Periodic solution for ϕ-Laplacian neutral differential equation

Lemma 2.1

([14]) Let X and Z be Banach spaces with norm ∥⋅∥_X and ∥⋅∥_Y, respectively, and Ω ⊂ X be an open and bounded set with origin θ ∈ Ω. Suppose that M : X ∩ domM → Z is a quasi – linear operator and

is an M-compact mapping. In addition, if

1. Mx ≠ N_λ x, λ ∈ (0, 1), x ∈ ∂ Ω,

2. deg{JQN, Ω ∩ kerM, 0} ≠ 0,

where N = N₁, then the abstract equation Mx = Nx has at least one solution in Ω.

Lemma 2.2

(see [15]) If |c| ≠ 1, then the operator (Ax)(t) := x(t) − cx(t − τ) has a continuous inverse A⁻¹ on the space C_T, and satisfying

Lemma 2.3

(see [16, 17]) The follow propositions are true:

1. Suppose c = −1, |τ| = (m/n)π, where m, n are coprime positive integers with m even, then A : L2π2→L2π2 , has a unique inverse A⁻¹ : L2π2→L2π2 satisfying

   ∥A−1∥≤1σ1,

   where σ1:=infk∈N|1−ce−ikτ|=infk∈N2(1+cos⁡kτ)12>0.

2. Suppose c = −1, |τ| = (m/n)π, where m, n are coprime odd positive integers, then A : L2π2+→L2π2+ , has a unique inverse A⁻¹ : L2π2+→L2π2+ satisfying

   ∥A−1∥≤1σ2,

   where σ2:=infk∈N1|1−ce−ikτ|=infk∈N12(1+cos⁡kτ)12>0.

3. Suppose c = −1, |τ| = (m/n)π, where m, n are coprime positive integers with m odd and n even, then A : L2π2−→L2π2− , has a unique inverse A⁻¹ : L2π2−→L2π2− satisfying

   ∥A−1∥≤1σ3,

   where σ3:=infk∈N2|1−ce−ikτ|=infk∈N22(1+cos⁡kτ)12>0.

4. Suppose c = 1, |τ| = (m/n)π, where m, n are coprime positive integers with m odd, then A : L2π2−→L2π2− , has a unique inverse A⁻¹ : L2π2−→L2π2− satisfying

   ∥A−1∥≤1σ4,

   where σ4:=infk∈N1|1−ce−ikτ|=infk∈N12(1+cos⁡kτ)12>0.

5. Suppose c = 1, |τ| = π, then A : L2π2−→L2π2− , has a unique inverse A⁻¹ : L2π2−→L2π2− satisfying

   ∥A−1∥≤1σ5,

   where σ5:=infk∈N1|1−ce−ikτ|=infk∈N12(1+cos⁡kτ)12=2>0.

Theorem 3.1

Assume that condition (A₁), (A₂) and |c| ≠ 1, Ω is an open bounded set in CT1 . Suppose the following conditions hold:

1. For each λ ∈ (0, 1) the equation

   (ϕ(Ax)′(t))′=λf(t,x(t),x′(t)) (3.1)

   has no solution on ∂Ω;

2. The equation

   F(a):=1T∫0Tf(t,x(t),x′(t))dt=0,

   has no solution on ∂Ω ∩ℝ;

3. The Brouwer degree

   deg⁡{F,Ω∩R,0}≠0.

Then (1.1) has at least one periodic solution on Ω.

Proof

In order to use Lemma 2.1 studying the existence of a periodic solution to (3.1), we set X := {x ∈ C[0, T] : x(0) = x(T)} and Z := C[0, T],

where dom M := {u ∈ X : ϕ(Au)′ ∈ C¹(ℝ, ℝ)}. Then ker M = ℝ. In fact

where c, c₁, c₂ are constant in ℝ. Since (Ax)(0) = (Ax)(T), then, we get ker M = {x ∈ X : (Ax)(t) ≡ c₂}. In addition,

So M is quasi-linear. Let

Clearly, dim X₁ = dim Z₁ = 1, and X = X₁ ⊕ X₂, P : X → X₁, Q : Z → Z₁, are defined by

For ∀ Ω ⊂ X define N_λ : Ω → Z by

We claim (I − Q)N_λ(Ω) ⊂ ImM = (I − Q)Z holds. In fact, for x ∈ Ω, we have

Hence, we have (I − Q)N_λ(Ω) ⊂ Im M.

Moreover, for any x ∈ Z, we have

So, we have (I − Q)Z ⊂ ImM. On the other hand, x ∈ ImM and ∫0T x(t)dt = 0, then we have x(t) = x(t) − ∫0T x(t)dt. Hence, we can get x(t) ∈ (I − Q)Z. Therefore, ImM = (I − Q)Z.

From QN_λ x = 0, we can get λT∫0T f(t, x(t), x′(t))dt = 0. Since λ ∈ (0, 1), then we have 1T∫0T f(t, x(t), x′(t))dt = 0. Therefore, we can get QNx = 0, then, (2.4) also holds.

Let J : Z₁ → X₁, J(x) = x, then J(0) = 0. Define R : Ω × [0, 1] → X₂, by Lemma 2.2, we know that there exists a continuous inverse operator A⁻¹ of neutral operator A such that

where a ∈ R is a constant such that

From Lemma 2.3 of [13], we know that a is uniquely defined by

where ã(x, λ) is continuous on Ω × [0, 1] and bounded sets of Ω × [0, 1] into bounded sets of ℝ.

From (3), we can find that

Now, for any x ∈ ∑_λ = {x ∈ Ω : Mx = N_λ x} = {x ∈ Ω : (ϕ(Ax)′(t))′ = λ f(t, x(t), x′(t))}, we have ∫0T f(t, x(t), x′(t))dt = 0, together with (3) gives

Take a = ϕ(Ax)’(0), from (Ax)′(t) = (Ax′)(t), then we can get

where a is unique, we see that

So, we have

which yields the second part of (2.4). Meanwhile, if λ = 0, the

where c₃ ∈ ℝ is a constant, so by the continuity of ã(x, λ) with respect to (x, λ), a = ã(x, 0) = ϕ(Ac)′(0) = θ. So,

which yields the first part of (2.4). Furthermore, we consider the following equation

In fact,

Integrating both sides of (3.5) over [0, s], we have

Therefore, we have

where a := ϕ(A(P + R))′(0). Then, we can get

Then, we have

since (A(P + R))′(s) = A(P + R)′(s). Hence, we have

since R(x, λ)(0) = 0. So, we can get

Thus, N_λ is M-compact on Ω. Obviously, the equation

can be converted to

where M and N_λ are defined by (3.2) and (3), respectively. As proved above,

is an M-compact mapping. From assumption (C₁), one find

and assumptions (C₂) and (C₃) imply that deg{JQN, Ω ∩ kerM, θ} is valid and

So by applications of Lemma 2.1, we see that (3.1) has a T-periodic solution. □

In the following, applying Lemma 2.3 and Theorem 3.1, we consider the existence of a periodic solution to (1.1) in the case that |c| = 1.

Theorem 3.2

Assume that conditions (A₁), (A₂) (C₁), (C₂) and (C₃) hold, Ω is an open bounded set in C2π1. Furthermore, suppose one of the following conditions holds:

1. c = −1 and |τ| = (m/n)π, with m, n are coprime positive integers with m even;

2. c = −1 and |τ| = (m/n)π, with m, n are coprime odd positive integers;

3. c = −1 and |τ| = (m/n)π, with m, n are coprime positive integers with m odd and n even;

4. c = 1 and |τ| = (m/n)π, with m, n are coprime positive integers with m odd;

5. c = 1 and |τ| = π.

Then (1.1) has at least one periodic solution on Ω.

Proof

We follow the same strategy and notation as in the proof of Theorem 3.1. Next, we consider R(x, λ)(t).

Case (i) c = −1 and |τ| = (m/n)π, with m, n are coprime positive integers with m even. Take T = 2π, from (3.3) and (3.4), applying Lemma 2.3, we know that there exist a continuous inverse operator A⁻¹ of neutral operator A in the case that c = −1 such that

where a ∈ R is a constant such that

Similarly, we can get Case (ii)-Case (v). This proves the claim and the rest of the proof of the theorem is identical to that of Theorem 3.1. □

Theorem 4.1

Suppose |c| ≠1, (A₁) and (A₂) hold. Assume that the following conditions hold:

1. There exists a constant D > 0 such that

   xg(t,x)>0,  ∀ (t,x)∈[0,T]×R, with |x|>D.

2. There exist two positive constants σ_*, σ^* such that σ_* ≤ |f(x(t))| ≤ σ^*, ∀ t ∈ ℝ.

3. There exist positive constants a, b, B such that

   |g(t,x(t))|≤a|x(t)|+b,   for |x(t)|>B and t∈R.

Then (4.1) has at least one solution with period T if σ∗−σ∗|c|+22(1+|c|)aT>0.

Proof

Consider the homotopic equation

Firstly, we will claim that the set of all T-periodic solutions of (4.2) is bounded. Let x(t) ∈ CT1 be an arbitrary T-periodic solution of (4.2). As (Ax)(0) = (Ax)(T), there exists a point t₀ ∈ (0, T) such that (Ax)′(t₀) = 0, while ϕ(0) = 0, and we see

Integrating both sides of (4.2) over [0, T], we have

From the mean value theorem, there is a constant ξ ∈ (0, T) such that

In view of (H₁), we obtain

Then, we have

Multiplying both sides of (4.2) by (Ax)′(t) and integrating over the interval [0, T], we get

Substituting ∫0T (ϕ(Ax)′(t))′(Ax)′(t)dt = 0, ∫0T f(x(t))x′(t)(Ax)′(t)dt = ∫0T f(x(t))(x′(t))²dt – c ∫0T f(x(t))x′(t)x′(t – τ))dt

Therefore, we have

From (H₂), we have

So, we have

Define

From (H₂) and the Hölder inequality, we have

where ∥e∥:=maxt∈[0,T]|e(t)|. Substituting ∫0T |x′(t – τ)|dt = ∫0T |x′(t)|dt into (4.7), and by applications of condition (H₃), we have

where ∥g_B∥ := max|x(t)|≤B |g(t, x(t))| and N1:=(1+|c|)(|gB|+b+∥e∥)T12. Substituting (4.5) into (4.8), we have