Abstract

Abstract

A direct method for the exact solution of loaded differential or Fredholm Integro-Differential Equations (IDEs) with nonlocal integral boundary conditions is proposed. We consider the abstract operator equations of the form Bu = Au - g Ψ( u ) = f with abstract nonlocal boundary conditions. In this paper we investigate the correctness of the equation Bu = f and its exact solution in closed form.

Full Text

Introduction In recent years the theory of loaded functional, differential and integro-differential equations (IDEs) has been advanced. These equations describe problems in optimal control, regulation of the layer of soil water and ground moisture, underground fluid and gas dynamics, mathematical biology, economics, ecology, and pure mathematics [1-7]. An equation is called loaded equation if it contains the solution function on a manifold with dimension less than the dimension of domain of this function [2]. For example, an ordinary loaded differential equation is represented by dy/dx = f(x, y) + ψ(y(xj )), x ∈ [0, 1], xj ∈ [0, 1], where xj are fixed points. Boundary value problems (BVPs) for differential and IDEs with nonlocal boundary conditions arise in various fields of mechanics, physics, biology, biotechnology, chemical engineering, medical science, finance and others (see [8-15]). Fredholm IDEs with nonlocal integral boundary conditions and ordinary differential operators, probably, first were considered by J.D. Tamarkin [16]. Nonlocal BVPs for the ordinary differential equations with integral boundary conditions were studied in [17-19]. Nonlocal BVPs for the partial differential equations were investigated in [8, 20-25]. Loaded nonlocal BVPs for ordinary differential equations were considered in [7, 26]. This work is a generalization of the papers [27-32], where integral boundary conditions have not been considered. The problems of the type arise naturally from A.A. Dezin, R.O. Oinarov extensions of linear operators [33, 34], which are not restrictions of a maximal operator, unlike the classical M.G. Krein, J.Von. Neuman extensions [35, 36] in Hilbert space and M. Otelbaev [37] in Banach space. We investigate the abstract operator equations of the form Bu = Au - gΨ(u) = f with abstract nonlocal boundary conditions D(B) = {u ∈ D(A) : Φ(u) = NF(Au)}, where B : X → Y, X, Y are Banach spaces, g ∈ Yn is a vector, are vector functionals, N is a matrix, A is a differential operator with finite-dimensional kernel and gΨ(u) is a loaded part of Bu = f or a Fredholm integro-differential operator defined on the space with graph norm If g = 0, we have a differential equation with nonlocal integral boundary conditions. The technique presented here allows us to find exact solutions in closed form for loaded Differential or Fredholm Integro-Differential Equations with separable kernels and integral boundary conditions. This technique is simple to use and can be easily incorporated to any Computer Algebra System (CAS). The essential ingredient in our approach is the extension of the main idea in [34]. The paper is organized as follows. In Section 1 we recall some basic terminology and notation about operators. In Section 2 we prove the main general results. Finally, in Section 3 we discuss some examples of loaded differential and IDEs which show the usefulness of our technique. 1. Terminology and notation Let X be a complex Banach space and X∗ its adjoint space, i.e. the set of all complex-valued linear and bounded functionals on X. We denote by f (x) the value of f on x. We write D(A) and R(A) for the domain and the range of the operator A, respectively. An operator A2 is said to be an extension of an operator A1, or A1 is said to be a restriction of A2, in symbol A1 ⊂ A2, if D(A2) ⊇ D(A1) and A1x = A2x, for all x ∈ D(A1). An operator A: X → Y is called closed if for every sequence xn in D(A) converging to x0 with Axn → f0, it follows that x0 ∈ D(A) and Ax0 = f0. A closed operator A0 : X → Y is called minimal if R(A0) ≠ Y and the inverse exists on R(A0) and is continuous. A closed operator A is called maximal if R(A) = Y and ker A ≠{0}. An operator is called correct if = Y and the inverse exists and is continuous. The problem is said to be correct when the operator is correct. If Ψi ∈ X∗, i = 1, ..., m, then we denote by Ψ = col(Ψ1, ..., Ψm) and Ψ(x) = col(Ψ1(x), ..., Ψm(x)). Let q = (q1, ..., qm) be a vector of Xm. We will denote by Ψ(q) the m×m matrix whose i, j-th entry is the value of functional Ψi on element qj and by Im and 0m the identity m×m and the zero m×m matrices, respectively. By we will denote the zero column vector. 2. Loaded differential and Fredholm IDE with nonlocal integral boundary conditions Lemma 2.1. Let the operator A: C[0, 1] → C[0, 1] be defined by Au(t) =u'(t), D(A) = C1[0, 1], . Then: (i) The operator A is closed and A2: C[0, 1] → C[0, 1] is determined by A2u(t) = u''(t), D(A2) = C2[0, 1] and closed too. (ii) For all u(t)∈C2[0, 1] the next inequality holds (1) Proof. (i)-(ii) Let x1, x2 ∈ [0, 1]. Then from the equality we obtain From the last relation follows (1). Now we will prove the closedness of A2. Let un(t) ∈ D(A2), un(t) → u(t), n → ∞ and A2un(t) = un''(t) → y(t), n → ∞. Put yn(t) = un''(t). Then (2) From (1) we get From this inequality, because of un(t) and un''(t) are fundamental sequences in C[0, 1], it follows that un'(t) is also a fundamental sequence in Banach space C[0, 1]. Then there exists a function v(t) ∈ C[0, 1] such that un'(t) → v(t), n → ∞. The last implies un'(0) → v(0), n → ∞. For n → ∞ from (2) we get It is evident that v(t) ∈ C1[0, 1] and v'(t) = y(t). From un(t) →u(t), un'(t) → v(t), n → ∞, since the operator A is closed, we get u(t) ∈ C1[0, 1] and u'(t) = v(t). But v(t) ∈ C1[0, 1] and v'(t) = y(t). Hence u(t) ∈ C2 [0, 1], u''(t) = y(t) and A2 is a closed operator. Lemma 2.2. Let A, B, C, D, E be n×n matrices and . Then (3) Proof. Let In this case |F| = |F-1| = 1. So |F G| =|F| |G| = = |G| and |F-1G| = |F-1| |G| = |G|. From the last relations we get (3). The next theorem easy follows from Theorem 3 [32]. Theorem 2.3. Let u(t) ∈ Wl2(a, b), hj (t) ∈ L2(a, b), j = 0, 1, ..., l, (ai, bi) ⊆ (a, b), i = 1, ..., n and the functional (4) Then Ψi(u) is linear and bounded on Wl2(a, b) and hence Ψi ∈[Cl[a, b]]*, i = 1, ..., n. We generalize Theorem 1 [32] without using the linear independence of components of the vector Ψ. Theorem 2.4. Let X, Y and Z be complex Banach spaces and : X → Y a linear correct operator with D ⊂ Z ⊆ X. Further let the components of the vector g = (g1, . . . , gn), be linearly independent in Y and the vector Ψ = col(ψ1, . . . , ψn) ∈ [Z∗]n. Then: (i) The operator B defined by Bu = - gΨ(u) = f, D(B) = D f ∈ Y. (5) is injective if and only if detW = det[In - Ψ-1g)] ≠ 0, (6) (ii) If B is injective, then for any f ∈ Y the unique solution of (5) is given by u = B-1f = -1f + -1g)[In - Ψ-1g)]-1Ψ-1f). (7) Proof. (i). Let det W ≠ 0. Since f, gi ∈ R i = 1, ..., m, by using (5) we have u - -1gΨ(u) = -1f, (8) Ψ(u) - Ψ(-1g)Ψ(u) = Ψ(-1f), [Im - Ψ(-1g)]Ψ(u) = Ψ(-1f), (9) Ψ(u) = [In - Ψ(-1g)]-1Ψ(-1f) for all f ∈ R(). (10) Since Ψ(u) is defined uniquely by (9) for all f ∈ R() = Y, then by substituting (10) into (8), we obtain the unique solution (7) of the problem (5) for all f ∈ Y. So B is an injective operator and R() = Y. Conversely. Let detW = 0. Then there exists a vector= col(c1, ..., cn) ≠ such that W= 0. For u0 = -1g ≠0 follows that Bu0 =u0 -gΨ(u0) = g - gΨ(-1g) = = g[In - Ψ(-1g)] = gW = g = 0. Hence u0 = -1g ∈ ker B ≠ {0}. Note that u0 ≠, because of g1, ..., gn are linearly independent and the operator is correct. So B is not injective. So the statement (i) holds. (ii). We have proved in (i) that in the case det W ≠ 0 Equations (8)-(10) hold. Substituting (10) into (8), we obtain the unique solution (7) of the problem (5) for all f ∈ Y. This means that R(B) = Y. The boundedness of B-1 on Y follows from the boundedness of the operator -1 and Ψ1, ..., Ψn. Hence B is correct. Remark 2.5. The linear independence of g1, ..., gn has been used only to prove the necessary condition for injectivity of the operator B. Now we generalize this theorem on the case of perturbed boundary conditions. Let X, Y be Banach spaces, A : X Y be a linear closed operator. We remind that for A, the graph norm associated with A, is defined by: ||x||XA = ||x||X + ||Ax||Y ∀x ∈ D(A) (11) and that the set XA =(D(A), || · ||XA) (12) is a Banach space. Note that usually in the previous theorem Z = XA. Theorem 2.6. Let X, Y be complex Banach spaces, A : X → Y a maximal linear operator with finite dimensional kernel z = (z1, ..., zm) which is a basis of ker A and a correct restriction of A defined by ⊂ A, D() = {u ∈ D(A) : Φ(u) = 0}. (13) Suppose also that the components of the vector functionals Φ = = col(Φ1, ..., Φm), Ψ = col(Ψ1, ..., Ψm), F = col(F1, ..., Fm), belong to X∗A, X∗A, Y∗, respectively, where Φ1, ..., Φm biorthogonal to z1, ..., zm and that the vector g = (g1, ..., gm) ∈ Ym and N is an m × m matrix, g1, ..., gm linearly independent. Then: (i) The operator B defined by Bu =Au - gΨ(u) = f, f ∈ Y, D(B) ={u ∈ D(A) : Φ(u) = NF(Au)}, (14) is injective if and only if det L2 = det[Im - Ψ(-1g) - Ψ(z)NF(g)] ≠ 0. (15) (ii) If B is injective, then B is correct and for all f ∈ Y the unique solution of (14) is given by u = B-1f =-1f + pΨ(-1f) +[z + pΨ(z)]NF(f), where (16) p =-1g + zNF(g)]L-12. (17) Proof. (i). Since z ∈ [ker A]m, Φ(z) = Im, the problem (14) is written as Bu = A(u - zNF(Au)) - gΨ(u) = f, f ∈ Y, D(B) = {u ∈ D(A) : Φ(u - zNF(Au)) = 0}, (18) Then u - zNF(Au) ∈ D() for every u ∈ D(B) and Bu = (u - zNF(Au)) - gΨ(u) = f, (19) u - zNF(Au) - -1gΨ(u) = -1f, (20) Ψ(u) - Ψ(z)NF(Au) - Ψ(-1g)Ψ(u) = Ψ(-1f), [Im - Ψ(-1g)]Ψ(u) - Ψ(z)NF(Au) = Ψ(-1f), (21) -F(g)Ψ(u) + F(Au) = F(f). (22) From (21), (22) we obtain (23) Denoting the matrix from the left by L and using the formula (3), where we take E = Ψ(z)N, we obtain Let det L2 ≠ 0 and u ∈ ker B. Then Bu = Au - gΨ(u) = 0, Φ(u) = NF(Au), det L ≠ 0. (24) Now using Equations (19)-(23), where f = 0 we get (25) From the above equation, since det L = ± det L2 ≠ 0 follows that Ψ(u) = , F(Au) = . Substituting these values into (24), we obtain Φ(u) = and Bu = u = 0. From u = 0, since is correct follows that u = 0. This proves that ker B = {0}. So B is injective. Conversely. Let det L2 = 0. Then there exists a vector = col(c1,..., cm) ≠ ≠ such that L2 = 0. Observe that he element u0 = -1g + zNF(g) ≠ 0, otherwise the components of g will be linearly dependent, which contradicts the hypothesis that g1, ..., gm are linearly independent. Note that u0 ∈ D(B), since Φ(u0) = NF(g), F(Au0) = F(g) and so Φ(u0)-NF(Au0) = NF(g)-NF(g)= = . We will show now that u0 ∈ ker B. Bu0 =Au0 - gΨ(u0) = g - gΨ( -1g + zNF(g)) = = g - gΨ(-1g) - gΨ(z)NF(g) = = g[Im - Ψ(-1g) - Ψ(z)NF(g)] = gL2 = g. Such u0 ∈ ker B. Hence u0 ∈ D(B) and u0 ∈ ker B. So ker B ≠ {0} and B is not injective. The statement (i) holds. (ii) Let det L2 ≠ 0. Then det L ≠ 0 and from (23) we obtain (26) We put Then [Im - Ψ(g)]K11 - Ψ(z)NK21 = Im, (27) [Im - Ψ(g)]K12 - Ψ(z)NK22 = 0m, (28) -F(g)K11 + K21 = 0m, (29) -F(g)K12 + K22 = Im. (30) From (30) we obtain K21 = F(g)K11 and from (27), (29), we get [Im - Ψ(g)]K11 - Ψ(z)NF(g)K11 = Im, [Im - Ψ(g) - Ψ(z)NF(g)]K11 = Im, K11 = K21 = F(g) From (28), since K22 = Im + F(g)K12, we have [Im - Ψ(g)]K12 - Ψ(z)N[Im + F(g)K12] = 0m, [Im - Ψ(g) - Ψ(z)NF(g)]K12 = Ψ(z)N, K12 = Ψ(z)N. The last equation and K22 = Im + F(g)K12 implies K22 = Im + + F(g)Ψ(z)N. So Now we rewrite (26) in the form (31) Then Ψ(u) = Ψf) + Ψ(z)NF(f), F(Au) =F(g) Ψf) + [Im + F(g) Ψ(z)N]F(f). Substituting these values into (20) we get the solution of the problem (14): u =f + zNF(g) Ψ(f) + [Im + Fg)Ψ(z)N]F(f)} - [Ψ(f) +Ψ(z)NF(f)], From the last equation for every f ∈ Y follows the unique solution (16) of (14). Because f in (16) is arbitrary, we obtain R(B) = Y. Since the operator and the functionals F1, ..., Fm, Ψ1, ... ,Ψm are bounded, from (16) follows the boundedness of B-1. Hence, the operator B is correct if and only if (15) holds and the unique solution of (14) is given by (16). The theorem is proved. From the previous theorem for g = 0 follows the next corollary which is useful for solving differential equations with integral boundary conditions. Corollary 2.7. Let X, Y be complex Banach spaces, the operators A, , the vector z = (z1, ..., zm) and vector functionals Φ = col(Φ1, ..., Φm), Ψ = col(Ψ1, ..., Ψm) and the matrix N are defined as in Theorem 2.6. Then: (i) The operator B defined by Bu = Au = f, f ∈ Y, D(B) = {u ∈ D(A) : Φ(u) = NF(Au)}. (32) is correct and for all f ∈ Y the unique solution of ( 32) is given by u = B-1f = f + zNF(f). (33) Remark 2.8. In applications we encounter operators B1 : X → Y of the form Au - g1Ψ1(u) - ... - gkΨk(u) = f, k ≤ m, Φ1(u) = ν11F1(Au) + ... + ν1mFm(Au) ... ... ... ... Φm(u) = νm1F1(Au) + ... + νmmFm(Au) (34) where X, Y are Banach spaces, Φi, Fi, Ψj, i = 1, ..., m, j = 1, ..., k are functionals. Then we are interested in knowing whether the operator B1 is B type operator defined by Bu =Au - gΨ(u) = f, f ∈ Y, D(B) = {u ∈ D(A) : Φ(u) = NF(Au)}, (35) and, therefore, Theorem 2.6 applies. For this purpose, we proceed as follows: 1.1. We check that m = k and show that the operator A is maximal, namely A is closed with dim ker A = m and R(A) = Y, 1.2. We find the basis z1, ..., zm of ker A and the Banach space XA = (D(A), || · ||XA), where ||u||XA = ||u||X + ||Au||Y, 1.3. We show that the operator defined by u = Au, D() = {u ∈ D(A) : Φ(u) = }, is correct and find the inverse operator , 1.4. We find the vector g, the element f, the m×m matrix N, the functional vectors Φ(u) = col(Φ1(u), ..., Φm(u)), Ψ(u) = col(Ψ1(u), ..., Ψm(u)), F(Au) = col(F1(Au), ..., Fm(Au)) and F(f) = col(F1(f), ..., Fm(f)), 1.5. We check the condition Φ(z) = Im, 1.6. We check the conditions Fi ∈ Y∗, Φi and Ψi ∈ X∗A, i = 1, ..., m. If all these conditions hold true then we apply Theorem 2.6. If one of these steps fails, then B1 is not identified as B-type operator and, therefore, the theory can not be applied. To prove that B is an injective and correct operator, we proceed as follows: 2.1. We calculate the vector g = ( g1, ...,gm), 2.2. We calculate the m×m matrices 2.2.4. L2 = Im - Ψ(g) - Ψ(z)NF(g). If the determinant det L2 ≠ 0, then B is an injective and correct operator. To find the solution of (34), we proceed as follows: 3.1. We calculate the inverse matrix and the element f, 3.2. We calculate the vectors Ψ(f) = col(Ψ1(f), ..., Ψm(f)), F(f) = col(F1(f), ..., Fm(f)), p =[g + zNF(g)] 3.3. We find the solution of (34) by u = B-1f = f + pΨ(f) +[z + pΨ(z)]NF(f). 4. If k < m, then we take as vector g the vector = ( ..., where = gi, i = 1, ..., k, = 0, i = k + 1, ..., m. As functional vector Ψ(u) we can take (u) = ..., (u)) with linearly independent elements, where (u) = ψi(u), i = 1, ..., k, and (u), i = k+1, ..., m are arbitrary functionals. 3. Examples Example 3.1. The next problem with loaded differential equation and nonlocal integral boundary conditions on C[0, 1] u''(t) - 4t u(1/2) - (2t + 1) u(1) = 1 - 5t, (36) (37) is correct and the unique solution of (36), (37) is given by u(t) = t2 - t + 1. (38) Proof. First we rewrite the boundary conditions (37) in the form (39) If we compare (36), (39) with (14), it is natural to take X = Y = C[0, 1], Au(t) = u (t) = u''(t), D(A) = C2[0, 1], The operator A, by Lemma 3.1, is closed. Furthermore A is a maximal operator, XA = (C2[0, 1], ||·||XA), where ||u(t)||XA = ||u(t)||+||u''(t)||. The set z = (1, t) constitute a basis of ker A. As the operator it is natural to take u (t) = Au(t) = u''(t), D() = {u(t) ∈ ∈ D(A) : u(0) = u'(0) = 0}. The initial problem u (t) = f(t) is correct and has the unique solution -1f(t) = ∫t0(t - x)f(x)dx. By comparing (36), (39) with (14), it is natural to take g1 = 4t, g2 = 2t + 1, g = (g1, g2) = (4t, 2t + 1), f = 1 - 5t, where Φ1(u) = u(0), Φ2(u) = u'(0), Ψ1(u) = u(1/2), Ψ2(u) = u(1), Then It is evident that the vector z = (1, t) is biorthogonal to (Φ1, Φ2). and that |F1(f)| ≤ ||f||, |F2(f)| ≤ 5||f|| for all f ∈ C[0, 1]. So F1, F2 ∈ C∗[0, 1] = X∗. Because of |Ψ(u)| = |u(t0)| ≤ ≤ ||u(t)|| ≤ ||u(t)|| + ||u''(t)|| = ||u(t)||XA, we conclude that Ψ1, Ψ2 ∈ [XA]∗. In the same way is proved that Φ1 ∈ [XA]∗. Now by using Inequality (1), for all u ∈ C2[0, 1] we obtain |Φ2(u)| = |u'(0)| ≤ ||u'(t)|| ≤ ||u''(t)|| +2||u(t)|| ≤ ≤ 2(||u''(t)|| +||u(t)||) = 2||u(t)||XA. This proves that Φ2 ∈ [XA]∗. The conditions 1.1-1.6 of Remark 2.8 hold true and so we can apply Theorem 2.6. Now we calculate Since det L2 ≠ 0, by Theorem 2.6, the problem (36), (37) is correct. The inverse matrix For f(t) = 1-5t we compute f(t) = F(f) = col(-11/12, -13/2), Ψ(f) = col(Ψ1(f), Ψ2(f))= (1/48, -1/3). From (17) we get p =[g + zNF(g)] = -1/303(4(28t3 + 43t2 - - 91t + 106), 196t3 - 2t2 - 334t + 439). From (16), by Theorem 2.6, we obtain the unique solution Example 3.2. The next problem with differential equation and nonlocal integral boundary conditions on C[0, 1] u''(t) = -π2cos(πt), (40) is correct and the unique solution of (40) is given by u(t) = cos(πt) + 3t + 2. (41) Proof. First we rewrite the boundary conditions (40) in the form (42) If we compare (40), (42) with (32), it is natural to take the spaces X, Y, XA, the operators A, the vectors z, Φ1, Φ2, F1 as in Example 3.1, Then Note that for all f(t) ∈ C[0, 1] it follows that |F2(f)| ≤ 5||f(t)|| and so F2 ∈ C∗[0, 1]. For f(t) = -π2cos(πt) we compute By Corollary 2.7, the problem (40) is correct. Next as in Example 3.1 is proved that F2 ∈ X∗. From (33), by Corollary 2.7, we obtain the unique solution which yields (41). Example 3.3. The next problem with loaded integro-differential equation and nonlocal integral boundary conditions on C[0, 1] (43) is correct and the unique solution of (43) is given by u(t) = 2 cos(πt) - 3 sin(πt). (44) Proof. First we rewrite the boundary conditions (43) in the form (45) If we compare (43), (4.5) with (14), it is natural to take the spaces X, Y, XA, the operators A, the vectors z, Φ1, Φ2 as in Ex. 3.1, g1 = 4π2cos(πt), g2 = -(π/6) sin(πt), f = 6π2cos(πt) + (3π2 - 1) sin(πt), Then Further, we take It is evident that |Fi(f)| ≤ ||f(t)|| for all f(t)∈C[0,1], i = 1, 2. Such Fi ∈ C∗[0, 1], i = 1,2. We proved in Ex. 3.1 that the functional defined by Ψ(u) = u(1) belongs to . So Ψ1 ∈ . From Theorem 2.3 it follows that Ψ2 ∈ C∗[0, 1]. Then for all u(t) ∈ XA we get |Ψ2(u)|≤k||u(t)||≤ k(||u(t)|| + ||u''(t)||) = k||u(t)||XA. So Ψ2 ∈ Further we calculate So Also we compute Since det L2 ≠ 0, by Theorem 2.6, the problem (43) is correct. For f(t) = = 6π2cos(πt) + (3π2-1) sin(πt) we compute Next we find Sustituting these values into (16), we obtain the unique solution (44).