Volume 13, MAIN ARTICLES

Transcription

1 Voume 13, MAIN ARTICLES THE BASIC BVPs OF THE THEORY OF ELASTIC BINARY MIXTURES FOR A HALF-PLANE WITH CURVILINEAR CUTS Bitsadze L. I. Vekua Institute of Appied Mathematics of Iv. Javakhishvii Tbiisi State University Phone: (+99532) , E-mai: amarabits@yahoo.com Abstract. The first and second boundary vaue probems of the theory of eastic binary mixtures for a transversay isotropic haf-pane with curviinear cuts are investigated. The sovabiity of a system of singuar integra equations is proved by using the potentia method and the theory of singuar integra equations. Key words: Eastic mixture, uniqueness theorem, potentia method, expicit soution. MSC 2000 : 74E30, 74G05. Introduction. In the present paper the first and second boundary vaue probems (BVPs) of eastic binary mixture theory are investigated for a transversay-isotropic haf pane with curviinear cuts. The boundary vaue probems of easticity for anisotropic media with cuts were considered in [1, 2]. In this paper we extend this resut to BVPs of eastic mixture for a transversayisotropic eastic body. Here we sha be concerned with the pane probem of eastic binary mixture theory (it is assumed that the second components u 2 and u 2 of the three-dimensiona partia dispacement vectors u (u 1, u 2, u 3) and u (u 1, u 2, u 3) are equa to zero, whie the components u 1, u 3, u 1, u 3 depend ony on the variabes x 1, x 3 ). A soution of the first BVP is sought in the form of a doube-ayer potentia whie a soution of the second BVP is sought in the form of a singe-ayer potentia. For the unknown density we obtain a system of singuar integra equations. Using the potentia method and the theory of singuar integra equations we rigorousy prove the sovabiity of system of singuar integra equations, corresponding to the boundary vaue probems. The basic homogeneous equations of statics of the transversay isotropic eastic binary mixtures theory in the case of pane deformation can be written in the form [3] C( x)u = ( C (1) ( x) C (3) ( x) C (3) ( x) C (2) ( x) ) U = 0, (1) where the components of the matrix C () ( x) = C () pq ( x) 2x2 are given in the

2 2 Buetin of TICMI form C pq () ( x) = C qp () ( x), = 1, 2, 3, p, q = 1, 2, C () 11 ( x) = c () 11 C () 12 ( x) = (c () 13 + c () 44 ) 2, C () 22 ( x) = c () 44 x 1 x 3 2 x c () 33 2 x x 2 1, + c () 44 2 x 2 3, c (k) pq -are constants, characterizing the physica properties of the mixture and satisfying certain inequaities caused by the positive definiteness of potentia energy. U = U T (x) = (u, u ) is four-dimensiona dispacement vector-function, u (x) = (u 1, u 3) and u (x) = (u 1, u 3) are partia dispacement vectors depending on the variabes x 1, x 3. Evrerywhere beow by T we denote transposition. Let D be the haf-pane x 3 < 0 with the boundary x 3 = 0 and suppose that the boundary of the haf-pane is fastened. Let s assume that in D we have p curviinear cuts, = a b, = 1, 2,..., p, which are simpe reativey nonintersecting open Lyapunov arcs having no common points, and do not intersect the boundary. The positive direction on is chosen from the point a to the point b. The norma on is direct to the right with respect to the positive motion of direction. = p. We suppose that D is fied with binary transversay-isotropic eastic mixture. We introduce the notation z = x 1 + ix 3, ζ k = y 1 + α k y 3, t k = t 1 + α k t 3, σ k = z k ς k, z k = x 1 + α k x 3, t = t 1 + t 3. The basic boundary vaue probems of static of the theory of eastic binary mixtures are formuated as foows: Probem 1. Find a reguar soution of the equation (1) in D, when the boundary vaues of the dispacement vector are given on both sides of the, = 1, 2,..., p and on the boundary x 3 = 0. Let s aso assume, that the principa vector of externa force acting on, stress vector and the rotation at infinity are zero. It is required to define the deformed state of the pane. If we denote by U + (U ) the imits of U on from the eft (right), then the boundary conditions of the probem wi take the form U + (t 0 ) = f + (t 0 ), U (t 0 ) = f (t 0 ), t 0, U = 0, x 3 = 0, (2) where f +, and f are the given functions on satisfying Höder s conditions on the cuts, having derivatives in the cass H (for the definitions of the casses H and H see[4]) and satisfying the foowing conditions on the ends a and b of f + (a ) = f (a ), f + (b ) = f (b ). Probem 2. Find a reguar soution of the equation (1) in D, when the stress vector is given on both sides of the, = 1, 2,..., p and the boundary x 3 = 0 is traction free. In addition it is assumed that the principa vector of externa force acting on, stress vector and the rotation at infinity are zero. The

3 Voume 13, boundary conditions can be written as foows: [T U] + (t 0 ) = f + (t 0 ), [T U] (t 0 ) = f (t 0 ), t 0, [T U] = 0, x 3 = 0, < x 3 < +, (3) where f +, and f are the given known vector-functions on of the Hoder cass H, which have derivatives in the cass H and satisfying at the ends a and b of, the conditions f + (a ) = f (a ), f + (b ) = f (b ). Therefore, it is interesting to study the behavior of the soution of the probem in the neighborhood of cuts. The first BVP for the haf-pane with curviinear cuts. We seek for a soution of the probem in the form [5], U(z) = 1 π Im n(t k z k ) E (k) [g(s) + ih(s)]ds+ + 1 π Im E (k) E n(t k z k ) [g(s) ih(s)]ds + p V (z), (4) where E (k) = pq A 1 4x4 denotes a specia matrix that reduces the first BVP to a Fredhom integra equation of second order, = pq 4x4. The eements of the matrix E (k) and the matrix A 1 are defined as foows A 1 = A i = 4 A A A A 24 1 A A A A 22 1 i, A(k) i = i, 11 = d k [c (2) 11 q 4 + αk 2t 11 + αk 4t 12 + c (2) 44 q 3 αk 6], 22 = d k [q 1 c (2) 44 + α 2 k t 44 + α 4 k t 42 + c (2) 33 q 4 α 6 k ], 33 = d k [c (1) 11 q 4 + α 2 k t 33 + α 4 k t 23 + c (1) 44 q 3 α 6 k ], A ((k)) 44 = d k [q 1 c (1) 44 + α 2 k t 55 + α 4 k t 52 + c (1) 33 q 4 α 6 k ], 12 = d k α k [v 12 + v 11 α 2 k + v 13α 4 k ], 14 = d k α k [w 11 + w 12 αk 2 + w 13αk 4 ], k = 3, 4, 23 = d k α k [v 22 α k + v 21 α 2 k + v 23α 4 k ], 34 = d k α k [w 24 + w 14 αk 2 + w 34a 4 k ], k = 3, 4, 5, 6, 24 = d k [ q 1 c (3) 44 + α 2 k t 66 + α 4 k t 62 c (3) 33 q 4 α 6 k ], 13 = d k [ q 4 c (3) 11 + α 2 k t 22 + α 4 k t 13 c (3) 33 q 3 α 6 k ],, (5)

4 4 Buetin of TICMI 1 = a 1 a 2 a 3 a 4 2, 2 = A 22 A 44 A 2 24, 1 = A 11 A 33 A 2 13, t 11 = c (2) 11 δ 22 + c (2) 44 q 4 α (3) c (3) 44 α (2) 13 α (2)2 t 12 = c (2) 44 δ 22 + c (2) 11 q 3 α (3) c (3) 33 α (2) 13 α (2)2 33, t 22 = c (3) 11 δ 22 c (3) 44 q 4 + α (1) 44 c (3) 44 (α (2) 13 + α (3)2 13 ) + α (2) t 13 = c (3) 44 δ 22 c (3) 11 q 3 + α (1) 33 c (3) 33 (α (2) 13 + α (3)2 13 ) + α (2) 33, t 52 = t 33 c (1) 11 δ 22 + c (1) 33 δ 11, t 62 = t 22 c (3) 11 δ 22 + c (3) 33 δ 11, t 42 = t 11 c (2) 11 δ 22 + c (2) 33 δ 11, t 44 = c (2) 44 δ 11 + c (2) 33 q 1 α (3) c (3) 11 α (2) 11, t 66 = c (3) 44 δ 11 c (3) 33 q 1 + α (2) 11 c (3) 11 (α (2) 13 + α (3)2 13 ) + α (1) 11, t 23 = c (1) 44 δ 22 + c (1) 11 q 3 α (1) c (3) 33 α (1) 33, t 33 = c (1) 11 δ 22 + c (1) 44 q 4 α (1) c (3) 44 α (1) t 55 = c (1) 44 δ 11 + c (1) 33 q 1 α (1) c (3) 11 α (1) 11, v 11 = α (2) 13 (α (2) 13 ) α (1) c (2) 33 ) α (2) 13 (c (3)2 33 ) +α (3) 13 (2c (2) 44 + c (2) w 12 = α (2) 13 (α (2) 13 ) α (3) 44 c (2) 44 + c (2) 2 33 ) +α (2) 33 ) + α (1) 44 + c (2) v 21 = α (3) 13 (α (2) 13 ) α (3) 44 c (1) c (1) 33 ) +α (2) 44 + c (1) 33 ) + α (1) w 14 = α (1) 13 (α (2) 13 ) + α (3) 13 (2c (1) c (1) 33 ) α (1) 13 (c (3)2 33 ) α (2) c (1) v 12 = α (1) 44 c (2) 11 + α (3) 44 ) α (2) v 13 = α (1) 44 c (2) 33 + α (3) 33 c (3) 33 c (2) 44 ) α (2) 33 c (3) w 11 = α (1) 44 c (2) 11 α (3) 44 ) + α (2) w 13 = α (1) 33 α (3) 33 c (2) 33 c (3) 44 ) + α (2) 44 c (1) 11, v 22 = α (1) 11 α (3) 44 ) + α (2) v 23 = α (1) 44 c (2) 33 α (3) 33 c (1) 33 c (3) 44 ) + α (2) 33 c (1) w 24 = α (1) 11 + α (3) 13 (c (3) 44 + c (1) 44 ) α (2)

5 Voume 13, w 34 = α (1) 33 + α (3) 33 c (3) 33 c (1) 44 ) α (2) 33 c (1) α k, k = 1, 2, 3, 4, are the roots of the characteristic equation ([6]). g and h are the unknown rea vectors from the Hoder cass that have derivatives in the cass H. We write V (z) = 1 π Im a (k) A (z (k) k b (k) )n(z k b (k) ) (z k a (k) )n(z k a (k) ) K, b (k) a (k) = Re a + α k Im a, b (k) = Re b + α k Im b. K (), = 1,..., p, are the unknown rea constant vectors to be defined ater on. The vector V (z) satisfies the foowing conditions: 1. V (z) has the ogarithmic singuarity at infinity V = 1 π Im ( n z k + 1)K + O(z 1 k ). 2. Under V is supposed a branch, which is uniquey defined on the cut pane aong. 3. V is continuousy continued function on from the eft and right, incuding the end points a and b, i. e., we have the equaities V + V + (a ) = V (a ), V + (b ) = V (b ), V = 2 Re 4 A τ (k) k a (k) b (k) It is easiy to check U = 0, x 3 = 0, and a (k) K, = 1,..., p. + (T U) ds + [(T U) + (T U) ]ds = 0. To define the unknown density, we obtain by virtue of (4)-(2), the foowing system of singuar integra equation of the norma type where 2g(τ) = f f + Re 1 π h(t)dt + 1 t t 0 π p A τ (k) k a (k) b (k) a (k) K, K(t 0, t)dt = Ω(t 0 ), t 0, (6) K(t 0, t) = ie θ +

6 6 Buetin of TICMI +Re 4 E (k) n(1 + λ t t 0 k ), Re E (k) E q t t 0 n(t q t k0 ), q=1 Ω(t 0 ) = 1 2 (f + + f ) 1 p (V + + V 2 ) 1 [E θ π (t) + Im E (k) (t) n(1 + λ t t 0 k ) t t 0 +Im 4 E (k) E q (t) n(t q t k0 )]g(t)ds, q=1 λ k = 1 + iα k, θ = arg(t t 0 ), t k0 = Re t 0 + α k Im t 0, t = y 1 + iy 3. 1 iα k Thus we defined vector g on. It is not difficut to verify, that g H, g H, g(a ) = g(b ) = 0, Ω H, Ω H. Formua (6) is a system of singuar integra equations of the norma type with respect to the vector h. The points a and b are nonsinguar ones, whie the summary index of the cass h 2p is equa to 4p (for the definition of the cass h 2p see [4]). If the soution of equation (6) on the cass h 2p exists, it wi satisfy Hoder s condition on, vanishing on the points a and b and having to derivatives in the cass H. Let s prove that the homogeneous equation corresponding to the system of equation (6) ony has a trivia soution h 0 in the cass h 2p. Let s assume the contrary. Let h 0 be nontrivia soution of the homogeneous system in the cass h 2p and construct the potentia U 0 (z) = 1 π Re n (t k z k ) E (k) [g(t) + ih(t)]ds + 1 π Im E (k) E n (t z k ) [g(t) ih(t)]ds. It is cear U + 0 = U 0 = 0, t and on the basis of uniqueness theorem U 0 = 0, z D. Therefore T U 0 = 0, z D, and (T U 0 ) + (T U 0 ) = 2A 1 h 0 = 0. Consequenty h 0 = 0, since h 0 (a ) = 0, that competes the proof. Thus the corresponding adoint homogeneous equation has 4p ineary independent soution σ, = 1,..., 4p in the adoined cass and the conditions of sovabiity are Ωσ ds = 0. From ast equaity we get 4p agebraic equation with respect to K. On the basis of the uniqueness theorem it is not difficut to prove that the ast system is sovabe.

7 Voume 13, The second BVP for the haf-pane with curviinear cuts. We seek for a soution of the above formuated second BVP for the haf-pane in the form [5] U(z) = 1 π Im L (k) il R(k)iL( T n (z k t )[g(s) ih(s)]ds), where g and h are unknown rea vector-functions. n (t k z k )[g(s) + ih(s)]ds R (k) = R pq (k) 4x4, p, q = 1, 2, 3, 4, R (k) 1 = α k (c (1) c(3) 44 3 ) + c(1) 44 2 R (k) 2 = α 1 k R(k) 1, R (k) 3 = α k (c (3) 44 1 R (k) 4 = α 1 k R(k) 3 + c(2) 44 3 ) + c(3) 44 2, = 1, 2, 3, 4, + c(3) 44 4, + c(2) 44 4, (7) are given by (5) and L (k)t L denotes the compex conugate matrix of L (k)t L, L L L = 1 0 L L L L , (8) 0 L L 22 1 pq L 11 = q 4 [a 44 B 1 + (b a 34 )A 1 + a 33 D 1 ], A 1 = B 0 m 3, B 1 = B o m 1, L 13 = q 4 [a 24 B 1 + ( b 33 + a 14 + a 23 )A 1 + a 13 D 1 ], C 1 = A! + B 1 m 2 α 1 α 2 α 3 α 4, L 22 = q 4 [a 44 C 1 + (b a 34 )B 1 + a 33 A 1 ], D 1 = A 1 m 2 B 1 α 1 α 2 α 3 α 4, L 24 = q 4 [a 24 C 1 + ( b 33 + a 14 + a 23 )B 1 + a 13 A 1 ], 1 = L 11 L 33 L 2 13, L 33 = q 4 [a 22 B 1 + (b a 12 )A 1 + a 11 D 1 ], 2 = L 22 L 44 L 2 24, L 44 = q 4 [a 22 C 1 + (b a 12 )B 1 + a 11 A 1 ], 2 = [b 4 (m 1 m 3 2 a 1 a 2 a 3 a 4 ) + q 4 m 0 ]q 4 B 0 > 0, m 0 = (a 11 a 44 + a 33 a 22 2a 13 a 24 ), 1 = a 1 a 2 a 3 a 4 2,

8 8 Buetin of TICMI m 1 = 4 α k, m 2 = α 1 α 2 + α 1 α 3 + α 1 α 4 + α 2 α 3 + α 2 α 4 + α 3 α 4, m 3 = α 1 α 2 α 3 + α 1 α 2 α 4 + α 1 α 3 α 4 + α 2 α 3 α 4, B 1 0 = (α 1 + α 2 )(α 1 + α 3 )(α 1 + α 4 )(α 2 + α 3 )(α 2 + α 4 )(α 3 + α 4 ). Between the coefficients a pq, b pp and c () pq there are the reations a 11 = c (2) 11 q 3 c (1) 33 c (2) c (2) 33 c (2) 33 c (2)2 13 > 0, a 12 = c (2) 13 c (3)2 13 ) c (1) 33 c (2) a 13 = c (3) 11 q 3 + c (2) 33 c (1) 13 + c (1) 33 c (2) 13 c (3) 33 (c (1) 13 + c (3)2 13 ), a 14 = c (3) 13 c (3)2 13 ) + c (1) 33 + c (2) 33 c (3) a 23 = c (3) 13 c (3)2 13 ) + c (1) 33 + c (2) 33 c (3) a 22 = c (2) 33 q 1 c (1) c (2) 11 c (2) 2 13 > 0, a 24 = c (3) 33 q 1 + c (2) 13 + c (2) 11 c (3) 11 (c (1) 13 + c (3)2 13 ), a 33 = c (1) 11 q 3 c (2) 33 c (1) c (1) 33 c (1) 33 c (3)2 13 > 0, a 34 = c (1) 13 c (3)2 13 ) c (2) 33 c (1) a 44 = c (1) 33 q 1 c (2) c (1) 11 c (1) 2 13 > 0, = (c (1) 11 a 11 + c (2) 11 a c (3) 11 a 13 ) q 1 q 3 + (c (1) 13 c (3)2 13 ) > 0, b = c () 44 q4 1 > 0, = 1, 2, 3. L (k) = αk 2L(k) 22 α k L (k) 22 αk 2L(k) 24 α k L (k) 24 α k L (k) 22 L (2) 22 α k L (k) 24 L (k) 24 αk 2L(k) 24 α k L (k) 24 αk 2L(k) 44 α k L (k) 44 α k L (k) 24 L (k) 24 α k L (k) 44 L (k) 44 L (k) 22 = q 4 d k [a 44 + α 2 k (b a 34 ) + a 33 α 4 k ], L (k) 24 = q 4 d k [a 24 + α 2 k ( b 33 + a 14 + a 23 ) + a 13 α 4 k ], L (k) 44 = q 4 d k [a 22 + α 2 k (b a 12 ) + a 11 α 4 k ]. For the stress vector we get T ( x, n)u(z) = 1 π Im 1 π Im L () il L (k) il[ n (z k t ) [g(t) ih(t)]ds]., n(t k z k ) [g(t) + ih(t)]ds+

9 Voume 13, It can be shown that the vector T U satisfies the condition (T U) = 0, automaticay; [T ( x, n)u(z)] = 1 π Im + 1 π Im L () il L (k) il[ g(t) ih(t) x 1 t ds] = 0. But on the arc ends, = 1, 2,..., p, we have [T ( x, n)u(z)] ± = g(t 0 ) + 1 π Im + 1 π Im L () il From ast equation we get where L (k) il[ g(t) + ih(t)) ds x 1 t n (t k0 t ) [g(t) ih(t)]ds] = f ± (t 0 ). 2g(τ) = f f +, 1 π Re P (k)[ P () n (t k0 t k ) h(t)dt n (t k0 t ) h(t)dt Ω(t 0 ) = 1 2 (f + f + ) 1 π Im P () n(t k0 t ) g(t)dt]. P (k) [ n (t k0 t k ) [g(t) + ih(t)]ds ] = Ω(t 0 ), n (t k0 t k ) g(t)dt We must require in addition that the soution h of the system of singuar integra equations (9) satisfies the condition h(t)ds = RL (f f + )ds, (10) (9) where 4 R (k)t = E + ir. We seek for the soution of the system (9) in the cass h 0 ([4]). Therefore, the tota index in the cass h 0 is 4p.

10 10 Buetin of TICMI Let s prove that the adoint homogeneous system corresponding to the system (9) has ony the trivia soution in the adoint cass. The adoint homogeneous system has the form 1 π Re n (t k t k0 ) ill (k) ν(t)ds 4 LiL () ill k n(t t k0 ) ν(t)ds = 0. (11) Note, that the soution of the system (11) wi be vector, satisfying the Hoder s conditions, vanishing at the end points a and b, and having derivative in the cass H ([4]). Mutipying the system (11) by nonsinguar matrix a = L 1 given by (8) and taking into account the identity all (k) = P (k) a, we obtain 1 π Re P (k) Re 4 P () P k n (t k t k0 ) aν(t)ds n(t t k0 ) aν(t)ds = 0. (12) Let ν 0 be the nontrivia soution of the system (12) and construct the potentia U 0 (z) = 1 π Re Re P () P k R(k)iL T n (t k z k ) aν(t)ds n (t t k ) aν(t)ds. Then at every point of, except may be a and b [T ( t, n)u 0 ] ± = 0 (13) and on the basis of the uniqueness theorem we have U 0 = 0. Finay, from equaity: U 0 + U0 = 2aν(t 0 ) = 0, t 0, it foows, that ν 0 = 0, which contradicts our assumption. Thus, the system (9) is sovabe in the cass h 0 for the arbitrary righthand side and the soution depended on the 4p arbitrary constants. These constants are fixed by the conditions (10), given the system of agebraic 4p inear equations. Let s prove that the determinant of this system is not zero. Indeed, et s take the homogeneous system, corresponding to the conditions f ± = 0. Supposing the soution K (0) 1,..., K (0) 4p nontrivia, we construct the potentia U 0 (z) = 1 π Re R(k)L T n (z k t k )h (0) ds, (14)

11 Voume 13, where h (0) is a inear combinations of soutions h () h (0) = 4p K () h (), and h () are ineary independent soutions of the homogeneous equation corresponding to (9). h () has satisfy the foowing condition h (0) ds = 0. Then the potentia (14) is reguar at infinity and by the uniqueness theorem h (0) = 0. But we have the foowing equaity ( u ) + ( u ) = Lh (0) = 0. Whence we concude that K () = 0, which contradict the assumption. Thus the sovabiity of the probem is proved. Acknowedgement. The designated proect has been fufied by financia support of Georgian Nationa Science Foundation (Grant N GNSF/ST06/3-033). Any idea in this pubications is possessed by the author and may not represent the opinion of Georgian Nationa Science Foundation itsef. R e f e r e n c e s [1] Sh. Zazashvii. Mixed Boundary Vaue Probem for an Infinite Pane with a Rectiinear Cut. Reports of Enarged Sessions of the Seminar of I. Vekua Inst. of App. Math., 4, No. 2 (1989), (in Russian). [2] Sh. Zazashvii. Contact Probem for two Anisotropic Haf-pane with a Cut aong the Contact Line. Bu. Acad. Sci. Georgia, 145, No. 2 (1992), (in Russian). [3] Ya. Ya. Rushchitski. Eements of Mixture Theory. Naukova Dumka, Kiev, 1991 (in Russian). [4] N. I. Muskheishvii. Singuar Integra Equations. Boundary Probems of the Theory of Functions and some their Appications in Mathematica Physics. 3rd ed. Nauka, Moscow, 1968 (in Russian); Engish Transation from 2nd Russian ed. (1946): P. Noordhoff, Groningen 1953, Corrected Reprint Dover Pubications, Inc., N. Y., [5] L. Bitsadze. The Basic BVPs of Statics of the Theory of Eastic Transversay Isotropic Mixtures. Reports Seminar of I. Vekua Inst. of App. Math. Reports, 26-27, No. 1-3 ( ), [6] L. Bitsadze. On Some Soutions of Statics of the Theory of Eastic Transversay Isotropic Mixtures. Reports of Enarged Session of the Seminar of I. Vekua Inst. of App. Math., 16, No. 2 (2001), Received ; revised ; accepted