GEORGIAN MATHEMATICAL JOURNAL: Vol. 4, No. 3, 1997, 279-302

ON THE RIEMANN–HILBERT PROBLEM IN THE DOMAIN WITH A NONSMOOTH BOUNDARY

V. KOKILASHVILI AND V. PAATASHVILI

Abstract. The following Riemann–Hilbert problem is solved: find an analytical function Φ from the Smirnov class E^{p}(D), whose angular boundary values satisfy the condition

Re[(a(t) + ib(t))Φ^{+}(t)] = f(t). The boundary Γ of the domain D is assumed to be a piecewise smooth curve whose nonintersecting Lyapunov arcs form, with respect to D, the inner angles with values ν_{k}π, 0 < ν_{k} ≤ 2.

Let Γ be a plane piecewise smooth closed Jordanian curve bounding the finite simply-connected domain D, 0 ∈ D; z = z(w) be the function conformally mapping the unit circle onto the domain D; w = w(z) be its inverse function. Denote by E^{p}(D) the Smirnov class, i.e., the set of analytic functions Φ in D for which

sup_{r∈(0,1)} ∫_{Γ_{r}} |Φ(z)|^{p}|dz| < ∞,

where p > 0, and Γ_{r} is the image of the circumference |w| = r for the conformal mapping of the unit circle U onto the domain D. As is well known, functions of the class E^{p}(D) have, almost everywhere on Γ, angular boundary values belonging to the class L^{p}(Γ), i.e., to the set of functions summable in power p on Γ with respect to the arc length measure.

Let the real measurable functions a, b and f be given on Γ, a and b being bounded and f being from the class L^{p}(Γ).

We shall solve the following Riemann–Hilbert problem: define a function Φ ∈ E^{p}(D), p > 1, whose boundary values satisfy, almost everywhere on Γ,

279 the boundary condition

1991 Mathematics Subject Classification. 35J25, 35J40, 35B65.

Key words and phrases. Riemann–Hilbert problem, Dirichlet problem, Smirnov class, angular boundary values, conformal mapping, piecewise smooth boundary..

1072-947X/97/0500-0279$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1997 Plenum Publishing Corporation

Re [(a(t) + ib(t))Φ^{+}(t)] = f(t), t ∈ Γ.  (1)

The Riemann–Hilbert problem and related singular integral equations under various assumptions for Γ and Φ were investigated by many authors (see, e.g., [1]–[20]); the case where Γ is a nonsmooth curve was also considered, (see, e.g., [10]–[20]). In the papers where the problem is treated in the above-given (or close to the above-given) formulation, certain restrictions are imposed on the angles under which there are arcs making up Γ. This is done with the aim of establishing the Fredholmian property of the problem or by the limited capacity of the investigation method used.

In this paper we investigate the problem for arbitrary piecewise Lyapunov curves without cusps and also for curves having cusps with the inner angle 2π. The solvability is discussed and when the problem is not solvable we give, as we think, an optimal and simple condition ensuring the existence of a solution. In all cases where solutions exist, they are constructed using Cauchy type integrals and the conformal mapping of a given domain onto the unit circle. In achieving this aim, we establish a two-weighted inequality for singular integrals with an optimal condition imposed on the pair of weights. Moreover, we give detailed consideration to the Dirichlet problem with respect to a harmonic function which is the real part of a function from E^{p}(D). By virtue of these results, problem (1) is solved by Muskhelishvili’s method, i.e., by reducing it to the linear conjugation problem [4], [5]. Though we have to solve the latter problem without making the traditional assumptions with respect to coefficients, we nevertheless succeed in creating the solvability situation and constructing solutions. In the conclusive part, the problem is considered in Smirnov weight classes.

We made essential use of the fact that when Γ is a piecewise Lyapunov curve and C is its angular point with an inner angle νπ, 0 < ν ≤ 2, then in the neighborhood of the point c = w(C) we have

z′(w) = (w − c)^{ν−1}z_{0}(w),  (2)

where z_{0}(w) is continuous and different from zero [21], ([7], Ch. I).

1^{0}. A Two-Weighted Estimate for Singular Integrals. For a 2πperiodic summable function f on (−π, π) we set

f̃(x) = ∫_{−π}^{π} \frac{f(y)}{e^{ix} − e^{iy}} dy.

It will be assumed that 1 < p < ∞ and the positive number α is so large that the function ψ(x) = x^{p−1} ln^{p} \frac{α}{x} increases on (0, π), α > eπ.

Theorem 1. Let 1 < p < ∞ and x_{0} ∈ (−π, π). Then there exists a constant M(p) > 0 such that the inequality

∫_{−π}^{π} |f̃(x)|^{p}|x − x_{0}|^{p−1}dx ≤

≤ M(p) ∫_{−π}^{π} |f(x)^{p}|x − x_{0}|^{p−1} ln^{p} \frac{α}{|x − x_{0}|} dx  (3)

holds for arbitrary f for which the integral on the right-hand side is finite.

Moreover, the power p on the right-hand side with the logarithm is sharp, that is, it cannot be replaced by any γ < p.

The proof will be based on the following Hardy type two-weight inequality.

Theorem A ([22]). Let 1 ≤ p < q < ∞ and the functions u, w defined on (0, π) be positive. Then for the equality

∫_{0}^{π} v(x)| ∫_{0}^{x} F (y)dy|^{p}dx ≤ N(p) ∫_{0}^{π} w(x)|F (x)|^{p}dx  (4)

with the constant N(p) not depending on F to hold, it is necessary and sufficient that the conditions

sup_{x>0} ( ∫_{x}^{π} v(y)dy)( ∫_{0}^{x} w^{1−p′}(y)dy)^{p−1} < ∞  (5)

(p′ = \frac{p}{p−1}) be fulfilled.

Proof of Theorem 1. It is assumed without loss of generality that x_{0} = 0. Note that if the integral on the right-hand side of (3) is finite, then, as readily follows from the Hölder inequality, the function f is summable on (−π, π) and therefore f̃(x) exists almost everywhere. Indeed,

∫_{−π}^{π} |f(x)|dx = ∫_{−π}^{π} |f(x)| |x|^{1− \frac{1}{p}} ln \frac{α}{|x|}|x|^{\frac{1}{p} −1} ln^{−1} \frac{α}{|x|} dx ≤

≤ ( ∫_{−π}^{π} |f(x)|^{p}|x|^{p−1} ln^{p} \frac{α}{|x|} dx)^{\frac{1}{p}} ( ∫_{−π}^{π} \frac{dx}{|x| ln^{p′} \frac{α}{|x|}} )^{\frac{1}{p′}} < ∞.

282 V. KOKILASHVILI AND V. PAATASHVILI

Further we have

∫_{−π}^{π} |f̃(x)|^{p}|x|^{p−1}dx = (p − 1) ∫_{−π}^{π} |f̃(x)|^{p}( ∫_{0}^{|x|}τ^{p−2}dτ)dx =

= (p − 1) ∫_{0}^{π} τ^{p−2}( ∫_{π>|x|>τ} |f̃(x)|^{p}dx)dτ ≤

≤ 2^{p−1}(p − 1)[ ∫_{0}^{π} τ^{p−2}( ∫_{π>|x|>τ} | ∫_{π>|y|> \frac{τ}{2}} \frac{f(y)}{e^{ix} − e^{iy}} dy|^{p}dx)dτ +

+ ∫_{0}^{π} τ^{p−2}( ∫_{π>|x|>τ} | ∫_{0<|y|< \frac{τ}{2}} \frac{f(y)}{e^{ix} − e^{iy}} dy|^{p}dx)dτ] =

= 2^{p−1}(p − 1)(I_{1} + I_{2}).

By Riesz’ theorem we conclude, that

I_{1} = ∫_{0}^{π} τ^{p−2}( ∫_{π>|x|>τ} (| ∫_{−π}^{π} \frac{f(y)χ\{y : π > |y| > \frac{τ}{2} \}}{e^{ix} − e^{iy}} dy|^{p}dx)dτ ≤

≤ R_{p} ∫_{0}^{π} τ^{p−2}[ ∫_{−π}^{π} (|f(y)|χ\{y : π > |y| > \frac{τ}{2} \})^{p}dy]dτ ≤

≤ R_{p} ∫_{0}^{π} τ^{p−2}( ∫_{π>|y|> \frac{τ}{2}} |f(y)|^{p}dτ).

By changing the integration order in the latter expression we obtain

I_{1} ≤ M_{1} ∫_{−π}^{π} |f(y)|^{p}( ∫_{0}^{2|y|}τ^{p−2}dτ)dy ≤
≤ M_{2} ∫_{−π}^{π} |f(y)|^{p}|y|^{p−1}dy ≤ M_{2} ∫_{−π}^{π} |f(y)|^{p}|y|^{p−1} ln^{p} \frac{α}{|y|} dy.  (6)

Let us now estimate I_{2}. For 0 < τ < π, π > |x| > τ, 0 < |y| < \frac{τ}{2} we have |x−y| ≤ |x|+|y| ≤ π + \frac{π}{2} = \frac{3π}{2} . Moreover, |x| ≤ |x−y|+|y| ≤ |x−y|+ \frac{τ}{2} ≤ |x − y| + \frac{1}{2}|x| and hence |x − y| ≥ \frac{1}{2}|x| > \frac{1}{2}τ. Also,

|e^{ix} − e^{iy}| = 2| sin \frac{x − y}{2} | ≥ \frac{2}{π} |x − y|

for \frac{1}{2}τ ≤ |x − y| ≤ π, and |e^{ix} − e^{iy}| ≥ 2 sin \frac{3π}{4} for π ≤ |x − y| ≤ \frac{3π}{2} . By virtue of all of the above inequalities we obtain

I_{2} ≤ ∫_{0}^{π} τ^{p−2}( ∫_{π>|x|>τ} (  ∫_{\{y: |y|< \frac{τ}{2} \}∩\{ \frac{τ}{2} <|x−y|<π\}} |f(y)| \frac{1}{|e^{ix} − e^{iy}|} dy)^{p}dx)dτ +

+ ∫_{0}^{π} τ^{p−2}( ∫_{π>|x|>τ} ( ∫_{\{y: |y|< \frac{τ}{2} \}∩\{π<|x−y|< \frac{3π}{2} \}} |f(y)| \frac{1}{|e^{ix} − e^{iy}|} dy)^{p}dx)dτ ≤

≤ M_{3}[ ∫_{0}^{π} τ^{p−2}( ∫_{π>|x|>τ} \frac{dx}{|x|^{p}} ( ∫_{|y|< \frac{τ}{2}} |f(y)|dy)^{p})dτ +

+ ∫_{0}^{π} τ^{p−2}( ∫_{π>|x|>τ} dx ∫_{|y|< \frac{π}{2}} |f(y)|dy)^{p}dτ].

Furthermore,

I_{2} ≤ M_{4} ∫_{0}^{π} \frac{1}{τ} ( ∫_{|y|< \frac{τ}{2}} |f(y)|dy)^{p}dτ +

+ M_{3} ∫_{0}^{π} τ^{p−2}( ∫_{π>|x|>τ} dx ∫_{|y|< \frac{π}{2}} |f(y)|dy)^{p})dτ = I_{21} + I_{22}. (7)

Let us verify whether condition (5) is fulfilled for the pair of weights v(τ) = \frac{1}{τ}, w(τ) = τ^{p−1} ln^{p} \frac{α}{τ}. We have

∫_{x}^{π} \frac{dτ}{τ} ( ∫_{0}^{x} \frac{1}{τ} ln^{p(1−p′)} \frac{α}{τ} dτ)^{p−1} = M_{5} ∫_{x}^{π} \frac{dτ}{τ} ( ∫_{0}^{x} \frac{d ln \frac{α}{τ}}{ln^{p′} \frac{α}{τ}} )^{p−1} =

= c ln \frac{π}{x} \frac{1}{ln \frac{α}{x}} ≤ M_{6}.

Therefore we can use Theorem A to estimate I_{21}. We obtain

I_{21} ≤ M_{7} ∫_{−π}^{π} |f(x)|^{p}|x|^{p−1} ln^{p} \frac{α}{|x|} dx.  (8)

Now we shall estimate I_{22} as

I_{22} ≤ M_{8} ∫_{0}^{π} τ^{p−2} ∫_{π>|x|>τ} dx( ∫_{|y|< \frac{|x|}{2}} |f(y)|dy)^{p}dτ ≤

≤ M_{9} ∫_{−π}^{π} |x|^{p−1}( ∫_{|y|< \frac{|x|}{2}} |f(y)|dy)^{p}dx ≤

≤ M_{10} ∫_{−π}^{π} |x|^{−1}( ∫_{|y|< \frac{|x|}{2}} |f(y)|dy)^{p}dx ≤

≤ M_{11} ∫_{−π}^{π} |f(x)|^{p}|x|^{p−1} ln^{p} \frac{α}{|x|} dx.  (9)

By (6), (8), and (9) we conclude that

∫_{−π}^{π} |f̃(x)|^{p}|x|^{p−1}dx ≤ M(p) ∫_{−π}^{π} |f(x)|^{p}|x|^{p−1} ln^{p} \frac{α}{|x|} dx.  (10)

It remains to show that in inequality (10) the power index p of the logarithm cannot be replaced by a smaller number. Let us assume the contrary. Let ε ∈ (0, 1). Fix the number t > 0 and set

f_{t}(y) = \begin{cases}\frac{α}{y} ln^{(p−ε)(1−p′)} \frac{1}{y} , & for 0 < y < \frac{t}{2} \\ 0, & for y\overline{∈}(0, \frac{t}{2}) \end{cases} .

By substituting the function f_{t} into inequality (10), where the power index p of the logarithm is replaced by p − ε, we obtain

∫_{−π}^{π} | ∫_{0}^{\frac{t}{2}} \frac{f_{t}(y)}{x − y} dy|^{p}|x|^{p−1}dx ≤ M ∫_{0}^{\frac{t}{2}} \frac{1}{y} ln^{(p−ε)p(1−p′)} \frac{α}{y} ln^{p−ε} \frac{α}{y} dy =

= M ∫_{0}^{\frac{t}{2}} \frac{1}{y} ln^{(p−ε)(1−p′)} ln \frac{α}{y} dy.

Hence

∫_{t}^{π} ( ∫_{0}^{\frac{t}{2}} \frac{f_{t}(y)}{x − y} dy)^{p}|x|^{p−1}dx ≤ M ∫_{0}^{\frac{t}{2}} \frac{1}{y} ln^{(p−ε)(1−p′)} \frac{α}{y} dy.  (11)

On the other hand, it is obvious that

∫_{t}^{π} ( ∫_{0}^{\frac{t}{2}} \frac{f_{t}(y)}{x − y} dy)^{p}|x|^{p−1}dx ≥ ∫_{t}^{π} \frac{1}{x}( ∫_{0}^{\frac{t}{2}} f_{t}(y)dy)^{p}dx.  (12)

By virtue of (11) and (12) we must have

∫_{t}^{π} \frac{dx}{x}( ∫_{0}^{\frac{t}{2}} f_{t}(y)dy)^{p} ≤ M ∫_{0}^{\frac{t}{2}} \frac{1}{y} ln^{(p−ε)(1−p′)} \frac{α}{y} dy,

i.e., the inequality

ln \frac{π}{t} ( ∫_{0}^{t} \frac{1}{y} ln^{(p−ε)(1−p′)} \frac{α}{y} dy)^{p−1} ≤ M  (13)

must be fulfilled for 0 < t < π. But this is impossible, since

( ∫_{0}^{t} \frac{1}{y} ln^{(p−ε)(1−p′)} \frac{α}{y} dy)^{p−1} ∼ ln^{ε−1} \frac{α}{y} .  (14)

We have therefore proved the validity of the last part of the theorem. We shall now formulate the theorem as needed for our further discussion:

Theorem 1′. Let 1 < p < ∞, γ be the unit circumference, c ∈ γ. Then the operator

T : f → T f, (T f)(ζ_{0}) = (ζ_{0} − c)^{\frac{1}{p′}} ∫_{γ} \frac{f(ζ)}{(ζ − c)^{\frac{1}{p′}} ln |ζ − c|(ζ − ζ_{0})} dζ

is continuous in L^{p}(γ).

2^{0}. Let Γ be a simple piecewise Lyapunov closed curve bounding the finite domain D. It is assumed that Γ = ∪_{k=1}^{n}Γ_{k}, where the non-intersecting Lyapunov arcs Γ_{k} meet at the points C so as to form, with respect to D, the inner angles

φ_{k} = ν_{k}π, 0 < ν_{k} ≤ 2, k = \overline{1, n}.  (15)

We shall investigate the following Dirichlet problem: In the domain D define a function u by the conditions

\begin{matrix}Δu = 0, u = Re Φ, Φ ∈ E^{p}(D) \\ u^{+}(t) = f(t), f ∈ L^{p}(Γ) \end{matrix} \} .  (16)

It will be sufficient for us to define a function Φ ∈ E^{p}(D) whose angular boundary values Φ^{∗}(t) satisfy the condition

Re[Φ^{+}(t)] = f(t).  (17)

The latter condition is the particular case of problem (1).

The function Φ(z(w)) is regular in the unit circle. In this case, Φ(z) ∈ E^{p}(D) iff the function Ψ(w) = \sqrt[p]{z′(w)}Φ(z(w)) belongs to the Hardy class H^{p} (see, e.g., [23], Ch. IX). Hence the solution of problem (17) in the class E^{p}(D) is equivalent to the solution of the boundary value problem

Re [ \frac{1}{\sqrt[p]{z′(ζ)}} Ψ^{+}(ζ)] = f(z(ζ)), |ζ| = 1  (18)

in the class H^{p}.

Following [4], [5], we introduce the function

Ω(w) = \begin{cases}Ψ(w), & |w| < 1 \\ \overline{Ψ(\frac{1}{\overline{w}} )}, & |w| > 1 \end{cases} .  (19)

It is not difficult to show that (see, e.g., [8], [19]) in a domain |w| > 1 the function Ω(w) − Ω(∞) is representable by a Cauchy integral. Therefore in a complex plane cut along the circumference γ = {ζ : |ζ| = 1} the function Ω(w) is representable by a Cauchy type integral with the constant principal part.

Let p ≥ 1 and n be an integer ≥ 0. We set

K_{p,n} = {Φ : ∃q, q(w) = a_{0} + a_{1}e + ⋯ + a_{n}w^{n}, φ ∈ L^{p}(γ),
Φ(w) = \frac{1}{2πi} ∫_{γ} \frac{φ(ζ)}{ζ − w} dζ + q(w)\}.  (20)

Denote K̃_{p} = K_{p,0} and

K_{p} = \{Φ : ∃φ ∈ L^{p}(γ), Φ(w) = \frac{1}{2πi} ∫_{γ} \frac{φ(ζ)}{ζ − w} dζ\}.  (21)

Thus, if Ψ ∈ H^{p} is a solution of problem (18), then the function Ω defined by (19) belongs to K̃_{p}. It is not difficult to find that (see, e.g., [5], §41)

Ω^{+}(ζ) = − \frac{\sqrt[p]{z′(ζ)}}{\overline{\sqrt[p]{z′(ζ)}}} Ω^{−}(ζ) + g(ζ),  (22)

where g(ζ) = 2f(z(ζ)\sqrt[p]{z′(ζ)}. We therefore conclude that if Ψ ∈ H^{p} is a solution of problem (18), then Ω ∈ K̃_{p} is a solution of the boundary value problem (22). The converse statement is not true. For the restriction of Ω(w) onto the unit circle U to give a solution of problem (18), it is necessary and sufficient that the equality

Ω_{∗}(w) = Ω(w),  (23)

where Ω_{∗}(w) = \overline{Ω(\frac{1}{\overline{w}})} (see [5], Ch. 5), be fulfilled for any |w| ≠ 1.

We have come to the problem: Define solutions Ω ∈ K̃_{p} of problem (22) satisfying the additional condition (23). If Ω is such a solution, then its restriction onto U will give the desired solution of problem (18), while the function

Φ(z) = (\sqrt[p]{w′(z)})^{−1}Ω(w(z)), z ∈ D,  (24)

will be a solution of the class E^{p}(D) of our main problem (17). Also, if Ω is a solution of the boundary value problem (22), then the function

\frac{1}{2}(Ω(w) + Ω_{∗}(w))  (25)

is a solution of problem (22) satisfying condition (23).

We introduce the notation G(ζ) = −\sqrt[p]{z′(ζ)}[\overline{\sqrt[p]{z′(ζ)}}]^{−1}. The boundary value problem

Ω^{+}(ζ) = G((ζ)Ω_{0}^{−}(ζ)  (22_{0})

will be called the homogeneous problem corresponding to the nonhomogeneous problem (22).

A function X defined on the set |w| ≠ 1 will be called a factor-function for G in the class K_{p} if it satisfies the conditions: (i) X(z) ∈ ∪_{n}K_{p,n}, [X(z)]^{−1} ∈ ∪_{n}K_{p′,n}; (ii) X^{+}(ζ)[X^{−}(ζ)]^{−1} = G(ζ); (iii) the operator

g → T g, (T g)(ζ_{0}) = \frac{X^{+}(ζ_{0})}{2πi} ∫_{γ} \frac{g(ζ)}{X^{+}(ζ)(ζ − ζ_{0})} dζ  (26)

is continuous in L^{p}(γ).

3^{0}. First, it is assumed that Γ has an angular point C with the inner angle νπ, 0 < ν ≤ 2π. We shall consider the following cases: (1) 0 < ν < p;

(2) p < ν < 2; (3) ν = p; (4) ν = 2.

4^{0}. The Case 0 < ν < p. Consider the function

X(w) = \begin{cases}−\sqrt[p]{z′(w)}, & |w| < 1 \\ \overline{\sqrt[p]{z′(\frac{1}{\overline{w}} )}}, & |w| > 1 \end{cases} .  (27)

In the case under consideration we have

−\frac{1}{p} < \frac{ν − 1}{p} < \frac{1}{p′} .  (28)

By virtue of (2) it is easy to establish that (see, e.g., [8]) X is a factorfunction for G in K_{p} and all solutions of problem (22) of the class K̃_{p} are given by the formula

Ω(w) = \frac{X(w)}{2πi} ∫_{γ} \frac{g(ζ)}{X^{+}(ζ)(ζ − w)} dζ + αX(w),  (29)

where α is an arbitrary constant.

Let us turn to condition (23). A general solution of problem (22_{0}) is given by the equality Ω_{0}(w) = αX(w). But

(αX)_{∗}(w) = \begin{cases}\overline{α}\sqrt[p]{z′(w)}, & |w| < 1 \\ −\overline{α}\overline{\sqrt[p]{z′(\frac{1}{\overline{w}} )}}, & |w| > 1 \end{cases} ,  (30)

while the equality (αX)_{∗} = αX implies \overline{α} = −α. Thus we have Re α = 0. Hence it follows that all solutions of problem (22_{0}) have the form α\sqrt[p]{z′(w)}, Re α = 0 and by equality (24) we conclude that for f = 0 problem (16) has only a trivial solution. Moreover, assuming additionally that α = 0 and using (25), from (29) we obtain the solution of (22) satisfying condition (23)

Ω(w) = \frac{1}{2}[\frac{X(w)}{2πi} ∫_{γ} \frac{g(ζ)}{X^{+}(ζ)(ζ − w)} dζ +

+ (\overline{\frac{X(\frac{1}{\overline{w}})}{2πi}})\overline{∫_{γ} \frac{g(ζ)}{X^{+}(ζ)(ζ − \frac{1}{w})} dζ}].  (31)

Since X(w(z)) = − \frac{1}{\sqrt[p]{w′(z)}}, X(1/\overline{w′(z)}) = 1/\overline{\sqrt[p]{w′(z)}}, from (24) and (31)

we have the equality

u(z) = Re [\frac{1}{2πi} ∫_{γ} \frac{f(z(ζ))}{ζ − w(z)} dζ + \frac{w(z)}{2πi} ∫_{γ} \frac{\overline{f(z(ζ))}}{ζ} \frac{dζ}{ζ − w(z)}]. (32)

Eventually, we find that for 0 < ν < p problem (16) is uniquely solvable for any f ∈ L^{p}(Γ) and its solution is given by (32).

5^{0}. The Case p < ν < 2. This time we shall investigate the function

X̃(w) = \begin{cases}− \frac{\sqrt[p]{z′(w)}}{w − c} , & |w| < 1 \\ \frac{\overline{\sqrt[p]{z′(\frac{1}{\overline{w}})}}}{w − c} , & |w| > 1 \end{cases} .  (˜27)

By virtue of (2) we have X̃ = O((w − c)^{\frac{ν−p−1}{p}}) in the neighborhood of the point c. Since −\frac{1}{p} < \frac{ν−p−1}{p} < \frac{1}{p′} (this is equivalent to the inequality p < ν < 2p which we have in the case under consideration), only the function

Ω(w) = \frac{X̃(w)}{2πi} ∫_{γ} \frac{g(ζ)}{ζ − w} dζ + (αw + β)X̃(β)

can be a solution of problem (22), and the function Ω_{0}(w) = (αw + β)X̃(w) a solution of problem (22_{0}). Let us see for which α and β condition (23) is fulfilled. We must have

−(\overline{α}\frac{1}{w} + \overline{β}) \frac{\sqrt[p]{z′(w)}}{\frac{1}{w} − \overline{c}} = \frac{(αw + β)\sqrt[p]{z′(w)}}{w − c} .

Since \overline{c} = c^{−1}, we obtain

−(\frac{\overline{α}}{w} + \overline{β})\frac{wc}{c − w} = \frac{αw + β}{w − c} .

Hence \overline{β}c − α = 0, \overline{α}c − β = 0. This gives us β = \overline{α}c. For arbitrarily chosen β we obtain α = \overline{β}c. Thus the function

u_{0}(z(w)) = Re [ \frac{1}{\sqrt[p]{z′(w)}} \frac{w\overline{β}c + β}{w − c} \sqrt[p]{z′(w)}] = Re [\frac{w\overline{β}c + β}{w − c} ]

will be a solution of the homogeneous Dirichlet problem in the circle. But if w = re^{iθ}, c = c_{1} + ic_{2} = e^{iθ_{c}}, β = λ + iμ, then

\frac{\overline{β}cw + β}{w − c} = \frac{(\overline{β}cw + β)(\overline{w} − \overline{c})}{|w − c|^{2}} = \frac{(\overline{β}cr^{2} − β\overline{c}) − (\overline{β}w − β\overline{w})}{|w − c|^{2}}

and, keeping in mind that Re[\overline{β}w − β\overline{w}] = 0 and assuming \overline{β}c = d + ie, we obtain

Re \frac{\overline{β}cw + β}{w − c} = Re \frac{\overline{β}cr^{2} − \overline{\overline{β}c}}{|w − c|^{2}} = Re \frac{(d + ie)r^{2} − (d − ie)}{|w − c|^{2}} = \frac{d(r^{2} − 1)}{|w − c|}

with d = Re \overline{β}c = Re[(λ − iμ)(c_{1} + ic_{2})] = λc_{1} + μc_{2}.

Thus

u_{0}(z(re^{iθ})) = \frac{(λc_{1} + μc_{2})(1 − r^{2})}{1 + r^{2} − 2r cos(θ − θ_{c})} ,

where λ, μ are arbitrary real constants. Clearly, λc_{1} + μc_{2} runs through the set of all real numbers so that

u_{0}(z(re^{iθ})) = M Re \frac{c + w}{c − w} ,

where M is an arbitrary real constant. For f = 0 a general solution of problem (16) is given by the equality

u_{0}(z) = M Re \frac{c + w(z)}{c − w(z)} .  (33)

A particular solution of the nonhomogeneous problem is obtained after replacing in it X(w) by X̃(w). Finally, we obtain a particular solution of the form

ũ(z) = Re [\frac{1}{w(z) − c}(\frac{1}{2πi} ∫_{γ} \frac{f(z(ζ))(ζ − c)}{ζ − w} dζ −

− \frac{cw^{2}}{2πi} ∫_{γ} \frac{\overline{f(z(ζ))}(\overline{ζ} − \overline{c})}{ζ(ζ − w)} dζ)].  (34)

Thus for p < ν < 2 the Dirichlet problem (16) is solvable for any f ∈ L^{p}(Γ) and has an infinite number of solutions given by the equality u(z) = u_{0}(z) + ũ(z), where u_{0} and ũ are defined by (33) and (34), respectively.

6^{0}. The Case ν = p. Let X be the function defined by (27). Then by (2) we have X(w) = O((w − c)^{\frac{1}{p′}}) (including the case ν = 2 = p). We shall first consider the homogeneous problem. The function F (w) = Ω(w)[X(w)]^{−1} satisfies the condition F^{+} = F^{−} and, in the domains |w| < 1 and |w| > 1, belongs to the set ∩_{δ<1}H^{δ}, where H^{δ} is the Hardy class (it is assumed that F ∈ H^{δ} in the domain |w| > 1 if F (\frac{1}{w}) ∈ H^{δ} in the domain |w| < 1). We shall show that F (w) is regular in the whole plane except the point c. Let ζ be an arbitrary point on γ, different from c. Let us take a pair of points ζ_{1} and ζ_{2} from the different sides of ζ and consider the domain S_{ζ}^{+} which is a sector of the circle U bounded by the radii passing through ζ_{1} and ζ_{2} and by the circumference arc γ(ζ_{1}, ζ_{2}) containing ζ. Since for the function Ω ∈ H^{p} we have by the Feyer–Riesz theorem (see, e.g., [24], p. 46)

∫_{0}^{1} |Ω(re^{iθ})|^{p}dr ≤ M ∫_{0}^{2π}|Ω(e^{iψ})|^{p}dψ,

it follows that Ω ∈ E^{p}(S_{ζ}^{+}). Hence we easily find that Ω ∈ E^{p}(S_{ζ}^{−}), where S_{ζ}^{−} is the domain bounded by the arc γ(ζ_{1}, ζ_{2}), by the continuation of the radii passing through ζ_{1} and ζ_{2}, and by the circumference arc |w| = 1 + η, η > 0. For ζ ≠ c points ζ_{1} and ζ_{2} can be chosen so that c would not lie on γ_{1}(ζ_{1}, ζ_{2}). Therefore the function [X(w)]^{−1} is bounded in the domains S_{ζ}^{±} and hence in these domains the function F (w) belongs to the Smirnov class and is represented by the Cauchy integral

\begin{aligned}F (w) & = \begin{cases}\frac{1}{2πi} ∫_{γ_{1}} \frac{F (ζ)}{ζ − w} dζ, & w ∈ S_{ζ}^{+} \\ 0, & w ∈ S_{ζ}^{−} \end{cases} , \\ F (w) & = \begin{cases}\frac{1}{2πi} ∫_{γ_{2}} \frac{F (ζ)}{ζ − w} dζ, & w ∈ S_{ζ}^{−} \\ 0, & w ∈ S_{ζ}^{+} \end{cases} , \end{aligned}  (35)

where γ_{1} and γ_{2} are the boundaries of the domains S_{ζ}^{+} and S_{ζ}^{−}, respectively. Within these boundaries there lies the arc γ(ζ_{1}, ζ_{2}) on which the integration on γ_{1} and γ_{2} is performed in the opposite directions and on which F^{+}(ζ) = F^{−}(ζ). Hence the function

F̃(w) = \frac{1}{2πi} ∫_{γ_{1}∪γ_{2}} \frac{F (ζ)}{ζ − w} dζ = \frac{1}{2πi} ∫_{γ_{3}} \frac{F (ζ)}{ζ − w} dζ,
γ_{3} = (γ_{1} ∪ γ_{2})\γ(ζ_{1}, ζ_{2}),

on the one hand, coincides in S_{ζ}^{+} ∪ S_{ζ}^{−} with the function F (w) (by virtue of (35)) and, on the other hand, is analytic inside γ_{3} and therefore in the neighborhood of the point ζ. Thus F is regular almost everywhere except the point c at which it may have only a first order pole, since otherwise it could not belong to the class H^{δ}, δ ∈ (0, 1). Therefore F (w) = α + \frac{β}{w−c} and only the function Ω_{0}(w) = αX(w) + β[w − c]^{−1}X(w) can be a solution of problem (22_{0}). But since in the neighborhood of c we have Ω_{0}(w) = O((w−c)^{− \frac{1}{p}}), Ω_{0} will belong to K̃_{p} iff β = 0, i.e., for ν = p the homogeneous problem has only a solution of the form Ω_{0}(w) = αX(w). For this solution to satisfy condition (23) we shall show, like we did in Subsection 4^{0}, that α = 0.

Returning to the nonhomogeneous problem, note that there are functions in L^{p}(γ) for which it is not solvable (see Subsection 8^{0} below),

We shall show now how using Theorem 1′ one can construct a solution of problem (22)–(23) for a wide subclass of functions f ∈ L^{p}(Γ).

Assuming that

g(ζ) ln(ζ − c) ∈ L^{p}(γ),  (36)

we obtain g(ζ) = \frac{g_{1}(ζ)}{ln(ζ−c)}, where g_{1} ∈ L^{p}(γ) and therefore by Theorem 1′ the function

\frac{(ζ_{0} − c)^{\frac{1}{p′}}}{2πi} ∫_{γ} \frac{1}{(ζ − c)^{\frac{1}{p′}} ln(ζ − c)} \frac{g_{1}(ζ)}{ζ − ζ_{0}} dζ = (ζ_{0} − c)^{\frac{1}{p′}} F (ζ)

belongs to L^{p}(γ). Hence due to the inclusion F ∈ H^{δ} and the relation X(w) = O((w − c)^{\frac{1}{p′}}) it follows that the function

Ω(w) = \frac{X(w)}{2πi} ∫_{γ} \frac{g(ζ)}{X^{+}(ζ)(ζ − w)} dζ

belongs to K̃_{p} (we have used here Smirnov’s theorem: if Φ ∈ H^{p_{1}} has boundary values Φ^{+} ∈ L^{p_{2}}, p_{2} > p_{1}, then Φ ∈ H^{p_{2}}). Condition (36) is equivalent to the condition f(t) ln |ζ(t)−ζ(c)| ∈ L^{p}(Γ), since arg(ζ(t)−ζ(c)) is the bounded function. But in the case under consideration

ζ(t) − ζ(c) = (t − C)^{\frac{1}{p′}} ζ_{0}(t), 0 < m < |ζ_{0}(t)| < M

(see [21] or [7]). Therefore ln |ζ(t) − ζ(c)| < const ln |t − C| and (36) is equivalent to the condition

f(t) ln |t − C| ∈ L^{p}(Γ).  (37)

If this condition is fulfilled, then the nonhomogeneous problem (16) is solvable and its solution is given by equality (32).

7^{0}. The Case ν = 2. Equality (2) implies z′(w) = (w − c)z_{0}(w), 0 < m < |z_{0}(w)| < M. For p < 2 we shall establish, like we did in Subsection 5^{0}, that Ω_{0}(w) = (αw + β)X̃(w), where X̃ is given by equality (˜27). Using corresponding arguments we again find that the homogeneous Dirichlet problem has an infinite number of solutions given by (33), while the function ũ given by (34) is a particular solution of the nonhomogeneous problem.

For p > 2 we have 2 = ν < p (see Subsection 4^{0}). If p = 2, then ν = p = 2 (see Subsection 6^{0}).

To summarize, we have: for ν = 2 the homogeneous Dirichlet problem has only a trivial solution if p ≥ 2. If however p < 2, then it has an infinite set of solutions given by equality (33). The nonhomogeneous problem is solvable for any f ∈ L^{p}(Γ) if p ≠ 2. When p = 2, for this problem to be solvable it is sufficient that condition (37) be fulfilled. In that case, a particular solution is constructed by (32) if p ≥ 2, and by (34) if p < 2.

8^{0}. An Example of the Function f_{0} ∈ L^{p}(Γ) for Which the Dirichlet Problem (16) Has No Solution if Γ Has an Angular Point with an Angle pπ.

To construct such a function we need to show first that a solution of the Dirichlet problem, if it exists, has the definite form. We rewrite condition (22) as follows:

(ζ − c)Ω^{+}(ζ) = (ζ − c)G(ζ)Ω^{−}(ζ) + g(ζ)(ζ − c).

It can be easily verified that if X is defined by equality (27), then the function F (w) = (w − c)[X(w)]^{−1} belongs to the class K_{p,1}. Therefore possible solutions of the class K̃_{p} of problem (22) lie in the set of functions

Ω̃(w) = \frac{X(w)}{w − c} \frac{1}{2πi} ∫_{γ} \frac{g(ζ)(ζ − c)}{X^{+}(ζ)(ζ − w)} dζ + \frac{αw + β}{w − c} X(w),  (38)

where α and β are arbitrary constants. Since \frac{αw+β}{w−c} X(w) = α + \frac{β+αc}{w−c} X(w) and the function αX(w) belongs to K̃_{p}, we conclude that Ω̃(w) will be a solution of the class K̃_{p} only if the function

Ω(w) = \frac{X(w)}{w − c} \frac{1}{2πi} ∫_{γ} \frac{g(ζ)(ζ − c)}{X^{+}(ζ)(ζ − w)} dζ + \frac{DX(w)}{w − c} ,  (39)

where D is some constant, belongs to this class. But Ω ∈ H^{δ}, δ < 1, and therefore Ω will belong to the class K̃_{p} if the function Ω^{+}(ζ) belongs to L^{p}(Γ). For this it is necessary and sufficient that the function

g̃(ζ_{0}) = \frac{X^{+}(ζ_{0})}{ζ_{0} − c} ∫_{γ} \frac{g(ζ)(ζ − c)}{X^{+}(ζ)(ζ − ζ_{0})} dζ + \frac{DX^{+}(ζ_{0})}{ζ_{0} − c}  (40)

belong to the class L^{p}(γ).

Let us now find a function g_{0} ∈ L^{p}(γ) for which g̃_{0} ∉ L^{p}(γ) no matter what the constant D is. We set c = 1 and assume that

g_{0}(ζ) = g_{0}(e^{iθ}) = \begin{cases}\frac{m_{n}X^{+}(ζ)}{ζ − 1} , & θ_{n} ≤ θ ≤ θ_{n+1} \\ 0, & θ ∈ (1, 2π) \end{cases} ,  (41)

where θ_{n} = \frac{1}{n}, m_{n} = \frac{1}{ln(n+1)}.

Since in the case under consideration |\frac{X^{+}}{ζ−1}| ≤ _{\frac{M}{|ζ−1|^{\frac{1}{p}}}} , we have

∫_{γ} |g_{0}|^{p}|dζ| ≤ ∑_{n=1}^{∞} m_{n}^{p} ∫_{θ_{n}}^{θ_{n+1}} \frac{dθ}{|ζ − 1|} ≤

≤ ∑_{n=1}^{∞} \frac{m_{n}^{p}}{\sqrt{(1 − cos \frac{1}{n})^{2} + sin^{2} \frac{1}{n}}} (\frac{1}{n} − \frac{1}{n + 1}) ≤

≤ const ∑_{n=1}^{∞} \frac{1}{n ln^{p}(n + 1)} < ∞.

Thus g_{0} ∈ L^{p}(γ). We shall show that g̃_{0} ∉ L^{p}(γ) no matter what the constant D is. Since

g̃_{0}(ζ) = \frac{X(ζ_{0})}{(ζ_{0} − c)}[ ∫_{γ} \frac{g_{0}(ζ)(ζ − c)}{X^{+}(ζ)(ζ − ζ_{0})} dζ + D]

and, by (2) \frac{X(ζ)}{ζ_{0}−c} = (ζ_{0} − c)^{− \frac{1}{p}} z_{0}(ζ_{0}), 0 < m < |z_{0}| < M, it is sufficient to prove the equality

lim_{ζ_{0}→1} ∫_{γ} \frac{g_{0}(ζ)(ζ − c)}{X^{+}(ζ)(ζ − ζ_{0})} dζ = ∞.  (42)

Given a number K > 0, we choose N such that ∑_{n=1}^{N} \frac{1}{n ln(n+1)} > K and assume that ζ_{0} = e^{iθ}, θ ∈ (2π − \frac{1}{N} , 2π). We have

| ∫_{γ} \frac{g_{0}(ζ)(ζ − c)}{X^{+}(ζ)(ζ − ζ_{0})} dζ| ≥ | ∑_{n=1}^{∞} m_{n} ln |\frac{ζ_{n} − ζ_{0}}{ζ_{n+1} − ζ_{0}} ||.  (43)

But

ln |\frac{ζ_{n} − ζ_{0}}{ζ_{n+1} − ζ_{0}} | = ln |\frac{sin \frac{1}{2}(\frac{1}{n} − θ)}{sin \frac{1}{2}(\frac{1}{n+1} − θ)}|

so that assuming that θ = 2π − α we obtain

ln |\frac{ζ_{n} − ζ_{0}}{ζ_{n+1} − ζ_{0}} | = ln |\frac{sin(\frac{α}{2} + \frac{1}{2n})}{sin(\frac{α}{2} + \frac{1}{2(n+1)})}| > 0, α ∈ (0, \frac{1}{N} ).

Now (43) implies

|g̃_{0}(ζ)| ≥ ∑_{n=1}^{∞} m_{n} ln |\frac{sin(\frac{α}{2} + \frac{1}{2n})}{sin(\frac{α}{2} + \frac{1}{2(n+1)})}| =

= ∑_{n=1}^{∞} m_{n} ln (1 + \frac{2 cos(\frac{α}{2} + \frac{1}{4n(n+1)}) sin \frac{1}{4n(n+1)}}{sin(\frac{α}{2} + \frac{1}{2(n+1)})} ) ≥

≥ ∑_{n=1}^{N} m_{n} \frac{2 cos(\frac{α}{2} + \frac{1}{4n(n+1)}) sin \frac{1}{4n(n+1)}}{sin(\frac{α}{2} + \frac{1}{2(n+1)})} ≥ m_{0}K,

where m_{0} = \frac{2}{π} cos \frac{2N+3}{4N(N+1)}. Thus relation (42) is proved.

Clearly, the function f_{0}(t) = g_{0}(w(t))[\sqrt[p]{w′(t)}]^{−1} belongs to L^{p}(Γ) and the corresponding Dirichlet problem is not solvable for it.

9^{0}. Formulation of the Result on Problem (16) in the Case of One Angular Point. Based on the arguments of Subsections 4^{0}–8^{0}, we come to a conclusion that the following theorem is valid.

Theorem 2′. Let D be a finite singly connected domain with a piecewise Lyapunov boundary, one angular point C with an inner angle νπ, 0 < ν ≤ 2. Then for the Dirichlet problem the following statements are true:

If 0 < ν < p, then the problem is uniquely solvable and its solution is given by equality (32). If p < ν ≤ 2, then it has an infinite number of solutions of the form

u(z) = ũ(z) + M Re \frac{c + w(z)}{c − w(z)} , c = w(C),

where w(z) is the function mapping conformally the domain D onto the unit circle, M is an arbitrary constant, and ũ is defined by equality (34). If ν = p, then the fulfillment of the condition f ∈ L^{p}(Γ) is not enough for the problem to be solvable. However, if the condition

f(t) ln |t − C| ∈ L^{p}(Γ)  (37)

is satisfied, it has a (unique) solution given by equality (34).

10^{0}. The Dirichlet Problem in a General Case. We shall formulate the result for a more general case which follows from Theorem 2′.

Theorem 2. Let D be a domain bounded by a simple piecewise Lyapunov closed curve with angular points C_{k}, k = \overline{1, n}, at which the inner angle values are equal to ν_{k}π, 0 < ν_{k} ≤ 2. Note that there are n_{1} angular points with values ν_{k} from the interval (p, 2] (it is assumed that (2, 2] = {2}). In that case, all solutions of the homogeneous problem (16) are given by the equality

u_{0}(z) = ∑_{\begin{matrix}k \\ ν_{k}∈(p,2] \end{matrix}} M_{k} Re \frac{c_{k} + w(z)}{c_{k} − w(z)} , c_{k} = w(C_{k}),  (44)

where M_{k} are arbitrary real constants. Generally speaking, the fulfillment of the condition f ∈ L^{p}(Γ) is not enough for a homogeneous problem to be solvable. However, if the condition

f(t) ln ∏_{\begin{matrix}k \\ ν_{k}=p \end{matrix}}(t − C_{k}) ∈ L^{p}(Γ)  (45)

is fulfilled, then the problem is solvable. In all cases in which a solution exists, it is given by the equality u(z) = u_{0}(z) + ũ(z), where u_{0}(z) is defined by equality (44), and

ũ(z) = Re [(\frac{1}{2πi} ∫_{γ} \frac{f(z(ζ))ρ(ζ)}{ζ − w(z)} dζ +

+(−1)^{n_{1}} \frac{w^{n_{1}+1}(z) ∏_{\begin{matrix}k \\ ν_{k}∈(p,2] \end{matrix}} c_{k}}{2πi} ∫_{γ} \frac{\overline{f(z(ζ))}\overline{ρ(ζ)}}{ζ(ζ − w(z))} dζ)\frac{1}{ρ(w(z))}],  (46)

where ρ(w) = ∏_{\begin{matrix}k \\ ν_{k}∈(p,2] \end{matrix}}(w − c_{k}) and ρ(w) ≡ 1, when \{ν_{k} : ν_{k} ∈ (p, 2]\} = ∅.

11^{0}. The Riemann–Hilbert Problem in a Domain whose Boundary Has One Angular Point. We make the same assumptions for Γ as in Subsection 3^{0}. It is assumed that the coefficients a(t) and b(t) are measurable and the function G(t) = (a(t) − ib(t))(a(t) + ib(t))^{−1} belongs to the class Ã_{p}(Γ), i.e., (1) inf |G| > 0, sup |G| < ∞; (2) for each point t ∈ Γ, except perhaps the points t_{k}, k = \overline{1, m}, there exists a neighborhood in which the values G lie in some sector with the vertex at the origin, whose angle is less than 2π[max(p, p′)]^{−1}; (3) in unilateral neighborhoods of points t_{k}, G satisfies the Hölder condition, at points t_{k} there exist limits G(t_{k}, ±), and values of the angles δ_{k} between the vectors G(t_{k}−) and G(t_{k}+) are such that \frac{2π}{p} < δ_{k} ≤ \frac{2π}{p′} for p > 2, \frac{2π}{p′} ≤ δ_{k} < \frac{2π}{p} for 1 < p < 2 and δ_{k} ≠ π for p = 2.

For such a function we define, depending on p, the function Θ_{p}(t) = arg_{p} G(t) and the integer number κ = κ(p; G) = \frac{1}{2π} [Θ_{p}]_{Γ} [13] ([φ]_{Γ} denotes an increment of the function when t passes over the curve Γ).

Passing to the circle, we come to a problem of defining the function Ω ∈ K̃_{p}(γ) by the set of conditions

\begin{matrix}Ω^{+}(ζ) = − ^{\begin{matrix}^{p} & \sqrt{z′(ζ)} \\ ^{p} & \sqrt{z′(ζ)} \end{matrix}} \frac{A(ζ)−iB(ζ)}{A(ζ)+iB(ζ)} Ω^{−}(ζ) + g(ζ), ζ ∈ Γ, \\ Ω_{∗}(w) = Ω(w), |w| ≠ 1, \end{matrix} \}  (47)

where A(ζ) = a(z(ζ)), B(ζ) = b(z(ζ)), g(ζ) = 2f(z(ζ))\sqrt[p]{z′(ζ)}[a(ζ) + iB(ζ)]^{−1}.

The function G_{γ}(ζ) = G(z(ζ)) belongs to Ã_{p}(γ). We need only to check whether the Hölder property takes place in the unilateral neighborhoods of points τ_{k} = w(t_{k}). To this end, we recall that in our case the function z(ζ) satisfies the Hölder condition with the index min(1, ν) [21]. By [13] G_{γ}(ζ) is factorized in K_{p}(γ) and its factor-function has the form

Y (w) = \begin{cases}exp (\frac{1}{2πi} ∫_{γ} \frac{Θ_{p}(ζ)}{ζ − w} dζ), & |w| < 1 \\ w^{−κ} exp (\frac{1}{2πi} ∫_{γ} \frac{Θ_{p}(ζ)}{ζ − w} dζ), & |w| > 1 \end{cases} .  (48)

Assuming that Γ has one angular point C ≠ t_{k}, k = \overline{1, m}, with an angle νπ, 0 < ν ≤ 2, we shall consider the cases: (i) 0 < ν < p; (ii) p < ν < 2; (iii) ν = p; (iv) ν = 2.

(i) If 0 < ν < p and the function X is given by (27), then the function

T (w) = AY (w)X(w),  (49)

where A is an arbitrary constant, will be the factor-function for G̃_{γ}(ζ) = −\sqrt[p]{z′(ζ)}[\overline{\sqrt[p]{z′(ζ)}}]^{−1}G_{γ}(ζ). By an appropriate choice of A [5] we can fulfill the equality T_{∗}(w) = w^{κ}T (w) and hence make the following conclusion: if κ ≥ 0, then the homogeneous problem corresponding to problem (1) has an infinite number of solutions given by the equality

Φ(z) = T (w(z))P_{κ}(w(z))[\sqrt[p]{w′(ζ)}]^{−1},

where P_{κ}(w) = a_{0} + a_{1}w + ⋯ + a_{κ}w^{κ} is an arbitrary polynomial whose coefficients satisfy the conditions c_{i} = \overline{c}_{κ−i}, i = \overline{1, κ}. The nonhomogeneous problem is solved unconditionally. If, however, κ < 0, then the homogeneous problem has only a trivial solution, while for the nonhomogeneous problem to be solvable it is necessary and sufficient that the conditions

∫_{Γ} w^{k}(t)\frac{f(t)w′(t)}{T^{+}(w(t))} dt = 0, k = 0, 1, …, |κ| − 1,  (50)

be fulfilled.

(ii) p < ν < 2. We set T̃(w) = AY (w)X̃(w), where Y and X̃ are given by equalities (48) and (˜27), respectively. By an appropriate choice of A we can fulfill the equality

T̃_{∗}(w) = w^{κ+1}T̃(w), |w| ≠ 1.

The result of (i) therefore remains in force if we replace κ and T by κ + 1 and T̃, respectively. We write the solvability conditions as

∫_{Γ} w^{k}(t)\frac{f(t)w′(t)}{T̃^{+}(w(t))} dt = 0, k = 0, 1, …, |κ| − 2.  (51)

(iii) ν = p. Applying a reasoning similar to that of Subsection 6^{0} we find that only the function

Ω_{0}(w) = Y (w)[P_{κ}(w) + \frac{D}{w − c}]X(w)

can be a solution of the homogeneous problem. But for ν = p we have X(w) = (w − c)^{\frac{1}{p′}} X_{0}(w), while the assumptions made for G_{γ} imply that |Y (w)| ≥ m_{0} > 0 in the neighborhood of the point c. Thus for the function Ω_{0} to belong to the class K̃_{p}(γ) we must set D = 0. So we obtain the same result for the homogeneous problem as in the case (i). The nonhomogeneous problem is not always solvable. Let the condition (37) be fulfilled for f. Then the function

Ω(w) = \frac{T (w)}{2πi} ∫_{γ} \frac{g(ζ)}{T^{+}(ζ)} \frac{dζ}{ζ − w}  (52)

will be a solution of problem (47) from the class K̃_{p}(γ) iff conditions (50) are fulfilled.

(iv) ν = 2. We have a different situation for p ≠ 2 and p = 2. When p < 2, we can show, as above, that the function Ω(w)[Y (w)X(w)]^{−1} is analytically continuable onto the entire plane except the point c and thus come to the conclusion that the general solution of the homogeneous problem has the form

Ω_{0}(w) = Y (w)X(w)(w − c)^{−1}P_{κ+1}(w)  (53)

(P_{κ+1}(w) ≡ 0 for κ + 1 < 0). The nonhomogeneous problem is solved unconditionally for κ ≥ −1. If, however, κ ≤ −2, then for this problem to be uniquely solvable it is necessary and sufficient that conditions (51) be fulfilled. In all cases one can easily write all solutions.

If p > 2, then 2 = ν < p and this case is considered in (i); if p = 2, the ν = 2 = p and this case is treated in (iii).

Now we are able to summarize the results for the general case.

12^{0}. The Riemann–Hilbert Problem in the General Case. We introduce the notation

ρ(w) = ∏_{\begin{matrix}k \\ ν_{k}∈(p,2] \end{matrix}}(w − c_{k})  (54)

and assume that (2, 2] = {2}. If there are no points C_{k} for which ν_{k} ∈ (p, 2], then it is assumed that ρ(w) = 1.

The results of Subsection 11^{0} give rise to

Theorem 3. Let the Riemann–Hilbert problem (1) from the class E^{p}(D) be considered in a finite simply connected domain D bounded by the curve Γ. It is assumed that :

(i) Γ is a simple piecewise Lyapunov curve with angular points C_{k}, k = \overline{1, n}, with the inner angles ν_{k}π, 0 < ν_{k} ≤ 2; n_{1} is the number of points at which ν_{k} ∈ (p, 2];

(ii) G(t) = (a(t) − ib(t))(a(t) + ib(t))^{−1} belongs to Ã_{p}, C_{k} ≠ t_{k}, and κ(p, G) = \frac{1}{2π} [arg_{p} G(t)]_{Γ} (see Subsection 11^{0});

(iii)

κ = κ(p, G) + n_{1},

T (w) = \begin{cases}−\frac{Y (w)\sqrt[p]{z′(w)}}{ρ(w)} , & |w| < 1 \\ \frac{\overline{Y (\frac{1}{\overline{w}})}\overline{\sqrt[p]{z′(\frac{1}{\overline{w}})}}}{ρ(w)} , & |w| > 1 \end{cases} .  (55)

Then:

(1) All solutions of the homogeneous problem are given by the equality

Φ_{0}(z) = AT (w(z))P_{κ}(w(z))[\sqrt[p]{w′(z)}]^{−1},  (56)

where, for κ ≥ 0, P_{κ}(w) = a_{0} +a_{1}w+⋯+a_{κ}w^{κ} is an arbitrary polynomial whose coefficients are related by the conditions c_{i} = \overline{c}_{κ−i}, i = \overline{1, n}, P_{κ}(w) ≡ 0 for κ < 0, and the constant A is uniquely defined by the equality

(AT)_{∗}(w) = w^{κ}AT (w).

(2) For the nonhomogeneous problem we have:

If κ ≥ 0 and

f(t) ln | ∏_{\begin{matrix}k \\ ν_{k}=p \end{matrix}}(t − C_{k})| ∈ L^{p}(Γ),  (57)

then the problem is solvable.

If κ < 0 and condition (57) is fulfilled, then for problem (1) to be solvable it is necessary and sufficient that

∫_{Γ} w^{k}(t)\frac{f(t)w′(t)}{T^{+}(w(t))} dt = 0, k = 0, 1, …, |κ| − 1.  (58)

In all these cases the solution is representable by the formula

Φ(z) = Φ̃(z) + Φ_{0}(z),

where Φ_{0} is given by (56), and

Φ̃(z) = \frac{T (w(z))}{2πi} ∫_{γ} \frac{f(z(ζ))}{T (z(ζ))} \frac{\sqrt[p]{z′(ζ)}}{ζ − w(ζ)} dζ +

+ (\overline{\frac{T (1/\overline{w(ζ)})}{2πi}})\overline{∫_{γ} \frac{f(z(ζ))\sqrt[p]{z′(ζ)}(\overline{w(z)} − \overline{ζ})}{(\overline{w(z)} − \overline{ζ})T (z(ζ))\overline{ζ} \overline{w(ζ)}} dζ}.  (59)

13^{0}. The Riemann–Hilbert Problem in the Smirnov Weighted Class. Let β be a real number, τ ∈ Γ and r(z) = (z − τ)^{β}, z ∈ D. An analytical function Φ(z) in the domain D will be said to belong to the class E^{p}(D; r) if Φ(z)r(z) ∈ E^{p}(D). When one considers problem (1) in this class, it is of special interest to investigate the case where τ coincides with the angular point C (it is assumed that there is only one such point). On rewriting the condition Re[(A(ζ) + iB(ζ))Φ^{+}(z(ζ))] = f(z(ζ)) as

Re [\frac{(A(ζ) + iB(ζ))\sqrt[p]{z′(ζ)r(z(ζ))}Φ^{+}(z(ζ))}{\sqrt[p]{z′(ζ)r(z(ζ))}}] = f(z(ζ)),  (60)

we reduce problem (1) to a problem of form (22) in which \sqrt[p]{z′(ζ)} is replaced by \sqrt[p]{z′(ζ)r(z(ζ))}. But r(z(ζ)) = (r(ζ) − z(c))^{β} and, since z(ζ) satisfies the Hölder condition with the index ν̃ = min(1, ν), we have \sqrt[p]{z′(ζ)r(z(ζ))} =

(ζ − c)^{\frac{ν−1}{p} +\frac{_{ν}β}{p}} r_{0}(ζ), r_{0} ≠ 0, in the neighborhood of c. Now, by setting ν_{β} = ν +ν̃β and assuming that ν_{β} ∈ (0, 2] we obtain analogues of Theorems 2 and 3 in which ν is replaced by ν_{β}. We shall not go into details and write only admissible values for β

max(−1, −ν) < β ≤ min (2 − ν, \frac{2 − ν}{ν} ).  (61)

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(Received 17.08.1995)

Authors’ address:

A. Razmadze Mathematical Institute Georgian Academy of Sciences 1, M. Aleksidze St., Tbilisi 380093 Georgia