# Periodic Solutions of a Class of Fourth-Order Superlinear Differential Equations

Authors: Yanyan Li; Yuhua Long
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

This paper deals with the periodic solutions of a class of fourth-order superlinear differential equations. By using the classical variational techniques and symmetric mountain pass lemma, the periodic solutions of a single equation in literature are extended to that of equations, and also, the cubic growth of nonlinear term is extended to a general form of superlinear growth.


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## Body

## 1. Introduction

The existence of periodic solutions of fourth-order differential equations has been studied by more and more researchers [1–6]. The application methods contain mainly Clark theorem [2–4], Cone theory [6], and so on.For a single equation, Tersian and Chaparova [2] study the existence of infinitely many unbounded solutions, using symmetric mountain pass lemma:
(1.1)uiv-pu″+a(x)u-b(x)u3=0,x∈R,u(0)=u(L)=0,u′′(0)=u′′(L)=0.It is a natural problem to wonder whether symmetric mountain pass lemma method may be applied not only to single equations but also to systems of differential equations. In this paper we study the existence of periodic solutions of the fourth-order equations, by making use of the classical variational techniques and symmetric mountain pass lemma(1.2)u(4)-cu′′+a(x)u-∂F(x,u,v)∂u=0,0<x<L,v(4)-dv′′+b(x)v-∂F(x,u,v)∂v=0,0<x<L,u(0)=u′′(0)=u(L)=u′′(L)=0,v(0)=v′′(0)=v(L)=v′′(L)=0.Through studying System (1.2),  (1.1) of the corresponding conclusions are extended.The paper is organized as follows. In Section2, we consider the result of System (1.2) under certain conditions. In Section 3, we prove the main result of this paper and give an example.


## 2. Main Result

In this paper, we state our main result. First we give the following list of assumptions on the parameters in System (1.2):(A)
a(x)>0,b(x)>0,c>-π2/L2,d>-π2/L2.(F1)
F is an even functional about (u,v). That is, F(x,-u,-v)=F(x,u,v) for every (u,v)∈R2.(F2)
There existsβ>2, as u2+v2≠0, we have
(2.1)u·∂F(x,u,v)∂u+v·∂F(x,u,v)∂v≥βF(x,u,v)>0foreveryx∈R.(F3)
F(x,u,v)=o(u2+v2) with respect to x consistently, as u2+v2→0.Denotea1=minx∈[0,L]a(x), a2=maxx∈[0,L]a(x), b1=minx∈[0,L]b(x), b2=maxx∈[0,L]b(x).From condition(A), we obtain ai>0,bi>0, when i=1,2.Remark 2.1.
Letz=(u,v)∈R2, then condition (F2) is transformed to
(2.2)(∇F(x,z),z)>βF(x,z)>0foreveryz≠0,
where (·,·) represents the usual inner product in R2.Remark 2.2.
From (F3), we obtain lim|z|→0F(x,z)/|z|2=0, where |·| represents normal norm in R2. Besides, from the continuity of F, we obtain F(x,0,0)=0.Our main result is as follows.Theorem 2.3.
Supposea(x),b(x), and F satisfy (A), (F1)–(F3). Then System (1.2) has infinitely many distinct pairs of solutions zn=(un,vn), which are critical points of the functional I:X→R, and I(zn)→∞ as n→∞.In this paper, the existence of periodic solutions of a single equation in System (1.1) are extended to the case of equations, and also the cubic growth of nonlinear term is extended to a general form of superlinear growth.


## 3. Variational Structure and the Proof of Result

In this section, we prove the main result stated in Section2.


### 3.1. Variational Structure

Denote(3.1)X(L)=(H2(0,L)∩H01(0,L))2.
Then X(L) is a Hilbert space. The norm is
(3.2)∥z∥2=∥u∥c2+∥v∥d2,
where
(3.3)∥u∥c={∫0L‍[|u′′(x)|2+c|u′(x)|2+a(x)|u(x)|2]dx}1/2,∥v∥d={∫0L‍[|v′′(x)|2+d|v′(x)|2+b(x)|v(x)|2]dx}1/2,z=(u,v)∈X(L). The corresponding inner product are
(3.4)〈z1,z2〉|=∫0L‍[(z1′′,z2′′)+c(u1′,u2′)+d(v1′,v2′)+a(x)(u1,u2)+b(x)(v1,v2)]dx,〈u1,u2〉|c=∫0L‍[(u1′′,u2′′)+c(u1′,u2′)+a(x)(u1,u2)]dx,〈v1,v2〉|d=∫0L‍[(v1′′,v2′′)+d(v1′,v2′)+b(x)(v1,v2)]dx.
For every z=(u,v)∈X(L), using Poincaré inequality [7], we obtain
(3.5)∫0L‍u2dx≤L2π2∫0L‍u'2dx,∫0L‍u'2dx≤L2π2∫0L‍u′′2dx.
Thus, we can define another norm ∥·∥1 in X(L). That is, for every z∈X(L),
(3.6)∥z∥1={∫0L‍|z′′(x)|2dx}1/2.
The inner product in X(L) as follows:
(3.7)〈z1,z2〉|1=∫0L‍(z1′′(x),z2′′(x))dx,z1,z2∈X(L).
The two different norms (3.2) and (3.6) are equivalent in X(L).In this section we consider System (1.2). The Fréchet derivative of I is given by the following:
(3.8)I(u,v)=12∫0L‍[u′′2+cu'2+a(x)u2+v′′2+dv'2+b(x)v2]dx-∫0L‍F(x,u,v)dx,
where z=(u,v)∈X(L).Remark 3.1.
In general, the growth ofF is limited by the differentiability of functional I, but we apply truncation techniques in [8]. First, introduce auxiliary functional and the auxiliary functional is Fréchet differentiable. Second, we use critical point theory to prove the existence of critical point of auxiliary functional, then prove the existence of the original equation. However, in order to avoid technical complexity, we assume directly functional I is Fréchet differentiable.In fact, for everyz=(u,v)∈X(L),z-=(u-,v-)∈X(L), we obtain
(3.9)〈I′(z),z-〉=〈Iu′(u,v),u-〉+〈Iv′(u,v),v-〉,
where
(3.10)〈Iu′(u,v),u-〉=∫0L‍[u′′u-′′+cu′u-′+a(x)uu--∂F(x,u,v)∂uu-]dx,〈Iv′(u,v),v-〉=∫0L‍[v′′v-′′+cv′v-′+a(x)vv--∂F(x,u,v)∂vv-]dx.
and Iu′(u,v),Iv′(u,v)∈[H2(0,L)∩H01(0,L)]*, I′(z)∈X(L)*.It is similar to the discussion of [8], the solutions of System (1.2) corresponds to the critical point of the functional I, so we need to discuss the critical point of functional I. In order to prove Theorem 2.3, we introduce below definition and lemma.Definition 3.2 (see [9]).
LetX be a real Banach space, I∈C1(X,R), I is a Fréchet continuously differentiable functional in X(L). I is said to be satisfying Palais-Smale (PS) condition if any sequence {un}⊂X for which {I(un)} is bounded and {I′(un)}→0 as j→∞, possesses a convergent subsequence.Lemma 3.3 (see [8]).
LetX be an infinite dimensional Banach space and (Xn)n be a sequence of finite dimensional subspaces of X such that dimXn=n,
(3.11)X1⊂X2⊂⋯⊂Xn⊂X,⋃n=1∞Xn¯=X.
Let I∈C1(X,R) be an even functional, I(0)=0, and I satisfy (PS) condition. Suppose that(A1)
there are constantsρ,α>0 such that I|∂Bρ≥α, and(A2)
for everyn there is an Rn>0 such that I≤0 on Xn∖BRn.
ThenI possesses infinitely many pairs of critical points with unbounded sequence of critical values.


### 3.2. The Proof of Result

Step 1 (FunctionalI satisfies (PS) condition).
Let{zn}={(un,vn)} be a (PS) sequence in X, that is, {I(zn)} is bounded and I′(zn)→0, as n→∞. Suppose that {zn} is unbounded in X, that is, ∥zn∥→∞ as n→∞. Since
(3.12)I(zn)+1γ∥I′(zn)∥∥zn∥≥I(zn)-1γ〈I′(zn),zn〉=1γ∥zn∥2,
it follows that
(3.13)I(zn)∥zn∥2+∥I′(zn)∥γ∥zn∥≥1γ,
where γ≥4. Letting n→∞ in (3.13), we have a contradiction with ∥zn∥→∞ as n→∞.
Therefore{zn} is a bounded sequence in X(L). Passing if necessary to a subsequence we may assume that {zn} is weakly convergent to a function z∈X(L), zn⇀z in X(L), and zn→z in C[(0,L)].
From the Lebesgue theorem,z∈X(L), zn⇀z in X(L), and zn→z in C[(0,L)], letting n→∞ in (3.9)
(3.14)〈I′(zn,zn)〉=∥zn∥2-∫0L‍∂F(x,un,vn)∂uundx-∂F(x,un,vn)∂vvndx,〈I′(zn,z)〉=〈zn,z〉-∫0L‍∂F(x,un,vn)∂uudx-∂F(x,un,vn)∂vvdx,
we obtain
(3.15)limn→∞∥zn∥2=∫0L‍∂F(x,u,v)∂uudx+∂F(x,u,v)∂vvdx=∥z∥2.
From (3.15) and z∈X(L), zn⇀z in X(L), we have ∥zn-z∥→0 as n→∞.Remark 3.4.
γ is the largest sum of the order of u and v.Step 2 (Geometric conditions).
Lete1=(1,0),e2=(0,1), then {e1,e2} constitutes a pair of standard orthogonal base in R2. Let us define X2m to be the subspace of X(L)(3.16)X2m=span{sinkπxLei,i=1,2,k=1,2,…,m},
for every m∈N. We have dimX2m=2m, X1⊂X2⋯⊂X2m⊂X,⋃n=1∞Xn¯=X.
For a given constantρ>0, define a bounded closed set K⊂X2m(3.17)K={z=(u,v)∈X2m∣z=∑k=1m‍[αksinkπxLe1+βksinkπxLe2],∑k=1m‍(αk2+βk2)=ρ2}.Define mappingH:X2m→R2m. For any z∈X2m, we obtain
(3.18)H(z)=(α1,β1,α2,β2,…,αm,βm)ρ.
It is clear that H is a linear odd mapping. For every z∈X2m, we have
(3.19)∥z∥12=∫0L‍[|u′′(x)|2+|v(x)′′|2]dx=π42L3∑k=1m‍k4(αk2+βk2).So(3.20)ρ2π42L3|H(z)|2≤|z|12≤ρ2(mπ)42L3|H(z)|2.
From (3.20), we obtain H is an odd homeomorphism from X2m to R2m. Then H is an odd homeomorphism from K to S2m-1, since H(K)=S2m-1.On one hand, from functional (3.8) and using Sobolev’s embedding theorem, we obtain
(3.21)I(z)≥12∥z∥2-ε|z|2L≥12∥z∥2-επ2L∥z∥2.
Thus condition (A1) is fulfilled if ε=L/4π2, ρ=∥z∥/2.On the other hand, as-F(x,u,v)<0, then there exists σ, such that -F(x,u,v)<-(1/4)σ∥z∥14.DenoteA(n)=(nπ/L)4+p(nπ/L)2+a. From functional (3.8), we obtain
(3.22)I(z)≤12∫0L‍(z′′2+pz'2+az2)dx-∫0L‍F(x,z)dx≤L4A(n)∥z∥12-∫0L‍F(x,z)dx≤L4A(n)∥z∥12-L4σ∥z∥14,
where p=max{c,d},a=max{a2,b2}. Here choosing Rn=∥z∥1≥A(n)/σ, we obtain
(3.23)I(z)≤0.
So (A2) holds. The proof of Theorem 2.3 is completed.Example 3.5.
In System (1.2), consider the problem:
(3.24)F(x,u,v)=p0(x)un+p1(x)un-1v+⋯+pi(x)un-ivi+⋯+pn-1(x)uvn-1+pn(x)vn,
where pi(x)≥0, but there exists at least one pi(x)≠0, n is an even and n≥4, i=0,1,2,…,n.It is obvious thatF(x,-u,-v)=F(x,u,v) and F(x,u,v)=o(u2+v2) as u2+v2→0.For the superlinear property, we calculate that(3.25)u·∂F(x,u,v)∂u+v·∂F(x,u,v)∂v=np0(x)un+(n-1)p1(x)un-1v+⋯+(n-i)pi(x)un-ivi+⋯+pn-1(x)uvn-1+p1(x)un-1v+⋯+ipi(x)un-ivi+⋯+(n-1)pn-1(x)uvn-1+npn(x)vn=nF(x,u,v)≥4F(x,u,v).
Therefore, there exists β=4>2, as u2+v2≠0, we have
(3.26)u·∂F(x,u,v)∂u+v·∂F(x,u,v)∂v≥4F(x,u,v)>0foreveryx∈R.SoF satisfies the conditions (F1)–(F3). We only choose a(x)>0,b(x)>0, c>-π2/L2,d>-π2/L2, then the condition (A) is satisfied. Therefore, System (1.2) has infinitely many distinct pairs of solutions by using Theorem 2.3.


## 3.1. Variational Structure

Denote(3.1)X(L)=(H2(0,L)∩H01(0,L))2.
Then X(L) is a Hilbert space. The norm is
(3.2)∥z∥2=∥u∥c2+∥v∥d2,
where
(3.3)∥u∥c={∫0L‍[|u′′(x)|2+c|u′(x)|2+a(x)|u(x)|2]dx}1/2,∥v∥d={∫0L‍[|v′′(x)|2+d|v′(x)|2+b(x)|v(x)|2]dx}1/2,z=(u,v)∈X(L). The corresponding inner product are
(3.4)〈z1,z2〉|=∫0L‍[(z1′′,z2′′)+c(u1′,u2′)+d(v1′,v2′)+a(x)(u1,u2)+b(x)(v1,v2)]dx,〈u1,u2〉|c=∫0L‍[(u1′′,u2′′)+c(u1′,u2′)+a(x)(u1,u2)]dx,〈v1,v2〉|d=∫0L‍[(v1′′,v2′′)+d(v1′,v2′)+b(x)(v1,v2)]dx.
For every z=(u,v)∈X(L), using Poincaré inequality [7], we obtain
(3.5)∫0L‍u2dx≤L2π2∫0L‍u'2dx,∫0L‍u'2dx≤L2π2∫0L‍u′′2dx.
Thus, we can define another norm ∥·∥1 in X(L). That is, for every z∈X(L),
(3.6)∥z∥1={∫0L‍|z′′(x)|2dx}1/2.
The inner product in X(L) as follows:
(3.7)〈z1,z2〉|1=∫0L‍(z1′′(x),z2′′(x))dx,z1,z2∈X(L).
The two different norms (3.2) and (3.6) are equivalent in X(L).In this section we consider System (1.2). The Fréchet derivative of I is given by the following:
(3.8)I(u,v)=12∫0L‍[u′′2+cu'2+a(x)u2+v′′2+dv'2+b(x)v2]dx-∫0L‍F(x,u,v)dx,
where z=(u,v)∈X(L).Remark 3.1.
In general, the growth ofF is limited by the differentiability of functional I, but we apply truncation techniques in [8]. First, introduce auxiliary functional and the auxiliary functional is Fréchet differentiable. Second, we use critical point theory to prove the existence of critical point of auxiliary functional, then prove the existence of the original equation. However, in order to avoid technical complexity, we assume directly functional I is Fréchet differentiable.In fact, for everyz=(u,v)∈X(L),z-=(u-,v-)∈X(L), we obtain
(3.9)〈I′(z),z-〉=〈Iu′(u,v),u-〉+〈Iv′(u,v),v-〉,
where
(3.10)〈Iu′(u,v),u-〉=∫0L‍[u′′u-′′+cu′u-′+a(x)uu--∂F(x,u,v)∂uu-]dx,〈Iv′(u,v),v-〉=∫0L‍[v′′v-′′+cv′v-′+a(x)vv--∂F(x,u,v)∂vv-]dx.
and Iu′(u,v),Iv′(u,v)∈[H2(0,L)∩H01(0,L)]*, I′(z)∈X(L)*.It is similar to the discussion of [8], the solutions of System (1.2) corresponds to the critical point of the functional I, so we need to discuss the critical point of functional I. In order to prove Theorem 2.3, we introduce below definition and lemma.Definition 3.2 (see [9]).
LetX be a real Banach space, I∈C1(X,R), I is a Fréchet continuously differentiable functional in X(L). I is said to be satisfying Palais-Smale (PS) condition if any sequence {un}⊂X for which {I(un)} is bounded and {I′(un)}→0 as j→∞, possesses a convergent subsequence.Lemma 3.3 (see [8]).
LetX be an infinite dimensional Banach space and (Xn)n be a sequence of finite dimensional subspaces of X such that dimXn=n,
(3.11)X1⊂X2⊂⋯⊂Xn⊂X,⋃n=1∞Xn¯=X.
Let I∈C1(X,R) be an even functional, I(0)=0, and I satisfy (PS) condition. Suppose that(A1)
there are constantsρ,α>0 such that I|∂Bρ≥α, and(A2)
for everyn there is an Rn>0 such that I≤0 on Xn∖BRn.
ThenI possesses infinitely many pairs of critical points with unbounded sequence of critical values.


## 3.2. The Proof of Result

Step 1 (FunctionalI satisfies (PS) condition).
Let{zn}={(un,vn)} be a (PS) sequence in X, that is, {I(zn)} is bounded and I′(zn)→0, as n→∞. Suppose that {zn} is unbounded in X, that is, ∥zn∥→∞ as n→∞. Since
(3.12)I(zn)+1γ∥I′(zn)∥∥zn∥≥I(zn)-1γ〈I′(zn),zn〉=1γ∥zn∥2,
it follows that
(3.13)I(zn)∥zn∥2+∥I′(zn)∥γ∥zn∥≥1γ,
where γ≥4. Letting n→∞ in (3.13), we have a contradiction with ∥zn∥→∞ as n→∞.
Therefore{zn} is a bounded sequence in X(L). Passing if necessary to a subsequence we may assume that {zn} is weakly convergent to a function z∈X(L), zn⇀z in X(L), and zn→z in C[(0,L)].
From the Lebesgue theorem,z∈X(L), zn⇀z in X(L), and zn→z in C[(0,L)], letting n→∞ in (3.9)
(3.14)〈I′(zn,zn)〉=∥zn∥2-∫0L‍∂F(x,un,vn)∂uundx-∂F(x,un,vn)∂vvndx,〈I′(zn,z)〉=〈zn,z〉-∫0L‍∂F(x,un,vn)∂uudx-∂F(x,un,vn)∂vvdx,
we obtain
(3.15)limn→∞∥zn∥2=∫0L‍∂F(x,u,v)∂uudx+∂F(x,u,v)∂vvdx=∥z∥2.
From (3.15) and z∈X(L), zn⇀z in X(L), we have ∥zn-z∥→0 as n→∞.Remark 3.4.
γ is the largest sum of the order of u and v.Step 2 (Geometric conditions).
Lete1=(1,0),e2=(0,1), then {e1,e2} constitutes a pair of standard orthogonal base in R2. Let us define X2m to be the subspace of X(L)(3.16)X2m=span{sinkπxLei,i=1,2,k=1,2,…,m},
for every m∈N. We have dimX2m=2m, X1⊂X2⋯⊂X2m⊂X,⋃n=1∞Xn¯=X.
For a given constantρ>0, define a bounded closed set K⊂X2m(3.17)K={z=(u,v)∈X2m∣z=∑k=1m‍[αksinkπxLe1+βksinkπxLe2],∑k=1m‍(αk2+βk2)=ρ2}.Define mappingH:X2m→R2m. For any z∈X2m, we obtain
(3.18)H(z)=(α1,β1,α2,β2,…,αm,βm)ρ.
It is clear that H is a linear odd mapping. For every z∈X2m, we have
(3.19)∥z∥12=∫0L‍[|u′′(x)|2+|v(x)′′|2]dx=π42L3∑k=1m‍k4(αk2+βk2).So(3.20)ρ2π42L3|H(z)|2≤|z|12≤ρ2(mπ)42L3|H(z)|2.
From (3.20), we obtain H is an odd homeomorphism from X2m to R2m. Then H is an odd homeomorphism from K to S2m-1, since H(K)=S2m-1.On one hand, from functional (3.8) and using Sobolev’s embedding theorem, we obtain
(3.21)I(z)≥12∥z∥2-ε|z|2L≥12∥z∥2-επ2L∥z∥2.
Thus condition (A1) is fulfilled if ε=L/4π2, ρ=∥z∥/2.On the other hand, as-F(x,u,v)<0, then there exists σ, such that -F(x,u,v)<-(1/4)σ∥z∥14.DenoteA(n)=(nπ/L)4+p(nπ/L)2+a. From functional (3.8), we obtain
(3.22)I(z)≤12∫0L‍(z′′2+pz'2+az2)dx-∫0L‍F(x,z)dx≤L4A(n)∥z∥12-∫0L‍F(x,z)dx≤L4A(n)∥z∥12-L4σ∥z∥14,
where p=max{c,d},a=max{a2,b2}. Here choosing Rn=∥z∥1≥A(n)/σ, we obtain
(3.23)I(z)≤0.
So (A2) holds. The proof of Theorem 2.3 is completed.Example 3.5.
In System (1.2), consider the problem:
(3.24)F(x,u,v)=p0(x)un+p1(x)un-1v+⋯+pi(x)un-ivi+⋯+pn-1(x)uvn-1+pn(x)vn,
where pi(x)≥0, but there exists at least one pi(x)≠0, n is an even and n≥4, i=0,1,2,…,n.It is obvious thatF(x,-u,-v)=F(x,u,v) and F(x,u,v)=o(u2+v2) as u2+v2→0.For the superlinear property, we calculate that(3.25)u·∂F(x,u,v)∂u+v·∂F(x,u,v)∂v=np0(x)un+(n-1)p1(x)un-1v+⋯+(n-i)pi(x)un-ivi+⋯+pn-1(x)uvn-1+p1(x)un-1v+⋯+ipi(x)un-ivi+⋯+(n-1)pn-1(x)uvn-1+npn(x)vn=nF(x,u,v)≥4F(x,u,v).
Therefore, there exists β=4>2, as u2+v2≠0, we have
(3.26)u·∂F(x,u,v)∂u+v·∂F(x,u,v)∂v≥4F(x,u,v)>0foreveryx∈R.SoF satisfies the conditions (F1)–(F3). We only choose a(x)>0,b(x)>0, c>-π2/L2,d>-π2/L2, then the condition (A) is satisfied. Therefore, System (1.2) has infinitely many distinct pairs of solutions by using Theorem 2.3.



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*Source: 469863-2012-10-10.xml*