# Existence of Positive Solutions for a Fourth-Order Periodic Boundary Value Problem

Authors: Yongxiang Li
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

The existence results of positive solutions are obtained for the fourth-order periodic boundary value problemu(4)−βu′′+αu=f(t,u,u′′), 0≤t≤1, u(i)(0)=u(i)(1),  i=0,1,2,3,  where  f:[0,1]×ℝ+×ℝ→ℝ+ is continuous, α,β∈ℝ, and satisfy 0<α<((β/2)+2π2)2, β>−2π2,(α/π4)+(β/π2)+1>0. The discussion is based on the fixed point index theory in cones.


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## Body

## 1. Introduction

This paper concerns the existence of positive solutions for the fourth-order periodic boundary value problem (PBVP)(1.1)u(4)(t)-βu′′(t)+αu(t)=f(t,u(t),u′′(t)),0≤t≤1,u(i)(0)=u(i)(1),i=0,1,2,3,
where α,β∈ℝ and f:[0,1]×ℝ+×ℝ→ℝ+ is continuous, ℝ+=[0,∞). PBVP (1.1) describes the deformations of an elastic beam in equilibrium state with periodic boundary condition. In the equation, the u′′ denotes the bending moment term which represents bending effect. Owing to its importance in physics, the existence of solutions to this problem has been studied by some authors, see [1–6]. In practice, only its positive solutions are significant. In this paper, we discuss the existence of positive solutions of PBVP (1.1).In [1, 2], Cabada and Lois obtained the maximum principles for fourth-order operator L4,αu=u(4)+αu in periodic boundary condition and then they proved the existence of solutions and the validity of the monotone method in the presence of lower and upper solutions for the periodic boundary problem (1.2)u(4)(t)=g(t,u(t)),0≤t≤1,u(i)(0)=u(i)(1),i=0,1,2,3.
In [3], the present author established a strongly maximum principle for operator L4u=u(4)-βu′′+αu in periodic boundary condition, and showed that if α,β satisfy the assumption
(1.3)0<α<(β2+2π2)2,β>-2π2,απ4+βπ2+1>0,
then L4 is strongly inverse positive in space(1.4)F4={u∈C4[0,1]∣u(i)(0)=u(i)(1),i=0,1,2;u(3)(0)≥u(3)(1)}.
As an application of this strongly maximum principle, the author considered the existence of positive solutions for the special fourth-order periodic boundary problem (1.5)u(4)(t)-βu′′(t)+αu(t)=g(t,u(t)),0≤t≤1,u(i)(0)=u(i)(1),i=0,1,2,3,
and obtained the following result.Theorem A.
Letg:[0,1]×ℝ+→ℝ+ be continuous and the assumption  (1.3) hold. If  g satisfies one of the following conditions
(G1)g0<α,g∞>α;
(G2)g0>α,g∞<α,
where(1.6)g0=lim⁡⁡inf⁡u→0+min⁡t∈[0,1](f(t,u)u),g0=lim⁡⁡sup⁡u→0+max⁡t∈[0,1](f(t,u)u),g∞=lim⁡⁡inf⁡u→+∞min⁡t∈[0,1](f(t,u)u),g∞=lim⁡⁡sup⁡u→+∞max⁡t∈[0,1](f(t,u)u),
then PBVP (1.5) has at least one positive solution.Based upon this strongly maximum principle, the authors of [4, 5] further consider the existence and multiplicity of positive solutions of PBVP (1.5). In [6], Bereanu obtained existence results for PBVP (1.5) by using the method of topological degree. However, all of these works are on the special equation (1.5), and few people consider the existence of positive solutions of PBVP (1.1) that nonlinearity f contains the bending moment term u′′. The purpose of this paper is to discuss the existence of positive solutions of PBVP (1.1).The strongly maximum principle implies that the fourth-order linear boundary value problem (LBVP)(1.7)L4u:=u(4)-βu′′+αu=0,0≤t≤1,u(i)(0)-u(i)(1)=0,i=0,1,2,u(3)(0)-u(3)(1)=1
has a unique positive solution Φ:[0,1]→(0,∞), see [3, Lemma 3]. This function has been introduced in [2, Lemma 2.1 and Remark 2.1]. Let I=[0,1], and set
(1.8)σ=min⁡t∈IΦ(t)max⁡t∈IΦ(t),M=max⁡t∈I|Φ′′(t)|min⁡t∈IΦ(t).
Let f:I×ℝ+×ℝ→ℝ+ be continuous. To be convenient, we introduce the notations(1.9)f0=lim⁡⁡inf⁡u→0+min⁡|v|≤Mu,t∈I⁡(f(t,u,v)u),f0=lim⁡⁡sup⁡u→0+max⁡|v|≤Mu,t∈I⁡(f(t,u,v)u),f∞=lim⁡⁡inf⁡u→+∞min⁡|v|≤Mu,t∈I⁡(f(t,u,v)u),f∞=lim⁡⁡sup⁡u→+∞max⁡|v|≤Mu,t∈I⁡(f(t,u,v)u).Our main result is as follows.Theorem 1.1.
Letf:[0,1]×ℝ+×ℝ→ℝ+ be continuous, and let the assumption  (1.3) hold. If  f satisfies one of the following conditions:
(F1)f0<α,f∞>α,
(F2)f0>α,f∞<α,
then PBVP (1.1) has at least one positive solution.Clearly, Theorem1.1 is an extension of Theorem A. Since that α is an eigenvalue of linear eigenvalue problem (1.10)u(4)-βu′′+αu=λu,
with periodic boundary condition, if one inequality in (F1) or (F2) of Theorem 1.1 is not true, the existence of solution to PBVP (1.1) cannot be guaranteed. Hence, (F1) and (F2) are the optimal conditions for the existence of the positive of PBVP (1.1).In Theorem1.1, the condition (F1) allows that f(t,u,v) may be superlinear growth on u and v, for example, f(t,u,v)=u2+v2, and the condition (F2) allows that f(t,u,v) may be sublinear growth on u and v, for example, f(t,u,v)=u2+v23.The proof of Theorem1.1 is based on the theory of the fixed point index in cones. Since the nonlinearity f of PBVP (1.1) contains u′′, the argument of Theorem A in [3] is not applicable to Theorem 1.1. We will prove Theorem 1.1 by choosing a proper cone of C2(I) in Section 3. Some preliminaries to discuss PBVP (1.1) are presented in Section 2.


## 2. Preliminaries

LetC(I) be the Banach space of all continuous functions on the unit interval I=[0,1] with the norm ∥u∥C=max0≤t≤ω|u(t)|. Let C+(I) denote the cone of all nonnegative functions in C(I). Generaly, for n∈ℕ, we use Cn(I) to denote the Banach space of the nth-order continuous differentiable functions on I with the norm ∥u∥Cn=∑k=1n∥u(k)∥C. In C2(I), we define a new norm by
(2.1)‖u‖C02=‖u‖C+‖u′′‖C.
Then ∥u∥C02 is equivalent to ∥u∥C2. In fact, for every u∈C2(I), it is clear that ∥u∥C02≤∥u∥C2. On the other hand, by the Lagrange mean-value theorem, there exists ξ∈(0,1) such that u(1)-u(0)=u′(ξ). For t∈I, we have(2.2)|u′(t)|≤|u′(t)-u′(ξ)|+|u′(ξ)|=|∫ξtu′′(s)ds|+|u(1)-u(0)|≤∫01|u′′(s)|ds+|u(1)|+|u(0)|≤‖u′′‖C+2‖u‖C≤2‖u‖C02.
Hence, ∥u′∥C≤2∥u∥C02. By this, we have
(2.3)‖u‖C2=‖u‖C+‖u′‖C+‖u′′‖C=‖u‖C02+‖u′‖C≤3‖u‖C02.Therefore, the norms∥u∥C02 and ∥u∥C2 are equivalent.Letα,β∈ℝ satisfy the assumption (1.3). For h∈C(I), we consider the fourth-order linear periodic boundary value problem (LPBVP) (2.4)u(4)(t)-βu′′(t)+αu(t)=h(t),0≤t≤1,u(i)(0)=u(i)(1),i=0,1,2,3.
Let Φ(t) be the unique positive solution of LBVP(1.7), and set (2.5)G(t,s)={Φ(t-s),0≤s≤t≤1,Φ(1+t-s),0≤t<s≤1.
By [3, Lemma 1], we have the following result.Lemma 2.1.
Letα,β∈ℝ satisfy the assumption  (1.3). Then for every  h∈C(I), LPBVP (2.4) has a unique solution  u(t) which is given by
(2.6)u(t)=∫01G(t,s)h(s)ds:=Sh(t),t∈R.
Moreover, S:C(I)→C4(I) is a linear bounded operator.Letσ and M be the positive constants given by (1.8). Choose a cone K in C2(I) by
(2.7)K={u∈C2(I)∣u(t)≥σ‖u‖C,|u′′(t)|≤M|u(t)|,t∈I}.
We have the following.Lemma 2.2.
Letα,β∈ℝ satisfy the assumption  (1.3). Then for every  h∈C+(I), the solution of LPBVP (2.4)  u=Sh∈K. Namely,  S(C+(I))⊂K.Proof.
Leth∈C+(I), u=Sh. For every t∈I, from (2.6) it follows that
(2.8)0≤u(t)=∫01G(t,s)h(s)ds≤max⁡t∈I⁡Φ(t)∫01h(s)ds,
which implies that
(2.9)‖u‖C≤max⁡t∈I⁡Φ(t)∫01h(s)ds.
By this and (2.6), we have
(2.10)u(t)=∫01G(t,s)h(s)ds≥min⁡t∈I⁡Φ(t)∫01h(s)ds=σmax⁡t∈I⁡Φ(t)∫01h(s)ds≥σ‖u‖C.
For t∈I, by the definition of G and Φ, we have
(2.11)u(t)=∫0tΦ(t-s)h(s)ds+∫t1Φ(1+t-s)h(s)ds.
Making derivation to both sides of this equality, we have
(2.12)u′′(t)=∫0tΦ′′(t-s)h(s)ds+∫t1Φ′′(1+t-s)h(s)ds,
from which it follows that
(2.13)|u′′(t)|≤max⁡t∈I⁡|Φ′′(t)|∫0th(s)ds+max⁡t∈I⁡|Φ′′(t)|∫t1h(s)ds=max⁡t∈I⁡|Φ′′(t)|∫01h(s)ds=Mmin⁡t∈I⁡Φ(t)∫01h(s)ds≤Mu(t),t∈I.
Therefore, u∈K. This means that S(C+(I))⊂K.For everyu∈K, since f:I×ℝ+×ℝ→ℝ+ is continuous, we see that F(u):=f(·,u(·),u′′(·))∈C+(I). By Lemma 2.2, w=S(F(u))∈K. Define an operator A:K→K by
(2.14)Au(t)=S(F(u))=∫01G(t,s)f(s,u(s),u′′(s))ds.
We have the following.Lemma 2.3.
A:K→K is a completely continuous operator.Proof.
LetD⊂K be a bounded set in C2(I). By the continuity of f:I×ℝ+×ℝ→ℝ+, F(D) is a bounded set in C(I). By the boundedness of the operator S:C(I)→C4(I), A(D)=S(F(D)) is a bounded set in C4(I). By the compactness of the embedding C4(I)↪C2(I), A(D) is a precompact set in C2(I). So A:K→K is completely continuous.By the definition ofS and K, the positive solution of PBVP (1.1) is equivalent to the nontrivial fixed point of A. We will find the nonzero fixed point of A by using the fixed point index theory in cones.We recall some concepts and conclusions on the fixed point index in [7, 8]. Let E be a Banach space, and let K⊂E be a closed convex cone in E. Assume Ω is a bounded open subset of E with boundary ∂Ω, and K∩Ω≠∅. Let A:K∩Ω¯→K be a completely continuous mapping. If Au≠u for any u∈K∩∂Ω, then the fixed point index i(A,K∩Ω,K) has definition. One important fact is that if i(A,K∩Ω,K)≠0, then A has a fixed point in K∩Ω. The following two lemmas are needed in our argument.Lemma 2.4 (see [8]).
LetΩ be a bounded open subset of  E with  θ∈Ω, and let  A:K∩Ω¯→K be a completely continuous mapping. If  λAu≠u for every u∈K∩∂Ω and 0<λ≤1, then  i(A,K∩Ω,K)=1.Lemma 2.5 (see [8]).
LetΩ be a bounded open subset of  E, and let  A:K∩Ω¯→K be a completely continuous mapping. If there exists an  e∈K∖{θ} such that  u-Au≠τe for every  u∈K∩∂Ω and  τ≥0, then  i(A,K∩Ω,K)=0.


## 3. Proof of the Main Result

Proof of Theorem1.1.
Choose the working spaceE=C2(I) with the norm ∥u∥C02. Let K be the closed convex cone in C2(I) defined by (2.7), and let A:K→K be the operator defined by (2.14). By Lemma 2.3 and the definition of K, the nonzero fixed of the operator A is the positive solution of PBVP (1.1). Let 0<r<R<+∞, and set
(3.1)Ω1={u∈C2(I)∣‖u‖C02<r},Ω2={u∈C2(I)∣‖u‖C02<R}.
We show that, if r is small enough and R large enough, the operator A has a fixed point in K∩(Ω2∖Ω¯1) in either case that (F1) holds or (F2) holds.Case 1.
Assume that (F1) holds.
Sincef0<α, by the definition of f0, we may choose ɛ∈(0,α) and δ>0, such that(3.2)f(t,u,v)≤(α-ɛ)u,t∈I,|v|≤Mu,0≤u≤δ.
Let r∈(0,δ). We prove that A satisfies the condition of Lemma 2.4 in K∩∂Ω1; namely, λAu≠u, for every u∈K∩∂Ω1 and 0<λ≤1. In fact, if there exist u0∈K∩∂Ω1 and 0<λ0≤1 such that λ0Au0=u0, then by the definition of A and Lemma 2.1, u0∈C4(I) satisfies the differential equation
(3.3)u0(4)(t)-βu0′′(t)+αu0(t)=λ0f(t,u0(t),u0′′(t)),t∈I,
and the periodic boundary condition
(3.4)u(i)(0)=u(i)(1),i=0,1,2,3.
Since u0∈K∩∂Ω1, by the definitions of K and Ω1, we have
(3.5)|u0′′(t)|≤Mu0(t),0<σ‖u0‖C≤u0(t)≤‖u0‖C02=r<δ,t∈R.
From this and (3.2), it follows that
(3.6)f(t,u0(t),u0′′(t))≤(α-ɛ)u0(t),t∈R.
By this inequality and (3.3), we have
(3.7)u0(4)(t)-βu0′′(t)+αu0(t)≤λ0(α-ɛ)u0(t)≤(α-ɛ)u0(t),∈I.
Integrating this inequality from 0 to 1 and using the periodic boundary condition (3.4), we obtain that
(3.8)α∫01u0(t)dt≤(α-ɛ)∫01u0(t)dt.
Since ∫01u0(t)dt≥σ∥u0∥C>0, form this inequality it follows that α≤α-ɛ, which is a contradiction. Hence, A satisfies the condition of Lemma 2.4 in K∩∂Ω1. By Lemma 2.4 we have
(3.9)i(A,K∩Ω1,K)=1.
On the other hand, sincef∞>α, by the definition of f∞, there exist ɛ1>0 and H>0 such that(3.10)f(t,u,v)≥(α+ɛ1)x,t∈I,|v|≤Mu,u≥H.
Choose R>max{(1+M/σ)H,δ}, and let e(t)≡1. Clearly, e∈K∖{θ}. We show that A satisfies the condition of Lemma 2.5 in K∩∂Ω2; namely, u-Au≠τe, for every u∈K∩∂Ω2 and τ≥0. In fact, if there exist u1∈K∩∂Ω2 and τ1≥0 such that u1-Au1=τ1e, since u1-τ1e=Au1, by definition of A and Lemma 2.1, u1(t)∈C4(I) satisfies the differential equation
(3.11)u1(4)(t)-βu1′′(t)+α(u1(t)-τ1)=f(t,u1(t),u1′′(t)),t∈I,
and the periodic boundary condition (3.4). Since u1∈K∩∂Ω2, by the definition of K, we have
(3.12)u1(t)≥σ‖u1‖C,|u1′′(t)|≤Mu1(t),t∈I.
By the second inequality of (3.12), we have
(3.13)‖u1‖C02=‖u1‖C+‖u1′′‖C≤‖u1‖C+M‖u1‖C=(1+M)‖u1‖C.
Consequently,
(3.14)‖u1‖C≥11+M‖u1‖C02.
By (3.14) and the first inequality of (3.12), we have
(3.15)u1(t)≥σ‖u1‖C≥σ1+M‖u1‖C02=σ1+MR>H,t∈I.
From this, the second inequality of (3.12), and (3.10), it follows that
(3.16)f(t,u1(t),u1′′(t))≥(α+ɛ1)u1(t),t∈I.
By this and (3.11), we have
(3.17)u1(4)(t)-βu1′′(t)+α(u1(t)-τ1)≥(α+ɛ1)u1(t),t∈I.
Integrating this inequality on I and using the periodic boundary condition (3.4), we get that
(3.18)α∫01u1(t)dt-ατ1≥(α+ɛ1)∫01u1(t)dt.
Since ∫01u1(t)dt≥σ∥u1∥C>0, from this inequality it follows that α≥α+ɛ1, which is a contradiction. This means that A satisfies the condition of Lemma 2.5 in K∩∂Ω2. By Lemma 2.4,
(3.19)i(A,K∩Ω2,K)=0.
Now, by the additivity of fixed point index, (3.9), and (3.19), we have(3.20)i(A,K∩(Ω2∖Ω¯1),K)=i(A,K∩Ω2,K)-i(A,K∩Ω1,K)=-1.
Hence, A has a fixed point in K∩(Ω2∖Ω¯1), which is the positive solution of PBVP (1.1).Case 2.
Assume that (F2) holds.
By the assumption off0>α and the definition of f0, there exist ɛ>0 and δ>0, such that(3.21)f(t,u,v)≥(α+ɛ)u,t∈I,|v|≤Mu,0≤u≤δ.
Let r∈(0,δ), and let e(t)≡1. We prove that A satisfies the hypothesis of Lemma 2.5 in K∩∂Ω1; namely, u-Au≠τe for every u∈K∩∂Ω1 and τ≥0. In fact, if there exist u0∈K∩∂Ω1 and τ0≥0 such that u0-Au0=τ0e, since u0-τ0e=Au0, by the definition of A and Lemma 2.1, u0(t)∈C4(I) satisfies the differential equation
(3.22)u0(4)(t)-βu0′′(t)+α(u0(t)-τ0)=f(t,u0(t),u0′′(t)),t∈I,
and the periodic boundary condition (3.4). Since u0∈K∩∂Ω1, by the definitions of K and Ω1, u0 satisfies (3.5). From (3.5) and (3.22), it follows that
(3.23)f(t,u0(t),u0′(t))≥(α+ɛ)u0(t),t∈I.
By this inequality and (3.22), we have
(3.24)u0(4)(t)-βu0′′(t)+α(u0(t)-τ0)≥(α+ɛ)u0(t),t∈I.
Integrating this inequality on I and using the periodic boundary condition (3.4), we have
(3.25)α∫01u0(t)dt-ατ0≥(α+ɛ)∫01u0(t)dt.
Since ∫01u0(t)dt≥σ∥u0∥C>0, from this inequality, it follows that α≥α+ɛ, which is a contradiction. Hence A satisfies the hypothesis of Lemma 2.5 in K∩∂Ω1. By Lemma 2.5,
(3.26)i(A,K∩Ω1,K)=0.
Sincef∞<α, by the definition of f∞, there exist ɛ1∈(0,α) and H>0 such that(3.27)f(t,u,v)≤(α-ɛ1)u,t∈I,|v|≤Mu,u≥H.
Choosing R>max{(1+M/σ)H,δ}, we show that A satisfies the condition of Lemma 2.4 in K∩∂Ω2; namely, λAu≠u, for every u∈K∩∂Ω2 and 0<λ≤1. In fact, if there exist u1∈K∩∂Ω2 and 0<λ1≤1 such that λ1Au1=u1, then by the definition of A and Lemma 2.1, u1∈C4(I) satisfies the differential equation
(3.28)u1(4)(t)-βu1′′(t)+αu1(t)=λ1f(t,u1(t),u1′′(t)),t∈I,
and the periodic boundary condition (3.4). Since u1∈K∩∂Ω2, by the definition of K, u1 satisfies (3.12). From (3.12), we can show that u1 satisfies (3.14). By (3.14) and the first inequality of (3.12), we have
(3.29)u1(t)≥σ‖u1‖C≥σ1+M‖u1‖C02=σ1+MR>H,t∈R.From this, the second inequality of (3.12), and (3.27), it follows that
(3.30)f(t,u1(t),u1′′(t))≤(α-ɛ1)u1(t),t∈I.
By this inequality and (3.28), we have
(3.31)u1(4)(t)-βu1′′(t)+αu1(t)≤λ1(α-ɛ1)u1(t)≤(α-ɛ1)u1(t),t∈I.
Integrating this inequality on I and using the periodic boundary condition (3.4), we obtain that
(3.32)α∫01u1(t)dt≤(α-ɛ1)∫01u1(t)dt.
Since ∫01u1(t)dt≥σ∥u1∥C>0, form this inequality it follows that α≤α-ɛ1, which is a contradiction. This means that A satisfies the condition of Lemma 2.4 in K∩∂Ω2. By Lemma 2.4,
(3.33)i(A,K∩Ω2,K)=1.
From (3.26) and (3.33), it follows that(3.34)i(A,K∩(Ω2∖Ω¯1),K)=i(A,K∩Ω2,K)-i(A,K∩Ω1,K)=1.
Hence, A has a fixed point in K∩(Ω2∖Ω¯1), which is the positive solution of PBVP (1.1).
The proof of Theorem1.1 is completed.Example 3.1.
Consider the superlinear fourth-order periodic boundary problem(3.35)u(4)-u′′+u=a1(t)u2+a2(t)(u′′)2,0≤t≤1,u(i)(0)=u(i)(1),i=0,1,2,3,
where a1,a2∈C(I) and a1(t),a2(t)>0 for t∈I. It is easy to verify that α=1 and β=1 satisfy the assumption (P). f(t,u,v)=a1(t)u2+a2(t)v2 satisfies the condition (F1), in which f0=0 and f∞=+∞. Hence, by Theorem 1.1, (3.35) has at least one positive solution.Example 3.2.
Consider the sublinear fourth-order periodic boundary problem(3.36)u(4)+u′′+u=b1(t)u+b2|u′′|,0≤t≤1,u(i)(0)=u(i)(1),i=0,1,2,3,
where b1,b2∈C(I) and b1(t),b2(t)>0 for t∈I. For PBVP (3.36), it is easy to verify that α=1 and β=-1 satisfy the assumption (1.3), and f(t,u,v)=b1(t)u+b2(t)|v| satisfies the condition (F2) with f0=+∞ and f∞=0. By Theorem 1.1, (3.36) has a positive solution.
Since (3.35) and (3.36) have nonlinear terms of u′′, which are not in the range considered by [1–6], the existence results in Example 3.1, and Example 3.2 cannot be obtained from [1–6].



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