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# Closing the assumptions

문서에서 Singularity formation of Euler equations (페이지 46-56)

### 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−δs+Me−(1+δ)s+Me−s≤M2emin{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≤cεδδ2 for sufficiently smallε.Then additionally, if we integrate τ˙ in t fort≤Tand useτ(−ε) =0,it gives

τ(t) = Z T

−ε

˙

τ(s)dt≤1

(δ2+1). (185a)

Closing (181d) and (181e)) 1. ||Zx||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˙es2xZ+∂xgZ)∂xZ+ (gZ+3x

2 + 1

1−τ˙es2Z)∂x2Z=FZ(1), (186a) where gZ= 1

1−τ˙(e2s(1−α

1+ακ−ξ˙) +1−α

1+αW). (186b)

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

2+ 1

1−τ˙e2sxZ+∂xgZ=3

2+e2sZx

1−τ˙+ (1−α)Wx

(1−τ˙)(1+α) (187a)

≥3

2−(1+2εδ2)(Me−δs+|1−α

1+α|||Wx||L). (187b)

We obtain the last inequality by1−1τ˙1

1−εδ2

≥1+2εδ2.(181b) have to be changed to||Wx||L≤1 for all s, becauseD>0 is necessary for sufficiently smallε,too. Then, since(1+2εδ2)(Mεδ+

|1−α1+α|)≤(1+2εδ2)(Mεδ+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,

||FZ(1)||L. e−s

1−τ˙(||Ax||L(||e2sW+κ||L+||Z||L) +||A||L(es2||Wx||L+||Zx||L)) (189a) .e−s(Me−(12+δ)s(2M+cκ+M) +M(es2+Me−(12+δ)s)) (189b)

.M2e32s. (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)||Le−(12)(s+logε)+M2e(1−δ)logεe−(12+δ)s (190a)

.(ε1−δ+M2ε1−δ)e−(12+δ)s, (190b)

because||Zx(−logε)||L

3

2||∂θz0||L≤ε

3

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

||Zx(s)||L≤ε

1

4e−(12)s. (191a)

2. ||Zxx||L ≤Me−(12+δ)s. By (113) withn=2,

(∂s+3+ 3

1−τ˙e2sxZ+2∂xgZ)∂x2Z+ (gZ+3x 2 + 1

1−τ˙e2sZ)∂x3Z=FZ(2). (192a) In this equation, the damping term D satisfies

D=3+ 3

1−τ˙es2xZ+2∂xgZ≥3−2(1+2εδ2)||Wx||L−3(1+2εδ2)es2||∂xZ||L, (193a) because∂xgZ=(1−(1−α)Wτ)(1+α)˙ x .Therefore, from (181b) and (181d),

D≥3−2(1+2εδ2)−3(1+2εδ2)Mεδ≥3

4 (194a)

for small enoughε.

On the other hand,

||FZ(2)||L=||∂x2FZ−∂x2gZxZ||L .||∂x2FZ||L+||∂x2gZ||L||∂xZ||L. (195a) Since ∂x2gZ = (1−(1−α)Wτ)(1+α)˙ xx , ||∂x2gZ||L .||Wxx||L is also satisfied. At this time, in order for the decay rate of ||Zxx|| to be e−(12+δ)s, we need an uniform bound of ||∂x2gZ||,because by (124),

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

||Wxx||L≤M ∀s. (196a)

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

||FZ(2)||L .M2e−s+Mε14e−(12)s.Mε14e−(12+δ)s. (197a) As we already know||∂x2Z(·,−logε)||L=||ε3θ2z0(·)||L ≤ε3 by (121a), we can apply the first maximum principle with the settingλD=34,F0=Mε14F =12+δ.It gives

||∂x2Z(s)||L.||∂x2Z(·,−logε)||Le34(s+logε)+Mε14e−(12+δ)s (198a) .ε

9

4e34s+Mε14e−(12)s.Mε14e−(12+δ)s ∀s≥ −logε. (198b) 3. ||Ax||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=FA(1)shows that

D=3

2+∂xgA= 3

2+ Wx+e2sZx

(1+α)(1−τ˙) ≥3

2−(1+2εδ2)

1+α (1+Mεδ)≥1

2+δ (199a)

for smallε,and

||FA(1)||L=||∂xFA||L =|| e−s

2(1+α)(1−τ˙)(−8AAx+2(e2sW+κ+Z)(es2W+Z)x (200a)

−2α(e2sW+κ−Z)(es2Wx−Zx))||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

1

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=FA(2).In this equation, D=3+2∂xgA=3+ 2(Wx+es2Zx)

(1+α)(1−τ˙)≥3−2(1+2εδ2)

1+α (1+Mεδ)≥1+δ, (202a) and

||FA(2)||L =||∂x2FA−∂x2gAxA||L.M2e−s+M||∂xA||L (203a) .M2e−s+Mε14e−(12)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 (∂s12)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 (∂s12)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)

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=−3Wxx2 (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+4Wx0))Wxxx0 =−W0Wxxxx0 −3(Wxx0)2. Using the initial data of W, the above equation is changed to∂sWxxx0 =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)

⇔∂sx+ (1+W˜x+2Wx)W˜x+ (3

2x+W)W˜xx=−WW˜ xx. (207c) The last step is established by(1+Wx)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˜xx+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=Vg,W˜xx=Vx

gx

gV

g .We find(1+2Wx)Vg+ (32x+W)(1gVxggx2V)+(terms of ˜W) turns to1g{(1+2Wx)−ggx(3x2 +W)}V+1g(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+8x6x22,if g satisfies−ggx =x(1+x2 2),the first damping term guarantee a positive lower bound 1g1+8x6x22.We can find suchg=1+xx22 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+xx22, 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+8x6x22,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+xx22. Then we have

|W˜x+ 2

x(1+x2)W˜ | ≤h(x) + 2

|x|(1+x2) Z |x|

0

|W˜x(x0,s)|dx0 (212a)

≤h(x) + 2

|x|(1+x2) Z |x|

0

h(x0)dx0. (212b)

The first inequality uses the fundamental theorem of calculus with ˜W(0) =0.Hence,D≥1+8x6x22−h(x)−

2

|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+xx2 2). Since |x|(1+x2 2)

R|x|

0

(x0)2

20(1+(x0)2)dx010(1+xx2 2), D≥ 1+8x6x22

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

R

|K(x,x0,s)|dx0≤1+x2 x2 |Wxx|

Z |x|

0

(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+xx22|Wx(x,−logε)−Wx(x)|,if we add the initial-assumption

|ε(∂θw0)(θ)−Wx(θ ε32

)| ≤

( θ

ε

3 2

)2 40(1+ ( θ

ε

3 2

)2), (214a)

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

According to lemma (6),Wx is compactly supported, while from the taylor series ofW atx=∞, Wx decays as |x| →∞. Therefore, (237a) is satisfied. Conclude from mλD = 201 921+8ll2 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.

∴|Wx(x,s)−Wx| ≤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,

|Wx(x,s)−Wx(s)| ≤ x2

20(1+x2) (217a)

turns into||Wx(·,s)||L ≤1.This is the end of proof (181b).

Remark) In this proof, we used bootstrap method to||V||,not||Wx||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)(−gx

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−ggx =x(1+x1 2).By solving the ODE, we getg=(1+x2)

1 2

x ,and if so,D11+xx22 is satisfied.

According to (217a),D2≥ −|x|(1+x1 2)R0|x|20(1+(x(x0)20)2)dx0−3201 1+xx22,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 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||Wxx(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.

||∂x2W(·,−logε)||L≤1 (221a)

||∂x4W(·,−logε)||L≤1 (221b) If so, the taylor series ofWxx gives||Wxx(x,−logε)|| ≤ |x|Wxxx(0,−logε) +|x|22||∂x4W||L ≤6|x|+x22, using (221b). Therefore by (221a),

||Wxx(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 Wxx as in the previous step. Wxx(x,s) =|x|Wxxx(0,s) + x22Wxxxx(x0,s). By the bootstrap assumptions, |Wxx(x,s)| ≤ 6|x|+x22M≤ 7|x|

(1+x2)12.Therefore, we obtain

||V˜||L(|x|<1

M)≤7. (223a)

Since mλD=142M12 ≥0=4F0,we conclude||V˜||L <12.Therefore,||Wxx||L < 12|x|

(1+x2)12

≤Mfor large enough M.

Closing (215a)) (215a) tells us (∂s+112 +5Wx)∂x4W+ (32x+W)∂x5W =−10WxxWxxx. In this equa- tion, the damping term D is automatically positive, since D= 112 +5Wx112 −5= 1214. Also,

||F||L ≤C||Wxxx||L by (196a). So, if we additionally assume

||Wxxx||L ≤Mk (224a)

for somek∈R+,||F||L ≤CMk≡F0. Therefore, by the first maximum principle,

||∂x4W||L≤ ||∂x4W(·,−logε)||Le14(s+logε)+4CMk. (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 ofWxxx, we have to use the second maximum principle because of the damping. (206c) shows∂sWxxx+ (4+4Wx)Wxxx+ (32x+W)Wxxxx=−3Wxx2,and the damping term sat- isfiesD≥4(1+Wx20(1+xx2 2))≥ 1+xx22.The last inequality uses the property 1+Wx4(1+x3x22).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|Wxxx(x,s)−Wxxx(0,s)| ≤ |x|M.It transforms into

|Wxxx(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|Wxxx(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≤ε52,||∂θ3w0||L ≤7ε−4,||∂θ4w0||L ≤ε112 3. |ε(∂θw0)(θ)−Wx( θ

ε

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˜ <(8c˜(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.∴|Wx(0,−logε)−Wx(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)| ≤ |Wx(x)|+min{

( θ

ε

3 2

)2 40(1+ ( θ

ε

3 2

)2), 1 2((θ

ε

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,CαHolder norm of w blows up as t→Twith a rate(T−t)1−3α2 .

Proof 17 W(x) = (−x2+ (271 +x42)12)13 −(2x+ (271 +x42)12)13 is known. Since Wx=Wx at x=0by the inequality(217a), and the fact that|x|1−α|Wx|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, Cα Holder norm of w blows up as t→T.

On the other hand, |w(θ,t)−w(θ−θ0|α0,t)|= |W(x,s)−W(x|x−x0|α 0,s)| is obtained by change of variables.

When x0=0,the right side of it shows

|W(x,s)|

|x|α ≤6|x|13−α=6|θ−x(t)|13−α(s(t)−t)12(3α−1). (230a) When x06=0, it gives

supx>x0|W(x,s)−W(x0,s)|

|x−x0|α .supx>x0 Rx

x0(1+y2)13dy

(x−x0)α .(s(t)−t)12(3α−1). (231a) Therefore, we can say that it blows up with a rate at most(T−t)1−3α2 .

### VI Appendix A : Bootstrap method and Maximum principle

문서에서 Singularity formation of Euler equations (페이지 46-56)

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