I'm currently working to solve the Black-Scholes model partial differential equation (it's a model for a.o. stock option prices). The Black-Scholes equation for a calloption C(S,t) is given by
$ \frac{∂C}{∂t}+\frac{1}{2} σ^2 S^2 \frac{∂^2 C}{∂S^2}+rS \frac{∂C}{∂S}-rC=0$
S is the stock price and t is time to expiration.
I am solving the equation by transforming it to the heat equation. I am having a little trouble with the first few transformation. The transformations are:
- $x=ln(\frac{S}{K})$ which gives $S=Ke^x$
- $τ=\frac{σ^2}{2} (T-t)$ which gives $t=T-\frac{2τ}{σ^2}$
The function becomes:
- $U(x,τ)= \frac{1}{K} C(S,t)=\frac{1}{K} C(Ke^x,T- \frac{2τ}{σ^2})$
These transformation applied to the partial differential equation above gives the following outcomes for the different terms. I found the following solution on the internet, my problem is that I don't really understand what they do here:
$\frac{∂C}{∂t}=K \frac{∂U}{∂τ} \frac{∂τ}{∂t}=\frac{-Kσ^2}{2} \frac{∂U}{∂τ},$
$\frac{∂C}{∂S}=K \frac{∂U}{∂x} \frac{∂x}{∂S}=\frac{K}{S} \frac{∂U}{∂x}=e^{-x} \frac{∂U}{∂x},$
$\frac{∂^2 C}{∂S^2}=\frac{-K}{S^2} \frac{∂U}{x}+\frac{K}{S} \frac{∂}{∂S} (\frac{∂U}{∂x})$ $=\frac{-K}{S^2} \frac{∂U}{∂x}+\frac{K}{S} \frac{∂}{∂x} (\frac{∂U}{∂x}) \frac{∂x}{∂S}$ $=\frac{-K}{S^2} \frac{∂U}{∂x}+\frac{K}{S^2} \frac{∂^2 U}{∂x^2}$ $=\frac{e^{-2x}}{K} (\frac{∂^2 U}{∂x^2} \frac{-∂U}{∂x}$)
Can someone help me out? I would really appreciate it!!!!
thanks in advance
edit: I now understand the first two but the third is a lot harder.