Multiple existence of sign changing solutions for a nonlinear elliptic problem

Nonlinear Analysis 68 (2008) 1043–1063 www.elsevier.com/locate/na

Multiple existence of sign changing solutions for a nonlinear elliptic problem Norimichi Hirano Department of Mathematics, Faculty of Engineering, Yokohama National University, Tokiwadai, Hodogayaku, Yokohama, Japan Received 28 June 2006; accepted 17 November 2006

Abstract Let N ≥ 3, 2 < p < 2∗ = 2N /(N − 2), ε > 0 and Ω be a bounded domain with a smooth boundary ∂Ω . Our purpose in this paper is to consider the multiple existence of sign changing solutions of the problem −ε2 ∆u + u = |u| p−2 u,

u ∈ H01 (Ω ).

c 2006 Elsevier Ltd. All rights reserved.

MSC: primary 35J65 Keywords: Semilinear elliptic problem; Sign changing solutions

1. Introduction Let N ≥ 3, 2 < p < 2∗ = 2N /(N − 2). In the present paper, we consider the multiple existence of sign changing solutions of the singularly perturbed problem  2 −ε ∆u + u = |u| p−2 u in Ω (Pε ) u=0 on ∂Ω where ε > 0 and Ω ⊂ R N is a bounded domain with a smooth boundary ∂Ω . In the last twenty years, the effect of the geometry and the topology of domain Ω on the existence and the multiplicity of positive solutions of problem (P) have been investigated by many authors (cf. [2,5,9]). On the other hand, little seems to be known about the effect of the geometry and the topology of domain Ω on the existence and multiplicity of sign changing solutions of problem (P) (cf. [7,14]). In [8], Clapp and Weth proved multiple existence results for the sign changing solutions of (P) under some symmetric conditions on Ω (cf. also [6]) for the critical case p = 2∗ . Recently, Bartsch and Weth [3] established multiple existence results for (Pε ) in the case where the nonlinear term |t| p−2 t is replaced a more general function, under topological assumptions on the space {(x, y) ∈ Ω × Ω , x 6= y} (cf. also [4]). Our purpose in the present paper is to show the multiple existence of sign changing solutions for problem (P) under conditions for the shape of the domain Ω . In this paper, instead of dealing with general cases, we restrict ourselves to the case in which the shape of E-mail address: [email protected]. c 2006 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.11.034

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Fig. 1.

Fig. 2.

the domain Ω is relatively simple. In the forthcoming paper, we will consider the case where the shape of the domain is complicated and the conditions for the shape are given by topological conditions. Our method employed here is a minimax argument in which we make use of results from differential topology (cf. [12]). To state our results, we need to give some notation. For each (x, r ) ∈ R N × R+ , we denote by Br (x) the open ball centered at x with radius r . ∂ Br (x) denotes the boundary of Br (x). For each x ∈ ∂Ω , we define a number H (x) ∈ (0, ∞] by  H (x) = sup r > 0 : Br (z) ∩ ∂Ω = {x} , for some z ∈ Ω c . For each r > 0, we put Ωr = {x ∈ Ω : d(x, ∂Ω ) ≥ r }, where d(x, A) denotes the distance of x ∈ R N from A ⊂ R N . For subsets A, B ⊂ R N , we write A ∼ = B when A and B are homotopic. Let h 0 > 0. We impose the following conditions: (Ω1 ) H (∂Ω ) = infx∈∂ Ω H (x) > h 0 ; (Ω2 ) there exists r0 , R0 > 0 such that Ωr ∼ = Ω for r ∈ (0, r0 ), and Ωr consists of two connected components for r ∈ (r0 , R0 ]. We can now state our main result. Theorem 1.1. If r0 /R0 is sufficiently small, there exists ε0 > 0 such that for each ε ∈ (0, ε0 ), problem (P) has at least 7 pairs of sign changing solutions which have exactly two nodal domains. Remark 1.1. The assumptions (Ω1 ), (Ω2 ) are satisfied by a dumbbell type domain (Fig. 1) or a cylindrical domain from which a cylinder is removed (Fig. 2) under suitable conditions. This paper is organized as follows. In Section 2, we give notation and preliminary results. In Section 3, we show the existence of three sign changing solutions of local minimal type. In each of Sections 4 and 5, we find a minimax type sign changing solution. In Section 6, we show the existence of two minimax type sign changing solutions different from those obtained in Sections 4 and 5. In Appendix there are collected the proofs of some technical lemmas. 2. Preliminaries Throughout the rest of this paper, we assume that conditions (Ω1 ) and (Ω2 ) hold. We also assume that ε0 ∈ e instead of Ω eε when we do not need to specify eε = 1 Ω . We write simply Ω (0, min {h 0 , 4r0 }) and ε ∈ (0, ε0 ). We put Ω ε the parameter ε. Then the problem (P) is equivalent to the problem  e −∆u + u = |u| p−2 u in Ω (P) e. u=0 on ∂ Ω  er = x ∈ Ω e : d(x, ∂ Ω e ) ≤ r for each r > 0. For q ∈ [1, ∞], we Then we will work on (P) instead of (Pε ). We put Ω e ). For each k ∈ N and d ∈ [0, 1), we put I k = [−1, 1]k and I k = I k \[−1+d, 1−d]k . denote by |·|q the norm of L q (Ω d

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For d ∈ [0, 1], we put b Id2 = Id2 × {0} ⊂ R3 . For each F ⊂ R3 and α > 0, we put α F = {αx : x ∈ F}. For subsets A, B of a Banach space X with A ⊂ B, we denote by ∂ B A the relative boundary of A in B (we simply write ∂ A when there is no risk of confusion about the set B). For two topological spaces A, B, we write A ∼ = B when A and B are e ). homotopy equivalent. We write A ≡ B when A and B are isomorphic. To simplify the notation, we put H = H01 (Ω R R 2 2 2 We denote by k · k the norm of H defined by kvk = Ω e (|∇v| + |v| )dx for v ∈ H . For u, v ∈ H , hu, vi = Ω e uv dx. For each u ∈ H and r > 0, Dr (u) denotes the open ball in H centered at u with radius r > 0, and supp u stands for the support of function u. For each function u ∈ H , we put u + (x) = max {u(x), 0} and u − (x) =Rmax {−u(x), 0} for e . Let m p > 0 be such that |z| p ≤ m p kzk for z ∈ H . For each u ∈ L p (R N ), we put b x ∈Ω u (x) = B1 (x) |u(y)| p dy for o n u |∞ and x ∈ R N , Ω (u) = x ∈ R N : b u (x) ≥ |b 2 R

Ω (u)

β0 (u) = R

|b u |∞ 2 )dx . |b u |∞ 2 )dx

x(b u (x) −

u (x) − Ω (u) (b

The mapping β0 is called the generalized barycenter (cf. [15,4]). We put β(u) = (β+ (u), β− (u)) = (β0 (u + ), β0 (u − ))

for each u ∈ L p (R N ).

For each a ∈ R, and a functional F : H → R, we denote by F a the level set F a = {v ∈ H : F a (v) ≤ a}. We also denote by F [a,b] the set F [a,b] = {v ∈ H : a ≤ F(v) ≤ b} for each a, b ∈ R. We define a functional I : H −→ R by  Z  1 1 I (u) = (|∇u|2 + |u|2 ) − |u| p dx for u ∈ H. (2.1) e 2 p Ω Then u ∈ H is a solution of (P) if and only if u is a critical point of I on H . We denote by K the set of critical points of I on H . We put n o  p M = v ∈ H \ {0} : kvk2 + |v|22 = |v| p and Ms = u ∈ M : u ± ∈ M . Then each nontrivial critical point of I is in M, and each sign changing critical point of I is in Ms . It is known that for each v ∈ H \ {0}, there exists a unique positive number tv such that tv v ∈ M (cf. [11,16]). We put N v = tv v for v ∈ H \ {0}. Then one can verify easily that N ∈ C 1 (H \ {0} , M). By a standard argument, we can define a semiflow Ψ : [0, ∞) × M −→ M associated with I satisfying Z n o 1 t min k∇ I (Ψ (s, v))k2 , 1 ds for v ∈ M and t ≥ 0 (2.2) I (Ψ (t, v)) − I (v) ≤ − 2 0 and

d

Ψ (t, v) ≤ 2 min {k∇ I (Ψ (t, v))k, 1}

dt

for v ∈ M and t ≥ 0.

(2.3)

For each u ∈ M, we define a mapping γ by p

γ (u) = (ku + k2 − |u + | p )/kuk2 .

(2.4)

Then it is clear that for each sign changing function u ∈ M, γ (u) = 0 if and only if u ∈ Ms . Here we recall that e replaced by R N has a positive radial solution, denoted by U0 . U0 is the unique positive smooth problem (P) with Ω  solution of (P) up to translation, and U0 satisfies I (U0 ) = c = min I (v) : v ∈ MR N , where MR N is the set M with H replaced by H 1 (R N ). It is known that c is not attained by a function v ∈ M, i.e., cΩ e = infv∈M I (v) > c. Here we fix a function α ∈ C ∞ (R, [0, 1]) such that α(t) = 1 for t ≤ 0 and α(t) = 0 for t ≥ 1. For each x ∈ R N ex,r = N (αx,r Ux ). Then one can see that and r > 1, we put Ux (·) = U0 (· − x), αx,r (·) = α(| · −x| + 1 − r ) and U e supp Ux,r = Br (x). We also have from the definition that ex,r ) > I (Ux ) I (U

all (x, r ) ∈ R N × R+

and

ex,r = Ux . lim U

r −→∞

(2.5)

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For each r > 0, we define a subset Λr ⊂ R N × R N by  er × Ω er ) : |x − y| > 2r . Λr = (x, y) ∈ int(Ω er(0) , Ω er(1) the two connected components of Ω er for each r > r0 /ε. We denote by Ω Lemma 2.1. There exists s0 ∈ (0, 1) and a mapping σ ∈ C([s0 , 1] × (Ms ∩ I 3c ), R+ ) satisfying the following conditions: (1) su + − σ (s, u)u − ∈ M for all (s, u) ∈ [s0 , 1] × (Ms ∩ I 3c ); (2) for each σ ∈ (s0 , 1), there exist m σ,1 , m σ,2 > 0 such that m σ,1 <

d I (su + − σ (s, u)u − ) < m σ,2 ds

for all (s, u) ∈ [s0 , σ ] × (Ms ∩ I 3c ).

Proof. Let u ∈ Ms ∩ I 3c . That is u ± ∈ M. Then for each s ∈ [0, 1], there exists a unique positive number t > 0 p p p such that (t 2 − t p )|u − | p = −(s 2 − s p )|u + | p . Then noting that ku ± k2 = |u ± | p , we have, putting t = σ (s, u), that s 2 ku + k2 + σ (s, u)2 ku − k2 = s p |u + | p + σ (s, u) p |u − | p .

(2.6)

That is su + − σ (s, u)u − ∈ M. We note that σ (s, u) ≥ 1. If we put m = ku − k2 /ku + k2 , then s 2 − s p = p (σ (s, u) p −σ (s, u)2 )m. We note that 1/2 ≤ m ≤ 2. In fact, recalling that u ± ∈ M and I (u ± ) = 21 ku ± k2 − 1p |u ± | p = p−2 ± 2 2 p ku k

≥ c, we have ku ± k2 ≥ 2 pc/( p − 2). On the other hand, we have ku ± k2 ≤ 4 pc/( p − 2) because u ∈ I 3c . Then 1/2 ≤ m ≤ 2 as claimed. We also have from (2.6) that 2s − ps p−1 = σ 0 (s)( pσ (s) p−1 − 2σ (s))m, where σ (s) = σ (s, u). Then since σ (s) ≥ 1, one can see σ 0 (s) =

2s − ps p−1 m<0 pσ (s) p−1 − 2σ (s)

(2.7)

on (s0 , 1), where s0 = (2/ p)1/( p−2) . We also have that for each s ∈ (s0 , 1), d I (su + − σ (s)u − ) = (2s − ps p−1 )ku + k2 + σ 0 (s)(σ (s) − σ (s) p−1 )ku − k2 . ds Then recalling that 1/2 ≤ m ≤ 2, assertion (2) follows directly from (2.7) and (2.8).

(2.8) 

We note that the same observation as in the proof of the lemma above yields that kuk2 ≤ C0 = 6cp/( p − 2) for all u ∈ M ∩ I 3c . For each u ∈ Ms ∩ I 3c , we put  (1 + (1 − s0 )s)u + − σ (1 + (1 − s0 )s, u)u − for s ∈ [−1, 0] π(s, u) = σ (1 − (1 − s0 )s, u)u + − (1 − (1 − s0 )s)u − for s ∈ [0, 1]. From the definition of σ and γ , we have that for each u ∈ Ms ∩ I 3c , γ (π(s, u)) = ((1 + (1 − s0 )s)2 − (1 + (1 − s0 )s) p )

ku + k2 >0 kπ(s, u)k2

for s ∈ [−1, 0]

and γ (π(s, u)) = −((1 − (1 − s0 )s)2 − (1 − (1 − s0 )s) p )

ku − k2 <0 kπ(s, u)k2

for s ∈ [0, 1].

Then it follows that s −→ γ (π(s, u))

is strictly monotone decreasing on [−1, 1]

(2.9)

and that there exists π0 such that γ (π(−s, u)) > π0 ,

γ (π(s, u)) < −π0

 1 , 1 × (Ms ∩ I 3c ). for all (s, u) ∈ 2 

(2.10)

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We also have by (2) of Lemma 2.1 that s −→ I (π(±s, u))

is decreasing on [0, 1] for u ∈ Ms .

(2.11)

For each σ ∈ (0, 1), we set Ms,σ = {π(s, v) : v ∈ Ms , |s| < σ }. Then Ms,σ is an open neighborhood of Ms in M. From the definition of π(·, ·), we have that each element u of Ms,1 is sign changing. Moreover one can see that for each σ ∈ (0, 1), m(σ ) = inf{ku − vk : u ∈ Ms,σ ∩ I 3c , v ∈ ∂M (Ms,1 ∩ I 3c )} > 0. For each (d, σ, δ) ∈ R+ × R+ × R+ , we denote by ρd,σ,δ the mapping ρd,σ,δ ∈ C([0, 1] × (Ms,1/2 ∩ I d+δ ), Ms,1 ∩ I d+δ ) satisfying the following conditions:  for all u ∈ Ms,1/2 ∩ I d+δ ρd,σ,δ (0, u) = u (ρ) ρ (1, u) ∈ Ms,1 ∩ I d−σ for all u ∈ Ms,1/2 ∩ I d+δ  d,σ,δ ρd,σ,δ (t, u) = u for all (t, u) ∈ [0, 1] × (Ms,1/2 ∩ I d−σ ). We also define a continuous mapping Q : Ms,1 −→ Ms by (2.12)

Qv = u where u ∈ Ms such that π(s, u) = v for some s ∈ (−1, 1). Then one can see that Q is continuous on Ms,1 . Moreover we have Lemma 2.2. For given δ > 0, there exists sδ > 0 such that if v ∈ Ms,1 ∩ I 3c and |γ (v)| < sδ , then kv − Qvk < δ

and

I (Qv) ≤ I (v) + δ.

Lemma 2.2 is a direct consequence of the definition of Q. Thus we omit the proof. Lemma 2.3. Let d ∈ (2c, 3c) such that there is no critical point of I in Ms with critical value d. Then there exists ε(d) > 0 such that for each σ ∈ (0, ε(d)), there exists a mapping ρd,σ,ε(d) ∈ C([0, 1] × (Ms,1/2 ∩ I d+ε(d) ), Ms,1 ∩ I d+ε(d) ) satisfying (ρ). Proof. Suppose that d ∈ (2c, 3c) satisfies the assumption. Then there exists ε 0 > 0 such that K∩(Ms ∩I [d−ε ,d+ε ] ) = φ. Then noting that each sign changing critical point of I is in Ms , we have that there exists µ ∈ (0, 1) such that 0 0 0 0 0 0 k∇ I (v)k ≥ µ for all v ∈ Ms,1 ∩I [d−ε ,d+ε ] . Let v ∈ Ms,1 ∩I [d−ε ,d+ε ] and t ≥ 0 be such that Ψ (t, v) ∈ I [d−ε ,d+ε ] . Then by (2.2) and (2.3), it follows that 0

I (Ψ (t, v)) − I (v) ≤ −

0

µ2 t 2

and  Z t 1/2 Z t d kΨ (s, v) − vk2 t 1/2 . Ψ (s, v) − v, Ψ (s, v) ds ≤ 2 dt 0 0 R 1/2 t Then 0 kΨ (s, v) − vk2 ds ≤ 2t 3/2 and kΨ (t, v) − vk2 ≤ 2

4 (I (v) − I (Ψ (t, v))). µ2 n 2 o Here we put ε(d) = min µ16 m( 12 ), ε0 . Let σ ∈ (0, ε(d)). If v ∈ Ms,1/2 ∩ I d+ε(d) and tv ≥ 0 such that I (Ψ (tv , v)) = kΨ (t, v) − vk ≤

d − σ , we have from the definition of ε(d) that kΨ (tv , v) − vk ≤ 12 m( 12 ). That is Ψ (tv , v) ∈ Ms,1 by the definition of m(·). Now we define the deformation by ρd,σ,ε(d) (t, v) = Ψ (tv t, v) for (t, v) ∈ [0, 1] × (Ms,1/2 ∩ I d+ε(d) ). Then ρd,σ,ε(d) has the desired properties. 

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3. Existence of local minima Throughout the rest of this paper, we put r = r0 /ε and R = R0 /ε. From (2) of Lemma 2.1 and the definition of π(·, ·), we can choose ε1 > 0 so small that   1 (3.1) π ± , v ∈ I 2c for all v ∈ Ms ∩ I 2c+ε1 . 2 Then we have Proposition 3.1. Let V ⊂ Ms ∩ I 2c+ε1 be a bounded open set in Ms such that  d = inf {I (v) : v ∈ V } < db = inf I (v) : v ∈ ∂Ms V .

(3.2)

Then there exists a critical point u ∈ V of I such that I (u) = d. Proof. Let V ⊂ Ms ∩ I 2c+ε1 satisfy the assumption. Assume that there exists σ0 ∈ (0, min{db − d, d − 2c}) such that V ∩ I d+σ0 contains no critical point of I . Then there exists δ > 0 such that U σ0 ∩ Ms ⊂ V and infv∈Uσ0 k∇ I (v)k > 0,

where Uσ = (∪v∈V ∩I d+σ Dδ (v)) ∩ M and U σ is the closure of Uσ for each σ > 0. Then by (2.11) and (2) of Lemma 2.1, we can choose σ1 ∈ (0, σ0 /2) so small that   1 1 . (3.3) π(s, u) ∈ Uσ0 /2 ∪ (M ∩ I d−σ1 ) for u ∈ V ∩ I d+σ1 and s ∈ − , 2 2 On the other hand, noting that infv∈Uσ0 k∇ I (v)k > 0, we can choose σ ∈ (0, σ1 ) so small that Ψ (t, v) ∈ Uσ0 ∪ (M ∩ I d−σ )

for t ≥ 0 and v ∈ Uσ0 /2 ∩ I d+σ .

Then in the same way as in the proof of Lemma 2.3, we can see that there exists a continuous mapping ρ ∈ C([0, 1] × ((Uσ0 /2 ∩ I d+σ ) ∪ (M ∩ I d−σ )), Uσ0 ∪ (M ∩ I d−σ )) such that  for all v ∈ (Uσ0 /2 ∩ I d+δ ) ∪ (M ∩ I d−σ ), ρ(0, v) = v ρ(t, v) = v for all (t, v) ∈ [0, 1] × (M ∩ I d−σ ),  d−σ ρ(1, v) ∈ U σ0 ∩ M ∩ I for all v ∈ Uσ0 /2 ∩ I d+σ . Now let u ∈ V ∩ I d+σ . Since d − σ > 2c, we have by (2.11) and (3.1) that there exist s1 < 0 and s2 > 0 such that I (π(s1 , u)) = I (π(s2 , u)) = d − σ and I (π(s, u)) > d − σ for s ∈ (s1 , s2 ). Then by (3.3), we have π((s1 , s2 ), u) ⊂ Uσ0 /2 . Then we have ρ(1, π([s1 , s2 ], u)) ⊂ U σ0 ∩ (M ∩ I d−σ ). Since γ (ρ(1, π(s1 , u))) = γ (π(s1 , u)) > 0 and γ (ρ(1, π(s2 , u))) = γ (π(s2 , u)) < 0, we have that there exists s ∈ (s1 , s2 ) such that γ (ρ(1, π(s, u))) = 0, i.e., ρ(1, π(s, u)) ∈ U σ0 ∩ Ms ⊂ V . Since ρ(1, π(s, u)) ∈ I d−σ , this contradicts (3.2). This completes the proof.  Remark 3.1. It is known that each global minimal point of Ms is a critical point of I (cf. [4,7]). We also note that the argument employed in Proposition 14 of [8] is valid for the proof of Proposition 3.1 above. e replaced by R N . Let s > 0. In the following, we use the same symbol I for the functional defined by (2.1) with Ω We put MR N \B1 ((s+1)x1 ) = {u ∈ H01 (R N \ B1 ((s + 1)x1 )) : h∇ I (u), ui = 0}. and  cs = inf I (u) : u ∈ MR N \B1 ((σ +1)x1 ) , β+ (u) = 0, 0 < σ ≤ s , where x1 = (1, 0, . . . , 0) ∈ R N . From the definition of cs , it is clear that cs is continuous and monotone decreasing as s −→ ∞. Moreover we have e and d(∂ Ω e , β+ (u)) = s. Then I (u) ≥ cs .(2): Lemma 3.1. (1): Let s > 0 and u ∈ M be such that u ≥ 0 on Ω lims−→∞ cs = c.

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e ) > 1. Let s and u satisfy the assumption. Let y ∈ ∂ Ω e Proof. (1) Since ε < h 0 , we have by (Ω1 ) that H (∂ Ω N e e be such that d(y, β+ (u)) = d(∂ Ω , β+ (u)) = s. We may assume, by translating and
rotating Ω in R , that e ) > 1, we have that B1 ((s + 1)x1 ) ⊂ Ω e c . Therefore we find that β+ (u) = 0 and y = sx1 . Then since H (∂ Ω e0,s ∈ MR N \B ((s+1)x ) for each u ∈ MR N \B1 ((s+1)x1 ) . Then by the definition of cs , I (u) ≥ cs follows. (2) Since U 1 1 e0,s ) = c, the assertion holds.  s > 0 and lims−→∞ I (U From (2) of the lemma above, we can take ε0 sufficiently small that cr0 /ε0 < c + ε1 .

(3.4)

Then from the definition of r, cr < c + ε1 . In the rest of this section, we assume that ε0 satisfies (3.4). We define a function R(·) by   c + cs e for s ≥ r. R(s) = sup R > 2r : I (U0,R ) ≥ 2 Here we put n o e (i) , β− (u) ∈ Ω e ( j) V i, j = u ∈ Ms : β+ (u) ∈ Ω 2r 2r

for i, j ∈ {0, 1} . (i, j)

Proposition 3.2. Assume that R > 2R(2r ). Then there exist u 0 critical point of I and (i, j)

I (u 0

i, j

) = c0 = min{I (u) : u ∈ V i, j }

(i, j)

∈ V i, j , i, j ∈ {0, 1}, such that each u 0

is a

for i, j ∈ {0, 1} .

Proof. By Proposition 3.1, it is sufficient to show that d = inf{I (u) : u ∈ V i, j } < inf{I (u) : u ∈ ∂Ms V i, j }

for i, j ∈ {0, 1},

and d < 2c + ε1 . Let i, j ∈ {0, 1} and u ∈ ∂Ms V i, j . We may assume without any loss of generality that e , β+ (u)) = 2r . Then by the definition, I (u + ) ≥ c2r . Since u − ∈ M, I (u − ) ≥ c e > c. Then we have d(∂ Ω Ω i. j e (i) ×Ω e ( j) ). I (u) > cΩ e +c2r for u ∈ ∂ V . On the other hand, since R > 2R(2r ), there exists (x, y) ∈ ΛR(2r ) ∩(Ω R(2r ) R(2r ) ex,R(2r ) − U ey,R(2r ) ∈ V i, j and Then from the definition of R(·), we have that U ex,R(2r ) − U ey,R(2r ) ) ≤ c + c2r < c + cr < 2c + ε1 . d ≤ I (U  (i, j) i, j . This completes the proof. Then c0 ≤ c + c2r < cΩ e + c2r ≤ inf I (u) : u ∈ ∂Ms V



Remark 3.2. It is known and easy to verify that each critical point u ∈ Ms with I (u) < 3c has exactly two nodal (0,0) (1,1) (0,1) (0,0) (1,1) domains (cf. [4]). From the definition of V i, j , u 0 6= ±u 0 , u 0 6= ±u 0 , ±u 0 . But it is not clear whether (0,1) (1,0) u0 6= ±u 0 . Then the number of distinct pairs of critical points guaranteed by Proposition 3.2 is three. 4. Minimax type critical points I e0,s ). Then In this section, we show the existence of minimax type critical points. For each s > 0, we put e cs = I (U we have e cs > c for s > 0 and lims−→∞ e cs = c. Here we choose ε0 such that e cr0 /4σ < c +

ε1 2

for all σ ∈ (0, ε0 ).

(4.1)

Since cr0 /ε0 ≤ cr0 /4ε0 ≤ e cr0 /4ε0 , we find that if (4.1) holds, then (3.4) holds. In the rest of this section, we assume that (4.1) holds and R > max {4R(r ), 2R(2r )}. From (4.1), we have that e cr 0 < c + ε21 for all r 0 ≥ r/4. From the (i) (i) (i) e xi ∈ Ω e , and yi ∈ Ω e definition, we can choose z i ∈ Ω R R/2 R/2 for i = 0, 1 such that d(x i , z i ) = R/2 and yi = 2z i − x i e e \ BR(r ) (y0 ))r/2 is connected. Then for i = 0, 1. Since Ωr/2 is connected and BR(r ) (y0 ) ⊂ B3R/4 (z 0 ), we have that (Ω

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e ) such that p0 (−1) = z 0 , p0 (1) = z 1 , d( p0 (t), ∂(Ω e \ BR(r ) (y0 ))) ≥ we can choose p0 ∈ C ∞ (I, Ω ∞ + next choose a function θ ∈ C (I, R ) such that r e \ BR(r ) (y0 )), p0 (t)) for t ∈ I. θ (−1) = θ (1) = R(r ), ≤ θ (t) ≤ d(∂(Ω 4

r 4

for t ∈ I. We

Now we define a mappings ω1 ∈ C ∞ (I, Ms ) by ep (t),θ(t) − U ey ,R(r ) ω1 (t) = U 0 0

for t ∈ I.

(4.2)

I (ω1 (−1)) = I (ω1 (1)) = 2e cR(r ) ≤ c + cr

(4.3)

From the definition of ω1 , we have

and I (ω1 (t)) ≤ e cR(r ) + e cr/4 ≤ 2e cr/4 < 2c + ε1 on I. Here we put Γ1 = {v ∈ C(I, Ms ∩ I 2c+ε1 ) : v(−1) = ω1 (−1), v(1) = ω1 (1)}

(4.4)

and cm,1 = inf sup I (v(t)). v∈Γ1 t∈I

(0)

e , β+ (ω1 (1)) = z 1 ∈ Since ω1 ∈ Γ1 by the definition, Γ1 6= ∅. On the other hand, noting that β+ (ω1 (−1)) = z 0 ∈ Ω R (1) (0) (1) e e e Ω R and Ω R ∩ Ω R = ∅, we have by (Ω2 ) that er 6= ∅ for all v ∈ Γ1 . β+ (v(I)) ∩ Ω Then maxt∈I I (v(t)+ ) ≥ cr for any v ∈ Γ1 . Then noting infv∈Ms I (v − ) ≥ cΩ e > c by the definition of c, we find i, j that cm,1 > c + cr > c0 , for i, j ∈ {0, 1}. We will see that there exists a critical point u m,1 of I with I (u m,1 ) = cm,1 . Lemma 4.1. π(·, ω1 (·)), π(·, ω1 (·))+ ∈ C 2 (I 2 , M). e ) and θ ∈ C ∞ (I, R+ ), we have from ep (t),θ(t) for t ∈ I. Since p0 ∈ C ∞ (I, Ω Proof. Here we put τ (t) = ω1+ (t) = U 0 ∞ e the definition of Ux,r that τ ∈ C (I, M). On the other hand, we have from (2.6), p

p

s 2 kτ (t)k2 + σ (s, ω1 (t))2 kω1 (t) − τ (t)k2 = s p |τ (t)| p + σ (s, ω1 (t)) p |ω1 (t) − τ (t)| p for t ∈ [0, 1]. Then one can see, noting that p > 2, that the mapping σ (·, ω1 (·)) ∈ C 2 (I 2 , R+ ). Then since  ey ,R(r ) for s ∈ [−1, 0] (1 + (1 − s0 )s)τ (t) − σ (1 + (1 − s0 )s, ω1 (t))U 0 π(s, ω1 (t)) = ey ,R(r ) for s ∈ [0, 1], σ (1 − (1 − s0 )s, ω1 (t))τ (t) − (1 − (1 − s0 )s)U 0 we obtain the assertion.



Lemma 4.2. (1): Let k ∈ N, m ∈ N ∪ {∞} , θ ∈ C(I k , R) and d ∈ [0, 1/2) be such that θ ∈ C m (Idk , R). Then for each σ ∈ (0, 1), there exists θ ∈ C m (I k , R) such that |θ (s, t) − θ (s, t)| < σ

on I k

and

θ (s, t) = θ (s, t)

k for (s, t) ∈ Id/2 .

2 , where e∪b (2): Let F ⊂ I 3 be a two dimensional C ∞ manifold such that ∂ F = ∂R2 I 2 × {0} and F = F I1/2 3 e ⊂ [−1/2, 1/2] . Let θ ∈ C(I × F, R) satisfy that the mapping s −→ θ (s, z) is monotone decreasing on I for each F z ∈ F. Then for each σ ∈ (0, 1), there exists θ ∈ C ∞ (I × F, R) such that

|θ − θ|∞ < σ s −→ θ (s, t)

2 on I × F, and for each z ∈ b I1/4 ,

is monotone decreasing on [−1 + σ, 1 − σ ].

Proof. We give a proof for (2). The assertion of (1) can be proved by a parallel argument. Let F and θ satisfy the assumption. Let ψ ∈ C0∞ (R3 ) be defined by ψ(x) = δ exp(1/(|x|2 − 1)) for |x| < 1 and ψ(x) = 0 for |x| ≥ 1, where

N. Hirano / Nonlinear Analysis 68 (2008) 1043–1063

δ is fixed so that

ψ = 1. We put   Z (s, t) − y b −3 ψ θ (y)dy θh (s, t) = h h R3 R

1051

R3

for (s, t) ∈ R × R2 and h > 0,

(4.5)

2 ⊂ R3 and b 2 ). Then θ ∈ C ∞ (R3 ) for all h > 0 and |b where b θ = θ on I × b I1/2 θ = 0 on R3 \ (I × b I1/2 θ − θh |∞ → 0, h 0 as h → 0 (cf. [10], Section 7.2). Then since s −→ θ (s, t) is decreasing on I for t ∈ F, we have from the definition 2 and h ∈ (0, 1). of ψ and (4.5) that s −→ θh (s, t) is decreasing on [−1 + h, 1 − h] for each t ∈ b I1/2 e⊂ Let σ ∈ (0, 1). Let E ⊂ I × F be a three dimensional manifold with a smooth boundary such that I × F 2 ∞ b E ⊂ (I × F) \ (I × I1/4 ). Then since C (E) is dense in C(E) (cf. Chapter 2 of [13]), we have that there exists e θ ∈ C ∞ (E) such that |e θ − θ|∞ < σ/2 on E. We extend e θ by putting e θ = 0 on (I × F) \ E. On the other hand, Let 2 . We put η ∈ C ∞ (I × F, [0, 1]) be such that η = 1 on E and η = 0 on I × b I1/4

θ (x) = η(x)e θ (x) + (1 − η(x))θh (x)

for x ∈ I × F.

Then since η(s, t) = ∈ (0, 1/2), s −→ θ (s, t) is decreasing on [−1+h, 1−h] 2 b for t ∈ I1/4 . Then the assertion follows by taking h sufficiently small.  2 , we obtain that for each h 0 on I × b I1/4

Proposition 4.1. If R > {4R(r ), 2R(2r )}, then there exists a critical point u m,1 ∈ Ms with I (u m,1 ) = cm,1 . Proof.  We assume that there exists no critical point of I in Ms with critical value cm,1 . Let σ = min ε(cm,1 ), (cm,1 − c − cr )/2 , where ε(·) is the mapping defined in Lemma 2.3. We can choose v ∈ Γ1 such that max I (v(t)) < cm,1 + σ. t∈I

Since v(t) ∈ Ms ∩ I 2c+ε1 on I by the definition of Γ1 , we have by (3.1) that    1 < 2c for all t ∈ [0, 1]. I π ± , v(t) 2 We also have by (2.11) that   s , v(i) ≤ I (π(0, v(i))) ≤ c + cr for i = 0, 1 and s ∈ I. I π 2 Therefore  s  , v(t) ≤ c + cr for (s, t) ∈ ∂I 2 . I π 2 On the other hand, recalling that  s  kπ s , ω1 (t)
+ k2 − |π s , ω1 (t)
+ | pp 2
2 γ π , v(t) = , 2 kπ 2s , ω1 (t) k2

(4.6)

(4.7)

we find by Lemma 4.1 that γ ◦ π( 2· , v(·)) ∈ C 2 (I 2 , R). By Lemma 2.3, there exists a continuous mapping ρ = ρcm,1 ,σ,ε(cm,1 ) satisfying (ρ) and we have   s  I ρ 1, π , v(t) ≤ cm,1 − σ on I 2 . (4.8) 2 Since c + cr < cm,1 − σ and (4.7) holds, we can choose d1 > 0 so small that  s  I π , v(t) ≤ cm,1 − σ for (s, t) ∈ Id21 . 2 Then  s  s  ρ 1, π , v(t) = π , v(t) on Id21 . 2 2

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Here we put ϕ(s, t) = ρ(1, π( 2s , v(t))) for (s, t) ∈ I 2 . By Lemma 4.2, we can choose γ ∈ C 2 (I 2 , R+ ) such that  γ (ϕ) = γ on Id21 /2 and |γ (ϕ) − γ |∞ ≤ min sσ/2 /2, π0 /2 , where sσ/4 is the number defined in Lemma 2.2. Since we have by (2.10) that  γ (−1, t) = γ (π(−1/2, v(t))) > π0 , and (4.9) γ (1, t) = γ (π(1/2, v(t))) < −π0 for all t ∈ I, the range R(γ ) of γ contains the interval [−π0 , π0 ]. Then since the regular value of γ ∈ C 2 (I 2 , R+ ) is dense in R(γ ) by Sard’s theorem (cf. 4.33 of [1]), we can find a regular value q ∈ R of γ such that |q| < min sσ/2 /2, π0 /2 . Then γ −1 (q) is a disjoint union of one dimensional submanifolds of I 2 . Since γ = γ (ϕ) on Id21 /2 and s −→ γ (π(s, v(t))) is monotone decreasing for t ∈ Id1 /2 by (2.9), we have that there exists a unique point s1 ∈ (−1, 1) and a unique one dimensional manifold {ξ(t) : t ∈ I} ⊂ γ −1 (q) ⊂ I 2 such that ξ(−1) = (s1 , −1) and   s 1 , v(−1) = q. γ (s1 , −1) = γ (ϕ(ξ(−1))) = γ π 2 Then the other end point ξ(1) of ξ must be in the boundary ∂I 2 \ {(s, −1) : s ∈ I}. We have by (4.9) that γ −1 (q) ∩ {(±1, t) : t ∈ I} = ∅. That is the other end ξ(1) of ξ must be in {(s, 1) : s ∈ I}. Therefore Q(ϕ(ξ(−1))) = ω1 (−1) and Q(ϕ(ξ(1))) = ω1 (1). On the other hand, noting that |γ (ϕ(ξ(t)))| ≤ |γ (ϕ(ξ(t))) − γ (ξ(t))| + |γ (ξ(t))| < sσ/2 , we have by Lemma 2.2 that I (Q(ϕ(ξ(t)))) ≤ I (ϕ(ξ(t))) + σ/2 ≤ cm,1 − σ/2

for all t ∈ I.

Then we find that Q(ϕ(ξ )) ∈ Γ1 . Since cm,1 ≤ maxt∈I I (Q(ϕ(ξ(t)))) by the definition, this is a contradiction.



5. Minimax type critical point II In this section, we show the existence of a critical point u m,2 ∈ Ms whose critical value is greater than cm,1 . Let e be the points defined in the previous section. We start from the following lemma. xi , yi and z i ∈ Ω Lemma 5.1. There exists ε2 ∈ (0, ε1 ) and r1 > 0 satisfying the following conditions: (i) For each positive function u ∈ M ∩ I c+ε2 , e , β+ (u)) ≥ r1 d(∂ Ω

and

eβ+ (u),s ) ≤ c + ε1 /2 I (U

for s ≥ r1 /4 :

(ii) For each u ∈ Ms ∩ I 2c+ε2 , β(u) ∈ Λr1 and eβ+ (u),s − U eβ− (u),s )))) < 2c + ε1 I (Q(N (αu + (1 − α)(U for all α ∈ [0, 1] and s ≥ r1 /4, where Q is the mapping defined in (2.12). Proof. (i): Suppose to the contrary that there exists a sequence of positive functions {u n } ⊂ M such that I (u n ) = c e , β+ (u n )) = s < ∞. By the same observation as in the proof of Lemma 3.1, we may assume and limn−→∞ d(∂ Ω that {u n } ⊂ MR N \B1 ((r +1)x1 ) and limn−→∞ d(∂ B1 ((r + 1)x1 ), β+ (u n )) = s. By the concentrate compactness lemma (cf. [17]), we have that there exists {xn } ⊂ R N such that limn−→∞ xn = x ∈ R N and limn−→∞ ku n − Uxn k = 0. But this is impossible because u n = 0 on B1 ((r + 1)x1 ). Then there exists ε2 > 0 and r1 > 0 such that for each e , β+ (u)) ≥ r1 . By choosing r1 sufficiently large, we have that I (U eβ+ (u),s ) ≤ c + ε1 /2 for s ≥ r1 u ∈ M ∩ I c+ε2 , d(∂ Ω holds. (ii): Let {u n } ⊂ Ms be such that limn−→∞ I (u n ) = 2c. Then we have again by the concentrate compactness − lemma that limn−→∞ ku + n − Uβ+ (u n ) k = limn−→∞ ku n − Uβ− (u n ) k = 0. Then d(β+ (u n ), β− (u n )) → ∞, as n → ∞. On the other hand, one can see from the definition of Q that for {rn } ⊂ R+ with limn−→∞ rn = ∞, eβ+ (u n ),rn − U eβ− (u),rn ))) − (U eβ+ (u n ),rn − U eβ− (u),rn )k = 0 lim kQ(N (αu n + (1 − α)(U

n−→∞

eβ+ (u n ),rn − U eβ− (u),rn )))) = 2c for α ∈ [0, 1]. Therefore the for α ∈ [0, 1] and then limn→∞ I (Q(N (αu n + (1 − α)(U assertion follows. 

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1053

Let ε0 > 0 be such that r0 > r1 ε0

and e cr0 /4σ < c +

ε2 2

for all σ ∈ (0, ε0 ).

(5.1)

Since ε2 < ε1 , we have that (5.1)  implies (4.1). Throughout this section, we assume (5.1). Then by the definition of r , cr ,e cr/2 < c + ε2 /2. We put we have that r > r1 and max e  ez ,R(r ) , v(1) = U ez ,R(r ) , v(t) ≥ 0 on I Γ0 = v ∈ C(I, M ∩ I c+ε1 ), v(−1) = U 0 1 and c p = inf sup I (v(t)). v∈Γ0 t∈I

er is connected, there exists p ∈ C(I, Ω er ) such that p(−1) = z 0 and p(1) = z 1 . We put v(·) = U ep(·),r (·) , Since Ω  e , p(t)) for t ∈ I. Then supt∈I I (v(t)) ≤ e where r (t) = max R(r ), d(∂ Ω cr < c + ε2 /2. That is v(·) ∈ Γ0 and then er cp ≤ e cr . On  the other hand, for each v ∈ Γ0 , there exists t0 ∈ I such that v(t0 ) ∈ Ω . Then we have that c p ≥ cr . cr ,e cr/2 < c + ε2 /2, we can choose δ0 > 0 so small that Since max e c p + δ0 <

2c + ε2 2

and e cR(r ) + δ0 < cr .

(5.2)

In the following, we will construct a function ω2 ∈ C(∂I 2 , Ms ) which is used for our minimax argument. We need the following lemma. Lemma 5.2. There exists R1 > {8R(r ), 2R(2r )} such that if R > R1 and u ∈ Γ0 ∩ C(I, M ∩ I c p +δ0 /2 ), then the following hold. (1) For j ∈ {0, 1}, there exists v0 j ∈ C(I, Ms ) such that ez ,R(r ) − U ey ,R(r ) , v0 j (1) = U ez ,R(r ) − U ey ,R(r ) , (i) v0 j (−1) = U j j 0 1 − (ii) I (v0 j (t)) ≤ c p + e cR(r ) + δ0 and I (v0 j (t) ) = e cR(r ) on I, (iii) β− (v0 j (t)) ∈ B R/4 (yi ) and β+ (v0 j (t)) = β+ (u(t)) on I. (2) For j ∈ {0, 1}, there exists v1 j ∈ C(I, Ms ) such that ex ,R(r ) − U ez ,R(r ) , v1 j (1) = U ex ,R(r ) − U ez ,R(r ) , (iv) v1 j (0) = U j j 0 1 + (v) I (vi j (t)) ≤ c p + e cR(r ) + δ0 and I (v1 j (t) ) = e cR(r ) on I, (vi) β+ (v1 j (t)) ∈ B R/4 (x j ) and β− (v1 j (t)) = β+ (u(t)) on I. The proof of Lemma 5.2 is given in Appendix. In the following, we assume that R > R1 . Then since v00 ∈ Γ1 , we have as a direct consequence from Lemma 5.2 that cm,1 ≤ c p + e cR(r ) + δ0 .

(5.3)

Let u ∈ Γ0 ∩ C(I, M ∩ I c p +δ0 /2 ) and vi j

⊂ C(I, Ms ) be the functions obtained in Lemma 5.2. The e2r ) connecting x0 and x1 , by putting path β+ (u(·)) connecting z 0 and z 1 can be extended to p1 ∈ C([−2, 2], Ω e (0) p1 (t) = −(1 + t)x0 + (2 + t)z 0 for t ∈ [−2, −1] and p1 (t) = (2 − t)z 0 + (t − 1)x1 for t ∈ [1, 2]. Then p1 (t) ∈ Ω 2r e (1) for t ∈ [1, 2]. We define p2 ∈ C([−2, 2], Ω e2r ) in the same way with xi replaced for t ∈ [−2, −1] and p1 (t) ∈ Ω 



i, j∈{0,1}

2r

by yi , i = 0, 1. Now we define a mapping ω2 ∈ C(∂I 2 , Ms ) by    1 1   e e U p1 (2t),R(r ) − U yi ,R(r ) for t ∈ [−1, 1] \ − ,   2 2 ω2 (t, 2i − 1) = 1 1   for t ∈ − , v0i (2t) 2 2

(5.4)

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and ω2 (2i − 1, t) =

   ex ,R(r ) − U ep (2t),R(r ) U i 2   v1i (2t)



1 1 for t ∈ [−1, 1] \ − ,   2 2 1 1 for t ∈ − , 2 2

 (5.5)

for i = 0, 1. From the definition of ω2 and (5.1), we have that for i ∈ {−1, 1}, e0,R(r ) ) = 2e I (ω2 (t, i)) = I (ω2 (i, t)) = 2I (U cR(r ) < 2cr < 2e cr < 2c + ε2 ,

for t ∈ I1/2 .

Then from the equality above and (ii), (v) of Lemma 5.2, we find I (ω2 (s, t)) ≤ c p + e cR(r ) + δ0 < 2c + ε2

for (s, t) ∈ ∂I 2 .

(5.6)

Now we define a set Γ2 by Γ2 = {( f, F) : F ⊂ I 3 is a two dimensional manifold with ∂ F = S, f ∈ C(F, Ms ∩ I 2c+ε1 ), with f (z) = ω2 (s, t) for z = (s, t, 0) ∈ S}, where S = ∂R2 I 2 × {0} ⊂ I 3 . We put cm,2 =

inf

sup I (v).

( f,F)∈Γ2 v∈ f (F)

We will show that cm,2 is a critical value of I . For technical convenience, we define a subset Γ2∞ of Γ2 below and work on Γ2∞ instead of Γ2 .  3    2  ( f, F) ∈ Γ2 : F is a C ∞ manifold such that F = F e∪b e ⊂ −1, 1 ,  I1/2 with F   2 2 Γ2∞ = .  2   1 1     and f satisfies f (ax) = f (x) for x ∈ ∂ , × {0} and a ∈ [1, 2]. 2 2 (∞)

(∞)

Lemma 5.3. cm,2 = cm,2 , where cm,2 = inf( f,F)∈Γ2∞ supv∈ f (F) I (v). The proof of Lemma 5.3 is given in Appendix. Lemma 5.4. cm,2 > cm,1 . Proof. First we put   c+ε1 ), v(−1) = ω2 (−1, τ−1 ), v(1) = ω2 (1, τ1 ) e0 = v ∈ C(I, M ∩ I Γ . for some τ−1 , τ1 ∈ I, and v(t) ≥ 0 on I e0 , we can define v ∈ Γ0 by For each e v∈Γ +   5   , −1 ω −2t − 2   2   +     3   ω −1, −1 + 4(1 + τ−1 ) +t    2 4 v (2t) v(t) = e +     3    − t ω 1, −1 + 4(1 + τ ) 2 1   4   +    5  ω2 −2t + , −1 2

t ∈ [−1, −3/4) t ∈ [−3/4, −1/2) t ∈ [−1/2, 1/2], t ∈ [1/2, 3/4] t ∈ [3/4, 1].

cR(r ) on I1/2 . Then since e cR(r ) ≤ cr ≤ c p , we have Then by (v) of Lemma 5.2, (5.4) and (5.5), I (v(t)) = e maxt∈I I (e v (t)) = maxt∈I I (v(t)). Then we find that c p = inf max I (e v (t)). e0 t∈I e v ∈Γ

(5.7)

N. Hirano / Nonlinear Analysis 68 (2008) 1043–1063

1055

Since s −→ cs is continuous and decreasing as s −→ ∞, and cR(r ) ≤ e cR(r ) < cr , we can choose r 0 ∈ (r, 2r ) such that e cR(r ) + δ0 < cr 0 < cr . We define a mapping ξ ∈ C(Ms , [0, 1]) by ξ(u) =

e (0) d(β− (u), Ω r0 ) (0) e 0 ) + d(β− (u), Ω e (1) d(β− (u), Ω 0 ) r

for u ∈ Ms .

(5.8)

r

e (0) e (1) From the definition, ξ(u) = 0 if β− (u) ∈ Ω r 0 and ξ(u) = 1 if β− (u) ∈ Ωr 0 . We also have that if β− (u) ∈ (0, 1), 0 er . To prove the assertion, it is sufficient to show that for each ( f, F) ∈ Γ2 , there exists p ∈ C(I, F) then β− (u) ∈ Ω such that e0 f ( p(·))+ ∈ Γ

and

er 0 β− ( f ( p(t))) ∈ Ω

on I.

(5.9)

In fact, if there exists p ∈ C(I, F) satisfying (5.9), then by (5.7), max I ( f ( p(t)+ )) ≥ c p .

(5.10)

t∈I

er 0 on I, we have I ( f ( p(t))− ) ≥ cr 0 for all t ∈ I. Therefore we find that On the other hand, since β− ( f ( p(t))) ∈ Ω there exists t0 ∈ I such that I ( f ( p(t0 ))) = I ( f ( p(t0 ))+ ) + I ( f ( p(t0 ))− ) ≥ c p + cr 0 . This implies that max I ( f (z)) ≥ I ( f ( p(t0 ))) ≥ c p + cr 0 z∈F

for any ( f, F) ∈ Γ2 .

cR(r ) + δ0 ≥ cm,1 . Then by the definition of r 0 and (5.3), cm,2 ≥ c p + cr 0 > c p + e Now we show that (5.9) holds for each ( f, F) ∈ Γ2 . By Lemma 5.3, we may assume that ( f, F) ∈ Γ2∞ , i.e., 2 with F 2 and χ = 1 on F. e∪b e ⊂ [− 1 , 1 ]3 . Let χ ∈ C ∞ (F, [0, 1]) be such that χ = 0 on b e We define a F=F I1/2 I1/4 2 2 mapping τ ∈ C(F, [0, 1]) by  1+t  2 (1 − χ (z)) + χ (z)(ξ( f (z))) for z = (s, t, 0) ∈ b I1/2 τ (z) = (5.11) 2 ξ( f (z)) e for z ∈ F. 2 , R). Let δ ∈ (0, 1/4). Then by the same argument as in the proof From the definition, we have that τ ∈ C ∞ (b I1/4 2 . Note τ ∈ C ∞ (F, [0, 1]) such that |e τ − τ |∞ < δ and e τ = τ on b I1/8 of Lemma 4.2, we can find a mapping e that ξ( f (z)) = ξ(ω2 (s, t)) for z = (s, t, 0) ∈ ∂I 2 × {0}, ξ(ω2 (−1, −1)) = 0 and ξ(ω2 (1, 1)) = 1. Then the range R(e τ ) contains the interval [0, 1]. Then we have by the density of regular values of e τ in the range R(e τ ) that there exists a regular value q ∈ R of e τ such that |q − 21 | < δ. Then we have that e τ −1 (q) ⊂ F is a disjoint 2 , we have union of one dimensional manifolds. Since e τ ((s, t)) = τ (z) = (1 + t)/2 for z = (s, t, 0) ∈ b I1/8 e τ −1 (q) ∩ {(s, t, 0) : t ∈ I} = (s, 2q − 1, 0) for s ∈ I1/8 . Therefore we have that there exists a unique one dimensional p (t) : t ∈ I} ⊂ e manifold {e τ −1 (q) ⊂ F such that e τ (−1, 2q − 1) = q = τ (e p (−1)). Then the other end e p (1) of e p must 2 be in (∂I × {0}) \ ({−1} × I × {0}).  2 e (1) By (iii) of Lemma 5.2, (5.4) and (5.5), we find that β− ( f (x)) ∈ Ω r 0 for x ∈ (s, t) ∈ ∂I × {0} : t ≥ 1/2 .  Therefore we have that ξ( f (z)) = 1 on (s, t) ∈ ∂I 2 × {0} : t ≥ 1/2 . Then recalling that f (z) = f (az) for z ∈ ∂[− 21 , 12 ]2 × {0} with a ∈ [1, 2], we have that ξ( f (z)) = 1 for z ∈ I × [ 21 , 1] × {0}. Then by the definition of τ ,

τ (z) = (1 − χ (z))

t +1 1+t 3 + χ (z)(ξ( f (z))) = (1 − χ (z)) + χ (z) ≥ 2 2 4

for z = (s, t, 0) with (s, t) ∈ I × [ 21 , 1]. On the other hand, by (vi) of Lemma 5.2, (5.4) and (5.5), we find that   2 2 e (0) β− ( f (z)) ∈ Ω r 0 for z ∈ (s, t) ∈ ∂I × {0} : t ≤ −1/2 , i.e., ξ( f (z)) = 0 on (s, t) ∈ ∂I × {0} : t ≤ −1/2 . Therefore we have 1+t 1 τ (z) = (1 − χ (z)) ≤ 2 4

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for z = (s, t, 0) with (s, t) ∈ I × [−1, − 12 ]. Therefore      1 1 ∪ , 1 = ∅. e τ −1 (q) ∩ (s, t, 0) : s ∈ I, t ∈ −1, − 2 2

(5.12)

Then e p (1) = (1, 2q − 1, 0). From (5.12), we have that there exists t1 , t2 ∈ (−1, 1) and τ1 , τ2 ∈ (− 21 , 12 ) such that e We put p(t) = e e p (t1 ) = (− 12 , τ1 , 0) and e p (t1 + (t2 − t1 ) 1+t p (t2 ) = ( 12 , τ2 , 0), and {e p (t) : t1 ≤ t ≤ t2 } ⊂ F. 2 ) for ∞ t ∈ I. Then by the definition of Γ2 , we have f ( p(−1))+ = f (e p (t1 ))+ = f (2e p (t1 ))+ = ω2 (−1, 2τ1 )+ e0 . On the other hand, noting that p(t) ∈ F e on (−1, 1), and f ( p(1))+ = ω2 (1, 2τ2 )+ . Therefore we find f ( p(·))+ ∈ Γ we have by the definition of τ that for t ∈ I, 1 1 ξ( f ( p(t))) − 1 = τ ( p(t)) − 1 ≤ |τ ( p(t)) − e τ ( p(t))| + q − < 2δ < . 2 2 2 2 0

er by the definition of ξ . This completes the proof. That is β− ( f ( p(t))) ∈ Ω



Proposition 5.1. If R > max {R1 , r1 }, then there exists a critical point u m,2 ∈ Ms of I such that I (u m,2 ) = cm,2 . Proof. The proof is by contradiction. We assume that there is no critical point in Ms with critical value cm,2 . We put σ = min (cm,2 − c p − e cR(r ) − δ0 )/2, ε(cm,2 ) , where ε(·) is the number defined in Lemma 2.3. Putting d = cm,2 , we have, by Lemma 2.3, that there exists a deformation retraction ρ = ρd,σ,ε(d) ∈ C([0, 1] × (Ms,1/2 ∩ I d+ε(d) ), Ms,1 ∩ I d+ε(d) ) satisfying (ρ). Here we choose ( f, F) ∈ Γ2∞ such that maxv∈F I ( f (v)) ≤ d + ε(d). We put   s ϕ(s, z) = ρ 1, π , f (z) for (s, z) ∈ I × F. 2 Then ϕ ∈ C(I × F, M) ∩ I cm,2 −σ . Recall that for z ∈ b I 2 , f (z) = f (az) = ω2 (az) where a > 0 is such that 1/2

az ∈ I 2 × {0}. Then we find by (5.6) that  s  2 , f (z) < c p + e cR(r ) + δ0 < cm,2 − σ on I × b I1/2 . (5.13) I π 2 Therefore from (3.1) and the inequality above, s  2 , f (z) for (s, z) ∈ (∂I × F) ∪ (I × b I1/2 ). (5.14) ϕ(s, z) = π 2 2 . We can find γ ∈ C ∞ (I × F, R) satisfying Here we choose δ > 0 such that |γ (ϕ(s, z))| > δ for (s, z) ∈ I1/2 × b I1/2  the assertion of (2) of Lemma 4.2 with θ, θ and σ replaced γ (ϕ), γ and  min sσ/2 , 1/2, δ/2 , where sσ/2 is the number defined in Lemma 2.2. That is |γ (ϕ) − γ |∞ ≤ min sσ/2 , 1/2, δ/2 . From (2.10) and (5.14), we have that the range R(γ ) of γ contains [−π0, π0 ]. Then since the regular value of γ is dense in R(γ ), we can find a regular value q ∈ R of γ such that |q| < min sσ/2 , 1/2, δ/2 . Then γ −1 (q) is disjoint union of two dimensional submanifolds of I × F. Since 1 |γ (±1, z)| ≥ |γ (ϕ(±1, z))| − |γ (±1, z) − γ (ϕ(±1, z))| ≥ π0 − > q, 2 we have that γ −1 (q) ∩ {(±1, z) : z ∈ F} = ∅. Therefore if ζ is a two dimensional submanifold in γ −1 (q) with boundary ∂ζ , then ∂ζ must be included in (−1, 1) × ∂I 2 × {0}. Note that by (2) of Lemma 4.2, the mapping 2 and I1/4 s −→ γ (s, z) is monotone decreasing on [−1/2, 1/2] for z ∈ b |γ (s, z)| ≥ |γ (ϕ(s, z))| − |γ (s, z) − γ (ϕ(s, z))| ≥ δ/2

2 for (s, z) ∈ I1/2 × b I1/2 .

2 , there exists a unique s ∈ (−1, 1) such that γ (s , z) = q. Then there exists a two Therefore for each z ∈ b I1/2 z q  −1 2 dimensional manifold ζq ⊂ γ (q) such that ∂ζq = (sz , z) : z ∈ ∂I × {0} ≡ S 1 . Since F is a C ∞ manifold, 2 × {0}) ⊂ I × F with we may think of I × F as a submanifold of I 3 and identify z = (σ, (s, t, 0)) ∈ I × (I1/2

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 (s, t, σ ) ∈ I 2 × I for (σ, t, s) ∈ I 3 . Then by this identification, ζ ⊂ I 3 and ∂ζq = (z, sz ) : z ∈ ∂I 2 . Here we put   2 × I) and l(z) = z ζ = l(z) : z ∈ ζq , where l(z) = (s, t, min 1, 4d((s, t), ∂I 2 ) σ ) for z = (s, t, σ ) ∈ ζ ∩ (I1/4 2 × I). Then ζ is a manifold with ∂ζ = S. From the definition of l(·), ϕ and Q, we can see that for z ∈ ζ \ (I1/4 −1 Q(ϕ(l (z))) = Q(ϕ(z)) for all z ∈ ζ . Moreover we have by Lemma 2.2 and the assumption that I (Q(ϕ(l −1 (x)))) ≤ I (ϕ(l −1 (x))) + σ/2 ≤ cm,2 − σ/2

for all x ∈ ζ.

Then (Q ◦ ϕ ◦ l −1 , ζ ) ∈ Γ2 . Since cm,2 ≤ maxz∈ζ I (Q(ϕ(l −1 (z)))), this is a contradiction.



6. Minimax type critical points III In this section, we will show the existence of critical points u 3,0 ∈ V 0,0 , u 3,1 ∈ V 1,1 such that u 3,0 6= (0,0) (1,1) ±u 0 , u 3,1 6= ±u 0 and I (u 3,0 ), I (u 3,1 ) < cm,1 . We start from the following lemma. Lemma 6.1. There exist positive numbers ε3 , r2 , and s1 ∈ (0, 1) such that β(u) ∈ Λr2 for all u ∈ Ms,s1 ∩ I 2c+ε3 . Proof. Let {sn } ⊂ R and {u n } ⊂ M be such that limn−→∞ sn = 0, u n ∈ Ms,sn for each n ≥ 1 and limn−→∞ I (u n ) = − ± 2c. Then we have that limn−→∞ I (u + n ) = limn−→∞ I (u n ) = c. Then we find limn→∞ ku n − Uβ± (u n ) k = 0, and then limn→∞ d(β+ (u n ), β− (u n )) = ∞. This implies that the assertion holds.  (0,0)

In the following, we show the existence of u 3,0 ∈ V 0,0 with u 3,0 6= ±u 0 by the same argument. Let R2 > max {r2 , R1 } be such that n ε3 o e for all σ ≥ R2 . cσ/2 < min c2r ,e cR(r ) , c + 2

. The existence of u 3,1 ∈ V 1,1 is proved

(6.1)

In the following, we assume that R > R2 . Here we put Γ3 = {( f, F) : F ⊂ I × S N −1 is an N − 1 dimensional manifold without boundary, f ∈ C(F, Ms ∩ I 2ec R/2 ∩ V 0,0 ) : β( f (F)) is not contractible in Λr2 } and c3 =

inf

sup I ( f (x)).

( f,F)∈Γ3 x∈F

e (0) fixed in Section 3. Let ι : S N −1 −→ ∂ B R/2 (z 0 ) be an isomorphism. Then we can define Let z 0 be the element of Ω R N −1 , Ms ) by a mapping ω ∈ C(S eι(x),R/2 − U e2z −ι(x),R/2 ω3 (x) = U 0

for x ∈ S N −1 .

Then we have Lemma 6.2. (ω3 , {0} × S N −1 ) ∈ Γ3 . Proof. The set {0} × S N −1 is an N − 1 dimensional manifold in [−1, 1] × S N −1 without boundary. For each (0, x) ∈ {0} × S N −1 , ω3 (x) ∈ Ms ∩ I 2ec R/2 . We also have by the definition that ω3 (S N −1 ) ⊂ V 0,0 . Since I (ω3 (x)) = 2e c R/2 < 2c + ε3 on S N −1 by (6.1), we have β(ω3 (S N −1 )) ⊂ Λr2 by Lemma 6.1. Then to show the assertion, it is sufficient to show that β(ω3 (S N −1 )) is not contractible in Λr2 . We define a mapping Π : Λr2 −→ S N −1 by Π (x, y) =

x−y |x − y|

for each (x, y) ∈ Λr2 .

Then from the definition of ω3 , Π (β(ω3 (·))) is an isomorphism on S N −1 . Then β(ω3 (S N −1 )) is not contractible in Λr2 .  Since cm,1 > c + cr > 2e cR(r ) > 2e c R/2 by (6.1), we find by Lemma 6.2 that c3 < cm,1 .

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N. Hirano / Nonlinear Analysis 68 (2008) 1043–1063 (0,0)

Lemma 6.3. If c3 = c0

(0,0)

, then there exist at least two critical points in V 0,0 which attain the local minimum c0

(0,0)

(0,0)

Proof. Suppose that u 0 is the unique local minimum of I in V 0,0 . We choose d > 0 so small that β(Dd (u 0 Ms ) is contractible in Λr2 . Let n o e c = inf I (u) : u ∈ V 0,0 \ Dd (u 0,0 ) . 0

.

)∩

(0,0) (0,0) Then by the assumption e c > c0 . If c3 = c0 , then by the definition of c3 , there exists ( f, F) ∈ Γ3 such that (0,0) (0,0) (0,0) supx∈F I ( f (x)) < c0 + δ, where δ ∈ (0,e c − c0 ). This implies that f (x) ∈ Ms ∩ Dd (u 0 ) for x ∈ F. Then β( f (F)) is contractible in Λr2 . This is a contradiction. 

Proposition 6.1. There exists R3 > R2 such that if R > R3 , there exists a critical point u 3,0 ∈ V 0,0 such that (0,0) u 3,0 6= ±u 0 and I (u 3,0 ) < cm,1 . (0,0)

Proof. By Lemma 6.3, to prove the assertion, it is sufficient to consider the case where c3 > c0 . Suppose (0,0) that c3 > c0 and that there is no critical point of I in I [c3 −σ0 ,cm,1 −σ0 ] ∩ V 0,0 , where σ0 > 0 such that (0,0) 0,0 . c0 < c3 − σ0 < c3 < cm,1 − σ0 . By the proof of Proposition 3.2, we have that I (u) ≥ cΩ e + c2r for u ∈ ∂Ms V Then we can choose s2 ∈ (0, s1 ) sufficiently small that e0,0 ∩ Ms,s I (u) ≥ c + c2r for all u ∈ ∂M V (6.2) 2 o n e (0) , β− (v) ∈ Ω e (0) . Recall that there is no critical point in Ms,1 \ Ms . Then by e0,0 = v ∈ M : β+ (v) ∈ Ω where V 2r 2r (2) of Lemma 2.1, we have that there exists m 1 > 0 such that o n inf k∇ I (u)k : u ∈ (Ms,s2 \ Ms,s2 /2 ) ∩ I 3c > m 1 . (6.3) We note that s2 and m 1 can be chosen independent of R. Then by (6.3), we can choose R3 > R2 and σ1 ∈ (0, σ0 ) such that for each σ ≥ R3 and v ∈ Ms,s2 /2 ∩ I 2ecσ/2 , there exists tv > 0 such that Ψ (t, v) ∈ Ms,s2

for t ∈ [0, tv ]

and

I (Ψ (tv , v)) < c3 − σ1

and    s 2 (6.4) I π ± , v ≤ c3 − σ1 for v ∈ Ms ∩ I 2ecσ/2 . 2 Now we assume that R > R3 . Then we can define, as in Lemma 2.3, a deformation retraction ρ ∈ C([0, 1] × (Ms,s2 /2 ∩ I 2ec R3 /2 ), Ms,s2 ∩ I 2ec R3 /2 ) satisfying  for all u ∈ Ms,s2 /2 ∩ I 2ec R3 /2 ρ(0, u) = u 0 c −σ (ρ ) ρ(1, u) ∈ Ms,s2 ∩ I 3 1 for all u ∈ Ms,s2 /2 ∩ I 2ec R3 /2  ρ(1, u) = u for all u ∈ Ms,s2 /2 ∩ I c3 −σ1 . We put   s 2 s, ω3 (x) ϕ(s, x) = ρ 1, π 2 Then we have by the definition of ρ, ϕ(s, x) ∈ I c3 −σ1

on I × S N −1 .

for (s, x) ∈ I × S N −1 .

(6.5)

Since ω3 (x) ∈ for x ∈ S N −1 , ϕ(s, x) ∈ Ms,s2 on I × S N −1 . Then we have by Lemma 6.1 that β(ϕ(s, x)) ∈ Λr2 e0,0 = ∅ for on I × S N −1 . We also have, from (6.2) and e c R/2 < c + c2r , that {ρ(t, π( s22 s, ω3 (x))) : t ∈ [0, 1]} ∩ ∂M V e0,0 . On the other hand, we have by (6.4), (s, x) ∈ I × S N −1 . Then by the definition of ρ and ϕ, ϕ(s, x) ∈ V   s   s  2 2 ϕ(±1, x) = ρ 1, π ± , ω3 (x) = π ± , ω3 (x) ∈ I c3 −σ1 ∩ M, for x ∈ S N −1 . (6.6) 2 2 I 2ec R/2

N. Hirano / Nonlinear Analysis 68 (2008) 1043–1063

Let π1 > 0 be such that   s  2 γ π − , v > π1 2

and

 s  2 γ π , v < −π1 2

for v ∈ Ms ∩ I 2c+ε3 .

1059

(6.7)

 We can choose e γ ∈ C ∞ (I × S N −1 , R) such that |γ (ϕ) − e γ | < min π1 /4, sσ1 /2 /2 , where sδ is the number γ in R(e γ ), we can choose a regular value defined in Lemma 2.2. By (6.7) and the density of regular values of e q ∈ R such that |q| < min π1 /4, sσ1 /2 /2 . Then e γ −1 (q) is a disjoint union of N − 1 dimensional submanifolds of  γ −1 (q), we find from (6.7) I × S N −1 . Since |γ (ϕ(v))| ≤ |e γ (v) − γ (ϕ(v))| + |e γ (v)| < min π1 /2, sσ1 /2 for v ∈ e that e γ −1 (q) ⊂ (−1, 1) × S N −1 . If all of the manifolds in e γ −1 (q) are contractible in (−1, 1) × S N −1 , we can find N −1 a path p ∈ C([0, 1], I × S ) connecting the two boundaries {−1} × S N −1 and {1} × S N −1 without intersecting −1 N −1 γ (q), i.e., p(0) ∈ {−1}× S e and p(1) ∈ {1}× S N −1 , and { p(t) : t ∈ [0, 1]}∩e γ −1 (q) = ∅. But this is impossible, −1 because e γ ( p(0)) > π1 , e γ ( p(1)) < −π1 , and then p(t) has to intersect the set e γ (q). Therefore there exists an N − 1 dimensional manifold ζ ⊂ e γ −1 (q) which is not contractible in I × S N −1 . We will show that (Q ◦ ϕ, ζ ) ∈ Γ3 . Since e0,0 for (s, x) ∈ I × S N −1 , Q ◦ (ϕ(ζ )) ⊂ V 0,0 . We also have β(Q ◦ (ϕ(ζ ))) ⊂ Λr because β ◦ ϕ(ζ ) ⊂ Λr . ϕ(s, x) ∈ V 2 2 We see that β ◦ ϕ(ζ ) is not contractible in Λr2 . We define a homotopy θ ∈ C([0, 1] × ζ, I × S N −1 ) such that for each (s, x) ∈ ζ ⊂ I × S N −1 and t ∈ [0, 1], θ (t, (s, x)) = θ1 (t, (z, x)), θ2 (t, (s, z)) = (s + (1 − s)t, x). Then θ (1, ζ ) ⊂ {1} × S N −1 ≡ S N −1 and θ (1, ζ ) is not contractible in {1} × S N −1 , because θ (0, ζ ) = ζ is not contractible in I × S N −1 . Then it is sufficient to show that β ◦ ϕ(θ (1, ζ )) is not contractible in Λr2 . Since s  2 β ◦π , ω3 (x) = (ι(x), 2z 0 − ι(x)) ∈ Λr2 for x ∈ S N −1 , 2 mapping β ◦ π( s22 , ω3 (·)) : S N −1 −→ β ◦ π( s22 , ω3 (S N −1 )) ⊂ Λr2 is an isomorphism. Since  (ι(x), 2z 0 − ι(x)) : x ∈ S N −1 is not contractible in Λr2 , we find from (6.6) that β ◦ (ϕ(θ (1, ζ ))) = β ◦ π( s22 , ω3 (θ2 (1, ζ ))) is not contractible in Λr2 . Therefore we have that β ◦ ϕ(ζ ) is not contractible in Λr2 . Then since β(Q(ϕ(ζ ))) = β(ϕ(ζ )), we obtain that β ◦ (Q(ϕ(ζ ))) is not contractible in Λr2 . On the other hand, recalling that |γ (ϕ(z))| < sσ1 /2 on e γ −1 (q), we have by Lemma 2.2 that I (Q(ϕ(z))) ≤ I (ϕ(z)) + σ1 /2 ≤ c3 − σ1 + σ1 /2 < c3 − δ1 /2

for z ∈ ζ.

Then we obtain that (Q ◦ ϕ, ζ ) ∈ Γ3 . But this is a contradiction, because supx∈ζ I (Q ◦ ϕ(x)) ≥ c3 has to hold. This completes the proof.  Acknowledgement The author expresses his hearty thanks to the referee for reading this article very carefully and giving valuable suggestions. Appendix Proof of Lemma 5.2. We will prove the case where i = 0 in (1). The other cases can be proved by the same way. We put Z e. αu (x) = (|∇u(z)|2 + |u(z)|2 )dz for u ∈ H and x ∈ Ω B2R(r ) (x)

 Put d0 = supu∈M∩I c+ε1 inf αu (x) : x ∈ B R/4 (y0 ) . We note that d0 −→ 0, as R −→ ∞. Recall that kuk2 ≤ C0 3c for each u ∈ M ∩ I c+ε1 . Then for given d > d0 , there exists kd ∈ N such that for each  u ∈ M ∩ I and for each e continuous function e α ∈ C(Ω ) with |αu − e α |∞ < d/2, there exist at most kd points x1 , x2 , . . . , xkd ⊂ B R/4 (y0 ) satisfying the following condition:  e α (x ) = d for each xi , d(xi , x j ) ≥ N −1/2 R(r ) for i 6= j, and (∗)  i x ∈ B R/4 (y0 ), e α (x) = d ⊂ ∪ B2R(r ) (xi ).

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Let u ∈ Γ0 with maxt∈I I (u(t)) ≤ c p + δ0 /2. Let d > d0 be a point in the range R(αu ) of αu. such that [d − σ, d + σ ] ⊂ R(αu ) for some σ ∈ (0, d). We will show that if R is sufficiently large, there exists a path e ) such that p(−1) = z 0 , p(1) = z 1 and α p(t) < 3d/2 on I. By the density of C ∞ (I × Ω e ) in C(I × Ω e ), p ∈ C ∞ (I, Ω σ ∞ e e we can choose e α ∈ C (I × Ω ) such that |αu(t) (x) − e α (t, x)| < 2 on I × Ω . Since regular values of e α are dense ez ,R(r ) and in the range R(e α ) of e α and d ∈ R(e α ), we may assume that d is the regular value of e α . Since u(−1) = U 0 d(z 0 , y0 ) = R/2 > 4R(r ), we find  (−1, B R/4 (y0 )) ⊂ A = (t, x) ∈ I × B R/4 (y0 ) : e α (t, x) ≤ d . Let K be a connected component of A such that (−1, B R/4 (y0 )) ⊂ K . We put K (t) = K ∩ ({t} × B R/4 (y0 )) for t ∈ I. Then mes|K (−1)| = mes|B R/4 (y0 )|, where mes|B| is the measure of B ⊂ R N . Note that (1, B R/4 (y0 )) ⊂ A.  e, e Then if (1, B R/4 (y0 )) ( K , we have B R/4 (y0 ) ∩ K (1) = ∅. Note that (t, x) ∈ (0, 1) × Ω α (t, x) = d is a smooth manifold, because d is the regular value of e α . Then we find that there exists t ∈ (−1, 1) such that 1 1 mes|B R/4 (y0 )| or mes|int(K (t))| < mes|B R/4 (y0 )| < mes|K (t)|. 2 2 Since |αu(t) (x) − e α (t, x)| < d/2, we have that ∂ B R/4 (y0 ) K (t) has to be covered by at most kd balls with radius 2R(r ) and satisfying (∗). But if R is sufficiently large, this is impossible. Therefore for R sufficiently large, (1, B R/4 (y0 )) ⊂ K (1). Then there exists a path p ∈ C ∞ (I, B R/4 (y0 )) such that p(−1) = p(1) = y0 , and p(t) ∈ A on I. By the definition of e α , we find that α p(t) < 3d/2 on I. We now define a function ζ ∈ C ∞ ([0, ∞), [0, 1]) such that ζ (t) = 0 for t ≤ R(r ) and ζ (t) = 1 for t ≥ 2R(r ). We put mes|K (t)| =

w(t, x) = ζ (|x − p(t)|)u(t, x)

e. for (t, x) ∈ I × Ω

Then w(t, x) = 0 on BR(r ) ( p(t)) for t ∈ [0, 1] and Z 2 kw(t) − u(t)k ≤ (|∇(ζ (|x − p(t)|)u(t, x))|2 + |ζ (|x − p(t)|)u(t, x)|2 )dx B2R(r ) ( p(t))

Z ≤C

B2R(r ) ( p(t))

(|∇u(t, x)|2 + |u(t, x)|2 )dx ≤ Cd,

where C > 0 is a constant depends only on ζ . Here we put ep(t),R(r ) v(t) = N w(t) − U

for t ∈ [0, 1].

ep(t),R(r ) for t ∈ [0, 1]. Recalling that u(−1) = U ez ,R(r ) , we find that Then v(t)+ = N w(t) and v(t)− = U 0 ez ,R(r ) − U ey ,R(r ) . We also have v(1) = U ez ,R(r ) − U ey ,R(r ) . That is v satisfies (i) of (1) with v00 = v. v(−1) = U 0 0 1 0 Since kN w(t) − u(t)k −→ 0, uniformly on [0, 1] as d −→ 0, we find that for R > 0 sufficiently large and d > 0 sufficiently small, I (v(t)+ ) ≤ I (u(t)) + δ0 /2 on I. Therefore (ii) of (1) holds with v00 = v. The first assertion of (iii) holds by the definition of v. We lastly show that β+ (u(t)) = β+ (v(t)) on I. To show that assertion, + (x) on Ω (u). From the shape of U , |U b0 |∞ > 1 it is sufficient to show that Ω (u) = Ω (v + ) and b u (x) = vc 0 (cf. [4]). Recall that if {u n } ⊂ M satisfies I (u n ) −→ c, then ku n − Uxn k −→ 0 for some {xn } ⊂ R N . Then we find that if ε2 is sufficiently small, |b z|∞ > 1 for all z ∈ M ∩ I c+ε2 . Then by taking d > 0 sufficiently + c+ε 2 small, we have v(t) ∈ I and then |b v (t)|∞ > 1 for all t ∈ [0, 1]. Here we recall that 2R(r ) > 1. In fact 2R(r ) > 4r > 4r0 /ε0 > 1 by the assumption that ε0 < 4r0 . Here we choose d > 0 so small that |tw − 1| < 21 and (tw m p ) p (3d/2) p < 1, where v = tw w = N w and m p is the constant defined in Section 2. Then we find [+ (x) ≤ (tw m p αv + (t) (x)) p/2 < (tw m p ) p/2 (3d/2) p/2 < 1 < |v|∞ on B2R(r )−1 ( p(t)) on I. Similarly we have that v(t) d u(t)(x) < 1 on B2R(r )−1 ( p(t)). On the other hand, noting B2R(r )−1 ( p(t)) ⊂ B2R(r ) ( p(t)) for t ∈ I, we find that [+ (x) = u(t)(x) d e \ B2R(r )−1 ( p(t)) on I by the definition of v. Therefore we find that Ω (v + ) = Ω (u) and v(t) on Ω [+ (x) = u(t)(x) d v(t) on Ω (v). Thus we find β+ (v(t)) = β+ (u(t)). This completes the proof.  (∞)

Proof of Lemma 5.3. We first see that if there exists ( f, F) ∈ Γ2 with supx∈F I ( f (x)) < 2c + ε1 , then cm,1 = cm,1 . 2 ⊂ I 3 . We will define a C ∞ manifold close e= 1F ∪b Let ( f, F) ∈ Γ2 with supx∈F I ( f (x)) < 2c + ε1 . We put F I3/4 4 e For given δ ∈ (0, 1/4), we define a mapping ψ ∈ C(I 3 , [0, 1]) by ψ(x) = [(δ − d(x, F)) e + /δ]2 for x ∈ I 3 . Let to F.

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N. Hirano / Nonlinear Analysis 68 (2008) 1043–1063 2 2 η ∈ C ∞ (I 3 , [0, 1]) be such that η(x) = 1 on I7/12 × I and η(x) = 0 on I 3 \ (I8/12 × I). We put

ϕ(s, t) = (1 − η(s, t))ψ(s, t) + η(s, t)(1 + t)

for (s, t) ∈ I 2 × I.

2 × I, [0, 2]). Then by (1) of Lemma 4.2, we can choose e Then ϕ ∈ C ∞ (I7/8 ϕ ∈ C ∞ (I 3 , [0, 2]) such that 2 × I. Since ψ = 1 on F e and ψ = 0 on I 2 × ∂I, we have by the |e ϕ − ϕ|C(I 3 ,R) < δ and e ϕ = ϕ on I1/2 density of regular values of e ϕ , we can choose a regular value q ∈ (1/2, 1] sufficiently close to 1. Then noting 2 2 × I) = that t −→ ϕ(s, t) is increasing on I for s ∈ I1/2 by the definition of ϕ, we find that e ϕ −1 ∩ (I1/2 n o 2 × I) = 2 ϕ −1 ∩ (I1/2 (s, q − 1) : s ∈ I1/2 . In particular e ϕ −1 ∩ (∂I 2 × I) is a one dimensional manifold  S1 = (s, q − 1) : s ∈ ∂I 2 ≡ S 1 . Then there exists a two dimensional C ∞ manifold ζq ⊂ e ϕ −1 (q) such that 2 , where e ∂ζq = S1 . By putting ζ = ζq − (q − 1), we find that ζ is of the form ζ = e ζ ∪b I1/2 ζ ⊂ [− 12 , 12 ]3 . e Ms ) by On the other hand, we define a function e f ∈ C( F,   f (4z) for z ∈ 1 F e 4 f (z) =  2 f (az) for z ∈ b I1/2 , where a > 0 with az ∈ ∂I 2 × {0}.

e f can be extended to a continuous function, again denoted by f , defined on I 3 with values in Ms . Then by choosing q sufficiently close to 1, we have supx∈ζ I ( e f (x)) < 2c + ε1 , i.e., we have that ( e f , ζ ) ∈ Γ2∞ . Therefore for given σ > 0, we have by choosing δ sufficiently small and by choosing q sufficiently close to 1 that sup I ( e f (x)) ≤ sup I ( f (x)) + σ. x∈ζ

x∈F (∞)

Since σ is arbitrary, we find that cm,2 = cm,2 . We next show that there exists ( f, F) ∈ Γ2 with supx∈F I ( f (x)) < 2c + ε1 . Note that I 2 is a two dimensional C ∞ manifold. Then to prove the assertion, it is enough to extend ω2 defined by (5.4) and (5.5) to a mapping e ω2 ∈ C(I 2 , Ms ) with supx∈I 2 I (e ω2 (x)) < 2c + ε1 . er /4 ) such that We first assume that there exist homotopies of mappings p± ∈ C([0, 1] × I 2 , Ω 1 p± (0, s, t) = β± (ω2 (s, t))

for (s, t) ∈ ∂I 2 ,

(7.1)

d( p+ (σ, s, t), p− (σ, s, t)) ≥ r1 /2

for (σ, s, t) ∈ [0, 1] × ∂I

d( p+ (1, s, t), p− (1, s 0 , t 0 )) ≥ r1 /2

for (s, t), (s 0 , t 0 ) ∈ I 2 .

2

(7.2)

and Let χ1 , χ2 ∈

C([0, 1]2 , [0, 1])

(7.3)

be such that

χ1 (s, t) = 0

on ∂I 2 ,

χ1 (s, t) = 1

χ2 (s, t) = 0

  5 5 2 on [−1, 1] \ − , 6 6 2

  5 5 2 on − , , 6 6 and

χ2 (s, t) = 1



4 4 on − , 6 6

2

.

We denote by P : [−2, 2]2 −→ I 2 the projection defined by P(z) = az for z ∈ [−2, 2]2 \ I 2 , where a > 0 is such that az ∈ ∂I 2 . Now we define e ω2 by ep+ (χ (s,t),P(2s,2t)),r /4 − U ep− (χ (s,t),P(2s,2t)),r /4 ))) e ω2 (s, t) = Q(N ((1 − χ1 (s, t))ω2 (P(2s, 2t)) + χ1 (s, t)(U 2 1 2 1 for (s, t) ∈ I 2 , where ω2 = ω2 on ∂I 2 and ω2 = 0 on I 2 \ ∂I 2 . Since χ1 = 0 on ∂I 2 and ω2 (P(2s, 2t)) = ω2 (s, t) on ∂I 2 , we find that e ω2 (s, t) = ω2 (s, t) on ∂I 2 . Then to complete the proof, it is sufficient to show that sup(s,t)∈I 2 I (e ω2 (s, t)) < 2c + ε1 . Since χ2 = 1 on [− 46 , 46 ]2 and (7.3) holds, we find that ep+ (χ (s,t),P(2s,2t)),r /4 − U ep+ (χ (s,t),P(2s,2t)),r /4 ) I (e ω2 (s, t)) = I (U 2 1 2 1  2 4 4 = 2cr1 /4 < 2c + ε1 on − , . 6 6

(7.4)

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N. Hirano / Nonlinear Analysis 68 (2008) 1043–1063

2 , we find from (7.2) that (7.4) holds On the other hand, noting that χ1 = 1 on [− 65 , 56 ]2 and P(2s, 2t) ∈ ∂I 2 on I1/2 2 and p (s, t) = β (ω (s, t)) for (s, t) ∈ ∂I 2 , we have by on [− 65 , 65 ]2 \ [− 46 , 46 ]2 . Finally, noting that χ2 = 0 on I5/6 ± ± 2 (ii) of Lemma 5.1 that

I (e ω2 (s, t)) < 2c + ε1

2 on I5/6 .

Thus we find I (e ω(s, t)) < 2c + ε1 on I 2 . That is (e ω, I 2 ) ∈ Γ2∞ . er /4 ) satisfying (7.1)–(7.3). Then to finish the proof, it is sufficient to show that there exist p± ∈ C([0, 1] × I 2 , Ω 1 We put p± (s, t) = β± (ω2 (s, t)) for ∂I 2 . By the definition of ω2 , we have that p+ (s, 1) = p+ (s, −1) = p− (±1, s)

on [−1/2, 1/2].

(7.5)

e , β± (ω2 (s, t))) Since I (ω2 (s, t)) < 2c + ε2 on ∂I 2 , we have by Lemma 5.1 that d(∂ Ω 2 d(β+ (ω2 (s, t)), β− (ω2 (s, t))) ≥ r1 on ∂I . Then we have that d( p+ (s, t), p− (s, t)) ≥ r1

and

e , p± (s, t)) ≥ r1 d(∂ Ω

on ∂I 2 .

≥

r1 and (7.6)

We also have from (iii) and (vi) of Lemma 5.2 that ∪ s∈[−1/2,1/2]

∪ s∈[−1/2,1/2]

p− (s, −1) ⊂ B R/4 (y0 ),

∪ s∈[−1/2,1/2]

p+ (−1, s) ⊂ B R/4 (x0 ),

∪ s∈[−1/2,1/2]

p− (s, 1) ⊂ B R/4 (y1 ),

(7.7)

p+ (1, s) ⊂ B R/4 (x1 ).

(7.8)

Then by (7.5)–(7.8), we can pull out the path p+ (1, ·) : [− 21 , 12 ] −→ Ms from the balls B R/4 (xi ), B R/4 (yi ), i ∈ {0, 1} by a continuous deformation. In fact, if s ∈ [−1/2, 1/2] and p+ (s, −1) ∈ B R/4 (y0 ), then we can move p+ (s, −1) out from B R/4 (y0 ) along the half-line {t p+ (s, −1) + (1 − t) p− (s, −1) : t ≥1}. If s ∈ [−1/2, 1/2] and p+ (s, −1) ∈ B R/4 (x0 ), then we can pull out p+ (s, −1) from B R/4 (y0 ) along the half-line t p+ (s, −1) + (1 − t)x00 : t ≥ 1 , where x00 ∈ B R/4 (x0 ) such that x00 6∈ { p− (±1, s) : s ∈ I}. In other cases, the procedure is the same. That is we can find a er ) such that e homotopy of mappings e p ∈ C([0, 1] × [−1/2, 1/2], Ω p (0, s) = p+ (s, 1) on [−1/2, 1/2], 1   1 1 (7.9) d(e p (σ, s), p− (s, ±1)) ≥ r1 , d(e p (σ, s), p+ (±1, s)) ≥ r1 for (σ, s) ∈ [0, 1] × − , 2 2 and e p (1, s) 6∈ (∪i=1,2 B R/4 (xi )) ∪ (∪i=1,2 B R/4 (yi )).

(7.10)

p replaced by p+ , that is Then we may assume that p+ (s, 1) satisfies (7.10) with e   1 1 p+ (s, ±1), p− (±1, s) 6∈ (∪i=1,2 B R/4 (xi )) ∪ (∪i=1,2 B R/4 (yi )) for s ∈ − , . 2 2

(7.11)

er continuously On the other hand, from (7.7), (7.8) and (7.11), we can deform the path p+ (−1, s) : I −→ Ω 1 e to the point x0 . That is we can define a homotopy of mappings e p ∈ C([0, 1] × I, Ωr1 ) such that e p (0, s) = p+ (−1, s), e p (1, s) = x0 and e p (σ, s) ∈ B R/4 (x0 ) for (σ, s) ∈ [0, 1] × I. By the same procedure, we may assume that p+ (−1, s) = x0 ,

p+ (1, s) = x1 ,

p+ (s, 1) = p+ (s, −1)

on I

(7.12)

p− (s, −1) = y0 ,

p− (s, 1) = y1 ,

p− (1, s) = p− (−1, s)

on I.

(7.13)

and

Since d(xi , yi ) ≥ R, we can deform the path p+ (·, 1) connecting x0 and x1 , and the path p− (·, 1) connecting y0 and y1 continuously to a path p1 and a path p2 , respectively, where p1 , p2 are paths such that p1 (−1) = x0 ,

p1 (1) = x1 ,

p2 (−1) = y0 ,

p2 (1) = y1

(7.14)

and er , p1 (s), p2 (t) ∈ Ω 1

d( p1 (s), p2 (t)) ≥ r1 /2.

for (s, t) ∈ I 2 .

(7.15)

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N. Hirano / Nonlinear Analysis 68 (2008) 1043–1063

Fig. 3.

er is connected, it is possible to deform p+ (1, ·), p− (1, ·) to paths satisfying (7.14) (See Fig. 3.) Since r1 < r and Ω er ) such that and (7.15). That is we can define homotopies of mappings p± ∈ C([0, 1] × I, Ω 1 p+ (0, s, ±1) = p+ (s, ±1), p+ (σ, −1, s) = x0 ,

p− (0, ±1, s) = p− (±1, s)

p+ (σ, 1, s) = x1 ,

d( p+ (σ, s, t), p− (σ, s, t)) ≥ r1 /2

on I,

p− (σ, s, −1) = y0 ,

on [0, 1] × ∂I

p− (σ, s, 1) = y1

on I,

2

and p+ (1, s, ±1) = p1 (s),

p− (1, ±1, s) = p2 (s)

on I.

Then we find that p± satisfies the desired properties (7.1)–(7.3). This completes the proof.



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