On a Class of Semilinear Elliptic Eigenvalue Problems in ℝ2

Proceedings of the Edinburgh Mathematical Society (2017) 60, 107–126 DOI:10.1017/S0013091516000080 ON A CLASS OF SEMILINEAR ELLIPTIC EIGENVALUE PROBLEMS IN R2 MARCELO F. FURTADO1 , EVERALDO S. MEDEIROS2 AND UBERLANDIO B. SEVERO2 1 Universidade de Brası́lia, Departamento de Matemática, 70910-900 Brası́lia-DF, Brazil ([email protected]) 2 Universidade Federal do Paraı́ba, Departamento de Matemática, 58051-900 João Pessoa-PB, Brazil ([email protected]; [email protected]) (Received 29 December 2013) Abstract We consider the semilinear problem −Δu − 21 (x · ∇u) = λf (u), x ∈ R2 , where λ is a positive parameter and f has exponential critical growth. We first establish the existence of a non-zero weak solution. Then, by assuming that f is odd, we prove that the number of solutions increases when the parameter λ becomes large. In the proofs we apply variational methods in a suitable weighted Sobolev space consisting of functions with rapid decay at infinity. Keywords: eigenvalue problem; Trudinger–Moser inequality; self-similar solutions; critical exponential growth equation 2010 Mathematics subject classification: Primary 35J50; 35B33; 58E05 1. Introduction In this paper we consider the semilinear elliptic eigenvalue problem −Δu − 12 (x · ∇u) = λf (u), x ∈ R2 , (Pλ ) where λ > 0 is a parameter and f is a continuous function satisfying some suitable conditions. We shall look for solutions in a weighted Sobolev space of functions having a rapid decay at infinity. As quoted by Haraux and Weissler [10] and Escobedo and Kavian [8], the above equation is closely related to the study of self-similar solutions for the heat equation. More specifically, if we consider the evolution equation ut − Δu = |u|p−1 u on (0, ∞) × RN and we seek solutions of the form u(t, x) = t−1/(p−1) ω(t−1/2 x), a straightforward calculation shows that ω : RN → R needs to satisfy −Δω − 21 (x · ∇ω) = 1 ω + |ω|p−1 ω p−1 c 2016 The Edinburgh Mathematical Society  on RN . 107 108 M. F. Furtado, E. S. Medeiros and U. B. Severo For higher dimensions, N  3, there are some results concerning the above equation and its variants obtained by replacing the right-hand side of the equality by more general nonlinearities g(ω) (see [3, 5, 8, 9, 13, 14] and references therein). In a large class of such results the authors used variational techniques in such a way that the range of the power p is limited from above by the critical Sobolev exponent 2N/(N − 2). In dimension 2 the role of the critical exponent is played by the limiting exponent α0 for the Trudinger–Moser inequality in H 1 (R2 ), which is α0 = 4π. Here we are interested in the case in which the function f has the maximal growth, which allows us to deal with (Pλ ) variationally. More precisely, in dimension 2 the maximal growth is related to the so-called Trudinger–Moser inequality, namely,  2 sup eαu dx  C(α) uH 1 (Ω)1 Ω 0 for all α  4π and Ω ⊂ R2 bounded. Motivated by this inequality, in [1, 6] the authors introduced the following notion of criticality in dimension 2: we say that f : R → R has exponential critical growth if there exists α0 > 0 such that  0 if α > α0 , f (s) (1.1) lim 2 = |s|→+∞ eαs +∞ if α < α0 . We assume throughout this paper that the nonlinearity f is continuous and satisfies this growth condition. Problems involving critical exponential growth in bounded domains Ω ⊂ R2 have been studied by many authors (see [6, 7, 15, 16] and references therein). When dealing with problems on the entire space we need a version of the Trudinger–Moser inequality for unbounded domains. It asserts that  2 sup (eαu − 1) dx  C(α) uH 1 (R2 )1 R2 for all α  4π (see [4, 17] and references therein). Since we intend to apply variational methods, we first observe that (Pλ ) can be rewritten in a divergence form. Indeed, if we set  2 |x| k(x) := exp , x ∈ R2 , 4 a direct calculation shows that (Pλ ) is equivalent to − div(k(x)∇u) = λk(x)f (u), x ∈ R2 . (1.2) Due to the presence of the weight k(x) we are not able to use the usual Sobolev spaces. As quoted in [8], the natural space to look for rapid decay solutions is the space X defined ∞ as the closure of Cc,rad (R2 ) with respect to the norm u :=  R2 1/2 . k(x)|∇u| dx 2 On a class of semilinear elliptic eigenvalue problems in R2 109 As we will see, the space X is compactly immersed into the Lebesgue spaces with weight k. Furthermore, with the aid of the Trudinger–Moser inequality, we prove that the embed2 ding of X into the Orlicz space LA (R2 ), where A is the N -function A(t) = eαt − 1, is continuous, although it is not compact. In order to perform our minimax approach we need to impose some suitable assumptions on the behaviour of f . More precisely, we shall assume the following conditions: (f1 ) f (s) = o(s) as s → 0, (f2 ) there exists θ > 2 such that 0  θF (s) := θ  s f (t) dt  sf (s) for all s ∈ R, 0 (f3 ) the following limit holds: f (s)s = +∞. |s|→+∞ F (s) lim In our first result we establish the existence of a non-trivial weak solution for problem (Pλ ) when λ > 0 is arbitrary. More precisely, we have the following theorem. Theorem 1.1. Suppose that f satisfies (f1 )–(f3 ) and (f4 ) there exists β0 > 0 such that lim inf |s|→∞ f (s)s 4  β0 > min λα0 eα0 s2  1 exp r2    r4 r2 + :r>0 . 4 256 Then, for any λ > 0, problem (Pλ ) has at least one non-zero weak solution. We remark that our first theorem can be applied for the model nonlinearity 2 f (s) = (es − 1)sβ0 , 1 + s2 s ∈ R. Indeed, this function satisfies conditions (f1 )–(f4 ) for α0 = 1 and λβ0 > 3. In the proof of Theorem 1.1 we shall use the mountain pass theorem. After having established all the properties of the space X we need to overcome the difficulty of dealing with the unboundedness of the domain. Actually, this implies some problems in handling with Palais–Smale sequences. Since the embedding of X in the Orlicz space LA (R2 ) is not compact, we need to prove some technical convergence results as well as perform some careful estimates of the minimax level of the associated functional. We also emphasize the necessity of establishing a version of a convergence result due to Lions [11] for our variational setting (see Lemma 2.6). Condition (f3 ), which essentially has already appeared in [15,19], is crucial in order to get some of the required convergence results. Although it appears to be quite restrictive, 110 M. F. Furtado, E. S. Medeiros and U. B. Severo the model functions with critical exponential growth satisfy (f3 ). Moreover, this condition is implied by (f3 ) there exist constants R0 , M0 > 0 such that 0 < F (s)  M0 f (s) for all |s|  R0 , which has been used in many papers (see [1, 6, 7], for instance). Concerning the correct localization of the minimax level, we use the technical condition (f4 ) and adapt some ideas presented in [15] by performing some careful estimates of a slight modification of the Green functions considered by Moser [12]. It is worthwhile mentioning that our condition (f4 ) is more general than the analogous one considered in [15], since here the number β0 may be finite. Actually, a numerical computation shows that the relation between α0 and β0 in condition (f4 ) is satisfied if λα0 β0 > 9. In our last result we introduce more symmetry into the problem and show that the value of the parameter λ > 0 affects the number of solutions. Theorem 1.2. Suppose that the odd function f satisfies (f1 )–(f3 ) and (f5 ) there are μ0 , s0 > 0 and θ0 > 2 such that F (s)  μ0 |s|θ0 for all |s|  s0 . Then, for any given m ∈ N, there exists Λm > 0 such that problem (Pλ ) has at least m pairs of non-zero weak solutions provided that λ > Λm . The proof of this last theorem will be done as a byproduct of the compactness condition established in the proof of Theorem 1.1 and an application of the symmetric mountain pass theorem. Notice that, in this symmetric setting, we replace the condition at infinity (f4 ) by (f5 ), which provides the control of F at the origin. As far as we know, there are no multiplicity results concerning problems like (Pλ ) when the nonlinearity has critical growth. The paper is organized as follows. In § 2 we present some useful inequalities concerning the space X. In § 3 we establish the variational setting of our problem and prove a local compactness result for the associated functional. The main results are proved in § 4. 2. Some properties of the space X In this section we present some technical results. Throughout the paper we write u instead of R2 u(x) dx. ∞ (R2 ) We recall that X denotes the Hilbert space obtained as the completion of Cc,rad with respect to the norm  1/2 2 u := k(x)|∇u| , which is induced by the inner product u, v :=  k(x)(∇u · ∇v). On a class of semilinear elliptic eigenvalue problems in R2 111 For each p  2 we also consider the weighted Lebesgue space Lp (R2 , k) of all radial functions u : R2 → R such that 1/p  up := k(x)|u|p < ∞. Lemma 2.1. The space X is compactly embedded into Lp (R2 , k) for any p ∈ [2, +∞). ∞ Proof . For any given u ∈ Cc,rad (R2 ) we have that    |∇(k(x)1/2 u)|2 = k(x)|∇u|2 + ∇(k(x)1/2 u2 )∇(k(x)1/2 ). Integrating by parts we get    1 |∇(k(x)1/2 u)|2 = k(x)|∇u|2 − k(x)u2 (Δθ + 12 |∇θ|2 ), 2 (2.1) where θ(x) := |x|2 /4. Since Δθ(x) + 12 |∇θ(x)|2 = 1 + 18 |x|2  1, it follows that  2 k(x)u  2  k(x)|∇u|2 . (2.2) By density, we have the same inequality for any u ∈ X. This establishes the continuous embedding X ֒→ L2 (R2 , k). If p > 2 and u ∈ X, we can use (2.1) and (2.2) to get    1 k(x)u2  2u2 . (|∇(k(x)1/2 u)|2 + k(x)u2 )  k(x)|∇u|2 + 2 Thus, we conclude that k 1/2 u ∈ H 1 (R2 ). Hence, we can use (2.1) again to infer that    1 2 1/2 2 k(x)|∇u|  |∇(k(x) u)| + k(x)|u|2 2  1 1/2 2 k(x)|u|2 = k uH 1 (R2 ) − 2  Cp  k(x)p/2 |u|p 2/p − 1 2  k(x)|u|2 , where Cp > 0 is related to the embedding H 1 (R2 ) ֒→ Lp (R2 ). Since k(x)  1 and p  2, we have that k(x)p/2  k(x). It follows from (2.2) that Cp  p k(x)|u| 2/p  Cp  and therefore X ֒→ Lp (R2 , k) for p  2. p/2 k(x) p |u| 2/p 2  k(x)|∇u|2 , (2.3) 112 M. F. Furtado, E. S. Medeiros and U. B. Severo It was proved in [8, Proposition 11] that the embedding X ֒→ L2 (R2 , k) is compact. For the p > 2 case, we take a sequence (un ) ⊂ X such that un ⇀ 0 weakly in X. Fix p̃ > p and consider τ ∈ (0, 1) such that p = (1 − τ )2 + τ p̃. Hölder’s inequality with exponents 1/(1 − τ ) and 1/τ provides   k(x)|un |p = k(x)(1−τ ) |un |(1−τ )2 k(x)τ |un |τ p̃   2 k(x)|un | 2(1−τ )  cun 2 1−τ   p̃ k(x)|un | τ . Up to a subsequence, we have that un → 0 in L2 (R2 , k). The above expression implies that un → 0 in Lp (R2 , k).  Remark 2.2. As a byproduct of the above calculations we see that X ֒→ H 1 (R2 ). Indeed, for any u ∈ X, it holds that   (|∇u|2 + |u|2 )  k(x)(|∇u|2 + |u|2 )  cu2 . We also quote for future reference that, in view of the second inequality of (2.3), for any r  1 there exists C = C(r) such that  1/r  k(x)r |u|2r  C k(x)|∇u|2 for all u ∈ X. (2.4) We shall need the following variant of a well-known radial lemma of Strauss [18]. Lemma 2.3. There exists c0 > 0 such that, for all v ∈ X, it holds that |v(x)|  c0 |x|−1/2 e−|x| 2 /8 v for all x ∈ R2 \ {0}. ∞ (R2 ). Let r = |x| and ϕ : [0, +∞) → Proof . It suffices to prove the lemma for v ∈ Cc,rad R be such that ϕ(r) = v(|x|). We have that  ∞ ϕ(r)2 = −2 ϕ(s)ϕ′ (s) ds 2  r ∞ e−s 2 r r −1 −r 2 /4 e /4 −1 s  |ϕ(s)||ϕ′ (s)|es 2 /4 ∞ (ϕ(s)2 + ϕ′ (s)2 )es s ds 2 /4 s ds r  c1 r−1 e−r Since X ֒→ L2 (R2 , k), we get 2 /4  (k(x)|∇v|2 + k(x)v 2 ). ϕ(r)2  cr−1 e−r and the lemma follows. 2 /4 v2 ,  On a class of semilinear elliptic eigenvalue problems in R2 113 In order to make the proof of our results more direct and effective, we state a technical lemma. Lemma 2.4. Suppose that G ∈ C(R, R) satisfies 2 G(s)  c1 s4 (eαs − 1) for all s ∈ R, with c1 , α > 0. Then there exists c > 0 such that, for any R > 1 and u ∈ X, it holds that  2 2 c k(x)G(u) dx  u4 (eαc0 u − 1), R BR (0)c where c0 > 0 comes from Lemma 2.3. Proof . It follows from the monotone convergence theorem that   2 k(x)|u|4 (eαu − 1) dx k(x)G(u) dx  c1 BR (0)c BR (0)c +∞ = c1 j=1 αj j!  k(x)|u|2j+4 dx. BR (2.5) (0)c By using Lemma 2.3 we can estimate the last integral above as   2 e(|x| /4)(1−j−2) |x|−j−2 dx k(x)|u|2j+4 dx  (c0 u)2j+4 BR (0)c BR (0)c  2π(c0 u)2j+4 = 2π(c0 u)2j+4   ∞ s−j−2 s ds R 1 jRj 2π (c0 u)2j+4 , R where we have used that j  1 and R > 1. The above expression and (2.5) provide  ∞ 2π (αc20 u2 )j k(x)G(u)dx  c1 (c0 u)4 R j! BR (0)c j=1 = 2 2 c u4 (eαc0 u − 1) R with c := 2πc1 c40 > 0, and this completes the proof.  We now recall the Trudinger–Moser inequality for the whole space R2 , the proof of which can be found, for instance, in [4, 17]. Lemma 2.5 (Trudinger–Moser inequality). If α > 0 and v ∈ H 1 (R2 ), then (e − 1) ∈ L1 (R2 ). Moreover, if ∇vL2 (R2 )  1, vL2 (R2 )  M < ∞ and α  4π, then there exists a constant C = C(M, α) such that  2 (eαv − 1)  C(M, α). (2.6) αv 2 114 M. F. Furtado, E. S. Medeiros and U. B. Severo In what follows we establish a version of a result due to Lions (see [11, Theorem I.6]) for the whole space R2 and considering our functional space. Lemma 2.6. Let (vn ) in X with vn  = 1 and suppose that vn ⇀ v weakly in X with v < 1. Then for each 0 < p < 4π(1 − v2 )−1 , up to a subsequence, we have that  2 sup (epvn − 1) < ∞. n∈N Proof . We first notice that, if a, b, ε > 0, Young’s inequality implies that   1 b a2 = (a − b)2 + b2 + 2ε(a − b)  (1 + ε2 )(a − b)2 + c 1 + 2 b2 . ε ε Hence, we can use Young’s inequality again to get     2 1 1 pvn p(1+ε2 )(vn −v)2 p(1+1/ε2 )v 2 (e − 1)  e e − − ′ γ γ   ′ 2 2 2 2 1 1 (eγp(1+ε )(vn −v) − 1) + ′ (eγ p(1+1/ε )v − 1),  γ γ where γ > 1 and 1/γ + 1/γ ′ = 1. Since the last integral above is finite, it suffices to prove that  2 2 sup (eγp(1+ε )(vn −v) − 1) < ∞. n∈N Since vn ⇀ v and vn  = 1, we conclude that lim vn − v2 = 1 − v2 < n→∞ 4π . p Thus, we can take 0 < α < 4π and choose γ > 1 and ε > 0 small in such a way that γp(1 + ε2 )vn − v2 < α < 4π. (2.7) We now set un := (vn − v)/vn − v and notice that, since |∇un |2  un 2 = 1, we have that ∇un L2 (R2 )  1. Moreover, it follows from (2.2) that un L2 (R2 )  2. Hence, we can invoke Lemma 2.5 and (2.7) to obtain a positive constant C(M, β) such that   2 2 2 γp(1+ε2 )(vn −v)2 (e − 1) = (eγp(1+ε )vn −v un − 1)  2  (eαun − 1)  C(M, β), and the lemma is proved.  On a class of semilinear elliptic eigenvalue problems in R2 115 3. A local compactness condition Let α > α0 and let q  1. By using the critical growth of f we obtain lim |s|→+∞ f (s) = 0. |s|q−1 (eαs2 − 1) This and (f1 ) imply that, for any given ε > 0, it holds that 2 max{|f (s)s|, |F (s)|}  21 εs2 + c1 |s|q (eαs − 1) for all s ∈ R. (3.1) Given u ∈ X we can set q = 2 to obtain    2 k(x)F (u)  c2 k(x)|u|2 + c3 k(x)|u|2 (eαu − 1). If we apply the inequality (1 + t)r  1 + tr with t = es − 1  0, we get (es − 1)r  (ers − 1) for all r  1, s  0. (3.2) It follows from Hölder’s inequality that  2 k(x)|u|2 (eαu − 1)   k(x)2 |u|4  c4 u2 < +∞,  1/2   2 (eαu − 1)2 1/2 2 (e2αu − 1) 1/2 where we have used (2.4) and Lemma 2.5 to conclude that the last term is finite. All together the above estimates imply that the functional  2 1 Iλ (u) := 2 u − λ k(x)F (u), u ∈ X, is well defined. Standard calculations show that Iλ ∈ C 1 (X, R) with derivative given by   Iλ′ (u)ϕ = k(x)∇u · ∇ϕ − λ k(x)f (u)ϕ for all u, ϕ ∈ X, and therefore the critical points of Iλ are precisely the weak solutions of (1.2). Remark 3.1. Since the value of λ > 0 is not relevant in Theorem 1.1, we shall assume, until § 4.2, that λ = 1. To simplify notation we write simply I to denote I1 . We devote the rest of this section to the proof of a compactness condition for the functional I. We follow here some arguments developed in [15]. First, we recall that (un ) ⊂ X is said to be a (PS)c sequence (where ‘PS’ stands for ‘Palais–Smale’) for the functional I if I(un ) → c and I ′ (un ) → 0. We say that I satisfies the (PS)c condition if any (PS)c sequence has a convergent subsequence. 116 M. F. Furtado, E. S. Medeiros and U. B. Severo Lemma 3.2. Suppose that f satisfies (f2 ) and (f3 ). If (un ) ⊂ X is a (PS)c sequence for I, then, up to a subsequence, we have that (i) un ⇀ u weakly in X with I ′ (u) = 0,   (ii) lim k(x)F (un ) = k(x)F (u), n→∞ (iii) lim sup R→∞  k(x)f (un )un = 0. BR (0)c Proof . Let (un ) ⊂ X be such that I ′ (un ) → 0 and I(un ) → c. By using (f2 ) and a standard argument we can check that (un ) is bounded in X. Hence, up to a subsequence, un ⇀ u weakly in X and un (x) → u(x) almost everywhere (a.e.) in R2 . Recalling that I ′ (un )un = on (1), using the above expression, (f2 ) and the boundedness of (un ), we obtain c1 > 0 such that     1 (3.3) k(x)f (un )un  c1 , k(x) f (un )un − F (un )  c1 . θ The first estimate above, k(x)  1 and condition (f2 ) again imply that    |f (un )un | = f (un )un  k(x)f (un )un  c1 . Thus, it follows from a convergence result due to de Figueiredo et al . (see [6, Lemma 2.1]) ∞ (R2 ), we can take the limit in that f (un ) → f (u) in L1loc (R2 ). Hence, given ϕ ∈ Cc,rad ′ ′ ′ I (un )ϕ to conclude that I (u)ϕ = 0. By density, I (u) = 0. Let M > 0 be such that un , u  M . Given R, ε > 0, we can use (3.1) with q = 4, Lemma 2.4 and the embedding X ֒→ L2 (R2 , k) to get   2 2 ε c3 c k(x)|un |2 + un 4 (eαc0 un  − 1)  c2 ε + , k(x)F (un )  2 R R BR (0)c with c2 , c3 > 0 depending only on M . It follows that  lim sup k(x)F (un )  c2 ε. R→∞ Moreover, since k(x)F (u) ∈ L1 (R2 ), it holds that  lim sup k(x)F (u) = 0. R→∞ BR (0) (3.5) BR (0)c For any fixed R > 0 we claim that   lim k(x)F (un ) dx = n→∞ (3.4) BR (0)c k(x)F (u) dx. BR (0) If this is true, we can use the above convergence, (3.4) and (3.5) to get (ii). (3.6) On a class of semilinear elliptic eigenvalue problems in R2 117 In order to prove (3.6) we first notice that, by (f3 ), we can obtain θ0 , R0 > 0 such that θ0 F (s)  f (s)s for any |s| > R0 . Moreover, we can suppose that θ0 also satisfies (θc1 )/(θ0 − θ) < ε. If An := {|un |  R0 }, we can use the second inequality in (3.3) and (f2 ) to obtain  θc1  k(x)(f (un )un − θF (un )) dx An   = (θ0 − θ) k(x)(f (un )un − θ0 F (un )) dx, k(x)F (un ) dx + An An and therefore it follows that  k(x)F (un ) dx  An θc1 < ε. θ0 − θ (3.7) Applying Egoroff’s theorem we obtain a measurable set A ⊂ BR (0) such that |A| < ε and un (x) → u(x) uniformly on BR (0) \ A. Hence,  BR (0) k(x)(F (un ) − F (u)) dx   k(x)F (un ) dx +  k(x)F (u) dx + on (1). (3.8) A A 2 Given α > α0 , it follows from the growth condition of f that F (s)  c4 eαs for all s ∈ R and some c4 > 0. Thus, given γ > 1, we can use Hölder’s inequality to get  A 1/γ k(x)F (u) dx  M1 c4 |A|  e αγ ′ u2 1/γ ′  c5 ε1/γ (3.9) with 1/γ + 1/γ ′ = 1. On the other hand, using (3.7), Lebesgue’s theorem and the above expression, we obtain    k(x)F (un ) dx k(x)F (un ) dx + k(x)F (un ) dx  A A∩{|un |<Rθ } A∩An ε+  k(x)F (u) dx + on (1) A∩{|un |<Rθ }  ε + c5 ε1/γ + on (1). Since ε > 0 is arbitrary, the convergence in (3.6) follows from (3.8), (3.9) and the above inequality. In order to prove (iii) it suffices to use the estimate for |f (s)s| in (3.1) with q = 4 and proceed as in the proof of (3.4).  Now we are ready to prove our compactness result. Proposition 3.3. Suppose that f satisfies (f2 ) and (f3 ). Then the functional I satisfies the (PS)d condition for any 0 < d < (2π)/α0 . 118 M. F. Furtado, E. S. Medeiros and U. B. Severo Proof . Let (un ) ⊂ X be such that I ′ (un ) → 0 and I(un ) → d < 2π/α0 . According to the previous lemma we have that un ⇀ u weakly in X, with I ′ (u) = 0, and moreover k(x)F (un ) → k(x)F (u). We shall consider two possible cases. Case 1 (u = 0). In this case we have that k(x)F (u) = 0, and therefore    4π 2 lim un  = 2 d + k(x)F (u) = 2d < . n→∞ α0 Since I ′ (un )un = on (1), in order to prove that un → 0 it is enough to check that  k(x)f (un )un = 0. lim n→∞ By using (3.1) with q = 3, we get  ε k(x)f (un )un  un 22 + c1 Dn , 2 where Dn :=  2 k(x)|un |2 |un |(eαun − 1). Since the embedding X ֒→ L2 (R2 , k) is compact, we have that un 2 → 0. Hence, it is enough to verify that Dn → 0. By taking ri > 1, i = 1, 2, 3, such that 1/r1 +1/r2 +1/r3 = 1 and r2 > 2, we can use Hölder’s inequality to get   1/r3 1/r1 2 2 un Lr2 (R2 ) (eαr3 un  (un /un ) − 1) k(x)r1 |un |2r1 Dn  c2  1/r3 2 2 ,  c3 un 2 on (1) (eαr3 un  (un /un ) − 1) (3.10) where we have used (2.4), (3.2) and the compactness of X ֒→ Lr2 (R2 ). Since un 2 → γ < 4π/α0 , we can choose r3 close to 1 and α > α0 close to α0 in such a way that r3 αun 2  γ̃ < 4π. It follows from Lemma 2.5 that the last term in the parentheses in (3.10) is bounded. Hence, Dn → 0 and the proposition is proved in the first case. Case 2 (u = 0). We are going to verify that   k(x)f (un )un = k(x)f (u)u. lim n→∞ (3.11) If this is true, we obtain on (1) = I ′ (un )un = un 2 −  k(x)f (u)u + on (1) = un 2 − u2 + I ′ (u)u + on (1). Since I ′ (u) = 0, we conclude that un  → u and the proposition follows from the weak convergence of un . On a class of semilinear elliptic eigenvalue problems in R2 119 It remains to check (3.11). In view of item (iii) of the previous lemma, and since lim supR→∞ BR (0)c k(x)f (u)u = 0, we need only verify that, for any R > 0, the following holds:   lim k(x)f (u)u dx. (3.12) k(x)f (un )un dx = n→∞ BR (0) BR (0) With this aim we first notice that, as in the first case, we have    lim un 2 = 2 d + k(x)F (u) = 2(d + d0 ) > 0, n→∞ (3.13) where d0 := k(x)F (u). We may suppose that un  = 0 for all n  n0 , and therefore vn := un /un  is well defined. The weak convergence of (un ) and (3.13) imply that vn ⇀ v := u 2(d + d0 ) weakly in X. It follows from (f2 ) that I(u)  0. Thus, we can take α > α0 such that d < I(u) + 2π/α. Hence, 4π/α 1 − v2 < 2(d + d0 ) and we can use (3.13) to obtain p0 > 0 such that αun 2 < p0 < (4π)/(1 − v2 ). We now choose q > 1 sufficiently close to 1 in such way that 4π , 1 − v2 αqun 2 < p < with p = p0 q. It follows from Lemma 2.6 that   2 2 2 sup (eαqun  vn − 1)  sup (epvn − 1) = c4 < ∞. n∈N n∈N Up to a subsequence, we have that un → u strongly in Ls (BR (0)) for any s  1. Hence, there exists Ψs ∈ L1 (BR (0)) such that |un (x)|s  Ψs (x) a.e. in BR (0). Hence, by using that k ∈ L∞ (BR (0)), (3.1) with q = 1, Hölder’s inequality and the above expression, for any measurable subset A ⊂ BR (0), we get  k(x)f (un )un dx  c5  A A  c5  2 |un | dx + c6 Ψ2 dx + c7  c5 A 1  A A   Ψ2 dx + c8  A A 1/q 1/q′   αu2n q (e − 1) dx |un | dx q′ A 1/q′   1/q 2 2 Ψq′ dx (eαqun  vn − 1) 1/q′ Ψq′ dx . Since Ψs ∈ L (BR (0)) and the set A ⊂ BR (0) is arbitrary, we conclude that the first integral above is uniformly small provided that the measure of A is small. Hence, the set {k(x)f (un )un } is uniformly integrable, and Vitali’s theorem implies that k(x)f (un )un →  k(x)f (u)u in L1 (BR (0)). This establishes (3.12) and concludes the proof. 120 M. F. Furtado, E. S. Medeiros and U. B. Severo 4. Proof of the main results We start this section with the following technical result. Lemma 4.1. If u ∈ X, β > 0, q > 0 and u  M with βM 2 < 4π, then there exists C = C(β, M, q) > 0 such that  2 k(x)|u|2+q (eβu − 1)  Cu2+q . Proof . Let ri > 1, i = 1, 2, 3, be such that 1/r1 + 1/r2 + 1/r3 = 1 and qr2  2. Hölder’s inequality implies that 1/r1  1/r3   q 2+q βu2 r1 2r1 βu2 r3 k(x)|u| (e uLqr2 (R2 ) − 1)  . (e − 1) k(x) |u| Using the embedding X ֒→ Lqr2 (R2 ), (2.4) and (3.2) we get  1/r3  2+q βu2 2+q βr3 u2 . − 1)  C(q)u (e − 1) k(x)|u| (e (4.1) By choosing r3 close to 1, we can suppose that α := βr3 M 2 < 4π. Thus,    2 2 2 2 (eβr3 u − 1)  (eβr3 M (u/u) − 1) = (eαv − 1), with v := u/u. Arguing as in the proof of Lemma 2.6, we obtain C(M, β) > 0 such that  2 (eβr3 u − 1)  C(M, β). This and (4.1) conclude the proof.  4.1. Proof of Theorem 1.1 In this section we prove our existence result. Since we have already proved a local compactness result, the key point is the correct localization of the mountain pass level. This will be done using the following result, the proof of which will be postponed to the last section of the paper. Proposition 4.2. Suppose that f satisfies (f2 ) and (f4 ). Then there exists v ∈ X with compact support such that max I(tv) < t0 2π . α0 (4.2) If we assume the above proposition, we can prove Theorem 1.1 as follows: by using (3.1) with q > 2, Lemma 4.1 and (2.2), we obtain   2 ε 2 2 1 k(x)|u| − c1 k(x)|u|2+(q−2) (eαu − 1) I(u)  2 u − 2  21 (1 − c2 ε)u2 − c3 uq , (4.3) On a class of semilinear elliptic eigenvalue problems in R2 121 with c1 , c2 , c3 > 0. Since q > 2 and ε > 0 is arbitrary, it is standard to obtain α, ρ > 0 such that I(u)  α > 0 for all u ∈ ∂Bρ (0). Moreover, from (f2 ) we obtain c4 , c5 > 0 such that F (s)  c4 |s|θ − c5 for all s ∈ R. (4.4) If v is given by Proposition 4.2 and A ⊂ R2 is its support, it follows from the above inequality that, for any t > 0, t2 I(tv)  v2 − c4 tθ 2  A k(x)|v|θ dx − c5 |A|. Since θ > 2, we conclude that I(tv) → −∞ as t → ∞. Hence, I(t0 v) < 0 for some t0 > 0 large enough. The above calculations show that I has the mountain pass geometry, and therefore we can define the minimax level cM := inf max I(γ(t)), γ∈Γ t∈[0,1] where Γ := {γ ∈ C([0, 1], X) : γ(0) = 0, γ(1) = t0 v}. The definition of cM and (4.2) imply that 2π cM  max I(tv) < . t0 α0 It follows from Proposition 3.3 and the mountain pass theorem that I has a non-zero critical point, and the theorem is proved.  4.2. Proof of Theorem 1.2 In order to prove our multiplicity result we shall use the following version of the symmetric mountain pass theorem (see [2]). Theorem 4.3. Let E be a real Banach space, and let J ∈ C 1 (E, R) be an even functional satisfying J(0) = 0 and (J1 ) there are constants ρ, α > 0 such that J|∂Bρ (0)  α; (J2 ) there is A > 0 and a finite-dimensional subspace V of E such that max J(u)  A. u∈V If the functional J satisfies the (PS)d condition for 0 < d < A, then it possesses at least dim V pairs of non-zero critical points. 122 M. F. Furtado, E. S. Medeiros and U. B. Severo Given m ∈ N, we are going to apply this abstract result with E = X and J = Iλ and V := span{ψ1 , . . . , ψm }, ∞ 2 where {ψi }m i=1 ⊂ C0 (R ) is a collection of smooth functions with disjoint supports. We notice that, for any λ > 0, all the results proved in the last sections for the functional I = I1 also hold for Iλ . Thus, arguing as in the proof of Theorem 1.1 we can easily check that the even functional Iλ satisfies (J1 ). By using the local condition (f5 ) and (f3 ) we can obtain μ > 0 such that F (s)  μ|s|θ0 for all s ∈ R. Since V is finite dimensional, all the norms in these spaces are equivalent. This fact and the above inequality yield, for any u ∈ V , Iλ (u)  12 u2 − λcm uθ0 , where cm > 0 depends only on m. If we consider the maximum value of the map t → (1/2)t2 − λcm tθ0 on [0, +∞), the above expression, θ0 > 2 and a straightforward calculation imply that max I(u)  Am u∈V    2/(θ0 −2) θ0 /(θ0 −2)  1 1 1 := − cm λ2/(2−θ0 ) , 2 θ0 cm θ0 cm and therefore Iλ verifies (J2 ). Since 2/(2−θ0 ) < 0, we have that limλ→+∞ Am = 0. Hence, there exists Λm > 0 such that Am < (2π)/α0 for any λ > Λm . In view of Proposition 3.3, for any λ > Λm we can apply Theorem 4.3 to obtain m pairs of non-zero critical points of Iλ . This finishes the proof.  4.3. Proof of Proposition 4.2 We devote this section to the proof of Proposition 4.2 by supposing, once again, that λ = 1 and I = I1 . The general case follows from obvious modifications. We consider a small modification to the sequence of scaled truncated Green functions considered by Moser (see [12]). More specifically, we define for n > 1, ⎧   −1/2 ⎪ r ⎪ ⎪ k log n ⎪ ⎪ ⎪ n ⎨   1 r M̃n (x) := √ −1/2 k(x) log 2π log n ⎪ ⎪ |x| ⎪ ⎪ ⎪ ⎪ ⎩0 if |x|  r/n, if r/n  |x| < r, if |x|  r, with r > 0 fixed. Notice that M̃ n ∈ H 1 (R2 ) and supp(M̃ n ) = B̄ r (0). Moreover, the following lemma holds. On a class of semilinear elliptic eigenvalue problems in R2 123 Lemma 4.4. There exist D = D(r) > 0 and a sequence (dn ) ⊂ R, which also depends on r, such that D − dn , M̃n 2 = 1 + log n with limn→∞ dn log n = 0. In particular, lim M̃n 2 = 1. (4.5) n→∞ Proof . We set An := Br (0) \ Br/n (0) and notice that ∇M̃ n is zero outside the set An , and   x −|x|2 /8 −1/2 1 + 4 x log(r/|x|) , ∇M̃n (x) = −e (2π log n) |x|2 x ∈ An . Hence, we can compute  k(x)|∇M̃n |2 =    1 |x|2 1 2 1 log + (r/|x|) + log(r/|x|) dx 2 2π log n An |x|2 16   r  1 s3 1 = + log2 (r/s) + 21 s log(r/s) ds log n r/n s 16   r4 r2 1 + − Γr,n,1 − Γr,n,2 log n + = log n 8 512 with Γr,n,1 r2 := 8   2 log n 1 + 2 , n2 n Γr,n,2   1 r4 8 log2 n 4 log n + + 4 . := 512 n4 n4 n If we now set r2 r4 , dn := (log n)−1 (Γr,n,1 + Γr,n,2 ), + 8 512 we get the conclusions of the lemma. D := (4.6)  We now normalize the Green function and consider the function Mn defined by Mn := M̃n . M̃n  Since Mn  = 1, Proposition 4.2 is a direct consequence of the following lemma. Lemma 4.5. Suppose that f satisfies (f2 ) and (f4 ). Then there exists n ∈ N such that   2  t 2π − k(x)F (tMn ) < max . (4.7) t0 2 α0 124 M. F. Furtado, E. S. Medeiros and U. B. Severo Proof . Since Mn has compact support, we can argue as in the proof of Theorem 1.1 to conclude that the function  t2 gn (t) := − k(x)F (tMn ), t  0, 2 goes to −∞ as t → +∞. Hence, it attains its global maximum at a point tn > 0 such that gn′ (tn ) = 0, that is,  t2n = k(x)tn Mn f (tn Mn ) dx. (4.8) Br (0) Suppose, by contradiction, that the lemma is false. Then for any n ∈ N it holds that  2π t2n − k(x)F (tn Mn )  , 2 α0 and therefore t2n  4π α0 for all n ∈ N. (4.9) We claim that (tn ) ⊂ R is bounded. Indeed, let β0 > 0 be given by (f4 ) and let 0 < ε < β0 . Condition (f4 ) provides R = R(ε) > 0 such that sf (s)  (β0 − ε) exp(α0 s2 ) for all |s|  R. (4.10) The definition of Mn , (4.9) and M̃n  → 1 imply that, for any large values of n, it holds that  2 log n −r 2 /8 tn Mn (x)  e  R for all x ∈ Br/n (0). α0 It follows from (4.8), (4.10), the above expression, k(x)  1 and the definition of Mn that  t2n  k(x)tn Mn f (tn Mn ) dx Br/n (0)  (β0 − ε) = (β0 − ε)  exp(α0 (tn Mn )2 ) dx Br/n (0)  exp Br/n (0)  2 2 e−r /(4n 2π M̃n 2 log n α0 t2n )  dx. This, (4.5), the equation 1/n2 = exp(−2 log n) and direct calculation provide     −r2 /(4n2 ) α0 2 e tn − 1 log n . t2n  (β0 − ε)πr2 exp 2 M̃n 2 4π (4.11) Since exp(s)  s, we can invoke Lemma 4.4 and the above expression to conclude that (tn ) is bounded. By going to a subsequence, we may use (4.9) to get t2n → γ  4π/α0 . Since e−r 2 /(4n2 )M̃n −2 → 1, On a class of semilinear elliptic eigenvalue problems in R2 125 we can take the limit in (4.11) to conclude that γ > 4π/α0 cannot occur. Hence, lim t2n = n→∞ 4π . α0 By using (4.9) and (4.11) again, we get   2 2 −2 t2n  (β0 − ε)πr2 exp (M̃n 2 − e−r /(4n ) ) log n . M̃n 2 (4.12) (4.13) It follows from Lemma 4.4 and L’Hopital’s rule that (M̃n 2 − e−r 2 /(4n2 ) ) log n = (1 − e−r 2 /(4n2 ) ) log n + D − dn log n = D + on (1). 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