Existence of Minimal and Maximal Solutions for a Quasilinear Elliptic Equation With Integral Boundary Conditions

ELECTRONIC JOURNAL OF QUALITATIVE THEORY OF DIFFERENTIAL EQUATIONS, ISSN : 0927-0256, DOI : 10.1016/j.commatsci.2010.10.025, Issue : 3, Volume : 50, pp. 880–885, January 2011.


Introduction
This work is concerned with the construction of the minimal and the maximal solutions of the following nonlinear boundary value problem where ϕ p (y) = |y| p−2 y, p > 1, f : [0, 1] × R 2 → R and g i : [0, 1] → R + are a continuous functions (i = 1, 2 ) and a 0 and a 1 are two positive real numbers.
It is well know that the method of upper and lower solutions coupled with monotone iterative technique has been used to prove existence of solutions of nonlinear boundary value problems by various authors ( see [3], [6], [9], [16] and [17]).
The purpose of this work is to show that it can be applied successfully to problems with integral boundary conditions of type (1).Our results improve and generalize those obtained in [6], [7] and [9].
The plan of this paper is as follows: In section 2, we give some preliminary results that will be used throughout the paper.In section 3, we state and prove our main result.Finally in section 4, we give an example to illustrate our results.

Preliminary results
In this section, we give some preliminary results that will be used in the remainder of this paper.
We consider the following problem where F : [0, 1] × R → R is a continuous function, h : [0, 1] × R → R is a continuous function and strictly increasing in its second variable, a 2 and a 3 are a positive real numbers, a ∈ R and b ∈ R.

Lemma 1 (Weak comparison principle).
Let u 1 , u 2 are such that Proof.Assume that there exists A similar argument holds if x 0 = 1.If x 0 ∈ (0, 1), we have then since ϕ p is strictly increasing, we obtain that But on this point, we have Which means that, Since u 1 (x 0 ) > u 2 (x 0 ) and the function h is strictly increasing in its second variable, we obtain that Which is a contradiction.

Definition 2.1:
We say that α is a lower solution of (2 Now, if moreover F is a bounded function, then we have the following result. Theorem 2 Suppose that α and β are lower and upper solutions of problem (2) Proof.Using a proof similar to that of Theorem 1 in [25], we can prove that the problem (2) admits at least one solution and by Lemma 1, it follows that this problem admits a unique solution.Now, we consider the following problem where f : [0, 1] × R 2 → R is a continuous function, g i : [0, 1] → R + are a continuous functions (i = 0, 1 ) and a 0 and a 1 are two positive real numbers.
Definition 2.6: Wintner condition relative to the pair u and u, if there exist C ≥ 0 and a functions for all (x, u, v) ∈ D, where We have the following result 4) and ( 5) in D. Then there exists a constant K > 0, such that every solution of problem (3 By the mean value theorem, there exists x 0 ∈ (0, 1) such that EJQTDE, 2011 No. 6, p. 5 We put by definition Now, we are going to prove that Suppose, on the contrary that there exists Then by the continuity of u ′ , we can choose x 2 ∈ [0, 1] verifying one of the following situations: Assume that the case i) holds.The others can be handled in a similar way.Since u is a solution of the problem (3) and by the Nagumo-Wintner condition (4), we have , for all x ∈ (x 0 , x 2 ) .(7) Since L ≤ η and ϕ p is increasing, we have Now if we put s = ϕ p (u ′ (x)), we obtain that EJQTDE, 2011 No. 6, p. 6 Then by ( 7) and ( 8), it follows that Which a contradiction with (6).

Main result
In this section, we state and prove our main result.
On the nonlinearity f , we shall impose the following condition: H) There exists a continuous function h : R → R strictly increasing such that s → f (x, s, z) + h (s) is increasing for all x ∈ [0, 1] and all z ∈ R.
The main result of this work is: Theorem 4 Let u and u be a lower and upper solution solution respectively for problem (3) and such that u ≤ u in [0, 1] .Assume that H) is satisfied and the Nagumo-Wintner conditions relative to u and u holds.Then the problem (3) has a maximal solution u * and a minimal solution u * such that for every solution u of (3 For the proof of this theorem, we need a preliminary lemma.Let w, w Let δ (v) := max (−K, min (v, K)), for all v ∈ R, where K is the constant defined in the proof of lemma 3. Then the function δ is continuous and bounded.
In fact, we have δ (v) = v for all v such that |v| ≤ K; and |δ (v)| ≤ K for all v ∈ R.
We consider the following problems: and Lemma 5 Let w and w be a lower and upper solution respectively for problem (3).Assume that H) is satisfied and the Nagumo-Wintner conditions relative to u and u holds.Then there exists a unique solution u and u of ( 9) and (10) such that u ≤ w ≤ u ≤ u ≤ w ≤ u.

Proof.
The proof will be given in several steps.
Step 1: w is a lower solution of (9).Proof: Let x ∈ (0, 1) , we have This means that, Now since w is a lower solution of (3) and u ≤ w ≤ u in [0, 1], then by using a proof similar to that of lemma 3, we prove that w ′ 0 ≤ K. Hence δ (w ′ ) = w ′ and we obtain that EJQTDE, 2011 No. 6, p. 8 On the other hand, we have That is, Similarly, we have Then by ( 11), ( 12) and ( 13), it follows that w is a lower solution of (9).
Step 2: w is an upper solution of (9).Proof: Let x ∈ (0, 1), we have Now by using a proof similar to that of lemma 3, we prove that w ′ 0 ≤ K. Hence δ (w ′ ) = w ′ and we obtain that Also, we have and Then by ( 14), ( 15) and ( 16), it follows that w is an upper solution of (9).
By Steps 1 and 2 and since the functions (x, is a bounded continuous function and u → h (u) is continuous and strictly increasing, then by theorem 2, it follows the existence of a unique solution u of ( 9) such that w ≤ u ≤ w.
Similarly, we can prove the existence and uniqueness of a solution to (10), which we call u such that w ≤ u ≤ w.EJQTDE, 2011 No. 6, p. 9 Finally, by using a proof similar to that of lemma 1, we prove that u ≤ u in [0, 1] .

Proof. of Theorem 4
The proof will be given in several steps.We take u 0 = u, u 0 = u and define the sequences (u n ) n≥1 , (u n ) n≥1 by Step 1: For all n ∈ N, we have Proof: EJQTDE, 2011 No. 6, p. 10 Since u and u are lower and upper solutions of problem (3), then by lemma 5, it follows that ii) Assume for fixed n > 1, we have and we show that Let x ∈ (0, 1), we have Since u n−1 ≥ u n and using the hypothesis H), we obtain Then by ( 17) and ( 18), it follows that Now by using a proof similar to that of lemma 3, we can prove that n and we obtain that On the other hand, we have That is, and That is, Then by ( 19), ( 20) and ( 21), it follows that u n is an upper solution of (3).
Similarly, we can prove that u n is a lower solution of (3).Then by lemma 5, there exists a unique solution u n+1 and u n+1 of (P n+1 ) and Hence, we have Step 2: The sequence (u n ) n∈N converge to a maximal solution of (3).Proof: By Step 1 and since u ′ n 0 ≤ K, for all n ∈ N, it follows that the sequence (u n ) n∈N is uniformly bounded in C 1 ([0, 1]) .Now let ε 1 > 0 and t, s ∈ [0, 1] such that t < s, then for each n ∈ N, we have where , we obtain p is an increasing homeomorphism from R onto R, we deduce from EJQTDE, 2011 No. 6, p. 12 that the sequence (u ′ n ) n∈N is equicontinuous on [0, 1] .Hence by the Arzéla-Ascoli theorem, there exists a subsequence u nj of (u n ) n∈N which converges in C 1 ([0, 1]) . Let

But by
Step 1 the sequence (u n ) n∈N is decreasing and bounded from below, then the pointwise limit of this sequence exists and it is denoted by u * .Hence, we have u = u * and moreover, the whole sequence converges in C 1 ([0, 1]) to u * .Let x ∈ (0, 1) , we have Now, as n tends to +∞, we obtain that Also, we have Hence, the dominated convergence theorem of Lebesgue implies that Thus, we obtain Also, by the dominated convergence theorem of Lebesgue, we have and EJQTDE, 2011 No. 6, p. 13 By ( 22), ( 23) and (24), it follows that u * is a solution of the following problem Now using a proof similar to that of lemma 3, we prove that u * ≤ K. Hence δ (u * ′ ) = u * ′ and consequently u * is a solution of (3).Now, we prove that if u is another solution of (3 Since u is a lower solution of (3), then by Step 1, we have Letting n → +∞, we obtain that Which means that u * is a maximal solution of problem (3).
Step 3: The sequence (u n ) n∈N converges to a minimal solution of (3).Proof: The proof is similar to that of Step 2, so it is omitted.
The proof of our result is complete.

Application
In this section, we apply the previous result to the following problem where 0 < k 1 < p − 1, k 2 > k 1 , λ 1 , and are a positive real parameters and To study this problem, we need first consider the following problem: where k 3 > 0 and k 3 = p − 1.
It is not difficult to see that the equation G (ρ) = 1 2 admits a unique solution.
Hence the problem (27) admits a unique positive solution.
Theorem 7 Assume that λ 1 > 1, then the problem (26) admits a maximal solution u * and a minimal solution u * .
This implies that the problem (26) admits a maximal solution u * and a minimal solution u * .EJQTDE, 2011 No. 6, p. 16

1 2 .
Defining u via equation (29), we can prove that problem (27) admits a unique positive solution u, with max B (k, l) is the Euler beta function defined by B (k, l) = 1 0