Change of variables in Black and Scholes equation

Question

Consider the Black and Scholes equation $$V_t - \frac{1}{2}x^2 \sigma^2 V_{xx} - \rho xV_x + \rho V = 0 \quad \text{ in } (0,\infty)\times \mathbb{R},$$ with initial condition $$V(0,x) = V_0(x).$$

What changes of variables reduce this equation to $V_t - \frac{1}{2}V_{xx} = 0$? How does the initial condition change?

Answer 1

The Black-Scholes PDE is $V_t - \frac{1}{2}x^2 \sigma^2 V_{xx} - \rho xV_x + \rho V = 0 $.

Taking $U(t,y) = e^{\rho t}V(t,e^y)$ we have

$$U_t - \frac{\sigma^2}{2}U_{yy} - (\rho - \sigma^2/2)U_y = 0.$$

Next, taking $W(t,z) = U(t, \sigma z - (\rho - \sigma^2/2)t)$ we have

$$W_t - \frac{1}{2} W_{zz} = 0.$$

Putting it together, we have the transformation

$$W(t,z) = e^{\rho t}V\left(\, t,\, \exp[\, \sigma z - (\rho - \sigma^2/2)t)\, ]\, \right),$$

and the initial condition is

$$W(0,z) = V(0,e^{\sigma z})= V_0(e^{\sigma z}).$$