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# Relaxation Dynamics of Point Vortices

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Reviewed: 22 September 2021 Published: 09 November 2021

DOI: 10.5772/intechopen.100585

From the Edited Volume

## Vortex Dynamics - From Physical to Mathematical Aspects

Edited by İlkay Bakırtaş and Nalan Antar

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## Abstract

We study a model describing relaxation dynamics of point vortices, from quasi-stationary state to the stationary state. It takes the form of a mean field equation of Brownian point vortices derived from Chavanis, and is formulated by our previous work as a limit equation of the patch model studied by Robert-Someria. This model is subject to the micro-canonical statistic laws; conservation of energy, that of mass, and increasing of the entropy. We study the existence and nonexistence of the global-in-time solution. It is known that this profile is controlled by a bound of the negative inverse temperature. Here we prove a rigorous result for radially symmetric case. Hence E/M2 large and small imply the global-in-time and blowup in finite time of the solution, respectively. Where E and M denote the total energy and the total mass, respectively.

### Keywords

• point vortex
• quasi-equilibrium
• relaxation dynamics

## 1. Introduction

Our purpose is to study the system

ωt+ωψ=ω+βωψinΩ×0T,ων+βωψν∂Ω=0,ωt=0=ω0xE1

with

Δψ=ωinΩ,ψ∂Ω=0,β=ΩωψΩωψ2,E2

where ΩR2 is a bounded domain with smooth boundary ∂Ω, ν is the outer unit normal vector on ∂Ω, and

=x1x2,=x2x1,x=x1x2.E3

The unknown ω=ωxtR stands for a mean field limit of many point vortices,

ωxtdx=i=1Nαiδxitdx.E4

It was derived, first, for Brownian point vortices by [1, 2], with β=βt standing for the inverse temperature. Then, [3, 4] reached it by the Lynden-Bell theory [5] of relaxation dynamics, that is, as a model describing the movement of the mean field of many point vortices, from quasi-stationary state to the stationary state. This model is consistent to the Onsager theory [6, 7, 8, 9, 10, 11, 12] on stationary states and also the patch model proposed by [13, 14], that is,

ωxt=i=1Npσi1Ωitx,E5

where Np, σi, and Ωit denote the number of patches, the vorticity of the i-th patch, and the domain of the i-th patch, respectively [15, 16, 17].

This chapter is concerned on the one-sided case of

ω0=ω0x>0.E6

If this initial value is smooth, there is a unique classical solution to (1)(4) local in time, denoted by ω=ωxt, with the maximal existence time T=Tmax0+. More precisely, the strong maximum principle to (1) guaranttes

ω=ωxt>0onΩ¯×0T.E7

Then, the Hopf lemma to the Poisson equation in (2) ensures

ψν∂Ω<0,E8

and hence the well-definedness of

β=ΩωψΩωψ2.E9

We confirm that system (1)(3) satisfies the requirements of isolated system of thermodynamics. First, the mass conservation is derived from (1) as

ddtΩω=0,E10

because

νψ∂Ω=0E11

holds by (2). Second, the energy conservation follows as

12ddtψ22=ψψt=ωtψ=ωψψω+βωψψ=ωψβΩωψ2=0E12

by (1) and (2), because

ψψ=0,E13

where , denotes the L2 inner product. Third, the entropy increasing is achieved, writing (1) as

ωt=ωψ+logω+βψ,νlogω+βψ∂Ω=0.E14

In fact, it then follows that

Ωωtlogω+βψ=Ωωψlogψ+βψωlogω+βψ2dxE15

with

Ωωψlogω+βψ=Ωωψ=∂ΩωνψΩωψ=0E16

from (11) and

==0.E17

Since

Ωωtlogω=ddtΩωlogω1,Ωωtψ=12ddtψ22=0,E18

We thus end up with the mass conservation

M=Ωω,E19

the energy conservation

E=ψ22=ψω,E20

and the entropy increasing

ddtΩωlogω1=Ωωlogω+βψ20.E21

Henceforth, C>0 stands for a generic constant. In the previous work [4] we studied radially symmetric solutions and obtained a criterion for the existence of the solution global in time. Here, we refine the result as follows, where B01 denotes the unit ball.

Theorem 1 Let

Ω=B01,ω0=ω0r,ω0r<0,0<r=x1.E22

Then there is C0>0 such that

C0ω023Eω¯T=+,ωtC,t0,E23

where

ω¯=minΩ¯ω0>0.E24

Theorem 2 Under the assumption of (22) there is δ0>0 such that

EM2<δ0T<+.E25

Remark 1 Since

ω023=Ωω023/2ω¯2/3Ωω04/33/2=ω¯Ωω04/33/2ω¯Ω1/2Ωω02=ω¯Ω1/2M2E26

the assumption (23) implies

EM2C0Ω1/2.E27

Therefore, roughly, the conditions E/M21 and E/M21 imply T=+ and T<+, respectively.

Remark 2 The assumption (22) implies

β=βt<0,0t<T,E28

and then we obtain Theorem 1. In other words, the conclusion of this theorem arises from (28), without (22).

Remark 3 Since

EM2=Ωψ2Ωω2E29

it holds that

EM2=c22,c=Δ1ω0Ωω0=ψ0∂Ωψ0ν,E30

where ψ0=Δ1ω0.

The system (1)(4) thus obays a profile of the micro-canonical ensemble. In a system associated with the canonical ensemble, the inverse temperature β is a constant in (1) independent of t, with the third equality in (2) elimiated:

ωt+ωψ=ω+βωψ,ων+βωψν∂Ω=0,ωt=0=ω0x>0Δψ=ω,ψ∂Ω=0.E31

Then there arise the mass conservation

ddtΩω=0,E32

and the free energy decreasing

ddtΩωlogω1+β2ψ2dx=Ωωlogω+βψ20.E33

The system (31) without vortex term,

ωt=ω+βωψ,ων+βωψν∂Ω=0,ωt=0=ω0x>0Δψ=ω,ψ∂Ω=0.E34

is called the Smoluchowski-Poisson equation. This model is concerned on the thermodynamics of self-gravitating Brownian particles [18] and has been studied in the context of chemotaxis [19, 20, 21, 22, 23]. We have a blowup threshold to (34) as a consequence of the quantized blowup mechanism [19, 23]. The results on the existence of the bounded global-in-time solution [24, 25, 26] and blowup of the solution in finite time [27] are valid even to the case that β is a function of t as in β=βt. provided with the vortex term ωψ on the right-hand side. We thus obtain the following theorems.

Theorem 3 It holds that

βtδ,ω01<8πδ1T=+,ωtCE35

in (31), where δ>0 is arbitrary.

Theorem 4 It holds that

βtδ,ω01>8πδ1ω0>0,ω01>8πδ1suchthatT<+E36

in (31), where δ>0 is arbitrary.

Remark 4 In the context of chemotaxis in biology, the boundary condition of ψ is required to be the form of Neumann zero. The Poisson equation in (34) is thus replaced by

Δψ=ω1ΩΩω,ψν∂Ω=0E37

or

Δψ+ψ=ω,ψν∂Ω=0E38

by [28] and [29], respectively. In this case there arises the boundary blowup, which reduces the value 8π in Theorems 3–4 to 4π. The value 8π in Theorems 3–4, therefore, is a consequence of the exclusion of the boundary blowup [30]. This property is valid even for (37) or (38) of the Poisson part, if (22) is assumed.

Remark 5 The requirement to ω0 in Theorem 4 is the concentration at an interior point, which is not necessary in the case of (22). Hence Theorems 3 and 4 are refined as

βtδ,ω01<8πδ1T=+,ωtCE39

and

βtδ,ω01>8πδ1T<+,E40

if (22) holds in (35). The main task for the proof of Theorems 1 and 2, therefore, is a control of β=βt in (1).

This paper is composed of four sections and an appendix. Section 2 is devoted to the study on the stationary solutions, and Theorems 1 and 2 are proven in Sections 3 and 4, respectively. Then Theorem 4 is confirmed in Appendix.

## 2. Stationary states

First, we take the canonical system (31) with β independent of t. By (32) and (33), its stationary state is defined by

logω+βψ=constant,ω=ωx>0,Ωω=M.E41

Then it holds that

ω=MeβψΩeβψE42

and hence

Δψ=MeβψΩeβψ,ψ∂Ω=0.E43

There arises an oredered structure arises in β<0, as observed by [11], as a consequence of a quantized blowup mechanism [19, 20, 31]. In the micro-canonical system (1) and (2), the value β in (43) has to be determined by E besides M.

Equality (21), however, still ensures (41) and hence (42) in the stationary state even for (1)(3). Writing

v=βψ,μ=βMΩeβψ,E44

we obtain

Δv=μevinΩ,v∂Ω=0,EM2=v22Ωvν2E45

by (30) and (43).

This system is the stationary state of (1) and (2) introduced by [4]. The first two equalities

Δv=μev,v∂Ω=0E46

comprise a nonlinear elliptic eigenvalue problem and the unknown eigenvalue μ is determined by the third equality,

EM2=v22Ωvν2.E47

The elliptic theory ensures rather deailed features of the set of solutions to (46). Here we note the following facts [31].

1. There is μ¯=μΩ>0 such that the problem (46) does not admit a solution for μ>μ¯.

2. Each μ0 admits a unique solution.

3. Each 0<δ<μ¯ admits a constant C=Cδ>0 such that vC for any solution v=vx.

4. There is a family of solutions μv such that μ0 and v+.

We show the following theorem, consistent to Theorem 2.

Theorem 5 If Ω=B01R2, there is δ>0 such that any solution vμ to (45) admits

EM2δ.E48

Proof: If μ=0, it holds that v=0. We have ±v>0 exclusively in Ω, provided that ±μ>0, respectively. By the elliptic theory [32], therefore, any solution v to (46) is radially symmetric as in v=vr, r=x. We have, furthermore, ±vr<0 in 0<r1, if ±μ>0, respectively.

Then it holds that ψ=ψr, and hence

1rrψrr=ωin0<r1,ψr=1=0E49

by (42) and (43), which implies

rψrr=0rsds>0,0<r1.E50

We thus obtain μ0, in particular.

If μ<0 we have β>0 by (44), and therefore, ψr>0 in 0<r1 by vr>0 there. It is a contradiction, and hence μ>0. In this case, the solution v=vr to (46) is explicit [31]. The numbers of the solution is 0, 1, and 2, according to μ>2, μ=2, and 0<μ<2, respectively, and if 0<μ2 the solutions v=v± are given as

v±r=log8γ±r1+γ±r22,γ±=4μ1μ4±1μ21/2.E51

In fact, we have γ+=γ for μ=2.

This solution is parametrized by

σ=Ωμev08π.E52

Hence each 0<σ<8π admits a unique solution vμ to (46), and v=v+ and v=v according as σ4π and σ4π, respectively. It holds also that μ0 if either σ8π or σ0. Thus we have only to confirm that E/M2 is bounded, both as σ8π and σ0.

As σ8π, we have

v=v+x4log1xlocallyuniformlyonΩ¯\0E53

and hence

v22+,∂Ωvν8π,E54

which implies

limσ8πEM2=+.E55

As σ0, on the other hand, we have

v=vx0uniformlyinΩ¯.E56

Since μ0, furthermore, there arises that

γ=γ=4μ1μ41μ21/2=μ1+o1.E57

It holds also that

vr=log8γμ2log1+μr2E58

and hence

vrr=4μr1+μr22=4μr1+o1uniformlyonΩ¯.E59

Then, (59) implies

v22=2π01vr2rdr=2π16μ201r3dr1+o1=8πμ21+o1E60

as well as

∂Ωvν2=16μ22π1+o1.E61

It thus follows that

limσ0EM2=14E62

and hence the conclusion.

## 3. Proof of Theorem 1

The first observation is the following lemma.

Lemma 1 Under the assumption of (22), it holds that

β=βt<0,ωrrt<0,0<r1,0t<T.E63

Proof: We have (7) and hence

ψrrt<0,0<r1,0t<TE64

by (49), which implies, in particular,

β=ωψΩωψ2<0E65

at t=0 by (22).

Since ω=ωrt and ψ=ψrt, we obtain ψ=0, and hence

ωt=ωrr+1rωr+βψrωrβω2E66

by (1). Then z=ωr satisfies

zt=zrr1r2z+1rzr+βψrrz+βψrzr2βωz,0<r1,0t<Tzr=0=0,zt=0=ω0rr<0,0<r1E67

and

z=βωψr,r=1,0t<T.E68

Putting

mt=min∂Ωzt=ωrtr=1,E69

we obtain m0<0 from the assumption. If there is 0<t0< such that

mt<0,0t<t0<T,mt0=0,E70

we obtain zrt>0 for 0t<t0, 0<r1, and t=t0, 0<r<1 by the strong maximum principle. By (64), we have (65) for 0tt0, that is,

β=01ψrzrdr01ωψr2rdr<0,0tt0,E71

and hence

z=βωψr<0r=1,t=t0,E72

a contradiction. It holds that z=ωr<0 for 0t<T, r=1, and hence

β=01ψrωrrdr01ωψr2rdr<0,0t<T.E73

The proof of Theorem 3 relies on the fact

βC,Ωωlogω1CT=+,ωtC.E74

This property is known for the Smoluchoski-Poisson equation (34), but the proof is valid even to (31) with vortex term. Having (21), therefore, we have to provide the inequality βC.

The inequality β<0, on the other hand, is sufficient for the following arguments.

Lemma 2 If β0, 0t<T, it holds that

ωω¯minΩ¯ω0>0onΩ¯×0T.E75

Proof: Since (17) we obtain

ωt+ψω=Δω+βψω+βΔψ=Δω+βψωβω2Δω+βψωinΩ×0TE76

with

ων=βωψν>0on∂Ω×0TE77

by (8). Then the result follows from the comparison theorem.

Lemma 3 Under the assumption of the previous lemma, there is C0=C0Ω>0 such that

C0ω023Eω¯ωt2ω02,βtαω022Eω¯,0t<T.E78

Proof: Using (11) and (17), we obtain

Ωωψω=Ωωωψ=12Ωω2ψ=12Ωω2ψ=0.E79

Hence (1) with (2) implies

12ddtω22+ω22=βΩωψω=β2ψω2=β2∂Ωω2ψν+β2Δψω2β2ω33E80

by β0 and (88). Since

Ωωψ=∂Ωωψν+ΩωΔψΩω2E81

follows from (8), furthermore, it holds that

β=ΩωψΩωψ2ω¯1ω22ψ22=1Eω¯ω22.E82

Then ineqality (80) induces

12ddtω22+ω2212Eω¯ω22ω33.E83

Here we use the Gagliardo-Nirenberg inequality (see (4.16) of [19]) in the form of

ω33CωH1ω22=Cω22ω2+ω2,E84

to obtain

12ddtω22+ω22CEω¯ω24ω2+ω212ω22+C28Eω¯2ω28+C2Eω¯ω25E85

and hence

ddtω22+ω22CEω¯ω25CEω¯ω23+1.E86

Then, Poincaré-Wirtinger’s inequality ensures

ddtω22+μω22CEω¯CEω¯ω26+ω23ω22,E87

where μ=μΩ>0 is a constant.

Writing

yt=CEω¯ω23,E88

we obtain

ddtω22+μω22y2+yω22,E89

and therefore, if

y2+y<μ/2E90

holds at t=0, it keeps to hold that

ddtω220E91

and (90) for 0t<T. Then, we obtain

ωt2ω02,0t<T,E92

and hence

βtω022Eω¯=α,0t<TE93

by (82).

The condition y0<μ2 means

C0ω02Eω¯E94

for C0>0 sufficiently large, and hence we obtain the conclusion.

Proof of Theorem 1: By the parabolic regularity, it suffices to show that

ωtC,0t<TE95

under the assumption. We have readily shown

ωt2C,0βtC,0t<TE96

by Lemma 3. Then, the conclusion (95) is obtained similarly to (34). See [26] for more details.

In fact, we have

Ωωψωp=Ωωψωp=pΩωpψω=pp+1Ωψωp+1=pp+1Ωωp+1ψ=0E97

for p>0 by (11) and (34). Then it follows that

1p+1ddtΩωp+1+4pp+12ωp+1222=βΩωψωp=βpp+1Ωψωp+1βpp+1Ωωp+1Δψ=βpp+1Ωωp+1CΩωp+2E98

by β<0 and (8). Then, Moser’s iteration scheme ensures (95) as in [33].

## 4. Proof of Theorem 2

We begin with the following lemma.

Lemma 4 Under the assumption of (22), it holds that

βtδ,0t<T,M=ω01>8πδT<+E99

in (31), where δ>0 is a constant.

Proof: We have ω=ωrt and ψ=ψrt for r=x under the assumption, which implies ψ=0. Then we obtain

ωψ=ωψ=0E100

by (17). It holds also that

ωψ=ωψrxr=xrωψr+xrωψr=1rωψr+ωψrr=1rrωψrr,E101

and therefore, there arises that

ωt=1rrωr+βrωψrr,ωr+βωψrr=1=0.E102

from (31).

Then (102) implies

ddt01ωr3dr=01ωtr3dr=01rωr+βrωψrrr2dr=012r2ωr+βωψrdr=2r2ωr=0r=1+0142βωψrr2dr.E103

Here we use (50) derived from the Poisson part of (31), that is,

rψrrt=Art0rstds.E104

Putting

λ=01ωrdr=M2π,E105

we obtain

ddt01ωr3dr=2ωr=1+4λ+2β01AArdr=2ωr=1+4λ+βA2r=0r=1=2ωr=1+4λ+βλ2<4λβ+M8π4λδ+M8π.E106

Since δ+M8π<0, therefore, T=+ is impossible, and we obtain T<+.

Lemma 5 Under the assumption (22), there is δ>0 such that

EM2<δ,βt0,0t<Tβt1CE1/20t<T.E107

Proof: First, Lemma 1 implies

ωωωr=1.E108

Second, we have

Ωψω=∂Ωψνω+ΩΔψω=ω∂Ωψν+ω22=ωΩΔψ+ω22=ω22ωM,E109

and hence

β=ΩψωΩωψ2=ω22ωMΩωψ2.E110

Here, we use the Gagliardo-Nirenberg inequality in the form of

w42Cw2wH1,E111

which implies

Ωωψ2ω2ψ42Cω2ψ2ψH1CE1/2ω22E112

by the elliptic estimate of the Poisson equation in (2),

ψH2Cω2.E113

We have, on the other hand,

ωMMEΩωψ2E114

by (110), and therefore,

β1CE1/2EM12CE1/2,E115

provided that

EM2<12C2.E116

Then the conclusion follows.

Proof of Theorem 2: By Lemma 5, there is δ0> such that

EM2<δβ1CE1/2δ1,E117

and then, Lemma 4 ensures

M>8πδ1T<+.E118

The assumption in (118) means

EM2<18πc2,E119

and hence we obtain the conclusion.

## Acknowledgments

This work was supported by JSPS Grand-in-Aid for Scientific Research 19H01799.

This theorem is valid to the general case of Ω and ω0 without (22). We assume δ=1 without loss of generation, so that

β1.E120

We follow the argument [27] concerning (34) with the Poisson part replaced by (42) or (43). Thus we have to take case of the vortex term ωψ, time varying β=βt, and the Dirichlet boundary condition in (31).

We recall the cut-off function used in [34] (see also Chapter 5 of [19]). Hence each x0Ω¯ and 0<R1 admit φ=φx0,RC2Ω¯ with

φν∂Ω=0,0φ1,φ=1inΩBx0R/2,φ=0inΩ\Bx0R,E121

and

φCR1φ1/2,2φCR2φ1/2.E122

In more details, we take a cut-off function, denoted by ψ, satisfying (121), using a local conformal mapping, and then put φ=ψ4.

Let

φC2Ω¯,φν∂Ω=0.E123

be given. First, we have

ddtΩωφ=Ωωψφω+βωψφdx=Ωωψφ+ωΔφβωψφdxE124

by (11). It holds that

Ωωψφ=Ω×ΩωxtxGxxφxωxtdxdx=ΩΩωxtφx0,2RxxGxxφxωxtdxdx+Ω×Ωωxt1φx0,2RxxGxxφxωxtdxdx=I+II.E125

Let, furthermore, x0Ω and 0<R1 in the above equality. Then,

φ=xx02φx0,.RE126

satisfies the requirement (123).

It holds that

φ=2xx0φx0,R+xx02φx0,RE127

and hence

φCxx0φx0,R+xx0R1φx0,R1/2Cxx0φx0,R1/2.E128

We obtain, furthermore,

xx02R,xx0RxxR,E129

and hence

xGxxCR1E130

in this case. Then it follows that

IICR1MΩxx0φx0,R1/2ωxtdxCR1M3/2A1/2,E131

where

A=Ωxx02φx0,Rω.E132

We have, on the other hand,

I=Ω×Ωωxtφx0,2RxxGxxφxωxtdxdx=12ΩΩφx0,2RxφxxGxx+φx0,2RxφxxGxxωω,E133

where G=Gxx is the Green’s function to

Δψ=ω,ω∂Ω=0E134

and

ωω=ωxtωxtdxdx.E135

Here we use the local property of the Green’s function

Gxx=Γxx+Kxx,KC2Ω¯×ΩC2Ω×Ω¯,E136

where

Γx=12πlog1xE137

stands for the fundamental solution to Δ.

Let

ρx0,R2(x,x=φx0,2RxφxxKxx+φx0,2RφxxKx.x.E138

Since (128) implies

φx0,2RxφxCφx0,2Rxxx0φx0,R1/2xCxx0φx0,R1/2x,E139

it holds that

ρx0,R1xxCxx0φx0,R1/2x+xx0φx0,R1/2x.E140

Then, we obtain

I=12Ω×Ωρx0,R0xxωω+IIIE141

with

IIICM3/2A1/2CR1M3/2A1/2,E142

where

ρx0,R0xx=Γxxφx0,2Rxφxφx0,2Rxφx.E143

Here, we have

Γx=x2πx2,E144

and therefore,

ρx0,R0xx=ρx0,R2xx+ρx0,R3xxE145

fo

ρx0,R2xx=12πxxxx2φx0,2RxφxφxE146
ρx0,R3xx=12πxxxx2φx0,2Rxφx0,2Rxφx.E147

Since (128) implies

ρx0,R3xxCR1φxCR1xx0φx0,R1/2x,E148

there arises that

I=12Ω×Ωρx0,R2xxωω+IV,E149

with

IVCR1M3/2A1/2,E150

similarly.

We have, furthermore,

φxφx=2xxφx0,Rx+2xx0φx0,Rxφx0,Rx+xx02φx0,Rxφx0,Rx+xx02xx02φx0,Rx,E151

and hence

ρx0,R2xx=1πφx0,2Rxφx0,Rx+ρx0,R4xx+ρx0,R5xx+ρx0,R6xxE152

with

ρx0,R4xxCxx1φx0,2Rxxx0φx0,Rxφx0,RxCR1xx0φx0,2Rx,E153
ρx0,R5xxCxx1φx0,2Rxxx02φx0,Rxφx0,RxCR2xx02φx0,2RxCR1xx0φx0,2Rx,E154

and

ρx0,R6xxCxxφx0,2Rxxx02xx02φx0,RxCR1xx0+xx0φx0,Rxφx0,2RxCR1xx0φx0,Rx+R1xx0φx0,2RxE155

by

xx02xx02=xxx+x2x0xxxx0+xx0.E156

The residual terms are thus treated similarly, and it follows that

I+12πΩωφx0,RΩωφx0,2RCR1M3/2A1/2,E157

which results in

Ωωψφ+12πΩωφx0,RΩωφx0,2RCR1M3/2A1/2.E158

We can argue similarly to the vortex term in (124). This time, from

Γxx=0E159

it follows that

ΩωψφCR1M3/2A1/2.E160

Concerning the principal term of (124), we use

Δφ=4φx0,R+4xx0φx0,R+xx02Δφx0,R.E161

From

xx0φx0,RCR1xx0φx0,R1/2E162

and

x02Δφx0,RCR2xx02φx0,R1/2CR1xx0φx0,R1/2,E163

it follows that

ΩωΔφ4Ωωφx0,RCΩR1xx0φx0,RωCR1M1/2A1/2.E164

Let M1=Mx0,R and M2=Mx0,2R for

Mx0,R=Ωωφx0,R.E165

Then, using (120), we end up with

Inequalilty (166) implies T<+ if A01, as is observed by [27] (see also Chapter 5 of [19]). Here we describe the proof for completeness.

The first observation is the monotoniity formula

ddtΩωφCM+M2φC1,E167

derived from (124) and the symmetry of the Green’s function: Gxx=Gxx. The proof is the same as in (34) and is omitted.

Second, we put I1=Ix0,R and I2=Ix0,2R for

Ix0,R=Ωxx02ωφx0,R.E168

Then it holds that

M2M1R<xx0<2Rφx0,2Rω2R1Ωxx0φx0,2Rω2M1/2R1I21/2E169

and

A2=A1+Ωxx02φx0,2Rφx0,RωA1+4R2Ωφx0,2Rφx0,Rω,E170

which implies

dA1dt4M1M122π+CR1M3/2+M1/2A11/2+CM3/2+M1/2Ωφx0,2Rφx0,Rω1/2.E171

Here, we use (167) to ensure

ddt(4M1M22πCM+M2R2E172

and

ddtΩφx0,2Rφx0,RωCM+M2R2.E173

Then, it follows that

4M1M122π4M10M1022π+CBaR1t1/2E174

and

Ωφx0,2Rφx0,RωΩφx0,2Rφx0,Rω0+CBaR1t1/22R2A20+CBaR1t1/2E175

for

B=M3/2+M1/2,as=s2+s.E176

Thus we obtain

dA1dt4M10M1022π+CR1BA11/2+CBA201/2+CBaR1t1/2=J0+CBaR1t1/2+CBR1A11/2E177

for

J=4M1M124π+CBR1A21/2.E178

Assume M10>8π, and put

4δ=4M10M1022π<0.E179

Let, furthemore,

1R2Ωxx02φx0,2Rω0η.E180

Now we define s0 by

CBas0=δE181

in (177), and take 0<η1 such that

ηδs02.E182

Then, if R and T0 satisfy R2T0=ηδ1, it holds that

A10R2η<2δT0.E183

Making 0<η1, furthermore, we may assume

J0+CBR1A101/24δ+CBR1A201/24δ+CBη1/23δ,E184

which results in

dA1dtJ0+CBaR1T01/2+BR1A1t1/2
=J0+δ+CBR1A11/2,0t<T0,E185

provided that TT0.

A continuation argument to (184)(185) guarantees

dA1dt2δ,0t<T0,E186

and then we obtain

A1T0A102δT0<0E187

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