# Solvability of a Fourth-Order Boundary Value Problem with Integral Boundary Conditions

Authors: Hui Li; Libo Wang; Minghe Pei
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We investigate the existence of solutions and positive solutions
for a nonlinear fourth-order differential equation with integral boundary conditions of the
formx(4)(t)=f(t,x(t),x′(t),x′′(t),x′′′(t)), t∈[0,1], x(0)=x′(1)=0, x′′(0)=∫01h(s,x(s),x′(s),x′′(s))ds, x′′′(1)=0, where f∈C([0,1]×ℝ4), h∈C([0,1]×ℝ3). By using a fixed point theorem due to D.
O'Regan, the existence of solutions and positive solutions for the previous boundary value
problems is obtained. Meanwhile, as applications, some examples are given to illustrate
our results.


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## Body

## 1. Introduction

It is well known that fourth-order boundary value problems (BVPs) arise in a variety of different areas of the flexibility mechanics and engineering physics and thus have been extensively studied; for instance, see [1–29] and references therein. Boundary value problems with integral boundary conditions appear in heat conduction, thermoelasticity, chemical engineering underground water flow, and plasma physics; see [12, 14, 21, 24, 26, 29] and references therein.Motivated by the previous works and [30], in this paper, we consider fully nonlinear fourth-order differential equation
(1)x(4)(t)=f(t,x(t),x′(t),x′′(t),x′′′(t)),t∈[0,1],
subject to the integral boundary conditions
(2)x(0)=x′(1)=0,x′′(0)=∫01h(s,x(s),x′(s),x′′(s))ds,x′′′(1)=0,
as well as its simplified form
(3)x(4)(t)=f(t,x(t)),t∈[0,1],
with the integral boundary conditions
(4)x(0)=x′(1)=0,x′′(0)=∫01h(s,x(s))ds,x′′′(1)=0,
where f and h are continuous functions.We notice that ifh≡0 in problems (1), (2) and (3), (4), then the models are known as the one endpoint simply supported and the other one sliding clamped beam. The study of this class of problems was considered by some authors via various methods; we refer the reader to the papers [2, 5, 8, 11, 22].The aim of this paper is to establish the existence results of solutions and positive solutions for problems (1), (2) and (3), (4), respectively. By positive solution, we mean a solution x(t) such that x(t)>0 for t∈(0,1]. Our main tool is the fixed point theorem due to D. O'Regan [31].


## 2. Preliminary

In this section, we present some lemmas which are needed for our main results.LetC[0,1] denote the Banach space of real-valued continuous functions on [0,1] with the norm ||x||0:=maxt∈[0,1]|x(t)|. Cn[0,1] is the Banach space of n times continuously differentiable functions defined on [0,1], with the norm ||x||Cn:=max{||x(i)||0,i=0,1,…,n}.Throughout this paper, we always assume thatf:[0,1]×ℝ4→ℝ=(-∞,+∞) (or [0,1]×ℝ→ℝ) and h:[0,1]×ℝ3→ℝ (or [0,1]×ℝ→ℝ) are continuous.We consider a priori bound of solutions of the following one-parameter family of boundary value problem:(5)x(4)(t)=λf(t,x(t),x′(t),x′′(t),x′′′(t)),t∈[0,1],(6)x(0)=x′(1)=0,x′′(0)=λ∫01h(s,x(s),x′(s),x′′(s))ds,x′′′(1)=0,
where 0≤λ≤1. Simple computations lead to the following lemma.Lemma 1.
BVP (5), (6) with λ=0 has only the trivial solution, and the corresponding Green function G(t,s) exists and is given by
(7)G(t,s)={(s-12s2)t-t36,0≤t≤s≤1;(s-12s2)t-t36+16(t-s)3,0≤s≤t≤1.Now, define a linear operatorL:C4[0,1]→C[0,1]×ℝ4 by
(8)(Lx)(t)=(x(4)(t),x(0),x′(1),x′′(0),x′′′(1)),aaaaaaaaaaaaaaaaaaaalaaaaaallllat∈[0,1].Then, we can easily show thatL is a Fredholm operator with index zero, and its inverse L-1:C[0,1]×ℝ4→C4[0,1] is given by
(9)L-1(y,a,b,c,d)(t)=∫01‍G(t,s)y(s)ds+a+bt+cg(t)+d(16t3-12t),t∈[0,1],
where g(t)=(1/2)t2-t.Define Nemytskii operatorsF:C3[0,1]→C[0,1] induced by f as
(10)(Fx)(t)=f(t,x(t),x′(t),x′′(t),x′′′(t)),t∈[0,1],
and H:C3[0,1]→C[0,1] induced by h as
(11)(Hx)(t)=h(t,x(t),x′(t),x′′(t)),t∈[0,1].Also, define an operatorT:C3[0,1]→C[0,1]×ℝ4 as
(12)(Tx)(t)=((Fx)(t),0,0,∫01‍(Hx)(s)ds,0).Simple computations yield the following lemma.Lemma 2.
BVP (5), (6) is equivalent to the abstract equation
(13)x=λL-1Tx
in C3[0,1]; that is, x∈C4[0,1] is a solution of BVP (5), (6) if and only if x∈C3[0,1] is a solution of the integral equation
(14)x(t)=λ[∫01‍G(t,s)(Fx)(s)ds+∫01‍φ(t,x(s))ds],
where φ(t,x(s))=g(t)(Hx)(s).Let us denote the operatorsP1, P2 as
(15)(P1x)(t)=∫01G(t,s)(Fx)(s)ds,(16)(P2x)(t)=∫01φ(t,x(s))ds.Then,L-1T can be written as
(17)L-1T=P1+P2.Now, we can easily give some properties of the Green functionG(t,s) and g(t) by direct computation.Lemma 3.
LetG(t,s) be as in Lemma 1 and g(t)=(1/2)t2-t. Then,(1)
0≤G(t,s)≤max0≤t,s≤1G(t,s)=1/3, t,s∈[0,1];
(18)0≤(∂∂t)G(t,s)={s-(12)s2-(12)t2,0≤t≤s≤1;s-ts,0≤s≤t≤1;(19)0≥(∂2∂t2)G(t,s)={-t,0≤t≤s≤1;-s,0≤s≤t≤1;(2)
g(t)≤0, g′(t)≤0, g′′(t)=1, g′′′(t)=0, t∈[0,1];(3)
||g||0=1/2, ||g′||0=1, ||g′′||0=1, ||g′′′||0=0, ||g||C3=1.Lemma 4.
Suppose that(i)
for each fixed(t,x2,x3)∈[0,1]×ℝ2, f(t,x0,x1,x2,x3) is nondecreasing in x0 and x1;(ii)
there exists a constantM>0 such that for |x|>M, t∈[0,1],
(20)xf(t,-x,-x,x,0)>0;(iii)
there existβ∈(0,1) and nondecreasing continuous function σ:[0,+∞)→[0,+∞) such that σ(u)≤βu for u>0, and
(21)|h(t,x0,x1,x2)-h(t,y0,y1,y2)|≤σ(max{|xi-yi|,i=0,1,2})
for all (t,x0,x1,x2),(t,y0,y1,y2)∈[0,1]×ℝ3.
Then, any solutionx=x(t) of BVP (5), (6) satisfies
(22)|x(i)(t)|≤M+r,i=0,1,2,t∈[0,1],
where r=(1-β)-1||h(·,0,0,0)||0.Proof.
Let us first show that(23)|x′′(t)|≤M+r,∀t∈[0,1].
Note that ifλ=0 in (5) and (6), then BVP (5), (6) has only the trivial solution, and thus (23) holds. Hence, we may assume that λ∈[0,1]. Suppose now that (23) is not true. Then, there exists t0∈[0,1] such that |x′′(t0)|>M+r. Let
(24)K:=|x′′(t1)|=maxt∈[0,1]|x′′(t)|.
Then,K>M+r, and from x(0)=x′(1)=0, we have
(25)|x(i)(t)|≤K,i=0,1,∀t∈[0,1].
It is easy to see thatt1∈[0,1]. In fact, if t1=0, then |x′′(0)|=K. From (6) and (iii), it follows that for some ζ∈[0,1],
(26)K=|x′′(0)|=λ|∫01‍h(s,x(s),x′(s),x′′(s))ds|≤|h(ζ,x(ζ),x′(ζ),x′′(ζ))|≤|h(ζ,x(ζ),x′(ζ),x′′(ζ))-h(ζ,0,0,0)|+|h(ζ,0,0,0)|≤βK+(1-β)r<βK+(1-β)K=K,
which is a contradiction, and thus t1∈(0,1]. Furthermore, by definition of t1 and (6), we have x′′′(t1)=0. Hence, from assumptions (i) and (ii) and (25), we have
(27)x′′(t1)x(4)(t1)=λx′′(t1)f(t1,x(t1),x′(t1),x′′(t1),0)≥λx′′(t1)f(t1,-x′′(t1),-x′′(t1),mmmmllmkklx′′(t1),0)>0.
We may assume thatx′′(t1)>0; then, x(4)(t1)>0. Thus, by the continuity of x(4)(t) on [0,1], there exists δ>0 such that x(4)(t)>0 for t∈(t1-δ,t1]⊂[0,1]. Since x′′′(t1)=0, it follows that x′′′(t)<0 for t∈(t1-δ,t1]; namely, x′′(t) is decreasing on (t1-δ,t1], which contradicts the fact that x′′(t) attains its positive maximum value at t=t1. In summary, inequality (23) is true, which implies from x(0)=x′(1)=0 that
(28)|x(i)(t)|≤M+r,i=0,1,t∈[0,1].
This completes the proof of the lemma.Remark 5.
In Lemma4, if condition (i) is replaced by
(i′), there exists a constantM>0 such that whenever |x2|>M and all (t,x0,x1)∈[0,1]×ℝ2,
(29)x2f(t,x0,x1,x2,0)>0;
then, the conclusion of Lemma 4 remains true.The following fixed point result due to D. O'Regan plays a crucial role.Lemma 6 (see [31]).
LetU be an open set in a closed, convex set C of a Banach space E. Assume that 0∈U, P(U¯) is bounded, and P:U¯→C is given by P=P1+P2, where P1:U¯→E is continuous and completely continuous and P2:U¯→E is a nonlinear contraction. Then, either(A1)
P has a fixed point in U¯, or(A2)
there is a pointu∈∂U and λ∈(0,1) with u=λP(u).


## 3. Main Results

Firstly in this section, we state and prove our existence results of solutions for BVP (1), (2).Theorem 7.
Suppose that(i)
for each fixed(t,x2,x3)∈[0,1]×ℝ2, f(t,x0,x1,x2,x3) is nondecreasing in x0 and x1;(ii)
there exists a constantM>0 such that for |x|>M, t∈[0,1],
(30)xf(t,-x,-x,x,0)>0;(iii)
there existβ∈(0,1) and nondecreasing continuous function σ:[0,+∞)→[0,+∞) such that σ(u)≤βu for u>0, and
(31)|h(t,x0,x1,x2)-h(t,y0,y1,y2)|≤σ(max{|xi-yi|,i=0,1,2})
for all (t,x0,x1,x2),(t,y0,y1,y2)∈[0,1]×ℝ3;(iv)
f(t,x0,x1,x2,x3) satisfies the Nagumo condition; that is, there exists a positive-valued continuous function Φ(s) on [0,+∞) with ∫0+∞‍(sds/Φ(s))=+∞ such that
(32)|f(t,x0,x1,x2,x3)|≤Φ(|x3|)
for all (t,x0,x1,x2,x3)∈[0,1]×[-M-r,M+r]3×ℝ, where
(33)r=(1-β)-1∥h(·,0,0,0)∥0.
Then, BVP (1), (2) has at least one solution.Proof.
Letx be a possible solution of BVP (5), (6). We now show that
(34)|x′′′(t)|≤N,t∈[0,1],
where N:=max{N0,2(M+r)} and ∫M+rN0‍(sds/Φ(s))=2(M+r)+1.
Suppose that (34) is not true. Then, there exists t1∈[0,1) such that |x′′′(t1)|>N. Since x′′′(1)=0, then there exists ξ, η (t1<ξ<η<1) such that
(35)|x′′′(ξ)|=N,|x′′′(η)|=M+r,M+r<|x′′′(t)|<N,∀t∈(ξ,η).
Therefore,x′′′(t) is positive or negative on (ξ,η) by the continuity of the x′′′(t). Hence, from assumption (iv), the definition of N, and Lemma 4, we can get the following contradiction:
(36)2(M+r)+1=∫M+rN0‍sdsΦ(s)≤∫M+rN‍sdsΦ(s)≤|∫ηξ‍x′′′(t)x(4)(t)dtΦ(|x′′′(t)|)|≤∫ξη‍|x′′′(t)|dt=|x′′(η)-x′′(ξ)|≤2(M+r).
Therefore, inequality (34) holds.
LetΩ:={x∈C3[0,1]:||x||C3<N+1=:R}. It follows easily from the properties of the Green function and the continuity of f that the operator P1:Ω¯→C3[0,1] is completely continuous.
We now show thatP2:Ω¯→C3[0,1] is a nonlinear contraction. In fact, from assumption (iii), we have
(37)|(Hx)(t)-(Hy)(t)|≤σ(∥x-y∥C3),mmmmlmt∈[0,1],∀x,y∈C3[0,1].
Consequently, from Lemma3, we have
(38)|φ(t,x(s))-φ(t,y(s))|=|g(t)(Hx)(s)-g(t)(Hy)(s)|≤∥g∥0σ(∥x-y∥C3)≤12β∥x-y∥C3,t,s∈[0,1],mmmmmlml∀x,y∈C3[0,1].
Similarly, for allx,y∈C3[0,1],
(39)|∂φ(t,x(s))∂t-∂φ(t,y(s))∂t|≤∥g′∥0σ(∥x-y∥C3)≤β∥x-y∥C3,t,s∈[0,1],|∂2φ(t,x(s))∂t2-∂2φ(t,y(s))∂t2|≤∥g′′∥0σ(∥x-y∥C3)≤β∥x-y∥C3,t,s∈[0,1],|∂3φ(t,x(s))∂t3-∂3φ(t,y(s))∂t3|=0,t,s∈[0,1].
Hence,(40)∥P2x-P2y∥C3=max{∥(P2x)(i)-(P2y)(i)∥0,i=0,1,2,3}≤β∥x-y∥C3,∀x,y∈C3[0,1].
Since all possible solutions of BVP (5), (6) satisfy ||x||C3≤N<R, it follows that there is no x∈∂Ω and λ∈(0,1) such that x=λL-1Tx. We conclude that (A2) of Lemma 6 does not hold. Consequently, L-1T=P1+P2 has a fixed point, which is a solution of BVP (1), (2). This completes the proof of the theorem.Remark 8.
In Theorem7, if condition (i) is replaced by(i′)
there exists a constantM>0 such that whenever |x2|>M and all (t,x0,x1)∈[0,1]×ℝ2,
(41)x2f(t,x0,x1,x2,0)>0,
then the conclusion of Theorem 7 remains true.Remark 9.
In Theorem7, if f≥0≥h and f(t,0,0,0,0)≢0, then all the solutions of BVP (1), (2) are monotone and positive. This is clear because by Lemma 3 we have x=P1x+P2x≥0, x′=(P1x)′+(P2x)′≥0 and x′′=(P1x)′′+(P2x)′′≤0.
Next, we consider the existence of solutions and positive solutions for BVP (3), (4).Theorem 10.
Suppose that(i)
there existsΨ:[0,+∞)→[0,+∞) being continuous and nondecreasing such that
(42)|f(t,x)|≤Ψ(|x|),∀(t,x)∈[0,1]×ℝ;(ii)
there existsβ∈(0,2) such that
(43)|h(t,x)-h(t,y)|≤β|x-y|,∀(t,x,y)∈[0,1]×ℝ2;(iii)
there existsγ>0 such that
(44)Ψ(γ)+(3/2)h0γ<3(1-12β),
where h0=max0≤t≤1|h(t,0)|.
Then, BVP (3), (4) has at least one solution.Proof.
It is easy to see thatx∈C4[0,1] is a solution of BVP (3), (4) if and only if x∈C[0,1] is a solution of the integral equation (14) with λ=1. Moreover, P1 is completely continuous, and P2 is a nonlinear contraction.
It follows from (15), (i), and Lemma 3 that ∀x∈C[0,1],
(45)|(P1x)(t)|≤∫01‍G(t,s)Ψ(|x(s)|)ds≤13Ψ(∥x∥0),t∈[0,1].
Also, (16), (ii), and Lemma 3 yield
(46)|(P2x)(t)|≤∫01‍|g(t)(Hx)(s)|ds≤∥g∥0∫01‍(|h(s,x(s))-h(s,0)|+|h(s,0)|)ds≤12(β∥x∥0+h0),∀x∈C[0,1],t∈[0,1].
From (14), (45), and (46), we have that all possible solutions of x=λ(P1+P2)x satisfy
(47)|x(t)|≤13Ψ(∥x∥0)+12(β∥x∥0+h0),t∈[0,1].
LetΩ:={x∈C[0,1]:||x||0<γ}. Then, Ω is open in C[0,1], 0∈Ω, and (P1+P2)(Ω¯) is bounded. Suppose that x∈∂Ω and λ∈(0,1) satisfy x=λ(P1+P2)x. Then, ||x||0=γ, and (i) and (47) lead to
(48)γ≤13Ψ(γ)+12(βγ+h0);
that is,
(49)Ψ(γ)+(3/2)h0γ≥3(1-12β),
which contradicts (iii). Hence, (A2) of Lemma 6 does not hold, and consequently L-1T=P1+P2 has a fixed point which is a solution of BVP (3), (4). This completes the proof of the theorem.Remark 11.
In Theorem10, if f≥0≥h and f(t,0)≢0, then all the solutions of BVP (3), (4) are monotone and positive.Now, we consider BVP (3), (4) with linear boundary conditions as
(50)h(t,x(t))=l(t)x(t),t∈[0,1],
where l(·)∈C[0,1]. Define
(51)K(t,s)=g(t)l(s),(t,s)∈[0,1]×[0,1].Then,(52)(P2x)(t)=∫01K(t,s)x(s)ds.Theorem 12.
Suppose that(i)
K0:=(1/2)∫01‍|l(s)|ds<1;(ii)
there existsΨ:[0,+∞)→[0,+∞) being continuous and nondecreasing such that
(53)|f(t,x)|≤Ψ(|x|),∀(t,x)∈[0,1]×ℝ,limsupρ→+∞Ψ(ρ)ρ<3(1-K0).
Then, the nonlinear fourth-order differential equation (3) with boundary conditions
(54)x(0)=x′(1)=0,x′′(0)=∫01l(s)x(s)ds,x′′′(1)=0
has at least one solution.Proof.
Notice that the existence of solutions of BVP (3), (54) is equivalent to the existence of fixed points of operator equation
(55)x=P1x+P2x.
As a linear operator onC[0,1], from (52) and (i), we get ||P2||=K0<1, which implies that I-P2 is invertible and its inverse is given by
(56)(I-P2)-1=∑n=0∞P2nwith∥(I-P2)-1∥≤11-K0.
Hence, we see from (55) that x is a solution of BVP (3), (54) if and only if x is a fixed point of the completely continuous operator S=(I-P2)-1P1.
Let us show that there existsρ*>0 such that any solution x of operator equation x=λSx (λ∈(0,1)) satisfies ||x||0<ρ*. In fact, any solution x of x=λSx (λ∈(0,1)) satisfies
(57)x(t)=λ(I-P2)-1∫01G(t,s)f(s,x(s))ds,
and, hence, by (ii) and Lemma 3, we have
(58)∥x∥0≤13(1-K0)Ψ(∥x∥0).
The conditionlimsupρ→+∞(Ψ(ρ)/ρ)<3(1-K0) implies that there exists ρ*>0 such that (Ψ(ρ)/ρ)<3(1-K0) for all ρ≥ρ*; that is,
(59)ρ>13(1-K0)Ψ(ρ),∀ρ≥ρ*.
Comparing (58) and (59), we see that ||x||0<ρ*.
Now, letΩ={x∈C[0,1]:||x||0<ρ*}. By the Leray-Schauder continuation theorem S has a fixed point in Ω, which is a solution of BVP (3), (54). This completes the proof of the theorem.Remark 13.
In Theorem12, if f≥0≥l and f(t,0)≢0, then all the solutions of BVP (3), (54) are monotone and positive since all the solutions of BVP (3), (54) satisfy
(60)x(t)=(∑n=0∞‍P2n)∫01G(t,s)f(s,x(s))ds≥0on[0,1],
and thus
(61)x′=(P1x)′+(P2x)′≥0,x′′=(P1x)′′+(P2x)′′≤0.Finally, we give some examples to illustrate our results.Example 14.
Consider the fourth-order boundary value problem(62)x(4)(t)=t2(x(t))3+12(x′(t))5+x′′(t)e(x′′(t))2+(x′′′(t))2,t∈[0,1],(63)x(0)=x′(1)=0,x′′(0)=12∫011+(x′′(s))2ds,x′′′(1)=0.
Let(64)f(t,x0,x1,x2,x3)=t2x03+12x15+x2ex22+x32aaaaaaaaaaaaaaaaaaaaaaaaaaaaon[0,1]×ℝ4,h(t,x0,x1,x2)=121+x22on[0,1]×ℝ3,σ(u)=12uon[0,+∞).
It is easy to check that all the assumptions in Theorem7 are satisfied. Hence, BVP (62), (63) has at least one solution.Example 15.
Consider the fourth-order boundary value problem(65)x(4)(t)=q(t)(x(t))2+1,t∈[0,1],(66)x(0)=x′(1)=0,x′′(0)=32∫01ln11+(x(s))2ds,x′′′(1)=0,
where q(t)∈C([0,1],[0,(1/8)]).
Let(67)f(t,x)=q(t)x2+1on[0,1]×ℝ,h(t,x)=32ln11+x2on[0,1]×ℝ,Ψ(x)=18x2+1on[0,+∞),β=32,γ=3.
It is easy to check that all the assumptions in Theorem10 and Remark 11 are satisfied. Hence, BVP (65), (66) has at least one monotone positive solution.Example 16.
Consider the fourth-order boundary value problem(68)x(4)(t)=q(t)(x(t))23+1,t∈[0,1],(69)x(0)=x′(1)=0,x′′(0)=∫01(s2-4s)x(s)ds,x′′′(1)=0,
where q(t)∈C([0,1],[0,+∞)).
Let(70)f(t,x)=q(t)x23+1on[0,1]×ℝ,Ψ(x)=(maxt∈[0,1]q(t))x23+1on[0,+∞),l(t)=t2-4ton[0,1].
It is easy to check that all the assumptions in Theorem12 and Remark 13 are satisfied. Hence, BVP (68), (69) has at least one monotone positive solution.



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*Source: 782363-2013-03-24.xml*