OF GROUND STATE IN AXIALLY SYMMETRIC DOMAINS

OF GROUND STATE IN AXIALLY SYMMETRIC DOMAINS

TSUNG-FANG WU Received 9 October 2003

We letΩ(r) be the axially symmetric bounded domains which satisfy some suitable con- ditions, then the ground-state solutions of the semilinear elliptic equation inΩ(r) are nonaxially symmetric and concentrative on one side. Furthermore, we prove the neces- sary and sufficient condition for the symmetry of ground-state solutions.

1. Introduction

Let N≥2 and 2< p <2^∗, where 2^∗=2N/(N−2) for N≥3 and 2^∗= ∞ forN=2.

Consider the semilinear elliptic equation

−∆u+u= |u|^p−2u inΩ,

u=0 on∂Ω, (1.1)

whereΩis a domain inR^N. WhenΩis a bounded domain inR^N being convex in the z_idirection and symmetric with respect to the hyperplane{z_i=0}, the famous theorem by Gidas, Ni, and Nirenberg [6] (or see Han and Lin [7]): ifuis a positive solution of (1.1) belonging toC²(Ω)∩C(Ω), thenuis axial symmetric inzi.However, the axially symmetry of positive solution generally fails ifΩis not convex in thez_i direction. For instance, Dancer [5], Byeon [2,3], and Jimbo [8] proved that (1.1) in axially symmetric dumbbell-type domain has nonaxially symmetric positive solutions. Wang and Wu [13]

and Wu [15] showed the same result in a finite strip with hole. In this paper, we want to show that the symmetry and concentration behavior of ground-state solutions in axially symmetric bounded domainsΩ(r) (will be defined later), where the domainsΩ(r) are different from those of Dancer [5], Byeon [2,3], Jimbo [8], and are extensions of Wang and Wu [13] and Wu [15]. The definition of ground-state solution of (1.1) is stated as follows. Consider the energy functionalsa,b, andJinH0¹(Ω),

a(u)=

Ω

|∇u|²+u², b(u)=

Ω|u|^p, J(u)=1

2a(u)−1

pb(u). (1.2)

Copyright©2004 Hindawi Publishing Corporation Abstract and Applied Analysis 2004:12 (2004) 1019–1030 2000 Mathematics Subject Classification: 35J20, 35J25, 35J60 URL:http://dx.doi.org/10.1155/S1085337504404023

It is well known that the solutions of (1.1) are the critical points of the energy func- tionalJ. Consider the minimax problem

αΓ(Ω)= inf

γ∈Γ(Ω)max

t∈[0,1]Jγ(t), (1.3)

where

Γ(Ω)=

γ∈C[0, 1],H0¹(Ω)|γ(0)=0,γ(1)=e, (1.4) J(e)=0 ande=0.We call a non zero critical pointuofJ inH0¹(Ω) withJ(u)=αΓ(Ω) a ground-state solution. It follows easily from the mountain pass theorem of Ambrosetti and Rabinowitz [1] that such a ground-state exists. We remark that the ground-state solutions of (1.1) can also be obtained by the Nehari minimization problem

α0(Ω)= inf

v∈M0(Ω)J(v), (1.5)

whereM0(Ω)= {u∈H0¹(Ω)\{0}|a(u)=b(u)}. Note thatM0(Ω) contains every nonzero solution of (1.1) andαΓ(Ω)=α0(Ω) (see Willem [14] and Wang [12]).

Now, we consider the following assumptions of an axially symmetric unbounded do- mainΩ. For the generic pointz=(x,y)∈R^N−1×R,

(Ω1)Ωis a y-symmetric (axially symmetric) domain ofR^N, that is, (x,y)∈Ωif and only if (x,−y)∈Ω;

(Ω2)Ωis separated by ay-symmetric bounded domainD, that is, there exist two dis- joint subdomainsΩ1andΩ2ofΩsuch that

(x,y)∈Ω2if and only if (x,−y)∈Ω1,

Ω\D=Ω1∪Ω2; (1.6)

(Ω3) equation (1.1) inΩdoes not admit any solution u∈H0¹(Ω) such that J(u)= α0(Ω).

Now, we give some examples. The infinite strip with hole: Ω =A^r\ω, whereA^r= B^N−1(0;r)×Randω⊂A^ris ay-symmetric bounded domain, andΩ = {(x,y)∈R^N−1× R||x|²<|y|+ 1}. Clearly,Ω andΩ satisfy (Ω1) and (Ω2). Furthermore, by Lien, Tzeng, and Wang [9, Lemma 2.5], ifΩis a ball-up domain inR^N, then (1.1) inΩdoes not admit any solutionu∈H0¹(Ω) such that J(u)=α0(Ω). Thus, the domainΩ satisfies (Ω3).

Moreover, along the same line of the proof of Lien, Tzeng, and Wang [9, Lemma 2.5], we obtainα0(Ω)=α0(A^r). ByLemma 2.8, the domainΩ satisfies (Ω3) (or see Wang [12, Example 2.13 and Proposition 2.14]).

LetΩ(r)=Ω∩B^N(0;r) be ay-symmetric bounded domain and letΩ⁺t = {(x,y)∈Ω| y > t}andΩ⁻_t = {(x,y)∈Ω|y < t}, then our first main result is the following theorem.

Theorem1.1. Suppose thatΩsatisfies(Ω1),(Ω2), and(Ω3). Then, for eachε >0and l≥0there exists anr(ε,l)>0such that forr >r(ε, l), ifvis a ground-state solution of (1.1) inΩ(r), then either_Ω⁺_l |v|^p< εor_Ω⁻

l |v|^p< ε.

Note that, if we takeε=(p/(p−2))α0(Ω) andl=0, then there exists anr0>0 such that forr > r0, every ground-state solution of (1.1) inΩ(r) is noty-symmetric. Then, we have the following result.

Corollary1.2. Letε=(p/(p−2))α0(Ω)andl=0, then there exists anr0>0such that forr > r0, (1.1) inΩ(r)has at least three positive solutions of which one isy-symmetric and the other two are noty-symmetric.

ByTheorem 1.1, for eachε >0 andl≥0 there exists anm0∈Nsuch that for eachm≥ m0, (1.1) inΩ(m) has a ground-state solutionvmthat satisfies_Ω⁺

l|vm|^p< εor_Ω⁻₋

l|vm|^p<

ε. Then, we have the following results.

Theorem1.3. (i)The sequence{v_m}is a (PS)_α0(Ω)-sequence inH0¹(Ω)forJ;

(ii)vm0weakly inL^p(Ω)and inH0¹(Ω)asm→ ∞.

ByTheorem 1.1, the ground-state solutions of (1.1) inΩ(r) are noty-symmetric for larger. In this motivation, we consider the positive ground-state solutions of the follow- ing equation:

−∆u+u= f(u) inΘ,

u=0 on∂Θ, (1.7)

whereΘis ay-symmetric bounded domain and the nonlinear term f is usually assumed to satisfy the following conditions:

(f1) f(−t)= −f(t) and f(t)=o(|t|) neart=0;

(f2) there exist two constantsθ∈(0, 1/2) andC0>0 such that 0< F(u)≡_u

0 f(s)ds≤ θu f(u) for allu≥C0;

(f3)|f(t)| ≤C|t|^qfor some 1< q <(N+ 2)/(N−2) ifN >2, 1< q <∞ifN=2 and for larget;

(f4)∂²f /∂t²(t)≥0 fort=0.

f(t)= |t|^p−2tis a typical example. Under the conditions (f1) through (f3), the def- inition of ground-state solutions of (1.7) is similar to the minimax problem (1.3). Here, we modify the proof of Chern and Lin [4] to get the following results.

Theorem1.4. Let v∈C²(Θ)∩C(Θ)be a positive ground-state solutions of (1.7) inΘ.

Then, there exists az0∈{y=0} ∩Θsuch that(∂v/∂y)(z0)=0if and only ifvisy-symmetric.

Corollary1.5. Ifvis a positive ground-state solution of (1.1) inΩ(r)as inCorollary 1.2 andzcis a critical point ofv, thenzc∈ {/ y=0} ∩Ω. In particular, either(∂v/∂y)(z)<0or (∂v/∂y)(z)>0for allz∈ {y=0} ∩Ω.

2. Preliminaries

We define they-symmetric domains andy-symmetric functions as follows.

Definition 2.1. (i)Ωisy-symmetric provided thatz=(x,y)∈Ωif and only if (x,−y)∈ Ω;

(ii) letΩbe ay-symmetric domain inR^N. A functionu:Ω→Risy-symmetric (axi- ally symmetric) ifu(x,y)=u(x,−y) for (x,y)∈Ω.

Throughout this paper, letΩbe ay-symmetric domain inR^N,H_s(Ω) theH¹- closure of the space {u∈C0^∞(Ω)|u is y-symmetric} and letX(Ω) be either the whole space H0¹(Ω) or they-symmetric Sobolev spaceHs(Ω). Then,Hs(Ω) is a closed linear subspace ofH0¹(Ω). LetH_s⁻¹(Ω) be the dual space ofHs(Ω).

We define the Palais-Smale (PS) sequences, (PS)-values and (PS)-conditions inX(Ω) forJas follows.

Definition 2.2. We define the following:

(i) forβ∈R, a sequence{un}is a (PS)_β-sequence inX(Ω) forJ ifJ(un)=β+o(1) andJ(un)=o(1) strongly inX⁻¹(Ω) asn→ ∞;

(ii)β∈Ris a (PS)-value inX(Ω) forJif there is a (PS)_β-sequence inX(Ω) forJ;

(iii)Jsatisfies the (PS)_β-condition inX(Ω) if every (PS)_β-sequence inX(Ω) forJcon- tains a convergent subsequence.

By Willem [14], for anyβ∈R, a (PS)_β-sequence inX(Ω) forJis bounded. Moreover, a (PS)-valueβshould be nonnegative.

Lemma2.3. Let β∈Rand{un} be a (PS)_β-sequence inX(Ω)forJ, then there exists a positive numberc(β)such thatu_nH¹≤c(β)for largen. Furthermore,

aun

=bun

+o(1)= 2p

p−2β+o(1) (2.1)

andβ≥0. Moreover,c(β)can be chosen so thatc(β)→0asβ→0.

Now, we consider the Nehari minimization problem αX(Ω)= inf

u∈M(Ω)J(u), (2.2)

whereM(Ω)= {u∈X(Ω)\{0} |a(u)=b(u)}. Note thatM(Ω) contains every nonzero solution of (1.1) inΩ,αX(Ω)>0 and ifu0∈M(Ω) achievesαX(Ω), thenu0is a positive (or negative) solution of (1.1) inΩ(see [13,14]). Moreover, we have the following useful lemma, whose proof can be found in [13, Lemma 7].

Lemma2.4. Let{u_n}be inX(Ω). Then,{u_n}is a (PS)_αX(Ω)-sequence inX(Ω)forJif and only ifJ(un)=αX(Ω) + o(1)anda(un)=b(un) + o(1).

We denote

(i)αX(Ω) byα0(Ω) forX(Ω)=H0¹(Ω) andαX(Ω) byαs(Ω) forX(Ω)=Hs(Ω), (ii)M(Ω) byM0(Ω) forX(Ω)=H0¹(Ω) andM(Ω) byM_s(Ω) forX(Ω)=Hs(Ω).

Remark 2.5. By the principle of symmetric criticality (see [11]), we have every (PS)_β- sequence inX(Ω) forJis a (PS)_β-sequence inH0¹(Ω) forJ.

LetΩbe any unbounded domain andξ∈C^∞([0,∞)) such that 0≤ξ≤1 and ξ(t)=





0 fort∈[0, 1],

1 fort∈[2,∞). (2.3)

Let

ξ_n(z)=ξ 2|z|

n

. (2.4)

Then, we have the following results whose proof can be found in [15].

Proposition2.6. Equation (1.1) inΩdoes not admit any solutionu0such thatJ(u0)= αX(Ω)if and only if for each (PS)_αX(Ω)-sequence{un}inX(Ω)forJ,there exists a subse- quence{un}such that{ξnun}is also a (PS)_αX(Ω)-sequence inX(Ω)forJ.

Proposition2.7. Jdoes not satisfy the (PS)_αX(Ω)-condition inX(Ω)forJif and only if there exists a (PS)_αX(Ω)-sequence{u_n}inX(Ω)forJsuch that{ξ_nu_n}is also a (PS)_αX(Ω)-sequence inX(Ω)forJ.

LetΩ1Ω2, clearlyαX(Ω1)≥αX(Ω2). Then, we have the following useful results.

Lemma2.8. LetΩ1Ω2andJ:X(Ω2)→Rbe the energy functional. Suppose thatαX(Ω1)

=αX(Ω2). Then, the following hold:

(i)equation (1.1) in Ω1 does not admit any solution u0∈X(Ω1) such that J(u0)= α_X(Ω1);

(ii)Jdoes not satisfy the (PS)_αX(Ω2)-condition.

The proof is given by Wang and Wu [13, Lemma 13].

By the Rellich compact theorem,Jsatisfies the (PS)_αX(Ω)-condition inX(Ω) ifΩis a bounded domain.

Lemma2.9. Let Ωbe a bounded domain in R^N. Then, the (PS)_αX(Ω)-condition holds in X(Ω)forJ. Furthermore, (1.1) inΩhas a positive solutionu0such thatJ(u0)=αX(Ω).

3. Concentration behavior We need the following results.

Lemma3.1. LetΩbe an unbounded domain. Then, αX

Ω(r)αX(Ω) asr ∞. (3.1)

Proof. SinceΩ(r) is a bounded domain for allr >0, by Lemmas2.8and2.9, we have αX(Ω(r)) is monotone decreasing asr is monotone increasing andαX(Ω(r))> αX(Ω).

Thus, there exists ad0≥α_X(Ω) such that αX

Ω(r)d0 asr ∞. (3.2)

Claim thatd0≤α_X(Ω). Let{u_n}be a (PS)_αX(Ω)-sequence inX(Ω) forJ. ByLemma 2.3, there exists ac >0 such that

Ω

∇un²+u²_n≤c,

Ω

un^p≤c (3.3)

for alln∈N. Thus, for eachn∈N, there exists a sequence{r_n}such thatr_n>0 with r_n ∞asn→ ∞and

Ω∩{|z|≥rn}

∇un²+u²_n< 1 n,

Ω∩{|z|≥rn}

un^p<1

n. (3.4)

Now, defineηrn(z)=η(2|z|/rn), whereη∈C^∞_c ([0,∞)) satisfies 0≤η≤1 and

η(t)=





1 fort∈[0, 1],

0 fort∈[2,∞). (3.5)

Then,ηrnun∈X(Ω). From (3.4), we obtain

aη_rnu_n=au_n+o(1),

bη_rnu_n=bu_n+o(1). (3.6)

By the routine computations, there exists a sequence{s_n} ⊂R⁺such thata(s_nη_rnu_n)= b(s_nη_rnu_n),s_n=1 +o(1) and

Js_nη_rnu_n=Jη_rnu_n+o(1)=α_X(Ω) +o(1), (3.7)

that is,snηrnun∈M(Ω(rn)) andJ(snηrnun)≥αX(Ω(rn))=d0+o(1). Takingn→ ∞, we

getα_X(Ω)≥d0. Therefore,α_X(Ω)=d0.

LetΩ⁺_t = {(x,y)∈Ω|y > t}andΩ⁻_t = {(x,y)∈Ω|y < t}. Then, we have the follow- ing result.

Lemma3.2. Suppose that the domainΩsatisfies(Ω1),(Ω2), and(Ω3). Then, for eachε >0 andl≥0, there exists aδ(ε,l)>0such that ifu∈M0(Ω)andJ(u)< α0(Ω) +δ(ε,l), then either_Ω⁺

l|u|^p< εor_Ω⁻₋

l|u|^p< ε.

Proof. If not, there existc >0,l0≥0, and{u_n} ⊂M0(Ω) such thatJ(u_n)=α0(Ω) +o(1),

Ω⁺l0

u_n^p≥c,

Ω⁻−l0

u_n^p≥c. (3.8)

By Lemma 2.4, {u_n} is a (PS)_α0(Ω)-sequence inH0¹(Ω) for J. Now, Ω satisfies condi- tion (Ω3). ByProposition 2.6, there exists a subsequence{u_n}such that{ξ_nu_n}is also

a (PS)_α0(Ω)-sequence inH0¹(Ω) forJ, whereξ_nis as in (2.4). Letv_n=ξ_nu_n. We obtain Jv_n=α0(Ω) + o(1),

J v_n=o(1) inH⁻¹(Ω). (3.9)

SinceΩis a y-symmetric domain inR^N separated by a bounded domain, there exists a n0> l0such thatv_n=0 inΩ(n0) forn >2n0, andΩ\Ω(n0)=Ω1∪Ω2, whereΩ1=Ω⁺_n0

andΩ2=Ω⁻−n0.Moreover,vn=v¹_n+v²_n, where

v_nⁱ(z)=



v_n(z) forz∈Ωi

0 forz /∈Ωi

fori=1, 2. (3.10)

Then,v_nⁱ ∈H0¹(Ωi) anda(vⁱ_n)=b(v_nⁱ) +o(1). By (3.9), we obtain J vⁱ_n=o(1) strongly inH⁻¹Ωi

fori=1, 2. (3.11)

Assume that

Jvⁱ_n=ci+ o(1) fori=1, 2. (3.12) Since J(vn)=J(v¹_n) +J(v_n²)=α0(Ω) + o(1), we have c1+c2=α0(Ω). Since ci are (PS)- values inH0¹(Ωi) forJ, byLemma 2.3,c_i≥0 and

c1

2p p−2

=

Ω⁺_l0

v_n^1p+o(1)=

Ω⁺_l0

un^p+o(1),

c2

2p p−2

=

Ω⁻_−l0

v²_n^p+o(1)=

Ω⁻_−l0

un^p+o(1).

(3.13)

By (3.8), we havec_i>0 fori=1, 2. We have that α0(Ω)=c1+c2≥α0

Ω1

+α0

Ω2

, (3.14)

which contradicts the fact thatα0(Ω)≤α0(Ωi) fori=1, 2.

Now, we begin to show the proof ofTheorem 1.1. ByLemma 3.1, for eachε >0 and l≥0, there exists aδ(ε,l)>0 such that if u∈M0(Ω) and J(u)< α0(Ω) +δ(ε,l), then

Ω⁺_l |u|^p < ε or _Ω⁻₋

l|u|^p< ε. Moreover, by Lemma 3.2, there exists an r > 0 such that

α0(Ω(r))< α0(Ω) +δ(ε) for all r >r. Thus, if v is a ground-state solution of (1.1) in H0¹(Ω(r)) for r >r, then v∈M0(Ω(r))⊂M0(Ω), J(v)< α0(Ω) +δ(ε) and either

Ω⁺_l |v|^p< εor_Ω⁻₋

l|v|^p< ε.

Now, we begin to show the proof ofTheorem 1.3.

(i) ByLemma 3.1, we haveJ(vm)=α0(Ω(m))=α0(Ω) +o(1). Sincevm∈M0(Ω(m))⊂ M0(Ω), fromLemma 2.4we can conclude that {v_m}is a (PS)_α0(Ω)-sequence inH0¹(Ω) forJ.

(ii) Letv∈L^q(Ω), where 1/ p+ 1/q=1. Then, for eachε >0 there exists anl >0 such that

(Ω(l))^c|v|^q< ε^q. (3.15) ByTheorem 1.1, there exists anm0> lsuch that

Ω(l)

vm^q< ε^p ∀m > m0. (3.16)

Thus, for eachε >0 there exists anm0such that

Ωvmv=

(Ω(l))^cvmv+

Ω(l)vmv≤

(Ω(l))^c

vm^p1/ p

(Ω(l))^c|v|^q 1/q

+

Ω(l)

v_m^p 1/ p

Ω(l)|v|^q 1/q

≤ c1+c2

ε ∀m > m0,

(3.17)

wherec1=((2p/(p−2))α0(Ω)) andc2= v_L^q. This implies thatv_m0 weakly inL^p(Ω) asm→ ∞. Sincevmis a solution of (1.1) inΩ(m), we have

Ω(m)∇v_m∇ϕ+v_mϕ=

Ω(m)

v_m^p−2v_mϕ ∀ϕ∈H0¹

Ω(m). (3.18)

First, we need to show for eachε >0 andϕ∈C_c¹(S) there exists anm0such that

Ω(m)∇vm∇ϕ+vmϕ < ε ∀m > m0 (3.19) forϕ∈C_c¹(Ω). LetK=suppϕ, thenK⊂Ωis compact and there exists anm1such that K⊂Ω(m) for allm≥m1. FromTheorem 1.4, for eachε >0 there existl0>0 andm0such thatϕ∈H0¹(Ω(m)),

(Ω(l⁰))^c|ϕ|^p=0,

Ω(l⁰)

vm^p< ε^{(p−1)/ p} ∀m > m0. (3.20)

We obtain

Ω(m)

v_m^p−2v_mϕ=

(Ω(l0))^c

v_m^p−2v_mϕ+

Ω(l0)

v_m^p−2u¹_mϕ

≤

(Ω(l⁰))^c

vm^p(p−1)/ p

(Ω(l⁰))^c|ϕ|^p 1/ p

+

Ω(l0)

v_m^p

(p−1)/ p

Ω(l0)|ϕ|^p 1/ p

≤cε,

(3.21)

Ω∇vm∇ϕ+

Ωvmϕ=

Ω(m)∇vm∇ϕ+

Ω(m)vmϕ

=

Ω(m)

vm^p−2vmϕ ∀m > m0.

(3.22)

We have that

Ω∇vm∇ϕ+

Ωvmϕ≤cε ∀m > m0. (3.23) Sinceα0(Ω(m+ 1))< α0(Ω), there exists aC >0 such that vmH¹≤C. Thus, for each ε >0 andψ∈H0¹(Ω), there exists aϕ∈C¹_c(Ω) such that

ψ−ϕ_H¹< ε

C. (3.24)

From (3.23) and (3.24), we can conclude that for eachε >0 andψ∈H0¹(Ω) there exists anm0>0 such that

vm,ψ_H¹=

vm,ψ−ϕ_H¹+vm,ϕ_H¹

≤Cψ−ϕH¹+vm,ϕ_H¹

< ε+cε form > m0.

(3.25)

This implies thatvm0 weakly inH0¹(Ω).

4. Symmetry

Now, we begin to show the proof ofTheorem 1.4. Let vbe a ground-state solution of (1.7) inΘand letz^∗=(x,−y) be the reflection point ofz=(x,y) with respect to the hyperplaneT:= {y=0}. First, we claim that either

v(z)≥vz^∗ ∀z∈Θ⁺ (4.1)

or

v(z)≤vz^∗ ∀z∈Θ⁺, (4.2)

whereΘ⁺is one of half domainΘ\T. If not, then the following two sets A+=

z∈Θ⁺|v(z)> vz^∗, (4.3)

A−=

z∈Θ⁺|v(z)< vz^∗, (4.4)

are nonempty. Letw(z)=v(z)−v(z^∗) forz∈Θ⁺. Then,wsatisfies

∆w−w+f_vζ(z)w=0, inΘ⁺,

w=0, in∂Θ⁺, (4.5)

whereζ(z) is betweenv(z) andv(z^∗). Let A^∗₋=

z^∗|z∈A−

. (4.6)

Ford >0, we define a function

ud(z)=











w(z) ifz∈A+, dwz^∗ ifz∈A^∗₋,

0 otherwise.

(4.7)

Since_A+wφ1>0 and_A−wφ1<0, there exists a constantd0>0 such that

Θud0φ1=

A+

wφ1+d0

A−

wφ1=0, (4.8)

whereφ1is the first positive eigenfunction of the following eigenvalue problem:

∆−1 +fv

ζ(z)φ+λφ=0 inΘ,

φ=0 on∂Θ. (4.9)

Letλ2be the second eigenvalue of (4.9). Sincevis a ground-state solution of (1.7), by the same method of the proof of Theorem 2.11 in [10], we haveλ2is nonnegative. Moreover, by (4.3)–(4.7), we have

∆ud−ud+fv

ζ(z)ud>0 forz∈A+,

∆ud−ud+fv

ζ(z)ud<0 forz∈A^∗₋,

∆ud−ud+fv

ζ(z)ud=0 otherwise.

(4.10)

Therefore, from (4.8) and (4.10), we have 0>

Θ−u_d(z)∆ud(z)−u_d+ f_vζ(z)u_d(z)dz

=

Θ

∇u_d(z)²+u²_d−f_vζ(z)u²_d(z)dz

≥λ2

Θu²_d(z)dz≥0,

(4.11)

a contradiction. This proves inequalities (4.1) and (4.2). By (4.1) and (4.2), we may as- sumew(z)≥0 for allz∈Θ⁺, ifw(z)>0 for somez∈Θ⁺. Sincewsatisfies (4.5), by using the strong maximum principle, we havew >0 inΘ⁺. Similarly, ifw(z)≤0 andw(z)<0 for somez∈Θ⁺, we havew <0 inΘ⁺. Suppose thatw(z)>0 for allz∈Θ⁺. Then, from (4.5) and applying the Hopf Lemma, we have

∂w

∂(−y) z0

= −2∂v

∂y z0

<0. (4.12)

Similarly, ifw(z)<0 for allz∈Θ⁺, we have (∂v/∂)y(z0)<0, this contradicts the fact that (∂v/∂)y(z0)=0. Therefore,w(z)=0 for allz∈Θ⁺orv(x,y)=v(x,−y) for all (x,y)∈Θ.

The converse is obvious.

Acknowledgment

The author is partially supported by the National Science Council of Taiwan.

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Tsung-Fang Wu: Center for General Education, Southern Taiwan University of Technology, Tainan 71005, Taiwan

E-mail address:[email protected]