Spectrum of a Hamiltonian which is a perturbation of Laplacian

Let $ \Delta =\frac{\partial^2}{\partial x_1^2}+\frac{\partial^2}{\partial x_2^2}+\frac{\partial^2}{\partial x_3^2}$ be the Laplacian on $ \mathbb{R}^3$ . Consider an operator $ H$ on complex valued functions on $ \mathbb{R}^3$ $ $ H\psi=\Delta\psi(x) +i\sum_{p=1}^3A_p(x)\frac{\partial \psi(x)}{\partial x_p} +B(x)\psi(x),$ $ where $ A_i,B$ are smooth real valued functions.

I am looking for a precise result of the following approximate form: (1) if $ A_i$ and $ B$ are ‘small’ then the discrete spectrum of $ H$ is non-positive. (2) If $ A_i,B$ are ‘large’ then the discrete spectrum of $ H$ contains necessarily a positive element.

Positive and negative powers of small parameter in perturbation problem

I’d like to perturbatively handle an eigenvalue problem similar to this: $ $ \lambda f = (\hat{H} + 1/\epsilon^2 \hat{V} + \epsilon {W}) f, $ $ where $ f$ is a function, $ \lambda$ is an eigenvalue, $ \epsilon$ is a small parameter, and the rest are linear (differential) operators. The problem is, that if one writes a series for the eigenvalue and the eigenfunction, $ $ f = f_0 + \epsilon f_1 + \epsilon^2 f_2 + …\ \lambda = \lambda_0 + \epsilon \lambda_1 + \epsilon^2 \lambda_2 + …, $ $ one will get e.g. $ $ \lambda_0 f_0 = \hat{H} f_0 + \hat{V} f_2\ \lambda_1 f_0 + \lambda_0 f_1 = \hat{H} f_1 + \hat{W} f_0 + \hat{V} f_3\ … $ $ i.e. the different orders of the series start to mix. Is there a way to develop a systematic perturbation theory for this case?

What is an equilibrium called when a perturbation only moves it slightly?

In game theory, a equilibrium is:

Stable, if a perturbation form any of the players returns the equilibrium back to it’s original state

Unstable, if a perturbation moves the equilibrium away from the original state

Semi-stable if some of the players perturbations are acceptable.

What do you call an equilibrium if a perturbation merely moves it slightly.

Consider the following game: Two players must state a number. Each of them gets a payoff of $ 1 if they state the same number, otherwise there is no payoff. (1,1) is an equilibrium. If one of the players chooses a different strategy, say 1.01. Then the Equilibrium is (1.01, 1.01). This does not seem unstable – neither does this seem stable. What is this type of equilibrium called?

Perturbation of the adiabatic limit

Let $ (M,g_M)$ be a closed oriented Riemannian manifold that has a fibration structure $ $ Y \rightarrow M \overset{\pi}{\rightarrow} B $ $ where $ (Y,g_Y)$ and $ (B,g_B)$ are closed Riemannian manifolds such that $ \pi$ is a Riemannian submersion.

Now we define $ g_{\epsilon}=\epsilon^{-2}\pi^*g_B+g_Y$ and $ \tilde g_{\epsilon}=g_{\epsilon}+\alpha_{\epsilon}$ for some error term $ \alpha_{\epsilon}=O(\epsilon^{\tau-2})$ for some $ \tau>0$ . If we denote the signature operators on $ M$ with respect to $ g_{\epsilon}$ and $ \tilde g_{\epsilon}$ by $ A_{\epsilon}$ and $ \tilde A_{\epsilon}$ respectively. Can we conclude that

$ $ \lim_{\epsilon \to 0} \eta(A_{\epsilon})= \lim_{\epsilon \to 0} \eta(\tilde A_{\epsilon}), $ $ if the former exists?

Perturbation theory compact operator

Let $ K$ be a compact self-adjoint operator on a Hilbert space $ H$ such that for some normalized $ x \in H$ and $ \lambda \in \mathbb C:$

$ \Vert Kx-\lambda x \Vert \le \varepsilon.$

It is well-known that this implies that $ d(\sigma(K),\lambda) \le \varepsilon.$

However, I am wondering whether this implies also something about $ x.$

For example, it seems plausible that $ x$ cannot be orthogonal to the direct sum of eigenspaces of $ K$ with eigenvalues that are close to $ \lambda.$

In other words, are there any non-trivial restrictions on $ x$ coming from the spectral decomposition of $ K$ ?

Resolvent estimate of compact perturbation of self-adjoint operator

Let $ T$ be a selfadjoint operator on Hilbert space $ H$ . Then we know that there is a resolvent estimation $ $ \left\lVert (\lambda-T)^{-1}\right\rVert = \frac{1}{dist(\lambda,\sigma(T))}, \ \lambda \in \rho(T).$ $ But what if we want to consider the compact perturbation, that is, operator $ A = T+ D,$
where $ D$ is a compact operator on $ H$ . Is there any known results on the resolvent estimate? Does anyone know some related references ? Thank you very much.

Uniform inequality for an analytic perturbation

Let $ T$ be a bounded linear operator acting on a complex Banach space. Suppose that $ T$ has spectral radius strictly less than $ 1$ . If we introduce an analytic perturbation to $ T$ , $ s\mapsto T_s$ for $ |s|<\epsilon$ (with $ T_0 = T$ ) then by upper-semicontinuity of the spectrum, assuming that $ \epsilon$ is sufficiently small, the spectral radius of each $ T_s$ is less than $ \rho$ for some $ \rho <1$ . My question is the following:

Is it possible to find $ K>0$ and $ \rho <\rho ‘<1$ such that $ $ \|T_s^n\|\le K (\rho ‘)^n,$ $ for all $ |s|<\epsilon$ and $ n$ ?

This certainly seems like it should be true but I can’t find a proof – I think I’m missing something obvious. Any help would be greatly appreciated – cheers!

Stability of the second-order hyperbolic PDE add a perturbation

Here we assume that $ u=u^{\varepsilon}(x,t)$ satisfies the following hyperbolic PDE problem,

$ $ \partial^2_t u^{\varepsilon}-\partial_x \big(a(x)\partial_x u^{\varepsilon} \big)+\varepsilon \partial_x u^{\varepsilon} =0,~(x\in\mathbb{R},t>0),$ $ $ $ u^{\varepsilon}(x,0)=0,\quad \partial_t u^{\varepsilon}(x,0)=F_0(x).$ $

Suppose that $ a(x)>0$ is a smooth even periodic function, i.e., $ a(x+1)=a(x),~a(-x)=a(x)$ . The initial condition $ F_0(x)\in C^{\infty}( \mathbb{R})\cap L^2(\mathbb{R})$ .

If $ \varepsilon=0$ , we denote the above solution as $ u^0(x,t)$ . Our question is when $ \varepsilon>0$ , for any $ t\in[0,\frac1{\varepsilon}]$ , could we get the following estimate under the same initial condition $ $ \|u^{\varepsilon}(x,t)\|_{L^2(\mathbb{R})}\le C \sup_{\tau\in[0,\frac{1}{\varepsilon}]}\|u^0(x,\tau)\|_{L^2(\mathbb{R})} ~?$ $

Where $ C$ is independent of $ \varepsilon$ . When $ a(x)$ is a constant, namely $ \mathcal{L}=-\partial_x \big(a(x)\partial_x\big)$ is Laplacian, we can prove the above result by using Fourier tranform, Plancherel theorem and Gronwall inequality. How about the case of a general elliptic operator, is the result also the same? If so, does Riesz transform work here?