Abstract
In this work the Neumann boundary value problem for a nonhomogeneous polyharmonic equation is studied in a unit ball. Necessary and sufficient conditions for solvability of this problem are found. To do this we first reduce the Neumann problem to the Dirichlet problem for a different nonhomogeneous polyharmonic equation and then use the Green function of the Dirichlet problem.
MSC: 35J40, 35J30, 35A01.
Keywords:
nonhomogeneous polyharmonic equation; the Neumann problem; the necessary and sufficient conditions for solvability1 Introduction
Let be a unit ball, be a unit sphere and m be a positive integer.
Consider on the domain Ω the following Neumann boundary value problem:
where ν is the unit outer normal vector to sphere ∂Ω, and are given functions; we always suppose that these functions are sufficiently smooth, and from here on we do not pay any attention to their smoothness.
A function is called a solution of problem (1), (2) if it satisfies (1), (2) in a classical sense.
It is well known (see, for example, [1]) that even in case the considered problem (1), (2) does not have solutions for arbitrary (even, as we supposed, smooth) functions and ; in the case of the Poisson equation, the necessary and sufficient solvability condition for the Neumann problem is
In the paper [2] by Kanguzhin and Koshanov, in particular, it is shown that in case the necessary and sufficient condition for solvability of problem (1), (2) has the form
where is the scalar product in and .
The authors of the paper [3] presented this condition in a different form, which could be easily verified,
In the above paper [2] the authors found a solvability condition for Neumann problem (1), (2) for arbitrary m as well (see [2], Theorem 4.2). This condition follows from equality to zero of the determinant of an matrix, one column of which consists of integrals , and is a constant. Note that the equation which one has as a result is very difficult to verify.
The main goal of the present paper is to find a solvability condition for problem (1), (2) in a more simple form. It should be noted that in our study of problem (1), (2) the Green function of the Dirichlet problem for equation (1) is essentially used. In the paper [4] a similar method was used in the solution of the boundary value problem for the Poisson equation with the boundary operator of fractionalorder.
The paper is organized as follows. In the next section we study the properties of some integrodifferential operators, which we then use throughout the paper. In Section 3 we investigate the Dirichlet problem for a polyharmonic equation, making use of the explicit form of the Green function found in [57]. Then, in the following section, reducing Neumann problem (1), (2) to the considered Dirichlet problem, we give the necessary and sufficient solvability condition for problem (1), (2) with homogeneous boundary conditions. In the same way we consider in Section 5 the Neumann boundary value problem for the homogeneous equation with nonhomogeneous boundary conditions. Finally, in Section 6 we study problem (1), (2) in the general case. To present the necessary and sufficient conditions for solvability, here we apply the Almansi formula for constructing solutions to the Dirichlet problem.
2 Properties of some integrodifferential operators
Let be a sufficiently smooth function in Ω. Consider the following operators:
where and is a constant. Note that the operator is not defined for functions with .
Note that in the class of harmonic functions in a ball the properties of operators and with have previously been studied in the paper [8].
Lemma 1Letbe a smooth function. Then for anyone has
Hence, equality (5) and the second equality of (4) are proved.
As we noted above, if , then the expression is defined. Now apply the operator to this expression. Then
But due to equality (7) and the condition , the last expression is equal to . Hence, equality (6) is proved. The first equality in (4) can be proved in the same way. □
The following statement can be proved by a direct calculation.
Lemma 2Letbe a smooth function. Then for anyone has
Corollary 1Letbe a smooth function. Then for anyone has
3 Some properties of the solutions of the Dirichlet problem
Let be a solution of the Dirichlet problem
It is known (see, for example, [57]) that if is a sufficiently smooth function, then the solution of problem (10) exists, it is unique and has the form
where is the Green function of Dirichlet problem (10).
We make use of the following explicit form of the Green function [5]:
if n is odd, or even and , then
where
where
Lemma 3Letin Dirichlet problem (10), and letbe the unique solution of this problem. Thenif and only if
Proof Let be the solution of problem (10). Then it has the form (11). To use the explicit form of the Green function, we shall deal only with the case n is odd or even and , the other cases being exactly similar. So, if , then from (11) one has
If we denote and , then the last integral can be rewritten as
Now we consider the inner integral . Noting that , we introduce the following two integrals:
Integrating by part in the integral , we obtain
Therefore
where
It is not hard to prove by induction that
Now let us suppose that (13) holds true for some j and prove it for . We have
Thus equality (13) holds true for any . In particular,
where
Therefore
and going back to the Cartesian coordinate system, we have
□
4 The Neumann problem with homogeneous boundary conditions
In this section we study problem (1), (2) with homogeneous boundary conditions.
Theorem 1Letbe sufficiently smooth. Then the necessary and sufficient solvability condition for Neumann problem (1), (2) has the form (12).
If a solution exists, then it is unique up to a constant and can be represented as
whereis the solution of Dirichlet problem (10) with the righthand side, which satisfies the additional condition.
Proof Let a solution of problem (1), (2) exist and let be this solution. We apply an operator to a function and denote . Now we obtain the conditions for the function .
Obviously, . If we apply the operator to , then by virtue of (8) we have
Further, since
then
It is not hard to verify that for any and all one has [9]
Therefore from homogeneous conditions (2) we finally have
Thus, if is a solution of problem (1), (2) with homogeneous boundary conditions, then the function is the solution of Dirichlet problem (10) with the righthand side
Moreover, the function satisfies the condition and according to Lemma 3, the necessary condition for this is (12). Hence, if a solution of problem (1), (2) exists, then it is necessary for condition (12) to be satisfied.
Now we prove that if condition (12) is satisfied, then the solution of problem (1), (2) with homogeneous boundary conditions exists.
Indeed, if (12) is satisfied, then according to Lemma 3 the solution of Dirichlet problem (10) with exists and .
Therefore, we may apply the operator to and consider the function . It is not hard to show that this function is the solution of problem (1), (2).
Indeed, by virtue of (9) one has
Since , then one can show as above that the function satisfies all homogeneous boundary conditions. □
5 The Neumann problem for the homogeneous equation
In the present section we consider Neumann problem (1), (2) with .
Let A be the following matrix
where , . Note that if . Denote by , , the determinant of the matrix obtained from A by deleting the elements of the first column and the jth row. Obviously, .
Let be a solution of the following Dirichlet problem with sufficiently smooth boundary functions :
Theorem 2Letand, , be sufficiently smooth functions. Then the necessary and sufficient solvability condition for Neumann problem (1), (2) has the form
If a solution exists, then it is unique up to a constant and can be represented as
whereis the solution of Dirichlet problem (15) with boundary functions, , , which satisfies the additional condition.
Proof Let a solution of problem (1), (2) exist and let be this solution. We apply an operator to a function and denote . Now we prove that the function is the solution of Dirichlet problem (15) with the additional condition .
From the properties of the operator we have , . By virtue of the following formula [9]
We rewrite these conditions in a more convenient form. To do this we first consider the last two of them:
We multiply expression (19) by and sum to (18). Then, making use of (17), we obtain
Further, by repeating this argument for all , we get
Thus, if is the solution of Neumann problem (1), (2), then the function will be the solution of Dirichlet problem (15) with the additional condition . Note that, under the conditions of Theorem 2, the solution of problem (15) exists and it is unique (see, for example, [10]).
Next we find the conditions to the boundary functions , which guarantee the equality . Making use of the Almansi formula (see, for example, [11], p.188) we write the solution of problem (15) as
where are harmonic functions in the ball Ω. Obviously, if and only if .
Substituting function (20) into the boundary condition of (15) and integrating over the sphere, taking into account the equalities
we get the system of equations
where
and . The matrix of this system is matrix A, defined by (14). As we noted above, . By reducing to the Vandermonde determinant, it is not hard to find the value of this determinant; one has .
Making use of Cramer’s rule, we find from the above system of equations: , where is the determinant of the following matrix
Obviously,
where determinants , , are defined above. Therefore, the equality holds if and only if . But by the definition of , this condition is equivalent to (16).
Thus, if the solution of the considered Neumann problem exists, then necessarily condition (16) holds.
We now prove the converse, i.e., if (16) holds, then the solution of the Neumann problem exists.
Let be the solution of Dirichlet problem (15). If condition (16) holds, then and we may consider the function
and prove that this function is in fact the solution of the Neumann problem.
Indeed, after changing of variable , the last integral can be written as
Therefore,
In the subsequent discussions, we use formulas (17) and (21) and assume that . So, we have
Further, for the second derivative one has
Then
Hence
Using the same argument, we have for any
Consequently,
Therefore, finally we have by induction
□
6 The Neumann problem in the general case
In this final section we consider Neumann problem (1), (2) in the case when both the equation and the conditions are nonhomogeneous.
Let the solution of problem (1), (2) exist and denote this solution by . Apply the operator to and put . Then the function is a solution of the following Dirichlet problem:
where , . Moreover, by definition satisfies the additional condition . Since functions and are sufficiently smooth, then the solution of problem (22) exists and it is unique.
Next we find the conditions to functions and , which guarantee the equality . To do this we present as
where and are the considered above solutions of the corresponding Dirichlet problems (10) and (15). Obviously, if and only if .
We represent the functions and , according to (11) and (20), in the form
Then
where
It is not hard to see that using the formula we can simplify and obtain
Since , the condition has the form
Thus, we proved the following statement on the necessary and sufficient solvability condition for the general Neumann boundary value problem.
Theorem 3Let, andbe sufficiently smooth. Then the necessary and sufficient solvability condition for Neumann boundary value problem (1), (2) has the form (23).
If a solution exists, then it is unique up to a constant and can be represented as
whereis the solution of Dirichlet problem (22) with the righthand side, which satisfies the additional condition.
Example 1 Let us consider the biharmonic equation, i.e., .
In this case
Then
Therefore the solvability condition has the form
i.e., this condition coincides with the result of the paper [3].
Example 2 Let us consider the socalled threeharmonic equation, i.e., .
In this case
Then
Therefore the solvability condition has the form
Remark 1 Let . Obviously, if , then . Therefore, if we suppose that and , , then all the considered boundary value problems have solutions and these solutions are unique (see, for example, [2]).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Both authors completed the paper together. Both authors read and approved the final manuscript.
Acknowledgements
This work has been supported by the MON Republic of Kazakhstan under Research Grant 0830/GF2 and the Ministry of Higher and Secondary Special Education of Uzbekistan under Research Grant F4FAF010.
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