Solvability of Some Integral Equations in Banach Space and Their Applications to the Theory of Viscoelasticity

Abstract

An integral equation of Volterra type with additional compact operator in Banach space is considered. A special case is an integral equation of contact problem that arises in theory of viscoelasticity of mixed Fredholm and Volterra type with spectral parameter depending on time. In case the initial value of the parameter coincides with some isolated point of the spectrum of compact operator, the conditions of solvability are established.

1. Introduction

We consider in an arbitrary complex Banach space 𝐁 the following integral equation:^𝑡₀𝐾(𝑡,𝑠)𝑢(𝑠)𝑑𝑠+𝐴𝑢(𝑡)−𝜆(𝑡)𝑢(𝑡)=𝑓(𝑡),𝑡≥0, (1.1) where 𝑢∶[0,+∞)→𝐁 is unknown function, 𝐴∶𝐁→𝐁 is a linear compact operator, 𝐾∶𝑄→ℂ is the kernel, 𝑓∶[0,+∞)→𝐁 is a given function, and 𝜆∶[0,+∞)→ℂ is a function which we may interpret as spectral parameter. We indicated above by 𝑄 the set𝑄=(𝑡,𝑠)∈ℝ²∶0≤𝑠≤𝑡<∞. (1.2) The main example is the following integral equation:^𝑡₀𝐾(𝑡,𝑠)𝑢(𝑥,𝑠)𝑑𝑠+_Ω𝑅𝑥,𝑦𝑢𝑦,𝑡𝑑𝑦−𝜆(𝑡)𝑢(𝑥,𝑡)=𝑓(𝑥,𝑡), (1.3) where 𝑥∈Ω and 𝑡≥0, which we consider in Banach spaces 𝐁=𝐿_𝑝(Ω) or 𝐁=𝐶(Ω). We suppose that the set Ω⊂ℝ^𝑛 is measurable in Lebesgue sense. The equations of such type are known as partial integral equations and were first considered by Salam [1] (see also [2], and books [3, 4]). The equation (1.3) arises in the theory of viscoelasticity [5] (see also [6]). The kernels 𝐾(𝑡,𝑠) and 𝑅(𝑥,𝑦) are connected with some elastic creeping base and 𝜆(𝑡) is the given value which describes the elastic properties of deformable body. We may refer also to work [7], where the more general integral equations in Hilbert space were considered. The main purpose of the present paper is to find conditions of solvability of (1.1) in case where 𝜆(0) coincides with some isolated pole of the resolvent 𝑅_𝜆(𝐴)=(𝐴−𝜆𝐼)⁻¹.

2. The Conditions for Solvability on the Spectrum

We suppose that 𝜆(𝑡) is a continuous function. Denote by Λ(𝑡) the range of the function 𝜆(𝑠) on the interval [0,𝑡]Λ(𝑡)={𝜆(𝑠)∶0≤𝑠≤𝑡}. (2.1) It is clear that because of continuity of the function 𝜆(𝑡) the set Λ(𝑡) for every 𝑡≥0 is closed. We say that 𝑢∈𝐶_𝐁[0,∞) if 𝑢 is an abstract function 𝑢∶[0,+∞)→𝐁, which is continuous on the half-line [0,∞), and set‖𝑢‖_𝑡=sup_{0≤𝑠≤𝑡}‖𝑢(𝑠)‖,𝑡≥0. (2.2) Denote by 𝜎(𝐴) the spectrum of a compact operator 𝐴 and consider for 𝜆∉𝜎(𝐴) the resolvent 𝑅_𝜆(𝐴)=(𝐴−𝜆𝐼)⁻¹ of the operator 𝐴. In case where Λ(𝑡)∩𝜎(𝐴)=⊘ for all 𝑡≥0 it is not difficult to show that (1.1) has continuous solution 𝑢(𝑡) for any continuous function 𝑓(𝑡). The problem is more complicated when Λ(𝑡) has a common point with spectrum of 𝐴 and that is the main idea of our consideration. Note that the case where 𝐴 is self-adjoint operator in Hilbert space was studied in [7–9]. In this paper we assume that 𝜆(0) coincides with one of the points 𝜆₀≠0 of the spectrum of the operator 𝐴. From the mechanical point of view it means that the initial state of the considered system coincides with resonance. The problem is how to change the function 𝜆(𝑡) for 𝑡>0 in order to get the existence and uniqueness of the solution. It was proved in [7] that the answer is almost obvious: 𝜆(𝑡) must go away from the spectrum as fast as possible. We suppose that 𝜆(𝑡) has 𝑚 continuous derivatives on the half line 𝑡≥0, where 𝑚 will be chosen below. The main assumption is the following:𝜆^′(0)≠0. (2.3) It is necessary to add some conditions to establish the one-valued solvability of (1.1). It was shown in [7] that one of these conditions is 𝐾(0,0)=0. Indeed, if 𝐾(0,0)≠0 then there exists the function 𝜆(𝑡), which satisfies condition (2.3), however the homogeneous equation^𝑡₀𝐾(𝑡,𝑠)𝑢(𝑠)𝑑𝑠+[𝐴−𝜆(𝑡)𝐼]𝑢(𝑡)=0 (2.4) has nontrivial solution.

2.1. Example

Let 𝜇≠0 be a simple eigenvalue of the operator 𝐴 and 𝑢_𝜇 be the corresponding eigenvector:𝐴𝑢_𝜇=𝜇𝑢_𝜇. (2.5)

Set 𝑢(𝑡)=𝑢_𝜇=const≠0. Then^𝑡₀𝐾(𝑡,𝑠)𝑢(𝑠)𝑑𝑠+[𝐴−𝜆(𝑡)𝐼]𝑢(𝑡)=𝑘(𝑡)+𝜇−𝜆(𝑡)𝑢_𝜇, (2.6) where𝑘(𝑡)=^𝑡₀𝐾(𝑡,𝑠)𝑑𝑠. (2.7)

Set𝜆(𝑡)=𝜇+𝑘(𝑡). (2.8)

In case where 𝐾(0,0)≠0 this function satisfies condition (2.3) since𝜆^′(0)=𝐾(0,0)≠0. (2.9)

It is clear that 𝑢(𝑡) is a solution of the homogeneous equation (2.4).

Let 𝑝 be a natural number. We assume that the kernel 𝐾(𝑡,𝑠) is defined on the whole plane ℝ², has all partial derivatives of order ≤𝑝, which are continuous on the plane ℝ² and the following conditions:𝜕^𝑖+𝑗𝐾𝜕𝑡^𝑖𝜕𝑠^𝑗(0,0)=0,0≤𝑖+𝑗≤𝑝−1, (2.10) are fulfilled. For example, 𝐾(𝑡,𝑠)=(𝑡±𝑠)^𝑝𝐹(𝑡,𝑠), where 𝐹(𝑡,𝑠) is an arbitrary smooth enough function on the plane ℝ².

Let 𝐴^∗∶𝐁^∗→𝐁^∗ be the adjoint operator, where 𝐁^∗ is a conjugate space (the space of linear continuous functionals 𝑣∶𝐁→ℂ). As usual, we setker𝐴^∗=𝑣∈𝐁^∗∶𝐴^∗𝑣=0. (2.11)

Further, we say that 𝑢⟂ker𝐴^∗ if 𝑢∈𝐁 and 𝑣(𝑢)=0 for all 𝑣∈ker𝐴^∗.

According to Fredholm-Riesz-Schauder theory, every nonzero point 𝜆₀ of the spectrum of 𝐴 is an eigenvalue anddimker𝐴−𝜆₀𝐼=dimker𝐴^∗−𝜆₀𝐼≥1. (2.12)

It is well known that the resolvent of the compact operator is meromorphic function and the nonzero eigenvalues of this operator coincide with poles of this function (see, e.g., [10], Chapter VIII, Section 8).

3. Solvability on the Spectrum

Definition 3.1. Let 𝑚 be a natural number and 𝜆₀ be an eigenvalue of the compact linear operator 𝐴. We say that 𝜆₀≠0 is the isolated point of the spectrum of the operator 𝐴 of the type 𝑚 if 𝜆₀ is the pole of the resolvent 𝑅_𝜆(𝐴)=(𝐴−𝜆𝐼)⁻¹ of the order 𝑚, that is, there exist 𝐶>0 and 𝛿>0 so that 𝑅_𝜆(𝐴≤𝐶)𝜆−𝜆₀^𝑚,0<𝜆−𝜆₀<𝛿. (3.1)

We apply also the additional restrictions on the function 𝑓 and its derivatives at the point 𝑡=0 and prove the following result. In what follows we assume that the condition (2.10) with some natural 𝑝 is fulfilled.

We say that the function 𝜆∶[0,+∞)→ℂ belongs to the space 𝐶^𝑚[0,∞), if this function has 𝑚 continuous derivatives for 𝑡≥0.

Theorem 3.2. Suppose that 𝜆∈𝐶^𝑚[0,∞) satisfies condition (2.3). Let 𝜆(0) be an isolated point of the spectrum 𝜎(𝐴) of the type 𝑚≤𝑝 and 𝜆(𝑡)∉𝜎(𝐴) for all 𝑡>0. If the function 𝑓 has continuous derivatives of order ≤ 𝑚 on the half-line 𝑡≥0, and the following conditions: 𝑓^(𝑘)(0)⟂ker𝐴^∗−𝜆₀𝐼,𝑘=0,1,…,𝑚−1, (3.2) are fulfilled, then the continuous solution of (1.1) exists and is unique.

To prove Theorem 3.2 we consider the following auxiliary equation:^𝑡₀𝐾(𝑡,𝑠)𝑅_𝜆(𝑠)(𝐴)𝑣(𝑠)𝑑𝑠+𝑣(𝑡)=𝑓(𝑡),𝑡≥0, (3.3) which is equivalent to (1.1). At first we find the estimate of resolvent 𝑅_𝜆(𝑡) near to point 𝜆₀=𝜆(0).

Lemma 3.3. Let 𝜆(0)=𝜆₀ be an isolated point of the spectrum 𝜎(𝐴) of the type 𝑚 and 𝜆(𝑡)∉𝜎(𝐴) for all 𝑡>0. If the condition (2.3) is fulfilled then for all 𝑇>0 the following inequality: 𝑅_𝜆(𝑡)(𝐴≤𝐶)^𝑇𝑡^𝑚,0<𝑡≤𝑇, (3.4) is valid.

Proof. It is not difficult to show that the estimate 𝜆(𝑡)−𝜆₀≥𝑐(𝑇)𝑡,0≤𝑡≤𝑇, (3.5) follows from assumption (2.3). According to condition (3.1), for all 𝜆 near to 𝜆₀ the estimate 𝑅_𝜆(𝑡)(𝐴≤𝐶)𝜆(𝑡)−𝜆₀^𝑚 (3.6) is valid, and (3.4) follows from (3.5) and (3.6).

Lemma 3.4. Let the conditions (2.10) be fulfilled. Then for 𝑚≤𝑝 the following equality 𝐾(𝑡,𝑠)=𝑡^𝑚𝐾_𝑚(𝑡,𝑠), (3.7) is valid, where the function 𝐾_𝑚(𝑡,𝑠) is bounded on every compact set 𝐸⊂𝑄.

Proof. The assertion follows from equality 𝐾(𝑡,𝑠)=_{𝑖+𝑗≤𝑚}1𝜕𝑖!𝑗!^𝑖+𝑗𝐾(0,0)𝜕𝑡^𝑖𝜕𝑠^𝑗𝑡^𝑖𝑠^𝑗+𝑅_𝑚()=𝑡,𝑠_{𝑖+𝑗=𝑚}1𝜕𝑖!𝑗!^𝑖+𝑗𝐾(0,0)𝜕𝑡^𝑖𝜕𝑠^𝑗𝑡^𝑖𝑠^𝑗+𝑅_𝑚(𝑡,𝑠)=𝑡^𝑚𝐾_𝑚(𝑡,𝑠). (3.8) We are taking into account that 0≤𝑠≤𝑡 and, hence, 𝑡^𝑖𝑠^𝑗=𝑡^𝑚⋅𝑠𝑡^𝑗=𝑂𝑡^𝑚,𝑖+𝑗=𝑚. (3.9)

Lemma 3.5. Let the function 𝑔 be continuous on the open half-line (0,+∞) and bounded on the closed half-line [0,+∞). Then for every 𝑇>0 the solution of the equation ^𝑡₀𝑠𝑡^𝑚𝐾(𝑡,𝑠)𝑅_𝜆(𝑠)(𝐴)𝑤(𝑠)𝑑𝑠+𝑤(𝑡)=𝑔(𝑡),0≤𝑡≤𝑇, (3.10) exists, is unique, continuous, and bounded on the interval (0,𝑇].

Proof. According to Lemma 3.4, we may state that the function 𝐾_𝑚(𝑡,𝑠)=𝑡^−𝑚𝐾(𝑡,𝑠) (3.11) is bounded for 0≤𝑠≤𝑡≤𝑇. Set 𝐵(𝑡)=𝑡^𝑚𝑅_𝜆(𝑡)(𝐴)for𝑡>0,𝐵(0)=0. (3.12)
Then, (3.10) takes the form ^𝑡₀𝐾_𝑚(𝑡,𝑠)𝐵(𝑠)𝑤(𝑠)𝑑𝑠+𝑤(𝑡)=𝑔(𝑡),0≤𝑡≤𝑇. (3.13)
Note that, according to Lemma 3.3, ‖𝐵(𝑡)𝑤(𝑡)‖≤𝐶‖𝑤(𝑡)‖,0≤𝑡≤𝑇. (3.14)
Now it is clear that the integral operator on the left side (3.13) is quasi-nilpotent. Hence (3.13) has required unique solution, which is given by Neumann series.

Lemma 3.6. Let 𝑓(𝑡)=𝑡^𝑚𝑔(𝑡), where the function 𝑔 is continuous on the half-line [0,∞). Then the continuous solution of (1.1) exists and is unique.

Proof. Set 𝑣(𝑡)=𝑡^𝑚𝑤(𝑡), (3.15) where 𝑤(𝑡) is a solution to (3.10). Then 𝑣(𝑡) is a solution to (3.3). Hence, the function 𝑢(𝑡)=𝑅_𝜆(𝑡)(𝐴)𝑣(𝑡) is the required solution of (1.1).
For an arbitrary 𝑇>0 this solution exists on the interval [0,𝑇], and because of uniqueness we may state that this solution belongs to 𝐶_𝐁[0,∞).

Lemma 3.7. Let 𝑓_𝑘∈𝐁 and 𝑓_𝑘⟂ker𝐴^∗−𝜆₀𝐼,𝑘=0,1,…,𝑚−1. (3.16)
Then there exist the elements 𝜃_𝑘𝑗∈𝐁, 0≤𝑘≤𝑚−1, 0≤𝑗≤𝑚−𝑘−1, so that 𝜃_𝑘𝑗⟂ker𝐴^∗−𝜆₀𝐼 (3.17) and the functions 𝑣_𝑘(𝑡)=𝑡^𝑘𝜃_𝑘0+𝑡𝜃_𝑘1+𝑡²𝜃_𝑘2+⋯+𝑡^{𝑚−𝑘−1}𝜃_{𝑘,𝑚−𝑘−1} (3.18) which satisfy the following equations: [𝐴−𝜆(𝑡)𝐼]𝑣_𝑘(𝑡)=𝑡^𝑘𝑓_𝑘+𝑡^𝑚𝛽_𝑘(𝑡), (3.19) where the functions 𝛽_𝑘(𝑡) are continuous on the half-line [0,+∞) and 𝛽_𝑘(𝑡)⟂ker𝐴^∗−𝜆₀𝐼,𝑡≥0. (3.20)

Proof. We construct the elements 𝜃_𝑘𝑗 using the back induction and begin with the case 𝑘=𝑚−1. It is clear that we may choose element 𝜃_𝑚−1⟂ker(𝐴^∗−𝜆₀𝐼) so that 𝐴−𝜆₀𝐼𝜃_𝑚−1=𝑓_𝑚−1. (3.21)
Set 𝑣_𝑚−1(𝑡)=𝑡^𝑚−1𝜃_𝑚−1. (3.22)
Then [𝐴−𝜆(𝑡)𝐼]𝑣_𝑚−1(𝑡)=𝐴−𝜆₀𝐼𝑣_𝑚−1(𝑡)−𝜆(𝑡)−𝜆₀𝑣_𝑚−1(𝑡)=𝑡^𝑚−1𝐴−𝜆₀𝐼𝜃_𝑚−1−𝜆(𝑡)−𝜆₀𝑡^𝑚−1𝜃_𝑚−1=𝑡^𝑚−1𝑓_𝑚−1+𝑡^𝑚𝛽_𝑚−1(𝑡), (3.23) where 𝛽_𝑚−1(𝑡)𝜆(𝑡)=−−𝜆₀𝑡𝜃_𝑚−1𝛽,𝑡>0,_𝑚−1(0)=−𝜆^′(0)𝜃_𝑚−1. (3.24)
It is clear that 𝛽_𝑚−1(𝑡) satisfies condition (3.20).
Now assume that Lemma 3.3 is valid for some 𝑘<𝑚 and prove it for 𝑘−1, that is, show that there exist 𝜃_{𝑘−1,𝑗} so that the function 𝑣_𝑘−1(𝑡)=𝑡^𝑘−1𝜃_𝑘−1,0+𝑡𝜃_𝑘−1,1+𝑡²𝜃_𝑘−1,2+⋯+𝑡^𝑚−𝑘𝜃_{𝑘−1,𝑘} (3.25) satisfies equation [𝐴−𝜆(𝑡)𝐼]𝑣_𝑘−1(𝑡)=𝑡^𝑘−1𝑓_𝑘−1+𝑡^𝑚𝛽_𝑘−1(𝑡). (3.26)
If we change 𝑓_𝑘 to 𝑓_𝑘−1 (3.18) then, we may state that there exists the function 𝑢_𝑘(𝑡) so that 𝑢_𝑘(𝑡)=𝑡^𝑘𝜃_𝑘0+𝑡𝜃_𝑘1+𝑡²𝜃_𝑘2+⋯+𝑡^{𝑚−𝑘−1}𝜃_{𝑘,𝑚−𝑘−1},[𝐴−𝜆(𝑡)𝐼]𝑢_𝑘(𝑡)=𝑡^𝑘𝑓_𝑘−1+𝑡^𝑚𝛽_𝑘(𝑡). (3.27)(3.28)
Set for 𝑡≥0𝑣_𝑘−1(𝑡)=𝑡^𝑘−1𝜃_𝑘0+𝑡𝜃_𝑘1+𝑡²𝜃_𝑘2+⋯+𝑡^{𝑚−𝑘−1}𝜃_{𝑘,𝑚−𝑘−1}+𝑡^𝑚−𝑘𝜃^∗, (3.29) where 𝜃^∗ will be defined below. Taking into account (3.27) we may write for 𝑡>0𝑣_𝑘−1(𝑡)=1𝑡𝑢_𝑘(𝑡)+𝑡^𝑚−1𝜃^∗. (3.30)
Then, according to (3.28), [𝐴−𝜆(𝑡)𝐼]𝑣_𝑘−1(𝑡)=1𝑡[𝐴−𝜆(𝑡)𝐼]𝑢_𝑘(𝑡)+𝑡^𝑚−1[𝐴−𝜆(𝑡)𝐼]𝜃^∗=𝑡^𝑘−1𝑓_𝑘−1+𝑡^𝑚−1𝛽_𝑘(𝑡)+𝑡^𝑚−1[𝐴−𝜆(𝑡)𝐼]𝜃^∗=𝑡^𝑘−1𝑓_𝑘−1+𝑡^𝑚−1𝛽_𝑘(𝑡)+𝑡^𝑚−1𝐴−𝜆₀𝐼𝜃^∗−𝑡^𝑚−1𝜆(𝑡)−𝜆₀𝜃^∗. (3.31)
If we choose 𝜃^∗⟂ker(𝐴^∗−𝜆₀𝐼) so that 𝐴−𝜆₀𝐼𝜃^∗=−𝛽_𝑘(0), (3.32) then [𝐴−𝜆(𝑡)𝐼]𝑣_𝑘−1(𝑡)=𝑡^𝑘−1𝑓_𝑘−1+𝑡^𝑚−1𝛽_𝑘(𝑡)−𝛽_𝑘(0)−𝑡^𝑚−1𝜆(𝑡)−𝜆₀𝜃^∗=𝑡^𝑘−1𝑓_𝑘−1+𝑡^𝑚𝛽_𝑘−1(𝑡), (3.33) where 𝛽_𝑘−1(𝑡)=𝛽_𝑘(𝑡)−𝛽_𝑘(0)𝑡−𝜆(𝑡)−𝜆₀𝑡𝜃^∗𝛽for𝑡>0,_𝑘−1(0)=𝛽^′_𝑘(0)−𝜆^′(0)𝜃^∗. (3.34)
It is clear that the function 𝑣_𝑘−1(𝑡) satisfies both conditions (3.25) and (3.26).

Now, let us give the proof of Theorem 3.2.

Proof. Set 𝑓_𝑘=𝑓^(𝑘)(0).𝑘! (3.35)
According to (3.2) these elements satisfy conditions (3.16) of Lemma 3.7. Let 𝑣_𝑘(𝑡) be the solutions of (3.19) from Lemma 3.7. Set 𝑣(𝑡)=^𝑚−1_𝑘=0𝑣_𝑘(𝑡). (3.36)
Then, according to Lemma 3.7, [𝐴−𝜆(𝑡)𝐼]𝑣(𝑡)=^𝑚−1_𝑘=0𝑡^𝑘𝑓^(𝑘)(0)𝑘!+𝑡^𝑚𝛽(𝑡), (3.37) where 𝛽(𝑡)∈𝐶[0,∞). Hence, ^𝑡₀𝐾(𝑡,𝑠)𝑣(𝑠)𝑑𝑠+[𝐴−𝜆(𝑡)𝐼]𝑣(𝑡)=^𝑚−1_𝑘=0𝑓^(𝑘)(0)𝑡𝑘!^𝑘+𝑡^𝑚ℎ(𝑡), (3.38) where ℎ(𝑡)=𝛽(𝑡)+^𝑡₀𝐾_𝑚(𝑡,𝑠)𝑣(𝑠)𝑑𝑠 (3.39) and the kernel 𝐾_𝑚(𝑡,𝑠) is defined by (3.7).
Set 𝑓(𝑡)=𝑓(𝑡)−^𝑡₀𝐾(𝑡,𝑠)𝑣(𝑠)𝑑𝑠−[𝐴−𝜆(𝑡)𝐼]𝑣(𝑡). (3.40)
According to (3.38), 𝑓(𝑡)=𝑓(𝑡)−^𝑚−1_𝑘=0𝑓^(𝑘)(0)𝑡𝑘!^𝑘−𝑡^𝑚ℎ(𝑡)=𝑡^𝑚𝑔(𝑡), (3.41) where, obviously, 𝑔∈𝐶[0,∞) and 𝑔(0)=𝑓^(𝑚)(0)(0).𝑚!−ℎ (3.42)
Consider the following equation: ^𝑡₀𝐾(𝑡,𝑠)𝑤(𝑠)𝑑𝑠+[𝐴−𝜆(𝑡)𝐼]𝑤(𝑡)=𝑓(𝑡), (3.43) where 𝑓(𝑡)=𝑡^𝑚𝑔(𝑡). According to Lemma 3.6, the solution 𝑤(𝑡) exists and is unique.
Recall that according to definition (3.40), the function 𝑣(𝑡) satisfies equation ^𝑡₀𝐾(𝑡,𝑠)𝑣(𝑠)𝑑𝑠+[𝐴−𝜆(𝑡)]𝑣(𝑡)=𝑓(𝑡)−𝑓(𝑡). (3.44)
Now we can prove the existence of the solution by setting 𝑢(𝑡)=𝑣(𝑡)+𝑤(𝑡). (3.45)
Uniqueness follows from Lemma 3.6.

3.1. Remark

Note that assumption (3.2) of Theorem 3.2 is important. If this is not satisfied then existence of the continuous solution is not guaranteed.

In order to show this, it is enough to consider the finite-dimensional operator with corresponding Jordan matrix whose diagonal and subdiagonal elements are equal to 1 and all other elements are equal to 0. The next example makes clear this statement.

3.2. Example

Let us consider (1.3) for Ω=[0,𝜋] and for kernel𝑅𝑥,𝑦=^𝑚_𝑘=1𝑔_𝑘𝑦sin𝑘𝑥,0≤𝑥≤𝜋,0≤𝑦≤𝜋, (3.46) where𝑔_𝑘𝑦=2𝜋sin𝑘𝑦+sin(𝑘−1)𝑦. (3.47)

Define the matrix ‖𝑎_𝑗𝑘‖ with elements𝑎_𝑗𝑘=^𝜋₀𝑔_𝑗(𝑥)sin𝑘𝑥𝑑𝑥. (3.48)

It is clear that𝑎_𝑗𝑗=1,𝑗=1,2,…,𝑚,𝑎_{𝑗(𝑗−1)}=1,𝑗=2,3,…,𝑚, (3.49) and 𝑎_𝑗𝑘=0 otherwise.

Let 𝑢(𝑥) be the solution to the equation(𝐴−𝜆𝐼)𝑢(𝑥)=𝑓(𝑥), (3.50) or𝜆𝑢(𝑥)=^𝑚_𝑘=1𝑢,𝑔_𝑘sin𝑘𝑥−𝑓(𝑥), (3.51) where, as usual,(𝑢,𝑣)=^𝜋₀𝑢(𝑥)𝑣(𝑥)𝑑𝑥. (3.52)

Set𝑐_𝑘=𝑢,𝑔_𝑘,𝑓_𝑘=𝑓,𝑔_𝑘. (3.53)

Then (3.51) takes the form𝜆𝑐_𝑗=^𝑚_𝑘=1𝑎_𝑗𝑘𝑐_𝑘−𝑓_𝑗,𝑗=1,2,…,𝑚. (3.54)

We suppose that 𝜆≠0 and 𝜆≠1. Then the solution of this algebraic system is𝑐_𝑘=−^𝑘_𝑗=1(𝜆−1)^{𝑗−𝑘−1}𝑓_𝑗,𝑘=1,2,…,𝑚. (3.55)

Hence, the solution of the integral equation (3.51) equals𝑢(𝑥)=1𝜆^𝑚_𝑘=1𝑐_𝑘1sin𝑘𝑥−𝜆𝑓(𝑥). (3.56)

Assume that𝑓(𝑥,𝑡)=𝜑(𝑡)sin𝑥. (3.57)

Then, obviously,𝑓₁(𝑡)=𝑓₂(𝑡)(𝑡),𝑓=𝜑₃=𝑓₄=⋅⋅⋅=𝑓_𝑚=0. (3.58)

In this case, according to (3.55),𝑐_𝑘𝜑(𝑡)=−(𝜆−1)^𝑘−𝜑(𝑡)(𝜆−1)^𝑘−1𝜆=−(𝜆−1)^𝑘𝜑(𝑡),𝑘=1,2,…,𝑚. (3.59)

Thus, the function𝑢(𝑥,𝑡)=−𝜑(𝑡)^𝑚_𝑘=1sin𝑘𝑥(𝜆−1)^𝑘−𝜑(𝑡)𝜆sin𝑥 (3.60) satisfies (3.50) with right-hand side 𝑓(𝑥,𝑡)defined by (3.57).

Now it is easy to construct (1.3) so that its right-hand side satisfies conditions (3.2) for 𝑘=1,2,…,𝑚−2 and does not satisfy this condition for 𝑘=𝑚−1, and because of that this equation has no continuous solution.

For example, in the simplest case 𝐾(𝑡,𝑠)≡0 we may set 𝜆(𝑡)=1+𝑡 and 𝜑(𝑡)=𝑡^𝑚−1. The function (3.57) equals 𝑓(𝑥,𝑡)=𝑡^𝑚−1sin𝑥 and satisfies conditions (3.2) for all 𝑘 except 𝑘=𝑚−1. According to (3.60), there exists only one solution𝑢()1𝑥,𝑡=−𝑡^𝑚_𝑘=1𝑡^𝑚−𝑘𝑡sin𝑘𝑥−^𝑚−11+𝑡sin𝑥, (3.61) and it is clear that this solution does not belong to 𝐶[0,∞).

3.3. Remark

We may note that the estimate of the solution may be obtained using the properties of some entire functions just as in [11].

Acknowledgment

The author is grateful to Professor Shavkat Alimov for his valuable assistance.