1 INTRODUCTION

System of equations of the first order

$$\frac{\partial u_{i}}{\partial x_{i}}=\sum_{k=1}^{n}a_{ik}(x_{1},\dots,x_{n})u_{k}+f_{i}(x_{1},\dots,x_{n}),\quad i=1,\dots,n,$$
(1)

was studied by a number of authors (see, for instance, [1–6] and references in these papers). It is of interest, in particular, in connection with differential equations of mixed type. The most number of publications concerns the case \(n=2\).

The author prposed [7] a version of the ariemann method for systems of differential equations both with simple and multiple characteristics.

The Darboux problem for hyperbolic equations and systems is of great interest. This problem for equations of second order with two independent variables is considered by a number of authors. We mention here the works [8, p. 228–233], [9–14]. There exist also investigations of the Darboux problem for hyperbolic equations of the third order [15, 16].

2 THE GOURSAT PROBLEM

We consider the following system of equations

$$\begin{cases}u_{1x}=a_{11}(x,y)u_{1}+a_{12}(x,y)u_{2}+f_{1}(x,y),\\ u_{2y}=a_{21}(x,y)u_{1}+a_{22}(x,y)u_{2}+f_{2}(x,y),\end{cases}$$
(2)

in rectangle \(D=\{x_{0}<x<x_{1}\), \(y_{0}<y<y_{1}\}\). Here \(a_{11}\), \(a_{12}\), \(a_{21}\), \(a_{22}\), \(f_{1}\), \(f_{2}\in C(\overline{D})\). A solution \(u\) is called regular in \(D\) if \(u_{1},\)\(u_{2},\)\(u_{1x},\)\(u_{2y}\in C(D)\).

The Goursat problem: Find regular solution of system (2) in domain \(D\) which is continuously extendable on the boundary of domain \(D\) and satisfies boundary conditions

$$u_{1}(x_{0},y)=\varphi(y),\quad u_{2}(x,y_{0})=\psi(x),$$
$$\varphi(y)\in C([y_{0},y_{1}]),\quad\psi(x)\in C([x_{0},x_{1}]).$$
(3)

The Goursat problem has unique solution [].

A linear transformation of the desired functions reduces (2) to the case

$$a_{11}\equiv a_{22}\equiv 0.$$
(4)

In what follows we conider these identities as fulfilled.

Chekmarev [] by means of the successive approximations method obtained solution of the Goursat problem (2), (3) in the form

$$u_{1}(x,y)=\varphi(y)+\int\limits_{y_{0}}^{y}\varphi(\tau)L(x,y,x_{0},\tau)d\tau+\int\limits_{x_{0}}^{x}\psi(t)K(x,y,t,y_{0})dt+\int\limits_{x_{0}}^{x}f_{1}(t,y)dt$$
$${}+\int\limits_{x_{0}}^{x}\int\limits_{y_{0}}^{y}(L(x,y,t,\tau)f_{1}(t,\tau)+K(x,y,t,\tau)f_{2}(t,\tau))d\tau dt,$$
(5)
$$u_{2}(x,y)=\psi(x)+\int\limits_{y_{0}}^{y}\varphi(\tau)M(x,y,x_{0},\tau)d\tau+\int\limits_{x_{0}}^{x}\psi(t)N(x,y,t,y_{0})dt+\int\limits_{y_{0}}^{y}f_{2}(x,\tau)d\tau$$
$${}+\int\limits_{x_{0}}^{x}\int\limits_{y_{0}}^{y}(M(x,y,t,\tau)f_{1}(t,\tau)+N(x,y,t,\tau)f_{2}(t,\tau))d\tau dt,$$
(6)

where \(L\), \(K\), \(M\), \(N\) are determined by coefficients \(a_{12}\), \(a_{21}\) as uniformly convergent series, and satiffy relations

$$\displaystyle K(x,y,t,\tau)=a_{12}(t,y)+\int\limits_{t}^{x}a_{12}(\xi,y)N(\xi,y,t,\tau)d\xi,$$
$$\displaystyle L(x,y,t,\tau)=\int\limits_{t}^{x}a_{12}(\xi,y)M(\xi,y,t,\tau)d\xi,$$
$$\displaystyle M(x,y,t,\tau)=a_{21}(x,\tau)+\int\limits_{t}^{y}a_{21}(x,\eta)L(x,\eta,t,\tau)d\eta,$$
$$\displaystyle N(x,y,t,\tau)=\int\limits_{\tau}^{y}a_{21}(x,\eta)K(x,\eta,t,\tau)d\eta.$$
(7)

There exists other approach to solving of described aboveproblems. We can apply to system (2) the Riemann method. Rewrite (2) in vector-matrix form

$$L(\mathbf{U})=\mathbf{F},\quad L(\mathbf{U})\equiv\mathbf{A}\mathbf{U}_{x}+\mathbf{B}\mathbf{U}_{y}-\mathbf{C}\mathbf{U},\quad\mathbf{U}=\mathop{\textrm{colon}}\nolimits(u_{1},u_{2}),$$
$$\mathbf{A}=\begin{pmatrix}1&0\\ 0&0\\ \end{pmatrix},\quad\mathbf{B}=\begin{pmatrix}0&0\\ 0&1\end{pmatrix},\quad\mathbf{C}=\begin{pmatrix}0&a_{12}\\ a_{21}&0\end{pmatrix},$$
$$\mathbf{F}=\mathop{\textrm{colon}}\nolimits(f_{1},f_{2}).$$
(8)

Introduce Riemann matrix \(\mathbf{R}=\mathop{\textrm{colon}}\nolimits(\mathbf{R}_{1},\mathbf{R}_{2})\), where vectors \(\mathbf{R}_{i}(x,y,\xi,\eta)=(r_{i1},r_{i2})\), \(i=\overline{1,2}\), are solutions of systems

$$\begin{cases}\displaystyle r_{i1}(x,y)=\delta_{i1}-\int\limits_{\xi}^{x}a_{21}(\alpha,y)r_{i2}(\alpha,y)d\alpha,\\ \displaystyle r_{i2}(x,y)=\delta_{i2}-\int\limits_{\eta}^{y}a_{12}(x,\beta)r_{i1}(x,\beta)d\beta,\end{cases}$$
(9)

\(\delta_{ij}\) is Kronecker symbol. The solutions of systems (9) exist and are unique in the class of continuous functions. We differentiate (9), and see that system (9) is equivalent to two Goursat problems

$$\begin{cases}r_{i1x}=-a_{21}(x,y)r_{12},\quad r_{i2y}(x,y)=-a_{12}(x,y)r_{11},\\ r_{i1}|_{x=\xi}=\delta_{i1},\quad r_{i2}|_{y=\eta}=\delta_{i2},\quad i=1,2.\end{cases}$$
(10)

Clearly, function \(R\) with respect to the first pair of variables satisfies the conjugated to (2) system

$$L^{\ast}(\mathbf{V})=0,\quad L^{\ast}(\mathbf{V})\equiv-(\mathbf{V}\mathbf{A})_{x}-(\mathbf{V}\mathbf{B})_{y}-\mathbf{V}\mathbf{C}.$$

We integrate identity

$$\mathbf{R}L(\mathbf{U})=(\mathbf{R}\mathbf{A}\mathbf{U})_{x}+(\mathbf{R}\mathbf{B}\mathbf{U})_{y}$$
(11)

over rectangle \(D_{1}=\{x_{0}<x<\xi\), \(y_{0}<y<\eta\}\), \((\xi,\eta)\in D\), and obtain

$$\int\limits_{y_{0}}^{\eta}r_{11}(\xi,y,\xi,\eta)u_{1}(\xi,y)dy+\int\limits_{x_{0}}^{\xi}r_{12}(x,\eta,\xi,\eta)u_{2}(x,\eta)dx=\int\limits_{y_{0}}^{\eta}r_{11}(x_{0},y,\xi,\eta)u_{1}(x_{0},y)dy$$
$${}+\int\limits_{x_{0}}^{\xi}r_{12}(x,y_{0},\xi,\eta)u_{2}(x,y_{0})dx+\iint\limits_{D_{1}}(r_{11}(x,y,\xi,\eta)f_{1}(x,y)+r_{12}(x,y,\xi,\eta)f_{2}(x,y))dxdy,$$
(12)
$$\int\limits_{y_{0}}^{\eta}r_{21}(\xi,y,\xi,\eta)u_{1}(\xi,y)dy+\int\limits_{x_{0}}^{\xi}r_{22}(x,\eta,\xi,\eta)u_{2}(x,\eta)dx=\int\limits_{y_{0}}^{\eta}r_{21}(x_{0},y,\xi,\eta)u_{1}(x_{0},y)dy$$
$${}+\int\limits_{x_{0}}^{\xi}r_{22}(x,y_{0},\xi,\eta)u_{2}(x,y_{0})dx+\iint\limits_{D_{1}}(r_{21}(x,y,\xi,\eta)f_{1}(x,y)+r_{22}(x,y,\xi,\eta)f_{2}(x,y))dxdy.$$
(13)

We use the properties of the Riemann matrix, differentiate (12) with respect to variable \(\eta\) and (13) with respect to \(\xi\), and obtain solutions of the Goursat problem in the following form:

$$u_{1}(\xi,\eta)=\varphi(\eta)+\int\limits_{y_{0}}^{\eta}r_{11\eta}(x_{0},y,\xi,\eta)\varphi(y)dy+\int\limits_{x_{0}}^{\xi}r_{12\eta}(x,y_{0},\xi,\eta)\psi(x)dx+\int\limits_{x_{0}}^{\xi}f_{1}(x,\eta)dx$$
$${}+\int\limits_{y_{0}}^{\eta}\int\limits_{x_{0}}^{\xi}(r_{11\eta}(x,y,\xi,\eta)f_{1}(x,y)+r_{12\eta}(x,y,\xi,\eta)f_{2}(x,y))dxdy,$$
(14)
$$u_{2}(\xi,\eta)=\psi(\xi)+\int\limits_{y_{0}}^{\eta}r_{21\xi}(x_{0},y,\xi,\eta)\varphi(y)dy+\int\limits_{x_{0}}^{\xi}r_{22\xi}(x,y_{0},\xi,\eta)\psi(x)dx+\int\limits_{y_{0}}^{\eta}f_{2}(\xi,y)dy$$
$${}+\int\limits_{y_{0}}^{\eta}\int\limits_{x_{0}}^{\xi}(r_{21\xi}(x,y,\xi,\eta)f_{1}(x,y)+r_{22\xi}(x,y,\xi,\eta)f_{2}(x,y))dxdy.$$
(15)

The formulas (5), (6), (7), (9), (14), (15) imply relations

$$L(\xi,\eta,x,y)=r_{11\eta}(x,y,\xi,\eta),\quad K(\xi,\eta,x,y)=r_{12\eta}(x,y,\xi,\eta),$$
$$M(\xi,\eta,x,y)=r_{21\xi}(x,y,\xi,\eta),\quad N(\xi,\eta,x,y)=r_{22\xi}(x,y,\xi,\eta).$$

3 THE DARBOUX PROBLEM

We consider the Darboux problem:Find regular solution of system (2) in domain \(D_{0}=\{(x,y):0<y<x<T\}\), which is continuable on the boundary of domain \(D\) and satisfies boundary conditions

$$u_{1}(y,y)=\lambda(y),u_{2}(x,0)=\mu(x),$$
$$\lambda(y)\in C([0,T]),\quad\mu(x)\in C([0,T]).$$
(16)

Solution of the Darboux problem exists, and it is unique [3, p. 26–29].

We fix a point \(P(\xi,\eta)\in D_{0}\), and consider corresponding quadrangle \(D_{p}=\{(x,y):y<x<\xi,0<y<\eta\}\subset D_{0}\) with vertices at points \(O(0,0)\), \(P_{1}(\eta,\eta)\), \(P(\xi,\eta)\), \(P_{3}(\xi,0)\), rectangle \(D_{1}=\{(x,y):\eta<x<\xi,0<y<\eta\}\) with vertices at points \(P\), \(P_{1}\), \(P_{2}(\eta,0)\), \(P_{3}\), and triangle \(D_{2}=\{(x,y):y<x<\eta,0<y<\eta\}\) with vertices at points \(O\), \(P_{1}\), \(P_{2}\) (see Fig. 1).

figure 1

Fig. 1

Define Riemann–Hadamard matrix for the Darboux problem \(H(x,y,\xi,\eta)\). Let

$$H(x,y,\xi,\eta)=\begin{cases}R(x,y,\xi,\eta),\quad(x,y)\in D_{1},\\ V(x,y,\xi,\eta),\quad(x,y)\in D_{2},\end{cases}$$
(17)

where \(R\) is the defined above Riemann matrix, and matrix \(\mathbf{V}=\mathop{\textrm{colon}}\nolimits(\mathbf{V}_{1},\mathbf{V}_{2})\), with vectors \(\mathbf{V}_{i}(x,y,\xi,\eta)=(\bar{r}_{i1},\bar{r}_{i2})\), \(i=\overline{1,2}\), needs to be specified in domain \(D_{2}\).

We require that \(V_{1}\), \(V_{2}\) be solutions the Darboux problems in domain \(D_{2}\) for systems

$$\bar{r}_{i1x}=-a_{21}(x,y)\bar{r}_{12},\quad\bar{r}_{i2y}(x,y)=-a_{12}(x,y)\bar{r}_{11},\quad i=1,2,$$
(18)

with conditions relatively

$$\bar{r}_{11}(\eta,y,\xi,\eta)=r_{11}(\eta,y,\xi,\eta),\quad\bar{r}_{12}(y,y,\xi,\eta)=0$$
(19)

for vector \(V_{1}\), and

$$\bar{r}_{21}(\eta,y,\xi,\eta)=r_{21}(\eta,y,\xi,\eta),\quad\bar{r}_{22}(y,y,\xi,\eta)=0$$
(20)

for vector \(V_{2}\).

We integrate identity (11) over rectangle \(D_{1}\):

$$\int\limits_{0}^{\eta}r_{11}(\xi,y,\xi,\eta)u_{1}(\xi,y)dy+\int\limits_{\eta}^{\xi}r_{12}(x,\eta,\xi,\eta)u_{2}(x,\eta)dx=\int\limits_{0}^{\eta}r_{11}(\eta,y,\xi,\eta)u_{1}(\eta,y)dy$$
$${}+\int\limits_{\eta}^{\xi}r_{12}(x,0,\xi,\eta)u_{2}(x,0))dx+\iint\limits_{D_{1}}(r_{11}(x,y,\xi,\eta)f_{1}(x,y)+r_{12}(x,y,\xi,\eta)f_{2}(x,y))dxdy,$$
(21)
$$\int\limits_{0}^{\eta}r_{21}(\xi,y,\xi,\eta)u_{1}(\xi,y)dy+\int\limits_{\eta}^{\xi}r_{22}(x,\eta,\xi,\eta)u_{2}(x,\eta)dx=\int\limits_{0}^{\eta}r_{21}(\eta,y,\xi,\eta)u_{1}(\eta,y)dy$$
$${}+\int\limits_{\eta}^{\xi}r_{22}(x,0,\xi,\eta)u_{2}(x,0))dx+\iint\limits_{D_{1}}(r_{21}(x,y,\xi,\eta)f_{1}(x,y)+r_{22}(x,y,\xi,\eta)f_{2}(x,y))dxdy.$$
(22)

Integration of identity (11) over triangle \(D_{2}\) gives

$$-\int\limits_{0}^{\eta}\bar{r}_{12}(x,0,\xi,\eta)u_{2}(x,0)dx+\int\limits_{0}^{\eta}\bar{r}_{11}(\eta,y,\xi,\eta)u_{1}(\eta,y)dy+\int\limits_{\eta}^{0}(\bar{r}_{11}(y,y,\xi,\eta)u_{1}(y,y)$$
$${}-\bar{r}_{12}(y,y,\xi,\eta)u_{2}(y,y))dy+\iint\limits_{D_{2}}(\bar{r}_{11}(x,y,\xi,\eta)f_{1}(x,y)+\bar{r}_{12}(x,y,\xi,\eta)f_{2}(x,y))dxdy,$$
(23)
$$-\int\limits_{0}^{\eta}\bar{r}_{22}(x,0,\xi,\eta)u_{2}(x,0)dx+\int\limits_{0}^{\eta}\bar{r}_{21}(\eta,y,\xi,\eta)u_{1}(\eta,y)dy+\int\limits_{\eta}^{0}(\bar{r}_{21}(y,y,\xi,\eta)u_{1}(y,y)$$
$${}-\bar{r}_{22}(y,y,\xi,\eta)u_{2}(y,y))dy+\iint\limits_{D_{2}}(\bar{r}_{21}(x,y,\xi,\eta)f_{1}(x,y)+\bar{r}_{22}(x,y,\xi,\eta)f_{2}(x,y))dxdy.$$
(24)

We add (21) with (23) and (22) with (24), and obtain

$$\int\limits_{0}^{\eta}r_{11}(\xi,y,\xi,\eta)u_{1}(\xi,y)dy+\int\limits_{\eta}^{\xi}r_{12}(x,\eta,\xi,\eta)u_{2}(x,\eta)dx$$
$${}+\int\limits_{0}^{\eta}\bar{r}_{11}(\eta,y,\xi,\eta)u_{1}(\eta,y)dy+\int\limits_{\eta}^{0}(\bar{r}_{11}(y,y,\xi,\eta)u_{1}(y,y)-\bar{r}_{12}(y,y,\xi,\eta)u_{2}(y,y))dy$$
$${}=\int\limits_{0}^{\eta}r_{11}(\eta,y,\xi,\eta)u_{1}(\eta,y)dy+\int\limits_{\eta}^{\xi}r_{12}(x,0,\xi,\eta)u_{2}(x,0))dx$$
$${}+\int\limits_{0}^{\eta}\bar{r}_{12}(x,0,\xi,\eta)u_{2}(x,0)dx+\iint\limits_{D_{1}}(r_{11}(x,y,\xi,\eta)f_{1}(x,y)+r_{12}(x,y,\xi,\eta)f_{2}(x,y))dxdy$$
$${}+\iint\limits_{D_{2}}(\bar{r}_{11}(x,y,\xi,\eta)f_{1}(x,y)+\bar{r}_{12}(x,y,\xi,\eta)f_{2}(x,y))dxdy,$$
(25)
$$\int\limits_{0}^{\eta}r_{21}(\xi,y,\xi,\eta)u_{1}(\xi,y)dy+\int\limits_{\eta}^{\xi}r_{22}(x,\eta,\xi,\eta)u_{2}(x,\eta)dx$$
$${}+\int\limits_{0}^{\eta}\bar{r}_{21}(\eta,y,\xi,\eta)u_{1}(\eta,y)dy+\int\limits_{\eta}^{0}(\bar{r}_{21}(y,y,\xi,\eta)u_{1}(y,y)-\bar{r}_{22}(y,y,\xi,\eta)u_{2}(y,y))dy$$
$${}=\int\limits_{0}^{\eta}r_{21}(\eta,y,\xi,\eta)u_{1}(\eta,y)dy+\int\limits_{\eta}^{\xi}r_{22}(x,0,\xi,\eta)u_{2}(x,0))dx$$
$${}+\int\limits_{0}^{\eta}\bar{r}_{12}(x,0,\xi,\eta)u_{2}(x,0)dx+\iint\limits_{D_{1}}(r_{21}(x,y,\xi,\eta)f_{1}(x,y)+r_{22}(x,y,\xi,\eta)f_{2}(x,y))dxdy$$
$${}+\iint\limits_{D_{2}}(\bar{r}_{21}(x,y,\xi,\eta)f_{1}(x,y)+\bar{r}_{22}(x,y,\xi,\eta)f_{2}(x,y))dxdy.$$
(26)

It follows from (9), (19), (20) that \(r_{11}(\xi,y,\xi,\eta)=1\), \(r_{12}(x,\eta,\xi,\eta)=0\), \(r_{21}(\xi,y,\xi,\eta)=0\), \(r_{22}(x,\eta,\xi,\eta)=1\), \(\bar{r}_{12}(y,y,\xi,\eta)=0\), \(\bar{r}_{22}(y,y,\xi,\eta)=0\), and

$$\bar{r}_{11}(\eta,y,\xi,\eta)-r_{11}(\eta,y,\xi,\eta)=0,\quad\bar{r}_{21}(\eta,y,\xi,\eta)-r_{21}(\eta,y,\xi,\eta)=0.$$

Consequently, (25) and (26) are representable as

$$\int\limits_{0}^{\eta}u_{1}(\xi,y)dy=F_{1}(\xi,\eta),\quad\int\limits_{\eta}^{\xi}u_{2}(x,\eta)dx=F_{2}(\xi,\eta),$$
(27)

where \(F_{1}(\xi,\eta)\), \(F_{2}(\xi,\eta)\) are completely determined by the Riemann–Hadamard matrix \(H\) and boundary data of the Darboux problem.

We differentiate the first equality (27) in \(\eta\), and second in \(\xi\), and obtain explicit solution of the Darboux problem in terms of the Riemann–Hadamard matrix.