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Sorry as really basic question. Chapter 8 of Wilmott introduces Q Finance the BS equation is transformed into the heat equation. Firstly by using $ V(S,t) \rightarrow \mathrm{e}^{-r(T - t)}U(S,t) $ and then $ \tau = T - t $

Resulting in: $$ \frac{\partial U}{\partial \tau} = \frac{1}{2}\sigma^2\frac{\partial^2U}{\partial \xi^2} + (r - \frac{1}{2}\sigma^2)\frac{\partial U}{\partial \xi} $$

The final change of variables used is $ x = \xi + (r - \frac{1}{2}\sigma^2)\tau $ which results in the heat equation in terms of $ x $ and $ \tau $.

Could someone please tell me how exactly this variable change reduces the above equation to the heat equation? I seem to be getting $ \frac{1}{2}\sigma^2\frac{\partial^2 U}{\partial x^2} = 0 $. After applying the chain rule for each of the terms.

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    $\begingroup$ You are missing one change of variable at least. You need to take the log of S to make the equation with constant coefficients $\endgroup$ – Ezy Dec 31 '18 at 11:22
  • $\begingroup$ Sorry I should have said $ \xi = log(S) $ so that part has been done. It's just the last step. $\endgroup$ – bjp93 Dec 31 '18 at 11:41
  • $\begingroup$ This is just chain rule $\endgroup$ – Ezy Dec 31 '18 at 12:01
  • $\begingroup$ Yeah think I've not had enough sleep haha. But wouldn't $ \frac{\partial U}{\partial \tau} = \frac{\partial U}{\partial x} \frac{\partial x}{\partial \tau} = (r - \frac{\sigma^2}{2}) \frac{\partial U}{\partial x}$ ?? Which then gives the second derivative of x as being equal to zero? $\endgroup$ – bjp93 Dec 31 '18 at 12:49
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    $\begingroup$ you mixed yourself up. Just write it as a new function $V(\tau,x):=U(\tau,x-(...)\tau)$ and take the $\tau$ derivative on both sides to get the heat equation for $V$ $\endgroup$ – Ezy Dec 31 '18 at 13:37

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