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?
