Multiple periodic solutions for asymptotically linear Duffing equations with resonance (II)

J. Math. Anal. Appl. 397 (2013) 156–160

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Multiple periodic solutions for asymptotically linear Duffing equationswith resonance (II)✩

Keqiang Li a,∗, Shangjiu Wang b, Yonggang Zhao a

a Department of Mathematics, Henan Normal University, Xinxiang, Henan 453000, PR Chinab School of Mathematics and Information Science, Shaoguan University, Shaoguan, Guangdong 512005, PR China

a r t i c l e i n f o

Article history:Received 22 April 2012Available online 24 July 2012Submitted by J. Mawhin

Keywords:Asymptotically linear Duffing equationwith resonance

Multiple periodic solutionIndex theoryMorse theoryThree-critical-points theorem

a b s t r a c t

We investigate the existence ofmultiple periodic solutions of asymptotically linear Duffingequations with resonance on the left side of the first eigenvalue using index theory and thethree-critical-points theorem and obtain a new result.

© 2012 Elsevier Inc. All rights reserved.

1. Introduction and the main results

In this paper, we consider the existence of multiple periodic solutions of the Duffing equation:

x′′+ f (t, x) = 0, (1.1)

x(1)− x(0) = 0 = x′(1)− x′(0), (1.2)

where f : [0, 1] × R → R is a continuous differential function satisfying

(H′

1)

1

0f (t,+∞)dt < 0 <

1

0f (t,−∞)dt,

where f (t,+∞) = lim supx→+∞ f (t, x) and f (t,−∞) = lim infx→−∞ f (t, x). Let f ′(t, x) denote the derivative of f withrespect to x. Our main result is the following theorem.

Theorem 1.1. Let (H′

1) hold. Assume that f satisfies the following conditions:

(H′

2) There exists a constant r > 0 such that f (t,x)x ≤ 0 for |x| ≥ r.

(H′

3) f (t, 0) = 0, i(f ′(t, 0)) ≥ 1, and ν(f ′(t, 0)) = 0.

Then (1.1)–(1.2) has at least two nontrivial solutions.

✩ Partially supported by the National Natural Science Foundation of China (11171157).∗ Corresponding author.

E-mail address: likeqiang000@sina.com (K. Li).

0022-247X/$ – see front matter© 2012 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2012.07.048

K. Li et al. / J. Math. Anal. Appl. 397 (2013) 156–160 157

Remark 1.1. This paper is an extension to the work of [1], where the author studied the existence of multiple periodicsolutions under the assumption that 0 ≤

f (t,x)x < λ(t) for |x| ≥ r , with i(λ) = 1 and ν(λ) = 0. Here, we only allow that

resonance happens for |x| ≥ r on the left side of the first eigenvalue. This case cannot be solved by degree arguments. Weobtain results that also cannot be obtained by degree arguments. We will use the three-critical-points theorem to proveTheorem 1.1.

Here for any a ∈ L∞[0, 1], i(a) and v(a) denote its index and the nullity of the associated linear Duffing equations (see

[2,3,1] for reference). In [1], the author discussed the index theory for the linear Duffing equation and gave somepropositionsand examples of indexes. In [2], an index for second-order linear Hamiltonian systems was defined. And in [3], an indexfor more general linear self-adjoint operator equations was developed. In [4–7], an index theory for symplectic paths wasdefined by Conley et al.. More applications of the index theory can be found in [8–13]. As in [12], throughout this paper, fora1, a2 ∈ L∞

[0, 1], wewrite a1 ≤ a2 if a2(t)−a1(t) ≥ 0, for a.e. t ∈ [0, 1]; wewrite a1 < a2 if a1 ≤ a2, and a2(t)−a1(t) > 0holds on a subset of [0, 1] with nonzero measure.

It is well known that the linear periodic boundary value problem

x′′+ λx = 0, x(1)− x(0) = 0 = x′(1)− x′(0) (1.3)

has the eigenvaluesλ = 4π2m2, m ∈ N, withmultiplicity 2 form ≥ 1. Sowe regard the problem (1.1) and (1.2) as resonancefor |x| ≥ r > 0 on the left side of the first eigenvalue of (1.3) under assumption (H′

1). Resonance problems have been paidmuch attention during the last few decades since the celebrated work [14] appeared.

There are many results for (1.1)–(1.2) in the literature. It is well known [15] that under non-resonant conditions,

(2kπ)2 + δ ≤f (t, x)

x≤ (2(k + 1)π)2 − δ, as |x| > r > 0, k ∈ N,

(1.1)–(1.2) has at least one solution. The resonant conditions in [16,17]:

(2kπ)2 ≤f (t, x)

x≤ (2(k + 1)π)2, as |x| > r > 0, k ∈ N,

are not enough for the existence of solutions of (1.1)–(1.2). An additional condition called the (LL) condition, like (H′

1), isusually needed. The three papers [15–17] are about the existence of solutions.

By the discussion above, our theorem is a new result. Solutions of (1.1)–(1.2) correspond to critical points of the C2

functional ϕ:

ϕ(x) =12

1

0|x′(t)|2dt −

1

0F(t, x(t))dt, ∀x ∈ E, (1.4)

where F(t, u) = u0 f (t, s)ds, u ∈ R, and E := {x ∈ H1

[0, 1]|x(0) = x(1)} with the inner product and norm given by

⟨x, y⟩ :=

01[xy + x′y′

]dt, ∥x∥E :=

1

0[|x|2 + |x′

|2]dt

12

, ∀x, y ∈ E.

The derivatives of ϕ are

⟨ϕ′(x), y⟩ =

1

0x′y′dt −

1

0f (t, x)ydt, ∀x, y ∈ E, (1.5)

⟨ϕ′′(x)y, z⟩ =

1

0y′z ′dt −

1

0f ′(t, x)yzdt, ∀x, y, z ∈ E. (1.6)

By f (t, 0) = 0 in (H′

3), it is obvious that (1.1)–(1.2) has a trivial solution x = 0. We are looking for nontrivial solutionsfor (1.1)–(1.2), for which it turns out that the existence of nontrivial solutions of (1.1)–(1.2) depends on the interplay of thebehaviors of f near infinity and near zero with the eigenvalues of (1.3).

We conclude this section by giving a review of the index theory that will be used below.For any a ∈ L∞

[0, 1], consider the following equations:x′′(t)+ a(t)x = 0,x(0)− x(1) = x′(0)− x′(1) = 0, (1.7)

and

qa(x, y) =

1

0[x′(t)y′(t)− a(t)x(t)y(t)dt], ∀x, y ∈ E. (1.8)

From [2,3,1], we have the following proposition.

158 K. Li et al. / J. Math. Anal. Appl. 397 (2013) 156–160

Proposition 1.2. For any a ∈ L∞[0, 1],

(1) E can be divided into three parts:

E = E+(a)⊕ E0(a)⊕ E−(a)

such that qa is positive definite, null and negative definite on E+(a), E0(a) and E−(a) respectively. Furthermore, E0(a) and E−(a)are finitely dimensional. We call ν(a) := dim E0(a) and i(a) := dim E+(a) the nullity and index respectively.(2) ν(a) is the dimension of the solution subspace of (1.7), and i(a) =

s<0 ν(a + s).

2. Proof of Theorem 1.1

The proof of Theorem 1.1 will depend on the following lemmas. Let X be a Hilbert space and ψ ∈ C1(X,R). As in [18],let K = {x ∈ X |ψ ′(x) = θ}, which is a set of all critical points of ψ . When ψ ∈ C2(X,R) and p ∈ K , we have that ψ ′′(p)is a self-adjoint operator. We call the dimension of negative space corresponding to the spectral decomposition the Morseindex of p and denote it by m−(ψ ′′(p)), and set m0(ψ ′′(p)) = dim ker ψ ′′(p). If ψ ′′(p) has a bounded inverse we say that pis non-degenerate. By (1.6), the following remark is known ([2], Chapter 9).

Remark 2.1. (1) The Morse index of p is equal to i(b)where b = f ′(t, p).(2) p is non-degenerate if and only if ν(b) = 0.

The following lemma is Theorem 3.1 in [19], Chapter 5.

Lemma 2.1. Assume that f ∈ C2(X, R), where X is a Hilbert space, satisfies the (PS) condition and is lower semi-bounded.Meanwhile, suppose that x0 is a non-degenerate non-minimum critical point of f with finite Morse index. Then f has at leastthree critical points.

Wewill use Lemma 2.1 to prove Theorem 1.1. The critical points of ϕ defined by (1.4) correspond to the solutions of (1.1)–(1.2),so we need to verify that ϕ satisfies the assumptions of Lemma 2.1. We will see that ϕ is lower semi-bounded and satisfies the(PS) condition in Lemmas 2.2 and 2.3 below, respectively.

Lemma 2.2. Under assumption (H′

2), ϕ defined by (1.4) is lower semi-bounded.

Proof. Since ϕ ∈ C2(E, R), to prove that ϕ is lower semi-bounded, it is sufficient to prove that ϕ is lower bounded. Since fis continuous, we can assume that there exists a constantM (>0) such that r

0f (t, s)ds

< M and 0

−rf (t, s)sds

< M.

Meanwhile, by (H′

2), it is obvious that

f (t, x) ≤ 0, as x ≥ r and f (t, x) ≥ 0, as x ≤ −r.

So at once, we get the results that x

rf (t, s)ds ≤ 0, as x ≥ r and

−r

xf (t, s)ds ≥ 0, as x ≤ −r.

What is more, it is also obvious that when x ≥ r , x

0f (t, s)ds =

r

0f (t, s)ds +

x

rf (t, s)ds ≤ M (2.1)

and when x ≤ −r , x

0f (t, s)ds =

−r

0f (t, s)ds +

x

−rf (t, s)ds ≤ M. (2.2)

So by (2.1) and (2.2), we obtain 1

0F(t, x)dt =

1

0

x

0f (t, s)ds

dt

=

{t;|x(t)|<r}

x

0f (t, s)dsdt +

{t;x(t)≥r}

x

0f (t, s)dsdt +

{t;x(t)≤−r}

x

0f (t, s)dsdt

≤

{t;|x(t)|<r}

x

0f (t, s)dsdt + 2M

≤ C1, (2.3)

K. Li et al. / J. Math. Anal. Appl. 397 (2013) 156–160 159

where C1 > 0 is a constant. So by (2.3), we have

ϕ(x) ≥ −C1.

So ϕ is lower bounded. The proof is completed. �

Lemma 2.3. Under assumptions (H′

1), the functional ϕ satisfies the (PS) condition.

Proof. To verify that ϕ satisfies the (PS) condition, we assume that {xn} ⊂ E, s.t. {ϕ(xn)} is bounded, and ϕ′(xn) → 0 asn → ∞. Then there exists a convergent subsequence. We claim that ∥xn∥C is bounded, where ∥x∥C := maxt∈[0,1] |x(t)|. Ifnot, assume ∥xn∥C → +∞. Set yn =

xn(t)∥xn∥C

. By (2.3) and multiplying by ∥xn∥−2C on both sides of (1.4), we have

12

1

0|y′

n(t)|2dt = ∥xn∥−2

C ϕ(xn)+ ∥xn∥−2C

1

0F(t, xn)dt ≤ C2,

where C2 (>0) is a constant. Then ∥yn∥E is bounded. So we can assume that yn ⇀ y0 in E and yn → y0 in C[0, 1]. Noticethat ∥yn∥C = 1; thus ∥y0∥C = 1. By (1.4), (1.5), and (H′

2), we get

1∥xn∥C

⟨ϕ′(xn), yn⟩ =

1

0|y′

n(t)|2dt −

10 f (t, xn(t)yn(t)dt)

∥xn∥C

≥

1

0|y′

n(t)|2dt − ∥xn∥−1

C

{t;|xn(t)|≤r}

f (t, xn(t))yn(t)dt. (2.4)

We know that lim supn→∞

10 |y′

n(t)|2dt ≥

10 |y′(t)|2dt as yn ⇀ y in L2[0, 1]. By (2.4), the following holds: 1

0|y′

0(t)|2dt ≤ 0.

Furthermore, we get y′

0(t) = 0 for a.e. t ∈ (0, 1). So we have y0 = c, c = ±1. If y0 = 1, under hypothesis yn =xn(t)∥xn∥C

, theequation xn(t) = yn(t)∥xn∥C holds. Since yn → y0 = 1 in C[0, 1], it is obvious that xn(t) → +∞ uniformly for t ∈ [0, 1] asn → ∞. So by ϕ′(xn) → 0, and the equation below:

⟨ϕ′(xn), y⟩ =

1

0x′

n(t)y′(t)dt −

1

0f (t, xn(t))y(t)dt, ∀y ∈ E,

letting y = 1, we have 10 f (t, xn(t))dt → 0 as n → ∞. By Fatou’s Lemma, we have

0 = lim supn→∞

1

0f (t, xn(t))dt ≤

1

0lim supn→∞

f (t, xn(t))dt

=

1

0f (t,+∞)dt

which is a contradiction to assumption (H′

1). If y0 = −1, under hypothesis yn =xn(t)∥xn∥C

, the equation xn(t) = yn(t)∥xn∥C

holds. Since yn → y0 = −1 in C[0, 1], it is obvious that xn(t) → −∞ uniformly for t ∈ [0, 1] as n → ∞. So by ϕ′(xn) → 0,and the equation below:

⟨ϕ′(xn), y⟩ =

1

0x′

n(t)y′(t)dt −

1

0f (t, xn(t))y(t)dt, ∀y ∈ E,

letting y = 1, we have 10 f (t, xn(t))dt → 0 as n → ∞. By Fatou’s Lemma, we obtain

0 = lim infn→∞

1

0f (t, xn(t))dt ≥

1

0lim infn→∞

f (t, xn(t))dt

=

1

0f (t,−∞)dt,

which is also a contradiction to assumption (H′

1). The (PS) condition is verified. �

The last lemma is used to verify ϕ satisfies the remaining assumptions in Lemma 2.1.

Lemma 2.4. Let (H′

3) hold. Then x = 0 is a non-degenerate non-minimum critical point with finite Morse index.

160 K. Li et al. / J. Math. Anal. Appl. 397 (2013) 156–160

Proof. By (1.6), we have

⟨ϕ′′(0)u, v⟩ =

1

0u′v′dt −

1

0f ′(t, 0)uvdt, ∀u, v ∈ E. (2.5)

Firstly, by (1.5) and f (t, 0) = 0, it is obvious that x = 0 is a critical point. Secondly, since ν(f ′(t, 0)) = 0, by (2) inRemark 2.1, we have that x = 0 is non-degenerate. Thirdly, it is obvious that x = 0 is non-minimum. Indeed, if 0 is aminimum of ϕ, there exists a neighborhood Br1 of 0, with radius r1 > 0, such that (ϕ′′(0)x, x) ≥ 0, ∀x ∈ Br1 . Meanwhile, byi(f ′(t, 0)) ≥ 1 and (1) in Proposition 1.2, there exists a subspace E∗(⊂ E)with dimension greater than or equal 1 such that(ϕ′′(0)z, z) < 0,∀z ∈ E∗ ∩ (Br1 \ {0}) (⊂ Br1). This is a contradiction. Finally, by (2.5), (1.6), (1) in Proposition 1.2, and (1)in Remark 2.1, x = 0 has a finite Morse index. This completes the proof. �

After giving the preliminary work, now we can prove Theorem 1.1.

Proof of Theorem 1.1. Firstly, by Lemmas 2.2 and 2.3, we know that ϕ is lower semi-bounded and satisfies the (PS)condition. Secondly, by Lemma 2.4 and (1) in Remark 2.1, we know that x = 0 is a non-degenerate non-minimum criticalpoint of ϕ with finite Morse index i(f ′(t, 0)) ≥ 1. Therefore, all assumptions of Lemma 2.1 are satisfied by ϕ. So we obtain atleast three distinct critical points and (1.1)–(1.2) has accordingly at least three distinct solutions. Besides the trivial solutionx = 0, there also exist two nontrivial solutions of (1.1)–(1.2). This completes the proof. �

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