2
$\begingroup$

As we know that skew-symmetricity means

$A=-A^\top$

where $A\in\mathbb{R}^{n\times n }$. But recently I came across an inequality that states,

$A+A^\top\preceq0$ can also be considered as an approximation towards skew symmetricity. (This particular approximation can be widely found out while dealing with port hamiltonian systems for partial differential equations/a straight example would be when this acts a necessary assumption for formulating the port hamiltonian of a counter heat flow model represented in terms of PDE). The PDE is described as,

$\frac{\partial}{\partial{t}}\begin{pmatrix}\theta_1\\\theta_2\end{pmatrix}=\begin{pmatrix}-v_1&0\\0&v_2\end{pmatrix}\frac{\partial}{\partial \zeta}\begin{pmatrix}\theta_1\\\theta_2\end{pmatrix}-\begin{pmatrix}h_1&-h_1\\-h_2&h_2\end{pmatrix}\begin{pmatrix}\theta_1\\\theta_2\end{pmatrix}\\ ~~~ ~~~~~~~~~~~:=B\frac{\partial}{\partial \zeta}\begin{pmatrix}\theta_1\\\theta_2\end{pmatrix}+A\begin{pmatrix}\theta_1\\\theta_2\end{pmatrix}$

where $\theta_1$ and $\theta_2$ represents the temperature at space position $\zeta\in(a,b)$ and time $t>0$, respectively. The positive constants $v_1,v_2,h_1$ and $h_2$ are physical parameters and the boundary conditions are,

$\theta_1(a,t)+k_1\theta_2(a,t)=0, k_2\theta_1(b,t)+\theta_2(b,t)=0$

But quite frankly I am not sure how this approximation could be made, can anyone help me with this.

$\endgroup$
  • 1
    $\begingroup$ Could you link to the specific context, or tell us a little bit more about it? What is the specific role of $A$ in this PDE? $\endgroup$ – Omnomnomnom Jun 29 '16 at 8:44
  • 1
    $\begingroup$ This sounds rather strange. It would mean that a negative definite symmetric matrix $A$ qualifies as being "approximately skew symmetric". $\endgroup$ – Andreas Cap Jun 29 '16 at 13:01
  • $\begingroup$ @Omnomnomnom You can find this inequality at "An Introduction to Infinite-Dimensional Linear Systems Theory" by Ruth F. Curtain, Hans Zwart. ISBN: 978-1-4612-8702-5 (Print) 978-1-4612-4224-6 (Online). $\endgroup$ – Raaja Jun 29 '16 at 15:38
  • $\begingroup$ @Andreas Cap Even I have the same doubt. $\endgroup$ – Raaja Jun 29 '16 at 15:40
  • $\begingroup$ @Omnomnomnom Moreover the $A$ in PDE describes the mapping of the system, a rather clear explanation could be found out in the previously mentioned reference. $\endgroup$ – Raaja Jun 29 '16 at 20:52
1
$\begingroup$

It seems that $A$ dictates the evolution of the system over time. My best guess is that the relevant piece of information here is

$A + A^T = 0 \implies$ $A$ has imaginary eigenvalues $\implies$ energy is conserved

On the other hand, $A + A^T \preceq 0 \implies$ $A$ has eigenvalues with a non-positive real part $\implies $ energy decays with a steady state solution corresponding to purely imaginary eigenvalues.

At the very least, all of this is true when $B$ is taken to be zero, and I'm assuming something along these lines holds in general. It might also be the case that the decay aspect gives some stability to the solution so that numerical methods converge nicely; hence, a system with energy decay might be used as a nicer approximation of the actual system.

$\endgroup$
  • $\begingroup$ Again, this is all a guess based on intuition from the Schrodinger equation, but maybe it helps. $\endgroup$ – Omnomnomnom Jun 29 '16 at 22:10
  • $\begingroup$ It seems correct...and yes it kind of holds when I formulate my energy space in $\mathbb{L}^2$. Thanks. $\endgroup$ – Raaja Jun 30 '16 at 6:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.