In this paper, the authors give a new version of the sub-super solution method and prove the existence of positive solution for a (p, q)-Laplacian system under weak assumptions than usually made in such systems. In particular, nonlinearities need not be monotone or positive.
The authors prove that the sub-super solution method can be proved by the Shcauder fixed-point theorem and use the method to prove the existence of a positive solution in elliptic systems, which appear in some problems of population dynamics.
The results complement and generalize some results already published for similar problems.
The result is completely new and does not appear elsewhere and will be a reference for this line of research.
1. Introduction
Consider the following (p1, p2)-Laplacian system,
Ω is an open bounded domain of with smooth boundary ∂Ω. For i = 1, 2, is the pi-Laplacian operator, pi > 1, μi is a positive parameter and is a continuous function.
Many authors have been interested by the problem (1) in different ways [1–4]. The sub-super solution method, given in [5] by using a monotony argument, is the principal tool used to prove the existence of solution of the problem (1) in [1, 3, 4]. Recently, a new version of the method of the sub-super solution is given to prove the existence of solution for the (p(x), q(x))-Laplacian systems by using the Schaefer’s fixed-point theorem [6].
Our main contribution in this article is, in first, to give a new version of the sub-super solution method based on Schauder’s famous fixed-point theorem and, in second, use the method to prove the existence of a positive solution of problem (1) under the continuity assumptions on functions F and G. The functions F and G need not to be nondecreasing as in [1, 4].
Recall that the sub-super solution method is a topological method, which does not require strong regularity assumptions as the variational method.
2. Preliminaries and main results
We start by the definition of sub-super solution of the problem (1).
We say that is a pair of sub-super solution of the problem (1) if they satisfy
a.e in Ω and on ∂Ω for i = 1, 2.
, ,
, .
[u, v] = {z : u(x) ≤ z(x) ≤ v(x), a.e. ∈ Ω}.
For i = 1, 2, assume that Fi is continuous in . Then, if there exists a pair of sub-super solution of (1) in the sense of Definition (2.1), system (1) has a positive weak solution .
Consider, for i = 1, 2, the truncation operators, defined by
Let be the Nemytskii operator defined by
Then, is -bounded. By the dominate convergence theorem and the continuity of Fi, we conclude the continuity of , and we have .
Now, fix , there exists a unique pair solution of the problem,
Therefore, we can define the operator by S(z1, z2) = (w1, w2), where (w1, w2) is the unique solution of problem (3).
S is a compact operator. Indeed, let (z1,n, z2,n) be a bounded sequence in and (w1,n, w2,n) = S(z1,n, z2,n), then
If the test function ϕi = wi,n, by the Sobolev embedding theorem, there exists some constants Ki such tat
Then, (w1,n, w2,n) is bounded in . By the compact embedding, there exists a convergent sub-sequence of (w1,n, w2,n) in . So, S is compact.
From (4), there exists Li > 0 such that
for some . By the Schauder fixed-point theorem, in , there exists a unique such that S(u1, u2) = (u1, u2).
Finally, (u1, u2) is a solution of problem (1) if, and only if, T1(u1) = u1 and T2(u2) = u2, which means that and . We need to prove that , , and . Let us prove, for example, that . The same argument works for the others cases.
Let . Since is a sub-solution, then for and v = T2(u2), we have,
and as (u1, u2) is a solution of (6),
Then,
Remark that in Ω+, . So,
Therefore, by the monotonicity of the p1-Laplacian, in Ω+. Then, in Ω. In the same way, we get in Ω. Then, and and T2(u2) = u2. Finally, (u1, u2) is a solution of the problem (1). □
3. Applications
Consider system (1) and assume that for i = 1, 2,
A.1 ∃ Ci, αi, βi > 0, such that ,
A.2 F1(x, 0, 0) + F2 (x, 0, 0) > 0 a.e. x ∈ Ω.
Then,
If max(αi, βi) < pi − 1, for i = 1, 2, then ∀μi > 0, there exists a weak positive solution of problem (1).
If min(αi, βi) ≥ pi − 1, for i = 1, 2, then, there exists positive numbers such that the problem (1) has at least a weak positive solution (u1, u2) such that ‖ui‖∞ ≤ ‖ei‖∞. ei is the unique solution of problem (7).
If αi + βi < pi − 1 and αj + βj ≥ pj − 1. j = 2 if i = 1 and j = 1 if i = 2. Then there exists such that problem (1) has at least a weak positive solution ∀μi > 0 and .
By A.2, (0, 0) is a sub-solution, but not a solution, of problem (1). By Theorem 2.1, we need to find a super solution of problem. Let ei be the unique positive solution of the Dirichlet boundary condition problem,
In the sub-linear case, as max(αi, βi) < pi − 1, for i = 1, 2, there exists K > 1, large enough, such that
In the super-linear case, for i = 1, 2, min(αi, βi) ≥ pi − 1. Put and let . Then,
(e1, e2) is a super-solution, and we deduce that the problem (1) admits a weak positive solution (u1, u2), such that 0 < ‖ui‖∞ ≤ ‖ei‖∞.Consider the sub-super linear case, 0 < α1 + β1 < p1 − 1 and α2 + β2 ≥ p2 − 1 and put . Let μ1 > 0 and . Then, there exists K, large enough, such that
So, (Ke1, e2) is as super solution of problem (P).
By the same argument, if α1 + β1 ≥ p1 − 1 and 0 < α2 + β2 < p2 − 1. Put . For all and μ2 > 0, there exists K, large enough, such that
So, (e1, Ke2) is as super solution; hence, the problem (1) admits a weak positive solution (u1, u2). The proof of Theorem 3.1 is complete.
□
The second point of Theorem 3.1 is of great importance because no restriction was made on the growth of nonlinearities but only on the parameter μi which must not be large.
The hypothesis A.2 plays an essential role in the way that we need only to find a super-solution. In what follows, we provide an example without the hypothesis A.2 of Theorem (3.1). We will see that the assumptions on nonlinearities become more restrictive.
For i = 1, 2, assume that ∃ 0 < ci ≤ Ci, 0 < αi, βi < pi − 1, ,
Then, ∀μi > 0, there exists a weak positive solution of the problem (1).
According to the proof of Theorem 3.1, there exists K, large enough, such that (Ke1, Ke2) is a super-solution of the problem (1). Then, we need only to find a sub-solution. Let , ϕi is the principal eigenfunction (positive) associated with the principal eigenvalue of pi-Laplacian operator such that ‖ϕi‖∞ = 1,
is as sub-solution of the problem (1). We need to have
As ‖ϕi‖∞ = 1, it is enough to have (δ1 = α1 and δ2 = β2). This is possible for ɛ, small enough, ɛ < 1 and for all μi > 0. To end the proof, we need to verify that for a small ɛ and K large. By the maximum principle, we have
□
4. Examples
In this section, we use Theorem 3.1 and solve some elliptic (p1, p2)-Laplacian systems studied in some published articles see [7].
Consider the following (p1, p2)-Laplacian system
4.1 The case where (0, 0) is a sub-solution
Assume, for i = 1, 2, that ai(x) and bi(x) are continuous and nonnegative in , a1 or a2 not identically null. Then, (0, 0) is a sub-solution of problem (P). Taking into account Theorem 3.1 and its proof, we get the following propositions,
Problem (P) has a positive weak solution provided that for i = 1, 2,
μi > 0 if 0 < αi + βi < pi − 1 (The sub-linear case),
if αi + βi ≥ pi − 1. Ci = max(‖ai‖∞, ‖bi‖∞) (The super linear case) and
μi > 0 and if αi + βi < pi − 1 and αj + βj ≥ pj − 1. j = 2 if i = 1 and j = 1 if i = 2 (The sub-super linear case).
We have to look for a super solution of our problem.
- (1)
Since 0 < αi + βi < pi − 1, for ∀μi > 0, there exists K such that for i = 1, 2, . Then, we have
In the same way, we show that , ∀u ∈ [0, Ke1]. So, (Ke1, Ke2) is a super solution of problem (P), and then the problem has a weak positive solution.
- (2)
Consider the super linear case, for i = 1, 2, αi + βi ≥ pi − 1. Let , (e1, e2) is a super solution of problem (P). Indeed, we have
In the same way, we obtain , ∀u ∈ [0, e1]. We conclude that problem (P) has a positive weak solution in [0, e1] × [0, e2].
- (3)
Finally, the third point, the sub-super linear case, follows from the two previous ones.
□
4.2 The case where (0, 0) is a trivial solution
We examine only the case where p1 = p2 = p. (0, 0) is a trivial solution of (P) if ai, i = 1, 2, are identically null. Our goal is to find a positive solution of problem (P). To do this, we need to find a pair of positive sub-super solution. Nevertheless, we have to add some more assumptions. Assume that, for i = 1, 2, bi is positive continuous in . So, ci ≤ ‖bi‖∞ ≤ Ci for some positive constants ci and Ci. The problem becomes
Assume that, for i = 1, 2, 0 < αi + βi < p − 1. Then, the problem (P1) has a positive weak solution ∀μi > 0.
According to Theorem 2.1, we need to find a pair of sub-super solution of the problem (P1). Assume that for i = 1, 2, 0 < αi + βi < p − 1, then ∀μi > 0, we can choose 0 < ɛ < 1, such that for i = 1, 2. Fix such ɛ and choose such that for i = 1, 2. Then, (ɛϕ1, ɛϕ1) and (Ke1, Ke1) is a pair of sub-super linear solution of the problem (P1) (‖ϕ1‖∞ = 1). Indeed, we have, by the maximum principle, ɛϕ1 ≤ Ke1 because
∀(u, v) ∈ [ɛϕ1, Ke1] × [ɛϕ1, Ke1], we have
and
The problem (P1) has a weak positive solution in the set [ɛϕ1, Ke1] × [ɛϕ1, Ke1]. The proof is complete. □
