Open access peer-reviewed chapter - ONLINE FIRST

Relaxation Dynamics of Point Vortices

By Ken Sawada and Takashi Suzuki

Submitted: July 5th 2021Reviewed: September 22nd 2021Published: November 9th 2021

DOI: 10.5772/intechopen.100585

Downloaded: 24

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 ΩR2is a bounded domain with smooth boundary ∂Ω, νis the outer unit normal vector on ∂Ω, and

=x1x2,=x2x1,x=x1x2.E3

The unknown ω=ωxtRstands 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=βtstanding 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 Ωitdenote 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 L2inner 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>0stands 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 B01denotes the unit ball.

Theorem 1Let

Ω=B01,ω0=ω0r,ω0r<0,0<r=x1.E22

Then there isC0>0such that

C0ω023Eω¯T=+,ωtC,t0,E23

where

ω¯=minΩ¯ω0>0.E24

Theorem 2Under the assumption of(22)there isδ0>0such that

EM2<δ0T<+.E25

Remark 1Since

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 conditionsE/M21andE/M21implyT=+andT<+, respectively.

Remark 2The 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 3Since

EM2=Ωψ2Ωω2E29

it holds that

EM2=c22,c=Δ1ω0Ωω0=ψ0∂Ωψ0ν,E30

whereψ0=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 βis 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 βis a function of tas in β=βt. provided with the vortex term ωψon the right-hand side. We thus obtain the following theorems.

Theorem 3It holds that

βtδ,ω01<8πδ1T=+,ωtCE35

in(31), whereδ>0is arbitrary.

Theorem 4It holds that

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

in(31), whereδ>0is arbitrary.

Remark 4In the context of chemotaxis in biology, the boundary condition ofψis 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 value8πin Theorems 3–4 to4π. The value8πin 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 5The requirement toω0in 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=βtin (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.

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2. Stationary states

First, we take the canonical system (31) with βindependent 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 βin (43) has to be determined by Ebesides 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 μis 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>0such that the problem (46) does not admit a solution for μ>μ¯.

  2. Each μ0admits a unique solution.

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

  4. There is a family of solutions μvsuch that μ0and v+.

We show the following theorem, consistent to Theorem 2.

Theorem 5IfΩ=B01R2, there isδ>0such that any solutionvμto(45)admits

EM2δ.E48

Proof:If μ=0, it holds that v=0. We have ±v>0exclusively in Ω, provided that ±μ>0, respectively. By the elliptic theory [32], therefore, any solution vto (46) is radially symmetric as in v=vr, r=x. We have, furthermore, ±vr<0in 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<0we have β>0by (44), and therefore, ψr>0in 0<r1by vr>0there. It is a contradiction, and hence μ>0. In this case, the solution v=vrto (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<μ2the solutions v=v±are given as

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

In fact, we have γ+=γfor μ=2.

This solution is parametrized by

σ=Ωμev08π.E52

Hence each 0<σ<8πadmits a unique solution vμto (46), and v=v+and v=vaccording as σ4πand σ4π, respectively. It holds also that μ0if either σ8πor σ0. Thus we have only to confirm that E/M2is bounded, both as σ8πand 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.

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3. Proof of Theorem 1

The first observation is the following lemma.

Lemma 1Under 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=0by (22).

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

ωt=ωrr+1rωr+βψrωrβω2E66

by (1). Then z=ωrsatisfies

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

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

we obtain zrt>0for 0t<t0, 0<r1, and t=t0, 0<r<1by 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<0for 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 2Ifβ0,0t<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 3Under the assumption of the previous lemma, there isC0=C0Ω>0such 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 β0and (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>0is 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<μ2means

C0ω02Eω¯E94

for C0>0sufficiently 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>0by (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<0and (8). Then, Moser’s iteration scheme ensures (95) as in [33].

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4. Proof of Theorem 2

We begin with the following lemma.

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

βtδ,0t<T,M=ω01>8πδT<+E99

in(31), whereδ>0is a constant.

Proof:We have ω=ωrtand ψ=ψrtfor r=xunder 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 5Under the assumption(22), there isδ>0such 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.

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Acknowledgments

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

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This theorem is valid to the general case of Ωand ω0without (22). We assume δ=1without 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<R1admit 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, x0Ωand 0<R1in 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=Gxxis 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,Rand M2=Mx0,2Rfor

Mx0,R=Ωωφx0,R.E165

Then, using (120), we end up with

dAdt4M1M122π+CR1M3/2+M1/2A1/2+CM2M1.E166

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,Rand I2=Ix0,2Rfor

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 s0by

CBas0=δE181

in (177), and take 0<η1such that

ηδs02.E182

Then, if Rand T0satisfy 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

by (183), a contradiction.

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Ken Sawada and Takashi Suzuki (November 9th 2021). Relaxation Dynamics of Point Vortices [Online First], IntechOpen, DOI: 10.5772/intechopen.100585. Available from:

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