I found this question on the physics side of our network and tried in vain to solve it. It's better suited here, so I'll post it.
We have a vibrating string. One side of the string is fixed, and the other side of the string vibrates by the force $F(t) = A \sin\omega t$.
Wave equation in 1D:
$u_{tt}=k^2u_{xx}$
Boundary conditions:
$u(0, t) = 0,u(l, t) = A \sin \omega t$
$u(x, 0) = u_t(x, 0) = 0$
My attempt:
The $\sin \omega t$ term was always going to worry me but I ran into problems well before that!
If we use separation of variables with Ansatz:
$$u(x,t)=X(x)T(t)$$
and separation constant $-m^2$, we obtain two ODEs:
$$\ddot{X}+m^2F=0$$
and:
$$\ddot{T}+m^2k^2G=0$$
The second has the classic solution:
$$T(t)=A\cos(mkt)+B\sin(mkt)$$
Using the boundary condition: $u(x,0)=u_t(x,0)=0$:
$$u(x,0)=0\implies T(0)=0$$
$$0=A\cos 0+B\sin 0\implies 0=A+0\implies A=0$$
So $T(t)=B\sin(mkt)$.
And for the second one:
$$u_t(x,0)=0\implies \dot{T}(0)=0$$
$$\dot{T}(0)=Bmk\cos 0=0 \implies B=0$$
Or $mk=0$. Either way this makes:
$$T(t)=0$$
A commenter mused "You need to account for resonant frequencies where nonzero solutions are possible" but I'm far from sure how to go about that.
Best Answer
We can search for a solution via the Laplace transform. The inverse laplace transform is what is going to be cumbersome, and I believe we might be able to extract a series solution of sorts from it.
We would have
$$s^2 U(x,s) - su(x,0)-u_t(x,0) = k^2 \partial_{xx}U(x,s)$$
Which leads to the Boundary Value Problem in $x$,
$$\begin{cases} k^2 U''(x,s) - s^2(x,s) = 0 \\ U(0,s)=0 \\ U(L,s)=A\displaystyle\frac{\omega}{\omega^2 + s^2} \end{cases}$$
This has the solution $$U(x,s) = A\left(\frac{\omega}{\omega^2 + s^2}\right)\text{csch}{\left(\frac{Ls}{k}\right)}\text{sinh}{\left(\frac{sx}{k}\right)}$$
and so,
$$u(x,t) = \frac{1}{2 \pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}U(x,s)e^{st}\,ds$$
In order to do this, consider the integral
$$\int_{\Gamma}U(x,z)e^{zt}\,dz$$
Where $\Gamma$ is the contour taken from the vertical line $x=c$, connected to a circular contour with radius $R$ in the first quadrant going around and connecting back to $x=c$ in the fourth quadrant.
Since our integral is made up of exponentials, it's easy to bound and show that it vanishes as $R\rightarrow \infty$ (I avoided this computation, but it would be nice if someone could verify my suspicion.). Further, we have simple poles at $z = \pm i \omega$ and $z=\frac{kn\pi i}{L}$ for $n \in \mathbb{Z}$. There is no pole at $z=0$ as it is a removable singularity.
Thus,
$$u(x,t) = \text{Res}(U(x,z)e^{z t},i\omega) +\text{Res}(U(x,z)e^{z t},-i\omega) +\sum_{n\in \mathbb{Z}}\text{Res}(U(x,z)e^{z t},\frac{k n \pi i}{L}) $$
To which we have the following from the residues:
$\text{Res}(U(x,z)e^{z t},i\omega) = \frac{1}{2 \pi i}(A e^{i t \omega} \csc{\frac{L \omega}{k}}\sin{\frac{\omega x}{k}})$
$\text{Res}(U(x,z)e^{z t},-i\omega) = -\frac{1}{2 \pi i}(A e^{-i t \omega} \csc{\frac{L \omega}{k}}\sin{\frac{\omega x}{k}})$
$\sum_{n\in \mathbb{Z}}\text{Res}(U(x,z)e^{z t},\frac{k n \pi i}{L}) = \frac{AkL\omega}{i} \sum_{n \in \mathbb{Z}} \left(\frac{(-1)^n}{(kn\pi)^2-(L\omega)^2} \right)\sin{\frac{n \pi x}{L}}e^{\frac{k n t \pi i}{L}}$
So that
$u(x,t) = A\csc{\left(\frac{L\omega}{k}\right)}\sin{(t\omega)}\sin{\left(\frac{\omega x}{k}\right)} + \frac{AkL\omega}{i}\sum_{n\in \mathbb{Z}}\left(\frac{(-1)^n}{(kn\pi)^2-(L\omega)^2} \right)\sin{\left(\frac{n \pi x}{L}\right)}e^{\left(\frac{k n t \pi i}{L}\right)}$
Now, noticing the summation in the index actually simplifies..we get
$u(x,t) = A\csc{\left(\frac{L\omega}{k}\right)}\sin{(t\omega)}\sin{\left(\frac{\omega x}{k}\right)} + 2AkL\omega\sum_{n=1}^{\infty}\left(\frac{(-1)^n}{(kn\pi)^2-(L\omega)^2} \right)\sin{\left(\frac{n \pi x}{L}\right)}\sin{\left(\frac{k t n \pi}{L}\right)}$
Here's a gif of the string in action for $A=L=k=\omega=1$