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

$\begin{array}{rcl}M& =& |\sum _{k=1}^{\mathrm{\infty}}{a}_{k}({x}_{0}){J}_{\mu}^{\alpha -\beta}[v][{t}_{k}({x}_{0})]|\le \sum _{k=1}^{\mathrm{\infty}}|{a}_{k}({x}_{0})||{J}_{\mu}^{\alpha -\beta}[v]({t}_{k}({x}_{0}))|\\ \le & \sum _{k=1}^{\mathrm{\infty}}|{a}_{k}({x}_{0})|\frac{1}{\mathrm{\Gamma}(\alpha -\beta )}{\int}_{0}^{1}{s}^{\mu -1}{|lns|}^{(\alpha -\beta -1)}\left|v(s{t}_{k}({x}_{0}))\right|\phantom{\rule{0.2em}{0ex}}ds.\end{array}$

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 $P(x,y)=\frac{1}{{\omega}_{n}}\frac{1-{|x|}^{2}}{{|x-y|}^{n}}$ 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. □