I have been searching for a solution online, but cannot find one that fits the B.C. and I.C. for this wave equation. I read through this PDF, page 7; although I had similar conditions I just obtained trivial solutions.
Now, the system is
$\left\{\begin{array}{ll}u_{tt}(x,t)-c^2 u_{xx}(x,t)=0,\quad 0<x<L,\quad t>0\\u(0,t)=0,\quad u_{x}(L,t)=A\cos(\Omega t),\quad t>0\\u(x,0)=0,\quad u_{t}(x,0)\quad 0<x<L\end{array}\right.$
As usual I use separation of variables and then obtain the solutions for $X(x)$ and $T(t)$
$\left\{\begin{array}{ll}X(x)=B\cos(\omega _{1}x)+C\sin(\omega _{1}x)\\T(t)=D\cos(\omega _{2}t)+E\sin(\omega _{2}t)\\\end{array}\right.$
$\omega ^{2} _{1}=\lambda / c^2,\quad \omega ^{2}_{2}=\lambda,\quad \lambda>0$.
I use the I.C. and I obtain $D=E=0$. This is wrong. Have I used the wrong method solving this system?
Best regards//
Best Answer
You need to subtract off the boundary conditions before you can apply separation of variables. Try a solution of the form
$$ u(x,t) = Ax\cos(\Omega t) + v(x,t) $$
where the boundary function was obtained from $f(x)A\cos(\Omega t)$ such that $f(0)=0$ and $f'(L) = 1$
Then $v(x,t)$ satisfies
\begin{cases} v_{tt} - c^2v_{xx} = \Omega^2 Ax\cos(\Omega t)\\ v(0,t) = v_x(L,t) = 0 \\ v(x,0) = -Ax \\ v_t(x,0) = 0 \end{cases}
The equation is no longer homogeneous but you can still decompose into eigenfunctions by solving
\begin{cases} X''(x) + \lambda^2 X(x) = 0 \\ X(0) = X'(L) = 0 \end{cases}
Then we have
$$ v(x,t) = \sum_{n=0}^\infty T_n(t) \sin\left(\frac{(2n+1)\pi}{2L} x\right) $$
Plugging into the equation gives
$$ \sum_{n=0}^\infty \left[ T_n''(t) + \frac{c^2(2n+1)^2\pi^2}{4L^2} T_n(t) \right] \sin\left(\frac{(2n+1)\pi}{2L} x\right) = \Omega^2 Ax \cos(\Omega t) $$
Decompose the RHS (and also the initial condition) into it's corresponding series
$$ x = \sum_{n=0}^\infty b_n \sin\left(\frac{(2n+1)\pi}{2L} x\right) $$
where
$$ b_n = \frac{\int_0^L x \sin\left(\frac{(2n+1)\pi}{2L} x\right)\ dx}{\int_0^L \sin^2\left(\frac{(2n+1)\pi}{2L} x\right)\ dx} $$
You'll get a family of IVPs in $T_n(t)$
\begin{cases} T_n'' + \dfrac{c^2(2n+1)^2\pi^2}{4L^2} T_n(t) = b_n\Omega^2 A \cos(\Omega t) \\ T_n(0) = -b_nA \\ T_n'(0) = 0 \end{cases}