Abstract
A Hilbert-type boundary value problem in the classes of quasi-harmonic functions is considered. Quasi-harmonic functions are regular solutions of an elliptic differential equation form $\frac{\partial ^2 W}{\partial z\partial \overline z} + \frac{n(n + 1)}{(1 + z \overline z)^2}W = 0$, where $\frac{\partial }{\partial z} = \frac{1}{2} ( \frac{\partial }{\partial x} - i\frac{\partial }{\partial y})$ , $\frac{\partial }{\partial \overline z} = \frac{1}{2} (\frac{\partial }{\partial x} + i\frac{\partial }{\partial y})$ , and n is a given positive integer. Using the fact that a circle is an analytic curve, we have developed an explicit method for finding solutions of the Hilbert homogeneous boundary value problem for quasi-harmonic functions in circular domains. The principal logic of this method consists of two stages. At stage one we are using a representation of quasi-harmonic function via analytic function and its derivatives to reduce the problem to the classical Hilbert problem for some auxiliary analytic function in the circular domain. A solution Φ( z ) for this problem will be used at stage two, when we solve the linear differential Euler equation of order n with the right-hand side Φ( z ). General solution for the problem can be explicitly expressed in terms of the solution of the Euler equation. Moreover, we have established that the solvability for the considered boundary-value problem depends essentially on whether a unit circumference is the carrier of boundary conditions or a non-unit circle
Highlights
Что равенство (32) есть краевое условие однородной задачи Гильz →t∈Lr берта относительно аналитической в круге Tr+ = {z : z < r} функции Ψ(z)
Using the fact that a circle is an analytic curve, we have developed an explicit method for finding solutions of the Hilbert homogeneous boundary value problem for quasi-harmonic functions in circular domains
Сравнивая изложенные выше решения однородной задачи Г10 в единичном круге T + = {z : z < 1} и в круге { Tr+ = z : z < r} , r ≠ 1, замечаем, что существенное различие в картинах разрешимости этой задачи в указанных двух случаях возникает из-за того, что однородное дифференциальное уравнение вида (34) в круге Tr+ = {z : z < r} , r ≠ 1, не имеет (нетривиальных) решений, принадлежащих классу A(Tr+ ) ∩ H (1) (Lr ) , а в единичном круге, т.е
Summary
Тогда (в силу представления (5)) общее решение искомой однородной задачи Г10 при χ ≥ 0 можно задавать формулой: ( ) ∫ ∫ W (z) = C1 + z−1Φ(z) + Г10 в классе квазигармонических функций первого рода, необходимо и достаточно, чтобы функция h(t) удовлетворяла условиям (17) и (20), причем при выполнении указанных условий, общее решение задачи Г10 можно задавать формулой (22).
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More From: Bulletin of the South Ural State University series "Mathematics. Mechanics. Physics"
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