www.elsevier.com/locate/na
Four positive solutions of semilinear elliptic equations in
exterior domains
Huei-Li Lin
a,∗, Hwai-Chiuan Wang
b, Tsung-Fang Wu
c,1aCenter for General Education, Chang Gung University, 259, Wen Hwa 1st Road, Kwei Shan, 33375, Tao-Yuan, Taiwan bDepartment of Applied Mathematics, Hsuan Chuang University, Hsinchu, Taiwan
cCenter for General Education, Southern Taiwan University of Technology, Tainan, Taiwan
Received 22 November 2005; accepted 7 July 2006
Abstract
In this paper, assuming that h is nonnegative and khkL2 > 0, we prove that if khkL2 is sufficiently small, then there are at least
four positive solutions of the semilinear elliptic equation in RN\D, where D is a C1,1bounded domain in RN. c
2006 Elsevier Ltd. All rights reserved.
Keywords:Four solution; Semilinear elliptic equations; Category; Minimax method
1. Introduction
Consider the semilinear elliptic equation −1u + u = |u|p−2u + h(z) in Ω;
u ∈ H01(Ω), (1)
where Ω = RN\D, D is a C1,1bounded domain in RN, 2< p < 2∗=2N/(N − 2) for N ≥ 3. Let
d(p, α) = (p − 2) 1 p −1 p−1p−2 2 p p −2 12 α(Ω)12, (2)
whereα(Ω) is defined below. In this paper, we first consider h(z) ≥ 0 and 0 < khkL2 < d(p, α). Associated with
Eq.(1), we define the functionals a, b, and Jh, for u ∈ H01(Ω),
a(u) = Z Ω |∇u|2+u2; b(u) = Z Ω |u|p; ∗Corresponding author.
E-mail address:[email protected](H.-L. Lin).
1 Present address: Department of Applied Mathematics, National University of Kaohsiung, Kaohsiung, Taiwan. 0362-546X/$ - see front matter c 2006 Elsevier Ltd. All rights reserved.
Jh(u) = 1 2a(u) − 1 pb(u+) − Z Ω hu.
By Rabinowitz [14, Proposition B.10.], a, b, and Jhare of C2. For h = 0, we consider the semilinear elliptic equation
−1u + u = |u|p−2u in Ω ;
u ∈ H01(Ω), (3)
and the energy functional J(u) = 12a(u) − 1pb(u+). Bahri–Lions [5] showed that Eq.(3)admits at least one positive
solution u in RN, and Kwong [12] proved that the solution u is unique.
Suppose that h is nonnegative, small and exponential decay, Zhu [18] and Hsu–Wang [10] proved that Eq.(1)
admits at least two positive solutions in RN, an exterior strip domain Ar\D, respectively. Without the condition of exponential decay, Cao–Zhou [8] and Hirano [9] proved that Eq. (1)admits at least two positive solutions in RN. About the existence of two positive solutions, Jeanjean [11] and Adachi–Tanaka [2] studied a more general equation −1u + u = g(z, u) + h (z) in RN under some conditions. Moreover, Adachi–Tanaka [1] asserted that there are at least four positive solutions of Eq.(4)
−1u + u = p(z)|u|p−2u + h(z) in RN;
u ∈ H1(RN), (4)
where p(z) 6≡ 1 and p(z) ≥ 1 − C exp(− (2 + δ) |z|) for some C, δ > 0. In this paper, we study the idea of category in Adachi–Tanaka [1] to show that there exist at least four positive solutions of Eq.(1)in Ω .
For 1 ≤ l ≤ N − 1, let z = (x, y) ∈ RN −l × Rl. Consider the exterior strip domain Ar\D, where Ar = {(x, y) ∈ RN | |x |< r}. We also obtain the same result in an exterior strip domain.
2. Existence of (PS)-Sequences
We define the Palais–Smale (denoted by (PS)) sequences, (PS)-values, and (PS)-conditions in H01(Ω) for Jh as
follows.
Definition 1. (i) Forβ ∈ R, a sequence {un} is a(PS)β-sequence in H01(Ω) for Jh if Jh(un) = β + o(1) and
Jh0(un) = o(1) strongly in H−1(Ω) as n → ∞;
(ii) β ∈ R is a (PS)-value in H01(Ω) for Jhif there is a(PS)β-sequence in H01(Ω) for Jh;
(iii) Jh satisfies the (PS)β-condition in H01(Ω) if every (PS)β-sequence in H01(Ω) for Jh contains a convergent
subsequence. Lemma 2. Let u ∈ H1
0(Ω) be a critical point of Jh, then u is a nonnegative solution of Eq.(1). Moreover, if u 6≡0
or h 6≡0, then u is positive in Ω .
Proof. Suppose that u ∈ H01(Ω) satisfies hJh0(u), ϕi = 0 for any ϕ ∈ H01(Ω), that is, Z Ω ∇u∇ϕ + uϕ = Z Ω u+p−1ϕ + hϕ for any ϕ ∈ H01(Ω).
Thus, u is a weak solution of −1u+u = u+p−1+h(z) in Ω. Since h ≥ 0, by the maximum principle, u is nonnegative.
If u 6≡ 0 or h 6≡ 0, we have that u is positive in Ω . Let
Mh= {u ∈ H01(Ω)\{0} | u ≥ 0 and hJ 0
h(u) , ui = 0} and αh(Ω) = inf u∈Mh
Jh(u).
Denote by M0=M, J0(u) = J(u) and α0(Ω) = α (Ω).
Applying the same argument of Chen–Wang [7], we have the following lemmas.
Lemma 4. (i) For each nonnegative u ∈ H01(Ω)\{0}, there exists a su > 0 such that suu ∈M;
(ii) Letβ > 0 and {un} be a sequence in H01(Ω)\{0} for J such that J(un) = β + o(1) and a(un) = b(u+n) + o(1).
Then there is a sequence {sn}in R+such that sn=1 + o(1), {snun}inM and J(snun) = β + o(1).
Lemma 5. If u ∈ H01(Ω)\{0} and u ≥ 0, then a(u)p2 b(u) !p−21 ≥ 2 p p −2 12 α(Ω)12.
Proof. ApplyingLemma 4.
By Lien–Tzeng–Wang [13], we have the following lemma.
Lemma 6 (Palais–Smale Decomposition Lemma for J ). Let {un} be a(PS)β-sequence in H01(Ω) for J. Then there
are a subsequence {un}, a positive integer l, sequenceszin
∞
n=1in R
N, functions u in H1
0(Ω), and w
i 6=0 in H1(RN)
for1 ≤ i ≤ l such that z i n → ∞ for 1 ≤ i ≤ l; −4u + u = |u|p−2u inΩ ; −4wi +wi = w i p−2 wi in RN; un=u + l X i =1 wi(· − zi n) + o(1) strongly in H 1(RN); J(un) = J(u) + l X i =1 J(wi) + o(1).
In addition, if un≥0, then u ≥ 0 andwi ≥0 for 1 ≤ i ≤ l.
Lemma 7 (Palais–Smale Decomposition Lemma for Jh). Let {un} be a(PS)β-sequence in H01(Ω) for Jh. Then there
are a subsequence {un}, a positive integer l, sequenceszin
∞
n=1in R
N, functions u in H1
0(Ω), and w
i 6=0 in H1(RN)
for1 ≤ i ≤ l such that z i n → ∞ for1 ≤ i ≤ l; −4u + u = |u|p−2u + h(z) in Ω; −4wi +wi = w i p−2 wi in RN; un=u + l X i =1 wi(· − zi n) + o(1) strongly in H1(RN); Jh(un) = Jh(u) + l X i =1 J(wi) + o(1).
In addition, if un≥0, then u ≥ 0 andwi ≥0 for 1 ≤ i ≤ l.
Proof. See Zhu–Zhou [19].
Defineψ(u) = hJh0(u) , ui = a(u)−b (u+)−RΩhu. Then if 0< khkL2 < d(p, α), we have the following lemma.
Lemma 8. For each u ∈ Mh, we have hψ0(u), ui = a(u) − (p − 1) b(u) 6= 0.
ByLemma 8, we write Mh=M+h ∪M − h, where M+h = {u ∈Mh |a(u) − (p − 1) b(u) > 0}, M−h = {u ∈Mh |a(u) − (p − 1) b(u) < 0}. Define α+ h (Ω) = inf u∈M+h Jh(u); α−h (Ω) = inf u∈M−h Jh(u).
By Wang–Wu [17], we have the following lemma.
Lemma 9. {un} is a (PS)α(Ω)-sequence in H01(Ω) for J if and only if J (un) = α(Ω) + o (1) and a (un) =
b u+n + o(1). In particular, every minimizing sequence {un}inM of α(Ω) is a (PS)α(Ω)-sequence in H01(Ω) for J.
For each nonnegative u ∈ H01(Ω)\{0}, we write
tmax=
a(u) (p − 1) b (u)
p−21 > 0.
Lemma 10. For each nonnegative u ∈ H01(Ω)\{0}, we have the following results.
(i) There is a unique number t−=t−(u) > tmax> 0 such that t−u ∈M−h and Jh t−u = maxt ≥tmaxJh(tu);
(ii) t−(u) is a continuous function;
(iii) M−h =nu ∈ H01(Ω)\{0} | u ≥ 0 and kuk1
H 1 t−kuku H 1 =1o; (iv) If R
Ωhu > 0, then there is a unique number 0 < t
+ = t+(u) < t
max such that t+u ∈ M+h and
Jh(t+u) = min0≤t ≤t− Jh(tu).
Proof. See Tarantello [15] and Cao–Zhou [8].
Lemma 11. (i) For each u ∈ M+h, we haveRΩhu> 0 and Jh(u) < 0. In particular, αh(Ω) ≤ α+h (Ω) < 0;
(ii) Jhis coercive and bounded below onMh.
Proof. (i) For each u ∈ M+h, a(u) − (p − 1) b(u) > 0 and a(u) = b (u) + RΩhu. Then Z
Ω
hu = a(u) − b(u) > (p − 2) b(u) > 0. Hence Jh(u) = 1 2 − 1 p b(u) − 1 2 Z Ω hu< p −2 2 p b(u) − p −2 2 b(u) = −(p − 1) (p − 2) 2 p b(u) < 0. (ii) By Tarantello [15, p. 288].
Lemma 12. Let u be in Mhsuch that Jh(u) = min v∈Mh
Jh(v) = αh(Ω). Then
(i) RΩhu> 0;
(ii) u is a solution of Eq.(1)inΩ . Proof. (i) ByLemma 11(i), we have
0> αh(Ω) = Jh(u) = 1 2 − 1 p a(u) − 1 − 1 p Z Ω hu. Thus,R Ωhu> 0.
(ii) ByLemma 8, hψ0(v), vi 6= 0 for each v ∈ Mh. Since Jh(u) = minv∈Mh Jh(v), by the Lagrange multiplier
theorem, there is aλ ∈ R such that Jh0(u) = λψ0(u) in H−1(Ω). Then we have 0 = h Jh0(u), ui = λhψ0(u), ui.
Thus,λ = 0 and Jh0(u) = 0 in H−1(Ω). Therefore, u is a solution of Eq.(1)in Ω with Jh(u) = αh(Ω).
By Cao–Zhou [8], we have the following two lemmas.
Lemma 13. Given u ∈ Mh, then there exist aδ > 0 and a differentiable functional l : B (0; δ) ⊂ H01(Ω) → R+
such that l(0) = 1, l (v) (u − v) ∈ Mhforv ∈ B (0; δ) and
hl0(v) , ϕi (l,v)=(1,0) =hψ
0(u), ϕi
hψ0(u), ui forϕ ∈ C ∞ c (Ω).
Lemma 14. (i) There exists a(PS)αh(Ω)-sequence {un} inMhfor Jh;
(ii) There exists a(PS)α+
h(Ω)-sequence {un} inM
+ h for Jh;
(iii) There exists a(PS)α−
h(Ω)-sequence {un} inM
− h for Jh.
3. Existence of the first solution
By Lemma 14(i), there is a (PS)αh(Ω)-sequence {un} in Mh for Jh. Then we have the following (PS)αh(Ω)
-condition.
Lemma 15. Let {un} ⊂ Mh be a(PS)αh(Ω)-sequence for Jh. Then there exist a subsequence {un} and a nonzero
u0 ∈ H01(Ω) such that un → u0 strongly in H01(Ω). Moreover, u0 is a positive solution of Eq. (1) such that
Jh(u0) = αh(Ω).
Proof. ByLemma 11(ii), {un} is bounded in H01(Ω). Take a subsequence {un} and u0∈ H01(Ω) such that un * u0
weakly in H01(Ω). Then we have that u0is a nonzero solution of Eq.(1)in Ω . Since
Jh(un) = 1 2a(un) − 1 pb(un) − Z Ω hun=αh(Ω) + o (1), hJh0(un) , uni =a(un) − b (un) − Z Ω hun=o(1), we obtain 1 2 − 1 p a(un) − 1 − 1 p Z Ω hun=αh(Ω) + o (1).
Since the functional a is weakly lower semicontinuous andRΩhun →
R
Ωhu0as n → ∞, then Jh(u0) = αh(Ω).
Let pn=un−u0. By the Br´ezis–Lieb Lemma, we get
Jh(pn) = 1 2a(pn) − 1 pb(pn) − Z Ω hpn = 1 2a(un) − 1 2a(u0) − 1 pb(un) + 1 pb(u0) − Z Ω hun+ Z Ω hu0+o(1) = Jh(un) − Jh(u0) + o (1) = o (1). (5)
By the Br´ezis–Lieb Lemma,RΩhpn=o(1) and u0is a solution of Eq.(1), so
hJh0(pn) , pni =a(pn) − b (pn) − Z Ω hpn =a(un) − a (u0) − b (un) + b (u0) − Z Ω hun+ Z Ω hu0+o(1) = hJh0(un) , uni − hJh0(u0) , u0i =o(1). (6)
Thus, by(5)and(6)RΩhpn=o(1), we have
p −2
2 p a(pn) = o (1), that is,
un →u0 strongly in H01(Ω).
Moreover, u0is a solution of Eq.(1)such that Jh(u0) = αh(Ω). ByLemma 2, u0is positive in Ω .
We prove that u0is the unique critical point of Jhin B(r0) in the following lemma.
Lemma 16. Let r0= 1 p−1 p−21 2 p p−2 12 α (Ω)1 2. Then (i) M+ h ⊂B(r0) = {u ∈ H 1 0(Ω) | kukH1 < r0};
(ii) Jh(u) is strictly convex in B (r0).
Proof. (i) If u ∈ M+h, then a(u) > (p − 1) b(u) and a(u) = b (u) + RΩhu. Thus, a(u) < 1 p −1a(u) + khkL2kukH1. This implies kukH1 < p − 1 p −2 khkL2 < p − 1 p −2 (p − 2) 1 p −1 p−1p−2 2 p p −2 12 α (Ω)12 = 1 p −1 p−21 2 p p −2 12 α (Ω)12 =r0. (ii) We know Jh00(u) (v, v) = a (v) − (p − 1) Z Ω |u|p−2v2 for allv ∈ H01(Ω). Thus, byLemma 5, we obtain
Jh00(u) (v, v) ≥ a (v) − (p − 1) kukLp−2p kvk2Lp ≥ a(v) − (p − 1) a(u) p−2 2 p − 2 2 p p−22 α (Ω)−(p−2)22 p × a(v) p − 2 2 p p−2p α (Ω)−(p−2)p ≥ a(v) 1 −(p − 1) 2 p p −2α (Ω) 2− p2 kukp−2 H1 > 0 for u ∈ B(r0)\{0}.
Thus, Jh00(u) is positive definite for u ∈ B (r0) and Jhis strictly convex in B(r0).
ByLemma 15, there exists a solution u0 ∈ Mhof Eq.(1)such that Jh(u0) = αh(Ω). Furthermore, we have the
following lemma.
Lemma 17. (i) u0∈M+h and Jh(u0) = αh+(Ω) = αh(Ω);
(ii) u0is the unique critical point of Jh(u) in B (r0), where r0is defined as inLemma16;
Proof. (i) ByLemma 12(i),RΩhu0> 0. We claim that u0∈M+h. Otherwise, if u0∈M−h, then byLemma 10, there
exists a unique t−(u
0) = 1 > t+(u0) > 0 such that t+(u0)u0∈M+h and
αh(Ω) ≤ α+h (Ω) ≤ Jh(t+(u0)u0) < Jh(t−(u0)u0) = αh(Ω),
which is a contradiction. Since u0 ∈M+h,α+h (Ω) ≤ Jh(u0) = αh(Ω) ≤ αh+(Ω), that is, Jh(u0) = αh+(Ω) =
αh(Ω).
(ii) By part (i) andLemma 16. (iii) See Cao–Zhou [8, p. 452]. 4. Existence of the other two solutions
By Cao–Zhou [8, Proposition 3.1] and Palais–Smale DecompositionLemma 7, we have the following restricted (PS)β-condition.
Lemma 18. (i) If {un} is a(PS)β-sequence in H01(Ω) for Jhwithβ < αh(Ω)+α (Ω), then there exist a subsequence
{un} and a nonzero u0in H01(Ω) such that un → u0 strongly in H01(Ω) and Jh(u0) = β. Moreover, u0 is a
positive solution of Eq.(1)inΩ ;
(ii) If {un} ⊂M−h is a(PS)β-sequence in H01(Ω) for Jhwith
αh(Ω) + α (Ω) < β < α−h (Ω) + α (Ω),
then there exist a subsequence {un} and a nonzero u0 ∈ M−h such that un → u0 strongly in H01(Ω) and
Jh(u0) = β. Moreover, u0is a positive solution of Eq.(1)inΩ .
By Chen–Chen–Wang [6, Proposition 1], we have the following lemma.
Lemma 19. Let u be a positive solution of Eq.(1)inΩ . Then for anyε > 0, there are positive constants cε and R such that D ⊂ BN(0; R) and
u(z) ≥ cεexp(− (1 + ε) |z|) for |z| ≥ R.
It is well-known that there is a positive radially symmetric smooth solution ¯u of Eq. (3) in RN such that J( ¯u) = α(RN). Recall the facts
(i) for anyε > 0, there exist constants C0, C00> 0 such that for all z ∈ RN
u(z) ≤ C0exp(−|z|) and |∇u(z)| ≤ C00exp(− (1 − ε) |z|);
(ii) for anyε > 0, there exists a constant Cε> 0 such that u(z) ≥ Cεexp(− (1 + ε) |z|) for all z ∈ RN.
For such R inLemma 19, letψR : RN →[0, 1] be a C∞-function on RN such that 0 ≤ψR ≤1, |∇ψR| ≤ c and
ψR(z) =1 for |z| ≥ R + 1;0 for |z| ≤ R.
We define
vz(z) = ψR(z) ¯u (z − z) for z ∈ RN.
Clearly,vz(z) ∈ H01(Ω).
In order to proveLemma 24, we need the following lemmas.
Lemma 20. (i) a (vz) = b (vz) + o (1) = p−22 p α(RN) + o(1) as |z| → ∞;
(ii) J (vz) = α (Ω) + o (1) = α(RN) + o (1) as |z| → ∞;
(iii) vz * 0 weakly in H01(Ω) as |z| → ∞.
Lemma 21. Let E be a domain in RN. If f : E → R satisfies Z E f(z)e σ|z| dz< ∞ for some σ > 0, then Z E f(z)e−σ|z−z|dz eσ |z|= Z E f(z)eσ hz,zi |z| dz + o(1) as |z| → ∞.
Proof. Sinceσ |z| ≤ σ |z| + σ |z − z|, we have f(z)e −σ |z−z| eσ|z| ≤ f(z)e σ|z| .
Since −σ |z − z| + σ |z| = σhz|z|,zi +o(1) as |z| → ∞, then the lemma follows from the Lebesque dominated convergence theorem.
Lemma 22. For t ≥ 0, we have the following inequalities. (i) (1 + t)q≥1 + tq+ q
q−1tq−1where q ≥2;
(ii) (1 + t)q≥1 + tq+qt where q ≥2;
(iii) (1 + t)q≥1 + tq+qt +q−2q tq−1where q ≥3;
(iv) If t ≤ c for some c> 0, then (1 + t)q≥1 + tq+qt + A(c)t2where2< q < 3 and A(c) > 0. Proof. (i) Let f(t) = (1 + t)q−1 − tq− q
q−1t
q−1for t ≥ 0 and q ≥ 2. Then f(0) = 0, and
f0(t) = q[(1 + t)q−1−tq−1−tq−2].
Since q ≥ 2, we get(1 + t)q−1=(1 + t)q−2+t(1 + t)q−2≥tq−2+tq−1. Thus, f0(t) ≥ 0. (ii) The proof is similar to(i).
(iii) Let g(t) = (1 + t)q−1 − tq−qt −q−2q tq−1for t ≥ 0 and q ≥ 3. Then g(0) = 0, and by (i), we obtain g0(t) = q (1 + t)q−1−tq−1−1 −q −1 q −2t q−2 ≥0. (iv) Let h(t) = (1 + t)q−tq. Then h(0) = 1,
h0(t) = q[(1 + t)q−1−tq−1]and h0(0) = q, h00(t) = q (q − 1) [(1 + t)q−2−tq−2]> 0.
Since t ≤ c for some c> 0, applying the Taylor theorem, we have (1 + t)q−tq−1 − qt ≥ q(q − 1) 2 [(1 + c) q−2−cq−2]t2. ByLemma 22, we obtain (a + b)q≥aq+bq+qaq−1b + q q −2ab q−1 for q ≥ 3 and a, b ≥ 0, (7) and
(a + b)q≥aq+bq+qaq−1b + A(c)aq−2b2 for 2< q < 3 and b/a ≤ c. (8)
Lemma 23. (i) There exists a number t0> 0 such that for 0 ≤ t < t0and eachvz ∈H01(Ω), we have
Jh(u0+tvz) < Jh(u0) + α(Ω);
(ii) There exist positive numbers l1and t1such that for any t> t1and |z| ≥ l1, we have
Proof. (i) Since Jhis continuous in H01(Ω) and {vz} is bounded in H01(Ω), there is a t0> 0 such that for 0 ≤ t < t0 and eachvz ∈H01(Ω) Jh(u0+tvz) < Jh(u0) + α (Ω). (ii) ByLemma 20, Jh(tvz) = t2 2 − tp p 2 p
p−2α (Ω) + o(1) as |z| → ∞. There is an l1> 0 such that for |z| ≥ l1
Jh(tvz) < t2 2 − tp p 2 p p −2α (Ω) + 1. Thus, there exists a t1> 0 such that
Jh(tvz) < 0 for any t > t1and |z| ≥ l1.
Lemma 24. There exists a number l0> 0 such that for |z| ≥ l0
sup
t ≥0
Jh(u0+tvz) < Jh(u0) + α(RN) = αh(Ω) + α(Ω),
where u0is the local minimum inLemma17.
Proof. ByLemma 23, we only need to show that there exists an l0> 0 such that for |z| ≥ l0
sup
t0≤t ≤t1
Jh(u0+tvz) < Jh(u0) + α(RN).
Since u0is a positive solution of Eq.(1)in Ω , then
hu0, tvziH1 = Z Ω t u0p−1vz+htvz dz.
For t0≤t ≤ t1, since supt ≥0J(tu) = α(RN) and 0 ≤ ψR ≤1, we obtain
Jh(u0+tvz) = 1 2ku0+tvzk 2 H1 − 1 p Z Ω(u0 +tvz)p− Z Ω h(u0+tvz) = Jh(u0) +J(tvz) + hu0, tvziH1+ 1 p Z Ω u0p+(tvz)p−(u0+tvz)p−htvz = Jh(u0) + J (tvz) − 1 p Z Ω(u 0+tvz)p−u0p−(tvz)p−pu0p−1(tvz) ≤ Jh(u0) + α(RN) + t2 2 Z RN |∇ψR|2[u(z − z)]2dz +t2 Z RN |∇ψR| |∇u(z − z)| u (z − z) dz +t p p Z RN 1−ψRp [u(z− z)]pdz −1 p Z RN(u0 +tvz)p−u0p−(tvz)p−pu0p−1(tvz).
Since the support of 1 −ψRp is bounded, then Z RN 1 −ψRp [u(z − z)]pdz ≤ C1exp(−p |z|). (9) Similarly, we have Z supp(∇ψR) |∇ψR|2[u(z − z)]2dz ≤ C2exp(−2 |z|), (10)
and Z supp(∇ψR) |∇ψR| |∇u(z − z)| u (z − z) dz ≤ C3exp(− (2 − ε) |z|). (11) (i) For 3 ≤ p< 2∗, by(7) Z RN(u0 +tvz)p≥ Z RN u0p+(tvz)p+pu0p−1(tvz) + p p −2u0(tvz) p−1.
Thus, byLemma 21, for |z| ≥ R1
Z
RN
u0vzp−1dz ≥ c1exp(− min {1, p − 1} (1 + ε) |z|) ≥ c1exp(− (1 + ε) |z|). (12)
Choosingε < 1/2 and using(9)–(12)we have for |z| ≥ R10 ≥ R1
sup
t0≤t ≤t1
Jh(u0+tvz) < Jh(u0) + α(RN).
(ii) For 2< p < 3, byLemma 22(ii), we get (I ) = (u0+tvz)p−u0p−(tvz)p−pu p−1 0 (tvz) ≥ 0. Then Z RN(I ) dz ≥ Z { R≤|z|≤2R} (I ) dz. (13)
Since max {vz(z) /u0(z) | R ≤ |z| ≤ 2R} ≤ c < ∞ for each z ∈ RN, by(8)
Z { R≤|z|≤2R} (u0+tvz)p≥ Z { R≤|z|≤2R} u0p+(tvz)p+pu0p−1(tvz) + A(c)u0p−2(tvz)2.
Thus, byLemma 21, for |z| ≥ R2
Z
{ R≤|z|≤2R}
u0p−2v2zdz ≥ c2exp(− min {2, p − 2} (1 + ε) |z|)
≥ c2exp(− (p − 2) (1 + ε) |z|). (14)
Choosingε < (4 − p) / (p − 1) and using(9)–(11),(13)and(14), we have for |z| ≥ R02≥ R2
sup
t0≤t ≤t1
Jh(u0+tvz) < Jh(u0) + α(RN).
Let l0=max{R10, R 0
2}, we complete the proof.
For n ∈ N, we define
vn(z) = ψR(z) ¯u (z − nz),
where z is a unit vector in RN. Clearly,vn∈ H01(Ω).
Remark 1. There exists a positive integer n0such that for n ≥ n0
sup
t ≥0
Jh(u0+tvn) < Jh(u0) + α (Ω) uniformly in z,
where u0is the local minimum inLemma 17.
In the following, we study the idea of Adachi–Tanaka [1]. For c ∈ R, we denote [ Jh≤c] =u ∈ M−h | Jh(u) ≤ c ⊂ M−h.
We try to show for a sufficiently smallσ > 0
To prove(15), we need some preliminaries. Recall the definition of Lusternik–Schnirelman category.
Definition 25. (i) For a topological space X , we say a non-empty, closed subset A ⊂ X is contractible to a point in Xif and only if there exists a continuous mapping
η : [0, 1] × A → X such that for some x0∈X
η (0, x) = x for all x ∈ A, and η (1, x) = x0 for all x ∈ A. (ii) We define cat(X) = min (
k ∈ N | there exist closed subsets A1, . . . , Ak ⊂Xsuch that
Aj is contractible to a point in X for all j and k [ j =1 Aj =X ) .
When there do not exist finitely many closed subsets A1, . . . , Ak ⊂ Xsuch that Aj is contractible to a point in X
for all j and ∪kj =1Aj =X, we say cat(X) = ∞.
We need the following two lemmas.
Lemma 26. Suppose that X is a Hilbert manifold and Ψ ∈ C1(X, R). Assume that there are c0∈ R and k ∈ N,
(i) Ψ (x) satisfies the (PS)c-condition for c ≤ c0;
(ii) cat ({x ∈ X | Ψ (x) ≤ c0}) ≥ k.
ThenΨ(x) has at least k critical points in {x ∈ X; Ψ (x) ≤ c0}.
Proof. See Ambrosetti [3, Theorem 2.3].
Lemma 27. Let N ≥ 1, SN −1= {z ∈ RN | |z| =1}, and let X be a topological space. Suppose that there are two continuous maps
F : SN −1→X, G : X → SN −1
such that G ◦ F is homotopic to the identity map of SN −1, that is, there exists a continuous mapζ : [0, 1] × SN −1→ SN −1such that
ζ (0, z) = (G ◦ F) (z) for each z ∈ SN −1,
ζ (1, z) = z for each z ∈ SN −1.
Then
cat(X) ≥ 2.
Proof. See Adachi–Tanaka [1, Lemma 2.5]. Let A1= u ∈ H01(Ω)\{0} u ≥0 and 1 kukH1 t− u kukH1 > 1 ∪ {0} A2= u ∈ H01(Ω)\{0} u ≥0 and 1 kukH1 t− u kukH1 < 1 . From Tarantello [15], we have the following results.
Lemma 28. (i) A\M−h =A1∪A2, where A = {u ∈ H01(Ω) | u ≥ 0};
(ii) M+h ⊂A1;
(iii) there exist t0> 1 and n1≥n0such that u0+t0vn ∈ A2for each n ≥ n1, where n0is defined as inRemark1;
(iv) there exists a sequence {sn} ⊂(0, 1) such that u0+snt0vn∈M−h for each n ≥ n1;
(v) αh−< αh(Ω) + α (Ω).
Proof. (i) ByLemma 10(iii). (ii) For each u ∈ M+h, we have
1< tmax(u) < t−(u) =
1 kukH1 t− u kukH1 ,
then M+h ⊂A1. In particular, u0∈ A1, where u0is defined as inLemma 17.
(iii) There is a constant c> 0 such that 0 < t− u0+tvn
ku0+tvnkH 1 < c for each t ≥ 0 and each n ∈ N. Otherwise, there
exist a sequence {tn} and a subsequence {vn} such that t−
u0+tnvn ku0+tnvnkH 1 → ∞as n → ∞. Letwn = ku0u+t0+tnvnnvkn H 1 . We claim that b(wn) is bounded below away from zero.
Case (a): there is a subsequence {tn}such that tn=c0+o(1) as n → ∞, where c0> 0. ByLemma 20, we have
a(vn) = b (vn) + o (1) = 2 p p −2α (Ω) + o (1). Thus, b(wn) = 1 u0 tn +vn p H1 Z Ω u0 tn +vn p ≥ b(vn) 2p−1 u0 tn p H1 + kvnk p H1 = 2 p p−2α (Ω) 2p−1 ku0kH 1p c0p + 2 p p−2α (Ω) p2 +o(1). Case (b): tn→ ∞as n → ∞. The proof is similar to Case (a).
Case(c): there is a subsequence {tn}such that tn=o(1) as n → ∞. ByLemma 20, we have
ku0+tnvnk2H1 = ku0k2H1+t 2 nkvnk2H1+2tnhvn, u0iH1 = ku0k2H1+o(1). Thus, b(wn) ≥ 1 ku0+tnvnkHp1 Z Ω u0p = 1 ku0kpH1 Z Ω u0p+o(1). Since t−(wn) wn∈M−h ⊂Mh, we have Jh t−(wn) wn = 1 2t −(w n)2− 1 pt −(w n)pb(wn) − t−(wn) Z Ω hwn →−∞ as n → ∞.
However, Jhis bounded below on Mh, which is a contradiction. Let
t0= p −2 2 pα (Ω) c 2−a(u 0) 12 +1,
then ku0+t0vnk2H1 =a(u0) + t02 2 p p −2 α (Ω) + o(1) > c2+o(1) ≥ t− u 0+t0vn ku0+t0vnkH1 2 +o(1).
Thus, there is an n1≥n0, where n0is defined as inRemark 1, such that, or n ≥ n1,
1 ku0+t0vnkH1 t− u 0+t0vn ku0+t0vnkH1 < 1, or u0+t0vn∈ A2.
(iv) Define a pathγn(s) = u0+st0vnfor s ∈ [0, 1] and each n ≥ n1where t0> 1, then
γn(0) = u0∈ A1, γn(1) = u0+t0vn ∈ A2. Since kuk1 H 1 t−kuku H 1
is a continuous function for nonzero u andγn([0, 1]) is connected, there exists a sequence
{sn} ⊂(0, 1) such that u0+snt0vn∈M−h.
(v) By part (iv) andRemark 1, α−
h ≤ Jh(u0+snt0vn) < Jh(u0) + α (Ω) = αh(Ω) + α (Ω).
For n ≥ n1, we define a map Fn:SN −1→H01(Ω) by
Fn(z) (z) = u0(z) + snt0vn(z) for z ∈ SN −1,
wherevn(z) = ψR(z) u (z − nz). Then we have
Lemma 29. There exists a sequence {σn} in R+such that
Fn(SN −1) ⊂ [Jh≤αh(Ω) + α (Ω) − σn].
Proof. ByLemma 28(iv) andRemark 1, we have that u0+snt0vn∈Mh−and Jh(u0+snt0vn) ≤ αh(Ω) + α(Ω) − σn
for each n ≥ n1, the conclusion holds.
For c> 0, we define bc(u) = Z Ω cup; Ic(u) = 1 2a(u) − 1 pbc(u+) ; MIc = {u ∈ H 1 0(Ω)\{0} | hI 0 c(u), ui = 0}.
Recall that there exist a unique t−=t−(u) > 0 and a unique t1=t1(u) > 0 such that t−u ∈M−h and t1u ∈M. Let Σ = {u ∈ H01(Ω) | u ≥ 0 and kukH1 =1}. Then we have the following results.
Lemma 30. (i) For each u ∈ Σ , there exists a unique number tc(u) > 0 such that tc(u)u ∈ MIcand
max t ≥0 Ic(tu) = Ic t c(u) u = 1 2− 1 p bc(u) − 2 p−2;
(ii) For each nonnegative u ∈ H01(Ω) and 0 < µ < 1, we have (1 − µ) I 1 1−µ(u) − 1 2µkhk 2 L2 ≤ Jh(u) ≤ (1 + µ) I 1 1+µ(u) − 1 2µkhk 2 L2;
(iii) For each u ∈ Σ and 0< µ < 1, we have (1 − µ)p−2p J(t1u) − 1 2µkhk2L2 ≤ Jh t−u ≤(1 + µ) p p−2 J(t1u) − 1 2µkhk2L2;
(iv) αh−> 0 for sufficiently small khkL2.
Proof. (i) For each u ∈ Σ , let f(t) = Ic(tu) =12t2−1ptpbc(u), then f (t) → −∞ as t → ∞, f0(t) = t −tp−1bc(u)
and f00(t) = 1 − (p − 1) tp−2bc(u). Let
tc(u) = 1
bc(u)
p−21 > 0. Then f0(tc(u)) = 0, tc(u) u ∈ MIc and
tc(u)2 f00 tc(u) = a tc(u)u − (p − 1) bc tc(u)u
=(2 − p) tc(u)2a(u) < 0.
Thus, there exists a unique tc(u) > 0 such that tc(u)u ∈ MIcand
max t ≥0 Ic(tu) = Ic t c(u) u = 1 2 − 1 p bc(u) −p−22 . (ii) Forµ ∈ (0, 1), we get
Z Ω hudz ≤ kukH1khkL2 ≤ µ 2 kuk 2 H1+ 1 2µkhk 2 L2.
Thus, for each nonnegative u ∈ H01(Ω), then 1 −µ 2 kuk 2 H1− 1 p Z Ω up− 1 2µkhk 2 L2 ≤ Jh(u) ≤ 1 +µ 2 kuk 2 H1− 1 p Z Ω up+ 1 2µkhk 2 L2.
(iii) Applying part (ii), we have that for each u ∈ Σ (1 − µ) I 1 1−µ t c1u − 1 2µkhk 2 L2 ≤Jh t−u ≤(1 + µ) I 1 1+µ t c2u − 1 2µkhk 2 L2, where tc1u ∈M I 1 1−µ and tc2u ∈M I 1 1+µ
. By part (i), then
I 1 1−µ t c1u = 1 2− 1 p b 1 1−µ(u) − 2 p−2 =(1 − µ)p−22 1 2 − 1 p b(u)−p−22 =(1 − µ)p−22 J(t1u). Similarly, I 1 1+µ(t
c2u) = (1 + µ)p−22 J(t1u). Hence, (iii) holds.
(iv) We know thatα (Ω) > 0 and applying part (ii) to obtain α− h ≥(1 − µ) p p−2 α (Ω) − 1 2µkhk 2 L2.
The following lemma is a key lemma to prove our main result.
Lemma 31. There exists a numberδ0> 0 such that if u ∈ M and J(u) ≤ α (Ω) + δ0, then
Z RN z |z| |∇u|2+u2dz 6= 0.
Proof. On the contrary, there exists a sequence {un} in M such that J(un) = α (Ω) + o (1) and
Z RN z |z| |∇un|2+u2n dz = 0.
ByLemma 9, {un} is a(PS)α(Ω)-sequence in H01(Ω) for J. Since inf
v∈M(Ω)J(v) = α (Ω) = α(R
N) is not achieved.
Let u be the unique positive solution of Eq.(3)in RN. It follows from the DecompositionLemma 6that there exists a sequence {zn} in RNsuch that |zn| → ∞ as n → ∞ and
un(z) = u(z − zn) + o (1) strongly in H1(RN).
Suppose the subsequence zn
|zn| → z0as n → ∞, where z0is a unit vector in R
N. Then by the Lebesgue dominated
theorem, we have 0 = Z RN z |z| |∇un|2+u2n dz = Z RN z + zn |z + zn| |∇u|2+u2dz + o(1) = 2 p p −2 α(RN)z 0+o(1), which is a contradiction.
Lemma 32. There exists a number d0> 0 such that for khkL2 < d0, we have
Z RN z |z| |∇u|2+u2dz 6= 0, for u ∈[ Jh< αh(Ω) + α (Ω)].
Proof. For u ∈ [ Jh< αh(Ω) + α(Ω)], then u/ kukH1 ∈ Σ . There exists a t1 > 0 such that t1u/ kukH1 ∈ M. By
Lemma 30(ii), we have for anyµ ∈ (0, 1) and khkL2 < d1(µ)
J t1u kukH1 ≤(1 − µ)− p p−2 Jh(u) + 1 2µkhk2L2 . Sinceαh(Ω) < 0, we have, for u ∈ [Jh< αh(Ω) + α (Ω)],
J t1u kukH1 ≤(1 − µ)− p p−2 α (Ω) +21µkhk2L2 .
Takeµ ∈ (0, 1) such that there exists d1(µ) > d0> 0 such that for khkL2 < d0
J t1u
kukH1
≤α (Ω) + δ0. Since t1u/ kukH1 ∈M, byLemma 31
Z RN z |z| ∇ t1u kukH1 2 + t1u kukH1 2! dz 6= 0, or Z RN z |z| |∇u|2+(u)2dz 6= 0.
For khkL2 < d0, we obtain Z RN z |z| |∇u|2+ |u|2dz 6= 0
for all u ∈ [ Jh< αh(Ω) + α (Ω)]. Now, we define
G :[ Jh < αh(Ω) + α (Ω)] → SN −1 by G(u) = Z RN z |z| |∇u|2+ |u|2dz Z RN z |z| |∇u|2+ |u|2dz . We have
Lemma 33. For n ≥ n1and khkL2 < d0, the map
G ◦ Fn:SN −1→SN −1
is homotopic to the identity. Proof. We define ζn(θ, z) : [0, 1] × SN −1→ SN −1 by ζn(θ, z) = G((1 − 2θ) Fn(z) + 2θu (z − nz)) for θ ∈ [0, 1/2) ; G u z − n 2(1 − θ)z forθ ∈ [1/2, 1) ; z forθ = 1.
First, we claim that limθ→1−ζn(θ, z) = z and lim
θ→1 2
−ζn(θ, z) = G (u (z − nz)).
(a) limθ→1−ζn(θ, z) = z: since
Z RN z |z| ∇u z − n 2(1 − θ)z 2 +u z − n 2(1 − θ)z 2! dz = Z RN z +2(1−θ)n z z + n 2(1−θ)z |∇u(z)|2+u(z)2dz = 2 p p −2 α(RN)z + o(1) as θ → 1−, then limθ→1−ζn(θ, z) = z. (b) limθ→1 2 −ζn(θ, z) = G (u (z − nz)): since k(1 − 2θ) Fn(z) + 2θu (z − nz)kH1 = ku(z − nz)kH1+o(1) as θ → 1 2 − , by the continuity of G, we obtain limθ→1
2
−ζn(θ, z) = G (u (z − nz)). Thus, ζn(θ, z) ∈ C([0, 1] × SN −1, SN −1) and
ζn(0, z) = G (Fn(z)) for all z ∈ SN −1,
ζn(1, z) = z for all z ∈ SN −1,
provided n ≥ n1and khkL2 < d0. This completes the proof.
Lemma 34. Assume that h(z) ≥ 0 and 0 < khkL2 < min {d(p, α), d0}, where d0is defined as inLemma32. Jh(u)
has at least two critical points in [ Jh < αh(Ω) + α (Ω)].
Proof. ApplyingLemmas 27and33, we have for sufficiently large n ≥ n1and khkL2 < d0,
cat([Jh≤αh(Ω) + α (Ω) − σn]) ≥ 2.
ByLemmas 18and26, Jh(u) has at least two critical points in [Jh < αh(Ω) + α (Ω)].
5. Existence of the fourth solution
Sinceαh−> 0 for sufficiently small khkL2, we define
Kh(u) = sup t ≥0
Jh(tu) = Jh(t−u) > 0,
where t−u ∈Mh. We observe that if khkL2 is sufficiently small, Bahri–Li’s minimax argument [4] also works for Kh.
Let
Γ = {γ ∈ C(Br(0), Σ) | γ |∂ Br(0)=ψR(z) ¯u (z − y) / kψR(z) ¯u (z − y)kH1} for large r = |y|,
where Σ = {u ∈ H01(Ω) | u ≥ 0 and kukH1 =1}. Then we define
γh(Ω) = inf γ ∈Γy∈RsupN Kh(γ (y)); γ0(Ω) = inf γ ∈Γy∈RsupN K0(γ (y)).
ByLemma 30(iii), for 0< µ < 1, we have (1 − µ)p−2p γ 0(Ω) − 1 2µkhk 2 L2 ≤γh(Ω) ≤ (1 + µ) p p−2γ 0(Ω) − 1 2µkhk 2 L2. (16) Lemma 35. α (Ω) < γ0(Ω) < 2α (Ω).
Proof. Bahri–Li [4] proved that Eq.(3)admits at least one positive solution u in Ω and J(u) = γ0(Ω) < 2α (Ω).
Lien–Tzeng–Wang [13] proved that Eq.(3) does not have a positive ground state solution in Ω . Hence,α (Ω) < γ0(Ω) < 2α (Ω).
Lemma 36. There exists a number d00 > 0 such that if 0 < khkL2 < d00, then
αh(Ω) + α(Ω) < γh(Ω) < α−h(Ω) + α(Ω).
Moreover, there exists a positive solution u of Eq.(1)inΩ such that Jh(u) = γh(Ω).
Proof. ByLemma 30(iii), we also have that for 0< µ < 1 (1 − µ)p−2p α(Ω) − 1 2µkhk 2 L2 ≤α − h(Ω) ≤ (1 + µ) p p−2 α(Ω) − 1 2µkhk 2 L2.
For anyε > 0, there exists a d1(ε) > 0 such that if khkL2 < d1(ε), then
α(Ω) − ε < α−
h(Ω) < α(Ω) + ε.
Thus,
2α(Ω) − ε < α−
h(Ω) + α(Ω) < 2α(Ω) + ε.
Applying(16), for anyδ > 0, there exists a d2(δ) > 0 such that if khkL2 < d2(δ), then
Fix a small 0 < ε < (2α (Ω) − γ0(Ω))/2, since α (Ω) < γ0(Ω) < 2α (Ω), choosing a δ > 0 such that for
khkL2 < d00 =min{d1, d2}, we get
αh(Ω) + α(Ω) < α(Ω) < γh(Ω) < 2α(Ω) − ε < αh−(Ω) + α(Ω).
Therefore, by Lemmas 2 and18(ii), we obtain that there exists a positive solution u of Eq. (1) in Ω such that Jh(u) = γh(Ω).
We can conclude the following theorem.
Theorem 37. Assume that h(z) ≥ 0 and 0 < khkL2 < min{d(p, α), d0, d00}, where d0is defined as inLemma32and
d00 is defined as inLemma36. Then there are at least four positive solutions of Eq.(1)inΩ .
Proof. Therefore, byLemmas 2,15,34and36, we have that Eq.(1)has at least four positive solutions in Ω . References
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