- #1

perishingtardi

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- 1

## Homework Statement

By changing variables from [itex](S,t,V)[/itex] to [itex](x,\tau,u)[/itex] where

[tex]\tau = T - t,[/tex]

[tex]x = \ln\left(\frac{S}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)(T-t),[/tex]

[tex]u=e^{r\tau}V,[/tex]

where [itex]r, \sigma, \tau, K[/itex] are constants, show that the Black-Scholes equation

[tex]\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[/tex]

reduces to the diffusion equation

[tex]\frac{\partial u}{\partial \tau} - \frac{\sigma^2}{2}\frac{\partial^2 u}{\partial x^2}=0.[/tex]

## Homework Equations

Chain rule.

## The Attempt at a Solution

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