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.


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?


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