SOLVING A DUAL INTEGRAL EQUATION INVOLVING FOURIER TRANSFORMS ENCOUNTERED IN A CRACK PROBLEMS FOR FRACTURE ELASTIC MATERIALS NGUYEN VAN NGOC AND MA DINH TREN Abstract. The aim of the present work is to consider a mixed bound- ary value problem in a strip related to cracks of fracture materials. By the Fourier transform, the problem are reduced to studying a dual inte- gral equations on one edge of the strip. The uniqueness and existence theorems of solution of dual integral equation are established in appro- priate Sobolev spaces. A method for reducing the dual integral equation to a hypersingular integral equation of second order is also proposed. AMS Subject Classification: 45H05, 42A38, 46F05, 46F10, 47G30. Key words: Biharmonic equation, mixed boundary value problems of mathmatical physics, dual integral equations. 1. Introduction During the last decade, numerous publics about crack problems of fracture materials (see, [1], [2]). These problems often are reduced to hypersingular integral equations [3], [4]. Formal technique for solving such problems have been developed exten- sively, but their solvability problems have been not considered. The aim of the present work is to consider a mixed boundary value prob- lem of the linear elastic fracture mechanics in a strip by the dual integral equations method [6], [8]. The problem may be intepreted as a plane crack defect on the interval −a x a, y = 0. A siminary problem for the half-plane worse considered in [1]-[2]. For investigating this problem, we use the pseudo-differemtial equations approach [7] for investigation on solvability of a dual integral equations involving Fourier transform, which are medial equations between the initial mixed boundary value problems and hypersingular integral equations . Our work is constructed as follows: In Section 2 we formulate the mixed boundary value problem for the linear elastic fracture materials in a strip and reduce it to a dual integral equation involving Fourier transforms. Sec- tion 3 is intended for the solvability of dual integral equation with complex inccreasing symbol in appropriate Sobolev spaces. Finally in the last sec- tion, Section 4 we present a manners reducing the dual integral equation involving Fourier transforms to a hypersingular integral equation of second order. 13 .\WXF{QJWUuQKNKRDKeF3K?Q, 7UuQJ9LKeF7KQJ/RQJ 7UuQJ9LKeF7KQJ/RQJ 2. Formulations of the problem 2.1. Formulation of the mixed boundary value problem. We study the solution Φ(x, y) of a boundary-value problem for the biharmonic equa- tion ∂ 2 Φ ∂x2 + ∂2Φ ∂y 2 + β ∂Φ ∂x = 0, β = const = 0 (2.1) in the Π = {(x, y) : −∞ < x < ∞, 0 < y < h}. Let R be a real axis, Ω = (−a, a) denotes certain bounded interval and Ω = R \ Ω. Consider the following mixed boundary value problems. To find solution Φ(x, y) of the equation (2.1) in the strip Π that satisfies the boundary conditions αΦ y=h + γ ∂Φ ∂y y=h = g(x), x ∈ R, (2.2) − ∂Φ ∂y y=0 = f(x), x ∈ Ω, Φ y=0 = 0, x ∈ Ω , (2.3) where α, γ are certain real constants such that α 2 + γ 2 > 0. This problem may be intepreted as a crack problem on the interval −a x a, y = 0. 2.2. Reduction to a dual integral equation. We shall solve the for- mulated problems by the method of Fourier transforms and reduce it to a system of dual equations involving inverse Fourier transforms. For a suitable function f(x), x ∈ R( for example, f (x) ∈ L 1 (R)), direct and inverse Fourier transforms are defined by the formulas f(ξ) = F [f](ξ) = ∞ −∞ f(x)e ixξ dx, (2.4) ˘ f(ξ) = F −1 [f](ξ) = 1 2π ∞ −∞ f(x)e −ixξ dx. (2.5) The Fourier transforms of tempered generalized functions may be found, for example, in [5, 9]. Taking the Fourier transform with respect to the variable x for the equa- tion (2.1), we obtain ∂ 2 ˆ Φ(ξ, y) ∂y 2 − (ξ 2 + βiξ)Φ(ξ, y) = 0, (2.6) where Φ(ξ, y) = F x Φ(ξ, y) = A(ξ) cosh(λy) + B(ξ) sinh(λy), (2.7) where A(ξ), B(ξ) are arbitrary functions of the variable ξ, and [Φ(x, y)](ξ) is the Fourier transform with respect to x of the function Φ(x, y). The general solution of the differential equation (2.6) is taken in the form NGUYEN VAN NGOC AND MA DINH TREN 14 .\WXF{QJWUuQKNKRDKeF3K?Q, 7UuQJ9LKeF7KQJ/RQJ 7UuQJ9LKeF7KQJ/RQJ λ = λ(ξ) = 1 √ 2 ξ 4 + β 2 ξ 2 + ξ 2 + i √ 2 sgn(βξ) ξ 4 + β 2 ξ 2 − ξ 2 , (2.8) where the sgn(.) is defined as sgn(η) = 1, η > 0, 0, η = 0, −1, η < 0. The value ˆ Φ(0, y) is understood in the sense Φ(0, y) = lim ξ→0 ˆ Φ(ξ, y). We introduce notations u(ξ) := Φ(ξ, 0) = A(ξ), u(x) := F −1 [u](x), (2.9) where u(ξ) is an unknown function. Using boundary condition (2.2) and the relation (2.9), we express un- known functions A(ξ) and B(ξ) in terms of u(ξ) and g(ξ). We have A(ξ) = u(ξ), (2.10) B(ξ) = g(ξ) α sinh(λh) + γλ cosh(λh) − α cosh(λh) + γλ sinh(λh) α sinh(λh) + γλ cosh(λh) u(ξ). (2.11) From (2.7), (2.10) and (2.11), we have Φ(ξ, y) = u(ξ) cosh(λy) + [g(ξ) − u(ξ)(α cosh(λh) + γλ sinh(λh))] sinh(λy) α sinh(λh) + γλ cosh(λh) = u(ξ) α sinh λ(h − y) + γλ cosh λ(h − y) α sinh(λh) + γλ cosh(λh) + g(ξ) sinh(λy) α sinh(λh) + γλ cosh(λh) , . (2.12) In order to determine the unknown function u(ξ), we use the condition (2.3). Sastifying the mixed conditions (2.3), we obtain the dual integral equation with respec to u(ξ) = F [u](ξ) : F −1 L(ξ)u(ξ) (x) = f(x), x ∈ Ω, F −1 [u(ξ)](x) = 0, x ∈ Ω , (2.13) where L(ξ) = λ(ξ)[α + γλ(ξ) tanh(λh)] α tanh(λh) + γλ(ξ) , (2.14) f(x) = f(x) + F −1 G(ξ)g(ξ) , (2.15) G(ξ) = λ(ξ) α sinh(λh) + γλ(ξ) cosh(λh) . (2.16) SOLVING A DUAL INTEGRAL EQUATION 15 .\WXF{QJWUuQKNKRDKeF3K?Q, 7UuQJ9LKeF7KQJ/RQJ 7UuQJ9LKeF7KQJ/RQJ Denote µ = Reλ = 1 √ 2 ξ 4 + β 2 ξ 2 + ξ 2 , (2.17) ν = Imλ = 1 √ 2 sgn(βξ) ξ 4 + β 2 ξ 2 − ξ 2 , (2.18) D = α 2 (cosh 2µh − cos 2νh) + γ 2 (µ 2 + ν 2 )(cosh 2µh + cos 2νh) + 2γα(µ sinh 2µh + ν sin 2νh). (2.19) We have L(ξ) = α 2 (µ sinh 2µh + ν sin 2νh) + γ 2 (µ 2 + ν 2 )(µ sinh 2µh − ν sin 2νh) D + 2αγ(µ 2 cosh 2µh + ν 2 cos 2νh) D + i α 2 (µ sinh 2µh + ν sin 2νh) D + γ 2 (µ 2 + ν 2 )(ν sinh 2µh + µ sin 2νh) + 2αγµν(cosh 2µh − cos 2νh) D . (2.20) 3. Solvability of the dual integral equations Our aim in this Section is to establish the solvability of the dual series equation (??) in some appropriate Sobolev spaces. 3.1. Sobolev spaces. Let H s := H s (R)(s ∈ R) be the Sobolev-Slobodeskii space defined as a closure of the set C ∞ o (R) of infinitely differentiable func- tions with a compact support with respect to the norm [5,9] ||u|| s := ∞ −∞ (1 + |ξ|) 2s |ˆu(ξ)| 2 dξ 1/2 < ∞, ˆu = F [u]. (3.1) The space H s is Hilbert with the following scalar product (u, v) s := ∞ −∞ (1 + |ξ|) 2s u(ξ)v(ξ)dξ. (3.2) Let Ω be an interval in R. The subspace of H s (R) consisting of functions u(x) with supp u ⊂ Ω is denoted by H s o (Ω) [5] while the space of functions v(x) = pu(x), where u ∈ H s (R) and p is the restriction operator to Ω is denoted by H s (Ω). The norm in H s (Ω) is defined by ||v|| H s (Ω) = inf l ||lv|| s , where the infimum is taken over all possible extensions lv ∈ H s (R). Theorem 3.1. Let Ω ⊂ R, u ∈ H s (Ω), f ∈ H −s (Ω) and let lf be an extension of f from Ω to R belonging H −s (R). Then the integral [f, u] := (lf, u) o := ∞ −∞ lf (ξ) u(ξ)dξ (3.3) dos not depend on the choice of the extension lf. Therefore, this formula defines a linear continuous functional on H s o (Ω). Conversely, for every linear NGUYEN VAN NGOC AND MA DINH TREN 16 .\WXF{QJWUuQKNKRDKeF3K?Q, 7UuQJ9LKeF7KQJ/RQJ 7UuQJ9LKeF7KQJ/RQJ continuous functional Φ(u) on H s o (Ω) there exists an element f ∈ H −s (Ω) such that Φ(u) = [f, u] and ||Φ|| = ||f || H −s (Ω) . Lemma 3.1. Let Ω be a bounded subset of intervals in R. Then the imbed- ding H s (Ω) into H s−ε (Ω) is compact if and only if ε > 0. Proof. See [10]. 3.2. Pseudo-differential operators. Consider pseudodifferential opera- tors of the form [5] (Au)(x) := F −1 [A(ξ)u(ξ)](x), (3.4) where u = F [u], u ∈ H s (R). The function A(ξ) is called the symbol of the pseudodifferential operator Au. Definition 3.1. Let α ∈ R. We say that the function a(ξ) belongs to the class σ α (R), if |A(ξ)| C 1 (1 + |ξ|) α , ∀ξ ∈ R, (3.5) and belongs to the class σ α + (R), if C 2 (1 + |ξ|) α A(ξ) C 1 (1 + |ξ|) α , ∀ξ ∈ R, (3.6) where C 1 and C 2 are some positive constants. We shall need the following Lemma 3.2. Assume that A(ξ) ∈ σ α (R), u(ξ) ∈ H s (R), A(ξ)u(ξ) ∈ S (R). Then (Au)(x) ∈ H s−α (R), where (Au)(x) is defined by the equation (3.4). For the symbols of the dual integral equations (2.13) the following affir- mations hold Lemma 3.3. Let L(ξ) be determined by the formula (2.14). Then i) L(ξ) = O |ξ| + i 2 sign(βξ) , |ξ| → ∞, (3.7) ii) L(ξ) ∈ σ 1 (R) ∩ C ∞ (R), (3.8) iii) ReL(ξ) > 0, ∀ξ ∈ R, ξ = 0, αγ ≥ 0 (α 2 + γ 2 > 0), (3.9) iv) G(ξ) ∈ σ −p (p >> 1). (3.10) By virtue of (3.8) and (3.10), using Lemma 3.2 we have. Theorem 3.2. Suppose that g(x) ∈ H 1 2 (R) and u(x) := Φ(x, 0) ∈ H 1 2 (R). (3.11) Then we get F −1 [L(ξ)u(ξ)](x) ∈ H − 1 2 (R), F −1 [G(ξ)g(ξ)](x) ∈ H t (R), ∀t > 0. (3.12) Due to Theorem 3.2, besides the assumptions in (3.11), we suppose that f(x) ∈ H −1/2 (Ω). (3.13) Then in virtue of (2.15) and (3.11) we have f(x) ∈ H −1/2 (Ω). SOLVING A DUAL INTEGRAL EQUATION 17 .\WXF{QJWUuQKNKRDKeF3K?Q, 7UuQJ9LKeF7KQJ/RQJ 7UuQJ9LKeF7KQJ/RQJ 3.3. Solvability of dual integral equations (2.13). Denote (Lu)(x) = pF −1 [L(ξ)u(ξ)](x), x ∈ Ω (3.14) and rewite the (2.13) in the form (Lu)(x) = f(x), x ∈ Ω., u ∈ H 1/2 ◦ (Ω). (3.15) Our purpose now is to establish the uniqueness and existen theorems of solution for the pseudodifferential operator (3.15) in the space H 1/2 o (Ω). In virtue of Lemma 3.2 and 3.8, we have L : H 1/2 o (Ω) → H −1/2 (Ω). (3.16) Then we suppose the following assumption f(x) ∈ H −1/2 (Ω). (3.17) Theorem 3.3. (Uniqueness). Let g(x) and f (x) satisfy the conditions (3.11) and (3.17), respectively, and αγ ≥ 0(α 2 + γ 2 > 0). Then equation (3.15) has at most one solution with respect to u = F −1 [ˆu] ∈ H 1/2 o (Ω). We introduce notations L 1 (ξ) = Reλ(ξ) + 1, (3.18) L o (ξ) = iImλ(ξ) −1 + L(ξ) −λ(ξ). (3.19) It is not difficult show that L(ξ) − λ(ξ) = λ(ξ) [α − γλ][1 − tanh(λh)] α tanh(λh) + γλ ∈ σ −p (∀p > 1), L 1 (ξ) ∈ σ 1 + (R), L o (ξ) ∈ σ 0 (R) and L(ξ) = L 1 (ξ) + L o (ξ). (3.20) (3.21) Lemma 3.4. Then the scalar and product in H α/2 (R) can be defined by the formulas (u, v) L 1 ,1/2 = ∞ −∞ L 1 (ξ)F [u](ξ)F [v](ξ)dξ, , (3.22) ||u|| L 1 ,1/2 = ∞ −∞ L 1 (ξ)|F [u](ξ)| 2 dξ 1/2 . (3.23) Proof. The proof is based on the relations (3.1), (3.2) and (3.6). Theorem 3.4. (Existepnce). Let f(x) belong to H −1/2 (Ω), g(x) ∈ H 1/2 (R), and αγ ≥ 0 (α 2 + γ 2 > 0). Then the dual equation (3.15) has a unique solution u = F −1 [ˆu] ∈ H 1/2 o (Ω). NGUYEN VAN NGOC AND MA DINH TREN 18 .\WXF{QJWUuQKNKRDKeF3K?Q, 7UuQJ9LKeF7KQJ/RQJ 7UuQJ9LKeF7KQJ/RQJ 4. Reduction to a hypersingular integral equation The aim of this Section is to propose a method for reducing the dual integral equation (2.13) to an hypersingular integral equation. Definition 4.1. The Holder space C m,α (c, d), where (c, d) is an interval on R, and m ≥ 0 an integer, consists of those functions on (c, d) having continuous derivatives up to order m are Holder continuous with exponent α, where 0 < α 1. The Cauchy principle value (CPV) of integrals are defined following [??, ??] Definition 4.2. If φ(x) ∈ C 0,α (c, d), then − d c dt t − x = lim ε→+0 x−ε c dt t − x + d x+ε dt t − x = log d − x x − c , c < x < d, (4.1) − d c φ(t) t − x dt = lim ε→+0 |t−x|≥ε φ(t) − φ(x) t − x dt + φ(x) |t−x|≥ε dt t − x (4.2) = d c φ(t) − φ(x) t − x dt + φ(x) log d − x x − c , c < x < d (4.3) Finite part (HFP) integral is defined by [4]: Definition 4.3. If φ(x) ∈ C 1,α (c, d), then = d c φ(t) (t − x) 2 dt (c < x < d) := (4.4) lim ε→+0 x−ε c φ(t) (t − x) 2 dt + d x+ε φ(t) (t − x) 2 dt − 2φ(x) ε (4.5) = lim ε→+0 φ(c) c − x − φ(d) d − x + x−ε c φ (t) t − x dt + d x+ε φ (t) t − x dt (4.6) + φ(x − ε) + φ(x + ε) ε − 2φ(x) ε (4.7) = φ(c) c − x − φ(d) d − x + φ (x) log d − x x − c + d c φ (t) − φ (x) t − x dt. (4.8) Theorem 4.1. [9]. Let S and E be the spaces of generalized functions of slow growth and of generalized functions with compact support in R n , respectively. Let f, ∈ S and g ∈ E . Then their convolution f ∗ g ∈ S , and its Fourier transform can be calculated from the formular F [f ∗ g](ξ) = F [f](ξ)F [g](ξ) = f(ξ)g(ξ). (4.9) From the (4.9) we have F −1 [ fg](x) = (f ∗ g)(x). (4.10) SOLVING A DUAL INTEGRAL EQUATION 19 .\WXF{QJWUuQKNKRDKeF3K?Q, 7UuQJ9LKeF7KQJ/RQJ 7UuQJ9LKeF7KQJ/RQJ From the (4.9) we have F −1 [ fg](x) = (f ∗ g)(x). (4.11) Represent the symbol L(ξ) of the dual integral equation (2.13) in the form L(ξ) = L ∞ (ξ) + K(ξ), (4.12) where L ∞ (ξ) = |ξ|+ iβ 2 |ξ| ξ , (4.13) K(ξ) = L(ξ) − L ∞ (ξ). (4.14) We have F −1 [L ∞ (ξ)](x) = 1 2π − 2 x 2 + β x , . (4.15) Using (4.11)-(4.15) we can prove the following Theorem. Theorem 4.2. The dual integral equation (2.13) with respct to u(ξ) is equiv- alent to the following hypersingular integral equation − 1 π = a −a u(t) (x − t) 2 dt + β 2π − a −a u(t) x − t dt + 1 2π a −a k(x −t)u(t)dt = f(x), (4.16) where x ∈ (−a, a), u(x) = F −1 [u](x) = Φ(x, +0), k(x) = F −1 [K(ξ)], (4.17) References [1] Y S. Chan, A. C., Fannjiang,G.H. Paulino, Integral Equations with Hypresingular Kernel- Theory and Applications to Fracture Mechanics, it INt J. of Eng. Sci., 41, (2003) 683-720. [2] Y S. Chan, G.H. Paulino, A. C., Fannjiang, The crack problem for nonhomogeneous materials under antiplane shear loading-A displacement based formulation, Int. J. Solids Struct, 38 (2001) 2989-3005. [3] L.K. Lifanov, I. N. Poltavskii and G. N. Vaniko, HYpersingular Integral Equations and Their Applications, CRC, 2004. [4] P. A. Martin, Exact solution of a simple hypersingular integral equation, Fournal of Integral Equations and Applications bf 4(1992) 197-204. [5] G. I. Eskin, Boundary Value Problems for Elliptic Pseudodifferential Equations, Nauka, Moscow, 1973 (in Russian). [6] Mandal B.N., Advances in dual integral equations, Chapman & Hall / CRC Press, Boca Raton, 1999, 226 p. [7] Nguyen Van Ngoc, On the solvability of dual integral equations involving Fourier Transforms, Acta Math. Vietnamica, 13(2)(1988), 21-30. [8] Ia. S. Ufliand, Method of Dual Equations in Problems of Mathematical Physics, Leningrad, Nauka, 1977 (in Russian). [9] V. S. Vladimirov , Generalized Functions in Mathematical Physics, Moscow, Mir, 1979 (in Russian). [10] L. R. Volevich and B. P. Panekh , Some spaces of generalized functions and imbedding theorems, Uspekhii Math. Nauk, 20 (1) (1965), 3-74(in Russian). NGUYEN VAN NGOC AND MA DINH TREN 20 .\WXF{QJWUuQKNKRDKeF3K?Q, 7UuQJ9LKeF7KQJ/RQJ 7UuQJ9LKeF7KQJ/RQJ Nguyen Van Ngoc Hanoi Institute of Mathematics 18 Hoang Quoc Viet Road P.O. Box 10307, BoHo, Hanoi, Vietnam E-mail address: nvngoc@math.ac.vn SOLVING A DUAL INTEGRAL EQUATION 21 .\WXF{QJWUuQKNKRDKeF3K?Q, 7UuQJ9LKeF7KQJ/RQJ 7UuQJ9LKeF7KQJ/RQJ . SOLVING A DUAL INTEGRAL EQUATION INVOLVING FOURIER TRANSFORMS ENCOUNTERED IN A CRACK PROBLEMS FOR FRACTURE ELASTIC MATERIALS NGUYEN VAN NGOC AND MA DINH TREN Abstract. The aim of the. constructed as follows: In Section 2 we formulate the mixed boundary value problem for the linear elastic fracture materials in a strip and reduce it to a dual integral equation involving Fourier transforms. . investigation on solvability of a dual integral equations involving Fourier transform, which are medial equations between the initial mixed boundary value problems and hypersingular integral equations . Our