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Schrodinger operators, operators on manifolds, general differential operators, numerical studies, integral operators, discrete models, resonances, non-self-adjoint operators, random operators/matrices
9
votes
3
answers
1k
views
Functions of pseudodifferential operators
Suppose I have a self-adjoint pseudo-differential operator $A$ on $\mathbb{R}^n$ and a continuous function $f$ (possibly bounded, or Schwartz, or compactly supported) on its spectrum. Then I can consi …
4
votes
1
answer
361
views
Functional Calculus of closed operators
I learned that there is a holomorphic functional calculus for closed operators: If $T$ is a closed operator on a Hilbert space, and $f$ is a function that is holomorphic on some open subset $\Omega$ o …
4
votes
0
answers
112
views
Determinant of quotient of unbounded operators
I have been trying to prove this for a while but failed so far.
Let $A$ and $B$ are two positive, self-adjoint operators with compact resolvent on a Hilbert space $H$ defined on the same dense domai …
3
votes
Spectrum of Mathieu equation
The Wikipedia article (http://en.wikipedia.org/wiki/Mathieu_function) is quite extensive, I believe, and in general there is NO way to explicitly calculate either eigenvalues or eigenfunctions of the …
1
vote
Asymptotic of the heat kernel
You just need to check that
$$ \int_M H_k(x, y) \mathrm{d}y = 1 + O(t^{k+1})$$
for all $x \in M$ and for all $k$. For example, this follows directly from the method of stationary phase (just take a ge …
6
votes
1
answer
1k
views
Lower bounds on matrix eigenvalues
Let $A$ be a real $n\times n$ matrix and let $\mu_1, \dots, \mu_n$ the (generalized, complex) eigenvalues of $A$. Assume that
$$ 0 < \alpha < \mathrm{Re}(\mu_1) < \dots < \mathrm{Re}(\mu_n).$$
I am i …
3
votes
Accepted
The first eigenvalue of the Schrödinger operator is simple.
Roughly, the trick is not to view $L$ as an operator on $L^2$, but on $C^0$
I will use the following version of Krein-Rutmann which is proven in "Du, Yihong: Order Structure and Topological Methods i …
2
votes
2
answers
320
views
Smooth dependence of the spectrum on the operator
I would like to know if there are theorems that state under which circumstances spectra of operator families depend smoothly on the parameter.
To clarify, suppose I have a 1-parameter family $T_h$ of …
0
votes
Perturbative solution to an Eigenvalue Problem with a continuous spectrum
I find some statements your post quite curious.
If you look at the Laplace operator on $L^2(-\infty,\infty)$, it has a complete set of generalized eigenvectors, namely $\cos(\sqrt{\lambda}x)$, $\sin …
1
vote
Can a self-adjoint operator have a continuous set of eigenvalues?
The resolvent set is the set of all $\zeta \in \mathbb{C}$ for which $T-\zeta$ is invertible (which means especially that the Range is all of $H$). The spectrum $\Sigma$ is the complement of the resol …