Transforming Black–Scholes into the Heat Equation

Summary

The document gives intuition for the variable changes used to transform the Black–Scholes partial differential equation into the heat equation. The response breaks the spatial transformation into two steps: first express the underlying price in log-moneyness, then shift that coordinate to remove the drift in the log-price process. Because geometric Brownian motion becomes Brownian motion in log coordinates, this change simplifies the equation.

The time reversal and exponential rescaling of the option value are also identified as parts of the standard reduction, but the answer focuses on the spatial change and does not derive the full transformed equation. It offers a conceptual explanation rather than a worked proof, and points to further discussion for alternative reductions. The question about a Jacobian is not directly addressed; the transformation is a change of variables for a differential equation, not a probability-density transformation.

Key ideas

• Taking the logarithm of the underlying price turns geometric Brownian motion into a process with linear drift in log-price.
• Shifting log-moneyness by the drift removes that drift from the spatial coordinate.
• Time reversal and rescaling the option value help reduce the Black–Scholes equation to the heat equation.
• The response gives intuition for the coordinate changes but does not derive the full transformation or directly answer the Jacobian question.

Tags

# Intuition behind the change of variable of BS into Heat Equation


# Intuition behind the change of variable of BS into Heat Equation












$$\tau=T-t$$ $$u=Ce^{r\tau}$$ $$x=\ln(\frac{S}{K})+(r-\frac{1}{\sigma^2})\tau$$ The first transform is reverse in time. The second is reversely discounted call price. But I find it hard to see the intuition behind the third transformation. And I am also wondering why this is not a one-to-one transform of time, call price, and price of the underlying. Why is 'multiplying by a Jacobian' doesn't work in this case?

## Answer by Mark Joshi (score 2)

https://quant.stackexchange.com/a/33510

the third one is really two transformations:

$$ y = \log (S/K) $$ $$ x= y + (r-0.5\sigma^2)\tau $$

Move to log coordinates -- the stock followed geometric Brownian motion so it's log follows Brownian motion. So the equation should be simpler in log coordinates.

The second one is remove the drift. The log has drift $r-0.5\sigma^2$ so removing that changes to coords in which the log stock is driftless.

(there are actually two ways to do the reduction to the heat equation, see Concepts and Practice etc by me)

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