Solution to the Diffusion equation using a two dimensional fourier transform

[knowlewj01]

Homework Statement

at t=0, x=0, a schoolboy sets off a stink bomb halfway down a corridor that is long enough to be considered infinite. The dispersion of the particles obey the modified diffusion formula:

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

1) Express the density in terms of its Fourier transform and substitute into the above equation, also write out the dirac delta function as a product of the integral forms.
2) find the function \tilde{\rho}(p,\omega)

there is another part but i need to understand this bit before i make n attempt on it

Homework Equations

Inverse Fourier Transform:

f(x,t) = \frac{1}{2\pi}\int^\infty_{-\infty} dp \int^\infty_{-\infty} d\omega e^{ipx - i\omega t} \tilde{f}(p,\omega)

Dirac Delta Function:

\delta (x-x_0) = \frac{1}{2\pi}\int^\infty_{-\infty} e^{ip(x-x_0) dp

The Attempt at a Solution

having made a substitution for \rho(x,t) for the above equation and using the delta functions for x and t i get:

\int^\infty_{-\infty} dp \int^\infty_{-\infty} d\omega e^{ipx - i\omega t} \left(Dp^2 - i\omega\right) \tilde{\rho}(p,\omega) = \frac{K}{2\pi}\int^\infty_{-\infty} dp \int^\infty_{-\infty} d\omega e^{ipx - i\omega t}

From here, I am not sure how to get the function \tlide{\rho}(p,\omega)
is there some property of Fourier transforms or delta functions that i am forgetting?
Thanks

 

[knowlewj01]

equating the arguments of the integrals:

for t=0:

(Dp^2-i\omega)\tilde{\rho}(p,\omega) = \frac{K}{2\pi}

for t>0:

(Dp^2-i\omega)\tilde{\rho}(p,\omega) = 0i think this might be the next step. There is also a hint on the question paper:

this function could contain a term in the form:
f(p)\delta(Dp^2-i\omega)

where f(p) is an arbitary function, in this case you shhould assume f(p)=0

I don't fully follow the argument behind this

 

[I like Serena]

There seems to be a slight mistake in your calculation.
I believe that your factor K/2pi should actually be 2piK.

 

[knowlewj01]

ah sorry, missed a 1/2pi factor on the delta function when i typed it up, the product of the two d(x) d(t) gives a 1/4pi^2 on the right hand side, which cancels with 1/2pi on the left to leave K/2pi.

 

[I like Serena]

ah sorry, missed a 1/2pi factor on the delta function when i typed it up, the product of the two d(x) d(t) gives a 1/4pi^2 on the right hand side, which cancels with 1/2pi on the left to leave K/2pi.

Shouldn't the delta function have a factor of 1/sqrt(2pi) [edit]to be consistent with the f(x,t) transform[/edit]?

 

[knowlewj01]

in my notes it says that you get a factor of 1/2pi for each dimension in the delta function, ill have another read of it.

 

[knowlewj01]

for t>0:

(Dp^2-i\omega)\tilde{\rho}(p,\omega) = 0

On second thought i don't think this is right, this is just one possible solution. i think the previos one just needs to be rearranged and modified in some way.

\tilde{\rho}(p,\omega) = \frac{K}{2\pi}\frac{1}{Dp^2-i\omega}

\tilde{\rho}(p,\omega) needs to be chosen such that the origonal PDE is satisfied. not sure how to go about this

 

[I like Serena]

On second thought i don't think this is right, this is just one possible solution. i think the previos one just needs to be rearranged and modified in some way.

\tilde{\rho}(p,\omega) = \frac{K}{2\pi}\frac{1}{Dp^2-i\omega}

\tilde{\rho}(p,\omega) needs to be chosen such that the origonal PDE is satisfied. not sure how to go about this

Isn't this already the solution?
With your equation for F(rho)(p,w), the integral equation is satisfied, which means the original diffusion equation is satisfied.

The only question would be whether this is a "complete" solution, since there might be another solution.

 

[I like Serena]

in my notes it says that you get a factor of 1/2pi for each dimension in the delta function, ill have another read of it.

With Fourier, each and every time one needs to "chose" where to put the 2pi factor.
There are basically 3 conventions:
1. Within the exponential (which means using frequency f instead of angular velocity w)
2. Equally divided between the transform and the inverse transform
3. Only in either the transform or the inverse transform

In a 2-dimensional transform we have a factor (1/2pi)^2 that must be divided.
In a 1-dimensional transform we have a factor (1/2pi) that must be divided.

In your 2-dimensional transform the factor is 1/2pi for the inverse transform, implying you would have the same factor 1/2pi for the 2-dimensional transform.

So for transform of the dirac-function we would get 1/sqrt(2pi) to be consistent.
And for a 2-dimensional transform (of the product of 2 dirac-functions) we would get 1/2pi.

 

[knowlewj01]

ok, so if i don't ignore this:

for t>0:

(Dp^2-i\omega)\tilde{\rho}(p,\omega) = 0

a solution is possible where:

Dp^2 = i\omega

can rewrite this using the delta function:

\tilde{\rho}(p,\omega) = \delta\left(Dp^2-i\omega\right) \tilde{f}(p)

would the complete solution be the sum of this one and my previous one?:

\tilde{\rho}(p,\omega) = \delta\left(Dp^2-i\omega\right) \tilde{f}(p) + \frac{K}{2\pi}\frac{1}{Dp^2-i\omega}