N−LAPLACIAN EQUATIONS IN RN WITH CRITICAL GROWTH

We study the existence of nontrivial solutions to the following problem: { u ∈ W 1,N (R ), u ≥ 0 and −div(| ∇u |N−2 ∇u) + a (x) | u |N−2 u = f(x, u) in R (N ≥ 2), where a is a continuous function which is coercive, i.e., a (x) → ∞ as | x |→ ∞ and the nonlinearity f behaves like exp ( α | u |N/(N−1) ) when | u |→ ∞.


Introduction
In this paper, we apply a mountain pass type argument to prove the existence of nontrivial weak solutions to the following class of semilinear elliptic problems in R N (N ≥ 2) , involving critical growth: where a : R N → R is a continuous function satisfying a (x) ≥ a 0 , ∀x ∈ R N , and such that a (x) → ∞ as | x |→ ∞.It is assumed that the nonlinearity f : R N × R → R is also continuous and f (x, 0) ≡ 0. Thus, u ≡ 0 is a solution of (1) and has critical growth, i.e., f behaves like exp α | u | N/(N −1) when | u |→ ∞.
For N = 2, problems of this type, that is, involving the Laplacian operator and critical growth in the whole R 2 , have been considered by Cao in [10] and by Cao and Zhengjie in [11], under the decisive hypothesis that the function a be a constant.For that purpose, they used the concentration-compactness principle of P. L. Lions.
Recently, Rabinowitz in [29], among other results, obtained a nontrivial solution to the problem −∆u + a (x) u = f (x, u) in R N , under the assumption that a is coercive and that the potential F (x, u) = u 0 f (x, s)ds is superquadratic and f (x, u) has subcritical growth, that is, | f (x, u) |≤ b 1 + b 2 | u | s , where s ∈ (1, (N + 2)/(N − 2)).This result was extended by Costa [15] to a class of potentials F (x, u) which are nonquadratic at infinity.Miyagaki, in [24], has treated this problem for N ≥ 3, involving critical Sobolev exponent, namely for f ) and λ > 0. In [3], this result was generalized by Alves to the p−Laplacian operator.
In this paper, we complement the results mentioned above by establishing sufficient conditions for the existence of nontrivial solutions to (1).To treat variationally this class of problems, with we introduce a Trudinger-Moser type inequality.On the other hand, to overcome the lack of compactness that has arisen from the critical growth and the unboundedness of the domain, we use some recent ideas from [16,19] together with a compact imbedding result essentially given by the coerciveness of a (cf.[15]).
We would also like to mention that problems involving the Laplacian operator with critical growth range in bounded domains of R 2 have been investigated, among others, by [2,5,6,9,16,17,22,23,30].We refer to [1,19,26] for semilinear problems with critical growth for the N −Laplacian in bounded domains of R N .Now we shall describe the conditions on the functions a and f.Namely, for a we suppose that: (a 1 ) there exists a positive real number a 0 such that a On the other hand, motivated by a Trudinger-Moser type inequality (cf.Lemma 1 below), we assume the following growth condition on the nonlinearity f (x, u), for some constants α 0 , b 1 , b 2 > 0, where Moreover, f is assumed to satisfy the following conditions: (f 2 ) there is a constant µ > N such that, for all x ∈ R N and u > 0, As usual, W 1,N (R N ) denotes the Sobolev space of functions in L N (R N ) such that their weak derivatives are also in L N (R N ) with the norm and we consider the subspace E ⊂ W 1,N (R N ) given by Since a(x) ≥ a 0 > 0, we clearly see that the Banach space E is a continuously embedded in W 1,N R N and, moreover, The main result of this paper is the following Theorem 1. Suppose (a 1 ) − (a 2 ) and (f 1 ) − (f 4 ) are satisfied.Furthermore, assume that Then the problem (1) has a nontrivial weak solution u ∈ E.

Remark 1.
The assumption (a 2 ) implies that the Banach space E is compactly immersed in L q if N ≤ q < ∞.We observe that this compact embedding result is used here only to prove that the Palais-Smale sequence obtained by mountain pass type argument converges to a weak nontrivial solution.
Therefore, the same device can be applied when we have some assumption which implies a compact embedding result as the cited above.For instance, when the function a is a radially symmetric function, that is, [4,18]).

A Trudinger-Moser inequality
Let Ω be a bounded domain in R N (N ≥ 2).The Trudinger-Moser inequality (cf.[25,31]) asserts that and that there exists a constant C(N ) which depends on N only, such that where Inspired of this inequality and based on the related results [7,10,11,12,13,14], we get the following result.
then there exists a constant C = C(N, M, α), which depends only on N, M and α, such that Proof.We may assume u ≥ 0, since we can replace u by | u | without causing any increase in the integral of the gradient.Since we shall use Schwarz symmetrization method, we recall briefly some of theirs basic properties (cf.[20,27]).Let 1 ≤ p ≤ ∞ and u ∈ L p (R N ) such that u ≥ 0. Thus, there is a unique nonnegative function u * ∈ L p (R N ), called the Schwarz symmetrization of u, such that it depends only on and there exists R λ > 0 such that {x : u * ≥ λ} is the ball B (0, R λ ) of radius R λ centered at origin.Moreover, suppose that G : [0, +∞) → [0, +∞) is a continuous and increasing function such that G (0) = 0.Then, we have Thus, we can write and, for a real number r > 1 to be determined, we have Let us recall two elementary inequalities.Using the fact that the function h : (0, +∞) → R given by h (t) = [(t + 1) If γ and γ are positive real numbers such that γ + γ = 1, then for all ε > 0, we have because g : [0, +∞) → R, given by g (t) = t γ − εt, is bounded.
Let v (x) = u * (x) − u * (rx 0 ) where x 0 is some fixed unit vector in R N .Notice that v ∈ W 1,N 0 (B (0, r)) .Here, B(0, r) denotes the ball of radius r centered at the origin of R N .Now, from ( 5) and ( 6), we have, respectively, and hence, which, in view of Trudinger-Moser inequality, implies, Furthermore, taking ε > 0 such that (1 + ε) α < α N , we obtain ( 8) , where in the last inequality we have used Radial Lemma A.IV in [8]: On the other hand, we have ( 9) Now, notice that the estimate together with Radial Lemma, lead to (10) Finally, (7), ( 9) and (10) imply the existence of the integral in (3).Furthermore, in the case that α < α N and u which, in combination with (8), implies (4).

The variational formulation
First, we observe that since we are interested in obtaining nonnegative solutions, it is convenient to define (11) for some constants α 1 , b 3 > 0. Thus, by lemma 1, we have F (x, u) ∈ L 1 (R N ) for all u ∈ W 1,N R N .Therefore, the functional I : E → R given by is well defined.Furthermore, using standard arguments (cf.Theorem A.VI in [8]) as well as the fact that for any given strong convergent sequence (u n ) in W 1,N R N there is a subsequence (u n k ) and there exists h Consequently, critical points of the functional I are precisely the weak solutions of problem (1).Here, like in [29,24,19], we are going to use a Mountain-Pass Theorem without a compactness condition such as the one of Palais-Smale type.This version of Mountain -Pass Theorem is a consequence of Ekeland's variational principle (cf.[21]).In the next two lemmas we check that the functional I satisfies the geometric conditions of the Mountain-Pass Theorem (cf.[28]).
Lemma 2. Assume that (a 1 ), (f 1 ) , (f 2 ) and (f 3 ) are satisfied.Then for any u ∈ W 1,N R N −{0} with compact support and u ≥ 0, we have Proof.Let u ∈ W 1,N R N − {0} with compact support and u ≥ 0. By (f 2 ) and (f 3 ) there are positive constants c, d such that Thus, Lemma 3. Suppose that (a 1 ), (f 1 ) and (f 5 ) hold.Then there exist α, ρ > 0 such that Proof.From (f 5 ), there exist ε, δ > 0 in such a way that | u |≤ δ implies for all x ∈ R N .On the other hand, for q > N, by (f 1 ), there are positive constants for all x ∈ R N .These two estimates yield, In what follows we make use of the inequality to be proved later, assuming that u E ≤ M holds, where M is sufficiently small.Under the assumption we have just done, by means of ( 2) and the continuous imbedding E → L N (R N ), we achieve Thus, since ε > 0 and q > N, we may choose α, ρ > 0 such that Now, let us obtain inequality (12).As it has been done in the proof of lemma 1, we use shall the method of symmetrization.Letting R(β, u) where σ is a number to be determined later.Using the Hölder inequality, we obtain where 1/r + 1/s = 1.Now, proceeding as in the proof of lemma 1, we obtain On the other hand, the Radial Lemma leads to Finally, the combination of estimates ( 13) and ( 14) leads to (12).
In order to get a more precise information about the minimax level obtained by the Mountain Pass Theorem, let us consider the following sequence of nonnegative functions Finally, passing to limits, using (21) and the latter fact we obtain where d 0 = lim inf n→∞ d n is a nonnegative number.Thus, which implies  In view of lemmas 2 and 3 we can apply the Mountain-Pass Theorem to obtain a sequence (u n ) ⊂ E such that I(u n ) → c > 0 and I (u n ) → 0, that is, for all v ∈ E, where ε n → 0 as n → ∞.Furthermore, by lemma 4, the level c is less than 1 N ( α N α 0 ) N −1 ).From now on, we shall be working in order to prove that (u n ) converges to a weak nontrivial solution u of problem (1).From ( 22), ( 23) and (f 2 ), Now, using the same argument as in Proposition 2.1 of [15], in view of Sobolev's Theorem together with conditions (a 1 ) − (a 2 ), the Banach space E is compactly immersed in L q if N ≤ q < ∞, (cf.[3]).Therefore, up to r), is the ball B[0, r] of radius r centered at zero, R N | ∇ M n (x, r) | N dx = 1 and since d n ≥ 0. Using the change of variable τ = log r s /(ζ n log n), with ζ n = M n E , we have, by straightforward computation, r/n≤|x|≤r exp .