# Applying Convolution to a PDE with a Fourier Transform

• member 428835
No need for a Fourier transform, a convolution, or anything.In summary, the conversation discusses using a Fourier transform to solve a partial differential equation with initial conditions and boundary conditions. The solution involves using a Gaussian function as the inverse Fourier transform of the transformed equation, with the final step being to apply the convolution theorem. However, it is noted that the solution can also be obtained by using a change of variables and the method of characteristics.
member 428835

## Homework Statement

$$u_{xx}=u_t+u_x$$ subject to ##u(x,0)=f(x)## and ##u## and ##u_x## tend to 0 as ##x\to\pm\infty##.

## Homework Equations

Fourier Transform

## 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?

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)##.

Orodruin said:
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)##.
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?

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

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?

... 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...)

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

Orodruin said:
... 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...)
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?

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.

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?

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.

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 by a moderator:
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
$$G(x,t) = \frac{1}{\sqrt{4\pi t}} e^{-\frac{(x-t)^2}{4t}}.$$

## What is Convolution Fourier Transform?

The Convolution Fourier Transform is a mathematical operation that combines two mathematical functions to produce a third function. It is used to analyze the frequency components of a signal and is commonly used in signal processing and image processing applications.

## How does Convolution Fourier Transform work?

The Convolution Fourier Transform works by multiplying the Fourier transforms of two functions and then taking the inverse Fourier transform of the product. This process measures the overlap between the two functions and produces a new function that represents their combined frequency components.

## What are the applications of Convolution Fourier Transform?

Convolution Fourier Transform has a wide range of applications, including audio and image compression, noise reduction, and image filtering. It is also used in fields such as radar and sonar signal processing, medical imaging, and data analysis.

## What are the advantages of using Convolution Fourier Transform?

One of the main advantages of Convolution Fourier Transform is its ability to analyze complex signals and extract important information about their frequency components. It also allows for efficient processing of signals and images, making it a valuable tool in various scientific and engineering fields.

## Are there any limitations to Convolution Fourier Transform?

While Convolution Fourier Transform is a powerful tool, it does have some limitations. One limitation is that it assumes the functions being convolved are linear. It also requires a significant amount of computing power and is not ideal for real-time processing of signals with large data sets.

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