Are solutions coordinate invariant? In the case of electromagnetism, we can solve the sorceless wave equation in Cartesian coordinates ($x$,$y$,$z$) getting plane waves as solutions: $$ u(x) = A(x-ct) + B(x+ct) $$ and actually I am not entirely sure how to treat $y$ and $z$ here..
But if we go to Spherical polar coordinates, the solutions have an extra $\frac{1}{r}$ factor: $$ u(r) = \frac{1}{r}C(r-ct) + \frac{1}{r}D(r+ct) $$.
Physically, how is it possible that we get two different behaviours from the same original equation? Shouldn't coordinate transformations leave the solution invariant (or at least covariant)?
 A: The two solutions are different because they have different boundary conditions. In the first case, the equation is indeed
$$ \frac{\partial^2u}{\partial t^2} = c^2 \nabla^2 u = c^2 \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) u. $$
Here though we specify $u(t,x=x_0)$ to be some value independent of $y$ or $z$. That is, the source is an infinite plane. $y$ and $z$ drop out, and we are left with
$$ \frac{\partial^2u}{\partial t^2} = c^2 \frac{\partial^2u}{\partial x^2}, $$
which leads to the general solution you give.
In the second case, we rewrite the Laplacian in spherical coordinates:
$$ \frac{\partial^2u}{\partial t^2} = c^2 \nabla^2 u = \frac{1}{r^2} \left(\frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r} + \frac{1}{\sin\theta} \frac{\partial}{\partial\theta} \sin\theta \frac{\partial}{\partial\theta} + \frac{1}{\sin^2\theta} \frac{\partial^2}{\partial\phi^2}\right) u. $$
In order for this to reduce to the form for which you have a solution,
$$ \frac{\partial^2u}{\partial t^2} = \frac{c^2}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial u}{\partial r}\right), $$
we must have a boundary condition that does not depend on $\theta$ or $\phi$ -- i.e. $u(t,r=r_0)$ is specified. This corresponds to setting the field on a sphere rather than a plane.
You could use Cartesian coordinates and a spherical boundary condition, in which case transforming your solution into spherical coordinates at the end would produce the expected result. And similarly for spherical coordinates and a plane boundary. But this only works if the boundary conditions are physically the same, and symmetry in $y$ and $z$ is not the same as symmetry in $\theta$ and $\phi$. 
A: Note that you haven't actually found the general solution in spherical coordinates... What you have there is a solution known as a spherical wave, which describes a set of spherically symmetric wave fronts that diverges from (or converges towards) the origin $r=0$.
However, in general, a wave could also be a function of the angles $\theta$ and $\phi$, which breaks the spherical symmetry above. For example, if you want to describe a plane wave propagating along the $x$-axis, then clearly your wave amplitude should be a function of $x = r \cos \theta \sin \phi$, so you need the $\theta$ and $\phi$ dependence to describe plane waves.
