[Math] When solving the wave equation by separation of variables, is the separation constant always negative

partial differential equationswave equation

I am currently using Stanley F. Farlow's books Partial Differential Equations for Scientists and Engineers to prepare for an upcoming exam. On page 155 he illustrates how to solve the finite vibrating string problem. He employs the standard separation of variables technique by assuming a solution of the form $u(x,t)=X(x)T(t)$, justifying the existence of a separation constant $\lambda$ and separating the PDE into two second order ODEs. He then states:

… It will be left as an
exercise for the reader to verify that only negative values of
$\lambda$ give a feasible (nonzero and bounded) solution

Two pages later he writes:

…So we're done, the solution is

$$\boxed{u(x,t)=\sum_0^{\infty}\sin{(\frac{n \pi
x}{L}})[a_n\sin{(\frac{n \pi \alpha t}{L}})+b_n\cos{(\frac{n \pi
\alpha t}{L}})]}$$

Is it always the case that only negative values of $\lambda$ give feasible solutions or are there finite vibrating string problems where only positive or zero solutions are viable? Will the solution to the finite vibrating string problem always have the form of the equation I posted above?

Best Answer

The separation constants (eigenvalues) depend on the boundary conditions more than on the PDE: they are the same for wave and heat equations. The question is when there exists a nontrivial solution of $X'' = \lambda X $ with given (homogeneous) boundary conditions. Here is a summary of what happens:

  1. Dirichlet condition $\implies \lambda <0$
  2. Neumann condition $\implies \lambda \le 0$
  3. Mixed condition (Dirichlet at one end, Neumann at the other) $\implies \lambda <0$
  4. Robin conditions, $X'(a)=c_1X(a)$ and $X'(b)=c_2 X(b)$, are more complicated:
    • if $c_1 \le 0$ and $c_2\ge 0$ (dissipation at the boundary), then $\lambda\le 0$
    • if $c_1 >0 $ and $c_2<0$ (absorption at the boundary), then there is a positive eigenvalue
    • if $c_1,c_2 $ have the same sign, then the outcome also depends on their size.

I would not dismiss unbounded conditions as unfeasible. For example, the Neumann condition says the string is not being held. So, if given positive initial velocity it will just keep on moving forever, which is correctly described by an unbounded solution such as $u(x,t)=t$.

Related Question