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Volume 9, Issue 1

# Sign-changing multi-bump solutions for the Chern-Simons-Schrödinger equations in ℝ2

Zhi Chen
/ Xianhua Tang
/ Jian Zhang
• Corresponding author
• School of Mathematics and Statistics, Hunan University of Technology and Business, Changsha, 410205 Hunan, P. R. China
• Email
• Other articles by this author:
Published Online: 2019-09-20 | DOI: https://doi.org/10.1515/anona-2020-0041

## Abstract

In this paper we consider the nonlinear Chern-Simons-Schrödinger equations with general nonlinearity

$−Δu+λV(|x|)u+h2(|x|)|x|2+∫|x|∞h(s)su2(s)dsu=f(u),x∈R2,$

where λ > 0, V is an external potential and

$h(s)=12∫0sru2(r)dr=14π∫Bsu2(x)dx$

is the so-called Chern-Simons term. Assuming that the external potential V is nonnegative continuous function with a potential well Ω := int V–1(0) consisting of k + 1 disjoint components Ω0, Ω1, Ω2 ⋯, Ωk, and the nonlinearity f has a general subcritical growth condition, we are able to establish the existence of sign-changing multi-bump solutions by using variational methods. Moreover, the concentration behavior of solutions as λ → +∞ are also considered.

MSC 2010: 35J20; 58E50

## 1 Introduction and main results

In this paper we are interested in the following nonlinear Schrödinger system with the gauge field

$iD0ϕ+(D1D1+D2D2)ϕ+g(ϕ)=0,∂0A1−∂1A0=−Im(ϕ¯D2ϕ),∂0A2−∂2A0=Im(ϕ¯D1ϕ),∂1A2−∂2A1=−12|ϕ|2,$(1.1)

where i denotes the imaginary unit, $\begin{array}{}{\mathrm{\partial }}_{0}=\frac{\mathrm{\partial }}{{\mathrm{\partial }}_{t}},{\mathrm{\partial }}_{1}=\frac{\mathrm{\partial }}{{\mathrm{\partial }}_{{x}_{1}}},{\mathrm{\partial }}_{2}=\frac{\mathrm{\partial }}{{\mathrm{\partial }}_{{x}_{2}}}\end{array}$ for (t, x1, x2) ∈ ℝ1+2, ϕ : ℝ1+2 → ℂ is the complex scalar field, Aκ : ℝ1+2 → ℝ is the gauge field and Dκ = κ + iAκ is the covariant derivative for κ = 0, 1, 2. This model (1.1) was first proposed and studied in [22, 23, 24], and is sometimes called the Chern-Simons-Schrödinger equations. The two-dimensional Chern-Simons-Schrödinger equations is a nonrelativistic quantum model describing the dynamics of a large number of particles in the plane, which interact both directly and via a self-generated electromagnetic field. Moreover, it describes an external uniform magnetic field which is of great phenomenological interest for applications of Chern-Simons theory to the quantum Hall effect.

As usual in Chern-Simons theory, system (1.1) is invariant under gauge transformation

$ϕ→ϕeiχ, Aκ→Aκ−∂κχ$

for any arbitrary C function χ. The existence of standing wave solutions for system (1.1) with power type nonlinearity, that is, g(u) = λ|u|p–1u (p > 1 and λ > 0), has been investigated recently by a number of authors. For example, see [6, 7, 20, 25, 32] and the references therein. The standing wave solutions of system (1.1) have the following form

$ϕ(t,x)=u(|x|)eiωt, A0(t,x)=k(|x|),A1(t,x)=x2|x|2h(|x|), A2(t,x)=−x1|x|2h(|x|),$(1.2)

where ω > 0 is a given frequency, u, k, h are real valued functions depending only on |x|. Note that the ansatz (1.2) satisfies the Coulomb gauge condition 1A1 + 2A2 = 0. Inserting the ansatz (1.2) into the system (1.1), Byeon et al.[6] got the following nonlocal semilinear elliptic equation

$−Δu+ωu+h2(|x|)|x|2+∫|x|∞h(s)su2(s)dsu=λ|u|p−2u in R2,$(1.3)

where

$h(s)=hu(s)=12∫0sru2(r)dr=14π∫Bsu2(x)dx.$

Mathematically, equation (1.3) is not a pointwise identity as the appearance of the Chern-Simons term

$h2(|x|)|x|2+∫|x|∞h(s)su2(s)dsu.$

Hence problem (1.3) is called a nonlocal problem and is quite different from the usual semi-linear Schrödinger equation. From the variational point of view, the nonlocal term causes some mathematical difficulties that make the study of problem (1.3) more interesting.

Following [6], equation (1.3) possesses a variational structure, that is, the standing wave solutions are obtained as critical points of the energy functional associated to (1.3) defined by

$L(u)=12∫R2(|∇u|2+ωu2)dx+12∫R2u2|x|2∫0|x|s2u2(s)ds2dx−λp∫R2|u|pdx,$(1.4)

u$\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2), where $\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2) := {uH1(ℝ2)| u(x) = u(|x|)}. In [6, 10, 20, 29, 30, 33, 39, 40, 45], the critical points of 𝓛 are found by using variational methods. It is shown that the value p = 4 is critical for this problem. Indeed, for p > 4, it is known that the energy functional is unbounded from below and satisfies a mountain-pass geometry. In a certain sense, in this case the local nonlinearity dominates the nonlocal term. However the existence of a solution is not so direct, since for p ∈ (4, 6) the (PS)-condition is not known to hold. In the spirit of [34], this problem is bypassed in [6] by using a constrained minimization taking into account the Nehari-Pohozaev manifold.

A special case is p = 4: in this case, solutions have been explicitly found in [6, 7] as optimizers of a certain inequality. An alternative approach would be to pass to a self-dual equation, which leads to a Liouville equation that can be solved explicitly. For more information on the self-dual equations, see [14, 24]. For the case p ≥ 4, [29] and [30] proved the existence, multiplicity, quantitative property and asymptotic behavior of normalized solutions with prescribed L2-norm.

The situation is different if p ∈ (2, 4), solutions are found in [6] as minimizers on a L2-sphere. Later, the results has been extended by Pomponio and Ruiz [32] by investigating the geometry of the energy functional under the different range of frequency ω. Moreover, Pomponio and Ruiz in [33] also studied the bounded domain case for p ∈ (2, 4). By using singular perturbation arguments based on a Lyapunov-Schmidt reduction, they obtained some results on boundary concentration of solutions. Wan and Tan [40] studied the existence and multiplicity of standing waves for asymptotically linear nonlinearity case, and see [44] for the sublinear case. Cunha et al.[10] obtained a multiplicity result when the nonlinearity satisfies the general hypotheses introduced by Berestycki and Lions [8]. For more results about the initial value problem, well-posedness, existence and blow-up, scattering and uniqueness results for some nonlocal problems, we refer readers to [5, 11, 12, 19, 21, 26, 27, 42, 43] and references therein.

When p > 6, by using the symmetric mountain pass theorem, Huh [20] obtained the existence of infinitely many radially symmetric solutions for equation (1.3). Recently, this result has been extended to more general nonlinearity model by Zhang et al.[45]. Besides, Deng et al.[16] and Li et al.[31] investigated the existence and asymptotic behavior of radial sign-changing solutions by using constraint minimization method and quantitative deformation lemma for equation (1.3). Liu et al.[28] obtained a multiplicity result of sign-changing solutions via a novel perturbation approach and the method of invariant sets of descending flow.

Very recently, for the general 6-superlinear nonlinearity case, Tang et al.[39] considered the following nonlocal Schrödinger equation with the gauge field and deepening potential well

$−Δu+λV(|x|)u+h2(|x|)|x|2+∫|x|∞h(s)su2(s)dsu=f(u) in R2,$(1.5)

where the potential V is a continuous function satisfies

• (V1′$\begin{array}{} V'_{1} \end{array}$)

V(|x|) ∈ C(ℝ2) and V(|x|) ≥ 0 on ℝ2;

• (V2)

there exists a constant b > 0 such that the set Vb := {x ∈ ℝ2|V(|x|) < b} is nonempty and has finite measure;

• (V3)

there is bounded symmetric domain Ω such that Ω = int V–1(0) with smooth boundary ∂Ω and Ω̄ = V–1(0).

Under some suitable conditions on the nonlinearity f, the second and third authors proved the existence and multiplicity of solutions (possibly positive, negative or sign-changing) by using mountain pass theorem. Moreover, the concentration behavior of these solutions on the set Ω as λ → +∞ are also studied.

Involving the Chern-Simons-Schrödinger equations with potential wells, there is only the work [39] so far. As described above, the shape of solutions obtained in [39] may be single-bump. However, nothing is known for the existence of multi-bump type solutions. Motivated by the above facts, we intend in the present paper to study the existence of sign-changing multi-bump solutions for equation (1.5) with deepening potential well. To the best of our knowledge, it seems that such a problem was not considered in literature before. In order to state our statements, for the potential V we need to assume that the following conditions besides (V2) and (V3),

• (V1)

VC1(ℝ2) and V(|x|) ≥ 0 on ℝ2;

• (V4)

there are k + 1 disjoint open bounded components Ω0, Ω1, Ω2, ⋯, Ωk (k ≥ 2) such that Ω = int V–1(0) = $\begin{array}{}{\cup }_{i=0}^{k}\end{array}$Ωi and dist(Ωi, Ωj) > 0 for ij, i, j = 0, 1, 2, ⋯, k, where Ω0 = {x ∈ ℝ2, r0 = 0 ≤ |x| ≤ $\begin{array}{}{r}_{0}^{\prime }\end{array}$}, Ωi = {x ∈ ℝ2, ri ≤ |x| ≤ $\begin{array}{}{r}_{i}^{\prime }\end{array}$}.

Moreover, we suppose that the nonlinearity f satisfies

• (f1)

fC(ℝ, ℝ), and there exist constants C > 0 and q0 ∈ (4, +∞) such that

$|f(t)|≤C(|t|+|t|q0−1);$

• (f2)

f(t) = o(t) as t → 0;

• (f3)

$\begin{array}{}\underset{t\to \mathrm{\infty }}{lim}\frac{F\left(t\right)}{{t}^{6}}=+\mathrm{\infty }\end{array}$ where F(t) = $\begin{array}{}{\int }_{0}^{t}\end{array}$f(s)ds;

• (f4)

the function $\begin{array}{}\frac{f\left(t\right)}{{t}^{5}}\end{array}$ is increasing on (0, +∞) and decreasing on (–∞, 0).

Before stating our results we first need to introduce some notations. Throughout this paper, we define

$Hλ:=u∈Hr1(R2)∣∫R2λV(|x|)u2dx<+∞$

with the norm

$∥u∥λ2=∫R2|∇u|2+λV(|x|)u2dx.$

Clearly, the embedding Hλ$\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2) is continuous due to (V1) and (V2). We will give the proof later. Define the energy functional 𝓘λ : Hλ → ℝ by

$Jλ(u)=12∫R2(|∇u|2+λV(|x|)u2)dx+12∫R2u2|x|2∫0|x|s2u2(s)ds2dx−∫R2F(u)dx,$(1.6)

Then, our hypotheses imply that the functional 𝓘λC1(Hλ, ℝ), and for any u, φHλ, we have

$〈Jλ′(u),φ〉=∫R2∇u∇φ+λV(|x|)uφdx+∫R2h2(|x|)|x|2uφdx +∫R2u2|x|2∫0|x|s2u2(s)ds∫0|x|su(s)φ(s)dsdx−∫R2f(u)φdx.$(1.7)

Obviously, critical points of 𝓘λ are the weak solutions for equation (1.5). Furthermore, if uHλ is a solution of (1.5) and u± ≠ 0, then u is a sign-changing solution of (1.5), where

$u+(x)=max{u(x),0} and u−(x)=min{u(x),0}.$

Our main result can be stated as follows.

#### Theorem 1.1

Suppose that (V1)-(V4) and (f1)-(f4) hold. Then, for any non-empty subset T ⊂ {0, 1, 2, 3, …, k} with

$T=T1∪T2∪T3 and Ti∩Tj=∅ for i≠j, i,j=1,2,3.$(1.8)

There exists a constant ΛT > 0 such that for λ > ΛT, equation (1.5) has a sign-changing multi-bump solution uλ, which possesses the following property: for any sequence {λn} with λn → +∞ as n → ∞, there is a subsequences {uλni} converges strongly to u in $\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2), where u$\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2) is a nontrivial solution of the equation

$−Δu+(h2(|x|)|x|2+∫|x|rk′h(s)su2(s)ds)u=f(u), x∈ΩT=∪i∈TΩi,u=0 x∈R2∖ΩT.$(1.9)

Moreover, u|Ωi is positive for iT1, u|Ωi is negative for iT2, and u|Ωi changes sign exactly once for iT3.

The motivation of the present paper arises from the study of the local Schrödinger equations with deepening potential well

$−Δu+(λV(x)+a(x))u=f(x,u) in RN.$(1.10)

We remark that conditions (V1)-(V3) have been first introduced by Bartsch, Pankov and Wang [9] in studying the Schrödinger equation (1.10). They obtained some results on the existence of multiple solutions, and also showed that the solutions concentrated at the bottom of the potential well as λ → +∞. The existence and characterization of the solutions for equation (1.10) were considered in [1, 2, 15, 35] under conditions (V1)-(V4). For example, Ding and Tanaka [15] first constructed the existence of multi-bump positive solutions uλ for (1.10), and they also proved that, up to a subsequence, uλ converges strongly in H1(ℝN) to a function u, which satisfies u = 0 outside ΩT and u|Ωi is the positive least energy solution to the equation

$−Δu+a(x)u=f(x,u), u∈H01(Ωi).$(1.11)

Inspired by [15], Alves [1] and Sato and Tanaka [35] investigated the sign-changing multi-bump solutions to equation (1.10) independently. Later, Alves and Pereira [2] obtained a similar result for the critical growth case. We must point out that the equation (1.11) (called limit equation of (1.10)) plays an important role in the study of multi-bump solutions for equation (1.10). Because the positive, negative and sign-changing solutions of (1.11) are used as building bricks to construct the multi-bump solutions of (1.10) by using of gluing techniques. Recently, there are some works focused on study of multi-bump solutions and ground states solutions for other nonlocal problems. For instance, see [18, 36, 37] for Schrödinger-Kirchhoff equation, [3, 13, 38] for Schrödinger-Poisson system, and [4] for Choquard equation and so on.

From the commentaries above, it is quite natural to ask if the results in [1, [35] still hold for the Chern-Simons-Schrödinger equations. Unfortunately, we can not draw a similar conclusion in a straight way. Since problem (1.5) is a nonlocal one as the appearance of the Chern-Simons term, then the solutions of problem (1.5) need their global information. Thus, we cannot use the same arguments explored in [1, 35] to solve the corresponding limit equation (1.9) separately on each Ωi. This result in the effective methods used [1, 35] for local Schrödinger equations cannot be applied to nonlocal problem (1.5) directly, there arises a technical problem one should overcome. As we all know, in order to obtain the existence of sign-changing multi-bump solutions for problem (1.5), we need to use the existence and some properties of the least energy sign-changing solutions of limit equation (1.9). However, it is not easy to prove the existence and sign properties of the least energy solution for limit equation (1.9). Hence, the first step is to consider the limit equation (1.9) and to look for the existence of least energy sign-changing solution that is nonzero on each component Ωi, iT.

Our result on the limit equation (1.9) can be stated as follows.

#### Theorem 1.2

Suppose that (f1)-(f4) hold, then, for any non-empty subset T, equation (1.9) has a nontrivial solution u with u|Ωi is positive for iT1, u|Ωi is negative for iT2, and u|Ωi changes sign exactly once for iT3. Moreover, u is the least energy solution among all solutions with those sign properties.

To prove our results, some arguments are in order. First, to obtain the least energy sign-changing solution solutions with prescribed sign properties for limit equation (1.9), we first prove the set 𝓜T (see (2.1)) is nonempty and then we seek minimizers of the energy functional on 𝓜T. Observe that 𝓜T is not a C1-manifold, we will take advantage of constraint minimization method and quantitative deformation lemma to obtain the existence of minimizers of the energy functional. Second, we will use the penalization technique explored by del Pino and Felmer in [17] to cut off the nonlinearity f, then, to control the order of growth of nonlinearity f outside the potential well ΩT. In such a way, we build a modification of the energy functional associated to (1.5) and give some energy relations of problems (1.5) and (1.9) which play a key role in getting the critical point of (1.5) (see Lemma 4.3). Moreover, in order to show a critical point associated to the modified functional is indeed a solution to the original problem, we also need give a delicate L-estimation for the solutions of the modified problem. Finally, we study a special minimax value of the modified functional, which is crucial for proving Theorem 1.1. Furthermore, via a rather precise analysis of deformation flow to the modified functional, we prove the existence of sign-changing multi-bump solutions for (1.5).

The paper is organized as follows. In Section 2, we give some preliminary lemmas and the proof of Theorem 1.2. In Section 3, we define a penalization problem and modified functional, and give a L-estimation for the solutions of the modified problem. In Section 4, we study a special minimax value of the modified functional. At last, we give the proof of Theorem 1.1 in Section 5.

Throughout the sequel, we denote the usual Lebesgue space with norms $\begin{array}{}\parallel u{\parallel }_{p}={\left({\int }_{{\mathbb{R}}^{2}}|u{|}^{p}dx\right)}^{\frac{1}{p}}\end{array}$ by Lp(ℝ2), where 1 ≤ p < ∞, and C denotes different positive constant in different place.

## 2 Mixed type sign-changing solutions to limit problem

In this section, we study the existence of solutions for the limited equation (1.9) with prescribed sign properties. Firstly, we restrict the nonlinearity (x, t) = 0 if

$(i) t<0 and x∈Ωi for i∈T1; or (ii) t>0 and x∈Ωi for i∈T2.$

Denoted

$HT={u∈Hr1(R2)∣u=0 in x∈R2∖ΩT}$

with the norm

$∥u∥T2=∫ΩT|∇u|2dx.$

Now we define the energy functional JT corresponding to limit problem (1.9) on HT

$JT(u):=12∫ΩT|∇u|2dx+12∫ΩTu2|x|2(∫0|x|s2u2(s)ds)2dx−∫ΩTF~(x,u)dx, u∈HT,$

and the set

$MT={u∈HT∣〈JT′(u),ui〉=0 for i∈T,ui+≠0 for i∈T1,ui−≠0 for i∈T2 and 〈JT′(u),ui±〉=0,ui±≠0 for i∈T3},$(2.1)

where ui := u|Ωi and T = T1T2T3 ⊂ {0, 1, 2, 3, …, k} satisfies (1.8), and (x, u) = $\begin{array}{}{\int }_{0}^{u}\end{array}$(x, t)dt. Let

$m=infu∈MTJT(u).$

If m is attained by u0 ∈ 𝓜T and $\begin{array}{}{J}_{T}^{\prime }\end{array}$(u0) = 0, then u0 be called a the least energy sign-changing solution of limit problem (1.9).

Without loss of generality, we consider the case T1 = {1}, T2 = {2} and T3 = {3} for simplify. In this case, ΩT = $\begin{array}{}{\cup }_{i=1}^{3}\end{array}$Ωi with dist(Ωi, Ωj) > 0 for ij, i, j = 1, 2, 3. To simplify the notations, we use Ω, 𝓜, H to denote the sets ΩT, 𝓜T, HT respectively. Moreover, we define the functional on H as follows

$J(u):=12∫Ω|∇u|2dx+12∫Ωu2|x|2(∫0|x|s2u2(s)ds)2dx−∫ΩF~(x,u)dx, u∈H.$(2.2)

The functional J is well-defined and belongs to C1(H, ℝ). Moreover, for any u, φH, we have

$〈J′(u),φ〉=∫Ω∇u∇φdx+∫Ωh2(|x|)|x|2uφdx+∫Ωu2|x|2(∫0|x|s2u2(s)ds)(∫0|x|su(s)φ(s)ds)dx −∫Ωf~(x,u)φdx.$(2.3)

Clearly, critical points of J are the weak solutions for limit problem (1.9).

We use constraint minimizer on 𝓜 to seek a critical point of J with nonzero component. We first check that the set 𝓜 is nonempty in H.

#### Lemma 2.1

Assume that (f1)-(f4) hold. For uH with $\begin{array}{}{u}_{1}^{+}\ne 0,{u}_{2}^{-}\ne 0\end{array}$ and $\begin{array}{}{u}_{3}^{±}\ne 0,\end{array}$ then there exists a unique 4-tuple (s1, s2, s3, s4) ∈ (ℝ+)4 such that

$s1u1+s2u2+s3u3++s4u3−∈M.$

#### Proof

For uH with $\begin{array}{}{u}_{1}^{+}\ne 0,{u}_{2}^{-}\ne 0\end{array}$ and $\begin{array}{}{u}_{3}^{±}\ne 0,\end{array}$ we define

$F(u)=(F1(u),F2(u),F3(u),F4(u)),$(2.4)

where

$Fi(u)=J′(u)ui, i=1,2,J′(u)(u3+), i=3,J′(u)(u3−), i=4.$

Obviously, t1u1 + t2u2 + t3$\begin{array}{}{u}_{3}^{+}\end{array}$ + t4$\begin{array}{}{u}_{3}^{-}\end{array}$ ∈ 𝓜 if and only if

$F(t1u1+t2u2+t3u3++t4u3−)=0.$(2.5)

Next, we obtain the desired results by proving two claims.

#### Claim 1

For uH with u1 ≠ 0, u2 ≠ 0 and $\begin{array}{}{u}_{3}^{±}\end{array}$ ≠ 0, there exists at least one solution for (2.5).

Firstly, there exists a unique 1 > 0 such that

$F1(t¯1u1+t2u2+t3u3++t4u3−)=0.$(2.6)

for fixed (t2, t3, t4) ∈ (ℝ+)3. In fact, we define

$k(t):=t2∫Ω|∇u1|2dx+∫Ω(∫0|x|s2(t2u12(s)+t22u22(s)+t32u3+2(s)+t42u4+2(s))ds)2t2u12|x|2dx+∫Ωt2u12+t22u22+t32u3+2+t42u4+2|x|2(∫0|x|s2(t2u12(s)+t22u22(s)+t32u3+2(s)+t42u4+2(s))ds)(∫0|x|st2u12(s)ds)dx−∫Ωf~(x,tu1)tu1dx.$

By (f2), (f3) and (f4), it is easy to see that k(t) > 0 for t > 0 small enough and k(t) < 0 for t > 0 large enough. Hence, there exists 1 > 0 such that k(1) = 0. Moreover, by (f4) we can deduce that 1 is unique. Thus, we can get a function η1 : (ℝ+)3 → (0, +∞) defined by

$η1(t2,t3,t4)=t¯1,$

such that F1(1u1 + t2u2 + t3$\begin{array}{}{u}_{3}^{+}\end{array}$ + t4$\begin{array}{}{u}_{3}^{-}\end{array}$) = 0.

By the same arguments as above, we can define function ηi : (ℝ+)3 → (0, +∞), i = 2, 3, 4, given by

$η2(t1,t3,t4)=t¯2, η3(t1,t2,t4)=t¯3, η4(t1,t2,t3)=t¯4,$

satisfying F2(t1u1 + 2u2 + t3$\begin{array}{}{u}_{3}^{+}\end{array}$ + t4$\begin{array}{}{u}_{3}^{-}\end{array}$) = 0, F3(t1u1 + t2u2 + 3$\begin{array}{}{u}_{3}^{+}\end{array}$ + t4$\begin{array}{}{u}_{3}^{-}\end{array}$) = 0, and F4(t1u1 + t2u2 + t3$\begin{array}{}{u}_{3}^{+}\end{array}$ + 4$\begin{array}{}{u}_{3}^{-}\end{array}$) = 0.

Moreover, the functions ηi : (ℝ+)3 → (0, +∞), i = 1, 2, 3, 4, have the following three properties:

1. For any (a1, a2, a3, a4) ∈ (ℝ+)4, we have ηi(ai|a1, a2, a3, a4) > 0;

2. ηi are continuous on [0, +∞)3;

3. If $\begin{array}{}{a}_{max}^{i}\end{array}$ large enough, then

$a¯i=ηi(ai|a1,a2,a3,a4)

where (ai|a1, a2, a3, a4) := (a1, …, ai–1, ai+1, …, a4).

By the property (iii), there exists M1 > 0 such that ηi(ai|a1, a2, a3, a4) ≤ $\begin{array}{}{a}_{max}^{i}\end{array}$ for $\begin{array}{}{a}_{max}^{i}\end{array}$ > M1. From the property (i), we get

$M2:=maxi{maxηi(ai|a1,a2,a3,a4):amaxi≤M1,i=1,2,3,4}>0.$

Thus, M0 := max{M1, M2} > 0. For any (a1, a2, a3, a4) ∈ [0, M0]4, it follows from (iii) that ηi(ai|a1, a2, a3, a4) ≤ M0.

Hence, we can define L : [0, M0]4 → [0, M0]4 by

$L(a1,a2,a3,a4):=(a¯1,a¯2,a¯3,a¯4),$

where āi = ηi(ai| a1, a2, a3, a4) satisfying F1(ā1u1 + a2u2 + a3$\begin{array}{}{u}_{3}^{+}\end{array}$ + a4$\begin{array}{}{u}_{3}^{-}\end{array}$) = 0, F2(a1u1 + ā2u2 + a3$\begin{array}{}{u}_{3}^{+}\end{array}$ + a4$\begin{array}{}{u}_{3}^{-}\end{array}$) = 0, F3(a1u1 + a2u2 + ā3$\begin{array}{}{u}_{3}^{+}\end{array}$ + a4$\begin{array}{}{u}_{3}^{-}\end{array}$) = 0, F4(a1u1 + a2u2 + a3$\begin{array}{}{u}_{3}^{+}\end{array}$ + ā4$\begin{array}{}{u}_{3}^{-}\end{array}$) = 0.

Obviously, L(a1, a2, a3, a4) is continuous on [0, M0]4. Now, by applying the Brouwer Fixed Point Theorem, there exists (s1, s2, s3, s4) ∈ [0, M0]4 such that

$L(s1,s2,s3,s4)=(s1,s2,s3,s4).$

Thus, (s1, s2, s3, s4) is a solution of (2.5).

#### Claim 2

(s1, s2, s3, s4) obtained by Claim 1 is the unique solution of (2.5). To show Claim 2, the proof will be carried out in two cases.

• Case 1

u ∈ 𝓜. In this case, the 4-tuple (1, 1, 1, 1) is a solution of (2.5). We prove that (1, 1, 1, 1) is the unique solution of (2.5) in (ℝ+)4. Indeed, suppose (a1, a2, a3, a4) be any other solution, then

$Fi(a1u1+a2u2+a3u3++a4u3−)=0, i=1,2,3,4.$

Without loss of generality, we suppose a1 = max{a1, a2, a3, a4}. That is

$a12∫Ω|∇u1|2dx+∫Ω(∫0|x|s2(a12u12(s)+a22u22(s)+a32u3+2(s)+a42u4+2(s))ds)2a12u12|x|2dx +∫Ωa12u12+a22u22+a32u3+2+a42u4+2|x|2(∫0|x|s2(a12u12(s)+a22u22(s)+a32u3+2(s)+a42u4+2(s))ds) (∫0|x|sa12u12(s)ds)dx=∫Ωf~(x,a1u1)a1u1dx.$

Since u ∈ 𝓜, we have

$∫Ω|∇u1|2dx+∫Ω∫0|x|s2(u12(s)+u22(s)+u3+2(s)+u4+2(s))ds2u12|x|2dx +∫Ωu12+u22+u3+2+u4+2|x|2(∫0|x|s2(u12(s)+u22(s)+u3+2(s)+u4+2(s))ds) (∫0|x|su12(s)ds)dx=∫Ωf~(x,u1)u1dx.$

So we obtain

$1a14−1∫Ω|∇u1|2dx≥∫Ω(f~(x,a1u1)(a1u1)5−f~(x,u1)(u1)5)u16dx.$

By (f1)-(f4), it imply that a1 = max{a1, a2, a3, a4} ≤ 1. By the same arguments, we can easily conclude that min{a1, a2, a3, a4} ≥ 1. Consequently, the 4-tuple (1, 1, 1, 1) is the unique solution of (2.5).

• Case 2

u ∉ 𝓜. If (u1, u2, u3, u4) ∉ 𝓜, then by Claim 1, we know that (2.5) has a solution (s1, s2, s3, s4). Assume that $\begin{array}{}\left({s}_{1}^{\prime },{s}_{2}^{\prime },{s}_{3}^{\prime },{s}_{4}^{\prime }\right)\end{array}$ also be a solution. Then we have

$s1′s1s1u1+s2′s2s2u2+s3′s3s3u3++s4′s4s4u3−∈M.$

Since s1u1 + s2u2 + s3$\begin{array}{}{u}_{3}^{+}\end{array}$ + s4$\begin{array}{}{u}_{3}^{-}\end{array}$ ∈ 𝓜, by Case 1 we get

$s1′s1=s2′s2=s3′s3=s4′s4=1.$

Thus, (s1, s2, s3, s4) is the unique solution of (2.5) in (ℝ+)4.

#### Lemma 2.2

Assume that (f1)-(f4) hold, then the unique 4-tuple (t1, t2, t3, t4) ∈ (ℝ+)4 obtained by Lemma 2.1 is the unique maximum point of the function φ : (ℝ+)4 → ℝ defined by

$φ(s1,s2,s3,s4)=J(s1u1+s2u2+s3u3++s4u3−).$

#### Proof

From the proof of Lemma 2.1, we know that (t1, t2, t3, t4) is the unique critical point of φ in (ℝ+)4. By the assumption (f3), we deduce that φ(t1, t2, t3, t4) → –∞ as |(t1, t2, t3, t4)| → ∞, so it is sufficient to check that a maximum point cannot be achieved on the boundary of (ℝ+)4. Choose (s1, s2, s3, s4) ∈ (ℝ+)4, without loss of generality, we may assume that s1 = 0. But, it is obviously that

$l(s)=φ(s,s2,s3,s4)=J(su1+s2u2+s3u3++s4u3−)=12∫Ω|∇(su1+s2u2+s3u3++s4u3−)|2dx+12∫Ω(su1+s2u2+s3u3++s4u3−)2|x|2(∫0|x|s2(su1(s)+s2u2(s)+s3u3+(s)+s4u3−(s))2ds)2dx−∫ΩF~(x,su1+s2u2+s3u3++s4u3−)dx$

is an increasing function with respect to s if s > 0 is small enough, (0, s2, s3, s4) is impossible to be a maximum point of φ.

#### Lemma 2.3

Assume that (f1)-(f4) hold, let uH with $\begin{array}{}{u}_{1}^{+}\end{array}$ ≠ 0, $\begin{array}{}{u}_{2}^{-}\end{array}$ ≠ 0 and $\begin{array}{}{u}_{3}^{±}\end{array}$ ≠ 0 such that, Fi(u) ≤ 0 for i = 1, 2, 3, 4, where Fi are given by (2.4). Then the unique 4-tuples (t1, t2, t3, t4) ∈ (ℝ+)4 obtained by Lemma 2.1 satisfied (t1, t2, t3, t4) ∈ (0, 1]4.

#### Proof

Without loss of generality, we suppose that t1 = max{t1, t2, t3, t4}. Since t1u1 + t2u2 + t3$\begin{array}{}{u}_{3}^{+}\end{array}$ + t4$\begin{array}{}{u}_{3}^{-}\end{array}$ ∈ 𝓜, we get

$t12∫Ω|∇u1|2dx+∫Ω(∫0|x|s2(t12u12(s)+t22u22(s)+t32u3+2(s)+t42u4+2(s))ds)2t12u12|x|2dx+∫Ωt12u12+t22u22+t32u3+2+t42u4+2|x|2(∫0|x|s2(t12u12(s)+t22u22(s)+t32u3+2(s)+t42u4+2(s))ds)(∫0|x|st12u12(s)ds)dx=∫Ωf~(x,t1u1)t1u1dx.$(2.7)

At the same time, using Fi(u) ≤ 0, one has

$∫Ω|∇u1|2dx+∫Ω(∫0|x|s2(u12(s)+u22(s)+u3+2(s)+u4+2(s))ds)2u12|x|2dx +∫Ωu12+u22+u3+2+u4+2|x|2(∫0|x|s2(u12(s)+u22(s)+u3+2(s)+u4+2(s))ds) (∫0|x|su12(s)ds)dx≤∫Ωf~(x,u1)u1dx.$(2.8)

Therefore, (2.7) and (2.8) imply that

$(1t14−1)∫Ω|∇u1|2dx≥∫Ω(f~(x,t1u1)t15u15−f~(x,u1)u15)u16dx.$

By (f4), we obtain t1 ≤ 1. Thus we complete the proof.

Next, we consider the following constrained minimization problem m := infu∈𝓜 J(u).

#### Lemma 2.4

Assume that (f1)-(f4) hold, then m > 0 and m can be achieved by a function v ∈ 𝓜.

#### Proof

Let u ∈ 𝓜, then

$∫Ω|∇ui|2dx≤∫Ωf~(x,ui)uidx for i=1,2,$

and

$∫Ω|∇u3±|2dx≤∫Ωf~(x,u3±)u3±dx.$

Thus, by (f1), (f2) and Sobolev embedding theorem, there exists a positive constant C such that

$∥u1+∥,∥u2−∥,∥u3+∥,∥u3−∥≥C.$

Therefore,

$J(u)=J(u)−16∑i=13〈J′(u),ui〉=∑i=13(13∫Ω|∇ui|2dx+∫Ω16f~(x,ui)ui−F~(x,ui)dx)≥13∑i=13∫Ω|∇ui|2dx≥C,$

this implies that m > 0. Suppose that {vn} ⊂ 𝓜 satisfying

$limn→+∞J(vn)=m,$

we can easily to get that

$0

Passing to a subsequences, one has

$vn⇀v weakly in H,vn→v strongl in Lp(Ω) for 2≤p<∞,vn→v a.e. in Ω.$(2.9)

Assumptions (f1) and (f2) give

$∫Ωf~(x,v3±)v3±dx=limn→+∞∫Ωf~(x,vn,3±)vn,3±dx≥lim infn→+∞∥vn,3±∥2≥C1.$

Thus, $\begin{array}{}{v}_{3}^{±}\end{array}$ ≠ 0. By the same arguments, we conclude that $\begin{array}{}{v}_{1}^{+}\end{array}$ ≠ 0 and $\begin{array}{}{v}_{2}^{-}\end{array}$ ≠ 0. By Lemma 2.1, there exists a unique 4-tuple (t1, t2, t3, t4) ∈ (ℝ+)4 such that

$v¯:=t1v1+t2v2+t3v3++t4v3−∈M.$

On the other hand, due to {vn} ⊂ 𝓜, we have that

$Fi(v)≤0 for i=1,2,3.$

Thus, Lemma 2.3 deduces that

$(t1,t2,t3,t4)∈(0,1]4.$(2.10)

Thanks to the function sf(s) – 6F(s) is a non-negative function, increasing on (0, +∞), decreasing on (–∞, 0), it follow from (2.10) and Lemma 2.2, we have

$6m≤6J(v¯)=6J(v¯)−∑i=12〈J′(v¯),tivi〉−〈J′(v¯),t3v3+〉−〈J′(v¯),t4v4−〉=∑i=122ti2∥vi∥2+2t32∥v3+∥2+2t42∥v3−∥2+∑i=12∫Ω(f~(x,tivi)tivi−6F~(x,tivi))dx +∫Ω(f~(x,t3v3+)t3v3+−6F~(x,t3v3+))dx+∫Ω(f~(x,t4v3−)t4v3−−6F~(x,t4v3−))dx≤lim infn→+∞{∑i=122∥vn,i∥2+2∥vn,3+∥2+2∥vn,3−∥2+∑i=12∫Ω(f~(x,vn,i)vn,i−6F~(x,vn,i))dx +∫Ω(f~(x,vn,3+)vn,3+−6F~(x,vn,3+))dx+∫Ω(f~(x,vn,3−)vn,3−−6F~(x,vn,3−))dx}≤lim infn→+∞6J(vn)=6m.$

Thus, t1 = t2 = t3 = t4 = 1 and m is attained by v. Since the restriction on (x, u), we can get that v1 ≥ 0, v2 ≤ 0 and ($\begin{array}{}{v}_{3}^{±}\end{array}$) ≠ 0. So we prove the conclusion.

#### Proof of Theorem 1.2

We prove indirectly and assume that Φ′(v) ≠ 0. Then there exist δ > 0 and ϱ > 0 such that

$∥u−v∥≤3δ⇒∥Φ′(u)∥H∗≥ϱ.$(2.11)

For convenient, we define the function

$g(s1,s2,s3,s4)=s1v1+s2v2+s3v3++s4v3−, (s1,s2,s3,s4)∈D=(12,32)4.$

By Lemma 2.2, we have

$m¯:=max(s1,s2,s3,s4)∈∂DJ(g(s1,s2,s3,s4))(2.12)

For ε := min{(m)/2, ϱδ/8} and S := B(v, δ), then [41] yields a deformation ηC([0, 1] × H, H) such that

1. η(1, u) = u if J(u) < m – 2ε or J(u) > m + 2ε;

2. η(1, Jm0+εB(v, δ)) ⊂ Jmε;

3. J(η(1, u)) ≤ J(u), ∀ uH;

4. η(1, u) is a homeomorphism of H.

It is clear that

$max(s1,s2,s3,s4)∈D¯J(η(1,g(s1,s2,s3,s4)))(2.13)

We claim that η(1, g(D)) ∩ 𝓜 ∉ ∅. In fact, define

$h(s1,s2,s3,s4):=η(1,g(s1,s2,s3,s4))$

and

$Ψ0(s1,s2,s3,s4):=(1s1F1∘g,1s2F2∘g,1s3F3∘g,1s4F4∘g)(s1,s2,s3.s4),Ψ1(s1,s2,s3,s4):=(1s1F1∘h,1s2F2∘h,1s3F3∘h,1s4F4∘h)(s1,s2,s3.s4),$

Then the degree theory and Lemma 2.1 yield

$deg⁡(Ψ0,D,0)=1.$

It follows from (i), one has g = h on ∂D. Thus, we obtain

$deg⁡(Ψ1,D,0)=deg⁡(Ψ0,D,0)=1$

Thus, Ψ1(1, 2, 3, 4) = 0 for some (1, 2, 3, 4) ∈ D, this implies that η(1, g(D)) ∩ 𝓜 ≠ 0.

By (2.13) and the definition of m we reach a contradiction. Thus, J′(v) = 0. We can get vC2(Ω) by standard elliptic regularity theory. So v1 > 0, v2 < 0 by the Maximum Principle. At last, we show that v3 indeed has two nodal domains. Arguing by contradiction, we may assume that

$v3=w1+w2+w3,w1,w2,w3∈H01(Ω3),$

with wi ≠ 0, w1 ≥ 0, w2 ≤ 0 and suppt(wi) ∩ suppt(wj) = ∅, for ij, i, j = 1, 2, 3 and

$〈J′(v),wi〉=0, for i=1,2,3.$

Let u := w1 + w2, we see that u+ = w1 and u = w2. Setting

$v¯:=v1+v2+u∈H01(Ω).$

By Lemma 2.1, there exists a unique 4-tuple (1, 2, 3, 4) ∈ (ℝ+)4 such that

$t¯1v1+t¯2v2+t¯3u++t¯4u−∈M.$

Therefore

$J(t¯1v1+t¯2v2+t¯3u++t¯4u−)≥m.$(2.14)

Moreover, J′(v)wi = 0 implies that Fi() < 0 for i = 1, 2, 3, 4. Using Lemma 2.3, we deduce that (1, 2, 3, 4) ∈ (0, 1]4. At the same time,

$0=16〈J′(v),w3〉=16∫Ω|∇w3|2dx +16∫Ω(∫0|x|s2(v12(s)+v22(s)+w12(s)+w22(s)+w32(s))ds)2w32|x|2dx +16∫Ωv12+v22+w12+w22+w32|x|2(∫0|x|s2(v12(s)+v22(s)+w12(s) +w22(s)+w32(s))ds)(∫0|x|sw32(s)ds)dx−16∫Ωf~(x,w3)w3dx<12∫Ω|∇w3|2dx+16∫Ω(∫0|x|s2(v12(s)+v22(s)+w12(s)+w22(s)+w32(s))ds)2w32|x|2dx +16∫Ωv12+v22+w12+w22+w32|x|2(∫0|x|s2(v12(s)+v22(s)+w12(s) +w22(s)+w32(s))ds)(∫0|x|sw32(s)ds)dx−∫ΩF~(x,w3)dx$(2.15)

and

$J(t¯1v1+t¯2v2+t¯3u++t¯4u−)=J(⋅)−16∑i=12〈J′(⋅),t¯ivi〉+〈J′(⋅),t¯3u3+〉+〈J′(⋅),t¯4u4−〉=13∑i=12t¯i2∥vi∥2+13t¯32∥u3+∥2+13t¯42∥u3−∥2+16∑i=12∫Ω(f~(x,tivi)tivi−6F~(x,tivi))dx +16∫Ω(f~(x,t3u3+)t3u3+−6F~(x,t3v3+))dx+16∫Ω(f~(x,t4u3−)t4u3−−6F~(x,t4u3−))dx≤∑i=12(13∥vi∥2+16∫Ω(f~(x,vi)vi−6F~(x,vi))dx)+13∥u3+∥2+13∥u3−∥2 +16∫Ω(f~(x,u3+)u3+−6F~(x,u3+))dx+16∫Ω(f~(x,v3−)v3−−6F~(x,v3−))dx.$(2.16)

By using (2.14)-(2.16) and the fact that u+ = w1, u = w2, we have

$m≤J(t¯1v1+t¯2v2+t¯3u++t¯4u−)<∑i=12(13∥vi∥2+16∫Ω(f~(x,vi)vi−6F~(x,vi))dx)+13∥u+∥2+13∥u−∥2 +16∫Ω(f~(x,u+)u+−6F~(x,u+))dx+16∫Ω(f~(x,u−)u−−6F~(x,u−))dx +12∫Ω|∇w3|2dx+12∫Ω(∫0|x|s2(v12(s)+v22(s)+w12(s)+w22(s)+w32(s))ds)2w32|x|2dx +12∫Ωv12+v22+w12+w22|x|2(∫0|x|s2(v12(s)+v22(s)+w12(s) +w22(s))ds)(∫0|x|sw32(s)ds)dx +12∫Ωv12+v22+w12+w22+w32|x|2(∫0|x|s2w32(s)ds)2dx−∫ΩF~(x,w3)dx=J(v)=m,$(2.17)

which is a contradiction. So w3 = 0, and v3 has indeed two nodal domains.

## 3 Penalization of the nonlinearity and L∞-estimation

In this section, we will modify the functional Iλ by penalizing the nonlinearity f(u). It plays a key role in establishing the relation between mλ and m (will be defined later). On the other hand, we also give a delicate L-estimation for the critical points of the modified functional.

Since

$Ω=∪i=13Ωi and dist(Ωi,Ωj)>0 for i≠j, i,j=1,2,3,$

there exist open sets $\begin{array}{}{\mathit{\Omega }}_{i}^{\rho }\end{array}$ = {x ∈ ℝ2 : dist(x, Ωi) < ρ} for i = 1, 2, 3 with smooth boundary such that $\begin{array}{}\text{dist}\left({\mathit{\Omega }}_{i}^{\rho },{\mathit{\Omega }}_{j}^{\rho }\right)>0\end{array}$ for ij, i, j = 1, 2, 3. Denote $\begin{array}{}{\mathit{\Omega }}^{\rho }:={\cup }_{i=1}^{3}{\mathit{\Omega }}_{i}^{\rho }.\end{array}$ For open set Θ ⊂ ℝ2, we define

$Hλ(Θ):=u∈Hr1(Θ)|∫ΘV(|x|)u2dx<+∞ and u=0 in R2∖Θ$

with norm

$∥u∥λ,Θ2=∫Θ|∇u|2+λV(|x|)u2dx.$

By (V1) and (V2), there exists a positive constant ν0 such that

$ν0∫R2∖Ωρu2dx≤12∥u∥λ,R2∖Ωρ2 for all u∈Hλ(R2∖Ωρ).$(3.1)

Let a0 > 0 satisfy

$0

and , : ℝ → ℝ are the functions given by

$f~(s)=−f(−a0)a0s, if s<−a0,f(s), if |s|≤a0,f(a0)a0s, if s>a0,$

and

$F~(s)=∫0sf~(τ)dτ.$

Using the above notations, we denote

$g(|x|,s):=χΩρ(|x|)f(s)+(1−χΩρ(|x|))f~(s)$

and

$G(|x|,s):=∫0sg(|x|,t)dt=χΩρ(|x|)F(s)+(1−χΩρ(|x|))F~(s),$

where χΩρ denotes the characteristic function of the set Ωρ. We define the functional Φλ : Hλ → ℝ

$Φλ(u)=12∫R2|∇u|2+λV(|x|)u2dx+12∫R2u2|x|2∫0|x|s2u2(s)ds2dx−∫R2G(|x|,u)dx$(3.2)

and the critical points of Φλ are weak solutions of

$−Δu+λV(|x|)u+h2(|x|)|x|2+∫|x|+∞h(s)su2(s)dsu=g(|x|,u), x∈R2.$(3.3)

The next Proposition is about the asymptotic behavior of the critical points of Φλ as λ → +∞.

#### Proposition 3.1

Suppose λn → +∞ as n → ∞ and {uλn} ⊂ Hλn satisfying

$Φλn(uλn)→c and ∥Φλn′(uλn)∥Hλn∗→0.$

Then, up to a subsequence, there exists u$\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2) such that

1. unuλn → 0, consequently uλnu in $\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2);

2. u = 0 in2Ω and u is a solution to equation (1.9);

3. $\begin{array}{}{\mathit{\Phi }}_{{\lambda }_{n}}\left({u}_{{\lambda }_{n}}\right)\to J\left(u\right)=\frac{1}{2}{\int }_{\mathit{\Omega }}|\mathrm{\nabla }u{|}^{2}dx+\frac{1}{2}{\int }_{\mathit{\Omega }}\frac{{u}^{2}}{|x{|}^{2}}{\left({\int }_{0}^{|x|}\frac{s}{2}{u}^{2}\left(s\right)ds\right)}^{2}dx-{\int }_{\mathit{\Omega }}F\left(u\right)dx.\end{array}$

#### Proof

It is easy to know that {uλn} is bounded in Hλn(ℝ2) and hence {uλn} is bounded in $\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2). Passing to a subsequences, one has

$un⇀u weakly in Hr1(R2);un→u strongly in Lp(RN) for 2(3.4)

We prove (ii) firstly. Let m ∈ ℕ+, set Sm = {x ∈ ℝ2 : V(|x|) ≥ $\begin{array}{}\frac{1}{m}\end{array}$}, one has

$∫Smuλn2dx≤2mλn∫SmλnV(|x|)uλn2dx≤2mCλn,$(3.5)

and hence

$∫Smu2dx≤lim infn→∞∫Smuλn2dx=0.$

It implies that u ≡ 0 in Sm. So we prove the u ≡ 0 in ℝ2Ω̄. Now, for any φ$\begin{array}{}{C}_{0}^{\mathrm{\infty }}\end{array}$(Ω), since 〈$\begin{array}{}{\mathit{\Phi }}_{{\lambda }_{n}}^{\prime }\end{array}$(un), φ〉 = 0, we can deduce that u is a solution to equation (1.9).

Next, we prove uλnu in $\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2). For convenience, let

$Υ(u)=12∫R2u2|x|2(∫0|x|s2u2(s)ds)2dx$

and un = uλn. By virtue of 〈$\begin{array}{}{\mathit{\Phi }}_{{\lambda }_{n}}^{\prime }\end{array}$(un), unu〉 = 〈$\begin{array}{}{\mathit{\Phi }}_{{\lambda }_{n}}^{\prime }\end{array}$(u), unu〉 = 0 when n → +∞, we have

$(un,un−u)λn+〈Υ′(un),un−u〉=∫R2g(|x|,un)(un−u)dx,$(3.6)

$(u,un−u)λn+〈Υ′(u),un−u〉=∫R2g(|x|,u)(un−u)dx.$(3.7)

There, by Lemma 3.2 in [6], we have

$limn→∞(〈Y′(un),un−u〉−〈Y′(u),un−u〉)=0.$(3.8)

Using the standard argument, we can deduce that

$limn→∞∫R2g(|x|,un)(un−u)−g(|x|,u)(un−u)dx=0.$

Therefore, by (3.6), (3.7) and (3.8), we get

$limn→∞∥un−u∥λn2=0.$

On the other hand, the embedding Hλ$\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2) is continuous. Indeed, by (V1) and (V2), we can deduce that

$V(x)≥δ0>0, x∈R2∖Ωρ,$

so it is easy to get

$∫R2∖Ωρ|∇u|2+u2dx≤C∫R2|∇u|2+λV(|x|)u2dx.$

Thus we only need to show that

$∫Ωρ|∇u|2+u2dx≤C∫R2|∇u|2+λV(|x|)u2dx.$

We choose a cut-off function ΨC(ℝ2) such that 0 ≤ Ψ ≤ 1 in ℝ2, Ψ(x) = 1 for each xΩρ and Ψ(x) = 0 for x ∈ ℝ2Ω2ρ and |∇Ψ| < C. By Sobolev’s embedding inequality,

$∫Ωρu2dx≤∫Ω2ρ|uΨ|2dx≤C∫Ω2ρ|∇(uΨ)|2dx≤2C∫Ω2ρ|∇u|2dx+2C∫Ω2ρ∖Ωρu2dx≤C∫R2|∇u|2+λV(|x|)u2dx.$

So we get ∫2|∇u|2 + u2dxC2|∇u|2 + λV(|x|)u2dx. Thus we can get that

$∥un−u∥Hr1(R2)→0 as n→∞.$

Combining with (i), it is easy to prove the (iii).

The next Lemma is important which indicates that the critical points uλ of Φλ with bounded energy are the solutions of the original problem (1.9) if λ large enough.

#### Lemma 3.2

Fix M > 0, for any critical points uλ of Φλ(uλ) ≤ M. Then there exists Λ0 > 0 such that λΛ0, one has

$|uλ(x)|≤a0 for all x∈R2∖Ωρ.$(3.9)

#### Proof

We prove Lemma 3.2 by Moser’s iteration. By Proposition 3.1, it is easy to get that

$limλ→∞∫R2∖Ω|uλ|qdx=0, ∀ 2(3.10)

So, for any small η0 > 0 and λ large enough,

$∫R2∖Ω|uλ|6dx≤2η0.$(3.11)

Let ψ be a smooth cut-off function and β > 1, both of them will be specified later. For R > 0, we define

$uλR=R, if uλ>R,uλ, if |uλ|≤R,−R, if uλ

and multiply (3.3) by $\begin{array}{}{\psi }^{2}|{u}_{\lambda }^{R}{|}^{\beta -1}{u}_{\lambda },\end{array}$ then

$∫R2∇(ψ2|uλR|β−1uλ)∇uλdx+λ∫R2V(|x|)ψ2|uλR|β−1uλ2dx +∫R2(h2(|x|)|x|2+∫|x|+∞h(s)su2(s)ds)ψ2|uλR|β−1uλ2dx=∫R2g(|x|,uλ)ψ2|uλR|β−1uλdx.$(3.12)

That is

$∫R2ψ2|uλR|β−1|∇uλ|2dx+(β−1)∫R2ψ2|uλR|β−3uλRuλ∇uλR∇uλdx +2∫R2ψ|uλR|β−1uλ∇ψ∇uλdx+λ∫R2V(|x|)ψ2|uλR|β−1uλ2dx≤∫R2g(|x|,uλ)ψ2|uλR|β−1uλdx.$(3.13)

On the other hand, by Hölder’s inequality and Young’s inequality, one has

$∫R2ψ|uλR|β−1uλ∇ψ∇uλdx≤14∫R2ψ2|uλR|β−1|∇uλ|2dx+C∫R2|∇ψ|2|uλR|β−1uλ2dx.$

Note |g(|x|, u)u| ≤ |u|2 + C0|u|q0, so the inequality (3.13) leads to

$12∫R2ψ2|uλR|β−1|∇uλ|2dx+(β−1)∫R2ψ2|uλR|β−3uλRuλ∇uλR∇uλdx≤2C∫R2|∇ψ|2|uλR|β−1uλ2dx+∫R2ψ2uλ2|uλR|β−1dx+C0∫R2ψ2uλq0|uλR|β−1dx.$(3.14)

By sobolev imbedding theorem, we have

$S(p)∫R2(ψ|uλR|β−12uλ)pdx2p≤∫R2|∇(ψ|uλR|β−12uλ)|2dx≤(β+1)22∫R2ψ2|uλR|β−1|∇uλ|2dx+2∫R2|∇ψ|2|uλR|β−1uλ2dx,$(3.15)

where S(p) is imbedding constant. Using (3.14) and (3.15), one has

$S(p)∫R2(ψ|uλR|β−12uλ)pdx2p≤(2+2(β+1)2C)∫R2|∇ψ|2|uλR|β−1uλ2dx +C0(β+1)2∫R2ψ2uλq0|uλR|β−1dx+(β+1)2∫R2ψ2|uλR|β−1uλ2dx.$(3.16)

Now, for y ∈ ℝ2Ωρ and fix a r which $\begin{array}{}0 Then take the cut-off function ψ by

$ψ(x)=1, x∈B2r(y),0, R2∖B4r(y),$

and $\begin{array}{}0\le \psi \le 1,|\mathrm{\nabla }\psi |\le \frac{C}{r}.\end{array}$ Using Hölder’s inequality and (3.10), we have

$∫R2ψ2uλq0|uλR|β−1dx≤∫R2(ψ2|uλR|β−1uλ2)p2dx2p∫R2∖Ω(uλq0−2)pp−2dxp−2p≤S(p)2C0(β+1)2∫R2(ψ2|uλR|β−1uλ2)p2dx2p.$(3.17)

Combing (3.16) and (3.17), we have

$S(p)∫R2(ψ|uλR|β−12uλ)pdx2p≤(4+4(β+1)2C)∫R2|∇ψ|2|uλR|β−1uλ2dx +2(β+1)2∫R2ψ2|uλR|β−1uλ2dx.$(3.18)

Taking limit R → +∞ and β = 5 in (3.18), which implies that for any y ∈ ℝ2Ωρ,

$∫B2r(y)|uλ|3pdx2p≤C(r,p)∫B4r(y)|uλ|6dx.$(3.19)

Because p ∈ (2, +∞), we choose $\begin{array}{}p=\frac{3}{2}{q}_{0}-3>2\end{array}$ since q0 > 4, one has

$∫B2r(y)|uλ|92q0−9dx43q0−6≤C(r)∫B4r(y)|uλ|6dx.$(3.20)

Now, we use the above estimation combining with Moser’s iteration argument to complete the proof. Let $\begin{array}{}{Z}_{\lambda }=|{u}_{\lambda }^{R}{|}^{\frac{\beta -1}{2}}{u}_{\lambda },\end{array}$ where β > 1 will choose later, then (3.16) becomes

$S(6)∫R2(ψZλ)6dx13≤(2+2(β+1)2C)∫R2|∇ψ|2Zλ2dx +C0(β+1)2∫R2ψ2Zλ2uλq0−2dx+(β+1)2∫R2ψ2Zλ2dx,$(3.21)

where ψ is a cut-off function supported in B2r(y) with y ∈ ℝ3Ωρ and $\begin{array}{}r\le \frac{\rho }{4}.\end{array}$ By the Hölder’s inequality, we get

$∫R2ψ2Zλ2uλq0−2dx≤(∫R2(ψZλ)187dx)79(∫B2r(y)uλ92q0−9dx)29.$

Since $\begin{array}{}2<\frac{18}{7}<6,\end{array}$ thus, for any ϵ > 0,

$∥ψZλ∥1872≤ϵ∥ψZλ∥62+ϵ−12∥ψZλ∥22.$

By (3.21) and above estimate, it deduces that

$S(6)∫R2(ψZλ)6dx13≤(2+2(β+1)2C)∫R2|∇ψ|2Zλ2dx +C0C1(r)(β+1)2(ϵ∥ψZλ∥62+ϵ−12∥ψZλ∥22)+(β+1)2∫R2ψ2Zλ2dx,$(3.22)

where

$C1(r)=(∫B2r(y)uλ92q0−9dx)29≤(C(r)∫B4r(y)|uλ|6dx)q0−26≤[2η0C(r)]q0−26.$

Setting ϵ = S(6)[2C0C1(r)(β + 1)2]–1, we obtain from (3.22) that

$∫R2(ψZλ)6dx13≤(4+4(β+1)2C)S∫R2|∇ψ|2Zλ2dx+C2rq0−24(β+1)3∫R2ψ2Zλ2dx.$(3.23)

Now, for rr2 < r1 ≤ 2r, we choose ψ such that ψ ≡ 1 in Br2(y), ψ ≡ 0 in ℝ2Br1(y) and 0 ≤ ψ ≤ 1, $\begin{array}{}|\mathrm{\nabla }\psi |\le \frac{C}{{r}_{1}-{r}_{2}}.\end{array}$ Then we obtain that

$∥Zλ∥L6(Br2(y))≤C3(r1−r2)h32(rq0−24)−12∥Zλ∥L2(Br1(y)),$(3.24)

where h = (1 + β). Set

$L(p,r):=(∫Br(y)|uλ|pdx)1p.$

When R → +∞ in (3.24), then we have

$L(3h,r2)≤(C3(r1−r2))2hh3h(rq0−24)−1hL(h,r1).$(3.25)

Let h = hm = 6 ⋅ 3m, rm = r(1 + 2m) for m = 0, 1, 2, ⋯, by (3.25), we get

$L(6⋅3m+1,rm+1)=L(3hm,rm+1)≤(C3rm−rm+1)2hmhm3hm(rq0−24)−1hmL(hm,rm)=(2C3rq0+68632)13⋅3m(2⋅332)m3⋅3mL(hm,rm)≤⋯≤(2C3rq0+68632)13∑j=0∞3−j(2⋅332)13∑j=0∞j3−jL(6,2r).$(3.26)

Let m → ∞, we have

$supx∈Br(y)|u(x)|=lims→∞L(s,r)≤C4(r)(2η0)16.$(3.27)

We can choose Λ0, when λΛ0, we have $\begin{array}{}{C}_{4}\left(r\right)\left(2{\eta }_{0}{\right)}^{\frac{1}{6}}\le {a}_{0}.\end{array}$ So we can get

$∥uλn∥L∞(R2∖Ωρ)≤a0$

for λΛ0. Therefore we complete the proof.

## 4 A special minimax value for the modified functional

We investigate a special minimax value for the modified functional Φλ, which is used to get a key Lemma. We define a new functional

$J¯λ(u)=12∫Ωρ|∇u|2+λV(|x|)u2dx+12∫Ωρu2|x|2∫0|x|s2u2(s)ds2dx−∫ΩρF(u)dx$(4.1)

which is well defined and belongs to C1(Hλ(Ωρ), ℝ). We define set

$M¯={u∈Hλ(Ωρ)∣〈J¯′(u),ui〉=0 for i=1,2,u1+≠0,u2−≠0 and 〈J¯′(u),u3±〉=0,u3±≠0}$

and

$m¯λ:=infu∈M¯J¯λ(u).$

Similar to the proof of Section 2, we deduce that there exists λHλ(Ωρ) such that

$J¯λ(v¯λ)=m¯λ and J¯λ′(v¯λ)=0.$

#### Lemma 4.1

There holds that

1. 0 < λm, for all λ > 0;

2. λm, as λ → +∞.

#### Proof

The proof of (i) is trivial since u ∈ 𝓜 which also belongs to 𝓜̄ by zero extension.

Now we are going to prove (ii). Let {λn} be a sequence with λn → +∞. For each λn, there exists uλnHλn(Ωρ) with

$J¯λn(uλn)=m¯λn and J¯λn′(uλn)=0.$(4.2)

Since λnm, we can suppose {(uλn)} convergence (up to a subsequence) and $\begin{array}{}{\overline{J}}_{\phantom{\rule{negativethinmathspace}{0ex}}{\lambda }_{n}}^{\prime }\end{array}$(uλn) = 0. It is easy to know that there exists u$\begin{array}{}{H}_{0}^{1}\end{array}$(Ω) ∩ $\begin{array}{}{H}_{r}^{1}\end{array}$(Ω) ⊂ Hλ(Ωρ) such that

$uλn→u in Hλ(Ωρ) as n→+∞$

and (u|Ω1)+, (u|Ω2), (u|Ω3)± ≠ 0. Moreover,

$m¯λn=J¯λn(uλn)→J(u),$

and

$0=J¯λn′(uλn)→J′(u).$

By the definition of m, one has that

$limλn→+∞m¯λn=J(u)≥m.$

Using conclusion (i), we obtain that λnm as n → +∞.

In Section 2, we have known that there exists vH, that is

$v∈M, J(v)=m, J′(v)=0.$(4.3)

and v1 = v|Ω1 is positive, v2 = v|Ω2 is negative and v3 = v|Ω3 changes sign exactly once. At the same time, we can find two positive constants τ2 > τ1 such that

$τ1≤∥v1∥,∥v2∥,∥v3+∥,∥v3−∥≤τ2.$(4.4)

We define y0 : $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}\right]\end{array}$Hλ by

$y0(t1,t2,t3,t4):=t1v1+t2v2+t3v3++t4v3−$(4.5)

and

$mλ:=infy∈Σλmaxt∈[12,32]4Φλ(y(t)),$(4.6)

where

$Σλ:={y∈C([12,32]4,Hλ):∥y(t)∥λ≤6τ2+τ1,(y|Ω1ρ)+,(y|Ω2ρ)−,(y|Ω3ρ)±≠0 and y=y0 on ∂[12,32]4}.$(4.7)

It is easy to check y0Σλ, so Σλ ≠ ∅ and mλ is well defined.

The next Lemma is trivial by degree theory, so we omit the detail.

#### Lemma 4.2

For any yΣλ, there exists an 4-tuple $\begin{array}{}{t}^{\ast }=\left({t}_{1}^{\ast },{t}_{2}^{\ast },{t}_{3}^{\ast },{t}_{4}^{\ast }\right)\in D=\left(\frac{1}{2},\frac{3}{2}{\right)}^{4}\end{array}$ such that $\begin{array}{}〈{\overline{J}}_{\lambda }^{\prime }\left(y\left({t}^{\ast }\right){|}_{{\mathit{\Omega }}^{\rho }}\right),{y}_{1}^{+}\left({t}^{\ast }\right)〉=〈{\overline{J}}_{\lambda }^{\prime }\left(y\left({t}^{\ast }\right){|}_{{\mathit{\Omega }}^{\rho }}\right),{y}_{2}^{-}\left({t}^{\ast }\right)〉=0\end{array}$ and $\begin{array}{}〈{\overline{J}}_{\lambda }^{\prime }\left(y\left({t}^{\ast }\right){|}_{{\mathit{\Omega }}^{\rho }}\right),{y}_{3}^{±}\left({t}^{\ast }\right)〉=0\end{array}$ where $\begin{array}{}{y}_{i}\left(t\right)=y\left(t\right){\mid }_{{\mathit{\Omega }}_{i}^{\rho }}\end{array}$ for i = 1, 2, 3.

#### Lemma 4.3

There holds that

1. λmλm for all λ ≥ 1;

2. mλm as λ → +∞;

3. There exists ϵ0 > 0 such that Φλ(y(t)) < mϵ0 for all λ ≥ 0, yΣλ and t = (t1, t2, t3, t4) ∈ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$.

#### Proof

1. Since y0Σλ, we have

$mλ≤maxt∈[12,32]4Φλ(y0(t))=maxt∈[12,32]4J(y0(t))=m.$

Now, fixing $\begin{array}{}{t}^{\ast }\in \left(\frac{1}{2},\frac{3}{2}{\right)}^{4}\end{array}$ given by Lemma 4.2, it implies

$m¯λ≤J¯λ(y(t∗)|Ωρ).$

By the definition of g(|x|, u), we deduce that $\begin{array}{}|G\left(|x|,u\right)|\le \frac{{\nu }_{0}}{2}{u}^{2}\end{array}$ for x ∈ ℝ2Ωρ. By (3.1) we can get

$Φλ(y(t∗))≥J¯λ(y(t∗)|Ωρ).$

Therefore,

$maxt∈[12,32]4Φλ(y(t))≥J¯λ(y(t∗)|Ωρ)≥m¯λ, for each y∈Σλ.$

So mλλ.

2. it is obtained by Lemma 4.1 (ii) and Lemma 4.3 (i).

3. For t = (t1, t2, t3, t4) ∈ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$, we have

$Φλ(y(t))=J(y0(t)) for t=(t1,t2,t3,t4)∈∂[12,32]4.$

By Lemma 2.1, it is to get

$Φλ(y(t))

where ϵ0 is a small positive constant.

## 5 Proof of Theorem 1.1

In this section, we prove our main results. Define

$S:={u∈M∣J(u)=m}.$

Then we need further to study the properties of the set S.

#### Lemma 5.1

S is compact in H.

#### Proof

The proof is standard, we omit it immediately.

#### Lemma 5.2

Let d > 0 be a fixed number and let {un} ⊂ Sd be a sequence. Then, up to a subsequence, unu0 in Hλ as n → ∞, and u0S2d where

$Sd:={u∈Hλ:distλ(u,S)≤d}$

and distλ denotes the distance in Hλ.

#### Proof

Since S is compact in H, we can choose {ūn} ⊂ S satisfy

$distλ(un,u¯n)≤d.$

On the other hand, there exists ūS such that, up to a subsequence, ūnū in H. Hence, dist(un, u) ≤ d for n large enough. Thus {un} is bounded in Hλ. Up to a subsequence, unu0 weakly in Hλ. Since B2d(u) is weakly closed in Hλ, so u0B2d(u) ⊂ S2d.

#### Lemma 5.3

Let d ∈ (0, τ1), where τ1 is given by (4.4). Suppose that there exist a sequence λn > 0 with λn → ∞, and {un} ⊂ Sd satisfying

$limn→∞Φλn(un)≤m, limn→∞Φλn′(un)=0.$

Then, up to a subsequence, {un} converges strongly in $\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2) to an element uS.

#### Proof

Observe that, by limn→∞ Φλn(un) ≤ m and limn→∞ $\begin{array}{}{\mathit{\Phi }}_{{\lambda }_{n}}^{\prime }\end{array}$(un) = 0 we deduce that {∥uuλn} and {Φλn(un)} are bounded. Up to a subsequence, we may assume that Φλn(un) → cm. By Proposition 3.1, there exists u$\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2) such that unu in $\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2), u = 0 in ℝ2Ω and Φλn(un) → J(u). Moreover, u is a solution of equation (1.9). Next we prove that uS. Since {un} ⊂ Sd and d ∈ (0, τ1), we can deduce that (u|Ω1)+ ≠ 0, (u|Ω2) ≠ 0 and (u|Ω3)± ≠ 0. Indeed, if the conclusion is not correct, we can assume (u|Ω1)+ = 0, we can choose {ūn} ⊂ S satisfy

$distλ(un,u¯n)≤d,$

so

$τ1≤∥u¯n,1∥≤∥u¯n,1−un,1∥Hλn+∥un,1∥≤d,$

which implies that a contradiction. Hence, by Proposition 3.1 again, we get J′(u) = 0, u = 0 in ℝ2Ω. Then we get that J(u) ≥ m. At the same time, Φλn(un) → J(u) ≤ m, therefore uS.

#### Lemma 5.4

Let δ ∈ (0, τ1), where τ1 is given by (4.4). Then there exist constants 0 < σ < 1 and Λ1 > 0 such that $\begin{array}{}\parallel {\mathit{\Phi }}_{\lambda }^{\prime }\left(u\right){\parallel }_{{H}_{\lambda }^{\ast }}\ge \sigma \end{array}$ for any $\begin{array}{}u\in {\mathit{\Phi }}_{\lambda }^{m}\cap \left({S}^{\delta }\setminus {S}^{\frac{\delta }{2}}\right)\end{array}$ and λΛ1.

#### Proof

We prove it by contradiction. Suppose that there exist a number δ0 ∈ (0, τ1), a positive sequence {λj} with λj → 0, and a sequence of function $\begin{array}{}\left\{{u}_{j}\right\}\subset {\mathit{\Phi }}_{{\lambda }_{j}}^{m}\cap \left({S}^{{\delta }_{0}}\setminus {S}^{\frac{{\delta }_{0}}{2}}\right)\end{array}$ such that

$limj→+∞Φλj′(uj)=0.$

Up to a subsequence, we get {uj} ⊂ Sδ0 and limj→∞ Φλj(uj) ≤ m. By Lemma 5.3, we can deduce that there exists uS such that uju in Hλj(ℝ2). Therefore, distλj(uj, S) → 0 as j → +∞. This contradict the assumption that $\begin{array}{}{u}_{j}\notin {S}^{\frac{{\delta }_{0}}{2}}.\end{array}$

#### Lemma 5.5

There exist Λ2Λ1 and α > 0 such that for any λΛ2,

$Φλ(y0(t1,t2,t3,t4))≥mλ−α$

implies that y0(t1, t2, t3, t4) ∈ $\begin{array}{}{S}^{\frac{\delta }{2}}\end{array}$ for some δ ∈ (0, τ1).

#### Proof

We argue by contradiction. There exist λn → ∞, αn → 0 and $\begin{array}{}\left({t}_{1}^{\left(n\right)},{t}_{2}^{\left(n\right)},{t}_{3}^{\left(n\right)},{t}_{4}^{\left(n\right)}\right)\in \left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$ such that

$Φλ(y0(t1(n),t2(n),t3(n),t4(n)))≥mλn−αn and y0(t1(n),t2(n),t3(n),t4(n))∉Sδ2.$

We can choose a subsequence $\begin{array}{}\left({t}_{1}^{\left(n\right)},{t}_{2}^{\left(n\right)},{t}_{3}^{\left(n\right)},{t}_{4}^{\left(n\right)}\right)\to \left({\overline{t}}_{1},{\overline{t}}_{2},{\overline{t}}_{3},{\overline{t}}_{4}\right)\in \left[\frac{1}{2},\frac{3}{2}{\right]}^{4}.\end{array}$ Then we have

$J(y0(t¯1,t¯2,t¯3,t¯4))≥limn→∞(mλn−αn)=m.$

By the unique of 4-tuple, it is easy to have (1, 2, 3, 4) = (1, 1, 1, 1). It implies that

$limn→∞∥y0(t1(n),t2(n),t3(n),t4(n))−y0(1,1,1,1)∥=0.$

Since y0(1, 1, 1, 1) = vS, which contradicts the assumption.

Now, we define

$α0:=min{a2,ϵ02,18δσ2},$(5.1)

where δ, σ, α, ϵ0 are from Lemma 5.4, Lemma 5.5 and Lemma 4.3-(iii) respectively. Using Lemma 4.2, one has that there exists Λ3Λ2 such that

$|mλ−m|<α0 for all λ≥Λ3.$(5.2)

#### Lemma 5.6

There exists a critical point uλ of Φλ with uλSδ$\begin{array}{}{\mathit{\Phi }}_{\lambda }^{m}\end{array}$ for λΛ3.

#### Proof

We argue by contradiction. Fix a λΛ3, by the Lemma 5.3, we can assume that there exists 0 < ρλ < 1 such that ∥$\begin{array}{}{\mathit{\Phi }}_{\lambda }^{\prime }\end{array}$(u)∥ ≥ ρλ on Sδ$\begin{array}{}{\mathit{\Phi }}_{\lambda }^{m}\end{array}$. There exists a pseudo-gradient vector field Kλ in Hλ which is defined on a neighborhood Zλ of Sδ$\begin{array}{}{\mathit{\Phi }}_{\lambda }^{m}\end{array}$ such that for any uZλ there holds

$∥Kλ(u)∥≤2min{1,∥Φλ′(u)∥},〈Φλ′(u),Kλ(u)〉≥min{1,∥Φλ′(u)∥}∥Φλ′(u)∥.$

Define ψλ be a Lipschitz function on Hλ such that 0 ≤ ψλ ≤ 1, ψλ ≡ 1 on Sδ$\begin{array}{}{\mathit{\Phi }}_{\lambda }^{m}\end{array}$ and ψλ ≡ 0 on HλZλ. Define ξλ be a Lipschitz function on ℝ such that 0 ≤ ξλ ≤ 1, ξλ(t) ≡ 1 if |tmλ| ≤ $\begin{array}{}\frac{\alpha }{2}\end{array}$ and ξλ(t) ≡ 0 if |tmλ| ≥ α. Let

$eλ(u):=−ψλ(u)ξλ(Φλ(u))Kλ(u), if u∈Zλ,0, if u∈Hλ∖Zλ.$

Then there exists a global solution ηλ : Hλ × [0, +∞) → Hλ for the initial value problem

$ddθηλ(u,θ)=eλ(ηλ(u,θ)),ηλ(u,0)=u.$

We can deduce that ηλ has the following properties:

1. ηλ(u, θ) = u if θ = 0 or uHλZλ or |Φλ(u) – mλ| ≥ α;

2. $\begin{array}{}\frac{d}{d\theta }\end{array}$ηλ(u, θ)∥ ≤ 2;

3. $\begin{array}{}\frac{d}{d\theta }\end{array}$Φλ(ηλ(u, θ)) = 〈$\begin{array}{}{\mathit{\Phi }}_{\lambda }^{\prime }\end{array}$(ηλ(u, θ)), eλ(θλ(u, θ))〉 ≤ 0.

#### Assertion 1

For any (t1, t2, t3, t4) ∈ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$, there exists θ̄ = θ(t1, t2, t3, t4) ∈ [0, +∞) such that ηλ(y0(t1, t2, t3, t4), θ̄) ∈ $\begin{array}{}{\mathit{\Phi }}_{\lambda }^{{m}_{\lambda }-{\alpha }_{0}}.\end{array}$

Assuming by contradiction that there exists (t1, t2, t3, t4) ∈ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$ such that

$Φλ(ηλ(y0(t1,t2,t3,t4),θ))>mλ−α0$

for any θ ≥ 0. By Lemma 5.5, we get y0(t1, t2, t3, t4) ∈ $\begin{array}{}{S}^{\frac{\delta }{2}}\end{array}$. Note Φλ(y0(t1, t2, t3, t4)) ≤ mλ + α0, due the property (3) of ηλ,

$mλ−α0<Φλ(ηλ(y0(t1,t2,t3,t4),θ))<Φλ(y0(t1,t2,t3,t4))≤m≤mλ+α0.$

So we can deduce that ξλ(Φλ(ηλ(y0(t1, t2, t3, t4), θ))) ≡ 1. If ηλ(y0(t1, t2, t3, t4), θ) ∈ Sδ for all θ ≥ 0, so it imply that

$ψ(ηλ(y0(t1,t2,t3.t4),θ))≡1 and ∥Φλ′(ηλ(y0(t1,t2,t3,t4),θ))∥≥ρλ for all θ>0.$

It follows that

$Φλ(ηλ(y0(t1,t2,t3,t4),αρλ2))≤mλ+α2−∫0αρλ2ρλ2dt≤mλ−α2≤mλ−α0,$

which is a contradiction. Thus, there exists θ3 > 0 such that ηλ(y0(t1, t2, t3, t4), θ3) ∉ Sδ. Note that y0(t1, t2, t3, t4) ∈ $\begin{array}{}{S}^{\frac{\delta }{2}}\end{array}$, there exist 0 < θ1 < θ2 < θ3 such that

$ηλ(y0(t1,t2,t3,t4),θ1)∈∂Sδ2, ηλ(y0(t1,t2,t3,t4),θ2)∈∂Sδ$

and ηλ(y0(t1, t2, t3, t4), θ) ∈ Sδ$\begin{array}{}{S}^{\frac{\delta }{2}}\end{array}$ for all θ ∈ (θ1, θ2). By Lemma 5.4, one has

$∥Φλ′(ηλ(y0(t1,t2,t3,t4),θ))∥≥σ for all θ∈(θ1,θ2).$

Using the property (2) of ηλ we get that

$δ2≤∥ηλ(y0(t1,t2,t3,t4),θ2)−ηλ(y0(t1,t2,t3,t4),θ1)∥≤2|θ2−θ1|.$

This deduce that

$Φλ(ηλ(y0(t1,t2,t3,t4),θ2))≤Φλ(ηλ(y0(t1,t2,t3,t4),0))+∫0θ2ddθΦλ(ηλ(u,v,θ))dθ<Φy(y0(t1,t2,t3,t4))+∫θ1θ2ddθΦλ(ηλ(u,v,θ))dθ≤mλ+α0−σ2(θ2−θ1)≤mλ+α0−14δσ2≤mλ−α0,$

which is a contradiction. Therefore, we prove the assertion 1.

Now, we define

$Γ(t1,t2,t3,t4):=inf{θ≥0:Φλ(ηλ(y0(t1,t2,t3,t4),θ))≤mλ−α0}$

and

$y~(t1,t2,t3,t4):=ηλ(y0(t1,t2,t3,t4),Γ(t1,t2,t3,t4)).$

Then Φλ((t1, t2, t3, t4)) ≤ mλα0 for all (t1, t2, t3, t4) ∈ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$.

#### Assertion 2

(t1, t2, t3, t4) = ηλ(y0(t1, t2, t3, t4), Γ(t1, t2, t3, t4)) ∈ Σλ.

For any (t1, t2, t3, t4) ∈ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$, we have

$Φλ(y0(t1,t2,t3,t4))≤J(y0(t1,t2,t3,t4))

which implies that Γ(t1, t2, t3, t4) = 0. So (t1, t2, t3, t4) = y0(t1, t2, t3, t4) for (t1, t2, t3, t4) ∈ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$. We also need to prove ∥(t1, t2, t3, t4)∥ ≤ 6τ2 + τ1 for all $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$ and Γ(t1, t2, t3, t4) is continuous with respect to (t1, t2, t3, t4).

For any (t1, t2, t3, t4) ∈ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$, we have Γ(t1, t2, t3, t4) = 0 if Φλ(y0(t1, t2, t3, t4)) ≤ mλα0, so (t1, t2, t3, t4) = y0(t1, t2, t3, t4). By the definition of y0(t), we have ∥(t1, t2, t3, t4)∥ ≤ 6τ2 < 6τ2 + τ1.

On the other hand, if Φλ(y0(t1, t2, t3, t4)) > mλα0, it implies that y0(t1, t2, t3, t4) ∈ $\begin{array}{}{S}^{\frac{\delta }{2}}\end{array}$ and

$mλ−α0<Φλ(ηλ(y0(t1,t2,t3,t4),θ))

So one has

$ξλ(Φλ(ηλ(y0(t1,t2,t3,t4),θ)))≡1 for all θ∈[0,Γ(t1,t2,t3,t4)).$

Now, we will prove (t1, t2, t3, t4) ∈ Sδ. Otherwise, (t1, t2, t3, t4) ∉ Sδ, similar to the proof of assertion 1, we can find two constants 0 < θ1 < θ2 < Γ(t1, t2, t3, t4) such that

$Φλ(ηλ(y0(t1,t2,t3,t4),θ2))

It contradicts to the definition of Γ(t1, t2, t3, t4). Therefore

$y~(t1,t2,t3,t4)=ηλ(y0(t1,t2,t3,t4),Γ(t1,t2,t3,t4))∈Sδ.$

Thus, there exists uS, such that

$∥y~(t1,t2,t3,t4)∥≤∥u∥+τ1≤6τ2+τ1.$

To prove the continuity of Γ(t1, t2, t3, t4), we fix arbitrarily (t1, t2, t3, t4) ∈ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$. First, we assume that Φλ((t1, t2, t3, t4)) < mλα0. In this case, it is to see that Γ(t1, t2, t3, t4) = 0, which gives that Φλ(y0(t1, t2, t3, t4)) < mλα0. By the continuity of y0, there exists r > 0 such that for any (s1, s2, s3, s4) ∈ Br(t1, t2, t3, t4) ∩ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$, we have Φλ(y0(s1, s2, s3, s4)) < mλα0, so Γ(s1, s2, s3, s4) = 0, and hence Γ is continuous at (t1, t2, t3, t4). On the other hand, we assume that Φλ((t1, t2, t3, t4)) = mλα0. Similar to the proof of assertion 1, we can deduce that (t1, t2, t3, t4) = ηλ(y0(t1, t2, t3, t4), Γ(t1, t2, t3, t4)) ∈ Sδ, thus

$∥Φλ′(ηλ(y0(t1,t2,t3,t4),Γ(t1,t2,t3,t4)))∥≥ρλ>0.$

Therefore, for any w > 0, we have

$Φλ(ηλ(y0(t1,t2,t3,t4),Γ(t1,t2,t3,t4)+w))

Using the continuity of ηλ, there exists r > 0 such that

$Φλ(ηλ(y0(s1,s2,s3,s4),Γ(t1,t2,t3,t4)+w))

for any (s1, s2, s3, s4) ∈ Br(t1, t2, t3, t4) ∩ $\begin{array}{}\left[\frac{1}{2},\frac{3}{2}{\right]}^{4}\end{array}$. Thus, Γ(s1, s2, s3, s4) ≤ Γ(t1, t2, t3, t4) + w. It follows that

$0≤lim sup(s1,s2,s3,s4)→(t1,t2,t3,t4)Γ(s1,s2,s3,s4)≤Γ(t1,t2,t3,t4).$(5.3)

If Γ(t1, t2, t3, t4) = 0, we immediately obtain that

$lim(s1,s2,s3,s4)→(t1,t2,t3,t4)Γ(s1,s2,s3,s4)=Γ(t1,t2,t3,t4).$

If Γ(t1, t2, t3, t4) > 0, we can similarly deduce that

$Φλ(ηλ(y0(s1,s2,s3,s4),Γ(t1,t2,t3,t4)−w))>mλ−α0$

for any 0 < w < Γ(t1, t2, t3, t4).

By the continuity of ηλ again, we see that

$lim inf(s1,s2,s3,s4)→(t1,t2,t3,t4)Γ(s1,s2,s3,s4)≥Γ(t1,t2,t3,t4).$(5.4)

Combining (5.3) and (5.4), it is easy to see Γ is continuous at (t1, t2, t3, t4). This completes the proof of Assertion 2.

Thus, we have proved that (t1, t2, t3, t4) ∈ Σλ and

$max(t1,t2,t3,t4)∈[12,32]4Φλ(y~(t1,t2,t3,t4))≤mλ−α0$

which contradicts the definition of mλ. This completes the proof.

#### Proof of Theorem 1.1

We still prove it with T1 = {1}, T2 = {2} and T3 = {3}. By Lemma 5.6, when λ > Λ3, we can get that there exists a solution uλSδ$\begin{array}{}{\mathit{\Phi }}_{\lambda }^{m}\end{array}$ for equation (3.3). By Lemma 3.2, we can know that uλ is a solution of equation (1.5) when λ > Λ := max{Λ0, Λ3}. Moreover, combining with Lemma 5.3, uλuS (up to subsequence) strongly in $\begin{array}{}{H}_{r}^{1}\end{array}$(ℝ2). So, we complete the proof of Theorem 1.1.

## Acknowledgement

This work was supported by the NNSF (Nos. 11571370, 11601145, 11701173), by the Natural Science Foundation of Hunan Province (Nos. 2017JJ3130, 2017JJ3131), by the Excellent youth project of Education Department of Hunan Province (17B143, 17A113, 18B342), by the Hunan University of Commerce Innovation Driven Project for Young Teacher (16QD008), and by the Project of China Postdoctoral Science Foundation (2019M652790).

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Accepted: 2019-07-11

Published Online: 2019-09-20

Published in Print: 2019-03-01

Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 1066–1091, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496,

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