2005-Oujda International Conference on Nonlinear Analysis. Electronic Journal of Differential Equations, Conference 14, 2006, pp. 109–117. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

NONRESONANCE CONDITIONS FOR A SEMILINEAR WAVE EQUATION IN ONE SPACE DIMENSION

AOMAR ANANE, OMAR CHAKRONE, ABDELLAH ZEROUALI

Abstract. In this paper we study the existence of periodic weak solutions for semilinear wave equations in one space dimension in the case of nonresonance.

1. Introduction

In this paper we consider the existence of periodic solutions for the wave equation

\begin{matrix}□u = αu + βu_{x} − γu_{t} + g(x, t, u) + h(x, t) in Q, \\ u(x, t + 2π) = u(x, t) in ]0, π[×ℝ, \\ u(0, t) = u(π, t) = 0 ∀t ∈ ℝ, \end{matrix}  (1.1)

where Q =]0, π[×]0, 2π[, □ = \frac{∂^{2}}{∂t^{2}} − \frac{∂^{2}}{∂x^{2}} is the D’Alembertian, (α, β, γ) ∈ ℝ×ℝ×ℝ, h is a given function in L^{2}(Q), and g :]0, π[×ℝ × ℝ → ℝ is 2π-periodic in t and a Carathéodory function (i.e. measurable in (x, t) for each s ∈ ℝ and continuous in s for almost all (x, t) ∈ Q).

We are interested in the nonresonance for the problem (1.1) (i.e. in the condition for the function g such that there exist a solution u ∈ L^{2}(Q) for any given h ∈ L^{2}(Q)). We will assume that g satisfies the following conditions:

(C1) g(x, t, s) is nondecreasing in s;

(C2) for s ≠ r, (x, t) ∈ Q, we have

e^{β \frac{x}{2}} (\frac{g(x, t, r) − g(x, t, s)}{r − s}) ≥ \frac{β^{2}}{4} − α;

(C3) for all R > 0, there exists ϕ_{R} ∈ L^{2}(Q) such that a.e. (x, t) ∈ Q,

max_{|s|≤R} |g(x, t, s)| ≤ ϕ_{R}(x, t);

2000 Mathematics Subject Classification. 35J65, 35J25.

Key words and phrases. D’Alembertian; homotopy argument; eigenvalues.

\begin{matrix}c \\ ◯ \end{matrix}2006 Texas State University - San Marcos.

Published September 20, 2006.

(C4) a.e. (x, t) ∈ Q, we have

λ_{k} − \frac{γ^{2}}{4} + \frac{β^{2}}{4} − α < l(x, t) := lim inf_{|s|→+∞} \frac{g(x, t, s)}{s}

≤ lim sup_{|s|→+∞} \frac{g(x, t, s)}{s} := k(x, t)

< λ_{k+1} − \frac{γ^{2}}{4} + \frac{β^{2}}{4} − α,

where λ_{k} and λ_{k+1} are two consecutive eigenvalues of the D’Alembertian, and σ(□) denotes the spectrum of the D’Alembertian.

Problem (1.1) has been studied with conditions of resonances by several authors mention in particular: In the case α = β = γ = 0 and h = 0 Benaoum in [3, 4], Mustonen and Berkovits in [5, 6, 7, 8], and Brézis and Nirenberg in [12]. The case g(x, t, s) = g(s), has been studied by Mustonen and Berkovits in [9]. The case β = γ = 0 and α is a eigenvalue of the D’Alembertian operator (□), has been studied in [7]. The case β = 0 and α is a eigenvalue of the operator T defined by T u = □u + γu_{t}, where u_{t} = \frac{∂u}{∂t}, has been studied in [5, 12]. In the general case, Anane, Chakrone and Ghanim in [2]. In the case of nonresonance, the problem (1.1) has been studied by Mustonen and Berkovits in [6] and [10], and by Brezis and Nirenberg in [12] but only in particular cases. The situation that we consider here is marked by the presence of one term of transportation β∇u, what constitutes an extension of the cases studied by Mustonen and Berkovits in [6, 10]. In our work, we show (see Corollary 3.2) while using homotopy argument given by Mustonen and Berkovits in [6], and of analogous techniques developed by Anane and Chakrone in [1] for the Laplacian (Δ), that the problem (1.1) has at least a solution for all h ∈ L^{2}(Q).

2. Remarks and notation

Let δ, μ ∈ ℝ such that δ < μ, we introduce the following general hypothesis For a.e. (x, t) ∈ Q, we have

δ + \frac{β^{2}}{4} − α ≤≠ l(x, t) := lim inf_{|s|→+∞} \frac{g(x, t, s)}{s}

≤ lim sup_{|s|→+∞} \frac{g(x, t, s)}{s} := k(x, t)

≤≠ μ + \frac{β^{2}}{4} − α

(2.1)

The notation ≤≠ means that one has an large inequality on Q and strict on a set of measure different from zero.

Remark 2.1. (1) We denote by T u = □u + γu_{t}. Then (i) T is a densely defined closed linear operator with closed range. (ii) Im(T) = [ker(T)]^{⊥}. (ii) λ is a eigenvalue of the D’Alembertian if and only if λ − \frac{γ^{2}}{4} is a eigenvalue

of T (iii) If T_{0} is the restriction of the operator T on Im(T) = T (D(T)), with D(T)

is the domain of the operator T, then T_{0} has compact inverse.

For the proof of the remarks (i)–(iii), see [4].

(2) We put g̃(x, t, s) = (α − \frac{β^{2}}{4} )s + e^{\frac{β}{2} x}g(x, t, e^{− \frac{β}{2} x}s) and h̃(x, t) = e^{\frac{β}{2} x}h(x, t). Let N : L^{2}(Q) → L^{2}(Q),

N(u) = g̃(x, t, u)

be the Nemytskii operator generated by the function g̃. For r ∈ [0, 1], consider the operator H_{r} : D(T) ⊂ L^{2}(Q) → L^{2}(Q),

H_{r}(u) = T u − r(N(u) + h̃) − (1 − r)λu,

where δ < λ < μ.

If there exists R > 0, for all r ∈ [0, 1] and all u ∈ D(T),

with ∥u∥ = ( ∫_{Q} |u|^{2})^{1/2} = R, then H_{r}(u) ≠ 0.

(2.2)

(3) If (C1) and (C2) are verified, then g̃(x, t, s) is nondecreasing in s, thus the operator N is monotone. This statement and the following are easy to prove.

(4) Condition (C3) implies that for all R > 0 there exists ϕ̃_{R} ∈ L^{2}(Q) such that for a.e. (x, t) ∈ Q we have

max_{|s|≤R} |g̃(x, t, s)| ≤ ϕ̃_{R}(x, t).

(5) If (C4) is verified, then for a.e. (x, t) ∈ Q, we have

λ_{k} − \frac{γ^{2}}{4} < l̃(x, t) := lim inf_{|s|→+∞} \frac{g̃(x, t, s)}{s} ≤ lim sup_{|s|→+∞} \frac{g̃(x, t, s)}{s} := k̃(x, t) < λ_{k+1} − \frac{γ^{2}}{4}

(6) If (2.1) is satisfied, then for a.e. (x, t) ∈ Q, we have

δ ≤≠ l̃(x, t) := lim inf_{|s|→+∞} \frac{g̃(x, t, s)}{s} ≤ lim sup_{|s|→+∞} \frac{g̃(x, t, s)}{s} := k̃(x, t) ≤≠ μ

i.e. for all ε > 0 there exists a_{ε} ∈ L^{2}(Q) such that for a.e. (x, t) ∈ Q, and all s ∈ ℝ, we have

(l̃(x, t) − ε)s^{2} − a_{ε}(x, t)|s| ≤ sg̃(x, t, s) ≤ (k̃(x, t) + ε)s^{2} + a_{ε}(x, t)|s|.

(7) Under hypothesis (C3) and (2.1), there exists θ > 0 and η ∈ L^{2}(Q) such that a.e. (x, t) ∈ Q, and all s ∈ ℝ, we have

|g̃(x, t, s)| ≤ θ|s| + η(x, t).  (2.3)

Proposition 2.2. The problem (1.1) is equivalent to the problem

\begin{matrix}T v = g̃(x, t, v) + h̃(x, t) in Q, \\ v(x, t + 2π) = v(x, t) in ]0, π[×ℝ, \\ v(0, t) = v(π, t) = 0 ∀t ∈ ℝ, \end{matrix}  (2.4)

Proof. Assume that u is a solution of the problem (1.1). Let v = e^{\frac{β}{2} x}u, it is clear that v is 2π-periodic in t and v(0, t) = v(π, t) = 0 ∀t ∈ ℝ. On the other hand, we have

v_{x} = \frac{∂v}{∂x} = \frac{β}{2} e^{\frac{β}{2} x}u + e^{\frac{β}{2} x}u_{x},

v_{xx} = \frac{∂^{2}v}{∂x^{2}} = βe^{\frac{β}{2} x}u_{x} + \frac{β^{2}}{4} e^{\frac{β}{2} x}u + e^{\frac{β}{2} x}u_{xx},

v_{t} = \frac{∂v}{∂t} = e^{\frac{β}{2} x}u_{t}, v_{tt} = \frac{∂^{2}v}{∂t^{2}} = e^{\frac{β}{2} x}u_{tt};

thus

□v = v_{tt} − v_{xx}

= e^{\frac{β}{2} x}u_{tt} − βe^{\frac{β}{2} x}u_{x} − \frac{β^{2}}{4} e^{\frac{β}{2} x}u − e^{\frac{β}{2} x}u_{xx}

= −\frac{β^{2}}{4} v + e^{\frac{β}{2} x}(□u − βu_{x})

= (α − \frac{β^{2}}{4} )v − γv_{t} + e^{\frac{β}{2} x}(g(x, t, e^{− \frac{β}{2} x}v) + h(x, t)).

Hence T v = g̃(x, t, v) + h̃(x, t), and v is a solution of the problem (2.4). The reciprocal implication is demonstrated by an identical calculation. □

3. Main results

Theorem 3.1. Assume (C1), (C2), (C3) and (2.1). If (H_{r}) does not satisfy (2.2), then there exists m(x, t) ∈ L^{∞}(Q), v ∈ L^{2}(Q) \ {0} and (u_{n}) ⊂ L^{2}(Q) such that v is the nontrivial solution of the problem

\begin{matrix}T u = m(x, t)u in Q, \\ u(x, t + 2π) = u(x, t) in ]0, π[×ℝ, \\ u(0, t) = u(π, t) = 0 ∀t ∈ ℝ \end{matrix}  (3.1)

and

∥u_{n}∥ → +∞, \frac{u_{n}}{∥u_{n}∥} → v in L^{2}(Q),

δ ≤≠ m(x, t) ≤≠ μ a.e. in Q.

Corollary 3.2. Assume (C1), (C2), (C3) and (2.1). If there exist two consecutive eigenvalues of the D’Alembertian λ_{k} and λ_{k+1} such that 0 ≤ λ_{k} − \frac{γ^{2}}{4} < δ < μ < λ_{k+1} − \frac{γ^{2}}{4} , then problem (1.1) has at least one solution for all h ∈ L^{2}(Q).

Proof of theorem 3.1. As the proof is relatively long, we organize it in several lemmas. Suppose that (H_{r}) does not satisfy the estimate (2.2), then ∀n ∈ ℕ, there exist r_{n} ∈ [0, 1], and u_{n} ∈ D(T) with ∥u_{n}∥ = n such that

T u_{n} − r_{n}(N(u_{n}) + h̃) − (1 − r_{n})λu_{n} = 0  (3.2)

Let

v_{n} = \frac{u_{n}}{∥u_{n}∥}, g_{n}(x, t) = \frac{g̃(x, t, u_{n})}{∥u_{n}∥} a.e. in Q.

The sequence (v_{n}) is bounded in L^{2}(Q), then for subsequence v_{n} → v weakly in L^{2}(Q).

Lemma 3.3. Assume (2.3) and (3.2). (1) For a subsequence g_{n} → f weakly in L^{2}(Q). (2) v_{n} → v strongly in L^{2}(Q), in particular, ∥v∥ = 1, thus v ≠ 0.

Proof. (1) Dividing (2.3) by ∥u_{n}∥, we have

|g_{n}(x, t)| ≤ θ|v_{n}| + \frac{η(x, t)}{n},

thus

∥g_{n}∥ ≤ θ∥v_{n}∥ + \frac{∥η∥}{n} ≤ θ + \frac{∥η∥}{n},

hence g_{n} is bounded in L^{2}(Q), one deduces that for a subsequence g_{n} → f weakly in L^{2}(Q).

(2) Dividing by ∥u_{n}∥ in (3.2), we have

T v_{n} = r_{n}g_{n} + (1 − r_{n})λv_{n} + r_{n} \frac{h̃}{n}.

Which implies

v_{n} = (T_{0}^{−1})[r_{n}g_{n} + (1 − r_{n})λv_{n} + r_{n} \frac{h̃}{n}].

Since g_{n} → f weakly in L^{2}(Q) and v_{n} → v weakly in L^{2}(Q), then

r_{n}g_{n} + (1 − r_{n})λv_{n} + r_{n} \frac{h̃}{n} → rf + (1 − r)λv weakly in L^{2}(Q),

where r = lim_{n} r_{n}. The operator T_{0}^{−1} is compact, thus

v_{n} = (T_{0}^{−1})[r_{n}g_{n} +(1−r_{n})λv_{n} +r_{n} \frac{h̃}{n}] → (T_{0}^{−1})[rf +(1−r)λv] strongly in L^{2}(Q).

Therefore, v_{n} → (T_{0}^{−1})[rf + (1 − r)λv] = v strongly in L^{2}(Q). □

Lemma 3.4. Assume (2.3) and (III). Then f(x, t) = 0 a.e. in A = {(x, t) ∈ Q : v(x, t) = 0 a.e. in Q}.

Proof. Let ψ be the function

ψ(x, t) = sign(f(x, t))χ_{A}(x, t) a.e. in Q,

where χ_{A} is the indicatrice function. Since g_{n} → f weakly in L^{2}(Q), we have ∫_{Q} g_{n}ψ → ∫_{Q} fψ = ∫_{A} |f(x, t)|. On the other hand, as v_{n} → v, and using (2.3), we have

| ∫_{Q} g_{n}ψ| ≤ ∫_{Q} |g_{n}ψ| ≤ θ ∫_{Q} |v_{n}χ_{A}| + ∫_{Q} \frac{η(x, t)χ_{A}}{n} → θ ∫_{Q} |v|χ_{A} = 0,

thus ∫_{A} |f(x, t)| = 0 and f = 0 a.e. in A. □

We define the function

d(x, t) = \begin{cases}\frac{f(x,t)}{v(x,t)} & a.e. in Q \ A, \\ λ & a.e. in A. \end{cases}

Lemma 3.5. If one supposes (2.3) ,(3.2) and (2.1), then δ ≤ d(x, t) ≤ μ a.e. in Q.

Proof. We prove that δ ≤ d(x, t) a.e. in Q. We denote B = {(x, t) ∈ Q : δ(v(x, t))^{2} > v(x, t)f(x, t) a.e.}. It is sufficient to prove that meas B = 0. Under condition (2.1) (cf. Remark 2.1. No. 5), we have

(δ − ε)u_{n}^{2} − a_{ε}(x, t)|u_{n}| ≤ u_{n}g̃(x, t, u_{n}).

Dividing by ∥u_{n}∥^{2}, we get

(δ − ε)v_{n}^{2} − a_{ε}(x, t)\frac{|v_{n}|}{n} ≤ v_{n}g_{n}(x, t).

Multiplying by χ_{B} and integrating, we get

(δ − ε) ∫_{Q} v_{n}^{2}χ_{B} − ∫_{Q} \frac{a_{ε}(x, t)}{n}|v_{n}|χ_{B} ≤ ∫_{Q} v_{n}χ_{B}g_{n}(x, t).  (3.3)

Under conditions (2.3) and (3.2), g_{n} → f weakly in L^{2}(Q) and v_{n} → v strongly in L^{2}(Q). Going to the limit in (IV), we have

(δ − ε) ∫_{Q} |v(x, t)|^{2}χ_{B} ≤ ∫_{Q} v(x, t)f(x, t)χ_{B}.

Since ε is arbitrary, one concludes that

∫_{Q}[v(x, t)f(x, t) − δ|v(x, t)|^{2}]χ_{B} ≥ 0.

Therefore, by the definition of B, meas B = 0. By an analogous method, we prove that d(x, t) ≤ μ a.e. in Q. □

Lemma 3.6. If one supposes (2.3), (III) and (2.1), then

T v = m(x, t)v in Q,
v(x, t + 2π) = v(x, t) in ]0, π[×ℝ,
v(0, t) = v(π, t) = 0 ∀t ∈ ℝ,

where m(x, t) = rd(x, t) + (1 − r)λ and r = lim_{n} r_{n}.

Remark 3.7. It is easy to see that m(x, t) is 2π-periodic in t, and δ ≤ m(x, t) ≤ μ a.e. in Q.

Proof. In the proof of the Lemma 3.3, we have rf + (1 − r)λv = T v. From the definition of the function m, we have T v = mv. □

It remains to prove only the following lemma.

Lemma 3.8. If one supposes (2.3), (3.2) and (2.1), then

δ ≤≠ m(x, t) ≤≠ μ a.e. in Q.

Proof. We prove that m(x, t) ≤≠ μ a.e. in Q. (By analogous method, we prove that δ ≤≠ m(x, t) a.e. in Q). Suppose by contradiction that m(x, t) = μ a.e. in Q. Under assumption (2.1), we have

v_{n}g_{n} ≤ (k̃(x, t) + ε)v_{n}^{2} + \frac{a_{ε}|v_{n}|}{n},  (3.4)

where k̃(x, t) ∈ L^{∞}(Q) such that k̃(x, t) ≤≠ μ. By (V), we have

\begin{matrix}∫_{Q} r_{n}g_{n}v_{n} + (1 − r_{n})λv_{n}^{2} + r_{n} ∫_{Q} \frac{h̃v_{n}}{n} \\ ≤ ∫_{Q}[r_{n}(k̃(x, t) + ε) + (1 − r_{n})λ]v_{n}^{2} + r_{n} ∫_{Q}(h̃\frac{v_{n}}{n} + a_{ε} \frac{|v_{n}|}{n}). \end{matrix}  (3.5)

Under conditions (2.3) and (3.2), g_{n} → f weakly in L^{2}(Q) and v_{n} → v strongly in L^{2}(Q). Going to the limit in (V), we get

∫_{Q}[r(fv) + (1 − r)λv^{2}] ≤ ∫_{Q}[r(k̃(x, t) + ε) + (1 − r)λ]v^{2}.

By the definition of m, we have

∫_{Q}[r(fv) + (1 − r)λv^{2}] = ∫_{Q} m(x, t)v^{2} = ∫_{Q} μv^{2}.

Thus

∫_{Q} μv^{2} ≤ ∫_{Q}[r(k̃(x, t) + ε) + (1 − r)λ]v^{2}.

Since ε is arbitrary, we have

∫_{Q} μv^{2} ≤ ∫_{Q}[rk̃(x, t) + (1 − r)λ]v^{2}.

Hence

∫_{Q}[μ − rk̃(x, t) − (1 − r)λ]v^{2} ≤ 0.

Since k̃(x, t) ≤ μ a.e. in Q and λ < μ, we have μ − rk̃(x, t) − (1 − r)λ ≥ 0, then

∫_{Q}[μ − rk̃(x, t) − (1 − r)λ]v^{2} = 0.

Therefore, [μ − rk̃(x, t) − (1 − r)λ]v^{2} = 0 a.e. in Q. Since m(x, t) = μ a.e. in Q, by the definition of the function of d, (d(x, t) ≠ λ), we have meas A = 0 (i.e. v(x, t) ≠ 0 a.e. in Q). Thus, μ = rk̃(x, t) + (1 − r)λ a.e. in Q, this contradiction completes the proof. □

For the proof of Corollary 3.2 we will need the following two lemmas.

Lemma 3.9 ([6]). Assume (C1), (C2), (2.3), λ ∈ σ(T) and λ ≥ 0. Let h̃ ∈ L^{2}(Q), if there exist R > 0 such that

T u − r(N(u) + h̃) − (1 − r)λu ≠ 0, ∀u ∈ D(T), ∥u∥ = R, 0 ≤ r ≤ 1,

then problem (2.4) admits at least one solution u ∈ D(T) with ∥u∥ < R.

Proof. By (2.3), (C1) and (C2), N is continuous and monotone; therefore the result ensues while using by the homotopy studied in [6]. □

Lemma 3.10. If there exists two reals δ and μ such that

δ ≤ m(x, t) ≤ μ a.e. in Q with [δ, μ] ∩ σ(T) = ∅,  (3.6)

then the problem (3.1) has only the trivial solution.

Proof. Let c ∈ [δ, μ] be arbitrary with

\frac{max(|μ − c|, |δ − c|)}{dist(c, σ(T))} < 1

(for example, c = (δ + μ)/2). Then the operator T − cI is invertible and

∥(T − cI)^{−1}∥ = \frac{1}{dist(c, σ(T))}.

Hence for all u ∈ D(T),

∥T u − cu∥ ≥ dis(c, σ(T))∥u∥.

Assume now that u ∈ D(T) is a solution of the problem (3.1). Then ∥T u − cu∥ = ∥mu − cu∥. Therefore,

∥u∥ ≤ \frac{∥mu − cu∥}{dist(c, σ(T))}.

On the other hand, by (3.6),

|mu − cu| = |m − c||u| ≤ max(|μ − c|, |c − δ|)|u|

which implies ∥mu − cu∥ ≤ max(|μ − c|, |c − δ|)∥u∥. Thus

∥u∥ ≤ \frac{max(|μ − c|, |c − δ|)}{dist(c, σ(T))}∥u∥.

Since max(|μ − c|, |δ − c|)/ dist(c, σ(T)) < 1, it follows that u = 0. □

Proof of corollary 3.2. Suppose by contradiction that the problem (1.1) does not admit a solution. Thus by proposition 2.2, (2.4) does not admit a solution. Hence by lemma 3.9, the homotopy (H_{r}) does not satisfy the estimate (2.2). And by Theorem 3.1, there exists m(x, t) ∈ L^{∞}(Q), v ∈ L^{2}(Q) \ {0} such that v is the nontrivial solution of the problem (3.1) and δ ≤≠ m(x, t) ≤≠ μ a.e. in Q. Since 0 ≤ λ_{k} − \frac{γ^{2}}{4} < δ < μ < λ_{k+1} − \frac{γ^{2}}{4} , where λ_{k} − \frac{γ^{2}}{4} and λ_{k+1} − \frac{γ^{2}}{4} are two positive consecutive eigenvalues of T (cf. Remark 2.1 No. 1.ii), what is in contradiction with Lemma 3.10. Thus the proof is complete. □

Remark 3.11. (1) We have an analogous result, if in Corollary 3.2 λ_{k} and λ_{k+1} are two consecutive eigenvalues of the D’Alembertian such that λ_{k} − \frac{γ^{2}}{4} < δ < μ < λ_{k+1} − \frac{γ^{2}}{4} ≤ 0, while replacing the operator T by (−T).

(2) Note that if μ = 0, δ = 0 and γ = 0, we recover a result on the existence of the periodic solutions with conditions of non resonance of the problem

□u = g(x, t, u) + h(x, t) in Q,
u(x, t + 2π) = u(x, t) in ]0, π[×ℝ,
u(0, t) = u(π, t) = 0 ∀t ∈ ℝ

(3) Note that if μ = 0, and δ = 0, we recover a result on the existence of the periodic solutions with conditions of non resonance of the telegraph equation

□u = γu_{t} + g(x, t, u) + h(x, t) in Q,
u(x, t + 2π) = u(x, t) in ]0, π[×ℝ,
u(0, t) = u(π, t) = 0 ∀t ∈ ℝ

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[12] H. Brezis and L. Nirenberg, Characterization of the ranges of some nonlinear operators and applications to boundary value problems, Ann. Scuola Norm. Sup. Pisa (4) 5, pp. 225-326, 1978.

Aomar Anane Département de Mathématiques et Informatique, Faculté des Sciences, Université Mohammed 1er, Oujda, Maroc

E-mail address: anane@sciences.univ-oujda.ac.ma

Omar Chakrone Département de Mathématiques et Informatique, Faculté des Sciences, Université Mohammed 1er, Oujda, Maroc

E-mail address: chakrone@sciences.univ-oujda.ac.ma

Abdellah Zerouali Département de Mathématiques et Informatique, Faculté des Sciences, Université Mohammed 1er, Oujda, Maroc

E-mail address: abdellahzerouali@hotmail.com