1 INTRODUCTION

In 1857, B. Riemann (see, e.g. [29]) posed the following problem: to find (construct) a system of functions \(Y\left(z\right)=\left(y_{1}\left(z\right),\ldots,y_{m}\left(z\right)\right)^{T}\), satisfying three conditions:

  1. 1.

    functions are analytic everywhere in \(\widehat{\mathbb{C}}\) except at a finite number of points \(a_{1},a_{2},\ldots,a_{n}\);

  2. 2.

    the vector \(Y\left(z\right)\) possesses a linear transformation with a non-singular constant matrix \(V_{k}\) whenever \(z\) is orbiting around each singular point \(a_{k}\) \(\left(k=1,2,\ldots,n\right)\), i.e. \(Y\rightarrowtail V_{k}Y\), and \(V_{1}\cdot V_{2}\cdot\ldots\cdot V_{n}=E\), where \(E\) is the \(m\times m\) identity matrix;

  3. 3.

    the functions \(y_{1}\left(z\right),y_{2}\left(z\right),\ldots,y_{m}\left(z\right)\) possess power type asymptotics of a finite order at each singular point \(a_{k}\left(k=1,2,\ldots,n\right)\), i.e. \(|y_{j}(z)|\leq C|z-a_{k}|^{-\alpha}\), \(\forall z\), \(0<|z-a_{k}|<r_{k}\), \(a_{k}\not=\infty\) and \(|y_{j}(z)|\leq C|z|^{\alpha}\), \(\forall z\), \(|z|>r_{\infty}\) for an appropriate \(\alpha\geq 0\).

The matrices \(V_{1},V_{2},\ldots,V_{n}\) generate the so called monodromy group ([2, 5]). Riemann found a representation of the solution in the neighborhood of each particular singular point \(a_{k}\),

Riemann also pointed out that the functions \(y_{1},\ldots,y_{m}\) will be solutions of an \(m\)-th order complex differential equation with rational coefficients (see, e.g., [4]). In 1900 Hilbert included the problem of construction of a Fuchsian differential system into his list of mathematical problems for the twentieth century. This question is now known as the twenty-first Hilbert problem or as the Riemann–Hilbert problem, see [5, 8].

The Riemann–Hilbert problem can be formulated as the Riemann boundary value problem for analytic vector-functions ([24, 25, 31]). The vector-matrix Riemann boundary value problem with piecewise continuous algebraic coefficients was first solved in [10] by using Green’s function method. The solution of the Riemann-Hilbert problem by reduction to the Riemann boundary value problem was proposed in 1908 by Plemelj [27] (see also [28]).

For a long time it was thought that Plemelj had found a complete and positive answer to the question of existence of a complex differential equation with a given monodromy group. Therefore interest in this problem was moved to the effective construction of its solution, the most known are results by Lappo-Danilevsky (functions of matrices method), Röhrl (fibre bundles method) (see, e.g. [5, 3]). Erugin [7] considered the case of four singular points and showed, in particular, that the Riemann–Hilbert problem is related to Painlevé type differential equations (among the recent constructive results we can mention the paper [3]).

The gaps in Plemelj’s 1908 paper were first discovered by Kohn [21] and by Arnol’d and Ill’yashenko [2]. In the late 1980s Bolibrukh (see, e.g., [4]) found the first negative answer to the question. In fact, Plemelj’s conclusion is valid for the so-called ‘‘regular’’ (see, [4, p. 7]) variant of the Riemann–Hilbert problem. An extended description of the modern state of the Riemann-Hilbert problem as well as the main results by Bolibruch are presented in [4] and [5].

Although this problem has in general a negative solution, there are few cases when an existence of the positive solutions can be proved (for \(2\times 2\) Fuchsian systems and any number of singular points this is in Dekkers’ theorem [6]). Constructive procedures for solution to the Riemann–Hilbert problem is known only for few cases. Thus, the solution for the systems of two unknown functions (\(m=2\)) with three singular points (\(n=3\)) is given in [22]. We should also mention the paper [8], cf. [23, 30], which is devoted to the connection between the factorization of piecewise constant \(m\times m\) matrix functions with \(n\) jumps and the Riemann–Hilbert problem (cases \(m=2\), \(n=4\) and \(m=3\), \(n=3\) are treated there). Several constructive results related to the method developed in this paper are presented in [12–19, 20]. The problem of construction of the differential equation of Fuchsian type is highly related to the determination of accessor parameters which is not considered here, see [1].

The aim of our paper is to determine a procedure which makes the general case (small number of unknown functions and arbitrary number of singular points) constructively solvable. We propose a novel approach (logarithmization method) to the solution of the Riemann–Hilbert problem by bypassing of the commutativity assumption of the generator (see [4]). It generalizes Lappo-Danilevsky’s approach and is based on an exact representation for \(\ln(V_{1}\cdots V_{k})\). We limit our attention to the first nontrivial case for which the problem is not solved, namely, for vector functions \(Y\left(z\right)=\left(y_{1},y_{2}\right)^{T}\) (\(m=2\)), and five singular points (\(n=5\)). The procedure has been checked to ensure that it is universal, i.e. that it works after slight modifications in the case of six or more singular points and larger \(m\), but its algoritmisation is not yet done.

Within the paper we solve the following problems provided that the monodromy matrices are given. Each of these problems has a particular interest.

  1. 1.

    the logarithmization method is developed, i.e. the method for an exact representation of the logarithm of matrices, Sec. 3;

  2. 2.

    explicitly determined, by applying the logarithmization method, so called the differential matrix ([5]) of the system of differential equation, Sec. 4;

  3. 3.

    the above system is reduced to a Fuchsian differential equation of the second order and the fundamental system of its solutions is constructed in Subsec. 4.1;

  4. 4.

    the local solution to the Riemann-Hilbert problem in a vicinity of each singular point is determined in Subsec. 4.2;

  5. 5.

    the Riemann-Hilbert problem is solved globally by gluing the local solutions into a unique global solution, Sec. 5;

  6. 6.

    the partial indices of the piecewise constant matrix are found and a solution to the corresponding homogeneous Riemann boundary value problem with the piecewise constant matrix coefficient and five singular points is constructed, Sec. 6.

Solutions of these problems are realized in the case (\(m=2\), \(n=5\)). Moreover, intermediate results of Subsec. 3.1 allow us to apply the same technique in the cases \(m=2\), \(n=3\), and \(m=2,n=4\). The obtained solutions ([12–14]) coincide with those known in the literature. We believe that the general situation can be treated by using our method, it is a goal for our further study.

2 RIEMANN–HILBERT PROBLEM

2.1 Riemann–Hilbert Problem for Two Unknown Functions and Five Singular Points (\(m=2\), \(n=5\))

The Riemann–Hilbert problem in the case \(m=2,n=5\) consists in the determination of a vector function (or, equivalently, a system of two functions) \(Y(z)=(y_{1}(z),y_{2}(z))^{T}\), which is analytic in the extended complex plane \(\hat{\mathbb{C}}\) except at five (different) singular points \(a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{4}\), \(a_{5}\). This problem we consider in the following class (see, e.g. [24], cf. [6]): \(Y(z)\) is supposed to be integrable in neighbourhoods of \(a_{1},\ldots,a_{4}\) (more precisely, \(|y_{j}(z)|\leq C|z-a_{k}|^{-\alpha}\), \(0<|z-a_{k}|<r_{k}\)) and almost bounded in a neighbourhood of \(a_{5}=\infty\) (the latter means that \(Y(z)\) is either bounded or has a logarithmic singularity at \(a_{5}=\infty\)). Note that the problem in other classes may be unsolvable (cf. [24]).

Let \(\alpha_{k},\beta_{k}\) be eigenvalues of the matrices \(V_{k},k=1,\ldots,5\). We choose the branch of the logarithmic function in such a way that \(\rho_{k}=\frac{1}{2\pi i}\ln\alpha_{k},\sigma_{k}=\frac{1}{2\pi i}\ln\beta_{k}\) satisfy the conditions (which is possible since \(\Delta\) is a real number, see e.g. [4])

$$-1<{\textrm{Re}\ }\rho_{k}\leq 0,\quad-1<{\textrm{Re}\ }\sigma_{k}\leq 0,\quad\Delta=\sum\limits_{k=1}^{5}\left(\rho_{k}+\sigma_{k}\right),\quad-9\leq\Delta\leq 0.$$
(1)

Thus, the behaviour of the components of the solution at \(a_{5}=\infty\) is determined by the numbers \(\rho:=\rho_{5}+k_{1}\), \(\sigma:=\sigma_{5}+k_{2}\), where integer numbers \(k_{1},k_{2}\) are chosen in such a way that the so called Fuchs relation is satisfied,

$$\sum\limits_{k=1}^{4}\left(\rho_{k}+\sigma_{k}\right)+\rho+\sigma=1,$$
(2)

which is equivalent to the relation \(k_{1}+k_{2}=1-\Delta\). These numbers can be chosen as

$$k_{1}=\left[\frac{2-\Delta}{2}\right],\quad k_{2}=\left[\frac{1-\Delta}{2}\right]\quad\textrm{if}\quad\textrm{Re}\ \rho_{5}\leq\textrm{Re}\ \sigma_{5};$$
$$k_{1}=\left[\frac{1-\Delta}{2}\right],\quad k_{2}=\left[\frac{2-\Delta}{2}\right]\quad\textrm{if}\quad\textrm{Re}\ \sigma_{5}\leq\textrm{Re}\ \rho_{5}.$$
(3)

Since the symmetry of relation (2) we can consider without loss of generality that \(\textrm{Re}\ \rho\geq\textrm{Re}\ \sigma\).

It is known (see, e.g. [4]) that the solution to the Riemann–Hilbert problem can be represented in the locality of each singular point by the following form

$$Y(z)=\left(\begin{matrix}{y_{1}(z)}\\ {y_{2}(z)}\end{matrix}\right)=D_{k}\left(\begin{matrix}{\left(z-a_{k}\right)^{\rho_{k}}}&{u_{k}\left(z\right)}\\ {\left(z-a_{k}\right)^{\sigma_{k}}}&{v_{k}\left(z\right)}\end{matrix}\right),\quad 0<|z-a_{k}|<r_{k},\quad r_{k}>0,\quad k=1,2,3,4,$$
(4)
$$Y(z)=\left(\begin{matrix}{y_{1}(z)}\\ {y_{2}(z)}\end{matrix}\right)=D_{5}\left(\begin{matrix}{z^{-\rho}}&{u_{5}\left(z\right)}\\ {z^{-\sigma}}&{v_{5}\left(z\right)}\end{matrix}\right),\quad|z|>r_{\infty},\quad r_{\infty}>0,$$
(5)

where \(D_{k}\) \((k=1,2,3,4,5)\) transform the monodromy matrices \(V_{k}\) into a normal Jordan form, the functions \(u_{k}\left(z\right)\) are analytic in the neighborhoods of points \(a_{k}\), and the functions \(v_{k}\left(z\right)\) are either analytic if \(\rho_{k}\neq\sigma_{k}\), or have the form

$$v_{k}\left(z\right)=\frac{1}{2\pi i}\ln\left(z-a_{k}\right)u_{k}\left(z\right)+w_{k}\left(z\right)\quad\text{if}\quad\rho_{k}=\sigma_{k},\quad k=1,2,3,4,$$
(6)
$$v_{5}\left(z\right)=\frac{1}{2\pi i}\ln z\cdot u_{5}\left(z\right)+w_{5}\left(z\right)\quad\text{if}\quad\rho_{5}=\sigma_{5},$$
(7)

with \(w_{k}\left(z\right)\) being analytic in a neighborhood of the point \(a_{k}\), \(u_{k}\left(a_{k}\right)\neq 0\), \(w_{k}\left(a_{k}\right)\neq 0\), \(k=1,2,3,4,5\).

2.2 Relation to the Matrix Differential Equation

Let \(Y_{1}(z)=\left(\begin{matrix}y_{11}(z)\\ y_{21}(z)\end{matrix}\right)\), \(Y_{2}(z)=\left(\begin{matrix}y_{12}(z)\\ y_{22}(z)\end{matrix}\right)\) be two linear independent solutions to the Riemann–Hilbert problem. Then the matrix \(Y(z)=\left(\begin{matrix}y_{11}(z)&y_{12}(z)\\ y_{21}(z)&y_{22}(z)\end{matrix}\right)\) satisfies the following matrix differential equation with five singular points \(a_{1},a_{2},a_{3},a_{4},a_{5}=\infty\)

$$\frac{dY}{dz}=Y\sum_{k=1}^{4}\frac{U_{k}}{z-a_{k}},$$
(8)

where \(U_{k}\) are so called differential matrices (note that in general the differential equation obtained from the Riemann–Hilbert problem is not Fuchsian, i.e. not necessarily its differential matrix can have only first order singularities, see e.g. [6]). The constant matrices \(U_{k},k=1,2,3,4\) are similar to matrices \(W_{k}=\frac{1}{2\pi i}\ln V_{k}\) and matrix \(U_{5}=-\sum\limits_{k=1}^{4}U_{k}\) is similar to the matrix \(S=\left(\begin{matrix}-\rho&0\\ 0&1-\sigma\end{matrix}\right)\). Multiplying both sides of equation (8) by a nonsingular constant matrix \(C\) we arrive at the following differential equation

$$\frac{d\left(YC\right)}{dz}=\left(YC\right)\sum_{k=1}^{4}\frac{C^{-1}U_{k}C}{z-a_{k}}.$$
(9)

At the bypasses around the singular points \(a_{1},\ldots,a_{5}\) the matrix \(\left(YC\right)\) is transformed as follows \(\left(YC\right)\mapsto\left(V_{k}YC\right)\). Hence the columns of the matrix \(\left(YC\right)\) are linear independent solutions of the same Riemann–Hilbert problem and this matrix satisfies the differential equation (9) with differential matrices \(C^{-1}U_{k}C\).

Therefore the differential matrices of the Riemann–Hilbert problem can differ only in similarity transformation. A method of construction of the differential matrices in the case of the Riemann–Hilbert problem with two unknown functions is presented below in Sec. 4.

3 LOGARITHMIZATION METHOD

In this section, we describe in detail the logarithmization method for the product of matrices which allows us to get solution to the Riemann–Hilbert problem not assuming any commutativity of the given matrices. In order to show the essence of the method, we start with the case of the product of two and three matrices (Subsec. 3.1). Further development of the method is presented in the next subsection (Subsec. 3.2). For readers convenience we include the detailed description of the logarithmization method into Appendix at the end of the paper.

3.1 Logarithmization Method for the Product of two and Three Invertible Matrices

Let \(V_{1}\), \(V_{2}\) be a constant nonsingular square \(2\times 2\) matrices. The equality \(\ln\left(V_{1}\cdot V_{2}\right)=\ln V_{1}+\ln V_{2}\) is satisfied only if \(V_{1}\), \(V_{2}\) are permutation matrices. Now we establish a relationship between the logarithms of matrices \(V_{1},V_{2},V_{3}=V_{1}\cdot V_{2}\) for any (non-permutation) matrices.

Proposition 1. Let \(\alpha_{k}\) , \(\beta_{k}\) be the eigenvalues of the matrices \(V_{k},k=1,2,3\) and \(\rho_{k}=\frac{1}{2\pi i}\ln\alpha_{k}\) , \(\sigma_{k}=\frac{1}{2\pi i}\ln\beta_{k}\) be the eigenvalues of the matrices \(W_{k}=\frac{1}{2\pi i}\ln V_{k},k=1,2,3\) . Let the branches of the logarithmic functions be fixed in such a way that \(|\textrm{Re}\ (\rho_{k}-\sigma_{k})|<1\) , \(k=1,2\) , and the branches for \(\rho_{3}\) , \(\sigma_{3}\) be satisfied the following relation

$$\rho_{1}+\sigma_{1}+\rho_{2}+\sigma_{2}=\rho_{3}+\sigma_{3}.$$
(10)

Let \(\rho_{3}\not=\sigma_{3}\).

Then the matrix \(S_{3}=\left(\begin{matrix}\rho_{3}&0\\ 0&\sigma_{3}\end{matrix}\right)\) can be represented as a sum of two matrices

$$S_{3}=S_{1}+S_{2},$$
(11)

where

$$S_{1}=\begin{pmatrix}\frac{\rho_{1}\sigma_{1}-(\rho_{3}-\rho_{2})(\rho_{3}-\sigma_{2})}{\sigma_{3}-\rho_{3}}&\frac{\left[(\rho_{3}-\sigma_{1})(\sigma_{3}-\rho_{1})-\rho_{2}\sigma_{2}\right]a}{\sigma_{3}-\rho_{3}}\\ \frac{\rho_{2}\sigma_{2}-(\rho_{3}-\rho_{1})(\sigma_{3}-\sigma_{1})}{\left[\sigma_{3}-\rho_{3}\right]a}&\frac{(\sigma_{3}-\rho_{2})(\sigma_{3}-\sigma_{2})-\rho_{1}\sigma_{1}}{\sigma_{3}-\rho_{3}}\end{pmatrix},$$
(12)
$$S_{2}=\begin{pmatrix}\frac{\rho_{2}\sigma_{2}-(\rho_{3}-\rho_{1})(\rho_{3}-\sigma_{1})}{\sigma_{3}-\rho_{3}}&\frac{\left[\rho_{2}\sigma_{2}-(\rho_{3}-\sigma_{1})(\sigma_{3}-\rho_{1})\right]a}{\sigma_{3}-\rho_{3}}\\ \frac{(\rho_{3}-\rho_{1})(\sigma_{3}-\sigma_{1})-\rho_{2}\sigma_{2}}{\left[\sigma_{3}-\rho_{3}\right]a}&\frac{(\sigma_{3}-\rho_{1})(\sigma_{3}-\sigma_{1})-\rho_{2}\sigma_{2}}{\sigma_{3}-\rho_{3}}\end{pmatrix},$$
(13)

and \(a\) is an arbitrary constant. Here \(S_{k}\sim W_{k},k=1,2\) ( \(\sim\) stands for similarity relation which, in fact, preserves eigenvalues of the matrix).

Note that the logarithmization method can be applied to any Jordan form of the matrix. Let us show it at the construction of the differential matrix of the Riemann–Hilbert problem with three singular points. In this case it is suitable to use in of \(V_{3}=V_{1}V_{2}\). In this case the Fuchs relation becomes

$$\rho_{1}+\sigma_{1}+\rho_{2}+\sigma_{2}+\rho+\sigma=1.$$
(14)

By taken \(\rho_{3}=-\rho\) and \(\sigma_{3}=1-\sigma\) we can see that (14) coincides with (10). Thus

$$\left(\begin{matrix}-\rho&0\\ 0&1-\sigma\end{matrix}\right)=S_{1}+S_{2},\quad\textrm{or}\quad\left(\begin{matrix}0&0\\ 0&1\end{matrix}\right)=S_{1}+S_{2}+\left(\begin{matrix}-\rho&0\\ 0&-\sigma\end{matrix}\right).$$
(15)

If \(V_{3}^{0}=\left(\begin{matrix}\alpha_{3}&0\\ 0&\beta_{3}\end{matrix}\right)\) is the Jordan form of the matrix \(V_{3}\), then the matrices \(S_{3}=\left(\begin{matrix}\rho&0\\ 0&\sigma\end{matrix}\right)\) and \(S_{3}^{\prime}=\left(\begin{matrix}\rho&0\\ 0&\sigma-1\end{matrix}\right)\) are different branches of the multivalued function \(\frac{1}{2\pi i}\ln V_{3}^{0}\), and the matrix \(N=\left(\begin{matrix}0&0\\ 0&1\end{matrix}\right)\) is one of the branches of the multivalued function \(\frac{1}{2\pi i}\ln E\), moreover

$$N=S_{1}+S_{2}+S_{3}.$$
(16)

Let us show that representation (15) remains valid if \(\alpha_{3}=\beta_{3}\) and the Jordan form of the matrix \(V_{3}\) has the form \(V_{3}^{0}=\left(\begin{matrix}\alpha_{3}&0\\ 1&\beta_{3}\end{matrix}\right)\). Then by Lagrange-Sylvester formula [9] \(f(V_{3}^{0})=f(\alpha_{3})E+f^{\prime}(\alpha_{3})(V_{3}^{0}-\alpha_{3}E)\) we find \(S_{3}=\frac{1}{2\pi i}\ln V_{3}^{0}=\left(\begin{matrix}\rho&0\\ \delta&\rho\end{matrix}\right)\), where \(\delta=\frac{1}{2\pi i\alpha_{3}}\). Thus we have in (16) \(\left(\begin{matrix}-\rho&0\\ -\delta&1-\rho\end{matrix}\right)=S_{1}+S_{2}\). Matrix \(T=\left(\begin{matrix}1&0\\ \delta&1\end{matrix}\right)\) adduce the matrix in the left hand-side to the diagonal form \(\left(\begin{matrix}-\rho&0\\ 0&1-\rho\end{matrix}\right)=S_{1}^{\prime}+S_{2}^{\prime}\), where \(S_{1}^{\prime}\), \(S_{2}^{\prime}\) have the form (12), (13), respectively, with \(\rho_{3}=-\rho\), \(\sigma_{3}=1-\rho\). Therefore in what follows we can use logarithmization method for any Jordan form of \(2\times 2\) matrices.

Corollary 1. If \(\rho_{3}=\rho_{1}+\rho_{2}\) , \(\sigma_{3}=\sigma_{1}+\sigma_{2}\) , i.e the matrices \(S_{1}\) and \(S_{2}\) can be reduced to a triangular form by a unique similarity transformation, then the matrix \(S\) may be presented in the simple form

$$S_{3}=\left(\begin{matrix}\rho_{1}&c\\ 0&\sigma_{1}\end{matrix}\right)+\left(\begin{matrix}\rho_{2}&-c\\ 0&\sigma_{2}\end{matrix}\right)\quad\textrm{or}\quad S_{3}=\left(\begin{matrix}\rho_{1}&0\\ c&\sigma_{1}\end{matrix}\right)+\left(\begin{matrix}\rho_{2}&0\\ -c&\sigma_{2}\end{matrix}\right).$$

Proposition 2. Let \(\alpha_{k},\beta_{k}\) be the eigenvalues of the matrices \(V_{k},k=1,2,3,V_{4}=V_{1}V_{2}V_{3}\) and \(\rho_{k}=\frac{1}{2\pi i}\ln\alpha_{k},\sigma_{k}=\frac{1}{2\pi i}\ln\beta_{k}\) be the eigenvalues of the matrices \(W_{k}=\frac{1}{2\pi i}\ln V_{k},k=1,2,3,4\) . Let the branches of the logarithmic functions be fixed as before. Let \(\rho_{4}\not=\sigma_{4}\) . Then the Jordan form \(S_{4}=\left(\begin{matrix}{\rho_{4}}&{0}\\ {0}&{\sigma_{4}}\end{matrix}\right)\) of the logarithm of the product of three nonsingular \(2\times 2\) matrices can be represented by a sum of logarithms of three matrices which are similar to \(W_{k},k=1,2,3,\) in one of the following form:

$$\left(\begin{matrix}{s_{1}}&{c}\\ {\frac{\gamma_{1}}{c}}&{\rho_{1}+\sigma_{1}-s_{1}}\end{matrix}\right)+\left(\begin{matrix}{s_{2}}&{-\left(1+\tau\right)c}\\ {\frac{-\gamma_{2}}{\left(1+\tau\right)c}}&{\rho_{2}+\sigma_{2}-s_{2}}\end{matrix}\right)+\left(\begin{matrix}{s_{3}}&{\tau c}\\ {\frac{\gamma_{3}}{\tau c}}&{\rho_{3}+\sigma_{3}-s_{3}}\end{matrix}\right),$$
(17)
$$\left(\begin{matrix}{s_{1}}&{\frac{\gamma_{1}}{d}}\\ {d}&{\rho_{1}+\sigma_{1}-s_{1}}\end{matrix}\right)+\left(\begin{matrix}{s_{2}}&{\frac{-\gamma_{2}}{\left(1+\tau\right)d}}\\ {-\left(1+\tau\right)d}&{\rho_{2}+\sigma_{2}-s_{2}}\end{matrix}\right)+\left(\begin{matrix}{s_{3}}&{\frac{\gamma_{3}}{\tau d}}\\ {\tau d}&{\rho_{3}+\sigma_{3}-s_{3}}\end{matrix}\right),$$
(18)

\(s_{k}=\frac{1}{\sigma_{4}-\rho_{4}}\left[\rho_{4}\left(\sigma_{4}-\rho_{k}-\sigma_{k}\right)+\rho_{k}\sigma_{k}-\tau_{k}\right]\), \(\gamma_{k}=-\left(s_{k}-\rho_{k}\right)\left(s_{k}-\sigma_{k}\right)\), \(\tau_{k}\) are defined by the formulas \(\tau_{3}=\rho_{12}\sigma_{12}\), \(\tau_{1}=\rho_{23}\sigma_{23}\), \(\tau_{2}=\rho_{13}\sigma_{13}=\sum_{k=1}^{4}\rho_{k}\sigma_{k}-\tau_{1}-\tau_{3}\), \(\tau\) is a solution to algebraic equation equation \(\gamma_{1}\tau^{2}+\left(\gamma_{1}+\gamma_{3}-\gamma_{2}\right)\tau+\gamma_{3}=0\), \(c\) and \(d\) are arbitrary constants.

Remark 1. Representations (17) and (18), are, in general, equivalent (are equal up to a similarity transformation by a diagonal matrix). If the matrix \(D_{4}\) transforms \(V_{4}\) to its Jordan form, and any of the matrices \(V_{k}\) \(\left(k=1,2,3\right)\) are transformed to their triangular forms, then \(\gamma_{k}=0\) and we can choose the representation that corresponds to the form of the triangular matrix (upper or lower triangular form).

Note that as in Proposition 1 we can show that the logarithmization method can be applied to any Jordan form of the matrix \(W_{4}\).

Corollary 2. If \(\gamma_{1}=-\left(s_{1}-\rho_{1}\right)\left(s_{1}-\sigma_{1}\right)=0\), then \(c_{1}d_{1}=0\). The following simpler matrix representations are then possible for \(S_{4}\):

$$S_{4}=\left(\begin{matrix}{\rho_{1}}&{0}\\ \frac{\gamma_{2}-\gamma_{3}}{c}&{\sigma_{1}}\end{matrix}\right)+\left(\begin{matrix}{s_{2}}&{-c}\\ \frac{-\gamma_{2}}{c}&{\rho_{2}+\sigma_{2}-s_{2}}\end{matrix}\right)+\left(\begin{matrix}{s_{3}}&{c}\\ \frac{\gamma_{3}}{c}&{\rho_{3}+\sigma_{3}-s_{3}}\end{matrix}\right),$$
$$S_{4}=\left(\begin{matrix}{\rho_{1}}&\frac{\gamma_{2}-\gamma_{3}}{d}\\ {0}&{\sigma_{1}}\end{matrix}\right)+\left(\begin{matrix}{s_{2}}&\frac{-\gamma_{2}}{d}\\ {-d}&{\rho_{2}+\sigma_{2}-s_{2}}\end{matrix}\right)+\left(\begin{matrix}{s_{3}}&\frac{\gamma_{3}}{d}\\ {d}&{\rho_{3}+\sigma_{3}-s_{3}}\end{matrix}\right),$$
$$S_{4}=\left(\begin{matrix}{\rho_{1}}&{0}\\ {0}&{\sigma_{1}}\end{matrix}\right)+\left(\begin{matrix}{s_{2}}&\frac{-\gamma_{3}}{d}\\ {-d}&{\rho_{2}+\sigma_{2}-s_{2}}\end{matrix}\right)+\left(\begin{matrix}{s_{3}}&\frac{\gamma_{3}}{d}\\ {d}&{\rho_{3}+\sigma_{3}-s_{3}}\end{matrix}\right).$$

Similar representations occur in the cases \(\gamma_{2}=0\), \(\gamma_{3}=0\).

3.2 Logarithmization Method for the Product of Four Invertible Matrices

Let us present the matrix \(S_{5}=\left(\begin{matrix}-\rho&0\\ 0&1-\sigma\end{matrix}\right)\) as a sum of four matrices \(S_{5}=S_{1}+S_{2}+S_{3}+S_{4}\), where \(S_{k}\sim W_{k}=\frac{1}{2\pi i}\ln V_{k}\), \(k=1,2,3,4\). For this, we rewrite the matrix \(V_{5}:=V_{1}\cdot V_{2}\cdot V_{3}\cdot V_{4}\) as the following products of three matrices

$$V_{5}=V_{1}\cdot V_{2}\cdot(V_{3}\cdot V_{4})=V_{1}\cdot V_{2}\cdot V_{34},$$
(19)
$$V_{5}=(V_{1}\cdot V_{2})\cdot V_{3}\cdot V_{4}=V_{12}\cdot V_{3}\cdot V_{4},$$
(20)

and apply formulas (17), (18). For further analysis we need to know not only the eigenvalues \(\alpha_{k},\beta_{k}\) of the monodromy matrices \(V_{k}\), but also the eigenvalues of the products \(V_{1}V_{2}\), \(V_{2}V_{3}\), \(V_{3}V_{4}\), \(V_{1}V_{2}V_{3}\), \(V_{2}V_{3}V_{4}\) of these matrices and their logarithms.

We denote by \(\alpha_{k,k+1}\), \(\beta_{k,k+1}\) the eigenvalues of \(V_{k,k+1}=V_{k}V_{k+1}\), \(k=1,2,3,\) and by \(\alpha_{k,k+1,k+2}\), \(\beta_{k,k+1,k+2}\) the eigenvalues of \(V_{k,k+1,k+2}=V_{k}V_{k+1}V_{k+2}\), \(k=1,2.\) We determine the following parameters \(\rho_{k,k+1}=\frac{1}{2\pi i}\ln\alpha_{k,k+1}\), \(\sigma_{k,k+1}=\frac{1}{2\pi i}\ln\beta_{k,k+1}\), \(\rho_{k,k+1,k+2}=\frac{1}{2\pi i}\ln\alpha_{k,k+1,k+2}\), \(\sigma_{k,k+1,k+2}=\frac{1}{2\pi i}\ln\beta_{k,k+1,k+2},\) in such a way that branches of the logarithmic functions are fixed according to the conditions

$$\rho_{k,k+1}+\sigma_{k,k+1}=\rho_{k}+\rho_{k+1}+\sigma_{k}+\sigma_{k+1},\quad|\textrm{Re}\ (\rho_{k,k+1}-\sigma_{k,k+1})|<1,\quad k=1,2,3,$$
$$\rho_{k,k+1,k+2}+\sigma_{k,k+1,k+2}=\rho_{k}+\rho_{k+1}+\rho_{k+2}+\sigma_{k}+\sigma_{k+1}+\rho_{k+2},$$
$$|\textrm{Re}\ (\rho_{k,k+1,k+2}-\sigma_{k,k+1,k+2})|<1,\quad k=1,2.$$

Proposition 3. Let \(\alpha_{k},\beta_{k}\) be the eigenvalues of the matrices \(V_{k},k=1,2,3,4,\) and \(\rho_{k}=\frac{1}{2\pi i}\ln\alpha_{k}\), \(\sigma_{k}=\frac{1}{2\pi i}\ln\beta_{k}\) be the eigenvalues of the matrices \(W_{k}=\frac{1}{2\pi i}\ln V_{k}\), \(k=1,2,3,4\). Let the branches of the logarithmic functions are fixed as above. Let \(\rho_{5}\not=\sigma_{5}\).

Then the Jordan form \(S_{5}=\left(\begin{matrix}{\rho_{5}}&{0}\\ {0}&{\sigma_{5}}\end{matrix}\right)\) of the logarithm of the product of four nonsingular matrices of the second order can be represented by a sum of logarithms of three matrices which are similar to \(W_{k},k=1,2,3,4\):

$$S_{5}=\left(\begin{matrix}s_{1}&c\\ \frac{\gamma_{1}}{c}&\rho_{1}+\sigma_{1}-s_{1}\end{matrix}\right)+\left(\begin{matrix}s_{2}&-(1+\tau_{1})c\\ -\frac{\gamma_{2}}{(1+\tau_{1})c}&\rho_{3}+\sigma_{3}-s_{3}\end{matrix}\right)$$
$${}+\left(\begin{matrix}s_{3}&\tau_{1}(1+\tau_{2})c\\ \frac{\gamma_{3}}{\tau_{1}(1+\tau_{2})c}&\rho_{3}+\sigma_{3}-s_{3}\end{matrix}\right)+\left(\begin{matrix}s_{4}&-\tau_{1}\tau_{2}c\\ -\frac{\gamma_{4}}{\tau_{1}\tau_{2}c}&\rho_{4}+\sigma_{4}-s_{4}\end{matrix}\right),$$
(21)

where \(\tau_{1},\tau_{2}\) are solutions to \(\gamma_{1}\tau_{1}^{2}+(\gamma_{1}+\gamma_{34}-\gamma_{2})\tau_{1}+\gamma_{34}=0\) , \(\gamma_{12}\tau_{2}^{2}+(\gamma_{12}+\gamma_{4}-\gamma_{3})\tau_{2}+\gamma_{4}=0\) , respectively, and \(c\) is an arbitrary constant.

If we replace \(\tau_{1}\) by \(\frac{1}{\tau_{1}}\) and take into account that \(\gamma_{12}=\gamma_{34}\), then we arrive at another presentation of the matrix \(S_{5}\):

$$S_{5}=\left(\begin{matrix}s_{1}&\tau_{1}c\\ \frac{\gamma_{1}}{\tau_{1}c}&\rho_{1}+\sigma_{1}-s_{1}\end{matrix}\right)+\left(\begin{matrix}s_{2}&-(1+\tau_{1})c\\ -\frac{\gamma_{2}}{(1+\tau_{1})c}&\rho_{3}+\sigma_{3}-s_{3}\end{matrix}\right)$$
$${}+\left(\begin{matrix}s_{3}&(1+\tau_{2})c\\ \frac{\gamma_{3}}{(1+\tau_{2})c}&\rho_{3}+\sigma_{3}-s_{3}\end{matrix}\right)+\left(\begin{matrix}s_{4}&-\tau_{2}c\\ -\frac{\gamma_{4}}{\tau_{2}c}&\rho_{4}+\sigma_{4}-s_{4}\end{matrix}\right)$$
$${}=S_{1}^{\prime}+S_{2}^{\prime}+S_{3}^{\prime}+S_{4}^{\prime},$$
(22)

as well as at the equivalent presentation

$$S_{5}=\left(\begin{matrix}s_{1}&\frac{\gamma_{1}}{\tau_{1}d}\\ \tau_{1}d&\rho_{1}+\sigma_{1}-s_{1}\end{matrix}\right)+\left(\begin{matrix}s_{2}&-\frac{\gamma_{2}}{(1+\tau_{1})d}\\ -(1+\tau_{1})d&\rho_{3}+\sigma_{3}-s_{3}\end{matrix}\right)$$
$${}+\left(\begin{matrix}s_{3}&\frac{\gamma_{3}}{(1+\tau_{2})d}\\ (1+\tau_{2})d&\rho_{3}+\sigma_{3}-s_{3}\end{matrix}\right)+\left(\begin{matrix}s_{4}&-\frac{\gamma_{4}}{\tau_{2}d}\\ -\tau_{2}d&\rho_{4}+\sigma_{4}-s_{4}\end{matrix}\right)$$
$${}=S_{1}^{\prime\prime}+S_{2}^{\prime\prime}+S_{3}^{\prime\prime}+S_{4}^{\prime\prime},$$
(23)

where \(c\) and \(d\) are arbitrary constants,

$$s_{k}=\frac{\rho(\sigma+\rho_{k}+\sigma_{k}-1)+\omega_{k}-\omega^{\ast}_{k}}{1+\rho-\sigma},\quad k=1,2,3,4,$$
$$\omega^{\ast}_{1}=\omega_{234},\quad\omega^{\ast}_{2}=\omega_{134},\quad\omega^{\ast}_{3}=\omega_{124},\quad\omega^{\ast}_{4}=\omega_{123},$$

and \(\tau_{1},\tau_{2}\) are solutions of the following equations, respectively,

$$\gamma_{12}\tau_{1}^{2}+(\gamma_{12}+\gamma_{1}-\gamma_{2})\tau_{1}+\gamma_{1}=0,\quad\gamma_{12}\tau_{2}^{2}+(\gamma_{12}+\gamma_{4}-\gamma_{3})\tau_{2}+\gamma_{4}=0.$$
(24)

It follows from (22) that the above presentation of the matrix \(S\) is determined uniquely up to similarity transformation via a diagonal matrix. Therefore, the matrices in (22) and (23) are differential matrices of the equation (8). It can be shown that the statement remains valid for another Jordan form of the matrix \(W_{5}\).

4 FUCHSIAN DIFFERENTIAL EQUATION WITH FIVE SINGULAR POINTS

4.1 Construction of the Differential Equation

Let us now construct the second order differential equation of the Fuchsian class, related to the following matrix differential equation

$$\frac{dY}{dz}(z)=Y(z)\sum\limits_{k=1}^{4}\frac{S_{k}}{z-a_{k}},\quad\textrm{or}\quad\frac{dY}{dz}=Y(z)S(z),$$
(25)

with respect to the unknown matrix \(Y(z)=\left(y_{ij}(z)\right)\) and matrices \(S_{k}\) as presented in (21), where \(S(z)=\left(s_{ij}(z)\right)\). Thus

$$s_{11}(z)=\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}},s_{12}(z)=\sum\limits_{k=1}^{4}\frac{c_{k}}{z-a_{k}},s_{21}(z)=\sum\limits_{k=1}^{4}\frac{d_{k}}{z-a_{k}},s_{22}(z)=\sum\limits_{k=1}^{4}\frac{\rho_{k}-\sigma_{k}-s_{k}}{z-a_{k}},$$
$$\sum\limits_{k=1}^{4}c_{k}=0,\sum\limits_{k=1}^{4}d_{k}=0,s_{k}^{\prime}=\rho_{k}-\sigma_{k}-s_{k},s_{k}s_{k}^{\prime}-c_{k}d_{k}=\rho_{k}\sigma_{k}.$$

Matrix equation (25) can be written in the following scalar form

$$\begin{cases}y_{11}^{\prime}=s_{11}y_{11}+s_{21}y_{12},\\ y_{12}^{\prime}=s_{12}y_{11}+s_{22}y_{12},\\ y_{21}^{\prime}=s_{11}y_{21}+s_{21}y_{22},\\ y_{22}^{\prime}=s_{12}y_{21}+s_{22}y_{22}.\end{cases}$$
(26)

By changing the first equation of (26) to express \(y_{12}\),

$$y_{12}=\frac{1}{s_{21}}\left(y_{11}^{\prime}-s_{11}y_{11}\right)$$
(27)

and substituting it into the second equation, we arrive at the following differential equation:

$$y^{\prime\prime}-\left(s_{11}+s_{22}+\frac{s_{21}^{\prime}}{s_{21}}\right)y^{\prime}+\left(s_{11}s_{22}-s_{12}s_{21}-s_{11}^{\prime}+\frac{s_{21}^{\prime}}{s_{21}}s_{11}\right)y=0.$$
(28)

It follows from the third and the fourth equations of (26) that the function \(y_{21}\) is also a solution to equation (28).

Now we reformulate equation (28), using the forms of the matrices \(S_{k}\) in (21). Note that \(\frac{s_{21}^{\prime}}{s_{21}}=\left(\ln s_{21}\right)^{\prime}=\left(\ln\sum\limits_{k=1}^{4}\frac{d_{k}}{z-a_{k}}\right)^{\prime}\). By direct calculation we obtain

$$\frac{s_{21}^{\prime}}{s_{21}}=\frac{1}{z-b_{1}}+\frac{1}{z-b_{2}}+\sum\limits_{k=1}^{4}\frac{1}{z-a_{k}}.$$

where \(b_{1},b_{2}\) are the roots of the following quadric equation

$$\left(\sum\limits_{k=1}^{4}a_{k}d_{k}\right)z^{2}-\left(\sum\limits_{k=1}^{4}a_{k}d_{k}(a_{1}+a_{2}+a_{3}+a_{4}-a_{k})\right)z-a_{1}a_{2}a_{3}a_{4}\left(\sum\limits_{k=1}^{4}\frac{d_{k}}{a_{k}}\right)=0.$$
(29)

Since \(d_{1}=d\tau_{1}\), \(d_{2}=-d(1+\tau_{1})\), \(d_{3}=d(1+\tau_{2})\), \(d_{4}=-d\tau_{2}\) we can rewrite (29) in the form

$$\left(a_{3}-a_{2}+\tau_{1}(a_{1}-a_{2})+\tau_{2}(a_{3}-a_{4})\right)z^{2}-\left((a_{3}-a_{2})(a_{1}+a_{4})+\tau_{1}(a_{1}-a_{2})(a_{3}+a_{4}\right.$$
$${}\left.+\tau_{2}(a_{3}-a_{4})(a_{1}+a_{2})\right)z+\left((a_{3}-a_{2})a_{1}a_{4}+\tau_{1}(a_{1}-a_{2})a_{3}a_{4}+\tau_{2}(a_{3}-a_{4})a_{1}a_{2}\right)=0.$$

Further, we have

$$s_{11}+s_{22}+\frac{s_{21}^{\prime}}{s_{21}}=\sum\limits_{k=1}^{4}\frac{\rho_{k}+\sigma_{k}-1}{z-a_{k}}+\frac{1}{z-b_{1}}+\frac{1}{z-b_{2}},$$
$$s_{11}s_{22}-s_{12}s_{21}=\frac{s_{1}s_{1}^{\prime}-c_{1}d_{1}}{(z-a_{1})^{2}}+\frac{s_{1}s_{2}^{\prime}+s_{1}^{\prime}s_{2}-c_{1}d_{2}-c_{2}d_{1}}{(z-a_{1})(z-a_{2})}+\cdots$$
$${}=\frac{\rho_{1}\sigma_{1}}{(z-a_{1})^{2}}+\frac{(s_{1}+s_{2})(s_{1}^{\prime}+s_{2}^{\prime})-(c_{1}+c_{2})(d_{1}+d_{2})-s_{1}s_{1}^{\prime}-s_{2}s_{2}^{\prime}+c_{1}d_{1}+c_{2}d_{2}}{(z-a_{1})(z-a_{2})}+\cdots$$
$${}=\frac{\rho_{1}\sigma_{1}}{(z-a_{1})^{2}}+\frac{\omega_{12}-\rho_{1}\sigma_{1}-\rho_{2}\sigma_{2}}{(z-a_{1})(z-a_{2})}+\cdots=\sum\limits_{k=1}^{4}\frac{\omega_{k}}{(z-a_{k})^{2}}+\sum\limits_{k,j=1,k\not=j}^{4}\frac{\omega_{kj}-s_{kj}-\omega_{k}-\omega_{j}}{(z-a_{k})(z-a_{j})},$$
$$-s_{11}^{\prime}+\frac{s_{21}^{\prime}}{s_{21}}s_{11}=\left(\frac{1}{z-b_{1}}+\frac{1}{z-b_{2}}\right)\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}-\sum\limits_{k,j=1,k\not=j}^{4}\frac{s_{k}+s_{j}}{(z-a_{k})(z-a_{j})}.$$

It follows from Proposition 3 that

$$s_{1}+s_{2}=\frac{[\rho(\sigma+\rho_{12}+\sigma_{12}-1)+\omega_{12}-\omega_{34}]}{1+\rho-\sigma}=s_{12}.$$

Analogously

$$s_{k}+s_{j}=\frac{[\rho(\sigma+\rho_{kj}+\sigma_{kj}-1)+\omega_{kj}-\omega_{kj}^{\ast\ast}]}{1+\rho-\sigma}=s_{kj},$$
(30)

where

$$\omega_{12}^{\ast\ast}=\omega_{34},\quad\omega_{34}^{\ast\ast}=\omega_{12},\quad\omega_{14}^{\ast\ast}=\omega_{23},\quad\omega_{13}^{\ast\ast}=\omega_{24},\quad\omega_{23}^{\ast\ast}=\omega_{14},\quad\omega_{24}^{\ast\ast}=\omega_{13}.$$
(31)

Substituting the obtained formulas into (28), we arrive at the following form for the solution of the Riemann–Hilbert problem with five singular points

Theorem 1. Second order differential equation of the Fuchsian type, with five singular points and a given monodromy group, can be presented in the form

$$y^{\prime\prime}(z)+\left(\sum\limits_{k=1}^{4}\frac{1-\rho_{k}-\sigma_{k}}{z-a_{k}}-\frac{1}{z-b_{1}}-\frac{1}{z-b_{2}}\right)y^{\prime}(z)+\left(\sum\limits_{k=1}^{4}\frac{\omega_{k}}{(z-a_{k})^{2}}{}\right.$$
$${}\left.+\sum\limits_{k,j=1,k\not=j}^{4}\frac{\omega_{kj}-s_{kj}-\omega_{k}-\omega_{j}}{(z-a_{k})(z-a_{j})}+\left(\frac{1}{z-b_{1}}+\frac{1}{z-b_{2}}\right)\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}\right)y(z)=0.$$
(32)

4.2 Local Solution to the Differential Equation

In a neighbourhood of each of the singular points \(a_{1},\ldots,a_{5}\), equation (32) has two independent solutions (see e.g. [11], cf. [7])

$$u_{k}(z)=(z-a_{k})^{\rho_{k}}\sum\limits_{n=0}^{\infty}B_{n}^{(k)}(z-a_{k})^{n},$$
$$v_{k}(z)=(z-a_{k})^{\sigma_{k}}\sum\limits_{n=0}^{\infty}C_{n}^{(k)}(z-a_{k})^{n},\quad{\textrm{whenever}}\quad\rho_{k}\not=\sigma_{k},$$
$$v_{k}(z)=\frac{1}{2\pi i}\ln(z-a_{k})u_{k}(z)+(z-a_{k})^{\rho_{k}}w_{k}(z),\quad{\textrm{whenever}}\quad\rho_{k}=\sigma_{k},k=1,\ldots,4,$$
(33)
$$u_{5}(z)=z^{-\rho}\sum\limits_{n=0}^{\infty}B_{n}^{(5)}z^{-n},$$
$$v_{5}(z)=z^{-\sigma}\sum\limits_{n=0}^{\infty}C_{n}^{(5)}z^{-n},\quad{\textrm{whenever}}\quad\rho\not=\sigma$$
$$v_{5}(z)=\frac{1}{2\pi i}\ln zu_{k}(z)+z^{-\rho}w_{5}(z),\quad{\textrm{whenever}}\quad\rho=\sigma.$$
(34)

Here the functions \(w_{k}(z)\) are analytic in the vicinity of the singular points \(a_{k},k=1,\ldots,5\), and the coefficients \(B_{n}^{(k)},C_{n}^{(k)}\) are defined from recurrence relations, after substitution into equation (32). In the vicinity of points \(b_{1}\) and \(b_{2}\), equation (32) has two linear independent solutions

$$u_{b_{k}}(z)=\sum\limits_{n=0}^{\infty}B_{n}^{(b_{k})}(z-b_{k})^{n},\quad v_{b_{k}}(z)=(z-b_{k})^{2}\sum\limits_{n=0}^{\infty}C_{n}^{(b_{k})}(z-b_{k})^{n},\quad k=1,2.$$
(35)

The point \(b_{1}\) and/or \(b_{2}\) might coincide with one of the singular points \(a_{k}(k=1,\ldots,4)\). If the matrix \(S_{1}\) is triangular, then \(\gamma_{1}=0\) and \(d_{1}=0\). Then if by linear transformation the point \(a_{1}\) corresponds to \(0\), then \(z=0\) is a root of equation (29) and \(b_{1}=a_{1}\) whenever \(\gamma_{1}=0\).

Now we present the elements \(y_{12},y_{22}\) of the matrix \(Y(z)\) in terms of the elements \(y_{11},y_{21}\). Substituting expressions for \(s_{21},s_{11}\) into (32) we then obtain

$$y_{12}=\frac{1}{dr}\frac{\prod\limits_{k=1}^{4}(z-a_{k})}{(z-b_{1})(z-b_{2})}\left(y_{11}^{\prime}-\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}y_{11}\right)=\frac{1}{d}\frac{p(z)}{q(z)}\left(y_{11}^{\prime}-\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}y_{11}\right),$$

where

$$p(z)=\prod\limits_{k=1}^{4}(z-a_{k}),\quad q(z)=r(z-b_{1})(z-b_{2})=\left(a_{3}-a_{2}+\tau_{1}(a_{1}-a_{2})+\tau_{2}(a_{3}-q_{4})\right)z^{2}$$
$${}-\left((a_{3}-a_{2})(a_{1}+a_{4})+\tau_{1}(a_{1}-a_{2})(a_{3}+a_{4})+\tau_{2}(a_{3}-a_{4})(a_{1}+a_{2})\right)z$$
$${}+\left((a_{3}-a_{2})a_{1}a_{4}+\tau_{1}(a_{1}-a_{2})a_{3}a_{4}+\tau_{2}(a_{3}-a_{4})a_{1}a_{2}\right).$$

Analogously,

$$y_{22}=\frac{1}{dr}\frac{\prod\limits_{k=1}^{4}(z-a_{k})}{(z-b_{1})(z-b_{2})}\left(y_{21}^{\prime}-\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}y_{21}\right)=\frac{1}{d}\frac{p(z)}{q(z)}\left(y_{21}^{\prime}-\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}y_{21}\right).$$

Therefore, if \(u_{k},v_{k}\) is a fundamental system of solutions to differential equation (32) in a neighbourhood of the point \(a_{k},k=1,\ldots,5,\) then the solution \(Y(z)\) to the matrix equation (25) in this neighbourhood can be written in the form

$$Y(z)=D_{k}\left(\begin{matrix}u_{k}&\frac{p(z)}{q(z)}\left(u_{k}^{\prime}-\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}u_{k}\right)\\ v_{k}&\frac{p(z)}{q(z)}\left(v_{k}^{\prime}-\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}v_{k}\right)\end{matrix}\right)\left(\begin{matrix}1&0\\ 0&d\end{matrix}\right)=X(z)\left(\begin{matrix}1&0\\ 0&d\end{matrix}\right),$$
(36)

where \(D_{k}\) are the matrices transforming matrices \(V_{k},k=1,\ldots,5,\) to their normal Jordan forms. The order \(h\) of the determinant of the matrix \(X(z)\) at infinity is

$$h=\textrm{Re}\ \rho+(\textrm{Re}\ \sigma+1)-2=\textrm{Re}\ \rho+\textrm{Re}\ \sigma-1,$$

and the order of the first column of the matrix \(X(z)\) is \(h_{1}=\min\left\{\textrm{Re}\ \rho,\textrm{Re}\ \sigma\right\}=\textrm{Re}\ \rho\). Since \(s_{1}+s_{2}+s_{3}+s_{4}=-\rho\), then the order of the second column of \(X(z)\) is \(h_{2}=\min\left\{\textrm{Re}\ \rho,\textrm{Re}\ \sigma-1\right\}=\textrm{Re}\ \sigma-1\). Hence \(h=h_{1}+h_{2}\). Therefore we arrive at the following result (cf. [24]).

Theorem 2. Let \(u_{k},v_{k},k=1,\ldots,5,\) be a linear independent solutions of equation (32) in a neighbourhood of the corresponding singular point \(a_{k}\) , represented in one of the forms (33), (34). Let \(p(z)=\prod\limits_{k=1}^{4}(z-a_{k}),q(z)=r(z-b_{1})(z-b_{2})\) and \(b_{1},b_{2}\) be the roots of the quadric equation (29). Then the matrix

$$X(z)=D_{k}\left(\begin{matrix}u_{k}&\frac{p(z)}{q(z)}\left(u_{k}^{\prime}-\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}u_{k}\right)\\ v_{k}&\frac{p(z)}{q(z)}\left(v_{k}^{\prime}-\sum\limits_{k=1}^{4}\frac{s_{k}}{z-a_{k}}v_{k}\right)\end{matrix}\right)$$
(37)

is a local solution to the Riemann–Hilbert problem (i.e. solves equation (25)) in a neighbourhood of each singular point \(a_{k},k=1,\ldots,5\) . The matrix \(X(z)\) meets the following conditions:

a) \(\det X(z)\not=0,\forall z\not=a_{k},k=1,\ldots,5\);

b) the columns of \(X(z)\) belong to a chosen class of functions, specifically they are supposed to be integrable in a neighbourhood of \(a_{1},\ldots,a_{4}\) and almost bounded in a neighbourhood of \(a_{5}=\infty\);

c) the order of \(\det X(z)\) at infinity is equal to the sum of the orders of its columns.

5 GLOBAL SOLUTION TO THE RIEMANN–HILBERT PROBLEM

The obtained solution exists not only in a neighbourhood of each singular point. An analytic continuation of a local solution in a neighbourhood of a certain singular point does not necessarily become the solution at another singular point. Thus, in order to obtain a solution over the whole complex plane (i.e. the global solution), we need to relate the local representations to each other (i.e. assemble these representations into a unique analytic solution). Note that analytic continuation of the local solution corresponding \(a_{5}\) coincides with the starting local solution corresponding to \(a_{1}\) due to fulfillment of the Fuchs relation. Possible direct calculation can be provided too.

The solution (37) to equation (25) in a neighbourood of each singular point \(a_{k}\) is determined up to two constants, since the matrix \(D_{k}\) that transforms \(V_{k}\) to a normal Jordan form is not defined uniquely and has the form

$$D_{k}=\tilde{D}_{k}T_{k},\quad{\textrm{where}}\quad T_{k}=\begin{cases}\left(\begin{matrix}\delta_{k}&0\\ 0&\varepsilon_{k}\end{matrix}\right),\quad\textrm{if}\quad\alpha_{k}\not=\beta_{k},\\ \left(\begin{matrix}\delta_{k}&0\\ \varepsilon_{k}&\frac{\delta_{k}}{\alpha_{k}}\end{matrix}\right),\quad\textrm{if}\quad\alpha_{k}=\beta_{k},\end{cases}$$
(38)

where \(\tilde{D}_{k}\) are fixed matrices transforming \(V_{k}\) to a normal Jordan forms, and the constants \(\delta_{k},\varepsilon_{k}\) are subjects for further determination. The fundamental system of solutions to differential equation (32) is also determined up to two constant multipliers. Without loss of generality we fix values of the coefficients \(B_{0}^{(k)},C_{0}^{(k)},k=1,\ldots,5,\) in (33), (34), e.g. \(B_{0}^{(k)}=1,C_{0}^{(k)}=1,k=1,\ldots,5\). Since equation (32) has only two linear independent solutions, the solutions \(u_{k},v_{k}\) and \(u_{k+1},v_{k+1}\) satisfy the following linear relations:

$$\left(\begin{matrix}u_{k}(z)\\ v_{k}(z)\end{matrix}\right)=\Lambda_{k}\left(\begin{matrix}u_{k+1}(z)\\ v_{k+1}(z)\end{matrix}\right),$$
(39)

where \(\Lambda_{k}\left(\lambda_{ij}\right)\) are constant nonsingular matrices whose elements can be found via the formulas

$$\Lambda_{k}=\left(\begin{matrix}u_{k}(z_{0})&u_{k}^{\prime}(z_{0})\\ v_{k}(z_{0})&v_{k}^{\prime}(z_{0})\end{matrix}\right)\left(\begin{matrix}u_{k+1}(z_{0})&u_{k+1}^{\prime}(z_{0})\\ v_{k+1}(z_{0})&v_{k+1}^{\prime}(z_{0})\end{matrix}\right)^{-1},$$

and \(z_{0}\) is a fixed point in a common part of the convergence domains for the corresponding series. Local solutions in neighbourhoods of points \(a_{k}\) and \(a_{k+1}\) are analytic continuation of each other iff \(D_{k}\left(\begin{matrix}u_{k}(z)\\ v_{k}(z)\end{matrix}\right)=D_{k}\Lambda_{k}\left(\begin{matrix}u_{k+1}(z)\\ v_{k+1}(z)\end{matrix}\right)=D_{k+1}\left(\begin{matrix}u_{k+1}(z)\\ v_{k+1}(z)\end{matrix}\right)\) or, in notation from (38) \(\tilde{D}_{k}T_{k}=\tilde{D}_{k+1}T_{k+1}\). Denoting \(M_{k}\left(\mu_{ij}^{(k)}\right)=\tilde{D}_{k+1}^{-1}\tilde{D}_{k}\), we define \(\delta_{k},\varepsilon_{k}\) from the system \(T_{k}\Lambda_{k}=M_{k}T_{k+1},k=1,\ldots,4\). In particular, if \(\rho_{1}\not=\sigma_{1}\) and \(\rho_{2}\not=\sigma_{2}\), then such a system has the form

$$\left(\begin{matrix}\delta_{1}&0\\ 0&\varepsilon_{1}\end{matrix}\right)\left(\begin{matrix}\lambda_{11}^{(1)}&\lambda_{12}^{(1)}\\ \lambda_{21}^{(1)}&\lambda_{22}^{(1)}\end{matrix}\right)=\left(\begin{matrix}\mu_{11}^{(1)}&\mu_{12}^{(1)}\\ \mu_{21}^{(1)}&\mu_{22}^{(1)}\end{matrix}\right)\left(\begin{matrix}\delta_{2}&0\\ 0&\varepsilon_{2}\end{matrix}\right).$$
(40)

The solvability condition for system (40) has the form

$$\lambda_{11}^{(1)}\lambda_{22}^{(1)}\mu_{12}^{(1)}\mu_{21}^{(1)}=\lambda_{12}^{(1)}\lambda_{21}^{(1)}\mu_{11}^{(1)}\mu_{22}^{(1)}.$$
(41)

If matrices \(V_{1}\) and \(V_{2}\) are transformed by the same similarity transformation to the triangular form, then relation (41) holds automatically. If not then we may rewrite (41) in the form

$$\frac{\lambda_{11}^{(1)}\lambda_{22}^{(1)}}{\lambda_{12}^{(1)}\lambda_{21}^{(1)}}=\frac{\mu_{11}^{(1)}\mu_{22}^{(1)}}{\mu_{12}^{(1)}\mu_{21}^{(1)}}.$$
(42)

Since \(\rho_{1}\not=\sigma_{1}\) and \(\rho_{2}\not=\sigma_{2}\), we can then take matrices \(\tilde{D}_{k}\) in one of the forms

$$\tilde{D}_{k}=\left(\begin{matrix}\nu_{12}^{(k)}&\nu_{12}^{(k)}\\ \alpha_{k}-\nu_{11}^{(k)}&\beta-\nu_{11}^{(k)}\end{matrix}\right)\quad\textrm{or}\quad\tilde{D}_{k}=\left(\begin{matrix}\alpha_{k}-\nu_{22}^{(k)}&\beta-\nu_{22}^{(k)}\\ \nu_{21}^{(k)}&\nu_{21}^{(k)}\end{matrix}\right),\quad k=1,2;$$
$$M_{1}=\frac{1}{\nu_{12}^{(1)}(\beta_{1}-\alpha_{1})}\left(\begin{matrix}(\beta_{1}-\nu_{11}^{(1)})\nu_{12}^{(2)}-(\alpha_{2}-\nu_{11}^{(2)})\nu_{12}^{(1)}&(\beta_{1}-\nu_{11}^{(1)})\nu_{12}^{(2)}-(\alpha_{2}-\nu_{11}^{(2)})\nu_{12}^{(1)}\\ -(\alpha_{1}-\nu_{11}^{(1)})\nu_{12}^{(2)}+(\alpha_{2}-\nu_{11}^{(2)})\nu_{12}^{(1)}&-(\alpha_{1}-\nu_{11}^{(1)})\nu_{12}^{(2)}+(\beta_{2}-\nu_{11}^{(2)})\nu_{12}^{(1)}\end{matrix}\right).$$

Hence by simple algebra we get

$$\mu_{11}^{(1)}\mu_{22}^{(1)}=\nu_{12}^{(2)}(\nu_{12}^{(1)})^{-1}(\beta_{1}-\alpha_{1})^{-2}(\alpha_{12}+\beta_{12}-\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}),$$
$$\mu_{12}^{(1)}\mu_{21}^{(1)}=\nu_{12}^{(2)}(\nu_{12}^{(1)})^{-1}(\beta_{1}-\alpha_{1})^{-2}(\alpha_{12}+\beta_{12}-\alpha_{1}\alpha_{2}-\beta_{1}\beta_{2}).$$

Hence condition (42) can be rewritten as

$$\frac{\lambda_{11}^{(1)}\lambda_{22}^{(1)}}{\lambda_{12}^{(1)}\lambda_{21}^{(1)}}=\frac{\alpha_{12}+\beta_{12}-(\alpha_{1}\beta_{2}+\alpha_{2}\beta_{1})}{\alpha_{12}+\beta_{12}-(\alpha_{1}\alpha_{2}+\beta_{1}\beta_{2})}.$$
(43)

We note that neither the numerator nor the denominator of the ratio in (43) is non-vanishing. Moreover, this ratio can not equal unity due to the non-singularity of matrices \(\Lambda_{1}\) and \(M_{1}\). Direct observation shows that the recurrent relations for coefficients \(B_{n}^{(1)}\) and \(C_{n}^{(1)}\) contain in the denominators only \(\rho_{1},\sigma_{1}\) and do not contain the other pairs \(\rho_{2},\sigma_{2};\rho_{3},\sigma_{3};\rho_{4},\sigma_{4};\ldots,\rho,\sigma\). Analogously, recurrent relations for coefficients \(B_{n}^{(2)}\) and \(C_{n}^{(2)}\) contain in the denominators only \(\rho_{2},\sigma_{2}\) and do not contain the other pairs \(\rho_{1},\sigma_{1};\rho_{3},\sigma_{3};\rho_{4},\sigma_{4};\ldots,\rho,\sigma\). Therefore, the function \(f(\rho_{1},\sigma_{1};\rho_{2},\sigma_{2};\rho_{3},\sigma_{3};\rho_{4},\sigma_{4};\rho_{12},\sigma_{12};\ldots;\rho_{123},\sigma_{123};\ldots;\rho,\sigma)=\frac{\lambda_{11}^{(1)}\lambda_{22}^{(1)}}{\lambda_{12}^{(1)}\lambda_{21}^{(1)}}\) is an entire function with respect to all of its variables excluding \(\rho_{1},\sigma_{1};\rho_{2},\sigma_{2};\rho_{12},\sigma_{12}\) and does not include the values \(0,1,\infty\). Hence, the function \(f\) depends only on \(\rho_{1},\sigma_{1};\rho_{2},\sigma_{2};\rho_{12},\sigma_{12}\).

Substitute the following into (32): \(\rho_{3}=\sigma_{3}(\omega_{3}=0);\rho_{4}=\sigma_{4}(\omega_{4}=0);\) \(\rho_{34}=\sigma_{34}(\omega_{34}=0);\rho_{23}=\rho_{234}=\rho_{2};\sigma_{23}=\sigma_{234}=\sigma_{2}(\omega_{23}=\omega_{234}=\omega_{2}).\) The Fuchs relation (2) then becomes \(\rho_{12}+\sigma_{12}+\rho+\sigma=1\). Hence we have \(\rho=-\rho_{12},\sigma=1-\sigma_{12}\). It then follows \(\omega_{13}=\omega_{14}=\omega_{1}\), \(\omega_{23}=\omega_{2}\), \(\omega=\omega_{123}=\omega_{13}\). Therefore

$$s_{12}=\frac{[\rho(\sigma+\rho_{12}+\sigma_{12}-1)+\omega_{12}-\omega_{34}}{1+\rho-\sigma}=\rho_{12};\quad s_{3}=\frac{[\rho(\sigma+\rho_{3}+\sigma_{3}-1)+\omega_{3}-\omega_{124}}{1+\rho-\sigma}=0;$$
$$s_{4}=\frac{[\rho(\sigma+\rho_{4}+\sigma_{4}-1)+\omega_{4}-\omega_{123}}{1+\rho-\sigma}=0;$$
$$\gamma_{12}=-(s_{12}-\rho_{12})(s_{12}-\sigma_{12})=0,\quad\gamma_{34}=\gamma_{12}=0,\quad\gamma_{3}=\gamma_{4}=0.$$

From the first equation (24) we have \(\tau_{1}=\frac{\gamma_{2}}{\gamma_{2}-\gamma_{1}}\), moreover the second equation is satisfied identically. It then follows from representation (22) that in equation (29) we have \(d_{1}+d_{2}=0,d_{3}=d_{4}=0\), and that equation (29) can be written in the form

$$(a_{1}-a_{2})z^{2}-(a_{1}-a_{2})(a_{3}+a_{4})z+(a_{1}-a_{2})a_{3}a_{4}=0,$$

whose roots are \(z_{1}=a_{3},z_{2}=a_{4}\). Thus we have from equation (32) that \(b_{1}=a_{3}\), \(b_{2}=a_{4}\), \(s_{12}=\rho_{12}\), \(s_{13}=s_{1}+s_{3}=s_{1}\), \(s_{14}=s_{1}+s_{4}=s_{1}\), \(s_{23}=s_{2}+s_{3}=s_{2}\), \(s_{24}=s_{2}+s_{4}=s_{2}\), \(s_{34}=0\), and that equation (32) can be written in the form

$$y^{\prime\prime}+\left(\frac{1-\rho_{1}-\sigma_{1}}{z-a_{1}}+\frac{1-\rho_{2}-\sigma_{2}}{z-a_{2}}\right)y^{\prime}$$
$${}+\left(\frac{\rho_{1}\sigma_{1}}{(z-a_{1})^{2}}+\frac{\rho_{2}\sigma_{2}}{(z-a_{2})^{2}}+\frac{\rho_{12}\sigma_{12}-\rho_{1}\sigma_{1}-\rho_{2}\sigma_{2}}{(z-a_{1})(z-a_{2})}\right)y=0.$$
(44)

Equation (44) is the Riemann differential equation (see [26]). Its solution can be presented via the Gauss hypergeometric function (see [26]), \({}_{2}F_{1}(a,b;c;z)=\sum\limits_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}z^{n}\), (\((q)_{n}\) is the Pochhammer symbol, \((q)_{0}=1\), \((q)_{n}=q(q-1)\cdots(q-n+1)\)) with parameters \(a=\rho_{1}+\rho_{2}-\rho_{12},b=\rho_{1}+\rho_{2}-\sigma_{12},c=1+\rho_{1}-\sigma_{1}\).

From (40) it follows that we can take \(\delta_{1}=c\), \(\varepsilon_{1}=\frac{\mu_{21}^{(1)}\lambda_{11}^{(1)}}{\mu_{11}^{(1)}\lambda_{21}^{(1)}}c\), \(\delta_{2}=\frac{\lambda_{12}^{(1)}}{\mu_{12}^{(1)}}c\), \(\varepsilon_{2}=\frac{\lambda_{11}^{(1)}}{\mu_{11}^{(1)}}c\), where \(c\) is an arbitrary constant. Analogously, we can define connections between \(\gamma_{k}\), \(\delta_{k}\), \(\varepsilon_{k}\) in those cases when at least one of the matrices \(V_{k},k=1,2,\) is transformed to the triangular Jordan canonical form.

6 PARTIAL INDICES AND SOLUTION TO BOUNDARY VALUE PROBLEM

The Riemann–Hilbert problem can be formulated as the Riemann boundary value problem for analytic functions. To see this we draw a simple closed loop \(\Gamma\) through our singular points. Then bypassing the point \(a_{k}\) the following transformation yields \(Y^{+}\rightarrowtail V_{1}\cdot V_{2}\cdot\ldots\cdot V_{k}\cdot Y^{+}=Y^{-}\). Hence we arrive at the boundary condition

$$Y^{+}(t)=A(t)Y^{-}(t),\quad t\in{\Gamma}\setminus\{a_{1},a_{2},\ldots,a_{5}\},$$
(45)

where \(A(t)=A_{k}=\left(V_{1}V_{2}\cdots V_{k}\right)^{-1},t\in(a_{k},a_{k+1}),A_{n}=E\) (note that we are looking for the solution to (45) unbounded near each singular point as stated in condition (iii) of the Riemann problem).

We must now recall some definitions from the theory of boundary value problems corresponding to the considered situation (see [24, §126]). The matrix-function \(X(z)\), analytic outside the contour \(\Gamma\), is called the canonical matrix for boundary value problem (45) if it satisfies the following conditions:

  1. 1.

    \(X^{+}(t)=A(t)X^{-}(t)\), \(t\in{\Gamma}\setminus\{a_{1},a_{2},\ldots,a_{5}\}\);

  2. 2.

    \(\det X(z)\not=0\), \(\forall z\not=a_{1},a_{2},\ldots,a_{5}\);

  3. 3.

    the columns of the matrix \(X(z)\) belong to the chosen class of solutions, namely, are supposed to be integrable in a neighbourhood of \(a_{1},\ldots,a_{4}\) (i.e. satisfy power type asymptotics similar to that in (iii) of Introduction) and almost bounded in a neighbourhood of \(a_{5}=\infty\);

  4. 4.

    the order of the determinant of \(X(z)\) at infinity is equal to the sum of the orders at infinity of its columns.

Thus the columns of the canonical matrix are formed of the special system of the linear independent solutions to the boundary value problem (45). The order of the determinant \(\det X(z)\) of the canonical matrix at infinity is called the index (or general index) of the boundary value problem (45), and the integer parts of the orders of its columns are called the partial indices.

Theorem 3. Let \(u_{k},v_{k},k=1,\ldots,5,\) be linear independent solutions of equation (32) in a neighbourhood of the corresponding singular point \(a_{k}\) , represented by one of the forms (33), (34). Let \(p(z)=\prod\limits_{k=1}^{4}(z-a_{k}),q(z)=r(z-b_{1})(z-b_{2})\) and \(b_{1},b_{2}\) be the roots of the quadratic equation (29). Then the matrix (37) is the canonical matrix of boundary value problem (45).

The orders of the columns of (37) at infinity are equal

$$p_{1}=\min\{\textrm{Re}\ \rho,\textrm{Re}\ \sigma\}=\textrm{Re}\ \rho=\textrm{Re}\ \rho_{5}+\left[\frac{1-\Delta}{2}\right],$$
$$p_{2}=\min\{\textrm{Re}\ \rho,\textrm{Re}\ \sigma-1\}=\textrm{Re}\ \sigma-1=\textrm{Re}\ \rho_{5}+\left[\frac{-\Delta}{2}\right],$$

where \(\Delta=\sum\limits_{k=1}^{5}\rho_{k}+\sigma_{k}\).

The partial indices of boundary value problem (45) are \(\chi_{1}=[p_{1}]=\left[\frac{1-\Delta}{2}\right],\chi_{2}=[p_{2}]=\left[\frac{-\Delta}{2}\right]\), \(0\leq\chi_{1}\leq 5\), \(0\leq\chi_{2}\leq 4\), \(|\chi_{2}-\chi_{1}|\leq 1\).

The general index of boundary value problem (45) is equal \(\chi=\chi_{1}+\chi_{2}=-\Delta\), \(0\leq\chi\leq 9\).

Problem (45) has \(l=\chi+2\) linear independent solutions in the class of analytic vector-functions integrable in a neighbourhood of \(a_{1},\ldots,a_{4}\) (i.e. satisfy power type asymptotics similar to that in (iii) of Introduction) and almost bounded at a neighbourhood of \(a_{5}=\infty\) , which can be found via the formula

$$Y(z)=X(z)\left(\begin{matrix}P_{\chi_{1}}(z)\\ P_{\chi_{2}}(z)\end{matrix}\right),$$

where \(P_{\chi_{1}}(z)\), \(P_{\chi_{2}}(z)\) are polynomials of orders \(\chi_{1}\), \(\chi_{2}\) respectively.

Corollary 3. With the solution of the boundary value problem we can construct the solution of the factorization problem of the piecewise constant matrix \(A(t),t\in\Gamma\) . The partial indices of this factorization are \(\chi_{1},\chi_{2}\) . It follows from Theorem 3 that the partial indices are stable.

7 CONCLUSION

In this paper, we have proposed a method of construction of the solution of the Riemann–Hilbert problem in the case of several singular points. To avoid additional calculations we have restricted ourselves to the case of five singular points, but the method is universal and can be applied for larger numbers of points. During the course of the construction we have also determined the canonical matrix of the homogeneous boundary value problem with piecewise constant matrix coefficient. Thus the solvability conditions and solutions of the boundary value problem can be written in the standard way (see, e.g., [24]). Finally, the canonical matrix allows us to construct a solution for the factorization problem of the piecewise constant matrix (cf. [23, 8]).