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

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 00\\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

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//

• separation of variables won't work unless the boundary conditions are homogeneous – Dylan Feb 1 at 18:52
• Thanks again Dylan, any advice on what method to use? – SimpleProgrammer Feb 1 at 18:54
• I may type out an answer later. Here is the general method. Look under "Nonhomogeneous boundary conditions". – Dylan Feb 1 at 18:57
• Here's a different method using the characteristic lines that maybe easier computationally. – Dylan Feb 1 at 19:03
• I checked your links, although the "characteristic lines" solution looks elegant and sophisticated, I am more looking for some other method that let's me use separation of variables. Can I make a similar claim to the question you linked that u(x,t) = v(x,t) + w(t) so that my boundary conditions becomes homogeneous? – SimpleProgrammer Feb 1 at 19:53

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 $$T_n(t)$$

$$\begin{cases} T_n'' + \dfrac{c^2(2n+1)^2\pi^2}{4L^2} T_n(t) = b_n\Omega^2 A \cos(\Omega t) \\ T_n(0) = -b_nA \\ T_n'(0) = 0 \end{cases}$$