Use change of variables to reduce a PDE to $\frac{\partial v}{\partial r}=\frac{\partial^2v}{\partial x^2}+(k-1)\frac{\partial v}{\partial x}-kv$

Question

$V(S,t)$ is a smooth function that satisfies the partial differential equation

$\frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2V}{\partial S^2}+rS\frac{\partial V}{\partial S}-rV=0$

subject to $V(S,T)=max(S-K,0),V(0,t)=0,V(S,t)\sim S$ as $S\to\infty$. Here $r$, $\sigma\ge 0,K\ge 0$ are constants.

We aim to reduce the complicated equation by some change of variables: Let

$t=T-\tau/\frac{1}{2}\sigma^2,S=Ke^x,V=Kv(x,r)$

and we aim to show the equation becomes:

$\frac{\partial v}{\partial r}=\frac{\partial^2v}{\partial x^2}+(k-1)\frac{\partial v}{\partial x}-kv$

where $k=r/\frac{1}{2}\sigma^2$ and $v(x,0)=max(e^x-1,0)$

I know it will be messy and hard to type. But I am too stuck on the manipulations. Big thanks for anyone who partiently typing it and answer it for me.

Answer 1

I presume that you recognize that, using the "chain rule", $\frac{\partial V}{\partial t}= \frac{\partial V}{\partial T}\frac{\partial T}{\partial t}$. With $T= t- \frac{2\tau}{\sigma^2}$, $\frac{\partial T}{\partial t} = 1$ so $\frac{\partial V}{\partial t}= \frac{\partial V}{\partial T}$. And with $V= Kv$, $\frac{\partial V}{\partial t}= K\frac{\partial v}{\partial T}$.
Further, $\frac{\partial V}{\partial S}= \frac{\partial V}{\partial x}\frac{\partial x}{\partial S}= \frac{1}{\frac{\partial S}{\partial x}}\frac{\partial V}{\partial x}$. With $S= Ke^x$, $\frac{\partial S}{\partial x}= Ke^x$ so $\frac{\partial V}{\partial S}= \frac{1}{Ke^x}\frac{\partial V}{\partial x}= \frac{1}{e^x}\frac{\partial v}{\partial x}$. Do that again to get $\frac{\partial^2V}{\partial S^2}$.