Polynomial Solutions of the Boundary Value Problems for the Poisson Equation in a Layer

Abstract

In a multidimensional infinite layer bounded by two hyperplanes, the Poisson equation with the polynomial right-hand side is considered. It is shown that the Dirichlet boundary value problem and the mixed Dirichlet-Neumann boundary value problem with polynomial boundary conditions have a unique solution in the class of functions of polynomial growth and it solution is a polynomial. An algorithm for constructing this polynomial solution is given and examples are considered.

Exact Solutions to the Boundary-Value Problems for the Helmholtz Equation in a Layer With Polynomials in the Right-Hand Sides of the Equation and of the Boundary Conditions

“…As λ tends to zero, the functions p 2m (λ, y) go over into polynomials p 2m (y), which were considered in [2]. For example,…”

“…which is a solution of the Dirichlet-Neumann boundary value problem for the Poisson equation [2] ∆u(x, y) = x 2 y 2 , − ∞ < x < ∞, 0 < y < a, u(x, 0) = 0, u y (x, a) = 0, − ∞ < x < ∞.…”

“…In this paper, exact solutions are obtained in the form of quasipolynomials of the Dirichlet and Dirichlet-Neumann boundary value problems for the Helmholtz equation in a layer in the case when the right-hand side of the Helmholtz equation and the right-hand sides of the boundary conditions are polynomials. If the parameter of the Helmholtz equation tends to zero, then the Helmholtz equations turn into the Poisson equation, and the quasipolynomial solutions of the boundary value problems turn into polynomial solutions of the boundary value problems for the Poisson equation [2]. In the same way, exact polynomial solutions of boundary value problems for the Tricomi equation in a strip are obtained [3], [4].…”

“…Solutions u(x, y) of the boundary value problems (1), (2) and (1), (5) we will sought in the class of functions of slow growth with respect to the variable x for each fixed y from the interval (0, a), i.e. for ∀y ∈ (0, a) there exists m ≥ 0 such that…”

“…2m (λ, y) = (2m − 1)!! − 1 λ ∂ ∂λ m cosh ((a − y)λ) cosh(aλ) , and for exampleu 1 (x, y) = x cosh(λy) − x sinh(λa) sinh(λy) cosh(λa) , u 2 (x, y) = x 2 cosh(λy) − x 2 sinh(λa) sinh(λy) cosh(λa) + y sinh(λa) cosh(λy) λ cosh(λa) + a sinh(λy) λ − − y sinh(λy) λ − a sinh 2 (λa) sinh(λy) λ cos 2 (λa)As λ → 0, the functions u k (x, y) go over into polynomials that are solutions of the Dirichlet -Neumann boundary value problem for the Laplace equation[2]. For example,lim λ→0 u 0 (x, y) = 1, lim λ→0 u 1 (x, y) = x, lim λ→0 u 2 (x, y) = x 2 + 2ay − y If φ(x) = 0, ψ(x) = x l , x ∈ R, l ∈ N, then the solution to problem (19), (20) is the function…”