0
$\begingroup$

Consider the heat equation

$$\color{blue}{\begin{align} u_t&=ku_{xx}-bt^2u,\quad-\infty<x<\infty,\quad t>0,\\ u(x,0)&=\exp\left[-x^2\right]. \end{align}}$$

I am asked to solve it using the Fourier transform pair

$$\begin{align} \color{blue}{F(\omega)}&\color{blue}{=\frac1{2\pi}\int_{-\infty}^\infty f(x)e^{i\omega x}dx,}\\ \color{blue}{f(x)}&\color{blue}{=\int_{-\infty}^\infty F(\omega)e^{-i\omega x}d\omega.} \end{align}$$

This is what I ended up with:

$$ U_t(\omega,t)=-k\omega^2U(\omega,t)-bt^2U(\omega,t), $$

where the solution to this ODE is

$$ U(\omega,t)=c(\omega)\exp\left[-\frac{bt^3}3-k\omega^2t\right]. $$

Applying the initial condition to solve for $c$ yields

$$ c(\omega)=\frac{\exp\left[-\frac{\omega^2}4\right]}{2\sqrt\pi}. $$

Hence, the final solution is

$$ U(\omega,t)=\frac{\exp\left[-\frac{bt^3}{3}-\frac{\omega^2}{4}(4kt+1)\right]}{2\sqrt\pi}. $$

I wonder if this is correct. The reason why I ask this is that our professor hinted us that we use the following two transforms:

$$\begin{align} \mathcal F\left(\exp\left[-x^2\right]\right)(\xi)&\stackrel{?}{=}(2\pi)^{\frac12}\exp\left[-\frac{\xi^2}2\right],\\ \mathcal F(f(ax))(\xi)&=a^{-1}\mathcal F\left(\frac\xi a\right), \end{align}$$

which make no sense because I am not sure of the validity of the first, and I see no place where the second one could be of any help.

Is this perhaps a typo? Thanks in advance.

$\endgroup$
1
$\begingroup$

Based on your calculations, you have reached to the following stage $$ U(\omega,t)= \frac{1}{2\sqrt\pi} e^{-\frac{bt^3}{3}} e^{-\frac{\omega^2}{4}(4kt+1)}. $$

Now, all you need to do is to find the inverse Fourier transform w.r.t. $\omega$ to get $u(x,t)$,

$$ u(x,t)= \int_{-\infty}^\infty U(\omega, t)e^{-i\omega x}d\omega= \frac{1}{2\sqrt\pi} e^{-\frac{bt^3}{3}}\int_{-\infty}^\infty e^{-\frac{\omega^2}{4}(4kt+1)}e^{-i\omega x}d\omega $$

$$\implies u(x,t) = \frac{1}{2\sqrt\pi} e^{-\frac{bt^3}{3}}\int_{-\infty}^\infty e^{-\frac{a\,\omega^2}{4}}e^{-i\omega x}d\omega,$$

where $ a = 4kt+1. $

Can you find the last integral now?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.