1 PROBLEM STATEMENT

It is well known that the theory of fractional differential equations is one of the most frequently used directions in the theory of differential equations (see [1, 2]). In addition, fractional calculus is widely used in studying some problems of partial differential equations, as well as equations of mixed type with degenerations [3–7]. There are many papers (see, for example, [8–10]), in which the authors considered some classes of boundary value problems for nondegenerate and degenerate differential equations of mixed type with Gerasimov–Caputo and Riemann–Liouville fractional derivatives of the order \(0<\alpha\leqslant 1\). It should be also noted that some problems for partial differential equations with various integro-differential operators of fractional order were investigated in the works [11–25].

In this paper, we consider the questions of the existence and uniqueness of the solution to the problem for a mixed-type equation with two lines of degeneration, containing the Gerasimov–Caputo fractional derivative. So, we study a boundary value problem for the following parabolic-hyperbolic equation for \(0<\alpha<1\)

$$0=\begin{cases}{u_{xx}}-{}_{\mathbb{C}}D_{oy}^{\alpha}u,\quad y>0,\\ {(-y)^{m}}{u_{xx}}-{x^{n}}{u_{yy}},\quad y<0\end{cases}$$
(1)

with operators

$${}_{\mathbb{C}}D_{oy}^{\alpha}u=\frac{1}{{\Gamma(1-\alpha)}}\int\limits_{0}^{y}{{{(y-t)}^{-\alpha}}{u_{t}}(x,t)}dt$$
(2)

in the domain \(\Omega={\Omega_{1}}\cup{\Omega_{2}}\cup{I_{1}}\). The domain \({\Omega_{1}}\) is bounded by segments: \({A_{1}}A_{2}=\{(x,y):\ x=0,\,0<y<{h_{2}}\}\), \(A_{1}B_{1}=\{(x,y):\ y=0,\,0<x<{h_{1}}\}\), \(B_{1}B_{2}=\{(x,y):\ x={h_{1}},\,0<y<{h_{2}}\}\), \(A_{2}B_{2}=\{(x,y):\ y={h_{2}},\,0<x<{h_{1}}\}\) for \(y>0\), while \({\Omega_{2}}\) is characteristic triangle bounded by the segment \({A_{1}}{B_{1}}\) of the axes \(Ox\) and by two characteristics \({A_{1}}C:\ \frac{1}{q}{x^{q}}-\frac{1}{p}{(-y)^{p}}=0\), \({B_{1}}C:\ \,\frac{1}{q}{x^{q}}+\frac{1}{p}{(-y)^{p}}=1\) of the Eq. (1), emerging from points \({A_{1}}\left({0;0}\right)\), \(B_{1}\left({{h_{1}};0}\right)\) and intersecting at the point \(C\left({{{\left({\frac{q}{2}}\right)}^{{1\mathord{\left/{\vphantom{1q}}\right.\kern-1.2pt}q}}},\,-{{\left({\frac{p}{2}}\right)}^{{1\mathord{\left/{\vphantom{1p}}\right.\kern-1.2pt}p}}}}\right)\) for \(y<0\). Here \(2q=n+2\), \(2p=m+2\), \({h_{1}}={q^{{1\mathord{\left/{\vphantom{1q}}\right.\kern-1.2pt}q}}}\), \({h_{2}}>0\), \(m,\,n={\textrm{const}}>0,\>\ m>n\).

Let us introduce the following notations:

$${\Omega^{+}}=\Omega\cap(y>0),\quad{\Omega^{-}}=\Omega\cap(y<0),\quad{I_{1}}=\left\{{(x,y):\ 0<x<{h_{1}},\ y=0}\right\},$$
$${I_{2}}=\left\{{(x,y):\ x=0,\quad 0<y<{h_{2}}}\right\},\quad 2{\alpha_{1}}=n/(n+2),\quad 2{\beta_{1}}=m/(m+2)$$

and

$$0<{\alpha_{1}}<{\beta_{1}}<0.5,$$
(3)
$${F_{0x}}\left[{\begin{matrix}a,\quad b\\ c,\quad{x^{k}}\\ \end{matrix}}\right]f(x)=\frac{1}{{\Gamma(c)}}\int\limits_{0}^{x}{f(t)}{\big{(}{x^{k}}-{t^{k}}\big{)}^{c-1}}F\left({a,b,c,\frac{{{x^{k}}-{t^{k}}}}{{{x^{k}}}}}\right)k\,{t^{k-1}}dt,\quad k>0,\quad c>0,$$
(4)

where \(\Gamma(z)\) is Gamma function, \(F(a,b,c;z)\) is Gaussian hypergeometric function, \({F_{0x}}\left[{...}\right]\) is known operator (see, [5]).

In the domain \(\Omega\) for the Eq. (1) we study the following

Problem. To find the function \(u(x,y)\) defining from the class:

1) \(\Delta=\left\{{u(x,y):\ u(x,y)\in C(\bar{\Omega})\cap{C^{2}}({\Omega^{-}}),\ {u_{xx}}\in C\left({{\Omega^{+}}}\right),\ {}_{C}D_{oy}^{\alpha}u\in C\left({{\Omega^{+}}}\right)}\right\}\);

2) \(u(x,y)\) satisfies the Eq. (1) in the domains \({\Omega_{1}}\) and \({\Omega_{2}}\);

3) \({y^{1-\alpha}}{u_{y}}\in C({\Omega_{1}})\), \({u_{y}}\in C({\Omega_{2}})\), moreover, these functions are continuous up to the boundary \({A_{1}}{B_{1}}\). In addition, on \({A_{1}}{B_{1}}\) is fulfilled the following bonding condition

$$\lim\limits_{y\to+0}{y^{1-\alpha}}{u_{y}}(x,+0)={\lambda_{1}}(x){u_{y}}(x,-0)+{\lambda_{2}}(x)u(x,0)+{\lambda_{3}}(x),\,(x,0)\in{A_{1}}{B_{1}},$$
(5)

where the function \({\nu^{\pm}}(x)\) may have a singularity of order less than one as \(x\to 0\) and is bounded as \(x\to{h_{1}}\);

4) \(u(x,y)\) satisfies the boundary conditions

$$\left[{{\gamma_{1}}{u_{x}}(x,y)+{\gamma_{2}}u(x,y)}\right]\left|{{}_{{}_{{A_{1}}{A_{2}}}}}\right.={\varphi_{1}}(y),\quad 0\leqslant y\leqslant{h_{2}},$$
(6)
$$\left[{{\delta_{1}}{u_{x}}(x,y)+{\delta_{2}}u(x,y)}\right]\left|{{}_{{}_{{B_{1}}B_{2}}}}\right.={\varphi_{2}}(y),\quad 0\leqslant y\leqslant{h_{2}},$$
(7)
$$\frac{d}{{d\left({{x^{2q}}}\right)}}{\left({{x^{2q}}}\right)^{\frac{{1-{\alpha_{1}}-{\beta_{1}}}}{2}}}{F_{ox}}\begin{bmatrix}\frac{{{\alpha_{1}}+{\beta_{1}}-1}}{2},&\frac{{{\alpha_{1}}+{\beta_{1}}}}{2}\\ \quad{\beta_{1}},&{x^{2q}}\\ \end{bmatrix}\;{\left({{x^{2q}}}\right)^{\frac{{2{\alpha_{1}}-1}}{2}}}u\left[{\theta(x)}\right]$$
$${}=a(x){u_{y}}(x,0)+b(x),\quad\quad 0<x<{h_{1}},$$
(8)

where \({\gamma_{1}},\,{\gamma_{2}},\,{\delta_{1}},\,{\delta_{2}}={\textrm{const}}\) and \({\lambda_{i}}(x)\>(i=\overline{1,3})\), \({\varphi_{1}}(y),\,{\varphi_{2}}(y)\), \(\tilde{a}(x)=a({x^{1/2q}})\), \(\tilde{b}(x)=b({x^{1/2q}})\) are given functions, \({F_{0x}}\left[{...}\right]\) is generalized fractional integral operator [5] and

$$\theta(x)=\left({\frac{{{x^{q}}}}{2}}\right)^{1/q}-i\left({\frac{p}{q}\frac{{{x^{q}}}}{2}}\right)^{1/p}$$
(9)

is the point of intersection of the characteristics of the Eq. (1) emerging from the point \((x,0)\in{I_{1}}\) with characteristic \(AC\).

2 BASIC FUNCTIONAL RELATIONSHIP

In proving the uniqueness and existence of the solution to the problem, an important role is played some functional relations between \(\tau\,(x)\) and \(\nu\,(x)\), which have brought on \({I_{1}}\) from \({\Omega_{i}}\,(i=1,2)\). It is known that in domain \(\Omega_{2}^{-}\) the solution to the Cauchy problem for the Eq. (1) with initial value conditions \(u(x,-0)={\tau^{-}}(x)\), \(0\leqslant x\leqslant{h_{1}}\), \({u_{y}}(x,-0)={\nu^{-}}(x)\), \(0<x<{h_{1}}\) can by represented as [5]

$$u(x,y)=\frac{{\Gamma\left({\frac{1}{2}+{\beta_{1}}}\right)}}{{\sqrt{\pi}\Gamma({\beta_{1}})}}{2^{2{\beta_{1}}-1}}{\left({\frac{1}{q}{x^{q}}}\right)^{-{\alpha_{1}}}}\int\limits_{0}^{1}{{{\left[{\tfrac{1}{p}{{(-y)}^{p}}(2z-1)+\frac{1}{q}{x^{q}}}\right]}^{{\alpha_{1}}}}{{\left[{z(1-z)}\right]}^{{\beta_{1}}-1}}}$$
$${}\times{\tau^{-}}\left\{{\left[{\frac{q}{p}{{(-y)}^{p}}(2z-1)+{x^{q}}}\right]^{\frac{1}{q}}}\right\}\,F({\alpha_{1}},1-{\alpha_{1}},{\beta_{1}},\rho)dz$$
$${}-\frac{{(2p)^{1-2{\beta_{1}}}\Gamma\left({\frac{3}{2}+{\beta_{1}})}\right)}}{{\sqrt{\pi}\Gamma(1-{\beta_{1}})}}{\left({\frac{1}{p}{{(-y)}^{p}}}\right)^{1-2{\beta_{1}}}}{\left({\frac{1}{q}{x^{q}}}\right)^{-{\alpha_{1}}}}\int\limits_{0}^{1}{{{\left[{\frac{1}{p}{{(-y)}^{p}}(2z-1)+\frac{1}{q}{x^{q}}}\right]}^{{\alpha_{1}}}}{{\left[{z(1-z)}\right]}^{-{\beta_{1}}}}}$$
$${}\times\nu^{-}\left\{\left[{\frac{q}{p}{{(-y)}^{p}}\cdot\left({2z-1}\right)+{x^{q}}}\right]^{\frac{1}{q}}\right\}\,F({\alpha_{1}},1-{\alpha_{1}},{\beta_{1}},\rho)dz,$$
(10)

where \(\rho=\frac{{q{{(-y)}^{\frac{1}{p}}}z(1-z)}}{{{p^{2}}{x^{q}}\left[{\frac{1}{p}{{(-y)}^{p}}\left({2z-1}\right)+\frac{1}{q}{x^{q}}}\right]}}\).

Based on the solution to the Cauchy problem (10), (9), taking into account the property of the Gamma function, we have

$$u\left[{\theta\left(x\right)}\right]=u[\textrm{Re}\,\theta(x),\quad\textrm{Im}\theta(x)]={\gamma_{1}}{({x^{2q}})^{\frac{{2-{\alpha_{1}}-3{\beta_{1}}}}{2}}}F_{0x}\begin{bmatrix}\frac{{{\beta_{1}}-{\alpha_{1}}}}{2},&\frac{{{\alpha_{1}}+{\beta_{1}}-1}}{2}\\ {\beta_{1}},&{x^{2q}}\end{bmatrix}{({x^{2q}})^{\frac{{{\alpha_{1}}+{\beta_{1}}-2}}{2}}}\tau^{-}(x)$$
$${}-{\gamma_{2}}{({x^{2q}})^{\frac{{{\beta_{1}}-{\alpha_{1}}}}{2}}}F_{0x}\begin{bmatrix}\frac{{1-{\beta_{1}}-{\alpha_{1}}}}{2},&\frac{{{\alpha_{1}}-{\beta_{1}}}}{2}\\ 1-{\beta_{1}},&{x^{2q}}\end{bmatrix}{({x^{2q}})^{\frac{{{\alpha_{1}}-{\beta_{1}}-1}}{2}}}\nu^{-}(x),\quad(x,0)\in{I_{1}},$$
(11)

where \({\gamma_{1}}={\frac{{\Gamma(2{\beta_{1}})}}{{\Gamma({\beta_{1}})}}}{2^{{\alpha_{1}}-{\beta_{1}}}}\), \({\gamma_{2}}=\frac{{{2^{{\alpha_{1}}+3{\beta_{1}}-2}}\Gamma(2-2{\beta_{1}})}}{{\Gamma(1-{\beta_{1}})}}{\left({\frac{p}{q}}\right)^{1-2{\beta_{1}}}}\). Substituting (11) into (8), we obtain a functional relationship between \(\tau^{-}(x)\) and \(\nu^{-}(x)\) on the segment \({I_{1}}\), which have brought from the domain \(\Omega_{2}^{-}\):

$$\overline{a}(x)\nu^{-}(x)={\gamma_{1}}{\left({x^{2q}}\right)^{\frac{{1-2{\alpha_{1}}}}{2}}}\frac{d}{d{x^{2q}}}{\left({{x^{2q}}}\right)^{\frac{1-2{\beta_{\,1}}}{2}}}{F_{0x}}\left[\begin{matrix}{\alpha_{1}}+{\beta_{\,1}},&\frac{2\beta_{\,1}-1}{2}\\ {2{\beta_{\,1}},}&{x^{2q}}\end{matrix}\right]\left({x^{2q}}\right)^{\frac{2{\alpha_{\,1}}-1}{2}}\tau^{-}(x)$$
$${}-\left({{x^{2q}}}\right)^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}b(x),\quad 0<x<{h_{1}},$$
(12)

where \({\gamma_{1}}={\frac{{\Gamma(2{\beta_{1}})}}{{\Gamma({\beta_{1}})}}}{2^{{\alpha_{1}}-{\beta_{1}}}}\), \({\gamma_{2}}=\frac{{{2^{{\alpha_{1}}+3{\beta_{1}}-2}}\Gamma(1-2{\beta_{1}})}}{{\Gamma(1-{\beta_{1}})}}{\left({\frac{p}{q}}\right)^{1-2{\beta_{1}}}}\), \(\overline{a}(x)\equiv{\gamma_{2}}+{\left({{x^{2q}}}\right)^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}}a(x)\).

On the other hand, by the aid of given notation \(u(x,-0)={\tau^{-}}(x),\,0\leqslant x\leqslant{h_{1}}\), \({u_{y}}(x,-0)={\nu^{-}}(x),\,0<x<{h_{1}}\) and \(\mathop{\lim}\limits_{y\to+0}{y^{1-\alpha}}{u_{y}}(x,y)=\nu^{+}(x)\), \(0<x<{h_{1}}\) from the bond condition (5) we obtain

$${\nu^{+}}(x)={\lambda_{1}}(x){\nu^{-}}(x)+{\lambda_{2}}(x)\tau(x)+\lambda_{3}(x).$$
(13)

Taking (2), (13) and

$$\mathop{\lim}\limits_{y\to 0}D_{0y}^{\alpha-1}f(y)=\Gamma(\alpha)\mathop{\lim}\limits_{y\to 0}{y^{1-\alpha}}f(y)$$

into account, from the Eq. (1) as \(y\to+0\) we derive

$$\big{(}{\tau^{+}}(x)\big{)}^{\prime\prime}-\Gamma(\alpha){\nu^{+}}(x)=0.$$
(14)

3 UNIQUENESS OF SOLUTION TO THE PROBLEM

Theorem 1. If the following conditions are fulfilled: (3), \(\lambda_{i}(x)>0\,(i=1,2)\) and

$$\overline{a}(x)\equiv{\left({{x^{2q}}}\right)^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}}a(x)+{\gamma_{2}}>0,\,\forall(x,0)\in{\overline{I}_{1}},$$

then the solution of the problem is unique.

Proof. As usual, consider a homogeneous problem, i.e. suppose \({{\varphi_{1}}(y)\equiv{\varphi_{2}}(y)\equiv 0}\). We will prove that \(u(x,y)\equiv 0\). For this purpose, we will multiply the Eq. (14) by \(\tau(x)\) and integrate from 0 to \({h_{1}}\):

$$\Gamma(\alpha)\int\limits_{0}^{{h_{1}}}{\tau(x)\nu^{+}(x)dx}=\int\limits_{0}^{{h_{1}}}{{{\tau^{\prime\prime}}}(x)\tau(x)dx}.$$
(15)

Let be \({\gamma_{1}}={\delta_{1}}=0\). Then from the conditions (6) and (7) we have \(\tau(0)=\tau({h_{1}})=0\). Taking into account the last conditions and integrating by parts the right-hand side of the Eq. (15), by applying (13) we obtain

$$\Gamma(\alpha)\int\limits_{0}^{{h_{1}}}{\tau(x){\nu^{+}}(x)dx}=\int\limits_{0}^{{h_{1}}}{\tau(x)\lambda_{1}(x){\nu^{-}}(x)dx}+\int\limits_{0}^{{h_{1}}}\tau^{2}(x)\lambda_{2}(x)=-\int\limits_{0}^{{h_{1}}}{{{\left({\tau^{\prime}(x)}\right)}^{2}}dx\leqslant 0}.$$
(16)

Now we will prove that \(\int\limits_{0}^{{h_{1}}}{\lambda_{1}(x)\tau(x){\nu^{-}}(x)dx\geqslant 0}\). Using the formula (4), we make some simplifications in (12):

$${\nu^{-}}(x)=\frac{{{\gamma_{1}}{{\left({{x^{2q}}}\right)}^{\frac{{1-2{\alpha_{1}}}}{2}}}}}{{\overline{a}(x)\Gamma(2{\beta_{1}})}}\frac{d}{{d{x^{2q}}}}{\left({{x^{2q}}}\right)^{\frac{{1-2{\beta_{1}}}}{2}}}\int\limits_{0}^{x}{{{\left({{x^{2q}}-{t^{2q}}}\right)}^{2{\beta_{1}}-1}}}{\left({{t^{2q}}}\right)^{\frac{{2{\alpha_{1}}-1}}{2}}}$$
$${}\times{\tau^{-}}(t)F\left({{\alpha_{1}}+{\beta_{1}},\frac{{2{\beta_{1}}-1}}{2},2{\beta_{1}},\frac{{{x^{2q}}-{t^{2q}}}}{{{x^{2q}}}}}\right)d{t^{2q}}-\frac{{{{\left({{x^{2q}}}\right)}^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}}b(x)}}{{\overline{a}(x)}}.$$
(17)

Introducing the replacement \(t=x\,z\), after some simplifications we have

$${\nu^{-}}(x)=\frac{{2q{\gamma_{1}}\left({{\alpha_{1}}+{\beta_{1}}}\right){{\left({{x^{2q}}}\right)}^{\frac{{2{\beta_{1}}-1}}{2}}}}}{{\overline{a}(x)\Gamma(2{\beta_{1}})}}\int\limits_{0}^{1}{{{\left({1-{z^{2q}}}\right)}^{2{\beta_{1}}-1}}}{\left({{z^{2q}}}\right)^{\frac{{2{\alpha_{1}}+1}}{2}}}$$
$${}\times{\tau^{-}}(xz)F\left({{\beta_{1}}-{\alpha_{1}},\frac{{2{\beta_{1}}+1}}{2},2{\beta_{1}},1-{z^{2q}}}\right)dz+\frac{{2q\,{\gamma_{1}}{{\left({{x^{2q}}}\right)}^{\frac{{2{\beta_{1}}+1}}{2}}}}}{{\overline{a}(x)\Gamma(2{\beta_{1}})}}\int\limits_{0}^{1}{{{\left({1-{z^{2q}}}\right)}^{2{\beta_{1}}-1}}}{z^{2q}}\,\big{(}{\tau^{-}}(xz)\big{)}^{\prime}$$
$${}\times F\left({{\beta_{1}}-{\alpha_{1}},\frac{{2{\beta_{1}}+1}}{2},2{\beta_{1}},1-{z^{2q}}}\right)dz-\frac{{{{\left({{x^{2q}}}\right)}^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}}b(x)}}{{\overline{a}(x)}}.$$

Further, performing the reverse change \(s=x\,z\), we can obtain

$${\nu^{-}}(x)=\frac{{2q{\gamma_{1}}\left({{\alpha_{1}}+{\beta_{1}}}\right)}}{{\overline{a}(x)\Gamma(2{\beta_{1}})}}{\left({{x^{2q}}}\right)^{-{\alpha_{1}}-{\beta_{1}}}}$$
$${}\times\int\limits_{0}^{x}{{\frac{{\left({{s^{2q}}}\right)}^{2{\alpha_{1}}}}{\left({{x^{2q}}-{s^{2q}}}\right)}^{1-2{\beta_{1}}}}}{\tau^{-}}(s)F\left({{\alpha_{1}}+{\beta_{1}},\frac{{2{\beta_{1}}-1}}{2},2{\beta_{1}},\frac{{{x^{2q}}-{s^{2q}}}}{{{x^{2q}}}}}\right)ds$$
$${}+\frac{{2q{\gamma_{1}}}}{{\overline{a}(x)\Gamma(2{\beta_{1}})}}{\left({{x^{2q}}}\right)^{\frac{1}{2}-{\beta_{1}}}}\int\limits_{0}^{x}{{{\left({{s^{2q}}}\right)}^{{\alpha_{1}}+\frac{1}{2}}}{{\left({{x^{2q}}-{s^{2q}}}\right)}^{2{\beta_{1}}-1}}\big{(}{\tau}^{-}(s)\big{)}^{\prime}}$$
$${}\times\,F\left({{\alpha_{1}}+{\beta_{1}},\frac{{2{\beta_{1}}-1}}{2},2{\beta_{1}},\frac{{{x^{2q}}-{s^{2q}}}}{{{x^{2q}}}}}\right)ds-\frac{{{{\left({{x^{2q}}}\right)}^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}}b(x)}}{{\overline{a}(x)}}.$$
(18)

To complete the proof of the theorem, we need to apply the following lemma.

Lemma. If the function \(\tau(x)\) has a positive maximum (negative minimum) at the point \(x={x_{0}}\in(0,{h_{1}})\), then \({\nu^{-}}({x_{0}})>0\) (\({\nu^{-}}({x_{0}})<0\)).

Proof. Let the function \(\tau(x)\) has a positive maximum and \(b(x)\equiv 0\). Then from the last relation (18) we have

$${\nu^{-}}({x_{0}})=\frac{{2q{\gamma_{1}}\left({{\alpha_{1}}+{\beta_{1}}}\right)}}{{\overline{a}(x)\Gamma(2{\beta_{1}})}}\int\limits_{0}^{{x_{0}}}\frac{\big{(}{\tau}^{-}(t)\big{)}^{\prime}dt}{{\left({x_{0}^{2q}}\right)^{{\alpha_{1}}+{\beta_{1}}}}}$$
$${}\times\int\limits_{t}^{{x_{0}}}\frac{\left({{s^{2q}}}\right)^{2{\alpha_{1}}}}{{{{\left({x_{0}^{2q}-{s^{2q}}}\right)}^{1-2{\beta_{1}}}}}}F\left({{\alpha_{1}}+{\beta_{1}},\frac{{2{\beta_{1}}-1}}{2},2{\beta_{1}},\frac{{x_{0}^{2q}-{s^{2q}}}}{{x_{0}^{2q}}}}\right)ds+\frac{{2q{\gamma_{1}}}}{{\overline{a}(x)\Gamma(2{\beta_{1}})}}{\left({x_{0}^{2q}}\right)^{\frac{1}{2}-{\beta_{1}}}}$$
$${}\times\int\limits_{0}^{x}{\big{(}{\tau}^{-}(s)\big{)}^{\prime}{{\left({{s^{2q}}}\right)}^{{\alpha_{1}}+\frac{1}{2}}}{{\left({x_{0}^{2q}-{s^{2q}}}\right)}^{2{\beta_{1}}-1}}F\left({{\alpha_{1}}+{\beta_{1}},\frac{{2{\beta_{1}}-1}}{2},2{\beta_{1}},\frac{{x_{0}^{2q}-{s^{2q}}}}{{x_{0}^{2q}}}}\right)ds}.$$

It is obviously that

$${\gamma_{1}}>0,\quad\overline{a}(x)>0,\quad\Gamma(2{\beta_{1}})>0,\quad 0\leqslant\frac{{x_{0}^{2q}-{s^{2q}}}}{{x_{0}^{2q}}}\leqslant 1,\quad\int\limits_{0}^{{x_{0}}}{\tilde{\tau}^{\prime}(s)ds}=\int\limits_{0}^{{x_{0}}}\lim\limits_{{x_{0}}\to s}\frac{{\tilde{\tau}({x_{0}})-\tilde{\tau}(s)}}{{{x_{0}}-s}}ds>0.$$

Then, by virtue of (3), from the last relation we obtain \({\nu^{-}}({x_{0}})>0\). In a similar way, one can prove that at the given point it has a negative minimum \({\nu^{-}}\left({{x_{0}}}\right)<0\). Lemma is proved. \(\Box\)

Based on the lemma just proved above, we conclude that

$$\Gamma(\alpha)\int\limits_{0}^{{h_{1}}}{{\lambda_{1}}(x)\,\tau(x)\,{\nu^{-}}(x)dx+\Gamma(\alpha)\int\limits_{0}^{{h_{1}}}{{\lambda_{2}}(x)\,\tau^{2}(x)\geqslant 0}}$$

in the case \({\lambda_{i}}(x)\geqslant 0\) \((i=1,2)\). So, from the formula (16) follows that \({\nu^{-}}(x)\equiv\tau(x)\equiv 0\). Consequently, by virtue of the solution of the first boundary value problem, for the Eq. (1) ([9, 26]), (6) and (7) we derive \(u(x,y)\equiv 0\) in \(\overline{\Omega}^{+}\). From the solution of the Cauchy problem (10) we obtained \(u(x,y)\equiv 0\) in the closed domain \(\overline{\Omega}^{-}\). Theorem 1 is proved. \(\Box\)

4 EXISTENCE OF SOLUTION OF THE PROBLEM

Theorem 2. If the conditions of the Theorem 1 are satisfied and

$${\varphi_{1}}(y),\,{\varphi_{2}}(y)\in C\left({{\overline{I}}_{2}}\right)\cap{C^{1}}\left({{I_{2}}}\right),\quad a(x)\in C({\overline{I}_{1}})\cap{C^{2}}({I_{1}}),$$
(19)
$$b(x)\in{C^{2}}({I_{1}}),\quad{\lambda_{i}}(x)\in C[0,h_{1}]\cap C^{2}(0,h_{1}),\quad i=1,2,3,$$
(20)

then the solution to the problem exists.

Proof. By virtue of (13), from the Eq. (14) we obtained

$$\big{(}\tau^{+}(x)\big{)}^{\prime\prime}-\Gamma(\alpha)\,{\lambda_{2}}(x)\,\tau(x)=f(x),$$
(21)

where

$$f(x)=\Gamma(\alpha)\,{\lambda_{1}}(x)\,{\nu^{-}}(x)+\Gamma(\alpha){\lambda_{3}}(x).$$
(22)

The solution of the problem (21) with condition

$$\gamma_{1}\tau^{\prime}(0)+\gamma_{2}\tau(0)={\varphi_{1}}(0),\quad\delta_{1}\tau^{\prime}({h_{1}})+\delta_{2}\tau(h_{1})={\varphi_{2}}(0)$$
(23)

we write as

$${\tau^{+}}(x)=\Gamma(\alpha)\int\limits_{0}^{{h_{1}}}{G(x,t)\,{\lambda_{1}}(t)\,\nu(t)dt}+\Gamma(\alpha)\int\limits_{0}^{{h_{1}}}G(x,t){\lambda_{3}}(t)dt,$$
(24)

where \(G(x,t)\) is Green function of the problem (21)–(23).

The presentation (24) is functional relationship between \({\tau^{+}}(x)\) and \({\nu^{+}}(x)\), brought from the area \({\Omega_{1}}\) to \({I_{1}}\). By replacing \(x={({q^{2}}x^{\prime})^{1/2q}},\quad t={({q^{2}}t^{\prime})^{1/2q}}\) in (12) and (24), we obtain

$${\tilde{\nu}^{-}}(x)=\frac{{{\gamma_{1}}{q^{1-2{\beta_{1}}}}}}{{\overline{a}(x)\Gamma(2{\beta_{1}})}}{x^{\frac{{1-2{\alpha_{1}}}}{2}}}\frac{d}{{dx}}{x^{\frac{{1-2{\beta_{1}}}}{2}}}\int\limits_{0}^{x}{\frac{t^{\frac{{1-2{\alpha_{1}}}}{2}}}{{(x-t)}^{2{\beta_{1}}-1}}}F\left({{\alpha_{1}}+{\beta_{1}},\;\frac{{2{\beta_{1}}-1}}{2},\;2{\beta_{1}};\;\frac{{x-t}}{x}}\right)\tilde{\tau}^{-}(t)dt$$
$${}-\frac{1}{{\tilde{\overline{a}}(x)}}{({q^{2}}x)^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}}\tilde{b}(x),$$
(25)
$${\tilde{\tau}^{+}}(x)=\Gamma(\alpha)\int\limits_{0}^{1}{\tilde{G}(x,t){{\tilde{\lambda}}_{1}}(t)}{\tilde{\nu}^{+}}(t)dt+\Gamma(\alpha)\int\limits_{0}^{1}{\tilde{G}(x,t)\,{{\tilde{\lambda}}_{3}}(t)dt},$$
(26)

where

$$\tilde{\tau}^{\pm}(x)=\tau^{\pm}\left[{{{(q^{2}x)}^{1/2q}}}\right],\quad\tilde{\nu}^{\pm}(x)=\nu^{\pm}\left[{{{(q^{2}x)}^{1/2{q}}}}\right],$$
$$\tilde{f}(x)=f\left[{{{(q^{2}x)}^{1/2{q}}}}\right],\quad\tilde{\overline{a}}(x)=\overline{a}\left[{{{(q^{2}x)}^{1/2{q}}}}\right],$$
$$\tilde{b}(x)=b\left[{{{(q^{2}x)}^{1/2{q}}}}\right],\quad{\tilde{\lambda}_{1}}(x)={\lambda_{1}}\left[{{{(q^{2}x)}^{1/2{q}}}}\right],\quad\tilde{G}(x,t)=G\left[{{{({q^{2}}x)}^{1/2q}},{{({q^{2}}t)}^{1/2q}}}\right].$$

By excluding \(\tilde{\tau}^{+}(x)\) from the relationship (25) and (26), we derived

$${\tilde{\nu}^{-}}(x)=\frac{{{\gamma_{1}}\Gamma(\alpha)q^{1-2{\beta_{1}}}}}{{\tilde{\overline{a}}(x)\Gamma(2{\beta_{1}})}}{x^{\frac{{1-2{\alpha_{1}}}}{2}}}\frac{d}{{dx}}{x^{\frac{{1-2{\beta_{1}}}}{2}}}\int\limits_{0}^{x}{\frac{t^{\frac{{2{\alpha_{1}}-1}}{2}}}{{(x-t)}^{1-2{\beta_{1}}}}}F\left({{\alpha_{1}}+{\beta_{1}},\;\frac{{2{\beta_{1}}-1}}{2},\;2{\beta_{1}};\;\frac{{x-t}}{x}}\right)dt$$
$${}\times\left[{\int\limits_{0}^{1}{\tilde{G}(t,z)}{\tilde{\lambda}_{1}}(z)\,\tilde{\nu}^{-}(z)dz+\int\limits_{0}^{1}{\tilde{G}(t,z)\,{{\tilde{\lambda}}_{3}}(z)dz}}\right]-\frac{1}{{\tilde{\overline{a}}(x)}}{(q^{2}x)^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}}\tilde{b}(x).$$
(27)

The expression (27) we will write in the form of Fredholm integral equation of the second kind

$${\tilde{\nu}^{-}}(x)=\int\limits_{0}^{1}{\tilde{S}(x,z)}{\tilde{\lambda}_{1}}(z)\tilde{\nu}^{-}(z)dz+\tilde{F}(x),$$
(28)

where

$$\tilde{S}(x,z)=\frac{{{\gamma_{1}}{\Gamma(\alpha)}q_{1}^{1-2{\beta_{1}}}}}{{\tilde{\overline{a}}(x)\Gamma(2{\beta_{1}})}}{x^{\frac{{1-2{\alpha_{1}}}}{2}}}$$
$${}\times\frac{d}{{dx}}{x^{\frac{{1-2{\beta_{1}}}}{2}}}\int\limits_{0}^{x}\frac{t^{\frac{{2{\alpha_{1}}-1}}{2}}}{{{(x-t)}^{1-2{\beta_{1}}}}}\,\,F\left({{\alpha_{1}}+{\beta_{1}},\;\frac{{2{\beta_{1}}-1}}{2},\;2{\beta_{1}};\;\frac{{x-t}}{x}}\right)\tilde{G}(t,s)dt,$$
(29)
$$\tilde{F}(x)=\frac{{{\gamma_{1}}{\Gamma(\alpha)}q_{1}^{1-2{\beta_{1}}}}}{{\tilde{\overline{a}}(x)\Gamma(2{\beta_{1}})}}{x^{\frac{{1-2{\alpha_{1}}}}{2}}}\frac{d}{{dx}}{x^{\frac{{1-2{\beta_{1}}}}{2}}}\int\limits_{0}^{x}\frac{t^{\frac{{2{\alpha_{1}}-1}}{2}}}{{{(x-t)}^{1-2{\beta_{1}}}}}\,F\left({{\alpha_{1}}+{\beta_{1}},\;\frac{{2{\beta_{1}}-1}}{2},\;2{\beta_{1}};\;\frac{{x-t}}{x}}\right)dt$$
$${}\times\int\limits_{0}^{1}{{\tilde{G}(t,s)\,{{\tilde{\lambda}}_{3}}(s)ds}}-\frac{1}{\tilde{\overline{a}}(x)}{(q^{2}x)^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}}\tilde{b}(x).$$
(30)

By replacing \(t=x\mu\) in (29), we have

$$\tilde{S}(x,z)=\frac{{{\gamma_{1}}{\Gamma(\alpha)}q_{1}^{1-2{\beta_{1}}}}}{{\tilde{\overline{a}}(x)\Gamma(2{\beta_{1}})}}{x^{\frac{{1-2{\alpha_{1}}}}{2}}}$$
$${}\times\frac{d}{{dx}}{x^{{\alpha_{1}}+{\beta_{1}}}}\int\limits_{0}^{1}{\frac{\mu^{\frac{{2{\alpha_{1}}-1}}{2}}}{{(1-\mu)}^{1-2{\beta_{1}}}}}\,\tilde{G}(x\,\mu,s)F\left({{\alpha_{1}}+{\beta_{1}},\;\frac{{2{\beta_{1}}-1}}{2},\;2{\beta_{1}};\;1-\mu}\right)d\mu,$$
(31)
$$\tilde{F}(x)=\frac{{{\gamma_{1}}{\Gamma(\alpha)}q_{1}^{1-2{\beta_{1}}}}}{{\tilde{\overline{a}}(x)\Gamma(2{\beta_{1}})}}{x^{\frac{{1-2{\alpha_{1}}}}{2}}}\frac{d}{{dx}}{x^{{\alpha_{1}}+{\beta_{1}}}}\int\limits_{0}^{1}{\frac{\mu^{\frac{{2{\alpha_{1}}-1}}{2}}}{{(1-\mu)}^{1-2{\beta_{1}}}}}\,F\left({{\alpha_{1}}+{\beta_{1}},\;\frac{{2{\beta_{1}}-1}}{2},\;2{\beta_{1}};\;1-\mu}\right)d\mu$$
$${}\times\int\limits_{0}^{1}{{\tilde{G}(t,s)\,{{\tilde{\lambda}}_{3}}(s)ds}}-\frac{1}{{\tilde{\overline{a}}(x)}}{(q^{2}x)^{\frac{{1-{\alpha_{1}}+{\beta_{1}}}}{2}}}\tilde{b}(x).$$
(32)

After differentation the expressions (31) and (32) on \(x\), by virtue of (3), (19), we conclude that the kernel of the integral Eq. (28) and the right-hand side admit the estimates

$$\left|{\tilde{S}(x,z)}\right|\leqslant c_{1}x^{\frac{{2{\beta_{1}-1}}}{2}},\quad\left|{\tilde{F}(x)}\right|\leqslant c_{2}x^{\frac{{2{\beta_{1}-1}}}{2}}.$$
(33)

Since the integral Eq. (28) is a Fredholm integral equation of the second kind with a weak singularity, by virtue of (33) and (3), we deduce that the solvability of the Eq. (28) follows from the uniqueness of the solution to the problem. So, the solution of the Eq. (28) can be represented as

$${\nu^{-}}(x)=\tilde{f}(x)+\int\limits_{0}^{1}{\Re(x,z)\tilde{f}(z)dz}$$
(34)

and belongs to the class \(\nu^{-}(x)\in{C^{2}}(0,{h_{1}})\). Moreover, the function \(\nu^{-}(x)\) may have a singularity of order less than \(\frac{{1-2{\beta_{1}}}}{2}\) as \(x\to{h_{1}}\). But, this function as \(x\to 0\) is bounded, where \(\Re(x,z)\) is resolvent of the kernel \(S(x,z)\).

Now \(\nu^{-}(x)\) is known. So, from (13) we can find \(\nu^{+}(x)\) for \(\lambda_{1}(x)\neq 0\). Further, from the relation (24) we can determine \(\tau^{+}(x)=\tau^{-}(x)=\tau(x)\) in the class \(\tau\in C({\overline{I}_{1}})\cap{C^{2}}({I_{1}})\). We restore the solution of the problem in the domain \({\Omega^{+}}\) as the solution of the first boundary value problem (for \(\gamma_{1}=\delta_{1}=0\)) [9, 27]:

$$u(x,y)=\int\limits_{0}^{y}{{G_{\xi}}(x,y,0,\eta)\psi(\eta)d\eta}-\int\limits_{0}^{y}{{G_{\xi}}(x,y,1,\eta)\varphi(\eta)d\eta}+\int\limits_{0}^{1}{{G_{0}}(x-\xi,y)\tau(\xi)d\xi},$$

where

$${G_{0}}(x-\xi,y)=\frac{1}{{\Gamma(1-\alpha)}}\int\limits_{0}^{y}{{\eta^{-\alpha}}G(x,y,\xi,\eta)}d\eta,$$

\(G(x,y,\xi,\eta)=\frac{{{{(y-\eta)}^{\frac{\alpha}{2}-1}}}}{2}\sum\limits_{n=-\infty}^{\infty}{\left[{e_{1,{\alpha\mathord{\left/{\vphantom{\alpha 2}}\right.\kern-1.2pt}2}}^{1,{\alpha\mathord{\left/{\vphantom{\alpha 2}}\right.\kern-1.2pt}2}}\left({-\frac{{\left|{x-\xi+2n}\right|}}{{{{(y-\eta)}^{{\alpha\mathord{\left/{\vphantom{\alpha 2}}\right.\kern-1.2pt}2}}}}}}\right)-e_{1,{\alpha\mathord{\left/{\vphantom{\alpha 2}}\right.\kern-1.2pt}2}}^{1,{\alpha\mathord{\left/{\vphantom{\alpha 2}}\right.\kern-1.2pt}2}}\left({-\frac{{\left|{x+\xi+2n}\right|}}{{{{(y-\eta)}^{{\alpha\mathord{\left/{\vphantom{\alpha 2}}\right.\kern-1.2pt}2}}}}}}\right)}\right]}\) is Green’s function of the first boundary value problem for the Eq. (1) in the domain \({\Omega^{+}}\) [27], \(e_{1,\delta}^{1,\delta}(z)=\sum\limits_{n=0}^{\infty}{\frac{{{z^{n}}}}{{n!\Gamma(\delta-\delta n)}}}\) is Wright type function [26].

The solution of the second and mixed boundary value problem for the Eq. (1) in the domain \({\Omega^{+}}\), can also be restored following the work of Pskhu [27]. Theorem 2 is proved. \(\Box\)