MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Bifurcation theory
Question Bank - Set 2
Liberty University
Question 1
Question
Consider the differential equation: dx
dt =r−x2, where ris a constant parameter.
For what values of rdoes this equation have a bifurcation?
Solution
Step 1: To find bifurcation points, set dx
dt = 0 and solve for x.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the equilibrium points x=√rand x=
−√rby analyzing the sign of d2x
dt2.
To do this, differentiate dx
dt =r−x2with respect to x:
d
dx dx
dt =d
dx(r−x2)
d2x
dt2=−2x
Step 3: Plug x=√rinto d2x
dt2=−2x.
d2x
dt2=−2√r
<0for r > 0
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
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Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
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Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
17
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
18
Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
19
Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
20
Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
21
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
5
Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
10
Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
17
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
18
Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
19
Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
20
Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
21
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
5
Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
10
Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
17
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
18
Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
19
Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
20
Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
21
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
5
Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
10
Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
17
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
18
Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
19
Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
20
Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
21
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
5
Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
10
Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
17
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
18
Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
19
Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
20
Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
21
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
5
Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
10
Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
17
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
18
Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
19
Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
20
Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
21
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
5
Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
10
Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
17
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
18
Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
19
Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
20
Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
21
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
5
Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
10
Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
17
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
18
Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
19
Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
20
Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
21
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
5
Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
10
Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
17
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
18
Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
19
Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
20
Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
21
Step 4: Plug x=−√rinto d2x
dt2=−2x.
d2x
dt2= 2√r
>0for r > 0
Step 5: Analyze the sign of d2x
dt2near x= 0, which is the nontrivial solution.
d2x
dt2= 0
This information is not enough to determine the stability. Further analysis is needed.
Therefore, the bifurcation occurs at r= 0.
Question 2
Question
Consider the differential equation dy
dx =λy −y3, where λis a real parameter.
1. Find all the critical points of this system.
2. Determine the stability of each critical point for λ < 0,λ= 0, and λ > 0.
Solution
1. To find the critical points of the system, we set dy
dx =λy −y3= 0 and
solve for y.
λy −y3= 0
y(λ−y2) = 0
y= 0 or y2=λ
So the critical points are y= 0 and y=±√λ.
2. Next, we determine the stability of each critical point for different values
of λ.
Case 1: λ < 0
For λ < 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Substitute y= 0 into the equation dy
dx =λy −y3. The
derivative is d(0)
dx =λ·0−03= 0. Since the derivative is 0, we
consider the linear approximation:
d2y
dx2=−3y2
y=0
= 0
Since the second derivative is zero, we have an inconclusive test for
stability at y= 0.
2
• At y=√λand y=−√λ: Substitute y=±√λinto the equation
dy
dx =λy −y3. The derivative is d(±√λ)
dx =λ(±√λ)−(±√λ)3= 0.
Since the derivative is 0, we again consider the linear approximation:
d2y
dx2=−3y2
y=±√λ
=−3λ
Since the second derivative is negative for λ < 0, the critical points
y=±√λare stable.
Case 2: λ= 0
For λ= 0, the critical points are y= 0 and y=±√0 = 0. The analysis
at y= 0 is the same as in Case 1, resulting in an inconclusive test for
stability. Since y= 0 is a repeated root, the stability of this critical point
cannot be determined from the linearization near y= 0.
Case 3: λ > 0
For λ > 0, the critical points are y= 0 and y=±√λ.
• At y= 0: Similar to the previous cases, we consider the linear ap-
proximation near y= 0:
d2y
dx2=−3y2
y=0
= 0
The inconclusive test for stability at y= 0indicatesthatthestabilitycannotbedeterminedaty=0f orλ >
0.
• At y=√λand y=−√λ: The linear approximation near y=±√λ
is the same as in Case 1. The second derivative is negative, indicating
the critical points y=±√λare stable for λ > 0.
Question 3
Question
Consider the logistic map given by the recursive formula xn+1 =rxn(1 −xn)
where ris a bifurcation parameter and x0is the initial condition.
Given that the logistic map exhibits chaotic behavior when r≈3.57, deter-
mine the value of rat which a period-3 orbit first appears.
Solution
Step 1: To find the value of rat which a period-3 orbit first appears, we need
to set up the conditions for a period-3 orbit in the logistic map. For a period-3
orbit, we require the following equilibria: x1, x2,and x3such that x2=f(x1),
x3=f(x2), and x1=f(x3), where f(x) = rx(1 −x).
3
Step 2: Let’s denote x1, x2,and x3as x,f(x), and f(f(x)) respectively.
Then, we have the following equations:
x=rf(x)(1 −f(x))
f(x) = r(f(x))(1 −f(x))(1 −r(f(x))(1 −f(x)))
f(f(x)) = r(f(f(x)))(1 −f(f(x)))
Step 3: By solving the above equations simultaneously, we can find the values
of xthat satisfy the conditions for a period-3 orbit.
Step 4: Substitute f(x) = rx(1 −x)into the equations and solve for x. This
may result in a non-linear equation that can be solved using numerical methods.
Step 5: Once we have the solutions for x, plug them back into the logistic
map f(x) = rx(1 −x)to find the corresponding values of rthat produce a
period-3 orbit.
Step 6: By following the steps above, we can determine the value of rat
which a period-3 orbit first appears in the logistic map.
Question 4
Question
Consider the differential equation dy
dt =ky2−y.
(a) Determine all equilibrium points of the system.
(b) Use the method of linear stability analysis to classify the stability of each
equilibrium point.
(c) For what values of the parameter kdoes a bifurcation occur in the sys-
tem?
Solution
(a) To find the equilibrium points of the system, we set dy
dt = 0 and solve for y:
ky2−y= 0
y(ky −1) = 0
y= 0 or ky −1 = 0
y= 0 or y=1
k
So, the equilibrium points are y= 0 and y=1
k.
(b) To classify the stability of each equilibrium point, we consider the sign
of dy
dt in the vicinity of each point.
For y= 0:dy
dt = 0 −0 = 0
This indicates that y= 0 is a non-hyperbolic equilibrium point.
4
For y=1
k:
dy
dt =k1
k2
−1
k= 1 −1
k
At y=1
k, the derivative is positive for k < 1and negative for k > 1. Therefore,
y=1
kis a stable equilibrium point for k < 1and unstable for k > 1.
(c) A bifurcation in the system occurs at the critical point k= 1. At k= 1,
the stability of the equilibrium point y=1
kchanges from stable to unstable.
Question 5
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
(a) Find all the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which the system
undergoes a pitchfork bifurcation.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r·x−x3= 0
x(r−x2) = 0
So, the equilibrium points are x= 0 and x=±√r.
(b) To determine the values of rfor which the system undergoes a pitchfork
bifurcation, we examine the behavior of the equilibrium points at x= 0 and
x=±√r.
At x= 0, the stability of the equilibrium point can be determined by looking
at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x= 0. This occurs when r= 0.
At x=±√r, the stability of the equilibrium points can be determined by
looking at the sign of d
dx (r·x−x3):
d
dx(r·x−x3) = r−3x2
For a pitchfork bifurcation, we need the sign of d
dx (r−3x2)to change at
x=±√r. This occurs when r= 0.
Therefore, the system undergoes a pitchfork bifurcation at r= 0.
5
Question 6
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
1. Find the equilibrium solutions of the system.
2. Determine the stability of each equilibrium solution based on the value of
r.
3. Sketch a bifurcation diagram showing the stability of equilibrium solutions
as a function of r.
Solution
1. Find the equilibrium solutions of the system.
Setting dx
dt = 0, we have:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = 0
x(r−x) = 0
So, the equilibrium solutions are x= 0 and x=r.
2. Determine the stability of each equilibrium solution based on
the value of r.
To determine the stability, we examine the sign of d2x
dt2near each equilibrium
solution.
For x= 0, we have:
d2x
dt2=r
Since the sign of d2x
dt2depends on the value of r, we will analyze it further in
step 3.
For x=r, we have:
d2x
dt2=−2r2
Thus, x=ris a stable equilibrium when r < 0and an unstable equilibrium
when r > 0.
3. Sketch a bifurcation diagram showing the stability of equilib-
rium solutions as a function of r.
When r < 0, the equilibrium solution at x= 0 is stable, and the one at
x=ris unstable. As rincreases past 0, the stability of the equilibrium solutions
changes; when r= 0, the equilibrium at x=rbecomes stable, and the one at
x= 0 becomes unstable. This change indicates a bifurcation point.
The bifurcation diagram can be sketched as follows:
6
rStability of Equilibrium Solutions
r < 0 0 stable, runstable
r= 0 0 unstable, rstable
r > 0 0 unstable, runstable
Question 7
Question
Consider the differential equation dy/dt =ry −y3. Determine the equilibrium
solutions of the equation and investigate their stability using bifurcation theory.
Solution
Step 1: To find the equilibrium solutions, set dy/dt =ry −y3equal to 0and
solve for y:
ry −y3= 0
y(r−y2) = 0
This gives us two equilibrium solutions: 1. y= 0 2. y=±√r
Step 2: To determine the stability of the equilibrium solutions, we need to
compute the derivative of dy/dt with respect to y:
d
dy ry −y3=r−3y2
Step 3: Substitute the equilibrium solutions into the derivative to analyze
stability: 1. For y= 0:
r−3(0)2=r
Since rcan be positive, negative, or zero, y= 0 is a non-hyperbolic equilibrium.
2. For y=√r:
r−3(√r)2=r−3r=−2r
Since −2ris negative for positive r,y=√ris a stable equilibrium. 3. For
y=−√r:
r−3(−√r)2=r−3r=−2r
Similarly, y=−√ris also a stable equilibrium.
Therefore, the equilibrium solutions are y= 0,±√r, and all are stable for
r > 0.
Question 8
Question
Consider the differential equation dy
dt =r−y2, where ris a constant parameter.
For what values of rdoes the bifurcation diagram of the equation have two
stable fixed points and one unstable fixed point?
7
Solution
Step 1: Find the fixed points by setting dy
dt = 0: Setting dy
dt =r−y2= 0, we
get y2=r. So, the fixed points are at y=√rand y=−√r.
Step 2: Examine the stability of the fixed points by analyzing the sign of
d2y
dt2at each fixed point. Taking the derivative of dy
dt with respect to y, we get
d2y
dt2=−2y. Substitute the fixed points y=√rand y=−√rinto d2y
dt2: At
y=√r,d2y
dt2=−2√rand at y=−√r,d2y
dt2= 2√r.
Step 3: Identify the regions where the fixed points are stable or unstable
based on the signs of d2y
dt2. For two stable fixed points and one unstable fixed
point, we need d2y
dt2>0for the fixed points at y=−√rand y=√r, and
d2y
dt2<0for the fixed point between them. So, we need 2√r > 0for y=−√r
and y=√rto be stable, which implies r > 0. Also, we need −2√r < 0for the
fixed point between them to be unstable, which also implies r > 0.
Therefore, the bifurcation diagram of the equation has two stable fixed points
and one unstable fixed point when r > 0.
Question 9
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−xy, dy
dt =−y+y2−2xy
where ris a parameter. Determine the critical points of the system and inves-
tigate their stability for different values of r.
Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =dy
dt = 0 and solve for xand y.
rx −x2−xy = 0 and −y+y2−2xy = 0
Factoring xfrom the first equation and yfrom the second equation, we get:
x(r−x−y) = 0 and y(y−1−2x) = 0
This gives us the critical points (0,0),r
3−1
3,0, and r−1
2,1−r
2.
Step 2: Study the stability at each critical point
Let’s investigate the stability of each critical point by linearizing the system
about each critical point.
8
•For the critical point (0,0):
Linearizing the system around (0,0), we have the Jacobian matrix:
J=r−x
−2y2y−1
(0,0)
=r0
0−1
The eigenvalues are λ1=rand λ2=−1.
If r > 0, the eigenvalues have opposite signs, so the critical point (0,0) is
a saddle point. If r < 0, the eigenvalues are both negative, so the critical
point is stable.
•For the critical point r3−1
3,0:
Linearizing the system around r
3−1
3,0and simplifying, we find the
eigenvalues and classify the critical point’s stability based on the sign of
r.
•For the critical point r−12,1−r
2:
Linearizing the system around r−1
2,1−r
2and simplifying, we find the
eigenvalues and classify the critical point’s stability with respect to differ-
ent values of r.
Therefore, by finding the critical points and investigating their stability for
different values of r, we can understand the dynamics of the given system of
differential equations.
Question 10
Question
Consider the differential equation dx
dt =αx3−βx −γ, where α, β, γ are positive
constants. Investigate the possible bifurcation scenarios for this equation as α
varies. Show the critical values of αat which bifurcations occur.
Solution
To investigate the possible bifurcation scenarios for the given differential equa-
tion as αvaries, we need to find the critical values of αat which bifurcations
occur.
Step 1: Find the equilibrium points Setting dx
dt = 0, we find the equi-
librium points:
0 = αx3−βx −γ
This gives us the equilibrium points x=−β
3α+C
α, where C=3
qβ3
27α3+γ
α.
Step 2: Analyze the equilibrium points We need to analyze the behav-
ior of the equilibrium points as αvaries.
9
Case 1: β3<27α3γIn this case, there is one real equilibrium point and
two complex conjugate equilibrium points. A supercritical pitchfork bifurcation
occurs at α=27γ
β21/3.
Case 2: β3= 27α3γIn this case, there is one real equilibrium point and
one double real equilibrium point. A subcritical pitchfork bifurcation occurs at
α=27γ
β21/3.
Case 3: β3>27α3γIn this case, there are three real equilibrium points. A
transcritical bifurcation occurs at α=27γ
β21/3.
Therefore, the critical values of αat which bifurcations occur are α=
27γ
β21/3.
Question 11
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Deter-
mine the critical points and classify their stability as a function of r.
Solution
Step 1: To find the critical points, we set dx
dt = 0:
rx −x3= 0
Step 2: Factor out xfrom the equation:
x(rx −x2) = 0
Step 3: Set each factor to zero:
x= 0 or rx −x2= 0
Step 4: For x= 0, the derivative becomes d2x
dt2=r. Thus the critical point
at x= 0 is a saddle for r= 0.
Step 5: For rx −x2= 0, rearrange the equation:
x(rx −x) = 0
x(1 −x) = 0
Step 6: This gives critical points at x= 0 and x= 1. Now, we investigate
the stability at these points.
Step 7: For x= 0, the linearization gives dx
dt =rx, so the stability changes
based on the sign of r. When r < 0,x= 0 is stable, and when r > 0,x= 0 is
unstable.
10
Step 8: For x= 1, the linearization gives dx
dt =r−3, so x= 1 is stable when
r < 3and unstable when r > 3.
Step 9: In summary, the critical point at x= 0 is stable for r < 0and
unstable for r > 0, while the critical point at x= 1 is stable for r < 3and
unstable for r > 3.
Question 12
Question
Consider the differential equation given by dy
dt =ry −y3, where ris a constant.
Determine the bifurcation points and classify the type of bifurcation that occurs
at each point.
Solution
Step 1: Find the bifurcation points by setting dy
dt =ry −y3equal to zero and
solving for y.
0 = ry −y3
y3=ry
y(y2−r) = 0
So, the bifurcation points are y= 0 and y=√r.
Step 2: Examine the behavior of the system near each bifurcation point to
determine the type of bifurcation.
• For y= 0: Let’s analyze the behavior of the system near y= 0. Define
f(y) = ry −y3.
For y < 0:f(y) = ry −y3>0since −y3>0.
For 0< y < √r:f(y) = ry −y3>0since ry > y3.
For y > √r:f(y) = ry −y3<0since ry < y3.
Since the signs of f(y)changes from positive to negative as ypasses
through y= 0, a transcritical bifurcation occurs at y= 0.
• For y=√r: Let’s analyze the behavior of the system near y=√r. Define
f(y) = ry −y3.
For y < √r:f(y) = ry −y3<0since ry < y3.
For y > √r:f(y) = ry −y3>0since ry > y3.
Since the signs of f(y)changes from negative to positive as ypasses
through y=√r, a saddle-node bifurcation occurs at y=√r.
11
Question 13
Question
Consider the differential equation dy
dx =ry2−y, where ris a parameter.
For what values of rdoes this differential equation exhibit bifurcation be-
havior? Describe the type of bifurcation that occurs at each critical value of
r.
Solution
Step 1: Find the equilibrium points by setting dy
dx = 0:
ry2−y= 0 =⇒y(r−1) = 0
So, the equilibrium points are y= 0 and y=1
r.
Step 2: Identify the critical values of rwhere bifurcation occurs: Bifurcation
occurs when the equilibrium points change stability. This happens when the
derivative of the right-hand side of the differential equation with respect to y,
i.e., 2ry −1, evaluated at the equilibrium points is zero.
Evaluating at y= 0:2r(0) −1 = −1
Evaluating at y=1
r:2r1
r−1 = 1 −1 = 0
So, bifurcation occurs at r= 1.
Step 3: Describe the type of bifurcation at r= 1: - For r < 1: The equi-
librium point y= 0 is stable. - For r > 1: The equilibrium point y= 0
becomes unstable and a stable equilibrium point appears at y=1
r, indicating a
transcritical bifurcation.
Therefore, the differential equation dy
dx =ry2−yexhibits a transcritical
bifurcation at r= 1.
Question 14
Question
Consider the one-dimensional dynamical system defined by the differential equa-
tion dx
dt =rx −x3, where ris a parameter. Investigate the bifurcation behavior
of this system as rvaries.
Solution
To analyze the bifurcation behavior of the system, we first need to find the
critical points by setting dx
dt = 0:
rx −x3= 0
Factoring out an xgives:
x(rx −x2) = 0
12
So, the critical points are x= 0 and x=r. Next, we analyze the stability
of these critical points based on the sign of d2x
dt2:
Case 1: x= 0
Evaluate d2x
dt2at x= 0:
d2x
dt2=r
For r < 0,d2x
dt2is negative, indicating a stable critical point at x= 0.
Case 2: x=r
Evaluate d2x
dt2at x=r:
d2x
dt2= 2r
For r > 0,d2x
dt2is positive, indicating an unstable critical point at x=r.
Therefore, as rvaries: - For r < 0, the system has a stable critical point at
x= 0. - For r > 0, the system has an unstable critical point at x=r.
This indicates a bifurcation at r= 0, where the stability of the system
changes.
Question 15
Question
Consider the differential equation dx
dt =µx −x3, where µis a real parameter.
For what values of µdoes this equation exhibit a bifurcation? Determine
the bifurcation points and classify their stability.
Solution
Step 1: Find the equilibria
To find the equilibria, we set dx
dt = 0:
0 = µx −x3
x(µ−x2) = 0
So the equilibria are at x= 0 and x=±√µ.
Step 2: Analyze stability at the equilibria
Let’s examine the stability of the equilibria:
- At x= 0: Substitute x= 0 into dx
dt :
d(0)
dt =µ·0−03= 0
Since the derivative is zero, we have to perform further analysis.
- At x=õ: Substitute x=õinto dx
dt :
d(õ)
dt =µ√µ−µ= 0
13
Hence, the equilibrium x=õis stable.
- At x=−√µ: Substitute x=−√µinto dx
dt :
d(−√µ)
dt =−µ√µ−µ=−2µ√µ
Since the derivative is nonzero, we can affirm that the equilibrium x=−√µis
unstable.
Thus, the bifurcation occurs when x=√µand x=−√µ. The equilibrium
x=õchanges stability at this point.
Question 16
Question
Consider the system of differential equations given by:
dx
dt =rx −x2−αy −x2y, dy
dt =βy −γx2y
where r, α, β, γ > 0are parameters. Determine the bifurcation points of the
system.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0 and dy
dt = 0:
rx −x2−αy −x2y= 0, βy −γx2y= 0
Step 2: Solve the system of equations to find the equilibrium points (x∗, y∗).
Step 3: Linearize the system by computing the Jacobian matrix at the
equilibrium points:
J=r−2x∗−α−x∗
−2γx∗y∗β−γx∗2
Step 4: Calculate the determinant and trace of the Jacobian matrix to
determine the stability of the equilibrium points.
Step 5: Set the determinant equal to zero to find the bifurcation points.
Solve for the critical values of the parameters that lead to bifurcations.
Step 6: Analyze the eigenvalues of the Jacobian matrix at the bifurcation
points to determine the type of bifurcation (saddle node, transcritical, etc.).
Step 7: After determining the bifurcation points and types, provide a thor-
ough analysis of the system’s behavior near each bifurcation point.
Question 17
Question
Consider the differential equation given by dx
dt =rx−x3, where ris a parameter.
Determine the critical points and examine their stability as rvaries.
14
Solution
Step 1: Find the critical points by setting dx
dt = 0.
rx −x3= 0
x(r−x2) = 0
So, the critical points are x= 0 and x=±√r.
Step 2: Examine the nature of the critical points.
For x= 0, the linear approximation of the differential equation near x= 0
is dx
dt =rx. - If r < 0, the critical point x= 0 is stable. - If r > 0, the critical
point x= 0 is unstable.
For x=±√r, we have two cases:
a) x=√r: The linear approximation near x=√ris dx
dt =r√r−r=
r√r−r3
2. - If r < 0, the critical point x=√ris unstable. - If r > 0, the critical
point x=√ris stable.
b) x=−√r: The linear approximation near x=−√ris dx
dt =−r√r+r=
r−r3
2. - If r < 0, the critical point x=−√ris stable. - If r > 0, the critical
point x=−√ris unstable.
Question 18
Question
Consider the nonlinear dynamical system given by dx
dt =rx −x3, where ris a
real parameter. Study the fixed points and stability of the system as rvaries.
Solution
Step 1: Find the fixed points by setting dx
dt = 0.
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
The fixed points are x= 0,x=r.
Step 2: Study the stability of the fixed points by linearizing the system. Let
f(x) = rx −x3, then the linearized system is given by dx
dt =f′(x0)(x−x0),
where x0is a fixed point.
For x= 0, we have f′(0) = r, so the linearized system is dx
dt =r·xwhich is
a linear system with a stable fixed point at the origin.
For x=r, we have f′(r) = 0, so the linearized system is dx
dt = 0 which does
not give us information about the stability of x=r.
Step 3: Explore the bifurcation points where the stability of the system
changes. The bifurcation points occur when the stability of a fixed point
changes. In this case, the stability changes occur when r= 0.
15
Thus, for r < 0,x= 0 is a stable fixed point. For r > 0,x= 0 becomes
unstable and x=rbecomes a stable fixed point.
Therefore, the dynamical system exhibits a transcritical bifurcation at r= 0.
Question 19
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Find the fixed points of the system.
(b) Determine the stability of each fixed point as rvaries.
(c) Sketch a bifurcation diagram showing the stability of the fixed points as
a function of r.
Solution
(a) To find the fixed points, we set dx
dt = 0:
r−x2= 0 =⇒x=±√r
So the fixed points are x=√rand x=−√r.
(b) To determine the stability of each fixed point, we calculate the derivative
of dx
dt at each fixed point:
d
dx(r−x2) = −2x
For x=√r:
d
dx(r−x2)
x=√r
=−2√r < 0
Therefore, the fixed point x=√ris stable.
For x=−√r:
d
dx(r−x2)
x=−√r
= 2√r > 0
Therefore, the fixed point x=−√ris unstable.
(c) The bifurcation diagram indicates the stability of fixed points as a func-
tion of r. On the r−xplane, we mark a stability change with a dashed line.
For r < 0, there are no fixed points. For r= 0, there is a stable node at the
origin. For 0< r < 1, the fixed point moves to x=√r. For r > 1, the fixed
point at x=√rbecomes unstable, and there are no fixed points.
Question 20
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the bifurcation points of this equation and classify their stability.
16
Solution
Step 1: Find the equilibrium points by setting dx
dt to zero:
r−x2= 0
x2=r
x=±√r
Step 2: Calculate the derivative of dx
dt with respect to xto determine the
stability of the equilibrium points.
d
dx(r−x2) = −2x
Step 3: Analyze the stability at the equilibrium points. For x=−√r:
d
dx(r−(−√r)2) = −2(−√r) = 2√r > 0
Therefore, at x=−√r, the equilibrium point is unstable.
For x=√r:
d
dx(r−(√r)2) = −2(√r) = −2√r < 0
Therefore, at x=√r, the equilibrium point is stable.
Step 4: Determine the bifurcation points. Bifurcation occurs when the sta-
bility changes. Since the stability changes at x= 0 (from stable to unstable as
rmoves from negative to positive), x= 0 is a bifurcation point.
Step 5: Classify the stability at the bifurcation point. For x= 0:
d
dx(r−02) = −2(0) = 0
Since the derivative is zero, we cannot classify the stability at the bifurcation
point using linear stability analysis. Higher-order analysis would be required to
determine the stability at the bifurcation point.
Question 21
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter. Deter-
mine the critical points of the system and sketch the bifurcation diagram as r
varies.
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Solution
Step 1: Find the critical points
To find the critical points, we set dx
dt =r−x2= 0 and solve for x:
r−x2= 0
x2=r
x=±√r
So the critical points are x=±√r.
Step 2: Analyze the bifurcation
Let’s analyze the behavior of the system as rvaries:
1. When r < 0, there are no critical points as ris negative. The phase line
will have one steady-state solution at x= 0. 2. When r= 0, the critical points
are at x= 0. The phase line will have two steady-state solutions at x= 0. 3.
When r > 0, the critical points are at x=±√r. The phase line will have three
steady-state solutions at x=−√r,x= 0, and x=√r.
Therefore, the bifurcation diagram will show the change in the number of
steady-state solutions as rvaries.
Question 22
Question
Consider the differential equation dy
dt =r(1 −y2), where ris a parameter rep-
resenting the rate of growth.
(a) Determine the equilibrium solutions of the differential equation.
(b) Use bifurcation theory to analyze the behavior of the system as rvaries.
(c) Sketch a bifurcation diagram illustrating the equilibrium solutions as r
varies.
Solution
(a) To find the equilibrium solutions of the differential equation, we set dy
dt = 0:
r(1 −y2) = 0.
This implies y=±1. So, the equilibrium solutions are y= 1 and y=−1.
(b) Next, we analyze the behavior of the system as rvaries.
For r > 0, the equilibrium points y=±1are stable.
For r < 0, the equilibrium points y=±1are unstable.
This change in stability at r= 0 is a bifurcation point.
(c) To sketch a bifurcation diagram, we plot the equilibrium points y=±1
on the y-axis and indicate that they switch stability at r= 0.
The diagram will have the equilibrium points connected by a solid line,
indicating stability, and dashed lines near r= 0 to show the change in stability.
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Question 23
Question
Consider the differential equation dy
dt =ry −y3where ris a constant parameter.
1. Determine the equilibrium solutions of the differential equation.
2. Use a bifurcation diagram to classify the equilibrium solutions as stable
or unstable as rvaries.
3. Determine the critical values of rat which bifurcations occur.
Solution
1. Equilibrium Solutions: To find the equilibrium solutions, we set dy
dt = 0:
ry −y3= 0
y(r−y2) = 0
So, the equilibrium solutions are y= 0 and y=±√r.
2. Bifurcation Diagram: To determine the stability of the equilibrium
solutions as rvaries, we need to analyze the sign of dy
dt near each equilibrium
point. For y= 0,dy
dt =ry −y3. When r < 0,dy
dt is positive, resulting in an
unstable equilibrium at y= 0. When r > 0,dy
dt is negative, resulting in a stable
equilibrium at y= 0. For y=±√r,dy
dt =ry −y3. For r < 0, both ±√rare
stable equilibrium points. For r > 0,±√rare unstable equilibrium points.
3. Critical Values of r: Bifurcations occur when the stability of the
equilibrium points changes, i.e., when the sign of dy
dt changes near an equilibrium
point. From the analysis above, the critical values of rat which bifurcations
occur are r= 0 and r=−1.
Question 24
Question
Consider the logistic map given by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnrepresents the population proportion in generation n.
The bifurcation diagram for this map shows the values of xnas rvaries. For a
certain value of r, the logistic map exhibits a period-3 bifurcation.
If the logistic map has a period-3 orbit at r= 3.2, find the three fixed points
of the logistic map associated with this period-3 orbit.
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Solution
Step 1: Calculate the fixed points of the logistic map.
To find the fixed points, we set xn+1 =xn:
rx(1 −x) = x
rx −rx2=x
rx2−(r+ 1)x= 0
x(rx −(r+ 1)) = 0
Therefore, the fixed points are x= 0 and x=r+ 1
r.
Step 2: Determine the period-3 orbit points.
For a period-3 orbit, the logistic map must satisfy the conditions:
x1=x4, x2=x5, x3=x6
Substitute the logistic map equation into these conditions:
rx(1 −x) = r3x(1 −x)(1 −r2x(1 −x))
This simplifies to:
r=r3(1 −r2x(1 −x))
1 = r2(1 −r2x(1 −x))
1 = r2−r4x(1 −x)
r4x2−r2x+ 1 = 0
By solving this quadratic equation, we can find the values of xfor a period-3
orbit. Substituting r= 3.2into the equation gives:
3.24x2−3.22x+ 1 = 0
Solve the quadratic equation and find the three distinct values of xfor r=
3.2.
Question 25
Question
Consider the differential equation dy
dt =r−y2, where ris a parameter.
a) Find the equilibrium solutions of the differential equation in terms of r.
b) Use bifurcation theory to determine the values of rfor which bifurcations
occur, and classify the type of bifurcation that occurs at each critical value of
r.
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Solution
a) To find the equilibrium solutions, we set dy
dt =r−y2equal to 0 and solve for
y:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
b) To determine the values of rfor which bifurcations occur, we need to find
the critical values of rat which the equilibrium solutions change stability. We
do this by considering the derivative of dy
dt =r−y2with respect to y:
d
dy (r−y2) = −2y
At the equilibrium points y=√rand y=−√r, we have:
d
dy (r−y2)
y=√r=−2√r
d
dy (r−y2)
y=−√r= 2√r
For a bifurcation to occur, the stability of the equilibrium solutions must change.
This happens when the derivative with respect to ychanges sign at the critical
values.
Setting −2√r= 0 gives r= 0, which is a critical point. At r= 0, the
derivative changes sign from negative to positive. Therefore, a bifurcation occurs
at r= 0. This is a transcritical bifurcation.
In conclusion, at r= 0, a transcritical bifurcation occurs.
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