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!

 

[paranormal]

The solution you have obtained for T(x,t) using the inverse Fourier transform is not entirely correct. The integral you have evaluated is actually the inverse Fourier transform of the function e^(-Dk^2t + ikx), not T(x,t). The correct way to evaluate the integral is by using the properties of the Fourier transform, specifically the property that states e^(-a^2x^2) transforms to sqrt(pi/a^2) e^(-k^2/a^2). Using this property, we can rewrite the integral as follows:

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

From here, we can use the inverse Fourier transform property to obtain the final expression for T(x,t):

T(x,t) = T_0 \sqrt{\frac{\pi}{Dt}} e^{\frac{-x^2}{4Dt}} + \sum_{m=1}^{\infty} T_m cos(\frac {m \pi x}{L}) \sqrt{\frac{\pi}{Dt}} e^{-(\frac{m\pi}{L})^2Dt}

This is the correct solution for T(x,t) using the inverse Fourier transform. It is important to note that the factor of sqrt(1/t) is present in the coefficients of the cosine terms as well, as it should be in order for the solution to be correct.