1D Wave equation mixed boundary conditions and I.C.

Question

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//

Answer 1

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}

Answer 2

Equivalently, without doing the "relevement" with the BC, you can consider what happens physically to your system: you continuously excite the field with your harmonic boundary condition, which makes a forced mode $u_f$ appear. This mode must satisfy your BC and the wave equation, and take the form of a stationnary wave looking like $$u_f(x,t)=\frac{A}{\sin(\frac{\Omega L}{c})}\sin(\frac{\Omega}{c}x)\cos(\Omega t)$$ Then this perturbation $u_f$ acts like an initial condition for your field which behaves according to its own eigenmodes $u_i$ for $i\in\mathbb{N}$.

We can also write $u_m$ the part of the solution comprised of all the eigenmodes, so $u_m=\sum_i u_i$. These modes are solutions of your equation for homogeneous BC, and you find the adapted IC by considering at a given time $t_0$ (usually the time where the first emitted wave reaches the scattering boundary, and where the system gets the "drum" behavior and the eigenmodes, so something like $t_0=L/c$), and look for $u_f(x,0)$ and $\partial_t u_f(x,0)$.

Then, writting $u$ as the sum $u=u_f+u_m=u_f+\sum_i u_i$, you have: $$u_m(x,t_0)=u(x,t_0)-u_f(x,t_0)$$ and $$\partial_t u_m(x,t_0)=\partial_t u(x,t_0)-\partial_t u_f(x,0)$$.

Since before $t_0$ the emitted wave is not scattered, it is only a propagated wave of the form $A\cos(\Omega(t-x/c))$, so you know it at $t_0$ together with its derivative. Then you have "IC" for $u_m$ at $t_0$, apply a time shift $t=T+t_0$ and then get the IC of your new problem at $T=0$, with homogeneous Dirichlet BC, and then apply separation of variables, Fourier series expansion, using the Fourier expansion of the IC, and identifying term by term, as explained in the answer before.

At then end you apply the reverse $T=t-t_0$ and write $u$ as the sum $$u=u_f+\sum_i u_i$$