Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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