Question About Converting Black Scholes Differential Equation to Heat Equation

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I'm reading a book about converting Black Scholes equation to heat equation and I highlighted in bold for my doubt, 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 $\frac{\partial^2 V}{\partial S^2}=\frac{\partial }{\partial S}(\frac{\partial V}{\partial S})=\frac{\partial }{\partial S}(\frac{1}{S}\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S}\frac{\partial }{\partial S}(\frac{\partial V}{\partial y})=-\frac{1}{S^2}\frac{\partial V}{\partial y}+\frac{1}{S^2}\frac{\partial^2 V}{\partial y^2}$,

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 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?