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Solving the 'modified diffusion equation' using fourier transform

  • Thread starter mudkip9001
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Homework Statement



A the density of a gas [tex]\rho[/tex] obeys the modified diffusion equation

[tex]\frac{\partial \rho(x,t)}{\partial t}-D\frac{\partial^2 \rho(x,t)}{\partial x^2}=K\delta(x)\delta(t)[/tex]

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

B) Find the function [tex]\widetilde{\rho}(p,\omega)[/tex]
[Note: It might occur to you that this function could contain a term of the form [tex]f(p)\delta(Dp^2-i\omega )[/tex], where [tex]f(p)[/tex] is an arbitrary function. If so, you should assume that [tex]f(p)=0[/tex]. 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:

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

and

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

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

[tex]\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][/tex]

...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.
 
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Answers and Replies

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

I have also considered taking [tex]\frac{\partial^2 }{\partial p \partial \omega}[/tex] to both sides to get a partial differential equation, but i don't think that will work, since it's a definate integral, and you would still have an integral on the r.h.s from the chain rule.
 
  • #3
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i'm sorry, this is nought but a shameless bump, can't diguise it as anything else. Any hint would be appreciated.
 
  • #4
marcusl
Science Advisor
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You can't take the term [tex](Dp^2-i\omega)[/tex] out of the integral on the left since both variables are operated on by the integrals. Then maybe the hint can be used.
 
  • #5
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I stared at this for so long, can't believe I didn't realize that...
 

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