Evaluating solutions on an elliptic problem in a gravitational gauge field theory

Journal of Functional Analysis 265 (2013) 1240–1263

www.elsevier.com/locate/jfa

Evaluating solutions on an elliptic problem in a gravitational gauge field theory

Jann-Long Chern

, Sze-Guang Yang

aDepartment of Mathematics, National Central University, Chung-Li 32001, Taiwan bNational Center for Theoretical Sciences, National Tsing Hua University, Hsinchu 30043, Taiwan

Received 23 May 2012; accepted 28 May 2013 Available online 17 June 2013

Communicated by F.-H. Lin

Abstract

An elliptic equation arising from the study of static solutions with prescribed zeros and poles of the Ein- stein equations coupled with the classical sigma model and an Abelian gauge field, is considered. We classify the solutions and establish the uniqueness of radially symmetric solutions. We also complete a clas- sification of symmetric solutions of an elliptic equation on the sphere.

©2013 Elsevier Inc. All rights reserved.

Keywords:Nonlinear elliptic equation; A gravitational gauge field theory; Uniqueness and classification of symmetric solutions

1. Introduction

In this paper we study the equation

u=2e^η

e^u−1 e^u+1

+4π L j=1

δ_pj −4π M j=1

δ_qj inR², (1)

✩ Work partially supported by National Science Council of Taiwan.

* Corresponding author at: Department of Mathematics, National Central University, Chung-Li 32001, Taiwan.

E-mail address:[email protected](J.-L. Chern).

0022-1236/$ – see front matter ©2013 Elsevier Inc. All rights reserved.

http://dx.doi.org/10.1016/j.jfa.2013.05.041

whereδ_pis the Dirac distribution concentrated atp∈R²andηis the function given by

e^η(x)=g₀

e^u(x) (e^u(x)+1)²

L j=1

|x−p_j|⁻² M j=1

|x−q_j|⁻² 8π G

. (2)

Hereg0>0 is an arbitrary constant,Gis Newton’s gravitational constant,p1, p2, . . . , p_Land q1, q2, . . . , qM are the locations of vortices and antivortices of unit charges. This model is ini- tiated by Yang[8,10]on the coexistence of cosmic strings and antistrings in an Abelian gauge field theory originating with Schroers[6,7]in which the classical sigma model is coupled with a gauge field.

In the theory, with a Minkowski spacetime of signature(+ − − −), the action density of the gauged sigma model is given by

L= −1

4g^μμg^ννF_μνF_μν+1

2g^μν(D_μφ)·(D_νφ)−1

2(n·φ)²,

where the field configuration φ=(φ₁, φ₂, φ₃) is a spin vector which maps R² into the unit 2-sphereS²,nis the north pole ofS²,Fμν=∂μAν−∂νAμ the electromagnetic field induced from the Abelian gauge fieldAμ(μ, ν=0,1,2,3 witht=x0),Dμφ=∂μφ+Aμ(n×φ)is the gauge-covariant derivative, and the gauge group is the subgroup ofO(3)that leavesninvariant.

Through a stereographic projection from the south polenofS²into the complex plane, we set ψ=ψ1+iψ2, ψ1= φ₁

1+φ₃, ψ2= φ₂ 1+φ₃.

So that with the substitution A_μ → −A_μ and the new gauge-covariant derivative given by D_μψ=∂_μψ+iAμψ, we arrive at an Abelian gauge theory with the action density

L= −1

4g^μμg^ννFμνF_μν+ 2

(1+ |ψ|²)²g^μν(Dμψ )(Dνψ)−1 2

1− |ψ|² 1+ |ψ|²

2

. (3)

With the coupling of gravity, the equations of motion of the field-theoretic model(3)are G_μν+Λg_μν= −8π GTμν, (4)

√1

|g|D_μ √

|g|g^μν (1+ |ψ|²)²D_νψ

=f, (5)

√1

|g|∂_μ

g^μνg^μν |g|F_νν

=j^μ, (6)

whereG_μν is the Einstein tensor,Gis Newton’s gravitational constant,Λis the cosmological constant,T_μν is the energy-momentum tensor defined by

T_μν= −g^μνF_μμF_νν+ 2

(1+ |ψ|²)²(D_μψ D_νψ+D_μψ D_νψ )−g_μνL, (7) andf,j^μare the force and current density terms respectively given by

f =(1− |ψ|²−2g^μνDμψ Dνψ) (1+ |ψ|²)³ ψ, j^μ= 2i

(1+ |ψ|²)²g^μν(ψ D_νψ−ψD_νψ).

Letx₀be the temporal component and (xi),i=1,2,3, the spatial coordinates. To looking for static the field configurations which depend only on spatial variable x₁,x₂, we assumeA₀= A₃=0 and the metric tensor is of the form

gμν=diag

1,−e^η,−e^η,−1

, (8)

withηbeing a function ofx₁, x₂only. In this manner, the energy densityH=T₀₀has the repre- sentation

H=e^−η(F12+J12)+1 2

e^−ηF12−1− |ψ|² 1+ |ψ|²

2

+ 2e^−η

(1+ |ψ|²)²|D1ψ+iD2ψ|², (9) in whichJj k=∂jJk−∂kJj with a new current density

J_k= i

1+ |ψ|²(ψ D_kψ−ψD_kψ ), k=1,2.

As a consequence of(9), one can obtain the following Bogomol’nyi equations

D₁ψ+iD2ψ=0, (10)

F₁₂=e^η1− |ψ|²

1+ |ψ|². (11)

Note that the metric(8)implies the Einstein tensorGμν assumes the form

G₀₀= −G₃₃= −K_g, G_μν=0 for other(μ, ν)pairs, (12) whereK_gis the Gauss curvature of the 2-surface(R², e^ηδ_{j k})which can be given by

K_g= −1 2e^−ηη

with=∂₁²+∂₂²being the Laplace operator on R². From(12)we see that the cosmological constantΛ=0 and the Einstein equations(4)are boiled down to a single equation

K_g=8π GH, (13)

which relates the Gauss curvature to the energy density. The sets{p_j}^L_j=1,{q_j}^M_j=1of zeros and poles ofψare the sites for strings and antistrings to reside. Using the substitutionu=log|ψ|², Eqs. (10) and (11)is then reduced to Eq.(1). A solution pair(ψ, A) of(10) and (11)can be constructed fromuvia the standard procedure

ψ (z)=exp 1

2u(z)+iθ (z)

, θ (z)= L j=1

arg(z−p_j)− M j=1

arg(z−q_j), A₁(z)= −Re

2i∂logψ (z)

, A₂(z)= −Im

2i∂logψ (z) . Hence the energy densityHcan be rewritten by

e^ηH=

log 1+e^u

−1 2u

+2π

L j=1

δ_pj+2π M j=1

δ_qj inR².

It follows from(13)that the factore^ηis explicitly determined by the expression(2).

The solutions of(1)can be evaluated by a functionalβ that associates each solutionuwith the value

β(u)= −1 π

R²

e^η

e^u−1 e^u+1

dx.

We remark thatβdescribes the value of magnetic flux in the model corresponding to a solutionu, and on the other hand, it characterizes the asymptotic behavior ofunear infinity, that is (via a potential analysis[2]),

β=2(L−M)− lim

|x|→∞

u(x)

log|x|. (14)

Here arise the following interesting problem. For a prescribed valueλ∈R, we inquire whether there exists any solution usuch that β(u)=λ. And, if such solution exists, we wish to seek for its uniqueness. In this paper, we concentrate our study on radially symmetric solutions and on the cases either (i) L >0, M=0 or (ii) L=0, M >0. In either cases, we assume that the locations{p_j}^L_i=1(or{q_j}^M_j=1) of vortices (or antivortices) cluster at the origin. Specifically, lettingL=N >0 andM=0, Eq.(1)is reduced to

u= −|x|^−2aNh e^u

+4π N δ0 inR², (15)

where

h(t)= −2g0

t (t+1)²

a t−1 t+1

, a=8π G.

The functionalβ, depending onu(x)=u(r)withr= |x|, is given by

β(u)= ∞ 0

r^−2aN+1h e^u(r)

dr. (16)

Here are our consequences.

Theorem 1.1.Letube a radially symmetric solution of Eq.(15). Then the following statements are true.

(a) If aN <1, then β(u)∈E:=(−∞,^8(aN_a⁻¹⁾)∪ {2N} ∪(⁴_a,∞). Conversely, for any pre- scribed value λ∈E, there exists a unique (radially symmetric) solution u_λ such that β(uλ)=λ. In addition,uλhas the asymptotic behavior

u_λ(r)=(2N−λ)logr+O(1) asr→ ∞;especially, whenλ=2N,u_λ(r)→0asr→ ∞.

(b) IfaN=1, thenβ(u)≡constant∈ {0,2N,4N}.

(c) IfaN2, then every solutionusatisfies thatu(r)→ ∞asr→ ∞. Moreover, 0< β(u) <4

a.

Note.WhenL=0 andM=N >0 in Eq.(1), the problem is read as the following:

u= −|x|^−2aNh e^u

−4π N δ0 inR². (17)

Note thath(t⁻¹)= −h(t). Takingu→ −uinterchanges Eq.(17)with(15). Clearly,β(−u)=

−β(u). So we have mirror consequences ofTheorem 1.1for this case.

As an interesting aspect of this theory, the model can be performed an Abelian gauge field over a compact Riemann surface[10], in which there arises a nonlinear elliptic equation on the unit sphereS²about locating vortices and antivortices at the north and south poles ofS². In fact, denoting the north and south poles respectively bynands, here is the equation:

⎧⎪

⎪⎨

⎪⎪

⎩

gu=2e^η

e^u−1 e^u+1

+4π N δn+4π P δs, _g

η

16π G+log 1+e^u

−1 2u

= K₀

8π G−2π|N|δ_n−2π|P|δ_s,

(18)

where_g is the Laplace–Beltrami operator on(S², g)with is the standard metricg,K₀ is the associated (constant) Gauss curvature andN, P are integers for which positive or negative integer represents the corresponding number of prescribed zeros or poles ofurespectively. We remark that, as a result in[10], a necessary condition for the existence of a solutionusolving Eq.(18)is that the Newton gravitational constantGsatisfies the quantization condition

4π G

|N| + |P|

=1. (19)

To classify the symmetric solutionuof(18)with respect tonands, here are the known conse- quences:

Theorem A.(See[10].) Consider symmetric solutions of(18).

(i) Existence:If|N| = |P|>0, then there is a(symmetric)solution.

(ii) Nonexistence:If|N|>0,P =0orN =0,|P|>0, then there is no solution;if|N| = |P| withN P >0, then there is no solution.

There remains an unsolved case inTheorem A, namely|N| = |P|withN P <0. To look for the answer, we note that the symmetric case of(18)can be equivalently put into the framework of Eq.(15) associated with a prescribed asymptotic behavior at infinity. In fact, since(18) is invariant under the transformation(η, u)→(η,−u), it suffices to consider the caseN >0,P <0 and|N|<|P|. Note that, through a stereographic projection from the south pole ofS², we are looking for a radially symmetric solutionu=u(r),r= |x|, satisfying

⎧⎪

⎪⎨

⎪⎪

⎩

u=2e^η

e^u−1 e^u+1

+4π N δ0 inR²,

|xlim|→∞

u(r)

logr = −2P ,

(20)

with

e^η=λr^−2aN e^u

(e^u+1)² a

, a=8π G,

whereλ >0 is a constant. In view of(19),a(|N| + |P|)=2, we haveaN <1. So that, if there is a solutionusolving(20), then byTheorem 1.1, it follows that

−2P ∈

−∞,2N−4 a

∪ {0} ∪

−6N+8 a,∞

.

Since 2N−(4/a) <0, it must hold that−2P >−6N+(8/a), or equivalently, 4

a =2

|N| + |P|

>−4N+8 a >4

a,

yielding a contradiction. Therefore, we complete the classification inTheorem Awith the fol- lowing corollary:

Corollary 1.If|N| = |P|withN P <0, then Eq.(18)admits no symmetric solution.

Therefore,A symmetric solution of Eq.(18)exists only when the zeros or poles located at the north and south poles ofS²are balanced, in the sense that there are equal numbers of zeros or poles clustered atnands.

Remark.In the case aN =1 inTheorem 1.1, the constant associated with the value of β is dependent (only) on the given of the parameterg₀in the expression(2). For example, as one may see in[9, Theorem 11.3.7]with the choiceg₀=4^aN, we conclude thatβ=2N. On the other hand, in view of the context of(20)withP = ±N, the existence results inTheorem Aindicate that bothβ=0 andβ=4N are also possible provided suitableg₀is selected.

The paper is organized as follows. We divided the proof ofTheorem 1.1into several aspects.

In Section2we include a preliminary classification of solutions. In Section3we establish the uniqueness of solutions in the caseaN <1. In Section4we identify the range ofβ(u)and carry out the main theorem.

2. Preliminaries

To make sense of the solution classification, we associate the radially symmetric solutions of Eq.(15)with real numbers in the sense thatu(r)=u(r;s)withs∈Rsatisfies

⎧⎪

⎪⎨

⎪⎪

⎩

u(r)+1

ru(r)= −r^−2aNh e^u(r)

, r= |x|>0, u(r)=2Nlogr+s+o(1), u(r)=2N

r +o(1) asr→0.

(21)

Sinceh:R⁺→Ris a bounded function,ucannot blow up at a finiter. LetΛbe the solution set of(21). We divideΛinto three classes that

S⁺=

u∈Λ: lim

r→∞u(r)= ∞

; S^∗=

u∈Λ: lim

r→∞u(r)=0 ; S⁻=

u∈Λ: lim

r→∞u(r)= −∞

. Then the set of real numbers is split up into the components:

J⁺=

s∈R: u(r;s)is of the classS⁺

; J^∗=

s∈R: u(r;s)is of the classS^∗

; J⁻=

s∈R: u(r;s)is of the classS⁻ .

Note that J⁺∪J^∗∪J⁻=R. In the following we conclude some elementary facts associated with these settings.

(P1) By the continuous dependence ofuons,J⁺andJ⁻are open sets whose boundary satisfies

∂J⁺∪∂J⁻⊂J^∗.

(P2) By the maximum principle, we see that (i) ifu∈S^∗∪S⁻, thenu <0;

(ii) ifu∈S^∗∪S⁺, thenu>0.

Theβ(u)defined in(16)is treated as a function ofswhich is given by

β(s)=β u(·, s)

= ∞ 0

r^−2aN+1h e^u(r;s)

dr.

(P3) According to Eq.(21), we have

ru(r)=2N− r 0

τ^−2aN+1h e^{u(τ )}

dτ, (22)

from which it is not hard to see that|β|<∞, limr→∞ru(r)=2N−β and

u(r)=(2N−β)logr+O(1), r→ ∞. (23)

Clearly,β(u)=2N provided u∈S^∗. Assume s∈J⁻. Since e^u(r;s) vanishes asr→ ∞, we extract constants, r0>0 such thate^auh(e^u)for allrr0. By(23),

∞ r₀

r^−2aN+1e^au(r)dr ∞ r₀

r^−2aN+1h e^u(r)

dr <∞.

So that

β(s) > 2

a, s∈J⁻. (24)

Similarly, ifs∈J⁺,e^−u(r;s) vanishes asr→ ∞. Hence applying the facth(t⁻¹)= −h(t)we conclude that

β(s) <4N−2

a, s∈J⁺. (25)

A Pohozaev-type identity associated with Eq. (21)is established as follows. By multiply- ing(21)by the factorτ u(τ )and taking integration both sides over(0, r),

ru(r)2

−4N²= −2 r 0

τ^−2aN+2h e^{u(τ )}

u(τ ) dτ

= −2 r 0

τ^−2aN+2 d dτH

e^{u(τ )} dτ

= −2

τ^−2aN+2H

e^{u(τ )}τ=r τ=0

+4(1−aN ) r 0

τ^−2aN+1H e^{u(τ )}

dτ, (26)

in whichH:R⁺→Ris given by

H (t )=Hσ(t)= t σ

h(ζ ) ζ dζ,

whereσ may denote either 0 or∞regarding the following context. Denote

I_r = r 0

τ^−2aN+1h e^{u(τ )}

dτ.

From(22),I_r =2N−ru(r). UsingI_r in substitution forru(r)in(26), we get

I_r

I_r−4 a

= −2

τ^−2aN+2H

e^{u(τ )}τ=r

τ=0+4(1−aN ) r 0

τ^−2aN+1G u(τ )

dτ, (27)

whereG(ζ )=H (e^ζ)−(1/a)h(e^ζ). Note that

G(ζ )= −2e^(a+1)ζ (e^ζ+1)^2a+2

e^ζ−e^μ0

, e^μ0=1+1

a. (28)

Therefore,Gis increasing in the interval(−∞, μ₀), decreasing in(μ₀,∞)and attains its maxi- mum atζ=μ₀>0.

Theorem 2.1. AssumeaN <1. Let u=u(r;s) be the solution of(21) withs∈R. Then the following statements are true.

(i) Ifs∈J⁻, thenβ(s) >_a⁴. Moreover,β(s_j)→ ∞whenever{s_j} ⊂J⁻ands_j→∂J⁻. (ii) Ifs∈J⁺, thenβ(s) <0. Moreover,β(s_j)→ −∞whenever{s_j} ⊂J⁺ands_j→∂J⁺. Proof. Assumeu∈S⁻. Recall thatu <0 andu(r)→ −∞asr→ ∞. Consider

G(ζ )=H₀ e^ζ

−(1/a)h e^ζ

=

exp(ζ )

0

h(t )

t dt−(1/a)h e^ζ

. (29)

Clearly,G(−∞)=0. Thus, by(28),G(u)is positive. On the other hand, applying L’Hospital’s Rule,

rlim→∞r^−2aN+2H₀ e^u(r)

= 1

2(aN−1) lim

r→∞r^−2aN+2h e^u

ru

=0,

in which the estimates(23) and (24)are taken into account. Lettingr→ ∞in(27), we get

β

β−4 a

=4(1−aN ) ∞ 0

r^−2aN+1G u(r)

dr >0. (30)

So thatβ >4/a, as desired. Similarly, in the case u∈S⁺ whereu(r)→ ∞as r→ ∞, we consider

G(ζ )=H_∞ e^ζ

−(1/a)h e^ζ

=

exp(ζ )

∞

h(t)

t dt−(1/a)h e^ζ

.

Note thatG(∞)=0. Sinceh(ζ⁻¹)= −h(ζ ),

G(−∞)= − 1

0

+ ∞ 1

h(ζ ) ζ dζ= −

1 0

− 1 0

h(ζ ) ζ dζ=0.

From(28)again,G(u)is positive. By(23), (25)and L’Hospital’s Rule,

rlim→∞r^−2aN+2H_∞ e^u(r)

= 1

2(1−aN ) lim

r→∞r^−2aN+2h e^−u

ru

=0.

In view of(25) and (30),β <0. To complete the first conclusion, we assume, contrarily, that there exist a sequences_j∈J⁻andC >0 such thats_j→s^∗∈∂J⁻and|β(s_j)|< C. Let

u_j(r)=u(r;s_j), u^∗(r)=u r;s^∗

.

Note thats^∗∈J^∗. Forr >0, we haveuj−u^∗L^∞(0,r)→0 assj→s^∗. SinceIr has a uniform bound with variousrby means of(27), it is possible to extractC >¯ 0 such that

0<

r 0

τ^−2aN+1H₀ e^{u∗(τ )}

dτ <C,¯ (31)

regardless of the selection of r. This indicates r^−2aN+1H₀(e^u∗(r))is integrable over (0,∞);

however, it is impossible becauser^−2aN+1=r⁻¹⁺for some >0 and

rlim→∞H₀ e^u∗(r)

= 1 0

h(ζ ) ζ dζ >0.

Therefore,β(sj)→ ∞wheneversj→∂J⁻andsj∈J⁻. The second conclusion for the asymp- totic behavior ofβ(s)comes from the same argument by usingH_∞in place ofH0in(31). We omit the details here. 2

Letu(r)=u(r;s)solve(21). Denote

ϕ(r)=ϕ(r;s)=∂u

∂s(r;s).

Clearly,ϕ satisfies

⎧⎨

⎩

ϕ(r)+1

rϕ(r)=Q(r)ϕ(r), r >0, ϕ(0)=1, ϕ(0)=0,

(32)

whereQ(r)= −r^−2aNh(e^u(r))e^u(r). Especially whenaN=1, the functionw(r)=ru(r)sat- isfies(32)withw(0)=2N. Thus,w=2N ϕ. This leads to the following theorem.

Theorem 2.2.IfaN=1, thenβ(s)≡constant∈ {0,2N,4N}.

Proof. By definition,

2N ϕ(r)=w(r)=ru(r)+u(r)= −r^−2aN+1h e^u(r)

, u(r)=u(r;s). (33) From property (P2) before, we have (i) if s∈J⁻∪J^∗, thenϕ(·, s) <0; (ii) ifs∈J⁺, then ϕ(r;s) <0 forr∈(0, r0)andϕ(r;s) >0 forr∈(r0,∞)whereu(r0)=0. On the other hand, in the light of(23) and (26), it is not hard to see that

β(s)=0 fors∈J⁺; β(s)=2N fors∈J^∗; β(s)=4N fors∈J⁻. Note thatw(r;s)=ru(r;s)→2N−β(s)asr→ ∞. Hence

rlim→∞ϕ(r;s)= lim

r→∞w(r;s)/2N=1,0,−1,

fors∈J⁺, J^∗, J⁻respectively. So|ϕ(r;s)|1 for all(r, s). Lets₁, s₂∈R. Then u(r;s₁)−u(r;s₂)

∂u

∂s(r,·)

|s₁−s₂||s₁−s₂|, r >0,

indicating that the functionu(·, s₁)−u(·, s₂)must be bounded. In view of the asymptotic behav- ior characterized in(23), it follows thatβ(s₁)=β(s₂). Therefore the theorem is proved. 2 Theorem 2.3.LetaN2. Assumeuis a solution of Eq.(21). Thenusatisfies thatu(r)→ ∞ asr→ ∞. Moreover,

0< β(u) < 4 a. Proof. From(27), we have

β

β−4 a

=4(1−aN ) ∞ 0

r^−2aN+1G u(r)

dr <0. (34)

Obviously, (21)admits noS⁻ type solution because of(23) and (34). Assume there exists a solutionu^∗∈S^∗. Note thatβ(u^∗)=2Nandru^∗(r)→0 asr→ ∞. From(26),

N² aN−1=

∞ 0

τ^−2aN+1H e^{u(τ )}

dτ

>1 a

∞ 0

τ^−2aN+1h e^{u(τ )}

dτ=2N a ,

indicatingaN <2. This contradicts our assumption. So there is no solution ofS^∗type. There- fore, nothing butS^∗ type solutions can exist in this case; furthermore, the range ofβ follows readily from(34). 2

3. Uniqueness of the evaluation withaN <1

The following theorem shows the monotonicity ofβ(s), by which the setsJ⁻,J^∗andJ⁺are identified.

Theorem 3.1.IfJ^∗is not empty, then there iss^∗∈Rsuch thatJ⁻=(−∞, s^∗),J^∗= {s^∗}and J⁺=(s^∗,∞). Moreover,β(s) >0fors∈R− {s^∗}and

lim

s→s^∗−

β(s)= ∞, lim

s→s^∗+

β(s)= −∞.

We now anticipate the required pieces in advance of assemblingTheorem 3.1. We apply the techniques introduced in[1]. Recall the linear equation(32)we mention before. Let us rewrite it as follows:

⎧⎨

⎩

ϕ(r)+1

rϕ(r)=f e^u(r)

q(r)ϕ(r), r >0, ϕ(0)=1, ϕ(0)=0,

(35)

wheref (t )= −at²+(2a+2)t−aand q(r)=2g0r^−2aN

e^u (e^u+1)²

a 1 (e^u+1)²

>0.

Obviously, the quadratic polynomialf has two positive zeros att=T1, T2given by T₁=a+1−√

2a+1

a , T₂=a+1+√

2a+1

a ;

f is increasing in(−∞, a+1)and decreasing in(a+1,∞). To figure out the behavior ofϕ, we compare it with the functionwc(r)=ru(r)+c, wherecis a constant yet to be determined.

From(21) and (35),wcsatisfies the equation w_c(r)+1

rw_c(r)=f e^u(r)

q(r)w_c(r)+q(r)p_c e^u(r)

, r >0, (36)

wherep_cis the quadratic polynomial given by

Fig. 1.cis a function oftwithpc(t)=p(c, t)=0.

p_c(t)=2(1−aN )(t+1)(t−1)−cf (t)

=a(c+m)t²−2c(a+1)t+a(c−m), m=2(1−aN )/a >0. (37) Ifc= −m, the functionp_c(t)has only one zero att=a/(a+1). Ifc∈R\{−m}, thenp_c(t) admits two distinct zeros. As shown in Fig. 1,p_c(t)=p(c, t ) can be treated as a function of (c, t)∈R²and the level set {(c, t): p_c(t)=0}consists of three disjoint curves containing the pointsA,BandC in the diagram respectively. They divide the plane into four regions denoted by I, II, III and IV, in whichp_c(t)is positive in regions I, III and is negative in II, IV. From(37), the level setp_c(t)=0 definescas a function oft, namely

c(t )=am(t+1)(t−1) f (t ) fort∈R− {T1, T2}. Note that

lim

t→ξ⁻

c(t )= ∞, lim

t→ξ⁺

c(t)= −∞, lim

t→−∞c(t)= lim

t→∞c(t)= −m, whereξ denotesT1,T2. Now we combine Eqs.(35) and (36)as follows. Since

d dr

r

ϕw_c−w_cϕ

=ϕ rw_c

−w_c rϕ

, by taking integration on both sides of the formula, we get the identity

r2

r1

rP_cϕ dr=r₂

ϕ(r₂)w_c(r₂)−w_c(r₂)ϕ(r₂)

−r₁

ϕ(r₁)w_c(r₁)−w_c(r₁)ϕ(r₁) , (38)

whereP_c(r)=q(r)p_c(e^u(r)).

Lemma 3.1.Letϕbe the solution of(35)withu(r)=u(r;s).

(a) Ifs∈J^∗, thenϕis positive.

(b) Ifs∈J⁺, thenϕchanges sign at most twice.

(c) Ifs∈J⁻, thenϕchanges sign either once or twice.

Proof. Letu(r)=u(r;s). We perform the proof case by case withsin the following.

Case1. Fors∈J^∗, we assume contrarily thatz₁>0 is the first zero ofϕfor whichϕ(z₁)=0 andϕ(r) >0 in (0, z1). We considerc=0 in(38). Note that w(r)=w₀(r)=ru(r) >0 for allr. Sincee^u<1 andp₀(t) <0 in(0,1), we haveP₀(r)negative in(0, z1). So that

0>

z₁

0

rP0ϕ dr= −z1w(z1)ϕ(z1) >0, (39)

which is a contradiction. Therefore,ϕ(r;s)does not change sign provideds∈J^∗.

Case2. Lets∈J⁺. Assumeϕ changes sign at least three times. Letzbe the second zero ofϕ.

Pickr₁,r₂with 0< r₁< z < r₂, such that

ϕ(r) <0 forr∈(r₁, z),

ϕ(r) >0 forr∈(z, r₂), (40)

andφ(r)has local minimum and maximum atr=r₁, r₂respectively. Letσ₁=e^u(r1). From(35), we have

f (σ₁)q(r₁)ϕ(r₁)=ϕ(r₁)0,

indicatingf (σ₁)0. Thus eitherσ₁T₁orσ₁T₂, where recall thatT₁, T₂are the zeros of f withT₁< T₂. Now we show, in the following, that either of these situations for σ₁ cannot occur.

(2-1) Assumeσ₁T₁. Letz₁be the first zero ofϕ. Clearly,z₁< r₁. SinceT₁<1 andu(r) decreases asr decreases, it follows thate^u(r)<1 for allr∈(0, z1). Consider(38)and letw= w_c=0.P₀(r)is negative for allr∈(0, z1)becausep₀(t) <0 fort∈(0,1). Therefore, from the same expression as(39), we obtain a contradiction.

(2-2) Assumeσ₁T₂. Letζ=e^u(z)andσ₂=e^u(r2). Clearly, 1< T₂σ₁< ζ < σ₂.

By virtue of the behavior ofc(t ), as shown inFig. 1, we letc <−mbe the constant such that pc(ζ )=0,pc(t) >0 fort∈(σ1, ζ )andpc(t) <0 fort∈(ζ, σ2). Hence

Pc(r) >0 forr∈(r1, z),

P_c(r) <0 forr∈(z, r₂). (41)

With our selection ofr₁,r₂,cand in view of(40) and (41), the left-hand side (LHS) of(38)is negative, whereas the right-hand side (RHS) is positive. That is a contradiction. Here we also make use of(33), concludingw_c >0 providede^u>1. The second assertion of this lemma is concluded.

Case3. Supposes∈J⁻. We have to exclude the situations thatϕ >0 as well as thatϕchanges sign more than twice. Recall that e^u<1 and w_c<0 for anyS⁻ solution u. We divide the argument into the following subcases.

(3-1) Assumeϕis positive all the time. Then eitherϕ is bounded orϕ(r)→ ∞asr→ ∞.

Consider the identity(38)withc=0. Letw=ru. From(21), (24) and (35),w(r)approaches to 2N−β <0 andϕ(r)=O(logr)asr→ ∞. Moreover, from(23), (24) and (33), we have rw=O(r⁻)at infinity for some >0. Becausep0(t) <0 fort∈(0,1),

0>

∞ 0

rP₀ϕ dr= lim

r→∞r

ϕ(r)w(r)−w(r)ϕ(r)

=(β−2N ) lim

r→∞rϕ(r)0, which is impossible.

(3-2) Assume ϕ changes sign more than twice. Let z be the second zero ofϕ. Pick 0<

r1< z < r2such that ϕ satisfies(40)and has a local minimum and a local maximum at r= r₁, r₂respectively. Letz₁be the first zero ofϕ. Clearly,z₁< r₁. Letw=ru. Sinceϕ(z₁) <0, from(38),

0>

z₁

0

rP0ϕ dr= −z1w(z1)ϕ(z1),

implying that

u(z₁) <0. (42)

Henceu(r) <0 for allrz1by the property ofS⁻solutions. Furthermore, from(35), f

e^u(r1)

q(r₁)ϕ(r₁)=ϕ(r₁)0,

which reveals thate^u(r1)T₁. The numbersσ₁:=e^u(r1),ζ :=e^u(z)andσ₂:=e^u(r2)thus satisfy 0< σ₂< ζ < σ₁T₁.

From the c–t diagram shown in Fig. 1, there exists a (unique) constant c (> m) such that p_c(ζ )=0,p_c(t) >0 fort∈(σ₂, ζ )andp_c(t) <0 fort∈(ζ, σ₁). Hence

Pc(r) <0 forr∈(r1, z),

P_c(r) >0 forr∈(z, r₂). (43)

By (40) and (43) with such choice ofr₁, r₂, c, the LHS of (38)is positive while the RHS is negative; it is impossible. The proof is completed. 2

Lemma 3.2.limr→∞rϕ(r;s)=0for anys∈R.

Proof. Let ϕ(r)=ϕ(r;s) be the solution of (35) with u(r)=u(r;s) dependent on s∈R. The conclusion for s∈J^∗ follows readily from (35),Lemma 3.1and the fact that e^u(r)→1 asr→ ∞. In fact, pick a sufficiently larger₀>0, we have

rϕ(r)=r₀ϕ(r₀)− r r₀

q(τ )ϕ(τ )

e^{u(τ )}−T₁

e^{u(τ )}−T₂ τ dτ

r0ϕ(r0):=C (44)

forr > r0, hereT1<1< T2. We complete the proof for other cases by contradiction. Assume

rlim→∞rϕ(r)=0.

From the assumption, it is necessary thatϕis bounded. Recall that for anyS⁻orS⁺solutionu, the functionwcgiven in(36)is also a bounded function; it holds thatrw_c(r)→0 asr→0. Now we consider the casess∈J⁻ands∈J⁺in the following.

Case1. Lets∈J⁻. Note thatw_c<0 forS⁻solution. According toLemma 3.1, there are two possible situations as follows.

(1-1) Ifϕ changes sign twice, the proof is the same as the subcase (3-2) ofLemma 3.1, just with the modificationr2= ∞. The details are omitted here.

(1-2) Assumeϕchanges sign only once. Letz1be the zero ofϕ. Letc1be the number such that wc₁(z1)=0. Observe that forc=m,pmhas zeros at 0 and at some numberσ >1; especially, p_mis negative in(0,1); seeFig. 1for illustration. So by(38)we have

0>

z₁

0

rPmϕ dr= −z1wm(z1)ϕ(z1),

which implies thatw_m(z₁) <0 and hencec₁> m. As a result,p_c1 has two positive zerosθ₁, θ₂ withθ₁<1< θ₂. In particular,

p_c1>0 in(0, θ1),

pc₁<0 in(θ1,1). (45)

Now we claim that

P_c1(z₁) <0. (46)

Indeed, if P_c1(z₁)0 or equivalently p_c1(e^u(z1))0, then e^u(z1)θ₁ and thus by (42), e^u(r)< θ₁for allr > z₁. HenceP_c1(r)is positive for allr > z₁. So we have

0<

∞ z₁

rPc₁ϕ dr= lim

r→∞r ϕ(r)w_c

1(r)−wc₁(r)ϕ(r)

=0,

which is a contradiction. On the other hand, sincee^u(r)→0 asr→0,∞, we haveP_c1>0 near the origin and infinity. This together with (46)indicates that there arer₁, r₂withr₁< r₂such thatz₁∈(r₁, r₂)and

Pc₁(r) <0 forr∈(r1, r2),

Pc₁(r) >0 forr∈(0, r1)∪(r2,∞). (47) Letw=w_c=0=ru. Note that

r

ϕ(r)w(r)−w(r)ϕ(r)

= r 0

τ P0(τ )ϕ(τ ) dτ:=Θ(r). (48)

Clearly,Θ(r) <0 in(0, z1)and>0 in(z₁,∞), and in addition, limr→∞Θ(r)=0. So thatΘ is negative all the time. Hence for anyr∈R− {z₁},

w ϕ

(r) <0. (49)

Set

w(r₁)

ϕ(r₁) =K₁, w(r₂) ϕ(r₂)=K₂. From(49)we conclude that

K₁ϕ(r) < w(r) forr∈(0, r1), K₁ϕ(r) > w(r) forr∈(r₁, z₁);

K₂ϕ(r) > w(r) forr∈(z₁, r₂),

K₂ϕ(r) < w(r) forr∈(r₂,∞). (50) Recall thatz₁∈(r₁, r₂). So that from(47), lettingχ (r)=r^2aNq(r),

r^2−2aN−r₁^2−2aN

χ (r)p_c1 e^u(r)

<0 forr∈(0, z1), r^2−2aN−r₂^2−2aN

χ (r)p_c1 e^u(r)

>0 forr∈(z₁,∞). (51) Since w(z1) <0 via (42), we have w(r2) <0 and thus K2>0. On the other hand, letting b >0 be the site at whichuattains its maximum, from(45)–(46), we havePc₁(b) <0, because e^u(b)> e^u(z1)> θ1. Henceb∈(r1, z1). Note thatuis positive in(0, b)and negative in(b,∞).

In particular,w(b)=0. So thatK₁>0. Recall thatϕ(z₁)=w_c1(z₁)=0 by definition. So that, applying(38),

b 0

+

z1

b

rP_c1ϕ dr=

z1

0

rP_c1ϕ dr=0,