Fourier Transform Applied to Differential Equations

Fourier Transforms can also be applied to the solution of differential equations. To introduce this idea, we will run through an Ordinary Differential Equation (ODE) and look at how we can use the Fourier Transform to solve a differential equation. Consider the ODE in Equation [1]: [Equation 1] We are looking for the function y(t) that satisfies Equation [1] above. We know that we can take the Fourier Transform of a function, so why not take the fourier transform of an equation? It turns out there is no reason we can't. And since the Fourier Transform is a linear operation, the time domain will produce an equation where each term corresponds to the a term in the frequency domain. Taking the Fourier Transform of Equation [1], we get Equation [2]: [Equation 2] Now, if you recall the differentiation property of the Fourier Transform, we note that derivatives in time become simple multiplication in the Frequency domain: [Equation 3] Hence, Equation [2] becomes: [Equation 4] Equation [4] is a simple algebraic equation for Y(f)! This can be easily solved. This is the utility of Fourier Transforms applied to Differential Equations: They can convert differential equations into algebraic equations. Equation [4] can be easiliy solved for Y(f): [Equation 5] In general, the solution is the inverse Fourier Transform of the result in Equation [5]. For this case though, we can take the solution farther. Recall that the multiplication of two functions in the time domain produces a convolution in the Fourier domain, and correspondingly, the multiplication of two functions in the Fourier (frequency) domain will give the convolution in the time domain. Hence, Equation [5] becomes: [Equation 6] Equation [6] might not look helpful, but note that we already know the inverse Fourier Transform for the left-most inverse Fourier transform in the second line of [6]: it's one half of the two-sided decaying exponential function. Hence, we can start to simplify equation [6]: [Equation 7] ... and we've just derived the solution for the differential equation [1]. Give yourselves a round of applause. Now for the fine print. When we went from Step 1 to Step 2, we assumed the Fourier Transform for y(t) existed. This is a non-trivial assumption. You may recall from your differential equations class that the solution should also contain the so-called homogeneous solution, when g(t)=0: [Equation 8] The "total" solution is the sum of the solution we obtained in equation [7] and the homogeneous solution y_h of equation [8]. So why does the homogeneous solution not come out of our method? The answer is simple: the non-decaying exponentials of equation [8] do not have Fourier Transforms. That is, if you try to take the Fourier Transform of exp(t) or exp(-t), you will find the integral diverges, and hence there is no Fourier Transform. This is a very important caveat to keep in mind. In the next section, we'll look at applying Fourier Transforms to partial differential equations (PDEs).