Statement in textbook: inhomogeneous wave equation has finite solutions only if RHS is orthogonal to homogeneous solutions Summary: Brillouin states that an inhomogeneous hyperbolic PDE has a finite solution only if the RHS is orthogonal to the homogenous solutions

Hi, I've been reading Brillouin's 'Wave Propagation in Periodic Structures'.
About the following equation
$$\nabla^2u_1+\frac{\omega^2_0}{V_0^2}u_1 = R(r)$$
Brillouin states that "it is well known that such an equation possesses a finite solution only if the right-hand term is orthogonal to all solutions of the homogeneous equation:"
$$\iint_{\text{all space}} u_1^*R(r) dr = 0$$
This is not a property of hyperbolic PDEs I've come across before. I wasn't able to find anything in my PDE textbooks. Would anyone be able to suggest why this is the case? I would be very appreciative.
 A: Let $v$ be a solution of the homogeneous equation:
$$\nabla^2v+\frac{\omega_0^2}{V_0^2}v=0 \tag{1}$$
Let's take the inhomogeneous equation
$$\nabla^2u_1+\frac{\omega_0^2}{V_0^2}u_1=R(r) \tag{2}$$
We multiply both sides of $(2)$ by a solution $v$ of the homogeneous eq. and take the integral over all space:
$$\int v\nabla^2u_1 dr + \int \frac{\omega_0^2}{V_0^2}vu_1dr=\int v R(r) dr  \tag{3} $$
We concentrate on the term $\int v\nabla^2u_1 dr$. Expanding the nabla operator, it becomes
$$ \int \left( v\partial_x^2u_1+ v\partial_y^2u_1 + v\partial_z^2u_1 \right) dr  \tag{4}$$
If $v$, $u_1$ and their derivatives go to zero at infinity, then two successive integrations by parts yield (I consider only the x-derivative term for simplicity): $$\int v\partial_x^2u_1 dr=-\int (\partial_xv) (\partial_xu_1) dr=\int (\partial_x^2v)u_1 dr \tag{5}$$
So we can write
$$\int v\nabla^2u_1 dr = \int u_1\nabla^2v dr  \tag{6}$$
Now equation $(3)$ can be rewritten by means of $(6)$ as
$$\int u_1\nabla^2v dr + \int \frac{\omega_0^2}{V_0^2}vu_1dr=\int v R(r) dr \tag{7} $$
The left side of $(7)$ can be written as
$$\int u_1 \left( \nabla^2v + \frac{\omega_0^2}{V_0^2}v \right) dr  \tag{8} $$
We recognize in the expression in brackets in $(8)$ the left side of $(1)$. Then the expression in brackets is zero, and $(8)$ as a whole is consequently zero. Then $(7)$ becomes
$$0=\int v R(r) dr \tag{9} $$
or, switching sides,
$$\int v R(r) dr=0  \tag{10} $$
which means that $R$ is orthogonal to the solutions $v$ of the homogeneous equation.
