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.
Best Answer
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?