2005-Oujda International Conference on Nonlinear Analysis. Electronic Journal of Differential Equations, Conference 14, 2006, pp. 83–94. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu (login: ftp)

ON A PROBLEM OF LOWER LIMIT IN THE STUDY OF NONRESONANCE WITH LERAY-LIONS OPERATOR

AOMAR ANANE, OMAR CHAKRONE, MOHAMMED CHEHABI

Abstract. We prove the solvability of the Dirichlet problem

Au = f(u) + h in Ω,

u = 0 on ∂Ω

for a given h, under a condition involving only the asymptotic behaviour of the potential F of f, where A is a Leray-Lions operator.

1. Introduction and statement of results

This paper concerns the existence of solutions to the problem

\begin{matrix}Au = f(u) + h in Ω, \\ u = 0 on ∂Ω \end{matrix}  (1.1)

where Ω is a bounded domain of ℝ^{N}, N ≥ 1, A is an operator of the form A(u) = − ∑_{i=1}^{N} \frac{∂}{∂x_{i}} A_{i}(∇u), f is a continuous function from ℝ to ℝ and h is a given function on Ω. Also we consider the problem

\begin{matrix}−Δ_{p}u = f(u) + h in Ω \\ u = 0 on ∂Ω \end{matrix}  (1.2)

where Δ_{p} denotes the p-Laplacian Δ_{p}u = div(|∇u|^{p−2}∇u), 1 < p < ∞.

A classical result, essentially due to Hammerstein [9] asserts that if f satisfies a suitable polynomial growth restriction connected with the Sobolev imbeddings and if

lim sup_{x→±∞} \frac{2F (s)}{|s|^{2}} < λ_{1},  (1.3)

then problem (1.2) with p = 2 is solvable for any h. Here F denotes the primitive F (s) = ∫_{0}^{s} f(t)dt and λ_{1} is the first eigenvalue of −Δ on H_{0}^{1}(Ω). Several improvements of this result have been considered in the recent years.

2000 Mathematics Subject Classification. 35J60, 35P30.

Key words and phrases. Leray-Lions operator; Nonresonance; fist eigenvalue.

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

Published September 20, 2006.

In 1989, the case N = 1 and p = 2 was considered in [7]. It was shown there that (1.2) with p = 2 is solvable for any h ∈ L^{∞}(Ω) if

lim inf_{x→±∞} \frac{2F (s)}{|s|^{2}} < λ_{1}.  (1.4)

When N ≥ 1 and p = 2, showed later in [8] that (1.2) is solvable for any h ∈ L^{∞}(Ω) if

lim inf_{s→±∞} \frac{2F (s)}{|s|^{2}} < (\frac{π}{2R(Ω)})^{2},  (1.5)

where R(Ω) denotes the radius of the smallest open ball B(Ω) containing Ω. This result was extended to the p-Laplacian case in [5] where solvability of (1.2) was derived under the condition

lim inf_{s→±∞} \frac{pF (s)}{|s|^{p}} < (p − 1) \{\frac{1}{R(Ω)} ∫_{0}^{1} \frac{dt}{(1 − t^{p})^{1/p}} \}^{p} .  (1.6)

Note that this condition reduces to (1.5) when p = 2.

The question now naturally arises whether (p − 1)\{\frac{1}{R(Ω)} ∫_{0}^{1} \frac{dt}{(1−t^{p})^{1/p}} \}^{p} can be

replaced by λ_{1} in (1.6), where λ_{1} denotes the first eigenvalue of −Δ_{p} on W_{0}^{1,p}(Ω) (cf[1]).

Observe that for N > 1 and p = 2, (\frac{π}{2R(Ω)})^{2} < λ_{1}, and a similar strict inequality holds when 1 < p < ∞. In [2], it was showed that the constants in (1.5) and (1.6) can be improved a little bit.

Denote by l(Ω) the length of the smallest edge of an arbitrary parallelepiped containing Ω. If

lim inf_{s→±∞} \frac{pF (s)}{|s|^{p}} < C_{p}(l)  (1.7)

where C_{p}(l) = (p − 1)\{\frac{2}{l(Ω)} ∫_{0}^{1} \frac{dt}{(1−t^{p})^{1/p}} \}^{p} then for any h ∈ L^{∞}(Ω) the problem

(1.2) has a solution u ∈ W_{0}^{1,p}(Ω) ∩ C^{1}(Ω).

Observe that for N = 1, C_{p} = λ_{1} the first eigenvalue of −Δ on Ω =]0, l(Ω)[. In particular, C_{2} = (\frac{π}{l} )^{2}, and it recovers the result of [7]. It is clear that (1.7) is a

weaker hypothesis than (1.6). The difference between (1.7) and (1.6) is particularly important when Ω is a rectangle or a triangle. However C_{p}(l) < λ_{1} when N > 1, and the question raised above remains open.

In this paper we investigate the question of replacing Δ_{p} by the operator of the form

A(u) = − ∑_{i=1}^{N} \frac{∂}{∂x_{i}} A_{i}(∇u).

We assume the following hypotheses:

(A0) For all i ∈ {1, 2, …, N}, A_{i} : ℝ^{ℕ} → ℝ is continuous.

(A1) there exists (c, k) ∈ (]0, +∞[)^{2} such that |A_{i}(ξ)| ≤ c|ξ|^{p−1} + K for all

ξ ∈ ℝ^{ℕ}, and all i ∈ {1, 2, …, N}. (A2) (a) ∑_{i=1}^{N}(A_{i}(ξ) − A_{i}(ξ′))(ξ_{i} − ξ′_{i}) > 0 for all ξ ≠ ξ′ ∈ ℝ^{N};

(b) for all i ∈ {1, 2, …, N}, the function defined by r_{i}(s) = A_{i}(0, …, 0, s, 0, …, 0) for s ∈ ℝ is odd;

(c) for each i ∈ {1, 2, …, N}, there exists a_{i} ∈]0, +∞[ such that lim_{s→+∞} r_{i}(s)/s^{p−1} = a_{i};

(d) for each i ∈ {1, 2, …, N}, r_{i} ∈ C^{1}(ℝ^{∗}) and lim_{s→0} sr′_{i}(s) = 0;

(e) for all i ∈ {1, 2, …, N}, A_{i}(ξ) = 0 for all ξ ∈ ℝ^{N} such that ξ_{i} = 0.

Remark 1.1. (1) The hypothesis (A2)(d) is in particular satisfied if we suppose that for i ∈ {1, …, N}, r_{i} ∈ C^{1}(ℝ^{∗}) and there exists q_{i}, 1 < q_{i} < ∞, there exists η_{i} > 0, there exists (a, b) ∈ ℝ^{2}, such that for all |s| < η_{i}, |r′_{i}(s)| ≤ a|s|^{q_{i}−2} + b.

(2) The assumption (A2)(d) is an hypothesis of homogenization at infinity for the operator A.

Definition 1.2. For i ∈ {1, 2, …, N}, we define

l_{i}(s) = \frac{1}{p − 1}[sr_{i}(s) − ∫_{0}^{s} r_{i}(t)dt] ∀s ∈ ℝ.

Proposition 1.3. Assume (A0), (A1) and (A2). Then: (1) The operator A : W_{0}^{1,p}(Ω) → W^{−1,p′}(Ω) is defined, strictly monotone and

⟨Au, v⟩ = ∑_{i=1}^{N} ∫_{Ω} A_{i}(∇u)\frac{∂v}{∂x_{i}} dx ∀u, v ∈ W_{0}^{1,p}(Ω).

(2) For each i ∈ {1, 2, …, N}, the function r_{i} : ℝ → ℝ is continuous, strictly increasing and r_{i}(0) = 0.

(3) For each i ∈ {1, 2, …, N}, the function l_{i} satisfies (i) l_{i} is even, continuous and l_{i}(0) = 0; (ii) lim_{s→+∞} \frac{l_{i}(s)}{s^{p}} = \frac{a_{i}}{p}

(iii) l_{i} ∈ C^{1}(ℝ) and l′_{i}(s) = \begin{cases}\frac{sr′_{i}(s)}{p−1} & if s ≠ 0 \\ 0 & if s = 0. \end{cases}

(iv) l_{i} is strictly increasing in ℝ^{+}.

Proof. (1) By (A0), (A1), it is clear that the operator A is defined from W_{0}^{1,p}(Ω) to W^{−1,p′}(Ω), we have

⟨Au, v⟩ = ∑_{i=1}^{N} ∫_{Ω} A_{i}(∇u)\frac{∂v}{∂x_{i}} dx ∀u, v ∈ W_{0}^{1,p}(Ω)

and by (A1)(a), we verify easily that A is strictly monotone.

(2) Let i ∈ {1, …, N}. By (A0) and (A2)-(b), r_{i} is continuous and r_{i}(0) = 0 , in the end r_{i} is strictly increasing. Indeed, let (s, s′) ∈ ℝ^{2} such that s ≠ s′, we have

(r_{i}(s) − r_{i}(s′))(s − s′) = ∑_{i=1}^{N}(A_{i}(ξ) − A_{i}(ξ′))(ξ_{i} − ξ′_{i}) > 0

where ξ = (0, …, s, … 0) and ξ′ = (0, …, s′, … 0)

(3)(i) By the foregoing, the function l_{i} is even, continuous and l_{i}(0) = 0 for every i ∈ {1, …, N}

(3)(ii) We show first that

lim_{s→+∞} \frac{1}{s^{p}} ∫_{0}^{s} r_{i}(t)dt = \frac{a_{i}}{p} .  (1.8)

Let ε > 0, by (A2)(c), there exists η_{ε} = η such that |r_{i}(s) − a_{i}s^{p−1}| ≤ εs^{p−1} for all s ≥ η.

Integrating from η to s, we obtain

| ∫_{0}^{s} r_{i}(t)dt − ∫_{0}^{η} r_{i}(t)dt − \frac{a_{i}}{p} [s^{p} − η^{p}]| ≤ \frac{ε}{p}[s^{p} − η^{p}].

Dividing by s^{p} and letting n → +∞, we obtain

lim_{s→+∞} |\frac{1}{s^{p}} ∫_{0}^{s} r_{i}(t)dt − \frac{a_{i}}{p} | = 0

i.e (1.8) holds. Writing

\frac{l_{i}(s)}{s^{p}} = \frac{1}{p − 1}\{\frac{r_{i}(s)}{s^{p−1}} − \frac{1}{s^{p}} ∫_{0}^{s} r_{i}(t)dt\}.

By (1.8) and (A2)(c), we have lim_{s→+∞} \frac{l_{i}(s)}{s^{p}} = \frac{a_{i}}{p}

(3)(iii) Since r_{i} ∈ C^{1}(ℝ^{∗}), we have l_{i} ∈ C^{1}(ℝ^{∗}) and l′_{i}(s) = \frac{1}{p−1}sr′_{i}(s) for every s ≠ 0. On the other hand, for s ≠ 0, since r_{i} is increasing and odd, we have

|\frac{l_{i}(s)}{s}| = \frac{1}{p − 1}|r_{i}(s) − \frac{1}{s}| ∫_{0}^{s} r_{i}(t)dt ≤ \frac{2}{p − 1}r_{i}(|s|).

It results that l′_{i}(0) exists and l′_{i}(0) = 0. By (A2)-(d) we obtain lim_{s→0} l′_{i}(s) = lim_{s→0} sr′_{i}(s). This proves that l_{i} ∈ C^{1}(ℝ).

(3)(iv) is a consequence of (3)(iii) □

Example 1.4. We give at first some examples for operators A satisfying the hypothesis (A0), (A1) and (A2). (1) Let

Au = −Δ_{p}u = − ∑_{i=1}^{N} \frac{∂}{∂x_{i}} (|∇u|^{p−2} \frac{∂u}{∂x_{i}})

Then we have A_{i}(ξ) = |ξ|^{p−2}ξ_{i} for every ξ = (ξ_{i}) ∈ ℝ^{N}. r(s) = r_{i}(s) = |s|^{p−2}s for every s ∈ ℝ and every i ∈ {1, …, N}. l(s) = l_{i}(s) = \frac{1}{p}|s|^{p} for every s ∈ ℝ and every i ∈ \{1, …, N\}.

(2) Let

Au = −Δ_{p}u − Δ_{q}u = − ∑_{i=1}^{N} \frac{∂}{∂x_{i}} (|∇u|^{p−2} \frac{∂u}{∂x_{i}} + |∇u|^{q−2} \frac{∂u}{∂x_{i}})

where 1 < q < p < +∞. Then we have A_{i}(ξ) = |ξ|^{p−2}ξ_{i} + |ξ|^{q−2}ξ_{i} for every ξ = (ξ_{i}) ∈ ℝ^{N}. r(s) = r_{i}(s) = |s|^{p−2}s + |s|^{q−2}s for every s ∈ ℝ and every i ∈ {1, …, N}. l(s) = l_{i}(s) = \frac{1}{p}|s|^{p} + \frac{q−1}{q(p−1)}|s|^{q} for every s ∈ ℝ and every i ∈ \{1, …, N\}.

(3) Let

Au = −Δ_{p,ε}u = − ∑_{i=1}^{N} \frac{∂}{∂x_{i}} [ε + |∇u|^{2})^{\frac{p−2}{2}} \frac{∂u}{∂x_{i}} ] ,

where ε > 0. Then we have A_{i}(ξ) = (ε + |ξ|^{2})^{\frac{p−2}{2}} ξ_{i} for every ξ = (ξ_{i}) ∈ ℝ^{N}. r(s) = r_{i}(s) = (ε + |s|^{2})^{\frac{p−2}{2}} s for every s ∈ ℝ and every i ∈ \{1, …, N\}. l(s) = l_{i}(s) = (ε + |s|^{2})^{\frac{p−2}{2}} (\frac{s^{2}}{p} − \frac{ε}{p(p−1)}) + \frac{1}{p(p−1)}ε^{\frac{p}{2}} for every s ∈ ℝ and every i ∈ {1, …, N}.

2. Proof of Main Theorem

We consider the Dirichlet problem (1.1) where Ω is a bounded domain of ℝ^{N}, N ≥ 1, f is a continuous function from ℝ to ℝ and h ∈ L^{∞}(Ω).

Denote by [AB] the smallest edge of an arbitrary parallelepiped containing Ω. Making an orthogonal change of variables, we can always suppose that AB is parallel to one of the axis of ℝ^{N}. So Ω ⊂ P = ∏_{j=1}^{N}[a_{j}, bj] with, for some i, |AB| = b_{i} − a_{i} = min_{1≤j≤N}{b_{j} − a_{j}}, a quantity which we denote by b − a.

Denote by l = l_{i}, r = r_{i}, F the primitive F (s) = ∫_{0}^{s} f(t)dt and

C_{p} = (p − 1)\{\frac{2}{b − a} ∫_{0}^{1} \frac{dt}{(1 − t^{p})^{\frac{1}{p}}} \}^{p}.

Theorem 2.1. Assume

lim inf_{s→±∞} \frac{F (s)}{l(s)} < C_{p}.  (2.1)

Then for any h ∈ L^{∞}(Ω), the problem (1.1) has a solution u ∈ W_{0}^{1,p}(Ω) ∩ L^{∞}(Ω) in the weak sense; i.e

∑_{i=1}^{N} ∫_{Ω} A_{i}(∇u)\frac{∂φ}{∂x_{i}} dx = ∫_{Ω} f(u)φ + ∫_{Ω} hφ ∀φ ∈ W_{0}^{1,p}(Ω).

Definition 2.2. An upper solution for (1.1) is defined as a function β : \overline{Ω} → ℝ such that

• β ∈ C^{1}(\overline{Ω})

• A(β) ∈ C(\overline{Ω})
• A(β)(x) ≥ f(β(x)) + h(x) a e x in Ω.

A lower solution α is defined by reversing the inequalities above.

Lemma 2.3. Assume that (1.1) admits an upper solution β and a lower solution α with α(x) ≤ β(x) in Ω. Then (1.1) admits a solution u ∈ W_{0}^{1,p}(Ω) ∩ C^{1}(Ω), with α(x) ≤ u(x) ≤ β(x) in Ω.

Proof. Let

f̃(x, s) = \begin{cases}f(β(x)) & if s ≥ β(x), \\ f(s) & if α(x) ≤ s ≤ β(x), \\ f(α(x)) & if s ≤ α(x) \end{cases}

for every (x, s) ∈ \overline{Ω} × ℝ such that f̃ is bounded and continuous in \overline{Ω} × ℝ, then the problem

\begin{matrix}Au = f̃(x, u) + h(x) in Ω \\ u = 0 on ∂Ω \end{matrix}  (2.2)

admits a solution u ∈ W_{0}^{1,p}(Ω) in the weak sense, indeed the operator A is strictly monotone, so we can use the result of Lions [10] concerned the pseudomonotones operators.

We claim that α(x) ≤ u(x) ≤ β(x) in Ω, which clearly implies the conclusion. To prove the first inequality, one multiplies (2.2) by w = u − u_{α}, where u_{α}(x) =

max(u(x), α(x)), integrates by parts and uses the fact that α is a lower solution we obtain ⟨A(u) − A(u − w), w⟩ ≤ 0, which implies w = 0 (since A is strictly monotone). □

Lemma 2.4. Let a < b and M > 0, and assume

lim inf_{s→+∞} \frac{F (s)}{l(s)} < C_{p}.  (2.3)

then there exists β_{1} ∈ C^{1}(I) such that (r(β′_{1}(t)))′ ∈ C(I) and

−(r(β′_{1}(t)))′ ≥ f(β_{1}(t)) + M ∀t ∈ I,
β_{1}(t) ≥ 0 ∀t ∈ I

where I = [a, b].

Lemma 2.5. Assume

lim inf_{s→−∞} \frac{F (s)}{l(s)} < C_{p}.  (2.4)

then there exists α_{1} ∈ C^{1}(I) such that (r(α′_{1}(t)))′ ∈ C(I) and

−(r(α′_{1}(t)))′ ≤ f(α_{1}(t)) − M ∀t ∈ I
α_{1}(t) ≤ 0 ∀t ∈ I

where I = [a, b].

Accepting for a moment the conclusion of these two lemmas, let us turn to the Proof of Theorem 2.1. By Lemma 2.3 it suffices to show the existence of an upper solution and a lower solution for (1.1). Let us describe the construction of the upper solution (that of the lower solution is similar).

Let M > ∥h∥_{∞} and i ∈ {1, 2, …, N} such that b = b_{i}, a = a_{i}. By Lemma 2.4 there exists β_{1} : I → ℝ such that β_{1} ∈ C^{1}(I), (r(β′_{1}(t)))′ ∈ C(I) and

−(r(β′_{1}(t)))′ ≥ f(β_{1}(t)) + M ∀t ∈ I,
β_{1}(t) ≥ 0 ∀t ∈ I.

Writing β(x) = β_{1}(x_{i}) for all x = (x_{i}) ∈ \overline{Ω}, it is clear that β ∈ C^{1}(\overline{Ω}), A(β(x)) = A(β_{1}(x_{i})) ∈ C(\overline{Ω}), and we have by (A2)(e):

A(β(x)) = − ∑_{j=1}^{n} \frac{∂}{∂x_{j}} A_{j}(∇β(x))

= −\frac{∂}{∂x_{i}} (r_{i}(β′_{1}(x_{i})))

= −(r(β′_{1}(x_{i})))′

≥ f(β_{1}(x_{i})) + M

= f(β(x)) + M

≥ f(β(x)) + h(x) a.e.x ∈ Ω

The proof of Theorem 2.1 is thus complete.

Next, we present the proof of Lemma 2.4. The proof of Lemma 2.5 follows similarly.

First case. Suppose inf_{s≥0} f(s) = −∞. Then there exists β ∈ ℝ^{∗}+ such that f(β) < −M, and the constant function β provides a solution to the problem in Lemma 2.4.

Second case. Suppose now inf_{s≥0} f(s) > −∞. Let K > M such that inf_{s≥0} f(s) > −K + 1. Thus f(s) + K ≥ 1 for all s ≥ 0. Define g : ℝ → ℝ by

g(s) = \begin{cases}f(s) + K & if s ≥ 0 \\ f(0) + K & if s < 0 \end{cases}

and denote G(s) = ∫_{0}^{s} g(t)dt for all s in ℝ. It is easy to see that g(s) ≥ 1 for all s in ℝ and that

0 ≤ lim inf_{s→+∞} \frac{G(s)}{l(s)} = lim inf_{s→+∞} \frac{F (s)}{l(s)} < C_{p}.

Now it is clearly sufficient to prove the existence of a function β_{1} : I → ℝ such that β_{1} ∈ C^{1}(I), (r(β′_{1}(t)))′ ∈ C(I) and

−(r(β′_{1}(t)))′ = g(β_{1}(t)) ∀t ∈ I
β_{1}(t) ≥ 0 ∀t ∈ I

For this purpose we will need the following four Lemmas.

Lemma 2.6. Let 0 < c < ∞ and t ∈]0, 1[, then

lim_{α→+∞} \frac{α}{l^{−1}(c(l(α) − l(αt)))} = \frac{1}{c^{1/p}(1 − t^{p})^{\frac{1}{p}}}

In particular by, Fatou Lemma,

\frac{1}{c^{1/p}} ∫_{0}^{1} \frac{dt}{(1 − t^{p})^{1/p}} ≤ lim inf_{α→+∞} ∫_{0}^{1} \frac{αdt}{l^{−1}(c(l(α) − l(αt)))}

Proof. Denote s(α) = \frac{α}{l^{−1}(c(l(α)−l(αt)))} and \frac{a_{i}}{p} = d. By Proposition 1.3 (3)(ii), we have

lim_{s→+∞} \frac{s^{1/p}}{l^{−1}(s)} = d^{1/p}.

On the other hand,

lim_{α→+∞}[c(l(α) − l(αt))] = +∞,

and more generally,

lim_{α→+∞} \frac{l(α) − l(αt)}{α^{p}} = d(1 − t^{p}) > 0 .

Writing

s(α) = \frac{1}{[\frac{c(l(α)−l(αt))}{α^{p}}]^{1/p}} \frac{[c(l(α) − l(αt))]^{1/p}}{l^{−1}(c(l(α) − l(αt)))}

Letting n → +∞ and by the three limits above, we have

lim_{α→+∞} s(α) = \frac{1}{c^{1/p}(1 − t^{p})^{1/p}}

□

Lemma 2.7. For d > 0, define

τ_{G}(d) = ∫_{0}^{d} \frac{ds}{l^{−1}[\frac{G(d)−G(s)}{p−1}]} .

Then

lim sup_{d→+∞} τ_{G}(d) ≥ ( ∫_{0}^{1} \frac{dt}{(1 − t^{p})^{1/p}} )(\frac{1}{p − 1} lim inf_{s→+∞} \frac{G(s)}{l(s)})^{1/p}.

In particular (2.3) implies lim sup_{d→+∞} τ_{G}(d) > (b − a)/2.

Proof. Let ρ be a positive number such that lim inf_{s→+∞} \frac{G(s)}{l(s)} < ρ < C_{l}. Then lim sup_{s→+∞}[ρl(s) − G(s)] = +∞. Let w_{n} be the smallest number in [0, n] such that

max_{0≤s≤n} K(s) = K(w_{n}) where K(s) = ρl(s) − G(s); it is easily seen that (w_{n}) is increasing with respect to n. Since ρl(s)−G(s) < ρl(w_{n})−G(w_{n}) for all s ∈ [0, w_{n}[, we have \frac{G(w_{n})−G(s)}{p−1} < \frac{ρ}{p−1}(l(w_{n}) − l(s)) for all s ∈ [0, w_{n}[, since l : [0, +∞[→ [0, +∞[ is an increasing homeomorphism, we have

\frac{1}{l^{−1}[\frac{ρ}{p−1}(l(w_{n}) − l(s))]} < \frac{1}{l^{−1}[\frac{1}{p−1}(G(w_{n}) − G(s))]}.

Integrating from 0 to w_{n} and changing variable s = uw_{n} in the first member of inequality, we obtain

∫_{0}^{1} \frac{w_{n}}{l^{−1}[\frac{ρ}{p−1}(l(w_{n}) − l(w_{n}s))]}ds ≤ τ_{G}(w_{n}).

Letting n → +∞, we obtain

lim inf_{n→+∞} ∫_{0}^{1} \frac{w_{n}}{l^{−1}[\frac{ρ}{p−1}(l(w_{n}) − l(w_{n}s))]}ds ≤ lim sup_{n→+∞} τ_{G}(w_{n}).

By Lemma 2.6, it results

lim sup_{d→+∞} τ_{G}(d) ≥ [ ∫_{0}^{1} \frac{dt}{(1 − t^{p})^{1/p}} ][\frac{ρ}{p − 1}]^{\frac{−1}{p}} .

Letting ρ → lim inf_{s→+∞} \frac{G(s)}{l(s)}, the Lemma is proved. □

Lemma 2.8. Let d > 0 and consider the mapping T_{d} defined by

T_{d}(u) = d − ∫_{a}^{t} r^{−1}([ ∫_{a}^{τ} g(u(s))ds]^{1/(p−1)})dτ

in the Banach space C(I). Then T_{d} has a fixed point.

Proof. Clearly by Ascoli’s theorem T_{d} is compact. The proof of Lemma 2.8 uses an homotopy argument based on the Leray Schauder topological degree. So T_{d} will have a fixed point if the following condition holds:

There exists ρ > 0 such that (I−λT_{d})(u) ≠ 0 for all u ∈ ∂B(0, ρ) for all λ ∈ [0, 1], where ∂B(0, ρ) = {u ∈ C(I); ∥u∥_{∞} = ρ}.

To prove that this condition holds, suppose by contradiction that for all n = 1, 2, … there exists u_{n} ∈ ∂B(0, n), λ_{n} ∈ [0, 1] such that: u_{n} = λ_{n}T_{d}(u_{n}). The latter relation implies

u_{n} = λ_{n}d − λ_{n} ∫_{a}^{t} r^{−1}([ ∫_{a}^{τ} g(u(s))ds]^{\frac{1}{p−1}} )dτ  (2.5)

Therefore, u_{n} ∈ C^{1}(I) and we have successively

\begin{matrix}u′_{n}(t) = −λ_{n}r^{−1}([ ∫_{a}^{τ} g(u(s))ds]^{\frac{1}{p−1}} ) < 0 ∀t ∈]a, b], \\ u′_{n}(a) = 0, \end{matrix}  (2.6)

(r[\frac{u′_{n}(t)}{λ_{n}}])′ ∈ C(I) and

−(r(\frac{u′_{n}(t)}{λ_{n}}))′ = g(u_{n}(t)) ∀t ∈ I.  (2.7)

Note that by (2.6), u′_{n}(t) < 0 in ]a, b], so that u_{n} is decreasing. Hence, for n > d, u_{n}(b) = −n. Multiplying the equation (2.7) by u′_{n}(t), we obtain

−λ_{n}(l(\frac{u′_{n}(t)}{λ_{n}}))′ = \frac{1}{p − 1} \frac{d}{dt}G(u_{n}(t)).  (2.8)

Indeed

(l(\frac{u′_{n}(t)}{λ_{n}}))′ = [l(r^{−1}(r(\frac{u′_{n}(t)}{λ_{n}})))]′

= (l ∘ r^{−1})′(r(\frac{u′_{n}(t)}{λ_{n}}))(r(\frac{u′_{n}(t)}{λ_{n}}))′

= \frac{1}{p − 1} \frac{u′_{n}(t)}{λ_{n}}(r(\frac{u′_{n}(t)}{λ_{n}}))′

By (2.8), we have

λ_{n}(l(\frac{u′_{n}(t)}{λ_{n}})) = \frac{1}{p − 1}(G(λ_{n}d) − G(u_{n}(t))

and

−\frac{u′_{n}(t)}{λ_{n}l^{−1}[\frac{G(λ_{n}d)−G(u_{n}(t))}{(p−1)λ_{n}}]} = 1.

Integrating from a to b and changing variable s = u_{n}(t) (u_{n}(a) = λ_{n}d and u_{n}(b) = −n), we obtain

∫_{−n}^{λ_{n}d} \frac{ds}{λ_{n}l^{−1}[\frac{G(λ_{n}d)−G(s)}{(p−1)λ_{n}}]} = b − a

i.e.

∫_{0}^{λ_{n}d} \frac{ds}{λ_{n}l^{−1}[\frac{G(λ_{n}d)−G(s)}{(p−1)λ_{n}}]} = b − a + ∫_{0}^{−n} \frac{ds}{λ_{n}l^{−1}[\frac{G(λ_{n}d)−G(s)}{(p−1)λ_{n}}]} ≥ 0

Since G(s) = sg(0) for s ≤ 0 and changing variable s = −u, we obtain

0 ≤ (b − a) − ∫_{0}^{n} \frac{ds}{λ_{n}l^{−1}[\frac{G(λ_{n}d)−sg(0)}{(p−1)λ_{n}}]}  (2.9)

Denote by l(u) = \frac{G(λ_{n}d)−G(s)}{(p−1)λ_{n}} such that l′(u)du = \frac{g(0)}{(p−1)λ_{n}} ds and ds = \frac{λ_{n}}{g(0)}r′(u)udu

for u ≠ 0 and denote α_{n} = l^{−1}[\frac{G(λ_{n}d)}{(p−1)λ_{n}} ] and β_{n} = l^{−1}[\frac{(G(λ_{n}d)+ng(0))}{(p−1)λ_{n}}]. By (2.9), we obtain

0 ≤ (b − a) − ∫_{α_{n}}^{β_{n}} \frac{r′(u)}{g(0)}du

= (b − a) − \frac{1}{g(0)}r\{l^{−1}[\frac{G(λ_{n}d) − ng(0)}{(p − 1)λ_{n}}]\} + \frac{1}{g(0)}r\{l^{−1}[\frac{G(λ_{n}d)}{(p − 1)λ_{n}} ]\}.

Since

\frac{G(λ_{n}d) − ng(0)}{(p − 1)λ_{n}} ≥ \frac{ng(0)}{(p − 1)}, \frac{G(λ_{n}d)}{(p − 1)λ_{n}} ≤ \frac{d}{p − 1} max_{0≤s≤d} |g(s)|

and r ∘ l^{−1} is increasing, it results that

0 ≤ (b − a) − \frac{1}{g(0)}r\{l^{−1}[\frac{ng(0)}{(p − 1)λ_{n}} ]\} + \frac{1}{g(0)}r\{l^{−1}[\frac{d}{p − 1} max_{0≤s≤d}|g(s)|]\}.

Letting n → +∞, we get a contradiction. Let us denote by u_{d} ∈ C(I) a fixed point of the mapping T_{d} of Lemma 2.8 □

Lemma 2.9. There exists d > 0 such that u_{d}(t) ≥ 0 for all t ∈ [a, \frac{a+b}{2}[.

Proof. We know that u_{d} is decreasing and that u_{d}(a) = d for all d > 0. Let us distinguish two cases.

First if there exists d > 0 such that u_{d}(b) ≥ 0, then the conclusion of Lemma 2.9 clearly follows. So we can assume that u_{d}(b) < 0 for every d > 0. Since u_{d}(a) = d > 0, there exists δ_{d} ∈]a, b[ such that u_{d}(δ_{d}) = 0. It is clear that u_{d}(t) ≥ 0 for all t ∈ [a, δ_{d}[, and so it is sufficient to show that lim sup_{d→+∞} δ_{d} > \frac{a+b}{2}. Processing as in the proof of Lemma 2.8, we obtain

−u′_{d}(t)\{l^{−1}(\frac{G(d) − G(u_{d}(t))}{p − 1})\}^{−1} = 1.

Integrating from a to δ_{d} and changing variable s = u_{d}(t), one gets

τ_{G}(d) = ∫_{0}^{d} \frac{ds}{l^{−1}[\frac{G(d)−G(s)}{p−1}]} = δ_{d} − a,

consequently

lim sup_{d→+∞} δ_{d} > a + \frac{b − a}{2} = \frac{a + b}{2}

□

Proof of Lemma 2.4 continued. . Denoting u_{d}(t) by u(t), we have u ∈ C^{1}(I), (r(u′))′ ∈ C(I) and

−(r(u′))′ = g(u(s)) ∀t ∈ I,

u(t) ≥ 0 ∀t ∈ [a, \frac{a + b}{2}[,

u′(a) = 0.

Define a function β_{1} from [a, b] to ℝ by

β_{1}(t) = \begin{cases}u(\frac{3a+b}{2} − t) & if t ∈ [a, \frac{a+b}{2}], \\ u(t − \frac{b−a}{2}) & if t ∈ [\frac{a+b}{2}, b]. \end{cases}

We will show that this function β fulfills the conditions of Lemma 2.4. To see this it is sufficient to show that

(a) β_{1} is nonnegative in [a, b],

(b) β_{1} ∈ C^{1}([a, b]),

(c) (r(β′_{1}(t)))′ ∈ C([a, b]) and −(r(β′_{1}(t)))′ = g(β_{1}(t)) for all t ∈ [a, b]. Proof of (a). If a ≤ t ≤ \frac{a+b}{2}, then a ≤ \frac{3a+b}{2} − t ≤ \frac{a+b}{2}, and if \frac{a+b}{2} ≤ t ≤ b, then a ≤ t − \frac{b−a}{2} ≤ \frac{a+b}{2}, so that the conclusion follows from the sign of u on [a, \frac{a+b}{2}].

Proof of (b). β_{1} ∈ C^{1}([a, \frac{a+b}{2}[), β_{1} ∈ C^{1}(]\frac{a+b}{2}, b]), and moreover \frac{d}{dt^{+}} β_{1}(\frac{a+b}{2}) = u′(a) = 0 and \frac{d}{dt^{−}} β_{1}(\frac{a+b}{2}) = u′(a) = 0.

Proof of (c). We know that −(r(u′(t)))′ = g(u(t)) for t ∈ [a, b], therefore

−(r(u′(t)) = ∫_{a}^{t} g(u(s))ds.

If a ≤ t ≤ \frac{a+b}{2} then a ≤ \frac{3a+b}{2} − t ≤ \frac{a+b}{2}, which gives

β_{1}(t) = u(\frac{3a + b}{2} − t) and β′_{1}(t) = −u′(\frac{3a + b}{2} − t).

We obtain

−(r(u′(\frac{a + b}{2} − t)) = r(β′_{1}(t)).

The change of variable u = \frac{3a+b}{2} − s yields

∫_{a}^{\frac{3a+b}{2} −t} g(u(s))ds = ∫t^{\frac{a+b}{2}} g(u(\frac{3a + b}{2} − s))ds,

hence

r(β′_{1}(t)) = ∫t^{\frac{a+b}{2}} g(β_{1}(s))ds ∀t ∈ [a, \frac{a + b}{2}]

and

−(r(β′_{1}(t)))′ = g(β_{1}(t)) ∀t ∈ [a, \frac{a + b}{2}]

The proof is similar for all t ∈ [\frac{a+b}{2}, b]. □

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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

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

E-mail address: chehb md@yahoo.fr