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Solving the Black-Scholes PDE via the heat equation

Question

Starting from the Black-Scholes PDE Vt+12σ2S2VSS+rSVS−rV=0V_t+\tfrac12\sigma^2S^2V_{SS}+rSV_S-rV=0 for a European call with strike KK and expiry TT, use a change of variables that turns it into the one-dimensional heat equation, solve that, and transform back to obtain the closed-form call price.

Short answer

C=S N(d1)−Ke−r(T−t)N(d2)C=S\,N(d_1)-Ke^{-r(T-t)}N(d_2), with d1,2=ln⁡(S/K)+(r±σ2/2)(T−t)σT−td_{1,2}=\frac{\ln(S/K)+(r\pm\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}}. Put S=KexS=Ke^x, τ=12σ2(T−t)\tau=\tfrac12\sigma^2(T-t) and V=Keαx+βτu(x,τ)V=Ke^{\alpha x+\beta\tau}u(x,\tau) with α,β\alpha,\beta chosen to kill the first-order and zeroth-order terms; then uτ=uxxu_\tau=u_{xx}. Convolving the transformed payoff with the Gaussian heat kernel and undoing the substitutions yields the two normal-CDF terms.

Adapted from Zhou, A Practical Guide to Quantitative Finance Interviews, sec. 6.1 Option Pricing, 'Black-Scholes formula C', p.145; question and answer rewritten by SelfMadeQuant.com.

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