# Black-Scholes to Diffusion Initial Condition

I'm having troubles with the transformation from the Black-Scholes PDE and transforming it to the diffusion equation. I read this other stackexchange post (Here) and I understand most of the process, except where they changed the initial condition.

I got $$$$\begin{split} u(x,0) &= e^{r\tau}C(S,T)\\ &=e^{r\tau}\text{max}(S-K,0)\\ &=e^{r\tau}\text{max}(e^y-K,0)\\ &=e^{r\tau}\text{max}(e^{x-(r-\sigma^2/2)\tau)}-K,0)\\ &=\text{max}(e^{x+\sigma^2\tau/2}-e^{r\tau}K,0)\\ \end{split}$$$$

Which is different from their equation of: $$$$u(x,0) = u_0(x) = \text{max}(e^{\frac{1}{2}(a+1)x}-e^{\frac{1}{2}(a-1)x},0)$$$$

Where $$a=2r/\sigma^2$$

I would comment on the other post, however I don't have enough 'reputation' and this is a very specific question that I can't find elsewhere. Apparently it's in the textbook referenced in the original post, but the particular page referenced isn't freely available.

• In Dewynne's et al. derivation, there is an extra step (involving going from "v" to "u") that is not shown in the stackexchange post you linked... Commented Dec 19, 2018 at 21:41

$$u(x,0)=e^{\frac{k-1}{2}x}v(x,0)$$
and $$v(x,0)=max(e^{x}-1,0)$$ Hence $$u(x,0)=max(e^{\frac{k+1}{2}x}-e^{\frac{k-1}{2}x},0)$$