GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 3, 1999, 213-232

BOUNDARY VALUE PROBLEMS OF THE THEORY OF ANALYTIC FUNCTIONS WITH DISPLACEMENTS

R. BANTSURI

Abstract. Integral representations are constructed for functions holomorphic in a strip. Using these representations an effective solution of Carleman type problem is given for a strip.

Introduction

In studying some problems of the theory of elasticity and mathematical physics there arise boundary value problems of the theory of analytic functions for a strip [1, 2, 3, 4] when a linear combination of function values is given at a point t of the lower strip boundary and at a point t + a of the upper boundary.

We refer this problem to Carleman type problem for a strip. To solve this problem, in §1 we construct integral representations which play the same role in its solution as a Cauchy type integral plays in solving a linear conjugation problem. In §2 a solution is obtained for a Carleman type problem for a strip with continuous coefficients and in §3 a solution is given for a Carleman type problem for a strip with a coefficient polynomially increasing at infinity. In §4 a conjugation problem with a displacement is investigated.

When the coefficient is a meromorphic function, a Carleman type homogeneous problem was solved by E. W. Barens in [5] by means of Euler’s gamma-functions (provided that the poles and zeros of the coefficient are known). Later various particular cases were studied in [1] and [6].

§ 1. Integral Representations of Holomorphic Functions in a Strip

Let a function Φ(z), z = x + iy, be holomorphic in a strip {a < y < b, −∞ < x < ∞}, continuous in a closed strip {a ≤ y <≤ b, −∞ < x < ∞}

213 and satisfy the condition Φ(z)e^{μ|z|} → 0 for |z| → ∞, μ ≥ 0. The class of functions satisfying these conditions will be denoted by A_{a}^{b}(μ).

1991 Mathematics Subject Classification. 30E25.

Key words and phrases. Carleman type problem, conjugation problem with a displacement.

1072-947X/99/0500-0213$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1997 Plenum Publishing Corporation

Let

Φ_{k}(z) ∈ A_{0}^{β}(μ_{k}), μ_{k} < \frac{πβ[3 + (−1)^{k}]}{2(α^{2} + β^{2})}, k = 1, 2,  (1.1)

where α and β are real numbers, β > 0. Then the following formulas are valid:

Φ_{1}(z) = \frac{1}{2a} ∫_{−∞}^{+∞}\frac{Φ_{1}(t) + Φ_{1}(t + a)}{sinh p(t − z)}dt, 0 < I_{m}z < β,  (1.2)

Φ_{2}(z)= \frac{cosh pz}{2a} ∫_{−∞}^{+∞}\frac{Φ_{2}(t)−Φ_{2}(t+a)}{cosh pt sinh p(t−z)}dt+Φ_{2}(\frac{a}{2}), 0<I_{m}z <β, (1.3)

where p = \frac{πi}{a}, a = α + iβ.

The above formulas are obtained using the theorem on residues.

If Φ_{k}(z) has a form

Φ_{k}(z) = Ψ_{k}(z) + ∑_{j=1}^{n} A_{j}(z − \frac{a}{2})^{−j}, Ψ_{k}(z) ∈ A_{0}^{β}(μ_{k}), k = 1, 2,

then we shall have

Φ_{1}(z)= \frac{1}{2a} ∫_{−∞}^{+∞}\frac{Φ_{1}(t)+Φ_{1}(t+a)}{sinh p(t−z)}dt−∑_{j=1}^{n} \frac{(−p)^{j}A_{j}}{j!} (\frac{1}{cosh pz} )^{(j−1)},

0 < I_{m}z < β,  (1.4)

Φ_{2}(z) = \frac{cosh pz}{2a} ∫_{−∞}^{+∞}\frac{Φ_{2}(t) − Φ_{2}(t + a)}{cosh pt sinh p(t − z)}dt − ∑_{j=1}^{n} \frac{A_{j}(−p)^{j}}{j!} (tanh pz)^{(j−1)} +

+Φ_{2}(\frac{a}{2}), 0 < I_{m}z < β.  (1.5)

Let further F_{k}, k = 1, 2, be functions given on the real axis L and having the form F_{k}(x) = f_{k}(x)e^{μ_{k} |x|}, f_{k}(±∞) = 0, where f_{k} are functions satisfying the Hölder condition everywhere on L, μ_{k} are numbers satisfying inequality (1.1).

Consider the integrals

Φ_{1}(z) = \frac{1}{2a} ∫_{−∞}^{+∞}\frac{F_{1}(t)dt}{sinh p(t − z)}, 0 < I_{m}z < β,  (1.6)

Φ_{2}(z) = \frac{cosh pz}{2a} ∫_{−∞}^{+∞}\frac{F_{2}(t)dt}{cosh pt sinh p(t − z)}, 0<I_{m}z <β.  (1.7)

It is obvious that the functions are holomorphic in a strip 0 < y < β.

Using the Sohotski–Plemelj formulas we can show that the boundary values of Φ_{1} and Φ_{2} are expressed by the formulas

Φ_{1}(t_{0}) = \frac{F_{1}(t_{0})}{2} + \frac{1}{2a} ∫_{−∞}^{+∞}\frac{F_{1}(t)dt}{sinh p(t − t_{0})},

Φ_{1}(t_{0} + a) = \frac{F_{1}(t_{0})}{2} − \frac{1}{2a} ∫_{−∞}^{+∞}\frac{F_{1}(t)dt}{sinh p(t − t_{0})};

(1.8)

Φ_{2}(t_{0}) = \frac{F_{2}(t_{0})}{2} + \frac{cosh pz}{2a} ∫_{−∞}^{+∞}\frac{F_{2}(t)dt}{cosh pt sinh p(t − t_{0})},

Φ_{2}(t_{0} + a) = −\frac{F_{2}(t_{0})}{2} + \frac{cosh pt_{0}}{2a} ∫_{−∞}^{+∞}\frac{F_{2}(t)dt}{cosh pt sinh p(t − t_{0})}. (1.9)

It follows from Plemelj–Privalov’s theorem that that the boundary values of Φ_{1} and Φ_{2} satisfy the Hölder condition on a finite part of the boundary.

Let us investigate the behavior of these functions in the neighbourhood of a point at infinity. We begin by considering the case with μ_{k} = 0, k = 1, 2.

Rewrite formula (1.6) as

Φ_{1}(z) = \frac{1}{2a} ∫_{−∞}^{+∞}[\frac{1}{sinh p(t − z)} − \frac{a}{p} \frac{1}{(t − z)(t + a − z)}]F_{1}(t)dt +

+ \frac{1}{2ap} ∫_{−∞}^{+∞}\frac{F_{1}(t)}{t − z}dt − \frac{1}{2ap} ∫_{−∞}^{+∞}\frac{F_{1}(t)dt}{t + a − z} , 0 < I_{m}z < β.

Here the first term is holomorphic in the closed strip 0 ≤ I_{m}z ≤ β and tends to zero at infinity. The second and the third term are analytic in the strip 0 < I_{m}z < β, vanish at infinity and their boundary values satisfy the Hölder condition, including points at infinity [7].

Therefore Φ_{1} ∈ A_{0}^{β}(0). Now let us consider the function Φ_{2}(z). Rewrite formula (1.7) as

Φ_{2}(z) = \frac{1}{2a} ∫_{−∞}^{+∞}\frac{(cosh pz − cosh pt)F_{2}(t)dt}{cosh pt sinh p(t − z)} + \frac{1}{2a} ∫_{−∞}^{+∞}\frac{F_{2}(t)dt}{sinh p(t − z)}.

As we have shown, the second term here belongs to the class A_{0}^{β}(0). Denote the first term by I and rewrite it as

I = −\frac{1}{2a} ∫_{−∞}^{+∞}\frac{sinh \frac{p}{2}(t + z)F (t)}{cosh pt sinh \frac{p}{2}(t − z)}dt =

= −\frac{1}{2a} ∫_{−∞}^{+∞}\frac{sinh \frac{p}{2}(2x − τ + iy)F_{2}(x − τ)}{cosh p(x − τ) cosh \frac{p}{2}(τ + iy)}dτ =

= −\frac{1}{2a}( ∫_{−∞}^{0} + ∫_{0}^{+∞})\frac{sinh \frac{p}{2}(2x − τ + iy)F_{2}(x − τ)}{cosh p(x − τ) cosh \frac{p}{2}(τ + iy)}dτ.

Let x > 0. Then the first integral will be bounded in the strip 0 ≤ I_{m}z < β, since 2x − τ < 2(x − τ).

Rewrite the second integral as

\frac{1}{2a} ∫_{0}^{+∞}\frac{sinh \frac{p}{2}(2x − τ + iy)F_{2}(x − τ)}{cosh p(x − τ) cosh \frac{p}{2}(τ + iy)}dτ =

= (\frac{1}{2a} ∫_{−∞}^{0} +\frac{1}{2a} ∫_{0}^{x} )\frac{sinh \frac{p}{2}(x + t + iy)F_{2}(t)dt}{cosh pt cosh \frac{p}{2}(x − t + iy)}.

The first term is bounded, since x + t < x − t. The second term can be written in the form

\frac{1}{2a} ∫_{0}^{x} \frac{sinh \frac{p}{2}([(x − t + iy) + 2t]F_{2}(t)dt}{cosh pt cosh \frac{p}{2}(x − t + iy)} = \frac{1}{2a} ∫_{0}^{x} tanh \frac{p}{2}(x − t + iy)F_{2}(t)dt +

+\frac{1}{2a} ∫_{0}^{x} tanh ptF_{2}(t)dt.  (1.10)

Since the function tanh \frac{p}{2}z = tanh \frac{|p|^{2}}{2π} (β + αi)z is holomorphic in the strip 0 ≤ I_{m}z ≤ δ < β and | tanh \frac{p}{2}z| → 1, the estimate

|Φ_{2}(z)| < |Φ_{0}(x)| + ε|x|  (1.11)

holds for the function Φ_{2} when x are large in the closed strip 0 ≤ I_{m}z ≤ δ. Φ_{0}(x) is bounded for x > 0 and ε < 0 is an arbitrarily small number. A similar estimate is also true for the case x < 0. In the same manner one can obtain an estimate of the form (1.11) in the strip 0 < δ ≤ I_{m}z ≤ β provided that the function Φ_{2}(z) is represented as

Φ_{2}(z) = \frac{1}{2a} ∫_{−∞}^{+∞}\frac{cosh pz + cosh pt}{cosh pt sinh p(t − z)}F_{2}(t)dt − \frac{1}{2a} ∫_{−∞}^{+∞}\frac{F_{2}(t)dt}{sinh p(t − z)}.

Now let us consider the case with μ_{k} > 0, k = 1, 2. Rewrite (1.6) as follows:

Φ_{1}(z) = \frac{1}{2a} ∫_{−∞}^{+∞}\frac{cosh μ_{1}tφ_{1}(t)dt}{sinh p(t − z)}, φ_{1}(t) ≡ f_{1}(t)/ cosh μ_{1}t.

It is obvious that φ_{1}(t) satisfies the Hölder condition in the neighbourhood of a point at infinity.

We write the function Φ_{1} in the form

Φ_{1}(z) = \frac{1}{2a} ∫_{−∞}^{+∞}\frac{φ_{1}(t) cosh[μ_{1}(t − z) + μ_{1}z]}{sinh p(t − z)}dt =

= \frac{cosh μ_{1}z}{2a} ∫_{−∞}^{+∞}\frac{cosh μ_{1}(t − z)}{sinh p(t − z)}φ_{1}(t)dt + \frac{sinh μ_{1}z}{2a} ∫_{−∞}^{+∞}\frac{sinh(t − z)μ_{1}}{sinh p(t − z)} φ_{1}(t)dt.

Since μ_{1} < πβ/(α^{2} + β^{2}) = Re p, we have Φ_{1} ∈ A_{0}^{β}(μ_{1}). Taking this into account and applying the arguments used in investigating the behavior of the function Φ_{2}(z) in the case with F_{2}(±∞) = 0, we show that

Φ_{2} ∈ A_{0}^{β}(β_{2}).

Let us formulate the results obtained above.

Theorem 1. If the functions F_{k}(x)e^{−μ_{k} |x|}, (k = 1, 2), satisfy the Hölder condition everywhere on L and F_{k}(x)e^{−μ_{k} |x|} → 0 for |x| → +∞, where μ_{k} are some numbers satisfying inequality (1.1), then Φ_{k} ∈ A_{0}^{β}(μ_{k}) for μ_{1} ≥ 0, μ_{2} > 0, exp Φ_{2} ∈ A_{0}^{β}(ε) for μ_{2} = 0, where ε is an arbitrarily small positive number.

Formulas (1.8) and (1.9) imply

Φ_{1}(t) + Φ_{1}(t + a) = F_{1}(t), t ∈ (−∞, ∞),  (1.12)
Φ_{2}(t) − Φ_{2}(t + a) = F_{2}(t), t ∈ (−∞, ∞),  (1.13)

i.e., Φ_{1}(z) and Φ_{2}(z) defined by (1.6) and (1.7) are solutions of boundary value problems (1.12) and (1.13) of the class A_{0}^{β}(μ_{k}), k = 1, 2.

Clearly, if the function Φ_{2}(z) is a solution of problem (1.13), then the function W (z) = c + Φ_{2}(z) will also be a solution. We shall show that problems (1.12) and (1.13) do not have other solutions of the class A_{0}^{β}(μ_{k}), k = 1, 2. For this we should prove

Theorem 2. If F_{2}(t) ∈ L(−∞, ∞), then for a solution of problem (1.13) of the class A_{0}^{β}(0) to exist it is necessary and sufficient that the condition

∫_{−∞}^{∞} F_{2}(t)dt = 0

be fulfilled.

Proof. We can rewrite formula (1.7) as

Φ_{2}(z) = \frac{1}{2a} ∫_{−∞}^{∞} coth p(t − z)F_{2}(t)dt − \frac{1}{2a} ∫_{−∞}^{∞} tanh ptF_{2}(t)dt.  (1.14)

It is obvious that the limits of Φ_{2}(z) exist for x → ±∞, 0 ≤ y ≤ β, and

C + Φ_{2}(±∞ + iy) = ±\frac{1}{2a} ∫_{−∞}^{∞} F_{2}(t)dt − \frac{1}{2a} ∫_{−∞}^{∞} tanh ptF_{2}(t)dt + C. (1.15)

Taking

C = \frac{1}{2a} ∫_{−∞}^{∞} F_{2}(t) tanh pt dt

and setting

∫_{−∞}^{∞} F_{2}(t)dt = 0,  (1.16)

we find by virtue of (1.15) and (1.16) that the solution of problem (1.13) has the form

W (z) = \frac{1}{2a} ∫_{−∞}^{∞} coth p(t − z)F_{2}(t)dt  (1.17)

and belongs to the class A_{0}^{β}(0).

The necessity is proved by integrating equality (1.13) and applying the Cauchy theorem.

It remains to prove

Theorem 3. If the function φ ∈ A_{0}^{β}(\frac{πβ(3+λ)}{2(α^{2}+β^{2})}), λ = ±1, and satisfies the condition φ(z) = λφ(x + a), then it is constant and, for λ = −1, equal to zero.

Proof. Let λ = −1 and

Ψ(z) = \frac{φ(z)}{cosh pz} + φ(\frac{a}{2})\frac{a}{π(z − \frac{a}{2})}.  (1.18)

The function Ψ(z) ∈ A_{0}^{β}(0) and satisfies the condition

Ψ(x) − Ψ(x + a) = \frac{2a^{2}}{π} φ(\frac{a}{2})\frac{1}{x^{2} − a^{2}/4}.  (1.19)

Since Ψ(z) is a solution of problem (1.19) of the class A_{0}^{β}(0), the condition

\frac{2a^{2}}{π} φ(\frac{a}{2}) ∫_{−∞}^{∞} \frac{dx}{x^{2} − a^{2}/4} = 4aiφ(\frac{a}{2}) = 0

is fulfilled on account of Theorem 2. Thus Ψ(z) is a solution of the homogeneous problem

Ψ(x) − Ψ(x + a) = 0, −∞ < x < +∞.

If we introduce the function

Ψ_{1}(z) = \frac{Ψ(z) − Ψ(\frac{a}{2})}{cosh pz},

then we shall have

Ψ_{1}(x) + Ψ_{1}(x + a) = 0, −∞ < x < +∞.

By applying the Fourier transform to the latter equality we obtain

Ψ̂_{1}(1 + e^{iαt}) ≡ 0.

Hence we have Ψ̂_{1}(t) ≡ 0, Ψ_{1}(z) = 0. Therefore by (1.18) φ(z) = 0. We have thereby proved the theorem for λ = −1.

Let λ = 1. Then

φ(x) − φ(x + a) = 0.

The function

Ψ(z) = φ(z) − φ(\frac{3}{4}a)  (1.20)

also satisfies this condition and φ(\frac{3}{4}a) = 0.

We introduce the notation

Ψ_{0}(z) = \frac{Ψ(z)}{cosh 2pz} + \frac{a}{2π} \frac{Ψ(\frac{a}{4})}{z − \frac{a}{4}} .

Now, repeating the above arguments, we find that Ψ(\frac{a}{4}) = 0, i. e., Ψ_{0}(z) ∈ A_{0}^{β}(0) and satisfies the condition

Ψ_{0}(x) − Ψ_{0}(x + a) = 0.

But, as shown above, in that case Ψ_{0}(z) = Ψ(z) = 0 and therefore equality (1.20) implies

φ(z) = φ(\frac{3}{4}a),

which proves the theorem.

§ 2. A Carleman Type Problem with a Continuous Coefficient for a Strip

Let us consider the following problem: find a function Φ of the class A_{0}^{β}(μ) by the boundary condition

Φ(x) = λG(x)Φ(x + a) + F (x), −∞ < x < +∞,  (2.1)

where a = α + iβ, β > 0, μ < πβ(3 + λ)/2(α^{2} + β^{2}), F and G are the given functions satisfying the Hölder condition including a point at infinity, G ≠ 0 and F (±∞) = 0, G(−∞) = G(∞) = 1, the constant λ takes the value 1 or −1.

The integer number κ = \frac{1}{2π} [arg G(x)]_{−∞}^{+∞}, where [arg G(x)]_{−∞}^{+∞} denotes an increment of the function arg G(x) when x runs over the entire real axis from −∞ to ∞, is called the index of the function G(x). The index of G_{0}(x) = G(x)[(x − a/2)/(x + a/2)]^{κ} is equal to zero and therefore any branch of the function ln G_{0}(x) is continuous all over the real axis. We choose a branch that vanishes at infinity. By formulas (1.7) and (1.9) G(x) can be represented as

G(x) = \frac{X(x)}{X(x + a)},  (2.2)

where

X(z) = (z − \frac{a}{2})^{κ}X_{0}(z),

X_{0}(z) = exp (\frac{cosh pz}{2a} ∫_{−∞}^{+∞}\frac{ln G_{0}(t)dt}{cosh pt sinh p(t − z)}).

(2.3)

By virtue of Theorem 1 X_{0}(z) and [X_{0}(z)]^{−1} ∈ A_{0}^{β}(ε), where ε is an arbitrarily small positive number.

Using (2.2), we rewrite condition (2.1) as

\frac{Φ(x)}{X(x)} = λ\frac{Φ(x + a)}{X(x + a)} + \frac{F (x)}{X(x)}, −∞ < x < ∞.  (2.4)

The function Φ(z)/X(z) is holomorphic in the strip 0 < I_{m}z < β except perhaps of the point z = \frac{a}{2}, at which it may have a pole of order κ, for κ > 0, and satisfies the condition

(Φ(z)/X(z))e^{−μ|z|} → 0 for |x| → ∞ and 0 ≤ y ≤ β,

where 0 < μ < πβ(3 + λ)/2(α^{2} + β^{2}). By (1.4) and (1.5) condition (2.4) implies

Φ(z) = \frac{X(z)}{2a} ∫_{−∞}^{+∞}\frac{F (t)dt}{X(t) sinh p(t − z)} + X(z)φ_{1}(z) for λ = −1, (2.5)

Φ(z) = \frac{X(z) cosh pz}{2a} ∫_{−∞}^{+∞}\frac{F (t)dt}{X(t) cosh pt sinh p(t − z)} +

+X(z)φ_{2}(z) for λ = 1,  (2.6)

where

φ_{1}(z) = \begin{cases}0, & κ ≤ 0, \\ ∑_{k=1}^{κ−1} C_{k}(1/ cosh pz)^{(k)}, & κ > 0, \end{cases}  (2.7)

φ_{2}(z) = \begin{cases}0, & κ < 0, \\ ∑_{k=1}^{κ} C_{k}(tanh pz)^{(k)}, & κ ≥ 0, \end{cases}  (2.8)

C_{k} are arbitrary constants.

Let us investigate the behavior of the function

φ(z) = \frac{X(z)}{2a} ∫_{−∞}^{+∞}\frac{F (t)dt}{X(t) sinh p(t − z)}, 0 ≤ I_{m}z ≤ β,  (2.9)

in the neighbourhood of a point at infinity. The function X(z) can be represented as

X(z) = (z − \frac{a}{2})^{κ} exp Γ_{1}(z) · exp Γ_{2}(z),

where

Γ_{1}(z) = −\frac{1}{2a} ∫_{−∞}^{∞} \frac{sinh \frac{p}{2}(z + t) ln G_{0}(t)}{cosh pt cosh \frac{p}{2}(t − z)}dt,

Γ_{2}(z) = \frac{1}{2a} ∫_{−∞}^{∞} \frac{ln G_{0}(t)dt}{sinh p(t − z)}.

As we have shown, Γ_{2} ∈ A_{0}^{β}(0), i. e., (exp Γ_{2}(z) − 1) ∈ A_{0}^{β}(0).

By differentiating the function Γ_{1}(z) we obtain

Γ′_{1}(z) = \frac{1}{2a} ∫_{−∞}^{∞} \frac{ln G_{0}(t)dt}{cosh \frac{p}{2}(t − z)}, 0 ≤ I_{m}z ≤ β_{0} < β.

It is easy to verify that Γ′_{1}(z) → 0 for |z| → ∞ and therefore for any there is a number N such that

|Γ′_{1}(x + iy)| < ε for |x| > N, 0 ≤ y ≤ β_{0} < β.  (2.10)

We represent φ(z) as

φ(z) = \frac{X(z)}{2a} ∫_{−N}^{N} \frac{F (t)}{X(t)} · \frac{dt}{sinh p(t − z)} +

+\frac{X(z)}{2a}( ∫_{−∞}^{N} + ∫_{N}^{∞} )\frac{F (t)dt}{X(t) sinh p(t − z)}, 0 ≤ I_{m}z ≤ β_{0}.

It is easy to show that the first and the second term vanish for x → +∞. We shall show that the third term also tends to zero for x → +∞, 0 ≤ y ≤ β_{0} < β. This term will be denoted by I.

I = ∫_{N}^{∞} \frac{exp(Γ_{2}(z) − Γ_{2}(t)) exp(Γ_{1}(z) − Γ_{1}(t))(z − \frac{a}{2})^{κ}F (t)}{(t − \frac{a}{2})^{κ} sinh p(t − z)}dt.

Assume that κ ≥ 0 and represent the function (z − \frac{a}{2})^{κ} as

(z − \frac{a}{2})^{κ} = κ! ∑_{n=1}^{κ} \frac{(t − \frac{a}{2})^{κ−n}(z − t)^{n}}{(κ − n)! n!} + (t − \frac{a}{2})^{κ}.  (2.11)

Inequality (2.10) implies that

|Γ_{1}(z) − Γ_{1}(t)| ≤ | ∫_{t}^{z} Γ′(s)ds| ≤ ε|t − z|, t < N, κ > N, 0 ≤ y ≤ β_{0},

i.e., Re(Γ_{1}(t) − Γ_{1}(z)) − ε|t − z| < 0. Thus we have

| exp[Γ_{1}(t) − Γ_{1}(z)) − ε|t − z| − 1| < A|t − z|, t > N, x > N.

The latter inequality and formula (2.11) imply

|I| ≤ c ∑_{n=1}^{κ} ∫_{N}^{∞} \frac{e^{ε|x−t|}|x − t + iy|^{n}|F (t)|dt}{(κ − n)!n!| sinh p(x − t + iy)||t − \frac{a}{2}|^{n}} +

+c_{1} ∫_{N}^{∞} \frac{e^{ε|x−t|}[A|x − t| + (1 − e^{−ε|x−t|})]|F (t)|dt}{| sinh p(x − t + iy)|} + | ∫_{N}^{∞} \frac{F (t)dt}{2|a| sinh p(t − z)}|.

where, as shown above, the third term is the modulus of a function of the class A_{0}^{β}(0). Since ε is an arbitrarily small number and F (∞) = F (−∞) = 0, the first two terms are the convolutions of functions summable with functions tending to zero for → ∞. Therefore they tend to zero for x → ∞, 0 ≤ y ≤ β_{0} < β.

It can be shown in a similar manner that φ(z) → 0 for x → −∞, 0 ≤ y ≤ β_{0}, as well. It is not difficult to prove that the function φ(z) tends to zero for |x| → ∞, β_{0} ≤ I_{m}z ≤ β. When κ < 0, one can use the same reasoning to show that φ(z) → 0 for |x| → ∞, 0 ≤ y ≤ β, provided that z and t are exchanged in equality (2.11). Thus the function Φ represented by (2.5) tends to zero for |x| → +∞, 0 ≤ y ≤ β. Quite similarly, it is proved that for the function Φ defined by (2.6) we have Φ(z)e^{−ε|z|} → 0 for |x| → ∞, 0 ≤ y ≤ β.

For κ < 0 the function X(z) has a pole of order −κ at the point z = \frac{a}{2}. In that case the solution exists only if the following conditions are fulfilled:

∫_{−∞}^{∞} \frac{F (t)}{X(t)}(\frac{1}{cosh pt})^{(k)}dt=0, k =0, …, (−κ−1), for λ=−1, (2.12)

∫_{−∞}^{∞} \frac{F (t)}{X(t)}(\frac{e^{pt}}{cosh pt})^{(k)}dt = 0, k = 1, …, (−κ − 1), for λ = 1. (2.13)

The results obtained can be formulated as

Theorem 4. For λ = −1 and κ ≥ 0 problem (2.1) is solvable in the class A_{0}^{β}(0) and a general solution is given by (2.5) with formula (2.7) taken into account. If κ < 0, then the problem is solvable if condition (2.12) is fulfilled. In these conditions problem (2.1) has a unique solution in the class A_{0}^{β}(0) which is given by formula (2.5) for φ_{1} = 0.

Theorem 5. if λ = 1 and κ ≥ −1, the problem (2.1) is solvable in the class A_{0}^{β}(ε) and the solution is given by (2.6) with (2.8) taken into account; for κ < −1 the solution exists provided that condition (2.13) is fulfilled. If these conditions are fulfilled, then problem (2.1) has a unique solution in the class A_{0}^{β}(ε). This solution is given by (2.6), where φ_{2} = 0.

§ 3. A Carleman Type Problem with Unbounded Coefficients for a Strip

Problems of the elasticity theory can often be reduced to a Carleman type problem with coefficients polynomially increasing or decreasing at infinity. We shall consider such a case below.

We write the boundary condition of the problem in the form

Φ(x) = P_{n}(x)G(x)Φ(x + iβ) + F (x), −∞ < x < ∞,  (3.1)

where G(x) and F (x) satisfy the conditions discussed in §2, and P_{n}(x) is a polynomial without real zeros. Condition (3.1) can be rewritten as

Φ(x) = q[x^{2} + 4β^{2}]^{[ \frac{n}{2} ]}(2β − ix)^{δ(n)}G_{0}(x)Φ(x + iβ) + F (x), (3.2)

where δ(n) = 0 for even n and δ(n) = 1 for odd n; q is a complex number; G_{0}(x) is a Hölder class function including a point at infinity G_{0}(−∞) = G_{0}(∞) = 1.

As shown above, the function G_{0}(x) can be represented as

G_{0}(x) = \frac{X_{0}(x)}{X_{0}(x + iβ)}, −∞ < x < ∞,  (3.3)

where

X_{0}(z) = (z − \frac{iβ}{2} )^{κ} exp (\frac{cosh pz}{2iβ} ∫_{−∞}^{∞} \frac{ln [G_{0}(t)(\frac{t+iβ/2}{t−iβ/2})^{κ}]}{cosh pt sinh p(t − z)}dt). (3.4)

Write the function [x^{2} + 4β^{2}]^{[ \frac{n}{2} ]}(2β − ix)^{δ(n)} in form (3.3). We shall find solutions of the problems

X_{1}(x) = (2β + ix)X_{1}(x + iβ), −∞ < x < +∞,  (3.5)
X_{2}(x + iβ) = (2β − ix)X_{2}(x), −∞ < x < +∞.  (3.6)

Applying the Fourier transformation to conditions (3.5) and (3.6), we obtain the differential equations

(f_{1}(t)e^{βt})′ = (1 − 2βe^{βt})f_{1}(t), −∞ < t < +∞,
f′_{2}(t) = (2β − e^{−βt})f_{2}(t), −∞ < t < +∞,

where f_{1}(t) and f_{2}(t) denote the Fourier transforms of the functions X_{1}(x) and X_{2}(x).

By performing the reverse Fourier transformation of the solutions of these equations we obtain the solutions of problems (3.5) and (3.6):

X_{1}(z) = ∫_{−∞}^{+∞}exp ( − \frac{1}{β} e^{βt} + 3βz + itz)dt, 0 < I_{m}z < β, (3.7)

X_{2}(z) = ∫_{−∞}^{+∞}exp ( − \frac{1}{β} e^{−βt} − 2βt + itz)dt, 0 < I_{m}z < β. (3.8)

On substituting e^{βt} = βτ, we have

\begin{matrix}X_{1}(z) = β^{2}β^{\frac{iz}{β}} ∫_{0}^{∞}e^{−τ} τ^{2+ \frac{iz}{β}} dτ = β^{2}β^{\frac{iz}{β}} Γ(3 + \frac{iz}{β} ), \\ X_{2}(z) = β^{− \frac{iz}{β}} ∫_{0}^{∞}e^{−τ} τ^{1− \frac{iz}{β}} dτ = ββ^{− \frac{iz}{β}} Γ(2 − \frac{iz}{β} ). \end{matrix}  (3.9)

We introduce the notation

X_{3}(z) = [\frac{X_{1}(z)}{X_{2}(z)}]^{[ \frac{n}{2} ]}(X_{2}(z))^{−δ(n)}, 0 < I_{m}z < β.  (3.10)

Using Stirling’s formulas [11], we obtain from (3.9) and (3.10) the following representations of the functions X_{1}(z) and X_{2}(z) in the neighbourhood of a point at infinity:

|X_{1}(z)| = C_{1}(y)e^{− \frac{π}{2β} |x|}|x|^{\frac{5}{2} − \frac{y}{β}} (1 + O(\frac{1}{x})), 0 ≤ y ≤ β,

|X_{2}(z)| = C_{2}(y)e^{− \frac{π}{2β} |x|}|x|^{\frac{3}{2} + \frac{y}{β}} (1 + O(\frac{1}{x})), 0 ≤ y ≤ β,

where C_{1}(y), C_{2}(y) are the bounded functions that do not vanish.

By virtue of these formulas, for sufficiently large values of |z| (3.10) implies

|X_{3}(z)| = C(y)(|x|^{\frac{β−2y}{β}} )^{[ \frac{n}{2} ]}(e^{− \frac{π}{2β} |x|}|x|^{\frac{3}{2} + \frac{y}{β}} )^{−δ(n)}(1 + O(\frac{1}{x})). (3.11)

Using equalities (3.3) and (3.11), we rewrite condition (3.2) as

\frac{Φ(x)}{X(x)} − q \frac{Φ(x + iβ)}{X(x + iβ)} = \frac{F (x)}{X(x)}, −∞ < x < ∞,  (3.12)

where X(z) = X_{0}(z)X_{3}(z).

The function Φ(z)/X(z) is holomorphic in the strip 0 < I_{m}z < β except perhaps for the point z = iβ/2, where for κ > 0 it may have a pole of order not higher than κ, and satisfies the condition

(Φ(z)/X(z))e^{−μ|z|} → 0 for |z| → ∞, μ < \frac{π}{2β} + ε.

Write q in the form

q = \frac{X_{4}(x)}{X_{4}(x + iβ)}, X_{4}(z) = exp (\frac{iz}{β} ln q).

From (2.7) and (2.5) it follows that if q is not a real positive number, then a general solution of problem (3.1) is given by the formula

Φ(z) = \frac{X(z)}{2iβ} ∫_{−∞}^{∞} \frac{exp (\frac{π−δ+iγ}{β} (z − t))}{X(t) sinh p(t − z)}F (t)dt + X(z)φ(z), (3.13)

where γ = ln |q|, δ = arg q, 0 < δ < 2π.

φ(z) = ∑_{j=0}^{κ−1} C_{j} \frac{d^{j}}{dz^{j}} ( exp \frac{(π − δ + iγ)z}{β} / cosh pz).  (3.14)

For κ ≥ 0 the solution of problem (3.1) is given by formulas (3.13) and (3.14). Note that for κ ≤ 0 it is assumed that φ(z) ≡ 0. For κ < 0 the function X(z) has, at the point z = \frac{iβ}{2} , a pole of order −κ and therefore the bounded solution exists in the finite part of the strip only if the conditions φ(z) = 0;

∫_{−∞}^{∞} F (t)Ψ_{j}(t) = 0, Ψ_{j}(t) = \frac{d^{j}}{dt^{j}} (\frac{exp(\frac{δ−π−ij}{β})t}{cosh pt}),

j = 0, …, (−1 − κ),  (3.15)

are fulfilled. Thus, like in §2, one can easily prove that in the case of even n problem (3.1) has a solution Φ(z) ∈ A_{0}^{β}(0) for any δ ∈ (0, 2π), while in the case of odd n it has a solution Φ(z) ∈ A_{0}^{β}(\frac{π−2δ}{2β} + ε) for δ ∈ (0, \frac{π}{2} ]; Φ(z) ∈ A_{0}^{β}(0) for δ ∈ (\frac{π}{2} , \frac{3}{2}π); Φ(z) ∈ A_{0}^{β}(\frac{2δ−3π}{2β} + ε) for δ ∈ [\frac{3}{2}π, 2π), where ε > 0 is an arbitrarily small number.

When q > 0, by substituting

Φ(z) = X_{4}(x)Ψ(t)

condition (3.12) can be reduced to the condition

\frac{Ψ(x)}{X(x)} − \frac{Ψ(x + iβ)}{X(x + iβ)} = \frac{F (x)X_{4}(x)}{X(x)}, −∞ < x < ∞.  (3.16)

By virtue of formula (3.15) a general solution of problem (3.1) has the form

Φ(z) = \frac{X^{∗}(z)}{2iβ} ∫_{−∞}^{∞} \frac{F (t)dt}{X^{∗}(t) sinh p(t − z)} + X^{∗}(z)φ_{2}(z),  (3.17)

where X^{∗}(z) = X(z) cosh pzX_{4}(z),

φ_{2}(z) = \begin{cases}∑_{j=0}^{κ−1} C_{j} \frac{d^{j}}{dz^{j}} (tanh pz) + Cx, & for κ > 0, \\ C, & for κ = 0, \\ 0, & for κ ≤ −1, \end{cases}  (3.18)

C, C_{j}, j = 0, …, (κ − 1), are arbitrary constants. If κ < −1, then the solution exists only provided that the condition

∫_{−∞}^{∞} \frac{F (t)}{X^{∗}(z)} · \frac{d^{j}}{dt^{j}} (\frac{1}{cosh pt})dt = 0, j = 0, …, (−κ − 2),

is fulfilled.

One can prove that Φ(z) ∈ A_{0}^{β}(ε) for an even n and Φ(z) ∈ A_{0}^{β}(π/(2β)+ε) for odd n; here ε is a small positive integer.

Remark 1. Formulas (3.8) and (3.9) can be obtained by applying formulas (3.3) and (3.4).

Indeed, if in formula (3.4) G_{0}(t) is replaced by the function (2β − ix)^{−1}, then we shall have

X_{2}(z) = exp (\frac{cosh pz}{2iβ} ∫_{−∞}^{∞} \frac{ln i − ln(x + 2iβ)}{cosh px sinh p(x − z)}dx).  (3.19)

By the function ln z we understand ln z = ln |z| + arg z, −π < arg z < π. After rewriting ln(x + 2iβ) as

ln(x + 2iβ) = ∑_{k=0}^{n} [ ln(x + iβ(k + 2)) − ln(x + iβ(k + 3))] + ln(x + iβ(3 + n)) and substituting this expression into (3.19), by virtue of (1.3) we obtain

ω(z) = \frac{cosh pz}{2iβ} ∫_{−∞}^{∞} \frac{ln i − ln(x + 2iβ)}{cosh px sinh p(x − z)}dx =

= ∑_{k=0}^{n} [ ln(x + iβ(k + 2)) − ln (\frac{5iβ}{2} + kiβ)] +

+ \frac{cosh pz}{2iβ} ∫_{−∞}^{∞} \frac{ln(1 + n)β)}{cosh pt sinh p(t − z)}dt + O(\frac{1}{n}).

If we perform some simple transformations and calculate the latter integral by the formula

\frac{cosh pt}{2iβ} ∫_{−∞}^{∞} \frac{ln[(n + 1)β]dx}{cosh px sinh p(x − z)} = ln[(1 + n)β](\frac{iz}{β} + \frac{1}{2}).

then we shall have

ω(z) = ∑_{k=1}^{n} ln [(1 + \frac{ζ}{k} )e^{− \frac{ζ}{k}} ] − ζ( ln(n + 1) − ∑_{k=1}^{n} \frac{1}{k} ) − ln β^{ζ} −

−\frac{5}{2}( ln(n + 1) − ∑_{k=1}^{n} \frac{1}{k} ) + ln ζ + C_{n}, ζ = \frac{z + 2iβ}{iβ} .

Passing to the limit as n → +∞, by virtue of (3.19) we obtain

X_{2}(z) = Aζ ∏_{1}^{∞} (1 + \frac{ζ}{k} )e^{− \frac{ζ}{k}} e^{−cζ}β^{ζ} = AΓ(2 − \frac{iz}{β} )β^{2− \frac{iz}{β}} .

§ 4. On a Conjugation Boundary Value Problem with Displacements

As an application of the results obtained in §2 we shall consider one kind of a conjugation problem with displacements, when the boundary is a real axis. Denote by S^{+} and S^{−} the upper and the lower half-planes, respectively.

Consider the following problem:

Find a piecewise-holomorphic function bounded throughout the plane using the boundary condition

Φ^{+}(x) = G(x)Φ^{−}[α(x)] + f(x), −∞ < x < +∞,  (4.1)

where G(x) and f(x) are the given functions satisfying the Hölder condition, G(x) ≠ 0, G(∞) = G(−∞) = 1, f(+∞) = f(−∞) = 0,

α(x) = \begin{cases}x, & x < 0, \\ bx, & x ≥ 0, \end{cases}

b is a constant.

If we denote by κ the index of the function G(x), then G(x) can be represented as [7]

G(x) = \frac{X^{+}(x)}{X^{−}(x)}, X(z) = \begin{cases}exp ω(z), & z ∈ S^{+}, \\ (\frac{z + i}{z − i})^{κ} exp ω(z), & z ∈ S^{−}, \end{cases} (4.2)

ω(z) = \frac{1}{2πi} ∫_{−∞}^{+∞}\frac{ln G_{0}(t)dt}{t − z}, G_{0}(x) = G(x)(\frac{x + i}{x − i})^{κ}.

On putting the value of G(x) into (4.1), we obtain

\frac{Φ^{+}(x)}{X^{+}(x)} − \frac{Φ^{−}(α(x))}{X^{−}(x)} = \frac{f(x)}{X^{+}(x)}, −∞ < x < +∞.  (4.3)

For x < 0 condition (4.3) takes the form

\frac{Φ^{+}(x)}{X^{+}(x)} − \frac{Φ^{−}(x)}{X^{−}(x)} = \frac{f(x)}{X^{+}(x)}.  (4.4)

A general solution of problem (4.4) can be written as

Φ(z) = \frac{X(z)}{2πi} ∫_{−∞}^{0} \frac{f(t)dt}{X^{+}(t)(t − z)} + X(z)Φ_{0}(z).  (4.5)

The function Φ(z) is holomorphic on the plane cut along the positive semi-axis except perhaps for the neighbourhood of the point z = −i at which it has a pole of order κ for κ > 0.

For κ < 0 the function X(z) has a pole of order −κ at the point z = −i. Therefore for a bounded solution to exist it is necessary that the condition

Φ_{0}^{(k)}(−i) + \frac{k!}{2πi} ∫_{−∞}^{0} \frac{f(t)dt}{X^{+}(t)(t + i)^{k+1}} = 0, k = 0, 1, …, (−κ − 1), (4.6)

be fulfilled.

If we put the value of Φ(z) into (4.3), then we have

Φ_{0}^{+}(x) = G_{1}(x)Φ_{0}^{−}(bx) + f_{0}(x), 0 < x < ∞,  (4.7)

where G_{1}(x) = \frac{X^{−}(bx)}{X^{−}(x)}, f_{0}(x) = \frac{f(x)}{X^{+}(x)} − A^{+}(x) + G_{1}(x)A^{−}(bx),

A(z) = \frac{1}{2πi} ∫_{−∞}^{0} \frac{f(t)dt}{X^{+}(t)(t − z)}.

The function z = e^{ζ}, ζ = ξ + iη, maps the strip 0 < η < 2π onto the plane having a cut along the axis x > 0.

On introducing the notation Φ_{0}(e^{ζ}) = Ψ_{0}(ζ), 0 < η < 2π, we obtain

Φ_{0}^{+}(x) = Ψ_{0}(ξ), Φ_{0}^{−}(bx) = Ψ_{0}(ξ + ln b + 2πi), −∞ < ξ < +∞. (4.8)

Thus problem (4.7) is reduced to the problem considered in §2

Ψ_{0}(ξ) = G^{+}(ξ)Ψ_{0}(ξ + ln b + 2πi) + F_{0}(ξ), −∞ < ξ < +∞, (4.9)

where G^{+}(ξ) = G_{1}(e^{ξ}), F_{0}(ξ) = f_{0}(e^{ξ}), G^{∗}(−∞) = G^{∗}(∞) = 1,

J_{n}dG^{∗} = 0, F_{0}(+∞) = 0, F_{0}(−∞) = \frac{f(0)}{X^{+}(0)}.

Since for κ > 0 the function Φ_{0}(z) can have a pole of order κ at the point z = −i, we seek a solution Ψ_{0} of problem (4.9) in the class of functions satisfying the condition

Ψ_{0}(ζ)(\frac{ζ − \frac{3}{2}πi}{ζ + \frac{3}{2}πi})^{κ} ∈ A_{0}^{β}(μ), μ < \frac{4π^{2}}{4π^{2} + ln b}.  (4.10)

By virtue of formula (2.6) it is easy to show that a general solution of problem (4.9) is given by the formula

Ψ_{0}(ζ) = \frac{X^{∗}(ζ) cosh pζ}{2a} ∫_{−∞}^{+∞}\frac{F_{0}(t)dt}{X^{+}(t) cosh pt sinh p(t − ζ)} + X^{∗}(ζ)Ψ(ζ), (4.11)

where a = ln b + 2πi, p = \frac{πi}{a},

ψ(ζ) = \begin{cases}∑_{k=0}^{κ} c_{k} coth^{k} p(ζ − \frac{3}{2}πi), & κ ≥ 0, \\ c_{−1}, & κ = −1, \\ 0, & κ < −1, \end{cases}

X^{∗}(ζ) = exp (\frac{cosh pζ}{2a} ∫_{−∞}^{+∞}\frac{ln G^{∗}(t)dt}{cosh pt sinh p(t − ζ)}).

Returning to the variable z, we obtain

Ψ_{0}(ζ) = \frac{X_{0}(z)}{a} ∫_{0}^{+∞}\frac{t^{2p−1}f_{0}(t)dt}{(t^{2p} − z^{2p})X_{0}^{+}(t)} + X_{0}(z)(φ_{0}(z) − A), (4.12)

X_{0}(z) = exp (\frac{1}{a} ∫_{0}^{∞} \frac{ln G_{1}(t)t^{2p−1}}{t^{2p} − z^{2p}} dt), A = \frac{1}{a} ∫_{0}^{∞} \frac{t^{2p−1}f_{0}(t)}{(t^{2p} + 1)X_{0}^{∗}(t)}dt.

With (4.5) and (4.12) taken into account we conclude that a general solution of problem (4.1) has the form

Φ(z) = X(z)[\frac{1}{2πi} ∫_{−∞}^{0} \frac{f(t)dt}{X^{+}(t)(t − z)} + \frac{X_{0}(z)}{a} ∫_{0}^{∞} \frac{t^{2p−1}f_{0}(t)dt}{X_{0}^{+}(t)(t^{2p} − z^{2p})} +

+ X_{0}(z)(φ_{0}(z) − A)],  (4.13)

φ_{0}(z) = \begin{cases}∑_{k=0}^{κ} c_{k}(\frac{z^{2p} + (−i)^{2p}}{z^{2p} − (−i)^{2p}} )^{k}, & κ ≥ 0, \\ c_{−1}, & κ = −1, \\ 0, & κ < −1. \end{cases}  (4.14)

The function z^{2p} is holomorphic on the plane cut along the positive axis if by this function we mean the branch for which the limit as z → 1 from the upper half-plane is equal to 1 while t^{2p} denotes the function value, at the point t, of the upper edge of the cut.

For κ = −1 the function X_{0}(z) has a pole of first order at the point z = −i. In that case φ_{0}(z) = C_{−1} and X_{0}(−i) ≠ 0 and therefore the constant c_{1} can be chosen so that for z = −i the expression in square brackets on the right-hand side of (4.13) would vanish. Hence when κ ≥ −1 problem (4.1) has a bounded solution for an arbitrary right-hand side. When κ < −1, for a bounded solution to exist it is necessary and sufficient that the conditions

\frac{d^{k}}{dz^{k}} [\frac{1}{2πi} ∫_{−∞}^{0} \frac{f(t)dt}{X^{+}(t)(t − z)} + \frac{X_{0}(z)}{a} ∫_{0}^{∞} \frac{t^{2p−1}f_{0}(t)dt}{X_{0}^{+}(t)(t^{2p} − z^{2p})} − AX_{0}(z)] = 0,

z = −i, k = 1, …, −κ,

be fulfilled. Then the solution is given by formula (4.13).

For b = 1 we have p = \frac{1}{2}, X_{0}(z) ≡ 1, f_{0}(t) ≡ f(t) and formulas (4.13) and (4.14) give a solution of the conjugation problem.

Conjugation problems with displacements are investigated in [8–10] in the case with α′(t) belonging to the Hölder class.

References

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11. V. I. Smirnov, A course on higher mathematics vol. 3, part II. (Russian) Gostekhizdat, Moscow–Leningrad, 1949.

(Received 21.04.1997)

Author’s address:

A.Razmadze Mathematical Institute

Georgian Academy of Sciences

1, Aleksidze St., Tbilisi 380093

Georgia