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.

  • 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


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:

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

  • $\begingroup$ 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, 2022 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, 2022 at 0:18

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