I'm trying to find the Inverse Fourier Transform for the following heat equation.


  1. Solve the problem

$$u_t=ku_{xx}+\frac{1}{\sqrt{2kt}}e^{\frac{-x^2}{4kt}}, -\infty < x< \infty, t >0$$

with I.C $$u(x,0)=0$$


We need to take the Fourier Transform of the entire heat equation.


Then we need to multiply the integrating factor, $e^{4\pi^2\xi^2tk}$, throughout the whole equation


By reverse product rule, we have


note that $\hat{u}=\left(\sqrt{2\pi}t+c\right)e^{-4\pi^2k\xi^2t}$

Taking the inverse fourier transform,we have

$\sqrt{2 \pi}e^{-4 \pi^2 k \xi^2 t}$

errrrr... I know that the inverse fourier transform is

$\bar{g}(x) = \int^{-\infty}_{\infty} g(\xi)e^{i \xi x}\frac{1}{\sqrt{2\pi}}$

So would the $g(\xi)$ be $\sqrt{2 \pi}e^{-4 \pi^2 k \xi^2 t}$ and then I plug it into the inverse fourier transform equation?

$\bar{g}(x) = \int^{-\infty}_{\infty} \sqrt{2 \pi}e^{-4 \pi^2 k \xi^2 t}e^{i \xi x}\frac{1}{\sqrt{2\pi}}$

$\bar{g}(x) = \int^{-\infty}_{\infty} \sqrt{2 \pi}e^{-4 \pi^2 k \xi^2 t+i \xi x}\frac{1}{\sqrt{2\pi}}$

  • $\begingroup$ How did you go from $\hat{u}=(\sqrt{2\pi}t + c)\exp(-4\pi^{2}k\xi^{2}t)$ to $u= \sqrt{2\pi}\exp(-4\pi^{2}k\xi^{2}t)$ ? $\endgroup$ – mattos Dec 7 '14 at 2:27
  • $\begingroup$ I am not even sure... I just .. ughhh trying to grasp it but this is harder than the wave equation version. $\endgroup$ – usukidoll Dec 7 '14 at 3:03
  • $\begingroup$ I'll write up a post for you so we don't have to use the comments $\endgroup$ – mattos Dec 7 '14 at 3:06
  • $\begingroup$ alright...that's good $\endgroup$ – usukidoll Dec 7 '14 at 3:10

First of all, I would suggest you try the homogeneous heat equation first and learn how to do that question before trying the heat equation with source. Fourier Transforms can get quite tricky and it helps to practice with easier problems first before moving onto to higher dimensional/inhomogeneous problems. But that is another discussion.

Remember that the FT is defined by

$$ \hat{f}(\xi) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)\exp(-i\xi x) dx $$


$$ f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \hat{f}(\xi)\exp(i\xi x) d\xi $$

To transform back (above, the $\hat{f}$ I have used is equivalent to your $\hat{u}$), just substitute it into the integral. You have (remember $t$ is just a parameter and can be treated as a constant)

$$ u(t,x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \hat{u}(t,\xi)\exp(i\xi x) dx $$

Using your equation for $\hat{u}$ and noting that you also need to use your IC

$$ \begin{align} u(0,x) &= 0 \\ \implies \hat{u}(0,\xi) &= 0 \\ &= A(\xi) \\ \end{align} $$


$$ \hat{u}(t,\xi) = \sqrt{2\pi}t\exp(-4\pi^{2}k\xi^{2}t) $$

Try and give it a go from there, I'll be here to help so just comment below if you are still having difficulty. You will need to manipulate your integral a bit to get the solution.

(Note that I've used $A(\xi)$ instead of $c$ as when you integrate a function of two variables, the constant of integration is actually a function of the variable that wasn't integrated over; You integrated w.r.t $t$, hence the constant was a function of $\xi$).

  • $\begingroup$ Alternatively, you can use the variation of constants formula and the solution is given by $$ u(t,x) = \int_{\mathbb{R}} G(t,x,y)\phi(y) dy + \int_{0}^{t} \bigg(\int_{\mathbb{R}} G(t-s,x,y)f(s,y) dy \bigg) dt $$ where $f(s,y)$ is your source term and $G(t,x,y)$ is the heat kernel. $\endgroup$ – mattos Dec 8 '14 at 18:15

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.