1 INTRODUCTION

Degenerate partial differential equations occupy one of the central places in the theory of general partial differential equations and have numerous applications in various branches of science and technology. Partial differential equations of mixed type with degenerations have been systematically studied since the middle of the last century after the well-known works of F. I. Frankl, which are reflected in [1]. He showed applications of degenerate mixed type differential equations in solving problems of transonic and supersonic gas dynamics. Later, another applications were find of degenerate differential equations of mixed type in other fields of science and technology.

I. N. Vekua in [2] showed the importance of studying mixed-type differential equations in solving problems of the theory of infinitesimal bending of surfaces. The problem of the outflow of a supersonic jet from a vessel with flat walls reduces to the Tricomi problem for the Chaplygin equation. There are a number of works in which the problems of Tricomi, Gellerstedt, and Bitsadze are studied. It is easy to meet works, where new correct problems with bias are posed for equations of elliptic-hyperbolic and parabolic-hyperbolic types of the first kind, for which the line of degeneration is not a characteristic [3–19].

A feature of degenerate hyperbolic differential equations is that for this kind of equations the Cauchy problem with initial condition on the line of parabolic degeneracy does not always hold. For example, the Cauchy problem in the usual formulation may turn out to be unsolvable, if the hyperbolic equations degenerate along a line that is simultaneously a characteristic. Such type of differential equations are called degenerate equations of the second kind.

Note that the solution of the Cauchy problem with initial condition

$$\lim\limits_{y\to 0-}u(x,y)=\tau(x),\quad 0\leq x\leq 1,\quad\lim\limits_{y\to 0-}\frac{\partial u(x,y)}{\partial y}=\nu(x),\quad 0<x<1$$

for a differential equation

$$\left(-y\right)^{m}u_{xx}-u_{yy}=0,\quad-1<m<0$$
(1)

on the negative half-axis \(y<0\) has a representation [19, pages 259–260]:

$$u\left(x,y\right)=\gamma_{2}\int\limits_{0}^{1}\tau\left(z\right)t^{\beta}\left(1-t\right)^{\beta}dt+\frac{2\gamma_{2}}{\left(1+2\beta\right)\left(m+2\right)}\left(-y\right)^{\frac{m+2}{2}}\int\limits_{0}^{1}\tau^{\prime}\left(z\right)t^{\beta}\left(1-t\right)^{\beta}\left(2t-1\right)dt$$
$${}+\left[2\left(1-2\beta\right)\right]^{1-2\beta}\gamma_{1}y\int\limits_{0}^{1}\nu\left(z\right)t^{-\beta}\left(1-t\right)^{-\beta}dt,$$
(2)

where \(-1<2\beta<0\) for \(-1<m<0\) and

$$2\beta=\frac{m}{m+2},\quad\gamma_{1}=\left[2\left(1-2\beta\right)\right]^{2\beta-1}\frac{\Gamma\left(2-2\beta\right)}{\Gamma^{2}\left(1-\beta\right)},$$
$$\gamma_{2}=\frac{\Gamma\left(2+2\beta\right)}{\Gamma^{2}\left(1+\beta\right)},\quad z=x+\frac{2}{m+2}\left(-y\right)^{\frac{m+2}{2}}\left(2t-1\right).$$

Definition 1. The function \(u\left(x,y\right)\) (represented by (2)) on the negative half-axis \(y<0\) is called a generalized solution from the class \(\mathbb{R}_{2}\) to the Cauchy problem for differential equation (1), if \(\tau\left(z\right)\) is representable in the following integral form \(\tau\left(z\right)=\int\limits_{0}^{z}\left(z-t\right)^{-2\beta}T\left(t\right)dt,\) where \(\nu\left(z\right)\) and \(T\left(z\right)\) are continuous and integrable functions on the interval \(\left(0,1\right)\).

In [20] a generalized solution was obtained from the class \(\mathbb{R}_{2}\) to the Cauchy problem for a hyperbolic differential equation of the second kind. This generalized solution has a form

$$u\left(\xi,\eta\right)=\int\limits_{0}^{\xi}\left(\eta-t\right)^{-\beta}\left(\xi-t\right)^{-\beta}T\left(t\right)dt+\int\limits_{\xi}^{\eta}\left(\eta-t\right)^{-\beta}\left(t-\xi\right)^{-\beta}N\left(t\right)dt,$$

where

$$\xi=x-\frac{2}{m+2}\left(-y\right)^{\frac{m+2}{2}},\quad\eta=x+\frac{2}{m+2}\left(-y\right)^{\frac{m+2}{2}},\quad N\left(t\right)=\frac{1}{2\cos\pi\beta}T\left(t\right)-\gamma_{1}\nu\left(t\right).$$

Using this solution representation form, in [21–27] local and nonlocal boundary value problems for mixed differential equations of the second kind were studied. It is proved that such type of problems arise in studying some problems of mathematical biology [28] and physics [29].

In [30] for the differential equation \(u_{xx}+yu_{yy}+\left(\alpha+\frac{1}{2}\right)u_{y}=0\) built a special class of solutions in the case, when \(y<0,\quad\alpha\in\left(-\frac{1}{2},0\right)\cup\left(0,\frac{1}{2}\right)\).

In the hyperbolic part of the mixed domain, solvability of local and nonlocal boundary value problems are studied in the class \(\mathbb{R}_{2}\), and in the elliptic part of the domain the questions of classical solvability are studied. These problems are equivalently reduced to singular integral equations. Then by the regularization method they are reduced to study Fredholm integral equations of the second kind.

Note that boundary value problems for mixed differential and integro-differential equations in rectangular domains were studied in [31–34]. Boundary value problems with the Poincare–Tricomi condition for degenerate differential equations of elliptic and elliptic-hyperbolic types of the second kind have been studied, relatively little. We indicate here only the papers [35–36].

In this presented our paper by the aid of properties of generalized functions we study a boundary value problem with the Poincare–Tricomi condition for an elliptic-hyperbolic differential equation of the second kind.

2 FORMULATION OF THE PROBLEM

We consider the following mixed differential equation

$${\textrm{sgn}}y\left|y\right|^{m}u_{xx}+u_{yy}=0,\quad-1<m<0$$
(3)

in the domain \(D=D_{1}\cup D_{2}\), where \(D_{1}\) is domain bounded by a curve \(\sigma\) for \(y>0\) with the end points \(A(0,0)\), \(B(1,0)\) and with segment \(AB(y=0)\), \(D_{2}\) is domain bounded by segment \(AB\) and by characteristics

$$AC:x-\frac{2}{m+2}(-y)^{\frac{m+2}{2}}=0,\quad BC:x+\frac{2}{m+2}(-y)^{\frac{m+2}{2}}=1.$$

We introduce the notations

$$J=\left\{\left(x,y\right):\ 0<x<1,\ y=0\right\},\quad\partial D=\vec{\sigma}\cup\overline{AB},\quad 2\beta=\frac{m}{m+2}$$

for

$$-1<2\beta<0.$$
(4)

Problem \(\boldsymbol{PT}\). Find the function \(u(x,y)\), for which are true the following properties:

1) \(u(x,y)\in C\left(\overline{D}\right)\cup C^{1}\left(D\cup\sigma\cup J\right)\), there \(u_{x}\) can goes to infinity of order less than one units at the point \(A\left(0,0\right)\) and \(u_{y}\) can goes to infinity of order less than \(-2\beta\) at the point \(B\left(1,0\right)\);

2) function \(u(x,y)\in C^{2}\left(\overline{D}_{1}\right)\) is a regular solution of the differential equation (3) in the domain \(D_{1}\), and is generalized solution from the class \(\mathbb{R}_{2}\) in the domain \(D_{2}\);

3) is fulfilled the gluing condition \(u_{y}(x,-0)=-u_{y}(x,+0);\)

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

$$\left.\left\{\delta(s)A_{s}[u]+\rho(s)u\right\}\right|_{\sigma}=\varphi(s),\quad 0<s<l,\quad A_{s}\left[u\right]=y^{m}\frac{dy}{ds}\frac{\partial u}{\partial x}-\frac{dx}{ds}\frac{\partial u}{\partial y},$$
(5)
$$\left.u(x,y)\right|_{AC}=\psi(x),\quad 0\leq x\leq\frac{1}{2},$$
(6)

where \(\frac{dx}{ds}=-\cos(n,y)\), \(\frac{dy}{ds}=\cos(n,x)\), \(n\) is external normal to the curve \(\sigma\), \(l\) is a length of the entire curve \(\sigma\), \(s\) is the arc length of the curve \(\sigma\), measured from point \(B\left(1,0\right)\); \(\delta(s)\), \(\rho(s)\), \(\varphi(s)\), \(\psi(x)\) are given sufficiently smooth functions and \(\psi(x)\in C^{1}\left[0,\frac{1}{2}\right]\cap C^{2}\left(0;\frac{1}{2}\right)\), \(\varphi(l)=\psi(0)=0\).

We note that if we set \(\delta(s)=0\) (\(\rho(s)=0\)), then the problem \(PT\)coincided with the problem \(T\left(T_{N}\right)\) for elliptic-hyperbolic equation of the first kind (see [37, pp. 177–185]). Therefore, we will assume in our work that \(\delta(s)\neq 0\), \(\rho(s)\neq 0\).

We assume that the curve \(\sigma\) satisfies the following conditions:

1) functions \(x(s)\) and \(y(s)\) are parametric equations of curve \(\sigma\), have continuous derivatives \(x^{\prime}(s)\), \(y^{\prime}(s)\) (these derivatives are nonzero simultaneously) and have second derivatives, which satisfy the Hölder condition of the order \(\kappa\left(0<\kappa<1\right)\) on the segment \(0\leq s\leq l\);

2) in the neighborhood of the end points of the curve \(\sigma\) the inequality is true:

$$\left|\frac{dx}{ds}\right|\leq Cy^{m+1}(s),\quad x(l)=y(0)=0,\quad x(0)=1,\quad y(l)=0,\quad C={\textrm{const}}.$$
(7)

3 UNIQUENESS OF THE SOLUTION OF THE PROBLEM \(PT\)

Theorem 1. If (4) and the following conditions are fulfilled

$$\delta(s)\rho(s)\geq 0,\quad 0\leq s\leq l,$$
(8)
$$\lim\limits_{y\to 0}(-y)^{\frac{m}{2}}u^{2}(1,y)=0.$$
(9)

Then the solution of the problem \(PT\) is unique in the domain \(D\).

Proof. We prove the theorem by the method of energy integrals. Let \(u\left(x,y\right)\) be a twice continuously differentiable solution of equation (3) in the domain

$$\overline{D}^{\varepsilon_{1},\varepsilon_{2}}\subset D,\quad D^{\varepsilon_{1},\varepsilon_{2}}=D_{1}^{\varepsilon_{1},\varepsilon_{2}}\cup D_{2}^{\varepsilon_{1},\varepsilon_{2}},$$

where \(D_{1}^{\varepsilon_{1},\varepsilon_{2}}\) is domain with border

$$\partial D_{1}^{\varepsilon_{1},\varepsilon_{2}}=A_{\varepsilon_{2}}^{\varepsilon_{1}}B_{\varepsilon_{2}}^{\varepsilon_{1}}\cup\sigma_{\varepsilon_{1}}\quad\left(A_{\varepsilon_{2}}^{\varepsilon_{1}}B_{\varepsilon_{2}}^{\varepsilon_{1}}:\ y=\varepsilon_{2}\right)$$

strictly lying in the domain \(D_{1}\), while \(D_{2}^{\varepsilon_{1},\varepsilon_{2}}\) is domain bounded with lines

$$A_{\varepsilon_{2}}^{\varepsilon_{1}}B_{\varepsilon_{2}}^{\varepsilon_{1}}:y=-\varepsilon_{2},\quad A_{\varepsilon_{2}}^{\varepsilon_{1}}C_{\varepsilon_{1}}:x-\frac{2}{m+2}(-y)^{\frac{m+2}{2}}=\varepsilon_{1},\quad B_{\varepsilon_{2}}^{\varepsilon_{1}}C_{\varepsilon_{1}}:x+\frac{2}{m+2}(-y)^{\frac{m+2}{2}}=1-\varepsilon_{1},$$

where \(\varepsilon_{1},\varepsilon_{2}\) are small positive numbers.

In the domain \(D_{2}\) differential equation (3) takes the form \((-y)^{m}u_{xx}-u_{yy}=0.\) It is easy to check that the following identity holds:

$$u\left[(-y)^{m}u_{xx}-u_{yy}\right]=\frac{\partial}{\partial x}\left[(-y)^{m}uu_{x}\right]-\frac{\partial}{\partial y}\left[uu_{y}\right]-(-y)^{m}u_{x}^{2}+u_{y}^{2}.$$

Integrating this identity over the domain \(D_{2}^{\varepsilon_{1},\varepsilon_{2}}\), we derive

$$0=\iint\limits_{D_{2}^{\varepsilon_{1},\varepsilon_{2}}}u\left[(-y)^{m}u_{xx}-u_{yy}\right]dxdy=\iint\limits_{D_{2}^{\varepsilon_{1},\varepsilon_{2}}}\left\{\frac{\partial}{\partial x}\left[(-y)^{m}uu_{x}\right]-\frac{\partial}{\partial y}\left[uu_{y}\right]\right\}dxdy$$
$${}+\iint\limits_{D_{2}^{\varepsilon_{1},\varepsilon_{2}}}\left[u_{y}^{2}-(-y)^{m}u_{x}^{2}\right]dxdy.$$
(10)

Applying Green formula [35] to the first integral on the right-hand side of (10), we obtain

$$0=\iint\limits_{D_{2}^{\varepsilon_{1},\varepsilon_{2}}}u\left[(-y)^{m}u_{xx}-u_{yy}\right]dxdy=\int\limits_{A_{\varepsilon_{2}}^{\varepsilon_{1}}C^{\varepsilon_{1}}\cup CB_{\varepsilon_{2}}^{\varepsilon_{1}}\cup B_{\varepsilon_{2}}^{\varepsilon_{1}}A_{\varepsilon_{2}}^{\varepsilon_{1}}}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right]$$
$${}+\iint\limits_{D_{2}^{\varepsilon_{1},\varepsilon_{2}}}\left[u_{y}^{2}-(-y)^{m}u_{x}^{2}\right]dxdy.$$

Calculating the first integral on the right-hand side of the last equality, taking into account the condition (6) on the characteristic \(AC\), we obtain

$$\int\limits_{A_{\varepsilon_{2}}^{\varepsilon_{1}}C^{\varepsilon_{1}}\cup CB_{\varepsilon_{2}}^{\varepsilon_{1}}\cup B_{\varepsilon_{2}}^{\varepsilon_{1}}A_{\varepsilon_{2}}^{\varepsilon_{1}}}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right]=\int\limits_{A_{\varepsilon_{2}}^{\varepsilon_{1}}C^{\varepsilon_{1}}}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right]$$
$${}+\int\limits_{CB_{\varepsilon_{2}}^{\varepsilon_{1}}}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right]+\int\limits_{B_{\varepsilon_{2}}^{\varepsilon_{1}}A_{\varepsilon_{2}}^{\varepsilon_{1}}}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right]$$
$${}=\int\limits_{C^{\varepsilon_{1}}}^{B_{\varepsilon_{2}}^{\varepsilon_{1}}}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right]+\int\limits_{1-\varepsilon_{1}}^{\varepsilon_{1}}u(x,-\varepsilon_{2})u_{y}(x,\varepsilon_{2})dx.$$

Consequently, we have

$$\int\limits_{C^{\varepsilon_{1}}}^{B_{\varepsilon_{2}}^{\varepsilon_{1}}}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right]+\int\limits_{1-\varepsilon_{1}}^{\varepsilon_{1}}u(x,-\varepsilon_{2})u_{y}(x,\varepsilon_{2})dx+\iint\limits_{D_{2}^{\varepsilon_{1},\varepsilon_{2}}}\left[(-y)^{m}u_{x}^{2}+u_{y}^{2}\right]dxdy=0.$$

Passing to the limits as \(\varepsilon_{1}\to 0\) and \(\varepsilon_{2}\to 0\), taking the gluing condition \(u_{y}(x,-0)=-u_{y}(x,+0)\) into account we obtain

$$\int\limits_{0}^{1}\tau(x)\nu(x)dx=-\iint\limits_{D_{2}}\left[u_{y}^{2}-(-y)^{m}u_{x}^{2}\right]dxdy-\int\limits_{C}^{B}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right].$$

On the characteristic \(BC\) we have \(dx=(-y)^{\frac{m}{2}}dy\). Therefore we have

$$\int\limits_{C}^{B}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right]=\int\limits_{C}^{B}(-y)^{\frac{m}{2}}u\left[u_{x}dx+u_{y}dy\right]=\int\limits_{C}^{B}(-y)^{\frac{m}{2}}udy.$$
(11)

Integrating in parts the last integral of (11) with conditions \(\left.u\right|_{AC}=0\) and (9), we obtain

$$\int\limits_{C}^{B}(-y)^{\frac{m}{2}}udu=\frac{m}{4}\int\limits_{C}^{B}(-y)^{\frac{m-2}{2}}u^{2}dy.$$

For \(-1<m<0\) from (11) yields

$$\int\limits_{C}^{B}u\left[(-y)^{m}u_{x}dy+u_{y}dx\right]=\int\limits_{C}^{B}(-y)^{\frac{m}{2}}udu=\frac{m}{4}\int\limits_{C}^{B}(-y)^{\frac{m-2}{2}}u^{2}du\leq 0.$$
(12)

Now we show that the first integral on the right-hand side of equality (10) is not positive. Passing to characteristic variables

$$\xi=x-\frac{2}{m+2}(-y)^{\frac{m+2}{2}},\quad\eta=x+\frac{2}{m+2}(-y)^{\frac{m+2}{2}},$$

we obtain

$$\iint\limits\limits_{D_{2}}\left[u_{y}^{2}-(-y)^{m}u_{x}^{2}\right]dxdy=-2\iint\limits_{\Delta_{1}}\left(\frac{m+2}{4}\right)^{\frac{m}{2}}\left(\eta-\xi\right)^{\frac{m}{m+2}}u_{\xi}u_{\eta}d\xi d\eta,$$
(13)

where \(\Delta_{1}=\left\{\left(\xi,\eta\right):0<\xi<1,\ \xi<\eta<1\right\}\) is image of the domain \(D_{2}\) in coordinates \(\left(\xi,\eta\right)\).

In domain \(\Delta_{1}\) the differential equation (3) takes the form \(u_{\xi\eta}-\frac{\beta}{\eta-\xi}(u_{\eta}-u_{\xi})=0.\) Multiplying both sides of this equation by \(u_{\eta}\), we have

$$u_{\xi}u_{\eta}=u_{\eta}^{2}-\frac{\eta-\xi}{\beta}u_{\eta}u_{\xi\eta}.$$
(14)

Substitute (14) in (13). As a result yields

$$\iint\limits_{D_{2}}\left[u_{y}^{2}-(-y)^{m}u_{x}^{2}\right]dxdy=-2\left(\frac{m+2}{4}\right)^{\frac{m}{m+2}}$$
$${}\times\left[\iint\limits\limits_{\Delta_{1}}(\eta-\xi)^{\frac{m}{m+2}}u_{\eta}^{2}d\xi d\eta-\frac{1}{\beta}\iint\limits\limits_{\Delta_{1}}(\eta-\xi)^{\frac{2(m+1)}{m+2}}u_{\eta}u_{\xi\eta}d\xi d\eta\right].$$

Integrating the last integral in parts, we have

$$\iint\limits_{D_{2}}\left[u_{y}^{2}-(-y)^{m}u_{x}^{2}\right]dxdy=\frac{2(m+2)}{m}\left(\frac{m+2}{4}\right)^{\frac{m}{m+2}}$$
$${}\times\left[\iint\limits\limits_{\Delta_{1}}(\eta-\xi)^{\frac{m}{m+2}}u_{\eta}^{2}d\xi d\eta+\left.(\eta-\xi)^{\frac{2(m+1)}{m+2}}u_{\eta}^{2}\right|_{\eta=\xi}\right].$$

Since the last term in the square bracket is zero for \(\eta=\xi\), then we obtain

$$\iint\limits_{D_{2}}\left[u_{y}^{2}-(-y)^{m}u_{x}^{2}\right]dxdy=\frac{2(m+2)}{m}\left(\frac{m+2}{4}\right)^{\frac{m}{m+2}}\iint\limits\limits_{\Delta_{1}}(\eta-\xi)^{\frac{m}{m+2}}u_{\eta}^{2}d\xi d\eta.$$

Here by virtue of

$$\iint\limits\limits_{\Delta_{1}}(\eta-\xi)^{\frac{m}{m+2}}u_{\eta}^{2}d\xi d\eta\geq 0$$

for \(-1<m<0\) yields

$$\iint\limits_{D_{2}}\left[u_{y}^{2}-(-y)^{m}u_{x}^{2}\right]dxdy\leq 0.$$
(15)

By virtue of inequalities (12) and (15), from (10) we obtain

$$\int\limits_{0}^{1}\tau(x)\nu(x)dx\geq 0.$$
(16)

Differential equation (3) in domain \(D_{1}\) has the form \(y^{m}u_{xx}+u_{yy}=0\). Integrating the following identity

$$u\left[y^{m}u_{xx}+u_{yy}\right]=\frac{\partial}{\partial x}\left[y^{m}uu_{x}\right]+\frac{\partial}{\partial y}\left[uu_{y}\right]-y^{m}u_{x}^{2}-u_{y}^{2}$$

over the domain \(D_{1}^{\varepsilon_{1},\varepsilon_{2}}\subset D_{1}\), we derive

$$0=\iint\limits\limits_{D_{1}^{\varepsilon_{1},\varepsilon_{2}}}u\left[y^{m}u_{xx}+u_{yy}\right]dxdy=\iint\limits\limits_{D_{1}^{\varepsilon_{1},\varepsilon_{2}}}\left\{\frac{\partial}{\partial x}\left[y^{m}uu_{x}\right]+\frac{\partial}{\partial y}\left[uu_{y}\right]\right\}dxdy-\iint\limits\limits_{D_{1}^{\varepsilon_{1},\varepsilon_{2}}}\left[y^{m}u_{x}^{2}+u_{y}^{2}\right]dxdy.$$

Applying Green formula [35] to the first integral on the right-hand side of the last equality, we obtain

$$0=\iint\limits_{D_{1}^{\varepsilon_{1},\varepsilon_{2}}}u\left[y^{m}u_{xx}+u_{yy}\right]dxdy=-\iint\limits\limits_{D_{1}^{\varepsilon_{1},\varepsilon_{2}}}\left[y^{m}u_{x}^{2}+u_{y}^{2}\right]dxdy+\int\limits_{\partial D_{1}^{\varepsilon_{1},\varepsilon_{2}}}u\left[y^{m}u_{x}dy-u_{y}dx\right].$$

Hence by virtue of \(dy=0\) on \(AB\) and \(dy=\cos(n,x)ds\), \(dx=-\cos(n,y)ds\), we obtain

$$0=-\iint\limits\limits_{D_{1}^{\varepsilon_{1},\varepsilon_{2}}}\left[y^{m}u_{x}^{2}+u_{y}^{2}\right]dxdy-\int\limits_{x_{1}}^{x_{2}}u(x,\varepsilon_{2})u_{y}(x,\varepsilon_{2})dx+\int\limits_{\sigma_{\varepsilon_{1}}}uA_{s}[u]ds,$$
(17)

where \(x_{1}\), \(x_{2}\) are abscissas of the intersection points of line \(y=\varepsilon_{2}\) with curve \(\sigma_{\varepsilon_{1}}\).

By virtue of condition 1) of the problem \(PT\), zero values \(\varphi(s)\equiv b(x)\equiv 0\) and (5), from (17) as \(\varepsilon_{1}\to 0\), \(\varepsilon_{2}\to 0\) we obtain

$$\iint\limits_{D_{1}}\left[y^{m}u_{x}^{2}+u_{y}^{2}\right]dxdy+\int\limits_{0}^{1}\tau(x)\nu(x)dx+\int\limits_{\sigma}\frac{\delta(s)\rho(s)}{\delta^{2}(s)}u^{2}ds=0.$$
(18)

By virtue of inequalities (8) and (16), from (18) implies that \(u_{x}=u_{y}=0\) in \(D_{1}\), i.e. \(u(x,y)={\textrm{const}}\) for all \((x,y)\in D_{1}\). Since, each term of equality (18) is non-negative, then \(u(x,y)=0\) on \(\overline{\sigma}\). By the aid of the Hopf principle [19, pages 44–48], we conclude that \(u(x,y)\equiv 0\) in \(\bar{D}_{1}\) for \(\delta(s)\neq 0\). Now from the uniqueness of the solution of Cauchy problem it follows that \(u(x,y)\equiv 0\) in \(\bar{D}_{2}\). Consequently, \(u(x,y)\equiv 0\) in \(\overline{D}\). This is ended the proof of uniqueness of the solution of the problem \(PT\). Theorem 1 is proved. \(\Box\)

Remark. Uniqueness of the solution of the problem \(PT\) for \(\rho(s)\neq 0\), \(\forall s\in\left[0,l\right]\) be proved by the principle of extremum.

4 BASIC FUNCTIONAL RELATIONSHIPS

We introduce the following notations:

$$u\left(x,0\right)=\tau\left(x\right),\quad\left(x,0\right)\in\overline{J},$$
(19)
$$\lim\limits_{y\to-0}\frac{\partial u\left(x,y\right)}{\partial y}=\nu^{-}\left(x\right),\quad\lim\limits_{y\to+0}\frac{\partial u\left(x,y\right)}{\partial y}=\nu^{+}\left(x\right),\quad\left(x,0\right)\in J.$$
(20)

In studying this problem \(PT\) an important role are played functional relations between \(\nu^{\pm}\left(x\right)\) and \(\tau\left(x\right)\), which were bring from elliptical and hyperbolic parts of the domain \(D\).

A generalized solution from class \(\mathbb{R}_{2}\) to the Cauchy problem with conditions (19), (20) for differential equation (3) in the domain \(D_{2}\) is given by the formula [37, p. 230, form. 27.5]:

$$u\left(\xi,\eta\right)=\int\limits_{0}^{\xi}\left(\eta-t\right)^{-\beta}\left(\xi-t\right)^{-\beta}T\left(t\right)dt+\int\limits_{\xi}^{\eta}\left(\eta-t\right)^{-\beta}\left(t-\xi\right)^{-\beta}N\left(t\right)dt,$$
(21)

where

$$\xi=x-\frac{2}{m+2}(-y)^{\frac{m+2}{2}},\quad\eta=x+\frac{2}{m+2}(-y)^{\frac{m+2}{2}},\quad\gamma_{2}=\left[2(1-2\beta)\right]^{2\beta-1}\frac{\Gamma(2-2\beta)}{\Gamma^{2}(1-\beta)},$$
(22)
$$N\left(t\right)=\frac{T\left(t\right)}{2\cos\pi\beta}-\gamma_{2}\nu^{-}\left(t\right),$$
(23)
$$\tau\left(x\right)=\int\limits_{0}^{x}\left(x-t\right)^{-2\beta}T\left(t\right)dt,$$
(24)

\(T\left(x\right)\) and \(\nu^{-}\left(x\right)\) are functions of continuous in \(\left(0,1\right)\) and integrable on \(\left[0,1\right]\), \(\tau\left(x\right)\) is zero of the order not less then \(-2\beta\) as \(x\to 0\). Putting \(\xi=0\) in (21) and taking into account (6), (22), we obtain

$$\psi(\eta)=\int\limits_{0}^{\eta}N(\zeta)\zeta^{-\beta}(\eta-\zeta)^{-\beta}d\zeta.$$

This is first kind Volterra integral equation with respect to \(N(\zeta)\). We set

$$\Phi(\zeta)=N(\zeta)\zeta^{-\beta}.$$
(25)

Then we have

$$\int\limits_{0}^{\eta}\Phi(\zeta)(\eta-\zeta)^{-\beta}d\zeta=\psi(\eta).$$
(26)

We apply a fractional order differential operator \(D_{0x}^{\alpha}f(x)\) to (26):

$$D_{0\eta}^{\beta-1}\Phi(\eta)=\frac{1}{\Gamma(1-\beta)}\psi(\eta).$$
(27)

We apply the differential operator \(D_{0\eta}^{1-\beta}\) to both sides of equation (27) taking into account \(D_{0\eta}^{1-\beta}D_{0\eta}^{\beta-1}\Phi\left(\eta\right)=\Phi\left(\eta\right)\). Then we have

$$\Phi(\eta)=\frac{1}{\Gamma(1-\beta)}D_{0\eta}^{1-\beta}\psi(\eta).$$
(28)

By direct verification it is easy to check that function (28) is a solution of the equation (26). Taking into account (25) and (28) from (23) we obtain the first functional relation between \(T(x)\) and \(\nu^{-}(x)\), which are bringing from the \(D_{2}\) to the domain \(J\):

$$T(\zeta)=\gamma_{3}\nu^{-}(\zeta)+\frac{2\cos\pi\beta}{\Gamma(1-\beta)}\zeta^{\beta}D_{0\eta}^{1-\beta}\psi(\eta),$$
(29)

where \(\gamma_{3}=2\gamma_{2}\cos\pi\beta\).

In the positive half-plane \(y>0\) the differential equation (3) takes the form

$$y^{m}u_{xx}+u_{yy}=0,\quad-1<m<0.$$
(30)

Consider the following auxiliary problem.

Problem \(\boldsymbol{PT}^{\mathbf{+}}\). Find in domain \(D_{1}\) a solution \(u(x,y)\in C(\overline{D}_{1})\cap C^{1}\big{(}D_{1}\cup\sigma\cup J\big{)}\cap C^{2}\big{(}D_{1}\big{)}\) of the equation (30), satisfying to the boundary value conditions (5) and (19).

Solution of the problem \(PT^{+}\) with conditions (5) and (19) for differential equation (3) in domain \(D_{1}\) exists, unique and has the form (see [37, page 179]):

$$u(x,y)=\int\limits_{0}^{1}\tau(\xi)\frac{\partial}{\partial\eta}G_{2}(\xi,0;x,y)d\xi+\int\limits_{0}^{l}\frac{\varphi(s)}{\delta(s)}G_{2}(\xi,\eta;x,y)ds,$$
(31)

where \(G_{2}(\xi,\eta;x,y)\) is Green function of the problem \(PT^{+}\) for equation (3) and

$$G_{2}(\xi,\eta;x,y)=G_{02}(\xi,\eta;x,y)+H_{2}(\xi,\eta;x,y),$$

\(G_{02}(\xi,\eta;x,y)\) is the Green function of the problem \(PT^{+}\) for equation (3) in normal domain \(D_{0}\), which bounded with segment \(\overline{AB}\) and normal curve

$$\sigma_{0}:\left(x-\frac{1}{2}\right)^{2}+\frac{4}{(m+2)^{2}}y^{m+2}=\frac{1}{4},$$
$$H_{2}\left(\xi,\eta;x,y\right)=G_{2}\left(\xi,\eta;x,y\right)-G_{02}\left(\xi,\eta;x,y\right)$$
$${}=\int\limits_{0}^{l}\lambda_{2}(s;\xi,\eta)\left\{A_{s}\left[G_{02}\left(\xi\left(s\right),\eta\left(s\right);x,y\right)\right]+\frac{\rho(s)}{\delta(s)}G_{02}\left(\xi\left(s\right),\eta\left(s\right);x,y\right)\right\}ds,$$
(32)

\(\lambda_{2}\left(s;\xi,\eta\right)\) is solution of the integral equation

$$\lambda_{2}\left(s;\xi,\eta\right)+2\int\limits_{0}^{l}\lambda_{2}\left(t;\xi,\eta\right)\left\{A_{s}\left[q_{2}\left(\xi\left(t\right),\eta\left(t\right);x(s),y(s)\right)\right]\right.$$
$${}+\left.\frac{\rho(s)}{\delta(s)}q_{2}\left(\xi\left(t\right),\eta\left(t\right);x(s),y(s)\right)\right\}dt=-2q_{2}(\xi(s),\eta(s);\xi,\eta),$$

\(q_{2}\left(x,y,x_{0},y_{0}\right)\) is fundamental solution of differential equation (3) and

$$q_{2}\left(x,y,x_{0},y_{0}\right)=k_{2}\left(\frac{4}{m+2}\right)^{4\beta-2}\left(r_{1}^{2}\right)^{-\beta}\left(1-\sigma\right)^{1-2\beta}F\left(1-\beta,1-\beta,2-2\beta;1-\sigma\right),$$
(33)
$$\left.\begin{aligned} \displaystyle{r^{2}}\\ \displaystyle{r_{1}^{2}}\end{aligned}\right\}=\left(x-x_{0}\right)^{2}+\frac{4}{\left(m+2\right)^{2}}\left(y^{\frac{m+2}{2}}\mp y_{0}^{\frac{m+2}{2}}\right)^{2},$$
$$\sigma=\frac{r^{2}}{r_{1}^{2}},\quad\beta=\frac{m}{2(m+2)}<0,\quad k_{2}=\frac{1}{4\pi}\left(\frac{4}{m+2}\right)^{2-2\beta}\frac{\Gamma^{2}\left(1-\beta\right)}{\Gamma\left(2-2\beta\right)},$$

\(F\left(a,b,c;z\right)\) is Gauss hypergeometric function.

We differentiate (31) with respect to \(y\). Then taking (32) and (33) into account as \(y\to 0\) we obtain the functional relation between \(\tau(x)\) and \(\nu(x)\), which are bringing from the domain \(D_{1}\) to \(J\):

$$\nu(x)=k_{2}\int\limits_{0}^{1}\left|t-x\right|^{2\beta-2}\tau(t)dt-k_{2}\int\limits_{0}^{1}\frac{\tau(t)dt}{\left(t+x-2tx\right)^{2-2\beta}}$$
$${}+\int\limits_{0}^{1}\tau(t)\frac{\partial^{2}H_{2}(t,0;x,0)}{\partial\eta\partial y}dt+\int\limits_{0}^{l}\chi(s)\frac{\partial q_{2}(t,\eta;x,0)}{\partial y}ds,$$
(34)

where \(\chi(s)\) is solution of the integral equation

$$\chi(s)+2\int\limits_{0}^{l}\chi(t)\left\{A_{s}\left[q_{2}\left(\xi(t),\eta(t);x(s),y(s)\right)\right]+\frac{\rho(s)}{\delta(s)}q_{2}\left(\xi(t),\eta(t);x(s),y(s)\right)\right\}dt=\frac{2\varphi(s)}{\delta(s)}.$$

Substituting (24) into 34 and taking into account the identities:

$$\int\limits_{0}^{x}(x-t)^{2\beta-2}\tau(t)dt=\frac{\Gamma\left(1+2\beta\right)\Gamma\left(1-2\beta\right)}{2\beta\left(2\beta-1\right)}D_{0x}^{1-2\beta}D_{0x}^{2\beta-1}T(x)=\frac{\pi T(x)}{(2\beta-1)\sin 2\pi\beta},$$
$$\int\limits_{x}^{1}(t-x)^{2\beta-2}\tau(t)dt=\frac{\Gamma\left(1+2\beta\right)\Gamma\left(1-2\beta\right)}{2\beta\left(2\beta-1\right)}D_{x1}^{1-2\beta}D_{0x}^{2\beta-1}T(x)$$
$${}=\frac{\pi\cot 2\beta\pi}{1-2\beta}T\left(x\right)+\frac{1}{1-2\beta}\int\limits_{0}^{1}\left(\frac{1-t}{1-x}\right)^{1-2\beta}\frac{T\left(t\right)}{t-x}dt,$$
$$\int\limits_{0}^{1}\left(t+x-2tx\right)^{2\beta-2}\tau(t)dt=\frac{1}{1-2\beta}\int\limits_{0}^{1}\left(\frac{1-t}{1-x}\right)^{1-2\beta}\frac{T(t)dt}{x+t-2xt},$$

we obtain the functional relationship between \(T(x)\) and \(\nu^{+}(x)\), which are bringing from \(D_{1}\) to domain \(J\):

$$\nu^{+}(x)=-\frac{\pi k_{2}\tan\beta\pi}{1-2\beta}T(x)+\frac{k_{2}}{1-2\beta}\int\limits_{0}^{1}\left(\frac{1}{x}\right)^{-2\beta}T(t)\left[\frac{1}{x-t}+\frac{1-2t}{x+t-2xt}\right]dt$$
$${}+\int\limits_{0}^{1}T(t)dt\int\limits_{0}^{t}\left(t-z\right)^{-2\beta}\frac{\partial^{2}H_{2}(z,0;x,0)}{\partial\eta\partial y}dz+\int\limits_{0}^{l}\frac{\partial q_{2}(t,\eta;x,0)}{\partial y}\chi(s)ds,\quad(x,0)\in J.$$
(35)

5 EXISTENCE OF THE SOLUTION OF PROBLEM \(PT\)

Definition 2 [38, pages 255–259]. We say that the solution \(\omega(z)\) of a singular integral equation

$$\omega(z)+\lambda\int\limits_{0}^{1}\frac{\omega(\zeta)d\zeta}{\zeta-z}-\int\limits_{0}^{1}K(z,\zeta)\omega(\zeta)d\zeta=F(z)$$

belongs to class \(h(0)\), if this function \(\omega(z)\) is bounded as \(z\to 0\) and unbounded as \(z\to 1\).

Theorem 2. If conditions (4) and (7) are satisfied, then a solution of problem \(PT\) exists in the domain \(D\).

Proof. Eliminating \(\nu^{-}(x)\) and \(\nu^{+}(x)\) from relations (29) and (35), taking into account (20) and the gluing condition \(\nu^{-}(x)=-\nu^{+}(x)\), we have

$$\tilde{T}(x)-\gamma_{4}\int\limits_{0}^{1}\left[\frac{1}{x-t}-\frac{1}{x+t-2xt}\right]\tilde{T}(t)dt-\int\limits_{0}^{1}K(x,t)\tilde{T}(t)dt=F(x),$$
(36)

where \(\gamma_{4}=\frac{\cos\beta\pi}{\pi\left(\sin\beta\pi-1\right)}\), \(\tilde{T}(x)=x^{1-2\beta}T(x)\),

$$K\left(x,t\right)=\frac{\gamma_{3}}{\sin\beta\pi-1}\left(\frac{x}{t}\right)^{1-2\beta}\int\limits_{0}^{t}\left(t-z\right)^{-2\beta}\frac{\partial^{2}H_{2}(z,0;x,0)}{\partial\eta\partial y}dz,$$
(37)
$$F(x)=\frac{2\cos\pi\beta}{\left(\sin\pi\beta-1\right)\Gamma(1-\beta)}x^{1-\beta}D_{0\eta}^{1-\beta}\psi(\eta)+\frac{\gamma_{3}x^{1-2\beta}}{\left(\sin\pi\beta-1\right)}\int\limits_{0}^{l}\frac{\partial q_{2}(t,\eta;x,0)}{\partial y}\chi(s)ds.$$
(38)

We study the kernel and the right side of the singular integral equation (36). For \(0<x<1\) and \(0<z<1\) it is true the inequality [37, page 181]:

$$\left|\frac{\partial^{2}H_{2}(z,0;x,0)}{\partial\eta\partial y}\right|<C_{1}(x+z-2xz)^{2\beta-1},$$
(39)

where \(C_{1}={\textrm{const}}\) in \(D_{1}\). By virtue of (39), from (37) we obtain

$$\left|K(x,t)\right|\leq C_{1}\frac{\gamma_{3}}{\sin\beta\pi-1}\left(\frac{x}{t}\right)^{1-2\beta}\left|\int\limits_{0}^{t}\left(t-z\right)^{-2\beta}(x+z-2xz)^{2\beta-1}dz\right|.$$
(40)

By changing the variable \(z=t\) \((1-\sigma)\) and using the integral representation of the hypergeometric function [37, § 2, form. 2.10], from (40) we obtain

$$\left|K(x,t)\right|\leq C_{1}\frac{\gamma_{3}}{\sin\beta\pi-1}\left(\frac{x}{x+t-2xt}\right)^{1-2\beta}\left|\int\limits_{0}^{1}\sigma^{-2\beta}\left[1-\frac{(1-2x)t}{x+t-2xt}\sigma\right]^{2\beta-1}d\sigma\right|$$
$${}\leq C_{1}\frac{\gamma_{3}}{\sin\beta\pi-1}\left(\frac{x}{x+t-2xt}\right)^{1-2\beta}\left|F\left(1-2\beta,1-2\beta,2-2\beta;\frac{t\left(1-2x\right)}{x+t-2xt}\right)\right|.$$
(41)

Since \(c-a-b=2-2\beta-2+4\beta=2\beta<0\), then by the aid of the formula:

$$F\left(a,b,c,z\right)=\left(1-z\right)^{c-a-b}F\left(c-a,c-b,c;z\right),\quad\left|\arg(1-z)\right|<\pi$$

and by the estimate

$$\left|F\left(a,b,c,z\right)\right|\leq\begin{cases}\textrm{const},\quad c-a-b>0,\quad 0\leq z\leq 1,\\ {\textrm{const}}\cdot(1-z)^{c-a-b},\quad c-a-b<0,\quad 0<z<1,\\ {\textrm{const}}\cdot\left[1+\ln(1-z)\right],\quad c-a-b=0,\quad 0<z<1\end{cases}$$
(42)

from (41) for \(0\leq t\leq 1\) we come to an estimate

$$\left|K(x,t)\right|\leq C_{1}C_{2}\frac{\gamma_{3}}{\sin\beta\pi-1}\left(\frac{x}{x+t-2xt}\right)^{1-2\beta}\left(\frac{x}{x+t-2xt}\right)^{2\beta}\leq\frac{C_{3}x}{x+t-2xt}.$$

Now we estimate the right-hand side of equation (36). Differentiating (33) with respect to \(y\), then putting \(y=0\), we obtain

$$\frac{\partial q_{2}\left(\xi,\eta;t,0\right)}{\partial y}=k_{2}\eta\left[\left(\xi-t\right)^{2}+\frac{4}{\left(m+2\right)^{2}}\eta^{m+2}\right]^{\beta-1}.$$
(43)

Substituting (43) into (38), we have

$$F(x)=\frac{2\cos\pi\beta}{\left(\sin\pi\beta-1\right)\Gamma(1-\beta)}x^{1-\beta}D_{0\eta}^{1-\beta}\psi(\eta)$$
$${}-\frac{\gamma_{3}x^{1-2\beta}}{1-\sin\pi\beta}\int\limits_{0}^{l}\left[\left(\xi(s)-t\right)^{2}+\frac{4}{\left(m+2\right)^{2}}\eta^{m+2}(s)\right]^{\beta-1}\eta(s)\chi(s)ds.$$
(44)

By virtue of properties of the functions \(\psi\left(x\right)\) and \(\varphi\left(s\right)\), it follows from (44) that the function \(F\left(x\right)\) has derivatives of any order in the interval \((0,1)\). Let us find out the behavior of the function \(F\left(x\right)\) and its derivative as \(x\to 0\) and as \(x\to 1\). Consider the expression

$$F_{1}(x)=\int\limits_{0}^{l}\left[\left(\xi(s)-t\right)^{2}+\frac{4}{\left(m+2\right)^{2}}\eta^{m+2}(s)\right]^{\beta-1}\eta(s)\chi(s)ds.$$
(45)

We estimate the expression (45). By virtue of \(\delta(s),\rho(s),\varphi(s)\in C\left[0,l\right]\), for enough small \(x>0\) there hold inequalities

$$\left|F_{1}\left(x\right)\right|\leq\int\limits_{l-\varepsilon}^{l}\left|\chi(s)\right|\left[\left(\xi-x\right)^{2}+\frac{4}{\left(m+2\right)^{2}}\eta^{m+2}\right]^{\beta-1}\eta ds+O(1)$$
$${}<C_{4}\int\limits_{l-\varepsilon}^{l}\left[\left(\xi-x\right)^{2}+\frac{4}{\left(m+2\right)^{2}}\eta^{m+2}\right]^{\beta-1}\eta ds+O(1).$$

Hence, by virtue of (26), for sufficiently small \(\varepsilon>0\) we obtain

$$\left|F_{1}\left(x\right)\right|<C_{5}\int\limits_{l-\varepsilon}^{l}\eta^{\frac{m}{2}}\left|\frac{d\eta}{ds}\right|\left[x^{2}+\frac{4}{\left(m+2\right)^{2}}\eta^{m+2}\right]^{-\frac{1}{2}-\beta}ds+O(1)$$
$${}<C_{6}\int\limits_{0}^{\delta}\left[x^{2}+\tilde{\eta}^{2}\right]^{-\frac{1}{2}-\beta}d\tilde{\eta}+O(1).$$
(46)

By the change \(\mu^{2}=\omega\) in (46), using the integral representation of the hypergeometric function [37, § 2, form. 2.10] and taking into account (42) we obtain

$$\left|F_{1}\left(x\right)\right|<\frac{\delta^{2}}{x^{2\beta+1}}\left|\frac{\Gamma(1,5)\Gamma(\beta)}{\Gamma(0,5+\beta)}\right|\left(\frac{\delta^{2}}{x^{2}}\right)^{-\frac{1}{2}}$$
$${}+\frac{\delta}{x^{2\beta+1}}\left|\frac{\Gamma(1,5)\Gamma(-\beta)}{\Gamma(0,5)\Gamma(1-\beta)}\right|(x^{2}+\delta^{2})^{-\beta}x^{2\beta+1}F\left(\beta,\frac{1}{2},1+\beta;\frac{x^{2}}{x^{2}+\delta^{2}}\right)$$
$${}<\frac{\delta}{x^{2\beta}}\left|\frac{\Gamma(1,5)\Gamma(\beta)}{\Gamma(0,5+\beta)}\right|+\left|\frac{\delta\Gamma(1,5)\Gamma(-\beta)}{\Gamma(0,5)\Gamma(1-\beta)}\right|(x^{2}+\delta^{2})^{-\beta}<C_{7}x^{-2\beta}.$$
(47)

If \(1-x\) is sufficiently small, by the similarly way we obtain

$$\left|F_{1}\left(x\right)\right|=C_{8}\left(1-x\right)^{-2\beta}.$$
(48)

Carrying out the same reasoning, we obtain

$$\left|F_{1}^{\prime}\left(x\right)\right|<C_{9}x^{-2\beta-1},\quad\left|F_{1}^{\prime}\left(x\right)\right|=C_{10}\left(1-x\right)^{-2\beta-1}.$$
(49)

By virtue of (47)–(49), from (44) we deduce that \(F(x)\in C(\overline{J})\cap C^{2}(J)\) and the function \(F^{\prime}(x)\) turns to infinity order less then \(2\beta+1\) as \(x\to 1\), and as \(x\to 0\) this derivative \(F^{\prime}(x)\) is bounded.

By the change of variables \(\zeta=\frac{t^{2}}{1-2t+2t^{2}},\) \(z=\frac{x^{2}}{1-2x+2x^{2}},\) we represent the equation (36) as follows

$$\omega(z)+\gamma_{4}\int\limits_{0}^{1}\frac{\omega(\zeta)d\zeta}{\zeta-z}\quad-\int\limits_{0}^{1}\overline{K}(z,\zeta)\omega(\zeta)d\zeta=\tilde{F}(z),$$
(50)

where

$$\omega(z)=(1-2x+2x^{2})\tilde{T}(x),\quad\tilde{F}(z)=(1-2x+2x^{2})F(x),$$
$$\overline{K}(z,\zeta)=\frac{1-2t+2t^{2}}{2t(1-t)(1-2x+2x^{2})}K(x,t)+\gamma_{4}\frac{(1-2x+2x^{2})(1-2t+2t^{2})}{(1-t)(t+x-2xt)},$$
$$x=\frac{\sqrt{z}}{\sqrt{z}+\sqrt{1-z}},\quad t=\frac{\sqrt{\zeta}}{\sqrt{\zeta}+\sqrt{1-\zeta}}.$$

Since \(1-\gamma_{4}^{2}\neq 0\), then equation (50) is a normal type equation. Its index is zero in the class \(h(0)\), i.e. in the class of functions, which bounded as \(z\to 0\) and unbounded as \(z\to 1\).

Thus, we studied the solution \(\omega(z)\) of singular integral equation (50) in the class \(h(0)\).

We reduce the singular integral equation (50) by the well-known Carleman–Vekua regularization method [38] to an equivalent Fredholm equation of the second kind, the solvability of which implies from the uniqueness of the solution of problem \(PT\).

From the equality \(\tilde{T}(x)=x^{1-2\beta}T(x)\) and \(\omega(z)=(1-2x+2x^{2})\tilde{T}(x)\) we find the function \(T(x)\), which is continuous in \((0,1)\) and integrable on \(\left[0,1\right]\).

Substituting \(T(x)\) into (24), we find \(\tau(x)\). Then from (29) and (35) we find \(\nu^{\pm}(x)\). When \(\tau(x)\) is known function, then the solution of the problem \(PT\) for differential equation (3) in the domain \(D_{1}\) we restore as a solution of the problem \(PT^{+}\) for differential equation (3) with conditions (5) and (19). The solution of the problem \(PT\) for differential equation (3) in the domain \(D_{2}\) we restore as a generalized solution of the Cauchy problem with conditions (19) and (20) for differential equation (3).

Thus, in the domain \(D\) a solution of the problem \(PT\) exists. Theorem 2 is proved. \(\Box\)