Hyperbolic, parabolic, elliptical PDE related to under-, critical- and overdamped in harmonic osciallation A damped harmonic oscillator has three cases for the damping: underdamped, critically damped and overdamped. With partial differential equations, I know the hyperbolic wave equation, the parabolic heat equation and the elliptical Laplace equation.
Since underdamping means oscillation and therefore waves and the heat diffusion looks like a critically damped fluid, I was wondering, if those have anything in common.
 A: OK, let's look at a general second order damped oscillator.  We have
$$\begin{align}
F&=-kx -bv\\
ma &= -kx -bv\\
0&= {\ddot x} + \beta {\dot x} + \omega^{2}x
\end{align}$$
We can solve this by guessing the solution $x = x_{0}e^{k\,t}$.  Substituting this into our differential equation, and dividing through by $x_{0}e^{k\,t}$, we get
$$k^{2} + \beta k + \omega^{2}=0$$
In general, we get two distinct solutions 
$$k = \frac{ -\beta \pm \sqrt{\beta^{2} - 4\omega^{2}}}{2}$$
In general, this is a complex number.  In the case where $2\omega > \beta$, the imaginary part of this number is nonzero, and you have an underdamped oscillation, since your solution will be of the form $x = x_{0}e^{\rm Re}e^{i {\rm Im}}$.  Note that the simple harmonic oscillator is a special case of this, where the real part is zero.
If $\beta > 2\omega$, then the imaginary part will be zero, the motion will be overdamped, and your solutions will be two different rates of exponential decay in the particle's motion.
In the case where the discriminant is zero, the motion will be critically damped, and your two linearly independent solutions will coincide.  Additional analysis will be required to find the second linearly independent solution.
The PDE case decompositions is almost identical, but more complicated--you chracterize the pde's by taking only the highest derivative terms and fourier transforming, and then analyzing the eigenvalues.  In the ODE case, you take the laplace transform and solve the polynomial.  Either way, you characterize the three solutions by finding a characteristic polynomial from the equations, and then characterize the types of growth you get.
