1
$\begingroup$

I am hoping to receive some thoughts on an analytical solution of the following equation, $$ u_{tt}-u_{xx}=\gamma u, $$ for $x\in (0,L)$ and $t\in (0,\infty)$, with initial conditions $$ u(0,x)=0, \qquad u_t(0,x)=0,$$ and boundary conditions $$ u(t,0)=f(t), \qquad u_x(t,L)+cu(t,L)=g(t),$$ where $\gamma$ and $c$ are known constants.

I have tried separation of variables, $u=T(t)X(x)$, but obtain only a trivial solution for $T$.

$\endgroup$
1
  • 1
    $\begingroup$ What did you expect? If there is no initial wave, what would cause one? $\endgroup$
    – user247327
    Jul 21, 2019 at 18:41

1 Answer 1

1
$\begingroup$

Using separation of variables $u(x,t) = X(x) T(t)$, we have $$ \frac{T''}{T} = \frac{X''}{X} + \gamma = -\lambda \, , $$ i.e. $$ {T''} + \lambda T = 0 \qquad\text{and}\qquad {X''} + (\lambda-\gamma) X = 0 $$ for which the Fourier series approach could be applied. However, this might not be an easy task in the case where the boundary conditions are time-dependent.

The problem at hand is very similar to those of a one-dimensional string with vibrating ends. A possible strategy would consist in searching the solution as a sum of eigenfunctions -- so-called normal mode decomposition or expansion. To find the normal modes, one solves the same problem with homogeneous boundary conditions ($f=0$, $g=0$), see e.g. this link and chap. 4 of (1). Note that this post is closely related.

Alternatively, one may use the method of characteristics for the system $$ \begin{aligned} u_t \qquad &= v\\ \varepsilon_t -v_x &= 0\\ v_t - \varepsilon_x &= \gamma u \end{aligned} $$ where $\varepsilon = u_x$, but this might not be straightforward due to the right-hand side. You may have a look at (1), chap. 12.

Lastly, one may solve the problem by using Laplace transforms (see e.g. chap. 13 of (1)).


(1) R. Habermann, Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, 5th ed. Pearson Education Inc., 2013.

$\endgroup$
1
  • $\begingroup$ Dear Harry49, thank you for the response. It was informative and helpful. $\endgroup$
    – lsbaker1
    Jul 27, 2019 at 18:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .