Let and satisfy the conditions from Section 2.
Consider the following problem in the domain Ω:
(18)
where , , and , i.e. α and β are not equal to zero simultaneously, .
A harmonic function from the class , satisfying condition (18) in the classical case, will be called a solution of problem (17)-(18).
It should be noted that problem (17)-(18) was investigated for the case of in [30].
Let us investigate uniqueness for the solution of problem (17)-(18). The following statement holds.
Lemma 6 Let , , , , be continuous functions satisfying the condition
(19)
and let a solution of problem (17)-(18) exist.
Then:
-
(1)
If
(20)
then the solution of problem (17)-(18) is unique.
-
(2)
If
(21)
then the solution of problem (17)-(18) is unique up to a constant summand.
Proof Let be the solution of problem (17)-(18) at .
Denote , .
Then if , then, by virtue of the maximum principle for harmonic functions [46], the inequality holds for any .
The boundary condition (18) at implies
Further, since , , , then , and for any , . Therefore .
Hence,
If now condition (19) is realized , then , and we obtain from this the contradiction .
Hence, if condition (19) holds, it is necessary that . Since , substituting the function into the boundary condition (18), for we have
The last equality is equivalent to the equality
We obtain from this the result that either or .
Thus, if conditions (19) and (20) are fulfilled, we obtain , i.e. .
If the conditions (21) are fulfilled, then any constant is a solution of the homogeneous problem (17)-(18). In fact, substituting into equation (18), we obtain
The lemma is proved. □
Now investigate existence of a solution of problem (17)-(18). Let and let be the Poisson kernel of the Dirichlet problem, and the area of the unit sphere.
Introduce the function
(22)
and consider the equation
(23)
The following statement holds.
Lemma 7 Let , , , be continuous functions satisfying the condition (19). Then:
-
(1)
If the condition (20) is realized, then problem (17)-(18) is uniquely solvable at any .
-
(2)
If the condition (21) is realized, then problem (17)-(18) is solvable if the following condition is realized:
(24)
where the function is a solution of equation (23), moreover the number of independent solutions of this equation under these conditions is equal to 1.
Proof Since is a harmonic function, a solution of problem (17)-(18) can be found in the form of the Poisson integral where is an unknown function. Substituting this function into the boundary condition (18), we obtain the integral equation with respect to the unknown function ,
(25)
Designate
Then equation (25) can be rewritten in the form of
(26)
To investigate the solvability of the integral equation (26), we study the properties of the kernel . We show that is a continuous function on .
In fact, since , we obtain for all , and therefore the function is continuous on . Further, the function has an integrable singularity, and that is why the function is continuous on . Then by virtue of the uniform convergence of the series , the kernel is also a continuous function on .
Hence, one can apply Fredholm theory to equation (26). Since in the case of and fulfillment of the condition (20), the solution of problem (17)-(18) can only be , for the integral equation (26) has only a trivial solution.
Hence, for any the solution of equation (26) exists, is unique, and belongs to the class . Using this solution, we construct the function which will satisfy all the conditions of problem (17)-(18).
If the condition (21) is valid, then satisfies the condition (18) at , i.e. the corresponding homogeneous equation (26) has the nonzero solution . Then the adjoint homogeneous equation has also a nonzero solution, and that is why in this case fulfillment of the condition (24) is necessary and sufficient for solvability of problem (17)-(18). The lemma is proved. □