# Nontrivial Solutions of a Fully Fourth-Order Periodic Boundary Value Problem

Authors: Haitong Li; Minghe Pei; Libo Wang
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We investigate the solvability of a fully fourth-order periodic boundary value problem of the formx(4)=f(t,x,x′,x′′,x′′′),x(i)(0)=x(i)(T),i=0,1,2,3, where f:[0,T]×R4→R satisfies Carathéodory conditions. By using the coincidence degree theory, the existence of nontrivial solutions is obtained. Meanwhile, as applications, some examples are given to illustrate our results.


---

## Body

## 1. Introduction

In this paper, we consider a fully nonlinear fourth-order periodic boundary value problem of the form(1)x(4)=f(t,x,x′,x′′,x′′′),
subject to the boundary conditions
(2)x(i)(0)=x(i)(T),i=0,1,2,3,
where f:[0,T]×R4→R=(-∞,+∞) satisfies Carathéodory conditions; that is, (i)
for a.e.t∈[0,T], the function f(t,x0,x1,x2,x3):R4→R is continuous;(ii)
for every(x0,x1,x2,x3)∈R4, the function f(t,x0,x1,x2,x3):[0,T]→R is measurable;(iii)
for eachρ>0, there is a real valued function hρ(·)∈L1[0,T] such that
(3)|f(t,x0,x1,x2,x3)|≤hρ(t),for a.e.t∈[0,T] and x02+x12+x22+x32<ρ2.It is well known that fourth-order periodic boundary value problems are important research topics which arise in a variety of different areas, such as nonlinear oscillations, fluid mechanical, and nonlinear elastic mechanical phenomena, and thus have been extensively studied; for instance, see [1–30] and references therein. However, most of the works in the above-mentioned references allow only having t, x or t, x, x′′ in the right-hand side nonlinear function f; see [2–11, 13, 15–18, 20–30]. The works on the fully nonlinear cases of which f contains explicitly t and every derivative of x up to order three have been quite rarely seen; see [1, 12, 14, 19].The aim of this paper is to establish the existence of solutions and nontrivial solutions for the fully nonlinear fourth-order PBVP (1), (2). Our main tool is the coincidence degree theory. The paper [31] motivated our study.


## 2. Preliminary

In this section, we present some lemmas which are needed for our main results.At first, we will briefly recall some notations that are needed for our discussion.LetX, Z be real Banach spaces. A linear mapping L:dom⁡L⊂X→Z will be called aFredholm mapping of index zero if the following two conditions hold:(i)
Im⁡L is a closed subspace of Z;(ii)
dimKerL=codimImL<+∞.LetL:dom⁡L⊂X→Z be a Fredholm mapping of index zero; then there exist continuous projectors P:X→X and Q:Z→Z such that
(4)Im⁡P=Ker⁡L,Ker⁡Q=Im⁡L,
so that
(5)X=Ker⁡L⊕Ker⁡P,Z=Im⁡L⊕Im⁡Q.
It follows that L|dom⁡L∩Ker⁡P:dom⁡L∩Ker⁡P→Im⁡L is invertible. We denote the inverse of that map by KP. Let Ω be an open bounded subset of X such that dom⁡L∩Ω≠∅; the map N:X→Z will be called L-compact on Ω¯, if QN:Ω¯→Z and KP(I-Q)N:Ω¯→X are compact.Lemma 1 (see [32]).
LetL:dom⁡L→Z be a linear Fredholm mapping of index zero and let Ω⊂X be an open bounded set. Let N:Ω¯→Z be L-compact on Ω¯ and let A:X→Z be L-completely continuous such that(i)
Ker⁡(L-A)={0};(ii)
for every(x,λ)∈(dom⁡L∩∂Ω)×(0,1),
(6)Lx-(1-λ)Ax-λNx≠0,and assume that0∈Ω. Then equation
(7)Lx=Nxhas at least one solution indom⁡L∩Ω¯.Lemma 2 (see [33]).
LetL:dom⁡L→Z be a linear Fredholm mapping of index zero and let Ω⊂X be an open bounded set. Let N:Ω¯→Z be L-compact on Ω¯ and the coincidence degree D[(L,N),Ω] is well defined. If there exists z∈Z with z≠0 such that
(8)Lx-Nx≠λz,∀(x,λ)∈(dom⁡L∩∂Ω)×[0,+∞),
then D[(L,N),Ω]=0.In the following, we take Banach spaceX=C3[0,T] with the norm ∥x∥C3=∑i=03‍∥x(i)∥∞, and Z=L1[0,T]. Define a linear map L:dom⁡L⊂X→Z by
(9)(Lx)(t)=x(4)(t),∀x∈dom⁡L,
where dom⁡L={x∈W4,1(0,T):x(i)(0)=x(i)(T),i=0,1,2,3} and W4,1(0,T) is the usual Sobolev space. It is easy to see that L is a Fredholm mapping of index zero. Also define a nonlinear map N:X→Z by
(10)(Nx)(t)=f(t,x,x′,x′′,x′′′),∀x∈X.Define two projectsP:X→X and Q:Z→Z as follows:
(11)Px=1T∫0T‍x(t)dt,∀x∈X;Qz=1T∫0T‍z(t)dt,∀z∈Z.LetG(t,s) be Green function for the homogeneous BVP
(12)x(4)(t)=0,0<t<T,∫0T‍x(t)dt=0,x(i)(0)=x(i)(T),i=0,1,2,3.
Then KP:Im⁡L→dom⁡L∩Ker⁡P can be given by
(13)(KPz)(t)=∫01‍G(t,s)z(s)ds,∀z∈Im⁡L.
Hence the map KP:Im⁡L→dom⁡L∩Ker⁡P is continuous. We note that if f:[0,T]×R4→R satisfying Carathéodory conditions, then N:X→Z is bounded and continuous by Lebesgue's dominated convergence theorem. Furthermore, N is L-compact on every bounded set Ω¯⊂X.


## 3. Main Results

Forh∈L1[0,T] and r∈(0,+∞) we put
(14)h0=exp⁡(2∫0T‍h(t)dt),r0=r+eh0T3;ɛ∈(0,min⁡{12r0T,(2πT)4}),r3=h0exp⁡(2ɛr0T);r2=r3T,r1=r2T.In order to introduce our main theorem, we need some lemmas.Lemma 3.
Leth∈L1[0,T] be a nonnegative function and r∈(0,+∞). Let r0, r1, r2, r3, and ɛ fulfil (14). Then for any x∈dom⁡L, the inequalities
(15)|x(t)|≤r0,|x′(t)|≤r1,|x′′(t)|≤r2,mmmmmmmmmmmmmmmmm∀t∈[0,T],(16)x(4)(t)sign⁡x′′′(t)≤h(t)|x′′′(t)|+ɛ|x(t)|mmmfora.e.t∈{t∈[0,T]:|x′′′(t)|≥1}
imply
(17)|x′′′(t)|<r3,∀t∈[0,T].Proof.
Sincex∈dom⁡L, there exists t0∈(0,T) such that
(18)x′′′(t0)=0.
We will show that
(19)|x′′′(t)|<r3,∀t∈[t0,T].
By contradiction, assume that there exists t1∈(t0,T] such that
(20)|x′′′(t1)|≥r3.
Then there exists t2∈(t0,t1) such that
(21)|x′′′(t2)|=1,|x′′′(t)|≥1,∀t∈[t2,t1].
There are two cases to consider.
Case
1. Consider x′′′(t)≥1 on [t2,t1]. In this case, integrating (16) from t2 to t1 and using (15) and (21) we infer that
(22)∫t2t1‍x(4)(t)x′′′(t)dt≤∫t2t1‍(h(t)+ɛr0)dt<∫0T‍h(t)dt+ɛr0T.
Thus x′′′(t1)<r3, which contradicts (20).
Case
2. Consider x′′′(t)≤-1 on [t2,t1]. Similar to Case 1, we have
(23)∫t2t1‍-x(4)(t)-x′′′(t)dt≤∫t2t1‍(h(t)+ɛr0)dt<∫0T‍h(t)dt+ɛr0T.
Thus, x′′′(t1)>-r3, which contradicts (20). Therefore (19) is true. Furthermore, from the fact x′′′(0)=x′′′(T) it follows that |x′′′(0)|<r3.
Finally, we show that(24)|x′′′(t)|<r3,∀t∈[0,t0].
Suppose on the contrary that there exists t3∈(0,t0) satisfying
(25)|x′′′(t3)|≥r3.
Then there exists t4∈(0,t3) such that
(26)|x′′′(t4)|=r3,|x′′′(t)|≥r3,∀t∈[t4,t3].
There are two cases to consider.
Case  1
.
′ Consider x′′′(t)≥r3 on [t4,t3]. Similar to Case 1, we have x′′′(t3)<r3, which contradicts (25).
Case  2
.
′ Consider x′′′(t)≤-r3 on [t4,t3]. Similar to Case 2, one has x′′′(t3)>-r3, which also contradicts (25). Hence (24) is true.
In summary, from (19) and (24) it follows that estimate (17) holds. This completes the proof of the lemma.Lemma 4.
Letr∈(0,+∞) and let h∈L1[0,T] be a nonnegative function. Let r0, r1, r2, r3, and ɛ fulfil (14). Then for any function x∈dom⁡L the inequalities
(27)|x′′′(t)|<r3,∀t∈[0,T],(28)μx(4)(t)sign⁡x(t)>0fora.e.t∈{t∈[0,T]:|x(t)|≥r}
imply
(29)|x(t)|<r0,|x′(t)|<r1,|x′′(t)|<r2,mmmmmmmmmmmmmmmmmm∀t∈[0,T].Proof.
For everyx∈dom⁡L, from (28), there exist t0,t1,t2∈[0,T] such that
(30)|x(t0)|<r,x′(t1)=0,x′′(t2)=0.
Integrating (27) by (14) and (30) we get
(31)|x′′(t)|<r3T=r2,|x′(t)|<r2T=r1on⁡[0,T],|x(t)|<r+r2T2<r0on⁡[0,T].
This completes the proof of the lemma.Now, we apply Lemma1 to establish the existence results of solutions for the fourth-order PBVP (1), (2).Theorem 5.
Assume that there existμ∈{-1,1}, r∈(0,+∞), and a nonnegative function h∈L1[0,T]. Suppose further that (H0)
f:[0,T]×R4→R satisfies the Carathéodory conditions;(H1)
if|x0|≥r, |x1|≤r1, |x2|≤r2, |x3|≤r3, then
(32)μf(t,x0,x1,x2,x3)sign⁡x0≥0fora.e.t∈[0,T];(H2)
if|x0|≤r0, |x1|≤r1,  |x2|≤r2,  |x3|≥1, then
(33)f(t,x0,x1,x2,x3)sign⁡x3≤h(t)|x3|fora.e.t∈[0,T],wherer0, r1, r2, r3 fulfil (14). Then PBVP (1), (2) has at least one solution x such that
(34)|x(i)(t)|≤ri,i=0,1,2,3,∀t∈[0,T].Proof.
Let(35)Ω={x∈X:|x(i)(t)|<ri,i=0,1,2,3,t∈[0,T]}.
Then x∈∂Ω iff there exist some j∈{0,1,2,3} such that
(36)max⁡0≤t≤T|x(j)(t)|=rj,|x(i)(t)|≤ri,i=0,1,2,3,mmmmmmmmmmmmmmmmmmmmt∈[0,T].
Now, we show that(37)Ker⁡(L-A)={0},
where A:X→Z, x↦μɛx. To do this, we assume that x(t) is the solution of the following periodic boundary value problem:
(38)x(4)(t)=μɛx(t),a.e.t∈(0,T),x(i)(0)=x(i)(T),i=0,1,2,3.
Integrating the equation as above on [0,T], we obtain
(39)∫0T‍x(t)dt=0.
Thus, by Wirtinger inequality,
(40)∥x(i)∥2≤T2π∥x(i+1)∥2,i=0,1,2,3.
Hence from (38) it follows that
(41)∥x∥2≤(T2π)4∥x(4)∥2≤ɛ(T2π)4∥x∥2.
If ∥x∥2>0, then, from ɛ<(2π/T)4, the following contradiction holds:
(42)∥x∥2<∥x∥2.
Therefore, Ker⁡(L-A)={0}.
Finally, we show that, for every(x,λ)∈(dom⁡L∩∂Ω)×(0,1),
(43)Lx-(1-λ)Ax-λNx≠0.
To do this, let λ∈(0,1) and let xλ∈Ω¯ be a solution of the following PBVP:
(44)x(4)=λf(t,x,x′,x′′,x′′′)+(1-λ)μɛx,x(i)(0)=x(i)(T),i=0,1,2,3.
Then xλ∉∂Ω. In fact, let
(45)|xλ(t)|≤r0,|xλ′(t)|≤r1,|xλ′′(t)|≤r2,mmmmmmmmmmmmmmmmmm∀t∈[0,T].
Then, by (33),
(46)xλ(4)sign⁡xλ′′′=λfsign⁡xλ′′′+(1-λ)μɛxλsign⁡xλ′′′≤h(t)|xλ′′′|+ɛ|xλ|,
for a.e. t∈{t∈[0,T]:|xλ′′′(t)|≥1}. Applying Lemma 3, we obtain
(47)|xλ′′′(t)|<r3,∀t∈[0,T].
Thus according to (32), we have
(48)μxλ(4)sign⁡xλ=λμfsign⁡xλ+μ2(1-λ)ɛ|xλ|>0,
for a.e. t∈{t∈[0,T]:|xλ(t)|≥r}. It follows from Lemma 4 that
(49)|xλ(t)|<r0,|xλ′(t)|<r1,|xλ′′(t)|<r2,mmmmmmmmmmmmmmmmmmi∀t∈[0,T].
Thus xλ∉∂Ω. This implies that condition (ii) of Lemma 1 is valid.
In summary, all conditions of Lemma1 are satisfied. Therefore the conclusion of Theorem 5 holds. This completes the proof of the theorem.Next, we establish the existence result of nontrivial solutions for the fourth-order PBVP (1), (2) by means of Lemma 2.Theorem 6.
Assume that all conditions in Theorem5 hold with the exception of (H1), which is replaced by the following:(H1′)
there exists a constantr*∈(0,r) such that if x0>-r*, |x1|≤r1, |x2|≤r2, |x3|≤r3, then
(50)μf(t,x0,x1,x2,x3)≥0fora.e.t∈[0,T],and ifx0<-r, |x1|≤r1, |x2|≤r2, |x3|≤r3, then
(51)μf(t,x0,x1,x2,x3)≤0fora.e.t∈[0,T].Then PBVP (1), (2) has at least one nontrivial solution x satisfying (34).Proof.
From the proof of Theorem5 and Lemma 1, it follows that Lx=Nx has a solution in
(52)Ω={x∈X:|x(i)(t)|<ri,i=0,1,2,3,t∈[0,T]},
and |D[(L,N),Ω]|=1. Without loss of generality, we assume that μ=1 and Br*={x:∥x∥C3<r*}⊂Ω. We also assume that Lx≠Nx for all x∈∂Br*.
Now we assert that(53)D[(L,N),Br*]=0.
In fact, suppose that there exist x0∈∂Br* and λ0>0 such that
(54)Lx0-Nx0=λ0.
Applying Q to both sides of above equality, it follows that
(55)-QNx0=λ0Q(1);
that is,
(56)-1T∫0T‍f(t,x0(t),x0′(t),x0′′(t),x0′′′(t))dt=λ0.
Notice that ∥x0∥C3=r* and Br*⊂Ω; we have
(57)x0(t)≥-∥x0∥∞≥-∥x0∥C3=-r*,∀t∈[0,T],|x0′(t)|≤r1,|x0′′(t)|≤r2,|x0′′′(t)|≤r3,mmmmmmmmmmmmmmmmmm∀t∈[0,T].
Consequently, from assumption (H1′) one has
(58)f(t,x0(t),x0′(t),x0′′(t),x0′′′(t))≥0fora.e.t∈[0,T].
This together with (56) it follows that
(59)λ0≤0,
which is a contradiction. This implies that
(60)Lx-Nx≠λ,∀x∈∂Br*,∀λ≥0.
Thus from Lemma 2 it follows that
(61)D[(L,N),Br*]=0.
Hence
(62)D[(L,N),Ω∖B¯r*]=D[(L,N),Ω]-D[(L,N),Br*]≠0.
Therefore Lx=Nx has a solution in Ω∖B¯r*; that is, PBVP (1), (2) has at least one nontrivial solution. This completes the proof of the theorem.Finally, we give some examples to illustrate our results.Example 7.
Consider the fourth-order periodic boundary value problem(63)x(4)(t)=λh(t)(x(t)+1)ex(t)x′(t)(x′′(t))2(x′′′(t)+c),mmmmmmmmmmmmmmmiia.e.t∈[0,T],x(i)(0)=x(i)(T),i=0,1,2,3,
where λ is a parameter, h∈L1[0,T] is nonnegative, and
(64)c∈(exp⁡(2∫0T‍h(t)dt+1),+∞)
is a constant.
Let(65)f(t,x0,x1,x2,x3)=λh(t)(x0+1)ex0x1x22(x3+c).
Then f:[0,T]×R4→R satisfies Carathéodory conditions. Taking any r∈(1,+∞), then r0, r1, r2, and r3 are well defined by (14).
Now, we assert that all conditions of Theorem5 are satisfied when
(66)λ∈[-1(r0+1)er0r1r22(r3+c),1(r0+1)er0r1r22(r3+c)].
In fact, without loss of generality, we can assume λ∈[-1/(r0+1)er0r1r22(r3+c),0]. In this case, we choose μ=-1. It is easy to see that, for every (x0,x1,x2,x3)∈[r,+∞)×R2×[-r3,r3],
(67)μf(t,x0,x1,x2,x3)sign⁡x0≥0,a.e.t∈[0,T],
and, for every (x0,x1,x2,x3)∈(-∞,-r]×R2×[-r3,r3],
(68)μf(t,x0,x1,x2,x3)sign⁡x0≥0,a.e.t∈[0,T].
Hence condition (H1) of Theorem 5 is satisfied. In addition, for |x0|≤r0, |x1|≤r1, |x2|≤r2, |x3|≥1, we have
(69)f(t,x0,x1,x2,x3)sign⁡x3≤|λ|h(t)(|x0|+1)e|x0x1|x22(|x3|+c)≤h(t)|x3|+cr3+c≤h(t)|x3|,a.e.t∈[0,T].
Therefore, condition (H2) of Theorem 5 is also satisfied. Hence, from Theorem 5, the fourth-order PBVP (63) has at least a solution x=x(t), provided
(70)λ∈[-1(r0+1)er0r1r22(r3+c),1(r0+1)er0r1r22(r3+c)].Example 8.
Consider the fourth-order periodic boundary value problem(71)x(4)(t)=λh(t)(x(t)+1)k|x′′′(t)|1+(x′(t))n+(x′′(t))n,a.e.t∈[0,T],x(i)(0)=x(i)(T),i=0,1,2,3,
where λ is a parameter, k,n∈N={1,2,…}, k is odd, n is even, and h∈L1[0,T] is a nonnegative function.
Let(72)f(t,x0,x1,x2,x3)=λh(t)(x0+1)k|x3|1+x1n+x2n.
Then f:[0,T]×R4→R satisfies Carathéodory conditions. We choose r∈(1,+∞); then r0 is well defined by (14).
Now, we assert thatf satisfies all conditions of Theorem 6 when
(73)λ∈[-(r0+1)-k,(r0+1)-k].
In fact, without loss of generality, we can assume that λ∈[0,(r0+1)-k]. Choose μ=1 and r*=1. Then it is easy to see that, for every (x0,x1,x2,x3)∈[-r*,+∞)×R3,
(74)μf(t,x0,x1,x2,x3)≥0,a.e.t∈[0,T],
and, for every (x0,x1,x2,x3)∈(-∞,-r]×R3,
(75)μf(t,x0,x1,x2,x3)≤0a.e.t∈[0,T].
On the other hand, for every (x0,x1,x2,x3)∈[-r0,r0]×R3,
(76)f(t,x0,x1,x2,x3)sign⁡x3≤λ|x0+1|k1+x1n+x2nh(t)|x3|≤11+x1n+x2nh(t)|x3|≤h(t)|x3|,a.e.t∈[0,T].
In summary, all conditions of Theorem 6 are satisfied. Therefore, from Theorem 6, the fourth-order PBVP (71) has at least one nontrivial solution x=x(t), provided λ∈[-(r0+1)-k,(r0+1)-k].



---

*Source: 895862-2014-08-06.xml*