# Solving DE using Fourier Transform

1. Jan 9, 2014

### unscientific

1. The problem statement, all variables and given/known data

Part (a): State inverse fourier transform. Show fourier transform is:
Part (b): Show fourier transform is:
Part (c): By transforming LHS and RHS, show the solution is:
Part(d): Using inverse fourier transform, find an expression for T(x,t)

2. Relevant equations

3. The attempt at a solution

Part(d)
$$T_{(x,t)} = \int_{-\infty}^{\infty} T e^{-Dk^2t}e^{ikx} dk$$
= $$\int_{-\infty}^{\infty} (T_0 + \sum_{m=1}^{\infty} Tm cos(\frac {m\pi x}{L}) e^{-Dk^2t + ikx} dk$$
= $$\int_{-\infty}^{\infty} T_0 e^{-Dk^2t + ikx} dk + \sum_{m=1}^{\infty} T_m cos(\frac {m \pi x}{L}) \int_{-\infty}^{\infty} e^{-Dk^2t + ikx} dk$$

Attempt at evaluating the integral, letting $$a^2 = \frac {1}{Dt}$$

$$\int_{-\infty}^{\infty} e^{-(\frac{k^2}{a^2} - ikx)} dx$$
$$= \int_{-\infty}^{\infty} e^{-\frac{({k - \frac{ixa^2}{2}})^2}{a^2}} e^{-\frac {x^2a^2}{4}} dk$$
$$= e^-{\frac{x^2a^2}{4}} \int_{-\infty - i\frac{x^2a^2}{4}}^{\infty - i\frac {x^2a^2}{4}} e^{- \frac{k^2}{a^2}} dk$$
$$= e^{\frac {-x^2a^2}{4}} \sqrt{\pi a^2} = \sqrt {\frac{\pi}{Dt}} e^{\frac {-x^2a^2}{4}}$$

But this appears to be wrong as their final expression do not have the factor of √(1/t) in their coefficients..

Last edited: Jan 9, 2014
2. Jan 9, 2014

### vanhees71

First you have to evaluate $\tilde{T}(k,0)$ from the initial condition which is given in the position domain as $T(x,0)$! Then you can plug it into the general solution and Fourier transform back!

3. Jan 9, 2014

### unscientific

Ha ha, that's a silly mistake I made!

So I must fourier transform the initial condition, then plug it back in, then inverse fourier transform everything?

4. Jan 10, 2014

Yes!