@MichaelRenardy I have edited to restrict to only those systems where spectrum does determine stability. Note that I am not asking for computation of the full spectrum, but rather concluding stability using a combination of analysis of the truncated system and some estimate of the neglected modes.

Axler is not good for first course, especially for a self-study. Go with Strang.

No, I meant your proof claiming that (3) is trivially satisfied. Lets take $v$ to be a constant vector field pointing in horizontal direction, parallel to x axis. Clearly (3) doesn't hold for most points on boundary of $M$.

Do you see why its true when M is a disk in a plane? Does your 'proof' given at end of your question hold in this case?

Unabomber has to be #1 on this list.

There are no videos on that page

I thought "pure math" when applied becomes applied math, no ?

I highly doubt anyone will accept unless the workshop is 2+ years away in future (due to COVID). You might have better luck organizing something online.

Isn't all this addressed by symbolic dynamics ? en.wikipedia.org/wiki/Symbolic_dynamics

Not sure if RMT is applicable, but here's a work related spectral graph theory to grain boundary modeling: Johnson, Oliver K., Jarrod M. Lund, and Tyler R. Critchfield. "Spectral graph theory for characterization and homogenization of grain boundary networks." Acta Materialia 146 (2018): 42-54.

@Tao.Zhou I believe you can look into Ref. [4] in the paper to see why the ground state has lowest eigenvalue. In general, such operators have a unique positive eigenfunction (i.e., all other eigenfunctions must be positive and negative somewhere) and it corresponds to lowest eigenvalue, thats why it is called ground (i.e., lowest) state

Seems like a Rayleigh quotient argument ? Q is the ground state, i.e. eigenfunction corresponding to lowest eigenvalue, which is equal to 0 here since $L_{-}Q=0$. Now apply rayleigh quotient thm to prove positivity.

see this: mathoverflow.net/questions/327863/…

Divergence controls the rate of expansion/contraction of volumes (or Lebesgue measures), think about case where initial data is uniform small ball around origin. If div=0, this ball will deform but volume will be conserved (e.g. it will become ellipse in the case I mentioned above). If div>0, volume will grow. Hence, a neccesary condition for all traj. to go to 0 is div<0, but certainly it is not sufficient. E.g. $\dot{x}=x$, $\dot{y}=-2y$.

Take a 2D example with 0 divergence. $\dot{x}=x$,$\dot{y}=-y$. See what you get.

I am saying the opposite, that is if div v=0, then ODE cannot decay to 0 for all initial conditions.. Just take tiny circle around origin and apply divergence thm. Since all traj. are going into that circle, the line integral will be non-zero, but the area integral is 0 if div.v=0.

V is invompressible if its divergence is 0 everywhere. Such flows cannot have sinks as required by your 2nd eq.