نتایج جستجو برای: self adjoint problem
تعداد نتایج: 1370929 فیلتر نتایج به سال:
The spectral problem (A+V (z))ψ = zψ is considered with A, a self-adjoint Hamiltonian of sufficiently arbitrary nature. The perturbation V (z) is assumed to depend on the energy z as resolvent of another self-adjoint operator A ′ : V (z) = −B(A ′ −z) −1 B *. It is supposed that operator B has a finite Hilbert-Schmidt norm and spectra of operators A and A ′ are separated. The conditions are form...
We consider a regular indefinite Sturm-Liouville problem with two self-adjoint boundary conditions, one being affinely dependent on the eigen-parameter. We give sufficient conditions under which a basis of each root subspace for this Sturm-Liouville problem can be selected so that the union of all these bases constitutes a Riesz basis of a corresponding weighted Hilbert space.
We study the behavior of solutions to the problem { ε (u′′ ε (t) +A1uε(t)) + u ′ ε(t) +A0uε(t) = fε(t), t ∈ (0, T ), uε(0) = u0ε, uε(0) = u1ε, as ε → 0, where A1 and A0 are two linear self-adjoint operators in a Hilbert space H. MSC: 35B25, 35K15, 35L15, 34G10 keywords: singular perturbations; Cauchy problem; boundary layer function.
for the differential equation in a Hilbert space H with the self-adjoint positive definite operator A is considered. The well-posedness of this problem in Hölder spaces without a weight is established. The coercivity inequalities for solutions of the boundary value problem for elliptic-parabolic equations are obtained.
Abstract For a given self-adjoint operator A with discrete spectrum, we completely characterise possible eigenvalues of its rank-one perturbations B and discuss the inverse problem reconstructing from spectrum.
A linear operator on a Hilbert space , in the classical approach of von Neumann, must be symmetric to guarantee self-adjointness. However, it can shown that symmetry could omitted by using criterion for graph and adjoint graph. Namely, S is densely defined closed if only . In more general setup, we consider relations instead operators prove this situation similar result holds. We give necessary...
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