Multiplicity of positive and nodal solutions for
semilinear elliptic equations in infinite strips
Tsung-fang Wu
Department of Applied Mathematics,
National University of Kaohsiung, Kaohsiung 811, Taiwan
Abstract
In this paper, we study the multiplicity of positive and nodal solutions of the fol-lowing semilinear elliptic equation:
−∆u + u = gµ(x) |u|p−2u in A, u ∈ H1 0(A) ,
where 2 < p < N −22N (N ≥ 3), the parameter µ ≥ 0, A = Θ × R is an infinite strip in RN and Θ is a bounded domain in RN −1. We assume that g
µ(x) = a (x) + µb (x) is a positive function, where the functions a and b are satisfying some suitable conditions.
Key words: Semilinear elliptic equation, Nehari manifold, Multiple solutions
1 Introduction
In this paper, we consider the multiplicity of positive and nodal solutions of the following semilinear elliptic equation:
−∆u + u = gµ(x) |u|p−2u in A, u ∈ H1 0(A) , (Eµ) where 2 < p < 2N
N −2 (N ≥ 3) , the parameter µ ≥ 0, A = Θ × R is an infinite
strip in RN and Θ is a bounded domain in RN −1. Let x = (x0, x
N) ∈ RN −1×R,
we assume that gµ(x) = a (x) + µb (x) , where the functions a and b satisfy
the following conditions:
(D1) a, b ∈ C³A´ and there are positive numbers q, r with r < min {p, q} such that
1 ≥ a (x) ≥ 1−c0exp (−q |xN|) for some c0 < 1 and for all x = (x0, xN) ∈ A
and
b (x) ≥ d0exp (−r |xN|) for some d0 > 0 and for all x = (x0, xN) ∈ A;
(D2) a (x) ≤ 1 on A with a strict inequality on a set of positive measure; (D3) b (x) → 0 and a (x) → 1 as |xN| → ∞ uniformly in x0 ∈ Θ.
It is well known that equation (Eµ) has infinitely many solutions in a
bounded domain (see [19]). Lions [15] proved that if gµ(x) ≥ lim|x|→∞gµ(x) =
1, then equation (Eµ) has a positive ground state solution in RN. Bahri–Li [1]
and Bahri-Lions [2] proved that there is at least one positive solution of equa-tion (Eµ) in RN when lim|x|→∞gµ(x) = 1 and gµ≥ 1−C exp (−δ |x|) for some
δ > 0. Zhu [25] has studied the multiplicity of solutions of equation (Eµ) in RN
as follows. Assume that N ≥ 5, gµ(x) ≥ lim|x|→∞gµ(x) = 1 and there exist
positive constants C, γ and R0such that gµ(x) ≥ 1+C/ |x|γfor |x| ≥ R0. Then equation (Eµ) has at least one positive solution and one nodal solution. Hsu
[14] has studied the multiplicity of solutions of equation (Eµ) in unbounded
cylinder domain Ω = ω × Rn as follows. Assume that ω ⊂ Rm a bounded
do-main, m+n ≥ 3(m ≥ 2 and n ≥ 1), gµ(y, z) ≥ lim|z|→∞gµ(y, z) = 1 and there
exist positive constants C, δ and R0 such that gµ(y, z) ≥ 1 + C exp (−δ |z|) for
|z| ≥ R0, uniformly for y ∈ ω. Then equation (Eµ) has at least one positive
solution and one nodal solution.
In this paper, we consider the multiplicity of positive and nodal solutions of equation (Eµ) in an infinite strip. Our results are the following theorems.
Theorem 1.1 Suppose that the functions a and b satisfy (D1) − (D3) . Then
there exists a positive number µ0 such that
(i) for each µ ∈ [µ0, ∞), equation (Eµ) has at least one positive solution;
(ii) for each µ ∈ (0, µ0) , equation (Eµ) has at least two positive solutions.
Theorem 1.2 Suppose that the functions a and b satisfy (D1) − (D3) and in
addition to r < 1. Then there exists a positive number µ0 such that for each µ ∈ (0, µ0) , equation (Eµ) has at least two positive solutions and four 2–nodal
solutions.
In the following sections, we give the proofs of Theorems 1.1, 1.2. We use the variational methods to find nontrivial solutions of equation (Eµ) . Associated
with the equation (Eµ) , we consider the energy functional Jµ in H01(A)
Jµ(u) = 1 2kuk 2− 1 p Z Agµ(x) |u| pdx,
where kuk =³RA|∇u|2+ u2dx´1/2 is the standard norm in H1
0(A) . It is well known that Jµ ∈ C2(H01(A) , R) and the solutions of equation (Eµ) are the
critical points of the energy functional Jµ in H01(A) (see Rabinowitz [20]). Consider the equation
−∆u + u = |u|p−2u in A, u ∈ H1 0(A) , (E∞)
and its associated energy functional J∞ defined by
J∞(u) = 1 2kuk 2− 1 p Z A|u| pdx, u ∈ H1 0 (A) . By [16], equation (E∞) has a positive ground state solution w
0 such that
J∞(w0) = α∞= inf
n
J∞(u) | u ∈ H01(A) \ {0} ,D(J∞)0(u) , uE = 0o.
This paper is organized as follows. In section 2, we give some notations and preliminaries. In section 3, we prove Theorem 1.1. In section 4, we prove Theorem 1.2.
2 Notations and Preliminaries
Throughout this paper, we denote by Sp the best Sobolev constant for the
imbedding of H1
0 (A) in Lp(A). In particular,
µZ
A|u|
pdx¶1/p ≤ S−12
p kuk for all u ∈ H01(A) \ {0} .
Now we define the Palais–Smale (simply by (PS)) sequences, (PS)–values and (PS)–conditions in H1
0 (A) for Jµ as follows.
Definition 2.1 (i) For β ∈ R, a sequence {un} is a (PS)β–sequence in H01(A)
for Jµ if Jµ(un) = β + o(1) and Jµ0(un) = o(1) strongly in H−1(A) as n → ∞;
(ii) β ∈ R is a (PS)–value in H1
0(A) for Jµ if there exists a (PS)β–sequence
in H1
0(A) for Jµ;
(iii) Jµ satisfies the (PS)β–condition in H01(A) if every (PS)β–sequence in
H1
0 (A) for Jµ contains a convergent subsequence.
Lemma 2.2 Let {un} is a (PS)β–sequence in H01(A) for Jµ. Then there exist
a subsequence {un} , a positive integer m, a sequence {yni} ∞
n=1 in R, functions
u0 ∈ H01(A) and wi ∈ H01(A) \ {0} for 1 ≤ i ≤ m such that
|yi n− yjn| → ∞ for 1 ≤ i, j ≤ m and i 6= j; −∆u0+ u0 = gµ(x) |u0|p−2u0 in A; −∆wi+ wi = |wi|p−2wi in A; un = u0+ Pm
i=1wi(x0, xN − yin) + o (1) strongly in H01(A) ;
Jµ(un) = Jµ(u0) +
Pm
i=1J∞(wi) + o (1) .
Now, we consider the minimization problems
αµ = inf u∈Mµ Jµ(u) ; θµ= inf u∈Nµ Jµ(u) , where Mµ= n u ∈ H01(A) \{0} | DJµ0 (u) , uE= 0o and Nµ= n u ∈ H01(A) | u± ∈ Mµ o .
Clearly, 2αµ ≤ θµ. Then we have the following results.
Lemma 2.3 Suppose that u0 is a local minimizer for Jµon Mµ. Then Jµ0 (u0) = 0 in H−1(A) .
Proof. Our proof is almost the same as that in Brown-Zhang [7, Theorem 2.3] (or see Binding-Drabek-Huang [5]).
Lemma 2.4 If µ = 0, then
(i) α0 = α∞ and equation (E0) does not admit any solution u0 such that
J0(u0) = α0;
(ii) θ0 = 2α∞ and equation (E0) does not admit any nodal solution v0 such
that J0(v0) = θ0.
Proof. (i) Similar to the method used in Chabrowski [8, p.38].
(ii) First, by Lien-Tzeng-Wang [16], equation (E∞) has a positive solution w
0 such that J∞(w
0) = α∞. Let ψ : A → [0, 1] be a C∞ function defined by
ψ (x0, x N) = 0, xN ≤ 0; 1, xN ≥ 1,
such that |∇ψ| ≤ c. Let ψ(i)(x0, x
N) = ψ ³ x0, (−1)i+1x N ´ and u(i) n (x0, xn) =
(−1)i+1ψ(i)(x0, x N) w0 ³ x0, x N − (−1)i+1n ´
for i ∈ {1, 2} and n ∈ N, then suppu(1)
n ∩ suppu(2)n = ∅. (2.1)
Similar to the method used in Wu [23, Lemma 2.8], we have
° ° °u(i)n ° ° °2 = Z A ¯ ¯ ¯u(i)n ¯ ¯ ¯pdx + o (1) = Z Aa (x) ¯ ¯ ¯u(i)n ¯ ¯ ¯pdx + o (1) and J0 ³ u(i) n ´ = J∞³u(i) n ´ + o (1) = J∞(w 0) + o (1) . By Wang-Wu [21, Lemma 7], nu(i)
n
o
are (PS)α∞–sequences in H01(A) for J∞.
Moreover, by routine computations, there exists a sequence {s(i)
n } ⊂ R+ such
that s(i)
n = 1 + o(1), {s(i)n u(i)n } ⊂ M0 and J0(s(i)n u(i)n ) = J∞(w0) + o(1). Let
vn = s(1)n u(1)n + s(2)n u(2)n , then by (2.1) we have vn ∈ N0 and
J0(vn) ≥ θ0 ≥ 2α0 = 2α∞. Moreover,
J0(vn) = J0(s(1)n u(1)n ) + J0(s(2)n u(2)n ) = 2J∞(w0) + o(1) = 2α∞+ o(1). Thus, θ0 = 2α∞. Next, we will show that equation (E0) does not admit any nodal solution v0 such that J0(v0) = θ0. Suppose otherwise, then equation (E0) has a nodal solution v0 such that J0(v0) = θ0. Clearly, v0± ∈ M0 and
J0 ³ v± 0 ´ ≥ α0 = α∞. Since θ0 = 2α∞, we have J0 ³ v± 0 ´ = α0 = α∞. Then v0± are ground state solutions of equation (E0) , which contradicts the part (i) . Proposition 2.5 (i) For each minimizing sequence {un} in M0 for J0, there
exists a subsequence {un} such that it is a (PS)α∞–sequence in H01(A) for J∞;
(ii) For each minimizing sequence {un} in N0 for J0 there exist subsequences
{u±
n} such that they are (PS)α∞–sequences in H01(A) for J∞. Proof. (i) By Lemmas 2.2, 2.4.
(ii) Let {un} be a minimizing sequence in N0 for J0. Then by Lemma 2.4, lim n→∞J0(un) = θ0 = 2α ∞ and J0(un) = J0 ³ u+ n ´ + J0 ³ u− n ´ ≥ 2α∞.
Thus, limn→∞J0(u±n) = α∞. Since u±n ∈ M0, we have {u±n} are minimizing
sequences in M0 for J0. By part (i) there exist subsequences {u±n} such that
they are (PS)α∞–sequences in H01(A) for J∞.
Now, we recall the generalized barycenter map (cf. Bartsch-Weth [6, The-orem 2.1] and Cerami-Passaseo [10]) given by Φ : Lp³RN´\ {0} → RN a
continuous map satisfying
Φ (u(z − ζ)) = ζ + Φ (u) and Φ³u ◦ A−1´= AΦ (u)
for every ζ ∈ RN, every orthogonal N×N matrix A and every u ∈ LP ³RN´\ {0} .
Since Lp(A) ⊂ Lp³RN´ and the infinite strip A is only translation invariant
and symmetric on xN−axis. So we may redefine a new generalized barycenter
map h : Lp(A) \ {0} → R such that for every ξ ∈ R and u ∈ Lp(A) \ {0} , we
have
h (u (x0, xN − ξ)) = ξ + h (u (x0, xN)) and h (u (x0, −xN)) = −h (u (x0, xN)) .
Then we have the following results.
Lemma 2.6 For each positive number L there exists a positive number δ (L)
such that for every u ∈ M0 with J0(u) ≤ α∞+ δ (L) , we have either h(u) > L
or h(u) < −L.
Proof. Suppose otherwise, then there exist L0 > 0 and a sequence {un} ⊂
M0 such that J0(un) = α∞+ o (1) and
−L0 ≤ h(un) ≤ L0. (2.2)
By Lemma 2.4 (i) and Proposition 2.5 (i) , we may assume that {un} is a
(PS)α∞(A)–sequence in H01(A) for J∞. Then by Lemmas 2.2, 2.4 there exist a
sequence {yn} ⊂ R with |yn| → ∞ and w0 is a solution of equation (E∞) such that
kun− u0(x0, xN − yn)k → 0 as n → ∞. (2.3)
Then by (2.3)
h(un) = h(w0(x0, xN − yn)) + o (1) = h(w0) + yn+ o (1)
and so |h(un)| → ∞ as n → ∞, this contradicts to (2.2) .
Lemma 2.7 For each positive number L there exists a positive number δ (L)
such that for every u ∈ N0 with J0(u) ≤ θ0+ δ (L) , we have (i) |h(u+) − h(u−)| > L;
(ii) the function u+ satisfies either h(u+) > L or h(u+) < −L; (iii) the function u− satisfies either h(u−) > L or h(u−) < −L.
Proof. (i) Suppose otherwise, then there exist L0 > 0 and a sequence
{un} ⊂ N0 such that J0(un) = θ0+ o (1) and
¯ ¯
¯h(u+n) − h(u−n)¯¯¯≤ L0. (2.4) Similar to the proof of Proposition 2.5 (ii) , we have {u±
n} are (PS)α0–sequences
in H1
are (PS)α∞–sequences in H01(A) for J∞. Clearly, {u±n} are bounded sets in H1
0 (A) . Then by using a similar argument as in Lien-Tzeng-Wang [16, Theo-rem 4.1], there exist R, C0 > 0 and {yn±} ⊂ R with |yn±| → ∞ such that
Z A−R,R−(0,y±n) ³ u± n ´2 dx ≥ C0 for all n ∈ N,
where A−R,R = {(x0, xN) ∈ A | |xN| < R} . Moreover, by Lien-Tzeng-Wang
[16, Theorem 4.1] and Chen-Chen-Wang [9], equation (E∞) has a positive
solution u+
0 and a negative solution u−0 such that u±0 are axially symmetric in
xN–axis and ° ° °u± n − u±0 ³ x0, x N − yn± ´° ° °→ 0 as n → ∞. (2.5)
Now we will show that |y+
n − y−n| → ∞ as n → ∞. Suppose otherwise, then
we can assume that y+
n − yn−→ y0 for some y0 ∈ R. Then by (2.5) 0 = Z A ¯ ¯ ¯u+ n ¯ ¯ ¯r ¯ ¯ ¯u− n ¯ ¯ ¯sdx = Z A ¯ ¯ ¯u+ 0 ³ x0, x N − yn+ ´¯¯ ¯r ¯ ¯ ¯u− 0 ³ x0, x N − yn− ´¯¯ ¯sdx + o (1) = Z A ¯ ¯ ¯u+ 0 (x0, xN) ¯ ¯ ¯r ¯ ¯ ¯u− 0 (x0, xN + yn− zn) ¯ ¯ ¯sdx + o (1) = Z A ¯ ¯ ¯u−0 (x0, xN) ¯ ¯ ¯r ¯ ¯ ¯u−0 (x0, xN + y0) ¯ ¯ ¯sdx + o (1) where r
p + sq = 1, which is a contradiction. Thus, |yn+− yn−| → ∞ as n → ∞.
Moreover, by (2.5) we can conclude that
h(u± n) = h(u±0 ³ x0, x N − yn± ´ ) + o (1) = h(u± 0) + y±n + o (1) . (2.6) This implies ¯ ¯ ¯h(u+n) − h(u−n) ¯ ¯ ¯→ ∞ as n → ∞ this contradicts to (2.4) .
(ii) and (iii) : By (2.6) we only need to show that
¯ ¯ ¯yn± ¯ ¯ ¯→ ∞ and ¯ ¯ ¯y−n ¯ ¯ ¯→ ∞ as n → ∞.
Since the proofs of two cases are the similar arguments, we only prove the case
|y+
n| → ∞ as n → ∞. Suppose otherwise, then {y+n} is a bounded sequence in
R. Without loss of generality, we may assume that y+
n → y0 for some y0 ∈ R. By (2.5) ° ° °u+ n − u+0 (x0, xN − y0) ° ° °→ 0 as n → ∞. Since {u+
n} is a (PS)α0–sequence in H01(A) for J0. This implies u+0 (x0, xN − y0) is a ground state solution of equation (E0) , which contradicts the result in Proposition 2.5 (i) . This completes the proof.
Lemma 2.8 (i) If u is a nontrivial nonnegative solution of equation (Eµ) and
(ii) If u is a nodal solution of equation (Eµ) and Jµ(u) < 3α∞, then u is a
2–nodal solution of equation (Eµ) .
Proof. The proof is similar to Proposition 3.1 in Furtado [13] (or see Bartsch-Weth [6]).
3 Proof of Theorem 1.1
In this section, we will establish the existence of positive solutions of equa-tion (Eµ) . First, let w0(x) be a xN−symmetric positive solution of
equa-tion (E∞) such that J∞(w
0) = α∞. Then by Chen-Chen-Wang [9], for any 0 < ε < 1 + θ1, there exist Aε > 0 and Bε > 0 such that
Aεφ1(x0) e− √ 1+θ1+ε|xN| ≤ w 0(x0, xN) ≤ Bεφ1(x0) e− √ 1+θ1−ε|xN| (3.1) for all (x0, x
N) ∈ A, where θ1 is the first eigenvalue and φ1 the corresponding first positive eigenfunction of the Dirichlet problem −∆φ = θφ in Θ, φ = 0 on ∂Θ. Let
wl(x) = w0(x0, xN − l) , l ∈ R.
Then we have the following result.
Proposition 3.1 For each µ > 0 there exists l0 > 0 such that sup
t≥0 Jµ(twl) < α
∞ for all |l| ≥ l
0.
In particular, αµ< α∞ for all µ > 0.
Proof. Since Jµ(twl) = 1 2ktwlk 2−1 p Z Agµ|twl| pdx < t2 2 kw0k 2 −tp p Z A−1,1 amin|w0|pdx, where A−1,1 = {(x0, xN) ∈ A | − 1 < xN < 1} and amin = inf {a (x) | x ∈ A} > 0.
Then we have Jµ(twl) → −∞ as t → ∞. Thus, there exists t1 > 0 such that
Jµ(twl) < α∞ for all t ≥ t1. (3.2) Moreover, Jµ(0) = 0 < α∞, Jµ ∈ C2(H01(A) , R) and kwlk2 = p−22p α∞ for all
l ∈ R, then there exists t2 > 0 such that
Therefore, we only need to show that there exists l0 > 0 such that for |l| > l0, sup t2≤t≤t1 Jµ(twl) < α∞. Since Jµ(twl) = J0(tw0) + 1 p Z A(1 − a) t pwp ldx − µ p Z Abt pwp ldx. (3.4)
By Brown-Zhang [7] and Willem [22], we know that
J0(tw0) ≤ α∞ for all l ∈ R. Thus, by (3.4) Jµ(twl) ≤ α∞+ tp p Z A(1 − a) w p ldx − µtp p Z Abw p ldx. (3.5)
By (3.1) , there exists B0 > 0 such that
w0(x0, xN) ≤ B0φ1(x0) e−|xN| for all (x0, xN) ∈ A. Setting C0 = C Ã min x∈A−1,1 wp0(x) ! > 0. By the condition (D1) Z Abw p ldx = Z Ab (x 0, x N + l) w0p(x) dx ≥ C0 Z A−1,1 b (x0, x N + l) dx ≥ C0d0exp (−r |l|) and Z A(1 − a) w p ldx ≤ Z Ac0exp (−q |xN|) B p 0φ1(x0) exp (−p |xN − l|) dx ≤ C exp (− min {p, q} |l|) .
Since r < min {p, q} and t2 ≤ t ≤ t1, we can find l0 > 0 such that
tp p Z A(1 − a) w p ldx < µ p Z Abw p
ldx for all |l| ≥ l0 and t ∈ [t2, t1] . (3.6) Thus, by (3.2) , (3.3) , (3.5) and (3.6), we obtain
sup
t≥0Jµ(twl) < α
∞ for all |l| ≥ l
0. This completes the proof.
Theorem 3.2 For each µ > 0, equation (Eµ) has a positive ground state
solution.
Proof. Analogously to the proof of Ni-Takagi [18], one can show that the Ekeland variational principle (see [12]), there exists a minimizing sequence
{un} ⊂ Mµ such that
Jµ(un) = αµ+ o (1) and Jµ0 (un) = o (1) in H−1(A) .
Since αµ < α∞, by Lemma 2.2 there exist a subsequence {un} and u0 ∈ Mµ
is a nonzero solution of equation (Eµ) such that
un → u0 strongly in H01(A) and Jµ(u0) = αµ.
Since Jµ(u0) = Jµ(|u0|) and |u0| ∈ Mµ. By Lemma 2.3 and the maximum
principle, we may assume that u0 is a positive solution of equation (Eµ) . This
completes the proof
For positive numbers L, δ, let
M0(δ) = {u ∈ M0 : Jµ(u) ≤ α∞+ δ} ; Mi(δ, L) = n u ∈ M0(δ) : (−1)ih (u) > L o for i = 1, 2.
By the consequence of Lemma 2.6, it is easy to prove the following result. Lemma 3.3 Let L and δ (L) be positive numbers as in Lemma 2.6. Then we
have
(i) Mi(δ, L) 6= ∅ for all i = 1, 2;
(ii) M0(δ) = M1(δ, L) ∪ M2(δ, L) ; (iii) M1(δ, L) ∩ M2(δ, L) = ∅.
Note that for each µ > 0 and u ∈ Mµ there is a unique
t0(u) = Ã kuk2 R Aa |u|pdx !1/(p−2) > 0 (3.7)
such that t0(u) u ∈ M0. Furthermore, we have the following results.
Lemma 3.4 (i) 1 < [t0(u)]p < (1 + µ kb/ak∞)p/(p−2) for all µ > 0 and u ∈
Mµ; (ii) αµ≥ p−2(1+µkb/ak∞) p−2 α∞ > 0 for all µ ∈ ³ 0,2kb/akp−2 ∞ ´ .
Proof. (i) Clearly,
t0(u) = Ã kuk2 R Aa |u|pdx !1/(p−2) > Ã kuk2 R Agµ(x) |u|pdx !1/(p−2) = 1.
Since kuk2 = Z Agµ(x) |u| pdx ≤ (1 + µ kb/ak ∞) Z Aa (x) |u| pdx. (3.8)
By (3.7) , (3.8) we have [t0(u)]p ≤ (1 + µ kb/ak∞)p/(p−2). (ii) For u ∈ Mµ. Since t0(u) u ∈ M0,
Jµ(u) = sup t≥0Jµ(tu) ≥ Jµ(t0(u) u) =1 2kt0(u) uk 2− 1 p Z Agµ(x) |t0(u) u| pdx ≥ Ã 1 2 − 1 + µ kb/ak∞ p ! Z Aa (x) |t0(u) u| pdx ≥p − 2 (1 + µ kb/ak∞) p − 2 α ∞ for all µ ∈ Ã 0, p − 2 2 kb/ak∞ ! . This implies αµ ≥ p − 2 (1 + µ kb/ak∞) p − 2 α ∞> 0 for all µ ∈ Ã 0, p − 2 2 kb/ak∞ ! .
This completes the proof. Let
M (µ) = {u ∈ Mµ: Jµ(u) ≤ α∞} ;
Mi(µ, L) =
n
u ∈ M (µ) : (−1)i+1h (u) > Lo for i = 1, 2.
By the proof of Proposition 3.1, for each l ∈ R with |l| ≥ l0 there exists a positive number t∗(l) such that t∗(l) wl ∈ Mµ and Jµ(t∗(l) wl) < α∞.This
implies Mi(µ, L) 6= ∅. Furthermore, we have the following result.
Lemma 3.5 Let L and δ (L) be positive numbers as in Lemma 2.6. Then
there exists a positive number µ0 such that for every µ ∈ (0, µ0) we have (i) M (µ) = M1(µ, L) ∪ M2(µ, L) ;
(ii) M1(µ, L) ∩ M2(µ, L) = ∅.
Proof. For u ∈ Mµ with Jµ(u) ≤ α∞, by (3.7) there exists t0(u) > 0 such that t0(u) u ∈ M0. Moreover,
Jµ(u) = sup t≥0Jµ(tu) ≥ Jµ(t0(u) u) = J0(t0(u) u) − µ [t0(u)] p p Z RNb (x) |u| pdx.
J0(t0(u) u) ≤ Jµ(u) + µ [t0(u)]p p Z RNb (x) |u| pdx ≤ α∞+ µ (1 + µ kb/ak∞) p/(p−2)kbk ∞ p S −p2 p kukpH1.
Moreover, we can find a positive numberc independent of µ such that kuke H1 ≤
e
c for all µ > 0 and for all u ∈ Mµ with Jµ(u) ≤ α∞. Therefore,
J0(t0(u) u) ≤ α∞+
µ (1 + µ kb/ak∞)p/(p−2)kbk∞
p S
−p2 p cep.
Then by Lemma 3.3 there exists a positive number µ1such that for µ ∈ (0, µ1) ,
J0(t0(u) u) < α∞+ δ (L) and either
h (t0(u) u) > L or h (t0(u) u) < −L
for all u ∈ Mµwith Jµ(u) ≤ α∞. Moreover, by Lemma 3.4 and the continuity
of h there exists a positive number µ0 ≤ µ1 such that for every µ ∈ (0, µ0) and u ∈ Mµ with Jµ(u) ≤ α∞ we have either h(u) > L or h(u) < −L. This
complete the proof.
Furthermore, we have the following result.
Lemma 3.6 Let µ0 > 0 as in Lemma 3.5. Then for each µ ∈ (0, µ0) , we have
that Mi(µ, L) is closed set for all i = 1, 2.
Proof. The proofs of cases “i = 1, 2” are the similar arguments. Therefore, we only need to prove the first case “i = 1”. Suppose that u0 is a limits point of M1(µ, L) , then h (u0) ≥ L and Jµ(u0) ≤ α∞. This implies u0 ∈ M (µ). If
h (u0) = L, then by Lemma 3.5 u0 ∈ M2(µ, L) . We obtain
L = h (u0) < −L
which is a contradiction. Thus, h (u0) < L, this implies u0 ∈ M1(µ, L) . There-fore, Mi(µ, L) are closed sets.
We need the following result.
Proposition 3.7 Let µ0 > 0 as in Lemma 3.5. Then for each µ ∈ (0, µ0) ,
there exist minimizing sequences nu(i)
n
o
⊂ Mi(µ, L) such that
Jµ
³
u(i)n ´= σ(i)µ + o (1) and Jµ0 ³u(i)n ´= o (1) in H−1(A) ,
where σ(i)
Proof. Analogously to the proof of Ni-Takagi [18], one can show that the Ekeland variational principle (see [12]), there exist minimizing sequences
n u(i) n o ⊂ Mi(µ, L) such that Jµ ³ u(i) n ´ = σ(i) µ + o (1) and Jµ0 ³ u(i) n ´ = o (1) in H−1(A) .
We will omit detailed proof here.
Theorem 3.8 Let µ0 > 0 as in Lemma 3.5. Then for each µ ∈ (0, µ0) ,
equation (Eµ) has positive solutions u(i)µ ∈ Mi(µ, L) such that Jµ
³ u(i) µ ´ = σ(i) µ .
Proof. By Propositions 3.7, there exist sequences nu(i)
n o ⊂ Mi(µ, L) such that Jµ ³ u(i) n ´ = σ(i) µ + o (1) and Jµ0 ³ u(i) n ´ = o (1) in H−1(A) . Since σ(i)
µ < α∞, by Lemmas 2.2, 3.6 there exist subsequences
n u(i) n o and u(i)
µ ∈ Mi(µ, L) are nonzero solutions of equation (Eµ) such that
u(i)n → u(i)µ strongly in H01(A). Since Jµ ³ u(i) µ ´ = Jµ ³¯¯ ¯u(i)µ ¯ ¯ ¯ ´ and ¯¯¯u(i)µ ¯ ¯
¯ ∈ Mi(µ, L) . By Lemmas 2.3 and the
maximum principle, we may assume that u(i)
µ are positive solutions of equation
(Eµ) .
We now begin to show the proof of Theorem 1.1: By Theorems 3.2, 3.8 we obtain proof.
Remark 3.1 By Chen-Chen-Wang [9] (or see Hsu [14, Proposition 3.4]), if
u0 is a positive solution of equation (Eµ) , then for any 0 < ε < 1 + θ1, there
exist Aε> 0 and R0 > 0 such that
u0(x0, xN) ≤ Aεφ1(x0) e−
√
1+θ1−ε|xN| for all (x0, x
N) ∈ A and |xN| ≥ R0,
where θ1 is the first eigenvalue and φ1 the corresponding first positive
eigen-function of the Dirichlet problem −∆φ = θφ in Θ, φ = 0 on ∂Θ.
4 Proof of Theorem 1.2
Let w0 be a positive ground state solution of equation (E∞) and wl(x) =
w0(x0, xN − l) for l ∈ R. Then we have the following results.
i = 1, 2 there exist l1 > 0, s∗(l) and t∗(l) such that for |l| ≥ l1
s∗(l) u(i)
µ − t∗(l) wl∈ Nµ
where u(i)
µ are positive solutions of equation (Eµ) as in Theorem 3.8 and 12 ≤
s∗(l) , t∗(l) ≤ 2.
Proof. The proof is similar to the that in [25] (or see [14, p. 728]) Lemma 4.2 We have (i)
° ° ° ° ³ s∗(l) u(i) µ − t∗(l) wl ´+ − u(i) µ ° ° ° °→ 0 as |l| → ∞; (ii) ° ° ° ° ³ s∗(l) u(i) µ − t∗(l) wl ´− + wl ° ° ° °→ 0 as |l| → ∞.
Proof. The proofs of all cases are the similar arguments. Therefore, we only need to prove the case “i = 1 and l → ∞”. Let ψ : A → [0, 1] be a C∞
function defined by ψ (x0, xN) = 0, xN ≤ 0; 1, xN ≥ 1,
such that |∇ψ| ≤ c and let ψ(i)l (x0, x
N) = ψ ³ x0, (−1)i(x N − l) ´ . First, we will show that ° ° ° ³ s∗(l) u(1) µ − t∗(l) wl ´ −³s∗(l) ψ(1) l/2u(1)µ − t∗(l) ψ (2) l/2wl ´° ° °→ 0 as l → ∞. (4.1) Since 1 2 ≤ s∗(l) , t∗(l) ≤ 2 and so ° ° ° ³ s∗(l) u(1) µ − t∗(l) wl ´ −³s∗(l) ψ(1) l/2u(1)µ − t∗(l) ψ (2) l/2wl ´°° ° ≤ s∗(l)°°°³1 − ψ(1) l/2 ´ u(1) µ ° ° ° H1 + t ∗(l)°°°³1 − ψ(2) l/2 ´ wl ° ° ° ≤ 2°°° ³ 1 − ψl/2(1)´u(1) µ ° ° ° H1 + 2 ° ° ° ³ 1 − ψl/2(2)´wl ° ° °.
Similar to the proof of theorem 2.4 in Benci-Cerami [3], we can conclude that
° ° ° ³ 1 − ψ(1)l/2´u(1) µ ° ° ° → 0 and ° ° ° ³ 1 − ψ(2)l/2´wl ° ° ° → 0 as l → ∞. (4.2)
Thus (4.1) holds. Moreover,
° ° ° ° ³ s∗(l) u(1) µ − t∗(l) wl ´± −³s∗(l) ψ(1) l/2u(1)µ − t∗(l) ψ (2) l/2wl ´±°° ° ° → 0 as l → ∞. (4.3) Since ³ s∗(l) ψl/2(1)u(1)µ − t∗(l) ψ(2)l/2wl ´+ = s∗(l) ψ(1)l/2u(1)µ , (4.4) ³ s∗(l) ψ(1) l/2u(1)µ − t∗(l) ψ (2) l/2wl ´− = −t∗(l) ψ(2) l/2wl (4.5)
and ³ s∗(l) u(1) µ − t∗(l) wl ´± ∈ Mµ (4.6)
for all l > maxn2h³u(1)
µ ´ , l1 o . By (4.3) − (4.6) , ° ° °s∗(l) ψ(1)l/2u(1)µ ° ° °= Z Agµ(x) ¯ ¯ ¯s∗(l) ψl/2(1)u(1)µ ¯ ¯ ¯p+ o (1) as l → ∞ (4.7) and ° ° °−t∗(l) ψl/2(2)wl ° ° °= Z Agµ(x) ¯ ¯ ¯−t∗(l) ψl/2(2)wl ¯ ¯ ¯pdx + o (1) = Z A ¯ ¯ ¯−t∗(l) ψl/2(2)wl ¯ ¯ ¯pdx + o (1) as l → ∞. (4.8) Moreover, by u(1) µ ∈ Mµ, wl ∈ M∞, (4.2) , (4.7) and (4.8) , we have s∗(l) = 1 + o (1) and t∗(l) = 1 + o (1) as l → ∞. (4.9)
Therefore, by (4.2) − (4.5) and (4.9) we can conclude that
° ° ° ° ³ s∗(l) u(1) µ − t∗(l) wl ´+ − u(1) µ ° ° ° ° → 0 as l → ∞ and ° ° ° ° ³ s∗(l) u(1) µ − t∗(l) wl ´− + wl ° ° ° °→ 0 as l → ∞.
This completes the proof.
Lemma 4.3 Let µ0 > 0 as in Lemma 3.5 and let l1 > 0 as in Lemma 4.1.
Then for each µ ∈ (0, µ0) and i = 1, 2 there exist el0 ≥ l1 such that for |l| >el0
sup 1 2≤s,t≤2 Jµ ³ su(i) µ − twl ´ < αµ+ α∞.
Proof. The proofs of all cases are the similar arguments. Therefore, we only need to prove the case “i = 1”. Since u(1)
µ is a positive solution of equation
Jµ ³ su(1) µ − twl ´ =1 2 ° ° °su(1)µ ° ° °2+1 2ktwlk 2 −st µZ A∇u (1) µ ∇wl+ u(1)µ wl ¶ −1 p Z Agµ ¯ ¯ ¯su(1)µ − twl ¯ ¯ ¯pdx ≤ Jµ ³ su(1) µ ´ + J∞(tw 0) − st Z Agµ ¯ ¯ ¯u(1) µ ¯ ¯ ¯p−2u(1) µ wl− 1 p Z Agµ|twl| pdx +1 pC1 Z Agµ · s³u(1) µ ´p−1 twl+ su(1)µ (twl)p−1 ¸ dx + 1 p Z A|twl| pdx ≤ αµ+ α∞− tpµ p Z Abw p ldx + tp p Z A(1 − a) w p ldx +Ce1 Z A ³ u(1)µ ´p−1wl+ (wl)p−1u(1)µ dx
Here, we have used the following inequality,
|c1− c2|p > cp1+ c22− C
³
cp−11 c2+ cp−12 c1
´
,
for any c1, c2 > 0 and some positive constant C. Then
sup 1/2≤s,t≤2 Jµ ³ su(1) µ − twl ´ ≤ αµ+ α∞− µ 2pp Z Abw p ldx + 2p p Z A(1 − a) w p ldx +Ce1 Z A ³ u(1) µ ´p−1 wl+ wlp−1u(1)µ dx.
(i) From the proof of Proposition 3.1, we have
Z Abw p ldx ≥ C0d0exp (−r |l|) and Z A(1 − a) w p ldx ≤ C exp (− min {p, q} |l|) .
(ii) By the Holder inequality,
Z A−1,1 ³ u(1) µ ´p−1 wldx ≤ ÃZ A−1,1 ³ u(1) µ ´p dx !(p−1)/pÃZ A−1,1 wpldx !1/p ≤ C1exp (− |l|) .
Applying Remark 3.1, there exists l2 > 0 such that for |l| ≥ l2
Z Ac −1,1 ³ u(1) µ ´p−1 wldx ≤ C2 Z
{y∈R | |y|>1}exp (− (p − 1) |xN|) exp (− |xN − l|) dxN
(iii) Similarly, we also obtain
Z
A−1,1
wlp−1u(1)
µ dx ≤ C3exp (− (p − 1) |l|) and there exists an l3 > 0 such that for |l| ≥ l3
Z
Ac −1,1
wlp−1u(1)
µ dx ≤ C4exp (− |l|) .
By (i) − (iii) and r < min {1, q} < p, we can choose el0 = max {l1, l2, l3} such that for |l| >el0 sup 1 2≤s,t≤2 Jµ ³ su(i)µ − twl ´ < αµ+ α∞.
For positive numbers L, δ, let N (µ) =nu ∈ Nµ : Jµ
³
u±´< α∞o; N1(µ, L) =
n
u ∈ N (µ) : h(u+) > L and h(u−) < −Lo; N2(µ, L) =
n
u ∈ N (µ) : h(u+) < −L and h(u−) > Lo;
N3(µ, L) =
n
u ∈ N (µ) : h(u+) > L, h(u−) > L and G (u) > Lo;
N4(µ, L) =
n
u ∈ N (µ) : h(u+) < −L, h(u−) < −L and G (u) > Lo,
where G (u) = |h(u+) − h(u−)| . By Lemma 4.2, there exists l
0 > 0 such that (−1)ihµ³s∗(l) u(i)µ − t∗(l) wl ´+¶ > L, sgn (l) hµ³s∗(l) u(i)µ − t∗(l) wl ´−¶ > L and G³s∗(l) u(i) µ − t∗(l) wl ´ > L
for all |l| > l0. This implies Ni(µ, L) 6= ∅ for all i = 1, 2, 3, 4. Thus, by Lemmas
2.4, 3.4, 4.1 and 4.3, we have 2p − 4 (1 + µ kb/ak∞) p − 2 α ∞≤ inf u∈Ni(µ,L) Jµ(u) < αµ+ α∞ (4.10)
for all µ ∈³0,2kb/akp−2
∞
´
. Furthermore, we have the following result.
Lemma 4.4 Let L and δ (L) be positive numbers as in Lemma 2.6 and let
µ0 > 0 as in Lemma 3.5. Then there exists a positive numberµe0 ≤ min
n
µ0,2kb/akp−2
∞
o
(i) Ni(µ, L) ∩ Nj(µ, L) = ∅ for all i 6= j;
(ii) N (µ) = ∪4
i=1Ni(µ, L) .
Proof. Similar to the argument in Lemma 3.5.
Similar to the argument in Lemma 3.6 we have the following result.
Lemma 4.5 For each µ ∈ (0,µe0) , we have that Ni(µ, L) is closed set for all
i = 1, 2, 3, 4.
Now we consider the minimization problems in Ni(µ, L) for Jµ:
θi,µ = inf u∈Ni(µ,L)
Jµ(u) for all i = 1, 2, 3, 4.
Then we have the following result.
Lemma 4.6 For each v0 ∈ Ni(µ, L) there exists a map Ψ : H01(A) → R2
such that
(i) Ψ³t1v0++ t2v0−
´
= (t1, t2) for t1, t2 ≥ 0; (ii) Ψ (u) = (1, 1) if and only if u ∈ Nµ.
Proof. Similar to the method used in Clapp-Weth [11, Lemma 13].
The next result is a variant of proposition 14 in Clapp-Weth [11], and its proof follows from the arguments of applying the Leray-Schauder continuation principle. Let µ0 = minnµe0,4kb/akp−2∞
o
and let
distH (u, B) = inf
n
ku − vk : v ∈ B ⊂ H01(A)o.
Then for each µ ∈ (0, µ0) and i ∈ {1, 2, 3, 4} we have the following results. Proposition 4.7 Let λ0 = αµ + α∞ − θi,µ. Then for each λ ∈ (0, λ0) and
² > 0 there exists u0 ∈ H01(A) such that (i) distH(u0, Ni(µ, L)) ≤ ²;
(ii) Jµ(u0) ∈ [θi,µ, θi,µ+ λ);
(iii) k∇Jµ(u0)k ≤ max n√ λ,λ ² o .
Proof. The proofs of all cases are the similar arguments. Therefore, we only need to prove the case “i = 1”. Fix v0 ∈ N1(µ, L) such that Jµ(v0) < θi,µ+ λ,
and fix d0 > 1 such that Jµ
³
d0v±0
´
≤ 0. Let Ψ : H1
0(A) → R2 be as in Lemma 4.6. We put K = [0, d0] × [0, d0] and define
η : K → H1
0 (A) , η (s1, s2) = s1v0++ s2v0−. Then Ψ ◦ η = id : K → K, in particular
Notice also that
Jµ(η (s1, s2)) ≤ Jµ(v0) < θi,µ+ λ for all (s1, s2) ∈ K. (4.12) We now choose a Lipschitz continuous function χ : R → R such that 0 ≤
χ ≤ 1, χ (s) = 1 for s ≥ 0 and χ (s) = 0 for s ≤ −1. Then since Jµ ∈
C2(H1
0 (A) , R) , there is a semiflow ϕ : [0, ∞) × H01(A) → H01(A) satisfying
∂ ∂tϕ (t, u) = −χ (J (ϕ (t, u))) ∇Jµ(ϕ (t, u)) , ϕ (0, u) = u.
We will frequently write ϕt in place of ϕ (t, ·) . Since µ < µ
0, Jµ ³ v± 0 ´ < α∞ and J µ ³ d0v±0 ´ ≤ 0, it follows that sup Jµ(η (∂K)) < α∞< 2p − 4 (1 + µ kb/ak∞) p − 2 α ∞. Hence, by (4.10) ³ ϕt◦ η´(∂K) ∩ N µ= ∅ for all t ≥ 0
and, by Lemma 4.6, this implies
³
Ψ ◦ ϕt◦ η´(y) 6= (1, 1) for all y ∈ ∂K, t ≥ 0.
Equality (4.11) and the global continuation principle of Leray-Schauder (see e.g. Zeider [24, p.629]) imply that there exists a connected subset Z ⊂ K×[0, 1] such that (1, 1, 0) ∈ Z; ϕt(η (s 1, s2)) ∈ Nµ for all (s1, s2, t) ∈ Z; Z ∩ (K × {1}) 6= ∅. We put e Z =nϕt(η (s1, s2)) ∈ Nµ: (s1, s2, t) ∈ Z o .
By inequality (4.12) and Lemma 3.5 (ii) , we obtain
e
Z ⊂ N1(µ, L) ∪ N2(µ, L) .
Since Z is connected, we obtain that Z ⊂ Ne 1(µ, L) . We now pick (s1, s2, 1) ∈
Z ∩ (K × {1}) and write
Then v2 ∈Z ⊂ Ne 1(µ, L) . We distinguish two case. Case 1. kϕt(v
1) − v2k ≤ ² for all t ∈ [0, 1] . We choose t0 ∈ [0, 1] with
° ° °∇Jµ ³ ϕt0(v 1) ´° ° °= min 0≤t≤1 ° ° °∇Jµ ³ ϕt(v 1) ´° ° °
and put u0 = ϕt0(v1) . Then
λ ≥ Jµ(v1) − Jµ(v2) = − Z 1 0 ∂ ∂tJµ ³ ϕt(v 1) ´ ds = Z 1 0 ° ° °∇Jµ ³ ϕt(v1) ´°° °2dt ≥ k∇Jµ(u0)k2.
Hence u0 has the desired properties.
Case 2. There exists t ∈ [0, 1] such that °°°ϕt(v1) − v2
° ° ° > ². Then let t1 = sup n t ≥ t :°°°ϕt(v1) − v2 ° ° °> µ o . We choose t0 ∈ [t1, 1] with ° ° °∇Jµ ³ ϕt0(v 1) ´° ° °= min t1≤t≤1 ° ° °∇Jµ ³ ϕt(v 1) ´° ° °
and put u0 = ϕt0(v1) . Then
µ ≤ Z 1 t1 ° ° ° ° ° ∂ ∂tϕ t(v 1) ° ° ° ° °dt ≤ Z 1 t1 ° ° °∇Jµ ³ ϕt(v1) ´°° °dt and λ ≥ Jµ ³ ϕt1(v 1) ´ − Jµ(v2) = Z 1 t1 ° ° °∇Jµ ³ ϕt(v 1) ´° ° °2dt ≥ k∇Jµ(u0)k Z 1 t1 ° ° °∇Jµ ³ ϕt(v1) ´°° °dt.
We conclude that k∇Jµ(u0)k ≤ λ². Thus, u0 has the desired properties. Corollary 4.8 For each µ ∈ (0, µ0) there exist sequences nu(i)
n o ⊂ H1 0(A) such that (i) distH ³ u(i) n , Ni(µ, L) ´ → 0; (ii) Jµ ³ u(i) n ´ → θi,µ < αµ+ α∞; (iii) J0
µ(u(i)n ) = o(1) strongly in H−1(A) .
We now begin to show the proof of Theorem 1.2: It follows from Corollary 4.8 that for each µ ∈ (0, µ0) and i ∈ {1, 2, 3, 4} we can find a sequencenu(i)
n
o
⊂ H1
Hsu [14], there exist a subsequencenu(i)
n
o
and u(i)0 ∈ H1
0(A) such that uin → ui0 strongly in H1
0 (A) . Moreover, u (i)
0 is a nontrivial solution of equation (Eµ) . By
Lemma 4.5, ui
0 ∈ Ni(µ, L) and so {ui0}i∈{1,2,3,4} are different. Since θi,µ < 2α∞,
by Lemma 2.8 {ui
0}i∈{1,2,3,4} are 2–nodal solutions of equation (Eµ) . Thus add
the result of Theorem 3.8, we can conclude that equation (Eµ) has at least
two positive solutions and four 2–nodal solutions.
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