# Convolution Fourier Transform

1. Apr 30, 2017

### joshmccraney

1. The problem statement, all variables and given/known data
$$u_{xx}=u_t+u_x$$ subject to $u(x,0)=f(x)$ and $u$ and $u_x$ tend to 0 as $x\to\pm\infty$.

2. Relevant equations
Fourier Transform

3. The attempt at a solution
Taking the Fourier transform of the PDE yields
$$(\omega^2-i\omega) F\{u\}= \partial_t F\{u\}\implies\\ F\{u\} = \exp(t/4)F\{f(\omega)\}\exp((\omega-i/2)^2t)\implies\\ u = \exp(t/4)\int_{-\infty}^\infty f(\omega)\exp((\omega-i/2)^2t)\exp(i\omega x) \, d\omega$$ From here I don't know how to apply convolution. Can anyone help?

2. Apr 30, 2017

### Orodruin

Staff Emeritus
What is the function that has $\exp((\omega-i/2)^2t)$ as its Fourier transform?

Edit: Also, $i^2 = -1$ ... You should have $\exp(-(\omega+i/2)^2t)$.

3. Apr 30, 2017

### joshmccraney

Thanks for the correction. And a Gaussian transforms another Gaussian I believe, which is why I casted it in that form. I know $$F\left( \sqrt{\frac{\pi}{\alpha}}\exp(-x^2/4\alpha)\right) = \exp(-\alpha\omega^2).$$ Then the inverse transform of $\exp(-(\omega-i/2)^2t)$ where $\alpha = 1/t$ should be $\sqrt{\pi t}\exp(-t(x-i/2)^2/4)$. What do you think?

4. Apr 30, 2017

### Orodruin

Staff Emeritus
I think you are guessing rather than performing the proper contour integral in the complex plane.

5. May 1, 2017

### joshmccraney

Yea I was looking at a formula that looked applicable. To work it out we'd have $$\int_{-\infty}^\infty \exp(-(\omega-i/2)^2t)\exp(i\omega x)\,d\omega = \int_{-\infty}^\infty \exp(-(\omega-i/2)^2t+i\omega x)\,d\omega\\ =\sqrt{\frac{\pi}{t}}\exp\left( -\frac{x(2t+x)}{4t} \right)$$
How does that look for the inverse Fourier transform?

6. May 1, 2017

### Orodruin

Staff Emeritus
... and multiplying this by the exponent you already extracted gives you? (In fact, it is preferable not to complete the square until you have the Fourier integral...)

7. May 1, 2017

### Orodruin

Staff Emeritus
You might also want to double check the normalisation of your Fourier transform and that of the table you are getting your transforms from.

8. May 1, 2017

### joshmccraney

Gotcha, so this gives us $$\sqrt{\frac{\pi}{t}}\exp\left( \frac{(t-x)^2}{4t} \right).$$ But how do I apply this to use convolution?

9. May 1, 2017

### Orodruin

Staff Emeritus
You should still double check the normalisation. And there is now a missing sign in the exponent.

Then use that the Fourier transform of a convolution is the product of the Fourier transforms.

10. May 1, 2017

### joshmccraney

Shoot, we should have $$\sqrt{\frac{\pi}{t}}\exp\left( -\frac{(t+x)^2}{4t} \right)\exp(-t/2).$$ Can you elaborate on what you mean by "use that the FOurier transform of a convolution is the product of the Fourier transforms?

11. May 1, 2017

### Orodruin

Staff Emeritus
Your conclusion in post 8 was closer. Note that the sign of the exponent with the t also changes due to your original sign error. Again, the best thing is to wait with completing the square until you have the inverse FT integral.

If you are not familiar with the convolution theorem, I suggest reading up on it. It should be explained in any introductory textbook on Fourier transforms. Or you can start on Wikipedia https://en.m.wikipedia.org/wiki/Convolution_theorem

Note the issues with normalisation.

12. May 1, 2017

### joshmccraney

Shoot, yea I made an algebraic error. For completeness the equation after transforming the PDE should read $$-(\omega^2+i\omega) F\{u\}= \partial_t F\{u\}\implies\\ F\{u\} = F\{f(\omega)\}\exp(-(\omega^2+i\omega)t)\implies\\ u = \int_{-\infty}^\infty F\{f(\omega)\}\exp(-(\omega^2+i\omega)t)\exp(i\omega x) \, d\omega\\ = \exp\left(- \frac{(x-t)^2}{4t} \right)\int_{-\infty}^\infty F\{f(\omega)\}\exp\left( -t\left(i\frac{x-t}{2t}-\omega\right)^2\right) \, d\omega$$ Proceeding (if this is correct), convolution seems to imply I should take the inverse Fourier transform of $F\{f(\omega)\}$ and $\exp\left( -t\left(i\frac{x-t}{2t}-\omega\right)^2\right)$ and multiply these together?

Last edited: May 1, 2017
13. May 1, 2017

### Orodruin

Staff Emeritus
Up to a factor of $1/2\pi$ depending on normalisation of the Fourier transform.

Note that the Green's function of the heat equation is
$$\frac{1}{\sqrt{4\pi t}} e^{-\frac{x^2}{4t}}.$$
The difference in your case is the appearance of the convection term. You can get rid of it using an appropriate change of variables $\tau = t$, $\xi = x-t$. This is also the procedure you would obtain using the method of characteristics. If you do this, you will get the standard heat equation $u_\tau = u_{\xi\xi}$. Reinserting your original coordinates into the resulting Green's function immediately gives
$$G(x,t) = \frac{1}{\sqrt{4\pi t}} e^{-\frac{(x-t)^2}{4t}}.$$