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Solving Differential Equations Using Fourier Transform 📂Fourier Analysis

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

ut=kuxx(−∞<x<∞) u_{t}=k u_{xx} (-\infty < x< \infty)

$$ u(x,0)=f(x) ( -\infty < x< \infty)
$$

For positive time tt, there are no boundary conditions. Assume uu and ff decrease sharply and converge to 00 when x→±∞x \rightarrow \pm \infty, i.e., assume a L1L^1 function. Then a Fourier transform exists. Applying the Fourier transform to the given differential equation for xx

F[ut](ξ, t)=kF[uxx](ξ, t) \mathcal{F} [u_{t}] (\xi,\ t) = k \mathcal{F}[u_{xx}] (\xi,\ t)

Applying the properties of the Fourier transform [f′]^=iξf^\hat{[f^{\prime}]}=i\xi \hat{f} to the right-hand side

F[ut](ξ, t)=−kξ2u^(ξ, t) \mathcal{F} [u_{t}] (\xi,\ t) = -k\xi^{2} \hat{u}(\xi,\ t)

If we spell out the left-hand side, it becomes ∫ute−iξxdx\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 uu is the same as the differentiation of the Fourier transform of uu. Hence, the given differential equation becomes the following simple ordinary differential equation:

∂u^∂t(ξ, t)=−kξ2u^(ξ, t) \dfrac{\partial \hat{u}}{\partial t}(\xi,\ t) = -k\xi^{2} \hat{u}(\xi,\ t)

Solving the above differential equation for a fixed ξ\xi gives

u^(ξ, t)=f^(ξ)e−kξ2t \hat{u}(\xi,\ t) = \hat{f} (\xi) e^{-k\xi^{2}t}

Taking the inverse Fourier transform on both sides gives

u(x, t)=12π∫f^(ξ)e−kξ2teiξxdξ \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 uu. 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)(d) F[f∗g]=f^g^\mathcal{F} [f \ast g]=\hat{f}\hat{g} gives

f∗g=F−1[f^g^] f \ast g=\mathcal{F}^{-1}[\hat{f} \hat{g}]

Let’s say that (eq1)\eqref{eq1}’s e−kξ2te^{-k\xi ^{2}t} is the Fourier transform of some function. Specifically, let’s call it F[Kt](ξ)=e−kξ2t\mathcal{F}[K_{t}] (\xi)=e^{-k\xi^{2} t}. Then equation (eq1)\eqref{eq1} becomes

u(x, t)=12π∫f^(ξ)Kt^(ξ)eiξxdξ=F−1[f^Kt^](x)=f∗Kt(x) \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 KtK_{t}. Taking the inverse transform on both sides of the initially defined equation gives

Kt(x)=F−1F[Kt](x)=F−1[e−kξ2t]=12π∫e−kξ2teiξxdξ=12π∫e−kξ2te−iξ(−x)dξ=12πF[e−kξ2t](−x)=14πkte−x2/4ktdx \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 uu gives

u(x, t)=f∗Kt(x)=∫f(y)Kt(x−y)dy=14πkt∫f(y)e−(x−y)2/4ktdy \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)u(x,\ 0)=f(x), then lim⁡t→0u(x, t)=lim⁡t→0f∗Kt(x)=f(x)\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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