MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Bifurcation theory
Question Bank - Set 1
Liberty University
Question 1
Question
Consider the differential equation dy
dt =ry −y2, where ris a parameter. Inves-
tigate the bifurcation behavior of this equation at r= 0.
Solution
Step 1: Find the equilibrium points of the differential equation by setting dy
dt =
0.
ry −y2= 0
y(r−y) = 0
Step 2: The equilibrium points are y= 0 and y=r.
Step 3: Determine the stability of each equilibrium point by examining the
sign of dy
dt in a small interval around each point.
For y= 0, if we choose y=ϵwhere ϵis a small positive number, then
dy
dt =rϵ −ϵ2<0for 0< ϵ < r. Thus, the equilibrium point y= 0 is stable for
r > 0.
For y=r, if we choose y=r+ϵ, then dy
dt =r(r+ϵ)−(r+ϵ)2=−ϵ2<0
for ϵsufficiently small. Thus, the equilibrium point y=ris stable for all r.
Step 4: Therefore, at r= 0, the equilibrium point y= 0 loses stability,
leading to a bifurcation in the system’s behavior.
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
10
Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
12
Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
14
Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
20
Question 2
Question
Consider the differential equation dx
dt =r·x−x3, where ris a real parameter.
1. Determine the equilibrium solutions of the system.
2. Use bifurcation theory to classify the stability of the equilibrium solutions
as rvaries.
Solution
1. Equilibrium Solutions:
To find the equilibrium solutions, we set dx
dt = 0:
r·x−x3= 0
Factoring out an x, we have:
x(r−x2) = 0
Setting each factor to zero gives us the equilibrium points:
x= 0 and x=±√r
2. Stability Analysis:
We analyze the stability of the equilibrium solutions by considering the
sign of the derivative d
dx (r·x−x3):
d
dx (r·x−x3) = r−3x2
For x= 0, the derivative is r.
• If r > 0, the equilibrium at x= 0 is unstable.
• If r < 0, the equilibrium at x= 0 is stable.
For x=±√r, the derivative is r−3r=−2r.
• If r > 0, the equilibrium at x=±√ris stable.
• If r < 0, the equilibrium at x=±√ris unstable.
Therefore, the equilibrium solutions at x= 0 are stable for r < 0and
unstable for r > 0, while the equilibrium solutions at x=±√rare stable
for r > 0and unstable for r < 0.
2
Question 3
Question
Consider the differential equation dy
dx =ry −y3where ris a real parameter.
1. Find the critical points of the system.
2. Determine the stability of each critical point as a function of r.
3. Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
1. To find the critical points, we set dy
dx = 0:
ry −y3= 0
y(r−y2) = 0
This gives us critical points at y= 0 and y=±√r.
2. To determine the stability of each critical point, we need to examine the
sign of d
dx (dy
dx )near the critical points.
• For y= 0, we have:
d
dx (dy
dx ) = r−3y2=r
Thus, the critical point y= 0 is stable for r > 0and unstable for r < 0.
• For y=±√r, we have:
d
dx (dy
dx ) = r−3y2= 2r
The critical points y=±√rare always unstable.
3. The bifurcation diagram is a plot of the critical points as rvaries, indi-
cating their stability.
• For r > 0:
–y= 0 is stable.
–y=±√rare unstable.
• For r < 0:
–y= 0 is unstable.
–y=±√rare unstable.
• Thus, the bifurcation diagram will show a bifurcation occurring at r= 0,
where the stability of the critical points changes.
3
Question 4
Question
Consider the system of differential equations given by:
dx
dt =r−x2−y2
dy
dt =−y+x2−y2
where ris a parameter. Determine the critical points of the system and classify
their stability for r > 0.
Solution
Step 1: Find the critical points
To find the critical points of the system, we set dx
dt =dy
dt = 0 and solve for x
and y.
Setting dx
dt = 0, we have:
r−x2−y2= 0
Setting dy
dt = 0, we have:
−y+x2−y2= 0
Solving these equations simultaneously, we find the critical points.
Step 2: Evaluate the critical points
By solving the system of equations, we find the critical points of the system.
By evaluating the stability of these critical points, we can classify their behavior.
Step 3: Linearize the system
For each critical point, we can linearize the system of differential equations
around that point by finding the Jacobian matrix and evaluating it at the critical
point.
Step 4: Determine stability
By examining the eigenvalues of the Jacobian matrix at each critical point,
we can determine the stability of the critical points. A positive real part of
the eigenvalues indicates instability, a negative real part indicates stability, and
complex eigenvalues indicate oscillatory behavior.
Question 5
Question
Consider the differential equation dx
dt =rx−x3, where ris a constant. Determine
the values of rfor which the equilibrium points of the system change stability
at the bifurcation point.
4
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0.
rx −x3= 0 =⇒x(rx −x2) = 0
In order for this equation to hold true, either x= 0 or rx −x2= 0.
Step 2: Find the equilibrium points when x= 0. If x= 0, then dx
dt =
rx −x3=r(0) −(0)3= 0. So, x= 0 is an equilibrium point.
Step 3: Find the equilibrium points when rx −x2= 0. Solving rx −x2= 0
for x, we get x(rx −x) = 0, which implies x(r−x) = 0. So, x= 0 or x=r.
Step 4: Analyze the stability of the equilibrium points. We need to differen-
tiate between the cases when x= 0 and when x=rto determine the stability
of the equilibrium points.
For x= 0, consider the sign of d2x
dt2at x= 0:
d2x
dt2=d
dt (rx −x3) = r−3x2
Substitute x= 0:d2x
dt2=r
The sign of d2x
dt2is positive for r > 0and negative for r < 0. Thus, the
equilibrium point x= 0 changes stability at r= 0.
For x=r, consider the sign of d2x
dt2at x=r:
d2x
dt2=r−3r2=r(1 −3r)
The sign of d2x
dt2is positive for 0< r < 1
3, negative for r > 1
3, and zero at
r= 0 and r=1
3. Thus, the equilibrium point x=rchanges stability at r= 0
and r=1
3.
Question 6
Question
Consider the differential equation given by dy
dx =ry −y3, where ris a parameter.
Determine the bifurcation points, classify their stability, and sketch the phase
portrait.
Solution
Step 1: To find the bifurcation points, we set dy
dx = 0 and solve for y.
dy
dx =ry −y3= 0
5
y(r−y2) = 0
This equation has bifurcation points at y= 0 and y=±√r.
Step 2: Next, we determine the stability of these bifurcation points. We
calculate the sign of d(dy
dx )
dy at each point.
ddy
dx
dy =r−3y2
At y= 0,d(dy
dx )
dy =r, so the stability depends on the value of r. At y=±√r,
d(dy
dx )
dy =r−3r=−2r. If r > 0, then the bifurcation points will be stable; if
r < 0, then they will be unstable.
Step 3: Finally, we sketch the phase portrait. For r > 0, the bifurcation
points at y=±√rwill be stable nodes, while for r < 0, they will be unstable
nodes.
The phase portrait will show the behavior of solutions near the bifurcation
points, which will help to understand the dynamics of the system.
Question 7
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
a) Determine the critical points of the system and classify their stability
based on the parameter r.
b) Sketch a bifurcation diagram showing how the stability of the critical
points changes as rvaries.
Solution
a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
Therefore, the critical points are x= 0 and x=±√r.
To classify their stability, we evaluate the sign of the derivative d
dx (rx −x3)
at each critical point:
For x= 0:d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=0
=r
6
Therefore, x= 0 is a critical point with stability determined by the sign of
r. - If r < 0,x= 0 is a stable node. - If r > 0,x= 0 is an unstable node.
For x=±√r:
d
dx (rx −x3) = r−3x2
d
dx (rx −x3)
x=±√r
=r−3r=−2r
Thus, x=±√rare saddle points for all r.
b) Now, we can sketch the bifurcation diagram with ras the parameter: -
For r < 0, the system has a stable node at x= 0. - For r > 0, the system has
an unstable node at x= 0. - The critical points at x=±√rremain as saddle
points for all r.
Question 8
Question
Consider the differential equation dy
dx =r−y2, where ris a constant parameter.
1. Determine the equilibrium solutions of the system.
2. Investigate the behavior of the equilibrium solutions as rvaries.
Solution
1. To find the equilibrium solutions, set dy
dx = 0:
r−y2= 0
Solving for ygives two equilibrium solutions:
y=±√r
2. To investigate the behavior of the equilibrium solutions as rvaries, we
will determine the values of rat which a bifurcation occurs. At r= 0, the
equilibrium solutions are at y= 0, indicating a saddle node bifurcation.
For r > 0, the equilibrium solutions are real and stable. However, at r= 0,
the equilibrium solutions become imaginary, leading to the bifurcation.
Hence, a bifurcation occurs at r= 0.
Question 9
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter.
7
a) Determine the critical points of the differential equation.
b) Use the parameter rto investigate the bifurcation behavior of the system.
c) Sketch the bifurcation diagram showing the qualitative behavior of the
solutions.
Solution
a) To find the critical points, we set dx
dt equal to zero and solve the resulting
equation:
rx −x3= 0
x(rx −x2) = 0
x(rx −x2) = x(r−x) = 0
So, we have x= 0 and x=ras the critical points.
b) To investigate the bifurcation behavior, we analyze the sign of dx
dt around
the critical points. For x= 0,dx
dt =r(0) −03= 0, which indicates that x= 0 is
a stable critical point for all r.
For x=r,dx
dt =rr −r3=r2−r3=r2(1 −r). - If 0< r < 1, then dx
dt >0,
meaning x=ris unstable. - If r > 1, then dx
dt <0, meaning x=ris stable.
c) The bifurcation diagram will have a stable critical point at x= 0 for all
values of r, and a bifurcation occurs at r= 1, where the stability of the critical
point at x=rchanges. When 0< r < 1, the critical point x=ris unstable,
and when r > 1, the critical point x=rbecomes stable.
Question 10
Question
Consider the logistic map defined by the equation xn+1 =rxn(1 −xn), where
ris a parameter and xnis the population proportion at time n. For certain
values of r, the logistic map exhibits bifurcation behavior.
Given that rranges from 2.4 to 4.0, determine the values of rfor which the
logistic map exhibits period-3 behavior. Recall that period-3 behavior refers
to when the population proportion oscillates among three values in a repeating
cycle.
Solution
Step 1: Start by considering the fixed points of the logistic map. The fixed points
occur when xn+1 =xn=x∗, which leads to the equation x∗=rx∗(1 −x∗).
Step 2: Solve for the fixed points. Setting xn=x∗in the logistic map
equation, we have xn+1 =rx∗(1 −x∗). Thus, the fixed points are the solutions
to x∗=rx∗(1 −x∗). Solving this equation gives us the fixed points x∗= 0 and
x∗= 1 −1
r.
8
Step 3: Determine the stability of the fixed points. To determine the stability
of the fixed points, we need to calculate the derivative of the logistic map at the
fixed points.
Step 4: Calculate the derivative of the logistic map at the fixed points. The
derivative of the logistic map is given by
f′(x) = r(1 −2x).
Step 5: Evaluate the derivative at the fixed points. Evaluate the derivative
at the fixed points: At x∗= 0,f′(0) = r. At x∗= 1 −1
r,f′1−1
r=
r1−21−1
r=−r.
Step 6: Analyze the stability of the fixed points. If |f′(0)|<1, then x∗= 0
is stable. If |f′1−1
r|<1, then x∗= 1 −1
ris stable.
Step 7: Identify the values of rthat lead to period-3 behavior. For period-3
behavior to occur, the logistic map must exhibit a period-doubling cascade that
results in a period-3 cycle. This occurs when the stable fixed point loses stability
and a new period-3 cycle emerges.
Step 8: Determine the values of r. By iterating the logistic map equation
for various values of rbetween 2.4 and 4.0, we can identify the values that lead
to period-3 behavior.
Therefore, the values of rfor which the logistic map exhibits period-3 be-
havior lie within the range of the period-doubling cascade.
Question 11
Question
Consider the differential equation dx
dt =r−x2, where ris a parameter.
(a) Determine the equilibrium points of the system.
(b) Use bifurcation theory to analyze how the equilibrium points change as
the parameter rvaries.
(c) Sketch the bifurcation diagram for the system.
Solution
(a) To find the equilibrium points of the system, we set dx
dt = 0:
r−x2= 0 =⇒x2=r=⇒x=±√r
So the equilibrium points are x=√rand x=−√r.
(b) To analyze how the equilibrium points change as rvaries, we look at
the critical points where the system behavior changes. The critical points occur
when dx
dt = 0 and the derivative with respect to ris also zero. Calculating the
derivative with respect to r, we have:
d
dr (r−x2) = 1 −2xdx
dr = 0 =⇒xdx
dr =1
2
9
Substitute x=√rand x=−√rto solve for rat the critical points. We get
r=1
4.
Therefore, the equilibrium points change at r=1
4.
(c) The bifurcation diagram can be sketched to visualize the changes in
stability of equilibrium points as rvaries. At r=1
4, a bifurcation occurs leading
to changes in the number and stability of equilibrium points.
This completes the analysis of the differential equation using bifurcation
theory.
Question 12
Question
Consider the differential equation dx
dt =r−x2, where ris a constant. Determine
the values of rfor which the equilibrium points are stable or unstable using
bifurcation theory.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0: Setting dx
dt = 0, we have:
r−x2= 0
x2=r
x=±√r
So, the equilibrium points are x=√rand x=−√r.
Step 2: Determine the stability of the equilibrium points using the derivative:
Compute the derivative of dx
dt with respect to x:
d
dx (r−x2) = −2x
Now evaluate the derivative at the equilibrium points x=√rand x=−√r.
At x=√r, the derivative is:
d
dx
x=√r
=−2√r
At x=−√r, the derivative is:
d
dx
x=−√r
= 2√r
Step 3: Determine the stability: - If the derivative at the equilibrium point
is negative, the equilibrium point is stable. - If the derivative at the equilibrium
point is positive, the equilibrium point is unstable.
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Since the derivative at x=√ris negative, the equilibrium point x=√ris
stable.
Since the derivative at x=−√ris positive, the equilibrium point x=−√r
is unstable.
Therefore, for stability: 1. x=√ris stable when r > 0. 2. x=−√ris
stable when r < 0.
Question 13
Question
Consider the differential equation dy
dt =r−y2, where ris a real parameter.
Determine the values of rfor which the equilibrium solutions of the system
undergo a bifurcation.
Solution
To find the values of rfor which bifurcation occurs, we need to first find the
equilibrium solutions of the system.
Step 1: Find the equilibrium solutions Setting dy
dt = 0, we have:
r−y2= 0
y2=r
y=±√r
So, the equilibrium solutions are y=√rand y=−√r.
Step 2: Analyze the behavior of equilibrium solutions To determine
the values of rfor which bifurcation occurs, we need to study the stability of the
equilibrium solutions. The stability can be analyzed by evaluating the derivative
d
dy (r−y2)at the equilibrium solutions.
For y=√r:
d
dy (r−y2) = −2√r
For y=−√r:
d
dy (r−y2) = 2√r
Step 3: Identify bifurcation points Bifurcation points occur when the
stability of the equilibrium solutions changes. Hence, bifurcation occurs when
the derivative d
dy (r−y2)at an equilibrium solution is zero.
Therefore, the bifurcation points occur when:
−2√r= 0 ⇒r= 0
2√r= 0 ⇒No solution since ris a real parameter
So, the system undergoes a bifurcation at r= 0.
11
Question 14
Question
Consider the system of differential equations given by
dx
dt =x−y+µx2,dy
dt =x+y+µy2,
where µis a real parameter.
Find and classify all bifurcation points for this system when µ= 0.
Solution
Step 1: Find the equilibrium points To find the equilibrium points, we set
dx
dt = 0 and dy
dt = 0:
x−y+µx2= 0, x +y+µy2= 0.
Solving these equations simultaneously gives us the equilibrium points.
Step 2: Calculate the Jacobian matrix The Jacobian matrix for this
system is given by
J(x, y) = 1+2µx −1
1 1 + 2µy.
Step 3: Find the eigenvalues at the equilibrium points We evaluate
the eigenvalues of the Jacobian matrix at the equilibrium points to determine
their stability.
Step 4: Analyze the bifurcation points For µ= 0, the equilibrium
points and their stability will change. Analyze the eigenvalues at this specific
parameter value to find the bifurcation points and their classification.
Question 15
Question
Consider the differential equation dx
dt =r·x−x3, where ris a parameter.
If r < 0, show that the equation undergoes a pitchfork bifurcation as rpasses
through 0.
Solution
Step 1: Find the equilibrium points by setting dx
dt = 0:
r·x−x3= 0
Step 2: Factor out xfrom the equation:
x(r−x2) = 0
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Step 3: Solve for x:
x= 0 or x2=r
Step 4: If r < 0, there is only one equilibrium point at x= 0. Next, analyze
the stability of this equilibrium point using the first derivative test:
Step 5: Take the derivative of dx
dt with respect to xto find d2x
dt2:
d2x
dt2=r−3x2
Step 6: Evaluate d2x
dt2at the equilibrium point x= 0:
d2x
dt2|x=0 =r
Step 7: If r < 0, the equilibrium point at x= 0 is stable, which means
the equation undergoes a pitchfork bifurcation as rpasses through 0. This
bifurcation results in the birth of two new equilibrium points at x=±√ras r
changes sign.
Question 16
Question
Consider the following differential equation:
dx
dt =rx −x3
where ris a parameter.
(a) Find the equilibrium points of the system.
(b) Use bifurcation theory to determine the values of rfor which a bifurcation
occurs at x= 0.
(c) Classify the bifurcation at x= 0.
Solution
(a) To find the equilibrium points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(x(r−x)) = 0
Therefore, the equilibrium points are x= 0 and x=r.
(b) To use bifurcation theory, we look at the Jacobian matrix of the system:
J=r−3x2
13
Evaluate the Jacobian at the equilibrium point x= 0:
J(0) = r
For a bifurcation to occur, we need the determinant of the Jacobian at x= 0
to be zero, and the trace to change sign as rcrosses a critical value. Therefore,
we set r= 0:
J(0) = 0
(c) Since rcrosses zero at r= 0, and the eigenvalue is zero, the bifurcation
at x= 0 is a transcritical bifurcation.
Question 17
Question
Consider the system of differential equations given by:
dx
dt =r·x−x2−xy
dy
dt =−y+x2
where ris a parameter.
Determine the values of rfor which the system exhibits a bifurcation.
Solution
Step 1: First, find the critical points of the system by setting dx
dt =dy
dt = 0.
r·x−x2−xy = 0 and −y+x2= 0
This gives us two critical points: (0,0) and (r, r2).
Step 2: Linearize the system around the critical points and find the eigen-
values of the resulting matrix. For the critical point (0,0), the linearized system
is given by:
0 0
0−1x
y
The eigenvalues are λ1= 0 and λ2=−1.
For the critical point (r, r2), the linearized system is given by:
r−r
2r−1x
y=0
0
The eigenvalues are the solutions to the characteristic equation λ2+λ−r= 0,
which are λ=−1±√1+4r
2.
Step 3: Determine the values of rfor which the system exhibits a bifurcation.
A bifurcation occurs when at least one of the eigenvalues becomes zero. This
happens when 1+4r= 0, i.e., r=−1
4.
Therefore, the system exhibits a bifurcation at r=−1
4.
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Question 18
Question
Consider the following differential equation with a parameter r:
dy
dt =r−y2
Find the values of rfor which the equilibrium points of the system change
stability.
Solution
In bifurcation theory, we look for values of the parameter rwhere a qualitative
change in behavior occurs. For this differential equation, we need to find the
values of rfor which the equilibrium points change stability.
Step 1: Find the equilibrium points To find the equilibrium points, we
set dy
dt = 0:
r−y2= 0 =⇒y=±√r
So the equilibrium points are y=√rand y=−√r.
Step 2: Determine the stability of the equilibrium points To deter-
mine the stability, we need to analyze the sign of dy
dt around the equilibrium
points.
For y=√r:
dy
dt =r−r= 0
For y=−√r:
dy
dt =r−r= 0
Both equilibrium points have zero derivative suggesting they are neither
stable nor unstable, and are linearly stable.
Step 3: Determine the stability change To determine when the stability
changes, we need to look at the second derivative of dy
dt with respect to y.
d2y
dt2
y=±√r=−2y
y=±√r=−2√ror −2√r
The stability changes at the values of rwhere the second derivative changes
sign. Since the second derivative is always negative, there is no change in sta-
bility for any value of r.
15
Question 19
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
(a) Find the critical points of the system and determine their stability for
r < 0.
(b) Determine the regions in the rx-plane where the system exhibits bista-
bility.
Solution
(a) To find the critical points, we set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
x(r−x2) = 0
Therefore, the critical points are x= 0 and x=±√r. To determine the
stability, we evaluate the sign of d2x
dt2at each critical point.
At x= 0:d2x
dt2=r−3x2=r
Hence, if r < 0, we have d2x
dt2<0at x= 0, meaning the critical point is
stable.
At x=±√r:
d2x
dt2=r−3x2=r−3r=−2r
Therefore, for r < 0, the critical points x= 0 and x=±√rare stable.
(b) The system exhibits bistability in the regions where the phase line inter-
sects the stability line twice. For r < 0, the stability line intersects the rx-plane
at x= 0 and x=±√r. Hence, the system exhibits bistability in the regions
−√r < x < 0and 0< x < √r.
Question 20
Question
Consider the differential equation dx
dt =rx −x3, where ris a real parameter.
Determine the critical points of the system and classify their stability using
bifurcation theory.
16
Solution
Step 1: To find the critical points, set dx
dt = 0:
rx −x3= 0
x(rx −x2) = 0
This gives us critical points at x= 0 and x=r.
Step 2: To classify the stability of these critical points, we need to analyze
the sign of the derivative of dx
dt near each critical point. For x= 0, calculate the
derivative: d2x
dt2=r−3x2
At x= 0,d2x
dt2=r, so: - If r > 0,x= 0 is unstable. - If r < 0,x= 0 is stable.
For x=r, calculate the derivative:
d2x
dt2=r−3r2=r(1 −3r)
- If 0< r < 1
3,x=ris stable. - If r > 1
3,x=ris unstable.
Therefore, the bifurcation occurs at r=1
3, changing the stability of the
critical point x=rfrom stable to unstable.
Question 21
Question
Consider the differential equation dy
dt =y2−1. Determine the equilibrium
solutions of the system and sketch the phase line with the equilibria labeled.
Identify the type of bifurcation that occurs at the bifurcation point.
Solution
Step 1: Find the equilibrium solutions by setting dy
dt = 0: Setting y2−1=0,
we find y=±1.
Step 2: Draw the phase line with the equilibria labeled:
Region y′sign Nature of Equilibrium
1y < −1 + Unstable
2−1< y < 1−Stable
3y > 1 + Unstable
Step 3: Identify the type of bifurcation: At the bifurcation point y=−1or
y= 1, we see a saddle-node bifurcation occurring where equilibria ±1collide
and disappear.
Therefore, the equilibrium solutions are y=−1and y= 1, and a saddle-
node bifurcation occurs at these points.
17
Question 22
Question
Consider the differential equation given by:
dx
dt =r−x2
where ris a parameter. Show that this differential equation undergoes a
saddle-node bifurcation at r= 0.
Solution
Step 1: Find the critical points of the differential equation by setting dx
dt = 0.
r−x2= 0
x2=r
x=±√r
Step 2: Determine the stability of the critical points using the sign of d2x
dt2.
d2x
dt2=−2x
For x=√r, we have d2x
dt2=−2√r. Since this is negative for r > 0,x=√r
is a stable critical point.
For x=−√r, we have d2x
dt2= 2√r. Since this is positive for r > 0,x=−√r
is an unstable critical point.
Step 3: Analyze the bifurcation at r= 0. (A) For r < 0: There are two
real critical points x=±√r, both of which are stable. (B) At r= 0: The two
critical points x=±0collide and vanish. (C) For r > 0: There are no real
critical points, indicating a change in the stability behavior.
Therefore, the differential equation undergoes a saddle-node bifurcation at
r= 0.
Question 23
Question
Consider the differential equation dx
dt =rx −x3, where ris a parameter. Inves-
tigate the bifurcation behavior of this system as rvaries.
18
Solution
To investigate the bifurcation behavior of the system, we will analyze the equi-
librium points and their stability as the parameter rvaries.
Step 1: Find the equilibrium points Setting dx
dt = 0, we have rx−x3= 0.
Factoring out an x, we get x(rx −x2) = 0. So the equilibrium points are x= 0
and x=±√r.
Step 2: Analyze the stability of equilibrium points - For x= 0:
Substitute x= 0 back into the differential equation to find dx
dt at x= 0. We
have dx
dt = 0 −0 = 0. The equilibrium point x= 0 is unstable. - For x=±√r:
Substitute x=√rback into the differential equation to find dx
dt at x=√r. We
have dx
dt =r√r−r=r(√r−1). Since ris a parameter, the stability of x=√r
depends on the value of r. We need to further analyze this case.
Step 3: Analyze the bifurcation behavior - When r= 0: The equilib-
rium points are x= 0,x=√0 = 0, and x=−√0 = 0. As we’ve seen, x= 0
is unstable. As rincreases from 0, the stability at x=√rchanges at r= 1.
For r > 1,x=√rbecomes a stable equilibrium point, while x=−√rremains
unstable. This indicates a pitchfork bifurcation at r= 1.
Therefore, the system exhibits a pitchfork bifurcation at r= 1, where a
stable equilibrium point emerges from x= 0 as rcrosses 1.
Question 24
Question
Consider the differential equation given by dy
dx =ry(1 −y), where ris a real
parameter.
1. Find the equilibrium solutions of the differential equation.
2. Use the equilibrium solutions to determine the bifurcation points of the
system.
Solution
1. Equilibrium solutions: Setting dy
dx = 0, we have ry(1 −y) = 0. The
equilibrium solutions are the values of ythat make this equation true. This
means y= 0 or y= 1 are the equilibrium solutions.
2. Bifurcation points: To find the bifurcation points, we substitute these
equilibrium solutions into the original differential equation. When y= 0,dy
dx =
r·0(1 −0) = 0. When y= 1,dy
dx =r·1(1 −1) = 0.
Therefore, the bifurcation points occur at y= 0 and y= 1 because the
derivative becomes zero at these points.
19
Question 25
Question
Consider the differential equation dy
dx =1
2y(4 −y). Determine the critical points
and classify their stability using bifurcation theory.
Solution
Step 1: Find the critical points by setting dy
dx = 0.
dy
dx =1
2y(4 −y) = 0
This equation is true when y= 0 or y= 4. So the critical points are y= 0 and
y= 4.
Step 2: Classify the stability at y= 0. For y= 0, we evaluate the sign
of dy
dx near y= 0. When y < 0,dy
dx is positive, indicating that solutions move
away from y= 0 (unstable). When 0< y < 4,dy
dx is negative, indicating that
solutions move towards y= 0 (stable).
Step 3: Classify the stability at y= 4. For y= 4, we evaluate the sign
of dy
dx near y= 4. When 4< y < ∞,dy
dx is positive, indicating that solutions
move away from y= 4 (unstable). When y < 4,dy
dx is negative, indicating that
solutions move towards y= 4 (stable).
Therefore, y= 0 is a saddle point, while y= 4 is a stable point.
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