On a class of Kirchhoff type problems

Arch. Math. 102 (2014), 127–139c© 2014 Springer Basel

0003-889X/14/020127-13

published online February 14, 2014DOI 10.1007/s00013-014-0618-4 Archiv der Mathematik

On a class of Kirchhoff type problems

Yisheng Huang and Zeng Liu

Abstract. In this paper we consider the following Kirchhoff type problem:

(K)

⎛⎝1 + λ

∫

R3

(|∇u|2 + V (y)u2dy)⎞⎠ [−Δu + V (x)u] = |u|p−2u, in R

3,

where p ∈ (2, 6), λ > 0 is a parameter, and V (x) is a given potential.Some existence and nonexistence results are obtained by using variationalmethods. Also, the “energy doubling” property of nodal solutions of (K)is discussed in this paper.

Mathematics Subject Classification (2010). 35J20, 35J60.

Keywords. Kirchhoff type problem, Ground state, Nodal solution,Nehari manifold.

1. Introduction. Recently, the existence of positive solutions, ground states,and semiclassical states of the following nonlinear Kirchhoff type problem⎛

⎝a + λb

∫

R3

(|∇u|2 + V (y)u2dy)⎞⎠ [−Δu + V (x)u] = f(x, u), in R

3 (1.1)

have been extensively studied by many authors, see [1,2,6,11,19] and the ref-erences therein, where, a and b are positive constants, λ > 0 is a parameter,V (x) ∈ C(R3, R+), and f ∈ C(R3 ×R, R). Problem (1.1) has a strong physicalmeaning. Indeed, replacing R

3 by a bounded domain Ω ⊂ R3 and setting λ = 1

and V (x) = 0, (1.1) becomes the following Kirchhoff Dirichlet problem:⎧⎨⎩

−(a + b

∫Ω

|∇u|2dy)Δu = f(x, u), in Ω,

u = 0, on ∂Ω,(1.2)

Supported by Natural Science Foundation of China (11071180, 11171247) and CPRI Projectof Jiangsu Province (CXZZ12 0802).

128 Y. Huang and Z. Liu Arch. Math.

which is related to the stationary analogue of the equation

utt −⎛⎝a + b

∫

Ω

|∇u|2dy

⎞⎠Δu = f(x, u), in Ω, (1.3)

that was first proposed by Kirchhoff in [7] describing the transversal oscilla-tions of a stretched string. For more details on the physical and mathematicalbackground of the problem (1.3), we refer the readers to the papers [2,7,8,13]and the references therein.

There are many works on the existence of nontrivial solutions to (1.2) byusing variational methods recently, see, for example, [5,13–16,20]. In particu-lar, by using the method of invariant sets of descent flow, the existence of nodalsolutions for (1.2) was obtained by several authors in [14,15,20]. However, asfar as we know, there seems to be no result on the existence of nodal solutionsfor problem (1.1). Therefore the main goal of this paper is to study the exis-tence of nodal solutions of (1.1). For simplicity of notations, we consider theproblem (1.1) with a = b = 1 and f(x, u) = |u|p−2u with p ∈ (2, 6):⎛

⎝1 + λ

∫

R3

(|∇u|2 + V (y)u2dy)⎞⎠ [−Δu + V (x)u] = |u|p−2u, in R

3. (1.4)

Moreover, to avoid involving too much details of checking the compactness,we simply assume that(V1) V (x) ∈ C(R3, R+) such that HV ⊂ H1(R3) and the embedding HV ↪→

Lq(R3)(q ∈ (2, 6)) is compact,where the weighted Sobolev space is defined as

HV := HV (R3) =

⎧⎨⎩u ∈ D1,2(R3) :

∫

R3

V (x)u2dx < ∞⎫⎬⎭

with the norm ‖u‖2V =

∫R3(|∇u|2 + V (x)u2)dx.

We remark that the function satisfying condition (V1) exists; for example,we may choose a function V ∈ C(R3, R+) such that V0 := inf{V (x) : x ∈R

3} > 0 and for every M > 0,

meas({x ∈ R3 : V (x) ≤ M}) < ∞,

then HV ⊂ H1(R3) and condition (V1) holds (cf. [3]).We say that a function u ∈ HV is a (weak) solution of (1.4) if u satisfies

that for all v ∈ HV ,

(1 + λ‖u‖2V )

∫

R3

(∇u∇v + V (x)uv)dx −∫

R3

|u|p−2uvdx = 0.

It is clear that (weak) solutions of (1.4) correspond to critical points of the C2

functional Eλ : HV → R defined as

Eλ(u) =12‖u‖2

V +λ

4‖u‖4

V − 1p

∫

R3

|u|pdx (1.5)

Vol. 102 (2014) On a class of Kirchhoff type problems 129

with

〈E′λ(u), v〉 = (1 + λ‖u‖2

V )∫

R3

(∇u∇v + V (x)uv)dx −∫

R3

|u|p−2uvdx.

If u ∈ HV is a solution of (1.4) with u± �= 0, then u is called a nodal solution of(1.4), where u+(x) = max{u(x), 0} and u−(x) = min{u(x), 0}. If u is a nodalsolution of (1.4) with Eλ(u) = inf{Eλ(v) : v is the nodal solution of (1.4)},then u is called the least energy nodal solution.

Let Sq be the Sobolev constant of the embedding HV ↪→ Lq(R3):

Sq = infu∈HV \{0}

‖u‖V

‖u‖q, q ∈ [2, 6]. (1.6)

For p ∈ (2, 4], we define

λ(p) =(p − 2

4 − p

)(4 − p

2Spp

) 2p−2

if p ∈ (2, 4); λ(4) = S−44 if p = 4. (1.7)

The first result in this paper is the following.

Theorem 1.1. Assume (V1) holds.(1) Let p ∈ (2, 4] and λ(p) be in (1.7).

(i) If λ > λ(p), then (1.4) has only the zero solution.(ii) There exists λ0 > 0 such that for each λ ∈ (0, λ0), (1.4) has two pos-

itive solutions provided p ∈ (2, 4) and one positive solution providedp = 4.

(2) Let p ∈ (4, 6) and λ > 0. Then problem (1.4) has a least energy nodalsolution with exactly two nodal domains.

Remark 1.2. Even though we assume that a = 1 and b = 1 in (1.4), our meth-ods for Theorem 1.1 are still valid for a > 0 and b > 0. Except the existence ofnodal solutions obtained in Theorem 1.1, (i) and (ii) of Theorem 1.1 respec-tively complement the corresponding results in [12] and [11], where Li and Ye[12] showed that (1.1) has only the zero solution provided f(x, u) = |u|p−2uwith p ∈ (2, 3], a > 1, b > 0 and λ ≥ 1/(4b(a − 1)S6

3), and only one positivesolution obtained in [11] for λ > 0 small.

Remark 1.3. In fact, if we only assume that HV ↪→ H1(RN ), then the con-clusion of Theorem 1.1 (i) also holds (cf. Section 2.1 and Remark 2.1 below).Moreover, we obtian the following estimate for λ(p) when p ∈ (2, 4):

λ(p) ≤ λ(p) := 8p−p

p−2 (p − 2)3(

4 − p

S2

) 2(4−p)p−2

S−44 . (1.8)

To find nodal solutions for the problem (1.4), motivated by [1,4,9], weconsider the following minimization problem:

m := infu∈N ±

Eλ(u),

where the set N ± is defined as

N ± := {u ∈ HV : u± �= 0, 〈E′λ(u), u+〉 = 0, 〈E′

λ(u), u−〉 = 0}. (1.9)


We observe that there is a typical approach (see e.g., [4,9]) to obtain thenodal solutions of (1.4) with λ = 0, i.e., the following semilinear Schrodingerequation:

− Δu + V (x)u = |u|p−2u, in R3. (1.10)

However, the arguments used in [4,9] cannot be applied directly to obtain thenodal solution of (1.4) since, in [4,9], the following decompositions are crucial:for u ∈ HV ,

〈E′0(u), u±〉 = 〈E′

0(u±), u±〉, E0(u) = E0(u+) + E0(u−), (1.11)

where E0 is the corresponding functional of (1.10) defined as

E0(u) =12

∫

R3

(|∇u|2 + V (x)u2)dx − 1p

∫

R3

|u|pdx.

While, in our case, namely λ > 0, if u ∈ HV with u± �= 0, then the functionalEλ corresponding to (1.4) satisfies

Eλ(u) = Eλ(u+) + Eλ(u−) +12‖u+‖2

V ‖u−‖2V ,

〈E′λ(u), u+〉 = 〈E′

λ(u+), u+〉 + ‖u+‖2V ‖u−‖2

V > 〈E′λ(u+), u+〉,

〈E′λ(u), u−〉 = 〈E′

λ(u−), u−〉 + ‖u+‖2V ‖u−‖2

V > 〈E′λ(u−), u−〉,

which mean that (1.11) does not hold anymore. Therefore, to achieve our goal,it is important to study the properties of N ± (see Lemma 2.2). However, in thisprocess, a careful analysis is necessary due to the interaction of the nonlocalterm.

Note that it was shown in [1] that (1.4) has a positive ground state solutionu ∈ N such that Eλ(u) = m > 0 if p ∈ (4, 6), where

N = {u ∈ HV \{0} : 〈E′λ(u), u〉 = 0}, m = inf

u∈NEλ(u). (1.12)

The other aim of this paper is to study the “energy doubling” property of thenodal solutions of (1.4), that is, the energy of each nodal solution of (1.4) islarger than two times the energy of the ground state solution. The “energydoubling” property was first given in [18] for the nodal solutions of (1.10)with V (x) = 1. Here, we give a general result on this aspect for the nonlinearKirchhoff type problem (1.4).

Theorem 1.4. Assume that p ∈ (4, 6), λ > 0, and (V1) is satisfied, then thereexists ε > 0 such that Eλ(v) ≥ 2m + ε for all v ∈ N ±. In particular, for eachnodal solution u of (1.4) in HV , we have

Eλ(u) ≥ 2m + ε.

We remark that Theorem 1.4 seems to be the first result on the “energydoubling” property of nodal solutions of Kirchhoff type problems no matter inR

3 or in bounded domains.Throughout this paper, we will use the following notations: Lp(R3) is the

usual Lebesgue space with the norm ‖u‖p =( ∫

R3 |u|pdx)1/p, H1(R3) is the


usual Sobolev space with the associated norm ‖u‖ =( ∫

R3(|∇u|2 + |u|2)dx)1/2,

C,C1, C2, . . . denote the positive (possibly different) constants.

2. Proof of Theorems 1.1 and 1.4.

2.1. The case p ∈ (2, 4].

Proof. Case (i). Arguing by contradiction, there exists u ∈ HV \{0} such that〈E′

λ(u), v〉 = 0 for all v ∈ HV . Then we have

〈E′λ(u), u〉 = ‖u‖2

V + λ‖u‖4V − ‖u‖p

p = 0. (2.1)

For t > 0, set

g(t) = E(tu) =t2

2‖u‖2

V +λt4

4‖u‖4

V − tp

p‖u‖p

p,

then we have

g′(t) = t‖u‖2V + λt3‖u‖4

V − tp−1‖u‖pp.

By (2.1), we know that g′(1) = 0. On the other hand, for all t > 0, g′(t) = tγ(t),where γ(t) = ‖u‖2

V + λt2‖u‖4V − tp−2‖u‖p

p. If p = 4, then

γ(t) ≥ ‖u‖2V + (λ − S−4

4 )t2‖u‖4V > 0 (2.2)

for all t > 0 since λ ≥ λ(4) = S−44 . If p ∈ (2, 4), let γ′(t) = 0, that is,

2λt‖u‖4V − (p − 2)tp−3‖u‖p

p = 0,

we obtain

t4−p =

(p − 2)‖u‖pp

2λ‖u‖4V

.

Note that p ∈ (2, 4), it is easy to see that

mint>0

γ(t) = γ(t) = ‖u‖2V + λt

2‖u‖4V − t

p−2‖u‖pp

= ‖u‖2V

(1 − 4 − p

2

(p − 22λ

) p−24−p

( ‖u‖p

‖u‖V

) 2p4−p

)

≥ ‖u‖2V

(1 − 4 − p

2

(p − 22λ

) p−24−p

S−2p4−pp

)

> 0 (2.3)

since λ > λ(p) = (p − 2)(4 − p)(4−p)/(p−2)(2Spp)2/(2−p). Thus, (2.2) and (2.3)

imply that g′(t) > 0 for all t > 0, which contradicts g′(1) = 0.Case (ii). First of all, there exist λ0 > 0 and v0 ∈ HV \{0} such that Eλ(v0) < 0for all λ ∈ (0, λ0) and p ∈ (2, 4] since E0(tv0) → −∞ as t → ∞. Moreover,Eλ(u) = 1

2‖u‖2V + o(1) if ‖u‖V small. Then Eλ satisfies the mountain pass

geometry(cf. [17]). Secondly, we see that if p ∈ (2, 4), then

Eλ(u) ≥ 12‖u‖2

V +λ

4‖u‖4

V − 1Sp

p‖u‖p

V → ∞

as ‖u‖V → ∞, namely, Eλ is coercive on HV . Finally, by standard arguments,we can show that Eλ satisfies the (PS) condition since (V1) holds. Therefore,


on one hand, it follows from the Ekeland variational principle that (1.4) hasa solution u ∈ HV \{0} such that Eλ(u) = inf{Eλ(v) : v ∈ HV \{0}} <0 provided λ ∈ (0, λ0) and p ∈ (2, 4), on the other hand, by applying themountain pass theorem (cf. [17]), we see that (1.4) has a mountain pass solutionprovided λ ∈ (0, λ0) and p ∈ (2, 4]. To obtain the positive solution, we mayconsider the following functional

E+λ (u) =

12‖u‖2

V +λ

4‖u‖4

V − 1p

∫

R3

|u+|pdx,

and repeat the above steps to obtain that (1.4) has two nontrivial nonnegativesolutions provided λ ∈ (0, λ0) and p ∈ (2, 4) and one nontrivial nonnegativesolution provided λ ∈ (0, λ0) and p = 4. Then it follows from the maximumprinciple that these nonnegative solutions are positive. �

Remark 2.1. Clearly, if we only assume that HV ↪→ H1(RN ) (that is, HV ↪→↪→Lp(RN ) may not hold), then the above proof of Theorem 1.1 (i) still hold.Moreover, from the Young inequality and (1.6), we see that for u ∈ HV \ {0}and p ∈ (2, 4)

‖u‖2p ≤ ‖u‖2α

2 ‖u‖2(1−α)4 ≤ min

ε>0

{(αε1/α

S22

+1 − α

S24ε1/(1−α)

)‖u‖2V

}

= 2αα(1 − α)1−αS−2α2 S

−2(1−α)4 ‖u‖2

V ,

which implies that S2p ≥ 2−1α−α(1 − α)−(1−α)S2α

2 S2(1−α)4 , where α = (4 −

p)/p ∈ (0, 1). Combining this and (1.7), we obtain (1.8).

2.2. The case p ∈ (4, 6). In this case, without loss of generality, we may assumethat λ = 1 and denote E := E1. We consider the following minimizationproblem

m = infu∈N ±

E(u),

where E and N ± are respectively given in (1.5) and (1.9) with λ = 1. For eachu ∈ N ±, we denote

h+(u) := E(u+) +14‖u−‖2

V ‖u+‖2V , h−(u) := E(u−) +

14‖u−‖2

V ‖u+‖2V ,

G+(u) := 〈E′(u), u+〉 =〈E′(u+), u+〉 + ‖u−‖2V ‖u+‖2

V = 0,

G−(u) := 〈E′(u), u−〉 =〈E′(u−), u−〉 + ‖u−‖2V ‖u+‖2

V = 0,(2.4)

then

h+(u) = h+(u) − 14G+(u) =

14‖u+‖2

V +(

14

− 1p

)‖u+‖p

p, (2.5)

h−(u) = h−(u) − 14G−(u) =

14‖u−‖2

V +(

14

− 1p

)‖u−‖p

p, (2.6)

E(u) = h+(u) + h−(u) = E(u) − 〈E′(u), u〉/4. (2.7)


The properties on the set N ± are given by the following lemma. We remarkthat N defined by (1.12) has similar properties (cf. [1]).

Lemma 2.2. (i) For each u ∈ HV with u± �= 0, there exists a unique(tu, su) ∈ R × R with tu, su > 0 such that tuu+ + suu− ∈ N ± with

E(tuu+ + suu−) = max{E(tu+ + su−) : t, s ≥ 0},

and Hgu(tu, su) is a negative definite matrix, where Hgu(t, s) is theHessian matrix of gu(t, s) := E(tu+ + su−).

(ii) There exists C > 0 such that ‖u±‖pp ≥ C > 0 for all u ∈ N ±. Further-

more, m > 0.(iii) If there exists u ∈ N ± such that E(u) = m, then u is a critical point of

(1.5) with exactly two nodal domains.

Proof. For (i), let u ∈ HV with u± �= 0, by the definition of gu(t, s), we knowthat gu ∈ C2(R2, R) and

gu(t, s) = E(tu+ + su−) = E(tu+) + E(su−) +t2s2

2‖u+‖2

V ‖u−‖2V ,

which implies that gu(t, s) > 0 for t, s > 0 small and gu(t, s) → −∞ as|(t, s)| → ∞. Noting that gu(t, s) = gu(|t|, |s|), hence there exist nonnegativereal numbers tu and su such that

gu(tu, su) = E(tuu+ + suu−) = max{E(tu+ + su−) : t, s ≥ 0},

so that, tuu+ + suu− ∈ N ±. Next, we claim that tu > 0 and su > 0. If not,without loss of generality, we may assume that su = 0, then we have

gu(tu, 0) ≥ gu(tu, s) ≥ gu(tu, 0) +s2

2‖u−‖2

V +s4

4‖u−‖4

V − |s|pp

‖u−‖pp

> gu(tu, 0)

if s > 0 is small enough since p > 4, this is a contradiction. Therefore, both tuand su are positive.

By a simple computation, we know that

Hgu(tu, su)

=[2t2u‖u+‖4

V − (p − 2)tp−2u ‖u+‖p

p 2tusu‖u+‖2V ‖u−‖2

V

2tusu‖u+‖2V ‖u−‖2

V 2s2u‖u−‖4

V − (p − 2)sp−2u ‖u−‖p

p

].

Noting that p ∈ (4, 6) and G±(tuu+ + suu−) = 0, therefore we have that

2t2u‖u+‖4V − (p − 2)tp−2

u ‖u+‖pp < 0, 2s2

u‖u−‖4V − (p − 2)sp−2

u ‖u−‖pp < 0

hold and detHgu(tu, su) > 0, that is, Hgu(tu, su) is a negative definite matrix.Thus we only need to prove the uniqueness of (tu, su). Assume that there

exists another (tu, su) with tu > 0, su > 0 such that tuu+ + suu− ∈ N ±.Similarly as the above, we see that the Hessian matrix Hgu(tu, su) is alsonegative definite. Therefore, (tu, su) is a local maximum point of gu. Recallthat (tu, su) is a global maximum point, we have gu(tu, su) ≥ gu(tu, su) > 0.Let

v+ = tuu+, v− = suu−, tu =tutu

, su =su

su, (2.8)


then v = v++v− = tuu++suu− ∈ N ± and tuv++suv− = tuu++suu− ∈ N ±,moreover,

gv(tu, su) = gu(tu, su) ≥ gu(tu, su) = gv(1, 1). (2.9)

Without loss of generality, we may assume that tu ≥ su > 0. It follows from(2.4) that G+(v) = 0 and G+(tuv+ + suv−) = 0, therefore, we obtain that

(1 − (tu)p−2)‖v+‖2V + ((tu)2 − (tu)p−2)‖v‖2

V ‖v+‖2V ≥ 0,

which means that tu ≤ 1 since p ∈ (4, 6), so that 0 < su ≤ tu ≤ 1. On theother hand, from (2.5), (2.6), and (2.7), we get that

gv(1, 1) = E(v) = h+(v) + h−(v)

= h+(tuv+ + suv−) + h−(tuv+ + suv−)

+1 − (tu)2

4‖v+‖2

V +(

14

− 1p

)(1 − (tu)p)‖v+‖p

p

+1 − (su)2

4‖v−‖2

V +(

14

− 1p

)(1 − (su)p)‖v−‖p

p

= gv(tu, su) +1 − (tu)2

4‖v+‖2

V +(

14

− 1p

)(1 − (tu)p)‖v+‖p

p

+1 − (su)2

4‖v−‖2

V +(

14

− 1p

)(1 − (su)p)‖v−‖p

p,

which, together with (2.9) and 0 < su ≤ tu ≤ 1, implies that su = tu = 1. Itfollows from (2.8) that tu = tu and su = su, therefore, (tu, su) is unique.

For (ii), arguing by contradiction, there exists {un} ⊂ N ± such that‖u+

n ‖pp → 0 or ‖u−

n ‖pp → 0 as n → ∞. Without loss of generality, we may assume

that ‖u+n ‖p

p → 0 as n → ∞. It follows from G+(un) = 0 that ‖u+n ‖2

V → 0 asn → ∞. On the other hand, again using the fact of G+(un) = 0, we have

‖u+n ‖2

V + ‖u+n ‖4

V + ‖u+n ‖2

V ‖u−n ‖2

V = ‖u+n ‖p

p ≤ C‖u+n ‖p

V ,

which implies that there exists C > 0 such that ‖u+n ‖2

V ≥ C, a contradiction.Therefore, there exists C > 0 such that ‖u±‖p

p > C for all u ∈ N ±. m > 0follows from (2.5), (2.6), and (2.7).

For (iii), the proof of this claim is similar to the one of [9, Lemma 2.5], seealso in [4,10]. For the sake of completeness, we give a proof in the Appendix.

�

Proof of Theorem 1.1 (2). Let {un} ⊂ N ± be a sequence such that E(un) →m > 0 as n → ∞. Up to a subsequence, we may assume that E(un) ≤ 2m forall n. Thus we have

2m ≥ E(un) = E(un) − 14〈E(un), un〉 =

14‖∇un‖2

V +(

14

− 1p

)∫

R3

|un|pdx,


which implies that∫

R3

(|∇un|2 + V (x)u2n)dx ≤ 8m,

∫

R3

|un|pdx ≤ 8p

p − 4m.

Moreover, by Lemma 2.2, there exists C > 0 such that ‖u±n ‖V ≥ C, ‖u±

n ‖pp ≥

C. Hence there exists u ∈ HV such that un ⇀ u and u±n ⇀ u± in HV as

n → ∞. By condition (V1), u±n → u± in Lq(R3) for q ∈ (2, 6) as n → ∞,

then u± �= 0. It follows from Lemma 2.2 that there exist t±∗ > 0 such thatt+∗ u+ + t−∗ u− ∈ N ±. Without loss of generality, we may assume that t+∗ ≥t−∗ > 0. Noting that {un} ⊂ N ±, then (2.4) holds, by the weak lower semi-continuity of the norm, we have

‖u+‖2V + ‖u+‖2

V ‖u‖2V ≤

∫

R3

|u+|pdx. (2.10)

On the other hand, since t+∗ u+ + t−∗ u− ∈ N ± and t+∗ ≥ t−∗ > 0, it holds

|t+∗ |2‖u+‖2V + |t+∗ |4‖u+‖2

V ‖u‖2V ≥ |t+∗ |p

∫

R3

|u+|pdx. (2.11)

Combining (2.10) and (2.11), we obtain that

(1 − |t+∗ |p−4)∫

R3

|u+|pdx ≥(

1 − 1|t+∗ |2

)‖u+‖2

V ,

which yields that t+∗ ≤ 1 since p ∈ (4, 6). Therefore, 0 < t−∗ ≤ t+∗ ≤ 1. Itfollows from (2.5), (2.6), and (2.7) that

m ≤ E(t+∗ u+ + t−∗ u−) = h+(t+∗ u+ + t−∗ u−) + h−(t+∗ u+ + t−∗ u−)≤ h+(u) + h−(u) ≤ lim

n→∞ E(un) = m,

which gives that E(t+∗ u+ + t−∗ u−) = m. Thus, by Lemma 2.2, we see thatt+∗ u+ + t−∗ u− ∈ N ± is a critical point of E. �

2.3. Proof of Theorem 1.4.

Proof. Without loss of generality, we may assume that λ = 1. Arguing bycontradiction, there exists {un} ⊂ N ± such that E(un) ≤ 2m + 1/n for all n.Clearly, {un} is bounded, going if necessary to a subsequence, we may assumethat un ⇀ u, u±

n ⇀ u± in HV for some 0 �= u ∈ HV as n → ∞ since, by thecondition (V1) and Lemma 2.2, there exists C > 0 such that

limn→∞ ‖u±

n ‖pp = ‖u±‖p

p ≥ C, lim infn→∞ ‖u±

n ‖2V ≥ C.

It follows that for n large enough

0 = 〈E′(un), u±n 〉 = 〈E′(u±

n ), u±n 〉 + ‖u+

n ‖2V ‖u−

n ‖2V ≥ 〈E′(u±

n ), u±n 〉 +

C

2,


which implies that 〈E′(u±n ), u±

n 〉 ≤ −C/2. On the other hand, from Lemma 2.2and (2.4), there exist {t±n } such that t±n u±

n ∈ N and t±n ∈ (0, 1), where N isdefined by (1.12). Therefore,

−C

2≥ 〈E′(u+

n ), u+n 〉 = 〈E′(u+

n ), u+n 〉 − 1

|t+n |p 〈E′(t+n u+n ), (t+n u+

n )〉

=(

1 − 1|t+n |p−2

)‖u+

n ‖2V +

(1 − 1

|t+n |p−4

)‖u+

n ‖4V ,

which means that t+n ≤ t+∗ < 1 holds for some t+∗ > 0 and n large enough.Similarly, there exists t−∗ with t−∗ < 1 such that t−n ≤ t−∗ < 1 for n largeenough. Thus, by using (2.5), (2.6), and (2.7), we have

2m +1n

≥ E(un) = h+(un) + h−(un)

= E(t+n u+n )+E(t−n u−

n )+1−|t+n |2

4‖u+

n ‖2V +

(14

− 1p

)(1−|t+n |p)‖u+

n ‖pp

+1 − |t−n |2

4‖u−

n ‖2V +

(14

− 1p

)(1 − |t−n |p)‖u−

n ‖pp

≥ 2m +1 − |t+∗ |2

4‖u+

n ‖2V +

(14

− 1p

)(1 − |t+∗ |p)‖u+

n ‖pp

+1 − |t−∗ |2

4‖u−

n ‖2V +

(14

− 1p

)(1 − |t−∗ |p)‖u−

n ‖pp

≥ 2m + C

for some C > 0, this is a contradiction. �

3. Appendix.

Proof of Lemma 2.2 (iii). We suppose that there exists u ∈ N ± such thatE(u) = m. Borrowing an argument in the proof of [9, Lemma 2.5], we firstshow that u must be a critical point of E. If not, then there exists v ∈ HV

such that

〈E′(u), v〉 = (1 + ‖u‖2V )

∫

R3

(∇u∇v + V (x)uv)dx −∫

R3

|u|p−2uvdx ≤ −1,

which implies that there exists ε0 ∈ (0, 1/8) such that

〈E′(tu+ + su− + σv), v〉 ≤ −12

for all |t − 1| ≤ ε0, |s − 1| ≤ ε0 and |σ| ≤ ε0. Let T = [1/2, 3/2] × [1/2, 3/2]and let η ∈ C(T, [0, 1]) be a cut-off function defined as

η(t, s) =

⎧⎨⎩

1, for |t − 1| ≤ ε0

2, |s − 1| ≤ ε0

2,

0, for |t − 1| ≥ ε0, or |s − 1| ≥ ε0.


For |t − 1| ≤ ε0, |s − 1| ≤ ε0, we have

E(tu+ + su− + ε0η(t, s)v)

= E(tu+ + su−) +

1∫

0

〈E′(tu+ + su− + θε0η(t, s)v), ε0η(t, s)v〉dθ

≤ E(tu+ + su−) − ε0η(t, s)2

.

If |t − 1| ≤ ε0/2 and |s − 1| ≤ ε0/2, then

E(tu+ + su− + ε0η(t, s)v) ≤ E(tu+ + su−) − ε0η(t, s)/2 ≤ m − ε0/2.

If |t−1| ≥ ε0/2 or |s−1| ≥ ε0/2, then by Lemma 2.2, there exists C ∈ (0, ε0/2)such that E(tu+ + su− + ε0η(t, s)v) ≤ E(tu+ + su−) ≤ m − C. Therefore,

sup(t,s)∈T

E(tu+ + su− + ε0η(t, s)v) ≤ m − C,

this means that tu+ + su− + ε0η(t, s)v /∈ N ± for all (t, s) ∈ T .On the other hand, for each ε : 0 ≤ ε ≤ ε0, let

φε(t, s) = tu+ + su− + εη(t, s)v and Φε(t, s) = (G+(φε(t, s)), G−(φε(t, s))),

then by Lemma 2.2 and the definition of η, we see that η(t, s) = 0, φε(t, s) =tu+ + su− and G±(φε(t, s)) = G±(tu+ + su−) �= 0 hold for all (t, s) ∈ ∂T and0 ≤ ε ≤ ε0. It follows that

deg(Φε0 , T, (0, 0)) = deg(Φ0, T, (0, 0)) = sign(det H ′0(1, 1)) = 1

since Φ0(t, s) = (G+(tu+ + su−), G−(tu+ + su−)) = (0, 0) if and only if t =s = 1 for (t, s) ∈ T and detH ′

0(1, 1) = det Hgu(1, 1) > 0, where

H ′0(t, s) =

∂(G+(tu+ + su−), G−(tu+ + su−))∂(t, s)

and detH ′0(t, s) is the determinant of H ′

0(t, s). This contradicts tu+ + su− +ε0η(t, s)v /∈ N ± for all (t, s) ∈ T . Therefore, u is a critical point of E.

Next we prove that u has exactly two nodal domains. If u(x) = u1(x) +u2(x) + u3(x) with ui �= 0, u1(x) ≥ 0, u2(x) ≤ 0, and meas(supp(ui) ∩supp(uj)) = 0 for i �= j, i, j = 1, 2, 3, then there exist t, s > 0 such thattu1+su2 ∈ N ±. Noting that 〈E′(u), ui〉 = 0 for i = 1, 2, 3, we have t, s ∈ (0, 1],therefore

m ≤ E(tu1 + su2) ≤ E(u) − 14‖u3‖2

V −(

14

− 1p

)‖u3‖p

p < m.

this is a contradiction, hence u has exactly two nodal domains. �

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Yisheng Huang and Zeng Liu

Department of Mathematics,Soochow University,Suzhou 215006, Jiangsu,People’s Republic of China

Yisheng Huang

e-mail: [email protected]

Zeng Liu

e-mail: [email protected]

Received: 21 November 2013

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On a class of Kirchhoff type problems