- #1

binbagsss

- 1,261

- 11

## Homework Statement

2. Homework Equations

3. The Attempt at a Solution [/B]- So with the (from what i interpret of the notes this is needed) the same boundary conditions when time is fixed, we can relate the 'fundamental problem'- the initial condition ##t=0## given by a delta function##c_f(c,t=0)=\delta(x)##, to a more general problem with any initial condition here, for a linear operator ##L## via:

##\int c_f (x-s)g(s) ds = c(x) ## (*)where ##c_f(x)## is the solution to the fundamental problem and ##g(x)=c(x,t=0)##

part a is fine but I will give my working if needed - basically just Fourier transform to turn a PDE into an ODE and Fourier transform the intial conditions also.

part b seems to be going very wrong:

- I think I am going wrong with the following conclusion from the quoted result:

Since ##\int cos(nx) sin (mx) =0 ## for any n, m, the non zero contribiution from the quoted result:

## F[cos(ax)](k)=\sqrt{\frac{\pi}{2}} (\delta(k-a)+\delta(k+a))##

##= \int (cos(-kx)-i sin (kx)) cos (ax) dx= \int cos (ax) (cos (kx)) dx ##

##= \int e^{ikx} cos (ax) dx ##

using that ##\int cos(nx) sin (mx) =0 ##

comes from the cos.cos term only, and so since cos is even function this result is the same for the inverse Fourier transform: ## F^{-1}[cos(ak)](x)=\sqrt{\frac{\pi}{2}} (\delta(x-a)+\delta(x+a))##

With this I am getting...

##c_f(x,t)=F^{-1}[\frac{1}{\sqrt{2\pi}}cos(\sqrt{D}kt)](x)=\frac{1}{2}(\delta(x-\sqrt{D}t)+\delta(x+\sqrt{D}t) ##

And therefore using (*) and the intial condition for ##c(x,t=0)## I have:

##c(x,t)= \frac{1}{2} \int\limits^{a}_{-a} \delta (x-s-\sqrt{D}t)+\delta(x-s+\sqrt{D}t) ds ##

which doesn't feel like it's on the right tracks...

Am I incorrect that ## F^{-1}[cos(ak)](x)=\sqrt{\frac{\pi}{2}} (\delta(x-a)+\delta(x+a))## and if so where did my reasoning go wrong?

Many thanks in advance.