Our goal - Singularity formation of Euler equations

Proposition 7 By changing the variables of 1D Burgers equation using self-similar variables and sub- stitutingτ˙=0,ξ˙ =c,κ=c, we get(∂s−¹₂)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)

(τ0−t)³2

,s(t) =

−log(τ₀−t), and the Burgers equation turns to u(θ,t) =e⁻²^sU(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˙ =e²^s(F_W⁰(s) + ^F

0,(2) W

W_xxx⁰ (s)) =0, so letκ≡c.(114b)impliesξ˙ =κ=c.

Next, we need^R_t^T

0 ||∂_θz||_L∞dt<∞.Since||∂_θz(·,t)||_L∞=||Z_x^∂x

∂ θ||_L∞ =e³²^s||Z_x(·,s)||_L∞, Z T

t₀

||∂_θz||L^∞dt=

Z −log(τ(T)−T)

−ε

e³²^s||Zx||L^∞

dt dsds=

Z −log(τ(T)−T)

−ε

e¹²^s

1−τ˙||Zx||L^∞ds. (118a) We therefore expert that

||Z_x||_L∞ ≤ce⁻⁽¹²^+δ^)s ∀s (119a) should be needed for any nonnegative valueδ.The value ofδ will be determined later.

Finally,^R_t^T

0 ||a(·,t)||_L∞(T)dt=^R−log(τ(T)−T)

−ε ||A(·,s)||_L∞(T)e^−s 1−τ˙ds. So,

||A||_L∞≤ce^s ∀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) =e^sW_x(0,s) =−e^s,∂_θwblows up ats→∞(t→T_∗). Also, we derive from (117a) the estimate ||∂_θw(·,t)||_L∞ =||e^sWx||_L∞ ≤ce^s,so when s is finite, it does not blow up. It means that∂_θwblows up first atT∗.

2. About P andω

To ensure thatT_∗is 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 impliessup_t∈[−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)| ≤ C_α(max{|w|,|z|,|a|})²,for short.

Let w(t)≡maxt₀≤s≤tw(s)andΠ(t) =max{|w(t)|,|z(t)|,|a(t)|}.By w(s)≤ |w(s₀)|+^R_s^s

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, sup_tΠ(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 sup_t∈[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|²≥ ^c₂^˜ 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=e⁻²^sW+κ, W =e²^s(w−κ)≤2Me^s² for sufficiently large M.∴||W(s)||_L∞

x ≤

2Me²^s.

(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))^α¹ ≥^ν₂⁰ >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=¹_r∂r(ru_θ)−¹_r∂_θur=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= ^ω_P⁰

0 satisfies _2M^c^˜ ≤ω0 ≤ ^2c+4

ν0 . Let φ⁰ = b,φ(−ε) =ω₀as a characteristic. Because of the equation∂tω+b∂_θω= _α^aω,

ω(φ(t)) =ω₀+ Z t

−ε

αω(φ)ds≤ω₀+ Z t

−ε

αω(φ)ds. (127a)

Hence, by Gronwall lemma,ω(φ(t))≤ω₀e^α²^(t+ε)≤^2c+4

ν0 e^2(t+ε)^α ≤2^2c+4_ν

0 .The last inequality holds for small enoughε.

Also, observe that−ω(φ(t)) =−ω₀+^R_−ε^t _α^a(−ω(φ))ds implies

−ω(φ(t))≤ −ω₀+| Z t

−ε

α(−ω(φ))ds| ≤ − c˜ 2M+|

Z t

−ε

α(−ω(φ))ds|. (128a) By Gronwall lemma, it shows−ω(φ(t))≤ −ω₀e^k(t), where k(t) =_α²(−ω(φ)) if ^R_−ε^t −ω(φ)>

0and k(t) =−²

α(−ω(φ))if^R_−ε^t −ω(φ)≤0.It implies ω(φ(t))≥ω₀e^k(t)≥ c˜

2Me^k(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))≥_4M^c^˜ .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=T∗according to(124).

∴|κ(t)| ≤c_κ ∀t<T∗ (131a) such c_κexists.

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|x²³W_x| → 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) =cx^k¹+dx^k²+

· · · with0≤k_i≤ ²₃.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)e^s²Zx

(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)e^s²Zx

(1+α)(1−τ)˙ )Wx. (134b) SinceW˜x= ^ν

x²³+f

andW˜xx=^ν^x^(x

3+f)−(²₃x⁻¹3+f⁰)ν

(x²³+f)² ,it turns into

∂sν+ (1+^W^˜^x₁₋^+2W_τ_˙ ^x+_(1+α)(1−^1−α _τ)_˙ e^s²Zx−(gW+^3x₂ +^W+W^˜₁₋_τ_˙ )

2 3x⁻¹³+f⁰

x²3+f )ν+ (gW+^3x₂ +^W^˜₁₋^+W_τ_˙ )νx= (x²³+ f)∂xF_W−(gW+^W+^˜₁₋^τW^˙_τ_˙ )(x²³+f)Wxx−(^τW₁₋^˙ _τ^x_˙+ ^(1−α)e

s 2

(1+α)(1−τ˙)Z_x)(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−α_1+αe²^sZ_x

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)e^s²

(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

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 f}⁰

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).

|K(x,x⁰,s)|dx⁰≤1

c(x²³ +f)|Wxx(x)|

Z |x|

1 (x⁰)²³ +f

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

0<Dlower(x)≤3

4Dupper(x) (138a)

in some domainΩ^cto apply the second maximum principle.

4D_upper−D_lower)(x) =3

4− 3

4(x²³+f)+3

2W_x− x²³ 2(x²³+f)(3

2+W x +1

x Z |x|

dx⁰

(x⁰)²³ +f) (139a)

−1

c(x²³+f)|Wxx| Z |x|

1 (x⁰)²³+f

dx⁰ (139b)

We can find an adequate value of f and corresponding domain ω to make D_bound ≡ ³₄Dupper− D_lower≥0inω^c,by drawing matlab graphs. Since ¹_c 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 ofD_boundwhen 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 ofD_bound 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 ³₄D_upper−D_lower ≥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| →∞, D_upper(x) =5x⁻²³+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∞(Ω^c)≤F₀<∞condition in the principle,||F₁(·,s)||_L∞and||F₂(·,s)||_L∞ should be studied. Let’s evaluate||F₁(·,s)||_L∞ first.

||(x²³+8)∂xF_W||_L∞ .e⁻^s²||(x²³+8)∂(AZ)||_L∞+e⁻²^s||(x²³+8)∂xA||_L∞||e⁻^s²W+κ||_L∞ (140a) +e^−s||A||_L∞(||(x²³ +8)Wx||_L∞) (140b) by(117b)and the definition of F_W.||(x²³+8)∂xF_W||_L∞.M²e^−δ^s+Me^−swill therefore be satisfied by the combination of (119a)and(124)if

||A_x||_L∞≤ce⁻⁽¹²^+δ^)s ∀s. (141a)

The second component of F₁satisfies

||(g_W+ τW˙

1−τ˙)(x²³+8)Wxx||.||g_W+ 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| →∞.

g_W= 1

1−τ˙e^s²(κ−ξ˙+1−α

1+αZ) (143a)

= F_W^0,(2)

W_xxx⁰ (s)+ 1−α

(1+α)(1−τ˙)e^s²(Z(x,s)−Z⁰(s)), (143b) and the last equality holds for(114b). Applying(119a), we then easily derive the estimates

|g_W(x,s)| ≤ (1−α)e^s²

(1+α)(1−τ˙)|Z(x,s)−Z⁰(s)|+|F_W^0,(2)|

W_xxx⁰ (s) (144a)

.|x|Me^−δs+||∂_xxF_W⁰||_L∞+e²^s||Z_xx⁰||_L∞||W_x⁰||_L∞

W_xxx⁰ (s) . (144b)

The Integration of the above equation gives|Z−Z⁰| ≤M|x|e⁻⁽¹²^+δ)s.To ensure that W_xxx⁰ (s)has a uniform lower bound, we need

|Wxxx(0,s)−Wxxx(0,s)|=|Wxxx(0,s)−6| ≤1. (145a) Additionally assume

||Z⁰_xx||_L∞.ce⁻⁽¹²^+δ^)s, (146a) then with(117a), they make|gW(x,s)|.|x|Me^−δs+M²e^−s+Me^−δs.Therefore, atΩ^c={|x| ≥2}, we obtain|g_W|.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)e²^sZ_x

(1+α)(1−τ˙))(x²³+8)Wx||_L∞ .e^−δs||W_x||_L∞+e²^s||Z_x||_L∞ ≤Me^−δ^s. (148a) In this process, we used the property Wx=O(|x|⁻²³)as|x| →∞. Therefore,

||F₁(·,s)||_L∞≤cM²e^−δ^s+cMe^−s≤e⁻^δ²^s. (149a) Similarly, we have

||F₂(·,s)||_L∞(|x|≥2)≤e⁻^δ²^s. (150a) Now, we will close the bootstrap assumptions.

Lemma 4 ||ν(·,s)||L^∞≤³₄.

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)≥D_upper(x∗(s))||ν(·,s)||_L∞ (151a)

≥Dlower(x∗(s))||ν(·,s)||L^∞ ≥ | Z

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) +^R₀^∞ν(x⁰,s)K(x,x⁰,s)dx⁰,the term D(x,s)νdominates^R₀^∞ν(x⁰,s)K(x,x⁰,s)dx⁰at x∗(s). Therefore,

||(∂sν+U(x,s)νx) + (D(x,s)ν− Z _∞

ν(x⁰,s)K(x,x⁰,s)dx⁰)||L^∞ ≤2e⁻^δs² (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

d_sν(x∗,s)≤2e⁻^δs². (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

d_s||ν(·,s)||_L∞≤2e⁻^δ²^s. (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

δε^δ² <3

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

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

|ν(x,s)|= (x²³+8)W˜x≤¹₂ ∀|x| ≤2.For this, we add the condition

|W_x−W_x|< 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)−W_x( θ ε

3 2

)| ≤ 1

2((^θ

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−W_x)(·,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) =e⁻^s²W(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

x⁰(1+y²)⁻¹³dy (x−x⁰)¹³

≤1. (162a)

Hence w∈C¹³.

Remark) Withα >¹₃,C^αnorm of w blows up att→T∗with 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)||_L∞≤⁴₃c+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=e^3s²∂xZ,so (119a) changes into|∂_θz| ≤Me^(1−δ)s.Sincee^−s=τ(t)−t=ε−^R_−ε^t (1−τ)dt˙ = RT∗

t (1−τ˙)dt,from (147a),

|τ˙(t)|.ε^δ (163a)

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

∴ |∂_θz| ≤C(1−ε^δ)^−(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^+δ^)s), (167a) by Fundamental Theorem of Calculus,||∂_θz||_L∞is 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^∞=e^s||Wx||L^∞ ande^s≥(1+ε^δ)⁻¹(T∗−t)⁻¹implies^R_t^T₀^∗||∂_θw(·,t)||L^∞= +∞.However, to ensure its blowup, we need^R|∂_θw(·,t)|= +∞,which will turn out to be wrong, soon.

Define natural lagrangian flowζx₀(t)as d

dtζ_θ₀(t) =z(ζ_θ₀(t),t) +1−α

1+αw(ζ_θ₀(t),t), ζ_θ₀(−ε) =θ₀. (168a) Proposition 9 As t→T∗,|∂_θw(ζθ0(t),t)|does not blow up at a non-integrable rate.

(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,^R_t^T^∗

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) Since^R_t^T^∗

1 t²3(t¹²+c)

dt<∞,our proof is done.

Proof of claim : ζ(t)−ξ(t) =

Z T∗

(ξ˙(t⁰)−z(ζ(t⁰),t⁰)−1−α

1+αw(ζ(t⁰),t⁰))dt (171a)

= Z _T_∗

κ(t⁰)dt⁰ (171b)

+ Z T∗

1−α

1+αZ⁰(s⁰)−Z((ζ(t⁰)−ξ(t⁰))e^3s

2 ,s⁰)dt⁰ (171c)

− Z T∗

1−α 1+αe⁻^s

2W((ζ(t⁰)−ξ(t⁰))e^3s

2 ,s⁰)dt⁰− Z T∗

(1−τ˙)e⁻^s

2 F_W^0,(2)(s⁰) W_xxx⁰ (s⁰) dt⁰.

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

|I₂(t)| ≤ Z T∗

|1−α

1+αz⁰(s⁰)−z((ζ(t⁰)−ξ(t⁰))e^3s

2 ,s⁰)|dt⁰ (172a)

≤(1−α

1+α2+2)(T∗−t) = 4

1−α(T∗−t). (172b)

|I₃(t)| ≤ Z T∗

1−α 1+α

|w(ζ(t⁰),t⁰)|dt⁰+ Z T∗

(1−τ)e˙ ⁻^s

2|F_W^0,(2)(s⁰)

W_xxx⁰ (s⁰) |dt⁰. (172c) In(172c), the first term is bounded by ⁴₃^1−α_1+αc(T∗−t)due to(124), and the second term is bounded by C(1+ε^δ)^R_t^T^∗(||∂xxF_W⁰||_L∞e⁻^s

2 +||Z_xx⁰||_L∞)dt⁰.So, if we additionally assume

||∂_xxF_W⁰||_L∞.M²e^−s⁰, (173a) we can simply conclude that the second term.(T∗−t)^(1+δ⁾.ε^δ(T∗−t).Thus

|ζ(t)−ξ(t)| ≥I₁− 4

1+α(T∗−t)−4 3

1−α

1+αc(T∗−t)−ε^δ(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) =^R_t^T^∗w(ξ(t⁰),t⁰)dt⁰≥₂^c^˜(T∗−t)is established by κ(t) =w(ξ(t),t). Hence, ₂^c^˜ >(4+⁴₃(1−α)c)_1+α¹ will make all the thing possible. We can rewrite the condition as(₈^c^˜(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)(e⁻^s²(AW)xx+κAxx)).So,(173a)is equivalent 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

||Z_xx||_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−α_1+αz,and it satisfies≤ _1+α² M≤_8ε^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×_8ε^π = ^π₄. Therefore all the values withθ ∈T\[−^3π₄,^3π₄ ]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 w₀=κ₀,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,π

2ε⁻

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)