I'm trying to understand what a correct procedure is for solving the following wave equation on a semi-infinite plane $-\infty < x < \infty$, $0 \le z < \infty$: $$ \begin{cases} \nabla^2 \Psi(x,z,t) - \dfrac{1}{c^2}\dfrac{\partial^2 \Psi(x,z,t)}{\partial t^2}=f(z,t)H(t)\sin (q_0x)\\ \dfrac{\partial^2 \Psi(x,0,t)}{\partial x\partial z} = 0\\ \dfrac{\partial^2 \Psi(x,0,t)}{\partial x^2}-\dfrac{\partial^2 \Psi(x,0,t)}{\partial z^2}=0\\ \Psi(x,z,0)=\dfrac{\partial \Psi(x,z,0)}{\partial t}=0 \end{cases}$$
where $H(t)$ is a Heaviside step function.
I started by separating the variables which leads to the solution:
$$\Psi(x,z,t)=T(t)(A\sin (kz)+B\cos(kz))(C\sin(qx)+D\cos(qx))$$
The first boundary conditions leads to $A=0$ and the second to $k=q$. To deal with the inhomogeneous part, one would generally do the expansion
$$\iint T_{qk}(t)\cos(kz)(C\sin(qx)+D\cos(qx))dqdk = \iint (C_s\sin(qx)+C_c\cos(qx))\cos(kz)$$
where $C_s$ and $C_c$ are the sine and cosine series coefficients of the source term. Now I'm struggling how to handle the fact that $k=q$. Moreover, I think that the inhomogeneous part of the equation implies that $q=q_0$, which if true, makes things much worse, since then we can't do any series expansion at all. I'm assuming I made an error somewhere in my reasoning and I would appreciate some help trying to figure this out.
Additionally, I can actually find the particular solution using the Fourier transform technique. If a take the Fourier transform of $\Psi$ and the source term with respect to $x$ and $t$, the PDE becomes an ODE with $z$ the only independent variable. Since the source term is well behaved in the Fourier space, I can find the solution as a sum of the homogeneous and particular solution, $\hat{\Psi} = \hat{\Psi}_{hom}+\hat{\Psi}_{par}$. To get the complete solution in real space, I need to take the inverse Fourier transform, but that can only be analytically solved for the particular term, since the integral for the homogeneous term is to complicated. Is there any way knowing the particular solution can help me figure out the complete solution?
Edit: As requested in the comments, I’m writing down the entire problem. I’m trying to solve the thermoelastic equations for the material displacement $\textbf{u}=(u_x(x,z,t),u_z(x,z,t))$
$$ \begin{cases} v_{44}^2\nabla^2\textbf{u}+(v_{11}^2-v_{44}^2)\nabla(\nabla\cdot\textbf{u})-\frac{\partial^2\textbf{u}}{\partial t^2}=\gamma\nabla T(x,z,t)\\ \frac{\partial u_x(x,0,t)}{\partial z}+\frac{\partial u_z(x,0,t)}{\partial x} = 0\\ 2v_{44}^2\frac{\partial u_z(x,0,t)}{\partial z}+(v_{11}^2-2v_{44}^2)(\nabla\cdot\textbf{u}(x,0,t))-\gamma T = 0 \end{cases} $$
First I made these boundary conditions homogeneous by taking $\textbf{u}=\textbf{v}+\textbf{w}$ where $\textbf{w}$ is chosen such that it satisfies the $-\gamma T$ part of the second boundary condition, so that $\textbf{v}$ only has to satisfy the homogeneous boundary conditions.
If we then express $\textbf{v}$ in terms of longitudinal ($\Phi$) and transverse ($\boldsymbol\Psi$) potentials (Helmholtz decomposition), $\textbf{v}=\nabla\Phi + \nabla\times\boldsymbol\Psi$, we can rewrite the equation as
$$\nabla(v_{11}^2\nabla^2\Phi-\frac{\partial^2 \Phi}{\partial t^2}-f(x,z,t)) + \nabla\times(v_{44}^2\nabla^2\boldsymbol\Psi-\frac{\partial^2\boldsymbol\Psi}{\partial t^2}-\textbf{g}(x,z,t))=0$$
where $\textbf{g}(x,z,t)$ only has a y component. With the additional requirement of the Helmholtz decomposition that $\nabla\cdot\mathbf{\Psi}=0$, we can assume that $\mathbf{\Psi}$ only has a y component too and can for the purpose of solving the equation be considered a scalar (here I’m following a similar derivation from a published paper where they did not justify this step in detail, so I’m not confident that it’s 100% true).
I rewrote the boundary conditions in terms of $\Phi$ and $\Psi$ and got
$$\frac{\partial^2\Phi(x,0,t)}{\partial x\partial z}+\frac{\partial^2\Psi(x,0,t)}{\partial x^2}-\frac{\partial^2\Psi(x,0,t)}{\partial z^2}=0$$
$$2v_{44}^2(\frac{\partial^2\Phi(x,0,t)}{\partial z^2}+\frac{\partial^2\Psi(x,0,t)}{\partial x\partial z})+(v_{11}^2-2v_{44}^2)(\frac{\partial^2\Phi(x,0,t)}{\partial x^2}+\frac{\partial^2\Phi(x,0,t)}{\partial z^2})=0$$
I assumed $\Phi$ and $\Psi$ will satisfy this both separately (maybe this is wrong?), so I got for $\Psi$ the condition I wrote above.
