Let {t}_{k} and {a}_{k}(x) satisfy the conditions from Section 2.
Consider the following problem in the domain Ω:
\mathrm{\Delta}u(x)=0,\phantom{\rule{1em}{0ex}}x\in \mathrm{\Omega},
(17)
v(x)\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x){J}_{\mu}^{\alpha \beta}[u]({t}_{k}(x))=f(x),\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega},
(18)
where \mu >0, 0\le \beta \le \alpha, and \alpha +\beta \ne 0, i.e. α and β are not equal to zero simultaneously, f(x)\in C(\partial \mathrm{\Omega}).
A harmonic function v(x) from the class {C}^{2}(\mathrm{\Omega})\cap C(\overline{\mathrm{\Omega}}), 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 \alpha =\beta in [30].
Let us investigate uniqueness for the solution of problem (17)(18). The following statement holds.
Lemma 6 Let {\mathrm{\Gamma}}_{k}\subset {\overline{\mathrm{\Omega}}}_{1}\subset \mathrm{\Omega}, k=1,2,\dots , {a}_{k}(x), k=0,1,2,\dots , be continuous functions satisfying the condition
\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x)\le {\mu}^{\alpha \beta},\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega},
(19)
and let a solution of problem (17)(18) exist.
Then:

(1)
If
\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x)\ne {\mu}^{\alpha \beta},\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega},
(20)
then the solution of problem (17)(18) is unique.

(2)
If
\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x)\equiv {\mu}^{\alpha \beta},\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega},
(21)
then the solution of problem (17)(18) is unique up to a constant summand.
Proof Let v(x) be the solution of problem (17)(18) at f(x)=0.
Denote M=v({x}_{0})={max}_{\partial \mathrm{\Omega}}v(x), {x}_{0}\in \partial \mathrm{\Omega}.
Then if v(x)\ne \text{const}, then, by virtue of the maximum principle for harmonic functions [46], the inequality v(x)<M holds for any x\in \mathrm{\Omega}.
The boundary condition (18) at f(x)=0 implies
\begin{array}{rcl}M& =& \sum _{k=1}^{\mathrm{\infty}}{a}_{k}({x}_{0}){J}_{\mu}^{\alpha \beta}[v][{t}_{k}({x}_{0})]\le \sum _{k=1}^{\mathrm{\infty}}{a}_{k}({x}_{0}){J}_{\mu}^{\alpha \beta}[v]({t}_{k}({x}_{0}))\\ \le & \sum _{k=1}^{\mathrm{\infty}}{a}_{k}({x}_{0})\frac{1}{\mathrm{\Gamma}(\alpha \beta )}{\int}_{0}^{1}{s}^{\mu 1}{lns}^{(\alpha \beta 1)}\leftv(s{t}_{k}({x}_{0}))\right\phantom{\rule{0.2em}{0ex}}ds.\end{array}
Further, since {t}_{k}:\partial \mathrm{\Omega}\to {\mathrm{\Gamma}}_{k}\subset {\overline{\mathrm{\Omega}}}_{1}\subset \mathrm{\Omega}, k=1,2,\dots , {\mathrm{\Gamma}}_{k}\ne \mathrm{\varnothing}, then {t}_{k}({x}_{0})\in \mathrm{\Omega}, and for any s\in [0,1], s{t}_{k}({x}_{0})\in \mathrm{\Omega}. Therefore v(s{t}_{k}({x}_{0}))<M.
Hence,
M<M\sum _{k=1}^{\mathrm{\infty}}{a}_{k}({x}_{0})\frac{1}{\mathrm{\Gamma}(\alpha \beta )}{\int}_{0}^{1}{s}^{\mu 1}{lns}^{\alpha \beta 1}\phantom{\rule{0.2em}{0ex}}ds=M{\mu}^{\beta \alpha}\sum _{k=1}^{\mathrm{\infty}}{a}_{k}({x}_{0}).
If now condition (19) is realized , then {\mu}^{\beta \alpha}{\sum}_{k=1}^{\mathrm{\infty}}{a}_{k}({x}_{0})\le 1, and we obtain from this the contradiction M<M.
Hence, if condition (19) holds, it is necessary that v(x)=C\equiv \text{const}. Since {J}_{\mu}^{\alpha \beta}[C]={\mu}^{\beta \alpha}\cdot C, substituting the function v(x)=C into the boundary condition (18), for f(x)=0 we have
C{\mu}^{\beta \alpha}\cdot C\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x)=0,\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega}.
The last equality is equivalent to the equality
C\cdot [1{\mu}^{\beta \alpha}\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x)]=0.
We obtain from this the result that either C=0 or {\sum}_{k=1}^{\mathrm{\infty}}{a}_{k}(x)={\mu}^{\alpha \beta}.
Thus, if conditions (19) and (20) are fulfilled, we obtain C=0, i.e. v(x)\equiv 0.
If the conditions (21) are fulfilled, then any constant is a solution of the homogeneous problem (17)(18). In fact, substituting v(x)\equiv C into equation (18), we obtain
C\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x){\mu}^{\beta \alpha}C=C[1{\mu}^{\beta \alpha}\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x)]=C[1{\mu}^{\beta \alpha}{\mu}^{\alpha \beta}]=0.
The lemma is proved. □
Now investigate existence of a solution of problem (17)(18). Let 0\le \beta \le \alpha and let P(x,y)=\frac{1}{{\omega}_{n}}\frac{1{x}^{2}}{{xy}^{n}} be the Poisson kernel of the Dirichlet problem, and {\omega}_{n} the area of the unit sphere.
Introduce the function
{P}_{\alpha \beta ,\mu}(x,y)=\{\begin{array}{cc}P(x,y),\hfill & \text{if}\alpha =\beta ,\hfill \\ \frac{1}{\mathrm{\Gamma}(\alpha \beta )}{\int}_{0}^{1}{s}^{\mu 1}{lns}^{\alpha \beta 1}P(sx,y)\phantom{\rule{0.2em}{0ex}}ds,\hfill & \text{if}\alpha \beta ,\hfill \end{array}
(22)
and consider the equation
\psi (x){\int}_{\partial \mathrm{\Omega}}[\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(y){P}_{\alpha \beta ,\mu}({t}_{k}(y),x)]\psi (y)\phantom{\rule{0.2em}{0ex}}d{S}_{y}=0.
(23)
The following statement holds.
Lemma 7 Let {\mathrm{\Gamma}}_{k}\subset {\overline{\mathrm{\Omega}}}_{1}\subset \mathrm{\Omega}, {a}_{k}(x), k=1,2,\dots , be continuous functions satisfying the condition (19). Then:

(1)
If the condition (20) is realized, then problem (17)(18) is uniquely solvable at any f(x)\in C(\partial \mathrm{\Omega}).

(2)
If the condition (21) is realized, then problem (17)(18) is solvable if the following condition is realized:
{\int}_{\partial \mathrm{\Omega}}f(x){\psi}_{0}(x)\phantom{\rule{0.2em}{0ex}}d{s}_{x}=0,
(24)
where the function {\psi}_{0}(x) is a solution of equation (23), moreover the number of independent solutions of this equation under these conditions is equal to 1.
Proof Since v(x) is a harmonic function, a solution of problem (17)(18) can be found in the form of the Poisson integral v(x)={\int}_{\partial \mathrm{\Omega}}P(x,y)\nu (x)\phantom{\rule{0.2em}{0ex}}d{s}_{x} where \nu (x) is an unknown function. Substituting this function into the boundary condition (18), we obtain the integral equation with respect to the unknown function \nu (x),
\nu (x){\int}_{\partial \mathrm{\Omega}}[\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x){P}_{\alpha \beta ,\mu}({t}_{k}(x),y)]\nu (y)\phantom{\rule{0.2em}{0ex}}d{S}_{y}=f(x),\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega}.
(25)
Designate
K(x,y)=\sum _{k=1}^{\mathrm{\infty}}{a}_{k}(x){P}_{\alpha \beta ,\mu}({t}_{k}(x),y).
Then equation (25) can be rewritten in the form of
\nu (x)+{\int}_{\partial \mathrm{\Omega}}K(x,y)\nu (y)\phantom{\rule{0.2em}{0ex}}d{S}_{y}=f(x),\phantom{\rule{1em}{0ex}}x\in \partial \mathrm{\Omega}.
(26)
To investigate the solvability of the integral equation (26), we study the properties of the kernel K(x,y). We show that K(x,y) is a continuous function on \partial \mathrm{\Omega}\times \partial \mathrm{\Omega}.
In fact, since {t}_{k}(x)\in {\overline{\mathrm{\Omega}}}_{1}\subset \mathrm{\Omega}, we obtain s{t}_{k}(x)y>0 for all x,y\in \partial \mathrm{\Omega}, and therefore the function P(s{t}_{k}(x),y) is continuous on \partial \mathrm{\Omega}\times \partial \mathrm{\Omega}. Further, the function {s}^{\mu 1}{lns}^{\alpha \beta 1} has an integrable singularity, and that is why the function {P}_{\alpha \beta ,\mu}({t}_{k}(x),y) is continuous on \partial \mathrm{\Omega}\times \partial \mathrm{\Omega}. Then by virtue of the uniform convergence of the series {\sum}_{k=1}^{\mathrm{\infty}}{a}_{k}(x), the kernel K(x,y) is also a continuous function on \partial \mathrm{\Omega}\times \partial \mathrm{\Omega}.
Hence, one can apply Fredholm theory to equation (26). Since in the case of f(x)=0 and fulfillment of the condition (20), the solution of problem (17)(18) can only be v(x)\equiv 0, for f(x)=0 the integral equation (26) has only a trivial solution.
Hence, for any f(x)\in C(\partial \mathrm{\Omega}) the solution of equation (26) exists, is unique, and belongs to the class C(\partial \mathrm{\Omega}). Using this solution, we construct the function v(x) which will satisfy all the conditions of problem (17)(18).
If the condition (21) is valid, then v(x)=C satisfies the condition (18) at f(x)\equiv 0, i.e. the corresponding homogeneous equation (26) has the nonzero solution v(x)\equiv C. 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. □