### 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}≤M^{2}e^{−}^{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
thanM^{2}. 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^{δ}^{2} 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^{(}^{δ}^{2}^{+1)}. (185a)

Closing (181d) and (181e))
1. ||Z_{x}||_{L}∞≤Me^{−(}^{1}^{2}^{+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^{s}^{2}∂xZ+∂xgZ)∂xZ+ (gZ+3x

2 + 1

1−τ˙e^{s}^{2}Z)∂_{x}^{2}Z=F_{Z}^{(1)}, (186a)
where gZ= 1

1−τ˙(e^{2}^{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˙e^{2}^{s}∂xZ+∂xgZ=3

2+e^{2}^{s}Zx

1−τ˙+ (1−α)Wx

(1−τ˙)(1+α) (187a)

≥3

2−(1+2ε^{δ}^{2})(Me^{−δs}+|1−α

1+α|||W_{x}||_{L}∞). (187b)

We obtain the last inequality by_{1−}^{1}_{τ}_{˙} ≥ ^{1}

1−ε^{δ}^{2}

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

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

2(1+α) >0, (188a)

D≥ ^{1}_{2}+δ. It corresponds toλD=^{1}_{2}+δ in the first maximum principle.

On the other hand,

||F_{Z}^{(1)}||L^{∞}. e^{−s}

1−τ˙(||Ax||L^{∞}(||e^{−}^{2}^{s}W+κ||L^{∞}+||Z||L^{∞}) +||A||L^{∞}(e^{−}^{s}^{2}||Wx||L^{∞}+||Zx||L^{∞})) (189a)
.e^{−s}(Me^{−(}^{1}^{2}^{+δ)s}(2M+c_{κ}+M) +M(e^{−}^{s}^{2}+Me^{−(}^{1}^{2}^{+δ)s})) (189b)

.M^{2}e^{−}^{3}^{2}^{s}. (189c)

We have used (181d) and (124) to ensure the second inequality. Therefore, those results corre-
spond toλF =^{3}_{2},F0=M^{2}in the maximum principle. Therefore, by (181a),

||Zx(s)||L^{∞} .||Zx(−logε)||L^{∞}e^{−(}^{1}^{2}^{+δ}^{)(s+logε)}+M^{2}e^{(1−δ)logε}e^{−(}^{1}^{2}^{+δ)s} (190a)

.(ε^{1−δ}+M^{2}ε^{1−δ})e^{−(}^{1}^{2}^{+δ)s}, (190b)

because||Z_{x}(−logε)||_{L}∞=ε

3

2||∂_{θ}z_{0}||_{L}∞≤ε

3

2.Then, takingεsufficiently small in terms of M, and
using the factδ ≤^{1}_{4},we obtain

||Z_{x}(s)||_{L}∞≤ε

1

4e^{−(}^{1}^{2}^{+δ}^{)s}. (191a)

2. ||Z_{xx}||_{L}∞ ≤Me^{−(}^{1}^{2}^{+δ)s}.
By (113) withn=2,

(∂s+3+ 3

1−τ˙e^{2}^{s}∂xZ+2∂xgZ)∂_{x}^{2}Z+ (gZ+3x
2 + 1

1−τ˙e^{2}^{s}Z)∂_{x}^{3}Z=F_{Z}^{(2)}. (192a)
In this equation, the damping term D satisfies

D=3+ 3

1−τ˙e^{s}^{2}∂xZ+2∂xg_{Z}≥3−2(1+2ε^{δ}^{2})||W_{x}||_{L}∞−3(1+2ε^{δ}^{2})e^{s}^{2}||∂_{x}Z||_{L}∞, (193a)
because∂xgZ=_{(1−}^{(1−α)W}_{τ)(1+α)}_{˙} ^{x} .Therefore, from (181b) and (181d),

D≥3−2(1+2ε^{δ}^{2})−3(1+2ε^{δ}^{2})Mε^{δ}≥3

4 (194a)

for small enoughε.

On the other hand,

||F_{Z}^{(2)}||L^{∞}=||∂_{x}^{2}FZ−∂_{x}^{2}gZ∂xZ||L^{∞} .||∂_{x}^{2}FZ||L^{∞}+||∂_{x}^{2}gZ||L^{∞}||∂xZ||L^{∞}. (195a)
Since ∂_{x}^{2}gZ = _{(1−}^{(1−α)W}_{τ)(1+α)}_{˙} ^{xx} , ||∂_{x}^{2}gZ||_{L}∞ .||W_{xx}||_{L}∞ is also satisfied. At this time, in order for the
decay rate of ||Z_{xx}|| to be e^{−(}^{1}^{2}^{+δ)s}, we need an uniform bound of ||∂_{x}^{2}g_{Z}||,because by (124),

||∂_{x}Z||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}∞ .M^{2}e^{−s}+Mε^{1}^{4}e^{−(}^{1}^{2}^{+δ}^{)s}.Mε^{1}^{4}e^{−(}^{1}^{2}^{+δ)s}. (197a)
As we already know||∂_{x}^{2}Z(·,−logε)||L^{∞}=||ε^{3}∂_{θ}^{2}z_{0}(·)||L^{∞} ≤ε^{3} by (121a), we can apply the first
maximum principle with the settingλD=^{3}_{4},F_{0}=Mε^{1}^{4},λF =^{1}_{2}+δ.It gives

||∂_{x}^{2}Z(s)||_{L}∞.||∂_{x}^{2}Z(·,−logε)||_{L}∞e^{−}^{3}^{4}^{(s+logε)}+Mε^{1}^{4}e^{−(}^{1}^{2}^{+δ)s} (198a)
.ε

9

4e^{−}^{3}^{4}^{s}+Mε^{1}^{4}e^{−(}^{1}^{2}^{+δ}^{)s}.Mε^{1}^{4}e^{−(}^{1}^{2}^{+δ)s} ∀s≥ −logε. (198b)
3. ||A_{x}||_{L}∞≤Me^{−(}^{1}^{2}^{+δ)s}

As in the case of Z, we will use (113) with n=1.(∂s+^{3}_{2}+∂xgA)∂xA+ (gA+^{3x}_{2})∂_{x}^{2}A=F_{A}^{(1)}shows
that

D=3

2+∂xgA= 3

2+ Wx+e^{2}^{s}Zx

(1+α)(1−τ˙) ≥3

2−(1+2ε^{δ}^{2})

1+α (1+Mε^{δ})≥1

2+δ (199a)

for smallε,and

||F_{A}^{(1)}||_{L}∞=||∂_{x}FA||_{L}∞ =|| e^{−s}

2(1+α)(1−τ˙)(−8AA_{x}+2(e^{−}^{2}^{s}W+κ+Z)(e^{−}^{s}^{2}W+Z)x (200a)

−2α(e^{−}^{2}^{s}W+κ−Z)(e^{−}^{s}^{2}W_{x}−Z_{x}))||_{L}∞ (200b)

.M^{2}e^{−s}. (200c)

We used (124) for the last inequality. Since||∂_{x}A(·,−logε)||_{L}∞≤ε^{3}^{2},the first maximum principle
implies that

||∂_{x}A(s)||_{L}∞ ≤ε

1

4e^{−(}^{1}^{2}^{+δ)s}. (201a)

So, the assumption is closed.

4. ||Axx||L^{∞} ≤Me^{−(}^{1}^{2}^{+δ)s}

By (113) with n=2,(∂s+3+2∂xgA)∂_{x}^{2}A+ (gA+^{3x}_{2})∂_{x}^{3}A=F_{A}^{(2)}.In this equation,
D=3+2∂xgA=3+ 2(Wx+e^{s}^{2}Zx)

(1+α)(1−τ˙)≥3−2(1+2ε^{δ}^{2})

1+α (1+Mε^{δ})≥1+δ, (202a)
and

||F_{A}^{(2)}||_{L}∞ =||∂_{x}^{2}F_{A}−∂_{x}^{2}g_{A}∂xA||_{L}∞.M^{2}e^{−s}+M||∂_{x}A||_{L}∞ (203a)
.M^{2}e^{−s}+Mε^{1}^{4}e^{−(}^{1}^{2}^{+δ}^{)s}.Mε^{1}^{4}e^{−(}^{1}^{2}^{+δ)s}. (203b)
Using (201a), we get the second inequality. Finally, by applying the first maximum principle, we
get

||∂_{x}^{2}A||_{L}∞.Mε^{1}^{4}e^{−(}^{1}^{2}^{+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−^{1}_{2})W+ (^{3}_{2}x+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−^{1}_{2})W+ (^{3}_{2}x+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,∂_{x}^{2}W(0,s) =0,∂_{x}^{3}W(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_{xx}^{2} (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_{x}^{0}))W_{xxx}^{0} =−W^{0}W_{xxxx}^{0} −3(W_{xx}^{0})^{2}.
Using the initial data of W, the above equation is changed to∂sW_{xxx}^{0} =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+ (^{3}_{2}x+W)Wxx=0,which is the differentiation of equation

−^{1}_{2}W+ (^{3}_{2}x+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˜_{x}W˜_{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}^{−}

gx

gV

g .We find(1+2Wx)^{V}_{g}+ (^{3}_{2}x+W)(^{1}_{g}Vx−^{g}_{g}^{x}_{2}V)+(terms of ˜W)
turns to^{1}_{g}{(1+2Wx)−^{g}_{g}^{x}(^{3x}_{2} +W)}V+^{1}_{g}(^{3}_{2}x+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+x}^{2} 2)(^{3x}_{2} +W)≥_{1+8x}^{6x}^{2}_{2},if g satisfies−^{g}_{g}^{x} =_{x(1+x}^{2} 2),the
first damping term guarantee a positive lower bound ^{1}_{g}_{1+8x}^{6x}^{2}2.We can find suchg=^{1+x}_{x}_{2}^{2} by solving the
ODE.

So, forV(x,s) =W˜x1+x^{2}

x^{2} ,we can rewrite the equation to

∂s( x^{2}

1+x^{2}V) + (1+W˜x+2Wx) x^{2}

1+x^{2}V+ (3

2x+W)( x^{2}

1+x^{2}V)x=−WW˜ xx. (209a)
Dividing by _{1+x}^{x}^{2}2, we obtain

∂sV+ (1+W˜x+2Wx+ 2
x(1+x^{2})(3

2x+W))V+ (3

2x+W)Vx=−1+x^{2}
x^{2} Wxx

Z x 0

V (x^{0})^{2}
1+ (x^{0})^{2}dx^{0}.

(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+x}^{2} 2)(^{3x}_{2} +W),U =^{3x}_{2} +W andF=0. First, we have to check that
there exists any positive lower bound of D. Since 1+2Wx+_{x(1+x}^{2} 2)(^{3x}_{2} +W)≥ _{1+8x}^{6x}^{2}_{2},split D into two
terms.

|D| ≥ |1+2Wx+ 2

x(1+x^{2})(3x

2 +W)| − |W˜x+ 2

x(1+x^{2})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}_{x}2^{2}. Then
we have

|W˜x+ 2

x(1+x^{2})W˜ | ≤h(x) + 2

|x|(1+x^{2})
Z |x|

0

|W˜x(x^{0},s)|dx^{0} (212a)

≤h(x) + 2

|x|(1+x^{2})
Z |x|

0

h(x^{0})dx^{0}. (212b)

The first inequality uses the fundamental theorem of calculus with ˜W(0) =0.Hence,D≥_{1+8x}^{6x}^{2}_{2}−h(x)−

2

|x|(1+x^{2})

R|x|

0 h(x^{0})dx^{0}.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 _{20}^{1},adequately.(we can change the number later, if it makes a contradiction.)
Therefore, it corresponds to h(x) = _{20(1+x}^{x}^{2} 2). Since _{|x|(1+x}^{2} 2)

R|x|

0

(x^{0})^{2}

20(1+(x^{0})^{2})dx^{0} ≤ _{10(1+x}^{x}^{2} _{2}_{)}, D≥ _{1+8x}^{6x}^{2}_{2} −

3x^{2}

20(1+x^{2}) ≥^{9}_{2} ^{x}^{2}

1+8x^{2}.∴at|x| ≥l,D≥ ^{9l}^{2}

2(1+8l^{2})≡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,x^{0},s) =−^{1+x}^{2}

x^{2} Wxx(x)1_{[0,x]}(x^{0})_{1+(x}^{(x}^{0}^{)}^{2}0)^{2}.Therefore,
Z

R

|K(x,x^{0},s)|dx^{0}≤1+x^{2}
x^{2} |Wxx|

Z _{|x|}

0

(x^{0})^{2}

1+ (x^{0})^{2}dx^{0}≤ 3x^{2}
1+8x^{2} ≤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)| ≤_{40}^{1} on|x| ≤l will be proved
below, and because|V(x,−logε)|= ^{1+x}_{x}2^{2}|W_{x}(x,−logε)−W_{x}(x)|,if we add the initial-assumption

|ε(∂_{θ}w_{0})(θ)−W_{x}(θ
ε^{3}^{2}

)| ≤

( ^{θ}

ε

3 2

)^{2}
40(1+ ( ^{θ}

ε

3 2

)^{2}), (214a)

we can conclude that|V(x,−logε)| ≤_{40}^{1}.

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 = _{20}^{1} ^{9}_{2}_{1+8l}^{l}^{2} 2 ≥0=4F_{0}
that||V(·,s)||_{L}∞(R)≤ _{80}^{3}.Since _{80}^{3} < _{20}^{1},we can understand our step as closing bootstrap assumption

||V(·,s)||_{L}∞(R)≤ _{20}^{1},and the proof is done on|x| ≥l.

Remark)|x| ≤l,wherel>0.

Write the taylor series ofWxat x=0 :Wx(x,s) =−1+^{x}_{2}^{2}Wxxx(0,s) +^{x}_{6}^{3}Wxxxx(x^{0},s)for some|x^{0}|<|x|.It
implies|Wx(x,s) +1−3x^{2}|=^{x}_{6}^{3}Wxxxx(x^{0},s).So, if we have the condition

||∂_{x}^{4}W||_{L}∞ ≤M, (215a)

|Wx(x,s) +1−3x^{2}| ≤ ^{M}_{6}x^{2}lholds. Let’s add it in our bootstrap assumption.

∴|W_{x}(x,s)−W_{x}| ≤M

6x^{2}l+15x^{4}+O(x^{6})≤x^{2}(Ml

6 +15l^{2})≤ x^{2}

40(1+x^{2}) (216a)
Hence,||V(·,s)||_{L}∞(R)≤ _{40}^{1} on|x| ≤l,too. Using the properties of the functionWx,

|W_{x}(x,s)−Wx(s)| ≤ x^{2}

20(1+x^{2}) (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+ (^{5}_{2}+3Wx)Wxx+ (^{3}_{2}x+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 termD_{1}and the otherD_{2}.
D={5

2+3Wx+ (−gx

g)(3

2x+W)}+{(−gx

g)(W−W) +3(Wx−Wx)} (219a)

≡D_{1}+D_{2}. (219b)

For the positivity ofD_{1}, we want to implement one of the properties ofW : ^{5}_{2}+3Wx+_{x(1+x}^{1} 2)(^{3}_{2}x+W)≥

x^{2}

1+x^{2} ∀x∈R.For this, g should satisfy−^{g}_{g}^{x} =_{x(1+x}^{1} 2).By solving the ODE, we getg=^{(1+x}^{2}^{)}

1 2

x ,and if
so,D_{1}≥_{1+x}^{x}^{2}_{2} is satisfied.

According to (217a),D_{2}≥ −_{|x|(1+x}^{1} _{2}_{)}^{R}_{0}^{|x|}_{20(1+(x}^{(x}^{0}^{)}^{2}0)^{2})dx^{0}−3_{20}^{1} _{1+x}^{x}^{2}2,so
D≥4

5
x^{2}

1+x^{2} ≥ 1

2M^{2} ≡λD (220a)

for all|x| ≥_{M}^{1}.In this point, D has no strictly positive lower bound if x goes to 0. Therefore, the first max-
imum 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+x^{2})^{1}^{2}.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 :

||∂_{x}^{2}W(·,−logε)||_{L}∞≤1 (221a)

||∂_{x}^{4}W(·,−logε)||_{L}∞≤1 (221b)
If so, the taylor series ofW_{xx} gives||W_{xx}(x,−logε)|| ≤ |x|W_{xxx}(0,−logε) +^{|x|}_{2}^{2}||∂_{x}^{4}W||_{L}∞ ≤6|x|+^{x}_{2}^{2},
using (221b). Therefore by (221a),

||W_{xx}(x,−logε)|| ≤min{6|x|+x^{2}

2,1} ≤ 7|x|

(1+x^{2})^{1}^{2}. (222a)
So we obtain||V˜(x,−logε)||_{L}∞≤7.

To satisfy the other condition ||V˜(x,s)||_{L}∞(|x|<_{M}^{1}), we will use the taylor series of W_{xx} as in the
previous step. Wxx(x,s) =|x|W_{xxx}(0,s) + ^{x}_{2}^{2}Wxxxx(x^{0},s). By the bootstrap assumptions, |W_{xx}(x,s)| ≤
6|x|+^{x}_{2}^{2}M≤ ^{7|x|}

(1+x^{2})^{1}^{2}.Therefore, we obtain

||V˜||_{L}∞(|x|<^{1}

M)≤7. (223a)

Since mλ_{D}=14_{2M}^{1}2 ≥0=4F_{0},we conclude||V˜||_{L}∞ <12.Therefore,||W_{xx}||_{L}∞ < ^{12|x|}

(1+x^{2})^{1}^{2}

≤Mfor large enough M.

Closing (215a)) (215a) tells us (∂s+^{11}_{2} +5Wx)∂_{x}^{4}W+ (^{3}_{2}x+W)∂_{x}^{5}W =−10W_{xx}Wxxx. In this equa-
tion, the damping term D is automatically positive, since D= ^{11}_{2} +5Wx ≥ ^{11}_{2} −5= ^{1}_{2} ≥ ^{1}_{4}. 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}≡F_{0}. Therefore, by the first maximum principle,

||∂_{x}^{4}W||L^{∞}≤ ||∂_{x}^{4}W(·,−logε)||L^{∞}e^{−}^{1}^{4}^{(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=^{3}_{4}.

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+ (^{3}_{2}x+W)Wxxxx=−3W_{xx}^{2},and the damping term sat-
isfiesD≥4(1+Wx−_{20(1+x}^{x}^{2} _{2}_{)})≥ _{1+x}^{x}^{2}_{2}.The last inequality uses the property 1+W_{x}≥ _{4(1+x}^{3x}^{2}_{2}_{)}.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≤M^{3}^{4}

2 (226a)

for|x| ≤ ^{1}

4M^{1}^{4}.So, the valuelbecomes ^{1}

4M^{1}^{4}.
Returning to |x| ≥ ^{1}

4M^{1}^{4},we obtain D≥ ^{1}

32M^{1}^{2}. |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||∂_{x}^{3}W||L^{∞} <M^{3}^{4}.

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. w_{0}(0) =κ_{0},∂_{θ}w_{0}(0) =−ε^{−1},∂_{θ}^{2}w_{0}(0) =0,∂_{θ}^{3}w_{0}(0) =6ε^{−4}

2. ||∂_{θ}w_{0}||_{L}∞≤ε^{−1},||∂_{θ}^{2}w_{0}||_{L}∞≤ε^{−}^{5}^{2},||∂_{θ}^{3}w_{0}||_{L}∞ ≤7ε^{−4},||∂_{θ}^{4}w_{0}||_{L}∞ ≤ε^{−}^{11}^{2}
3. |ε(∂_{θ}w_{0})(θ)−W_{x}( ^{θ}

ε

3 2

)| ≤min{

(^{θ}

ε 3 2

)^{2}
40(1+( ^{θ}

ε 3 2

)^{2}), ^{1}

2(( ^{θ}

ε 3 2

)^{2}^{3}+8)

}
4. ||z_{0}||_{C}^{n}+||a_{0}||_{C}^{n}≤1 ∀0≤n≤4.

5. supp(w0(θ)−κ0)∪supp(z0(θ))∪supp(a0(θ))⊆(−^{π}_{2},^{π}_{2}).

6. ˜c≤ ||w_{0}||L^{∞}≤c,where ˜c≥2(_{α}^{2}(^{ν}_{2})^{α}+2),_{α}^{4} <c,c˜ <(_{8}^{c}^{˜}(1+α)−1)×_{1−α}^{3} .

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 C^{4}satisfies 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))^{2}^{3})}, (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∂_{θ}w_{0}(0) =−ε^{−1},one result from the first condition.

Also, the third condition implies

|ε(∂_{θ}w_{0})(θ)| ≤ |W_{x}(x)|+min{

( ^{θ}

ε

3 2

)^{2}
40(1+ ( ^{θ}

ε

3 2

)^{2}), 1
2((^{θ}

ε

3 2

)^{2}^{3}+8)

} ≤1 ∀x∈R, (229a)

which means that the condition ||∂_{θ}w_{0}||_{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 >^{1}_{3},C^{α}Holder norm of w blows up as t→T_{∗}with a rate(T∗−t)^{1−3α}^{2} .

Proof 17 W(x) = (−^{x}_{2}+ (_{27}^{1} +^{x}_{4}^{2})^{1}^{2})^{1}^{3} −(_{2}^{x}+ (_{27}^{1} +^{x}_{4}^{2})^{1}^{2})^{1}^{3} 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 >^{1}_{3},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∗.

On the other hand, ^{|w(θ,t}_{|θ}^{)−w(θ}_{−θ}0|^{α}^{0}^{,t)|}= ^{|W}^{(x,s)−W(x}_{|x−x}0|^{α} ^{0}^{,s)|} is obtained by change of variables.

When x^{0}=0,the right side of it shows

|W(x,s)|

|x|^{α} ≤6|x|^{1}^{3}^{−α}=6|θ−x(t)|^{1}^{3}^{−α}(s(t)−t)^{1}^{2}^{(3α−1)}. (230a)
When x^{0}6=0, it gives

sup_{x>x}^{0}|W(x,s)−W(x^{0},s)|

|x−x^{0}|^{α} .sup_{x>x}^{0}
Rx

x^{0}(1+y^{2})^{−}^{1}^{3}dy

(x−x^{0})^{α} .(s(t)−t)^{1}^{2}^{(3α−1)}. (231a)
Therefore, we can say that it blows up with a rate at most(T∗−t)^{1−3α}^{2} .