The purpose of this paper is to investigate the existence of symmetric positive solutions for a class of fourth-order boundary value problem: u4(t)+βu′′(t)=f(t,u(t),u′′(t)), 0<t<1, u(0)=u(1)=∫01p(s)u(s)ds, u′′(0)=u′′(1)=∫01qsu′′(s)ds, where p,q∈L1[0,1], f∈C([0,1]×[0,∞)×(-∞,0],[0,∞)). By using a monotone iterative technique, we prove that the above boundary value problem has symmetric positive solutions under certain conditions. In particular, these solutions are obtained via the iteration procedures.

1. Introduction

The deformation of an elastic beam in equilibrium state, whose two ends are simply supported, can be described by a fourth-order ordinary differential equation BVP (short for boundary value problem). At present, two-point situation of fourth-order BVP has been studied by many authors, generally using the nonlinear alternatives of Leray-Schauder, the fixed point index theory, and the method of upper and lower solutions, monotone iteration; see [1–6].

Recently, problems with integral boundary value conditions arise naturally in thermal conduction problems [7], semiconductor problems [8], and hydrodynamic problems [9]. Hence, the existence results of positive solutions to this kind of problems have received a great deal of attentions. We refer the readers to [10–15].

In [13], Ma studied the following problem:
(1)u4t=htft,u,0<t<1,(2)u0=u1=∫01psusds,u′′0=u′′1=∫01qsu′′sds,
where p,q∈L1[0,1] and h and f are continuous. The existence of at least one symmetric positive solution is obtained by the application of the fixed point index in cones.

In [14], authors study the existence and nonexistence of symmetric positive solutions of the following fourth-order BVP:
(3)ϕpx′′t′′=wtft,xt,0<t<1,x0=x1=∫01gsxsds,ϕpx′′0=ϕpx′′1=∫01hsϕpx′′sds.
The argument was based on the fixed point theory in cones.

For fourth-order differential equation
(4)u4t+βu′′t=λft,ut,u′′t,0<t<1,
subject to boundary value conditions (2), author in [15] established the existence of positive solutions by the use of the Krasnoseliis fixed point theorem in cone.

The existing literature indicates that researches of fourth-order two point BVPs are excellent and methods are developed to be various. However, as to fourth-order BVPs with integral boundary value conditions, methods applied are relatively limited. Most of results are obtained by the use of fixed point theory in the cone or the fix point index theorem.

In this paper, we will apply the monotone iterative technique to the following fourth-order BVP with integral boundary conditions:
(5)u4t+βu′′t=ft,ut,u′′t,0<t<1,(6)u0=u1=∫01psusds,u′′0=u′′1=∫01qsu′′sds.
We do not assume that the upper and lower solutions to the boundary value problem should exist but construct the specific form of the symmetric upper and lower solutions. And we will construct successive iterative schemes for approximating solutions. In addition, it is worth stating that the first term of our iterative scheme is a simple function or a constant function. Therefore, the iterative scheme is feasible. Under the appropriate assumptions on nonlinear term, a new and general result to the existence of symmetric positive solution of BVP (5) and (6) is obtained.

We assume that the following conditions hold throughout the paper:

Given m(x)∈L1[0,1], let
(7)δ1=1-∫01mxdx;δ2=sinβ-∫01mxsinβxdx-∫01mxsinβ1-xdx
and δi≠0, i=1,2. Denoted by Hi(t,s), i=1,2, the Green’s function of the following problem:
(8)-u′′t+λiut=0,0<t<1,u0=u1=∫01msusds.
Then, careful calculation yields
(9)H1t,s=G1t,s+1δ1∫01G1s,xmxdx,G1t,s=t1-s,0⩽t⩽s⩽1,s1-t,0⩽s⩽t⩽1,H2t,s=G2t,s+sinβt+sinβ1-tδ2∫01G2s,xmxdx,G2t,s=sinβtsinβ1-sβsinβ,0⩽t⩽s⩽1,sinβssinβ1-tβsinβ,0⩽s⩽t⩽1.

Lemma 1 (see [<xref ref-type="bibr" rid="B15">15</xref>]).

Suppose that (S1)–(S3) hold. Then, for any g∈C[0,1], u solves the problem
(11)u4t+βu′′t=gt,0<t<1,u0=u1=∫01psusds,u′′0=u′′1=∫01qsu′′sds
if and only if u(t)=∫01∫01K1t,sK2s,τgτdτds, where
(12)K1t,s=G1t,s+ρ1∫01G1s,xpxdx,K2s,τ=G2s,τ+ρ2s∫01G2τ,xqxdx.

During the process of getting the above solution, we can also know
(13)u′′t=-∫01K2t,sgsds.

Lemma 2.

If (S2) is satisfied, the following results are true:

Ki(t,s)⩾0, for t,s∈[0,1], i=1,2;

Gi(1-t,1-s)=Gi(t,s),Gi(t,s)⩽Gi(t,t), for t,s∈[0,1], i=1,2.

Denote
(14)l1=∫01G1x,xpxdx;l2=∫01G2x,xqxdx;ht=G2t,t+ρ2tl2;l3=∫01htdt.
As β∈[π/2,π), it is easy to check that h(t)=h(1-t) and h′′(t)⩽0, for t∈[0,1]. Hence, from the symmetry and concavity of h(t), we have
(15)ht⩽h12⩽2∫01htdt=2l3.
In addition, for t,s∈[0,1], the following results hold:
(16)K1t,s=G1t,s+ρ1∫01G1s,xpxdx⩽G1t,t+ρ1∫01G1x,xpxdx=t1-t+ρ1l1,(17)K112,s⩽14+ρ1l1.
Further,
(18)K2s,τ=G2s,τ+ρ2s∫01G2τ,xqxdx⩽G2s,s+ρ2s∫01G2x,xqxdx=G2s,s+ρ2sl2=hs
and therefore
(19)K212,τ⩽h12.

We consider Banach space E=C2[0,1] equipped with the norm u=maxu∞,u′′∞, where u∞=supt∈[0,1]|u(t)|. In this paper, a symmetric positive solution u* of (5) means a function which is symmetric and positive on (0,1) and satisfies (5) as well as the boundary conditions (6).

In this paper, we always suppose that the following assumptions hold:

f(t,u,v)=f(1-t,u,v) for t∈[0,1], u∈[0,∞), v∈(-∞,0];

f(t,u1,v1)⩽f(t,u2,v2), for 0⩽u1⩽u2<+∞, 0⩽|v1|⩽|v2|<+∞;

f(t,0,0)≢0, for t∈[0,1].

Denote
(20)P=u∈E:ut⩾0,u′′t⩽0,ut=u1-t,t∈0,1u∈E:ut⩾0,u′′.
It is easy to see that P is a cone in E.

We define the operator T as follows:
(21)Tut=∬01K1t,sK2s,τfτ,uτ,u′′τdτds,00000000000000000000000000000000000000it∈0,1.
By the above argument, we know that, for any u∈E, Tu∈E and
(22)Tu′′t=-∫01K2t,τfτ,uτ,u′′τdτ,0000000000000000000000000000000t∈0,1.

Lemma 3.

If (S1)–(S3) are satisfied, T:P→P is completely continuous; that is, T is continuous and compact.

Proof.

For any u∈P, from (21) and (22), combining Lemma 2 and (S3), we know that (Tu)(t)⩾0 and Tu′′(t)⩽0 for t∈[0,1]. We now prove that Tu is symmetric about 1/2.

For t∈[0,1],
(23)Tu1-t=∬01K11-t,sK2s,τ·fτ,uτ,u′′τdτds=∬10G11-t,1-sG21-s,1-τ·f1-τ,u1-τ,u′′1-τ·d1-τd1-s+∫10∫01G11-t,1-s·ρ21-s∫01G2τ,xqxdx·fτ,uτ,u′′τdτd1-s+∬01ρ1∫01G1s,xpxdx·G2s,τfτ,uτ,u′′τdτds+∬01ρ1∫01G1s,xpxdx·G2s,τfτ,uτ,u′′τdτds=∬01G1t,sG2s,τ·fτ,uτ,u′′τdτds+∬01G1t,sρ2s∫01G2τ,xqxdx·fτ,uτ,u′′τdτds+∬01ρ1∫01G1s,xpxdx·G2s,τfτ,uτ,u′′τdτds+∬01ρ1∫01G1s,xpxdx·G2s,τfτ,uτ,u′′τdτds=Tut.
So, TP⊂P. The continuity of T is obvious. We now prove that T is compact. Let Ω⊂P be a bounded set. Then, there exists R such that
(24)Ω=u∈P:u⩽R.
For any u∈Ω, we have
(25)0⩽ft,ut,u′′t⩽maxft,u,u′′∣t∈0,1,u∈0,R,u′′∈-R,0=:M.
Therefore, from (17) and (18), we have
(26)Tu∞=Tu12=∬01K112,sK2s,τ·fτ,uτ,u′′τdτds⩽M∬0114+ρ1l1hsdτds=14+ρ1l1l3M,
and from (19), we have
(27)Tu′′∞=-Tu′′12=∫01K212,τfτ,uτ,u′′τdτ⩽M∫01h12dτ=Mh12.
So, Tu is uniformly bounded. Next we prove that Tu is equicontinuous.

For 0⩽t1⩽t2⩽s⩽1, we have
(28)Tut2-Tut1∞=∬01K1t2,s-K1t1,s·K2s,τfτ,uτ,u′′τdτds⩽M∬01K1t2,s-K1t1,s·K2s,τdτds=M∬01G1t2,s-G1t1,s·K2s,τdτds⩽M∬01t2-t1K2s,τdτds=N1t2-t1,
where N1=M∬01K2s,τdτds and
(29)Tu′′t2-Tu′′t1∞⩽M∫01K2t2,τ-K2t1,τdτ⩽M∫01G2t2,τ-G2t1,τdτ+Mρ2t2-ρ2t1·∬01G2τ,xqxdxdτ.
According to the Lagrange mean value theorem, we obtain that
(30)G2t2,τ-G2t1,τ=sinβ1-τβsinβsinβt2-sinβt1⩽1βsinββt2-βt1=1sinβt2-t1.
Similarly, we have
(31)ρ2t2-ρ2t1⩽2βρ2t2-t1.
Hence, there exists a positive constant N2 such that
(32)Tu′′t2-Tu′′t1∞⩽N2t2-t1.
And the similar results can be obtained for 0⩽s⩽t1⩽t2⩽1 and 0⩽t1⩽s⩽t2⩽1.

The Arzelà-Ascoli theorem guarantees that TΩ is relatively compact which means that T is compact.

3. Existence and Iterative of Solutions for BVP (<xref ref-type="disp-formula" rid="EEq1.3">5</xref>) and (<xref ref-type="disp-formula" rid="EEq1.4">6</xref>)Theorem 4.

Assume that (H1)–(H3) hold. If there exists two positive numbers a1<a such that
(33)supt∈0,1ft,a,a⩽a1,
where a and a1 satisfy
(34)a⩾max14+ρ1l1l3,2l3,h12a1.
Then, problem (5) and (6) has concave symmetric positive solution w*,v*∈P with
(35)w*∞⩽a,limn→∞Tnw0=w*,
where
(36)w0t=a1l3t1-t+ρ1l1,v*∞⩽a,limn→∞Tnv0=v*,
where
(37)v0t=0.

Proof.

We denote Pa={w∈P:w⩽a}. In what follows, we first prove that TPa⊂Pa.

Let w∈Pa; then 0⩽w(t)⩽supt∈0,1w(t)=w∞⩽a, |w′′(t)|⩽supt∈0,1|w′′(t)|=w′′∞⩽a.

By assumption (H2) and (33), for t∈[0,1], we have
(38)0⩽ft,wt,w′′t⩽a1.

For any w(t)∈Pa, by Lemma 3, we know that Tw∈P. According to (17), (18), and (33), we get
(39)Tw∞=Tw12=∬01K112,sK2s,τ·fτ,uτ,u′′τdτds⩽a1∬0114+ρ1l1hsdτds=a114+ρ1l1∫01hsds=a1l314+ρ1l1⩽a,
and from (19) and (33), we get
(40)Tw′′∞=-Tw′′12=∫01K212,τfτ,uτ,u′′τdτ⩽a1∫01K212,τdτ⩽a1∫01h12dτ=a1h12⩽a.

Hence, Tw⩽a. Thus, we get TPa⊂Pa. Let w0(t)=a1l3(t(1-t)+ρ1l1), for t∈[0,1]; then w0⩽a and w0(t)∈Pa. Let w1(t)=Tw0; then w1∈Pa. We denote
(41)wn+1=Twn=Tn+1w0n=0,1,2,….

From the definition of T, (16), (18), and (38), it follows that
(42)w1t=Tw0t=∬01K1t,sK2s,τ·fτ,w0τ,w0′′τdτds⩽a1∬01K1t,sK2s,τdτds⩽a1∬01t1-t+ρ1l1hsdτds=a1t1-t+ρ1l1∫01hsds=a1l3t1-t+ρ1l1=w0t.
On the other hand, from (15), (18), and (38), we have
(43)w1′′t=Tw0′′t=∫01K2t,τfτ,w0τ,w0′′τdτ⩽a1∫01K2t,τdτ⩽a1∫01htdτ=a1ht⩽2a1l3=w0′′t.
From (H2), it follows that
(44)w2t=Tw1t⩽Tw0t=w1t,0⩽t⩽1,w2′′t=Tw1′′t⩽Tw0′′t=w1′′t,0⩽t⩽1.
By induction,
(45)wn+1t⩽wnt,wn+1′′t⩽wn′′t,0⩽t⩽1n=0,1,2,….

Since TPa⊂Pa, we have wn∈Pa, n=0,1,2,…. From Lemma 3, T is completely continuous. We assert that {wn}n=1∞ has a convergent subsequence {wnk}k=1∞ and there exists w*∈Pa such that wnk→w*.

Let v0=0, t∈[0,1]; then v0∈Pa. Let v1=Tv0; then v1∈Pa; we denote
(46)vn+1=Tvn=Tn+1v0,n=0,1,2,….

Similarly to {wn}n=1∞, we assert that {vn}n=1∞ has a convergent subsequence {vnk}k=1∞ and there exists v*∈Pa, such that vnk→v*.

Since v1(t)=(Tv0)(t)=(T0)(t)∈Pa, we have
(47)v1t=Tv0t=T0t⩾0,0⩽t⩽1,v1′′t=Tv0′′t=T0′′t⩾0,0⩽t⩽1.
Hence,
(48)v2t=Tv1t⩾T0t=v1t,0⩽t⩽1,v2′′t=Tv1′′t⩾T0′′t=v1′′t,0⩽t⩽1.
By induction, vn+1⩾vn, |v′′n+1(t)|⩾|v′′n(t)|, 0⩽t⩽1, (n=0,1,2,…). Hence, we assert that vn→v*, Tv*=v*.

If f(t,0,0)≢0, 0⩽t⩽1, then the zero function is not the solution of BVP (5) and (6). Thus, max0⩽t⩽1|v*(t)|>0; we have
(49)v*⩾mint,1-tmax0⩽t⩽1v*t>0,0<t<1.

It is well known that the fixed point of operator T is the solution of BVP (5) and (6). Therefore, w* and v* are two positive, concave, and symmetric solutions of BVP (5) and (6).

Example 5.

Consider the following fourth-order boundary value problem with integral boundary conditions:
(50)u4t+π24u′′t=ft,ut,u′′t,0<t<1,u0=u1=∫01susds,u′′0=u′′1=∫01su′′sds,
where
(51)ft,u,v=130t1-t+12v2+lnu+2.

The calculation yields
(52)ρ1=2,ρ2=π-2π,l1=112,l2=1π2,l3=2ππ-2,ρ2t=sinπ/2t+sinπ/21-tρ2,ht=2πsinπ2tsinπ21-t+sinπ/2t+sinπ/21-tππ-2,h12=π-2+2ππ-2.
It is easy to check that assumptions (S1)–(S3) hold. Set a=12, a1=10. Then we can verify that conditions (H1) and (H2) and (33) are satisfied. Then applying Theorem 4, BVP (50) has two concave symmetric positive solutions w*,v*∈P with
(53)w*∞⩽10,limn→∞Tnw0=w*,
where
(54)w0t=60t1-t+103ππ-2.v*∞⩽10,limn→∞Tnv0=v*,
where
(55)v0t=0.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This research is supported by the Beijing Higher Education Young Elite Teacher Project (Project no. YETP0322) and Chinese Universities Scientific Fund (Project no. 2013QJ004).

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