One typically solves waves (fields) equations in Fourier space. For example, the 1D wave equation

$\frac{\partial^2\phi(x,t)}{\partial t^2}-\frac{\partial^2\phi(x,t)}{\partial x^2} = 0$

in Fourier space becomes

$(\omega^2-k^2)\tilde\phi(k,\omega) = 0$,

from which one writes the general solution

$\phi(x,t)=\int dk (A(k)e^{-i(\omega_kt-kx)}+B(k)e^{i(\omega_kt+kx)})$,

where $\omega_k^2 = k^2$.

But this suggests that the only time-independent solution (i.e. $\omega_k =0$) to the original differential equation is one with $k=0$, so that it is also independent of the spatial coordinate. But clearly, a straight line $\phi(x,t) = A x + B$ is also a good time-independent solution to the equation. Why does this procedure fail to see it?

It is true that the Fourier transform of $Ax$ is a functional, not a regular function. But so is the Fourier transform of the constant $B$, or of any of the $e^{-i\omega_kt+ikx}$ for that matter. So I don't understand why this procedure misses that solution functional, but not the other.

This question extends to the wave equation in higher dimensions too, where the dispersion relation is now $\omega_k^2 = \vec {k}\cdot\vec k$. There I am still missing all time-independent solutions linear in the spatial coordinates. Moreover, we typically write the general solution as an integral over real $\vec k$: $\phi(\vec x,t)=\int d^3\vec k (A(k)e^{-i(\omega_kt-\vec k\cdot \vec x)}+B(k)e^{i(\omega_kt+\vec k\cdot \vec x)})$. But if we allow $\vec k$ to have complex components, then there too we can have time-independent solutions with non-zero $\vec k$.

Is this simply saying that $\phi(\vec x,t)=\int d^3\vec k (A(k)e^{-i(\omega_kt-\vec k\cdot \vec x)}+B(k)e^{i(\omega_kt+\vec k\cdot \vec x)})$ is $not$ the general solution?


1 Answer 1


It seems to me that if you want time independent solutions, you have to all the way back to your wave equation before you start finding solutions. Your time derivatives are zero and you don't really have a wave equation any longer.


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