Exercise: Reducing Black-Scholes to the heat equation

The Black-Scholes PDE

V_t + \tfrac12\sigma^2 S^2 V_{SS} + rSV_S - rV = 0, \quad S > 0, \quad 0 \le t < T

can be reduced to the standard heat equation

u_\tau = \tfrac12\sigma^2 u_{xx}

by a sequence of changes of variables. This was Black, Scholes, and Merton's original path to the closed-form formula.

Tasks

Substitute $\tau = T - t$ (time to expiry). Show the PDE becomes $-V_\tau + \tfrac12\sigma^2 S^2 V_{SS} + rSV_S - rV = 0$ (note the sign flip).
Substitute $x = \ln S$ . Convert the derivatives: $V_S = V_x/S$ , $V_{SS} = (V_{xx} - V_x)/S^2$ . Show that the PDE becomes

V_\tau = \tfrac12\sigma^2 V_{xx} + (r - \tfrac12\sigma^2)V_x - rV.

The first-order term $(r - \tfrac12\sigma^2)V_x$ is an advection term. Use the substitution $V(x, \tau) = e^{ax + b\tau}u(x', \tau)$ with $x' = x + c\tau$ and well-chosen $a, b, c$ to eliminate it.

Specifically: choosing $c = -(r - \tfrac12\sigma^2)$ shifts the drift away; choosing $a = -(r - \tfrac12\sigma^2)/\sigma^2$ may also eliminate it. Pick one consistent set that gives

u_\tau = \tfrac12\sigma^2 u_{xx}.

Transform the terminal condition for a European call: $V(S, T) = (S - K)^+$ becomes $u(x', 0) = ?$ in the new variables.
The heat equation has fundamental solution $G(x, \tau) = (2\pi\sigma^2\tau)^{-1/2}\exp(-x^2/(2\sigma^2\tau))$ . Use it to write $u(x, \tau) = \int G(x - y, \tau)\,u(y, 0)\,dy$ , and back-transform to recover the Black-Scholes formula. (Details optional — outline the integral and identify the Gaussian integral that produces $\Phi(d_1)$ and $\Phi(d_2)$ .)

Hint

For part 3, try $V = e^{\alpha x + \beta\tau}u$ where $\alpha, \beta$ are chosen to cancel the $V_x$ and $V$ terms simultaneously. You'll get two algebraic equations in $\alpha, \beta$ .