2D Fourier Transform of a general function satisfying the wave equation

Question

A function $f(x,t)$ which satisfies the wave equation can be expressed generally as a function of a single argument $f(x-ct)$, where $c=\frac{\omega}{k}$. This is because you can express this function as an integral, according to Fourier Analysis: $$f(x-ct)=\int^\infty_{-\infty}C(r)e^{ir(x-ct)}dr \tag{1}$$ The wave equation is linear, which is why this continuous summation of wave solutions will also solve the equation.

The general function $\psi(x,t)$ can also be expressed using the 2-dimensional Fourier Transform:

$$\psi(x,t)=\int^\infty_{-\infty}\int^\infty_{-\infty}\beta(k,\omega)e^{i(kx-\omega t)}dk~d\omega \tag{2}$$

This is a general function which has no constraints. Substituting this form for $\psi(x,t)$ into the wave equation will give us the constraint $c=\frac{\omega}{k}$ for nonzero $\beta(k,\omega)$.

I would think that substituting the constraint $\omega=ck$ back into (2) should produce a solution in the same form as (1). However, when I do this:

\begin{align*} \psi(x,t)&=\int^\infty_{-\infty}\int^\infty_{-\infty}\beta(k,\omega)e^{i(kx-\omega t)}dk~d\omega\\ &=\int^\infty_{-\infty}\int^\infty_{-\infty}\beta(k)e^{ik(x-ct)}dk~(cdk)\\ &\stackrel{?}{=}\int^\infty_{-\infty}C(k)e^{ik(x-ct)}dk\\ \end{align*}

I'm not sure how to evaluate the integral such that the second line can lead to the third line. How should I proceed?

Answer 1

A function $f(x,t)$ which satisfies the wave equation can be expressed generally as a function of a single argument $f(x-ct)$, where $c=\frac{\omega}{k}$. This is because you can express this function as an integral, according to Fourier Analysis: $$f(x-ct)=\int^\infty_{-\infty}C(r)e^{ir(x-ct)}\mathrm dr \tag{1}$$

This isn't really right. The function you're describing is a solution of the wave equation, but it is not the most general solution of the wave equation in 1D.

The wave equation in 1D reads $$ \left[\frac{\partial^2}{\partial x^2} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right] f(x,t) = 0 $$ and its general solution is $$ f(x,t) = F(x-ct) + G(x+ct), $$ with components that propagate to the right and to the left. Each of those components satisfies a first-order PDE, $$ \left[\frac{\partial}{\partial x} \pm \frac{1}{c}\frac{\partial}{\partial t}\right] f(x,t) = 0, $$ but these are not the wave equation proper.

In terms of Fourier analysis, then, if you write your solution as $$ f(x,t)=\int^\infty_{-\infty}C(k)e^{i(kx-\omega t)}\mathrm dk \tag{1'} $$ then the dispersion relation requires you to set $\omega^2 = k^2 c^2$, so $\omega=\pm kc$, so in the end you need to write $$ f(x,t)=\int^\infty_{-\infty}\left[C(k)e^{i(kx-c|k| t)}+ D(k)e^{i(kx+c|k| t)}\right]\mathrm dk , \tag{1''} $$ where keeping $f(x,t)$ real requires you to set $D(-k) = C(k)^*$.

This is the form that extends to higher dimensionality, with the natural extension reading $$ f(\mathbf x,t)=\int\left[C(\mathbf k)e^{i(\mathbf k\cdot\mathbf x-c|\mathbf k| t)}+ D(\mathbf k)e^{i(\mathbf k\cdot\mathbf x+c|\mathbf k| t)}\right]\mathrm d\mathbf k . $$ You should be able to take it from there (but keep in mind that explicit solutions similar to $f(x,t) = F(x-ct) + G(x+ct)$ are not guaranteed to exist above 1D).