I'm reading a book about converting Black Scholes equation to heat equation and I highlighted in bold for those I have doubts, and really appreciate your advice on it.

Let $S$,$T$,$V$ denote underlying asset price, maturity and option price separately. Here is the convert process:

Let $y=lnS$ since $(S=e^y)$ and $\tau_t=T-t$,then $\frac{\partial V}{\partial t}=-\frac{\partial V}{\partial \tau_t}$,$\frac{\partial V}{\partial S}=\frac{\partial V}{\partial y}\frac{\partial y}{\partial S}=\frac{1}{S}\frac{\partial V}{\partial y}$ and $\frac{\partial^2 V}{\partial S^2}=\frac{\partial }{\partial S}(\frac{\partial V}{\partial S})=\frac{\partial }{\partial S}(\frac{1}{S}\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S}\frac{\partial }{\partial S}(\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S^2}\frac{\partial^2 V}{\partial y^2}$,

here is my first doubt: why $\frac{1}{S}\frac{\partial V}{\partial S}(\frac{\partial V}{\partial y})=\frac{1}{S^2}\frac{\partial^2 V}{\partial y^2}$ holds?

The Black Scholes equation $\frac{\partial V}{\partial t} + rS \frac{\partial V}{\partial S} + \frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}-rV = 0$

can be converted to

$-\frac{\partial V}{\partial \tau_t} + (r-\frac{1}{2}\sigma^2) \frac{\partial V}{\partial y} + \frac{1}{2}\sigma^2\frac{\partial^2 V}{\partial y^2}-rV = 0$

Let $u=e^{r\tau_t}V$,

the equation becomes

$-\frac{\partial u}{\partial \tau_t} + (r-\frac{1}{2}\sigma^2) \frac{\partial u}{\partial y} + \frac{1}{2}\sigma^2\frac{\partial^2 u}{\partial y^2} = 0$

Finally, let



$\tau=\tau_t$, then $\frac{\partial u}{\partial y}=\frac{\partial u}{\partial x}$


$\frac{\partial u}{\partial \tau_t}=\frac{\partial u}{\partial \tau}+(r-\frac{1}{2}\sigma^2)\frac{\partial u}{\partial x}$,

here is my second doubt: why $\frac{\partial u}{\partial y}=\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial \tau_t}=\frac{\partial u}{\partial \tau}+(r-\frac{1}{2}\sigma^2)\frac{\partial u}{\partial x}$ hold?

  • 1
    $\begingroup$ I hope this is a typo error, but $\partial^2V / \partial S^2$ does not equal to $\partial V/\partial S (\partial V/\partial S)$ $\endgroup$ Jan 3, 2020 at 11:40
  • $\begingroup$ @JónásBalázs, thanks a lot! I think it is a typo from the book and I corrected it in my question just now. $\endgroup$
    – M00000001
    Jan 3, 2020 at 15:38

1 Answer 1


The first part of your question:

  • $\frac{\partial y}{\partial S} = \frac{\partial ln S}{\partial S} = \frac{1}{S}$

  • $ \frac{\partial^2 V}{\partial S \partial y} = \frac{\partial}{\partial y} \frac{\partial V}{\partial S} = \frac{\partial}{\partial y} (\frac{\partial y}{\partial S}\frac{\partial V}{\partial y})= \frac{\partial}{\partial y} (\frac{1}{ S}\frac{\partial V}{\partial y}) = \frac{-1}{S^2} \frac{\partial S}{\partial y}\frac{\partial V}{\partial y} + \frac{1}{S}\frac{\partial^2 V}{\partial y^2} = \frac{-1}{S}\frac{\partial V}{\partial y} + \frac{1}{S}\frac{\partial^2 V}{\partial y^2} $

  • $ \frac{\partial^2 V}{\partial S^2} = \frac{\partial}{\partial S} (\frac{\partial V}{\partial y}\frac{\partial y}{\partial S}) = \\ \frac{\partial^2 V}{\partial S \partial y} \frac{\partial y}{\partial S} + \frac{\partial V}{\partial y} \frac{\partial^2 y}{\partial S^2} = \\ \frac{\partial^2 V}{\partial S \partial y} \frac{1}{S} - \frac{1}{S^2}\frac{\partial V}{\partial y} = \\ \frac{-1}{S^2}\frac{\partial V}{\partial y} + \frac{1}{S^2}\frac{\partial^2 V}{\partial y^2} -\frac{1}{S^2}\frac{\partial V}{\partial y} = \\ \frac{-2}{S^2}\frac{\partial V}{\partial y} + \frac{1}{S^2}\frac{\partial^2 V}{\partial y^2} $

The key for the second part is that $\frac{\partial x}{\partial y}$ is 1.

  • $\begingroup$ Thanks a lot for showing the process in such a clear way, really appreciate it! $\endgroup$
    – M00000001
    Jan 3, 2020 at 15:45

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