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This work is concerned with the Fredholm property of the second order differetial opertor associated to a class of boundary conditions. Several sufficient conditions will be proved along with constructing the generalized inverse for such operator. The result is a basic tool to analysis the boundary value problems at resonance for nonlinear perturbation of such operators. Keywords: Fredholm property, second order differential operator, continuation theorem, projector, generalized inverse
Fredholm alternative for the second order differential opertor associated to a class of boundary conditions Do Huy Hoanga , Le Xuan Truonga,∗ a Department of Mathematics and Statistics, University of Economics HoChiMinh city, 59C, Nguyen Dinh Chieu Str, District 3, HoChiMinh city, Vietnam Abstract This work is concerned with the Fredholm property of the second order differetial opertor associated to a class of boundary conditions. Several sufficient conditions will be proved along with constructing the generalized inverse for such operator. The result is a basic tool to analysis the boundary value problems at resonance for nonlinear perturbation of such operators. Keywords: Fredholm property, second order differential operator, continuation theorem, projector, generalized inverse 2010 MSC: 34B10, 34B15 1. Introduction The question of the solvability for nonlinear pertubation of differential operators associated to multi-point or nonlocal boundary conditions have been extensively studied. To indentify a few, we refer the reader to [1]-[12] and references therein. Almost such problems, written in operator form, is of the type x = N x, (1.1) where is a linear mapping between two Banach spaces X and Z, while N : X → Z is a nonlinear mapping. Studying problem (1.1) is often confronted with the difficulty that the relevant linearized operator is not invertible in suitable function spaces. There have been some methods to overcome this obstacle as the alternative method [1, 12], the pertubation method (the name was proposed by Kannan [7]) or continuation method of Mawhin [4],... One of important hypotheses of Mawhin’s continuation theorem that is the Fredholm of index zero property of linear operator L which mean that (i) ker has finite dimension, (ii) Im is a closed subset of Z and has finite codimension. (iii) ind = dim ker − codim Im = 0. In many papers (see [2, 3, 5], [8]-[11]), this property is often proved by constructing first a continuous projector Q : Z → Z which satisfies the condition ker Q = Im . And then it follows that Im has finite codimension as well as index of is equal to 0. Nevertheless it seems that the construction a such projector is difficult when dim ker is large ([8]-[11]). So looking for the sufficient conditions to ensure Fredholmness of is still quite limited. ∗ Corresponding author Email address: ❧①✉❛♥tr✉♦♥❣❅❣♠❛✐❧✳❝♦♠ (Le Xuan Truong ) Preprint submitted to Elsevier November 12, 2014 The goal of current paper is to study the differential operator x(t) = where X := C 1 [0, 1]; dom d = , Z := L 1 (0, 1); d t2 d x ∈ X , x ∈ Z and d2 x : dom ⊂ X → Z defined by (t) = x (t), endowed with their usual norms and Ax(0) + B x (0) = D 1 0 x(s)ds E x(1) + F x (1) = G 1 0 x(s)ds , with A, B, D and E, F , G are square matrices of order d. Our results mentioned two issues. The first is looking for conditions of coefficient matrices for which the operator is Fredholm of zero index (section 2) and the second is characterizing the set of all right-hand side functions y ∈ Z for which the equation x = y has at least one solution x ∈ dom (section 3). To the best of our knowledge, the above issues has not been developed in general cases of dimension of the kernel. Furthermore our method involves several new ideas and gives a unified method of attack for many boundary value problems at resonance. Previous paper dealt with one problem at a time whereas our method allows us to solve many problem at once. We end this section by noting that our results can be used to discuss the solvability of equation x(t) = f t, x(t), x (t) , t ∈ (0, 1), on dom by using the Mawhin’s continuation theorem. This can be done by standard arguments (see [4, 8, 11]). However we will not state here. 2. Fredholm property of the operator In the rest of paper we use the following notations is zero matrix of order d × d and θ is zero element of d . Id : the identity matrix of order d. ν z(t) := t (t 0 − s)ν−1 z(s)ds, for all z ∈ Z and ν ∈ {1, 2}. First it’s necessary to note that, for x ∈ dom So, when x ∈ dom , we can write x(t) = x(0) + x (0)t + 2 x(t). Ax(0) + B x (0) = D , the boundary conditions E x(1) + F x (1) = G x(0), x (0) where • = A− D E−G B − 12 D E + F − 12 G 2 T = ( x) , (2.2) 1 0 x(s)ds 1 0 x(s)ds are equivalent to (2.3) • :Z→ 2d is a continuous linear mapping defined by (z) = D 1 1 2 2 z(t)d t , G 0 0 z(t)d t − E z(1) + F z(1) , z ∈ Z. 2 1 (z) Therefore dom dom 1 (2.4) 2 (z) can be represented as follows = x(t) = x(0) + x (0)t + 2 x(0) x (0) x(t), t ∈ [0, 1] : = ( x) . Lemma 2.1. We have ker and Im ∼ = Ker = x ∈ X : x(t) = c0 + c1 t, t ∈ [0, 1], c0 , c1 ∈ ker = {z ∈ Z : (z) ∈ Im , }. Proof. The proof of this Lemma is straightforward and we will omit the details. Theorem 2.1. The operator • zero if dim (Im + Im • positive if dim (Im is Fredholm. Moreover the index of is ) = 2d. + Im ) < 2d. To prove this Theorem we need the following lemma Lemma 2.2. Let E, F be two vector spaces on field , be a linear operator from E into F . Assume that be a subspaces of F . If is an algebraic complement of −1 ( ) in E then is isomorphic to any algebraic complement of ∩ (E) in (E). Proof. Let be the restriction of on . Since −1 (0) ⊂ ∩ −1 ( ) = {0} it is easy to see that is an isomorphism from into ( ) which means that is isomorphic to ( ). So it is sufficient to show that ( ) ≡ ( ) is an algebraic complement of := ∩ (E) in (E). Indeed, for any y ∈ (E), there exist x 1 ∈ y= −1 ( ) and x 2 ∈ x1 + such that x2. Since x 1 ∈ and x 2 ∈ ( ) the above equality implies that (E) = + ( ). On the other hand, if y0 ∈ ∩ ( ) then y0 = (x 0 ), where x 0 ∈ −1 ( ) ∩ = {0}. Therefore y0 = (0) = 0. This implies ∩ ( ) = {0}. So (E) = ⊕ ( ). The proof of Lemma is complete. Proof of Theorem 2.1 Since is continuous from Z into 2d and Im is closed in 2d it is clear that Im is a closed subspace of Z. Further we have dim ker = dim ker < ∞. So it remains to show that codim Im = dim ker . Indeed, by using Lemma 2.2, if Z0 is an algebraic complement of Im in Z then Z0 isomorphic to any algebraic complement of Im ∩ Im in Im . So codim Im = dim (Z/ Im ) = dim Z0 = dim (Im / (Im = dim Im − dim (Im = dim (Im 3 + Im ∩ Im )) ∩ Im ) − dim Im ) This implies that codim Im < ∞ and so proof of Lemma is complete. is a Fredholm operator. The remains is evident. The Next we will provide some sufficient conditions for need the following lemma: Lemma 2.3. The image of can be defined by = Im Dα, Eβ + F γ + Gα ∈ Consequently, if Im G ⊂ Im E + Im F then Im Proof. Recall that (z) = 1 (z) 1 (z), 2 (z) 1 t = D = G d × d d : α, β, γ ∈ . = Im D × (Im E + Im F ). , where (t − s)z(s)ds, dt 0 1 2 (z) to be Fredholm of zero index. First we 0 1 t (t − s)z(s)ds − E dt 0 0 1 (1 − s)z(s)ds − F z(s)ds. 0 0 Clearly it’s suffit to prove the inclusion supset (⊃). Let (ξ, ζ) be an element of the set d Dα, Eβ + F γ + Gα ∈ d × d : α, β, γ ∈ . Then we can write ξ = Dα1 , and ζ = Eα2 + F α3 + Gα1 , where αi = αi1 , αi2 , ..., αid ∈ consider the function z(t) = z1 (t), z2 (t), ..., zd (t) , t ∈ [0, 1], d . We where z j (t) = a j + b j t + c j t 2 with the coefficients a j , b j , c j be choosen such that 1 dt 0 t (t 0 − s)z(s)ds = α1 j , 1 (1 − s)z j (s)ds 0 1 z (s)ds 0 j = −α2 j , = −α3 j . It is note that by some simple calculations we can see that the above system of linear equation has 1 t a unique solution. Hence we can deduce that ξ = D 0 d t 0 (t − s)z(s)ds = 1 (z) and 1 ζ=G 1 t which implies (ξ, ζ) ∈ Im . So = Im 2 (z), 0 0 0 z(s)ds = (1 − s)z(s)ds − F (t − s)z(s)ds − E dt 0 1 Dα, Eβ + F γ + Gα ∈ d × d : α, β, γ ∈ d . Now assume that Im G ⊂ Im E+Im F . If (ξ, ζ) ∈ Im D×(Im E + Im F ) then there exist α, β, γ ∈ such that ξ = Dα and ζ = Eβ + F γ. On the other hand, it’s clear that there are α1 , α2 ∈ d d such that Eα1 + F α2 = Gα. So we can write ξ = Dα and ζ = E β − α1 + F γ − γ2 + Gα, which implies (ξ, ζ) ∈ Im has been proved. . Hence it is easy to see that Im 4 = Im D × (Im E + Im F ). The Lemma Corollary 2.1. The operator holds is Fredholm of zero index provided that one of following conditions (a) det(D) = 0 and det(E) = 0, (b) det(D) = 0 and det(F ) = 0, (c) det(D) = 0 and det(E + F ) = 0, Proof. Using lemma 2.3 it’s easy to prove that if one of above conditions holds then Im So by theorem 2.1 is a Fredholm of zero index operator. Corollary 2.2. The operator hold (a) 2B = D, det E + F − G 2 = 2d . is Fredholm of zero index provided that one of following conditions = 0, and Im(A − D) + Im D = d , (b) det(A − D) = 0 and one of determinants det(E), det(F ), det(E + F ) is not equal to zero, (c) det(2B − D) = 0 and one of determinants det(E), det(F ), det(E + F ) is not equal to zero. Proof. (a) In this case it’s clear that Im 2.3 we get Im + Im and so dim (Im + Im = Im(A− D)× d . By combining this equality and lemma d = (Im(A − D) + Im D) × ) = 2d. From theorem 2.1 we deduce that + Im (b) We will check that Im = d × d . Indeed, for u, v ∈ is Fredholm of zero index. d , the system (A − D)x + B − 21 D y + Dα = u (E − G)x + E + F − 12 G y + Gα + Eβ + F γ = v has at least one solution defined by x = (A − D)−1 u, y = α = 0 and if det(E) = 0, β = E −1 (v − (E − G)x) , γ = 0, −1 β = 0, γ = F (v − (E − G)x) , if det(F ) = 0, β = γ = (E + F )−1 (v − (E − G)x) , if det(E + F ) = 0. This show that (u, v) ∈ Im + Im zero index by using theorem 2.1. which implies Im + Im = d × d . So is Fredholm of (c) This case can be proved similarly. Remark 2.1. By using lemma 2.3 and the analysis of matrix conditions which are similar the corollaries 2.1 and 2.2. we can obtain some various sufficient 3. The generalized inverse of operator Assume that dim (Im + Im ) = 2d. It follows from Theorem 2.1 that there exist continuous projectors P : X → X and Q : Z → Z such that Im P = Ker , KerQ = Im Moreover the restriction of on dom Im . Then the generalized inverse of , X = Ker ⊕ KerP, Z = Im ⊕ Im Q. ∩ ker P, P , is an isomorphism from dom is defined by P,Q = P−1 (I − Q). 5 ∩ ker P onto The construction such projectors as well as the generalized inverse of study the linear equation x = y, is very important to on dom as well as the correspondent nonlinear boundary value problems at resonance by using the Mawhin’s continuation theorem (see [4, 8, 10, 11] and references therein). In the following we will present a general way to construct the projectors Q, P and pseudo inverse P,Q . For this aim we denote the orthogonal complement of Im ∩ Im in Im by Σ and assume that ei : i = 1, 2, ..., m is an its basis. Denoted by the matrix whose jth column is e j . It is well known that −1 T T P = is the orthogonal projection matrix with Im P = Σ. 2 In order to construct the projector Q we first note that if z = c0 + c1 t + c2 t then c0 c1 , where c2 D/6 D/24 D/60 = G/6 − E/2 − F G/24 − E/6 − F /2 G/60 − E/12 − F /3 (z) = By some lengthy calculations we can prove the equality Im = Im . On the other hand it’s clear that the restriction ∗ of on the orthogonal complement ker ⊥ of ker is an isomorphism from ker ⊥ onto Im . For i = 1, 2, ..., m, we put ωi (t) = ω0i + ω1i t + ω2i t 2 , where ω0i , ω1i , ω2i = and consider the subspace Z0 of Z which is spanned by the vectors ωi dim Z0 = dim (Im ∩ Im )⊥ = dim Im We can also show that Z0 is an complement of Im z ∈ Im ⇔ − dim (Im −1 ∗ ei , m . i=1 Then it’s evident that ) = dim ker ∩ Im . in Z. Moreover, z, ei = 0, ∀i ∈ {1, 2, ..., m}. Now we define the linear map Q : Z → Z by m Q(z) = λi ωi (t), ∀z ∈ Z, (3.5) i=1 where (λ1 , λ2 , ..., λm ) is the unique solution of the system linear equation not difficult to prove that Q is a continuous projector on Z and Im Q = Z0 , ker Q = Im and Z = Im −1 P . Now we shall construct the projector P and the map doinverse of . For x ∈ X we put P x = (1, t) I2d − + T =P (z). It’s ⊕ Im Q. + Let x(0), x (0) m i=1 λi ei T , is the Moore-Penrose pseu(3.6) Here if α, β ∈ d × d then the notation (1, t) α, β stand for α + β t. Since I2d − + is an orthogonal projector on ker it’s not difficult to check that P is a continuous projector on X . Furthermore we have Im P = ker , X = ker ⊕ ker P. 6 Lemma 3.1. The operator ∩ ker P is invertible and its inverse on dom −1 P z(t) + = (1, t) (z)) T + ( 2 z(t), t ∈ [0, 1], −1 P Moreover, there exists positive constant C such that −1 P z Proof. Let z ∈ Im z ∈ Im is defined by . (3.7) satisfies the following estimate ≤ C z , ∀z ∈ Im . (3.8) then it follows from (3.7) that −1 P z(0), ( −1 P z Hence it follows that P we obtain −1 P z ∈ dom T = ( T . (3.9) ∈ ker P. Moreover by combining (3.7) and (3.9) (0) x(t) = (1, t) for all t ∈ [0, 1]. Since x ∈ dom (z) T + = −1 P is well-defined. −1 P z(t) = z(t), for all and so −1 P + (0) = −1 P z −1 P z On the other hand it’s clear that x ∈ dom ∩ ker P then we have x(0), x (0) −1 P z) = θ , that is, −1 P z(0), This show that −1 P + ( (z) = t ∈ [0, 1] and z ∈ Im 2 x) + (z). . Moreover, if (3.10) x(t) ∩ ker P it follows that x) −1 P By combining (3.10), (3.11) we get + and x(0), x (0) T = x(0), x (0) T . (3.11) x(t) = x(t), ∀t ∈ [0, 1]. Finally, by using (3.7) and the continuity of of Lemma is complete. we can obtain the estimate (3.8) and the proof The main result is presented as below. Its proof is directly corollary of above analysis. Theorem 3.1. Let = z∈Z: (i) no solution if y ∈ / (z), ei = 0, ∀i ∈ {1, 2, ..., m} . The the equation x = y has , (ii) at least one solution defined by x = −1 P ( y) if h ∈ . 4. Examples This section presents some examples which is to illustrate our results. Example 4.1 (Second order differential operators with Sturm-Liouville type conditions). Let d = 1 and A = α, B = β, E = γ, F = δ and D = G = 0, where α, β, γ, δ are real numbers which satisfy the conditions γ2 + δ2 = 0 and α(γ + δ) = βγ. We have α β and z = 0, −γ 2 z(1) − δ 1 z(1) . = γ γ+δ Without loss of generality we can assume γ = 0. Then it’s not difficult to prove that Im = {0} × , Ker = − γ+δ γ ξ, ξ : ξ ∈ 7 and Im = ζ, α γ ζ :ζ∈ . This implies dim (Im + Im ) = 2 and so Next we note that Σ = Im = {0} × on Im . On the other hand, since has a basis {e1 = (0, 1)} and P 0 0 0 −ρ1 −ρ2 −ρ3 = with ρ1 = γ + δ, ρ2 = 2 γ 6 + δ 2 , ρ3 = γ 6 and −1 ∗ : {0} × ⊥ → ker formula δ 2 is the identity mapping , , we can show easily that = λ ρ1 , ρ2 , ρ3 : λ ∈ , defined by −1 ∗ (0, ζ) Hence ω1 (t) = − + ⊥ ker is Fredholm operator of zero-index by theorem (2.1). ρ1 + ρ2 t + ρ3 t 2 = −ζ ρ12 ρ1 , ρ2 , ρ3 , ∀ζ ∈ . + ρ22 + ρ32 , for all t ∈ [0, 1]. So the projector Q be defined by following ρ12 + ρ22 + ρ32 γ 2 z(1) + δ 1 z(1) Qz(t) = ρ12 + ρ22 + ρ32 + Now it’s not difficult to show that ρ1 + ρ2 t + ρ3 t 2 . 0 γ−1 0 0 = + and I2 − = 0 γ−1 (γ + δ) 0 1 . By using (3.6) we can define the projector P by P x(t) = x (0) − Moreover, since (z) = (0, 0) for all z ∈ Im γ+δ γ + t , ∀t ∈ [0, 1]. it follows from lemma 3.1 that t −1 P z(t) (t − s)z(s)ds, t ∈ [0, 1], z ∈ Im = . 0 From above analyses we get Claim 1. The equation x = y has at least one solution defined by t −1 P y(t) x(t) = = (t − s) y(s)ds, t ∈ [0, 1] 0 if and only if 1 γ 1 (1 − s) y(s)ds + δ 0 y(s)ds = 0. 0 Example 4.2 (Second order differential operators with nonlocal boundary conditions). Let A= D = F = and B = E = G = Id . 1 Id 1 1 and (z) = 0, 1 = s2 z(s)ds − Id 2 0 2 2 In this case we have 1 z(s)ds . It is easy 0 to prove that Im = {θ } × d , Ker = d × {θ } 8 and Im = {(2ξ, ξ) : ξ ∈ d }. Since dim (Im zero-index. + Im ) = 2d it follows from Theorem 2.1 that the operator Next it’s easy to see that Σ = Im simple calculations we obtain and P = ⊥ ker and −1 ∗ : {θ } × d ⊥ → ker = is the identity mapping on Im I I − 3d α, defined by − 15d 3α α ∈ , 8 5 3d + 4800 1889 2 (z) 1+ d :α∈ , =− 3 1 α, α, α . So by using (3.5) we 1889 8 5 3 1 −1 ∗ (θ , α) get the formula of projector Q as follow Qz(t) = − . Further by some , I − 8d is Fredholm of 8 t+ 4800 5 t 2 , t ∈ [0, 1]. we can deduce that P x(t) = x(0) for all t ∈ [0, 1]. Id Further it follows easily from Lemma 3.1 that On the other hand since = t −1 P z(t) (t − s)z(s)ds, t ∈ [0, 1], z ∈ Im = . 0 Finally we obtain the following result Claim 2. The equation if x = y has at least one solution given by x(t) = t (t − s) y(s)ds 0 if and only 1 1 2 y(s)ds = 0. s y(s)ds − 0 0 Example 4.3. Let following matrices Λ1 = Consider the operator In this case we have 1 0 0 2 Λ2 = , on dom 1 2 3 4 , Λ3 = 1/2 2 2 3 . with A = E = G = Λ1 , B = Λ2 , F = Λ3 , D = . 1 0 1 2 0 0 2 3 4 0 = and (z) = , where z(t) = z1 (t), z2 (t) 0 0 1 2 φ(z) 0 0 2 4 ψ(z) and φ(z) = 1 1 1 2 s z1 (s)ds − 2 1 z1 (s)ds − 2 0 0 1 0 z2 (s)ds, 0 1 s2 z2 (s)ds − 2 ψ(z) = 1 z1 (s)ds − 4 0 It’s not difficult to show that • ker and = {λ(0, −1, 2, −1) : λ ∈ }, 9 z2 (s)ds. 0 • Im = (α, β, γ, 2γ) : α, β, γ ∈ • Im = {(0, 0)} × 2 , . Since Im ∩ Im = (0, 0, γ, 2γ) : γ ∈ Fredholm of zero index. it follows that dim (Im + Im ) = 4 and so is Construction of the projector Q: • The orthogonal complement of Im ∩ Im in Im is Σ = of this subspace is e1 = (0, 0, −2, 1) , and so we get 0 0 0 0 0 0 0 0 P = 4 2 . 0 0 − 5 5 1 2 0 0 −5 5 • 0 0 0 0 = − 5 −2 6 −2 − 11 3 • ker ⊥ −1 ∗ −1 ∗ . A basis 0 0 − 32 17 − 15 has a basis as follow 1 • 0 0 0 0 0 0 7 − 38 −1 − 30 −1 − 74 − 23 0, 0, −2γ, γ : γ ∈ 5 3 7 2 = − , −2, − , −1, − , − , 6 8 30 3 : {(0, 0)} × 2 0, 0, α, β = → ker ⊥ ω1 (t) = 11 7 2 17 , −1, − , − , − 3 4 3 15 . defined by 1204545600 119586041 • the formula of projector Q: = −2, − 2 α− 623952000 119586041 β 1+ − 623952000 119586041 α+ 328352400 119586041 Qz(t) = − 52 φ(z) + 15 ψ(z) ω1 (t), for all z ∈ Z, where − 624976800 119586041 286479600 119586041 + − 25815600 7034473 274954500 119586041 343127520 − 119586041 t+ 235604880 119586041 t 2. Construction of the projector P: It is noted that 1 0 −1 0 0 1/2 −3/2 0 + = . 0 0 1 0 0 0 0 0 For x(t) = (x 1 (t), x 2 (t)), t ∈ [0, 1] we have P x(t) = x 2 (0) 0 1 + −2 1 t . The generalized inverse: By the same arguments as above examples we have P z(t) = φ(z) −1 −3/2 + 1 0 t t + (t − s)z(s)ds, t ∈ [0, 1], z ∈ Im 0 10 . β 2 Claim 3. The equation x = y with y(t) = y1 (t), y2 (t) , has at least one solution given by t x(t) = φ( y)(−1 + t) + 0 (t − s) y1 (s)ds t − 32 φ( y) + 0 (t − s) y2 (s)ds , t ∈ [0, 1] if and only if 1 s2 y2 (s) − y1 (s) ds = 0. 0 Acknowledgments This work was completed while the second author were visiting the Vietnam Institute for Advanced Study in Mathematics (VIASM). References [1] L. Cesari, Functional analysis, nonlinear differential equations and the alternative method, Proc. Conf., Mich. State Univ., East Lansing, Mich., 1975, pp. 1-197. [2] W. Feng and J. R. L. 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R E Gaines, J Mawhin, Coincidence Degree and Nonlinear Differential Equations, vol 568, Lecture Notes in Math., Springer-Verlag, Berlin, 1977 [5] C P Gupta, Existence theorems for a second order m-point boundary value problem at resonance, Int J Math & Math Sci 18(4), (1995), 705-710 [6] V .A Il’in, E.I Moiseev, Nonlocal boundary value problem of the first kind for a Sturm–Liouville operator, J Differential. ..Claim 3 The equation x = y with y(t) = y1 (t), y2 (t) , has at least one solution given by t x(t) = φ( y)(−1 + t) + 0 (t − s) y1 (s)ds t − 32 φ( y) + 0 (t − s) y2 (s)ds , t ∈ [0, 1] if and only if 1 s2 y2 (s) − y1 (s) ds = 0 0 Acknowledgments This work was completed while the second author were visiting the Vietnam Institute for Advanced Study in Mathematics (VIASM) References [1] L Cesari, Functional ... several new ideas and gives a unified method of attack for many boundary value problems at resonance Previous paper dealt with one problem at a time whereas our method allows us to solve many... for conditions of coefficient matrices for which the operator is Fredholm of zero index (section 2) and the second is characterizing the set of all right-hand side functions y ∈ Z for which the. .. at resonance, J Math Anal Appl 294 (2004), 147-157 [10] X Zhang, M Feng, W Ge, Existence result of second- order differential equations with integral boundary conditions at resonance, J Math Anal