SOLVING A DUAL INTEGRAL EQUATION INVOLVING FOURIER TRANSFORMS ENCOUNTERED IN A CRACK PROBLEMS FOR FRACTURE ELASTIC MATERIALS

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.. A method for redu

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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 6 x 6 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 equaSec-tion 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

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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

∂2Φ

∂x2 +

∂2Φ

∂y2 + β∂Φ

∂x = 0, β = const 6= 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

Ω0 = 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





−∂Φ

∂y

y=0 = f (x), x ∈ Ω, Φ

y=0 = 0, x ∈ Ω0,

(2.3)

where α, γ are certain real constants such that α2+ γ2> 0

This problem may be intepreted as a crack problem on the interval −a 6

x 6 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) ∈ L1(R)), direct and inverse Fourier transforms are defined by the formulas

b

f (ξ) = F [f ](ξ) =

Z ∞

−∞

˘

f (ξ) = F−1[f ](ξ) = 1

2π

Z ∞

−∞

f (x)e−ixξdx (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

∂2Φ(ξ, y)ˆ

∂y2 − (ξ2+ βiξ)Φ(ξ, y) = 0, (2.6) where bΦ(ξ, y) = Fx

b 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

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λ = λ(ξ) = √1

2

q p

ξ4+ β2ξ2+ ξ2+√i

2sgn(βξ)

q p

ξ4+ β2ξ2− ξ2, (2.8) where the sgn(.) is defined as

sgn(η) =





1, η > 0,

0, η = 0,

−1, η < 0

The value ˆΦ(0, y) is understood in the sense

b Φ(0, y) = lim

ξ→0 ˆ Φ(ξ, y)

We introduce notations

b u(ξ) := bΦ(ξ, 0) = A(ξ), u(x) := F−1[u](x),b (2.9) whereu(ξ) is an unknown function.b

Using boundary condition (2.2) and the relation (2.9), we express un-known functions A(ξ) and B(ξ) in terms ofu(ξ) andb g(ξ) We haveb

α sinh(λh) + γλ cosh(λh) −

α cosh(λh) + γλ sinh(λh)

α sinh(λh) + γλ cosh(λh)bu(ξ) (2.11) From (2.7), (2.10) and (2.11), we have

b

Φ(ξ, y) =u(ξ) cosh(λy) +b [bg(ξ) −u(ξ)(α cosh(λh) + γλ sinh(λh))] sinh(λy)b

α sinh(λh) + γλ cosh(λh)

=bu(ξ)α sinh λ(h − y) + γλ cosh λ(h − y)

α sinh(λh) + γλ cosh(λh) +bg(ξ) sinh(λy)

α sinh(λh) + γλ cosh(λh),

(2.12)

In order to determine the unknown function u(ξ), we use the conditionb (2.3) Sastifying the mixed conditions (2.3), we obtain the dual integral equation with respec to bu(ξ) = F [u](ξ) :

(

F−1L(ξ)bu(ξ)(x) =f (x), x ∈ Ω,e

F−1[bu(ξ)](x) = 0, x ∈ Ω0, (2.13) where

L(ξ) = λ(ξ)[α + γλ(ξ) tanh(λh)]

e

f (x) = f (x) + F−1G(ξ)bg(ξ), (2.15)

α sinh(λh) + γλ(ξ) cosh(λh). (2.16)

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µ = Reλ = √1

2

q p

ν = Imλ = √1

2sgn(βξ)

q p

ξ4+ β2ξ2− ξ2, (2.18)

D = α2(cosh 2µh − cos 2νh) + γ2(µ2+ ν2)(cosh 2µh + cos 2νh)

We have

L(ξ) = α

2(µ sinh 2µh + ν sin 2νh) + γ2(µ2+ ν2)(µ sinh 2µh − ν sin 2νh)

D +2αγ(µ

2cosh 2µh + ν2cos 2νh)

hα2(µ sinh 2µh + ν sin 2νh)

D +γ

2(µ2+ ν2)(ν sinh 2µh + µ sin 2νh) + 2αγµν(cosh 2µh − cos 2νh)

D

i (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 Hs := Hs(R)(s ∈ R) be the Sobolev-Slobodeskii space defined as a closure of the set Co∞(R) of infinitely differentiable func-tions with a compact support with respect to the norm [5,9]

||u||s:=

hZ ∞

−∞

(1 + |ξ|)2s|ˆu(ξ)|2dξ

i1/2

< ∞, u = F [u].ˆ (3.1) The space Hs is Hilbert with the following scalar product

(u, v)s:=

Z ∞

−∞

(1 + |ξ|)2su(ξ)b bv(ξ)dξ (3.2) Let Ω be an interval in R The subspace of Hs(R) consisting of functions u(x) with supp u ⊂ Ω is denoted by Hos(Ω) [5] while the space of functions v(x) = pu(x), where u ∈ Hs(R) and p is the restriction operator to Ω is denoted by Hs(Ω) The norm in Hs(Ω) is defined by

||v||Hs(Ω)= inf

l ||lv||s, where the infimum is taken over all possible extensions lv ∈ Hs(R)

Theorem 3.1 Let Ω ⊂ R, u ∈ Hs(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:=

Z ∞

−∞

b

lf (ξ) du(ξ)dξ (3.3) dos not depend on the choice of the extension lf Therefore, this formula defines a linear continuous functional on Hs

o(Ω) Conversely, for every linear

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continuous functional Φ(u) on Hos(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 Hs(Ω) into Hs−ε(Ω) is compact if and only if ε > 0

3.2 Pseudo-differential operators Consider pseudodifferential opera-tors of the form [5]

(Au)(x) := F−1[A(ξ)u(ξ)](x),b (3.4) whereu = F [u], u ∈ Hb 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 σα(R), if

|A(ξ)| 6 C1(1 + |ξ|)α, ∀ξ ∈ R, (3.5) and belongs to the class σ+α(R), if

C2(1 + |ξ|)α 6 A(ξ) 6 C1(1 + |ξ|)α, ∀ξ ∈ R, (3.6) where C1 and C2 are some positive constants

We shall need the following

Lemma 3.2 Assume that A(ξ) ∈ σα(R),u(ξ) ∈ Hb s(R), A(ξ)bu(ξ) ∈ S0(R) Then (Au)(x) ∈ Hs−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

2sign(βξ)

iii) ReL(ξ) > 0, ∀ξ ∈ R, ξ 6= 0, αγ ≥ 0 (α2+ γ2 > 0), (3.9)

By virtue of (3.8) and (3.10), using Lemma 3.2 we have

Theorem 3.2 Suppose that

g(x) ∈ H12(R) and u(x) := Φ(x, 0) ∈ H12(R) (3.11) Then we get

F−1[L(ξ)bu(ξ)](x) ∈ H−12(R), F−1[G(ξ)bg(ξ)](x) ∈ Ht(R), ∀t > 0 (3.12) Due to Theorem 3.2, besides the assumptions in (3.11), we suppose that

Then in virtue of (2.15) and (3.11) we have ef (x) ∈ H−1/2(Ω)

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3.3 Solvability of dual integral equations (2.13) Denote

(Lu)(x) = pF−1[L(ξ)u(ξ)](x), x ∈ Ωb (3.14) and rewite the (2.13) in the form

( (Lu)(x) = ef (x), x ∈ Ω.,

Our purpose now is to establish the uniqueness and existen theorems of solution for the pseudodifferential operator (3.15) in the space Ho1/2(Ω)

In virtue of Lemma 3.2 and 3.8, we have

L : Ho1/2(Ω) 7→ H−1/2(Ω) (3.16) Then we suppose the following assumption

Theorem 3.3 (Uniqueness) Let g(x) and f (x) satisfy the conditions (3.11) and (3.17), respectively, and αγ ≥ 0(α2+ γ2 > 0) Then equation (3.15) has at most one solution with respect to u = F−1[ˆu] ∈ Ho1/2(Ω)

We introduce notations

Lo(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),

L1(ξ) ∈ σ1+(R), Lo(ξ) ∈ σ0(R)

and

L(ξ) = L1(ξ) + Lo(ξ) (3.20)

(3.21) Lemma 3.4 Then the scalar and product in Hα/2(R) can be defined by the formulas

(u, v)L1,1/2=

Z ∞

−∞

L1(ξ)F [u](ξ)F [v](ξ)dξ, , (3.22)

||u||L1,1/2=

Z ∞

−∞

L1(ξ)|F [u](ξ)|2dξ

1/2

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) ∈ H1/2(R), and αγ ≥ 0 (α2 + γ2 > 0) Then the dual equation (3.15) has a unique solution u = F−1[ˆu] ∈ Ho1/2(Ω)

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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 Cm,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 6 1

The Cauchy principle value (CPV) of integrals are defined following [??,

??]

Definition 4.2 If φ(x) ∈ C0,α(c, d), then

−

Z d

c

dt

t − x = limε→+0

nZ x−ε c

dt

t − x +

Z d x+ε

dt

t − x

o

= logd − x

x − c, c < x < d,

(4.1)

−

Z d

c

φ(t)

t − xdt = limε→+0

nZ

|t−x|≥ε

φ(t) − φ(x)

t − x dt + φ(x)

Z

|t−x|≥ε

dt

t − x

o (4.2)

=

Z 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) ∈ C1,α(c, d), then

=

Z d

c

φ(t)

lim

ε→+0

nZ x−ε c

φ(t) (t − x)2dt +

Z d x+ε

φ(t) (t − x)2dt −2φ(x)

ε

o

(4.5)

= lim

ε→+0

nφ(c)

c − x −

φ(d)

d − x+

Z x−ε c

φ0(t)

t − xdt +

Z d x+ε

φ0(t)

t − xdt (4.6) + φ(x − ε) + φ(x + ε)

2φ(x) ε

o

(4.7)

= φ(c)

c − x −

φ(d)

d − x + φ

0(x) logd − x

x − c +

Z d c

φ0(t) − φ0(x)

t − x dt. (4.8) Theorem 4.1 [9] Let S0 and E0 be the spaces of generalized functions

of slow growth and of generalized functions with compact support in Rn, respectively Let f, ∈ S0 and g ∈ E0 Then their convolution f ∗ g ∈ S0, and its Fourier transform can be calculated from the formular

F [f ∗ g](ξ) = F [f ](ξ)F [g](ξ) = bf (ξ)bg(ξ) (4.9) From the (4.9) we have

F−1[ bfbg](x) = (f ∗ g)(x) (4.10)

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From the (4.9) we have

F−1[ bfbg](x) = (f ∗ g)(x) (4.11) Represent the symbol L(ξ) of the dual integral equation (2.13) in the form

where

L∞(ξ) = |ξ| + iβ

2

|ξ|

We have

F−1[L∞(ξ)](x) = 1

2π

− 2

x2 +β x

Using (4.11)-(4.15) we can prove the following Theorem

Theorem 4.2 The dual integral equation (2.13) with respct tou(ξ) is equiv-b alent to the following hypersingular integral equation

−1

π=

Z a

−a

u(t) (x − t)2dt + β

2π−

Z a

−a

u(t)

x − tdt +

1 2π

Z a

−a k(x − t)u(t)dt = f (x), (4.16) where

x ∈ (−a, a), u(x) = F−1[bu](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).

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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

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[4] P A Martin, Exact solution of a simple hypersingular integral equation, Fournal of Integral Equations and... 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... 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

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