EXISTENCE OF RADIAL POSITIVE SOLUTIONS FOR A QUSILINEAR NON-POSITONE PROBLEM IN A BALL

Abstract

In this paper, we prove existence of radial positive solutions for the following boundary value problem

1. Introduction

Let us consider the the existence of radial positive solutions of the problem

where λ > 0, Ω denotes a ball in RN ; f (0) < 0, f has more than one zero and is not strictly increasing entirely on [0,∞). △pu = div(|∇u|p−2∇u) (1 < p ≤ N) is the p-Laplacian operator of u.

The problem (1.1) arises in the theory of quasiregular and quasiconformal mappings or in the study of non-Newtonian fluids. In the latter case, the quantity p is a characteristic of the medium.Media with p > 2 are called dilatant fluids and these with p < 2 are called pseudoplastics(see[18,19]). If p = 2, they are Newtonian fluid. When p ≠ 2, the problem becomes more complicated certain nice properties inherent to the case p = 2 seem to be lost or at least difficult to verify. The main differences between p = 2 and p ≠ 2 can be founded in [9,11].

In recent years, the asymptotic behavior, existence and uniqueness of the positive solutions for the quasilinear eigenvalue problems:

where λ > 0; p > 1;Ω ∈ RN, N ≥ 2 have been considered by a number of authors, see [5-15,20-24,26-28] and the references therein. In [11], Guo and Webb proved existence and uniqueness results of (1.2) for λ large when f ≥ 0, (f(x)=xp−1)′ < 0 for x > 0 and f satisfies some p-sublinearity conditions at 0 and ∞, generalizing a result in [11] where Ω is a ball. When p = 2, uniqueness results for semilinear equations were obtained in [29,30] where the assumption (f(x)=x)′ < 0 is required only for large x. Similar results for systems were discussed in [31]. Related results for the superlinear case when f ≥ 0 can be found in [26,32]. When p = 2, f(0) < 0, f(s) has only one zero and Ω being a unit ball or an annulus in RN the related results have been obtained by Castro and Shivaji [2], Arcoya and Zertiti[1]. The case when f(0) < 0 and p = 2 was treated in [33], in which uniqueness of positive solution to single equation of (1.1) for λ large was established for sublinear f. See also [34] where this result was extended to the case when Ω is any bounded domain with convex outer boundary.

In this paper, we study this problem for p ≠ 2, f(0) < 0 and Ω being a unit ball in RN. It extends and complements previous results in the literature [1].

The paper is organized as follows. In section 2, we recall some facts that will be needed in the paper and give the main results. In section 3, we give the proofs of the main results in this paper.

 

2. Main results

We consider radial solution of (1.1), then, the existence of radial positive solutions of (1.1) is equivalent to the existence of positive solutions of the problem

where Ω is the unit ball of RN and λ > 0. Here f : [0,+∞) → R satisfies the following assumptions:

(H1) f ∈ C1([0,+∞),R) such that f′ ≥ 0 on [β,+∞), where β is the greatest zero of f;

(H2) f(0) < 0;

(H3) where p−1 < q < p∗−1, p∗ = ∞ if p ≥ N;

(H4) For some k ∈ (0, 1), where

Remark 2.1. We note that in hypothesis (H1), there is no restriction on the function f(u) for 0 < u < β.

Remark 2.2. If f satisfies (H1), any nonnegative solution u of (2.1) is positive in Ω, radial symmetric and radially decreasing, that is

By a modification of the method given in [1], we obtain the following results.

Theorem 2.1. Let assumptions (H1)-(H4) be satisfied. Then there exists a positive real number λ0 such that if λ ∈ [0, λ0], problem (1.1) has at least one radial positive solution which is decreasing on [0,1].

The proof of the theorem is based on the following preliminaries and four Lemmas.

Lemma 2.2. Let u(r) be a solution of (2.1) in (r1,r2) ⊂ (0,∞) and let a be an arbitrary constant, then for each r ∈ (r1,r2) we have

Remark 2.3. The identity of Pohozaev type was obtained by Ni and Serrin [6].

By a modification of the method given in [1], we first introduce the notations and the following preliminaries. Let F be defined as and θ denotes the greatest zero of F.

From (H4), we have such that

Given d ∈ R, λ ∈ R, we define

By Lemma 2.2, we show the following Pohozaev identity on (r0, r1)

Moreover, for d ≥ γ, there exists t0 such that

Next from (H1), we obtain that f is nondecreasing on [kd, d] ⊂ (β,+∞), and from (2.1) we have

Integrating on [0, t0], which implies

where

Hence, taking r0 = 0,r1 = t0 in (2.4), and using (2.5)-(2.6), we find

where

Lemma 2.3. There exists λ1 > 0 such that if λ ∈ (0, λ1), then u(r, γ, λ) ≥ β, for ∀r ∈ [0, 1].

Proof. Let r∗ = sup{0 ≤ r ≤ 1 : u(r, γ, λ) ≥ β}. For u is decreasing on [0, r∗], then β ≤ u(r, γ, λ). ≤ u(0, r, λ) = γ, ∀r ∈ [0, r∗]

Moreover, since f ≥ 0 on [β,+∞) and we obtain

Then for we have

Next, by using the mean value theorem and (2.8), there exists such that

Assume that r∗ < 1, we have

which contradicts the definition of r∗. Then, the lemma is proved for r∗ = 1. □

Lemma 2.4. There exists λ2 > 0 such that for λ ∈ (0, λ2)

Proof. From Lemma 2.2, we have the following Pohozaev identity on (r, t0)

Extending f by f(x) = f(0) < 0, for x ∈ (−∞, 0], then there exists B < 0 such that

For sufficiently large γ, from (H4), we deduce

By (2.7) and (2.9), we get

Then, there exists λ2 such that

Hence, for all λ ∈ (0, λ2) and r ∈ [0, 1],H(t) > 0, ∀d ≥ γ. This also implies that u(r, d, λ)2 + u′(r, d, λ)2 > 0, for all t ∈ [0, 1] and all d ≥ γ. □ .

Lemma 2.5. For r ∈ [0, 1], there exists d ≥ γ such that u(r, d, λ) < 0.

Proof. By contradiction, let d ≥ λ, we assume that u(r, d, λ) ≥ 0 for ∀r ∈ [0, 1].

Let is decreasing on (0, r)}. Define ω be the solution of the following equation:

where δ is chosen such that the first zero of ω is and ω satisfies r ∈ (0; 1).

From (H3), there exists d0 ≥ γ such that

Since

Let v = dω,

Then, we obtain

Suppose u(r, d, λ) ≥ d0 for all from (2.12),

On the other hand, from the quality of we know that then

From (2.13)-(2.15), we obtain

On the other hand, since

which is contradiction with (2.16).

Hence, there exists such that

And since d0 ≥ γ > β, there exists such that

Now, we consider t0 defined in (2.5), also

On [0, t0], from (H1), F is nondecreasing on [β,+∞) and u(r, d, λ) ≥ kd ≥ β ∀r ∈ (0, t0]. We have

On the other hand, since then

Hence, by (2.10) we get

From (H4), (2.18), (2.19)

Therefore, there exists d1 ≥ d0 such that for d ≥ d1, we get

By (2.17), (2.18)

Which implies

The mean value theorem gives us such that

hence and since there exists T ∈ (0, 1) such that u(T, d, λ) < 0, which contradicts with the assuming, the lemma is proved. □

 

3. Proof of the Main Results

The proof of Theorem 2.1. Let λ0 = min{λ1, λ2}, for ∀λ ∈ (0, λ0). Define From Lemma 2.3, we obtain that the set {d ≥ γ : u(r, d, λ) ≥ 0, ∀r ∈ (0, 1]} is nonempty. From Lemma 2.5 implies that

Then we claim that is the solution of problem (1.1). Moreover, the solution satisfies the following properties:

For (i). By contradiction, if there exists 0 ≤ R1 < 1 such that from Lemma 2.4, then we can suppose

Hence from and we find there exists R2 ∈ (R1, 1) such that which contradicts with the definition of

So for all r ∈ [0, 1).

For (ii). By contradiction,we assume then from (i) there exists η such that for ∀r ∈ (0, 1), moreover, there exists δ > 0 such that for ∀t ∈ (0, 1], which is a contradiction with the definition of (ii) is proved.

For (iii). From (2.1), Taking into account Lemma 2.3 and (H1), we have for ∀λ ∈ (0, λ0), u(r, γ, λ) ≥ β and f(s) > 0, for ∀s ∈ (β,+∞), which implies for ∀λ ∈ (0, λ1),

So (iv) is also proved.