Solving Differential Equations Using Fourier Transform
Description
Fourier series and Fourier transforms emerged to solve heat equations. Of course, they can be used to solve other differential equations too, provided the conditions are met. Fourier series, in particular, are used in quantum physics to calculate the energy of particles through the Schrödinger equation. Many physics students use it without knowing it’s a Fourier series, but they understand what it is when told. Depending on the conditions of the given differential equation, it is determined whether to use Fourier transform or Fourier series. Fourier series are used when the range given in the problem is finite, and Fourier transforms are used when the range is infinite.
Solution
Suppose we are given the following heat equation
$$ u_{t}=k u_{xx} (-\infty < x< \infty) $$
$$
u(x,0)=f(x) ( -\infty < x< \infty)
$$
For positive time $t$, there are no boundary conditions. Assume $u$ and $f$ decrease sharply and converge to $0$ when $x \rightarrow \pm \infty$, i.e., assume a $L^1$ function. Then a Fourier transform exists. Applying the Fourier transform to the given differential equation for $x$
$$ \mathcal{F} [u_{t}] (\xi,\ t) = k \mathcal{F}[u_{xx}] (\xi,\ t) $$
Applying the properties of the Fourier transform $\hat{[f^{\prime}]}=i\xi \hat{f}$ to the right-hand side
$$ \mathcal{F} [u_{t}] (\xi,\ t) = -k\xi^{2} \hat{u}(\xi,\ t) $$
If we spell out the left-hand side, it becomes $\int u_{t}e^{-i\xi x}dx$, assuming it is permissible to swap the order of integration and differentiation due to favorable conditions. In general, it’s not possible to swap the order of differentiation and integration at will, but in these types of problems, it almost always holds, so it’s okay not to worry too much about it. Thus, the Fourier transform of the differentiation of $u$ is the same as the differentiation of the Fourier transform of $u$. Hence, the given differential equation becomes the following simple ordinary differential equation:
$$ \dfrac{\partial \hat{u}}{\partial t}(\xi,\ t) = -k\xi^{2} \hat{u}(\xi,\ t) $$
Solving the above differential equation for a fixed $\xi$ gives
$$ \hat{u}(\xi,\ t) = \hat{f} (\xi) e^{-k\xi^{2}t} $$
Taking the inverse Fourier transform on both sides gives
$$ \begin{equation} u(x,\ t) =\dfrac{1}{2\pi}\int \hat{f}(\xi) e^{-k\xi^{2} t}e^{i\xi x} d\xi \label{eq1} \end{equation} $$
This is called the Fourier integral formula for $u$. This expression will be simplified using the properties of Fourier transform and convolution. Taking the inverse transform on both sides of the property of Fourier transform $(d)$ $\mathcal{F} [f \ast g]=\hat{f}\hat{g}$ gives
$$ f \ast g=\mathcal{F}^{-1}[\hat{f} \hat{g}] $$
Let’s say that $\eqref{eq1}$’s $e^{-k\xi ^{2}t}$ is the Fourier transform of some function. Specifically, let’s call it $\mathcal{F}[K_{t}] (\xi)=e^{-k\xi^{2} t}$. Then equation $\eqref{eq1}$ becomes
$$ \begin{align*} u(x,\ t) &= \dfrac{1}{2\pi} \int \hat{f}(\xi) \hat{K_{t}}(\xi) e^{i\xi x} d\xi \\ &= \mathcal{F}^{-1}[\hat{f}\hat{K_{t}}] (x) \\ &= f \ast K_{t}(x) \end{align*} $$
Now it’s time to find $K_{t}$. Taking the inverse transform on both sides of the initially defined equation gives
$$ \begin{align*} K_{t}(x)= \mathcal{F}^{-1} \mathcal{F}[K_{t}] (x) &= \mathcal{F}^{-1} \left[ e^{-k\xi ^{2} t} \right] \\ &= \dfrac{1}{2\pi} \int e^{-k\xi^{2} t}e^{i\xi x} d\xi \\ &= \dfrac{1}{2\pi} \int e^{-k\xi^{2} t}e^{-i\xi (-x)}d\xi \\ &= \dfrac{1}{2\pi} \mathcal{F} \left[e^{-k\xi^{2}t} \right] (-x) \\ &= \dfrac{1}{\sqrt{4\pi kt}}e^{-x^{2}/4kt}dx \end{align*} $$
The final formula can be easily obtained with the Fourier transform formula of the Gaussian function. Therefore, inserting this into $u$ gives
$$ \begin{align*} u(x,\ t) &= f \ast K_{t}(x) \\ &= \int f(y) K_{t}(x-y) dy \\ &= \dfrac{1}{\sqrt{4 \pi kt}} \int f(y) e^{-(x-y)^{2}/4kt}dy \end{align*} $$
Given the condition of the initial differential equation, if $u(x,\ 0)=f(x)$, then $\lim \limits_{ t\rightarrow 0} u(x,\ t)=\lim \limits_{ t\rightarrow 0} f \ast K_{t}(x)=f(x)$ would be a fitting solution for the above problem. Of course, it can actually hold and be proven.
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