POSITIVE SOLUTIONS FOR A CLASS OF INFINITE ...

LE MATEMATICHE Vol. LXVIII (2013) – Fasc. II, pp. 159–166 doi: 10.4418/2013.68.2.12

POSITIVE SOLUTIONS FOR A CLASS OF INFINITE SEMIPOSITONE PROBLEMS INVOLVING THE p-LAPLACIAN OPERATOR M. CHOUBIN - S. H. RASOULI - M. B. GHAEMI - G. A. AFROUZI

We discuss the existence of a positive solution to the infinite semipositone problem c −∆ p u = au p−1 − buγ − f (u) − α , x ∈ Ω, u = 0, x ∈ ∂ Ω, u where ∆ p is the p-Laplacian operator, p > 1, γ > p−1, α ∈ (0, 1), a, b and c are positive constants, Ω is a bounded domain in RN with smooth boundary ∂ Ω, and f : [0, ∞) → R is a continuous function such that f (u) → ∞ as u → ∞. Also we assume that there exist A > 0 and β > p − 1 such that f (s) ≤ Asβ , for all s ≥ 0. We obtain our result via the method of sub- and supersolutions.

1.

Introduction

We consider the positive solution to the boundary value problem ( c −∆ p u = au p−1 − buγ − f (u) − α , x ∈ Ω, u u = 0, x ∈ ∂ Ω,

(1)

where ∆ p denotes the p-Laplacian operator defined by ∆ p z = div(|∇z| p−2 ∇z), p > 1, γ > p − 1, α ∈ (0, 1), a, b and c are positive constants, Ω is a bounded Entrato in redazione: 29 ottobre 2012 AMS 2010 Subject Classification: 35J25, 35J55. Keywords: Positive solution, Infinite semipositone, Sub and supersolutions.

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M. CHOUBIN - S. H. RASOULI - M. B. GHAEMI - G. A. AFROUZI

domain in RN with smooth boundary ∂ Ω, and f : [0, ∞) → R is a continuous function. We make the following assumptions: (H1) f : [0, +∞) → R is a continuous function such that lims→+∞ f (s) = ∞. (H2) There exist A > 0 and β > p − 1 such that f (s) ≤ Asβ , for all s ≥ 0. In [8], the authors have studied the equation −∆u = g(u) − (c/uα ) with Dirichlet boundary conditions,where g is nonnegative and nondecreasing and limu→∞ g(u) = ∞. The case g(u) := au − f (u) has been study in [7], where f (u) ≥ au − M and f (u) ≤ Auβ on [0, ∞) for some M, A > 0, β > 1 and this g may have a falling zero. In this paper, we study the equation −∆ p u = au p−1 − buγ − f (u) − (c/uα ) with Dirichlet boundary conditions. Our result in this paper include the result of [7] in the case p = 2 (Laplacian operator), where say in Remark 2.2. Let F(u) := au p−1 −buγ − f (u)−(c/uα ), then limu→0+ F(u) = −∞ and hence we refer to (1) as an infinite semipositone problem. In fact, our result in this paper is on infinite semipositone problems involving the p-Laplacian operator. In recent years, there has been considerable progress on the study of semipositione problems (F(0) < 0 but finite) (see [1],[2],[5]). We refer to [6], [7], [8] and [9] for additional results on infinite semipositone problems. We obtain our result via the method of sub- and supersolutions([3]).

2.

The main result

We shall establish the following result. p Theorem 2.1. Let (H1) and (H2) hold. If a > ( p−1+α ) p−1 λ1 , then there exists positive constant c∗ := c∗ (a, A, p, α, β , γ, Ω) such that for c ≤ c∗ , problem (1) has a positive solution, where λ1 be the first eigenvalue of the p-Laplacian operator with Dirichlet boundary conditions.

Remark 2.2. Theorem 2.1 was established in [7] for the case p = 2 (the Laplacian operator), f (u) := g(u) − buγ , where the function g satisfy the following assumptions: • g(u) ≈ buθ for some θ > γ. • There exist A > 0 and β > 1 such that g(u) ≤ Auβ , for all u ≥ 0. • There exist M > 0 such that g(u) ≥ au − M, for all u ≥ 0. In fact, the function f satisfy the hypotheses of Theorem 2.1 in this paper (Since limu→∞ (g(u)/buθ ) = 1, hence limu→∞ f (u) = ∞) and g satisfy the hypotheses of Theorem 2.1 in [7], where (1) changes to equation −∆u = au − g(u) − (c/uα ) with Dirichlet boundary conditions.

POSITIVE SOLUTIONS FOR A CLASS OF INFINITE SEMIPOSITONE . . .

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Proof of Theorem 2.1. We shall establish Theorem 2.1 by constructing positive sub-supersolutions to equation (1). From an anti-maximum principle (see [4, pages 155-156]), there exists σ (Ω) > 0 such that the solution zλ of ( −∆ p z − λ z p−1 = −1, x ∈ Ω, z = 0, x ∈ ∂ Ω, for λ ∈ (λ1 , λ1 + σ ) is positive in Ω and is such that ∂∂νz < 0 on ∂ Ω, where ν is outward normal vector on ∂ Ω. Fix λ ∗ ∈ (λ1 , min{λ1 + σ , ( p−1+α ) p−1 a}) and p let 1

1 n (p/p − 1 + α) p−1 γ−p+1 a − ( p ) p−1 λ ∗ γ−p+1 p−1+α , K := min , γ p−(p−1)(α−1) p(γ−p+1) p−1+α p−1+α 2b kzλ ∗ k∞ 3b kzλ ∗ k∞ 1

(p/p − 1 + α) p−1 β −p+1 a − ( p ) p−1 λ ∗ 1 o β −p+1 p−1+α , β p−(p−1)(α−1) p(β −p+1) 2A kzλ ∗ k∞ p−1+α 3A kzλ ∗ k∞p−1+α p

Define ψ = Kzλp−1+α . Then ∗ ∇ψ = K(

1−α p )zλp−1+α ∇zλ ∗ ∗ p−1+α

and − ∆ p ψ = − div(|∇ψ| p−2 ∇ψ) −α p 1−α p p p−1+α (p−1)(1−α) p−1+α p−1 p−1 = −K ( ) ( p−1+α )zλ ∗ |∇zλ ∗ | + zλ ∗ ∆ p zλ ∗ p−1+α (p−1)(1−α) −α p p p p−1+α p−1+α ∗ p−1 ∗ |∇z | ) p−1 ( (p−1)(1−α) ) z (1 − λ z ) = −K p−1 ( + z λ p−1+α λ∗ λ∗ λ∗ p−1+α    |∇z ∗ | p  p(p−1) (p−1)(1−α) p λ p−1+α (p−1)(1−α) p−1 p−1 ∗ p−1+α =K ( ) λ z ∗ − zλ ∗ − ( p−1+α ) αp  λ  p−1+α zλp−1+α ∗ Let δ > 0, µ > 0, m > 0 be such that |∇zλ ∗ | p ≥ m in Ωδ and zλ ∗ ≥ µ in Ω \ Ωδ , where Ωδ := {x ∈ Ω : d(x, ∂ Ω) ≤ δ }. Let o n p 1 p p−1 ∗ p p ) p−1 ( (p−1)(1−α) ) m , µ a − ( ) λ . c∗ := K p−1+α min ( p−1+α p−1+α 3 p−1+α p Let x ∈ Ωδ and c ≤ c∗ . Since ( p−1+α ) p−1 λ ∗ < a, we have

K p−1 (

p−1 p(p−1) p p ) p−1 λ ∗ zλp−1+α < a Kzλp−1+α . ∗ ∗ p−1+α

(2)

162


From the choice of K , we have γ p−(p−1)(α−1) 1 p ( ) p−1 ≥ bK γ−p+1 kzλ ∗ k∞ p−1+α 2 p−1+α β p−(p−1)(α−1) 1 p ( ) p−1 ≥ AK β −p+1 kzλ ∗ k∞ p−1+α 2 p−1+α

(3) (4)

and by (3),(4),(H2), we know that γ (p−1)(1−α) p p 1 ) p−1 zλ ∗p−1+α ≤ −b Kzλp−1+α − K p−1 ( ∗ 2 p−1+α β (p−1)(1−α) p p p 1 p−1 p−1+α p−1+α p−1+α p−1 − K ( ) zλ ∗ ≤ −A Kzλ ∗ ≤ − f Kzλ ∗ 2 p−1+α Since |∇zλ ∗ | p ≥ m in Ωδ , from the choice of c∗ we have p p p−1 p−1 (p − 1)(1 − α) |∇zλ ∗ | −K ( ) αp p−1+α p−1+α zλp−1+α ∗ (p − 1)(1 − α) m p p ) p−1 ≤ −K p−1 ( αp p−1+α p−1+α zλp−1+α ∗ c ≤ − α . p Kzλp−1+α ∗ Hence for c ≤ c∗ , combining (2), (5), (6) and (7) we have p −∆ p ψ = K p−1 ( ) p−1 p−1+α    p p(p−1) (p−1)(1−α) ∗ |∇z | (p − 1)(1 − α) λ × λ ∗ zλp−1+α − zλ ∗p−1+α − ∗ αp  p−1+α p−1+α  zλ ∗ p(p−1) (p−1)(1−α) p p 1 ) p−1 λ ∗ zλp−1+α ) p−1 zλ ∗p−1+α − K p−1 ( ∗ p−1+α 2 p−1+α (p−1)(1−α) 1 p − K p−1 ( ) p−1 zλ ∗p−1+α 2 p−1+α p (p − 1)(1 − α) |∇zλ ∗ | p − K p−1 ( ) p−1 αp p−1+α p−1+α zλp−1+α ∗ p−1 γ p p p c p−1+α p−1+α − b Kz − f Kz − ≤ a Kzλp−1+α α p ∗ λ∗ λ∗ p−1+α Kzλ ∗ c = aψ p−1 − bψ γ − f (ψ) − α , x ∈ Ωδ . ψ

= K p−1 (

(5) (6)

(7)


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Next in Ω \ Ωδ , for c ≤ c∗ from the choice of c∗ and K, we know that p−1 c 1 p−1 p p ∗ ≤ K zλ ∗ a − ( ) λ , Kα 3 p−1+α

(8)

and p−1 1 p ∗ ≤ a−( ) λ bK z 3 p−1+α p(β −p+1) p−1 p 1 ∗ β −p+1 p−1+α a−( ) λ . ≤ AK zλ ∗ 3 p−1+α γ−p+1

p(γ−p+1) p−1+α λ∗

(9) (10)

By combining (8), (9) and (10) we have p ) p−1 −∆ p ψ = K p−1 ( p−1+α    p p(p−1) (p−1)(1−α) (p − 1)(1 − α) |∇zλ ∗ | × λ ∗ zλp−1+α − zλ ∗p−1+α − ∗ αp   p−1+α zλp−1+α ∗ p(p−1) p ) p−1 λ ∗ zλp−1+α ∗ p−1+α 3 p−1 1 p 1 p−1 ∗ p = αp ∑ K ( ) λ zλ ∗ p−1+α i=1 3 zλp−1+α ∗ p(γ−p+1) 1 n 1 p−1 p c 1 p−1 p γ−p+1 p−1+α ≤ αp K zλ ∗ a − α + K zλ ∗ a − bK zλ ∗ 3 K 3 zλp−1+α ∗ o p(β −p+1) 1 a − AK β −p+1 zλ ∗p−1+α + K p−1 zλp ∗ 3 γp βp pp−1 c ≤ aK p−1 zλp−1+α − bK γ zλp−1+α − AK β zλp−1+α − ∗ ∗ ∗ αp K α zλp−1+α ∗ p−1 γ p p p c ≤ a Kzλp−1+α − b Kzλp−1+α − f Kzλp−1+α − α p ∗ ∗ ∗ Kzλp−1+α ∗ c = aψ p−1 − bψ γ − f (ψ) − α , x ∈ Ω \ Ωδ . ψ

≤ K p−1 (

Thus ψ is a positive subsolution of (1). From (H1) and γ > p − 1, it is obvious that z = M where M is sufficiently large constant is a supersolution of (1) with z ≥ ψ. Thus Theorem 2.1 is proven.

164 3.


An extension to system (11)

In this section, we consider the extension of (1) to the following system:  c1  −∆ p u = a1 u p−1 − b1 uγ − f1 (u) − α , x ∈ Ω,    v  c2 −∆ p v = a2 v p−1 − b2 vγ − f2 (v) − α , x ∈ Ω,  u    u = 0 = v, x ∈ ∂ Ω,

(11)

where ∆ p denotes the p-Laplacian operator, p > 1, γ > p − 1, α ∈ (0, 1), a1 , a2 , b1 , b2 , c1 and c2 are positive constants, Ω is a bounded domain in RN with smooth boundary ∂ Ω, and fi : [0, ∞) → R is a continuous function for i = 1, 2. We make the following assumptions: (H3) fi : [0, +∞) → R is a continuous functions such that lims→+∞ fi (s) = ∞ for i = 1, 2. (H4) There exist A > 0 and β > p − 1 such that fi (s) ≤ Asβ , i = 1, 2, for all s ≥ 0. We prove the following result by finding sub-super solutions to infinite semipositone system (11). p Theorem 3.1. Let (H3) and (H4) hold. If min{a1 , a2 } > ( p−1+α ) p−1 λ1 , then there exists positive constant c∗ := c∗ (a1 , a2 , b1 , b2 , A, p, Ω) such that for

max{c1 , c2 } ≤ c∗ , problem (11) has a positive solution. Proof. Let σ be as in section 2, a˜ = min{a1 , a2 } and b˜ = max{b1 , b2 }. Choose λ ∗ ∈ (λ1 , min{λ1 + σ , ( p−1+α ) p−1 a}). ˜ Define p 1

1 n (p/p − 1 + α) p−1 γ−p+1 a˜ − ( p ) p−1 λ ∗ γ−p+1 p−1+α K := min , , γ p−(p−1)(α−1) p(γ−p+1) p−1+α p−1+α ˜ ˜ 2b kzλ ∗ k∞ 3b kzλ ∗ k∞ 1

(p/p − 1 + α) p−1 β −p+1 a˜ − ( p ) p−1 λ ∗ 1 o β −p+1 p−1+α , , β p−(p−1)(α−1) p(β −p+1) p−1+α p−1+α 2A kzλ ∗ k∞ 3A kzλ ∗ k∞ and n o p−1 ∗ p p p , 1 µ p a˜ − ( c∗ := K p−1+α min ( p−1+α ) p−1 (p−1)(1−α) m ) λ . p−1+α 3 p−1+α By the same argument as in the proof of theorem 2.1, we can show that (ψ1 , ψ2 ) p

p

:= (Kzλp−1+α , Kzλp−1+α ) is a positive subsolution of (11) for max{c1 , c2 } ≤ c∗ . ∗ ∗


165

Also it is easy to check that constant function (z1 , z2 ) := (M, M) is a supersolution of (11) for M large. Further M can be chosen large enough so that (z1 , z2 ) ≥ (ψ1 , ψ2 ) on Ω. Hence for max{c1 , c2 } ≤ c∗ , (11) has a positive solution the proof is complete.

Acknowledgements The authors are extremely grateful to the referees for their helpful suggestions for the improvement of the paper.

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M. CHOUBIN Department of Mathematics Faculty of Basic Sciences Payame Noor University Tehran, Iran e-mail: [email protected] SAYYYED H. RASOULI Department of Mathematics Faculty of Basic Sciences Babol University of Technology Babol, Iran e-mail: [email protected] MOHAMMAD B. GHAEMI Department of Mathematics Iran University of Science and Technology Narmak, Tehran, Iran e-mail: [email protected] GHASEM A. AFROUZI Department of Mathematics Faculty of Basic Sciences Mazandaran University Babolsar, Iran e-mail: [email protected]