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?

  • 3
    $\begingroup$ you cannot integrate twice in the same variable! You are not actually imposing the constrain, just messing with integration variables. $\endgroup$
    – fqq
    Jul 12, 2020 at 14:52
  • $\begingroup$ How does the second integral disappear? I thought of factoring the inner integral out, but I'm left to integrate $\int^\infty_{-\infty} dk$. $\endgroup$ Jul 12, 2020 at 14:56

1 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).

  • $\begingroup$ Thank you for your correction. Does equation (1') come from a Fourier Transform of a general function $f(x,t)$? If so, why is the integral done with respect to $k$ while there is also $exp(i\omega t)$ in the integrand? Shouldn't it be a double integral, with the second integration with respect to $\omega$? $\endgroup$ Jul 12, 2020 at 15:40
  • 1
    $\begingroup$ @HexiangChang Yes, that is indeed the case. If you want to work rigorously, then you do a double (spatial+temporal) Fourier transform, $$f(x,t) = \iint \tilde f(k,\omega) e^{i(kx-\omega t)}\mathrm dk\mathrm d\omega,$$ and then you apply the wave equation and you show (nontrivially) that it requires $(c^2k^2-\omega^2)\tilde f(k,\omega) = 0$, so therefore $\tilde f(k,\omega) = C(k)\delta(\omega+ck) + D(k)\delta(\omega-ck)$. $\endgroup$ Jul 12, 2020 at 16:01
  • $\begingroup$ How exactly does this $(c^2k^2-\omega^2)\tilde f(k,\omega) = 0$, lead to this $\tilde f(k,\omega) = C(k)\delta(\omega+ck) + D(k)\delta(\omega-ck)$? I'm aware of a similar equation in quantum mechanics for the position eigenstate, but in that context there is a constraint on it's norm to be 1. $\endgroup$ Jul 12, 2020 at 16:46
  • $\begingroup$ The requirement is $(c^2k^2-\omega^2)\tilde f(k,\omega) = 0$ $\forall k,\omega$. The only way this can happen is if $\tilde f(k,\omega)$ is zero everywhere that $c^2k^2-\omega^2 \neq 0$. If you want something in full rigour in terms of products of distributions, then it's probably a complicated argument (as you need to rule out derivatives of the delta function), but the heuristics are quite clear. $\endgroup$ Jul 12, 2020 at 16:57
  • $\begingroup$ @EmilioPisanty you could include your comment in the answer, I think that was the core of the question. $\endgroup$
    – fqq
    Jul 13, 2020 at 7:46

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