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Consider the wave equation $$ y_{tt} = \Delta y - \epsilon y_t $$

on $\Omega\subset R^n$, with Dirichlet boundary conditions. Where $\epsilon >0$.

Is it possible to find an explicit value $\sigma=\sigma(\epsilon ) > 0$ such that the solution $(y,y_t)$ verifies the estimate

$$ \|(y,y_t)\| \le M e^{-\sigma t} $$

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If you diagonalize the Laplacian with mode functions $f_i$ with Dirichlet boundary conditions on $\Omega$, $\Delta f_i = -k_i^2 f_i$, then a simple calculation shows that a solution of the form $f_i e^{-i\omega_i t}$ will have a frequency of the form $$ \omega_i = \pm\frac{\sqrt{4k_i^2-\epsilon^2}}{2} - \frac{i\epsilon}{2} . $$ As you can see, the imaginary part of the frequency is always the same, so (modulo some technical details about expressing any solution in terms of mode functions) your decay constant is $\sigma = \epsilon/2$.

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    $\begingroup$ For the above analysis you need $\epsilon$ to be sufficiently small compared to the first eigenvalue of the Laplacian. Consider the case $\Omega = [0,\pi]\subset \mathbb{R}$, and $\epsilon = 4$ which is more than twice the first eigenvalue. One checks that $$ y(x,t) = e^{(-2 + \sqrt{3})t} \sin(x)$$ is a solution and decays strictly slower than $e^{-2t}$ ... See en.wikipedia.org/wiki/Damping#Over-damping_.28.CE.B6_.3E_1.29 also for the classical harmonic oscillator analogue. $\endgroup$ – Willie Wong Apr 23 '13 at 9:19
  • $\begingroup$ Of course. I intuitively considered $\epsilon$ to be "sufficiently small". $\endgroup$ – Igor Khavkine Apr 23 '13 at 10:20

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