# Solving Black-Scholes PDE using Laplace transform

I'm trying to obtain the Laplace transform of Call option price with repect to time to maturity under the CEV process.

The well known Black scholes PDE is given by $$\frac{1}{2}\sigma(x)^2x^2\frac{\partial^2}{\partial x^2}C(x,\tau)+\mu x\frac{\partial}{\partial x}C(x,\tau)-rC(x,\tau)-\frac{\partial}{\partial \tau}C(x,\tau)=0.$$ where the initial condition $C(x,0)=max(x-K,0)$ and $\sigma(x)=\delta x^\beta$.

Taking the Laplace transform with respect to $\tau$, we obtain the following ODE: $$\frac{1}{2}\delta x^{2\beta+2}\frac{\partial^2}{\partial x^2}\hat{C}(x,\lambda)+\mu x\frac{\partial}{\partial x}\hat{C}(x,\lambda)-(\lambda+r)\hat{C}(x,\lambda)=-max(x-K,0).$$ where $\hat{C}(x,\lambda)=\int_0^\infty e^{-\lambda \tau}C(x,\tau)d\tau$

and the initial condition is transformed to $$\hat{C}(x,\lambda)=\int_0^\infty e^{-\lambda \tau}C(x,0) d\tau=max(x-K,0)/\lambda$$(is this right??? it seems wrong..)

Then, $\hat{C}(x,\lambda)$ can be analytically formulated by the case $x>K$ and $x\leq K$.

How to get explicit formula for $\hat{C}(x,\tau)$? I can't proceed from this stage.

I know one paper, "(2001 Dmitry) Pricing and Hedging Path-Dependent Options under the CEV", related to this question. However, there's big jumps for me to understand readily. Could you explain it step by step?

• It's perfectly legitimate to use the Laplace transform (and vonjd's linked paper does a fine job), but I've personally always preferred to solve the PDE by changing variables until the PDE turns into the standard diffusion equation. – Brian B Sep 18 '14 at 15:39
• Welcome to Quant Stackexchange :-) Thank you for your interesting first question. If the answer was helpful you could upvote and accept it :-) – vonjd Sep 21 '14 at 6:38