# Solving wave equations with Fourier transform: where are the time-independent solutions?

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?