Solving the 'modified diffusion equation' using fourier transform

[mudkip9001]

Homework Statement

A the density of a gas $$\rho$$ obeys the modified diffusion equation

$$\frac{\partial \rho(x,t)}{\partial t}-D\frac{\partial^2 \rho(x,t)}{\partial x^2}=K\delta(x)\delta(t)$$

A) Express $$\rho$$ in terms of its 2D Fourier transform $$\widetilde{\rho}(p,\omega)$$ and express the right hand side as a Fourier type integral.

B) Find the function $$\widetilde{\rho}(p,\omega)$$
[Note: It might occur to you that this function could contain a term of the form $$f(p)\delta(Dp^2-i\omega )$$, where $$f(p)$$ is an arbitrary function. If so, you should assume that $$f(p)=0$$. If this doesn't occur to you, then don't worry about it!]

The Attempt at a Solution

A) I'm pretty sure I can do this:

$$\rho(x,t)=\frac{1}{2\pi}\iint_{-\infty}^{\infty}dp\; d\omega\; \left [e^{i(px-\omega t)}\widetilde{\rho}(p,\omega) \right ]$$

and

$$\delta(x)\delta(t)=\frac{1}{4\pi^2}\iint_{-\infty}^{\infty}dp\; d\omega\; \left [e^{i(px-\omega t)} \right]$$

B) Sticking the results from A) into the diffusion equation, taking the derivatives and rearranging:

$$\iint_{-\infty}^{\infty}dp\; d\omega\; \left [e^{i(px-\omega t)}\widetilde{\rho}(p,\omega) \right]=\frac{K}{2\pi(Dp^2-i\omega)}\iint_{-\infty}^{\infty}dp\; d\omega\; \left [e^{i(px-\omega t)} \right]$$

...and here i get stuck. how do i solve this? I have noticed that the two sides are conspicuously similar, but I'm not sure what conclusions I can make from that.

Edit: the choice of exponentials for the transform (negative for the time) is the convention that was recomended in lectures.

 

[mudkip9001]

I'm tempted to simply say $$\widetilde{\rho}(p,\omega)=1$$, and $$\frac{K}{2\pi (Dp^2-i\omega)}=1$$, but that wouldn't give much diffusion, so I assume that's not the answer...

I have also considered taking $$\frac{\partial^2 }{\partial p \partial \omega}$$ to both sides to get a partial differential equation, but i don't think that will work, since it's a definite integral, and you would still have an integral on the r.h.s from the chain rule.

 

[mudkip9001]

i'm sorry, this is nought but a shameless bump, can't diguise it as anything else. Any hint would be appreciated.

 

[marcusl]

You can't take the term $$(Dp^2-i\omega)$$ out of the integral on the left since both variables are operated on by the integrals. Then maybe the hint can be used.

 

[mudkip9001]

I stared at this for so long, can't believe I didn't realize that...