Age Diffusion Theory and Fourier Transforms

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ajhunte
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I am attempting to solve a second order differential, but I am have never done anything like this. I was told that it was a good Idea to think about Fourier transforms.

[tex]\frac{d^{2}q}{dx^{2}}=\frac{dq}{dt}[/tex]

Boundary Conditions:
[tex]q(+/-\infty,t)=0[/tex]
[tex]q(x,0)=S_{0}\delta(x)[/tex]

Apparently the final solution is:

[tex]q(x,t)=\frac{S_{0}exp[\frac{-x^{2}}{4t}]}{\sqrt{4(\pi)t}}[/tex]

If you were wondering the problem statement:
Determine the slowing down density established by a monoenergetic plane source at the origin of an infinite moderating medium as given by age-diffusion theory.
 
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This is one of the simplest PDE you can solve via Fourier Transform which accounts for two physical problem of interest, i mean heat equation and Schroedinger equation.
We start with PDE

[tex] \partial^2_x u(x,t) = \partial_t u(x,t)[/tex]

First start with projecting the equation on Fourier space (with respect to x), we get

[tex] - k^2 \hat{u}(k,t) = \partial_t \hat{u}(k,t)[/tex]

This is a first order ODE in t for unknown [tex]\hat{u}(k,t)[/tex] its solution can be found by separation of variables and reads

[tex] \hat{u}(k,t) = f(k) e^{- t k^2}[/tex]

for some unknown f(k) constant in time. To recover [tex]u(x,t)[/tex] we simply use inverse Fourier transform

[tex] u(x,t) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} \hat{u}(k,t) e^{ i k x} dk = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} f(k) e^{- t k^2} e^{ i k x} dk<br /> [/tex]

thus

[tex] u(x,0) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} f(k) e^{ i k x} dk[/tex]

hence f(k) is just the Fourier transform of the initial condition of the PDE. In our case

[tex] f(k) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} u(x,0) e^{- i k x} dx = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} S_0 \delta(x) e^{- i k x} dx = \frac{S_0}{\sqrt{2 \pi}} [/tex]

So we find the solution by evaluating the integral

[tex] u(x,t) = \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^{\infty} \frac{S_0}{\sqrt{2 \pi}} e^{-t k^2} e^{i k x} dk = \frac{S_0}{2 \pi} \int_{- \infty}^{\infty} e^{-t k^2 + i k x} dk = \frac{S_0}{2 \pi} \int_{- \infty}^{\infty} e^{-t ( k^2 + i \frac{k x} {t} + \frac{i^2 x^2} {4 t^2} - \frac{i^2 x^2} {4 t^2} ) } dk = \frac{S_0 e^{- \frac{x^2} {4 t}}} {2 \pi} \int_{- \infty}^{\infty} e^{- t ( k + \frac{i x} {2 t} )^2 } dk =[/tex]
[tex] = \frac{S_0 e^{- \frac{x^2} {4 t}}} {2 \pi} \int_{- \infty - \frac{i x} {2 t} }^{\infty - \frac{i x} {2 t} } e^{- t p^2 } dp = \frac{S_0 e^{- \frac{x^2} {4 t}}} {2 \pi} \sqrt{\frac{\pi}{t}} = \frac{S_0} {\sqrt{4 \pi t}} e^{- \frac{x^2} {4 t}}[/tex]

Hope this can help, ask if some steps are not clear. bye.