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\documentclass[12pt,a4paper]{report} \usepackage[T1]{fontenc} \usepackage[latin9]{inputenc} \usepackage{units} \usepackage{amsmath} \usepackage{amssymb} \linespread{2} \providecommand{\tabularnewline}{\\} \usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry} \makeatother \usepackage[english]{babel} \begin{document} \section*{Analytic Approximation} To find an approximation to for small $\tau$ we follow the method outlined by Howison. We let $\tau=\epsilon t^{\prime}$$\qquad0<\epsilon\ll1$$\qquad$ and assume the solution can be expanded as a series. \subsection*{Option Pricing Under CEV Models} \section*{Call Option} \section{find the call inner?? } \section{find the call outer ??} $\qquad dS=S\mu dv+\sigma S^{\gamma}dX\:\gamma>1$ $V_{t}+\dfrac{\sigma^{2}S^{2\gamma}}{2}V_{SS}+rSV_{S}-rV$ subject to \[ \begin{aligned}V(S,T) & =max(S-K,0)\\ V(0,t) & =0\\ V(\infty,t) & \sim S\left(S-Ke^{-r(T-t)}\right) \end{aligned} \] $\tau=T-t$ \[ V_{\tau}-\dfrac{\sigma^{2}S^{2\gamma}}{2}V_{SS}-rSV_{S}+rV=0 \] \[ \begin{aligned}V_{\tau}-\dfrac{\sigma^{2}S^{2\gamma}}{2}V_{SS}- & rSV_{S}+rV=0\\ & V(S,0)=max(S-K,0)\\ & V(0,\tau)=0\\ S\rightarrow\infty\ & V(S,\tau)\sim S\\ & \qquad(\rightarrow S-Ke^{-r\tau}) \end{aligned} \] To find an approximation for small $\tau$ let $\tau=\epsilon t^{\prime}$$\qquad0<\epsilon\ll1$$\qquad$and assume a solution of the form \[ C(S,\tau)=\sum_{i=0}^{\infty}\epsilon^{i}C_{i}(S,t^{\prime}) \] Show \[ \dfrac{\partial C}{\partial S}=\sum_{i=0}^{\infty}\epsilon^{i}\dfrac{\partial C_{i}}{\partial S},\qquad\dfrac{\partial^{2}C}{\partial S^{2}}=\sum_{i=0}^{\infty}\epsilon^{i}\dfrac{\partial^{2}C_{i}}{\partial S^{2}} \] \[ \begin{aligned}\dfrac{\partial C}{\partial t^{\prime}} & =\sum_{i=0}^{\infty}\epsilon^{i}\dfrac{\partial C_{i}}{\partial t^{\prime}} \end{aligned} \] So DE is \begin{equation} C_{t^{\prime}}-\dfrac{\epsilon\sigma^{2}S^{2\gamma}}{2}V_{SS}-rS\epsilon C_{S}+r\epsilon C=0\label{eq:**} \end{equation} Sub into (\ref{eq:**}) \begin{equation} \left[\dfrac{\partial C_{0}}{\partial t^{\prime}}+\epsilon\dfrac{\partial C_{1}}{\partial t^{\prime}}+\cdots\right]-\dfrac{\epsilon^{2}S^{2\gamma}}{2}\left[\dfrac{\partial^{2}C_{0}}{\partial S^{2}}+\epsilon\dfrac{\partial^{2}C_{1}}{\partial S^{2}}+\cdots\right]-rS\epsilon\left[\dfrac{\partial C_{0}}{\partial S}+\epsilon\dfrac{\partial C_{1}}{\partial S}+\cdots\right]+r\epsilon\left[C_{0}+\epsilon C_{1}+\cdots\right]\label{eq:***} \end{equation} equating coefficients of $\varepsilon^{0}$ we get \[ \dfrac{\partial C_{0}}{\partial t^{\prime}}=0\Rightarrow C_{0}=C_{0}(S) \] At $\tau=0$ \begin{align*} C_{0}(S,0) & =max(S-K,0)\\ \Rightarrow\quad C_{0}(S,t^{\prime}) & =max(S-K,0) \end{align*} Equating each coefficient of $\epsilon^{\prime}$ in \ref{eq:***} we get \[ \dfrac{\partial C_{1}}{\partial t^{1}}-\dfrac{\partial^{2}S^{2\gamma}}{2}\dfrac{\partial^{2}C_{0}}{\partial S^{2}}-rS\dfrac{\partial C_{0}}{\partial S}+rC_{0}=0 \] \begin{align*} C_{0} & =S-k & S>K\\ & 0 & S<K \end{align*} we get for $S>K$ \[ \dfrac{\partial C_{1}}{\partial t^{1}}=rS-r(S-K)=rK \] \[ \Rightarrow\quad C_{1}=rKt^{\prime}\qquad S>K \] and for $S<K\qquad C_{0}=0$ so $\dfrac{\partial C_{1}}{\partial t^{1}}=0\quad C_{1}=0$ So we have \[ C_{0}(S,t^{\prime})+\epsilon C_{1}(S,t^{1})\begin{cases} =S-K+rKt^{\prime}\epsilon & K-S\ll\epsilon K\\ =0 & K-S\gg\epsilon K \end{cases} \] This is the outer solution to $O(\epsilon)$. However, this solution is not differentiable at $S=K$. So we introduce the inner variable \[ x=\dfrac{K-S}{\epsilon^{\frac{1}{2}}K} \] and scale $C$ to $\epsilon^{\frac{1}{2}}KQ$. \section*{Find put use formal put and call parity??} \section*{ put option} \[ {\displaystyle C(t)-P(t)=S(t)-K\cdot B(t,T)} \] again, this is not differentiable at $S=K$. So we introduce the inner variable \[ x=\dfrac{S-K}{\epsilon^{\frac{1}{2}}K} \] \[ \begin{aligned}S & =x\epsilon^{\frac{1}{2}}K+K\\ & =K\left(1+x\epsilon^{\frac{1}{2}}\right) \end{aligned} \] and uscale $P$ to $\epsilon^{\frac{1}{2}}KQ$ Show (by taking partial derivative) that the DE becomes \[ \dfrac{K}{\sqrt{\epsilon}}\dfrac{\partial Q}{\partial t^{1}}=\dfrac{\sigma^{2}}{2}\left[K\left(1+x\epsilon^{\frac{1}{2}}\right)\right]^{2\gamma}\times\dfrac{1}{\sqrt{\epsilon}K}\dfrac{\partial^{2}Q}{\partial x^{2}}+rK\left(1+x\epsilon^{\frac{1}{2}}\right)\dfrac{\partial P}{\partial S}-r\epsilon^{\frac{1}{2}}KQ \] \[ \Rightarrow\quad\dfrac{\partial Q}{\partial t^{1}}=\dfrac{\sigma^{2}}{2}K^{2\gamma-2}\left(1+x\epsilon^{\frac{1}{2}}\right)^{2\gamma}\dfrac{\partial^{2}Q}{\partial x^{2}}+r\sqrt{\epsilon}(1+x\sqrt{\epsilon})\dfrac{\partial Q}{\partial x^{2}}+r\sqrt{\epsilon}\left(1+x\sqrt{\epsilon}\right)\dfrac{\partial Q}{\partial x}-r\epsilon Q \] which needs to be solved subject to \[ Q(x,0)=max(-x,0) \] \[ \lim_{x\rightarrow\infty}Q(x,t^{1})=0 \] \[ \lim_{x\rightarrow-\infty}Q(x,t^{1})=-x-rt^{1}\epsilon^{\frac{1}{2}}+O(\epsilon^{\frac{3}{2}}) \] for a put \[ p(s,0)=max(k-s,0) \] \[ p(\infty,\tau)=0 \] \[ p(0,\tau)= k e^{-r\tau} \] \[ s \to 0 \] \[ x \to \frac{-1}{\epsilon^{\frac{1}{2}}} \to -\infty \epsilon is small \\ \] For a $\quad x\rightarrow-\infty\quad(\Rightarrow S\rightarrow0\quad x\rightarrow-\nicefrac{1}{\sqrt{\epsilon}}$ \[ \begin{aligned}{and} & \begin{aligned}\mbox{} & P\rightarrow Ke^{-r\tau}\end{aligned} =K(1-r^{\tau}+)\\ & \Rightarrow Q=\dfrac{P}{\sqrt{\epsilon}V}\rightarrow Ke^{-rCt^{1}}\\ & Q\Rightarrow\dfrac{K\left(1-r\epsilon t^{1}+\cdots\right)}{\sqrt{\epsilon}K}\\ & Q\rightarrow\dfrac{1}{\sqrt{\epsilon}}-rt^{1}\sqrt{\epsilon}+O(\epsilon^{\frac{3}{2}})\\ & Q\rightarrow-x-rt^{1}\sqrt{\epsilon}+O(\epsilon^{\frac{3}{2}}) \end{aligned} \] now expand \[ Q(\gamma,\tau)=\sum_{i=0}^{\infty}\epsilon^{\nicefrac{i}{2}}Q_{i}(x,t^{1}) \] so into DE get \[ \begin{aligned}\dfrac{\partial}{\partial t^{1}}\left[Q_{0}+\epsilon^{\frac{1}{2}}Q_{1}+\cdots\right]= & \dfrac{\sigma^{2}}{2}K^{2\gamma-2}\left(1+x\epsilon^{\nicefrac{1}{2}}\right)^{2\gamma}\dfrac{\partial^{2}}{\partial x^{2}}\left[Q_{0}+\epsilon^{\nicefrac{1}{2}}Q_{1}+\cdots\right]\\ & +r\sqrt{\epsilon}\left[1+x\sqrt{\epsilon}\right]\frac{\partial}{\partial x}\left[Q_{0}+\epsilon^{\nicefrac{1}{2}}Q_{1}+\cdots\right]\\ & -r\epsilon\left[Q_{0}+\epsilon^{\nicefrac{1}{2}}Q_{1}+\cdots\right] \end{aligned} \] \[ \left(1+\alpha\right)^{p}=1+p\cdot\alpha+\dfrac{p(p-1)}{2!}\alpha^{2}+\cdots \] so \[ \left(1+x\epsilon^{\nicefrac{1}{2}}\right)^{2\gamma}=1+2\gamma x\epsilon^{\nicefrac{1}{2}}+\dfrac{(2\gamma)(2\gamma-1)}{2}x^{2}\epsilon+\cdots \] equating term of O(1) i.e $\epsilon^{0}$term: \begin{equation} \dfrac{\partial Q_{0}}{\partial t^{1}}=\dfrac{\sigma^{2}}{2}K^{2\gamma-2}\dfrac{\partial^{2}Q_{0}}{\partial x^{2}}\label{eq:1} \end{equation} to be solved subject to \begin{equation} \begin{aligned} & Q_{0}(x,0)=max(-x,0)\\ & \lim_{x\rightarrow\infty}Q_{0}(x,t^{1})=0\\ & \lim_{x\rightarrow-\infty}Q_{0}(x,t^{1})=-\lambda \end{aligned} \end{equation} To solve this, we use a similarity reduction transformation \[ \begin{aligned}\mbox{Let }z & =e^{\alpha}x\\ s & =e^{\beta}t^{1}\\ \psi_{0} & =e^{c}Q_{0}\left(e^{-\alpha}z,e^{-\beta}t^{1}\right) \end{aligned} \] Find $\alpha,\beta,C$ so that PDE is unchanged \[ \dfrac{\partial Q_{0}}{\partial t^{1}}=\dfrac{\sigma^{2}}{2}k^{2\gamma-2}\dfrac{\partial^{2}Q_{0}}{\partial X^{2}}\rightarrow Q_{0t^{1}}=\dfrac{\sigma^{2}}{2}k^{2\gamma-2}Q_{0xx} \] \begin{itemize} \item $\begin{aligned}Z & =e^{\alpha}X\rightarrow x=e^{-\alpha}Z\\ S & =e^{\beta}t^{\prime}\rightarrow t^{\prime}=e^{\beta}S\\ \psi_{0} & =e^{c}Q_{0}\rightarrow Q_{0}=e^{-c}\psi_{0} \end{aligned} $ \item $\begin{aligned}Q_{0t^{\prime}}=\dfrac{\partial Q_{0}}{\partial t^{\prime}} & =\dfrac{\partial}{\partial t^{\prime}}\left(e^{-c}\psi_{0}\right)\\ & =e^{c}\dfrac{\partial\psi_{0}}{\partial t^{\prime}}=e^{-c}e^{\beta}\dfrac{\partial\psi_{0}}{\partial S^{\prime}}=e^{(\beta-c)}\dfrac{\partial\psi_{0}}{\partial S} \end{aligned} $ \item $\begin{aligned}Q_{0x}=\dfrac{\partial Q_{0}}{\partial x} & =\dfrac{\partial}{\partial x}\left(e^{-c}\psi_{0}\right)\\ & =e^{-c}\dfrac{\partial\psi_{0}}{\partial x}=e^{-c}e^{\alpha}\dfrac{\partial\psi_{0}}{\partial Z}=e^{(\alpha-c)}\dfrac{\partial\psi_{0}}{\partial Z} \end{aligned} $ \item $\begin{aligned}Q_{0xx}=\dfrac{\partial^{2}Q_{0}}{\partial x^{2}} & =\dfrac{\partial}{\partial x}\left(e^{(\alpha-c)}\dfrac{\partial\psi_{0}}{\partial Z}\right)\\ & =e^{(\alpha-c)}\dfrac{\partial}{\partial x}\dfrac{\partial\psi_{0}}{\partial Z}=e^{(\alpha-c)}e^{\alpha}\dfrac{\partial^{2}\psi_{0}}{\partial Z^{2}}=e^{(2\alpha-c)}\dfrac{\partial^{2}\psi_{0}}{\partial Z^{2}} \end{aligned} $ \end{itemize} $e^{(\beta-c)}\dfrac{\partial\psi_{0}}{\partial S}=\dfrac{\sigma^{2}}{2}K^{2\gamma-2}e^{(2\alpha-c)}\dfrac{\partial^{2}\psi_{0}}{\partial Z^{2}}$ so for the equation to be invariant, we need $\beta=2\alpha$ $c$ is arbitrary, so let $c=0$ so our transformation is $\begin{aligned}X & =e^{-\alpha}Z\\ t^{\prime} & =e^{-2\alpha}S\\ Q_{0} & =\psi_{0} \end{aligned} $ \begin{itemize} \item $Q_{0}\left(e^{-\alpha}Z,e^{-\beta}t^{\prime}\right)=f(\eta)\rightarrow\eta=\dfrac{e^{-\alpha}}{e^{-\frac{\beta}{2}}}\dfrac{Z}{\sqrt{t^{\prime}}}$ \end{itemize} $\begin{aligned}\mbox{or lets say }\eta & =e^{\left(-\alpha+\nicefrac{\beta}{2}\right)}\dfrac{Z}{\sqrt{t^{\prime}}}\\ & =e^{\left(-\alpha+\nicefrac{\beta}{2}\right)}\cdot e^{\alpha}X\cdot e^{\nicefrac{\beta}{2}}S^{-\nicefrac{1}{2}}\\ & =e^{\beta}XS^{-\frac{1}{2}}=e^{\nicefrac{\beta}{2}}\dfrac{x}{\sqrt{t^{\prime}}}\\ & =f\left(e^{\nicefrac{\beta}{2}}\dfrac{x}{\sqrt{t^{\prime}}}\right) \end{aligned} $ $\begin{aligned}Q_{t^{\prime}} & =f^{\prime}(\eta)\dfrac{\partial\eta}{\partial t^{\prime}}\\ & =f^{\prime}(\eta)\dfrac{-e^{\nicefrac{\beta}{2}}X}{2t^{\prime^{\nicefrac{\beta}{2}}}} \end{aligned} $ $Q_{0x}=f^{\prime}(\eta)\dfrac{\partial\eta}{\partial x}=f^{\prime}(\eta)\times\dfrac{e^{\nicefrac{\beta}{2}}}{\sqrt{t^{\prime}}}$ $Q_{0xx}=\dfrac{\partial}{\partial x}\left[f^{\prime}(\eta)\times\dfrac{1}{\sqrt{t}}\right]=\dfrac{e^{\nicefrac{\beta}{2}}}{\sqrt{t}}f^{\prime\prime}(\eta)\times\dfrac{\partial\eta}{\partial x}=\dfrac{e^{\beta}}{t}f^{\prime}(\eta)$ sub into $Q_{0_{t^{\prime}}}=\dfrac{\sigma^{2}}{2}K^{2\gamma-2}Q_{0_{xx}}$ $\dfrac{-e^{\nicefrac{\beta}{2}}X}{2t^{\prime^{\nicefrac{3}{2}}}}f^{\prime}(\eta)=\dfrac{\sigma^{2}}{2}K^{2\gamma-2}\dfrac{e^{\beta}}{t}f^{\prime\prime}(\eta)$ $\therefore\dfrac{-e^{\nicefrac{\beta}{2}}X}{2t^{\prime^{\nicefrac{1}{2}}}}f^{\prime}(\eta)=\dfrac{\sigma^{2}}{2}K^{2\gamma-2}f^{\prime\prime}(\eta)$ $\therefore\dfrac{-e^{\nicefrac{\beta}{2}}\eta}{2}f^{\prime}(\eta)=\dfrac{\sigma^{2}}{2}K^{2\gamma-2}f^{\prime\prime}(\eta)$ \[ \begin{aligned}S & =x\epsilon^{\frac{1}{2}}K+K\\ & =K\left(1+x\epsilon^{\frac{1}{2}}\right) \end{aligned} \] and uscale $P$ to $\epsilon^{\frac{1}{2}}KQ$ Show (by taking partial derivative) that the DE becomes \[ \dfrac{K}{\sqrt{\epsilon}}\dfrac{\partial Q}{\partial t^{1}}=\dfrac{\sigma^{2}}{2}\left[K\left(1+x\epsilon^{\frac{1}{2}}\right)\right]^{2\gamma}\times\dfrac{1}{\sqrt{\epsilon}K}\dfrac{\partial^{2}Q}{\partial x^{2}}+rK\left(1+x\epsilon^{\frac{1}{2}}\right)\dfrac{\partial P}{\partial S}-r\epsilon^{\frac{1}{2}}KQ \] \[ \Rightarrow\quad\dfrac{\partial Q}{\partial t^{1}}=\dfrac{\sigma^{2}}{2}K^{2\gamma-2}\left(1+x\epsilon^{\frac{1}{2}}\right)^{2\gamma}\dfrac{\partial^{2}Q}{\partial x^{2}}+r\sqrt{\epsilon}(1+x\sqrt{\epsilon})\dfrac{\partial Q}{\partial x^{2}}+r\sqrt{\epsilon}\left(1+x\sqrt{\epsilon}\right)\dfrac{\partial Q}{\partial x}-r\epsilon Q \] which needs to be solved subject to \[ Q(x,0)=max(-x,0) \] \[ \lim_{x\rightarrow\infty}Q(x,t^{1})=0 \] \[ \lim_{x\rightarrow-\infty}Q(\gamma,t^{1})=-x-rt^{1}\epsilon^{\frac{1}{2}}+O(\epsilon^{\frac{3}{2}}) \] For a $\quad x\rightarrow-\infty\quad(\Rightarrow S\rightarrow0\quad x\rightarrow-\nicefrac{1}{\sqrt{\epsilon}}$ \[ \begin{aligned}{and} & \begin{aligned}\mbox{} & P\rightarrow Ke^{-r\tau}\end{aligned} =K(1-r^{\tau}+)\\ & \Rightarrow Q=\dfrac{P}{\sqrt{\epsilon}V}\rightarrow Ke^{-rCt^{1}}\\ & Q\Rightarrow\dfrac{K\left(1-r\epsilon t^{1}+\cdots\right)}{\sqrt{\epsilon}K}\\ & Q\rightarrow\dfrac{1}{\sqrt{\epsilon}}-rt^{1}\sqrt{\epsilon}+O(\epsilon^{\frac{3}{2}})\\ & Q\rightarrow-x-rt^{1}\sqrt{\epsilon}+O(\epsilon^{\frac{3}{2}}) \end{aligned} \] now expand \[ Q(x,\tau)=\sum_{i=0}^{\infty}\epsilon^{\nicefrac{i}{2}}Q_{i}(x,t^{1}) \] so into DE get \[ \begin{aligned}\dfrac{\partial}{\partial t^{1}}\left[Q_{0}+\epsilon^{\frac{1}{2}}Q_{1}+\cdots\right]= & \dfrac{\sigma^{2}}{2}K^{2\gamma-2}\left(1+x\epsilon^{\nicefrac{1}{2}}\right)^{2\gamma}\dfrac{\partial^{2}}{\partial x^{2}}\left[Q_{0}+\epsilon^{\nicefrac{1}{2}}Q_{1}+\cdots\right]\\ & +r\sqrt{\epsilon}\left[1+x\sqrt{\epsilon}\right]\frac{\partial}{\partial x}\left[Q_{0}+\epsilon^{\nicefrac{1}{2}}Q_{1}+\cdots\right]\\ & -r\epsilon\left[Q_{0}+\epsilon^{\nicefrac{1}{2}}Q_{1}+\cdots\right] \end{aligned} \] \[ \left(1+\alpha\right)^{p}=1+p\cdot\alpha+\dfrac{p(p-1)}{2!}\alpha^{2}+\cdots \] so \[ \left(1+x\epsilon^{\nicefrac{1}{2}}\right)^{2\gamma}=1+2\gamma x\epsilon^{\nicefrac{1}{2}}+\dfrac{(2\gamma)(2\gamma-1)}{2}x^{2}\epsilon+\cdots \] equating term of O(1) i.e $\epsilon^{0}$term: \begin{equation} \dfrac{\partial Q_{0}}{\partial t^{1}}=\dfrac{\sigma^{2}}{2}K^{2\gamma-2}\dfrac{\partial^{2}Q_{0}}{\partial x^{2}}\label{eq:1-1} \end{equation} to be solved subject to \begin{equation} \begin{aligned} & Q_{0}(x,0)=max(-x,0)\\ & \lim_{x\rightarrow\infty}Q_{0}(x,t^{1})=0\\ & \lim_{x\rightarrow-\infty}Q_{0}(x,t^{1})=-\lambda \end{aligned} \end{equation} To solve this, we use a similarity reduction transformation $\begin{aligned}x & =e^{a}x^{\star}\\ t^{\prime} & =e^{b}t^{\prime\star}\\ Q_{0} & =e^{c}Q_{0}^{\star} \end{aligned} $ sub into \[ \dfrac{\partial Q_{0}}{\partial t}=\left(\dfrac{\sigma^{2}K^{2\gamma-2}}{2}\right)\dfrac{\partial^{2}Q_{0}}{\partial x^{2}} \] \[ e^{c-b}\dfrac{\partial Q_{0}^{\star}}{\partial t^{\prime\star}}=\alpha\dfrac{\partial^{2}Q_{o}^{\star}}{\partial x^{\star}}e^{(c-2a)} \] so need $b=2a$ so far $c$ is arbitrary. As $x\rightarrow-\infty\rightarrow x^{\star}\rightarrow-\infty$ need $Q_{0}\rightarrow-x\qquad e^{c}Q_{0}^{\star}\rightarrow-e^{a}x^{\star}$ To be univariant need $a=c$ so if $a=\nicefrac{1}{2},\,c=\nicefrac{1}{2},\,b=1$ $\begin{aligned}x & =e^{\nicefrac{1}{2}}x^{\star}\\ t^{\prime} & =e(t^{\prime})^{\star}\\ Q_{0} & =e^{\nicefrac{1}{2}}Q_{0}^{\star} \end{aligned} $ so $Q_{0}=\sqrt{t^{\prime}}f(\xi)\quad\xi=\dfrac{x}{\sqrt{t^{\prime}}}$ This leads to \[ \sigma^{2}K^{2\gamma-2}\dfrac{d^{2}f}{d\xi^{2}}+\xi\dfrac{df}{d\xi}-f=0 \] Solving subject to BCS gives \[ Q_{0}=\dfrac{\sqrt{t^{\prime}}\cdot\sigma K^{\alpha-1}}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2t^{\prime}\sigma^{2}K^{2\gamma-2}}}-\dfrac{x}{2}erfc\left(\dfrac{x}{\sqrt{2t^{\prime}}\cdot\sigma K^{\gamma-1}}\right) \] Then put $x=\dfrac{S-K}{\sqrt{\epsilon}K}\qquad\tau=\epsilon t^{\prime}$ \[ P=\epsilon^{\nicefrac{1}{2}}KQ_{0}(S,\tau) \] To go to $Q_{1}$ Equate terms of $O(\epsilon^{\nicefrac{1}{2}})$ \[ \dfrac{\partial Q_{1}}{\partial t^{\prime}}=\dfrac{\sigma^{2}K^{2\gamma-2}}{2}\left[\dfrac{\partial^{2}Q_{1}}{\partial x^{2}}+2\gamma x\dfrac{\partial^{2}Q_{0}}{\partial x^{2}}\right]+r\dfrac{\partial Q_{0}}{\partial x} \] \[ \dfrac{\partial Q_{1}}{\partial t^{\prime}}=\dfrac{\sigma^{2}K^{2\gamma-2}}{2}\dfrac{\partial^{2}Q_{1}}{\partial x^{2}}+\sigma^{2}K^{2\gamma-2}\gamma x\dfrac{\partial^{2}Q_{0}}{\partial x^{2}}+r\dfrac{\partial Q_{0}}{\partial x} \] \[ \begin{aligned}\mbox{Subject to}\quad & Q_{1}(x,0)=0\\ & \lim_{x\rightarrow\infty}Q_{1}(x,t^{\prime})=0\\ & \lim_{x\rightarrow\infty}Q_{1}(x,t^{\prime})=-rt^{\prime} \end{aligned} \] PDE is linear, so $Q_{1}=Q_{11}+Q_{12}$ where \[ \dfrac{\partial Q_{11}}{\partial t^{\prime}}=\dfrac{\sigma^{2}K^{2\gamma-2}}{2}\dfrac{\partial^{2}Q_{11}}{\partial x^{2}}+r\dfrac{\partial Q_{0}}{\partial x} \] and $Q_{12}$ satisfies \[ \dfrac{\partial Q_{12}}{\partial t^{\prime}}=\dfrac{\sigma^{2}K^{2\gamma-2}}{2}\dfrac{\partial^{2}Q_{1}}{\partial x^{2}}+x\sigma^{2}K^{2\gamma-2}\gamma\dfrac{\partial^{2}Q_{0}}{\partial x^{2}} \] For ${Q_{11}}$ \begin{equation} \dfrac{\partial Q_{1}}{\partial t^{\prime}}=\dfrac{\sigma^{2}K^{2\gamma-2}}{2}\left[\dfrac{\partial^{2}Q_{1}}{\partial x^{2}}+r\dfrac{\partial Q_{0}}{\partial x}\right]\label{eq:paer2(1)} \end{equation} the hit $v_{t}=\frac{1}{2}v_{xx}+u$\\ let $t^{\prime\prime}=\alpha t^{\prime}\qquad$ what is $\alpha?$ \[ \dfrac{\partial Q_{11}}{\partial t^{\prime}}=\dfrac{\partial Q_{1}}{\partial t^{\prime\prime}}\times\dfrac{\partial t^{\prime\prime}}{\partial t^{\prime}}=\alpha\dfrac{\partial Q_{1}}{\partial t^{\prime\prime}} \] So in ($\ref{eq:paer2(1)}$) \[ \alpha\dfrac{\partial Q_{11}}{\partial t^{\prime\prime}}\times\dfrac{1}{\sigma^{2}K^{2\gamma-2}}=\dfrac{1}{2}\left[\dfrac{\partial^{2}Q_{1}}{\partial x^{2}}\right]+\dfrac{r}{\sigma^{2}K^{2\gamma-2}}\dfrac{\partial Q_{0}}{\partial x} \] pick $\alpha=\sigma^{2}K^{2\gamma-2}$ so coeff of $\dfrac{\partial Q_{1}}{\partial t^{\prime\prime}}=1$ Then \[ \dfrac{\partial Q_{11}}{\partial t^{\prime\prime}}=\dfrac{1}{2}\dfrac{\partial^{2}Q_{1}}{\partial x^{2}}+\dfrac{r}{\sigma^{2}K^{2\gamma-2}}\dfrac{\partial Q_{0}}{\partial x} \] now using hint, solution is \[ Q_{11}=t^{\prime\prime}\times\dfrac{r}{\sigma^{2}K^{2\gamma-2}}\dfrac{\partial Q_{0}}{\partial x} \] gn terms of original variable \[ Q_{11}=\sigma^{2}K^{2\gamma-2}t^{\prime}\times\frac{r}{x}\frac{\partial Q_{0}}{\partial x} \] then a particular solution is $\Rightarrow v=\frac{1}{2}t{u_{xx}}$\\ For $Q_{{12}}$ \[ Q_{12}=r^{2}K^{2\gamma-2}t^{\prime}\times\gamma\dfrac{\partial Q_{0}}{\partial x^{2}} \] the hit $v_{t}=\frac{1}{2}v_{xx}+xu$ \[ t^{\prime\prime}=\alpha t^{\prime}\qquad\alpha=\sigma^{2}K^{2\gamma-2} \] \[ \dfrac{\partial Q_{12}}{\partial t^{\prime}}=\dfrac{\partial Q_{1}}{\partial t^{\prime\prime}}\cdot\dfrac{\partial t^{\prime\prime}}{\partial^{\prime}}=\alpha\dfrac{\partial Q_{1}}{\partial t^{\prime\prime}} \] \[ \alpha\dfrac{\partial Q_{12}}{\partial t^{\prime\prime}}\times\dfrac{1}{\alpha}=\dfrac{1}{2}\dfrac{\partial^{2}Q_{1}}{\partial x^{2}}+\dfrac{\alpha}{\alpha}\gamma K\dfrac{\partial Q_{0}}{\partial x} \] \[ \dfrac{\partial Q_{12}}{\partial t^{\prime\prime}}=\dfrac{1}{2}\dfrac{\partial^{2}Q_{1}}{\partial x^{2}}+X\gamma\dfrac{\partial^{2}Q_{0}}{\partial x} \] \[ v_{t}=\frac{1}{2}u_{xx}+Xu \] \[ \dfrac{\partial Q_{12}}{\partial t^{\prime\prime}}=\dfrac{1}{2}\dfrac{\partial^{2}Q_{1}}{\partial x^{2}}+X\gamma\dfrac{\partial Q_{0}}{\partial x} \] then a particular solution is $\Rightarrow v=Xt^{\prime}u+\frac{1}{2}t^{\prime\prime}u_{x}$\\ ${Q_{0}}$ definition\\ $Q_{0}=\dfrac{\sqrt{t^{\prime}}k^{\gamma-1}\sigma}{\sqrt{2\pi}}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2t^{\prime}\sigma^{2}}}-\dfrac{x}{2}erfc\left(\dfrac{x\cdot k^{1-r}}{\sigma\sqrt{2t^{\prime}}}\right)$ \[ \begin{aligned}erfc(z)=\dfrac{2}{\sqrt{\pi}}\int_{z}^{-\infty}e^{-t^{2}}dt & =1-erf(z)\\ & =1-\dfrac{2}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^{2}}dt \end{aligned} \] \[ erfc\left(\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right)=1-\dfrac{2}{\sqrt{\pi}}\int_{-\infty}^{\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}}e^{-t^{2}}dt \] \[ \begin{aligned}\mbox{let\qquad t} & =\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\\ dt & =\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}dx \end{aligned} \] \[ erfc(t)=1-\dfrac{2}{\sqrt{\pi}}\int e^{-t^{2}}\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}} \] \[ \left[erfc\left[\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right]\right]^{\prime}=\dfrac{-2}{\sqrt{\pi}}\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}e^{-\frac{x^{2}k^{2-2}}{2\sigma t^{\prime}}} \] Find derivative of ${Q_{0}}$\\ $\begin{aligned}\dfrac{dQ_{0}}{dx} & =-2x\dfrac{k^{2-2\gamma}}{2t^{\prime}\sigma^{2}}\dfrac{\sqrt{t^{\prime}}k^{\gamma-1}\sigma}{\sqrt{2\pi}}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2t^{\prime}\sigma^{2}}}-\frac{1}{2}erfc\left(\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right)+\dfrac{x}{2}\dfrac{2}{\pi}\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}}\\ & =\dfrac{-xk^{1-\gamma}}{\sqrt{2\pi t^{\prime}}\sigma}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2t^{\prime}\sigma^{2}}}-\frac{1}{2}erfc\left(\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right)+\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2\pi t^{\prime}}}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}}\\ \dfrac{dQ_{0}}{dx} & =-\dfrac{1}{2}erfc\left(\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right)\\ \dfrac{d^{2}Q_{0}}{dx^{2}} & =-\dfrac{1}{2}\left(\dfrac{-2}{\sqrt{\pi}}\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}}\right)\\ \dfrac{d^{2}Q_{0}}{dx^{2}} & =\dfrac{1}{\sqrt{\pi}}\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}}\\ \dfrac{d^{3}Q_{0}}{dx^{3}} & =\dfrac{1}{\sqrt{\pi}}\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\left(\dfrac{-2xk^{2-2\gamma}}{2\sigma^{2}t^{\prime}}\right)e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}}\\ \dfrac{d^{3}Q_{0}}{dx^{3}} & =\dfrac{-2x}{\sqrt{\pi}}\dfrac{k^{3-3\gamma}}{\sigma^{2}\left(\sqrt{2t^{\prime}}\right)^{3}}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}} \end{aligned} $\\ Then solved ${Q_{11}}$\\ $\begin{aligned}Q_{11} & =t^{\prime}\dfrac{dq_{0}}{dx}\\ Q_{11} & =-\dfrac{1}{2}t^{\prime}r\,erfc\left(\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right) \end{aligned} $\\ Then solved ${Q_{12}}$\\ $\begin{aligned}Q_{12} & =xt^{\prime}r\cdot\dfrac{d^{2}Q_{0}}{dx^{2}}+\dfrac{1}{2}t^{\prime\prime}\dfrac{d^{3}Q_{0}}{dx^{3}}\\ Q_{12} & =xt^{\prime}r\cdot\dfrac{1}{\sqrt{\pi}}\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}}+\dfrac{1}{2}t^{\prime\prime}\dfrac{-2x}{\sqrt{\pi}}\left(\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right)^{3}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}} \end{aligned} $\\ $\begin{aligned}Q_{12}& =\dfrac{x\gamma}{2\sqrt{2\pi}}\sqrt{t^{\prime}}\sigma k^{\gamma-1}e^{-\dfrac{x^{2}}{2t^{\prime}\sigma^{2}k^{2\gamma-2}}} \end{aligned} $\\ Then find $Q_1$\\ $\begin{aligned}Q_{1} & =Q_{11}+Q_{12}\\ Q_{1} & =-\dfrac{t^{\prime}r}{2}erfc\left(\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right)+\dfrac{xt^{\prime}r}{\sqrt{\pi}}\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}}+\dfrac{1}{2}t^{\prime\prime}\left(\dfrac{-2x}{\sqrt{\pi}}\right)\left(\dfrac{k^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right)^{3}e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}} \end{aligned} $ \\ $\begin{aligned}Q_{1}& =\dfrac{x\gamma}{2\sqrt{2\pi}}\sqrt{t^{\prime}}\sigma k^{\gamma-1}e^{-\dfrac{x^{2}}{2t^{\prime}\sigma^{2}k^{2\gamma-2}}} \end{aligned} $ \\ $Q=Q_{1}+\sqrt{\epsilon}Q_{0}$ $Q_{0}=\dfrac{\sqrt{t^{\prime}}k^{\gamma-1}}{\sqrt{2\pi}}\sigma e^{-\dfrac{x^{2}k^{2-2\gamma}}{2\sigma^{2}t^{\prime}}}-\dfrac{x}{2}erfc\left(\dfrac{xk^{1-\gamma}}{\sigma\sqrt{2t^{\prime}}}\right)$ $\begin{aligned}Q_{1}& =\dfrac{x\gamma}{2\sqrt{2\pi}}\sqrt{t^{\prime}}\sigma k^{\gamma-1}e^{-\dfrac{x^{2}}{2t^{\prime}\sigma^{2}k^{2\gamma-2}}} \end{aligned} $ \\ Assumption $ \begin{aligned}x=\dfrac{s-k}{k\sqrt{\epsilon}} \\ \qquad\tau=t^{\prime}\epsilon\\ \qquad t^{\prime\prime}=\gamma^{2}k^{2\gamma-2}\\ x=s-k\mbox{(special case)} & \qquad\mbox{if }\epsilon=\dfrac{1}{k^{2}}\\ \end{aligned} $ \\ $Q=Q_{0}+\sqrt{\epsilon}Q_{1}$\\ $\begin{aligned}P & =\sqrt{\epsilon}kQ\\ & =\sqrt{\epsilon}k\left[Q_{0}+\sqrt{\epsilon}Q_{1}\right]\\ P & =\sqrt{\epsilon}\cdot kQ_{0}+k\epsilon Q_{1}\\ P & =\sqrt{\epsilon}\cdot kQ_{0}+\left[-\dfrac{k\gamma\tau}{2}erfc\left(\dfrac{s-k}{\sigma^{2}\sqrt{\xi\tau}k^{\gamma}}\right)+\dfrac{(s-k)\gamma\sigma}{2\sqrt{2\pi}}\sqrt{\tau}k^{\gamma-1}e^{\frac{-x^{2}}{2t^{\prime}\sigma^{2}k^{2\gamma-2}}}\right] \end{aligned} $ let $Q_{0}=\dfrac{\sqrt{t^{\prime}}}{\sqrt{2\pi}}\sigma k^{\gamma-1}e^{\frac{-x^{2}}{2t^{\prime}\sigma^{2}k^{2\gamma-2}}}-\dfrac{x}{2}erfc\left(\dfrac{x}{\sqrt{2t^{\prime}\sigma}k^{\gamma-1}}\right)$ $Q_{0}=\sqrt{\dfrac{\tau}{\epsilon}}\cdot\dfrac{\sigma}{\sqrt{2\pi}}k^{\gamma-1}e^{\frac{-x^{2}}{2t^{\prime}\sigma^{2}k^{2\gamma-2}}}-\dfrac{x}{2}erfc\left(\dfrac{x}{\sqrt{2t^{\prime}}\sigma k^{\gamma-1}}\right)$ $\begin{aligned}k\sqrt{\epsilon}Q_{0} & =\dfrac{\sqrt{\tau}k^{\gamma}\sigma}{\sqrt{2\pi}}e^{\dfrac{-(s-k)^{2}}{k^{2}\epsilon2\frac{\tau}{\epsilon}\sigma^{2}k^{2\gamma-2}}}-\dfrac{(s-k)^{2}}{2k\sqrt{\epsilon}}erfc\left(\dfrac{s-k}{k\sqrt{\epsilon}\sqrt{\frac{2\tau}{\epsilon}\cdot\sigma k^{\gamma-1}}}\right)\\ & =\dfrac{\sqrt{\tau}k^{\gamma}\sigma}{\sqrt{2\pi}}e^{\dfrac{-(s-k)^{2}}{2\tau\sigma^{2}k^{2\gamma}}}-\dfrac{(s-k)^{2}}{2k\sqrt{\epsilon}}erfc\left(\dfrac{s-k}{\sigma\sqrt{2\tau}k^{\gamma}}\right) \end{aligned} $ sub from (2) into (1) $\begin{aligned}P & =\dfrac{\sqrt{\tau}k^{\gamma}\sigma}{\sqrt{2\pi}}e^{\dfrac{-(s-k)^{2}}{2\tau\sigma^{2}k^{2\gamma}}}-\dfrac{(s-k)^{2}}{2k\sqrt{\epsilon}}erfc\left(\dfrac{s-k}{\sigma\sqrt{2\pi}k^{\gamma}}\right)\\ & +\left[\dfrac{k\gamma\tau}{2}erfc\left(\dfrac{s-k}{\sigma^{2}\sqrt{2\tau}k^{\gamma}}\right)+\dfrac{(s-k)\gamma\sigma\sqrt{\tau}k^{\gamma-1}}{}e^{-\frac{(s-k)^{2}}{2\tau\sigma^{2}k^{2\gamma}}}\right]\\ \end{aligned} $ \\ $\begin{aligned}P & =\dfrac{\sqrt{\tau}k^{\gamma}\sigma}{\sqrt{2\pi}}\left[1+\dfrac{(s-k)}{k}\dfrac{\gamma^{2}}{2}\right]e^{-\frac{(s-k)^{2}}{2\tau\sigma^{2}k^{2\gamma}}}-\left[\dfrac{(s-k)^{2}}{2k\sqrt{\xi}}+\dfrac{k\gamma\tau}{2}\right]erfc\left(\dfrac{s-k}{\sigma\sqrt{2\pi}k^{\gamma}}\right)\\ P & =\dfrac{\sqrt{\tau}k^{\gamma-1}\sigma}{\sqrt{2\pi}}\left[k+\dfrac{(s-k)\gamma}{2}\right]e^{-\frac{(s-k)^{2}}{2\tau\sigma^{2}k^{2\gamma}}}-\left[\dfrac{(s-k)^{2}}{2k\sqrt{\xi}}+\dfrac{k\gamma\tau}{2}\right]erfc\left(\dfrac{s-k}{\sigma\sqrt{2\pi}k^{\gamma}}\right) \end{aligned} \\ $ \end{document}