On the solvability of multipoint boundary value problems for discrete systems at resonance

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1 On the solvability of multipoint boundary value problems for discrete systems at resonance Daniel Maroncelli, Jesús Rodríguez Department of Mathematics, Box 8205, North Carolina State University, Raleigh,NC , U.S.A Abstract In this paper we provide sufficient conditions for the existence of solutions to multipoint boundary value problems for discrete systems at resonance. Our results depend on the properties of the nonlinearities as well as on the solution space of the associated linear problem. Our approach is based on topological degree arguments in conjunction with the Lyapunov-Schmidt procedure. Keywords: Boundary value problems, Schauder s fixed point theorem, Lyapunov-Schmidt procedure, Topological degree theory, resonance 1. Introduction In this paper we analyze nonlinear, discrete, multipoint boundary value problems of the form subject to x(t + 1 = A(tx(t + f(t, x(t, (1 m B i x(i = 0. (2 Throughout our discussion, we assume that for each t N = {0, 1, 2, }, A(t is an n n invertible matrix. m is a fixed integer greater than 2, f : R n+1 R n is assumed to be continuous, each B i, i = 0,, m, is n n matrix and the augmented matrix [B 1 B 2 B m ] has full row rank. The dimension of the solution space for linear homogeneous problem x(t + 1 = A(tx(t (3 Corresponding Author addresses: dmmaronc@ncsu.edu (Daniel Maroncelli, rodrigu@math.ncsu.edu ( Jesús Rodríguez Preprint submitted to Elsevier February 6, 2013

2 subject to m B i x(i = 0 will play a critical role in our analysis. We will primarily be concerned with the case of resonance; that is, the case when the solution space to (3, (2 is nontrivial. In particalur, we will focus on the case in which the solution space has dimension greater than 1. In this regard, our results constitue a significant generalization of the ideas in [2, 3] where the solution space is assumed to be less than Preliminaries We will view the nonlinear boundary value problem (1, (2 as an operator problem. To do so we introduce appropriate spaces and operators. We define { } m X = φ : {0, 1, 2,, m} R n B i x(i = 0, and Z = {φ : {0, 1, 2,, m 1} R n }. We use the sup norm on both X and Z; that is, for φ X φ = and for ψ Z ψ = sup φ(t, t {0,1,2,,m} sup ψ(t. t {0,1,2,,} Here represents the standard Euclidean norm on R n. It is clear that X and Z become Banach spaces under these norms. We define a linear operator L: X Z by [Lx](t = x(t + 1 A(tx(t, and we introduce a nonlinear operator F : X Z defined by [Fx](t = f(t, x(t. 2

3 Remark 2.1. With the definitions above, it is clear that solving the nonlinear boundary value problem (1-(2 is equivalent to solving Lx = Fx. It is equally clear that the solution space of the linear homogeneous problem (3, (2 is given by the Ker(L. Let Φ(t = { I if t = 0 A(t 1A(t 2 A(0 if t = 1, 2, It is well known, see [1], that Φ is the principal fundamental matrix solution to (3. While analyzing the nonlinear boundary value problem (1, (2, it will be useful to have a characterization of the Im(L. Proposition 2.2. An element h, from Z, is contained in the Im(L if and only ( m T if B 1 Φ(1Φ 1 (1h(0 + + B m Φ(m Φ 1 (i + 1h(i Ker B i Φ(i. Proof. From the variation of parameters formula, see [1], we have Lx = h if and only if there exists an element x X such that t 1 x(t = Φ(tx(0 + Φ(t Φ 1 (i + 1h(i. Here 1 Φ 1 (i + 1h(i is taken to be 0. Imposing the boundary conditions, we have h Im(L if and only if there exists w R n such that ( m B i Φ(iw + B 1 Φ(1Φ 1 (1h(0 + + B m Φ(m Φ 1 (i + 1h(i = 0 ( m B 1 Φ(1Φ 1 (1h(0 + + B m Φ(m Φ 1 (i + 1h(i R B i Φ(i. ( m ( m T Using the fact that R B i Φ(i = Ker B i Φ(i, the result follows. 3

4 m Remark 2.3. It is now clear that L is invertible if and only if B i Φ(i is invertible. In this case, from the proof of proposition 2.2, we have that the unique solution to Lx = h is given by [ m ] 1 m j 1 x(t =Φ(t B i Φ(i B j Φ(jΦ 1 (i + 1h(i j=1 Φ 1 (i + 1h(i. t 1 + ( m T We let c 1, c 2,, c p denote a basis for the Ker B i Φ(i. If we define W = [c 1,..., c p ] and Ψ(t T : {0, 1, 2,, m 1} R n by Ψ T (t = m i=t+1 W T B i Φ(iΦ 1 (t + 1, we get the following corollary: Corollary 2.4. An element h, from Z, is contained in the Im(L if and only if Ψ T (ih(i = 0. We have, from remark 2.3, that the linear homogeneous problem (3, (2 has m a nontrivial soltution space whenever B i Φ(i is singular. It will be useful to have a discription of the solution space in this case. From remark 2.1, this is equivalent to finding a description of the Ker(L. Proposition 2.5. ( The solution space of the linear homogenous problem (3, m (2 and the Ker B i Φ(i have the same dimension. 4

5 Proof. Lx = 0 x(t + 1 = A(tx(t and x(t = Φ(tx(0 and m B i x(i = 0 m B i x(i = 0 there exists c R n such that x(t = Φ(tc and m B i Φ(ic = 0. ( m It follows that the map c Φ( c is an isomorphism from Ker B i Φ(i to Ker(L. If we choose vectors b 1,, b p, where p n, from R n which form a basis for Ker( m B iφ(i we may make the following definition: Definition 2.6. We define S(t to be the n p matrix whose ith column is S i (t := Φ(tb i. We now get the following, very useful, characterization of the Ker(L: a function x Ker(L if and only if x( = S( α for some α R p. Remark 2.7. For each α 0, S( α is a nonzero solution to (3; it follows that S(tα > 0. min t=0,1,2,,m 3. Main Results We now come to our first result regarding the nonlinear boundary value problem (1, (2. We will restrict our attention to the case of invertible L; that is, the case in which the linear homogeneous problem (3, (2 has only the trivial solution. We prove the existence of solutions by placing growth restrictions on the nonlinearity f. Theorem 3.1. Suppose that the only solution to (3, (2 is the trivial solution. Suppose further that there exists a function g : R n R + and a real number M such that for all t {0, 1, 2,, m} and s R n, f(t, s M s + g(s. If g(s g(s g(r when s r and lim = 0, then the nonlinear boundary value s s problem (1, (2 has a solution provided M is sufficiently small. 5

6 Proof. Define H : X X by m m j 1 [H(x](t =Φ(t [ B i Φ(i] 1 B j Φ(jΦ 1 (i + 1f(i, x(i j=1 Φ 1 (i + 1f(i, x(i. t 1 + It follows that there exists real numbers B 1 and B 2 such that H(x MB 1 x + B 2 g(x(β x, where β x is any point with x(β x = x. If MB 1 < 1, we may choose r sufficiently large such that for all s with s r, We define B = {x X : x r}. B 2 g(s < (1 MB 1 r. It is clear that H(B B. The existence of a solution is now a consequence of Brouwer s fixed point theorem. We now focus our attention on the resonant case. From proposition 2.5 we know this is equivalent to the matrix m B iφ(i being singular. We will analyze (1, (2 using a projection scheme known as the Lyapunov-Schmidt procedure. To do so we need projections onto the Ker(L and Im(L. In this regard, we choose to follow [4]. Let V : R n R n be the orthogonal projection onto Ker( m B iφ(i. Definition 3.2. Define P : X X by [P x](t = Φ(tV x(0 Definition 3.3. Define E : Z Z by [Eh](t] = h(t Ψ(t Ψ T (jψ(j 1 j=0 Ψ T (ih(i. The proofs that E and P are projections, as well a many other properties these maps, may be found in [4]. The proof of the following result may be found in [5]. 6

7 Proposition 3.4. Solving Lx = Fx is equivalent to solving the system where M p is L 1 (Ker(P dom(l. x = P x + M p EFx and (I EFx = 0 We now come to our main result concerning the nonlinear boundary value problem (1-(2. For convenience, we assume {b 1, b 2,, b p } and {c 1, c 2,, c p } have been chosen such that S( 1 and Ψ T ( 1. The following observation will play a critical role in what follows. Since the map taking (t, α S(tα is a continuous mapping, it attains its minimum on the compact set O := {0,, m} {α R p : α = 1}. From remark 2.7, we have that η = inf S(tα > 0. (t,α O Theorem 3.5. Suppose the following conditions hold: C1. The function f is sublinear ; that is, there exists real numbers M 1, M 2 and β, with 0 β < 1, such that for each t = 0,, m, we have f(t, x M 1 x β + M 2. C2. There exists real numbers k > 0, and γ such that for all M R there exists α(m such that if α > α(m and v M, with v R n, then, Ψ T (if(i, S(iα + v k. and α, Ψ T (if(i, S(iα + v γ > k 2. C3. There exists a c such that α(m cm, eventually. Then, there exists a solution to (1-(2. 7

8 Proof. We may assume c > 1. Since f is sublinear, we may choose constants B 1 and B 2 such that following hold: 1. M p EF (w B 1 w β + B Ψ T (if(i, w(i B 1 w β + B 2. ( c Choose ε such that ε η We may then choose R, large, such that and provided w R, and M p EF (w εr Ψ T (if(i, w(i εr, α(r cr. We define an operator H : R p Im(I P R p Im(I P by H(α, x = N 1 j=0 Ψ T (jf(j, S(jα + x(j x M p EF (S( α + x We use { a scaled max } norm on the space R p Im(I P ; that is, (α, x max = 1 max c α, x. We define Ω = B(0, R in (R p Im(I P, max. We will show that deg(h, Ω, 0 0. To this end, define Q : [0, 1] Ω R p Im(I P. by Q(λ, (α, x = N 1 (1 λα + λ Ψ T (jf(j, S(jα + x(j j=0 x λm p EF (S( α + x 8

9 Q is clearly a homotopy between I and H. We will show 0 / Q(λ, (Ω for each λ (0, 1. Now it is clear that (Ω = {(α, x : α = cr and x R or α cr and x = R}. Let (α, x be in (Ω and assume α cr and x = R. If follows that S( α + x α + x cr + R = (c + 1R Since the map x B 1x β + B 2 x is decreasing, it follows that M p EF (S( α + x ε(c + 1R. Thus, and x λm p EF (S( α + x x λ M p EF (S( α + x = R λε(c + 1R > 0 Now suppose (α, x is in (Ω with α = cr and x = R. Since we have that (1 λα + λ cr > R N 1 j=0 α = cr α(r ε(c + 1R N 1 Ψ T (jf(j, S(jα + x(j j=0 Ψ T (jf(j, S(jα + x(j min { } k2 + γ, k. 2 The proof is now complete by the invariance of degree under homotopy. 4. Example Consider x(t + 1 = A(tx(t + f(t, x(t 9

10 subject to m B i x(i = 0, where A = 0 1 0, B 1 = 0 0 0, B 2 = 1 0 2, B m = 0 0 0, and B i = 0 for i 1, 2, m (m t Since A is constant, we have that Φ(t = A t = It follows that m B i Φ(i = Thus, the solution space to the linear homogenous problem has dimension ( We choose 1, m 0 as a basis for Ker B i Φ(i. It follows that (t 1 S(t = (m 1 0 (m 1m Define E =

11 For m > 2, we have that E Ker(E is invertible. We denote the inverse simply by E 1. We now define f(x 1, x 2, x 3 = x x2 3 E 1 ( x 2 ln 1 + x x2 3 4 x 3 x x f 2 (x 1, x 2, x 3, 0 where f 2 is a continuous sublinear function. To simplify notation we introduce the following. For any vector v R 2, let ˆv R 3 denote the vector (0, v 1, v 2, where v = (v 1, v 2. Also, for any vector v R 3, let v R 2 denote the vector (v 2, v 3, where v = (v 1, v 2, v ( If we choose 0, m T 1 as a basis for Ker B i Φ(i, then 1 0 Ψ T (if(i, S(iα + v = (m 1 0 (m 1m f(ˆα + v = Ef(ˆα + v = (α + v 1 ln (1 + α + v 1 + α + v. (α + v 2 α + v Now, 1 (α + v 1 ln (1 + α + v 1 + α + v (α + v 2 α + v 2 = ( 2 1 ( (α + v 1 + α + v 2 1 ln 2 (1 + α + v + (α + v 2 2 α + v ( α + v α + v ln 2 (1 + α + v. 11

12 We also have that, α, Ψ T (if(i, S(iα + v = 1 ( 1 + α + v (α + v 2 1 ln (1 + α + v + (α + v 2 2 α + v α + v 1 + α + v ln (1 + α + v. If we choose a real number c, such that for all x R with x c and x 1 + x > 1 2 ln 1 + x > 1, then it is clear that k and α(m form theorem 3.5 may be chosen as 1 2 and c + M respectively. Since for any d > 1, c + M < dm eventually, the nonlinear boundary value problem has a solution. References [1] W.G. Kelley, A. Peterson, Difference Equations, Academic Press, New York, [2] D. Pollack, P. Taylor, Multipoint boundary value problems for discrete nonlinear systems at resonance, Int. J. Pure Appl. Math. 63 ( [3] J. Rodríguez, P. Taylor, Scalar discrete nonlinear multipoint boundary value problems, J. Math. Anal. Appl. 330 ( [4] J. Rodríguez, P. Taylor, Weakly nonlinear discrete multipoint boundary value problems, J. Math. Anal. Appl. 329 ( [5] J. Rodríguez, P. Taylor, Multipoint boundary value problems for nonlinear ordinary differential equations, Nonlinear Analysis 68 (