# PDE change of variables Black-Scholes equation

1. Oct 11, 2013

### perishingtardi

1. The problem statement, all variables and given/known data
By changing variables from $(S,t,V)$ to $(x,\tau,u)$ where
$$\tau = T - t,$$
$$x = \ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T-t),$$
$$u=e^{r\tau}V,$$
where $r, \sigma, \tau, K$ are constants, show that the Black-Scholes equation
$$\frac{\partial V}{\partial t} + \frac{\sigma^2}{2}S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0$$
reduces to the diffusion equation
$$\frac{\partial u}{\partial \tau} - \frac{\sigma^2}{2}\frac{\partial^2 u}{\partial x^2}=0.$$

2. Relevant equations
Chain rule.

3. The attempt at a solution
I know that $$\frac{\partial}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial}{\partial x} + \frac{\partial \tau}{\partial t}\frac{\partial}{\partial \tau}$$ and similarly for $\partial / \partial S$. I don't know what to do with the second-order derivative though. Since $\partial / \partial S$ turns out to be $\frac{1}{S}\frac{\partial}{\partial x}$, I reckoned that $$\frac{\partial^2 V}{\partial S^2} = \frac{1}{S}\frac{\partial}{\partial x} \left( \frac 1 S \frac{\partial V}{\partial x} \right),$$ but that seems to just make things more complicated.

2. Oct 11, 2013

### pasmith

Use
$$\frac{\partial^2 V}{\partial x^2} = S\frac{\partial }{\partial S}\left(S\frac{\partial V}{\partial S}\right) = S\frac{\partial V}{\partial S} + S^2 \frac{\partial^2 V}{\partial S^2}$$
to eliminate $S^2 \frac{\partial^2 V}{\partial S^2}$ to leave you with terms in $\partial V/\partial t$, $S\partial V/\partial S$, $V$ and ${\partial^2 V}/{\partial x^2}$. Then you can tidy up the first-order derivatives.