3
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ 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, 2018 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, 2018 at 11:41
  • $\begingroup$ This is just chain rule $\endgroup$
    – Ezy
    Dec 31, 2018 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, 2018 at 12:49
  • 1
    $\begingroup$ 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, 2018 at 13:37

1 Answer 1

3
$\begingroup$

The starting formulation of the Black-Scholes equation as found in the OP question:

$$ \frac{\partial U}{\partial \tau} = \frac{1}{2} \sigma^2 \frac{\partial^2 U}{\partial \xi^2} + \left(r - \frac{1}{2} \sigma^2 \right) \frac{\partial U}{\partial \xi} $$

This will be proven to be equivalent to the heat equation (the parabolic PDE) after a change of coordinates $(\xi, \tau) \rightarrow (x, \tau)$ defined as:

$$ \begin{align} x &= \xi + \left( r - \frac{1}{2} \sigma^2 \right) \tau\\ \tau &= \tau \end{align} $$

Use of the chain rule clarifies how first derivatives change when passing from a set of coordinates to the other:

$$ \begin{align} \frac{\partial}{\partial \xi (x, \tau)} (*) &= \overbrace{\frac{\partial x}{\partial \xi}}^{= 1} \frac{\partial}{\partial x} (*) + \overbrace{\frac{\partial \tau}{\partial \xi}}^{= 0} \frac{\partial}{\partial \tau} (*)\\ \frac{\partial}{\partial \tau (x, \tau)} (*) &= \underbrace{\frac{\partial x}{\partial \tau}}_{= r - \frac{1}{2} \sigma^2} \frac{\partial}{\partial x} (*) + \underbrace{\frac{\partial \tau}{\partial \tau}}_{= 1} \frac{\partial}{\partial \tau} (*) \end{align} $$

The second order derivative $\frac{\partial^2}{\partial \xi^2 (x, \tau)}$ needs also to be evaluated. Seen from above that $\frac{\partial}{\partial \xi (x, \tau)} = \frac{\partial}{\partial x}$, this is easily:

$$ \frac{\partial^2}{\partial \xi^2 (x, \tau)} (*) = \frac{\partial}{\partial \xi} \left( \frac{\partial}{\partial \xi} (*) \right) = \frac{\partial^2}{\partial x^2} (*) $$

Applying the above reformulations of $\frac{\partial}{\partial \xi (x, \tau)}$, $\frac{\partial}{\partial \tau (x, \tau)}$ and $\frac{\partial^2}{\partial \xi^2 (x, \tau)}$ to the Black-Scholes equation eliminates the first order derivative term and yields the classic heat equation:

$$ \begin{align} \require{cancel}\cancel{\left( r - \frac{1}{2} \sigma^2 \right) \frac{\partial U}{\partial x}} + \frac{\partial U}{\partial \tau} &= \frac{1}{2} \sigma^2 \frac{\partial^2 U}{\partial x^2} + \cancel{\left(r - \frac{1}{2} \sigma^2 \right) \frac{\partial U}{\partial x}} \qquad \qquad \Longrightarrow\\ \Longrightarrow \qquad \qquad \frac{\partial U}{\partial \tau} &= \frac{1}{2} \sigma^2 \frac{\partial^2 U}{\partial x^2} \end{align} $$

$\endgroup$
2
  • $\begingroup$ Why in the partial derivative $\frac{\partial}{\partial \tau(x,\tau)}(*)$ we have $\tau$ as a function of $x$ and $\tau$ instead of $x$ and $\xi$, i.e.: $\frac{\partial}{\partial \tau(x,\xi)}$? $\endgroup$ Feb 14 at 21:06
  • 1
    $\begingroup$ @userPrimeNumber What is of interest here is the change of coordinates $(\xi, \tau) \rightarrow (x, \tau)$, so $\tau$ has to be expressed in the plane $(x, \tau)$. $x$ is the new coordinate with the same meaning of $\xi$: the plane $(x, \xi)$ in this context is meaningless. $\endgroup$
    – Giogre
    Feb 15 at 0:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.