Open Access

Sub-super solutions for (p-q) Laplacian systems

Boundary Value Problems20112011:52

DOI: 10.1186/1687-2770-2011-52

Received: 13 August 2011

Accepted: 2 December 2011

Published: 2 December 2011

Abstract

In this work, we consider the system:

{ - Δ p u = λ [ g ( x ) a ( u ) + f ( v ) ] in Ω - Δ q v = λ [ g ( x ) b ( v ) + h ( u ) ] in Ω u = v = 0 on Ω , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equa_HTML.gif

where Ω is a bounded region in R N with smooth boundary ∂Ω, Δ p is the p-Laplacian operator defined by Δ p u = div (|u|p-2u), p, q > 1 and g (x) is a C1 sign-changing the weight function, that maybe negative near the boundary. f, h, a, b are C1 non-decreasing functions satisfying a(0) ≥ 0, b(0) ≥ 0. Using the method of sub-super solutions, we prove the existence of weak solution.

1 Content

In this paper, we study the existence of positive weak solution for the following system:
{ - Δ p u = λ [ g ( x ) a ( u ) + f ( v ) ] in Ω - Δ q v = λ [ g ( x ) b ( v ) + h ( u ) ] in Ω u = v = 0 on Ω , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equa_HTML.gif
(1)

where Ω is a bounded region in R N with smooth boundary ∂Ω, Δ p is the p-Laplacian operator defined by Δ p u = div(|u|p-2u), p, q > 1 and g(x) is a C1 sign-changing the weight function, that maybe negative near the boundary. f, h, a, b are C1 non-decreasing functions satisfying a(0) ≥ 0, b(0) ≥ 0.

This paper is motivated by results in [15]. We shall show the system (1) with sign-changing weight functions has at least one solution.

2 Preliminaries

In this article, we use the following hypotheses:

(Al) lim f M ( h ( s ) ) 1 q - 1 s p - 1 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_IEq1_HTML.gif as s → ∞, M > 0

(A2) lim f (s) = lim h (s) = ∞ as s → ∞.

(A3) lim a ( s ) s p - 1 = l i m b ( s ) s q - 1 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_IEq2_HTML.gif as s → ∞.

Let λ p , λ q be the first eigenvalue of -Δ p , -Δ q with Dirichlet boundary conditions and φ p , φ q be the corresponding positive eigenfunctions with ||φ p || = ||φ q || = 1.

Let m, δ, γ, μ p , μ q > 0 be such that
| φ p | p - λ p φ p m in Ω ¯ δ φ p μ p on Ω - Ω δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equ1_HTML.gif
(2)
and
{ | φ q | q - λ q φ q m in Ω ¯ δ φ p μ p on Ω - Ω δ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equ2_HTML.gif
(3)
Ω ¯ δ = { x Ω ; d ( x , Ω ) δ } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equb_HTML.gif
We assume that the weight function g(x) take negative values in Ω δ , but it requires to be strictly positive in Ω-Ω δ . To be precise, we assume that there exist positive constants β and η such that g(x) ≥-β on Ω ¯ δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_IEq3_HTML.gif and g(x) ≥ η on Ω-Ω δ . Let s0 ≥ 0 such that ηa(s) + f (s) > 0, ηb(s) + h(s) > 0 for s > s0 and
f 0 = max { 0 , - f ( 0 ) } , h 0 = max { 0 , - h ( 0 ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equc_HTML.gif
For γ such that γr-1t > s0; t = min {α p , α q }, r = min{p, q} we define
A = max [ γ λ p η a ( γ 1 p - 1 α p ) + f ( γ 1 q - 1 α q ) , γ λ q η b ( γ 1 q - 1 α q ) + h ( γ 1 p - 1 α p ) ] B = min [ m γ β a ( γ 1 p - 1 ) + f 0 , m γ β b ( γ 1 q - 1 ) + h 0 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equd_HTML.gif

where α p = p - 1 p μ p p p - 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_IEq4_HTML.gif and α q = q - 1 q μ q q q - 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_IEq5_HTML.gif.

We use the following lemma to prove our main results.

Lemma 1.1[6]. Suppose there exist sub and supersolutions (ψ1, ψ2) and (z1, z2) respectively of (1) such that (ψ1, ψ2) ≤ (z1, z2). then (1) has a solution (u, v) such that (u, v) [(ψ1, ψ2), (z1, z2)].

3 Main result

Theorem 3.1 Suppose that (A1)-(A3) hold, then for every λ [A, B], system (1) has at least one positive solution.

Proof of Theorem 3.1 We shall verify that (ψ1, ψ2) is a sub solution of (1.1) where
ψ 1 = γ 1 p - 1 p - 1 p φ p p p - 1 ψ 2 = γ 1 q - 1 q - 1 q φ q q q - 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Eque_HTML.gif
Let W H 0 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_IEq6_HTML.gif with w ≥ 0. Then
Ω | ψ 1 | p - 2 ψ 1 w d x = γ Ω ( λ p φ p p - | φ p | p ) w d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equ3_HTML.gif
(4)
Now, on Ω ¯ δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_IEq7_HTML.gif by (2),(3) we have
γ ( λ p φ p p - | φ p | p ) - m γ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equf_HTML.gif
Since λB then
λ m γ β a ( γ 1 p - 1 ) + f 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equg_HTML.gif
thus
γ ( λ p φ p p - | φ p | p ) - m γ λ - β a ( γ 1 p - 1 ) - f 0 λ g ( x ) a ( γ 1 p - 1 ) - f 0 λ g ( x ) a p - 1 p γ 1 p - 1 φ p 1 p - 1 + f q - 1 q γ 1 q - 1 φ q 1 q - 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equh_HTML.gif
then by (4)
Ω ¯ δ | ψ 1 | p - 2 ψ 1 w d x Ω ¯ δ λ g ( x ) a p - 1 p γ 1 p - 1 φ p p p - 1 + f q - 1 q γ 1 q - 1 φ q q q - 1 w d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equi_HTML.gif
A similar argument shows that
Ω ¯ δ | ψ 2 | q - 2 ψ 2 w d x Ω ¯ δ λ g ( x ) b q - 1 q γ 1 q - 1 φ q 1 q - 1 + h p - 1 p γ 1 p - 1 φ p 1 p - 1 w d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equj_HTML.gif
Next, on Ω - Ω ¯ δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_IEq8_HTML.gif. Since λA, then
λ γ λ p η a γ 1 p - 1 α p + f γ 1 q - 1 α q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equk_HTML.gif
so we have
γ ( λ p φ p p - | φ p | p ) γ λ p λ η a γ 1 p - 1 α p + f γ 1 q - 1 α q λ [ g ( x ) a ( ψ 1 ) + f ( ψ 2 ) ] , Ω - Ω δ ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equl_HTML.gif
Then by (4) on we have
- Δ p ψ 1 λ [ g ( x ) a ( ψ 1 ) + f ( ψ 2 ) ] on Ω - Ω δ ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equm_HTML.gif
A similar argument shows that
- Δ q ψ 2 λ [ g ( x ) b ( ψ 2 ) + h ( ψ 1 ) ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equn_HTML.gif
We suppose that κ p and κ q be solutions of
- Δ p κ p = 1 in Ω κ p = 0 on Ω - Δ q κ q = 1 in Ω κ q = 0 on Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equo_HTML.gif

respectively, and μ' p = ||κ p ||κ, ||κ q ||κ = μ' q .

Let
( z 1 , z 2 ) = c μ p λ 1 p - 1 κ p , 2 h c λ 1 q - 1 1 q - 1 λ 1 q - 1 κ q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equp_HTML.gif

Let W H 0 1 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_IEq9_HTML.gif with w ≥ 0.

For sufficient C large
μ p p - 1 [ | | g | | a ( C λ 1 p - 1 ) + f ( ( 2 h ( C λ 1 p - 1 ) ) 1 q - 1 λ 1 q - 1 μ q ) ] C p - 1 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equq_HTML.gif
then
| z 1 | p - 2 z 1 w d x = λ C μ p p - 1 w d x λ | | g | | a ( C λ 1 p - 1 ) + f ( 2 h ( C λ 1 p - 1 ) ) 1 q - 1 λ 1 q - 1 μ q d x λ g ( x ) a ( C λ 1 p - 1 κ p μ p ) + f ( 2 h ( C λ 1 p - 1 ) ) 1 q - 1 λ 1 q - 1 κ q d x = [ g ( x ) a ( z 1 ) + f ( z 2 ) ] w d x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equr_HTML.gif
Similarly, choosing C large so that
| | g | | b 2 h C λ 1 p - 1 1 q - 1 λ 1 q - 1 μ q h C λ 1 p - 1 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equs_HTML.gif
then
| z 2 | q - 2 z 2 w d x = 2 λ h C λ 1 p - 1 w d x λ | | g | | b ( z 2 ) + h ( z 1 ) w d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-52/MediaObjects/13661_2011_Article_97_Equt_HTML.gif

Hence by Lemma (1.1), there exist a positive solution (u, v) of (1) such that (ψ1, ψ2) ≤ (u, v) ≤ (z1, z2).

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Science and Research Branch, Islamic Azad University
(2)
Department of Mathematics, Faculty of Mathematical Sciences University of Mazandaran

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Copyright

© Haghaiegh and Afrouzi; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.