Closing the assumptions - Singularity formation of Euler equations

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.

Therefore, (124) implies

|τ|˙ .Me^−δ^s+Me^−(1+δ^)s+Me^−s≤M²e⁻^min{δ,1}s (184a) for sufficiently large M, for all s. As c in (181a) was an undetermined constant, let such c is larger thanM². If we add the assumptionδ <1,then our proof of (181a) is done. Inequality (181a) gives the uniform bound of ˙τ,|τ˙| ≤ce^−δs≤cε^δ <ε^δ² for sufficiently smallε.Then additionally, if we integrate τ˙ in t fort≤T_∗and useτ(−ε) =0,it gives

τ(t) = Z T∗

−ε

τ(s)dt≤1

2ε⁽^δ²⁺¹⁾. (185a)

Closing (181d) and (181e)) 1. ||Z_x||_L∞≤Me⁻⁽¹²^+δ)s

We will use the first maximum principle in Appendix, to prove it. By (113) with n=1, (∂s+3

2+ 1

1−τ˙e^s²∂xZ+∂xgZ)∂xZ+ (gZ+3x

2 + 1

1−τ˙e^s²Z)∂_x²Z=F_Z⁽¹⁾, (186a) where gZ= 1

1−τ˙(e²^s(1−α

1+ακ−ξ˙) +1−α

1+αW). (186b)

In this equation, the damping term D satisfies : D=3

2+ 1

1−τ˙e²^s∂xZ+∂xgZ=3

2+e²^sZx

1−τ˙+ (1−α)Wx

(1−τ˙)(1+α) (187a)

≥3

2−(1+2ε^δ²)(Me^−δs+|1−α

1+α|||W_x||_L∞). (187b)

We obtain the last inequality by₁₋¹_τ_˙ ≥ ¹

1−ε^δ²

≥1+2ε^δ².(181b) have to be changed to||Wx||L^∞≤1 for all s, becauseD>0 is necessary for sufficiently smallε,too. Then, since(1+2ε^δ²)(Mε^δ+

|^1−α_1+α|)≤(1+2ε^δ²)(Mε^δ+1−2δ)≤1−δ for small enoughε,if δ≡ min{α,1}

2(1+α) >0, (188a)

D≥ ¹₂+δ. It corresponds toλD=¹₂+δ in the first maximum principle.

On the other hand,

||F_Z⁽¹⁾||L^∞. e^−s

1−τ˙(||Ax||L^∞(||e⁻²^sW+κ||L^∞+||Z||L^∞) +||A||L^∞(e⁻^s²||Wx||L^∞+||Zx||L^∞)) (189a) .e^−s(Me⁻⁽¹²^+δ)s(2M+c_κ+M) +M(e⁻^s²+Me⁻⁽¹²^+δ)s)) (189b)

.M²e⁻³²^s. (189c)

We have used (181d) and (124) to ensure the second inequality. Therefore, those results corre- spond toλF =³₂,F0=M²in the maximum principle. Therefore, by (181a),

||Zx(s)||L^∞ .||Zx(−logε)||L^∞e⁻⁽¹²^+δ^)(s+logε)+M²e^{(1−δ)logε}e⁻⁽¹²^+δ)s (190a)

.(ε^1−δ+M²ε^1−δ)e⁻⁽¹²^+δ)s, (190b)

because||Z_x(−logε)||_L∞=ε

2||∂_θz₀||_L∞≤ε

2.Then, takingεsufficiently small in terms of M, and using the factδ ≤¹₄,we obtain

||Z_x(s)||_L∞≤ε

4e⁻⁽¹²^+δ^)s. (191a)

2. ||Z_xx||_L∞ ≤Me⁻⁽¹²^+δ)s. By (113) withn=2,

(∂s+3+ 3

1−τ˙e²^s∂xZ+2∂xgZ)∂_x²Z+ (gZ+3x 2 + 1

1−τ˙e²^sZ)∂_x³Z=F_Z⁽²⁾. (192a) In this equation, the damping term D satisfies

D=3+ 3

1−τ˙e^s²∂xZ+2∂xg_Z≥3−2(1+2ε^δ²)||W_x||_L∞−3(1+2ε^δ²)e^s²||∂_xZ||_L∞, (193a) because∂xgZ=₍₁₋^(1−α)W_τ)(1+α)_˙ ^x .Therefore, from (181b) and (181d),

D≥3−2(1+2ε^δ²)−3(1+2ε^δ²)Mε^δ≥3

4 (194a)

for small enoughε.

On the other hand,

||F_Z⁽²⁾||L^∞=||∂_x²FZ−∂_x²gZ∂xZ||L^∞ .||∂_x²FZ||L^∞+||∂_x²gZ||L^∞||∂xZ||L^∞. (195a) Since ∂_x²gZ = ₍₁₋^(1−α)W_τ)(1+α)_˙ ^xx , ||∂_x²gZ||_L∞ .||W_xx||_L∞ is also satisfied. At this time, in order for the decay rate of ||Z_xx|| to be e⁻⁽¹²^+δ)s, we need an uniform bound of ||∂_x²g_Z||,because by (124),

||∂_xZ||already has the same decay rate. So, we need

||W_xx||_L∞≤M ∀s. (196a)

Let’s add this inequality as one of our bootstrap assumptions. Then we can conclude using (191a) that

||F_Z⁽²⁾||_L∞ .M²e^−s+Mε¹⁴e⁻⁽¹²^+δ^)s.Mε¹⁴e⁻⁽¹²^+δ)s. (197a) As we already know||∂_x²Z(·,−logε)||L^∞=||ε³∂_θ²z₀(·)||L^∞ ≤ε³ by (121a), we can apply the first maximum principle with the settingλD=³₄,F₀=Mε¹⁴,λF =¹₂+δ.It gives

||∂_x²Z(s)||_L∞.||∂_x²Z(·,−logε)||_L∞e⁻³⁴^(s+logε)+Mε¹⁴e⁻⁽¹²^+δ)s (198a) .ε

4e⁻³⁴^s+Mε¹⁴e⁻⁽¹²^+δ^)s.Mε¹⁴e⁻⁽¹²^+δ)s ∀s≥ −logε. (198b) 3. ||A_x||_L∞≤Me⁻⁽¹²^+δ)s

As in the case of Z, we will use (113) with n=1.(∂s+³₂+∂xgA)∂xA+ (gA+^3x₂)∂_x²A=F_A⁽¹⁾shows that

D=3

2+∂xgA= 3

2+ Wx+e²^sZx

(1+α)(1−τ˙) ≥3

2−(1+2ε^δ²)

1+α (1+Mε^δ)≥1

2+δ (199a)

for smallε,and

||F_A⁽¹⁾||_L∞=||∂_xFA||_L∞ =|| e^−s

2(1+α)(1−τ˙)(−8AA_x+2(e⁻²^sW+κ+Z)(e⁻^s²W+Z)x (200a)

−2α(e⁻²^sW+κ−Z)(e⁻^s²W_x−Z_x))||_L∞ (200b)

.M²e^−s. (200c)

We used (124) for the last inequality. Since||∂_xA(·,−logε)||_L∞≤ε³²,the first maximum principle implies that

||∂_xA(s)||_L∞ ≤ε

4e⁻⁽¹²^+δ)s. (201a)

So, the assumption is closed.

4. ||Axx||L^∞ ≤Me⁻⁽¹²^+δ)s

By (113) with n=2,(∂s+3+2∂xgA)∂_x²A+ (gA+^3x₂)∂_x³A=F_A⁽²⁾.In this equation, D=3+2∂xgA=3+ 2(Wx+e^s²Zx)

(1+α)(1−τ˙)≥3−2(1+2ε^δ²)

1+α (1+Mε^δ)≥1+δ, (202a) and

||F_A⁽²⁾||_L∞ =||∂_x²F_A−∂_x²g_A∂xA||_L∞.M²e^−s+M||∂_xA||_L∞ (203a) .M²e^−s+Mε¹⁴e⁻⁽¹²^+δ^)s.Mε¹⁴e⁻⁽¹²^+δ)s. (203b) Using (201a), we get the second inequality. Finally, by applying the first maximum principle, we get

||∂_x²A||_L∞.Mε¹⁴e⁻⁽¹²^+δ)s. (204a) We closed all the assumptions (181d) and (181e).

So far we have proven the inequalities forκandτ, and also found uniform upperbounds for Z and A.

The left bootstrap assumptions are about W. However, since finer bounds are needed for W than previous assumptions, we will focus on a toy problem to get an idea of the computation.

Previously in proposition 7, we saw that 1D modulus Burgers equation can be transformed into (∂s−¹₂)W+ (³₂x+W)Wx=0, the equation ofW,by substituting ˙τ=0,ξ˙=c,κ=c.Knowing that W in the 2D Euler system is a solution of Burgers equation atγ=3 with A=Z=0, and all the previous results : (131a), (181a), (124), (181d) and (181e) are satisfied in condition ˙τ=0,ξ˙=c, andκ=c,we will cover (∂s−¹₂)W+ (³₂x+W)Wx=0 as a toy problem of 2D Euler equation.

In fact, all the terms forτ,ξ,κare bounded by arbitrary small variable as time goes to the singularity time from (181a), so they would not affect the proof of bootstrap assumptions. Therefore, we can do exactly the same procedure to 2D Euler, as the proof in the self-similar Burgers equation we will see.

Let’s setW(0,s) =0,∂xW(0,s) =−1,∂_x²W(0,s) =0,∂_x³W(0,0) =6.

(The last one holds from (104), and all the other things are from the definition of self-similar variables.) The equation of W is

(∂s−1

2)W+ (3x

2 +W)Wx=0, (205a)

and its steady solution isW(x).

Derivated form of (205a) :

∂sWx+ (1+Wx)Wx+ (3

2x+W)Wxx=0 (206a)

∂sWxx+ (5

2+3Wx)Wxx+ (3

2x+W)Wxxx=0 (206b)

∂sWxxx+ (4+4Wx)Wxxx+ (3

2x+W)Wxxxx=−3W_xx² (206c)

∂sWxxxx+ (11

2 +5Wx)Wxxxx+ (3

2x+W)Wxxxxx=−10WxxWxxx (206d)

Closing (181c))

Expression by substituting 0 for the x variable of (206c) is(∂s+ (4+4W_x⁰))W_xxx⁰ =−W⁰W_xxxx⁰ −3(W_xx⁰)². Using the initial data of W, the above equation is changed to∂sW_xxx⁰ =0.ThereforeWxxx(0,s) =Wxxx(0,0) = 6.

Closing (181b))

Since the damping term of (206a)(D=1+Wx)cannot be said to have a uniform positive lower bound by the assumption||Wx||L^∞ ≤1 only, we will change the equation with other variable to have a positive damping term, using the prediction that W andWis close enough.

Letting ˜W=W−W,we may then write (206a) in the form

∂sWx+ (1+Wx)Wx+ (3

2x+W)Wxx=0. (207a)

⇔∂s(W˜_x+W) + (1+W˜_x+Wx)(W˜_x+Wx) + (3

2x+W)(W˜_xx+Wxx) =0. (207b)

⇔∂sW˜_x+ (1+W˜_x+2Wx)W˜_x+ (3

2x+W)W˜_xx=−WW˜ _xx. (207c) The last step is established by(1+W_x)Wx+ (³₂x+W)Wxx=0,which is the differentiation of equation

−¹₂W+ (³₂x+W)∂xW =0 by x.

Now, to evaluate the damping term of (207c), split the middle terms in two so that one contains only the functions we already know, likeW, and the other has the remainder, because the middle terms may become new damping term.

(1+W˜_x+2Wx)W˜_x+ (3

2x+W)W˜_xx⇒ {(1+2Wx)W˜_x+ (3

2x+W)W˜_xx}+{W˜_xW˜_x+W˜W˜_xx}. (208a) We do not know the exact value of the second term, but we want a situation where the first term has a sufficiently large positive lower bound to cover the second term so that there can be a positive lower bound of the total damping term.

For this, we will define a new variableV(x,s) =g(x,s)W˜_x with some functiong(x,s).Here, g will be defined later in the process of creating a positive lower bound. Let’s change (208a) to V’s one. We

can use the relations ˜Wx=^V_g,W˜xx=^V^x⁻

g .We find(1+2Wx)^V_g+ (³₂x+W)(¹_gVx−^g_g^x₂V)+(terms of ˜W) turns to¹_g{(1+2Wx)−^g_g^x(^3x₂ +W)}V+¹_g(³₂x+W)Vx+(terms of ˜W). The first term becomes a damping term of V, and the second is obviously not. But the third term, terms of ˜W, may include additional damping terms of V. Since we know 1+2Wx+_x(1+x² 2)(^3x₂ +W)≥_1+8x^6x²₂,if g satisfies−^g_g^x =_x(1+x² 2),the first damping term guarantee a positive lower bound ¹_g_1+8x^6x²2.We can find suchg=^1+x_x₂² by solving the ODE.

So, forV(x,s) =W˜x1+x²

x² ,we can rewrite the equation to

∂s( x²

1+x²V) + (1+W˜x+2Wx) x²

1+x²V+ (3

2x+W)( x²

1+x²V)x=−WW˜ xx. (209a) Dividing by _1+x^x²2, we obtain

∂sV+ (1+W˜x+2Wx+ 2 x(1+x²)(3

2x+W))V+ (3

2x+W)Vx=−1+x² x² Wxx

Z x 0

V (x⁰)² 1+ (x⁰)²dx⁰.

(210a) We will find the upperbound of V through the equation, and will change the bound into W’s one. For it, we would like to apply one of the maximum principles. According to the notation of the maximum principle,D=1+W˜x+2Wx+_x(1+x² 2)(^3x₂ +W),U =^3x₂ +W andF=0. First, we have to check that there exists any positive lower bound of D. Since 1+2Wx+_x(1+x² 2)(^3x₂ +W)≥ _1+8x^6x²₂,split D into two terms.

|D| ≥ |1+2Wx+ 2

x(1+x²)(3x

2 +W)| − |W˜x+ 2

x(1+x²)W˜|. (211a) We need a small upper bound of the second term. Let’s assume|W˜_x| ≤h(x), i.e. |V| ≤h(x)^1+x_x2². Then we have

|W˜x+ 2

x(1+x²)W˜ | ≤h(x) + 2

|x|(1+x²) Z |x|

|W˜x(x⁰,s)|dx⁰ (212a)

≤h(x) + 2

|x|(1+x²) Z |x|

h(x⁰)dx⁰. (212b)

The first inequality uses the fundamental theorem of calculus with ˜W(0) =0.Hence,D≥_1+8x^6x²₂−h(x)−

|x|(1+x²)

R|x|

0 h(x⁰)dx⁰.As we need some positiveλD≥0,we will find corresponding h(x). We are going to find a bound of V, but becauseF =0, the bound will be constant by the maximum principle. So, set the upperbound of V as ₂₀¹,adequately.(we can change the number later, if it makes a contradiction.) Therefore, it corresponds to h(x) = _20(1+x^x² 2). Since _|x|(1+x² 2)

R|x|

(x⁰)²

20(1+(x⁰)²)dx⁰ ≤ _10(1+x^x² ₂₎, D≥ _1+8x^6x²₂ −

3x²

20(1+x²) ≥⁹₂ ^x²

1+8x².∴at|x| ≥l,D≥ ^9l²

2(1+8l²)≡λDin the second maximum principle. At this point, we see that we cannot use the first maximum principle, because the lower bound goes to 0 as x decreases to 0.

Next, we haveK(x,x⁰,s) =−^1+x²

x² Wxx(x)1_[0,x](x⁰)_1+(x^(x⁰⁾²0)².Therefore, Z

|K(x,x⁰,s)|dx⁰≤1+x² x² |Wxx|

Z _|x|

(x⁰)²

1+ (x⁰)²dx⁰≤ 3x² 1+8x² ≤3

4D. (213a)

Hence K obeys the assumption (240a) of the second maximum principle. Also, ||F||=0 gives the condition (239a).

Lastly, we verify the bounds (241a). The first inequality, |V(x,s)| ≤₄₀¹ on|x| ≤l will be proved below, and because|V(x,−logε)|= ^1+x_x2²|W_x(x,−logε)−W_x(x)|,if we add the initial-assumption

|ε(∂_θw₀)(θ)−W_x(θ ε³²

)| ≤

( ^θ

3 2

)² 40(1+ ( ^θ

3 2

)²), (214a)

we can conclude that|V(x,−logε)| ≤₄₀¹.

According to lemma (6),W_x is compactly supported, while from the taylor series ofW atx=∞, Wx decays as |x| →∞. Therefore, (237a) is satisfied. Conclude from mλD = ₂₀¹ ⁹₂_1+8l^l² 2 ≥0=4F₀ that||V(·,s)||_L∞(R)≤ ₈₀³.Since ₈₀³ < ₂₀¹,we can understand our step as closing bootstrap assumption

||V(·,s)||_L∞(R)≤ ₂₀¹,and the proof is done on|x| ≥l.

Remark)|x| ≤l,wherel>0.

Write the taylor series ofWxat x=0 :Wx(x,s) =−1+^x₂²Wxxx(0,s) +^x₆³Wxxxx(x⁰,s)for some|x⁰|<|x|.It implies|Wx(x,s) +1−3x²|=^x₆³Wxxxx(x⁰,s).So, if we have the condition

||∂_x⁴W||_L∞ ≤M, (215a)

|Wx(x,s) +1−3x²| ≤ ^M₆x²lholds. Let’s add it in our bootstrap assumption.

∴|W_x(x,s)−W_x| ≤M

6x²l+15x⁴+O(x⁶)≤x²(Ml

6 +15l²)≤ x²

40(1+x²) (216a) Hence,||V(·,s)||_L∞(R)≤ ₄₀¹ on|x| ≤l,too. Using the properties of the functionWx,

|W_x(x,s)−Wx(s)| ≤ x²

20(1+x²) (217a)

turns into||W_x(·,s)||_L∞ ≤1.This is the end of proof (181b).

Remark) In this proof, we used bootstrap method to||V||,not||W_x||directly.

Closing (196a)) Through (206b),∂sWxx+ (⁵₂+3Wx)Wxx+ (³₂x+W)Wxxx=0.To rewrite this equation to another which has a positive damping, set ˜V(x,s) =g(x)Wxx(x,s).Then

(206b)⇔ 1

g∂sV˜ + (5

2+3Wx)V˜ g + (3

2x+W)(V˜

g)x=0 (218a)

⇔∂sV˜+ (5

2+3Wx+ (3

2x+W)(−g_x

g))V˜ + (3

2x+W)V˜x=0. (218b) Observe that, we can decompose the damping term into the sum of its well-known and remained parts.

We call the well-known termD₁and the otherD₂. D={5

2+3Wx+ (−gx

g)(3

2x+W)}+{(−gx

g)(W−W) +3(Wx−Wx)} (219a)

≡D₁+D₂. (219b)

For the positivity ofD₁, we want to implement one of the properties ofW : ⁵₂+3Wx+_x(1+x¹ 2)(³₂x+W)≥

x²

1+x² ∀x∈R.For this, g should satisfy−^g_g^x =_x(1+x¹ 2).By solving the ODE, we getg=^(1+x²⁾

1 2

x ,and if so,D₁≥_1+x^x²₂ is satisfied.

According to (217a),D₂≥ −_|x|(1+x¹ ₂₎^R₀^|x|_20(1+(x^(x⁰⁾²0)²)dx⁰−3₂₀¹ _1+x^x²2,so D≥4

5 x²

1+x² ≥ 1

2M² ≡λD (220a)

for all|x| ≥_M¹.In this point, D has no strictly positive lower bound if x goes to 0. Therefore, the first maximum principle is not suitable for this proof. We have to use the second one. Since|F|=0 andK=0, the conditions of them satisfy the second maximum principle. Then, how about ||V˜(x,−logε)||_L∞(R)? An initial bound ˜V(x,−logε)≤Cis equivalent to||W_xx(x,−logε)|| ≤C ^|x|

(1+x²)¹².Therefore, in order to have a constant bound on ˜V, additional initial assumptions about partial derivatives of W are required.

There are many other ways to make it possible, but this is the simplest case.

The additional initial condition :

||∂_x²W(·,−logε)||_L∞≤1 (221a)

||∂_x⁴W(·,−logε)||_L∞≤1 (221b) If so, the taylor series ofW_xx gives||W_xx(x,−logε)|| ≤ |x|W_xxx(0,−logε) +^|x|₂²||∂_x⁴W||_L∞ ≤6|x|+^x₂², using (221b). Therefore by (221a),

||W_xx(x,−logε)|| ≤min{6|x|+x²

2,1} ≤ 7|x|

(1+x²)¹². (222a) So we obtain||V˜(x,−logε)||_L∞≤7.

To satisfy the other condition ||V˜(x,s)||_L∞(|x|<_M¹), we will use the taylor series of W_xx as in the previous step. Wxx(x,s) =|x|W_xxx(0,s) + ^x₂²Wxxxx(x⁰,s). By the bootstrap assumptions, |W_xx(x,s)| ≤ 6|x|+^x₂²M≤ ^7|x|

(1+x²)¹².Therefore, we obtain

||V˜||_L∞(|x|<¹

M)≤7. (223a)

Since mλ_D=14_2M¹2 ≥0=4F₀,we conclude||V˜||_L∞ <12.Therefore,||W_xx||_L∞ < ^12|x|

(1+x²)¹²

≤Mfor large enough M.

Closing (215a)) (215a) tells us (∂s+¹¹₂ +5Wx)∂_x⁴W+ (³₂x+W)∂_x⁵W =−10W_xxWxxx. In this equation, the damping term D is automatically positive, since D= ¹¹₂ +5Wx ≥ ¹¹₂ −5= ¹₂ ≥ ¹₄. Also,

||F||_L∞ ≤C||W_xxx||_L∞ by (196a). So, if we additionally assume

||W_xxx||_L∞ ≤M^k (224a)

for somek∈R⁺,||F||_L∞ ≤CM^k≡F₀. Therefore, by the first maximum principle,

||∂_x⁴W||L^∞≤ ||∂_x⁴W(·,−logε)||L^∞e⁻¹⁴^(s+logε)+4CM^k. (225a)

Hence, we can close the assumption (215a) for sufficiently large M, ifk<1.Without losing of general- ity, let k=³₄.

Closing (224a)) In the case ofW_xxx, we have to use the second maximum principle because of the damping. (206c) shows∂sWxxx+ (4+4Wx)Wxxx+ (³₂x+W)Wxxxx=−3W_xx²,and the damping term sat- isfiesD≥4(1+Wx−_20(1+x^x² ₂₎)≥ _1+x^x²₂.The last inequality uses the property 1+W_x≥ _4(1+x^3x²₂₎.So, in some domain|x|>l,D is positive.

To find the exact value of the domain |x|>l,consider the|x| ≤l case, first. By the mean value theorem and (215a), we know|W_xxx(x,s)−W_xxx(0,s)| ≤ |x|M.It transforms into

|W_xxx(x,s)| ≤6+|x|M≤M³⁴

2 (226a)

for|x| ≤ ¹

4M¹⁴.So, the valuelbecomes ¹

4M¹⁴. Returning to |x| ≥ ¹

4M¹⁴,we obtain D≥ ¹

32M¹². |F|.1 by bootstrap assumption, and all the other conditions are proved by easy calculation. The only remained condition is the initial value. To apply the maximum principle, we need|W_xxx(x,−logε)| ≤ ^M

3 4

2 for allx∈R.It requires another initial condition,

|Wxxx(x,−logε)| ≤7. (227a)

Such number 7 doesn’t have much meaning. We can take any number larger than 6, since if not, it makes a contradiction between (181c). With our new condition, finally, we can say that||∂_x³W||L^∞ <M³⁴.

So, we have proved all the bootstrap assumptions. It was (181a), (181b), (181c), (196a), (215a), (224a), (181d) and (181e), in total. We have seen only a case of toy problem for (181b) (224a), but 2D Euler solution’s case is absolutely the same calculation, with modulus. This is the end of our bootstrap closing.

From now on, we will collect all the initial conditions that we used, and check whether they are rea- sonable. During the proof, the initial conditions (214a), (221a), (227a), (221b) was added. Therefore, our initial conditions are

1. w₀(0) =κ₀,∂_θw₀(0) =−ε⁻¹,∂_θ²w₀(0) =0,∂_θ³w₀(0) =6ε⁻⁴

2. ||∂_θw₀||_L∞≤ε⁻¹,||∂_θ²w₀||_L∞≤ε⁻⁵²,||∂_θ³w₀||_L∞ ≤7ε⁻⁴,||∂_θ⁴w₀||_L∞ ≤ε⁻¹¹² 3. |ε(∂_θw₀)(θ)−W_x( ^θ

3 2

)| ≤min{

(^θ

ε 3 2

)² 40(1+( ^θ

ε 3 2

)²), ¹

2(( ^θ

ε 3 2

)²³+8)

} 4. ||z₀||_Cⁿ+||a₀||_Cⁿ≤1 ∀0≤n≤4.

5. supp(w0(θ)−κ0)∪supp(z0(θ))∪supp(a0(θ))⊆(−^π₂,^π₂).

6. ˜c≤ ||w₀||L^∞≤c,where ˜c≥2(_α²(^ν₂)^α+2),_α⁴ <c,c˜ <(₈^c^˜(1+α)−1)×_1−α³ .

The second condition is the combination of (221a), (227a) and (221b). Also, the previous third condition was changed a little bit by adding the condition (214a). Everything else is the same as before.

Proposition 10 An open set of initial conditions in C⁴satisfies all the initial conditions that we need.

Proof 16 By the change of variables, we can show that third, fourth conditions can cover the fifth, sixth conditions. Since the third condition is equivalent to

|Wx(·,−logε)−Wx(·,−logε)| ≤min{ x(θ,−ε)²

40(1+ (x(θ,−ε))²), 1

2(8+ (x(θ,−ε))²³)}, (228a) when we substituteθ=0,the first term in the right hand side becomes 0.∴|W_x(0,−logε)−W_x(0)|=0.

It is the same as∂_θw₀(0) =−ε⁻¹,one result from the first condition.

Also, the third condition implies

|ε(∂_θw₀)(θ)| ≤ |W_x(x)|+min{

( ^θ

3 2

)² 40(1+ ( ^θ

3 2

)²), 1 2((^θ

3 2

)²³+8)

} ≤1 ∀x∈R, (229a)

which means that the condition ||∂_θw₀||_L∞ ≤ε⁻¹ is the corollary of the condition 3. Therefore, any open set satisfying the rest second conditions and the third one, which is transformed to satisfy the first condition, will satisfy all the initial conditions of W.

Proposition 11 Withα >¹₃,C^αHolder norm of w blows up as t→T_∗with a rate(T∗−t)^1−3α² .

Proof 17 W(x) = (−^x₂+ (₂₇¹ +^x₄²)¹²)¹³ −(₂^x+ (₂₇¹ +^x₄²)¹²)¹³ is known. Since W_x=W_x at x=0by the inequality(217a), and the fact that|x|^1−α|W_x|blows up at x=0holds for allα >¹₃,we can conclude that|x|^1−α|Wx|will blow up at x=0,too. Therefore, by integration, C^α Holder norm of w blows up as t→T∗.