This article discusses a few different types of singular integral equations of the convolution type with Carleman shift in class {0}. By using the theory of Fourier analysis, these equations under consideration are transformed into Riemann–Hilbert boundary value problems for analytic functions with shift and discontinuous coefficients. For such problems, we propose a method different from the classical ones, and we obtain the analytic solutions and the conditions of Noether solvability.

MSC2010 Classification: 45E10, 45E05, 30E25

1. Introduction

Singular integral equations (SIEs) and boundary value problems for analytic functions are well-known to be effective mathematical tools with a wide range of applications in physics, engineering mechanics, fracture mechanics, technology, and other fields. Many such mathematicians have done extensive study on the integral equations of the convolution type and SIEs. The general solutions and solvability conditions for Cauchy SIEs with translations were determined by Musknelishvilli [1] and Litvinchuk [2]. For various classes of SIEs with Carleman translation, Chuan and Tuan [3, 4] studied the problem and provided the existence of the solution and the methods of solution. Many mathematicians devote their time to researching some classes of SIEs with Carleman shift by the Fredholm operator theory; for example, see Karapetiants and Samko [5], Castro [6, 7], and Karlovich [8, 9]. Recent research by Li and Ren [10–14] yielded the explicit solutions and the Noether theory of solvability for several classes of SIEs with convolution kernels and constant coefficients. Pingrun studied SIEs of convolution type with reflection and translation shifts equations in [15]. Li et al. [16] solved convolution SIEs with reflection and translation displacements using the Riemann–Hilbert approach. Nagdy, Hashem, and Ebrahim [17] studied some classes of singular integral differential equations with reflection.

In this article, we will study some different types of SIEs of the convolution type with Carleman shift α(t) = a − t, α(t) : R → R, R = (−∞, ∞), a is a real constant and α²(t) = t. These equations will be solved using the Hilbert operator, convolution operator, and Fourier integral operator. Here, we use a novel method for solving equations that is distinct from those used in classical cases. Specifically, we first use the Fourier integral transform to convert the equations into a system of function equations, and then we transform the resulting equations into boundary value problems with both shift and discontinuous coefficients. The structure of this work is as follows. Section 2 is devoted mainly for the necessary definitions of the Fourier and Cauchy transforms. In Section 3, we use the Fourier transforms to convert SIEs with Carleman shift into Riemann boundary value problems and obtain the equation’s solutions. In Section 4, we consider dual SIEs with Carleman Shift. In Section 5, we consider Singular Integral Wiener Hope equation with Carleman shift.

2. Basic Definitions and Tools

This section is devoted mainly to the necessary definitions of the Fourier and Cauchy transforms.

Definition 1 ([17]). We say that F(x) is an element of a space with Holder continuous functions H on [−N, N], if any positive real integer r exists such that for any x₁, x₂ ∈ [−N, N], the condition holds.

Definition 2 ([17]). Suppose F(x) is a continuous function on the entire real domain. The function if the following conditions are satisfied:

i.
F(x) ∈ H on [−N, N] in the case of any large enough positive number N.
ii.
for any |x_i| > N(i = 1, 2), k > 0.

Definition 3 ([15]). Let the function F(x) satisfies the conditions:

(i) . (ii) F(x) ∈ L¹(R), where , then function F(x)∈{{0}}.

Definition 4 ([15]). The Fourier transform of function f(x) ∈ L¹(R) is denoted by the following equation:

and the inverse of the Fourier transform of function F(x) is defined by the following equation:

It is easy to see that . If F(x) ∈ {{0}}, then .

Now we define the Fourier transform of a function f(α(x)) ∈ L¹(R) by the following form:

by putting α(x) = a − x = t, we get the following equation:

where the inverse of the Fourier transform of a function e^isaF(−s) is defined by the following equation:

The Fourier transform can be extended to L²(R) [18].

Definition 5 ([18]). The convolution for any functions f(x), g(x) ∈ {0} is denoted by the following equation:

Definition 6 ([19]). The operator T of the Cauchy principal value integral is defined by the following equation:

Lemma 7. If F(x) belongs to {{0}}, then e^ixaF(−x) ∈ {{0}} and f(α(x)) ∈ {0}.

Proof. We prove :

(i)
For any two points x₁, x₂ ∈ [−N, N], we have the following equations:

(ii)
For |x_i| > N(i = 1, 2), we have the following equations:

From (i), (ii), and Definition 2, we obtain . It is easy to notice that e^ixaF(−x) ∈ L¹(R), hence from Definition 3, we obtain e^ixaF(−x) ∈ {{0}} and f(α(x)) ∈ {0}.

Lemma 8. If f(x) ∈ {0}, then and .

Proof. It is easy to prove that, if f(x) ∈ {0}, then .

Now, we will prove that

Since:

()

From the extended residue theorem [19], we have the following equation:

()

Putting Equation (2) into (1), we obtain the following equation:

()

Substituting by the Carleman shift α(y) = a − y in Equation (3), we obtain the following equation:

by putting a − y = t, we get the following equations:

Lemma 9. If f ∈ {0} and F(0) = 0, then (Tf(x)), (Tf(α(x))) ∈ {0}.

Proof. It is easy to prove that, if f ∈ {0} and F(0) = 0, then Tf(x) ∈ {0}.

Now, we will prove that Tf(α(x)) ∈ {0}.

Since f ∈ {0}, then F(−x) ∈ {{0}} ⊂ C(−∞, ∞). If F(0) = 0, we get −sgn(x)e^ixaF(−x) ∈ C(−∞, ∞). We now prove :

(i)
For any two points x₁, x₂ ∈ [−N, N], x₁x₂ > 0, we have the following equations:

For any two points x₁, x₂ ∈ [−N, N], x₁x₂ < 0, we have the following equations:

where r₀, r₁ are positive constants, if |x₂ − x₁| < 1, we take α = min{α₀, α₁}.

If |x₂ − x₁| ≥ 1, we take α = max {α₀, α₁}.

(ii)
For |x_i| > N(i = 1, 2), similar to the case for x₁, x₂ ∈ [−N, N], x₁x₂ > 0, in (i) it is easy to obtain the following equation:

From (i), (ii), and Definition 2, we obtain . It is easy to notice that −sgn(x)e^ixa F(−x) ∈ L¹(R), hence from Definition 3, we obtain −sgn(x)e^ixaF(−x) ∈ {{0}} and Tf(α(x)) ∈ {0}.

3. SIEs With Carleman Shift

In this section, we consider the following SIEs with Carleman shift:

()

where b, c ≠ 0 are constants. The functions k(x), d(x) are continuous known functions belonging to the class {0}, k(x) is an even function. The functions

Taking the Fourier transform to Equation (4), we get the following equation:

()

Equation (5) can be written in the following form:

()

where

()

The solution F(s) of Equation (6) satisfies the condition F(0) = 0, so that the function D(s) must be satisfied the condition D(0) = 0.

By putting −s in Equation (5) instead of s, we obtain the following equation:

()

Equation (8) can be written in the following form:

()

We now solve Equations (6) and (9) for the functions F(s), F(−s).

Let:

()

Equations (6) and (9) have a unique solution if and only if the condition K(s) ≠ −b must be satisfied, hence Δ(s) > 0 and we obtain the following equations:

()

By the inverse of the Fourier transform, the solution of Equation (4) is given by the following equation:

in class {0}.

Theorem 10. In Equation (4), suppose the functions k(x), d(x) ∈ {0}, the function k(x) is an even function and , then K(s), D(s) ∈ {{0}} and from Lemma 7, we have D(s) sgns ∈ {{0}}. Hence from Equation (11), the function . Then, Equation (4) has a unique solution in class {0}.

Example. In Equation (4), suppose

then:

K(s), D(s) ∈ {{0}}, so k(x), d(x) ∈ {0}. From Equation (11), we get the following equation:

Hence, .

4. Dual SIEs With Carleman Shift

In this section, we consider the following dual SIEs with Carleman shift:

()

where

b_l, c_l, d_l, q_l, n_l, m_l are real constants, and the functions k_l(x), h_l(x) ∈ {0}, l = 1, 2.

To solve Equation (12), we suppose ψ ∈ {0} and

()

We can rewrite Equation (12) in the following form:

()

By applying the Fourier transforms to Equation (14), we obtain the following equation:

()

where D(s), Ψ^±(s) are the Fourier transforms of d(x), ψ_±(x), respectively, and

Suppose that:

()

the function Ψ(z) is a holomorphic function in the upper and lower half-planes. Since the function Ψ(z) ∈ {{0}} and Ψ(±∞) = 0 then by Sokhotski formulae [1], we have the following equation:

()

where Ψ^± are the boundary values of Ψ. Putting Equation (17) in Equation (15), we obtain the following equation:

()

By solving Equation (18) and eliminating F(±s), we obtain the SIEs:

()

where

and

By putting −s instead of s in Equation (19), we obtain a system of SIEs of two-dimensional in class {0} in the following form:

()

where

The solution of Equation (19) is the solution of Equation (20), we define the function as follows:

From [1], we have the following equation:

()

Putting Equation (21) in Equation (20), we obtain the following equation:

()

where

Let:

()

From Equation (22), we obtain the following equation:

()

where

The functions , and det Y(s) ≠ 0.

Let μ = ind Y(s),

is the canonical solution matrix of the homogenous of Equation (24) [1, 17, 19]. Since U⁺(∞) = U⁻(∞) = U(∞) = 0, the general solution of Equation (24) is given by the following equation [1]:

()

where P(z) is a two-dimensional polynomial vector.

We denote

, the two components of

. Hence from [1, 19], Equation (25) can be written in the following form:

()

where

is a polynomial of degree μ_j−1.

We assume that μ₁ ≥ μ₂, then the solution U(z) of Equation (26) has the following results.

i.
When μ₁ ≥ μ₂ ≥ 0, U(∞) = 0, then U(z) has μ₁ + μ₂ arbitrary constants.
ii.
When μ₁ ≥ 0 > μ₂, U(∞) = 0, then U(z) has μ₁ arbitrary constants and the following conditions of solvability must be satisfied

.
iii.
When 0 > μ₁ ≥ μ₂, U(∞) = 0, then U(z) has the following solvability conditions

.

From Equation (25), we obtain U⁺(t) and U⁻(t), and by substituting them in Equation (21), we get Φ(s) the solution of Equation (20). Hence, we obtain the solution Ψ(s) of Equation (19).

By substituting Ψ(s) in Equation (15) and similar to Section 3. we obtain the solution F(s) of Equation (15). Therefore, is a solution of Equation (12).

Example. In Equation (15), suppose:

then by taking the Fourier transform to Equation (15), we obtain the following equations:

where

Similarly, as in solving Equations (6) and (9), we obtain the following equation:

Hence, we get the solution

5. Wiener–Hopf Equation With Carleman Shift

In this section, we consider the following Wiener–Hopf equation with Carleman shift:

()

where b, c, d, n, and m are real constants, and the functions k(x), h(x), d(x) ∈ {0} and their Fourier transforms are

belong to {{0}}.

By extending t to t ∈ R⁻ in Equation (27), it can be written in the following form:

()

where

()

By using Fourier transform in Equation (28), we obtain the following equation:

()

We replace s by −s in Equation (30) and obtain the following equation:

()

To solve Equations (30) and (31), we define the sectionally holomorphic function:

in the upper and lower half-planes, F(±∞) = 0, F(z) ∈ {{0}}.

Substituting by,

in Equations (30) and (31), respectively, we obtain the following system:

()

where

and

The conditions of solvability of Equation (32) are as follows:

Similarly, as in the method used in Section 4 to solve Equation (20), we can find the solution F(s) of Equation (32). By using the inverse Fourier transform, we obtain the solution of Equation (27).

Example. In Equation (27), suppose:

Hence, we have the following equation:

Let F(s) be the boundary value of a function analytic in (0, ∞), then F⁺(s) = F(s) and F⁻(s) = 0. Hence, Equations (30) and (31) has the following form:

(c sgn(s))F(s) + (b + c sgn(s) + K(s))F(−s) = D(s).

Similarly, as in the solution of example in Sections 3, we obtain the following equation:

Hence, .

6. Conclusions

Some classes of SIEs of the convolution type with Carleman shift were studied in this paper. We used the theory of Fourier transforms to convert SIEs Equations (4), (12), and (27) into Riemann boundary value problems and obtained the exact solutions in class {0}. These classes of equations have important applications in practical problems, such as elastic mechanics, heat conduction, and electrostatics. Hence, the study about these classes of SIEs has an important meaning not only in application but also in the theory of resolving the equation itself. Many problems, such as piezoelectric material, voltage, magnetic materials, and functional gradient materials, can often attribute the problem to finding solutions for these classes of equations. Here, our method is different from the ones for the classical Riemann–Hilbert boundary value problems, and it is novel and effective. Thus, this paper generalizes the theory of classical Riemann–Hilbert boundary value problems and SIEs.

Conflicts of Interest

The authors declare no conflicts of interest.

Funding

The funding for this submission was received by the Mathematics Department, Faculty of Science, Zagazig University, Zagazig, Egypt.

Open Research

Data Availability Statement

The data that support the findings of this study are available upon request from the corresponding author.

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Singular Integral Equations of Convolution Type With Carleman Shift

Abstract

1. Introduction

2. Basic Definitions and Tools

3. SIEs With Carleman Shift

4. Dual SIEs With Carleman Shift

5. Wiener–Hopf Equation With Carleman Shift

6. Conclusions

Conflicts of Interest

Funding

Open Research

Data Availability Statement

References

References

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Singular Integral Equations of Convolution Type With Carleman Shift

Abstract

1. Introduction

2. Basic Definitions and Tools

3. SIEs With Carleman Shift

4. Dual SIEs With Carleman Shift

5. Wiener–Hopf Equation With Carleman Shift

6. Conclusions

Conflicts of Interest

Funding

Open Research

Data Availability Statement

References

References

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