# The Existence of Least Energy Sign-Changing Solution for Kirchhoff-Type Problem with Potential Vanishing at Infinity

Authors: Ting Xiao; Canlin Gan; Qiongfen Zhang
Journal: Advances in Mathematical Physics
 (2021)

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## Abstract

In this paper, we study the Kirchhoff-type equation:−a+b∫ℝ3∇u2dxΔu+Vxu=Qxfu,inℝ3, where a, b>0, f∈C1ℝ3,ℝ, and V,Q∈C1ℝ3,ℝ+. Vx and Qx are vanishing at infinity. With the aid of the quantitative deformation lemma and constraint variational method, we prove the existence of a sign-changing solution u to the above equation. Moreover, we obtain that the sign-changing solution u has exactly two nodal domains. Our results can be seen as an improvement of the previous literature.


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## Body

## 1. Introduction

Consider the Kirchhoff-type equation(1)−a+b∫ℝ3∇u2dxΔu+Vxu=Qxfu,x∈ℝ3,where a, b>0, V,Q:ℝ3→ℝ+, and f:ℝ3→ℝ are continuous functions. Moreover, V and Q are vanishing at infinity.Similar to [1], we say V,Q∈Q, if Qx and Vx satisfy the following conditions:(VQ1)Vx, Qx>0 for any x∈ℝ3 and Q∈L∞ℝ3(VQ2) IfAnn⊂ℝ3 is a sequence of Borel sets such that their Lebesgue measures ∣An∣≤R for all n∈ℕ and some R>0, then
(2)limr→∞∫An∩Brc0Qx=0uniformlyinn∈ℕ.And one of the below conditions holds:(VQ3)Q/V∈L∞ℝ3 or(VQ4) there existsp∈2,6 such that
(3)QxVx6−p/4→0asx→+∞.Problem (1) is related to the following Kirchhoff equation
(4)−a+b∫ℝ3∇u2dxΔu+Vxu=fu,x∈ℝ3.Problem (4) is nonlocal due to the existence of ∫ℝ3∇u2dx which causes some mathematical analysis difficulties and makes problem (4) more interesting. The nonlocal operator ∫ℝ3∇u2dx comes from the Dirichlet problem
(5)−a+b∫Ω∇u2dxΔu=fu,x∈Ω,u=0,x∈∂Ω,where a, b>0, Ω=ℝN or Ω⊂ℝN is a bounded domain. Equation (5) is related to the following stationary analogue of the Kirchhoff-type equation:
(6)utt−a+b∫Ω∇u2dxΔu=fx,u,which was introduced by Kirchhoff [2] as generalization of the famous D’Alembert wave equation
(7)ρ∂2u∂t2−ρ0h+E2L∫0L∂u∂x2dx∂2u∂x2=fx,ufor free vibration of elastic strings.After Lion [3] investigated problem (5) involving an abstract framework, Kirchhoff-type equations have been extensively researched by many scholars. Hence, numerous interesting works to (5) or similar problems are obtained in the last decades. Please see [4–7] and the references therein.In particular, many scholars dedicated to searching sign-changing solutions to (4) and similar problems. Indeed, they obtained a lot of interesting results. For example, Li et al. [8] investigated the sign-changing solution to the following problem by using the constraint variational method
(8)−a+b∫ℝ3∇u2dxΔu+Vxu=fx,u,x∈ℝ3.They supposed thatfx,t satisfies the following conditions:(f1)fx,t=ot as t→0 uniformly in x∈ℝ3(f2)fx,t∈Cℝ3×ℝ,ℝ and ∣fx,t∣≤c∣t∣+tq−1 for some q∈2,2∗, where c>0 and 2∗=2N/N−2(f3)fx,t/t4→∞ as t→∞ uniformly in x∈ℝ3(f4)fx,t/t3 is an increasing function of t∈ℝ\0.In [9], by using a similar main method and theorem, Wang, Zhang, and Cheng proved that if V and f satisfy (f1), (f3)-(f4) and(V1)∃a>0 such that Vx≥a>0 for any x∈ℝ3(V2)lim∣x∣→+∞Vx=V∞≔supx∈ℝ3Vx<+∞(V3) there existR0>0 and ρ:R0,∞→0,∞ is a nonincreasing function such that
(9)limr→∞ρreδr=∞,(f5)limt→∞ft/t5=0, then (4) has a sign-changing solution with two nodal domains. It is worth pointing out that they considered the associated “limiting problem”
(10)−a+b∫ℝ3∣∇u2dxΔu+V∞u=fu,inℝ3,u∈H1ℝ3,with the corresponding energy functional
(11)J∞u=12∫ℝ3a∇u2+V∞u2dx+b4∫ℝ3∇u2dx2−∫ℝ3Fudx.Moreover, they obtainedc0≤c+c∞, which is crucial to show that Mb≠∅. Where
(12)c0≜infMbJu,c∞≜infMb∞J∞u,N∞≜u∈H1ℝ3\0:J∞′u,u=0,Mb≜u∈H1ℝ30:J′u,u+=J′u,u−=0,Ju=12∫ℝ3a∇u2+Vxu2dx+b4∫ℝ3∇u2dx2−∫ℝ3Fudx.For more interesting results, see [10–14].Whena=1, b=0, (4) reduces to
(13)−Δu+Vxu=fu,inℝ3.Recently, (13) and similar problems have also been received far-reaching research. For example, in [15], when Vx≡0, fx,u=fu, Chen considered (13) on a bounded domain Ω⊂ℝN and obtained the sign-changing solution to (13) under some sufficient hypothesis. Precisely, they assumed that fx,t satisfies the following conditions:(A1)f∈CΩ×ℝ, Fx,t≜∫0tfx,sds≥0 and fx,t=o∣t∣ as ∣t∣→0 uniformly in x∈Ω(A2)fx,t=V∞xt+f1x,t, V∞∈CΩ and f1x,t=o∣t∣ as ∣t∣→+∞ uniformly in x∈Ω(A3)t↦fx,t/∣t∣ is strictly increasing on −∞,0∪0,∞ for every x∈Ω(A4)F~x,t≔1/2fx,tt−Fx,t→+∞ as t→+∞ uniformly in x∈Ω(A5)infx∈ΩV∞x>μ≜infu∈Πmax∇u+22,∇u−22, where
(14)Π≔u∈H01Ω:u±≠0,∫Ωu±2dx=1.Besides, lettingλ1 be the first eigenvalue of −Δ. Then by (A5) and the definition of μ, they concluded λ≤μ<∞. By using variational methods, Liu and Wang [16] obtained the least energy nodal solution to (13) in ℝN, when Vx=λVx and N≥3, where λ is a parameter. They discussed two situations, the most important of which is that a nodal solution to (13) can be found for any λ∈ℝ+ under certain conditions, thanks to assumptions.(S1)Ω≜intV−10≠∅ and ∃Ω1, Ω2⊂Ω such that
(15)Ω¯1∩Ω¯2=∅,where Ω1 and Ω2 are bounded smooth domains.(S2)lim∣t∣→∞ft/t=L∈μ,+∞,where μ=maxμ1,μ2 and μi=inf∫Ωi∇u2dx:u∈H01Ω,∫Ωiu2dx=1, i=1, 2.It is worth noting that Liu and Wang [16] used the first eigenfunction of −Δ,H01Ωi, i=1,2, when they showed that the constraint set is nonempty. For researching on problem (4) under conditions similar to [16], see [13]. For more relevant results, see [17–21] and the references therein.In fact, not only the sign-changing solutions of Kirchhoff-type equation has received widespread attention but also other kinds of solutions. Particularly, many scholars have investigated positive solutions, ground state solutions, multiple solutions, etc., for the Kirchhoff equation. Please see the literature [22–25]. Moreover, in recent years, there are also a lot of works on the following Kirchhoff-Schro¨dinger-Poisson systems
(16)−a+b∫ℝ3∇u2dxΔu+Vxu+ϕu=fx,u,x∈ℝN,−Δϕ=u2,x∈ℝN.Readers can refer to [26–32]. These results are also helpful for us to study (1), since (1) and (16) have a great correlation.Motivated by above-mentioned results, we will establish the existence of sign-changing solution to (1). Throughout this paper, we will make the following hypothesis about f:(F1)limt→0ft/t3=0 if (VQ3) holds and limt→0ft/tp−1=0 for some p∈4,6 if (VQ4) holds(F2)limt→∞ft/t6=0(F3)limt→∞Ft/t4=∞, where Ft=∫0tfsds(F4)ft/t3 is nondecreasing for ∣t∣>0.Obviously, (F2)-(F3) are weaker than (f2)-(f3). In [9], the authors assume that f satisfies (F1)-(F4), but they have more other assumptions on V than this paper. This also makes the problem challenging and more interesting. Moreover, our hypothesis does not involve eigenvalue problems. Hence, our results can be regarded as the improvement and complementary of [15, 16]. In the paper, we use a direct method to obtain the sign-changing solution.In this paper, we consider our problem on the following space:(17)E≜u∈D1,2ℝ3:∫ℝ3Vxu2dx<∞with the norm
(18)∥u∥2=∫ℝ3a∇u2+Vxu2dx.SinceV,Q∈Q, the embedding E↪LQqℝ3 is compact for some q∈2,6 (see Lemma 1), where
(19)LQqℝ3=u:uismeasurableonℝ3and∫ℝ3Qxuqdx<+∞endowed with the norm
(20)uQ=∫ℝ3Qxuqdx1/q.It is clear thatE,∙ is a Banach space. Let ·p be the usual norm in Lpℝ3. Hence, the embedding E↪L6ℝ3 is continuous. Set
(21)S6=infu∈E,u≠0∥u∥2∫ℝ3u6dx1/3.Similar to [1], we can prove the following lemmas.Lemma 1.
LetV,Q∈Q. If (VQ3) occurs, then the embeddingE°LQqℝ3is continuous and compact. If (VQ4) satisfies, then the embeddingE°LQpℝ3is continuous and compact.Lemma 2.
Assume thatV,Q∈Q. Ifun⊂Eandun⇀uinE, then(22)∫ℝ3QxFundx→∫ℝ3QxFudx,∫ℝ3Qxfunundx→∫ℝ3Qxfuudx.Our main result is as follows.Theorem 3.
IfV,Q∈Qand (F1)-(F4) hold, then problem (4) has at least one least-energy sign-changing solutionu, which has precisely two nodal domains.Throughout this paper,C1, C2, ⋯ denote positive constants possibly different in different places.The rest of this paper is organized as follows: in Section2, some frameworks are demonstrated. In Section 3, the proof of the main result is given.


## 2. Preliminaries

It is no doubt that the weak solution for (1) corresponds to the critical point of the energy functional
(23)Iu=12∥u∥2+b4∫ℝ3∇u2dx2−∫ℝ3QxFudx.Meanwhile,I∈C1E,ℝ and we define
(24)I′u,v=∫ℝ3a∇u∇v+Vxuvdx+b∫ℝ3∇u2dx∫ℝ3∇u∇vdx−∫ℝ3Qxfuvdx.And ifu is a solution to (1) with u±≠0, then u is called the sign-changing solution of (1), where
(25)u+x=maxux,0,u−x=minux,0.Through a straightforward calculation, we obtain(26)Iu=Iu++Iu−+b2∫ℝ3∇u+2dx∫ℝ3∇u−2dx,I′u,u+=I′u+,u++b∫ℝ3∇u+2dx∫ℝ3∇u−2dx,I′u,u−=I′u−,u−+b∫ℝ3∇u+2dx∫ℝ3∇u−2dx.The only thing needed to obtain a sigh-changing solution to (1) is to find a minimizer of the energy of I over the following constraint:
(27)M=u∈H1ℝ3:u±≠0,I′u,u+=I′u,u−=0.For proving our result, we will adopt a regular process. First, we need to obtain the existence of a unique pairsu,tu such that suu++tuu−∈M and su,tu is the unique maximum point of Isu++tu−. That is to prove the following lemma.Lemma 4.
LetV,Q∈Q. Suppose that (F1)-(F4) hold, if  u∈Ewith  u±≠0, then there exists a unique pair  su,tuof positive numbers such that  suu++tuu−∈M  and(28)Isuu++tuu−=maxs,t≥0Isu++tu−.Proof.
Setu∈E with u±≠0. Letting
(29)gs,t=I′su++tu−,su+=∫ℝ3a∇su++tu−∇su++Vxsu++tu−su+dx+b∫ℝ3∇su++tu−2dx∫ℝ3∇su++tu−∇su+dx+b∫ℝ3∇su++tu−2dx∫ℝ3∇su++tu−∇su+dx−∫ℝ3Qxfsu++tu−su+dx=s2∥u+∥2+bs4∫ℝ3∣∇u+∣2dx2+bs2t2∫ℝ3∇u+∣2dx∫ℝ3∇u−2dx−∫ℝ3Qxfsu+su+dx,hs,t=I′su++tu−,tu−=∫ℝ3a∇su++tu−∇tu−+Vxsu++tu−tu−dx+b∫ℝ3∇su++tu−2dx∫ℝ3∇su++tu−∇tu−dx−∫ℝ3Qxfsu++tu−tu−dx=t2∥u−∥2+bt4∫ℝ3∇u−2dx2+bs2t2∫ℝ3∇u+2dx∫ℝ3∇u−2dx−Qx∫ℝ3ftu−tu−dx.
First, we will show that there exists0<r<R such that
(30)gr,t>0,gR,t<0,∀t∈r,R(31)hs,r>0,hs,R<0,∀s∈r,R.Case 1.
If (VQ2) holds, from (F1)-(F2), for eachε>0, there exists a Cε>0 such that
(32)ftt≤εt2+Cεt6,∀t∈ℝ.
Then by Sobolev imbedding theorem, one has(33)hs,t≥t2∥u−∥2+bt4∫ℝ3∇u−2dx2+bs2t2∫ℝ3∇u+2dx∫ℝ3∇u−2dx−ε∫ℝ3Qxtu−2dx−Cε∫ℝ3Qxtu−6dx≥t2∥u−∥2−C1εQxt2∥u−∥2−Cεt6∫ℝ3Qxu−6dx≥1−C1ε∥QV−1∥∞t2∥u−∥2−C2S6−3t6∥Qx∥∞∥u∥6.Case 2.
If (VQ4) holds, refer to the argument in [1], there exists R, Cp>0 such that
(34)∫BRc0Qxupdx≤ε∫BRc0Vxu2+u6dx,∀u∈E,ε∈0,Cp.
Then by (F1)-(F2), we deduce that(35)hs,t≥t2∥u−∥2−C3tp∫ℝ3Qxu−pdx−C4t6∫ℝ3Qxu−6dx.
According to (34), (VQ1) and Ho¨lder inequality, we obtain
(36)hs,t≥t2∥u−∥2−C4∥Q∥∞t6∫ℝ3u−6dx−C3εtp∫BRc0Vxu−2+u−6dx−C3tp∥Q∥6/6−pBR0∫BR0u−6dxp/6≥t2∥u−∥2−C4t6∥Q∥∞S6−3∥u−∥6−C3tpε∥u−∥2+C3S6−3tpε∥u−∥6+C3S6−p/2∥Q∥6/6−pBR0∥u−∥ptp.
From the above discussion andp>2, we can conclude that there is t1>0 enough small such that
(37)hs,t>0fort∈0,t1.
Due tou−≠0, then ∃K>0 such that
(38)measx∈ℝ3,u−>K>0.
Besides, it follows from (F3) that for allL>0, there exists M>0 such that
(39)fyy3≥L,foranyy∈M,∞.
Hence, ast satisfies tK>M, we get
(40)∫ℝ3Qxftu−tu−dx≥∫u−x>KQxftu−tu−3tu−4≥Lt4∫u−x>KQxu−4dx.
Assumes≤t. We have
(41)hs,t≤t2∥u−∥2+bs2t2∫ℝ3∇u+2dx∫ℝ3∇u−2dx+bt4∫ℝ3∇u−2dx2−Lt4∫u−x>KQxu−4dx≤t2∥u−∥2+b2s4∫ℝ3∇u+2dx2+b2t4∫ℝ3∇u−2dx2+bt4∫ℝ3∇u−2dx2−Lt4∫u−x>KQxu−4dx≤t2∥u−∥2+3b2t4∫ℝ3∇u−2dx2+b2t4∫ℝ3∇u+2dx2−Lt4∫u−x>KQxu−4dx.
And once again, ift≤s, we have
(42)gs,t≥∥u+∥2−C3tp∫ℝ3Qxu+pdx−C4t6∫ℝ3Qxu+6dx,or
(43)gs,t≥s2∥u+∥2−C4s6∥Q∥∞S6−3∥u+∥6−C3ε∥u+∥2+S6−3ε∥u+∥6+S6−p/2∥Q∥6/6−pBR0∥u+∥psp.
And(44)gs,t≤s2∥u+∥2+3b2s4∫ℝ3∇u+2dx2+b2s4∫ℝ3∇u−2dx2−Ls4∫u+x>KQxu+4dx.
Obviously, we can takeL large enough so that
(45)L∫u±x>KQxu±4dx≥3b∫ℝ3∇u±2dx2+b∫ℝ3∇u∓2dx2.
By (33), (34), and (41)–(45), we can deduce that (30) and (31) hold. Then from the point of view of Maranda’s theorem [2], there exists a pair su,tu of positive numbers such that
(46)gsu,tu=0,hsu,tu=0,i.e.,suu++tuu−∈M.
Next, we will prove the uniqueness ofsu,tu.
Without loss of generality, the only thing we need to do is to show thatsu which makes suu++tuu−∈M is unique. Assume that there exists another pair su1,tu such that su1u++tuu−∈M. Let su1<su, by the definition of M, we have
(47)su2∥u+∥2+bsu4∫ℝ3∇u+2dx2+bsu2tu2∫ℝ3∇u+2dx∫ℝ3∇u−2dx−∫ℝ3Qxfsuu+suu+dx=0,(48)bsu14∫ℝ3∇u+2dx2+bsu12tu2∫ℝ3∇u+2dx∫ℝ3∇u−2dx+su12∥u+∥2−∫ℝ3Qxfsu1u+su1u+dx=0.
Combining (47) and (48), we have
(49)1su2−1su12∥u+∥2+btu2∫ℝ3∇u+2dx∫ℝ3∇u−2dx=∫ℝ3Qxfsuu+suu+3−fsu1u+su1u+3u+4dx.
Thanks to (F4) andsu1<su, the right side of the above equality is positive, which is a contradiction since the left is negative. So su1 must be equal to su, that is, su is unique. Similarly, we can show the uniqueness of tu. In short, the pair su,tu is unique.
Next, we prove thatIsuu++tuu−=maxs,t≥0Isu++tu−.
LettingWs,t=Isu++tu−. It is obvious that W attains the unique critical value at su,tu in ℝ+×ℝ+. From (F3), we obtain
(50)Ws,t→−∞uniformlyas∣s,t∣→∞,which implies that a maximum point of W cannot be established on the boundary of ℝ+×ℝ+. If we may assume that s~,0 is a maximum of W, for t small enough, we can easily obtain that
(51)Wt′s~,t=Is~u++tu−t′=t∥u−∥2+t3∫ℝ3∣∇u−∣2dx2+s~2t∫ℝ3∇u+∣2dx∫ℝ3∇u−2dx−∫ℝ3Qxftu−u−dx>0.
Then fort small enough, Ws~,t is an increasing function with respect to t. Hence, according to our discussion, the pair s~,0 is not a maximum point of W in ℝ+×ℝ+.Remark 5.
For proving the uniqueness of the pairsu,tu, we can also discuss it in two situations. Case 1. u∈M. We need to show that su=tu=1. The process of proving in Case 1 shall be similar to the proof in Lemma 4. Case 2. u∉M. By Lemma 4, there exists a pair of positive numbers su,tu such that suu++tuu−∈M when u∉M. Assume that there exists another pair su′,tu′ such that su′u++tu′u−∈M. Set v≜suu++tuu− and v′≜su′u++tu′u−. Then, we have
(52)su′suv++tu′tuv−=su′u++tu′u−=v′∈M,which implies su=su′ and tu=tu′. Since v′=v∈M and by Case 1, we consider the minimization problem
(53)m≜infIu:u∈M.Lemma 6.
Assume that (F1)-(F4) hold andV,Q∈Q, then  m  is achieved.Proof.
Step 1.m>0.
For eachu∈M, we have I′u,u=0. From (33) and (36), we derive that
(54)∥u∥2≤ε∥QV−1∥∞∥u∥2+S6−3Cε∥Q∥∞∥u∥6or
(55)∥u∥2≤C3ε∥u∥2+S6−3ε∥u∥6+S6−p/2∥Q∥6/6−pBR0∥u∥p+C4∥Q∥∞S6−3∥u∥6.
Choosingε=min1/2C3,1/2∥QV−1∥∞. It is easy to verify that there exists δ>0 such that ∥u∥2>δ. According to (F4), we have
(56)Gx,t≜∫ℝ3Qxft−4Ftdx≥0,forallx∈ℝ3,t∈ℝ.
Then,(57)Iu=Iu−14I′u,u≥14∥u∥2+14Gx,u>0,i.e., m≥1/4δ>0.
Step 2.m is achieved.
Letun⊂M such that Iun→m. Then, un is bounded in E. Consequently, there exist u such that un⇀u in E. Due to un∈M, then I′un,un±=0. That is
(58)∥un±∥2+b∫ℝ3∇un±2dx2−∫ℝ3Qxfun±un±dx=0.
Similar to (57), we can find a γ>0 such that ∥un±∥>γ>0 for any n∈ℕ+. Moveover, according to (F1)-(F2), for every ρ>0, there is cρ such that
(59)ftt≤ρt2+ρt6+Cρtq,∀t∈ℝ.
Therefore, from (58), we conclude
(60)γ≤∥un±∥2<∫ℝ3Qxfun±un±≤ρ∫ℝ3Qxun±2dx+ρ∫ℝ3Qxun±6dx+Cρ∫ℝ3Qxun±qdx.
Based on the boundedness ofun, there is C7 such that
(61)γ≤ρC7+Cρ∫ℝ3Qxun±qdx.
SelectingC7=γ/2ρ, by (61), we have
(62)∫ℝ3Qxun±qdx≥γ2Cρ.
From (62) and the compactness of embedding E°Lqℝ3 for q∈2,6, then we have
(63)∫ℝ3Qxu±qdx≥γ2Cρ,i.e., u±≠0. According to the weak semicontinuity of the norm, we have
(64)∥u±∥2+b∫ℝ3∇u2dx∫ℝ3∇u±2dx≤limn→∞inf∥un±∥2+b∫ℝ3∇un2dx∫ℝ3∇un±2dx.
By (58), (64), and Lemma 2, we have
(65)∥u±∥2+b∫ℝ3∇u2dx∫ℝ3∇u±2dx≤∫ℝ3Qxfu±u±dx,i.e., I′u,u±≤liminfn→∞I′un,un±=0. Then, there exists su,tu such that suu++tuu−∈M. Suppose that su≥tu>0, then
(66)su2∥u+∥2+bsu4∫ℝ3∇u+2dx2+bsu4∫ℝ3∇u+2dx∫ℝ3∇u−2dx≥su2∥u+∥2+bsu4∫ℝ3∣∇u+∣2dx2+bsu2tu2∫ℝ3∇u+∣2dx∫ℝ3∇u−2dx=∫ℝ3Qxfsuu+suu+dx.
From (65) and (66), we have
(67)1su2−1∥u+∥2≥∫ℝ3Qxfsuu+suu+3−fu+u+3u+4dx.
Similar to the discussion in Lemma4, by (67), we obtain su≤1. Letting u¯=suu++tuu−. Then by Gx,t≥0, we obtain
(68)m≤Iu¯=Iu¯−14I′u¯,u¯=14∥suu+∥2+∥tuu−∥2+∫ℝ3Qxfsuu+suu+−4Fsuu+dx+14∫ℝ3Qxftuu−tuu−−4Ftuu−dx≤14∥u∥2+14∫ℝ3Qxfuu−4Fudx≤limn→∞infIun−14I′un,un=m.
(68) implies that su=tu=1. Then, u¯=u and Iu=m. The proof is completed.Lemma 7.
Assume (F1)-(F4) hold, thenuis a critical point ofI.Proof.
Due tou∈M, then I′uu+=I′uu−=0. By Lemma 4, for any s,t∈ℝ+,ℝ+ and s,t≠1,1, we have
(69)Isu++tu−<Iu++u−=m.
AssumeI′u≠0, then there exists μ>0 and θ>0 such that ∥I′u∥≥μ for every ∥u−v∥≤3θ and u∈E. Let D=1−σ,1+σ×1−σ,1+σ and φs,t≜su++tu−, s,t∈D, where σ=min1/2,θ2/∥u∥2. Then, from Lemma 4, we have
(70)m¯≜max∂DI∘φ<m.
ForS2θ=u∈E:∥u−v∥≤2θ, Lemma 6 in [2] shows that there is a deformation η such that
(i)
η1,u=u if u∉I−1c−2ε,c+2ε∩S2θ(ii)
η1,Ic+ε∩S⊂Ic−ε(iii)
Iη1,u≤Iu for all u∈E.
We claim that(71)maxs,t∈D¯Iη1,φs,t<m.
By Lemma4 and (iii), we have Iφs,t≤m<m+ε, i.e., φs,t∈Im+ε. Moveover, from the definition of φs,t, it follows that
(72)∥φs,t−u∥2≤s−12∥u+∥2+t−12∥u−∥2≤σ∥u∥2≤θ2,
which implies thatφs,t∈Sθ for all s,t∈D¯. So φs,t∈Im+ε∩Sθ. Then by (ii), we can obtain that (71) holds.
Next, we will show thatη1,φD∩M≠∅.
Lettingκs,t≜η1,φs,t,
(73)Ψ0s,t≜I′su++tu−,u+,I′su++tu−,u−≜ψ1s,t,ψ2s,t,Ψ1s,t≜1sI′κs,t,κ+s,t,1tI′κs,t,κ−s,t.
By direct calculation, we get(74)∂ψ1s,t∂s1,1=u+2+3b∫R3∇u+2dx2+b∫R3∇u+2dx∫R3∇u-2dx-∫R3Qxf′u+u+2dx,∂ψ2s,t∂t1,1=u-2+3b∫R3∇u-2dx2+b∫R3∇u+2dx∫R3∇u-2dx-∫R3Qxf′u-u-2dx,∂ψ1s,t∂t1,1=2b∫R3∇u+2dx∫R3∇u-2dx=∂ψ2s,t∂s.
From (F4), we have(75)f′tt2−3ftt>0,foranyt∈ℝ.∂ψ1s,t∂s1,1<2b∫ℝ3∇u+2dx2−2∫ℝ3Qxfu+u+dx,∂ψ2s,t∂t1,1<2b∫ℝ3∇u−2dx2−2∫ℝ3Qxfu−u−dx.
Then,(76)∂ψ1s,t∂s1,1≤−2∥u+∥2−2b∫ℝ3∇u+2dx∫ℝ3∇u−2dx,∂ψ2s,t∂t1,1≤−2∥u−∥2−2b∫ℝ3∇u+2dx∫ℝ3∇u−2dx.
Let(77)A≜∂ψ1s,t∂s∂ψ2s,t∂s∂ψ1s,t∂t∂ψ2s,t∂t,
thendetA>0. Therefore, 1,1 is the unique isolated zero point of Ψ0. Then by degree theorem, we have that degΨ0,D,0=1. From (70) and (i), we have κ=φ. Hence, degΨ1,D,0=degΨ0,D,0=1. Then, there is s0,t0∈D such that Ψ1s0,t0=0. Hence, η1,φs0,t0=Ψs0,t0∈M, which is a contradiction with (71).


## 3. Proof of Theorem 3

Proof of Theorem 3.
From Lemmas6 and 7, there is a u∈M such that Iu=m and I′u=0. Next, we will prove that u has exactly two nodal domains.
By a contradiction, letu=u1+u2+u3, ui≠0 and I′u,ui=0i=1,2,3, where
(78)u1≥0,u2≤0,Ω3∩Ω4=∅,u3Ω3∪Ω4=0,Ω3≔x∈Ω:u1x>0,Ω4≔x∈Ω:u2x<0.
Lettingv=u1+u2. Then v+=u1, v−=u2, that is, v±≠0. By Lemmas 4 and 6, there exists a unique pair sv,tv∈0,1×0,1 such that svv++tvv−∈M. Then,
(79)Isvv++tvv−≥m.
Therefore, we have(80)0=14I′u,u3=14∥u3∥2+b4∫ℝ3∇u32dx2+b4∫ℝ3∇u12dx∫ℝ3∇u32dx+b4∫ℝ3∇u2∣2dx∫ℝ3∇u32dx−14∫ℝ3Qxfu3u3dx<Iu3+b4∫ℝ3∇u12dx∫ℝ3∇u32dx+b4∫ℝ3∇u2∣2dx∫ℝ3∇u32dx.
On the other hand, we have(81)m≤Isvv++tvv−=Isvv++tvv−−14I′svv++tvv−,svv++tvv−≤14sv2∥u1∥2+14tv2∥u2∥2+14∫ℝ3Qxfsvu1svu1−4Fsvu1dx+14∫ℝ3Qxftvu2tvu2−4Ftvu2dx≤14∥u1∥2+14∫ℝ3Qxfu1u1−4Fu1dx+14∥u2∥2+14∫ℝ3Qxfu2u2−4Fu2dx=Iu1+u2−14I′u1+u2,u1+u2≤Iu1+Iu2+Iu3+b2∫ℝ3∇u12dx∫ℝ3∇u22dx+b2∫ℝ3∇u12dx∫ℝ3∇u32dx+b2∫ℝ3∇u2∣2dx∫ℝ3∇u32dx=Iu=m,which is impossible. Hence, u has two exactly two nodal domains. The proof is completed.



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*Source: 6690204-2021-01-31.xml*