Black-Scholes can be transformed to the heat equation (and vice versa).
I'm trying to do something with Green's functions on a half-space but I'm missing something which is probably obvious.
[$]u_{t}=ku_{xx}[$]
If we consider a problem on the half-space [$]x>0[$] and [$]t>0[$] with [$]u(x,0)=0[$] for [$]x>0[$] and [$]u(0,t)=f(t)[$] for [$]t>0[$], there's a Green's function solution
[$]u(x,t)=k\int_{0}^{t}\frac{x}{2\sqrt{\pi}}\frac{1}{(k(t-s))^{3/2}}\exp\left[-\frac{x^{2}}{4k(t-s)}\right]f(s)ds[$]
my question: when [$]x=0[$], pretty obviously we need this formula to reduce to [$]u(0,t)=f(t)[$] so somehow the Green's function reduces to a delta function in that limit. Anyone able to fill in the gaps in that statement?
I'm also going to need [$]u_{x}(0,t)[$], which I presume you can get by differentiating the formula for [$]u[$] above and then setting [$]x=0[$], but again there wlll be delta functions, so can someone point the way?