Problem With Critical Sobolev Exponent and With Potential on
Abstract
We consider the equation in D’, u = 0 on D′, where D′ is a geodesic ball with radius θ1, centered at the north pole, on , N ≥ 4, and q is a positive continuous function. We prove the existence of solutions that depends only on the behavior of the potential q near its minima.
1. Introduction
The function q is a positive differentiable potential in D′. The exponent 2∗ = 2N/(N − 2) is a critical Sobolev exponent in the sense of the embedding which is compact for all t < 2∗ and only continuous for t = 2∗. For more clarity, let us define the following spaces Lt(D′) = {f : D′⟶ℝ measurable function such that ∫D′|f|tdx < +∞} and with Diu as the partial derivative of u with respect to xi.
Much has been written about this problem on a bounded domain in ℝN, when the potential is constant, and many works have been devoted to study the relation between the geometry or the topology of the domain on the existence of nontrivial solutions. Pohozaev [1] had shown that the equation like (2) does not admit a positive solution if the domain is star-shaped. Bahri and Coron [2] have proved that the topology of the domain plays an important role in affirming or not the existence of solutions. Ding [3] obtained positive solutions on contractible domains. Molle and Passaseo [4] studied the problem for a perturbed domain; see also Li, Nie, and Zhang [5] and Li and Zou [6] for other types of critical Sobolev equations. Brezis and Nirenberg [7] proved that a linear perturbation of problem as (2) can generate the existence of solution in any bounded domain. Bandle and Benguria [9] considered the equation on S3 and obtained many interesting results. For a nonconstant potential, Hadiji and Yazidi [10] and Hadiji et al. [11, 12] have disturbed the operator Laplacian by considering a divergence operator with potential and have shown that the existence of solution depends on behavior of the potential function near its minima. In this work, we consider a problem with positive differentiable weight on SN with N ≥ 4. The dimension N = 3 will not be treated here, since it is a critical dimension and it is very delicate to obtain a complete result; see [13] for some numerical results. Alternatively, we show that the existence of solutions of (1) depends among other on the behavior of the potential q(.) near its minima. This gives an alternative to avoid adding some topological conditions to the domain or making some perturbations of the equation.
2. Main Results
Let .
Remark 1. The space H is not empty and S(q, D) is a finite quantity.
Indeed, let , where is a solution modulo, a constant of in ℝN (see [7 and 8] for more details), and φ ∈ C∞(D) is such that φ ≡ 1 ∈ B(x0, R) and φ ≡ 0 ∈ D\B(x0, (4/3)R) with R as a positive constant such that B(x0, 2R) ⊂ D.
Easily, we see that and . The main result of this work is the following theorem.
Theorem 1. Assume that D is a ball in ℝN, N ≥ 4, and q is a differentiable function satisfying (3) and (4). Then, for k ≥ 2, there exists a positive solution of Equation (2).
Let S be the best Sobolev constant for the injection of into . The proof of Theorem 1 is based on the following two lemmas.
Lemma 1. If S(q, D) < q(x0)S, then the minimizing problem (5) is achieved.
Proof 1. Let (uj) be a minimizing sequence of S(q, D), that is,
From (6), we write
Hence, (uj) is bounded in .
Then, for a subsequence that still denoted (uj), there exists such that
We claim that u ≠ 0.
Let vj = uj − u. Arguing by contradiction, suppose that u = 0; from (6) and (7), we have
By definition of S and x0, we have
This contradicts the hypothesis S(q, D) < q(x0)S; then, u ≠ 0.
By the definition of S(q, D), we have
Now, prove the inequality inverse.
Let vj = uj − u. The Brezis–Lieb lemma gives
Therefore,
By multiplying S(q, D), we get
Or
Then,
Therefore,
Since S(q, D) < q(x0)S, then
Combining (8) and (10), we obtain
Injecting (11) into (9) and using the fact that S(q, D) < q(x0)S, we conclude vj⟶0 strongly in and therefore uj⟶u strongly in .
Lemma 2.
- 1.
For N ≥ 4 and k > 2, we have
- 2.
For N ≥ 4 and k = 2, we have
- 3.
For N ≥ 4 and 0 < k < 2, we have
Proof 2. We make the change of variable w = ρ(N − 2)/2u in (5), and we claim that
Indeed, we have
Therefore,
We have
Or
Inserting (15) and (16) into (14), we obtain, after some computations,
Integrating (13) and using (17), we have
Since |x|2 ρ2 = 2 ρ − ρ2, we get, after some simplifications,
We estimate the energy at , where is a solution modulo, a constant of in ℝN and φ ∈ C∞(D) such that φ ≡ 1 ∈ B(x0, R) and φ ≡ 0 ∈ D\B(x0, (4/3)R) with R as a positive constant such that B(x0, 2R) ⊂ D.
We recall from [10],
We have from [7], after some modifications,
Using (4), some computations give
Looking at the previous estimation, we see that if we assume that β2 is small enough, then β2 < C. Also, since , then . Therefore, if , which satisfied since β2 is small enough. Consequently, the conclusion of Lemma 2 comes directly from (23).
Proof of Theorem 1. Summary.
Let be given by Lemma 1. We can assume that u ≥ 0 on D (otherwise, we replace u by |u|). Since u is a minimizer for S(q, D), therefore, there is a Lagrange multiplier μ = S(q, D) such that .
It follows that k u satisfies (2) for an appropriate constant . Since ρ and q are positive functions, then u > 0 by the strong maximum principle.
Conflicts of Interest
The authors declare no conflicts of interest.
Funding
The authors received no specific funding for this work.
Open Research
Data Availability Statement
Data are included in the article.
