Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 57, 1-29; http://www.math.u-szeged.hu/ejqtde/

Solvability of a one-dimensional quasilinear problem under nonresonance conditions on the

potential

Aboubacar MARCOS Institut de Mathématiques et de Sciences Physiques BP. 613, Porto-Novo (Bénin), e-mail:abmarcos@imsp-uac.org

Abstract Problem of the type −Δ_{p}u = f(u) + h(x) in (a, b) with u = 0 on {a, b} is solved under nonresonance conditions stated with respect to the first eigenvalue and the first curve in the Fučik spectrum of (−Δ_{p}, W_{0}^{1,p}(a, b)), only on a primitive of f.

Keywords: One-dimensional p-Laplacian, Fučik spectrum, Nonresonance, Timemapping, Degree theory. ^{1}

1 Introduction

This paper is mainly concerned with the following quasilinear two-point boundary value problem

(P) = \{ \begin{matrix}−(φ_{p}(u′))′ = f(u) + h(x)  in (a, b) \\ u = 0  on \{a, b\} \end{matrix}

where φ_{p} : ℝ ⟶ ℝ is defined by φ_{p}(s) = |s|^{p−2}s, with p ∈]1, +∞], f : ℝ ⟶ ℝ, is a continuous function and h ∈ L^{1}(a, b).

1AMS subject classification: 34B15

We denote by ∑ the set of couples of positive numbers (μ_{+}, μ_{−}) such that the homogeneous problem

(P∑) = \{ \begin{matrix}−(φ_{p}(u′))′ = μ_{+}φ_{p}(u^{+}) − μ_{−}φ_{p}(u^{−}) in (a, b) \\ u = 0 on \{a, b\} \end{matrix}

has a nontrivial solution u. Here u^{+} = max(u, 0), u^{−} = u^{+} − u. The set ∑ is called the Fučik spectrum of the p-Laplacian operator −Δ_{p} on W_{0}^{1,p}(a, b). Denote respectively by λ_{1} and λ_{2} the first and the second eigenvalue of −Δ_{p} on W_{0}^{1,p}(a, b). It is well known that ∑ is composed of two trivial lines λ_{1} × ℝ and ℝ × λ_{1}, and of a sequence of hyperbolic-like curves (cf [2],[6]).The first curve C_{1} passes through λ_{2} and is the set

C_{1} = \{(μ_{+}, μ_{−}) ∈ ℝ^{2}, 1/(μ_{+})^{1/p} + 1/(μ_{−})^{1/p} = \frac{b − a}{π_{p}}\}

where π_{p} = 2(p − 1)^{1/p} ∫_{0}^{1} \frac{ds}{(1−s^{p})^{1/p}}.

Let us denote by F the primitive of f defined by F (s) = ∫_{a}^{s} f(t) dt. In some previous works (see for instance [1],[3],[4], [10] ) many authors have proved the solvability of (P) when h ∈ L^{∞}(a, b) under various nonresonance assumptions on either the nonlinearity f, or on the primitive F, or on both f and F. As far as nonresonance conditions are considered at the right of λ_{1}, the Dolph-type condition:

λ_{1} < lim inf_{s→±∞} \frac{f(s)}{|s|^{p−2}s} ≤ lim sup_{s→±∞} \frac{f(s)}{|s|^{p−2}s} ≤ λ_{2}  (1)

is sufficient to yield solvability of (P) when h ∈ L^{∞}(a, b) ( See [1]). It was observed in a recent work in [3], that weaker conditions with respect to the first curve in the Fučik spectrum such as

λ_{1} < lim sup_{s→±∞} \frac{pF (s)}{|s|^{p}} ≤ lim sup_{s→±∞} \frac{f(s)}{|s|^{p−2}s} ≤ μ_{±}  (2)

and

lim inf_{\begin{matrix}s→+∞ \\ or \\ s→−∞ \end{matrix}} \frac{pF (s)}{|s|^{p}} < μ_{+}( or μ_{−})  (3)

coupled with

lim_{|s|→+∞} sgn(s)f(s) = +∞

yield the same conclusion. Adapting an example given in [5], one can observe that assumption (3) cannot be relaxed to

lim inf_{\begin{matrix}s→+∞ \\ or \\ s→−∞ \end{matrix}} \frac{f(s)}{|s|^{p−2}s} < μ_{+}( or μ_{−})

Our purpose in the present paper is to weaken nonresonance conditions (2) and (3) at the light of a recent contribution in [11] for p = 2. Indeed, in [11] the solvability of (P) when p = 2 occurs under assumptions such as

lim_{|s|→+∞} sgn(s)f(s) = +∞  (4)
lim inf_{s→±∞} \frac{2F (s)}{s^{2}} > μ_{1}  (5)

and

lim inf_{s→+∞} \frac{2F (s)}{s^{2}} = μ,  lim sup_{s→−∞} \frac{2F (s)}{s^{2}} = ν  (6)

where μ_{1} is the first eigenvalue of −Δ, on H_{0}^{1}(a, b) and (μ, ν) is such that \frac{π}{\sqrt{μ}} + \frac{π}{\sqrt{ν}} > b − a. It is worth noticing that the roles of s at infinity in (6) are interchangeable. Clearly, assumption such as (6 )improves (3) in the particular case of p = 2 and the question naturally arises to know whether similar assumption can be extended to the p-Laplacian. The aim of this work is to investigate such a problem and as a result of this investigation we have the following.

Theorem 1.1. Assume that

(h_{1}) lim_{|s|→+∞} sgn(s)f(s) = +∞

(h_{2}) lim inf_{s→±∞} \frac{pF (s)}{|s|^{p}} > λ_{1}

(h_{3}) lim inf_{s→+∞} \frac{pF (s)}{|s|^{p}} = μ,  lim sup_{s→−∞} \frac{pF (s)}{|s|^{p}} = ν
or lim inf_{s→−∞} \frac{pF (s)}{|s|^{p}} = μ,  lim sup_{s→+∞} \frac{pF (s)}{|s|^{p}} = ν

with 1/μ^{1/p} + 1/ν^{1/p} > \frac{b−a}{π_{p}} Then problem (P) is solvable for any h ∈ L^{1}(a, b).

As a consequence of our main result we have the following

Corollary 1.1. Assume that

lim_{|s|→+∞} sgn(s)f(s) = +∞

lim inf_{s→±∞} \frac{pF (s)}{|s|^{p}} > λ_{1}

and lim inf_{s→+∞} \frac{pF (s)}{|s|^{p}} < λ_{2},  lim sup_{s→−∞} \frac{pF (s)}{|s|^{p}} ≤ λ_{2}
or lim inf_{s→+∞} \frac{pF (s)}{|s|^{p}} ≤ λ_{2},  lim sup_{s→−∞} \frac{pF (s)}{|s|^{p}} < λ_{2}

Then problem (P) is solvable for any h ∈ L^{1}(a, b).

Needless to mention that the limits at ∞ are interchangeable. Thus, our result improves [3] in what concerns the conditions with respect to the first curve in the Fučik spectrum. The proof of Theorem 1-1 is given in section 4. Basically, it uses time-mapping estimates to yield the needed a-priori bounds for a suitable parametrized problem related to (P) and combines topological degree argument to conclude . Our section 2 is devoted to the establishment of general properties for quasilinear differential equations useful for the proof of our main result. In section 3 we have given new estimate results for the time-mapping related to the p-Laplacian and accordingly improved some estimate results stated in [10]. Those estimates play a central role in the proof of Theorem 1-1.

2 General properties

Here we give general results for a large class of parametrized quasilinear problem of the form

(Q_{γ}) = \{ \begin{matrix}−(φ_{p}(u′))′ = f̂(x, u, γ) + γh(x)  in (a, b), γ ∈ [0, 1] \\ u = 0  on \{a, b\} \end{matrix}

We assume that h ∈ L^{1}(a, b) and f̂: [a, b] × ℝ × [0, 1] ⟶ ℝ is a function satisfying the following:

(i) sign condition: sgn(s)f̂(x, s, γ) ≥ −c for a positive constant c, for a.e. x ∈ [a, b] and for γ ∈ [0, 1]; (ii) L^{1}-Carathéodory condition: f̂(x, ., γ) is continuous for a.e. x ∈ [a, b], γ ∈ [0, 1], f̂(., s, γ) is measurable for s ∈ ℝ and γ ∈ [0, 1]; moreover for each R > 0, there is Γ_{R} ∈ L^{1}(a, b) such that |f̂(x, s, γ)| ≤ Γ_{R}(x) for all |s| ≤ R, a.e. x ∈ [a, b], and for γ ∈ [0, 1]. Solutions to (Q_{γ}) are intended in the sense that u ∈ C^{1}[a, b], φ_{p}(u′) is absolutely continuous and u satisfies (Q_{γ}).

For any solution u of (Q_{γ}) we set here and henceforth the following Definition 2.1. We denote by x^{∗} the first point of maximum of u and x_{∗} the last point of minimum of u.

Definition 2.2. For any K such that 0 < K ≤ max u, we denote

α_{0} = max{x ∈ [a, x^{∗}), u(x) = 0}
β_{0} = min{x ∈ (x^{∗}, b], u(x) = 0}
α_{K} = min{x ∈ [α_{0}, x^{∗}], u(x) = K}
β_{K} = max{x ∈ [x^{∗}, β_{0},], u(x) = K}

Definition 2.3. For any K′ such that 0 > K′ ≥ min u, we denote

α′_{0} = max{x ∈ [a, x_{∗}), u(x) = 0}
β′_{0} = min{x ∈ (x_{∗}, b], u(x) = 0}
α′_{K′} = min{x ∈ [α′_{0}, x_{∗}], u(x) = K′}
β′_{K′} = max{x ∈ [x_{∗}, β′_{0},], u(x) = K′}

Writing the first equation in (Q_{γ}) in the planar system

φ_{p}(u′) = y(x) − γH̃(x)  (7)
y′(x) = −f̃(x, u(x), γ)  (8)

with f̃(x, s, γ) = f̂(x, s, γ) + c and H̃(x) = ∫_{a}^{x}(h(t) − c) dt, we derive the following.

Lemma 2.1. A positive constant L exists such that any solution u of (Q_{γ}) satisfying max u > L, fulfills the following conditions: there exist uniquely determined real numbers ρ and ρ̄, with α_{0} < ρ ≤ x^{∗} ≤ ρ̄ < β_{0} such that

(i)

y(x) > ∥H̃∥_{∞} on [α_{0}, ρ)
|y(x)| ≤ ∥H̃∥_{∞} on [ρ, ρ̄]
y(x) < −∥H̃∥_{∞} on (ρ̄, β_{0}]

(ii) u is strictly increasing on [α_{0}, ρ], strictly decreasing on [ρ̄, β_{0}] and

max u − L ≤ u(x) ≤ max u on [ρ, ρ̄]

If furthermore

lim_{s→+∞} f̂(x, s, γ) = +∞

uniformly for (γ ∈ [0, 1], and a.e. x ∈ [a, b]), then for any K > 0 such that K ≤ max u we have

lim_{K→+∞}(α_{K} − α_{K−L}) = lim_{K→+∞}(β_{K−L} − β_{K}) = 0

Remark 2.1. A dual version of Lemma (2.1)involving α′_{0}, β′_{0}, α′_{K′}, β′_{K′} can be obtained in the case that u is a solution of the planar system with min u < −L and

lim_{s→−∞} f̂(x, s, γ) = −∞

uniformly (for γ ∈ [0, 1] and a.e. x ∈ [a, b]). In this case f̃ and H̃ in the planar system are written f̃(x, s, γ) = f̂(x, s, γ) − c, H̃(x) = ∫_{a}^{x}(h(t) + c) dt.

Proof of Lemma 2.1 The proof of Lemma 2.1 in the particular case p = 2 is given in [11]. We give here the general case for any p > 1. So, let us consider α_{0}, β_{0}, x^{∗} as set in the definitions 2.1, 2.2. Since y′(x) = −f̃(x, u(x), γ), from the sign condition on f̂, we have that y is strictly decreasing on (α_{0}, β_{0}) and accordingly

y(α_{0}) > y(x^{∗}) > y(β_{0})

Moreover u′(x^{∗}) = 0, and then (7) yields

|y(x^{∗})| = |φ_{p}(0) + γH̃(x^{∗})| ≤ ∥H̃∥_{∞}

Let [ρ, ρ̄] ⊂ [α_{0}, β_{0}] be the maximal interval containing x^{∗} and such that

|y(x)| ≤ ∥H̃∥_{∞}  (9)

Clearly, for such an interval, part (i) of Lemma 2.1 holds. Since φ_{p} is a bijection on ℝ, one can write (7) on the form

u′(x) = φ_{p}^{−1}(y(x) − γH̃(x))

And then using the monotoniciity of φ_{p} and (9), we have for s ∈ [ρ, ρ̄]

|u′(x)| = |φ_{p}^{−1}(y(x) − γH̃(x))| ≤ φ_{p}^{−1}(2∥H̃∥_{∞})

Accordingly, for x′ and x′′ in [ρ, ρ̄], we get

|u(x′′) − u(x′)| ≤ | ∫_{x′}^{x′′} u′(x) dx| ≤ (b − a)φ_{p}^{−1}(2∥H̃∥_{∞}).

Consequently we get

∥u∥_{∞} − (b − a)φ_{p}^{−1}(2∥H̃∥_{∞}) ≤ u(x′′) ≤ ∥u∥_{∞} for all x′′ ∈ [ρ, ρ̄].

So by setting L = (b − a)φ_{p}^{−1}(2∥H̃∥_{∞}), we have part (ii) of the lemma. To deal with part (iii) of the lemma, we note that since lim_{s→+∞} f̂(x, s, γ) = +∞

uniformly for γ ∈ [0, 1], and a.e. x ∈ [a, b]), for any k > 0, one can choose v_{k} > 0 large enough such that

f̂(x, s, γ) ≥ k for all s ≥ v_{k}, γ ∈ [0, 1], and a.e. x ∈ [a, b].

Choose K with K ≥ v_{k} + L where L = (b − a)φ_{p}^{−1}(2∥H̃∥_{∞}). Let’s consider any solution of (Q_{γ}) such that max u ≥ K. Since K − L ≤ u(x) ≤ K when x ∈ [α_{K−L}, α_{K}], we have

f̃(x, u(x), γ) ≥ k, for all x ∈ [α_{K−L}, α_{K}], and γ ∈ [0, 1]  (10)

and then

y(x) = y(α_{K}) + ∫_{α_{K}}^{x} y′(t) dt
= ∫_{x}^{α_{K}} f̃(t, u(t), γ) dt on [α_{K−L}, α_{K}].

Since max u ≥ K, we have α_{K} ∈ [α_{0}, x^{∗}] and then by using part (i) of the lemma and condition (10) we have

y(x) ≥ −∥H̃∥_{∞} + k(α_{K} − x).

And then

u′(x) = φ_{p}^{−1}(y(x) − γH̃(x)) ≥ φ_{p}^{−1}(−2∥H̃∥_{∞} + k(α_{K} − x)) on [α_{K−L}, α_{K}].

Next, we derive from the integration of u′ on [α_{K−L}, α_{K}] the following inequality

L = u(α_{K}) − u(α_{K−L}) ≥ ∫_{α_{K−L}}^{α_{K}} φ_{p}^{−1}(−2∥H̃∥_{∞} + k(α_{K} − x)) dx  (11)

To go further with the integral in the left hand-side of (11), let us set

Ψ_{p}^{∗}(s) = ∫_{0}^{s} φ_{p}^{−1}(ξ) dξ.

A simple computation shows that φ_{p}^{−1}(s) = s|s|^{\frac{2−p}{p−1}} for all s ∈ ℝ and p ∈]1, +∞[ and then Ψ_{p}^{∗}(s) = \frac{p−1}{p}|s|^{\frac{p}{p−1}} for all s ∈ ℝ and p ∈]1, +∞[. Clearly Ψ_{p}^{∗} is an even strictly increasing function on ℝ_{+}. Let us denote by Ψ_{p}^{∗−1}(s) its positive inverse function on ℝ_{+}.

Ψ_{p}^{∗−1}(s) = (\frac{p}{p − 1})^{\frac{p−1}{p}} s^{\frac{p−1}{p}} , s ∈ ℝ_{+}.

On the other hand, the function

x ↦ Ψ_{p}^{∗}(−2∥H̃∥_{∞} + k(α_{K} − x))

is differentiable with respect to x and so

\frac{d}{dx}[Ψ_{p}^{∗}(−2∥H̃∥_{∞} + k(α_{K} − x))] = −kφ_{p}^{−1}(−2∥H̃∥_{∞} + k(α_{K} − x)) (12)

Combining (11)and (12), we get

L ≥ −\frac{1}{k}[Ψ_{p}^{∗}(−2∥H̃∥_{∞} + k(α_{K} − x))]_{α_{K−L}}^{α_{K}}

and then

kL + Ψ_{p}^{∗}(2∥H̃∥_{∞}) ≥ Ψ_{p}^{∗}(−2∥H̃∥_{∞} + k(α_{K} − α_{K−L})).

Now, using the positive inverse of Ψ_{p}^{∗}, one has

α_{K} − α_{K−L} ≤ \frac{Ψ_{p}^{∗−1}(kL + Ψ_{p}^{∗}(2∥H̃∥_{∞}) + 2∥H̃∥_{∞}}{k}  (13)

One can easily see that the right hand-side of inequality (13 ) is equivalent to k^{−1/p} at infinity. So when K tends to +∞, k tends to +∞, and then

lim_{K→+∞}(α_{K} − α_{K−L}) = 0.

A similar argument on [β_{K}, β_{K−L}] leads to

lim_{K→+∞}(β_{K−L} − β_{K}) = 0.

So Lemma (2.1) is proved. ■

Remark 2.2. When K = max u, then α_{K} = β_{K} = x^{∗} and then from (iii ) of Lemma(2.1) we have lim_{u^{∗}→+∞}(β_{max u−L} − α_{max u−L}) = 0.

Lemma 2.2. Let u be a changing sign solution of (Q_{γ}) for γ ∈ [0, 1] and let A, be a positive real number such that max u < A or min u > −A uniformly with respect to γ ∈ [0, 1]. Then a constant M ( depending only on A ) exists such that ∥u∥_{∞} < M.

Proof Let us consider only the case max u < A, the second case min u > −A of course can be proved similarly. Thus, suppose on the contrary that there exist a sequence (γ_{n}) ∈ [0, 1] denote (γ) for sake of simplicity of notation, and corresponding solutions u_{n} of (Q_{γ}), with max u_{n} < A and min u_{n} tending to −∞. Then, from the sign condition and the L^{1}-Carathéodory condition on f̂ we have

f̂(x, u_{n}, γ) ≤ cχ_{{u_{n}<0}} + Γ_{A}.χ_{{0≤u_{n}<A}} = Γ(x)

where for any set E, χ_{E} denote its characteristic function. Choose two points x_{n}^{∗} and x_{∗}_{n} such that u_{n}(x_{n}^{∗}) = max u_{n} and u_{n}(x_{∗}_{n}) = min u_{n}. We can suppose without loss of generality that x_{n}^{∗} > x_{∗}_{n}. Set ũ_{n} = u_{n} − ū_{n} with (x_{n}^{∗} − x_{∗}_{n})ū_{n} = ∫_{x_{∗}_{n}}^{x_{n}^{∗}} u_{n}(x) dx. Then after the multiplication of the first equation in (Q_{γ}) with ũ_{n} and its integration over [x_{∗}_{n}, x_{n}^{∗}], we have

∫_{x_{∗}_{n}}^{x_{n}^{∗}} |ũ′_{n}(x)|^{p} dx = ∫_{x_{∗}_{n}}^{x_{n}^{∗}}[f̂(x, u_{n}, γ)−Γ(x)]ũ_{n}(x) dx+∫_{x_{∗}_{n}}^{x_{n}^{∗}}[γh(x)+(x)]ũ_{n}(x) dx

≤ ||ũ_{n}||_{∞}{ ∫_{x_{∗}_{n}}^{x_{n}^{∗}}[−f̂(x, u_{n}, γ_{n}) + Γ(x)] dx + ||h||_{1} + ∥Γ∥_{1}}.

But

∫_{x_{∗}_{n}}^{x_{n}^{∗}}[−f̂(x, u_{n}, γ_{n}) + γh(x)] dx = 0

and then we have

∫_{x_{∗}_{n}}^{x_{n}^{∗}} |ũ′_{n}(x)|^{p} dx ≤ 2[||h||_{1} + ||Γ||_{1}]||ũ_{n}||_{∞}  (14)

From the Hölder inequality we have

(∫_{x_{∗}_{n}}^{x_{n}^{∗}} |ũ′_{n}(x)| dx)^{p} ≤ (b − a)^{p−1} ∫_{x_{∗}_{n}}^{x_{n}^{∗}} |ũ′_{n}(x)|^{p} dx.

Combining the above inequality and (14), we get

(max ũ_{n} − min ũ_{n}) ≤ 2(b − a)^{p−1}[||h||_{1} + ||Γ||_{1}]||ũ_{n}||_{∞}  (15)

Since ||ũ_{n}||_{∞}^{p} ≤ (max ũ_{n} − min ũ_{n}), inequality (15) yields

||u_{n}||_{∞} ≤ 2^{\frac{1}{p−1}} (b − a)[||h||_{1} + ||Γ||_{1}]^{\frac{1}{p−1}} .

So the sequence (ũ_{n}) is bounded and hence (u_{n}) is bounded. This is a contradiction to the fact that min u_{n} tends to −∞. ■

3 Time-mapping and auxiliary functions

3.1 Time-mapping estimates

Let’s consider the initial value problem

(I) = \{ \begin{matrix}−(φ_{p}(u′))′ = g(u)  on ℝ \\ u(0) = s,  u′(0) = 0 \end{matrix}

Where g : ℝ → ℝ is a continuous function satisfying sgn(s)g(s) > −c for c > 0 and G(s) → +∞. The function τ_{g} defined by

τ_{g}(s) = 2c_{p}sgn(s) ∫_{0}^{s} \frac{dξ}{[G(s) − G(ξ)]^{1/p}} for s in ℝ

with G(s) = ∫_{0}^{s} g(ξ) dξ, c_{p} = \frac{1}{p^{∗}^{1/p}} and p^{∗} = \frac{p}{p−1} is the time-mapping associated to (I). Under the assumptions sgn(s)g(s) > −c for c > 0 and G(s) → +∞ when |s| → +∞, τ_{g}(s) is well defined for |s| large enough. By adapting arguments developed in [12] for the case p = 2, one can easily derive that for s large enough (I) admits a periodic solution u_{s}with ||u_{s}||_{∞} = s and τ_{g}(s) is the value of the half period. Time-mapping enables to provide a-priori estimates for solutions of boundary value problems ( cf [7], [10], [11], [12]). Here, we give new results on the time-mapping estimates extending and even improving some results in [7], [10], [11], [12].

Lemma 3.1. Assume that there exist positive real numbers k^{±} and k_{±} such that

lim sup_{s→±∞} pG(s)/|s|^{p} = k^{±} (resp. lim inf_{s→±∞} pG(s)/|s|^{p} = k_{±})

then

lim inf_{s→±∞} τ_{g}(s) ≥ π_{p}/(k^{±})^{1/p} (resp. lim sup_{s→±∞} τ_{g}(s) ≤ π_{p}/(k_{±})^{1/p})

Proof. One can notice that under the assumption sgn(s)g(s) > −c for c > 0 and the fact that k^{±} k_{±} are greater than 0, G(s) → +∞ when |s| → +∞ so that τ_{g}(s) is well defined for s large enough. Let’s limit the proof of the lemma to the cases lim sup_{s→−∞} pG(s)/|s|^{p} = k^{−}, (resp. lim inf_{s→−∞} pG(s)/|s|^{p} = k_{−}), the other cases being

similar.

For s < 0 and for any ξ such that |s|^{p} > |ξ|^{p}, we have

lim sup_{s→−∞} \frac{pG(s)}{|s|^{p} − |ξ|^{p}}

= lim sup_{s→−∞} pG(s)/|s|^{p} × lim_{s→−∞} \frac{|s|^{p}}{|s|^{p} − |ξ|^{p}} = k^{−}

and then

lim sup_{s→−∞} [\frac{pG(s)}{|s|^{p} − |ξ|^{p}} − \frac{pG(ξ)}{|s|^{p} − |ξ|^{p}} ] = k^{−}.

So for ϵ > 0, there is a real number s_{0} < 0 such that for s < s_{0} we have

G(s) − G(ξ) ≤ 1/p(k^{−} + ϵ)(|s|^{p} − |ξ|^{p}).

Recalling the expression of τ_{g}(s) and taking into account inequality above, we get

τ_{g}(s) ≥ 2c_{p} ∫_{s}^{s_{0}} \frac{dξ}{[G(s) − G(ξ)]^{1/p}}

≥ \frac{2c_{p}p^{1/p}}{(k^{−} + ϵ)^{1/p}} ∫_{s}^{s_{0}} \frac{dξ}{[|s|^{p} − |ξ|^{p}]^{1/p}} .

Setting z = ξ/s, one has

τ_{g}(s) ≥ \frac{2(p − 1)^{1/p}}{(k^{−} + ϵ)^{1/p}} ∫_{0}^{1} \frac{dz}{[1 − z^{p}]^{1/p}} = \frac{π_{p}}{(k^{−} + ϵ)^{1/p}} for all s < s_{0} < 0.

Thus

lim inf_{s→−∞} τ(s) ≥ \frac{π_{p}}{(k^{−})^{1/p}} .

For the case lim inf_{s→−∞} pG(s)/|s|^{p} = k_{−}, we have ∀ϵ > 0, there is a real number s_{0} < 0 such that for s < s_{0}

G(s) − G(ξ) ≥ 1/p(k_{−} − ϵ)(|s|^{p} − |ξ|^{p}), for all s < ξ < 0. So, for ϵ sufficiently small such that k_{−} − ϵ > 0, we have

τ_{g}(s) ≤ \frac{2c_{p}p^{1/p}}{(k_{−} − ϵ)^{1/p}} ∫_{s}^{s_{0}} \frac{dξ}{[|s|^{p} − |ξ|^{p}]^{1/p}} .

And by a simple computation as previously done we get

lim sup_{s→−∞} τ_{g}(s) ≤ \frac{π_{p}}{(k_{−})^{1/p}} . ■

EJQTDE, 2009 No. 57, p. 11

3.2 Auxiliary functions related to the time-mapping

Let us consider the following parametrized problem

(P_{γ}) = \{ \begin{matrix}−(φ_{p}(u′))′ = g(u, γ) + γh(x)  in (a, b), γ ∈ [0, 1] \\ u = 0  on \{a, b\} \end{matrix}

where h ∈ L^{1}(a, b) and g(., γ) : ℝ ⟶ ℝ is a continuous function for any γ ∈ [0, 1]. (P_{γ}) is a particular type of problem (Q_{γ}) where the nonlinearity f̂ does not depend on x. We assume that g satisfies the following sign condition: lim_{|s|→+∞} sgn(s)g(s, γ) = +∞ uniformly with respect to γ ∈ [0, 1]. From this sign

condition, one can find a positive constant c such that sgn(s)g(s, γ) > c for all γ ∈ [0, 1]. Let us set

H̃(x) = ∫_{a}^{x}(h(t) − c) dt  G̃_{γ}(s) = ∫_{a}^{x} g̃(ξ, γ) dξ,

for each γ ∈ [0, 1], and where g̃(s, γ) = g(s, γ) + c for s ≥ 0 and g̃(s, γ) = g(s, γ) − c for s ≤ 0 The planar system equivalent to the first equation in (P_{γ}) is written

φ_{p}(u′) = y(x) − γH̃(x)  (16)
y′(x) = −g̃(u(x), γ)  (17)

for x ∈ (a, b) and γ ∈ [0, 1]. It is clear that the planar system (16), (17) is a particular case of the planar system (7), (8) and hence Lemma (2.1) is valid for any solution of (16), (17) as well. For any solution u of (16), (17), let us consider the function T_{ϵ} where ϵ = ±1 and defined by

T_{ϵ}(x) = \frac{p − 1}{p}|y(x) + ϵ||H̃||_{∞}|^{\frac{p}{p−1}} + G̃_{γ}(u(x)) on [a, b]

with γ ∈ [0, 1] and p > 1. One can easily see that

T′_{ϵ}(x) = y′(x)[|y(x) + ϵ||H̃||_{∞}|^{\frac{2−p}{p−1}} (y(x) + ϵ||H̃||_{∞}) − u′(x)].

Recalling part (i) of Lemma (2.1), we derive that: for ϵ = −1

T′_{−1}(x) = \{ \begin{matrix}y′(x)[(y(x) − ||H̃||_{∞})^{\frac{1}{p−1}} − u′(x)] on [α_{0}, ρ] \\ −y′(x)[(−y(x) + ||H̃||_{∞})^{\frac{1}{p−1}} + u′(x)] on [ρ, β_{0}] \end{matrix}

for ϵ = 1

T′_{1}(x) = \{ \begin{matrix}y′(x)[(y(x) + ||H̃||_{∞})^{\frac{1}{p−1}} − u′(x)] on [α_{0}, ρ̄] \\ −y′(x)[(−y(x) − ||H̃||_{∞})^{\frac{1}{p−1}} + u′(x)] on [ρ̄, β_{0}] \end{matrix}

So recalling again part (i) of Lemma (2.1), one can easily check that

T′_{−1}(x) ≥ 0 on [α_{0}, ρ]  T′_{1}(x) ≤ 0 on [ρ, β_{0}]  (18)

Accordingly we have

T_{−1}(x) ≤ T_{−1}(ρ) = G̃_{γ}(u(ρ)) ≤ G̃_{γ}(max u) on [α_{0}, ρ]  (19)
T_{1}(x) ≤ T_{1}(ρ̄) = G̃_{γ}(u(ρ̄)) ≤ G̃_{γ}(max u) on [ρ̄, β_{0}]

Taking into account the expressions of T_{−1}(x) and T_{1}(x) and recalling again (i) of Lemma 2.1, we get

u′(x) ≤ φ_{p}^{−1}[2||H̃||_{∞} + (p^{∗})^{\frac{p−1}{p}} (G̃_{γ}(max u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}}] on [α_{0}, ρ]
−u′(x) ≤ φ_{p}^{−1}[2||H̃||_{∞} + (p^{∗})^{\frac{p−1}{p}} (G̃_{γ}(max u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}}] on [ρ̄, β_{0}]

Next, by setting ξ = u(x), we get

(β_{0} − α_{0}) ≥ (β_{0} − ρ̄) + (ρ − α_{0}) ≥
c_{p} ∫_{0}^{u(ρ̄)} \frac{dξ}{φ_{p}^{−1}[2c_{p}^{p−1}||H̃||_{∞} + (G̃_{γ}(max u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}}]}  (20)

+c_{p} ∫_{0}^{u(ρ)} \frac{dξ}{φ_{p}^{−1}[2c_{p}^{p−1}||H̃||_{∞} + (G̃_{γ}(max u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}}]}

From Lemma (2.1), u(ρ) and u(ρ̄) are greater than max u − L and then writing max u = s in (20), one has

(β_{0} − α_{0}) ≥ 2c_{p} ∫_{0}^{s−L} \frac{dξ}{φ_{p}^{−1}[2c_{p}^{p−1}||H̃||_{∞} + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]}  (21)

for s > L. Considering the functions T_{−1} and T_{1} respectively on [α′_{0}, ρ′] and [ρ̄′, β′_{0}] where α′_{0}, ρ′, ρ̄′, β′_{0} are the equivalents of α_{0}, ρ, ρ̄, β_{0} in the dual version of Lemma (2.1), and arguing as above with min u playing the role of max u, we get

(β′_{0} − α′_{0}) ≥ (β′_{0} − ρ̄′) + (ρ′ − α′_{0}) ≥

c_{p} ∫_{u(ρ̄′)}^{0} \frac{dξ}{φ_{p}^{−1}[2c_{p}^{p−1}||H̃||_{∞} + (G̃_{γ}(min u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}}]}

+c_{p} ∫_{u(ρ′)}^{0} \frac{dξ}{φ_{p}^{−1}[2c_{p}^{p−1}||H̃||_{∞} + (G̃_{γ}(min u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}}]}

Next, writing min u = s and taking into account the fact that from the dual version of Lemma (2.1), u(ρ′) and u(ρ̄′) are lower than min u + L, one has

(β′_{0} − α′_{0}) ≥ 2c_{p} ∫_{s+L}^{0} \frac{dξ}{φ_{p}^{−1}[2c_{p}^{p−1}||H̃||_{∞} + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]}  (22)

for s < −L.

In conclusion we have

(β_{0} − α_{0}) + (β′_{0} − α′_{0}) ≥
2c_{p}σ ∑_{σ∈\{−1,1\}} ∫_{0}^{s−σL} \frac{dξ}{φ_{p}^{−1}[2c_{p}^{p−1}||H̃||_{∞} + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]}  (23)

for |s| > L. Thus T_{ϵ} provides lower estimates for the length of the intervals [α_{0}, β_{0}] and [α′_{0}, β′_{0}]. Let us now deal with upper estimates provide by T_{ϵ}. Going back to (18) and to the expressions of T_{1}(x) and T_{−1}(x) we derive that

T_{1}(x) ≥ T_{1}(x^{∗}) ≥ G̃_{γ}(max u) on [α_{0}, x^{∗}]
T_{−1}(x) ≥ T_{−1}(x^{∗}) ≥ G̃_{γ}(max u) on [x^{∗}, β_{0}]

and hence

u′(x) ≥ φ_{p}^{−1}[−2||H̃||_{∞} + (p^{∗})^{\frac{p−1}{p}} (G̃_{γ}(max u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}}]  (24)
on [α_{0}, x^{∗}]
−u′(x) ≥ φ_{p}^{−1}[−2||H̃||_{∞} + (p^{∗})^{\frac{p−1}{p}} (G̃_{γ}(max u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}}]  (25)
on [x^{∗}, β_{0}]

From part (ii) of Lemma (2.1), one has [α_{0}, α_{max u−L}] ⊂ [α_{0}, x^{∗}] and [β_{max u−L}, β_{0}] ⊂ [x^{∗}, β_{0}]. So let us consider inequalities in (24)and (25) respectively on [α_{0}, α_{max u−L}] and [β_{max u−L}, β_{0}] and let us assume that the following is satisfied (we will show farther in section 4 that such a condition is indeed satisfied under suitable condition ):

−2||H̃||_{∞} + (p^{∗})^{\frac{p−1}{p}} (G̃_{γ}(max u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}} > 0  (26)
on [α_{0}, α_{max u−L}] ∪ [β_{max u−L}, β_{0}].

Then, we derive after the change of variable ξ = u(x) in (24) and (25), that

(β_{0} − β_{max u−L}) + (α_{max u−L} − α_{0}) ≤
2c_{p} ∫_{0}^{max u−L} \frac{dξ}{φ_{p}^{−1}[−2c_{p}^{p−1}||H̃||_{∞} + (G̃_{γ}(max u) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]}.  (27)

Arguing in a similar way, one can show that

(β′_{0} − β′_{min u+L}) + (α′_{min u+L} − α′_{0}) ≤
2c_{p} ∫_{min u+L}^{0} \frac{dξ}{φ_{p}^{−1}[−2c_{p}^{p−1}||H̃||_{∞} + (G̃_{γ}(min u) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]}.  (28)

Now, let us set

T_{γ}(s) = 2c_{p}sgn(s) ∫_{0}^{s−sgn(s)L} \frac{dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]}

with |s| > L and K > 0.

τ_{γ}(s) = 2c_{p}sgn(s) ∫_{0}^{s} \frac{dξ}{[G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}}
with |s| > 0.

One can easily see that according to (21), (22)

(β_{0} − α_{0}) ≥ T_{γ}(max u) for max u > L  (29)
(β′_{0} − α′_{0}) ≥ T_{γ}(min u) for min u < −L  (30)

with K = 2||H̃||_{∞} On the other hand, the following lemma shows that τ_{γ}(s) is a good approximation of T_{γ}(s) for s large enough.

Lemma 3.2. Assume that lim_{|s|→+∞} g(s, γ) = +∞ uniformly with respect to γ. As-

sume that at least one of the functions T_{γ}(s) and τ_{γ}(s) is uniformly bounded with respect to γ. Then

lim_{s→±∞}[T_{γ}(s) − τ_{γ}(s)] = 0 uniformly with respect toγ.

Proof Without loss of generality, we can suppose that it is T_{γ}(s) which is bounded uniformly with respect to γ. Furthermore the proof will be given only for the case s → +∞, the case s → +∞ can be dealed similarly. So, let us consider

T_{γ}(s) = 2c_{p} ∫_{0}^{s−L} \frac{dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]}
for s > L
τ_{γ}(s) = 2c_{p} ∫_{0}^{s} \frac{dξ}{[G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}}
for s > 0.

We observe that K > 0 implies

φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] ≥ φ_{p}^{−1}[G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] = [G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}

for 0 < ξ < s, and thus τ_{γ}(s) ≥ T_{γ}(s) for s > L. So, it suffices to show that for any ϵ > 0, τ_{γ}(s) − T_{γ}(s) < ϵ for s large enough.

Since lim_{s→+∞} g̃(s, γ) = +∞ uniformly with respect to γ, for any A > 0 there exists a real number d > 0 such that

g̃(ξ, γ) ≥ A for ξ ≥ d and γ ∈ [0, 1]  (31)

and

G̃_{γ}(d) > G̃_{γ}(ξ) for 0 ≤ ξ < d and γ ∈ [0, 1].  (32)

Choose s such that s > d + L with L > 0,

τ_{γ}(s)−T_{γ}(s) ≥ I = 2c_{p} ∫_{0}^{s−L} \frac{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] − [G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}} dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}][G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}}

Let us split the integral I as follows

I = 2c_{p} [∫_{0}^{d} + ∫_{d}^{s−L}] = 2c_{p}[I_{1} + I_{2}].

Dealing with the first term of this decomposition, we get by using the monotonicity of φ_{p}^{−1}

I_{1} ≤ ∫_{0}^{d} \frac{dξ}{[G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}} ≤ ∫_{0}^{d} \frac{dξ}{[G̃_{γ}(s) − G̃_{γ}(d)]^{\frac{1}{p}}}

Tending s to infinity, we notice that the right-hand side integral tends to zero. So for s large enough we have

2c_{p}I_{1} < \frac{ϵ}{2}.

To deal with the second term I_{2},we write it as follows

I_{2} = 2c_{p} ∫_{0}^{s−L} \frac{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] − φ_{p}^{−1}[(G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}][G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}}

In order to estimate I_{2} the following inequalities will be useful.

Claim 1.

(i) A positive constant D exists such that for any real numbers a, b

|a − b|^{p} ≤ D(|a|^{p−2}a − |b|^{p−2}b)(a − b) for p ≥ 2

(ii) If a, b are non negative reals numbers then

(a + b)^{p} − b^{p} ≤ pa(a + b)^{p−1} for p > 1

and

(a + b)^{p} − b^{p} ≤ pab^{p−1} for 0 < p < 1

Proof. For the case (i), on can refer to [9]. In order to prove (ii), let us consider the function

r(y) = (y + b)^{p} for 0 ≤ y ≤ a.

Obviously r is derivable and its derivative function r′(y) = p(y + b)^{p−1} is increasing on [0,a]. So,

r(a) − r(0) = ar′(ξ) ≤ pa(a + b)^{p−1} for 0 < ξ < a

Thus (a + b)^{p} − b^{p} ≤ pa(a + b)^{p−1}. The second inequality in (ii) follows similarly and thus claim (1) is proved. ■

Now, let’s go ahead with the proof of the Lemma (3.2). Using (i) of claim (1) with a = φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] and b = φ_{p}^{−1}[(G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] we have

φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] − φ_{p}^{−1}[(G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] ≤ (DK)^{\frac{1}{p−1}}

for p ≥ 2. For 1 < p < 2, we have \frac{1}{p−1} > 1 and then we can apply the first inequality in (ii)

of claim (1) with \frac{1}{p−1} playing the role of p, a = K, b = (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}. Thus, we have

[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]^{\frac{1}{p−1}} − [(G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]^{\frac{1}{p−1}}

≤ \frac{1}{p − 1}K[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]^{\frac{2−p}{p−1}} ,

that is

φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] − φ_{p}^{−1}[(G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]

≤ \frac{1}{p − 1}K[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]^{\frac{2−p}{p−1}} for 1 < p < 2.

In conclusion: For p > 2,

2c_{p}I_{2} ≤ 2c_{p}(DK)^{\frac{1}{p−1}} ∫_{d}^{s−L} \frac{dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}][G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}}

≤ \frac{(DK)^{\frac{1}{p−1}}}{[L min g̃(ξ, γ)_{ξ∈[s−L,L]}]^{1/p}} × T_{γ}(s) ≤ \frac{(DK)^{\frac{1}{p−1}}}{[LA]}^{1/p}T_{γ}(s).

Since T_{γ} is uniformly bounded with respect to γ, by choosing A large enough we have 2c_{p}I_{2} ≤ ϵ/2. For 1 < p < 2

2c_{p}I_{2} ≤ \frac{2c_{p}K}{p − 1} ∫_{d}^{s−L} \frac{[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]^{\frac{2−p}{p−1}} dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}][G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}}

≤ \frac{2c_{p}K}{p − 1} ∫_{d}^{s−L} \frac{[K/(G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}} + 1]^{\frac{2−p}{p−1}} dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}][G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{p−1}{p}}}

≤ \frac{2c_{p}K}{p − 1}[\frac{K}{[L min g̃(ξ, γ)_{ξ∈[s−L,L]}]^{\frac{p−1}{p}}} + 1]^{\frac{2−p}{p−1}} × \frac{1}{[L min g̃(ξ, γ)_{ξ∈[s−L,L]}]^{1/p}} ×

∫_{d}^{s−L} \frac{dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]}

≤ \frac{K}{p − 1} [\frac{K}{(LA)^{\frac{p−1}{p}}} + 1]^{\frac{2−p}{p−1}} × \frac{1}{(LA)^{\frac{p−1}{p}}} × T_{γ}(s).

Here again, for A large enough, we have 2c_{p}I_{2} < ϵ/2 and finally, we get 2c_{p}(I_{1}+I_{2}) < ϵ for s large enough, that is τ_{γ}(s)−T_{γ}(s) < ϵ for s large enough. ■

An analogous of Lemma (3.2) holds when K is a negative real number. In order to state it let us start define

T̃_{γ}(s) = 2c_{p}sgn(s) ∫_{0}^{s−sgn(s)L} \frac{dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]} with |s| > L(33)

and K a negative real number.

τ̃_{γ}(s) = 2c_{p}sgn(s) ∫_{0}^{s−sgn(s)L} \frac{dξ}{[G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}} with |s| > L  (34)

It is worth noticing that according to (27) and (28)

(β_{0} − β_{max u−L}) + (α_{max u−L} − α_{0}) ≤ T̃_{γ}(max u − L)  (35)
(β′_{0} − β′_{min u+L}) + (α′_{min u+L} − α′_{0}) ≤ T̃_{γ}(min u + L)  (36)

with K = −2||H̃||_{∞}.

Lemma 3.3. Assume that lim_{|s|→+∞} g(s, γ) = +∞ uniformly with respect to γ and

that at least one of the functions T̃_{γ}(s) and τ̃_{γ}(s) is uniformly bounded with respect to γ. Moreover, suppose that the following condition is satisfied:

(c_{1}) K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}} > 0 for ξ ∈ [0, s − L], withs > 0

or

c_{2}) K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}} > 0 for ξ ∈ [s + L, 0], withs < 0.

Then respectively lim_{s→+∞}[T̃_{γ}(s)−τ̃_{γ}(s)] = 0 or lim_{s→−∞}[T̃_{γ}(s)−τ̃_{γ}(s)] = 0 uniformly with respect toγ.

Proof. The proof is not too different of that of Lemma (3.2).We will sketch it below. Suppose that T̃_{γ}(s) is uniformly bounded with respect to γ and let’s give the proof when s → +∞ (the other cases being similar).

Since T̃_{γ}(s) > τ̃_{γ}(s) for s > L, we shall just have to prove that for any ϵ > 0, T̃_{γ}(s) − τ̃_{γ}(s) < ϵ for s sufficiently large.

T̃_{γ}(s)−τ̃_{γ}(s) = 2c_{p} ∫_{0}^{s−L} \frac{([G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}} − φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]) dξ}{[G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}} φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]}

= 2c_{p} [∫_{0}^{d} + ∫_{d}^{s−L}] = 2c_{p}[Ĩ_{1} + Ĩ_{2}]

with d as in (31)and (32). And then

Ĩ_{1} ≤ ∫_{0}^{d} \frac{dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(d))^{\frac{p−1}{p}}]}.

So for s large enough, we have 2c_{p}Ĩ_{1} < ϵ/2. To estimate Ĩ_{2} in the case p ≥ 2, we proceed similarly as in the proof of Lemma (3.2) by using (i)of claim (1) to yield

φ_{p}^{−1}[(G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] − φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] ≤ (−DK)^{\frac{1}{p−1}}

and next

2c_{p}Ĩ_{2} ≤ \frac{(−DK)^{\frac{1}{p−1}}}{[LA]}^{1/p}τ̃_{γ}(s).

Hence 2c_{p}Ĩ_{2} ≤ ϵ/2 for s large enough and p ≥ 2. In the case 1 < p < 2, following the same way as in Lemma (3.2), we apply the (ii) of claim(1) by writing the numerator of I_{2} in the form

φ_{p}^{−1}[−K + K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}] − φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}]

and by setting a = −K, b = K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}; and then we obtain

2c_{p}Ĩ_{2} ≤ −2c_{p} \frac{K}{p − 1} ∫_{d}^{s−L} \frac{(φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}])^{2−p} dξ}{φ_{p}^{−1}[K + (G̃_{γ}(s) − G̃_{γ}(ξ))^{\frac{p−1}{p}}][G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}}

≤ −2c_{p} \frac{K}{(p − 1)(φ_{p}^{−1}[K + (L min g̃(ξ, γ)_{ξ∈[s−L,L]})^{\frac{p−1}{p}}])^{p−1}} ∫_{d}^{s−L} \frac{dξ}{[G̃_{γ}(s) − G̃_{γ}(ξ)]^{\frac{1}{p}}}

So

2c_{p}Ĩ_{2} ≤ −\frac{K}{(p − 1)(φ_{p}^{−1}[K + (L min g̃(ξ, γ)_{ξ∈[s−L,L]})^{\frac{p−1}{p}}])^{p−1}} τ̃_{γ}(s)

≤ −\frac{K}{(p − 1)(φ_{p}^{−1}[K + (LA)^{\frac{p−1}{p}}])^{p−1}} τ̃_{γ}(s)

where A is as in (31). Since τ̃_{γ}(s) is uniformly bounded with respect to γ, we have 2c_{p}Ĩ_{2} < ϵ/2 for s large enough and then T̃_{γ}(s) − τ̃_{γ}(s) < ϵ for s sufficiently large. ■

4 Proof of the Theorem.

Let us consider the following parametrized problem

(S_{γ}) = \{ \begin{matrix}−(φ_{p}(u′))′ = (1 − γ)θφ_{p}(u) + γ[f(u) + h(x)]  in (a, b) \\ u = 0  on \{a, b\} \end{matrix}

where γ ∈ [0, 1] and θ is such that λ_{1} < θ < min(μ_{−}, μ_{+}). Notice that the function defined by : (s, γ) ↦ (1 − γ)θφ_{p}(s) + γf(s) is a particular case of the function f̂ and accordingly under the assumptions of Lemma (2.1)(respectively Lemma (2.2)), the conclusions of Lemma (2.1)(respectively Lemma (2.2)) are also valid for (S_{γ}) as well. Under the assumptions (h_{1}), (h_{3}) of the theorem the following lemmas hold.

Lemma 4.1. Under assumption (h_{1}) and the first part of assumption (h_{3}), that is

lim_{|s|→+∞} sgn(s)f(s) = +∞

lim inf_{s→+∞} \frac{pF (s)}{|s|^{p}} = μ,  lim sup_{s→−∞} \frac{pF (s)}{|s|^{p}} = ν

with

1/μ^{1/p} + 1/ν^{1/p} > \frac{b − a}{π_{p}}

(i) there exists a sequence S_{n} → +∞ such that if u solves (S_{γ}) for some γ ∈ [0, 1] and u changes sign, then max u ≠ S_{n} for every n and every γ ∈ [0, 1]. (ii) When (h_{1}) and the second part of (h_{3}) hold, there exists a sequence T_{n} → −∞ such that if u solves (S_{γ}) for some γ ∈ [0, 1] and u changes sign, then min u ≠ T_{n} for every n and every γ ∈ [0, 1].

Proof. We prove only the first statement, the proof of the second one being similar. First let us denote by g the function :(s, γ) ↦ (1 − γ)θφ_{p}(s) + γf(s). According to (h_{1}) lim_{|s|→+∞} sgn(s)g(s, γ) = +∞ uniformly with respect to γ and hence there is

a positive constant c such that sgn(s)g(s, γ) ≥ −c for all s ∈ ℝ and all γ ∈ [0, 1]. Let G̃_{γ} be the primitive such that

G̃_{γ}(s) = \{ \begin{matrix}∫_{0}^{s}(g(t) + c) dt & for s > 0 \\ ∫_{0}^{s}(g(t) − c) dt & for s < 0. \end{matrix}

For such a G̃_{γ} we associate the function H̃ defined by

H̃(x) = \{ \begin{matrix}∫_{a}^{x}(h(t) − c) dt x ∈ [a, b] & when G̃_{γ} operates on [0, +∞[ \\ ∫_{a}^{x}(h(t) + c) dt x ∈ [a, b] & when G̃_{γ} operates on ] − ∞, 0]. \end{matrix}

From the first part of (h_{3}), we have

lim inf_{s→+∞} \frac{pG̃_{γ}(s)}{|s|^{p}} ≤ μ,  (37)
lim sup_{s→−∞} \frac{pG̃_{γ}(s)}{|s|^{p}} ≤ ν  (38)

Choose μ′ > μ such that the pair (μ′, ν) still lies below C_{1}, that is

1/μ′^{1/p} + 1/ν^{1/p} > \frac{b − a}{π_{p}}.

For such a μ′ we have

lim sup_{s→+∞}(μ′ \frac{s^{p}}{p} − G̃_{γ}(s)) = +∞

Hence, there exists an increasing sequence s_{n} → +∞ so that for each n

p(G̃(s_{n}) − G̃(s)) ≤ μ′(s_{n}^{p} − s^{p})  (39)

for all s ∈ [0, s_{n}[ with s_{n} > L = (b − a)φ_{p}^{−1}(2||H̃||_{∞})

Choose S_{n} as a tail sequence of the sequence s_{n} and suppose that with such a S_{n}, Lemma 4.1 is false. Then, one can find a subsequence of s_{n} still denoted by s_{n}, and solutions u_{n} of (S_{γ}) for γ = γ_{n} ∈ [0, 1] satisfying max u_{n} = s_{n} → +∞, and hence according of Lemma (2.2), min u_{n} → −∞. Let’s show below that such a sequence solutions leads to a contradiction. So, let’s consider the real numbers (α_{0}, ρ, ρ̄, β_{0}) and (α′_{0}, ρ′, ρ̄′, β′_{0}) corresponding to the sequence solutions u_{n} as respectively in Lemma (2.1) and in its dual version. For the sake of simplicity we will keep the notations, α_{0}, ρ, ρ̄, β_{0} and α′_{0}, ρ′, ρ̄′, β′_{0}, however those numbers depend on n . Recalling inequality (29), that is (β_{0} − α_{0}) ≥ T_{γ}(max u_{n}) for max u_{n} > L and using Lemma 3.2, we get ∀ϵ > 0, (β_{0} − α_{0}) ≥ τ_{γ}(S_{n}) − ϵ for S_{n} large enough. Next, combining (39) and inequality above, we get

(β_{0} − α_{0}) ≥ \frac{2c_{p}p^{1/p}}{μ′^{1/p}} ∫_{0}^{S_{n}} \frac{dξ}{[S_{n}^{p} − |ξ|^{p}]^{1/p}}

for S_{n} large enough. And finally

lim inf_{n→+∞}(β_{0} − α_{0}) ≥ \frac{π_{p}}{(μ′)^{1/p}}  (40)

To complete the proof, we need to estimate (β′_{0} − α′_{0}). Recalling (30), we have (β′_{0} − α′_{0}) ≥ T_{γ}(min u_{n}) for | min u_{n}| large enough and using again Lemma 3.2, we get

∀ϵ > 0, (β′_{0} − α′_{0}) ≥ τ_{γ}(min u_{n}) − ϵ

for , | min u_{n}| large enough. But, combining condition (38) and results in Lemma (3.1), we obtain

lim inf_{n→+∞}(β′_{0} − α′_{0}) ≥ lim inf_{n→+∞} τ_{γ}(min u_{n}) ≥ \frac{π_{p}}{(ν)^{1/p}}

(41)

So, putting (40), (41) together yields

b − a ≥ lim inf_{n→+∞}(β_{0} − α_{0}) + lim inf_{n→+∞}(β′_{0} − α′_{0}) > b − a.

This is a contradiction so Lemma 4.1 is proved. ■

The following lemma provides a-priori bounds for solutions of (S_{γ}) having a constant sign.

Lemma 4.2. Assume that

lim_{|s|→+∞} sgn(s)f(s) = +∞

and that condition (h_{2}) holds, that is lim inf_{|s|→+∞} \frac{pF (s)}{|s|^{p}} > λ_{1}.

Then, there are two constants K > 0, K′ < 0 such that there is no nonnegative solution or respectively no non-positive solution u of (S_{γ}) for some γ ∈ [0, 1] such that max u ≥ K or respectively min u ≤ K′.

Proof. We give the proof only in the case of non negative solution, the case of non positive solution being similar. Condition (h_{2}) gives

lim inf_{s→+∞} \frac{pG̃_{γ}(s)}{|s|^{p}} = k_{+} > λ_{1}  (42)

and accordingly from Lemma (3.1)

lim sup_{s→+∞} τ_{γ}(s) ≤ \frac{π_{p}}{k_{+}^{1/p}} < b − a.

But since τ̃_{γ}(s) ≤ τ_{γ}(s) for s > L we also have

lim sup_{s→+∞} τ̃_{γ}(s) ≤ \frac{π_{p}}{k_{+}^{1/p}} < b − a.  (43)

Suppose now that Lemma (4.2) is false, then we can find a sequence of non negative solutions (u_{n}) for some γ ∈ [0, 1] such that max u_{n} → +∞. Let us show that such an assertion is absurd. So let us write the length b − a = β_{0} − α_{0} as follows:

b − a = (β_{0} − β_{max u_{n}−L}) + (α_{max u−L} − α_{0}) + (β_{max u_{n}−L} − α_{max u−L}) (44)

Since lim_{s→+∞} sgn(s)g(s, γ) = +∞ uniformly with respect to γ, we have according to Lemma 2.1

lim_{n→+∞}(β_{max u_{n}−L} − α_{max u−L}) = 0  (45)

In order to use (35) for estimating (β_{0} − β_{max u_{n}−L}) + (α_{max u−L} − α_{0}), we will prove here that inequality (26) previously admitted, that is

−2||H̃||_{∞}+(p^{∗})^{\frac{p−1}{p}} (G̃_{γ}(max u)−G̃_{γ}(u(x)))^{\frac{p−1}{p}} > 0 on [α_{0}, α_{max u−L}]∪[β_{max u−L}, β_{0}]

is effectively achieved under (42). Indeed, under (42) we have the following: for ϵ > 0, there is a positive real number s_{0} > 0 such that for s > s_{0}

G̃_{γ}(s) − G̃_{γ}(ξ) ≥ 1/p(k_{+} − ϵ)(s^{p} − ξ^{p}) for 0 < ξ < s.

So for n sufficiently large we have max u_{n} − L > s_{0} and then

G̃_{γ}(max u_{n}) − G̃_{γ}(u_{n}(x)) ≥ 1/p(k_{+} − ϵ)((max u_{n})^{p} − (u_{n}(x))^{p})

for 0 < u_{n}(x) < max u_{n} − L. So, for ϵ sufficiently small, we get

G̃_{γ}(max u_{n}) − G̃_{γ}(u_{n}(x)) > 1/pλ_{1}((max u_{n})^{p} − (u_{n}(x))^{p})  (46)

for 0 < u_{n}(x) < max u_{n} − L. But 0 < u_{n}(x) < max u_{n} − L, implies

(max u_{n})^{p} − u_{n}^{p}(x) ≥ L^{p} with L = (b − a)φ_{p}^{−1}(2||H̃||_{∞})  (47)

Then, taking into account (46) and (47) we have

−2||H̃||_{∞} + (p^{∗})^{\frac{p−1}{p}} (G̃_{γ}(max u) − G̃_{γ}(u(x)))^{\frac{p−1}{p} ≥}

−2||H̃||_{∞} + (\frac{p^{∗}}{p})^{\frac{p−1}{p}} λ_{1}^{\frac{p−1}{p}} L^{p−1} =

−2||H̃||_{∞} + (\frac{1}{p − 1})^{\frac{p−1}{p}} (\frac{π_{p}}{b − a})^{p−1} × (b − a)^{p−1}[φ_{p}^{−1}(2||H̃||_{∞})]^{p−1} =

2||H̃||_{∞} (\frac{(π_{p})^{p−1}}{(p − 1)_{\frac{p−1}{p}}} − 1) .

Noticing that

π_{p} = 2(p − 1)^{1/p} ∫_{0}^{1} \frac{ds}{(1 − s^{p})^{1/p}} ≥ 2(p − 1)^{1/p}

for p > 1, we get

2||H̃||_{∞} (\frac{(π_{p})^{p−1}}{(p − 1)_{\frac{p−1}{p}}} − 1) ≥ 2||H̃||_{∞} (2^{p−1} − 1) > 0

for p > 1. In conclusion, it is proved that

−2||H̃||_{∞} + (p^{∗})^{\frac{p−1}{p}} (G̃_{γ}(max u) − G̃_{γ}(u(x)))^{\frac{p−1}{p}} > 0
on [α_{0}, α_{max u−L}] ∪ [β_{max u−L}, β_{0}]

which is of course inequality (26). Hence, we can estimate (β_{0} − β_{max u_{n}−L}) + (α_{max u−L} − α_{0}) by using (35) and then we have (β_{0} − β_{max u_{n}−L}) + (α_{max u_{n}−L} − α_{0}) ≤ T̃_{γ}(max u_{n}) with K equals −2||H̃||_{∞} in T̃_{γ}. Since inequality (26) implies condition (c_{1}) of Lemma (4.1) with K = −2||H̃||_{∞}, we have

(β_{0} − β_{max u_{n}−L}) + (α_{max u_{n}−L} − α_{0}) ≤ τ̃_{γ}(max u_{n}) + ϵ

for all ϵ > 0 and for max u_{n} large enough. Thus, combining (43),(44),(45) with the above inequality, one obtains

b − a ≤ lim sup_{n→+∞} τ̃_{γ}(max u_{n}) + ϵ < b − a + ϵ

for all ϵ > 0. This is a contradiction and then Lemma (4.2) is proved. ■

We are now ready to introduce the functional analysis framework in which invariance of topological degree property will be used to conclude the proof of the theorem. So, let us denote by

L : L^{1}(a, b) → C^{1}[a, b]

the operator which sends l ∈ L^{1}(a, b) on the unique solution of

(E_{1}) = \{ \begin{matrix}−(φ_{p}(u′))′ = l in (a, b) \\ u(a) = u(b) = 0 \end{matrix}

It is known that L is an odd and continuous operator (see [6]) and due to the compact embedding

J : C^{1}[a, b] → C^{0}[a, b], J ∘ L : L^{1}(a, b) → C^{0}[a, b]

is completely continuous. Moreover, for each γ ∈ [0, 1], denote by K_{γ} : C^{0}[a, b] → L^{1}(a, b) the operator defined by K_{γ}(u) = g(., u, γ) with K_{γ}(u)(x) = g(x, u, γ) = (1 − γ)θφ_{p}(u) + γ[f(u) + h(x)]. Clearly K_{γ} is continuous and map bounded sets into bounded sets, hence for each γ ∈ [0, 1], the operator

T_{γ} = J ∘ L ∘ K_{γ} : C^{0}[a, b] → C^{0}[a, b]

is completely continuous and its fixed points are exactly the solutions of (S_{γ}). Moreover,for γ = 0, K_{0} is an odd operator . Now let us built a suitable open bounded subset Ω of C^{0}[a, b] on which the degree of T_{γ}, γ ∈ [0, 1] is different of zero. The construction of Ω involves some constants provided by the different Lemmas (2.2), (4.1) (4.2). We consider first our theorem in the case that conditions (h_{1}), (h_{2}) and the first part of (h_{3}) are satisfied. So, choose a constant K according to Lemma (4.2). Next, for n large enough, choose an element S_{n} denoted S of the sequence (S_{n}) such that S > K. For any possible changing sign solution u of (S_{γ}) for some γ ∈ [0, 1] such that max u < S, Lemma (2.2) provides with positive real numbers such that min u > −M. Take −R = min(−M, K′) for a fixed M where K′ < 0 is chosen according to Lemma (4.2).

Ω = {u ∈ C^{0}[a, b], −R < u(x) < S, ∀x ∈ [a, b]}

The set Ω is such that T_{γ}(u) ≠ u for every u ∈ ∂Ω and γ ∈ [0, 1]. Hence, by the homotopy invariance of the topological degree

deg(I − T_{1}, Ω, 0) = deg(I − T_{0}, Ω, 0)

where I is the identity operator in C^{0}[a, b]. Claim 2. deg(I − T_{0}, Ω, 0) ≠ 0

Proof By definition of T_{γ}, γ ∈ [0, 1], u − T_{0}u = 0 if only if u is a solution of

(E_{2}) = \{ \begin{matrix}−(φ_{p}(u′))′ = θφ_{p}(u) in (a, b) \\ u(a) = u(b) = 0 \end{matrix}

But since λ_{1} < θ < min(μ, ν), u = 0 is the unique solution of (E_{2}). Moreover T_{0} is odd, therefore by the Borsuk theorem

deg(I − T_{0}, B_{r}, 0) ≠ 0

where B_{r} is the open ball of center O and raduis r in C^{0}[a, b]. Take r such that B_{r} ⊂ Ω, then 0 ∉ (I − T_{0})^{−1}(Ω̄ ∖ B_{r}) and from the excision property of the degree, we have deg(I − T_{0}, Ω, 0) = deg(I − T_{0}, B_{r}, 0) ≠ 0. ■

Consequently deg(I − T_{1}, Ω, 0) ≠ 0 and by the existence property of the topological degree T_{1} has a fixed point in Ω which in turn is precisely a solution of (P). In the context that it is the conditions (h_{1}), (h_{2}) and the second part of (h_{3}) which are satisfied, we construct our Ω with parameters S and R as follow: we fix −R = T_{n} with T_{n} < K′ < 0 for n large enough, where T_{n} is as in Lemma (4.1). Next, we choose S = max(M, K) for a fixed M where M is as in Lemma (2.2). A similar argument of topological degree as above yields again the solvability of (P) in this latter case. This completes the proof of the theorem. ■ Remark 4.1. It is worth noticing that the establishment of Lemma (2.1) does not involve the boundary conditions so that Lemma (2.1) can be usefully employed in dealing with other boundary conditions. The time-mapping estimates in Lemma (3.1) and the auxiliary functions estimates in Lemma (3.2) and Lemma (3.3) are many tools which can be combined with others in order to extend to the onedimensional p- Laplacian (p > 1), many others results previously obtained for the Laplacian.

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Acknowledgments: This work was carried out while the author was visiting the department of Mathematics of Université Libre de Bruxelles (ULB). The author thanks the department for its hospitality and Professor J.P. GOSSEZ for some very instructive discussions.

(Received August 7, 2008)