0
$\begingroup$

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//

$\endgroup$
  • $\begingroup$ separation of variables won't work unless the boundary conditions are homogeneous $\endgroup$ – Dylan Feb 1 at 18:52
  • $\begingroup$ Thanks again Dylan, any advice on what method to use? $\endgroup$ – SimpleProgrammer Feb 1 at 18:54
  • 1
    $\begingroup$ I may type out an answer later. Here is the general method. Look under "Nonhomogeneous boundary conditions". $\endgroup$ – Dylan Feb 1 at 18:57
  • 1
    $\begingroup$ Here's a different method using the characteristic lines that maybe easier computationally. $\endgroup$ – Dylan Feb 1 at 19:03
  • $\begingroup$ 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? $\endgroup$ – SimpleProgrammer Feb 1 at 19:53
1
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.