Electronic Journal of Qualitative Theory of Differential Equations Spec. Ed. I, 2009 No. 16, 1–11; http://www.math.u-szeged.hu/ejqtde/

A THIRD ORDER NONLOCAL BOUNDARY VALUE PROBLEM AT RESONANCE

Eric R. Kaufmann

Department of Mathematics & Statistics, University of Arkansas at Little Rock Little Rock, AR 72204 USA e-mail: erkaufmann@ualr.edu

Honoring the Career of John Graef on the Occasion of His Sixty-Seventh Birthday

Abstract

We consider the third-order nonlocal boundary value problem

u′′′(t) = f(t, u(t)), a.e. in (0, 1),
u(0) = 0, u′(ρ) = 0,
u′′(1) = λ[u′′],

where 0 < ρ < 1, the nonlinear term f satisfies Carathéodory conditions with respect to L^{1}[0, T], λ[v] = ∫_{0}^{1}v(t) dΛ(t), and the functional λ satisfies the resonance condition λ[1] = 1. The existence of a solution is established via Mawhin’s coincidence degree theory.

Key words and phrases: Coincidence degree theory, nonlocal boundary value problem, resonance. AMS (MOS) Subject Classifications: 34B10, 34B15

1 Introduction

We study the third-order nonlocal boundary value problem

u′′′(t) = f(t, u(t)), a.e. in (0, 1),  (1)
u(0) = 0, u′(ρ) = 0,  (2)
u′′(1) = λ[u′′],  (3)

where 0 < ρ < 1, the nonlinear term f satisfies Carathéodory conditions with respect to L^{1}[0, T], and λ[v] is a linear functional defined by the Riemann-Stieltjes integral λ[v] = ∫_{0}^{1}v(t) dΛ(t). Here Λ is a suitable monotonically increasing function on [0, 1]. We assume throughout that the functional λ satisfies

λ[1] = 1.  (4)

Due to the condition (4), the boundary value problem (1)-(3) can not be inverted, and as such, we say that the boundary value problem is at resonance. Recently, several authors have studied nonlocal boundary value problems at resonance, see for example [1, 2, 3, 9, 10, 11, 12, 13, 15, 16, 17, 18] and references therein. The literature is rich also with articles on nonlocal boundary value problems; see [4, 5, 6, 7, 8] and references therein. The primary motivation for this work is the article [8] due to Graef and Webb. In [8], the authors consider the existence of multiple positive solutions for the nonlocal boundary value problem

u′′′(t) = f(t, u(t)), t ∈ (0, 1),
u(0) = 0, u′(ρ) = 0,
u′′(1) = λ[u′′],

where ρ > 1/2 and λ[v] = ∫_{0}^{1}v(t) dΛ(t), as well as a generalization of this problem. To ensure that the boundary value problem is invertible, the authors impose the condition that λ[1] ≠ 1.

In Section 2 we give the background information from coincidence degree theory. So that the paper is self-contained, we state Mawhin’s coincidence theorem, [14], in this section. We also define appropriate mappings and projectors that will be used in the sequel. We state and prove our main result in Section 3.

2 Background

Let X and Z be normed spaces. A linear mapping L : dom L ⊂ X → Z is called a Fredholm mapping if ker L has a finite dimension, and Im L is closed and has finite codimension. The (Fredholm) index of a Fredholm mapping L is the integer, Ind L, given by Ind L = dim ker L − codim Im L.

For a Fredholm map of index zero, L : dom L ⊂ X → Z, there exist continuous projectors P : X → X and Q : Z → Z such that

Im P = ker L, ker Q = Im L, X = ker L ⊕ ker P, Z = Im L ⊕ Im Q,

and the mapping

L|_{dom L∩ker P} : dom L ∩ ker P → Im L

is invertible. The inverse of L|_{dom L∩ker P} is denoted by

K_{P} : Im L → dom L ∩ ker P.

The generalized inverse of L, denoted by K_{P,Q} : Z → dom L ∩ ker P, is defined by K_{P,Q} = K_{P} (I − Q).

If L is a Fredholm mapping of index zero, then for every isomorphism J : Im Q → ker L, the mapping JQ + K_{P,Q} : Z → dom L is an isomorphism and, for every u ∈ dom L,

(JQ + K_{P,Q})^{−1}u = (L + J^{−1}P)u.

Definition 2.1 Let L : dom L ⊂ X → Z be a Fredholm mapping, E be a metric space, and N : E → Z. We 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, we say that N is L-completely continuous if it is L-compact on every bounded E ⊂ X.

We will formulate the boundary value problem (1)-(3) as Lu = Nu, where L and N are appropriate operators. The existence of a solution to the boundary value problem will then be guaranteed by the following theorem due to Mawhin [14].

Theorem 2.1 Let Ω ⊂ X be open and bounded. Let L be a Fredholm mapping of index zero and let N be L-compact on \overline{Ω}. Assume that the following conditions are satisfied:

(i) Lu ≠ μNu for every (u, μ) ∈ ((dom L \ ker L) ∩ ∂Ω) × (0, 1);

(ii) Nu ∉ Im L for every u ∈ ker L ∩ ∂Ω;

(iii) deg_{B} (JQN|_{ker L∩∂Ω}, Ω ∩ ker L, 0) ≠ 0, with Q : Z → Z a continuous projector, such that ker Q = Im L and J : Im Q → ker L is an isomorphism.

Then the equation Lu = Nu has at least one solution in dom L ∩ \overline{Ω}.

We say that the map f : [0, T] × ℝ^{n} → ℝ satisfies Carathéodory conditions with respect to L^{1}[0, T] if the following conditions hold.

(i) For each z ∈ ℝ^{n}, the mapping t ↦ f(t, z) is Lebesgue measurable.

(ii) For a.e. t ∈ [0, T], the mapping z ↦ f(t, z) is continuous on ℝ^{n}.

(iii) For each r > 0, there exists α_{r} ∈ L^{1}([0, T], ℝ) such that for a.e. t ∈ [0, T] and for all z such that |z| < r, we have |f(t, z)| ≤ α_{r}(t).

Let AC[0, 1] denote the space of absolutely continuous functions on the interval [0, 1]. Define Z = L^{1}[0, 1] with norm ∥ · ∥_{1} and let

X = {u : u, u′, u′′ ∈ AC[0, 1] and u′′′ ∈ L^{1}[0, 1]}

be equipped with the norm ∥u∥ = max {∥u∥_{0}, ∥u′∥_{0}, ∥u′′∥_{0}}. Define the mapping L : dom L ⊂ X → Z, where

dom L = {u ∈ X : u satisfies (2) and (3)},

by

Lu(t) = u′′′(t), t ∈ [0, 1].

Define N : X → Z by

Nu(t) = f(t, u(t)), t ∈ [0, 1].

Lemma 2.1 The mapping L : dom L ⊂ X → Z is a Fredholm mapping of index zero.

Proof. Let g ∈ Z, and for t ∈ [0, 1], let

u(t) = a (t^{2}/2 − ρt) + ∫_{0}^{t}∫_{ρ}^{s}∫_{0}^{r}g(τ) dτ dr ds.  (5)

Then, u′′′(t) = g(t) for a.e. t ∈ [0, 1], u(0) = 0, and u′(ρ) = 0. Furthermore, if

∫_{0}^{1}g(s) ds − λ [∫_{0}^{t}g(s) ds] = 0,

then u satisfies the nonlocal boundary condition u′′(1) = λ[u′′]. If (5) is satisfied, then u ∈ dom L, and so,

{g ∈ Z : ∫_{0}^{1}g(s) ds − λ [∫_{0}^{t}g(s) ds] = 0} ⊆ Im L.

Now let g ∈ Im L. Then there exists a u ∈ dom L such that u′′′(t) = g(t) for a.e. t ∈ [0, 1]. We have,

u′′(t) = u′′(0) + ∫_{0}^{t}g(s) ds.  (6)

Apply the functional λ[v] to both sides of (6) to obtain,

λ[u′′] = u′′(0) + λ [∫_{0}^{t}g(s) ds] .  (7)

Also, evaluate (6) at t = 1.

u′′(1) = u′′(0) + ∫_{0}^{1}g(s) ds.  (8)

From the boundary condition u′′(1) = λ[u′′], and (7) and (8), we see that g satisfies

∫_{0}^{1}g(s) ds = λ [∫_{0}^{t}g(s) ds] .

Hence

Im L ⊆ {g ∈ Z : ∫_{0}^{1}g(s) ds − λ [∫_{0}^{t}g(s) ds] = 0} .

Since both inclusions hold, then

Im L = {g ∈ Z : ∫_{0}^{1}g(s) ds − λ [∫_{0}^{t}g(s) ds] = 0} .

Let φ ∈ C[0, 1] be such that φ(t) > 0 for t ∈ [0, 1] and

C ≡ ∫_{0}^{1}φ(t) dt − λ [∫_{0}^{t}φ(s) ds] ≠ 0.

Define Q_{1} : Z → ℝ by

Q_{1}g ≡ ∫_{0}^{1}g(s) ds − λ [∫_{0}^{t}g(s) ds] .

Now define the mapping Q : Z → Z by

(Qg)(t) = \frac{1}{C} (Q_{1}g)φ(t).  (9)

The mapping Q is a continuous linear mapping and

(Q^{2}g)(t) = (Q(Qg))(t)
= \frac{1}{C} (\frac{1}{C} Q_{1}g) (∫_{0}^{1}φ(t) dt − λ [∫_{0}^{t}φ(s) ds]) φ(t)
= (\frac{1}{C} Q_{1}g) φ(t) = (Qg)(t).

That is, Q^{2}g = Qg for all g ∈ Z. Furthermore, Im L = ker Q.

For g ∈ Z we have g−Qg ∈ ker Q = Im L and Qg ∈ Im Q. Hence Z = Im L+Im Q. Let g ∈ Im L ∩ Im Q. Since g ∈ Im Q, then there exists an η ∈ ℝ such that g(t) = ηφ(t), t ∈ [0, 1]. Since g ∈ Im L = ker Q, then

0 = Q_{1}g(t) = η (∫_{0}^{1}φ(t) dt − λ [∫_{0}^{t}φ(s) ds]) = ηC.

Since C ≠ 0, then η = 0 and so g(t) ≡ 0, t ∈ [0, 1]. Consequently, Z = Im L ⊕ ker Q. Note ker L = {c (t^{2}/2 − ρt) : c ∈ ℝ} ≅ ℝ, and so dim ker L = codim Im L = dim Im Q = 1. Thus, L is a Fredholm mapping on index zero. The proof is complete.

We are now ready to give the other projector employed in the proof of our main result. Define P : X → X by

(P u)(t) = u′′(0) (t^{2}/2 − ρt) , t ∈ [0, 1],  (10)

and note that ker P = {u ∈ X : u′′(0) = 0} and Im P = ker L. Since (P u)′′(t) = u′′(0), then (P^{2}u)(t) = (P u)(t), t ∈ [0, 1]. For all u ∈ X we have

u(t) = u′′(0) (t^{2}/2 − ρt) + (u(t) − u′′(0)(t^{2}/2 − ρt)),

and so, X = ker L ⊕ ker P.

Define the mapping K_{P} : Im L → dom L ∩ ker P by

K_{P} g(t) = ∫_{0}^{t}∫_{ρ}^{s}∫_{0}^{r}g(τ) dτ dr ds.

It follows that ∥K_{P} g∥_{0} ≤ \frac{1}{2}(1 − ρ)^{2}∥g∥_{1}, ∥(K_{P} g)′∥_{0} ≤ (1 − ρ)∥g∥_{1}, and ∥(K_{P} g)′′∥_{0} ≤ ∥g∥_{1}. As such, we have

∥K_{p}g∥ = max {∥K_{P} g∥_{0}, ∥(K_{P} g)′∥_{0}, ∥(K_{P} g)′′∥_{0}} ≤ ∥g∥_{1}.  (11)

If g ∈ Im L then,

(LK_{P})g(t) = \frac{d^{3}}{dt^{3}} ∫_{0}^{t}∫_{ρ}^{s}∫_{0}^{r}g(τ) dτ dr ds = g(t).

And if u ∈ dom L ∩ ker P then,

(K_{p}L)u(t) = ∫_{0}^{t}∫_{ρ}^{s}∫_{0}^{r}u′′′(τ) dτ dr ds
= u(t) − u(0) − tu′(ρ) − u′′(0) (t^{2}/2 − ρt)
= u(t).

Consequently, K_{P} = (L|_{dom L∩ker P})^{−1}. The generalized inverse of L is defined by

K_{P,Q}u(t) = ∫_{0}^{t}∫_{ρ}^{s}∫_{0}^{r}u(τ) − Qu(τ) dτ dr ds.

Lastly in this section, we show that N is L-compact.

Lemma 2.2 The mapping N : X → Z given by Nu(t) = f(t, u(t)) is L-completely continuous.

Proof. Let E ⊂ X be a bounded set and let r be such that ∥u∥ ≤ r for all u ∈ E. Since f satisfies Carathéodory conditions, there exists an α_{r} ∈ L^{1}[0, 1] such that for a.e. t ∈ [0, T] and for all z such that |z| < r we have |f(t, z)| ≤ α_{r}(t). Let M = ∫_{0}^{1}dΛ(t). Then,

∥QNu(t)∥_{1} ≤ \frac{1}{C} ∫_{0}^{1}∫_{0}^{1}|f(s, u(s))| ds + λ [∫_{0}^{t} |f(s, u(s))| ds] φ(t) dt
≤ \frac{1}{C} (∫_{0}^{1}∫_{0}^{1}α_{r}(s) dsφ(t) dt + ∫_{0}^{1} λ [∫_{0}^{t}α_{r}(s) ds] φ(t) dt)
≤ \frac{1}{C} ∥φ∥_{1} (∥α_{r}∥_{1} + ∫_{0}^{1}∫_{0}^{t} α_{r}(s) ds dΛ(t))
≤ \frac{1}{C} ∥φ∥_{1}∥α_{r}∥_{1}(1 + M).

Hence, QN(E) is uniformly bounded.

It is clear that the functions QN(u) are equicontinuous on E. By the Arzelà-Ascoli Theorem, QN(E) is relatively compact.

It can be shown that K_{P,Q}N(E) is relatively compact as well. As such, the mapping N : X → Z is L-completely continuous and the proof is complete.

3 Existence of Solutions

We will assume that the following conditions hold.

(H_{1}) There exists a constant c_{1} > 0 such that for all u ∈ dom L \ ker L satisfying |u′′(t)| > c_{1}, t ∈ [0, T], we have

QNu(t) ≠ 0, for all t ∈ [0, 1].

(H_{2}) There exist β, γ ∈ L^{1}[0, T], such that for all u ∈ ℝ and for all t ∈ [0, 1],

|f(t, u)| ≤ β(t) + γ(t)|u|.

(H_{3}) There exists a B > 0 such that for all c_{2} ∈ ℝ with |c_{2}| > B, either

c_{2} (∫_{0}^{1}f (s, c_{2}(s^{2}/2 − ρs)) ds − λ [∫_{0}^{t}f (s, c_{2}(s^{2}/2 − ρs)) ds]) < 0

or

c_{2} (∫_{0}^{1}f (s, c_{2}(s^{2}/2 − ρs)) ds − λ [∫_{0}^{t}f (s, c_{2}(s^{2}/2 − ρs)) ds]) > 0.

Theorem 3.1 Assume that conditions (H_{1})-(H_{3}) hold and that

1 − 2∥γ∥_{1} > 0.  (12)

Then the nonlinear periodic problem (1)-(3) has at least one solution.

Proof. Let Q : Z → Z and P : X → X be defined as in (9) and (10), respectively. We begin by constructing a bounded open set Ω that satisfies Theorem 2.1. To this end, define the set Ω_{1} as follows.

Ω_{1} = {u ∈ dom L \ ker L : Lu = μNu for some μ ∈ (0, 1)}.

Let u ∈ Ω_{1} and write u as u = P u + (I − P)u. Then

∥u∥ ≤ ∥P u∥ + ∥(I − P)u∥.  (13)

Since u ∈ Ω_{1} then (I −P)u ∈ dom L∩ker P = Im K_{P}, 0 < μ < 1 and Nu = \frac{1}{μ}Lu ∈ Im L. From (11) we have,

∥(I − P)u∥ = ∥K_{P} L(I − P)u∥ ≤ ∥L(I − P)u∥ = ∥Lu∥ < ∥Nu∥.  (14)

From (H_{2}) we see that ∥Nu∥ ≤ ∥β∥_{1} + ∥γ∥_{1}∥u∥, and so by (13) and (14), we obtain,

∥u∥ < ∥P u∥ + ∥β∥_{1} + ∥γ∥_{1}∥u∥.  (15)

Now, P u(t) = u′′(0) (t^{2}/2 − ρt). Since 0 < ρ < 1, then ∥P u∥ = |u′′(0)|. Also, since u ∈ Ω_{1} and ker Q = Im L, then QNu(t) = 0 for all t ∈ [0, 1]. By (H_{1}), there exists t_{0} ∈ [0, 1] such that |u′′(t_{0})| ≤ c_{1}. Now,

u′′(0) = u′′(t_{0}) − ∫_{0}^{t_{0}}u′′′(s) ds,

and so,

|u′′(0)| ≤ |u′′(t_{0})| + ∫_{0}^{t_{0}}|u′′′(s)| ds ≤ c_{1} + ∥Nu∥.

Consequently,

∥P u∥ = |u′′(0)| ≤ c_{1} + ∥β∥_{1} + ∥γ∥_{1}∥u∥.  (16)

By (15) and (16), we have for u ∈ Ω_{1},

∥u∥ ≤ c_{1} + 2∥β∥_{1} + 2∥γ∥_{1}∥u∥,

which, using (12), implies that

∥u∥ ≤ \frac{c_{1} + 2∥β∥}{1 − 2∥γ∥_{1}}.

The set Ω_{1} is bounded.

Next define the set Ω_{2} by

Ω_{2} = {u ∈ ker L : Nu ∈ Im L}.

Let u ∈ Ω_{2}. Since u ∈ ker L, then u(t) = c_{2} (t^{2}/2 − ρt) for some c_{2} ∈ ℝ. We also know that Nu ∈ Im L = ker Q and so,

0 = Q_{1}Nu = ∫_{0}^{1}f (s, c_{2}(s^{2}/2 − ρs)) ds − λ [∫_{0}^{t}f (s, c_{2}(s^{2}/2 − ρs)) ds] .

It follows from (H_{3}) that |c_{2}| < B and so Ω_{2} is bounded.

Before we define the set Ω_{3}, we must state our isomorphism, J : Im Q → ker L. Let

J(cφ(t)) = c(t^{2}/2 − ρt).

Suppose that the first part of (H_{3}) is satisfied. Then define

Ω_{3} = {u ∈ ker L := −μJ^{−1}u + (1 − μ)QNu = 0, μ ∈ [0, 1]}

and note that for each u ∈ Ω_{3},

μcφ(t) = (1 − μ)\frac{1}{C} (∫_{0}^{1}f (s, c_{2}(s^{2}/2 − ρs)) ds − λ [∫_{0}^{t}f (s, c_{2}(s^{2}/2 − ρs)) ds]) .

Suppose that μ = 1, then c = 0. If |c| > B, then

μc^{2}φ(t) = (1 − μ)\frac{c}{C} (∫_{0}^{1}f (s, c_{2}(s^{2}/2 − ρs)) ds
− λ [∫_{0}^{t}f (s, c_{2}(s^{2}/2 − ρs)) ds]) < 0.

In either case we get a contradiction and hence Ω_{3} is bounded.

If the second part of (H_{3}) is satisfied then define Ω_{3} by

Ω_{3} = {u ∈ ker L := μJ^{−1}u + (1 − μ)QNu = 0, μ ∈ [0, 1]} .

A similar argument as above shows that Ω_{3} is bounded.

Let Ω be an open and bounded set such that ∪_{i=1}^{3}\overline{Ω}_{i} ⊂ Ω. Then the assumptions (i) and (ii) of Theorem 2.1 are satisfied. By Lemma 2.1, L : dom L ⊂ X → Z is a Fredholm mapping of index zero. By Lemma 2.2, the mapping N : X → Z is Lcompletely continuous. We only need to verify that condition (iii) of Theorem 2.1 is satisfied.

We apply the invariance under a homotopy property of the Brower degree. Let

H(u, μ) = ±Id u + (1 − μ)JQNu.

If u ∈ ker L ∩ ∂Ω, then

deg_{B} (JQN|_{ker L∩∂Ω}, Ω ∩ ker L, 0) = deg_{B} (H(·, 0), Ω ∩ ker L, 0)
= deg_{B} (H(·, 1), Ω ∩ ker L, 0)
= deg_{B} ( ± Id, Ω ∩ ker L, 0)
≠ 0.

All the assumptions of Theorem 2.1 are fulfilled and the proof is complete.

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