I'm looking into the following waves equation problem:
A string is driven at $x=0$: $\psi(x=0,t)=a\cos(\omega_{d}t)$.
It is fixed at $x=L$: $\psi(x=L,t)=0$.
In the simple case when there is no damping, guessing a solution in the form of $$\psi(x,t)=A\sin(k_{d}x+\theta)\text{cos}(\omega_{d}t)$$ works and the solution gives $$\psi(x,t)=\frac{a}{\sin(k_{d}L)}\sin\left(k_{d}(L-x)\right)\cos(\omega_{d}t)$$.
I am trying to solve for the case in which the string is damped, and the string wave equation is given by $$\frac{\partial^{2}\psi(x,t)}{\partial t^{2}}=c^{2}\frac{\partial^{2}\psi(x,t)}{\partial x^{2}}-\Gamma\frac{\partial\psi(x,t)}{\partial t} .$$ I assume that the solution would also oscillate at frequency $\omega_d$, which means that $k=k_r+ik_i$ would have to be complex. However, I am unable to solve the waves equation with this assumption and I am unable to find a solution. I would be happy for some input.
I will write down my attempt at the solution:
$\psi(x,t)=A\sin(k_{d}x+\theta)\text{cos}(\omega_{d}t)$
Boundary condition at $x=0$:
$a\cos(\omega_{d}t)=A\sin(k_{d}x+\theta)\text{cos}(\omega_{d}t)\rightarrow\,\,\,a=A\sin(\theta)$
At $x=L$:
$0=A\sin(k_{d}L+\theta)\cos(\omega_{d}t)\rightarrow\theta_{r}+i\theta_{i}=n\pi-k_{r}L-ik_{i}L$
$A=\frac{a}{\sin(n\pi-k_{r}L-ik_{i}L)}$
$\psi(x,t)=\frac{a}{\sin(n\pi-k_{r}L-ik_{i}L)}\sin\left(k_{r}(x-L)+n\pi-ik_{i}\left(x-L\right)\right)\cos(\omega_{d}t)$
$\psi(x,t)=\frac{a}{\sin(k_{r}L+ik_{i}L)}\sin\left(k_{r}(L-x)-ik_{i}\left(L-x\right)\right)\cos(\omega_{d}t)$
At this point I assume I need to demand that the imaginary part is 0, but I get that this is not possible unless $k_i=0$, which is impossible. The math to get to this condition is ugly.