PDE change of variables Black-Scholes equation

perishingtardi
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Homework Statement


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.

Homework Equations


Chain rule.

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.
 
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perishingtardi said:

Homework Statement


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.

Homework Equations


Chain rule.


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.

Use
<br /> \frac{\partial^2 V}{\partial x^2} = S\frac{\partial }{\partial S}\left(S\frac{\partial V}{\partial S}\right) <br /> = 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.
 
Sorry to resurrect a very old thread...but can anyone provide more guidance on this? I'm also stuck on this question. Any help would be greatly appreciated!

I see how the second derivative above helps...but I'm not sure whether to start with V or u. How do I translate between the two?

Also can you help with the partial derivatives of V wrt S and t?
 
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