Abstract
The parqueting-reflection method is applied to a nonregular domain and the harmonic Green function for the half hexagon is constructed. The related Dirichlet problem for the Poisson equation is solved explicitly.
1. Introduction
The basic boundary value problems for the second-order complex partial differential equations are the harmonic Dirichlet and Neumann problems for the Laplace and Poisson equations. In order to find the solution in explicit or closed form diverse methods have been applied. In case a given domain is simply connected and has a piecewise smooth boundary the tools of complex analysis such as Schwarz reflection principle and conformal mapping serve perfectly. When a given domain is piecewise smooth polygonal and has corners the Schwarz-Christoffel formula can be used. Difficulties arise since the elliptic integrals appearing in the formula imply complicated computations and need to be solved numerically. As analogue to this formula, another method can be applied which gives the covering of the entire complex plane by reflection of the given domain at its boundary. The method is fully described in numerous papers of Begehr and other authors; see, for example [1–12]. Our aim is to find the solution of the Dirichlet boundary value problem for the Poisson equation through the Poisson integral formula. It is known that the Poisson kernel function is an analogue of the Cauchy kernel for the analytic functions and the Poisson integral formula solves the Dirichlet problem for the inhomogeneous Laplace equation. One way to obtain the Poisson kernel leads to the harmonic Green function which is to be constructed by use of the parqueting-reflection method.
In this paper we first consider the half hexagon domain and implement the parqueting-reflection method. The reflection points treated in a proper way help to construct the certain meromorphic functions needed to find the harmonic Green function and representation formula. The later one provides the solution to the harmonic Dirichlet problem which is shown in the last part.
2. Half Hexagon Domain and Poisson Kernel
We consider a polygonal domain with corner points. The half hexagon denoted as with four corner points at 2, , , and lies in the upper half plane. A point will later serve as a pole of the Green function. Its complex conjugate does not lie in . is reflected at the real axis so that the entire hexagon (Figure 1) is obtained. The pole is reflected onto which will later become a zero of a certain meromorphic function related to the Green function. The points and from are reflected again through all the sides of the hexagon, starting with the right upper side and continuing in a positive direction. The successive reflections of give the points, which will later become zeros of the meromorphic function mentioned above. They areReflection of the point defines the poles of the meromorphic function in the hexagons . These points in turn are reflected through the sides of the new hexagons, except for reflecting to the original hexagon . Hence each hexagon includes now 3 poles and 3 zeros. Continuation of these operations reveals that all the points have the same coefficients of rotation: , , , and displacement , . Note that reflection includes rotation and shifting and the points from one hexagon can be expressed through the points of another one. In general the points from the hexagons differ by displacements in the direction of the real and in the direction of the imaginary axes. Thus the main period is .
Obviously, the repeated reflections of the point are representable in different ways, using either of the pointswhich are connected by the relations and .
In general, all reflection points are either given byor bywhere such that .
We choose zeros as direct reflection of poles and poles as direct reflection of zeros. Then having a set of zeros and a set of poles, one can construct the Schwarz kernel for and treat the related Schwarz problem [9] and Riemann-Hilbert-type boundary value problem.
The half hexagon can be viewed as the complement of the intersection of four half planes. We define them by being the right-hand half plane which has the boundary line passing through the points 2 and , being the upper half plane with the border line through the points , being the left-hand half plane with the border line passing through the points and −2, and being the half plane which is below the real axis.
Let then , , , be the complementary half planes of those listed above. The Green functions of these half planes are, in fact, the Green functions for the complementary half planes . The outward normal derivatives of the Green function on the boundary is the Poisson kernel. The kernel provides the boundary condition in the Dirichlet problem.
The Poisson kernels can be found from the respective Green functions , , as described below.
For the half plane with the boundary described by the relation we havewhere , .
For the half plane the relation on the boundary is given as ; thenhere , .
The boundary of the half plane is described by andwhere , .
Finally, for the half plane with the boundary described by , we have
3. Green Representation Formula
The method of reflections helps to find the harmonic Green function; see [3–5]. The reflection points given in (3) or (4) are used to construct a meromorphic function:where , , or a functionwhere , . Here is considered as a parameter and is the variable.
For the boundary part , the line from to , a meromorphic function , is deduced from by rotating the variable and the parameter about the angle :which becomes 1 on the boundary , where .
The following lemmas will be needed to prove the Green representation formula below. The complete proofs of these lemmas are given in [9].
Lemma 1. The infinite productconverges, where , .
Lemma 2. The equalities hold for .
The proof of this equality is based on the fact that the functions , , since can be obtained from , have the same poles and zeros; see [9].
The Green function must satisfy the following conditions; see [13]:(10) is harmonic in ;(20) is harmonic in for any ;(30) for any ;and the additional properties:(40) , and in , ;(50) , and in , .By the properties ()–() the Green function is uniquely defined. Obviously, as defined above is harmonic in as is analytic in up to a single pole at . Adding gives a harmonic function of . The symmetry property () is a consequence from the properties ()–(). The harmonic Green function for the half hexagon is or, by the symmetry property,
Lemma 3. The function has vanishing boundary values on ; that is,
Theorem 4 (see [13]). Any can be represented aswhere is the arc length parameter on with respect to the variable and is the harmonic Green function for .
We consider now the different forms of the Green function and take the derivatives , .
For the right-hand side, a boundary , we choose the form (14) for , . Here the outward normal derivative is ; thensince , .
For the boundary part , a line between , on a real axis, the outward normal derivative is , ; thenFor the boundary part on the left-hand side of , we take form (15). The outward normal derivative is also here and ; thenFor the upper boundary part , a line joining the points , form (16) is valid. Here and ; then is
4. Harmonic Dirichlet Problem
The representation formula in Theorem 4 provides the solution to the Dirichlet problem for the Poisson equation.
At first the boundary behavior of the integral is to be studied. Let for
Lemma 5. For the function presented in (23) satisfies the relation where is any fixed point on .
Proof. Let be defined on different boundary parts and consider the boundary behavior when .
Case 1. If is taken on so that thenOn where , .
For formula (19) givesBecause , then . The limit in the following ratio as and gives For the other terms of the sum,which follows from the rearrangement of the indices in for certain . Thus Hence for on On and , for in (22), the formula becomesThis term is not singular for and the terms of the sum can be in general rewritten as for certain . ThereforeLetting , the sum (22) tends to 0.
Similarly, for the rest parts of the boundary , one can get that the sums in (21) and (20) tend to zero as we let . As a result for the case on the boundary .
Case 2. Let be from , where , .
On , , .
For the term in (22) isOn this boundary and or ; therefore Substituting the latter into (34) and considering , , givesFor , by Letting and since , the sum tends to 0. Then on . Similar computations on the boundary parts , , give that the sums (21) and (20) tend to zero as . Therefore, on the boundary part for Case 3. Let be defined on by .
On with , .
For in (21) the formula becomesSince , thenLetting , , and for the fractionFor the other terms of (21) andThereforefor . On the boundary parts , , in the same manner we can prove that the sums (19), (22), and (20) tend to zero as . Thus for this Case 3 is as follows:on the boundary part .
Case 4. Let be from , where .
Obviously, similar calculations on the boundary parts imply the related sums to be convergent to zero, except for the boundary part , where the boundary behavior is to be observed carefully.
On with for in formula (20) we haveHere ; then term (47) isand taking the limit in the second fraction for and since Again, the terms of the sum (20) are rewritten and it follows thatIf , , this sum (20) tends to 0 for . Thus on this boundary parton the boundary . Thus, equality (24) for the function is valid.
In the next lemma the boundary behavior of the function in the corner points , is observed. It is shown that the continuity of the function is preserved at all the corner points which are treated as an intersection of two lines through the boundary parts.
Lemma 6. If , then
The proof of this lemma is given in detail in [9]. We consider now the main theorem of this paper.
Theorem 7. The Dirichlet problem fo the Poisson equation is uniquely solvable in the Sobolev space bywhere .
Proof. We need to prove that (54) is a solution of the Poisson equation in problem (53). The property of the Pompeiu operator , described in [13, 14] as , provides a weak solution of and the derivativeIn order to construct the Pompeiu-type operator we consider the following term: Define a function which is analytic with respect to ; then , . Then, the equation is rewritten as This provides the solution to the differential equation in problem (53) in a weak sense. The boundary condition on the boundary holds because of Lemmas 5 and 6.
Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.