0
$\begingroup$

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?

$\endgroup$
2
  • 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 '20 at 14:52
  • $\begingroup$ How does the second integral disappear? I thought of factoring the inner integral out, but I'm left to integrate $\int^\infty_{-\infty} dk$. $\endgroup$ – Hexiang Chang Jul 12 '20 at 14:56
3
$\begingroup$

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).

$\endgroup$
5
  • $\begingroup$ 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 $\omega$? $\endgroup$ – Hexiang Chang Jul 12 '20 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 $\tilde f(k,\omega) = C(k)\delta(\omega+ck) + D(k)\delta(\omega-ck)$. $\endgroup$ – Emilio Pisanty Jul 12 '20 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. $\endgroup$ – Hexiang Chang Jul 12 '20 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. $\endgroup$ – Emilio Pisanty Jul 12 '20 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 '20 at 7:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.