# Black-Scholes PDE transformation

From "Mathematics of Financial Derivatives" by Wilmott, Howison and Dewynne, section 5.4, p76. How do I start making the transformations to get to the dimensionless equation? I.e. we start with the standard Black-Scholes PDE:

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

and the following transformations are applied: $$S=Ee^x, t = T - \tau/\frac{1}{2}\sigma^2, V=Ev(x,\tau)$$ to obtain: $${\frac {\partial v}{\partial \tau}} = {\frac {\partial ^{2}v}{\partial x^{2}}} +(k-1){\frac {\partial v}{\partial x}}-kv$$ where $$k = r/\frac{1}{2}\sigma^2$$.

How do I start with these transformations? Let's say I take the $$rS{\frac {\partial V}{\partial S}}$$ term and substitute for $$S$$ and $$V$$: $$rS{\frac {\partial V}{\partial S}} = r \times Ee^x \times\frac {\partial Ev(x,\tau)}{\partial Ee^x }$$ How can I proceed from here?

I think it simplifies by noting that

$$\dfrac{\partial}{\partial S} = \dfrac{\partial}{\partial (E e^x)} = \dfrac{\partial x}{\partial (E e^x)}\dfrac{\partial}{\partial x} = \left(\dfrac{\partial (E e^x)}{\partial x}\right)^{-1}\dfrac{\partial}{\partial x} = \dfrac{1}{E e^x}\dfrac{\partial}{\partial x} ,$$

and

$$\dfrac{\partial}{\partial t} = \dfrac{\partial \tau}{\partial t}\dfrac{\partial}{\partial \tau} = -\dfrac{2}{\sigma^2}\dfrac{\partial}{\partial \tau} .$$

Using those expressions, just substitute $$V$$ by the new variable and that should do it.

• Thanks, works really well. Does this technique have a name at all? Commented Apr 5, 2022 at 21:24
• You can find it at the third line here: en.wikipedia.org/wiki/Chain_rule Rule
– KT8
Commented Apr 6, 2022 at 6:58