Existence and multiplicity of positive solutions for a class of
nonlinear boundary value problems
Tsung-fang Wu1
Department of Applied Mathematics, National University of Kaohsiung, Kaohsiung 811, Taiwan
Abstract
In this paper, we study a class of nonlinear boundary value problems in RN+. By
means of minimax method and the Lusternik-Schnirelman category, the criteria of the existence, multiplicity and nonexistence of positive solutions are established. Keywords: Ljusternik-Schnirelmann category; Positive solutions; Nonlinear boundary value problems.
1. Introduction
Let Ω be a domain in RN, N ≥ 2, with smooth boundary ∂Ω and consider the following nonlinear boundary value problem:
−∆u = g (x, u) in Ω, (1.1)
∂u
∂n + f (x, u) = 0 in ∂Ω, (1.2)
where ∂n∂ is the outer unit normal derivative, g : Ω × R → R is a Carath´edory function and f : ∂Ω × R → R is a continuous function.
Equations of the type (1.1) arise in many and diverse contexts like differential geometry (e.g., in the scalar curvature problem and the Yamabe problem) [27], nonlinear elasticity [19], non-Newtonian fluid mechanics [20], glaciology [34], math-ematical biology [5], and elsewhere. As a result, questions concerning the solvability of problem (1.1) have received great attention, particularly after the famous work of Brezis and Nirenberg [13]. Among the vast number of results recorded in the litera-ture so far, the case which has been studied extensively concerns the class of positive or non-negative solutions under a variety of the nonlinear term g(x, u). However, an exhaustive review of the existing bibliography is beyond our present scope and the interested reader should consult the survey in [2], as well as the references cited therein.
Email address: [email protected] (Tsung-fang Wu )
In recent years, problem (1.1), (1.2) have become rather an active area of research; see for example [6, 7, 14, 16, 17, 18, 22, 25, 26, 35, 37, 40] and references therein. The existence of positive solutions of the problem in bounded domains is strongly depen-dent on a priori estimates of the solutions [29], so fewer results are known for N ≥ 2. On the other hand, many papers deal with the existence of positive solutions of the problems in unbounded domains. For example, in [16, 18], the authors considered the existence and nonexistence of positive solutions of the problem (1.1) and (1.2) in upper half-space of RN with g (x, u) = −u + |u|p−2u and f (x, u) = − |u|q−2u. They
gave the exact form of the solution when p = N −22N , q = 2(N −1)N −2 with N ≥ 3 in [18], and proved the existence of positive solutions for p ≥ N −22N , q ≥ 2(N −1)N −2 and the nonexis-tence of positive solutions for some cases of p and q in [16]. The nonexisnonexis-tence results of [16] in some sense can be regarded as an extension of the results in [26] where Hu considered the problem with g (x, u) = −u + a |u|p−2u and f (x, u) = − |u|q−2u in the exterior of a ball in the upper half-space of RN. In [40], the author
con-sider the multiplicity of positive solutions of the problem (1.1) and (1.2) in upper half-space of RN with g (x, u) = −u + |u|p−2u and f (x, u) = b (x) |u|q−2u, with 1 < q < 2 < p < 2N
N −2 and b is a sign-changing continuous function. In [17] the
authors proved that the number of sign-changing solutions strongly depends on the spatial dimension. For the existence and multiplicity of positive solutions by varia-tional methods, see [6, 7, 14, 22, 25, 35, 37].
In this paper, we consider the existence and multiplicity of positive solutions for the following nonlinear boundary value problem:
−∆u + u = gλ(x) |u|p−2u in RN+, ∂u ∂n + f (x) |u| q−2 u = 0 in ∂RN +, (Eλ) where 1 < q < min {2∗, p} , 2∗ = 2(N −1)N −2 if N ≥ 3, 2∗ = ∞ if N = 2 , 2 < p < 2∗ 2∗ = N −22N if N ≥ 3, 2∗ = ∞ if N = 2 , RN + = (x 0, x N) ∈ RN −1× R | xN > 0
is an upper half space in RN, the parameter λ ∈ R and gλ(x) = 1 + λa (x) . We
assume that the functions f and a satisfy the following conditions: (D1) f ∈ C ∂RN
+ \ {0} and there exists a positive number rf > 1 such that
0 ≤ f (x) ≤bc exp (−rf|x|) for somebc > 0 and for all x ∈ ∂R
N +;
(D2) a ∈ CRN +
and there exists a positive number ra < {rf, q} such that
a (x) ≥ c0exp (−ra|x|) for some c0 > 0 and for all x ∈ RN+
and
The following theorem is our main result.
Theorem 1.1. Suppose that the functions f and a satisfy the conditions (D1) and (D2) . Then there exists a positive number λ∗ such that equation (Eλ) has at least
three positive solutions for λ ∈ (0, λ∗) , and at least one positive solution for λ ∈
{0} ∪ [λ∗, ∞).
In the following sections, we proceed to prove Theorem 1.1. We use the variational methods to find positive solutions of equation (Eλ) . Associated with the equation
(Eλ) , we consider the energy functional Jλ in H1 RN+
Jλ(u) = 1 2kuk 2 H1 + 1 q Z ∂RN + f |u|qdσ −1 p Z RN+ gλ|u| p dx,
where dσ is the measure on the boundary and kukH1 =
R
RN+
|∇u|2+ u2dx1/2 is
the standard norm in H1 RN+ . It is well known that the solutions of equation (Eλ)
are the critical points of the energy functional Jλ in H1 RN+(see Rabinowitz [36]).
This paper is organized as follows. In section 2, we give some notations and preliminaries. In section 3, we establish the existence of a positive solution for λ = 0. In section 4, we establish the existence of a positive solution for λ > 0. In sections 5, 6, we prove Theorem 1.1.
2. Notations and Preliminaries
First, we define the Palais–Smale (simply (PS)–) sequences, (PS)–values, and (PS)–conditions in H1
RN+ for Jλ as follows.
Definition 2.1. (i) For β ∈ R, a sequence {un} is a (PS)β–sequence in H1 RN+
for Jλ if Jλ(un) = β + o(1) and Jλ0(un) = o(1) strongly in H−1 RN+ as n → ∞.
(ii) Jλ satisfies the (PS)β–condition in H1 RN+ if every (PS)β–sequence in H1 RN+
for Jλ contains a convergent subsequence.
As the energy functional Jλ is not bounded below on H1 RN+ , it is useful to
consider the functional on the Nehari manifold Nλ =u ∈ H1 RN+ \ {0} | hJ
0
λ(u) , ui = 0 .
Thus, u ∈ Nλ if and only if
kuk2H1 + Z ∂RN + f |u|qdσ − Z RN+ gλ|u| p dx = 0.
Lemma 2.2. The energy functional Jλ is coercive and bounded below on Nλ. Proof. If u ∈ Nλ, then Jλ(u) = p − 2 2p kuk 2 H1 + p − q pq Z ∂RN + f |u|qdσ ≥ p − 2 2p kuk 2 H1. (2.1)
Thus, Jλ is coercive and bounded below on Nλ.
Define ψλ(u) = kuk2H1 + Z ∂RN + f |u|qdσ − Z RN+ gλ|u|pdx. Then for u ∈ Nλ, hψ0λ(u) , ui = 2 kuk2H1 + q Z ∂RN + f |u|qdσ − p Z RN+ gλ|u| p dx = (2 − p) kuk2H1 + (q − p) Z ∂RN + f |u|qdσ < 0. Furthermore, we have the following result.
Lemma 2.3. Suppose that u0 is a local minimizer for Jλ on Nλ. Then Jλ0 (u0) = 0
in H−1 RN . Furthermore, if u0 is a non-trivial nonnegative function in Ω, then
u0 is a positive solution of equation (Eλ) .
Proof. By a similar argument to that in the proof of Brown and Zhang [15, Theorem 2.3] (or see Binding, Dr´abek and Huang [8]), we have Jλ0 (u0) = 0 in
H−1(Ω) , this implies that u0 is a weak solution of equation (Eλ) . Now, if u0 is a
non-trivial nonnegative function in Ω, then by the maximum principle, u0 is positive
in Ω.
To get a better understanding of the Nehari manifold, we consider the function mu : R+→ R defined by mu(t) = t2−qkuk2H1− t p−q Z RN+ gλ|u|pdx for t > 0.
Clearly, tu ∈ Nλ if and only if mu(t) +
R
∂RN + f |u|
q
dσ = 0 and mu btλ(u) = 0, where
b tλ(u) = kuk2H1 R RN+ gλ|u| p dx !p−21 > 0. Moreover, m0u(t) = t1−q " (2 − q) kuk2H1 − (p − q)t p−2 Z RN+ gλ|u|pdx # .
Thus, if q ≥ 2, then
m0u(t) < 0 for all t > 0,
which implies that mu is strictly decreasing on (0, ∞) with limt→0+mu(t) = ∞ and limt→∞mu(t) = −∞, and if q < 2, then mu has a unique critical point at
t =etλ(u) <btλ(u) , where
etλ(u) = (2 − q) kuk2H1 (p − q)R RN+ gλ|u| p dx !p−21 > 0 such that mu etλ(u) = kukqH1 p − 2 p − q 2 − q p − q 2−qp−2 kukp H1 R RN+ gλ|u| p dx !2−qp−2 > 0,
which implies that muis strictly increasing on (0,etλ(u)) and muis strictly decreasing
on (etλ(u) , ∞) with limt→∞mu(t) = −∞. Therefore, we can conclude that for each
u ∈ H1 RN+ \ {0} there exist 0 < etλ(u) < btλ(u) such that mu(t) > 0 for all
t ∈ 0,btλ(u) and mu is strictly decreasing on [etλ(u) , ∞) with mu(t) < 0 for all
t ∈ btλ(u) , ∞ . Moreover, we have the following lemma.
Lemma 2.4. Suppose that λ ≥ 0. Then for each u ∈ H1
RN+ \ {0} we have the
following.
(i) There is a unique tλ(u) ≥btλ(u) such that tλ(u) u ∈ Nλ. Furthermore,
Jλ(tλ(u) u) = sup t≥0
Jλ(tu) = sup t≥btλ(u)
Jλ(tu) . (2.2)
(ii) tλ(u) is a continuous function for u ∈ H1 RN+ \ {0} .
(iii) tλ(u) = kuk1 H1 tλ u kukH1 . (iv) Nλ = n u ∈ H1 RN+ \ {0} | 1 kukH1tλ u kukH1 = 1o. Proof. Fix u ∈ H1 RN+ \ {0} . (i) Let hu(t) = Jλ(tu) = t2 2 kuk 2 H1+ tq q Z ∂RN + f |u|qdσ − t p p Z RN+ gλ|u| p dx. Then h0u(t) = t kuk2H1 + tq−1 Z ∂RN + f |u|qdσ − tp−1 Z RN+ gλ|u| p dx = tq−1 mu(t) + Z ∂RN + f |u|qdσ ! .
Thus, by R ∂RN + f |u| q dσ ≥ 0, the equation mu(t) + R ∂RN + f |u| q dσ = 0 has a unique solution tλ(u) ≥btλ(u) , which implies that h0u(tλ(u)) = 0 and tλ(u) u ∈ Nλ.
More-over, hu is strictly increasing on (0, tλ(u)) and strictly decreasing on (tλ(u) , ∞).
Therefore, (2.2) holds.
(ii) By the uniqueness of tλ(u) and the extremal property of tλ(u) , we have tλ(u)
is a continuous function for u ∈ H1 RN+ \ {0} . (iii) Let v = kuku
H1. Then, by part (i), there is a unique tλ(v) > 0 such that tλ(v) v ∈ Nλ or tλ u kukH1 u
kukH1 ∈ Nλ. Thus, by the uniqueness of tλ(v) , we can
conclude that tλ(u) = kuk1H1tλ
u kukH1
. (iv) For u ∈ Nλ. By parts (i), (iii), tλ
u kukH1
u
kukH1 ∈ Nλ. Since u ∈ Nλ, we have
tλ u kukH1 1
kukH1 = 1, which implies that
Nλ ⊂ u ∈ H1 RN+ | 1 kukH1 tλ u kukH1 = 1 . Conversely, let u ∈ H1 RN+ such that 1 kukH1tλ u kukH1
= 1. Then, by part (iii) ,
tλ u kukH1 u kukH1 ∈ Nλ. Thus, Nλ = u ∈ H1 RN+ \ {0} | 1 kukH1 tλ u kukH1 = 1 . This completes the proof.
Now we consider the following elliptic problems: −∆u + u = |u|p−2 u in RN+, ∂u ∂n = 0 on ∂R N +. (E∞) and −∆u + u = |u|p−2 u in RN, lim|x|→∞u = 0. ( eE∞) Associated with the equations (E∞) and Ee∞
, we consider the energy functionals J∞ in H1 RN+ and Je∞ in H1 RN J∞(u) = 1 2 Z RN+ |∇u|2+ u2dx − 1 p Z RN+ |u|pdx and e J∞(u) = 1 2 Z RN |∇u|2+ u2dx −1 p Z RN |u|pdx,
respectively. Consider the minimizing problems: inf
u∈N∞J ∞
(u) = α∞ and inf
u∈ eN∞ e J∞(u) =αe∞, where N∞ =u ∈ H1 RN+ \ {0} | (J∞)0(u) , u = 0 and e N∞= u ∈ H1 RN \ {0} | e J∞ 0 (u) , u = 0 . It is known that equations (E∞) and Ee∞
have unique positive radial solutions w (x) and w (x) , respectively such that Je ∞(w) = α∞ and eJ∞(w) =e αe∞ (see [21, 28, 32, 33]). Without loess of generality, we may assume that
w (0) = max x∈∂RN + w (x) = max x∈RN + w (x) and e w (0) = max x∈∂RN + e w (x) = max x∈RNw (x) .e
Thus, we observe that solution w (x) can construct solutionw (x) of equatione
e E∞
by reflection with respect to ∂RN
+. Then αe
∞ = eJ∞(
e
w) = 2J∞(w) = 2α∞ (or see [21, p. 889]). For λ ≥ 0, similarly as in [10, 11, 30, 31], we have the following results. Proposition 2.5. Let {un} be a (PS)β–sequence in H1 RN+
for Jλ. Then there
exist a subsequence {un} , m,m ∈ N ∪ {0} , sequences {xe
i
n} ⊂ ∂RN+ and {xe
j
n} ⊂ RN,
function v0 ∈ H1 RN+ , 0 6= wi ∈ H1 RN+ , for 1 ≤ i ≤ m and 0 6=we
j ∈ H1
RN , for 1 ≤ j ≤m such thate
(i) |xin| → ∞ and |xi n− xkn| → ∞ as n → ∞, for 1 ≤ i 6= k ≤ m; (ii) |xej n| → ∞ and |ex j n−xe k n| → ∞ as n → ∞, for 1 ≤ j 6= k ≤m;e (iii) |xi n−ex j n| → ∞ as n → ∞, for 1 ≤ i ≤ m and 1 ≤ j ≤m;e (iv) −∆v0+ v0 = gλ(x) |v0|p−2v0 in RN+, ∂v0 ∂n + f (x) |v0| q−2 v0 = 0 on ∂RN+; (v) −∆wi+ wi = |wi|p−2wi in RN+, ∂wi ∂n = 0 on ∂R N +; (vi) −∆wej + e wj = | e wj|p−2 e wj in RN;
(vii) un = v0+ m P i=1 wi(· − xin) + e m P j=1e wj(· −xejn) + o(1) strongly in H1 RN ; (viii) kunk2H1( RN+) = kv0k 2 H1( RN+) + m P i=1 kwik2 H1( RN+) + e m P j=1 kwejk2 H1(RN)+ o(1); (viiii) Jλ(un) = Jλ(v0) + m P i=1 J∞(wi) + e m P j=1 e J∞(wej) + o(1); In addition, if un ≥ 0, then v0 ≥ 0, wi ≥ 0 and we
j ≥ 0 for each 1 ≤ i ≤ m and
1 ≤ j ≤m.e
Proof. Since {un} is a a (PS)β–sequence in H1 RN+ for Jλ, by Lemma 2.2, there
is a subsequence {un} and v0 in H1 RN+ such that
un* v0 weakly in H1 RN+
and in Lp RN+ (2.3)
and v0 is a solution of equation (Eλ) . Let ubn = un− v0. Then, by (2.3) , b
un * 0 weakly in H1 RN+
and in Lp RN+ .
Suppose that bun6→ 0 strongly in H1 RN+ (otherwise, the result is automatic). By
a similar argument to that in the proof of [11, Lemma 3.1], there exist δ > 0 and xn = (yn, zn) ∈ RN −1× R such that |xn| → ∞ and
Z BN(1)+x n |bun| 2 dx > δ,
where BN(1) = x ∈ RN | |x| < 1 . Moreover, we may assume that one of the following two cases occurs:
(a) {zn} is bounded;
(b) {zn} is unbounded.
Case (a) : Without loss of generality, we may assume that zn = 0. Set x1n = xn =
(yn, 0) . Then |x1n| → ∞ andbun(x + x
1
n) ∈ H1 RN+ for all n. Adopting the approach
employed in the proof of [10, Proposition II.1] (or see [22, Proposition 2.1]), there exists w1 ∈ H1 RN+ \ {0} such that b un x + x1n * w1 weakly in H1 RN+ and −∆w1+ w1 = |w1|p−2w1 in RN +, ∂wi ∂n = 0 on ∂R N +. Case (b) : Set xe1 n = xn = (yn, zn) . Then |xe 1 n| → ∞ and bun(x +ex 1 n) ∈ H1 RN for
all n. Again, using a similar procedure to that in the proof of [10, Proposition II.1] (or see [11, Lemma 3.1]), there exists we1 ∈ H1
RN \ {0} such that b
un x + x1n * w
1 weakly in H1
and
−∆we1+we1 = |we1|p−2we1 in RN.
Following the same lines of the proof in [10, Proposition II.1] (or see [11, Lemma 3.1]), we repeat the argument above, each iteration will likewise give rise to two cases and the procedure will terminate after some finite steps; the procedure will also lead us to conclude that (i) − (viiii) hold.
For λ ≥ 0, we define
αλ = inf u∈Nλ
Jλ(u) .
Then, by Proposition 2.5, we have the following compactness result. Corollary 2.6. Suppose that {un} is a (PS)β–sequence in H1 RN+
for Jλ with
0 < β < α∞+ min {α∞, αλ} and β 6= α∞. Then there exists a subsequence {un} and
a non-zero v0 in H1 RN+ such that un → v0 strongly in H1 RN+ and Jλ(v0) = β.
Furthermore, v0 is a non-zero solution of equation (Eλ) .
Proof. Sinceαe∞= 2α∞ and
β < α∞+ min {α∞, αλ} ≤ 2α∞,
we have m = 0, which implies thate kunk2H1( RN+) = kv0k 2 H1( RN+) + m X i=1 wi 2 H1( RN+) + o (1) and β = Jλ(v0) + m X i=1 J∞(wi).
Since 0 < β < α∞ + min {α∞, αλ} and β 6= α∞, by the uniqueness of positive
solutions of equation (E∞) , we conclude that m = 0. Thus, un → v0 strongly
in H1 RN+ and Jλ(v0) = β > 0, which implies that v0 is a non-zero solution of
equation (Eλ) .
3. Existence of positive solutions for λ = 0
Let w (x) be a positive radial solution of equation (E∞) such that J∞(w) = α∞. Then, by Gidas, Ni and Nirenberg [24] and Kwong [28], for any ε > 0, there exist positive numbers Aε and B0 such that
Aεexp (− (1 + ε) |x|) ≤ w (x) ≤ B0exp (− |x|) for all x ∈ RN+. (3.1)
Let y ∈ S =x0 ∈ RN −1 | |x0| = 1 and let z
0 = (δ0, 0, . . . , 0) ∈ RN −1, where
0 < δ0 =
minrf, q,p2 − 1
2 minrf, q,p2 + 1
Clearly,
1 − δ0 ≤ |y − z0| ≤ 1 + δ0 for all y ∈ S. (3.2)
Define
wy,l(x) = w (x − l (y, 0)) for l ≥ 0 and y ∈ S (3.3)
and
wz0,l(x) = w (x − l (z0, 0)) for l ≥ 0.
Clearly, wy,l and wz0,l are also least energy positive solutions of equation (E ∞) for
all l ≥ 0. Moreover, by Lemma 2.4 for each u ∈ H1
RN+ \ {0} there is a unique
t0(u) ≥bt0(u) such that t0(u) u ∈ N0. Then we have the following results.
Lemma 3.1. For each s0 ∈ (0, 1) there exist l (s0) > 0 and σ (s0) > 1 such that for
any l > l (s0) , we have bt p−2 0 (swy,l+ (1 − s) wz0,l) > σ (s0) sp−2+ (1 − s)p−2
for all y ∈ S and for all s ∈ (0, 1) with min {s, 1 − s} ≥ s0.
Proof. Since b tp−20 (swy,l+ (1 − s) wz0,l) = kswy,l+ (1 − s) wz0,lk 2 H1 R RN+ |swy,l+ (1 − s) wz0,l| p dx = s 2kwk2 H1 + (1 − s) 2 kwk2H1 + 2s (1 − s) hwy,l, wz0,li R RN+ |swy−z0,l+ (1 − s) w| p dx (3.4)
for all s ∈ [0.1] and for all y ∈ S. Moreover, by (3.1) , the triangle inequality and 1 − δ0 ≤ |y − z0| ≤ 1 + δ0 for all y ∈ S, (3.5)
we have hwy,l, wz0,li = Z RN+ wp−1wy−z0,ldx ≤ Bp0 Z |x|<(1+δ0)l exp (− (|x| + |x − l (z0− y, 0)|)) dx +B0p Z |x|≥(1+δ0)l exp (− (|x| + |x − l (z0− y, 0)|)) dx = Bp0lN Z |x|<(1+δ0) exp (−l (|x| + |x − (z0− y, 0)|)) dx +B0p Z |x|≥(1+δ0)l exp (− |x|) exp (− |x − l (z0− y, 0)|) dx ≤ Bp0lN Z |x|<(1+δ0) exp (−l (|x| + |x − (z0− y, 0)|)) dx +B0pexp (− (1 + δ0) l) Z |x|≥(1+δ0)l exp (− (|x − l (z0− y, 0)|)) dx ≤ Bp0lN Z |x|<(1+δ0) exp (−l (|y − z0|)) dx +B0pexp (− (1 + δ0) l) Z RN exp (− |x|) dx ≤ Bp0lNexp (− (1 − δ0) l) Z |x|<(1+δ0) 1dx + d0B p 0exp (− (1 + δ0) l) ≤ C0B0pl Nexp (−l (1 − δ
0)) for all l ≥ 1 and for all y ∈ S,
which implies that
lim
l→∞hwy,l, wz0,li = 0 uniformly in y ∈ S. (3.6)
By (3.1) , (3.5) and Br´ezis-Lieb lemma [12], for any s ∈ [0.1] we have lim l→∞ Z RN+ |swy−z0,l+ (1 − s) w| p − |sw y−z0,l| p dx = Z RN+ |(1 − s) w|pdx uniformly in y ∈ S. (3.7) Thus, by (3.4) , (3.6) and (3.7) , for any s ∈ [0.1]
lim l→∞bt p−2 0 (swy,l+ (1 − s) wz0,l) = s2+ (1 − s)2 kwk2H1 (sp+ (1 − s)p)R RN+ |w|pdx = s 2+ (1 − s)2 sp+ (1 − s)p uniformly in y ∈ S. (3.8)
Since s2 + (1 − s)2 sp−2+ (1 − s)p−2 sp+ (1 − s)p = 1 + s2(1 − s)p−2 + (1 − s)2sp−2 sp + (1 − s)p ≥ 1 + s 2 0(1 − s0)p−2+ (1 − s0)2sp−20 sp0+ (1 − s0) p (3.9)
for all s ∈ (0, 1) with min {s, 1 − s} > s0, by (3.8) and (3.9) , there exist l (s0) > 0
and σ (s0) > 1 such that for any l > l (s0) , we have
bt
p−2
0 (swy,l+ (1 − s) wz0,l) >
σ (s0)
sp−2+ (1 − s)p−2
for all y ∈ S and for all s ∈ (0, 1) with min {s, 1 − s} ≥ s0. This completes the
proof.
Proposition 3.2. There exists l1 > 0 such that for any l ≥ l1
sup
t≥0
J0(t [swy,l+ (1 − s) wz0,l]) < 2α ∞
for all y ∈ S,
where J0 = Jλ for λ = 0. Furthermore, there is a unique t0(swy,l+ (1 − s) wz0,l) > 0 such that
t0(swy,l+ (1 − s) wz0,l) [swy,l+ (1 − s) wz0,l] ∈ N0.
Proof. When s = 0 or 1, by a similar argument to that in the proof of Wu [40, Proposition 2], there exists et1 > 0 such that
max sup t≥0 J0(twy,l), sup t≥0 J0(twz0,l) ≤ α∞+et1C0 q exp (− min {rf, q} l) (3.10) for all y ∈ S, this implies that there exists el1 > 0 such that for any l > el1,
max sup t≥0 J0(twy,l), sup t≥0 J0(twz0,l) ≤ 3 2α ∞ for all y ∈ S. (3.11) Therefore, by J0 ∈ C2 H1 RN+ , R and (3.11) , there exist positive constants s0, el
such that for any l > el, sup
t≥0
J0(t [swy,l+ (1 − s) wz0,l]) < 2α ∞
for all y ∈ S and for all min {s, 1 − s} ≤ s0. In the following we always assume that min {s, 1 − s} ≥ s0. Since J0(t [swy,l+ (1 − s) wz0,l]) = t 2 2 s 2kwk2 H1 + (1 − s) 2 kwk2H1 + 2s (1 − s) hwy,l, wz0,li +t q q Z ∂RN + f (swy,l+ (1 − s) wz0,l) q dσ −t p p Z RN+ (swy,l+ (1 − s) wz0,l) p dx (3.13) ≤ t 2 2 s 2 + 2s (1 − s) + (1 − s)2 kwk2H1 +bct q2q−1 q (s q+ (1 − s)q ) Z ∂RN+ wqdσ − t p p max {s p, (1 − s)p}Z RN+ wpdx ≤ t2kwk2H1 + bct q2q−1 q Z ∂RN + wqdσ − t p p2p Z RN+ wpdx
for all 0 ≤ s ≤ 1 and y ∈ S. Then there exists t1 > 0 such that for any t ≥ t1,
J0(t [swy,l+ (1 − s) wz0,l]) < 2α ∞
for all 0 ≤ s ≤ 1 and for all y ∈ S. (3.14) Moreover, by Lemma 2.4 (i) and Lemma 3.1,
sup
t≥0
J0(t [swy,l+ (1 − s) wz0,l]) = sup t≥t2
J0(t [swy,l+ (1 − s) wz0,l]) (3.15)
for all y ∈ S, where t2 =
σ(s0) sp−2+(1−s)p−2
1/(p−2)
and σ (s0) > 1 is as in Lemma 3.1.
Thus, by (3.14) and (3.15) , we only need to show that there exists l1 ≥ el such that
for any l > l1, sup t2≤t≤t1 J0(t [swy,l+ (1 − s) wz0,l]) < 2α ∞ for all y ∈ S. (3.16) By lemma 2.1 in Bahri and Li [9], there exists Cp > 0, such that for any nonnegative
real numbers a, b,
Then, by (3.13) , (3.17) and Lemma 3.1, J0(t [swy,l+ (1 − s) wz0,l]) ≤ 2α∞+ s (1 − s)t2− tpsp−2− tp(1 − s)p−2 Z RN+ wy,lp−1wz0,ldx +t q 1 q Z ∂RN + f (swy,l+ (1 − s) wz0,l) q dσ + t p 1Cp p Z RN+ wy,lp/2wzp/2 0,ldx ≤ 2α∞− C02[σ (s0) − 1] Z RN+ wp−1y,l wz0,ldx +t q 1 q Z ∂RN + f (swy,l+ (1 − s) wz0,l) q dσ + t p 1Cp p Z RN+ wy,lp/2wzp/2 0,ldx (3.18)
for all y ∈ S, where we have used the result Z RN+ wp−1y,l wz0,ldx = hwy,l, wz0,li = Z RN+ wy,lwp−1z0,ldx. We first estimate R RN+ w p−1 y,l wz0,ldx. Setting C0 = min x∈BN +(eN,12) wp−1(x) > 0, where BN + eN,12 = x ∈ RN+ | |x − eN| ≤ 12 and eN = (0, . . . , 0, 1) ∈ RN. Then,
by (3.1) and (3.2) , for any ε > 0, Z RN+ wp−1y,l wz0,ldx = Z RN+ wp−1(x) w (x − l (z0− y, 0)) dx ≥ C0Aε Z BN +(eN,12) exp (− (1 + ε) |x − l (z0− y, 0)|) dx ≥ C0Aε Z BN +(eN,12) exp (− (1 + ε) |x| − l (1 + ε) |y − z0|) dx ≥ C0Aεexp (−l (1 + ε) |y − z0|) ≥ C0Aεexp (−l (1 + ε) (1 + δ0)) . (3.19)
From (3.2) we have Z RN+ wy,lp/2wzp/2 0,ldx ≤ B0p Z |x|<(1+δ0)l exp−p 2(|x| + |x − l (z0− y, 0)|) dx +B0p Z |x|≥(1+δ0)l exp−p 2(|x| + |x − l (z0− y, 0)|) dx ≤ B0plN Z |x|<(1+δ0) exp−p 2l (|x| + |x − (z0− y, 0)|) dx +B0pexp −(1 + δ0) pl 2 Z |x|≥(1+δ0)l exp−p 2(|x − l (y − z0, 0)|) dx ≤ B0plN Z |x|<(1+δ0) exp −pl 2 |y − z0| dx +bc0B0pexp −pl 2 |y − z0| ≤ C0B0pl Nexp −pl 2 |y − z0| ≤ C0B0plNexp − minnrf, q, p 2 o (1 − δ0) l for l ≥ 1. (3.20) By (D2) , we also have Z ∂RN + f (swy,l+ (1 − s) wz0,l) q dσ ≤ 2q−1 Z ∂RN+ f wy,lq dσ + Z ∂RN+ f wqz 0,ldσ ! ≤ Cb0B0plNexp (− min {rf, q} l |y − z0|) ≤ Cb0B0pl N exp − minnrf, q, p 2 o (1 − δ0) l for l ≥ 1. (3.21) Since 1 + δ0 = 1 + minrf, q,p2 − 1 2 minrf, q,p2 + 1 < min n rf, q, p 2 o 1 − minrf, q, p 2 − 1 2 minrf, q,p2 + 1 ! = minnrf, q, p 2 o (1 − δ0) ,
we may take 0 < ε << 1 such that
(1 + ε) (1 + δ0) < min n rf, q, p 2 o (1 − δ0) .
Then, by (3.18) − (3.21), there exists l1 ≥ max
n el, 1
o
such that (3.16) holds. There-fore, by (3.12) and (3.14) − (3.16), we can conclude that for any l > l1,
sup
t≥0
J0(t [swy,l+ (1 − s) wz0,l]) < 2α ∞
for all 0 ≤ s ≤ 1 and for all y ∈ S. Moreover, by Lemma 2.4, there is a unique t0(swy,l+ (1 − s) wz0,l) > 0 such that
t0(swy,l+ (1 − s) wz0,l) [swy,l+ (1 − s) wz0,l] ∈ N0. This completes the proof.
Theorem 3.3. We have α0 = inf u∈N0 J0(u) = inf u∈N∞J ∞ (u) = α∞.
Furthermore, equation (E0) does not admit any solution u0 such that J0(u0) = α0.
Proof. By Lemma 2.4, there is a unique t0(wy,l) > 0 such that t0(wy,l) wy,l ∈ N0
for all y ∈ S, that is
kt0(wy,l) wy,lk2H1 + Z ∂RN + f |t0(wy,l) wy,l|qdσ = Z RN+ |t0(wy,l) wy,l|pdx. Since Z ∂RN + f |wy,l| q dσ → 0 as l → ∞, and kwy,lk 2 H1 = Z RN+ |wy,l| p dx = 2p p − 2α ∞
for all l ≥ 0 and for all y ∈ S, we have t0(wy,l) → 1 as l → ∞. Thus,
lim
l→∞J0(t0(wy,l) wy,l) = liml→∞J ∞
(t0(wy,l) wy,l) = α∞ for all y ∈ S,
which implies that
α0 ≤ inf u∈N∞J
∞
(u) = α∞.
Let u ∈ N0. Then, by Lemma 2.4, J0(u) = supt≥0J0(tu) . Moreover, there is a
unique t∞> 0 such that t∞u ∈ N∞. Thus,
J0(u) ≥ J0(t∞u) ≥ J∞(t∞u) ≥ α∞ and so α0 ≥ α∞. Therefore, α0 = inf u∈N0 J0(u) = inf u∈N∞J ∞ (u) = α∞.
Next, we will show that equation (E0) does not admit any solution u0 such that
J0(u0) = α0. Suppose the contrary. Then we can assume that there exists u0 ∈ N0
such that J0(u0) = α0. Then, by Lemma 2.4 (i) , J0(u0) = supt≥0J0(tu0) . Moreover,
there is a unique t∞(u0) > 0 such that t∞(u0) u0 ∈ N∞. Thus,
α∞ = inf u∈N0 J0(u) = J0(u0) ≥ J0(t∞(u0) u0) ≥ α∞+ [t ∞(u 0)]q q Z ∂RN + f |u0|qdσ,
which implies that R
∂RN +
f |u0|qdσ = 0 and so u0 ≡ 0 in x ∈ ∂RN+ | f (x) 6= 0 ,
form the condition (D1) . Therefore, α∞ = inf
u∈N∞J ∞
(u) = J∞(tu0u0) .
Since |t∞(u0) u0| ∈ N∞ and J∞(|t∞(u0) u0|) = J∞(t∞(u0) u0) = α∞, by Willem
[39, Theorem 4.3] and the maximum principle, we can assume that tu0u0is a positive solution of (E∞) , this contradicts
u0 ≡ 0 in x ∈ ∂RN+ | f (x) 6= 0 .
This completes the proof.
By Theorem 3.3, equation (E0) does not admit any solution u0 such that J0(u0) =
α0 and α0 = inf u∈N0 J0(u) = inf u∈N∞J ∞(u) = α∞.
Furthermore, we have the following result.
Lemma 3.4. Suppose that {un} is a minimizing sequence for J0 in N0. Then
Z
∂RN +
f |un|qdσ = o (1) .
Furthermore, {un} is a (PS)α∞–sequence for J∞ in H1 RN+ .
Proof. For each n, there is a unique tn > 0 such that tnun ∈ N∞, that is
t2nkunk2H1 = t p n Z RN+ |un|pdx.
Then, by Lemma 2.4 (i) ,
J0(un) ≥ J0(tnun) = J∞(tnun) + tq n q Z ∂RN + f |un|qdσ ≥ α∞+t q n q Z ∂RN + f |un| q dσ.
Since J0(un) = α∞+ o (1) from Theorem 3.3, we have tq n q Z ∂RN + f |un|qdσ = o (1) .
We will show that there exists c0 > 0 such that tn > c0 for all n. Suppose the
contrary. Then we may assume tn → 0 as n → ∞. Since J0(un) = α∞+ o (1) , by
Lemma 2.2, we have kunk is uniformly bounded and so ktnunkH1 → 0 or J∞(tnun) →
0, and this contradicts J∞(tnun) ≥ α∞ > 0. Thus,
Z ∂RN + f |un| q dσ = o (1) ,
which implies that
kunk2H1 = Z RN+ |un|pdx + o (1) and J∞(un) = α∞+ o (1) .
Moreover, by Wang and Wu [38, Lemma 7], we have {un} is a (PS)α∞–sequence for J∞ in H1
RN+ .
For u ∈ H1 RN+, we define the center mass function from N0 to the unit ball of
RN −1 m(u) = 1 kukpLp(RN +) Z RN+ x0 |x0||u(x 0 , xN)| p dx0dxN.
Clearly, m is continuos from N0 to BN −1(0, 1) and |m (u)| < 1. Let
θ0 = inf {J0(u) | u ∈ N0, u ≥ 0, m(u) = 0} .
Then we have the following result.
Lemma 3.5. There exists ξ0 > 0 such that α∞< ξ0 ≤ θ0.
Proof. Suppose the contrary. Then there exists a sequence {un} ⊂ N0 and
m(un) = 0 for each n, such that J0(u) = α∞ + o (1) . By Lemma 3.4, we have
{un} is a (PS)α∞–sequence in H1 RN+ for J∞. By the concentration–compactness principle (see Lions [30, 31] or del Pino and Flores [22, proof of proposition 2.1]) and the fact that α∞ = αe∞/2 > 0, there exist a subsequence {un} , a sequence
{(x0
n, 0)} ⊂ ∂RN+, and a positive solution w ∈ H1 RN+ of equation (E∞) such that
Now we will show that |(x0n, 0)| → ∞ as n → ∞. Suppose the contrary. Then we may assume that {(x0n, 0)} is bounded and (x0n, 0) → (x00, 0) for some (x00, 0) ∈ ∂RN+. Thus, by (3.22) , Z ∂RN + f |un| q dσ = Z ∂RN + f (x) |w (x − (x0n, 0))|qdσ + o (1) = Z ∂RN + f (x + (x00, 0)) |w (x)|qdσ + o (1) , this contradicts the result of Lemma 3.4: R∂RN
+ f |un| q
dσ = o (1) . Hence we may assume that x0n
|x0
n| → e as n → ∞, where e ∈ S. Then, by (3.22) and the Lebesgue dominated convergence theorem, we have
0 = m(un) = kunk −p Lp(RN +) Z RN+ x0 |x0||un(x 0 , xN)| p dx0dxN = kwk−pLp(RN +) Z RN+ x0 + x0n |x0 + x0 n| |w(x0, xN)| p dx0dxN + o(1) = e + o(1) as n → ∞,
which is a contradiction. Therefore, there exists ξ0 > 0 such that α∞ < ξ0 ≤ θ0.
By Lemma 2.4 and Proposition 3.2, for each y ∈ S and l > l1there exists t0(wy,l) >
0 such that t0(wy,l) wy,l ∈ N0. Moreover, we have the following result.
Lemma 3.6. There exists l0 ≥ l1 such that for any l ≥ l0,
(i) α∞ < J0(t0(wy,l) wy,l) < ξ0 for all y ∈ S;
(ii) hm(t0(wy,l) wy,l), yi > 0, for all y ∈ S.
Proof. (i) Follow from (4.4) − (4.6) and Theorem 3.3. (ii) For x0 ∈ RN −1 with x0 + ly 6= 0, we have
( x 0+ ly |x0+ ly|, ly) = |x 0+ ly| − 1 |x0+ ly|(x 0+ ly, x0) ≥ |x0+ ly| − |x0| ≥ l |y| − 2 |x0| = l − 2 |x0| . Then hm(t0(wy,l) wy,l), yi = 1 l kwy,lk p Lp(RN +) Z RN+ x0 |x0|, ly |wy,l|pdx0dxN = 1 l kwkpLp(RN +) Z RN+ x0+ ly |x0+ ly|, ly |w|pdx0dxN ≥ 1 l kwkpLp(RN +) (l Z RN+ |w|pdx0dxN − 2 Z RN+ |x0| |w|pdx0dxN) = 1 − 2c0 l ,
where c0 = kwk −p Lp(RN +) R RN+ |x 0| |w|p
dx0dxN. Thus, there exists l0 ≥ l1 such that
hm(t0(wy,l) wy,l), yi ≥ 1 −
2c0
l > 0 for all l ≥ l0. This completes the proof.
In the following, we will use Bahri-Li’s minimax argument [9]. Let B =u ∈ H1 RN+ \ {0} | u ≥ 0 and kukH1 = 1 . We define
I0(u) = sup t≥0
J0(tu) : B → R.
Then, by Lemma 2.4 (iii), for each u ∈ H1 RN+ \ {0} there exists t0(u) = 1 kukH1 t0 u kukH1 > 0 such that t0(u) u ∈ N0 and
I0(u) = J0(t0(u) u) = J0 t0 u kukH1 u kukH1 (3.23) Next, we define a map h0 from S to B by
h0(y) = w (x − l (y, 0)) kw (x − l (y, 0))kH1 = wy,l kt0(wy,l) wy,lkH1 ,
where y ∈ S. Then, by (3.10) and (3.23) , for l > l0 sufficiently large, we have
I0(h0(y)) = J0(t0(wy,l) wy,l) < θ0 for all y ∈ S.
We define another map h∗ from BN −1(0, 1) to N 0 by
h∗(sy + (1 − s) z0) =
swy,l+ (1 − s) wz0,l kswy,l+ (1 − s) wz0,lkH1
where 0 ≤ s ≤ 1 and y ∈ S. It is clear that h∗|S = h0. It follows from Proposition
3.2 and (3.23) that
I0(h0(y)) = J0(t0(swy,l+ (1 − s) wz0,l) [swy,l+ (1 − s) wz0,l]) < 2α ∞
(3.24) for all y ∈ S. We next define a min-max value. Let
β0 = inf γ∈Γz∈BmaxN −1(0,1)I0(γ(z)), (3.25) where Γ =nγ ∈ CBN −1(0, 1), B | γ| S = h0 o . (3.26)
Lemma 3.7. We have
α∞< ξ0 ≤ θ0 ≤ β0 < 2α∞.
Proof. By Lemmas 3.5, 3.6, (3.24) and (3.23), we only need to show θ0 ≤ β0.
For any γ ∈ Γ, there exists t0(γ(z)) > 0 such that t0(γ(z)) γ(z) ∈ N0 and
t0(γ(z)) γ(z) = t0(wz,l) wz,l for all z ∈ S.
Consider the homotopy H(s, z) : [0, 1] × BN −1(0, 1) → R defined by
H(s, z) = (1 − s)m(t0(γ(z)) γ(z)) + sI(z),
where I denotes the identity map. Note that m(t0(γ(z)) γ(z)) = m(t0(wz,l) wz,l) for
all z ∈ S. By Lemma 3.6 (ii), H(s, z) 6= 0 for z ∈ S and s ∈ [0, 1]. Therefore, deg(m (t0(γ) γ) , BN −1(0, 1) , 0) = deg(I, BN −1(0, 1) , 0) = 1.
There exists z0 ∈ BN −1(0, 1) such that
m(t0(γ(z0)) γ(z0)) = 0.
Hence, for each γ ∈ Γ0, we have
θ0 = inf {J0(u) | u ∈ N0, u ≥ 0, m(u) = 0}
≤ I0(γ(z0))
≤ max
z∈BN −1(0,1)I0(γ(z)). This shows that θ0 ≤ β0.
Now, we are going to assert that the equation (E0) has a positive solution.
Theorem 3.8. Equation (E0) has a positive solution eu0 such that J0(eu0) = β0. Proof. By Lemma 3.7 and the minimax principle (see Ambrosetti and Rabinowitz [3]), there exists a sequence {un} ⊂ B such that
I0(un) = β0+ o (1) , kI0 0(un)kT∗ unB ≡ sup {I 0 0(un)φ | φ ∈ TunB, kφkH1 = 1} = o (1) as n → ∞, where α∞< β0 < 2α∞ and TunB =φ ∈ H 1 RN+ | hφ, uni = 0 . By an argument
similar to the proof of proposition 1.7 in Adachi and Tanaka [4], there exists t0(un) >
0 such that t0(un) un ∈ N0 and
J0(t0(un) un) = β0+ o (1) ,
J00(t0(un) un) = o (1) in H−1 RN+
as n → ∞.
Thus, by Corollary 2.6, Theorem 3.3 and the maximum principle, we can conclude that the equation (E0) has a positive solution ue0 such that J0(ue0) = β0.
4. Existence of positive solutions for λ > 0
By Lemma 2.4 for each u ∈ H1 RN+ \ {0} and λ > 0 there is a unique tλ(u) ≥
b
tλ(u) such that tλ(u) u ∈ Nλ. Let wy,l be as in (3.3) . Then we have the following
results.
Proposition 4.1. For each λ > 0, there exists bl1 = bl1(λ) > 0 such that for any
l ≥ bl1,
sup
t≥0
Jλ(twy,l) < α∞ for all y ∈ S.
Furthermore, there is a unique tλ(wy,l) > 0 such that tλ(wy,l) wy,l ∈ Nλ.
Proof. We have Jλ(twy,l) = t 2 2 kwy,lk 2 H1 + tq q Z ∂RN + f |wy,l| q dσ −t p p Z RN+ gλ|wy,l| p dx = t 2 2 kwk 2 H1 − tp p Z RN+ wpdx +t q q Z ∂RN + f wy,lq dσ − λt p p Z RN+ awpy,ldx (4.1) ≤ t 2 2 kwk 2 H1 +b ctq q Z ∂RN + wqdσ −t p p Z RN+ wpdx.
for all λ > 0. This implies that Jλ(twy,l) → −∞ as t → ∞ for all y ∈ S. Thus, there
exists t1 > 0 such that for any l ≥ 0,
Jλ(twy,l) < α∞ for all t ≥ t1 and for all y ∈ S. (4.2)
Moreover, Jλ(0) = 0 < α∞, Jλ ∈ C2 H1 RN , R and kwy,lk 2 H1 =
2p p−2α
∞ for all
l ≥ 0, this implies that there exists t2 > 0 such that for any l ≥ 0,
Jλ(twy,l) < α∞ for all 0 ≤ t ≤ t2 and for all y ∈ S. (4.3)
Moreover, by Brown and Zhang [15] and Willem [39], we know that J∞(tw) = t 2 2 kwk 2 H1 − tp p Z RN+ wpdx ≤ α∞ for all t > 0. (4.4) Thus, by (4.1) , Jλ(twy,l) ≤ α∞+ tq q Z ∂RN + f wy,lq dσ − λt p p Z RN+
awpy,ldx for all t > 0. (4.5)
By (4.2) and (4.3) , we only need to show that there exists bl1 > 0 such that for any
l > bl1,
sup
t2≤t≤t1
We set C0 = min x∈BN +(eN,12) wp(x) > 0, where BN + eN,12 = x ∈ RN+ | |x − eN| < 12 and eN = (0, . . . , 0, 1) ∈ RN. Then, by the conditions (D1) , Z RN awpy,ldx = Z RN a (x + l (y, 0)) wp(x) dx ≥ C0 Z BN +(eN,12) a (x + l (y, 0)) dx ≥ C0exp (−ral) .
Moreover, by (3.1) and the condition (D1) , Z ∂RN + f wy,lq dσ ≤ bcB0p Z ∂RN +
exp (−rf|x|) exp (−q |x − l (y, 0)|) dσ
≤ C0exp (− min {rf, q} l) (4.6)
Since ra < min {rf, q} and t2 ≤ t ≤ t1, we can find bl1 > 0 such that for any l > bl1,
tq q Z ∂RN + f wy,lq dσ < λt p p Z RN+
awy,lp dx for all y ∈ S and for all t ∈ [t2, t1] . (4.7)
Thus, by (4.2) − (4.5) and (4.7), we obtain that for any l > bl1
sup
t≥0
Jλ(twy,l) < α∞ for all y ∈ S.
Moreover, by Lemma 2.4, there is a unique tλ(wy,l) > 0 such that tλ(wy,l) wy,l ∈ Nλ.
This completes the proof.
Theorem 4.2. For each λ > 0, equation (Eλ) has a positive solution u0 such that
Jλ(u0) = αλ = inf u∈Nλ
Jλ(u) < α∞.
Proof. By analogy with the proof of Ni and Takagi [32], one can show that the Ekeland variational principle (see [23]), there exists a minimizing sequence {un} ⊂
Nλ such that
Jλ(un) = αλ+ o (1) and Jλ0 (un) = o (1) in H−1 RN+ .
Since infu∈NλJλ(u) < α
∞ from Proposition 4.1 (ii) , by Lemma 2.2 and Corollary
2.6 there exist a subsequence {un} and u0 ∈ Nλ is a nonzero solution of equation
(Eλ) such that
un→ u0 strongly in H1(RN+) and Jλ(u0) = αλ.
Since Jλ(u0) = Jλ(|u0|) and |u0| ∈ Nλ, by Lemma 2.3, we may assume that u0 is a
5. Existence of two positive solutions We need the following result.
Lemma 5.1. There exists d0 > 0 such that if u ∈ N0 and J0(u) ≤ α∞+ d0, then
Z RN+ x0 |x0| |∇u| 2 + u2 dx0dxN 6= 0.
Proof. Suppose the contrary. Then there exists sequence {un} ⊂ N0 such that
J0(u) = α∞+ o (1) and Z RN+ x0 |x0| |∇un| 2 + u2n dx0 dxN = 0.
Moreover, by Lemma 3.4, we have {un} is a (PS)α∞–sequence in H1 RN+ for J∞. By the concentration–compactness principle (see Lions [30, 31] or del Pino and Flores [22, proof of proposition 2.1]) and α∞ = αe∞/2, there exist a subsequence {un} , a
sequence {(x0n, 0)} ⊂ ∂RN+, and a positive solution w ∈ H1 RN+ of equation (E∞) such that
kun(x) − w (x − (x0n, 0))kH1 → 0 as n → ∞. (5.1) Now we will show that |(x0n, 0)| → ∞ as n → ∞. Suppose the contrary. Then we may assume that {(x0n, 0)} is bounded and (x0n, 0) → (x00, 0) for some (x00, 0) ∈ ∂RN
+. Thus, by (5.1) , Z ∂RN + f |un|qdσ = Z ∂RN + f (x) |w (x − (x0n, 0))|qdσ + o (1) = Z ∂RN + f (x + (x00, 0)) |w (x)|qdσ + o (1) ,
which contradicts the result of Lemma 3.4: R∂RN + f |un| q dσ = o (1) . Hence we may assume x0n |x0 n| → e as n → ∞, where e ∈ S = x 0 ∈ RN −1 | |x0| = 1 . Then, by the
Lebesgue dominated convergence theorem, we have 0 = Z RN+ x0 |x0| |∇un| 2 + u2n dx0dxN = Z RN+ x0+ x0n |x0+ x0 n| |∇w|2+ w2 dx + o (1) = 2p p − 2α ∞e + o (1) ,
which is a contradiction. This completes the proof.
For u ∈ Nλ, by Lemma 2.4, there is a unique t0(u) > 0 such that t0(u) u ∈ N0.
Let η0 = (1 + λ kak∞) 1
p−q. Then we have the following result.
Proof. Let u ∈ Nλ. Then we have kuk2H1 + Z ∂RN + f |u|qdσ = Z RN+ gλ|u| p dx. We distinguish two cases.
Case (A) : t0(u) < 1. Since θ0 > 1, we have
t0(u) < 1 < η0.
Case (B) : t0(u) ≥ 1. Since
[t0(u)] pZ RN+ |u|pdx = [t0(u)] 2 kuk2H1 + [t (u)] qZ ∂RN + f |u|qdσ ≤ [t0(u)]q kuk2H1 + Z ∂RN + f |u|qdσ ! = [t0(u)]q Z RN+ gλ|u|pdx ≤ [t0(u)] q (1 + λ kak∞) Z RN+ |u|pdx, we have t0(u) ≤ (1 + λ kak∞) 1 p−q = η 0.
This completes the proof.
By the proof of Proposition 4.1, there exist positive numbers tλ(wy,l) and bl1 such
that tλ(wy,l) wy,l ∈ Nλ and
Jλ(tλ(wy,l) wy,l) < α∞ for all l > bl1.
Then we have the following result.
Lemma 5.3. There exists a positive number λ0 such that for every λ ∈ (0, λ0) , we
have Z RN+ x0 |x0| |∇u| 2 + u2 dx0dxN 6= 0
for all u ∈ Nλ with Jλ(u) < α∞.
Proof. (i) Let u ∈ Nλ with Jλ(u) < α∞. Then, by Lemma 2.4 (i) , there exists
t0(u) > 0 such that t0(u) u ∈ N0. Moreover,
Jλ(u) = sup t≥0 Jλ(tu) ≥ Jλ(t0(u) u) = J0(t0(u) u) − λ [t0(u)] pZ RN+ a |u|pdx.
Thus, by Lemma 5.2 and the Sobolev inequality, J0(t0(u) u) ≤ Jλ(u) + λ [t0(u)]p
Z RN+ a |u|pdx < α∞+ λc0η0pkak∞kuk p H1 for some c0 > 0. (5.2) Moreover, by (2.1) , α∞> Jλ(u) ≥ p − 2 2p kuk 2 H1, which implies that
kukH1 <
2pα∞
p − 2 12
(5.3) for all u ∈ Nλ with Jλ(u) < α∞. Therefore, by (5.2) and (5.3) ,
J0(t0(u) u) < α∞+ λc0η0pkak∞
2pα∞
p − 2 p2
.
Let d0 > 0 be as in Lemma 5.1. Then there exists a positive number λ0 such that
for λ ∈ (0, λ0) ,
J0(t (u) u) < α∞+ d0. (5.4)
Since t0(u) u ∈ N0 and t0(u) > 0, by Lemma 5.1 and (5.4)
Z RN+ x0 |x0| |∇ (t0(u) u)| 2 + (t0(u) u) 2 dx0 dxN 6= 0,
which implies that
Z RN+ x0 |x0| |∇u| 2 + u2 dx0dxN 6= 0
for all u ∈ Nλ with Jλ(u) < α∞.
In the following, we use an idea of Adachi and Tanaka [4]. For c ∈ R+, we denote [Jλ ≤ c] = {u ∈ Nλ | u ≥ 0, Jλ(u) ≤ c} .
We then try to show for a sufficiently small σ > 0
cat ([Jλ ≤ α∞− σ]) ≥ 2. (5.5)
To prove (5.5) , we need some preliminaries. Recall the definition of Lusternik-Schnirelman category.
Definition 5.4. (i) For a topological space X, we say a non-empty, closed subset Y ⊂ X is contractible to a point in X if and only if there exists a continuous mapping
such that for some x0 ∈ X
ξ (0, x) = x for all x ∈ Y, and
ξ (1, x) = x0 for all x ∈ Y.
(ii) We define
cat (X) = min {k ∈ N | there exist closed subsets Y1, ..., Yk⊂ X such that
Yj is contractible to a point in X for all j and k
∪
j=1Yj = X
. When there do not exist finitely many closed subsets Y1, ..., Yk⊂ X such that Yj
is contractible to a point in X for all j and ∪k
j=1Yj = X, we say cat (X) = ∞.
We need the following two lemmas.
Lemma 5.5. Suppose that X is a Hilbert manifold and F ∈ C1(X, R) . Assume that there are c0 ∈ R and k ∈ N,
(i) F (x) satisfies the Palais–Smale condition for energy level c ≤ c0;
(ii) cat ({x ∈ X | F (x) ≤ c0}) ≥ k.
Then F (x) has at least k critical points in {x ∈ X; F (x) ≤ c0} .
Proof. See Ambrosetti [1, Theorem 2.3].
Let Sm−1 = {x ∈ Rm | |x| = 1} be a unit sphere in Rm for m ∈ N. Then we have
the following results.
Lemma 5.6. Let X be a topological space. Suppose that there are two continuous maps
Φ : Sm−1 → X, Ψ : X → Sm−1
such that Ψ ◦ Φ is homotopic to the identity map of Sm−1, that is, there exists a
continuous map ζ : [0, 1] × Sm−1 → Sm−1 such that
ζ (0, x) = (Ψ ◦ Φ) (x) for each x ∈ Sm−1, ζ (1, x) = x for each x ∈ Sm−1.
Then
cat (X) ≥ 2. Proof. See Adachi and Tanaka [4, Lemma 2.5].
For l > bl1, we may define a map Φλ,l : S(N −1)−1 → H1 RN+ by
Φλ,l(y) (x) = tλ(w (x − l (y, 0))) w (x − l (y, 0)) for y ∈ S(N −1)−1,
where tλ(w (x − l (y, 0))) w (x − l (y, 0)) is as in the proof of Proposition 4.1. Note
Lemma 5.7. There exists a sequence {σl} ⊂ R+ with σl → 0 as l → ∞ such that
Φλ,l S(N −1)−1 ⊂ [Jλ ≤ α∞− σl] .
Proof. By Proposition 4.1, for each l > bl1 we have
tλ(w (x − l (y, 0))) w (x − l (y, 0)) ∈ Nλ
and
sup
l>bl1
Jλ(tλ(w (x − l (y, 0))) w (x − l (y, 0))) < α∞ for all y ∈ S(N −1)−1.
Since S = S(N −1)−1 and Φ
λ,l S(N −1)−1 is compact,
Jλ(tλ(w (x − l (y, 0))) w (x − l (y, 0))) ≤ α∞− σl,
so that the conclusion holds. From Lemma 5.3, we define
Ψλ : [Jλ < α∞] → S(N −1)−1 by Ψλ(u) = R RN+ x0 |x0| |∇u| 2 + u2 dx0dx N R RN+ x0 |x0| |∇u| 2 + u2 dx0dx N .
Then we have the following results.
Lemma 5.8. Let λ0 > 0 be as in Lemma 5.3. Then for each λ ∈ (0, λ0) and there
exists bl0 ≥ bl1 such that for l > bl0, the map
Ψλ◦ Φλ,l : S(N −1)−1→ S(N −1)−1
is homotopic to the identity.
Proof. Let Σ = nu ∈ H1 RN+ \ {0} | R RN+ x0 |x0| |∇u| 2 + u2 dx0dxN 6= 0 o . We define Ψλ : Σ → S(N −1)−1 by Ψλ(u) = R RN+ x0 |x0| |∇u| 2 + u2 dx0dx N R RN+ x0 |x0| |∇u| 2 + u2 dx0dx N ,
an extension of Ψλ. Since w (x − l (y, 0)) ∈ Σ for all e ∈ S(N −1)−1and for l sufficiently
Ψλ(Φλ,l(y)) such that γ (s1) = Ψλ(wy,l) , γ (s2) = Ψλ(Φλ,l(y)) . By an argument
similar to that in Lemma 5.1, there exists a positive number bl0 ≥ bl1 such that for
l > bl0, w x − l (y, 0) 2 (1 − θ)
∈ Σ for all y ∈ S(N −1)−1 and θ ∈ [1/2, 1) .
We define ζl(θ, y) : [0, 1] × S(N −1)−1 → S(N −1)−1 by ζl(θ, y) = γ (2θ (s1 − s2) + s2) for θ ∈ [0, 1/2) ; Ψλ wx − 2(1−θ)l(y,0) for θ ∈ [1/2, 1) ; y for θ = 1.
Then ζl(0, y) = Ψλ(Φλ,l(y)) = Ψλ(Φλ,l(y)) and ζl(1, y) = y. First, we claim that
lim θ→1−ζl(θ, y) = y and lim θ→12− ζl(θ, y) = Ψλ(wy,l) . (a) lim θ→1−ζl(θ, y) = y : since Z RN+ x0 |x0| ∇ w x − l (y, 0) 2 (1 − θ) 2 + w x − l (y, 0) 2 (1 − θ) 2! dx0dxN = Z RN+ x0+ 2(1−θ)ly x 0+ ly 2(1−θ) |∇ [w (x)]|2+ [w (x)]2 dx0dxN = 2p p − 2 α∞y + o(1) as θ → 1−, then lim θ→1−ζl(θ, y) = y. (b) lim θ→12−
ζl(θ, y) = Ψλ(wy,l) : since Ψλ ∈ C Σ, S(N −1)−1 , we obtain lim θ→12−
ζl(θ, y) =
Ψλ(wy,l) .
Thus, ζl(θ, y) ∈ C [0, 1] × S(N −1)−1, S(N −1)−1 and
ζl(0, y) = Ψλ(Φλ,l(y)) for all y ∈ S(N −1)−1,
ζl(1, y) = y for all y ∈ S(N −1)−1,
provided l > bl0. This completes the proof.
Theorem 5.9. For each λ ∈ (0, λ0) , functional Jλ has at least two critical points
in [Jλ < α∞] . In particular, equation (Eλ) has two positive solutions u (1)
0 and u (2) 0
Proof. Applying Lemmas 5.6 and 5.8, we have for λ ∈ (0, λ0) ,
cat ([Jλ ≤ α∞− σl]) ≥ 2.
By Corollary 2.6 and Lemmas 5.5 and 5.7, Jλ(u) has at least two critical points in
[Jλ < α∞] . This implies, equation (Eλ) has two nontrivial nonnegative solutions u1
and u2 such that ui ∈ Nλ for i = 1, 2. Moreover, by the maximum principle, we
have ui > 0 in RN+.
6. Proof of Theorem 1.1
Given a positive real number r0 > max
n 2 p−2, q p−q o . Let Λ0 = min r0(p − 2) − 2 2 kak∞(r0+ 1) , r0(p − q) − q q kak∞(r0+ 1) , λ0 , where λ0 > 0 as in Lemma 5.3. Then we have the following results.
Lemma 6.1. We have 1 2(1 + λ kak∞) r0 − 1 p(1 + λ kak∞) r0+1−p − 2 2p > 0 and 1 q(1 + λ kak∞) r0 −1 p(1 + λ kak∞) r0+1−p − q pq > 0 for all λ ∈ (0, Λ0) . Proof. Let k (λ) = 1 2(1 + λ kak∞) r0 − 1 p(1 + λ kak∞) r0+1− p − 2 2p . Then k (0) = 0 and k0(λ) = r0 2 (1 + λ kak∞) r0−1kak ∞− r0 + 1 p (1 + λ kak∞) r0kak ∞ = kak∞(1 + λ kak∞)r0−1 r0 2 − r0+ 1 p (1 + λ kak∞) > 0 for all λ ∈ (0, Λ0) . This implies that k (λ) > 0 or
1 2(1 + λ kak∞) l0 − 1 p(1 + λ kak∞) r0+1− p − 2 2p > 0 for all λ ∈ (0, Λ0) . Similar to the argument we also have
1 q (1 + λ kak∞) r0 − 1 p(1 + λ kak∞) r0+1− p − q pq > 0 for all λ ∈ (0, Λ0) .
This completes the proof. We define
Iλ(u) = sup t≥0
Jλ(tu) : B → R.
Then we have the following result.
Lemma 6.2. For each λ ∈ (0, Λ0) and u ∈ B we have
(1 + λ kak∞)−r0
I0(u) ≤ Iλ(u) ≤ I0(u) .
Proof. Let u ∈ B. Then, by Lemmas 2.4 and 6.1, and (3.23) Iλ(u) = sup t≥0 Jλ(tu) ≥ Jλ(t0(u) u) ≥ 1 2 Z RN+ |∇t0(u) u|2 + (t0(u) u)2dx + 1 q Z ∂RN + f |t0(u) u|qdσ −(1 + λ kak∞) p Z RN+ |t0(u) u|pdx = 1 2 − 1 + λ kak∞ p Z RN+ |∇t0(u) u|2+ (t0(u) u)2dx + 1 q − 1 + λ kak∞ p Z ∂RN + f |t0(u) u| q dσ ≥ (1 + λ kak∞)−r0 J0(t0(u) u) = (1 + λ kak∞) −r0 I0(u) . Moreover,
Jλ(tu) ≤ J0(tu) ≤ I0(u) for all t > 0.
Then Iλ(u) ≤ I0(u) . This completes the proof.
We observe that if λ is sufficiently small, the minimax argument in Section 4 also works for Jλ. Let l > max
n l0, bl0
o
be very large and let βλ = inf
γ∈Γy∈BmaxN −1(0,1)Iλ(γ(y)).
where Γ is as in (3.26) . Then, by (3.25) and Lemma 6.2, for λ ∈ (0, Λ0), we have
(1 + λ kak∞)−r0
β0 ≤ βλ ≤ β0. (6.1)
Then we have the following result.
Theorem 6.3. There exists a positive number λ∗ ≤ Λ0 such that for λ ∈ (0, λ∗) ,
α∞< βλ < α∞+ αλ < 2α∞.
Furthermore, equation (Eλ) has a positive solution u (3)
0 such that Jλ
Proof. By Theorems 3.3 and 4.2, and Lemma 6.2, we also have that (1 + λ kak∞)−r0
α∞≤ αλ < α∞.
For any ε > 0 there exists a positive number λ1 ≤ Λ0 such that for λ ∈ 0, λ1 ,
α∞− ε < αλ < α∞.
Thus,
2α∞− ε < α∞+ αλ < 2α∞.
Applying (6.1) for any δ > 0 there exists a positive number λ2 ≤ Λ0 such that for
λ ∈ 0, λ2 ,
β0− δ < βλ ≤ β0.
Moreover, by Lemma 3.7,
α∞< β0 < 2α∞.
Fix a small 0 < ε < 2α∞− β0, choosing a δ > 0 such that for λ ∈ (0, λ∗) we get,
α∞< βλ < 2α∞− ε < α∞+ αλ < 2α∞,
where λ∗ = minλ1, λ2 . Similar to the argument in the proof of Theorem 3.8, we
can conclude that the equation (Eλ) has a positive solution u (3)
0 such that Jλ
u(3)0 = βλ. This completes the proof.
We can now complete the proof of Theorem 1.1: By Theorems 4.2 and 3.8, equa-tion (Eλ) has at least one positive solution for all λ ∈ [0, ∞). Moreover, by Theorems
5.9 and 6.3, there exists a positive number λ∗ such that for λ ∈ (0, λ∗) , equation
(Eλ) has three positive solutions u (1) 0 , u (2) 0 and u (3) 0 with 0 < Jλ u(i)0 < α∞ < Jλ u(3)0 for i = 1, 2. This completes the proof of Theorem 1.1.
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