How Do You Solve Differential Equations Using Fourier Transform?

[unscientific]

Homework Statement

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)

Homework Equations

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..

 

[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!

 

[unscientific]

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!

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?

 

[vanhees71]

Yes!