PDEs using Fourier Analysis II

Vivek Palaniappan
Sep 25, 2018 · 7 min read

In my previous post, PDEs using Fourier Analysis I, I investigated solving PDEs, in particular the wave equation, on a membrane using Fourier Series. In this article, I shall formulate a generalised method of solving PDEs, that is founded on the Fourier Transform. If you are unfamiliar with the Fourier Transform, check out my introduction here.

The method presented here works for initial value problems where the PDEs is solved with respect to some initial values known a priori. The workflow of this method is as follows:

  1. Transform the PDE into the frequency domain
  2. Solve the new PDE in frequency domain
  3. Apply the initial values (transformed)
  4. Invert the solution back into the time domain

General PDE

Consider the following general PDE

Where the bar over x denotes that it is a vector. Taking the Fourier Transform, with respect to x, of this equation

This is now merely an ODE in the time variable, and can be solved using standard ODE techniques.

Now, once the ODE is solved, we need to fit in the initial values g(x). For that, transform the initial values and plug it in

Now, in order to effectively find the inverse of this solution, we make use of convolution. Hence, we arrive at the final solution:

Now we will apply this method to the three broad categories of PDEs: Heat, Wave and Laplace.

Heat Equation

The heat equation is defined as the following PDE:

Now, applying the Fourier transform, and then solving the ODE, we get the following transformed solution

Now we label this solution as H(k,t), to make it easier to write and we don’t need to worry about deriving the inverse transform.

Applying our convolution trick, we get the following

Explicitly, we obtain the following integral

I have left out the solving of the ODE as an exercise for the reader to check and I feel it’s a good way to brush up on some differential equations skills.

Wave Equation

The wave equation we encountered before can now be written generally as

Notice that the wave equation requires two initial conditions to be satisfied. This is the a property of PDEs that are similar to the wave equation.

Now we use our standard trick to transform it and then solve the ODE.

Again I have left solving the ODE to the reader, but it is not too difficult I feel (you will get an ODE of the form y’+k²y=0). Now we assign the functions phi and psi to those solutions, resulting in

Applying our convolution

We get the explicit formula

We notice that the wave equation has two solutions due to its properties. It is useful to compare this to d’Alembert solution of the wave equation which results in waves moving in opposite directions.

Laplace Equation

Now we arrive at the famous and elegant Laplace equation. This PDE may seem simple and even a bit pointless to analyse, but surprisingly a lot of analysis of PDEs in general can be done using solutions of Laplace’s equation. Furthermore, the solutions of this equation led to the branch of mathematics called harmonic analysis. Laplace’s equation is defined as such:

Again, transforming and solving,

Letting the solution be P(k,y),

Convoluting, we obtain

One this to note is that the function P here is also the Poisson Kernel in n dimensions (stay tuned for an article on kernels and green’s functions for an elaboration). Writing the convolution explicitly, we obtain

Application: Smoke Spread

Consider the following problem:

Find the distribution of smoke in a square region of space and time, given that the initially there a number of point sources at known locations

In equations, the problem can be stated as

Where the initial condition is a sum of Dirac delta functions centered at the points where the sources are located.

Now, looking back at our solution for the general heat equation, we can get the Fourier transformed solution as

Now we need to find the inverse Fourier transform of our transformed solution.

This can evaluated with relative ease using the standard Gaussian integrals.

Now, we use the standard convolution to get our desired solution in the standard time domain.

This integral is surprisingly simple. Just take a moment to think about it. Dirac delta function is only non zero at the various xi points and therefore, when you integrate it over the entire space, only the values at those points will be reflected. Hence, our final solution takes the following form:

I think this is quite an elegant result. Let us break the expression down: the solution is the sum of Gaussians centered at the various points, with spread (variance), increasing with time. Also, there is a reciprocal square root term outside the summation that decreases the overall level of the smoke distribution. Isn’t that beautiful!

Now let us graph it for a for few sources.

Plotted using GeoGebra

The Gaussian makes it very easy to vary the number of sources and their properties, allowing for a clear understanding of the behavior of the heat equation under point source initial conditions.

Conclusion

This has been quite a general view of PDEs and how to use Fourier Transform to solve PDEs. This approach is very useful in solving any initial value problems, given that Fourier transforms of the initial conditions exist. However, the struggle always arises in solving the convolution integral. In the example we used, conveniently, the integral contained the Dirac delta function, which is very elementary to integrate and merely returns the integrand with different parameters. If the initial conditions very something other than delta functions, we would have a long day of integrating. This method of solving PDEs is privy to the method of Green’s functions and kernels, which I shall introduce at a latter date.

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Vivek Palaniappan

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Engineer Quant

Delve into engineering and quantitative analysis

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