Solving the 'modified diffusion equation' using fourier transform

In summary, the conversation discusses the modified diffusion equation for the density of a gas, which can be expressed in terms of its 2D Fourier transform and the right hand side can be written as a Fourier type integral. The function \widetilde{\rho}(p,\omega) is then found by substituting the results from A) into the diffusion equation and rearranging. The conversation ends with the question of how to solve the resulting equation, with the suggestion to use the term (Dp^2-i\omega) as a hint.
  • #1
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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  • #2
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
i'm sorry, this is nought but a shameless bump, can't diguise it as anything else. Any hint would be appreciated.
 
  • #4
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
I stared at this for so long, can't believe I didn't realize that...
 

Related to Solving the 'modified diffusion equation' using fourier transform

1. What is the 'modified diffusion equation'?

The modified diffusion equation is a mathematical equation used to describe the behavior of a diffusing substance in a medium with varying properties. It is a modification of the traditional diffusion equation to account for changes in diffusion coefficients or other parameters.

2. What is the Fourier transform?

The Fourier transform is a mathematical operation that decomposes a function into its individual frequency components. It is commonly used in signal processing and image processing to analyze and manipulate signals or images in the frequency domain.

3. How is the Fourier transform used to solve the modified diffusion equation?

The Fourier transform is used to convert the modified diffusion equation from the time domain to the frequency domain. This allows for a simpler and more efficient solution to the equation, as the transformed equation can be solved using basic algebraic operations.

4. What are the advantages of using the Fourier transform to solve the modified diffusion equation?

Using the Fourier transform to solve the modified diffusion equation allows for a more efficient and accurate solution compared to traditional methods. It also allows for the incorporation of boundary conditions and other constraints in a more straightforward manner.

5. Are there any limitations to using the Fourier transform to solve the modified diffusion equation?

One limitation is that the Fourier transform assumes the medium is infinite and homogeneous, which may not always be the case in practical applications. Additionally, the Fourier transform may not yield a solution for certain complex boundary conditions or discontinuous diffusion coefficients.

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