N −LAPLACIAN EQUATIONS IN RN WITH CRITICAL GROWTH JOÃO MARCOS B. DO Ó Abstract. We study the existence of nontrivial solutions to the following problem: u ∈ W 1,N (RN ), u ≥ 0 and −div(| ∇u |N −2 ∇u) + a (x) | u |N −2 u = f (x, u) in RN (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 |→ ∞. 1. 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 RN (N ≥ 2) , involving critical growth: (1) u ∈ W 1,N (RN ), u≥0 and ∇u) + a (x) | u |N −2 u = f (x, u) −div(| ∇u |N −2 in RN , where a : RN → R is a continuous function satisfying a (x) ≥ a0 , ∀x ∈ RN , and such that a (x) → ∞ as | x |→ ∞. It is assumed that the nonlinearity f : RN × 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 R2 , have been considered by Cao in [10] and by Cao and Zhengjie in [11], under the decisive hypothesis that the function 1991 Mathematics Subject Classification. 35A15, 35J60. Key words and phrases. Elliptic equations, p−Laplacian, critical growth, Mountain Pass theorem, Trudinger-Moser inequality. Work partially supported by CNPq/Brazil. Received: June 10, 1997. c 1996 Mancorp Publishing, Inc. 301 302 JOÃO MARCOS B. DO Ó 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 RN , under the as sumption that a is coercive and that the potential F (x, u) = 0u f (x, s)ds is superquadratic and f (x, u) has subcritical growth, that is, | f (x, u) |≤ b1 + b2 | 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 (x, u) = λ | u |q−1 u+ | u |p−1 u, where 1 < q < p ≤ (N − 2)/(N + 2) 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 N variationally this class of problems, with f behaving like exp α | u | N −1 when | u |→ ∞, 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 R2 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 RN . Now we shall describe the conditions on the functions a and f. Namely, for a we suppose that: (a1 ) there exists a positive real number a0 such that a (x) ≥ a0 , ∀x ∈ RN , (a2 ) a (x) → ∞ as | x |→ ∞. 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), (f1 ) the function f : RN × R → R is continuous and for all (x, u) ∈ RN × R, N | f (x, u) |≤ b1 | u |N −1 +b2 exp α0 | u | N −1 − SN −2 (α0 , u) , for some constants α0 , b1 , b2 > 0, where SN −2 (α0 , u) = N −2 k=0 N α0k | u | N −1 k . k! Moreover, f is assumed to satisfy the following conditions: (f2 ) there is a constant µ > N such that, for all x ∈ RN and u > 0, 0 ≤ µF (x, u) ≡ µ u 0 f (x, t)dt ≤ uf (x, u), N -LAPLACIAN EQUATIONS (f3 ) 303 there are constants R0 , M0 > 0 such that, for all x ∈ RN and u ≥ R0 , 0 < F (x, u) ≤ M0 f (x, u); (f4 ) N lim uf (x, u) exp −α0 | u | N −1 u→+∞ subsets of RN . ≥ β0 > 0 uniformly on compact As usual, W 1,N (RN ) denotes the Sobolev space of functions in LN (RN ) such that their weak derivatives are also in LN (RN ) with the norm . u N W 1,N = RN (| ∇u |N + | u |N )dx, and we consider the subspace E ⊂ W 1,N (RN ) given by E = {u ∈ W 1,N RN : endowed with the norm u . N E= RN RN a(x) | u |N dx < ∞} (| ∇u |N +a(x) | u |N )dx. see that the Banach space E is a continuously Since a(x) ≥ a0 > 0, we clearly embedded in W 1,N RN and, moreover, (2) λ1 (N ) = inf 0=u∈E u N E ≥ a0 > 0. u N LN The main result of this paper is the following Theorem 1. Suppose (a1 ) − (a2 ) and (f1 ) − (f4 ) are satisfied. Furthermore, assume that (f5 ) lim supu→0+ N F (x, u) < λ1 (N ) | u |N unif ormly in x ∈ RN . Then the problem (1) has a nontrivial weak solution u ∈ E. Remark 1. The assumption (a2 ) implies that the Banach space E is compactly immersed in Lq 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, a(x) = a(y) if | x |=| y | (cf. [4, 18]). 304 JOÃO MARCOS B. DO Ó 2. A Trudinger-Moser inequality Let Ω be a bounded domain in RN (N ≥ 2). The Trudinger-Moser inequality (cf. [25, 31]) asserts that N ∀u ∈ W01,N (Ω) , exp(α | u | N −1 ) ∈ L1 (Ω), ∀α > 0 and that there exists a constant C(N ) which depends on N only, such that u sup N Ω 1,N ≤1 W 0 exp(α | u | N −1 ) ≤ C(N ) | Ω |, if α ≤ αN , 1 −1 and wN −1 is the (N − 1)−dimensional where | Ω |= Ω dx, αN = N wNN−1 measure of the (N − 1)−sphere. Inspired of this inequality and based on the related results [7, 10, 11, 12, 13, 14], we get the following result. Lemma 1. If N ≥ 2, α > 0 and u ∈ W 1,N RN , then (3) RN N exp α | u | N −1 − SN −2 (α, u) < ∞. 1 −1 ≤ 1, u LN ≤ M < ∞ and α < αN = N wNN−1 , Moreover, if ∇u N LN then there exists a constant C = C(N, M, α), which depends only on N, M and α, such that (4) RN N exp α | u | N −1 − SN −2 (α, u) ≤ C (N, M, α) . 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 ∈ Lp (RN ) such that u ≥ 0. Thus, there is a unique nonnegative function u∗ ∈ Lp (RN ), called the Schwarz symmetrization of u, such that it depends only on | x |, u∗ is a decreasing function of | x |; for all λ > 0 | {x : u∗ (x) ≥ λ} |=| {x : u (x) ≥ λ} | and there exists Rλ > 0 such that {x : u∗ ≥ λ} is the ball B (0, Rλ ) of radius Rλ centered at origin. Moreover, suppose that G : [0, +∞) → [0, +∞) is a continuous and increasing function such that G (0) = 0. Then, we have ∗ RN G (u (x)) dx = Further, if u ∈ W01,p RN RN RN G (u (x)) dx. then u∗ ∈ W01,p RN | ∇u∗ |p (x) dx ≤ RN and | ∇u |p (x) dx. N -LAPLACIAN EQUATIONS Thus, we can write RN N exp α | u | N −1 305 −SN −2 (α, u)] = RN N exp α | u∗ | N −1 − SN −2 (α, u∗ ) , and, for a real number r > 1 to be determined, we have RN = |x|<r + ≤ N exp α | u∗ | N −1 − SN −2 (α, u∗ ) |x|≥r |x|<r N exp α | u∗ | N −1 − SN −2 (α, u∗ ) N exp α | u∗ | N −1 − SN −2 (α, u∗ ) N exp α | u∗ | N −1 + |x|≥r N exp α | u∗ | N −1 − SN −2 (α, u∗ ) . Let us recall two elementary inequalities. Using the fact that the function N N 1 h : (0, +∞) → R given by h (t) = [(t + 1) N −1 − t N −1 − 1]/t N −1 is bounded, we have a positive constant A = A(N ) such that N (5) N 1 N (u + v) N −1 ≤ u N −1 + Au N −1 v + v N −1 , ∀u, v ≥ 0. If γ and γ are positive real numbers such that γ + γ = 1, then for all ε > 0, we have uγ v γ ≤ εu + ε (6) − γγ v, ∀u, v ≥ 0, because g : [0, +∞) → R, given by g (t) = tγ − εt, is bounded. Let v (x) = u∗ (x) − u∗ (rx0 ) where x0 is some fixed unit vector in RN . Notice that v ∈ W01,N (B (0, r)) . Here, B(0, r) denotes the ball of radius r centered at the origin of RN . Now, from (5) and (6), we have, respectively, N N N 1 N | u∗ | N −1 =| v + u∗ (rx0 ) | N −1 ≤ v N −1 + Av N −1 u∗ (rx0 ) + u∗ (rx0 ) N −1 , 1 v N −1 u∗ (rx0 ) = ≤ N v N −1 1 N ε NN−1 v + A N u∗ (rx0 ) N −1 ε A 1 1−N N −1 N N u∗ (rx0 ) N −1 , and hence, N N N | u∗ | N −1 ≤ (1 + ε) v N −1 + K(ε, N )u∗ (rx0 ) N −1 , N 1 where K(ε, N ) = A N −1 ε 1−N + 1. Therefore, N ∗ N −1 ≤ |x|≤r exp α | u | N exp K(ε, N )u∗ (rx0 ) N −1 N N −1 , |x|≤r exp α | (1 + ε) v | 306 JOÃO MARCOS B. DO Ó which, in view of Trudinger-Moser inequality, implies, (7) N exp α | u∗ | N −1 < ∞, |x|≤r ∀u ∈ W 1,N RN , ∀α > 0. Furthermore, taking ε > 0 such that (1 + ε) α < αN , we obtain |x|≤r N exp α | u∗ | N −1 ≤ C (N ) (8) N wN −1 rN exp K(ε, N )u∗ (rx0 ) N −1 N NMN wN −1 rN exp ≤ C (N ) N wN −1 1 N −1 K(ε, N ) N r N −1 , ≤ 1 and u LN ≤ M , where in for all u ∈ W 1,N RN such that ∇u N LN the last inequality we have used Radial Lemma A.IV in [8]: ∗ −1 | u (x) |≤| x | |x|≥r (9) N 1 N ∀x = 0. ∗ N |u | |x|≥r | x | N −1 k dx = wN −1 = ∞ αk + k=N Now, notice that the estimate |x|≥r u∗ LN (RN ) , exp α | u∗ | N −1 − SN −2 (u∗ ) αN −1 = (N − 1)! N wN −1 On the other hand, we have 1 N ∞ N −1 t r N t N −1 k wN −1 N N −1 k − N k! N |x|≥r | u∗ | N −1 k . dt N rN − N −1 k ≤ wN −1 rN N r N −1 k , ∀k ≥ N, together with Radial Lemma, lead to ∞ αk k=N (10) k! N |x|≥r | u∗ | N −1 k ≤ wN −1 r N ∞ αk N k=N k! 1 N wN −1 u∗ LN (RN ) r NN−1 k . Finally, (7), (9) and (10) imply the existence of the integral in (3). Furthermore, in the case that α < αN and u LN ≤ M , if we choose r = M N wN −1 1 N , we have |x|≥r N exp α | u∗ | N −1 − SN −2 (u) ≤ N M N exp (αN ) , which, in combination with (8), implies (4). N -LAPLACIAN EQUATIONS 307 3. The variational formulation First, we observe that since we are interested in obtaining nonnegative solutions, it is convenient to define f (x, u) = 0, ∀(x, u) ∈ RN × (−∞, 0]. From (f1 ), we obtain for all (x, u) ∈ RN × R, N | F (x, u) |≤ b3 exp α1 | u | N −1 − SN −2 (α1 , u) , (11) 1 N for some constants α1 ,b3 > 0. Thus, by lemma 1, we have F (x, u) ∈ L (R ) for all u ∈ W 1,N RN . Therefore, the functional I : E → R given by 1 I(u) = u N E − N F (x, u)dx RN is well defined. Furthermore, using standard arguments (cf. Theorem A.VI in [8]) aswellas the fact that for any given strong convergent sequence (un) in W 1,N RN there is a subsequence (unk ) and there exists h ∈ W 1,N RN such that | unk (x) |≤ h(x) almost everywhere in RN , we see that I is a C 1 functional on E with I (u)v = (| ∇u |N −2 ∇u∇v+a(x) | u |N −2 uv)dx− RN f (x, u)vdx, ∀v ∈ E. RN 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. that (a1 ), (f1 ) , (f2 ) and (f3 ) are satisfied. Then for any Assume 1,N N u∈W R −{0} with compact support and u ≥ 0, we have I (tu) → −∞ as t → ∞. Proof. Let u ∈ W 1,N RN − {0} with compact support and u ≥ 0. By (f2 ) and (f3 ) there are positive constants c, d such that F (x, s) ≥ csµ − d, Thus, I (tu) ≤ ∀x ∈ supp (u) , ∀s ∈ [0, +∞). tN µ u N E −ct N uµ + d | supp (u) |, RN which implies that I (tu) → −∞ as t → ∞, since µ > N. Lemma 3. Suppose that (a1 ), (f1 ) and (f5 ) hold. Then there exist α, ρ > 0 such that I (u) ≥ α if u E = ρ. 308 JOÃO MARCOS B. DO Ó Proof. From (f5 ), there exist ε, δ > 0 in such a way that | u |≤ δ implies (λ1 (N ) − ε) | u |N N for all x ∈ RN . On the other hand, for q > N , by (f1 ), there are positive constants β, C = C(q, δ) such that | u |≥ δ implies F (x, u) ≤ N F (x, u) ≤ C | u |q exp β | u | N −1 − SN −2 (β, u) for all x ∈ RN . These two estimates yield, N (λ1 (N ) − ε) | u |N +C | u |q exp β | u | N −1 − SN −2 (β, u) , N N for all (x, u) ∈ R × R. In what follows we make use of the inequality F (x, u) ≤ (12) RN N | u |q exp β | u | N −1 − SN −2 (β, u) ≤ C(β, N ) u qE , 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 3→ LN (RN ), we achieve 1 (λ1 (N ) − ε) q u N u N E − LN −C u E N N (λ1 (N ) − ε) 1 q ) u N ≥ (1 − E −C u E . N λ1 (N ) I (u) ≥ Thus, since ε > 0 and q > N , we may choose α, ρ > 0 such that I (u) ≥ α if u E = ρ. Now, let us obtain inequality (12). As it has been done in the proof of lemma 1, weN use shall the method of symmetrization. Letting R(β, u) = N −1 exp β | u | − SN −2 (β, u), we have R(β, u) | u | dx = RN ∗ RN R(β, u∗ ) | u∗ |q dx RN and q ∗ q R(β, u ) | u | dx = ∗ |x|≤σ ∗ q R(β, u ) | u | dx+ |x|≥σ R(β, u∗ ) | u∗ |q dx, where σ is a number to be determined later. Using the Hölder inequality, we obtain |x|≤σ R(β, u∗ ) | u∗ |q dx ≤ ≤ |x|≤σ [exp(β |x|≤σ N | u∗ | N −1 )] | u∗ |q dx exp(βr | u∗ | N N −1 1 r ) |x|≤σ | u∗ |qs 1 s , where 1/r + 1/s = 1. Now, proceeding as in the proof of lemma 1, we obtain |x|≤σ N exp(βr | u∗ | N −1 )dx ≤ C(β, N ) N -LAPLACIAN EQUATIONS 309 N if u E ≤ M, where M is such that βrM N −1 < αN . Thus, using the continuous imbedding E 3→ Lqs (RN ), we have (13) |x|≤σ R(β, u∗ ) | u∗ |q dx ≤ C(β, N ) u qE . On the other hand, the Radial Lemma leads to N |x|≥σ ≤ | u∗ | N −1 k | u∗ |q dx 1 N N wN −1 ≤ 1 N N wN −1 ∗ u LN (RN ) NN−1 k |x|≥σ NN−1 k u∗ LN (RN ) N wN −1 ≤ wN −1 σ N 1 N LN (RN ) ≤ C(N, M ) u qE , for all k ≥ N , if we choose σ r = M0 (N )1/N M . |x|≥σ We also have that | u∗ |N | u∗ |q dx ≤ |x|≥σ N | x | N −1 kr 1 r |x|≥σ 1 ∗ qs |u | s N k N −1 u qLsq (RN ) σr dx 1 u∗ N | x | N −1 k |x|≥σ λ1 | u∗ |q N wN −1 1 N | u∗ |N r dx where u LN (RN ) ≤ M0 = 1 r |x|≥σ | u∗ |qs dx 1 s ≤ u∗ N u∗ qLqs (RN ) LN r (RN ) ≤ C (N, M ) u∗ qE , if u∗ qE ≤ M, via the continuous imbedding E 3→ W 1,N (RN ) 3→ LN r (RN ). Therefore, (14) |x|≥R RN (β, u∗ ) | u∗ |q dx ≤ C(N, M ) exp(β) u qE . 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 N −1 (log n) N 1 1 r n (x, r) = w− N N M N −1 log( | x | )/(log n) 0 if if | x |≤ r/n r/n ≤| x |≤ r if | x |≥ r. n (·, r) ∈ W 1,N (RN ), the support of M n (x, r), is the ball Notice that: M n (x, r) |N dx = 1 and B[0, r] of radius r centered at zero, RN | ∇M 310 JOÃO MARCOS B. DO Ó n (x, r) |N dx = O(1/ log n) as n → ∞. Moreover, let Mn (x, r) = |M n (x, r)/ M n E . Thus, it is not difficult to see that M RN N − 1 N −1 MnN −1 (x, r) = wN −1 log n + dn , ∀ | x |≤ r/n, (15) where dn is a bounded sequence of nonnegative numbers. Lemma 4. Suppose that (a1 ) and (f1 ) − (f5 ) hold true. Then there exists Mn (·, r) such that 1 αN N −1 ) . max{I(tMn ) : t ≥ 0} < ( N α0 Proof. Let r be a fixed positive real number such that 1 β0 > N r (16) N α0 N −1 . Suppose, by contradiction, that for all n we have 1 αN N −1 max{I(tMn ) : t ≥ 0} ≥ ( ) , N α0 where Mn (x) = Mn (x, r). In view of Lemma 2, given n there exists tn > 0 such that I (tn Mn ) = max{I (tMn ) : t ≥ 0}. So, tN 1 αN N −1 I (tn Mn ) = n − (17) F (x, tn Mn ) dx ≥ ( ) , N N α0 RN and using the fact that F (x, u) ≥ 0, we obtain αN N −1 tN (18) ) . n ≥( α0 d Since at t = tn , we have I (tMn ) = 0, it follows that dt tN n = (19) tn Mn f (x, tn Mn ) dx = RN |x|≤r tn Mn f (x, tn Mn ) dx. Now, using hypothesis (f5 ), given ε > 0 there exists Rε > 0 such that for all u ≥ Rε and for all | x |≤ r, N uf (x, u) ≥ (β0 − ε) exp(α0 | u | N −1 ). (20) From (19) and (20), for large n, we obtain, tN n ≥ (β0 − ε) r |x|≤ n wN −1 = (β0 − ε) N N exp(α0 | tn Mn | N −1 )dx N r n N − 1 N N −1 exp(α0 tn N −1 wN −1 log n + α0 tn N −1 dn ). Thus, 1 ≥ (β0 − ε) N wN −1 N α0 N log n NN−1 tn + α0 tn N −1 dn − N log tn − N log n]. r exp[ N αN N -LAPLACIAN EQUATIONS 311 Therefore, the sequence tn is bounded, since otherwise, up to subsequences, we would have N α0 N log n NN−1 tn + α0 tn N −1 dn − N log tn − N log n = +∞, n→∞ αN lim which leads to a contradiction. Moreover, by (18) and N tN n N wN −1 N α0 tn N −1 ≥ (β0 − ε) − 1)N log n + α0 tn N −1 dn ] r exp[( N αN it follows that tN n →( (21) αN N −1 ) , α0 as n → ∞. Now, in order to estimate (19) more precisely, we consider the sets An = {x ∈ B[0, r] : tn Mn ≥ Rε } and Bn = B[0, r] − An . From (19) and (20) we arrive at tN n ≥ (β0 − ε) |x|≤r − (β0 − ε) exp(α0 | tn Mn | N N −1 )dx + Bn tn Mn f (x, tn Mn ) dx N Bn exp(α0 | tn Mn | N −1 )dx. Notice that Mn (x) → 0 and the characteristic functions χBn → 1 for almost every x such that | x |≤ r. Therefore, the Lebesgue Dominated Convergence Theorem implies Bn tn Mn f (x, tn Mn ) dx → 0 and Bn exp(α0 | N tn Mn | N −1 )dx → wN −1 N r as n → ∞. N αN N −1 , Note also that, by (18), tN n ≥ ( α0 ) N |x|≤r = exp(α0 | tn Mn | N −1 )dx ≥ r |x|≤ n r |x|≤ n exp(αN | Mn | N N −1 )dx + N exp(αN | Mn | N −1 )dx = N |x|≤r exp(αN | Mn | N −1 )dx N r ≤|x|≤r n r |x|≤ n exp(αN | Mn | N −1 )dx, − 1 N −1 exp[αN ωN −1 log n + dn αN ] ωN −1 rN exp[N log n] exp[dn αN ] N nN ωN −1 rN ωN −1 rN exp[dn αN ] ≥ , = N N = 312 JOÃO MARCOS B. DO Ó since dn ≥ 0. Using the change of variable τ = log rs /(ζn log n), with n E , we have, by straightforward computation, ζn = M N r/n≤|x|≤r exp(αN | Mn | N −1 )dx N = wN −1 r ζn log n ζn−1 0 N exp[N log n(τ N −1 − ζn τ )]dτ → wN −1 rN as n → ∞ Finally, passing to limits, using (21) and the latter fact we obtain ( αN N −1 ωN −1 rN ) ≥ (β0 − ε) {exp[d0 αN ] − 1} α0 N +(β0 − ε)wN −1 rN , where d0 = lim inf n→∞ dn is a nonnegative number. Thus, αN N −1 ) ≥ (β0 − ε)wN −1 rN , ( α0 which implies 1 N N −1 , β0 ≤ N r α0 contradicting (16). 4. Proof of Theorem 1.1 In view of lemmas 2 and 3 we can apply the Mountain-Pass Theorem to obtain a sequence (un ) ⊂ E such that I(un ) → c > 0 and I (un ) → 0, that is, 1 N (22) F (x, un )dx → c, as n → ∞, u n E − N RN (23) | RN N −2 | ∇un | N −2 ∇un ∇v − a(x) | un | un v − RN f (x, un )v | ≤ εn v E , for all v ∈ E, where εn → 0 as n → ∞. Furthermore, by lemma 4, the level c is less than N1 ( ααN0 )N −1 ). From now on, we shall be working in order to prove that (un ) converges to a weak nontrivial solution u of problem (1). From (22), (23) and (f2 ), µ (µF (x, un ) − f (x, un )un )dx C + εn un E ≥ ( − 1) un E − N RN µ ≥ ( − 1) un N E, N which implies that un E ≤ C, RN f (x, un )un dx ≤ C, RN F (x, un )dx ≤ C. Now, using the same argument as in Proposition 2.1 of [15], in view of Sobolev’s Theorem together with conditions (a1 ) − (a2 ), the Banach space E is compactly immersed in Lq if N ≤ q < ∞, (cf. [3]). Therefore, up to N -LAPLACIAN EQUATIONS 313 subsequences, we have un : u weakly in E, un → u in Lq (RN ), ∀q ≥ N and un (x) → u(x) almost everywhere in RN . Moreover, arguing as in lemma 4 of [19], we get f (x, un ) → f (x, u) in L1 (B(0, R)), | ∇un |N −2 ∇un →| ∇u |N −2 ∇u (24) weakly in LN/(N −1) (B(0, R)) N , for all R > 0. Therefore, by (23), passing to the limit, RN | ∇u |N −2 ∇u∇ϕ − a(x) | u |N −2 uϕ dx = RN f (x, u)ϕdx for all ϕ ∈ C0∞ (RN ), that is, u is a weak solution of (1). Let us show that u is nontrivial. Assume, by contradiction, that u ≡ 0. Using the Generalized Lebesgue Dominated Convergence Theorem (cf. [20]), by (f3 ) and the first result in (24), we conclude that F (x, un ) → 0 in L1 (B(0, R)), for all R > 0. Thus, using (11), in view of Radial Lemma, we obtain F (x, un ) → 0 in L1 (RN ). This together with (22) imply (25) lim n→∞ RN | ∇un |N = N c, and hence given < > 0, we have ∇un N ≤ N c + <, for large n. Using c < LN 1 αN N −1 ( ) and choosing q > 1 sufficiently close to 1 and < sufficiently small, N α0 N we obtain qα0 ∇un LNN−1 < αN . Hence, by the same kind of argument as it has been done in the proof of lemma 1, we conclude that RN N exp α | un | N −1 − SN −2 (α, un ) ≤ C, ∀n, which, in combination with the Hölder inequality and (f1 ), implies that lim n→∞ RN | f (x, un ) |q dx = 0. Therefore, from (23) with v = un , we achieve lim n→∞ RN | ∇un |N = 0, which contradicts (25) since c > 0. Thus, u is nontrivial and the proof of our main result is complete. References [1] Adimurthi, Existence of positive solutions of the semilinear Dirichlet problem with critical growth for the N −Laplacian, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 17 (1990), 393–413. [2] Adimurthi and S. L. 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N -LAPLACIAN EQUATIONS 315 [29] P. H. Rabinowitz, On a class of nonlinear Schrödinger equations, Z. Angew. Math. Phys. 43 (1992), 270–291. [30] M.C. Shaw, Eigenfunctions of the nonlinear equation ∆u + νf (x, u) = 0 in R2 , Pacific J. Math. 129 (1987), 349–356. [31] N. S. Trudinger, On the imbedding into Orlicz spaces and some applications, J. Math. Mech. 17 (1967), 473–484. Departamento de Matemática Universidade Federal da Paraı́ba 58059.900 João Pessoa, Pb BRAZIL E-mail address: jmbo@mat.ufpb.br