Generalization of I . Vekua ’ s integral representations of holomorphic functions and their application to the Riemann – Hilbert – Poincaré problem

I. Vekua’s integral representations of holomorphic functions, whose m -th derivative (m ≥ 0) is Hölder-continuous in a closed domain bounded by the Lyapunov curve, are generalized for analytic functions whose m -th derivative is representable by a Cauchy type integral whose density is from variable exponent Lebesgue space Lp(·)(Γ;ω) with power weight. An integration curve is taken from a wide class of piecewise-smooth curves admitting cusp points for certain p and ω . This makes it possible to obtain analogues of I. Vekua’s results to the Riemann–Hilbert–Poincaré problem under new general assumptions about the desired and the given elements of the problem. It is established that the solvability essentially depends on the geometry of a boundary, a weight function ω(t) and a function p(t) .


Introduction
In many boundary value problems of function theory the boundary conditions contain not only the sought function, but also its derivatives up to certain order.Therefore it is useful to have formulas giving an integral representation of this holomorphic function.One form of such representations, quite convenient for applications, was proposed by I. Vekua ( [17], [18]).N. Muskhelishvili expounded them in his book, where they are called I. Vekua's integral representations (see [15, pp. 224-232]).This is I. Vekua's result.Theorem (I.Vekua).Let D + be a finite domain bounded by a simple closed Lyapunov curve Γ and Φ(z) be a holomorphic function in D + , whose derivative of order m is continuous in D + and the boundary belong to the Hölder's class H .Then, assuming that the origin is in D + , the function Φ is representable for m = 0 as and for m ≥ 1 as where ϕ(t) is a real function from the class H , and d is a real constant; ϕ(t) and d are defined uniquely with respect to Φ(z).
Subsequently, B. Khvedelidze [5] gave a generalization of this theorem to the case where a derivative of order m of the function Φ(z) is representable in D + by a Cauchy type integral with a density from the Lebesgue space L p (Γ; ω), where p > 1 and In that case, ϕ belongs to L p (Γ; ω).In [1] K. Aptsiauri generalized this result to the case ω(t) = 1 for a wide subclass of piecewise-smooth curves.I. Vekua used these representations for investigating quite a general boundary value problem, namely, the Riemann-Hilbert-Poincaré problem.An analogous investigation was carried out in [1] and [5].
In recent years, the attention of researchers has been attracted by Lebesgue spaces with a variable exponent p(t) and the boundary value problems where the boundary values of the sought functions belong to these spaces (see, for instance, [3]- [4], [6]- [14].Thus it is desirable to have representations (1)-( 2) of holomorphic functions Φ with boundary values having this property and for as wide as possible classes of curves Γ, functions p(t) and weights ω(t).
In this paper, we consider the question whether integrals of form ( 1)-( 2) can be used to present holomorphic functions whose m-th derivative is representable by a Cauchy type integral with a density from L p(•) (Γ; ω), where Γ is a piecewise-smooth curve.Relations are found between the values of the function p(t) at the angular points of the curve Γ, the values of these angles and the power exponents of weight functions, which provide the required representation.
Of the function p(t) it is only required that it satisfies the Log-Hölder's condition which we frequently encounter in the literature.For the weight function ω(t) it is assumed that α k ∈ − 1 p(t k ) , 1 p (t k ) , p (t) = p(t) p(t)−1 .The latter condition is necessary and sufficient for a singular operator with the Cauchy kernel to be continuous in the space L p(•) (Γ; ω) (see [13], [10]) without which we can hardly do in our consideration.As a result, the range of domains in which representations (1)-( 2) are valid has become much wider.This has been achieved thanks to the well studied Riemann-Hilbert problem in the considered classes.For this we had to extend the results obtained in [7], [8] for bounded domains to the case of unbounded domains.Also note that in [1], [5], [17]- [18] only the existence of ϕ and d such that (1), (2) holds is proved.Our approach enables us to indicate a more concrete construction of the function ϕ by means of the prescribed function Φ(z).
The representations obtained are used in investigating the Riemann-Hilbert-Poincaré problem under more general assumptions than in [1] and [5].
The statements we make in connection with this problem are seemingly analogous to the results in [1], [5], [17], [18], but, in fact, the relations between the values essentially depend on the geometry of the boundary Γ, multipliers |t−t k | α k from a weight function ω(t), discontinuity points of the principal coefficient from the boundary condition, as well as on the values of a function p(t) at the angular points of Γ and at points t k .
The results of the present paper were announced in [9].

Main definitions, the notation and auxiliary statements
1 0 .Let t = t(s), 0 ≤ s ≤ l , be an equation of a simple rectifiable curve Γ with respect to an arc abscissa.Let, further, p(t) be a positive measurable function on Γ, and ω(t) be a measurable, a.e.finite and nonzero function.We denote by L p(•) (Γ; ω) the set of measurable functions f on Γ for which then L p(•) (Γ; ω) is a Banach space with norm defined by (3).
2 0 .Definition 1.We say that a measurable function p on Γ belongs to the class P(Γ) if p − > 1 and, moreover, there exist positive constants A and ε such that for arbitrary t 1 and t 2 on Γ.The set of functions p for which p − > 1 and equality ( 4) is fulfilled for ε = 0 is denoted by P(Γ).It is obvious that P(Γ) ⊂ P(Γ).
at which the values of angles with respect to the domain D whose boundary is Γ are equal to πν k .If, additionally, Γ is a piecewise-Lyapunov curve, then we say that Γ is a curve of the class C Let D be a simply connected domain bounded by a simple rectifiable curve Γ, and the functions p and ω be such that L p(•) (Γ; ω) ⊂ L(Γ).We denote We denote by W p(•) (Γ) the set of weight functions ω for which the operator If Γ is a Carleson curve and p ∈ P(Γ), then the function ( 5) [13], [10].For the case of constant p we refer to [2].

Definition 3.
If m ≥ 0 is an integer number, then we denote by 5 0 .The problem of representation of holomorphic functions in terms of new assumptions reduces to the investigation of the Riemann-Hilbert problem in the class K p(•) D − (Γ; ω).In [7] and [8], this problem is solved in the class K p(•) D (Γ; ω) when D is a bounded domain with the boundary Γ.The case of an unbounded domain can be easily investigated by reducing it to the considered one.
In this subsection we recall the results from [7] and [8], and in Subsection 6 0 extend them to the case of an unbounded domain.
Let us consider the Riemann-Hilbert problem -Find a function Φ ∈ K p(•) D + (Γ; ω) for which the equality (6) Re (a(t is valid a.e. on Γ.We denote by B k , k = 1, λ, the discontinuity points of the function We assume Let us numerate the points of the set T ∪ A ∪ B as follows: (7) We set (τ ) = p(z(τ )) and write ( 8) For a real number x we set x = [x] + {x} , where [x] is an integer number and 0 ≤ {x} < 1.
Then, by assumptions I (or II) and ( 9) we have Proposition 1 ([7], [8]).If κ < −1 , then for problem (6) to be solvable in the class K be fulfilled, and their fulfillment implies that the problem has a unique solution where and is certainly solvable and all its solutions are given by the equality where P −1 (w) ≡ 0 , while for κ ≥ 0 is an arbitrary polynomial whose coefficients h k satisfy the relation Remark 1.If we search for a solution Φ(z) of problem ( 6) with the condition Φ(0) = 0 , then, as readily follows by the reasoning of [7], the above proposition remains valid when κ in it is replaced by κ − 1.Thus for κ < 0 the solvability conditions have the form and their fulfillment implies that there exists a unique solution Φ(z) = Ω(w(z)).
When κ ≥ 0 , the problem is solvable and all solutions are given by the equality Remark 2. The investigation of problem ( 6) reduces to that of the Riemann problem in the class [7] (and in [8]) is a canonical function for G(τ ).The order κ of the function Y (w) at infinity is the index of the Riemann problem (12) (or, which is the same, the index of the singular operator [11])).Problem ( 6) is solvable if and only if problem (12), too, is solvable.Note that when κ > 0 , the homogeneous problem (12) has a general solution depending on κ arbitrary complex constants, while the homogeneous problem (6) has a general solution depending on κ arbitrary real constants (distinguished by relation ( 11)).
6 0 .Let Γ be a curve of the class C 1 D − (A 1 , . . ., A i ; ν 1 , . . ., ν i ), p ∈ P(Γ) and the Riemann-Hilbert problem be considered in the class K p(•) D − (Γ; ω).This time the boundary condition is written in the form (13) Re (a(t We assume that the point z = 0 does not lie in D − .By transforming ζ = 1 z the domain D − becomes a finite domain which we denote by D + 1 .The function Φ(z) transforms to the function ).The weight function ω(t) transforms to the function and therefore conditions ( 5) are fulfilled for ω 1 (η).
The boundary condition ( 13) transforms to the condition ( 14) It can be easily verified that ).Thus we should solve problem (14) in the class K (Γ 1 ; ω 1 ) using the additional condition 1 with the condition ζ(0) = 0 , and w = ψ(ζ) be its inverse mapping.Besides, we denote .
For the domain D + 1 we define the numbers ω 1 k , δ 1 k and γ 1 k by analogy with ( 7), ( 8) and (10).The numbers δ 1 k are obtained from formulas (8) in which ν k is replaced by λ k (i.e., by 2 − ν k ), while when defining the numbers γ 1 k , (τ ) is understood as p( 1ζ(τ ) ). Define the function X 1 (w) and Q 1 (w) in accordance with the points w 1 k .Assume κ − = κ 1 0 + κ 1 1 .Now, applying the result from Remark 1 of Subsection 5 0 , we obtain then, assuming that κ − < 0 , for the problem to be solvable it is necessary and sufficient that the conditions is fulfilled, and when they are fulfilled, the problem has a unique solution where Ω * (w) = Ω 1 w , |w| = 1 and When κ − ≥ 0 , the problem is unconditionally solvable and all solutions are given by the equality where for the polynomial If we reject the condition Φ(∞) = 0 , then, assuming that κ − < −1, for problem (13) to be solvable in the class is necessary and sufficient that conditions (15) are fulfilled, where k = 0, |κ| − 2, and when they are fulfilled, there exists a unique solution given by equality (16).

I. Vekua's Integral Representation of Holomorphic
Functions from the Class K In this subsection we will prove (ii) ω be a power function of form (5); (iii) the point z = 0 lie in D + ; z = z(w) be a conformal mapping of the circle U onto the domain D − = CD + ; z(0) = ∞ and w = w(z) be its inverse mapping.Let a k = w(A k ), k = 1, i, τ k = w(t k ), k = 1, n, and the points of the set {a 1 , . . ., a i } ∪ {τ 1 , . . ., τ n } be numbered so that (18) then there exist a real function ϕ ∈ L p(•) (Γ; ω) and a real constant d such that The function ϕ and the constant d are defined in a unique manner.Before proceeding to the proof, we would like to note the particular cases of the theorem.Representations (20) and ( 21) are valid in the following cases: I. a) Γ is a smooth curve and p ∈ P(Γ) or Γ is a Lyapunov curve and Proof of Theorem 1.Since (20), (21) imply that d = Im Φ(0), we can assume that d = 0 .Let us prove the validity of equalities (20) and (21).Having these representations for d = 0 , from the equality Φ(z) = [Φ(z) − i Im Φ(0)] + i Im Φ(0) we will obtain the representations in the general case with d = Im Φ(0).
Let m = 0 and representation (20) hold.If t = t(s) is the equation of the curve Γ with respect to the arc abscissa, then ds = (t ) −1 dt = t dt (see [15, p. 225]).Therefore (20) can be rewritten in the form Combined with (22), this implies that Since Γ is a piecewise-smooth curve, and Φ + and ϕ belong to We have come to the Riemann-Hilbert problem with regard to the function N which we must solve in the class .
We will first show that representations (20) and (21) are unique.Both cases are treated in the same manner.Let, for instance, For z = 0 we have  Let us apply the results from Subsections 5 0 and 6 0 of the preceding section to problem (27).
In that case, Γ 1 is a curve of the class C 1

This function possesses discontinuity at the points
Furthermore, all discontinuity points of G(t) coincide with the angular points of Γ 1 .By assumption, on Γ there exist μ angular points coinciding with some points t k from the weight ω ; accordingly, on Γ 1 there exist μ angular points coinciding with some points τ k = 1 t k from the weight ω 1 .Furthermore, there are p = i − μ angular points not appearing among the singular points of the weight ω .Finally, in the weight ω (in the weight ω 1 ) there are M multipliers of the form (t )) be the inverse mapping.We have These points have been numbered according to (18).Then equalities (8) take form (19). Using these numbers, we define numbers γ k by means of equalities (10).Then with the condition F (0) = 0 ) take the form Hence we obtain where 1 (w)r(w) and Q 1 (w) is a rational function with zeros and poles on γ , r(w) is a power function, while X 1  1 ∈ β>0 E β (D + 1 ).Hence it follows that ] −1 and thereby Ψ m (ζ), too, belongs to E δ (D + 1 ), δ > 0.Moreover, since Γ is a piecewise-smooth curve without zero angles (since 0 < 2 − ν k ≤ 2 ), the construction of the functions X 1 and Q 1 implies that the function belongs to L(Γ 1 ) (see [7,Theorem 2] and [8]).Thus, according to Smirnov's theorem, Ψ ∈ E 1 (D + 1 ) and therefore I k,1 = 0. Let us write I k,2 in the form and therefore is holomorphic in the unit circle U and therefore [15, p. 143]).It is easy to verify that the functions By virtue of this fact we have Thereby I k = 0 and thus conditions (28) sufficient for problem (27) to be solvable are fulfilled.The solution is given by equality (16).

The Riemann-Hilbert-Poincaré Problem in the Class
where a k , H k , f are the given functions of Hölder's class, Γ is a Lyapunov curve bounding the finite domain D + , and the sought function Φ has a continuous derivative of order m in D + and with boundary values from H ( [17], [18]).In [5] this problem is considered when a k (t) are continuous and Φ (m) (z) is representable by a Cauchy type integral with a density from L p (Γ; ω), where p > 1 and ω is a power function.For ω = 1 , the problem is investigated for a wide subset of piecewise-smooth curves ( [1]).
Here we assume that for p, Γ and ω the conditions of Theorem 1 are fulfilled.We want to solve problem (30) in the class K the functions Φ (0) (z) = Φ(z), Φ (z), . . ., Φ (m−1) (z) are continuous in D + and absolutely continuous on Γ with respect to the arc abscissa (see, for instance, [16, p. 208]).Thus it is natural to assume that in condition (30) the coefficients a k (t), k = 0, m − 1, belong to L p(•) (Γ; ω).As to the coefficient a m (t), we should assume that it is bounded.However this is not enough.Following [17], [18], [15], we reduce the problem to a singular integral equation in the class L p(•) (Γ; ω), which is investigated in various conditions depending on the assumptions made for p, Γ and ω ([4], [11]).It is assumed for simplicity that a m (t) is piecewise-continuous on Γ and inf |a m (t)| > 0.
So, let Γ be a curve of the class , ω be the power function (5), the coefficients a 0 , a 1 , . . ., a m−1 belong to L p(•) (Γ; ω), p ∈ P(Γ), a m ∈ C( B 1 , . . ., B λ ) (i.e. a m is piecewise-continuous on Γ with discontinuity points B k ), and the operators It is required to find a function Φ ∈ K p(•) D + ,m (Γ; ω) for which equality (30) holds a.e. on Γ.Note that the compactness of the operators H k is provided, for instance, by the fulfillment of the conditions where A, λ ∈ [0, 1), are constants and s(t 0 , t) is the length of the smallest of two arcs connecting the points t 0 and t on Γ (see [12]).
Following [15, p. 233] this problem will sometimes be called Problem V.
2 0 .Since under the above assumptions the conditions of Theorem 1 are fulfilled, the sought solution Φ is representable by equality (20) for m = 0 and by equality (21) for m ≥ 1.
Assuming first that m ≥ 1 , we calculate the derivatives of the function Φ given by equality (21) and substitute them into (30).Thus we obtain (see [15, pp. 234-235]) that the function ϕ satisfies the condition where It is evident that where Moreover, (32) implies that the function A(t 0 ) can be written in the form (35) By virtue of our assumptions about Γ, the coefficients a 0 (t 0 ), . . ., a m−1 (t 0 ) and operators H k and using the above-mentioned result from [12] it is not difficult to establish that the operator T is compact in L p(•) (Γ; ω).
4 0 .Let us calculate the index of the operator N or, which is the same, of the operator N 0 (given by equality (33)).This is equivalent to the calculation of the index of the problem in the class K p(•) (Γ; ω) (see Remark 2 from Subsection 5 0 of Section 2).From ( 34) and ( 35) we obtain Denote by B k , k = 1, λ, the discontinuity points of this function and let b k = w(B k ).Among the points B k there are all points A k , k = 1, i.
Put s)   and write equality (36) in the form The function For problem (36) the index κ (see Remark 2 from 5 0 of Section 2), which depends on Γ, p, ω and the numbers u k , is defined in the class K p(•) (Γ; ω).Let for a pair of operators N and N the Noether theorems hold (see, for instance, [4], [11]).We can assume that the conditions of Theorem 11 from [11] are fulfilled.From this theorem it follows particulary the following statement.
Let ω = exp i 2 S Γ ϕ , where ϕ is a real piecewise-constant function, and B k,ϕ , k = 1, λ, are all discontinuity points of this function and Theorem 2. Let the conditions of Subsection 1 0 be fulfilled.Then for Problem V to be solvable in the class K p(•) D + ,m (Γ; ω) it is necessary and sufficient that for some real d the function where g 1 , . . ., g n are linearly independent solutions from the class L p (•) (Γ; ω −1 ) of the equation N g = 0, where N is the adjoint operator to the operator N .
In order that Problem V has a solution, for any right-hand part of f it is necessary and sufficient that n = 0 or n = 1 and in the latter case the solution g of the equation N g = 0 must satisfy the condition In both cases the homogeneous problem has κ + 1 linearly independent solutions (where κ ≥ −1 ).
If these conditions are misobserved, then: if (g k , σ) = 0 for any k = 1, n , then the homogeneous problem has κ + n linearly independent solutions, and if among the numbers (g k , σ) there is at least one nonzero number, then it has κ + n + 1 solutions.
If σ(t) = 0, then problem (30) is solvable for any right-hand part of f (t 0 ) if and only if n = 0; in that case the homogeneous problem has κ + 1 linearly independent solutions.
We omit the proof of this theorem because it is analogous to that of the respective I. Vekua's theorem ( [18], see also [15, pp. 238-240]).
of problem (38).In order to apply Proposition 1 given above to the case in which p(t) is replaced by p (t) and ω(t) by [ω(t)] −1 , we have to calculate the index κ = κ 0 + κ 1 of problem (38).Since for the considered Riemann-Hilbert problem a(t) + ib(t) = 1, we have κ 1 = 0.As to κ 0 , it is equal to the order at the point z = ∞ of the rational function Here the numbers γ k are defined from relations (10) in which the function (t) is replaced by the function (t) and the numbers α k by the numbers (−α k ).The number κ 0 is calculated by the equality where N (E) stands for the number of elements of the set E [7, Subsect.7].Problem (38) has κ 0 + 1 linearly independent solutions.We denote them by Ψ 0 (z), Ψ 1 (z), . . ., Ψ κ0 (z).

Let
Then from the definitions of the functions Ω * , Ψ 0 , . . ., Ψ κ0 it follows that for an arbitrary solution g of the equation N g = 0 there holds the equality By analogy with [15, p. 242] we call the function Ω(t 0 , z) a kernel.Now, following I. Vekua ( [18], see also [15, pp. 242-243]), condition (41) can be replaced by an equivalent condition Γ g(t)ω j (t) ds = 0, where ω j (t) (j = 0, 1, . . . ) can be understood as any of the following systems of functions: where z 0 , z 1 , . . . is an arbitrary sequence having a limit point in , where z 0 is an arbitrary fixed point in D − ; If k is a maximal number of linearly independent real functions g j (t) orthogonal to the kernel Ω(t, z) or to all elements of the sequence {ω j (t)} , then k is called the defect of the kernel or of the sequence {ω j (t)} .In the considered case, k coincides with n which is the number of linearly independent solutions in L p (•) (Γ; ω −1 ) of the equation N g = 0 .Hence the obtained results can be formulated as follows: For Problem V to have a solution for any right-hand part of f (t) it is necessary and sufficient that the defect of the sequence {ω j (t)} (or of the kernel Ω(t, z)) be equal to 0 or 1.Note that in the latter case the function g(t), which is orthogonal to all elements of the sequence {ω j (t)} , must not be orthogonal to the function σ(t).

Some Particular Cases
1 0 .Poincaré Problem.We will consider this problem formulated as follows.Find, in the domain D + , a harmonic function u from the set where [f ] Γ denotes an increment of the function f (t) when the point t performs one-time movement along the curve Γ.In that case the index of the operator N (see (31)) is calculated by the equality κ = κ 0 + κ 1 , where κ 1 = 2n and (recall that A k are the angular points of Γ and α k are power exponents from weight (5)).The kernel Ω(t 0 , z) is given by equalities (39)-(40), where Ω * (t j , z) = a(t 0 ) + ib(t 0 ) t 0 − z + γ(t 0 ) ln e 1 − t 0 z .
In that case for the Neumann problem we have κ(p ; 1) + 1 solvability conditions.Assuming that (9) holds and the solvability conditions are fulfilled, a general solution of problem (45) contains κ(p; 1) real constants, where κ(p; 1) = N {A k : p(A k ) < ν k } .