# Equivalent form of Black-Scholes Equation (to transform to heat equation)

I am trying to understand the transformation of the Black-Scholes equation to the one-dimensional heat equation from Joshi, M. (2011). The Concepts and practice of mathematical finance. 2nd ed. Cambridge, U.K.: Cambridge University Press, pp.119. . The author stated that the Black-Scholes equation:

$\frac{\partial C}{\partial t}(S,t) +rS \frac{\partial C}{\partial S}(S,t) + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 C}{\partial S^2}(S,t) -rC=0$

can be rewritten as:

$\frac{\partial C}{\partial t}(S,t) +(r-\frac{1}{2}\sigma^2) S \frac{\partial C}{\partial S}(S,t) + \frac{1}{2}\sigma^2 (S \frac{\partial}{\partial S})^2 C -rC=0$,

however I can't see how these two equations are equivalent, in particular, I don't understand how the additional $-\frac{1}{2} \sigma^2 S \frac{\partial C}{\partial S} (S,t)$ came about. I have tried expanding the squared differential operator as follows:

\begin{align*} & \frac{1}{2}\sigma^2(S \frac{\partial}{\partial S})^2 C(S,t) \\ &= \frac{1}{2} \sigma^2 \frac{\partial}{\partial S} [\frac{\partial}{\partial S} (S^2 C(S,t))] \\ &= \frac{1}{2}\sigma^2 \frac{\partial}{\partial S} [2S\cdot C(S,t) + S^2 \frac{\partial C(S,t)}{\partial S}] \\ &= \frac{1}{2}\sigma^2 [2C(S,t) + 2S \frac{\partial C(S,t)}{\partial S} + 2S \frac{\partial C(S,t)}{\partial S} + S^2 \frac{\partial^2 C(S,t)}{\partial S^2} \\ &= \frac{1}{2} \sigma^2 [2C(S,t) + 4S \frac{\partial C(S,t)}{\partial S}] + \frac{1}{2} \sigma^2 S^2 \frac{\partial ^2 C(S,t)}{\partial S^2} \end{align*}

Is there something that I missed out? Can anyone help me out?

Thanks.

• A warm welcome to Quant.SE and thank you for your question Commented Jan 8, 2017 at 9:35

I think you've cottoned on to the main question - that is,to see why $\frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{ \partial S^2 }$ is the same thing as $-\frac{1}{2} \sigma^2 S \frac{\partial C}{\partial S} + \frac{1}{2} \sigma^2 (S \frac{\partial}{\partial S})^2 C$. I think your confusion comes from dealing with the squared operator.

Writing down what the expression means, we have

$$(S \frac{\partial}{\partial S})^2 C= (S \frac{\partial}{\partial S}) (S \frac{\partial}{\partial S}) C$$

That is, apply $S \frac{\partial}{ \partial S}$ to $C$ first, then apply $S \frac{\partial}{\partial S}$ to the result. Note that we can't just switch the order of these symbols - this is not the same thing as $\frac{\partial}{\partial S} \frac{\partial}{\partial S} S^2$. (The derivative operator rarely commutes with anything - try evaluating this on a hypothetical option with constant value $C=1$) Expanding out the expression should give

$$(S \frac{\partial}{\partial S})^2 C = S \frac{\partial}{\partial S} (S \frac{\partial C}{\partial S})$$

and applying the product rule gives that this is equal to

$$S (\frac{\partial S}{\partial S} \frac{\partial C}{\partial S} + S \frac{\partial ^2 C}{\partial S ^2})$$

(Incidentally, we can see that it's exactly the product rule that makes the derivative fail to commute with multiplication by $S$). So

$$(S \frac{\partial}{\partial S})^2 C= S \frac{\partial C}{\partial S} + S^2 \frac{\partial ^2 C}{\partial S ^2}$$

Multiply this by $\frac{1}{2} \sigma^2$ and rearrange a little and we should have our result.