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# Our goal

문서에서 Singularity formation of Euler equations (페이지 34-46)

Proposition 7 By changing the variables of 1D Burgers equation using self-similar variables and sub- stitutingτ˙=0,ξ˙ =c,κ=c, we get(∂s12)U+ (32x+U)Ux =0 with W as the solution. Conversely, if we substitute w by a solution of 1D Burgers equation, z by 0 and a by 0 into(112)and(114), we get τ˙=0,ξ˙=c,κ=c. Therefore, in 2D Euler equation,τ˙=0,ξ˙ =c,κ=c is the condition equivalent to transforming the equation into 1D Burgers equation.

Proof 6 Assume that τ˙ =0,ξ˙ =c,κ =c. the self similar variables become x= θ−(θ0+ct)

0−t)32

,s(t) =

−log(τ0−t), and the Burgers equation turns to u(θ,t) =e2sU(x,s) +c.So we can rewrite Burgers equation to(−12+∂s)U+ (32x+U)Ux=0.

The only if direction can be proved by substituting A,Z=0. In this case,(114a)showsτ˙=0,(114c) showsκ˙ =e2s(FW0(s) + F

0,(2) W

Wxxx0 (s)) =0, so letκ≡c.(114b)impliesξ˙ =κ=c.

Next, we needRtT

0 ||∂θz||Ldt<∞.Since||∂θz(·,t)||L=||Zx∂x

∂ θ||L =e32s||Zx(·,s)||L, Z T

t0

||∂θz||Ldt=

Z −log(τ(T)−T)

−ε

e32s||Zx||L

dt dsds=

Z −log(τ(T)−T)

−ε

e12s

1−τ˙||Zx||Lds. (118a) We therefore expert that

||Zx||L ≤ce−(12)s ∀s (119a) should be needed for any nonnegative valueδ.The value ofδ will be determined later.

Finally,RtT

0 ||a(·,t)||L(T)dt=R−log(τ(T)−T)

−ε ||A(·,s)||L(T)e−s 1−τ˙ds. So,

||A||L≤ces ∀s (120a)

is also a necessary condition. (But later, it will be proved in(124).) (b) ∂θwblows up ast→T.(same as 1D Euler solution’s property.)

Since ∂θw(ξ(t),t) =esWx(0,s) =−es,∂θwblows up ats→∞(t→T). Also, we derive from (117a) the estimate ||∂θw(·,t)||L =||esWx||L ≤ces,so when s is finite, it does not blow up. It means that∂θwblows up first atT.

To ensure thatTis the smallest singularity time, we need the conditions∂θa,∂θz<∞ast→T. But before we show it, let’s check that our solutions satisfy the conditions of positive density and nonzero vorticity all the time.

(a) The density P is always positive

Proof 8 P= (α2(w−z))α1 impliessupt∈[−ε,T)||P(·,t)||L(T)≤(α2(||w(t)||L+||z(t)||L))α1. Therefore, if||w(t)||L ≤M and||z(t)||L≤M, we can conclude that||P||L ≤M for some M(α,κ0).

Also, since P(θ,t)≥(α2inft|w−z|)α1, in order to guarantee P(θ,t)>0,we need a uniform lower bound c of P that satisfies|w−z| ≥c>0.Here, we will define the following initial data conditions for the convenience of future calculations.

||z0||Cn+||a0||Cn ≤1 ∀0≤n≤4, (121a)

˜

c≤ ||w0||L≤c (121b)

Using these conditions, the other conditions of this proof are also proved.

Proposition 8 ||w(t)||L+||z(t)||L+||a(t)||L ≤M for large M(α,κ0),∀t∈[t0,T).

Proof 9 With the new notation ϕ ∈ {w,z,a}, we will write the equation (76) as ∂tϕ+ λ(w,z)ϕ0=Q(w,z,a)where Q: an explicit quadratic polynomial which obeys|Q(w,z,a)| ≤ Cα(max{|w|,|z|,|a|})2,for short.

0|Q|ds0 with s<T<ε, w(t)≤c+ε0|Π(t)|2 for some positive constant c and small enoughε0.

Similarly, we obtain z(t)≤1+ε0|Π(t)|2 and a(t)≤1+ε0|Π(t)|2. Combining the three inequalities gives the following expression

Π(t)≤(c+2) +3ε0Π2(t). (122a)

Since w,z,a are continuous and finite at s=s0, suptΠ(t)<∞for t∈[t0,t0+δ) where δ is a small enough value. In this process, we need the statement of Theorem 4. Therefore, let supt∈[t0,t0+δ)Π(t) ≤ c(c). Then by solving (122), we get z(t),a(t)≤c(c). It implies w(t)≤c+εw2(t) +ε(c(c))2 ∀t∈[t0,t0+δ).We can repeat this same process at t=t0+δ2, while t<T.Hence, by makingεsmall enough, we can say that w is less than a number that is slightly larger than c. In a similar way, z≤2, a≤2, w≥c˜−ε0|w|2c2˜ for sufficiently smallε0.Therefore, we conclude that

˜ c

2 ≤w(·,t)≤4

3c, ||z(·,t)||L≤2, ||a(·,t)||L ≤2 ∀t∈[−ε,T). (123a) In short,

||w(t)||L+||z(t)||L+||a(t)||L ≤4+4

3c≤M. (124a)

Lemma 2 Replace(124)with the equation for W,Z,A :

(W) Since w=e2sW+κ, W =e2s(w−κ)≤2Mes2 for sufficiently large M.∴||W(s)||L

x

2Me2s.

(Z,A) Since z=Z and a=A,||Z(s)||L+||A(s)||L ≤M∀s≥ −logε.

Through the results so far,|w−z| ≥c2˜−2for somec˜>4. Ifν0is defined to satisfy P(θ,t)≥ (α2(c2˜−2))α1ν20 >0,we get the information aboutc:˜

˜ c≥2(2

α(ν0

2)α+2). (125a)

With these variables, ν20 ≤P(θ,t)≤M.

(b) The vorticityωis nonzero for all time.

Proof 10 In 2D Euler equation, the vorticity isω=1rr(ruθ)−1rθur=2b(θ,t)−∂θa(θ,t).

So its initial data isω0=2b0−∂θa0=w0+z0−∂θa0.Using(121a),(121b)and(125), we get the relation

˜ c

2 ≤c˜−2≤ω0≤c+2. (126a)

Therefore, as mentioned in the proof of 4, ω0= ωP0

0 satisfies 2Mc˜ ≤ω02c+4

ν0 . Let φ0 = b,φ(−ε) =ω0as a characteristic. Because of the equation∂tω+b∂θω= αaω,

ω(φ(t)) =ω0+ Z t

−ε

a

αω(φ)ds≤ω0+ Z t

−ε

2

αω(φ)ds. (127a)

Hence, by Gronwall lemma,ω(φ(t))≤ω0eα2(t+ε)2c+4

ν0 e2(t+ε)α ≤22c+4ν

0 .The last inequality holds for small enoughε.

Also, observe that−ω(φ(t)) =−ω0+R−εt αa(−ω(φ))ds implies

−ω(φ(t))≤ −ω0+| Z t

−ε

2

α(−ω(φ))ds| ≤ − c˜ 2M+|

Z t

−ε

2

α(−ω(φ))ds|. (128a) By Gronwall lemma, it shows−ω(φ(t))≤ −ω0ek(t), where k(t) =α2(−ω(φ)) if R−εt −ω(φ)>

0and k(t) =−2

α(−ω(φ))ifR−εt −ω(φ)≤0.It implies ω(φ(t))≥ω0ek(t)≥ c˜

2Mek(t). (129a)

In this point, we showed thatω≥0, so k(t) =−2

α(−ω(φ)).As a result, for arbitrary small ε,ω(φ(t))≥4Mc˜ .Automatically,ω satisfies

1 M2 ≤ c˜

4M ν0

2 ≤ω(·,t)≤2(2c+4) ν0

M≤M2 ∀t∈[−ε,T) (130a) for large M(c,˜ c,ν0).

Lemma 3 Sinceκ=w(ξ(t),t),it is bounded until t=Taccording to(124).

∴|κ(t)| ≤cκ ∀t<T (131a) such cκexists.

3. About the solution’s singularity formation

From here on, we shall denoteW =W1, whereW1 was a solution of steady self-similar Burgers equation

−1

2W1+ (3x

2 +W1)∂xW1=0, (132a)

as we saw in Chapter 2. We regard W as a perturbed solution ofW, and the explicit form ofW is well-known. Since we can obtain|x23Wx| → 13 as|x| →∞through simple calculation, there is a possibility that W also satisfies|x23Wx| → 13.In other words,Wxhas a|x|23 decay rate as|x| →∞, uniformly in s.

Proof 11 We only need to prove||(x23+f(x))W˜x||L≤C whereW˜ =W−W , f(x) =cxk1+dxk2+

· · · with0≤ki23.Setν(x,s) = (x23+f(x))W˜x.The proof will use the maximum principle and bootstrap principle in Appendix.

Using the bootstrap method, after assuming H(t), we will prove C(t) which is stronger than H(t) and conclude that C(t) holds for all t. First,

||ν(x,s)||L≤1 (133a)

is our bootstrap assumption. From(112), we can find thatW˜xis the solution of (∂s+1+W˜x+2Wx

1−τ˙ + (1−α)es2Zx

(1+α)(1−τ)˙ )W˜x+ (gW+3x

2 + W

1−τ˙)W˜xx= (134a)

xFW−(gW+W˜ +τW˙

1−τ˙ )Wxx−(τW˙ x

1−τ˙ + (1−α)es2Zx

(1+α)(1−τ)˙ )Wx. (134b) SinceW˜x= ν

x23+f

andW˜xx=νx(x

2

3+f)−(23x13+f0

(x23+f)2 ,it turns into

sν+ (1+W˜x1−+2Wτ˙ x+(1+α)(1−1−α τ)˙ es2Zx−(gW+3x2 +W+W˜1−τ˙ )

2 3x13+f0

x23+f )ν+ (gW+3x2 +W˜1−+Wτ˙x= (x23+ f)∂xFW−(gW+W+˜1−τW˙τ˙ )(x23+f)Wxx−(τW1−˙ τx˙+ (1−α)e

s 2

(1+α)(1−τ˙)Zx)(x23+f)Wx. Therefore,

sν+ (1+W˜x+2Wx− 2x23 3(x23+f)(3

2+W˜ +W

x )− f0 x23 +f

(3x

2 +W˜ +W

1−τ˙ ))ν+ (135a) (gW+3x

2 + W

1−τ˙)νx= (135b)

−(τ˙(W˜x+2Wx) +1−α1+αe2sZx

1−τ˙ − 2x23

3(x23+f)( τW˙

(1−τ)x˙ +gW

x )− gWf0

x23+f)ν+ (135c)

(x23+f)∂xFW−(gW+ τW˙

1−τ˙)(x23+f)Wxx−(τW˙ x

1−τ˙+ (1−α)es2

(1+α)(1−τ˙)Zx)(x23+f)Wx− (135d) 1

1−τ˙(x23+f)Wxx

Z x 0

ν(x0) (x0)23+f

dx0. (135e)

We deliberately make the equation in the form of the last expression to apply the second maximum principle in Appendix A. To simply call the force term, let(135c)as F2and(135d)as F1. The final term gives K(x,x0,s) =−1−1˙

τ(x23+f)Wxx 1[0,x](x0)

(x0)23+f,which is the notation of a kernel in the second maximum principle.

We have a lower bound of D, which is a number we don’t know yet whether it is positive or not.

D=1+W˜x+2Wx− 2x23 3(x23 +f)(3

2+W˜ +W

x )− f0 x23+f(3x

2 +W˜ +W

1−τ˙ ) (136a)

≥1− 1

x23+f +2Wx− 2x23 3(x23+f)(3

2+W x +1

x Z x

0

dx0

(x0)23 +f)− f0 x23 +f(3x

2 +W˜ +W

1−τ˙ ) (136b)

≡Dupper(x). (136c)

Here, in the last term, we have x f0

x23+f

. Since it diverges at f(x) =−x23 if f06=0,let’s say f0≡0.

Next, we need to verify the condition(240a).

Z

R

|K(x,x0,s)|dx0≤1

c(x23 +f)|Wxx(x)|

Z |x|

0

1 (x0)23 +f

dx0≡Dlower(x). (137a) The first inequality holds for(117b). Now, we expect that,

0<Dlower(x)≤3

4Dupper(x) (138a)

in some domainΩcto apply the second maximum principle.

(3

4Dupper−Dlower)(x) =3

4− 3

4(x23+f)+3

2Wx− x23 2(x23+f)(3

2+W x +1

x Z |x|

0

dx0

(x0)23 +f) (139a)

−1

c(x23+f)|Wxx| Z |x|

0

1 (x0)23+f

dx0 (139b)

We can find an adequate value of f and corresponding domain ω to make Dbound34Dupper− Dlower≥0inωc,by drawing matlab graphs. Since 1c is larger than but sufficiently close to 1, we draw the graph with the setting c=1.

0 1 2 3 4 5 6 7 8 9 10

-0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2

X 2.06636 Y 0.0150911

Figure 2: The graph ofDboundwhen f=8

0 1 2 3 4 5 6 7 8 9 10

-0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2

X 2.06636 Y -0.00800797

Figure 3: The graph ofDbound when f=7 By testing the natural numbers that can fit in the place of f, we find that f=8 is a reasonable choice for 34Dupper−Dlower ≥0 where |x| ≥2. So, we will set our domainΩ={|x|<2} for the second maximum principle. However, there is no uniform lower boundλD,since as|x| →∞, Dupper(x) =5x23+O(|x|−1)goes to 0. Hence, we cannot use the principle directly, but we will use a similar idea.

To get||F(·,s)||L(Ωc)≤F0<∞condition in the principle,||F1(·,s)||Land||F2(·,s)||L should be studied. Let’s evaluate||F1(·,s)||L first.

||(x23+8)∂xFW||L .es2||(x23+8)∂(AZ)||L+e2s||(x23+8)∂xA||L||es2W+κ||L (140a) +e−s||A||L(||(x23 +8)Wx||L) (140b) by(117b)and the definition of FW.||(x23+8)∂xFW||L.M2e−δs+Me−swill therefore be satisfied by the combination of (119a)and(124)if

||Ax||L≤ce−(12)s ∀s. (141a)

The second component of F1satisfies

||(gW+ τW˙

1−τ˙)(x23+8)Wxx||.||gW+ τW˙ 1−τ˙

||||(x23+8)Wxx||. (142a) On the right side of the equation,||(x23 +8)Wxx|| ≤c for some constant c, since Wxx=O(|x|53) as|x| →∞.

gW= 1

1−τ˙es2(κ−ξ˙+1−α

1+αZ) (143a)

= FW0,(2)

Wxxx0 (s)+ 1−α

(1+α)(1−τ˙)es2(Z(x,s)−Z0(s)), (143b) and the last equality holds for(114b). Applying(119a), we then easily derive the estimates

|gW(x,s)| ≤ (1−α)es2

(1+α)(1−τ˙)|Z(x,s)−Z0(s)|+|FW0,(2)|

Wxxx0 (s) (144a)

.|x|Me−δs+||∂xxFW0||L+e2s||Zxx0||L||Wx0||L

Wxxx0 (s) . (144b)

The Integration of the above equation gives|Z−Z0| ≤M|x|e−(12+δ)s.To ensure that Wxxx0 (s)has a uniform lower bound, we need

||Z0xx||L.ce−(12)s, (146a) then with(117a), they make|gW(x,s)|.|x|Me−δs+M2e−s+Me−δs.Therefore, atΩc={|x| ≥2}, we obtain|gW|.12M|x|e−δs.To make time-decreasing of||1−τW˙ ˙

τ||,we also assume

|˙τ| ≤ce−δs. (147a)

Then by combining those two,||gW+1−τW˙ τ˙|| ||(x23 +8)Wxx|| ≤M2e−δs. Let’s move on to the next component.

||( τW˙

1−τ˙+ (1−α)e2sZx

(1+α)(1−τ˙))(x23+8)Wx||L .e−δs||Wx||L+e2s||Zx||L ≤Me−δs. (148a) In this process, we used the property Wx=O(|x|23)as|x| →∞. Therefore,

||F1(·,s)||L≤cM2e−δs+cMe−s≤eδ2s. (149a) Similarly, we have

||F2(·,s)||L(|x|≥2)≤eδ2s. (150a) Now, we will close the bootstrap assumptions.

Lemma 4 ||ν(·,s)||L34.

Proof 12 If this result is wrong, at some time s1>−log(ε),there must exist s0∈(−logε,s1), satisfying||ν(·,s)||L ≥ ||ν(·,s0)||L = 58 ∀s∈[s0,s1]because of the time continuity ofν. Let x(s)be a global maximum of|ν|for s∈[s0,s1).Then|x| ≥2. Otherwise, it becomes||ν|| ≤ 12, and a contradiction arises. (We will prove it soon.)

Let us consider the case whenν(x(s),s)is the global maximum forν.Then

D(x(s),s)ν(x(s),s)≥Dupper(x(s))||ν(·,s)||L (151a)

≥Dlower(x(s))||ν(·,s)||L ≥ | Z

R

K(x(s),x0,s)ν(x0,s)dx0|. (151b) So, in the equation∂sν+D(x,s)ν+U(x,s)νx=F1(x,s) +F2(x,s) +R0ν(x0,s)K(x,x0,s)dx0,the term D(x,s)νdominatesR0ν(x0,s)K(x,x0,s)dx0at x(s). Therefore,

||(∂sν+U(x,s)νx) + (D(x,s)ν− Z

0

ν(x0,s)K(x,x0,s)dx0)||L ≤2eδs2 (152a) by the inequalities(149a)and(150a). Since the second term is nonnegative at x(s)and the first term is equal to dsν(φ(x,s),s)for a characteristicφ which satisfies∂ φ∂s =U(x,s), we get

dsν(x,s)≤2eδs2. (153a)

Becauseν(x(s),s)is the global maximum or minimum forν,(we can do the same process for the global minimum) we conclude that

ds||ν(·,s)||L≤2eδ2s. (154a) In this process, we used a standard Rademacher theorem to define ds||ν(·,s)||L a.e. in s. Since the time-continuity of||ν||from Theorem(4)and the bootstrap bound Lemma(4)implies the lipschitz continuity of||ν||, the theorem implies that||ν||L is differentiable a.e. in time.

Applying one of the assumptions,||ν(·,s0)||L =58, by integration,

||ν(·,s)||L≤5 8+4

δεδ2 <3

4 ∀s>s0>−logε (155a) for sufficiently smallε. It contradicts the claim.

Therefore, we conclude that (133a) holds for x ∈Ωc, Now, let’s prove the same in |x| ≤2.

|ν(x,s)|= (x23+8)W˜x12 ∀|x| ≤2.For this, we add the condition

|Wx−Wx|< 1 2(x23+8)

∀|x| ≤2. (156a)

((156)is automatically proved by the result(217a), later.) On the other hand, we can argue that

|ν(x,−logε)|= (θ

2 3

ε +8)|Wx(x,−logε)−Wx( θ ε32

)|= (θ

2 3

ε +8)|ε(∂θw0)(θ)−Wx(θ ε32

)| ≤ 1 2 (157a)

is also satisfied in|x| ≤2.Therefore, we need one more initial data condition

|ε(∂θw0)(θ)−Wx( θ ε

3 2

)| ≤ 1

2((θ

ε

3 2

)23+8). (158a)

If all the conditions we need hold, we finally obtain(133a)for all x.

By the result before,||(x23+8)(Wx−Wx)(·,s)||L≤1.Thus||(x23+8)Wx||L≤1+||(x23+8)Wx||L, so we obtain

|Wx| ≤ 1

x23+8+|Wx| ≤ 2 x23

∀x∈R,s≥ −logε. (159a) The integration of (159a)with the condition W(0,s) =0∀s shows that

|W(x,s)| ≤6|x|13 ∀x∈R,s≥ −logε. (160a) Hence the conclusion is w∈L([−ε,T);C13(T)).Consider any two pointsθ6=θ0∈T.It implies x=θ−ξ(t)

(τ−t)32

6=x0=θ0−ξ(t)

(τ−t)32

.Then by w(θ,t) =es2W(x,s) +κ(t),

|w(θ,t)−w(θ0,t)|

|θ−θ0|13 =|W(x,s)−W(x0.s)|

|x−x0|13 . (161a)

(a) If x0=0(i.e. x6=0),

The right side of (161a)is equal to|W(x,s)|

|x|13 , so it is less than or equal to 6 by(160a). There- fore,|w(θ,t)−w(θ

0,t)|

|θ−θ0|13

≤c.It proves w∈C13.

(b) If x06=0 (general case)

By combining(117a)and(159a),|Wx(x,s)|.(1+x2)13.Then by fundamental theorem of calculus,

sup

x>x0

|W(x,s)−W(x0,s)|

|x−x0|13 .sup

x>x0

Rx

x0(1+y2)13dy (x−x0)13

≤1. (162a)

Hence w∈C13.

Remark) Withα >13,Cαnorm of w blows up att→Twith a rate(T−t)1−3α2 .It will be proved at proposition (11).

4. ∂θais bounded ast→T.

Since∂θa=w+z−ω,||∂θa(·,t)||L43c+2+2(2c+4)ν

0 M≤3M2 ∀t∈[−ε,T)for sufficiently large M, where we have employed (124).

5. ∂θzis bounded ast→T.

Our idea is to use characteristic method. For that, we need to check : finite initial data, finite speed of characteristics and finite force term. First, we will see its initial data is bounded or not.

θz=e3s2xZ,so (119a) changes into|∂θz| ≤Me(1−δ)s.Sincee−s=τ(t)−t=ε−R−εt (1−τ)dt˙ = RT

t (1−τ˙)dt,from (147a),

|τ˙(t)|.εδ (163a)

forδ is an fixed positive real number which is not determined yet, we will have a relation(1− εδ)(T−t).e−s.(1+εδ)(T−t).

∴ |∂θz| ≤C(1−εδ)−(1−δ)(T−t)−1+δM≤2M(T−t)−1+δ ∀t∈[−ε,T). (164a) We obtain the second inequality for small enoughε. Therefore, we conclude

||∂θz0||L≤1. (165a)

Now, we will check whether all terms corresponding to force and speed terms are finite or not.

Differentiating (76b) byθ, (∂t+ (z+1−α

1+αw)∂θ)(∂θz) = (166a)

−(∂θz+1−α

1+α∂θw)(∂θz)−1−2α

1+α a(∂θw)−3+2α

1+α a(∂θz) (166b)

− 1

1+α∂θa((1−2α)w+ (3+2α)z). (166c)

Note that by (124), (119a) and (141a), we can see that a,z,w and∂θaremain uniformly bounded inL.So, those terms do not affect the divergence of∂θz.Also,||∂θz0||L ≤1 says that if we have an additional condition

||∂θ2z||L ≤C(M) (i.e. ||∂x2Z||L≤Me−(12)s), (167a) by Fundamental Theorem of Calculus,||∂θz||Lis also bounded. Therefore,(∂θz)2does not affect its divergence. Therefore, only terms containing∂θware likely to affect the blowing up of∂θz.It is possible because||∂θw||L=es||Wx||L andes≥(1+εδ)−1(T−t)−1impliesRtT0||∂θw(·,t)||L= +∞.However, to ensure its blowup, we needR|∂θw(·,t)|= +∞,which will turn out to be wrong, soon.

Define natural lagrangian flowζx0(t)as d

dtζθ0(t) =z(ζθ0(t),t) +1−α

1+αw(ζθ0(t),t), ζθ0(−ε) =θ0. (168a) Proposition 9 As t→T,|∂θw(ζθ0(t),t)|does not blow up at a non-integrable rate.

Proof 13 |∂θw(ζ(t),t)|=es|Wx((ζ(t)−ξ(t))e3s2,s)|.T1−t(1+|ζ(t)−ξ(t)|

(T−t)32 )23. (a) In the case ofζ(T)6=ξ(T)

By continuity,|ζ(t)−ξ(t)| ≥c for t is sufficiently close to T.Then

|∂θw(ζ(t),t)| ≤ 1

T(1+ c

T32)23 = 1 (T32+c)23

≤1

c, (169a)

where T=T−t.Therefore,RtT

0 |∂θw(ζ(t),t)|dt<∞.So it does not blow up.

(b) In the case ofζ(T) =ξ(T)(Then,ζ(t)−ξ(t)→0as t→T)

Claim :∃c>0such that|ζ(t)−ξ(t)| ≥c(T−t) ∀t: sufficiently close to T. If the claim is true, then

|∂θw(ζ(T),T)|. 1

T−T(1+ c

(T−T)12)23 = 1

(T−T)23((T−T)12+c)23. (170a) SinceRtT

0

1 t23(t12+c)

dt<∞,our proof is done.

Proof of claim : ζ(t)−ξ(t) =

Z T

t

(ξ˙(t0)−z(ζ(t0),t0)−1−α

1+αw(ζ(t0),t0))dt (171a)

= Z T

t

κ(t0)dt0 (171b)

+ Z T

t

1−α

1+αZ0(s0)−Z((ζ(t0)−ξ(t0))e3s

0

2 ,s0)dt0 (171c)

Z T

t

1−α 1+αes

0

2W((ζ(t0)−ξ(t0))e3s

0

2 ,s0)dt0Z T

t

(1−τ˙)es

0

2 FW0,(2)(s0) Wxxx0 (s0) dt0.

(171d) Let I1=(171b), I2=(171c), I3=(171d). Thenζ(t)−ξ(t) =I1(t) +I2(t)−I3(t).

|I2(t)| ≤ Z T

t

|1−α

1+αz0(s0)−z((ζ(t0)−ξ(t0))e3s

0

2 ,s0)|dt0 (172a)

≤(1−α

1+α2+2)(T−t) = 4

1−α(T−t). (172b)

|I3(t)| ≤ Z T

t

1−α 1+α

|w(ζ(t0),t0)|dt0+ Z T

t

(1−τ)e˙ s

0

2|FW0,(2)(s0)

Wxxx0 (s0) |dt0. (172c) In(172c), the first term is bounded by 431−α1+αc(T−t)due to(124), and the second term is bounded by C(1+εδ)RtT(||∂xxFW0||Les

0

2 +||Zxx0||L)dt0.So, if we additionally assume

||∂xxFW0||L.M2e−s0, (173a) we can simply conclude that the second term.(T−t)(1+δ)δ(T−t).Thus

|ζ(t)−ξ(t)| ≥I1− 4

1+α(T−t)−4 3

1−α

1+αc(T−t)−εδ(T−t)>0 (174a) if I1>(4+43(1−α)c)1+α1 (T−t).Therefore, if only the last condition is proved, all proofs are completed.

Whenc˜≥2(α2(ν2)α+2)andν>0,I1(t) =RtTw(ξ(t0),t0)dt02c˜(T−t)is established by κ(t) =w(ξ(t),t). Hence, 2c˜ >(4+43(1−α)c)1+α1 will make all the thing possible. We can rewrite the condition as(8c˜(1+α)−1)1−α3 >c.

We have to combine the two conditions ofc,˜ c. To do this, let’s check whether both conditions can be satisfied at the same time. There is a possibility that a contradiction may arise from

˜ c

2≤w(t)≤43c.

˜ c 2<4

3(c˜

8(1+α)−1) 3 1−α

⇔(1−α)c˜ 8 <c˜

8(1+α)−1 (175a)

⇔ 4

α <c.˜ (175b)

Therefore, the conclusion of new conditions for c,c are˜ 4

α <c,˜ c<(c˜

8(1+α)−1)× 3

1−α. (176a)

This is the end of this chapter.

Lemma 5 Replacing(173a)with a simpler inequality.

Proof 14 ∂xxFW =− e

s 2

(1+α)(1−τ)˙ ((1−2α)(AZ)xx+ (3+2α)(es2(AW)xx+κAxx)).So,(173a)is equiva- lent to||Ax||L,||Axx||L,||Zx||L,||Zxx||L ≤Me−(12+δ)s.We have the first condition from(141a)and the third one from(119a), but the rest are new conditions. We will replace(146a)by

||Zxx||L ≤Me−(12+δ)s (177a)

because it is a tighter condition, and(173a)by

||Axx||L ≤Me−(12+δ)s. (178a)

Lemma 6 The first coordinate’s domain of W,Z,A can be extended toR.

Proof 15 According to the definition, the domain of W,Z,A are the same with the domain of θ in the sense of first coordinate. However, since we want to see W as a solution of Burgers equation atγ=3, hope to extend the domain intoR.

Luckily, (124) shows that the characteristic’s speed terms of (76) have an uniform upperbound depending on M. For example, the speed term of w is w+1−α1+αz,and it satisfies≤ 1+α2 M≤π.So it is possible to use the idea of finite speed of propagation.

If T<ε,the time from the initial to singularity time is less than 2ε.Then the maximum distance that the characteristics can reach is 2ε×π = π4. Therefore all the values withθ ∈T\[−4,4 ]is only affected by the initial data inT\[−π2,π2].It says that conversely, if the values of w0,z0,a0 onθ∈ T\[−π2,π2]are constant, it is easy to extend the domain. For simpler calculation, let w00,z0=a0=0 atT\[−π2,π2]. It implies w(θ,t) =w(π,t),z(θ,t) =z(π,t),a(θ,t) =a(π,t)atθ∈T\[−π2,π2], so the domain ofθ is extended toR.In other words, we need the condition

supp(w0(θ)−κ0)∪supp(z0(θ))∪supp(a0(θ))⊆(−π 2,π

2). (179a)

It is equivalent to

supp(W(x,−ε))∪supp(Z(x,−ε))∪supp(A(x,−ε))⊆(−π 2ε

3 2

3

2). (180a)

### V Proving all the equations in 4.3

Let’s rearrange the inequalities we needed for our goal and prove it with bootstrap method. Before doing that, we will recheck our initial data conditions.

문서에서 Singularity formation of Euler equations (페이지 34-46)

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