Transforming the Black–Scholes Equation into the Heat Equation

Summary

The document explains a time-variable substitution used to rewrite a zero-rate Black–Scholes form associated with the Bachelier model as a heat equation. It defines a new variable proportional to the remaining time to maturity, differentiates that variable with respect to calendar time, and applies the chain rule to replace the time derivative. Substitution into the original partial differential equation yields a diffusion equation in the new time variable and the underlying price coordinate.

The derivation illustrates how reversing time and rescaling it can clarify the connection between option-pricing equations and heat diffusion. It does not explain the role of the symbol eta mentioned in the question, specify boundary or terminal conditions, or show a verification beyond the algebraic substitution. Its scope is therefore a narrow PDE transformation; it does not derive a pricing formula or discuss model assumptions such as volatility behavior.

Key ideas

• A remaining-time variable can convert a terminal-value equation into a forward-time form.
• The chain rule relates calendar-time differentiation to differentiation in the transformed variable.
• Substituting the transformed derivative produces a heat-type diffusion equation.
• The derivation depends on the stated zero-rate equation and does not address boundary conditions.
• The response does not clarify the role of eta raised in the question.

Tags

# Black Scholes to Heat Equation


# Black Scholes to Heat Equation












Equation (2) was derived by setting r=0 in the Black-Scholes equation for the Bachelier model (1).

Can someone please help me understand all the steps for how we get from the heat equation under time reversal (2) to (3) and then show me how to verify that the equation still holds? I cannot understand what exactly using $\eta$ achieves. Thanks!

## Answer by ryc (score 3)

https://quant.stackexchange.com/a/57563

Let $n=\sigma^2(T-t)$

$$dn=-\sigma^2dt$$

$\frac{\partial {V}}{\partial {t}} = \frac{\partial {V}}{\partial {n}}\frac{d {n}}{d {t}}$

$\frac{\partial {V}}{\partial {t}} = -\sigma^2\frac{\partial {V}}{\partial {n}}$

since $$\frac{\partial {V}}{\partial {t}} = -\frac{1}{2}\sigma^2\frac{\partial^2 {V}}{\partial {S^2}} $$

so

$$-\sigma^2\frac{\partial {V}}{\partial {n}} = -\frac{1}{2}\sigma^2\frac{\partial^2 {V}}{\partial {S^2}} $$

$$\frac{\partial {V}}{\partial {n}} -\frac{1}{2}\frac{\partial^2 {V}}{\partial {S^2}} =0$$

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