# Question About Converting Black Scholes Differential Equation to Heat Equation

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

$$x=y+(r-\frac{1}{2}\sigma^2)\tau_t=lnS+(r-\frac{1}{2}\sigma^2)\tau_t$$

and

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

and

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

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?

• I hope this is a typo error, but $\partial^2V / \partial S^2$ does not equal to $\partial V/\partial S (\partial V/\partial S)$ Jan 3 '20 at 11:40
• @JónásBalázs, thanks a lot! I think it is a typo from the book and I corrected it in my question just now. Jan 3 '20 at 15:38

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.

• Thanks a lot for showing the process in such a clear way, really appreciate it! Jan 3 '20 at 15:45