A Class of Well Posed Damped PDEs

A large class of well posed PDEs is given by [45]

$\displaystyle {\ddot y} + 2\sum_{k=0}^M q_k \frac{\partial^{2k+1}y}{\partial x^{2k}\partial t} + \sum_{k=0}^N r_k\frac{\partial^{2k}y}{\partial x^{2k}} \protect$ (D.5)

Thus, to the ideal string wave equation Eq.$ \,$(C.1) we may add any number of even-order partial derivatives in $ x$, plus any number of mixed odd-order partial derivatives in $ x$ and $ t$, where differentiation with respect to $ t$ occurs only once. Because every member of this class of PDEs is only second-order in time, it is guaranteed to be well posed, as we now show.

To show Eq.$ \,$(D.5) is well posed [45], we must show that the roots of the characteristic polynomial equationD.3) have negative real parts, i.e., that they correspond to decaying exponentials instead of growing exponentials. To do this, we may insert the general eigensolution

$\displaystyle y(t,x) = e^{st+vx}$

into the PDE just like we did in §C.5 to obtain the so-called characteristic polynomial equation:

$\displaystyle s^2 + 2q(v)s + r(v) = 0


q(v) &\isdef & \sum_{k=0}^M q_k v^{2k}\\
r(v) &\isdef & \sum_{k=0}^N r_k v^{2k}

Let's now set $ v=jk$, where $ k=2\pi/\lambda$ is radian spatial frequency (called the ``wavenumber'' in acoustics) and of course $ j=\sqrt{-1}$, thereby converting the implicit spatial Laplace transform to a spatial Fourier transform. Since there are only even powers of the spatial Laplace transform variable $ v$, the polynomials $ q(jk)$ and $ r(jk)$ are real. Therefore, the roots of the characteristic polynomial equation (the natural frequencies of the time response of the system), are given by

$\displaystyle s = -q \pm \sqrt{q^2 - r}.

Next Section:
Proof that the Third-Order Time Derivative is Ill Posed
Previous Section:
Poles at