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Transforming the Black–Scholes PDE into Dimensionless Form

Article Quant Q&A · Author: userPrimeNumber
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Summary

The document shows how a change of variables converts the Black–Scholes partial differential equation into a dimensionless form. It defines log price relative to the strike, rescales time using volatility, and expresses option value relative to the strike. The key technique is applying the chain rule to rewrite derivatives with respect to the original price and time variables in terms of derivatives with respect to the transformed variables.

For price, the derivative operator becomes the transformed-price derivative divided by the original price; for time, the derivative operator gains a negative scale factor from the time substitution. Substituting these operators and the rescaled option value into the original equation yields the transformed PDE stated in the question. The answer supplies the derivative relationships but does not work through every substitution or discuss boundary conditions, so it is a concise starting point rather than a full derivation.

Key ideas

  • A logarithmic price variable can simplify the price dependence in the Black–Scholes equation.
  • Rescaling time by volatility converts the equation to a dimensionless form.
  • Use the chain rule to express derivatives in transformed variables before substituting into the PDE.
  • The time transformation introduces a negative scale factor, while the price derivative is divided by the original price.

Tags

OptionsDerivatives pricingStatistics
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# Black-Scholes PDE transformation


# 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?

## Answer by KT8 (score 1, accepted)

https://quant.stackexchange.com/a/70400

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.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.