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Planar vortices in a bounded domain with a hole

  • In this paper, we consider the inviscid, incompressible planar flows in a bounded domain with a hole and construct stationary classical solutions with single vortex core, which is closed to the hole. This is carried out by constructing solutions to the following semilinear elliptic problem

    {Δψ=λ(ψκ4πlnλ)p+,inΩ,ψ=ρλ,onO0,ψ=0,onΩ0,(1)

    where p>1, κ is a positive constant, ρλ is a constant, depending on λ, Ω=Ω0ˉO0 and Ω0, O0 are two planar bounded simply-connected domains. We show that under the assumption (lnλ)σρλ(lnλ)1σ for some σ>0 small, (1) has a solution ψλ, whose vorticity set {yΩ:ψ(y)κ+ρλη(y)>0} shrinks to the boundary of the hole as λ+.

    Citation: Shusen Yan, Weilin Yu. Planar vortices in a bounded domain with a hole[J]. Electronic Research Archive, 2021, 29(6): 4229-4241. doi: 10.3934/era.2021081

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  • In this paper, we consider the inviscid, incompressible planar flows in a bounded domain with a hole and construct stationary classical solutions with single vortex core, which is closed to the hole. This is carried out by constructing solutions to the following semilinear elliptic problem

    {Δψ=λ(ψκ4πlnλ)p+,inΩ,ψ=ρλ,onO0,ψ=0,onΩ0,(1)

    where p>1, κ is a positive constant, ρλ is a constant, depending on λ, Ω=Ω0ˉO0 and Ω0, O0 are two planar bounded simply-connected domains. We show that under the assumption (lnλ)σρλ(lnλ)1σ for some σ>0 small, (1) has a solution ψλ, whose vorticity set {yΩ:ψ(y)κ+ρλη(y)>0} shrinks to the boundary of the hole as λ+.



    In this paper, we consider a planar incompressible flow in a bounded smooth domain

    Ω=Ω0ˉO0,

    where O0, Ω0 are two bounded simply-connected open subsets of R2, such that ˉO0Ω0. A simple model describing this flow is

    {ψω=0,inΩ,Δψ=ω,inΩ,ψ=constant,onO0,ψ=0,onΩ0, (2)

    where ψ and ω are the stream function and the vorticity of this flow, respectively, and ψ:=(2ψ,1ψ). For a detailed presentation of this model, we refer the readers to [9].

    An existence result obtained by Smets and Schaftingen [15] via a variational method shows that (2) has a solution (ψλ,ωλ), such that ψλ=ρλ on O0,

    O0ψλν=0,

    and ωλ=λ(ψλqλ)p+, where 1<p<+, qλ=q+κ4πlnλ with κ>0 and q is a harmonic function in Ω. Moreover, as λ+, the total vorticity

    Ωωλκ,

    and the vorticity set {yΩ|ωλ(y)>0} shrinks to a point in Ω, which is a critical point of the Kirchhoff-Routh function corresponding to κ and q. For flows past obstacles, we refer the readers to [11,12,13,14] for other results.

    Note that in [15], the value ρλ of ψλ on O0 is as a lagrangian multiplier, which is unknown. In this paper, we assume that ρλ is a prescribed constant, and we remove the condition

    O0ψν=0.

    The first equation in (2) suggests that ψ and ω are functionally dependent. So, for simplicity, we consider the following elliptic equation

    {Δψ=λ(ψκ4πlnλ)p+,inΩ,ψ=ρλ,onO0,ψ=0,onΩ0, (3)

    where 1<p<+, κ>0 is a constant, and ρλ>0 is a constant, depending on λ.

    In this paper, we mainly focus on the solvability of (3), and the effect from the constant ρλ on the location of the vorticity set Ωλ:={yΩ|ψ(y)>κ4πlnλ}. We expect that for large ρλ>0, the vorticity set Ωλ concentrates near the boundary of O0, as λ+.

    Let η be the unique solution of the following problem

    Δη=0inΩ,η=1onO0,η=0onΩ0. (4)

    Making the change of ψ=lnλ4πu+ρλη, ε=λ12(lnλ4π)1p2 and λε=4πlnλρλ, (3) can be changed into

    {ε2Δu=(uκ+λεη)p+,inΩ,u=0,onΩ. (5)

    Now we want to find a solution to (3) by constructing a solution for (5), whose vorticity set is close to the boundary of O0. For this purpose, the following assumption can be imposed on ρλ:

    (Hλ) There is a small constant σ>0, such that

    (lnλ)σρλ(lnλ)1σ.

    It is easy to see that if ρλ satisfies the condition (Hλ), then λε satisfies the condition:

    (Hε) There are constants γ1 and γ2 with 0<γ1<γ2<1, such that

    1|lnε|γ2λε1|lnε|γ1.

    Before we state the main result, we give the following definition.

    Definition 1.1. Let f:RR be a continuous function. We call I=[a.b], ab a minimum interval of f, if f(t1)=f(t2) for any t1,t2I, and there is σ0>0 such that for 0<σ<σ0, f(aσ)>f(a) and f(b+σ)>f(b).

    Our main result of this paper can be stated as follows.

    Theorem 1.2. Let κ be a given positive number. Suppose that ρλ satisfies (Hλ). Then there is a constant C0>0, such that for any λ>C0, (3) has a solution ψλ, such that

    {yΩ|ψλ(y)>κ4πlnλ}BLλ(xλ),

    where xλΩ and L>0 is a constant independent of λ.

    Moreover, as λ+,

    dist(xλ,Ω)0,λΩ(ψλκ4πlnλ)p+κ.

    In particular, if I is a minimum interval of η(x(s))ν defined for x(s)Ω, then as λ+,

    dist(xλ,x(I))0.

    As a result of Theorem 1.2, we obtain a flow in Ω with single non-vanishing anti-clockwise vortex, which concentrates on the boundary of O0. Let us point out that we can also construct a planar Euler flow with single clockwise vortex, which nears the boundary of Ω0. This is carried out by considering the problem (3) with nonlinearity λ(ψκ4πlnλ)p+ replaced by λ(ψκ4πlnλ)p+.

    Theorem 1.2 is proved via the following theorem.

    Theorem 1.3. Let κ be a given positive number. Suppose that λε satisfies (Hε). Then there is a constant ε0>0, such that for any 0<ε<ε0, (5) has a solution uε, such that

    {yΩ|uε(y)κ+λεη(y)>0}BLε|lnε|p12(xε),

    where xεΩ and L>0 is a constant independent of ε. Moreover, as ε0,

    dist(xε,Ω)0.

    In particular, if there is a minimum interval I of η(x(s))ν defined for x(x)Ω, then as ε0,

    dist(xε,x(I))0.

    To prove Theorem 1.3, we will use a finite reduction argument as in [6,7,8,9]. Since we consider the vortex nears the boundary, it turns out that more delicate estimates are needed in the proof of Theorem 1.3 than those estimates in [6,7,8,9]. In particular, we need the estimates of the functions η and G near the boundary, where G is the Green's function for Δ in Ω with zero boundary condition, written as

    G(y,x)=12πln1|yx|H(y,x),x,yΩ, (6)

    and H(y,x) is the regular part of the Green's function. Recall that the Robin's function is defined by φ(x)=H(x,x).

    The stationary incompressible Euler equations have been studied by many authors, see for instance [1,3,4,5,6,7,8,9,10,15,16,17] and references therein. Roughly speaking, there are two commonly used methods to study the existence of the stationary incompressible Euler equations: the vorticity method and the stream-function method. The vorticity method was first established by Arnold [2] and further developed by Burton [4,5] and Tukington [16]. This argument roughly consists in maximizing the kinetic energy under a constrained sublevel set of ω. In this paper, we will use the stream-function method.

    This paper is organized as follows. In section 2, we construct approximate solutions for (5). We will carry out a reduction procedure in section 3 and the results of existence will be proved in section 4. In appendix A, we give the estimates for the radius of vortex core.

    In this section, following [10], we will construct an approximate solution for (5).

    For p>1, there is a unique solution ϕ for the following problem:

    Δϕ=ϕp,ϕ>0,ϕH10(B1(0)). (7)

    Moreover, ϕ is a radial function and satisfies

    B1(0)ϕp+1=π(p+1)2|ϕ(1)|2,B1(0)ϕp=2π|ϕ(1)|.

    Let R>0 be a large constant, such that for any xΩ, Ω⊂⊂BR(x). Now we consider

    {ε2Δu=(ua)p+,yBR(0),u=0,yBR(0), (8)

    where a>0 is a constant. Then, (8) has a solution Uε,a, which can be represented by

    Uε,a(y)={a+(εsε)2p1ϕ(|y|sε),|y|sε,aln|y|R/lnsεR,sε|y|R, (9)

    where and ϕ(y) is a radial solution of (7), and sε is a constant, such that Uε,aC1(BR(0)). So, sε is determined by

    (εsε)2p1ϕ(1)=alnsεR, (10)

    which gives an expansion for sε as follows

    sε=(|ϕ(1)|a)p12ε|lnε|p12(1+O(ln|lnε||lnε|)). (11)

    For any xΩ, define Uε,x,a(y)=Uε,a(yx). Because Uε,x,a does not vanish on Ω, we need to make a projection. Let PUε,x,a be the solution of

    {ε2Δu=(Uε,x,aa)p+,yΩ,u=0,yΩ.

    Then

    PUε,x,a(y)=Uε,x,a(y)alnRsεg(y,x), (12)

    where g(y,x)=lnR+2πH(y,x) and H(y,x) is the regular part of the Green's function G defined by (6).

    Take

    d1=r1|lnε|λε,d2=r2|lnε|λε, (13)

    where 0<r1<r2 are two fixed constants, which will be determined later, and set

    S=:(O0)d2¯(O0)d1, (14)

    where (O0)di={yΩ|dist(y,O0)<di} is the neighborhood of O0. Denote d(x)=dist(x,O0) and take ˆxO0 such that d(x)=|xˆx|. Define

    r(x)=|lnε|λεd(x). (15)

    Then we can rewrite S as follows

    S={xΩ:r1<r(x)<r2}. (16)

    In the rest of this paper, we assume that xS. We want to construct solutions for (5) of the form

    u=PUε,x,aε+ωε, (17)

    where ωε is a perturbation term. To make PUε,x,aε a good approximate solution, we need to require that aε and sε satisfy

    (εsε)2p1ϕ(1)=aεlnsεR,aε=κλεη(x)+aεlnRsεg(x,x). (18)

    Let us now show that (18) is solvable for ε>0 small. From the Taylor expansion we have

    η(x)=η(ˆx)η(ˆx)νd(x)+O(d2(x))=1η(ˆx)νd(x)+O(d2(x)),

    where ν is the outward unit normal of Ω. On the other hand, the following expansion for Robin's function is proved in [15] Appendix B:

    φ(x)=12πln12d(x)+O(d(x)).

    Then by our assumption on λε, as in Lemma 2.1 in [6], we can solve (18) to obtain aε(x) and sε(x). For simplicity, we use aε and sε instead of aε(x) and sε(x), respectively. Then for yBLsε(x), where L>0 is any fixed constant, we have

    PUε,x,aε(y)κ+λεη(y)=Uε,x,aε(y)κ+λεη(y)aεlnRsεg(y,x)=Uε,x,aε(y)aε+λεη(x),yxaεlnRsεg(x,x),yx+O(s1+σε), (19)

    where 0<σ<1 is a small constant.

    Remark 1. It is not difficult to get the following expansions:

    1lnRsε=1Aε+O(λε|lnε|2),aε=(1+lnR|lnε|)κλε+λεη(ˆx)νd(x)+κ|lnε|ln12d(x)+O(1|lnε|1+σ), (20)

    where σ>0 is a small constant and

    Aε=|lnε|p12ln|lnε|+lnR+p12lnκ|ϕ(1)|.

    Moreover, in view of φ(x)xi=O(1d(x)), we can prove that

    aεxi=O(λε),sεxi=O(λεsε),i=1,2. (21)

    By (9) and (21), we have the following expansion, which will be used in the rest of this paper,

    Uε,x,aε(y)xi={aε|ϕ(1)|1lnRsεϕ(|yx|sε)xiyi|yx|1sε+O(λε|lnε|),yBsε(x),aεlnRsεxiyi|yx|2+O(λεln|yx||lnε|),yΩBsε(x). (22)

    In this section, we reduce the problem of finding a solution for (5) of the form (17) to a finite dimension problem.

    Define

    Eε,x={u:uW2,p(Ω)H10(Ω),ΩuΔPUε,x,aεxidy=0,i=1,2},

    and

    Fε,x={u:uLp(Ω),ΩuPUε,x,aεxidy=0,i=1,2}.

    For any uLp(Ω), define the following projection

    Qεu=:u+2j=1bjε2ΔPUε,x,aεxj,

    where b1 and b2 are the constants such that QεuFε,x. Thus b1 and b2 should satisfy

    2j=1bjε2ΩPUε,x,aεxjPUε,x,aεxi=ΩuPUε,x,aεxi. (23)

    The existence of b1 and b2 can be obtained by the following estimate

    ε2ΩPUε,x,aεxjPUε,x,aεxi=pΩ(Uε,x,aεaε)p1+(Uε,x,aεxjaεxj)PUε,x,aεxi=c(δij+o(1))1|lnε|p+1, (24)

    where c>0 is a constant, δij=1, if i=j and δij=0, if ij.

    Set

    Lεu=ε2Δup(PUε,x,aεκ+λεη)p1+u. (25)

    We have the following result for the operator QεLε.

    Proposition 1. There are constants ε0>0 and σ0>0, such that for any ε(0,ε0), xS, uEε,x with QεLεu=0 in ΩBLsε(x) for some large L>0, then

    s22pεε2QεLεuLp(BLsε(x))σ0uL(Ω).

    As a consequence, QεLε is one to one and onto from Eε,x to Fε,x.

    Proof. Suppose to the contrary that there are εn0, xnSn, unEεn,xn with QεnLεnun=0 in ΩBLsεn(xn), and unL(Ω)=1, such that

    s22pεnε2nQεnLεnunLp(BLsεn(xn))1n. (26)

    First of all, we estimate b1,n and b2,n in the following formula:

    QεnLεnun=Lεnun+2j=1bj,nε2nΔPUεn,xn,aεnxj, (27)

    where b1,n and b2,n satisfy

    2j=1bj,nε2nΩPUεn,xn,aεnxjPUεn,xn,aεnxi=ΩPUεn,xn,aεnxiLεnun. (28)

    From (19), (21) and Lemma A.1, we have

    ΩPUεn,xn,aεnxiLεnun=ΩunLεnPUεn,xn,aεnxi=pΩ[(Uεn,xn,aεnaεn)p1+(Uεn,xn,aεnxiaεnxi)(PUεn,xn,aεnκ+λεnη)p1+PUεn,xn,aεnxi]un=pΩ[(Uεn,xn,aεnaεn)p1+Uεn,xn,aεnxi(Uεn,xn,aεnaεn+O(λεnsεn))p1+PUεn,xn,aεnxi]un+O(λεns2εn|lnεn|p1)=O(λεns2εn|lnεn|p1). (29)

    Then by (24), (28) and (29), we obtain

    bj,n=O(λεns2εn|lnεn|2). (30)

    Write

    ε2nΔun=p(PUεn,xn,aεnκ+λεnη)p1+un+fn, (31)

    where

    fn=QεnLεnun2j=1bj,nε2nΔPUεn,xn,aεnxj,

    Define

    ˜un(y)=un(sεny+xn),˜fn(y)=fn(sεny+xn).

    Then we have

    Δ˜un=ps2εnε2n(~PUεn,xn,aεnκ+λεn˜η)p1+˜un+s2εnε2n˜f2,n. (32)

    From (26), (30) and Lemma A.1, we find

    s2εnε2n˜fnLploc(R2)=on(1)+O(s22pεn2j=1bj,nΔPUεn,xn,aεnxjLp(Ω))=on(1). (33)

    Since the right hand side of (32) is bounded in Lploc(R2), ˜un is bounded in W2,ploc(R2). By the Sobolev embedding, ˜un is bounded in Cαloc(R2) for some α>0. So, we can assume that ˜un converges uniformly in any compact set of R2 to ˜uL(R2)C(R2). It is easy to check that ˜u satisfies

    Δ˜upwp1+˜u=0,inR2, (34)

    where

    w(y)={ϕ(|y|),|y|1,ϕ(1)ln|y|,|y|>1.

    So there exist constants b1 and b2 (see [10]), such that

    ˜u=b1wy1+b2wy2. (35)

    From unEεn,xn, we see that

    B1(0)ϕp1ϕyj˜u=0,j=1,2.

    So we get b1=b2=0. That is, ˜u0. Then we find

    un=on(1),inC(BLsεn(xn)). (36)

    By our assumption,

    QεnLεnun=0,inΩBLsεn(xn).

    We find that

    Δun=0,inΩBLsεn(xn).

    However, un=0 on Ω and un=on(1) on BLsεn(xn). By the maximum principle,

    un=on(1),inΩBLsεn(xn).

    So, we have proved that

    unL(Ω)=on(1),

    which contradicts unL(Ω)=1.

    Using the same argument as in Proposition 3.2 in [6], it is not difficult to prove that QεLε is one to one and onto from Eε,x to Fε,x. Therefore, we complete the proof.

    We now want to find a solution for (5) of the form PUε,x,aε+ω. Then ω should satisfy

    Lεω=lε+Rε(ω), (37)

    where

    lε=(PUε,x,aεκ+λεη)p+(Uε,x,aεaε)p+, (38)

    and

    Rε(ω)=(PUε,x,aε+ωκ+λεη)p+(PUε,x,aεκ+λεη)p+p(PUε,x,aεκ+λεη)p1+ω. (39)

    From (37), we see

    QεLεω=Qεlε+QεRε(ω). (40)

    Fot ωEε,x, using Proposition 1, (40) is equivalent to

    ω=Gεω=:(QεLε)1Qε(lε+Rε(ω)). (41)

    We have the following estimates for lε and Rε(ω).

    Lemma 3.1. It holds

    lεLp(BLsε(x))=O(λεs1+2pε|lnε|p1),

    and if ωL(Ω)=O(sε), then

    Rε(ω)Lp(BLsε(x))=O(s2pε|lnε|p2ω2L(Ω)).

    Proof. For any yBLsε(x), from (19), we have

    |lε|=|(PUε,x,aεκ+λεη)p+(Uε,x,aεaε)p+|=|(Uε,x,aεaε+O(λεsε))p+(Uε,x,aεaε)p+|C(λεsε(Uε,x,aεaε)p1++(λεsε)p)Cλεsε|lnε|p1.

    So we get

    lεLp(BLsε(x))s2pεlεL(BLsε(x))Cλεs1+2pε|lnε|p1.

    Similarly, using (19) and Lemma A.1, it ie east to see that

    Rε(ω)Lp(BLsε(x))Cω2L(Ω)(PUε,x,aεκ+λεη)p2+Lp(BLsε(x))Cs2pε|lnε|p2ω2L(Ω).

    So we complete the proof.

    Using Lemma 3.1, we can solve (41) in a standard way and obtain the following proposition.

    Proposition 2. There is ε0>0, such that for any ε(0,ε0) and xS, (40) has a unique solution ωε,xEε,x, with

    ωε,xL(Ω)=O(λεsε).

    Furthermore, ωε,x is a C1 map from xS to Eε,x.

    In this section, we will prove our main results.

    Define

    I(u)=ε22Ω|u|21p+1Ω(uκ+λεη)p+1+,

    and set

    F(x)=I(PUε,x,aε+ωε,x),

    where xS and ωε,x is found in Proposition 2. It is well-known that if x is a critical point of F, then PUε,x,aε+ωε,x is a solution of (5).

    Lemma 4.1. There holds:

    F(x)=I(PUε,x,aε)+O(λ2εs4ε|lnε|p1).

    Proof. Since ωε,xEε,x, then we have the following energy expansion

    I(PUε,x,aε+ωε,x)=I(PUε,x,aε)+I(PUε,x,aε),ωε,x+12I(PUε,x,aε)ωε,x,ωε,x+˜Rε(ωε,x)=I(PUε,x,aε)Ωlεωε,x+12Ωωε,xLεωε,x+˜Rε(ωε,x)=I(PUε,x,aε)Ωlεωε,x+12Ω(lε+Rε(ωε,x)2j=1bjε2ΔPUε,x,aεxj)ωε,x+˜Rε(ωε,x)=I(PUε,x,aε)12Ωlεωε,x+12ΩRε(ωε,x)ωε,x+˜Rε(ωε,x),

    where

    ˜Rε(ωε,x)=1p+1Ω[(PUε,x,aε+ωε,xκ+λεη)p+1+(PUε,x,aεκ+λεη)p+1+(p+1)(PUε,x,aεκ+λεη)p+ωε,xp(p+1)2(PUε,x,aεκ+λεη)p1+ω2ε,x].

    One sees

    |˜Rε(ωε,x)|CΩ(PUε,x,aεκ+λεη)p2+ω3ε,xCλ3εs5ε|lnε|p2.

    Then by Lemma 3.1, we can easily check that

    F(x)=I(PUε,x,aε)+O(λ2εs4ε|lnε|p1).

    Lemma 4.2. We have

    I(PUε,x,aε)=Cε+πκε2Aε(2λεη(ˆx)νd(x)+κ|lnε|ln12d(x))+O(ε2|lnε|2+σ),

    where σ>0 is a small constant, Aε is given in Remark 1, and Cε is a constant, depending on ε.

    Proof. We have

    ε2Ω|PUε,x,aε|2=Ω(ε2ΔPUε,x,aε)PUε,x,aε=Ω(Uε,x,aεaε)p+PUε,x,aε=Ω(Uε,x,aεaε)p+1++aεΩ(Uε,x,aεaε)p+aεlnRsεΩg(y,x)(Uε,x,aεaε)p+=π(p+1)2ε2(aεlnRsε)2+2πε2aεlnRsε(κλεη(x))+O(λεs3ε|lnε|p).

    On the other hand, by (19)

    Ω(PUε,x,aεκ+λεη)p+1+=Ω(Uε,x,aεaε+O(λεsε))p+1+=Ω(Uε,x,aεaε)p+1++O(λεs3ε|lnε|p)=π(p+1)2ε2(aεlnRsε)2+O(λεs3ε|lnε|p).

    Then we get that

    I(PUε,x,aε)=π(p1)4ε2(aεlnRsε)2+πε2aεlnRsε(κλεη(x))+O(λεs3ε|lnε|p).

    Therefore, since λε satisfies (Hε), it is not difficult from Remark 1 to see that

    I(PUε,x,aε)=Cε+πκε2Aε(2λεη(ˆx)νd(x)+κ|lnε|ln12d(x))+O(ε2|lnε|2+σ),

    where σ>0 is a small constant, Aε is given in Remark 1, and

    Cε=πε2Aε((p1)κ24Aε+(κλε)((1+lnR|lnε|)κλε)).

    Proof of Theorem 1.3. By Lemma 4.1 and Lemma 4.2, we have that for xS,

    F(x)=Cε+πκε2Aε(2λεη(ˆx)νd(x)+κ|lnε|ln12d(x))+O(ε2|lnε|2+σ). (42)

    Consider the following minimization problem

    min{F(x):xˉS}. (43)

    Let

    ρ(x)=2λεη(ˆx)νd(x)+κ|lnε|ln12d(x). (44)

    The Hopf lemma shows that for any yO0, η(y)ν>0. Define

    m1=κ2minyΩ{(η(y)ν)1},m2=κ2maxyΩ{(η(y)ν)1},

    and choose r1,r2R, such that 0<r1<m1m2<r2. Then there exists x0S, such that

    r(x0)=κ2(η(ˆx0)ν)1,

    where r(x) is given in (15). So

    ρ(x0)=2λεη(ˆx0)νd(x0)+κ|lnε|ln12d(x0)=κ|lnε|(1+ln12r(x0)+ln(λε|lnε|)).. (45)

    For xS, r(x)=r1, we have

    ρ(x)=2λεη(ˆx)νd(x)+κ|lnε|ln12d(x)=2|lnε|η(ˆx)νr1+κ|lnε|ln12r1+κ|lnε|ln(λε|lnε|)κ|lnε|(r1m2+ln12r1+ln(λε|lnε|))>h(x0), (46)

    if 0<r1<m1 small enough.

    Similarly, for xS, r(x)=r, we have

    ρ(x)κ|lnε|(r2m2+ln12r2+ln(λε|lnε|))>h(x0), (47)

    if r2m2 large enough.

    Therefore, from (42), (46) and (47), for any xS, we have

    F(x0)<F(x).

    Thus there is a minimum point xεS of F(x) in S, which is a critical point. As a result, uε=PUε,xε,aε+ωε,xε is a solution of (5).

    By our construction, we find

    Bsε(1Lsσε)(xε){yΩ:uε(y)κ+λεη(y)0}Bsε(1+Lsσε)(xε),

    and as ε0, dist(xε,O0)0.

    We remark that if there is a minimum interval I of η(x(s))ν for x(s)O0, then we can find a critical point xε of F, which nears x(I).

    In this appendix, we give the estimates for the radius of vortex core. The proofs of such results can be found in [6] Lemma A.1.

    In the following, we assume that xS, where S is given in (14).

    Lemma A.1. Let 0<σ<1 be a constant. Then there are ε0>0 and L>0 large enough, such that for any ε(0,ε0),

    PUε,x,aε(y)κ+λεη(y)>0,yBsε(1Lsσε)(x),

    while

    PUε,x,aε(y)κ+λεη(y)<0,yΩBsε(1+Lsσε)(x).

    Lemma A.2. Suppose that ω satisfies

    ωL(Ω)=O(sε).

    Then there is L>0 large enough, such that

    PUε,x,aε(y)+ω(y)κ+λεη(y)>0,yBsε(1Lsσε)(x),

    while

    PUε,x,aε(y)+ω(y)κ+λεη(y)<0,yΩBsε(1+Lsσε)(x).


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