A SYMMETRIC SOLUTION OF A MULTIPOINT BOUNDARY VALUE PROBLEM AT RESONANCE

We apply a coincidence degree theorem of Mawhin to show the existence of at least one symmetric solution of the nonlinear second-order multipoint boundary value problem u′′(t)= f (t,u(t),|u′(t)|), t ∈ (0,1), u(0)=∑i=1μiu(ξi), u(1− t)= u(t), t ∈ [0,1], where 0 < ξ1 < ξ2 < ··· < ξn ≤ 1/2, ∑n i=1μi = 1, f : [0,1]×R2 →R with f (t,x, y)= f (1− t,x, y), (t,x, y)∈ [0,1]×R2, satisfying the Carathéodory conditions.


Definitions and technical results
We study symmetric solutions of the multipoint nonlinear boundary value problem u (t) = f t,u(t), u (t) , t ∈ (0,1), (1.1) where and the inhomogeneous term satisfies (H0) f : If there is a μ i > 1, we assume, in addition, that n i=1 μ i ξ i 1 − ξ i = 0. (1.6) Due to the condition (1.4) the differential operator in the left side of (1.1) is not invertible.In the literature, boundary value problems of this type are referred to as problems at resonance.Boundary value problems at resonance have been studied by several authors including the most recent works [1][2][3][4][5][6][7][8][9]11].In the recent works [5,6,8,9], the inhomogeneous term is either a continuous function on [0,1] × R 2 or the sum of a continuous and a Lebesgue integrable functions.In this note, we merely require measurability of f in the first variable, continuity in the rest of variables for a. a. values of t, and, in addition, f being locally bounded by Lebesgue integrable functions for a. a. values of t.The above assumptions constitute the so-called Carathéodory conditions.
In this section, we provide the necessary background definitions and facts and state the key theorem due to Mawhin [10].In the second section, we provide additional assumptions on the inhomogeneous term and give the sufficient conditions of existence of at least one solution of (1.1)- (1.3).The emphasis in this note is on symmetric solutions at resonance.Definition 1.1.Let X and Z be normed spaces.A linear mapping L : domL ⊂ X → Z is called a Fredholm mapping if the following two conditions hold: (i) kerL has a finite dimension, (ii) ImL is closed and has a finite codimension.If L is a Fredholm mapping, its (Fredholm) index is the integer Ind L = dimkerL − codim Im L.
In this paper, we are concerned with a Fredholm mapping of index zero.From Definition 1.1, it follows that there exist continuous projectors P : X → X and Q : Z → Z such that and that the mapping is invertible.We denote the inverse of L| domL∩ker P by K P : ImL → dom L ∩ ker P. The generalized inverse of L denoted by K P,Q : Z → dom L ∩ ker P is defined by If L is a Fredholm mapping of index zero, then for every isomorphism J : ImQ → ker L, the mapping JQ + K P,Q : Z → dom L is an isomorphism and, for every u ∈ dom L, (1.9) Definition 1.2.Let L : domL ⊂ X → Z be a Fredholm mapping, let E be a metric space, and let N : E → Z be a mapping.Say that N is L-compact on E if QN : E → Z and K P,Q N : E → X are compact on E. In addition, say that N is L-completely continuous if it is Lcompact on every bounded E ⊂ X.
Nickolai Kosmatov 3 When the boundary value problem is shown to be equivalent to the abstract equation Lu = Nu, the existence of a solution will be guaranteed by the following theorem due to Mawhin [10,Theorem IV.13].
The following definition introduces the so-called Carathéodory conditions imposed on a map.Definition 1.4.Say that the map f : [0,1]×R n →R, (t,z) → f (t,z) satisfies the Carathéodory conditions with respect to L 1 [0,1] if the following conditions are satisfied: We introduce the Sobolev space by Define the mapping N : X → Z by Let u ∈ dom L and consider the linear equation (1.17) Conversely, if (1.17) holds for some g ∈ Z, we take the candidate of u ∈ dom L as given by (1.16) and establish that it is symmetric, absolutely continuous along with its derivative, u (t) = g(t) for a. a. t ∈ (0,1) and (1.2) is satisfied.In fact, we have 3) and (1.17) . (1.18) We recall the condition (1.6) and define the continuous linear mapping It is easy to see that Q 2 g = Qg for all g ∈ Z, that is, the mapping Q is idempotent.Observe also that (1.17) and (1.19) imply that ImL = ker Q.Take g ∈ Z in the form g = (g − Qg) + Qg so that g − Qg ∈ Im L and Qg ∈ R. If g ≡ c = 0, then, by (1.6), Qg = 0, which implies that ImL ∩ R = {0}.Hence Z = Im L ⊕ R. Now, Ind L = dimkerL − codim Im L = 0 and so L is a Fredholm mapping of index zero.
The continuous projector P : X → X is defined by (1.20) Note that the projectors P and Q are exact, that is, satisfy the relationships (1.7).Define K P : ImL → dom L ∩ ker P by Nickolai Kosmatov 5 Then K P g ∞ ≤ 2 g 1 and (K P g) ∞ ≤ 2 g 1 , and thus In fact if g ∈ Im L, then Also, if u ∈ dom L ∩ kerP, then (1.25) (since u ∈ ker P and u is symmetric, u(0) = u(1) = 0).Thus, we get that For convenience, we introduce a constant ( for all t ∈ [0,1] and k ∈ N, that is, the sequence {v k } is uniformly bounded on [0,1] and as such is equicontinuous on [0,1].Since {v k } is uniformly bounded and equicontinuous on [0,1], by Arzela-Ascoli theorem, it has a subsequence {v kl } that converges to some Nickolai Kosmatov 7 Consider the sequence {w kl } defined by Similar considerations apply to show that QN is continuous and that QN(E) is relatively compact.Now, since the mappings QN and K P,Q N are compact on an arbitrary bounded E ⊂ X, the mapping N : X → Z is L-completely continuous by Definition 1.2.
We take our isomorphism, J, to be the identity map Id : kerL → Im L, that is, Jc = c for c ∈ R. Set which, in either case, is a contradiction.If the other part of (H3) is satisfied, then we take and, again, obtain a contradiction.Thus, in either case u = c ≤ B for all u ∈ Ω 3 , that is, Ω 3 is bounded.
Let Ω be open and bounded such that 3 i=1 Ω i ⊂ Ω.Then the assumptions (i) and (ii) of Theorem 1.3 are fulfilled.By Definition 1.2, the mapping N is L-compact on Ω. Lemma 1.5 establishes that L is Fredholm of index zero.It only remains to verify that the third assumption of Theorem 1.3 applies.
× R 2 → R satisfies the Carathéodory conditions with respect to L 1 [0,1], there exists a Lebesgue integrable function α r such that for all k ∈ N and a.e.t 1.28) Lemma 1.6.The mapping N is L-completely continuous.Proof.Let E ⊂ X be bounded and {u k }⊂E.Define the sequence {v k } by v k (t)=K P,Q Nu k (t).Set r = sup u : u ∈ E .(1.29) Since the function f : [0,1] ∈ [0,1], Employing arguments similar to that for {v k } one can show that {w kl } is uniformly bounded and equicontinuous on [0,1].Hence {w kl } as a subsequence that converges to some w ∈ C[0,1].In fact, w(t) = v (t), t ∈ [0,1] and, thus, there is a subsequence of {v kl } that converges in C 1 [0,1].Therefore, the image of E under K P,Q N is relatively compact.Since the function f : [0,1] × R 2 → R satisfies the Carathéodory conditions with respect to L 1 [0,1], the continuity of K P,Q N on E follows from the Lebesgue dominated convergence theorem.