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Four positive solutions of semilinear elliptic equations in exterior domains
Nonlinear Analysis 67 (2007) 1129–1146 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,1 a Center for General Education, Chang Gung University, 259, Wen Hwa 1st Road, Kwei Shan, 33375, Tao-Yuan, Taiwan b Department of Applied Mathematics, Hsuan Chuang University, Hsinchu, Taiwan c Center 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 khk L 2 > 0, we prove that if khk L 2 is sufficiently small, then there are at least four positive solutions of the semilinear elliptic equation in R N \D, where D is a C 1,1 bounded domain in R N . 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−2 u + h(z) in Ω ; u ∈ H01 (Ω ),
(1)
where Ω = R N \D, D is a C 1,1 bounded domain in R N , 2 < p < 2∗ = 2N /(N − 2) for N ≥ 3. Let d( p, α) = ( p − 2)
1 p−1
p−1 p−2
2p p−2
1
2
1
α(Ω ) 2 ,
(2)
where α(Ω ) is defined below. In this paper, we first consider h(z) ≥ 0 and 0 < khk L 2 < d( p, α). Associated with Eq. (1), we define the functionals a, b, and Jh , for u ∈ H01 (Ω ), Z a(u) = |∇u|2 + u 2 ; ZΩ |u| p ; b(u) = Ω ∗ Corresponding author.
E-mail address: [email protected] (H.-L. Lin). 1 Present address: Department of Applied Mathematics, National University of Kaohsiung, Kaohsiung, Taiwan. c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.07.001
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Jh (u) =
Z
1 1 a(u) − b (u + ) − hu. 2 p Ω
By Rabinowitz [14, Proposition B.10.], a, b, and Jh are of C 2 . For h = 0, we consider the semilinear elliptic equation −1u + u = |u| p−2 u in Ω ; (3) u ∈ H01 (Ω ), and the energy functional J (u) = 21 a(u) − 1p b (u + ). Bahri–Lions [5] showed that Eq. (3) admits at least one positive solution u in R N , 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 R N , 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 R N . 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 R N 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−2 u + h (z) in R N ; (4) u ∈ H 1 (R N ), 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) ∈ R N −l × Rl . Consider the exterior strip domain Ar \D, where r A = {(x, y) ∈ R N | |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) For β ∈ R, a sequence {u n } is a (PS)β -sequence in H01 (Ω ) for Jh if Jh (u n ) = β + o(1) and Jh0 (u n ) = o(1) strongly in H −1 (Ω ) as n → ∞; (ii) β ∈ R is a (PS)-value in H01 (Ω ) for Jh if 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 ∈ H01 (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 Z p−1 ∇u∇ϕ + uϕ = u + ϕ + hϕ for any ϕ ∈ H01 (Ω ). Ω
Ω
p−1
Thus, u is a weak solution of −1u +u = u + +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
hJh0 (u) , ui = 0} and αh (Ω ) = inf Jh (u). u∈Mh
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 3. For each nonnegative u ∈ Σ = {u ∈ H01 (Ω ) | kuk H 1 = 1}, there exists a su > 0 such that su u ∈ M.
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Lemma 4. (i) For each nonnegative u ∈ H01 (Ω )\{0}, there exists a su > 0 such that su u ∈ M; (ii) Let β > 0 and {u n } be a sequence in H01 (Ω )\{0} for J such that J (u n ) = β + o(1) and a(u n ) = b(u + n ) + o(1). Then there is a sequence {sn } in R+ such that sn = 1 + o(1), {sn u n } in M and J (sn u n ) = β + o(1). Lemma 5. If u ∈ H01 (Ω )\{0} and u ≥ 0, then p
a(u) 2 b(u)
!
1 p−2
≥
2p p−2
Proof. Applying Lemma 4.
1
2
1
α(Ω ) 2 .
By Lien–Tzeng–Wang [13], we have the following lemma. Lemma 6 (Palais–Smale Decomposition Lemma for J ). Let {u n } be a (PS)β -sequence in H01 (Ω ) for J . Then there ∞ are a subsequence {u n }, a positive integer l, sequences z ni n=1 in R N , functions u in H01 (Ω ), and wi 6= 0 in H 1 (R N ) for 1 ≤ i ≤ l such that i z n → ∞ for 1 ≤ i ≤ l; −4u + u = |u| p−2 u in Ω ; p−2 −4wi + wi = wi wi in R N ; l X un = u + wi (· − z ni ) + o(1) strongly in H 1 (R N ); i=1
J (u n ) = J (u) +
l X
J (wi ) + o(1).
i=1
In addition, if u n ≥ 0, then u ≥ 0 and wi ≥ 0 for 1 ≤ i ≤ l. Lemma 7 (Palais–Smale Decomposition Lemma for Jh ). Let {u n } be a (PS)β -sequence in H01 (Ω ) for Jh . Then there ∞ are a subsequence {u n }, a positive integer l, sequences z ni n=1 in R N , functions u in H01 (Ω ), and wi 6= 0 in H 1 (R N ) for 1 ≤ i ≤ l such that i z n → ∞ for 1 ≤ i ≤ l; −4u + u = |u| p−2 u + h(z) in Ω ; p−2 −4wi + wi = wi wi in R N ; l X un = u + wi (· − z ni ) + o(1) strongly in H 1 (R N ); i=1
Jh (u n ) = Jh (u) +
l X
J (wi ) + o(1).
i=1
In addition, if u n ≥ 0, then u ≥ 0 and wi ≥ 0 for 1 ≤ i ≤ l. Proof. See Zhu–Zhou [19].
R Define ψ(u) = hJh0 (u) , ui = a(u)−b (u + )− Ω hu. Then if 0 < khk L 2 < d( p, α), we have the following lemma. Lemma 8. For each u ∈ Mh , we have hψ 0 (u), ui = a(u) − ( p − 1) b(u) 6= 0. Proof. By Lemma 5, Tarantello [15, Lemma 2.3] and Cao–Zhou [8].
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− By Lemma 8, we write Mh = M+ h ∪ Mh , where
M+ h = {u ∈ Mh | a (u) − ( p − 1) b(u) > 0}, M− h = {u ∈ Mh | a (u) − ( p − 1) b(u) < 0}. Define αh− (Ω ) = inf Jh (u).
αh+ (Ω ) = inf Jh (u);
u∈M− h
u∈M+ h
By Wang–Wu [17], we have the following lemma. Lemma 9. {u n } is a (PS)α(Ω ) -sequence in H01 (Ω ) for J if and only if J (u n ) = α(Ω ) + o (1) and a (u n ) = 1 b u+ n + o (1). In particular, every minimizing sequence {u n } in M of α(Ω ) is a (PS)α(Ω ) -sequence in H0 (Ω ) for J . For each nonnegative u ∈ H01 (Ω )\{0}, we write tmax =
a(u) ( p − 1) b (u)
1 p−2
> 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≥tmax Jh (tu); (ii) t − (u) isna continuous function; o u 1 1 − (iii) M− h = u ∈ H0 (Ω )\{0} | u ≥ 0 and kuk H 1 t kuk H 1 = 1 ; R (iv) If Ω hu > 0, then there is a unique number 0 < t + = t + (u) < tmax such that t + u ∈ M+ h and Jh (t + u) = min0≤t≤t − Jh (tu). Proof. See Tarantello [15] and Cao–Zhou [8]. R + Lemma 11. (i) For each u ∈ M+ h , we have Ω hu > 0 and Jh (u) < 0. In particular, αh (Ω ) ≤ αh (Ω ) < 0; (ii) Jh is coercive and bounded below on Mh . R Proof. (i) For each u ∈ M+ h , a(u) − ( p − 1) b(u) > 0 and a(u) = b (u) + Ω hu. Then Z hu = a(u) − b(u) > ( p − 2) b(u) > 0. Ω
Hence Z 1 1 p−2 p−2 1 Jh (u) = − b (u) − hu < b(u) − b(u) 2 p 2 Ω 2p 2 ( p − 1) ( p − 2) =− b(u) < 0. 2p
(ii) By Tarantello [15, p. 288].
Lemma 12. Let u be in Mh such that Jh (u) = min Jh (v) = αh (Ω ). Then v∈Mh R (i) Ω hu > 0; (ii) u is a solution of Eq. (1) in Ω . Proof. (i) By Lemma 11(i), we have Z 1 1 1 a(u) − 1 − 0 > αh (Ω ) = Jh (u) = − hu. 2 p p Ω R Thus, Ω hu > 0.
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(ii) By Lemma 8, hψ 0 (v), vi 6= 0 for each v ∈ Mh . Since Jh (u) = minv∈Mh Jh (v), by the Lagrange multiplier theorem, there is a λ ∈ R such that Jh0 (u) = λψ 0 (u) in H −1 (Ω ). Then we have 0 = hJh0 (u), ui = λhψ 0 (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 a δ > 0 and a differentiable functional l : B (0; δ) ⊂ H01 (Ω ) → R+ such that l (0) = 1, l (v) (u − v) ∈ Mh for v ∈ B (0; δ) and hψ 0 (u), ϕi hl 0 (v) , ϕi (l,v)=(1,0) = hψ 0 (u), ui
for ϕ ∈ Cc∞ (Ω ).
Lemma 14. (i) There exists a (PS)αh (Ω ) -sequence {u n } in Mh for Jh ; (ii) There exists a (PS)α + (Ω ) -sequence {u n } in M+ h for Jh ; h
(iii) There exists a (PS)α − (Ω ) -sequence {u n } in M− h for Jh . h
3. Existence of the first solution By Lemma 14(i), there is a (PS)αh (Ω ) -sequence {u n } in Mh for Jh . Then we have the following (PS)αh (Ω ) condition. Lemma 15. Let {u n } ⊂ Mh be a (PS)αh (Ω ) -sequence for Jh . Then there exist a subsequence {u n } and a nonzero u 0 ∈ H01 (Ω ) such that u n → u 0 strongly in H01 (Ω ). Moreover, u 0 is a positive solution of Eq. (1) such that Jh (u 0 ) = αh (Ω ). Proof. By Lemma 11(ii), {u n } is bounded in H01 (Ω ). Take a subsequence {u n } and u 0 ∈ H01 (Ω ) such that u n * u 0 weakly in H01 (Ω ). Then we have that u 0 is a nonzero solution of Eq. (1) in Ω . Since Z 1 1 Jh (u n ) = a (u n ) − b (u n ) − hu n = αh (Ω ) + o (1), 2 p ΩZ hu n = o (1), hJh0 (u n ) , u n i = a (u n ) − b (u n ) − Ω
we obtain Z 1 1 1 − a (u n ) − 1 − hu n = αh (Ω ) + o (1). 2 p p Ω R R Since the functional a is weakly lower semicontinuous and Ω hu n → Ω hu 0 as n → ∞, then Jh (u 0 ) = αh (Ω ). Let pn = u n − u 0 . By the Br´ezis–Lieb Lemma, we get Z 1 1 hpn Jh ( pn ) = a ( pn ) − b ( pn ) − 2 p Ω Z Z 1 1 1 1 = a (u n ) − a (u 0 ) − b (u n ) + b (u 0 ) − hu n + hu 0 + o (1) 2 2 p p Ω Ω = Jh (u n ) − Jh (u 0 ) + o (1) = o (1). (5) R By the Br´ezis–Lieb Lemma, Ω hpn = o (1) and u 0 is a solution of Eq. (1), so Z 0 hJh ( pn ) , pn i = a ( pn ) − b ( pn ) − hpn Ω Z Z = a (u n ) − a (u 0 ) − b (u n ) + b (u 0 ) − hu n + hu 0 + o (1) Ω
= hJh0 (u n ) , u n i − hJh0 (u 0 ) , u 0 i = o (1).
Ω
(6)
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R Thus, by (5) and (6) Ω hpn = o (1), we have p−2 a ( pn ) = o (1), 2p that is, un → u0
strongly in H01 (Ω ).
Moreover, u 0 is a solution of Eq. (1) such that Jh (u 0 ) = αh (Ω ). By Lemma 2, u 0 is positive in Ω .
We prove that u 0 is the unique critical point of Jh in B (r0 ) in the following lemma. 1 1 1 p−2 2p 2 1 Lemma 16. Let r0 = p−1 α (Ω ) 2 . Then p−2 1 (i) M+ h ⊂ B (r0 ) = {u ∈ H0 (Ω ) | kuk H 1 < r0 }; (ii) Jh (u) is strictly convex in B (r0 ).
R Proof. (i) If u ∈ M+ h , then a(u) > ( p − 1) b(u) and a(u) = b (u) + Ω hu. Thus, 1 a(u) + khk L 2 kuk H 1 . a(u) < p−1 This implies
p−1 khk L 2 < p−2 p−1 1 p−2 2 1 1 p−1 2p < α (Ω ) 2 ( p − 2) p−2 p−1 p−2 1 1 p−2 2 1 1 2p α (Ω ) 2 = r0 . = p−1 p−2
kuk H 1
(ii) We know Jh00 (u) (v, v) = a (v) − ( p − 1)
Z Ω
|u| p−2 v 2
for all v ∈ H01 (Ω ).
Thus, by Lemma 5, we obtain p−2
Jh00 (u) (v, v) ≥ a (v) − ( p − 1) kuk L p kvk2L p p−2 2 ( p−2)2 p−2 p − 2 − ≥ a (v) − ( p − 1) a(u) 2 α (Ω ) 2 p 2p p−2 p −( p−2) p − 2 × a(v) α (Ω ) p 2p 2− p 2 2 p p−2 kuk H 1 ≥ a(v) 1 − ( p − 1) α (Ω ) p−2 >0 Thus,
Jh00 (u)
for u ∈ B(r0 )\{0}.
is positive definite for u ∈ B (r0 ) and Jh is strictly convex in B (r0 ).
By Lemma 15, there exists a solution u 0 ∈ Mh of Eq. (1) such that Jh (u 0 ) = αh (Ω ). Furthermore, we have the following lemma. + Lemma 17. (i) u 0 ∈ M+ h and Jh (u 0 ) = αh (Ω ) = αh (Ω ); (ii) u 0 is the unique critical point of Jh (u) in B (r0 ), where r0 is defined as in Lemma 16; (iii) Jh (u 0 ) is a local minimum in H01 (Ω ).
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R − Proof. (i) By Lemma 12(i), Ω hu 0 > 0. We claim that u 0 ∈ M+ h . Otherwise, if u 0 ∈ Mh , then by Lemma 10, there + − + + exists a unique t (u 0 ) = 1 > t (u 0 ) > 0 such that t (u 0 )u 0 ∈ Mh and αh (Ω ) ≤ αh+ (Ω ) ≤ Jh (t + (u 0 )u 0 ) < Jh (t − (u 0 )u 0 ) = αh (Ω ), + + + which is a contradiction. Since u 0 ∈ M+ h , αh (Ω ) ≤ Jh (u 0 ) = αh (Ω ) ≤ αh (Ω ), that is, Jh (u 0 ) = αh (Ω ) = αh (Ω ). (ii) By part (i) and Lemma 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 Decomposition Lemma 7, we have the following restricted (PS)β -condition. Lemma 18. (i) If {u n } is a (PS)β -sequence in H01 (Ω ) for Jh with β < αh (Ω )+α (Ω ), then there exist a subsequence {u n } and a nonzero u 0 in H01 (Ω ) such that u n → u 0 strongly in H01 (Ω ) and Jh (u 0 ) = β. Moreover, u 0 is a positive solution of Eq. (1) in Ω ; 1 (ii) If {u n } ⊂ M− h is a (PS)β -sequence in H0 (Ω ) for Jh with αh (Ω ) + α (Ω ) < β < αh− (Ω ) + α (Ω ), 1 0 then there exist a subsequence {u n } and a nonzero u 0 ∈ M− h such that u n → u strongly in H0 (Ω ) and Jh (u 0 ) = β. Moreover, u 0 is a positive solution of Eq. (1) in Ω .
By Chen–Chen–Wang [6, Proposition 1], we have the following lemma. Lemma 19. Let u be a positive solution of Eq. (1) in Ω . Then for any ε > 0, there are positive constants cε and R such that D ⊂ B N (0; R) and u (z) ≥ cε exp (− (1 + ε) |z|)
for |z| ≥ R.
It is well-known that there is a positive radially symmetric smooth solution u¯ of Eq. (3) in R N such that J (u) ¯ = α(R N ). Recall the facts (i) for any ε > 0, there exist constants C0 , C00 > 0 such that for all z ∈ R N u(z) ≤ C0 exp (−|z|)
and
|∇u(z)| ≤ C00 exp (− (1 − ε) |z|);
(ii) for any ε > 0, there exists a constant Cε > 0 such that u(z) ≥ Cε exp (− (1 + ε) |z|)
for all z ∈ R N .
For such R in Lemma 19, let ψ R : R N → [0, 1] be a C ∞ -function on R N such that 0 ≤ ψ R ≤ 1, |∇ψ R | ≤ c and 1 for |z| ≥ R + 1; ψ R (z) = 0 for |z| ≤ R. We define vz (z) = ψ R (z) u¯ (z − z)
for z ∈ R N .
Clearly, vz (z) ∈ H01 (Ω ). In order to prove Lemma 24, we need the following lemmas. 2p Lemma 20. (i) a (vz ) = b (vz ) + o (1) = p−2 α(R N ) + o(1) as |z| → ∞; (ii) J (vz ) = α (Ω ) + o (1) = α(R N ) + o (1) as |z| → ∞; (iii) vz * 0 weakly in H01 (Ω ) as |z| → ∞.
Proof. It is similar to the proof of Wang [16, Lemma 30].
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Lemma 21. Let E be a domain in R N . If f : E → R satisfies Z f (z)eσ |z| dz < ∞ for some σ > 0, E
then Z f (z)e−σ |z−z| dz eσ |z| =
Z
f (z)eσ
hz,zi |z|
dz + o(1)
as |z| → ∞.
E
E
Proof. Since σ |z| ≤ σ |z| + σ |z − z|, we have f (z)e−σ |z−z| eσ |z| ≤ f (z)eσ |z| . Since −σ |z − z| + σ |z| = σ hz,zi |z| + o(1) as |z| → ∞, then the lemma follows from the Lebesque dominated convergence theorem. Lemma 22. For t ≥ 0, we have the following inequalities. q q−1 t where q ≥ 2; (i) (1 + t)q ≥ 1 + t q + q−1 q (ii) (1 + t) ≥ 1 + t q + qt where q ≥ 2; q q−1 (iii) (1 + t)q ≥ 1 + t q + qt + q−2 t where q ≥ 3;
(iv) If t ≤ c for some c > 0, then (1 + t)q ≥ 1 + t q + qt + A(c)t 2 where 2 < q < 3 and A(c) > 0. Proof. (i) Let f (t) = (1 + t)q − 1 − t q −
q q−1 q−1 t
for t ≥ 0 and q ≥ 2. Then f (0) = 0, and
f 0 (t) = q[(1 + t)q−1 − t q−1 − t q−2 ]. Since q ≥ 2, we get (1 + t)q−1 = (1 + t)q−2 + t (1 + t)q−2 ≥ t q−2 + t q−1 . Thus, f 0 (t) ≥ 0. (ii) The proof is similar to (i). q q−1 (iii) Let g(t) = (1 + t)q − 1 − t q − qt − q−2 t for t ≥ 0 and q ≥ 3. Then g(0) = 0, and by (i), we obtain q − 1 q−2 0 q−1 q−1 g (t) = q (1 + t) −t −1− t ≥ 0. q −2 (iv) Let h(t) = (1 + t)q − t q . Then h(0) = 1, h 0 (t) = q[(1 + t)q−1 − t q−1 ] and h 0 (0) = q, h 00 (t) = q (q − 1) [(1 + t)q−2 − t q−2 ] > 0. Since t ≤ c for some c > 0, applying the Taylor theorem, we have (1 + t)q − t q − 1 − qt ≥
q (q − 1) [(1 + c)q−2 − cq−2 ]t 2 . 2
By Lemma 22, we obtain (a + b)q ≥ a q + bq + qa q−1 b +
q abq−1 q −2
for q ≥ 3 and a, b ≥ 0,
(7)
for 2 < q < 3 and b/a ≤ c.
(8)
and (a + b)q ≥ a q + bq + qa q−1 b + A(c)a q−2 b2
Lemma 23. (i) There exists a number t0 > 0 such that for 0 ≤ t < t0 and each vz ∈ H01 (Ω ), we have Jh (u 0 + tvz ) < Jh (u 0 ) + α(Ω ); (ii) There exist positive numbers l1 and t1 such that for any t > t1 and |z| ≥ l1 , we have Jh (tvz ) < 0.
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Proof. (i) Since Jh is continuous in H01 (Ω ) and {vz } is bounded in H01 (Ω ), there is a t0 > 0 such that for 0 ≤ t < t0 and each vz ∈ H01 (Ω ) Jh (u 0 + tvz ) < Jh (u 0 ) + α (Ω ). 2 p 2p (ii) By Lemma 20, Jh (tvz ) = t2 − tp p−2 α (Ω ) + o(1) as |z| → ∞. There is an l1 > 0 such that for |z| ≥ l1 Jh (tvz ) <
t2 tp − 2 p
2p α (Ω ) + 1. p−2
Thus, there exists a t1 > 0 such that Jh (tvz ) < 0
for any t > t1 and |z| ≥ l1 .
Lemma 24. There exists a number l0 > 0 such that for |z| ≥ l0 sup Jh (u 0 + tvz ) < Jh (u 0 ) + α(R N ) = αh (Ω ) + α(Ω ), t≥0
where u 0 is the local minimum in Lemma 17. Proof. By Lemma 23, we only need to show that there exists an l0 > 0 such that for |z| ≥ l0 sup Jh (u 0 + tvz ) < Jh (u 0 ) + α(R N ).
t0 ≤t≤t1
Since u 0 is a positive solution of Eq. (1) in Ω , then Z p−1 hu 0 , tvz i H 1 = tu 0 vz + htvz dz. Ω
For t0 ≤ t ≤ t1 , since supt≥0 J (tu) = α(R N ) and 0 ≤ ψ R ≤ 1, we obtain Z Z 1 1 2 p h (u 0 + tvz ) = Jh (u 0 ) Jh (u 0 + tvz ) = ku 0 + tvz k H 1 − (u 0 + tvz ) − 2 p Ω Ω Z 1 p u + (tvz ) p − (u 0 + tvz ) p − htvz + J (tvz ) + hu 0 , tvz i H 1 + p Ω 0 Z 1 p p−1 = Jh (u 0 ) + J (tvz ) − (u 0 + tvz ) p − u 0 − (tvz ) p − pu 0 (tvz ) p Ω Z t2 |∇ψ R |2 [u(z − z)]2 dz ≤ Jh (u 0 ) + α(R N ) + 2 RN Z 2 |∇ψ R | |∇u (z − z)| u (z − z) dz +t RN Z tp p + 1− ψ R [u (z − z)] p dz p RN Z 1 p p−1 − (u 0 + tvz ) p − u 0 − (tvz ) p − pu 0 (tvz ). p RN p Since the support of 1 − ψ R is bounded, then Z p 1 − ψ R [u (z − z)] p dz ≤ C1 exp (− p |z|).
(9)
RN
Similarly, we have Z |∇ψ R |2 [u (z − z)]2 dz ≤ C2 exp (−2 |z|), supp(∇ψ R )
(10)
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and Z supp(∇ψ R )
|∇ψ R | |∇u (z − z)| u (z − z) dz ≤ C3 exp (− (2 − ε) |z|).
(i) For 3 ≤ p < 2∗ , by (7) Z Z p (u 0 + tvz ) ≥ RN
p
RN
p−1
u 0 + (tvz ) p + pu 0
(tvz ) +
(11)
p u 0 (tvz ) p−1 . p−2
Thus, by Lemma 21, for |z| ≥ R1 Z p−1 u 0 vz dz ≥ c1 exp (− min {1, p − 1} (1 + ε) |z|) ≥ c1 exp (− (1 + ε) |z|).
(12)
RN
Choosing ε < 1/2 and using (9)–(12) we have for |z| ≥ R10 ≥ R1 sup Jh (u 0 + tvz ) < Jh (u 0 ) + α(R N ).
t0 ≤t≤t1
(ii) For 2 < p < 3, by Lemma 22(ii), we get p
p−1
(I ) = (u 0 + tvz ) p − u 0 − (tvz ) p − pu 0
(tvz ) ≥ 0.
Then Z RN
(I ) dz ≥
Z
(I ) dz.
(13)
{R≤|z|≤2R}
Since max {vz (z) /u 0 (z) | R ≤ |z| ≤ 2R} ≤ c < ∞ for each z ∈ R N , by (8) Z Z p p−1 p−2 p u 0 + (tvz ) p + pu 0 (tvz ) + A(c)u 0 (tvz )2 . (u 0 + tvz ) ≥ {R≤|z|≤2R}
{R≤|z|≤2R}
Thus, by Lemma 21, for |z| ≥ R2 Z p−2 u 0 vz2 dz ≥ c2 exp (− min {2, p − 2} (1 + ε) |z|) {R≤|z|≤2R}
≥ c2 exp (− ( p − 2) (1 + ε) |z|). Choosing ε < (4 − p) / ( p − 1) and using (9)–(11), (13) and (14), we have for |z| ≥
(14) R20
≥ R2
sup Jh (u 0 + tvz ) < Jh (u 0 ) + α(R N ).
t0 ≤t≤t1
Let l0 = max{R10 , R20 }, we complete the proof.
For n ∈ N, we define vn (z) = ψ R (z) u¯ (z − nz), where z is a unit vector in R N . Clearly, vn ∈ H01 (Ω ). Remark 1. There exists a positive integer n 0 such that for n ≥ n 0 sup Jh (u 0 + tvn ) < Jh (u 0 ) + α (Ω )
uniformly in z,
t≥0
where u 0 is the local minimum in Lemma 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 ⊂ Mh . We try to show for a sufficiently small σ > 0 cat ([Jh ≤ αh (Ω ) + α (Ω ) − σ ]) ≥ 2.
(15)
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1139
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 X if 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 ⊂ X such that A j is contractible to a point in X for all j and
k [
) Aj = X .
j=1
When there do not exist finitely many closed subsets A1 , . . . , Ak ⊂ X such that A j is contractible to a point in X for all j and ∪kj=1 A j = X , we say cat (X ) = ∞. We need the following two lemmas. Lemma 26. Suppose that X is a Hilbert manifold and Ψ ∈ C 1 (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. Then Ψ (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, S N −1 = {z ∈ R N | |z| = 1}, and let X be a topological space. Suppose that there are two continuous maps F : S N −1 → X,
G : X → S N −1
such that G ◦ F is homotopic to the identity map of S N −1 , that is, there exists a continuous map ζ : [0, 1] × S N −1 → S N −1 such that ζ (0, z) = (G ◦ F) (z) ζ (1, z) = z
for each z ∈ S N −1 ,
for each z ∈ S N −1 .
Then cat (X ) ≥ 2. Proof. See Adachi–Tanaka [1, Lemma 2.5].
Let
H01 (Ω )\{0} u
A1 = u ∈ 1 A2 = u ∈ H0 (Ω )\{0} u
u 1 − ≥ 0 and t > 1 ∪ {0} kuk H 1 kuk H 1 1 u − ≥ 0 and t <1 . kuk H 1 kuk H 1
From Tarantello [15], we have the following results.
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1 Lemma 28. (i) A\M− h = A1 ∪ A2 , where A = {u ∈ H0 (Ω ) | u ≥ 0}; + (ii) Mh ⊂ A1 ; (iii) there exist t0 > 1 and n 1 ≥ n 0 such that u 0 + t0 vn ∈ A2 for each n ≥ n 1 , where n 0 is defined as in Remark 1; (iv) there exists a sequence {sn } ⊂ (0, 1) such that u 0 + sn t0 vn ∈ M− h for each n ≥ n 1 ; (v) αh− < αh (Ω ) + α (Ω ).
Proof. (i) By Lemma 10(iii). (ii) For each u ∈ M+ h , we have 1 t− 1 < tmax (u) < t (u) = kuk H 1 −
u , kuk H 1
then M+ h ⊂ A1 . In particular, u 0 ∈ A1 , where u 0 isdefined as in Lemma 17. u 0 +tvn − (iii) There is a constant c > 0 such that 0 < t ku 0 +tvn k H 1 < c for each t ≥ 0 and each n ∈ N. Otherwise, there 0 +tn vn 0 +tn vn → ∞ as n → ∞. Let wn = ku 0u+t . exist a sequence {tn } and a subsequence {vn } such that t − ku 0u+t n vn k H 1 n vn k 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. By Lemma 20, we have a (vn ) = b (vn ) + o (1) =
2p α (Ω ) + o (1). p−2
Thus, Z
1
p b (wn ) =
u0
tn + vn 1 Ω
u0 + vn tn
p
H
b (vn ) ≥ p
p u 2 p−1 tn0 1 + kvn k H 1 H 2p p−2 α (Ω )
=
2 p−1
ku 0 k p
c0
p H1
+
2p p−2 α (Ω )
p + o (1). 2
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 → ∞. By Lemma 20, we have ku 0 + tn vn k2H 1 = ku 0 k2H 1 + tn2 kvn k2H 1 + 2tn hvn , u 0 i H 1 = ku 0 k2H 1 + o(1). Thus, Z 1 p u p ku 0 + tn vn k H 1 Ω 0 Z 1 p u + o(1). = p ku 0 k H 1 Ω 0
b (wn ) ≥
Since t − (wn ) wn ∈ M− h ⊂ Mh , we have Jh t (wn ) wn −
Z 2 1 − p 1− − t (wn ) − t (wn ) b (wn ) − t (wn ) hwn = 2 p Ω →−∞ as n → ∞.
However, Jh is bounded below on Mh , which is a contradiction. Let 12 p − 2 2 t0 = c − a (u 0 ) + 1, 2 pα (Ω )
H.-L. Lin et al. / Nonlinear Analysis 67 (2007) 1129–1146
1141
then ku 0 + t0 vn k2H 1
2p α (Ω ) + o(1) = p−2 2 u 0 + t0 vn > c2 + o(1) ≥ t − + o(1). ku 0 + t0 vn k H 1 a (u 0 ) + t02
Thus, there is an n 1 ≥ n 0 , where n 0 is defined as in Remark 1, such that, or n ≥ n 1 , u 0 + t0 vn 1 − < 1, t ku 0 + t0 vn k H 1 ku 0 + t0 vn k H 1 or u 0 + t0 vn ∈ A2 . (iv) Define a path γn (s) = u 0 + st0 vn for s ∈ [0, 1] and each n ≥ n 1 where t0 > 1, then γn (0) = u 0 ∈ A1 , γn (1) = u 0 + t0 vn ∈ A2 . Since kuk1 1 t − kuku 1 is a continuous function for nonzero u and γn ([0, 1]) is connected, there exists a sequence H
H
{sn } ⊂ (0, 1) such that u 0 + sn t0 vn ∈ M− h. (v) By part (iv) and Remark 1, αh− ≤ Jh (u 0 + sn t0 vn ) < Jh (u 0 ) + α (Ω ) = αh (Ω ) + α (Ω ).
For n ≥ n 1 , we define a map Fn : S N −1 → H01 (Ω ) by Fn (z) (z) = u 0 (z) + sn t0 vn (z)
for z ∈ S N −1 ,
where vn (z) = ψ R (z) u (z − nz). Then we have Lemma 29. There exists a sequence {σn } in R+ such that Fn (S N −1 ) ⊂ [Jh ≤ αh (Ω ) + α (Ω ) − σn ]. Proof. By Lemma 28(iv) and Remark 1, we have that u 0 + sn t0 vn ∈ M− h and Jh (u 0 + sn t0 vn ) ≤ αh (Ω ) + α(Ω ) − σn for each n ≥ n 1 , the conclusion holds. For c > 0, we define Z cu p ; bc (u) = Ω
1 1 Ic (u) = a(u) − bc (u + ) ; 2 p M Ic = {u ∈ H01 (Ω )\{0} | hIc0 (u), ui = 0}. 1 Recall that there exist a unique t − = t − (u) > 0 and a unique t 1 = t 1 (u) > 0 such that t − u ∈ M− h and t u ∈ M. Let Σ = {u ∈ H01 (Ω ) | u ≥ 0 and kuk H 1 = 1}. Then we have the following results.
Lemma 30. (i) For each u ∈ Σ , there exists a unique number t c (u) > 0 such that t c (u)u ∈ M Ic and 1 1 − 2 c − bc (u) p−2 ; max Ic (tu) = Ic t (u) u = t≥0 2 p (ii) For each nonnegative u ∈ H01 (Ω ) and 0 < µ < 1, we have (1 − µ) I
1 1−µ
(u) −
1 1 khk2L 2 ≤ Jh (u) ≤ (1 + µ) I 1 (u) − khk2L 2 ; 1+µ 2µ 2µ
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(iii) For each u ∈ Σ and 0 < µ < 1, we have p
(1 − µ) p−2 J (t 1 u) −
p 1 1 khk2L 2 ≤ Jh t − u ≤ (1 + µ) p−2 J (t 1 u) − khk2L 2 ; 2µ 2µ
(iv) αh− > 0 for sufficiently small khk L 2 . Proof. (i) For each u ∈ Σ , let f (t) = Ic (tu) = 12 t 2 − 1p t p bc (u), then f (t) → −∞ as t → ∞, f 0 (t) = t −t p−1 bc (u) and f 00 (t) = 1 − ( p − 1) t p−2 bc (u). Let t (u) = c
1 bc (u)
1 p−2
> 0.
Then f 0 (t c (u)) = 0, t c (u) u ∈ M Ic and t c (u)
2
f 00 t c (u) = a t c (u)u − ( p − 1) bc t c (u)u 2 = (2 − p) t c (u) a (u) < 0.
Thus, there exists a unique t c (u) > 0 such that t c (u)u ∈ M Ic and max Ic (tu) = Ic t c (u) u t≥0 1 1 − 2 = − bc (u) p−2. 2 p (ii) For µ ∈ (0, 1), we get Z hudz ≤ kuk 1 khk 2 ≤ µ kuk2 1 + 1 khk2 2 . H L H L 2 2µ Ω Thus, for each nonnegative u ∈ H01 (Ω ), then Z Z 1 1 1+µ 1 1 1−µ kuk2H 1 − khk2L 2 ≤ Jh (u) ≤ kuk2H 1 − khk2L 2 . up − up + 2 p Ω 2µ 2 p Ω 2µ (iii) Applying part (ii), we have that for each u ∈ Σ (1 − µ) I
where t c1 u ∈ M I I
1 1−µ
1 1 khk2L 2 ≤ Jh t − u ≤ (1 + µ) I 1 t c2 u − khk2L 2 , t c1 u − 1+µ 2µ 2µ
1 1−µ
1 1−µ
t c1 u =
and t c2 u ∈ M I
1 1+µ
. By part (i), then
2 1 1 1 1 − 2 − 2 − b 1 (u) p−2 = (1 − µ) p−2 − b (u) p−2 1−µ 2 p 2 p 2
= (1 − µ) p−2 J (t 1 u). 2
Similarly, I
1 1+µ
(t c2 u) = (1 + µ) p−2 J (t 1 u). Hence, (iii) holds.
(iv) We know that α (Ω ) > 0 and applying part (ii) to obtain p
αh− ≥ (1 − µ) p−2 α (Ω ) − Thus, (iv) holds.
1 khk2L 2 . 2µ
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The following lemma is a key lemma to prove our main result. Lemma 31. There exists a number δ0 > 0 such that if u ∈ M and J (u) ≤ α (Ω ) + δ0 , then Z z |∇u|2 + u 2 dz 6= 0. R N |z| Proof. On the contrary, there exists a sequence {u n } in M such that J (u n ) = α (Ω ) + o (1) and Z z |∇u n |2 + u 2n dz = 0. R N |z| By Lemma 9, {u n } 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 R N . It follows from the Decomposition Lemma 6 that there exists a sequence {z n } in R N such that |z n | → ∞ as n → ∞ and u n (z) = u(z − z n ) + o (1)
strongly in H 1 (R N ).
Suppose the subsequence |zz nn | → z 0 as n → ∞, where z 0 is a unit vector in R N . Then by the Lebesgue dominated theorem, we have Z z |∇u n |2 + u 2n dz 0= N |z| ZR z + zn |∇u|2 + u 2 dz + o (1) = N |z + z n | R 2p = α(R N )z 0 + o (1), p−2 which is a contradiction.
Lemma 32. There exists a number d0 > 0 such that for khk L 2 < d0 , we have Z z |∇u|2 + u 2 dz 6= 0, R N |z| for u ∈ [Jh < αh (Ω ) + α (Ω )]. Proof. For u ∈ [Jh < αh (Ω ) + α(Ω )], then u/ kuk H 1 ∈ Σ . There exists a t 1 > 0 such that t 1 u/ kuk H 1 ∈ M. By Lemma 30(ii), we have for any µ ∈ (0, 1) and khk L 2 < d1 (µ) 1 t u 1 − p khk2L 2 . J ≤ (1 − µ) p−2 Jh (u) + kuk H 1 2µ Since αh (Ω ) < 0, we have, for u ∈ [Jh < αh (Ω ) + α (Ω )], 1 p t u 1 − p−2 2 khk L 2 . J ≤ (1 − µ) α (Ω ) + kuk H 1 2µ Take µ ∈ (0, 1) such that there exists d1 (µ) > d0 > 0 such that for khk L 2 < d0 1 t u J ≤ α (Ω ) + δ0 . kuk H 1 Since t 1 u/ kuk H 1 ∈ M, by Lemma 31 1 2 1 2 ! Z t u t u z ∇ + dz 6= 0, kuk H 1 kuk H 1 R N |z| or Z
z |∇u|2 + (u)2 dz 6= 0. R N |z|
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For khk L 2 < d0 , we obtain Z z |∇u|2 + |u|2 dz 6= 0 R N |z| for all u ∈ [Jh < αh (Ω ) + α (Ω )]. Now, we define G : [Jh < αh (Ω ) + α (Ω )] → S N −1 by Z
z |∇u|2 + |u|2 dz G(u) = R N |z|
Z
z 2 2 |∇u| + |u| dz . R N |z|
We have Lemma 33. For n ≥ n 1 and khk L 2 < d0 , the map G ◦ Fn : S N −1 → S N −1 is homotopic to the identity. Proof. We define ζn (θ, z) : [0, 1] × S N −1 → S N −1 by G ((1 − 2θ) Fn (z) + 2θu (z − nz)) n z ζn (θ, z) = G u z − 2 (1 − θ) z
for θ ∈ [0, 1/2) ; for θ ∈ [1/2, 1) ; for θ = 1.
First, we claim that limθ→1− ζn (θ, z) = z and limθ→ 1 − ζn (θ, z) = G (u (z − nz)). 2
(a) limθ →1− ζn (θ, z) = z: since 2 2 ! Z z n n dz ∇u z − 2 (1 − θ) z + u z − 2 (1 − θ) z R N |z| Z n z + 2(1−θ )z |∇u (z)|2 + u (z)2 dz = n R N z + 2(1−θ ) z 2p = α(R N )z + o(1) as θ → 1− , p−2
then limθ→1− ζn (θ, z) = z. (b) limθ→ 1 − ζn (θ, z) = G (u (z − nz)): since 2
k(1 − 2θ) Fn (z) + 2θu (z − nz)k H 1 = ku (z − nz)k H 1 + o(1)
as θ →
1− , 2
by the continuity of G, we obtain limθ → 1 − ζn (θ, z) = G (u (z − nz)). Thus, ζn (θ, z) ∈ C([0, 1] × S N −1 , S N −1 ) and 2
ζn (0, z) = G (Fn (z)) ζn (1, z) = z
for all z ∈ S N −1 ,
for all z ∈ S N −1 ,
provided n ≥ n 1 and khk L 2 < d0 . This completes the proof. Thus, we have
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Lemma 34. Assume that h (z) ≥ 0 and 0 < khk L 2 < min {d( p, α), d0 }, where d0 is defined as in Lemma 32. Jh (u) has at least two critical points in [Jh < αh (Ω ) + α (Ω )]. Proof. Applying Lemmas 27 and 33, we have for sufficiently large n ≥ n 1 and khk L 2 < d0 , cat ([Jh ≤ αh (Ω ) + α (Ω ) − σn ]) ≥ 2. By Lemmas 18 and 26, Jh (u) has at least two critical points in [Jh < αh (Ω ) + α (Ω )].
5. Existence of the fourth solution Since αh− > 0 for sufficiently small khk L 2 , we define K h (u) = sup Jh (tu) = Jh (t − u) > 0, t≥0
where Let
t −u
∈ Mh . We observe that if khk L 2 is sufficiently small, Bahri–Li’s minimax argument [4] also works for K h .
Γ = {γ ∈ C(Br (0), Σ ) | γ |∂ Br (0) = ψ R (z) u¯ (z − y) / kψ R (z) u¯ (z − y)k H 1 }
for large r = |y|,
where Σ = {u ∈ H01 (Ω ) | u ≥ 0 and kuk H 1 = 1}. Then we define γh (Ω ) = inf sup K h (γ (y)); γ ∈Γ y∈R N
γ0 (Ω ) = inf sup K 0 (γ (y)). γ ∈Γ y∈R N
By Lemma 30(iii), for 0 < µ < 1, we have p
(1 − µ) p−2 γ0 (Ω ) −
p 1 1 khk2L 2 ≤ γh (Ω ) ≤ (1 + µ) p−2 γ0 (Ω ) − khk2L 2 . 2µ 2µ
(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 < khk L 2 < d00 , then αh (Ω ) + α(Ω ) < γh (Ω ) < αh− (Ω ) + α(Ω ). Moreover, there exists a positive solution u of Eq. (1) in Ω such that Jh (u) = γh (Ω ). Proof. By Lemma 30(iii), we also have that for 0 < µ < 1 p
(1 − µ) p−2 α(Ω ) −
p 1 1 khk2L 2 ≤ αh− (Ω ) ≤ (1 + µ) p−2 α(Ω ) − khk2L 2 . 2µ 2µ
For any ε > 0, there exists a d1 (ε) > 0 such that if khk L 2 < d1 (ε), then α(Ω ) − ε < αh− (Ω ) < α(Ω ) + ε. Thus, 2α(Ω ) − ε < αh− (Ω ) + α(Ω ) < 2α(Ω ) + ε. Applying (16), for any δ > 0, there exists a d2 (δ) > 0 such that if khk L 2 < d2 (δ), then γ0 (Ω ) − δ < γh (Ω ) < γ0 (Ω ) + δ.
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Fix a small 0 < ε < (2α (Ω ) − γ0 (Ω ))/2, since α (Ω ) < γ0 (Ω ) < 2α (Ω ), choosing a δ > 0 such that for khk L 2 < 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 and 18(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 < khk L 2 < min{d( p, α), d0 , d00 }, where d0 is defined as in Lemma 32 and d00 is defined as in Lemma 36. Then there are at least four positive solutions of Eq. (1) in Ω . Proof. Therefore, by Lemmas 2, 15, 34 and 36, we have that Eq. (1) has at least four positive solutions in Ω .
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Four positive solutions of semilinear elliptic equations in exterior domains Huei-Li Lin a,∗ , Hwai-Chiuan Wang b , Tsung-Fang Wu c,1 a Center for General Education, Chang Gung University, 259, Wen Hwa 1st Road, Kwei Shan, 33375, Tao-Yuan, Taiwan b Department of Applied Mathematics, Hsuan Chuang University, Hsinchu, Taiwan c Center 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 khk L 2 > 0, we prove that if khk L 2 is sufficiently small, then there are at least four positive solutions of the semilinear elliptic equation in R N \D, where D is a C 1,1 bounded domain in R N . 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−2 u + h(z) in Ω ; u ∈ H01 (Ω ),
(1)
where Ω = R N \D, D is a C 1,1 bounded domain in R N , 2 < p < 2∗ = 2N /(N − 2) for N ≥ 3. Let d( p, α) = ( p − 2)
1 p−1
p−1 p−2
2p p−2
1
2
1
α(Ω ) 2 ,
(2)
where α(Ω ) is defined below. In this paper, we first consider h(z) ≥ 0 and 0 < khk L 2 < d( p, α). Associated with Eq. (1), we define the functionals a, b, and Jh , for u ∈ H01 (Ω ), Z a(u) = |∇u|2 + u 2 ; ZΩ |u| p ; b(u) = Ω ∗ Corresponding author.
E-mail address: [email protected] (H.-L. Lin). 1 Present address: Department of Applied Mathematics, National University of Kaohsiung, Kaohsiung, Taiwan. c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.07.001
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Jh (u) =
Z
1 1 a(u) − b (u + ) − hu. 2 p Ω
By Rabinowitz [14, Proposition B.10.], a, b, and Jh are of C 2 . For h = 0, we consider the semilinear elliptic equation −1u + u = |u| p−2 u in Ω ; (3) u ∈ H01 (Ω ), and the energy functional J (u) = 21 a(u) − 1p b (u + ). Bahri–Lions [5] showed that Eq. (3) admits at least one positive solution u in R N , 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 R N , 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 R N . 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 R N 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−2 u + h (z) in R N ; (4) u ∈ H 1 (R N ), 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) ∈ R N −l × Rl . Consider the exterior strip domain Ar \D, where r A = {(x, y) ∈ R N | |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) For β ∈ R, a sequence {u n } is a (PS)β -sequence in H01 (Ω ) for Jh if Jh (u n ) = β + o(1) and Jh0 (u n ) = o(1) strongly in H −1 (Ω ) as n → ∞; (ii) β ∈ R is a (PS)-value in H01 (Ω ) for Jh if 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 ∈ H01 (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 Z p−1 ∇u∇ϕ + uϕ = u + ϕ + hϕ for any ϕ ∈ H01 (Ω ). Ω
Ω
p−1
Thus, u is a weak solution of −1u +u = u + +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
hJh0 (u) , ui = 0} and αh (Ω ) = inf Jh (u). u∈Mh
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 3. For each nonnegative u ∈ Σ = {u ∈ H01 (Ω ) | kuk H 1 = 1}, there exists a su > 0 such that su u ∈ M.
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Lemma 4. (i) For each nonnegative u ∈ H01 (Ω )\{0}, there exists a su > 0 such that su u ∈ M; (ii) Let β > 0 and {u n } be a sequence in H01 (Ω )\{0} for J such that J (u n ) = β + o(1) and a(u n ) = b(u + n ) + o(1). Then there is a sequence {sn } in R+ such that sn = 1 + o(1), {sn u n } in M and J (sn u n ) = β + o(1). Lemma 5. If u ∈ H01 (Ω )\{0} and u ≥ 0, then p
a(u) 2 b(u)
!
1 p−2
≥
2p p−2
Proof. Applying Lemma 4.
1
2
1
α(Ω ) 2 .
By Lien–Tzeng–Wang [13], we have the following lemma. Lemma 6 (Palais–Smale Decomposition Lemma for J ). Let {u n } be a (PS)β -sequence in H01 (Ω ) for J . Then there ∞ are a subsequence {u n }, a positive integer l, sequences z ni n=1 in R N , functions u in H01 (Ω ), and wi 6= 0 in H 1 (R N ) for 1 ≤ i ≤ l such that i z n → ∞ for 1 ≤ i ≤ l; −4u + u = |u| p−2 u in Ω ; p−2 −4wi + wi = wi wi in R N ; l X un = u + wi (· − z ni ) + o(1) strongly in H 1 (R N ); i=1
J (u n ) = J (u) +
l X
J (wi ) + o(1).
i=1
In addition, if u n ≥ 0, then u ≥ 0 and wi ≥ 0 for 1 ≤ i ≤ l. Lemma 7 (Palais–Smale Decomposition Lemma for Jh ). Let {u n } be a (PS)β -sequence in H01 (Ω ) for Jh . Then there ∞ are a subsequence {u n }, a positive integer l, sequences z ni n=1 in R N , functions u in H01 (Ω ), and wi 6= 0 in H 1 (R N ) for 1 ≤ i ≤ l such that i z n → ∞ for 1 ≤ i ≤ l; −4u + u = |u| p−2 u + h(z) in Ω ; p−2 −4wi + wi = wi wi in R N ; l X un = u + wi (· − z ni ) + o(1) strongly in H 1 (R N ); i=1
Jh (u n ) = Jh (u) +
l X
J (wi ) + o(1).
i=1
In addition, if u n ≥ 0, then u ≥ 0 and wi ≥ 0 for 1 ≤ i ≤ l. Proof. See Zhu–Zhou [19].
R Define ψ(u) = hJh0 (u) , ui = a(u)−b (u + )− Ω hu. Then if 0 < khk L 2 < d( p, α), we have the following lemma. Lemma 8. For each u ∈ Mh , we have hψ 0 (u), ui = a(u) − ( p − 1) b(u) 6= 0. Proof. By Lemma 5, Tarantello [15, Lemma 2.3] and Cao–Zhou [8].
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− By Lemma 8, we write Mh = M+ h ∪ Mh , where
M+ h = {u ∈ Mh | a (u) − ( p − 1) b(u) > 0}, M− h = {u ∈ Mh | a (u) − ( p − 1) b(u) < 0}. Define αh− (Ω ) = inf Jh (u).
αh+ (Ω ) = inf Jh (u);
u∈M− h
u∈M+ h
By Wang–Wu [17], we have the following lemma. Lemma 9. {u n } is a (PS)α(Ω ) -sequence in H01 (Ω ) for J if and only if J (u n ) = α(Ω ) + o (1) and a (u n ) = 1 b u+ n + o (1). In particular, every minimizing sequence {u n } in M of α(Ω ) is a (PS)α(Ω ) -sequence in H0 (Ω ) for J . For each nonnegative u ∈ H01 (Ω )\{0}, we write tmax =
a(u) ( p − 1) b (u)
1 p−2
> 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≥tmax Jh (tu); (ii) t − (u) isna continuous function; o u 1 1 − (iii) M− h = u ∈ H0 (Ω )\{0} | u ≥ 0 and kuk H 1 t kuk H 1 = 1 ; R (iv) If Ω hu > 0, then there is a unique number 0 < t + = t + (u) < tmax such that t + u ∈ M+ h and Jh (t + u) = min0≤t≤t − Jh (tu). Proof. See Tarantello [15] and Cao–Zhou [8]. R + Lemma 11. (i) For each u ∈ M+ h , we have Ω hu > 0 and Jh (u) < 0. In particular, αh (Ω ) ≤ αh (Ω ) < 0; (ii) Jh is coercive and bounded below on Mh . R Proof. (i) For each u ∈ M+ h , a(u) − ( p − 1) b(u) > 0 and a(u) = b (u) + Ω hu. Then Z hu = a(u) − b(u) > ( p − 2) b(u) > 0. Ω
Hence Z 1 1 p−2 p−2 1 Jh (u) = − b (u) − hu < b(u) − b(u) 2 p 2 Ω 2p 2 ( p − 1) ( p − 2) =− b(u) < 0. 2p
(ii) By Tarantello [15, p. 288].
Lemma 12. Let u be in Mh such that Jh (u) = min Jh (v) = αh (Ω ). Then v∈Mh R (i) Ω hu > 0; (ii) u is a solution of Eq. (1) in Ω . Proof. (i) By Lemma 11(i), we have Z 1 1 1 a(u) − 1 − 0 > αh (Ω ) = Jh (u) = − hu. 2 p p Ω R Thus, Ω hu > 0.
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(ii) By Lemma 8, hψ 0 (v), vi 6= 0 for each v ∈ Mh . Since Jh (u) = minv∈Mh Jh (v), by the Lagrange multiplier theorem, there is a λ ∈ R such that Jh0 (u) = λψ 0 (u) in H −1 (Ω ). Then we have 0 = hJh0 (u), ui = λhψ 0 (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 a δ > 0 and a differentiable functional l : B (0; δ) ⊂ H01 (Ω ) → R+ such that l (0) = 1, l (v) (u − v) ∈ Mh for v ∈ B (0; δ) and hψ 0 (u), ϕi hl 0 (v) , ϕi (l,v)=(1,0) = hψ 0 (u), ui
for ϕ ∈ Cc∞ (Ω ).
Lemma 14. (i) There exists a (PS)αh (Ω ) -sequence {u n } in Mh for Jh ; (ii) There exists a (PS)α + (Ω ) -sequence {u n } in M+ h for Jh ; h
(iii) There exists a (PS)α − (Ω ) -sequence {u n } in M− h for Jh . h
3. Existence of the first solution By Lemma 14(i), there is a (PS)αh (Ω ) -sequence {u n } in Mh for Jh . Then we have the following (PS)αh (Ω ) condition. Lemma 15. Let {u n } ⊂ Mh be a (PS)αh (Ω ) -sequence for Jh . Then there exist a subsequence {u n } and a nonzero u 0 ∈ H01 (Ω ) such that u n → u 0 strongly in H01 (Ω ). Moreover, u 0 is a positive solution of Eq. (1) such that Jh (u 0 ) = αh (Ω ). Proof. By Lemma 11(ii), {u n } is bounded in H01 (Ω ). Take a subsequence {u n } and u 0 ∈ H01 (Ω ) such that u n * u 0 weakly in H01 (Ω ). Then we have that u 0 is a nonzero solution of Eq. (1) in Ω . Since Z 1 1 Jh (u n ) = a (u n ) − b (u n ) − hu n = αh (Ω ) + o (1), 2 p ΩZ hu n = o (1), hJh0 (u n ) , u n i = a (u n ) − b (u n ) − Ω
we obtain Z 1 1 1 − a (u n ) − 1 − hu n = αh (Ω ) + o (1). 2 p p Ω R R Since the functional a is weakly lower semicontinuous and Ω hu n → Ω hu 0 as n → ∞, then Jh (u 0 ) = αh (Ω ). Let pn = u n − u 0 . By the Br´ezis–Lieb Lemma, we get Z 1 1 hpn Jh ( pn ) = a ( pn ) − b ( pn ) − 2 p Ω Z Z 1 1 1 1 = a (u n ) − a (u 0 ) − b (u n ) + b (u 0 ) − hu n + hu 0 + o (1) 2 2 p p Ω Ω = Jh (u n ) − Jh (u 0 ) + o (1) = o (1). (5) R By the Br´ezis–Lieb Lemma, Ω hpn = o (1) and u 0 is a solution of Eq. (1), so Z 0 hJh ( pn ) , pn i = a ( pn ) − b ( pn ) − hpn Ω Z Z = a (u n ) − a (u 0 ) − b (u n ) + b (u 0 ) − hu n + hu 0 + o (1) Ω
= hJh0 (u n ) , u n i − hJh0 (u 0 ) , u 0 i = o (1).
Ω
(6)
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R Thus, by (5) and (6) Ω hpn = o (1), we have p−2 a ( pn ) = o (1), 2p that is, un → u0
strongly in H01 (Ω ).
Moreover, u 0 is a solution of Eq. (1) such that Jh (u 0 ) = αh (Ω ). By Lemma 2, u 0 is positive in Ω .
We prove that u 0 is the unique critical point of Jh in B (r0 ) in the following lemma. 1 1 1 p−2 2p 2 1 Lemma 16. Let r0 = p−1 α (Ω ) 2 . Then p−2 1 (i) M+ h ⊂ B (r0 ) = {u ∈ H0 (Ω ) | kuk H 1 < r0 }; (ii) Jh (u) is strictly convex in B (r0 ).
R Proof. (i) If u ∈ M+ h , then a(u) > ( p − 1) b(u) and a(u) = b (u) + Ω hu. Thus, 1 a(u) + khk L 2 kuk H 1 . a(u) < p−1 This implies
p−1 khk L 2 < p−2 p−1 1 p−2 2 1 1 p−1 2p < α (Ω ) 2 ( p − 2) p−2 p−1 p−2 1 1 p−2 2 1 1 2p α (Ω ) 2 = r0 . = p−1 p−2
kuk H 1
(ii) We know Jh00 (u) (v, v) = a (v) − ( p − 1)
Z Ω
|u| p−2 v 2
for all v ∈ H01 (Ω ).
Thus, by Lemma 5, we obtain p−2
Jh00 (u) (v, v) ≥ a (v) − ( p − 1) kuk L p kvk2L p p−2 2 ( p−2)2 p−2 p − 2 − ≥ a (v) − ( p − 1) a(u) 2 α (Ω ) 2 p 2p p−2 p −( p−2) p − 2 × a(v) α (Ω ) p 2p 2− p 2 2 p p−2 kuk H 1 ≥ a(v) 1 − ( p − 1) α (Ω ) p−2 >0 Thus,
Jh00 (u)
for u ∈ B(r0 )\{0}.
is positive definite for u ∈ B (r0 ) and Jh is strictly convex in B (r0 ).
By Lemma 15, there exists a solution u 0 ∈ Mh of Eq. (1) such that Jh (u 0 ) = αh (Ω ). Furthermore, we have the following lemma. + Lemma 17. (i) u 0 ∈ M+ h and Jh (u 0 ) = αh (Ω ) = αh (Ω ); (ii) u 0 is the unique critical point of Jh (u) in B (r0 ), where r0 is defined as in Lemma 16; (iii) Jh (u 0 ) is a local minimum in H01 (Ω ).
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R − Proof. (i) By Lemma 12(i), Ω hu 0 > 0. We claim that u 0 ∈ M+ h . Otherwise, if u 0 ∈ Mh , then by Lemma 10, there + − + + exists a unique t (u 0 ) = 1 > t (u 0 ) > 0 such that t (u 0 )u 0 ∈ Mh and αh (Ω ) ≤ αh+ (Ω ) ≤ Jh (t + (u 0 )u 0 ) < Jh (t − (u 0 )u 0 ) = αh (Ω ), + + + which is a contradiction. Since u 0 ∈ M+ h , αh (Ω ) ≤ Jh (u 0 ) = αh (Ω ) ≤ αh (Ω ), that is, Jh (u 0 ) = αh (Ω ) = αh (Ω ). (ii) By part (i) and Lemma 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 Decomposition Lemma 7, we have the following restricted (PS)β -condition. Lemma 18. (i) If {u n } is a (PS)β -sequence in H01 (Ω ) for Jh with β < αh (Ω )+α (Ω ), then there exist a subsequence {u n } and a nonzero u 0 in H01 (Ω ) such that u n → u 0 strongly in H01 (Ω ) and Jh (u 0 ) = β. Moreover, u 0 is a positive solution of Eq. (1) in Ω ; 1 (ii) If {u n } ⊂ M− h is a (PS)β -sequence in H0 (Ω ) for Jh with αh (Ω ) + α (Ω ) < β < αh− (Ω ) + α (Ω ), 1 0 then there exist a subsequence {u n } and a nonzero u 0 ∈ M− h such that u n → u strongly in H0 (Ω ) and Jh (u 0 ) = β. Moreover, u 0 is a positive solution of Eq. (1) in Ω .
By Chen–Chen–Wang [6, Proposition 1], we have the following lemma. Lemma 19. Let u be a positive solution of Eq. (1) in Ω . Then for any ε > 0, there are positive constants cε and R such that D ⊂ B N (0; R) and u (z) ≥ cε exp (− (1 + ε) |z|)
for |z| ≥ R.
It is well-known that there is a positive radially symmetric smooth solution u¯ of Eq. (3) in R N such that J (u) ¯ = α(R N ). Recall the facts (i) for any ε > 0, there exist constants C0 , C00 > 0 such that for all z ∈ R N u(z) ≤ C0 exp (−|z|)
and
|∇u(z)| ≤ C00 exp (− (1 − ε) |z|);
(ii) for any ε > 0, there exists a constant Cε > 0 such that u(z) ≥ Cε exp (− (1 + ε) |z|)
for all z ∈ R N .
For such R in Lemma 19, let ψ R : R N → [0, 1] be a C ∞ -function on R N such that 0 ≤ ψ R ≤ 1, |∇ψ R | ≤ c and 1 for |z| ≥ R + 1; ψ R (z) = 0 for |z| ≤ R. We define vz (z) = ψ R (z) u¯ (z − z)
for z ∈ R N .
Clearly, vz (z) ∈ H01 (Ω ). In order to prove Lemma 24, we need the following lemmas. 2p Lemma 20. (i) a (vz ) = b (vz ) + o (1) = p−2 α(R N ) + o(1) as |z| → ∞; (ii) J (vz ) = α (Ω ) + o (1) = α(R N ) + o (1) as |z| → ∞; (iii) vz * 0 weakly in H01 (Ω ) as |z| → ∞.
Proof. It is similar to the proof of Wang [16, Lemma 30].
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Lemma 21. Let E be a domain in R N . If f : E → R satisfies Z f (z)eσ |z| dz < ∞ for some σ > 0, E
then Z f (z)e−σ |z−z| dz eσ |z| =
Z
f (z)eσ
hz,zi |z|
dz + o(1)
as |z| → ∞.
E
E
Proof. Since σ |z| ≤ σ |z| + σ |z − z|, we have f (z)e−σ |z−z| eσ |z| ≤ f (z)eσ |z| . Since −σ |z − z| + σ |z| = σ hz,zi |z| + o(1) as |z| → ∞, then the lemma follows from the Lebesque dominated convergence theorem. Lemma 22. For t ≥ 0, we have the following inequalities. q q−1 t where q ≥ 2; (i) (1 + t)q ≥ 1 + t q + q−1 q (ii) (1 + t) ≥ 1 + t q + qt where q ≥ 2; q q−1 (iii) (1 + t)q ≥ 1 + t q + qt + q−2 t where q ≥ 3;
(iv) If t ≤ c for some c > 0, then (1 + t)q ≥ 1 + t q + qt + A(c)t 2 where 2 < q < 3 and A(c) > 0. Proof. (i) Let f (t) = (1 + t)q − 1 − t q −
q q−1 q−1 t
for t ≥ 0 and q ≥ 2. Then f (0) = 0, and
f 0 (t) = q[(1 + t)q−1 − t q−1 − t q−2 ]. Since q ≥ 2, we get (1 + t)q−1 = (1 + t)q−2 + t (1 + t)q−2 ≥ t q−2 + t q−1 . Thus, f 0 (t) ≥ 0. (ii) The proof is similar to (i). q q−1 (iii) Let g(t) = (1 + t)q − 1 − t q − qt − q−2 t for t ≥ 0 and q ≥ 3. Then g(0) = 0, and by (i), we obtain q − 1 q−2 0 q−1 q−1 g (t) = q (1 + t) −t −1− t ≥ 0. q −2 (iv) Let h(t) = (1 + t)q − t q . Then h(0) = 1, h 0 (t) = q[(1 + t)q−1 − t q−1 ] and h 0 (0) = q, h 00 (t) = q (q − 1) [(1 + t)q−2 − t q−2 ] > 0. Since t ≤ c for some c > 0, applying the Taylor theorem, we have (1 + t)q − t q − 1 − qt ≥
q (q − 1) [(1 + c)q−2 − cq−2 ]t 2 . 2
By Lemma 22, we obtain (a + b)q ≥ a q + bq + qa q−1 b +
q abq−1 q −2
for q ≥ 3 and a, b ≥ 0,
(7)
for 2 < q < 3 and b/a ≤ c.
(8)
and (a + b)q ≥ a q + bq + qa q−1 b + A(c)a q−2 b2
Lemma 23. (i) There exists a number t0 > 0 such that for 0 ≤ t < t0 and each vz ∈ H01 (Ω ), we have Jh (u 0 + tvz ) < Jh (u 0 ) + α(Ω ); (ii) There exist positive numbers l1 and t1 such that for any t > t1 and |z| ≥ l1 , we have Jh (tvz ) < 0.
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Proof. (i) Since Jh is continuous in H01 (Ω ) and {vz } is bounded in H01 (Ω ), there is a t0 > 0 such that for 0 ≤ t < t0 and each vz ∈ H01 (Ω ) Jh (u 0 + tvz ) < Jh (u 0 ) + α (Ω ). 2 p 2p (ii) By Lemma 20, Jh (tvz ) = t2 − tp p−2 α (Ω ) + o(1) as |z| → ∞. There is an l1 > 0 such that for |z| ≥ l1 Jh (tvz ) <
t2 tp − 2 p
2p α (Ω ) + 1. p−2
Thus, there exists a t1 > 0 such that Jh (tvz ) < 0
for any t > t1 and |z| ≥ l1 .
Lemma 24. There exists a number l0 > 0 such that for |z| ≥ l0 sup Jh (u 0 + tvz ) < Jh (u 0 ) + α(R N ) = αh (Ω ) + α(Ω ), t≥0
where u 0 is the local minimum in Lemma 17. Proof. By Lemma 23, we only need to show that there exists an l0 > 0 such that for |z| ≥ l0 sup Jh (u 0 + tvz ) < Jh (u 0 ) + α(R N ).
t0 ≤t≤t1
Since u 0 is a positive solution of Eq. (1) in Ω , then Z p−1 hu 0 , tvz i H 1 = tu 0 vz + htvz dz. Ω
For t0 ≤ t ≤ t1 , since supt≥0 J (tu) = α(R N ) and 0 ≤ ψ R ≤ 1, we obtain Z Z 1 1 2 p h (u 0 + tvz ) = Jh (u 0 ) Jh (u 0 + tvz ) = ku 0 + tvz k H 1 − (u 0 + tvz ) − 2 p Ω Ω Z 1 p u + (tvz ) p − (u 0 + tvz ) p − htvz + J (tvz ) + hu 0 , tvz i H 1 + p Ω 0 Z 1 p p−1 = Jh (u 0 ) + J (tvz ) − (u 0 + tvz ) p − u 0 − (tvz ) p − pu 0 (tvz ) p Ω Z t2 |∇ψ R |2 [u(z − z)]2 dz ≤ Jh (u 0 ) + α(R N ) + 2 RN Z 2 |∇ψ R | |∇u (z − z)| u (z − z) dz +t RN Z tp p + 1− ψ R [u (z − z)] p dz p RN Z 1 p p−1 − (u 0 + tvz ) p − u 0 − (tvz ) p − pu 0 (tvz ). p RN p Since the support of 1 − ψ R is bounded, then Z p 1 − ψ R [u (z − z)] p dz ≤ C1 exp (− p |z|).
(9)
RN
Similarly, we have Z |∇ψ R |2 [u (z − z)]2 dz ≤ C2 exp (−2 |z|), supp(∇ψ R )
(10)
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and Z supp(∇ψ R )
|∇ψ R | |∇u (z − z)| u (z − z) dz ≤ C3 exp (− (2 − ε) |z|).
(i) For 3 ≤ p < 2∗ , by (7) Z Z p (u 0 + tvz ) ≥ RN
p
RN
p−1
u 0 + (tvz ) p + pu 0
(tvz ) +
(11)
p u 0 (tvz ) p−1 . p−2
Thus, by Lemma 21, for |z| ≥ R1 Z p−1 u 0 vz dz ≥ c1 exp (− min {1, p − 1} (1 + ε) |z|) ≥ c1 exp (− (1 + ε) |z|).
(12)
RN
Choosing ε < 1/2 and using (9)–(12) we have for |z| ≥ R10 ≥ R1 sup Jh (u 0 + tvz ) < Jh (u 0 ) + α(R N ).
t0 ≤t≤t1
(ii) For 2 < p < 3, by Lemma 22(ii), we get p
p−1
(I ) = (u 0 + tvz ) p − u 0 − (tvz ) p − pu 0
(tvz ) ≥ 0.
Then Z RN
(I ) dz ≥
Z
(I ) dz.
(13)
{R≤|z|≤2R}
Since max {vz (z) /u 0 (z) | R ≤ |z| ≤ 2R} ≤ c < ∞ for each z ∈ R N , by (8) Z Z p p−1 p−2 p u 0 + (tvz ) p + pu 0 (tvz ) + A(c)u 0 (tvz )2 . (u 0 + tvz ) ≥ {R≤|z|≤2R}
{R≤|z|≤2R}
Thus, by Lemma 21, for |z| ≥ R2 Z p−2 u 0 vz2 dz ≥ c2 exp (− min {2, p − 2} (1 + ε) |z|) {R≤|z|≤2R}
≥ c2 exp (− ( p − 2) (1 + ε) |z|). Choosing ε < (4 − p) / ( p − 1) and using (9)–(11), (13) and (14), we have for |z| ≥
(14) R20
≥ R2
sup Jh (u 0 + tvz ) < Jh (u 0 ) + α(R N ).
t0 ≤t≤t1
Let l0 = max{R10 , R20 }, we complete the proof.
For n ∈ N, we define vn (z) = ψ R (z) u¯ (z − nz), where z is a unit vector in R N . Clearly, vn ∈ H01 (Ω ). Remark 1. There exists a positive integer n 0 such that for n ≥ n 0 sup Jh (u 0 + tvn ) < Jh (u 0 ) + α (Ω )
uniformly in z,
t≥0
where u 0 is the local minimum in Lemma 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 ⊂ Mh . We try to show for a sufficiently small σ > 0 cat ([Jh ≤ αh (Ω ) + α (Ω ) − σ ]) ≥ 2.
(15)
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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 X if 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 ⊂ X such that A j is contractible to a point in X for all j and
k [
) Aj = X .
j=1
When there do not exist finitely many closed subsets A1 , . . . , Ak ⊂ X such that A j is contractible to a point in X for all j and ∪kj=1 A j = X , we say cat (X ) = ∞. We need the following two lemmas. Lemma 26. Suppose that X is a Hilbert manifold and Ψ ∈ C 1 (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. Then Ψ (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, S N −1 = {z ∈ R N | |z| = 1}, and let X be a topological space. Suppose that there are two continuous maps F : S N −1 → X,
G : X → S N −1
such that G ◦ F is homotopic to the identity map of S N −1 , that is, there exists a continuous map ζ : [0, 1] × S N −1 → S N −1 such that ζ (0, z) = (G ◦ F) (z) ζ (1, z) = z
for each z ∈ S N −1 ,
for each z ∈ S N −1 .
Then cat (X ) ≥ 2. Proof. See Adachi–Tanaka [1, Lemma 2.5].
Let
H01 (Ω )\{0} u
A1 = u ∈ 1 A2 = u ∈ H0 (Ω )\{0} u
u 1 − ≥ 0 and t > 1 ∪ {0} kuk H 1 kuk H 1 1 u − ≥ 0 and t <1 . kuk H 1 kuk H 1
From Tarantello [15], we have the following results.
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1 Lemma 28. (i) A\M− h = A1 ∪ A2 , where A = {u ∈ H0 (Ω ) | u ≥ 0}; + (ii) Mh ⊂ A1 ; (iii) there exist t0 > 1 and n 1 ≥ n 0 such that u 0 + t0 vn ∈ A2 for each n ≥ n 1 , where n 0 is defined as in Remark 1; (iv) there exists a sequence {sn } ⊂ (0, 1) such that u 0 + sn t0 vn ∈ M− h for each n ≥ n 1 ; (v) αh− < αh (Ω ) + α (Ω ).
Proof. (i) By Lemma 10(iii). (ii) For each u ∈ M+ h , we have 1 t− 1 < tmax (u) < t (u) = kuk H 1 −
u , kuk H 1
then M+ h ⊂ A1 . In particular, u 0 ∈ A1 , where u 0 isdefined as in Lemma 17. u 0 +tvn − (iii) There is a constant c > 0 such that 0 < t ku 0 +tvn k H 1 < c for each t ≥ 0 and each n ∈ N. Otherwise, there 0 +tn vn 0 +tn vn → ∞ as n → ∞. Let wn = ku 0u+t . exist a sequence {tn } and a subsequence {vn } such that t − ku 0u+t n vn k H 1 n vn k 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. By Lemma 20, we have a (vn ) = b (vn ) + o (1) =
2p α (Ω ) + o (1). p−2
Thus, Z
1
p b (wn ) =
u0
tn + vn 1 Ω
u0 + vn tn
p
H
b (vn ) ≥ p
p u 2 p−1 tn0 1 + kvn k H 1 H 2p p−2 α (Ω )
=
2 p−1
ku 0 k p
c0
p H1
+
2p p−2 α (Ω )
p + o (1). 2
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 → ∞. By Lemma 20, we have ku 0 + tn vn k2H 1 = ku 0 k2H 1 + tn2 kvn k2H 1 + 2tn hvn , u 0 i H 1 = ku 0 k2H 1 + o(1). Thus, Z 1 p u p ku 0 + tn vn k H 1 Ω 0 Z 1 p u + o(1). = p ku 0 k H 1 Ω 0
b (wn ) ≥
Since t − (wn ) wn ∈ M− h ⊂ Mh , we have Jh t (wn ) wn −
Z 2 1 − p 1− − t (wn ) − t (wn ) b (wn ) − t (wn ) hwn = 2 p Ω →−∞ as n → ∞.
However, Jh is bounded below on Mh , which is a contradiction. Let 12 p − 2 2 t0 = c − a (u 0 ) + 1, 2 pα (Ω )
H.-L. Lin et al. / Nonlinear Analysis 67 (2007) 1129–1146
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then ku 0 + t0 vn k2H 1
2p α (Ω ) + o(1) = p−2 2 u 0 + t0 vn > c2 + o(1) ≥ t − + o(1). ku 0 + t0 vn k H 1 a (u 0 ) + t02
Thus, there is an n 1 ≥ n 0 , where n 0 is defined as in Remark 1, such that, or n ≥ n 1 , u 0 + t0 vn 1 − < 1, t ku 0 + t0 vn k H 1 ku 0 + t0 vn k H 1 or u 0 + t0 vn ∈ A2 . (iv) Define a path γn (s) = u 0 + st0 vn for s ∈ [0, 1] and each n ≥ n 1 where t0 > 1, then γn (0) = u 0 ∈ A1 , γn (1) = u 0 + t0 vn ∈ A2 . Since kuk1 1 t − kuku 1 is a continuous function for nonzero u and γn ([0, 1]) is connected, there exists a sequence H
H
{sn } ⊂ (0, 1) such that u 0 + sn t0 vn ∈ M− h. (v) By part (iv) and Remark 1, αh− ≤ Jh (u 0 + sn t0 vn ) < Jh (u 0 ) + α (Ω ) = αh (Ω ) + α (Ω ).
For n ≥ n 1 , we define a map Fn : S N −1 → H01 (Ω ) by Fn (z) (z) = u 0 (z) + sn t0 vn (z)
for z ∈ S N −1 ,
where vn (z) = ψ R (z) u (z − nz). Then we have Lemma 29. There exists a sequence {σn } in R+ such that Fn (S N −1 ) ⊂ [Jh ≤ αh (Ω ) + α (Ω ) − σn ]. Proof. By Lemma 28(iv) and Remark 1, we have that u 0 + sn t0 vn ∈ M− h and Jh (u 0 + sn t0 vn ) ≤ αh (Ω ) + α(Ω ) − σn for each n ≥ n 1 , the conclusion holds. For c > 0, we define Z cu p ; bc (u) = Ω
1 1 Ic (u) = a(u) − bc (u + ) ; 2 p M Ic = {u ∈ H01 (Ω )\{0} | hIc0 (u), ui = 0}. 1 Recall that there exist a unique t − = t − (u) > 0 and a unique t 1 = t 1 (u) > 0 such that t − u ∈ M− h and t u ∈ M. Let Σ = {u ∈ H01 (Ω ) | u ≥ 0 and kuk H 1 = 1}. Then we have the following results.
Lemma 30. (i) For each u ∈ Σ , there exists a unique number t c (u) > 0 such that t c (u)u ∈ M Ic and 1 1 − 2 c − bc (u) p−2 ; max Ic (tu) = Ic t (u) u = t≥0 2 p (ii) For each nonnegative u ∈ H01 (Ω ) and 0 < µ < 1, we have (1 − µ) I
1 1−µ
(u) −
1 1 khk2L 2 ≤ Jh (u) ≤ (1 + µ) I 1 (u) − khk2L 2 ; 1+µ 2µ 2µ
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(iii) For each u ∈ Σ and 0 < µ < 1, we have p
(1 − µ) p−2 J (t 1 u) −
p 1 1 khk2L 2 ≤ Jh t − u ≤ (1 + µ) p−2 J (t 1 u) − khk2L 2 ; 2µ 2µ
(iv) αh− > 0 for sufficiently small khk L 2 . Proof. (i) For each u ∈ Σ , let f (t) = Ic (tu) = 12 t 2 − 1p t p bc (u), then f (t) → −∞ as t → ∞, f 0 (t) = t −t p−1 bc (u) and f 00 (t) = 1 − ( p − 1) t p−2 bc (u). Let t (u) = c
1 bc (u)
1 p−2
> 0.
Then f 0 (t c (u)) = 0, t c (u) u ∈ M Ic and t c (u)
2
f 00 t c (u) = a t c (u)u − ( p − 1) bc t c (u)u 2 = (2 − p) t c (u) a (u) < 0.
Thus, there exists a unique t c (u) > 0 such that t c (u)u ∈ M Ic and max Ic (tu) = Ic t c (u) u t≥0 1 1 − 2 = − bc (u) p−2. 2 p (ii) For µ ∈ (0, 1), we get Z hudz ≤ kuk 1 khk 2 ≤ µ kuk2 1 + 1 khk2 2 . H L H L 2 2µ Ω Thus, for each nonnegative u ∈ H01 (Ω ), then Z Z 1 1 1+µ 1 1 1−µ kuk2H 1 − khk2L 2 ≤ Jh (u) ≤ kuk2H 1 − khk2L 2 . up − up + 2 p Ω 2µ 2 p Ω 2µ (iii) Applying part (ii), we have that for each u ∈ Σ (1 − µ) I
where t c1 u ∈ M I I
1 1−µ
1 1 khk2L 2 ≤ Jh t − u ≤ (1 + µ) I 1 t c2 u − khk2L 2 , t c1 u − 1+µ 2µ 2µ
1 1−µ
1 1−µ
t c1 u =
and t c2 u ∈ M I
1 1+µ
. By part (i), then
2 1 1 1 1 − 2 − 2 − b 1 (u) p−2 = (1 − µ) p−2 − b (u) p−2 1−µ 2 p 2 p 2
= (1 − µ) p−2 J (t 1 u). 2
Similarly, I
1 1+µ
(t c2 u) = (1 + µ) p−2 J (t 1 u). Hence, (iii) holds.
(iv) We know that α (Ω ) > 0 and applying part (ii) to obtain p
αh− ≥ (1 − µ) p−2 α (Ω ) − Thus, (iv) holds.
1 khk2L 2 . 2µ
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The following lemma is a key lemma to prove our main result. Lemma 31. There exists a number δ0 > 0 such that if u ∈ M and J (u) ≤ α (Ω ) + δ0 , then Z z |∇u|2 + u 2 dz 6= 0. R N |z| Proof. On the contrary, there exists a sequence {u n } in M such that J (u n ) = α (Ω ) + o (1) and Z z |∇u n |2 + u 2n dz = 0. R N |z| By Lemma 9, {u n } 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 R N . It follows from the Decomposition Lemma 6 that there exists a sequence {z n } in R N such that |z n | → ∞ as n → ∞ and u n (z) = u(z − z n ) + o (1)
strongly in H 1 (R N ).
Suppose the subsequence |zz nn | → z 0 as n → ∞, where z 0 is a unit vector in R N . Then by the Lebesgue dominated theorem, we have Z z |∇u n |2 + u 2n dz 0= N |z| ZR z + zn |∇u|2 + u 2 dz + o (1) = N |z + z n | R 2p = α(R N )z 0 + o (1), p−2 which is a contradiction.
Lemma 32. There exists a number d0 > 0 such that for khk L 2 < d0 , we have Z z |∇u|2 + u 2 dz 6= 0, R N |z| for u ∈ [Jh < αh (Ω ) + α (Ω )]. Proof. For u ∈ [Jh < αh (Ω ) + α(Ω )], then u/ kuk H 1 ∈ Σ . There exists a t 1 > 0 such that t 1 u/ kuk H 1 ∈ M. By Lemma 30(ii), we have for any µ ∈ (0, 1) and khk L 2 < d1 (µ) 1 t u 1 − p khk2L 2 . J ≤ (1 − µ) p−2 Jh (u) + kuk H 1 2µ Since αh (Ω ) < 0, we have, for u ∈ [Jh < αh (Ω ) + α (Ω )], 1 p t u 1 − p−2 2 khk L 2 . J ≤ (1 − µ) α (Ω ) + kuk H 1 2µ Take µ ∈ (0, 1) such that there exists d1 (µ) > d0 > 0 such that for khk L 2 < d0 1 t u J ≤ α (Ω ) + δ0 . kuk H 1 Since t 1 u/ kuk H 1 ∈ M, by Lemma 31 1 2 1 2 ! Z t u t u z ∇ + dz 6= 0, kuk H 1 kuk H 1 R N |z| or Z
z |∇u|2 + (u)2 dz 6= 0. R N |z|
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For khk L 2 < d0 , we obtain Z z |∇u|2 + |u|2 dz 6= 0 R N |z| for all u ∈ [Jh < αh (Ω ) + α (Ω )]. Now, we define G : [Jh < αh (Ω ) + α (Ω )] → S N −1 by Z
z |∇u|2 + |u|2 dz G(u) = R N |z|
Z
z 2 2 |∇u| + |u| dz . R N |z|
We have Lemma 33. For n ≥ n 1 and khk L 2 < d0 , the map G ◦ Fn : S N −1 → S N −1 is homotopic to the identity. Proof. We define ζn (θ, z) : [0, 1] × S N −1 → S N −1 by G ((1 − 2θ) Fn (z) + 2θu (z − nz)) n z ζn (θ, z) = G u z − 2 (1 − θ) z
for θ ∈ [0, 1/2) ; for θ ∈ [1/2, 1) ; for θ = 1.
First, we claim that limθ→1− ζn (θ, z) = z and limθ→ 1 − ζn (θ, z) = G (u (z − nz)). 2
(a) limθ →1− ζn (θ, z) = z: since 2 2 ! Z z n n dz ∇u z − 2 (1 − θ) z + u z − 2 (1 − θ) z R N |z| Z n z + 2(1−θ )z |∇u (z)|2 + u (z)2 dz = n R N z + 2(1−θ ) z 2p = α(R N )z + o(1) as θ → 1− , p−2
then limθ→1− ζn (θ, z) = z. (b) limθ→ 1 − ζn (θ, z) = G (u (z − nz)): since 2
k(1 − 2θ) Fn (z) + 2θu (z − nz)k H 1 = ku (z − nz)k H 1 + o(1)
as θ →
1− , 2
by the continuity of G, we obtain limθ → 1 − ζn (θ, z) = G (u (z − nz)). Thus, ζn (θ, z) ∈ C([0, 1] × S N −1 , S N −1 ) and 2
ζn (0, z) = G (Fn (z)) ζn (1, z) = z
for all z ∈ S N −1 ,
for all z ∈ S N −1 ,
provided n ≥ n 1 and khk L 2 < d0 . This completes the proof. Thus, we have
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Lemma 34. Assume that h (z) ≥ 0 and 0 < khk L 2 < min {d( p, α), d0 }, where d0 is defined as in Lemma 32. Jh (u) has at least two critical points in [Jh < αh (Ω ) + α (Ω )]. Proof. Applying Lemmas 27 and 33, we have for sufficiently large n ≥ n 1 and khk L 2 < d0 , cat ([Jh ≤ αh (Ω ) + α (Ω ) − σn ]) ≥ 2. By Lemmas 18 and 26, Jh (u) has at least two critical points in [Jh < αh (Ω ) + α (Ω )].
5. Existence of the fourth solution Since αh− > 0 for sufficiently small khk L 2 , we define K h (u) = sup Jh (tu) = Jh (t − u) > 0, t≥0
where Let
t −u
∈ Mh . We observe that if khk L 2 is sufficiently small, Bahri–Li’s minimax argument [4] also works for K h .
Γ = {γ ∈ C(Br (0), Σ ) | γ |∂ Br (0) = ψ R (z) u¯ (z − y) / kψ R (z) u¯ (z − y)k H 1 }
for large r = |y|,
where Σ = {u ∈ H01 (Ω ) | u ≥ 0 and kuk H 1 = 1}. Then we define γh (Ω ) = inf sup K h (γ (y)); γ ∈Γ y∈R N
γ0 (Ω ) = inf sup K 0 (γ (y)). γ ∈Γ y∈R N
By Lemma 30(iii), for 0 < µ < 1, we have p
(1 − µ) p−2 γ0 (Ω ) −
p 1 1 khk2L 2 ≤ γh (Ω ) ≤ (1 + µ) p−2 γ0 (Ω ) − khk2L 2 . 2µ 2µ
(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 < khk L 2 < d00 , then αh (Ω ) + α(Ω ) < γh (Ω ) < αh− (Ω ) + α(Ω ). Moreover, there exists a positive solution u of Eq. (1) in Ω such that Jh (u) = γh (Ω ). Proof. By Lemma 30(iii), we also have that for 0 < µ < 1 p
(1 − µ) p−2 α(Ω ) −
p 1 1 khk2L 2 ≤ αh− (Ω ) ≤ (1 + µ) p−2 α(Ω ) − khk2L 2 . 2µ 2µ
For any ε > 0, there exists a d1 (ε) > 0 such that if khk L 2 < d1 (ε), then α(Ω ) − ε < αh− (Ω ) < α(Ω ) + ε. Thus, 2α(Ω ) − ε < αh− (Ω ) + α(Ω ) < 2α(Ω ) + ε. Applying (16), for any δ > 0, there exists a d2 (δ) > 0 such that if khk L 2 < d2 (δ), then γ0 (Ω ) − δ < γh (Ω ) < γ0 (Ω ) + δ.
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Fix a small 0 < ε < (2α (Ω ) − γ0 (Ω ))/2, since α (Ω ) < γ0 (Ω ) < 2α (Ω ), choosing a δ > 0 such that for khk L 2 < 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 and 18(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 < khk L 2 < min{d( p, α), d0 , d00 }, where d0 is defined as in Lemma 32 and d00 is defined as in Lemma 36. Then there are at least four positive solutions of Eq. (1) in Ω . Proof. Therefore, by Lemmas 2, 15, 34 and 36, we have that Eq. (1) has at least four positive solutions in Ω .
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