Navier-Stokes Seminar

8/3/2019 Navier-Stokes Seminar

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Contents

Contents 1

1 Linear & Nonlinear Static NSE 31.1 Physical properties remarks . . . . . . . . . . . . . . . . . . . 31.2 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Shear flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Steady-state Stokes - Overview . . . . . . . . . . . . . . . . . 61.5 Steady-Stokes - Variational form . . . . . . . . . . . . . . . . 81.6 Wk,p, Hk, H12 . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.7 The Auxiliary space E() . . . . . . . . . . . . . . . . . . . . 101.8 Extension map . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.9 Trace and Stokes formula . . . . . . . . . . . . . . . . . . . . . 121.10 Preliminaries for Stokes Equation Existence . . . . . . . . . . 131.11 Leray projection operator . . . . . . . . . . . . . . . . . . . . . 151.12 Existence for Stokes Equation . . . . . . . . . . . . . . . . . . 151.13 Energy approach . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.14 Energy Method proof. . . . . . . . . . . . . . . . . . . . . . . . 191.15 Existence for Inhomogeous Stokes . . . . . . . . . . . . . . . . 201.16 Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.17 Navier-Stokes Existence . . . . . . . . . . . . . . . . . . . . . . 271.18 Existence - Galerkin . . . . . . . . . . . . . . . . . . . . . . . . 29

1


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CONTENTS 2

1.19 Variational Form for Inhomogeneous Navier-Stokes . . . . . 32

1.20 Uniqueness for Inhomogeneous Navier-Stokes . . . . . . . . . 331.21 Inhomogeneous NS Uniqueness . . . . . . . . . . . . . . . . . 341.22 Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.23 Regularity for NSE for n = 2, 3, 4, 5, 6, 7 . . . . . . . . . . . . . 43

2 Evolution NSE 552.1 Stokes operator . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.2 Function spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 562.3 Galerkin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592.4 Pass to limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 622.5 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

2.6 Stability Properties . . . . . . . . . . . . . . . . . . . . . . . . 662.7 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Denn nichts ist fur denMenschen als Menschen etwas

wert, was er nicht mitLeidenschaft tun kann.


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Chapter 1

Linear & Nonlinear StaticNSE

Friday, August 26 Overview of NSE

1.1 Physical properties remarks

These equations establish that changes in momentum (acceleration)of fluid particles are simply the product of changes in pressure anddissipative viscous forces (similar to friction) acting inside the fluid.These viscous forces originate in molecular interactions and dictatehow sticky (viscous) a fluid is. Thus, the Navier-Stokes equations area dynamical statement of the balance of forces acting at any givenregion of the fluid.

1.2 Derivation

F = Fint = where = normal + sheer strengthviscosity - 0 in ideal fluid

and

3


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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 4

(ij) = 13

normalize

p 0 00 p 00 0 p

+ T=

1

3 (pI3)gravity

+ tensor; material dependent

where we assume the tensor is symmetric.

For a Newtonian fluid,

stress tensor

u

y+ u

yT

velocity gradient - strain rate

where is proportional to viscosity, which varies with temperature.

For u = (u1, u2, u3) and p constant,ut + u u =

pI3 +

(u +

(u

)T

)???

uincompressibility

= 0

Calculating,

(u + (u)T) or xj

uixj

+uj

xi take transpose

= ui +

xiuj

xj div = 0 fluid incompressible

ut + u u + p = u. (NSE)

This is formally a coupled system in four unknowns (u1, u2, u3), p and2 eqns, and is therefore determinable. To uncouple, write (?) Difficulty: p is presumably a

function of 4 variables, but myeqn doesnt tell me; this makes

p difficult to handle.For fluid haveboundary conds., but for p ini-

tial condition


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Figure 1.1:Measuringshear

flow

p = div (u )uelliptic eqn

1.3 Shear flow

An element of solid has a preferred shape, to which it relaxes whenthe external forces on it are withdrawn (whereas a fluid does not).

Looking at a rectangular element ABCD, under the action of a shearforce F, the element assumes the shape ABCD. If the solid is per-


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Figure 1.2: Deformation of

solid (top) and fluid elements(bottom)

fectly elastic it goes back to its preferred shape ABCD when F iswithdrawn.

In contrast a fluid deforms continuously under the action of a shearforce (however small). Thus the element of the fluid ABCD con-fined between parallel plates deforms to shapes such as ABCD andABCD as long as the force F is maintained on the upper place.

How to Measure Viscosity

First measure the normal force that is exerted on the plate. From thiswe can determine pressure p:

F = This constant is to be determined

A u

ystrain rate: sort of like du

dy

measure area A, the strain rateuy

to determine experimentally

the pressure constant

1.4 Steady-state Stokes - Overviewu t=0= u0, u0 = 0.

In NSE (or just Stokes) the following boundary conditions are common-place:


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u

= 0 no slip boundary - fluid will have zero velocity relative to

the boundary.

u = 0or ( u) v = 0

vorticity - here no tang. component

Navier Slip boundary condition

A. Steady-State Stokes

u + p = 0 drop t.

p is harmonic,so p can bewild on boundary we have

this from h o l o m o r p h i c

f u n c t i o n s

u = 0u = 0

B. Evolution Stokes

ut u + p = 0

u = 0

u t=0= u0, u = 0

C. Steady NSE

u u u + p = 0

u = 0

u = 0


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1.5 Steady-Stokes - Variational form

Compare this to Lax-Milgram-type arguments for elliptic eqns. In fact,NS is almost elliptic:

u2ndorder function with divergent form

+

but have this termso we must find a different solution space.p = 0

For the first term

u u, where Cc

+ p, = 0 u + p, = 0,

For the second term,

p, Leibnitz rule= div (p)trace=0

p ( ) so

u,

p,

= 0

where we have assumed that the test function is divergence-free:

V = Cc () div = 0 .Reverse direction:

u, = 0 V

or

min u2 u V We calculate first variation to

get Euler-Langrange eqn

(above)

.

The Problem with the Space VThe space V = Cc () div = 0) is not a complete Vector Space! ! ! We need to complete this So how will we do this? dependson the space we want to close, i.e. H10 , L

2.


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Monday, August 29 Function spaces

Once we have the variational formulation of

u + p = 0 in

u = 0 on

we have to set up the corresponding function space. The usual Sobolevspace is deficient because it requires information about u, whereas here

only u is known. In the following, we consider only vector fields that aredivergence-free.

1.6 Wk,p, Hk, H12The Inner Product (, ) and the Pairing , (u, v) we can use the L2 inner product for two functions. u, v use

for those arguments that arent functions, for example (u, v) .Wk,p. Define Wk,p

(

)=

f Lp

(

) Ddf Lp

(

)

k, k 0, 1 p

fWk,p = k

DfpLp 1pwhere Wk,p(),Wk,p is a Banach space.Density theorem

We denote the space C() as the space of smooth functions up to theboundary. For p = 2, 1 p < , this space is dense in Wk,p().

Note that Cc Wk,p = Wk,p

0 and u = Du = 0

k 1 on . ?

H and V

For u Cc (), u = 0 H10 L2, define H = {u Cc u = 0}. Thisspace is not closed. After mollifying, define


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V =

u Cc

(

) u = 0

= H0 Here u H

10 and u = 0 preserve

their definitions. H = u Cc u = 0 and p H1() = L2

Here we have uk = Cc and

u

L2

uk Cc and uk = 0.

What is u = 0 here? In this space, u is = 0, remember thoughwe are taking distributional derivatives.

Since u = 0, can we talk about u . Ifu is a L2 function whosedistributional derivative is zero, then do we have enough information?No: If only u L2, then the boundary cant be defined.

We have partial information with u . This expression is welldefined when

u H12()

1.7 The Auxiliary space E()Define the auxiliary space E() =

u L2() udistributional divergence

L2().

ThenH1 E() L2.

Define (u, v)E() =

well defined each term is L2.

u + (div u)(div v)and

u

E() =

(

u, u

)E(). this function is onto but still not

1-1:ker 0 = H

10() = whole spaceNote that (E(),)E() is Hilbert (check).

Density theorem for E(). Assume C0,1. ThenCc () is dense in E().


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Proof. (...) (see s.31)

Trace theorem for E() . A trace theorem allows us to define u for u E().

To do this, we ask where the trace operator lives. The definition

0 H1() L2()

has the deficiency that not every L2 function can be the trace of a H1

function: this mapping is not onto. Replace this definition with

0 H1() H12 () D 12 . cant define fractional

derivatives pointwise. usefourier transform

Define Hs

(Rn

)

f

2

+

2

f

L2 =

f Rn R

1 +

2

s2

f

L2

. this is for flat case. not if domain

is nonflat. If have Lipschitz graph

=

(x, (x)

Lipschitz

) x Rn1

Consider

f(x) H12 (Rn1) = f(x, (x)) = Rn1 Rwhere f H

1

2 (). I push forward. is H1 so this isequivalentlocally Lipschitz Use Partition of unity and define globally.

1.8 Extension map

Define

H

1

2

()

H

1

().Findf F

by solving

F = 0

F = f.The solution of this Laplace equation gives F, the definition of the

extension map F. Note that f and F are reciprocals of each other:

FH1 cfH12 , n = 0 = id


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1.9 Trace and Stokes formula

For C1,1 there exists a linear continuous operator E() H 12 () require the (stronger) C1,1 regu-larity hereandH

1

2 () = H12 () dual space.Furthermore the following hold:

1. u = u u C1().2. Stokes formula holds:

div (u)+

u

u=

0 restrict

u restrict normal component

(u

) this allows us to integrate byparts in the space E().3. ker() = E0() and

E0() = Cc ()E()For Sobolev spaces we had that the whole component on the bound-

ary was zero; here, we have no information about the tangent com-ponent, just the normal component.

For the function u the normal component is well defined, but grad u isnot. Dx1, Dx2, . . . are not defined

which I need for the matrix rep-resenting the tangential compo-nents.


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Friday, September 2

1.10 Preliminaries for Stokes Equation

Existence

Hodge Decomposition

L2() = H H1 H2where H =

u L2

(

) div u = 0,

normal trace

= 0

H1 = u L2() u = grad p, p H1(), p = 0 no information on boundarydata for H1, but we do havep = 0.

H2 = u L2() u = grad p, p H10()Check:

1. H1 H2 = H H1 = H1H2 = {0} .2. H1, H2 H where H = u L2

(

), u = p,p H1

(

)

3. H1 H2 Stokes theorem can be relaxedonly the gradient need be in H1

0

with the space E().Check (3) :u = p, p = 0, (1.1)

v = q, q H10 . (1.2)

Stokes theorem applies:

(u, v

)=

(

well defined, p E()p , q

)well defined, q H1

0

+

(div p,q

)= (0(p)0Eq 1.2

, 0q)= 0


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4. Let u L2

(

)and p H10 solves

p = div u in

p = 0 on

By Lax-Milgram, there exists a solution.

div (p u) = 0u = (up) + pClaim: The term p u is divergence-free but perhaps not trace free( 0). To see this, write

u p

div -free

E() normal tracecan be defined

(up)well-defined

H12()

Let q H1 solve

q = 0 (1.3)

q= (u p) (1.4)

where must satisfy (u p), 1 = 0 div (up)=0

(the

above equation is solvable up to a constant. We write u H1R todenote this (modulo the constant).

Apply Lax-Milgram to solve (1.3,4) . Note: When applying Lax-Milgram with Neumann BC youhave to be careful (?)Thus

u=

u

p

qH+

H1

q+

pH2


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Figure 1.3: The HH plane

1.11 Leray projection operatorDefine P the bounded linear operator

P L2()H remember L2() = HH2which has the apriori estimate DuL2 uL2 where u u p q.

The Projection operator is used to interpret Stokes:

u + p = f in

div u = 0 in

u = 0 on .

Apply op on both sides of the equation:

Pu + P(p)H2

=

L2P fL2

Pu is Stokes operator,

denoted by A.

Later: Mild solutions KatoGiga, Fujita, (cf. Du Hamel formwhere we use P to kill pressure

p).

orAu =

f

Avantage: P eliminates thepressure term p.Disadvantage:1.12 Existence for Stokes Equation

Rn is bounded Lipschitz


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Use Leray projection operator to make p drop out:

u + p,v = f, v v V (uvint. by parts= (f, v) u V)

For the spaceV = u H10(), div u = 0 H10

we can write the inner product

(u, v

)V =

u v

and u2V = (u, u)V.We look for u V that satisfies

(u, v)V = (f, v) v Vwhere f V and f H1. Because u V , this u satis-

fies (A) incompressibility condi-tion (B) trace=0

This implies

(u + f, v

)D()= 0 v V.

+ fH1

=

L2p for some p D() Can recover p; see Notes last

class.

Summarize:

Fact u V solves (u, v)V = (f, v) for f V if and only ifu + p = f in

div u = 0 in

u = 0 on

for some p L2 V is read in the dual V


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1.13 Energy approach

Define the Energy functional E(u) = u2V 2(f, u). cf. Laplace Energy functionalu = f 12 u2 f, wherewe minimize the functional over

H10

; Here we minimize over V.

The variational problem u V solves (u, v)V = (f, v) for v V . isequivalent to

1. u minimizes E(u) over V.2. a(u, v) = (u, v)V u, v V.

bilinear form is (A) bounded (B)Coersive so we can apply Lax-Milgram


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Wednesday, September 7 Existence

Regularity

Theorem 1 For

u + p = f in V H10div u = f in V H10 you can enlarge the data here

so its in a larger space

u = 0.!

There exists a unique u V and unique p L2()R solving homoge-neous system with estimate

uV +pL2R uH1 .Proof. a(u, v) = (u, v)L2 u, v V.The following hold for a:

a(u, v)

uV vV boundednessa(u, v) u2V coersivityLax-Milgram gives that there exists a unique v V which satisfies

a(u, v) = f, v v VIt follows from Notes from last class that

u + p = f for some p D

Note that

p L2Poincaire

p + f + u H1Initially p is just a distribution. using info p L2.


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1.14 Energy Method proof.

The energy functional E is given by E(u) = u2L (f, v) v V.Claim u V solves (*) if and only if E(u) = minvVE(v).We know the minimization is always obtained.Pf. (u, v)L2 = (f, v) v V. Let

v = u w where w V.

Then(u, u)L2 = (f, u) (f, w) + (u, w)L2

Using that

(u, u)L2cs 2u2L2 + 2 w2L2

we have

(u, u)L2 = (f, u) (fw) + (u, w)L2 (f, u) (f, w) +

2u2L2 + 2 w2L2

12

E(u) = 2 u2L2 (fu)

2w2L2 (f, w) = 12 E(v)

In solving the Laplace equationusing Energy method, u H1

0

Here u V.

0 =d

dtE(u + tv) = 2(u, v)L2 2(f, v). (1.5)


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1.15 Existence for Inhomogeous Stokes

L

2

()R

means (f, 1)L2

f constant=

f1 = f= 0Theorem 2 For

(IS)

u p = f in divu = g in

u = on .

and Rn bounded, C2, let f H1, g L2, H12() and

g = u

divergence theorem=

!u H1 and ! p L2

R solving 1.7 and

uH1 +pL2 R fH1 +gL2 +fH12 Remember: A function f H10 div f L2The approach for the inhomogeneous is cook up new function that is

divergence-free and has zero trace. Note that f changes.

A Property of the divergence operator

The divergence operator H10()ontoL2() f = 0Define the restriction operator 0 H1

(

)H12

(

) u0 H1 such

that 0(u0) = first extend it.Because of the Compatibility Condition,

(div u0

gg ) = g

g

Compatibility Condition= 0

u1 H10 such that div u1 = g div u0 Somehow, u0,u1 kills g.

Let v=

u

u0

u1. Thendiv v = div u div u0 div u1

= g div u0 (g div u0)= 0.


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Trace

0(V) = 0(u) 0(u0) 0(u1)= = 0.

The Eq (IS) in v becomes

v = p = (u u0 u1) + p= f + (u0

H1

+ u1H1

) = fH1

Since u0 H1 and u1 H

1,

then u H1.

Then

v = 0 in

v = 0 on .

Now its time to prove that

H10()ontoL2() f = 0 (1.6)Not 1-1: add divergence-free function and we still have = f (no unique-

ness).The term p: The gradient operator L2

(

) H1 slash out constants so we can

have f+ c = f.

After takingout constants, this space is 1-1.

Adjoint:

, g = f, g (1.7)This equality 1.7 holds because div f, gintegrate by parts= div f, g1.16 Regularity

If Rn is bounded and Cm+

2 and u W2,, p W1,, 1 < < +.For u W2, , u is pointwise defined and p is well defined strong solution instead of weak

solution.If addi-

tionally f Wm,, g Wm+1, and () Wm+2 1(),u Wm+2,(), p Wm+1,


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and

uWm+2,

+pWm+1,

fWm,

+gWm+1

+W

m+2 1 ()+uL2

Question: Why do we have u W2,? Ans: Iff W,1 ??

Theorem 3 If f L2, g H1, H32, then

u W2,2 p W1,2.

Why? Suppose g = = 0 for the moment and = 0

u + p = f (1.8)

u = 0

u = 0

For this eq 1.16, which has only two terms on the lhs,

f L2 u L2 then automatically p W1,.

We must show, using finite quotients, that

(u, v) ()= (f, v) use u u(x+he)u(x)h

H1uniformly with respect to h. D2u L2.Approach: Use the finite difference quotient as test function, but have

to cut out.


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Friday, September 9 regularity for n = 2.

Theorem 4 (regularity for n = 2)

For f Wm,, g Wm+1,, Wm+21

,(), Cm+2 and g =

then Remember that (B) and (C) be-low are satisfied if and only if theCompatibility Condition is satis-fied.

u + p = f in

divu = g in (B)u = on(C)

has unique solution (modulo constant) such that u Wm+2,, p Wm+1,1

and

uWm+2,

+pWm+1, fWm, +gWm+ 1 +Wm+2 1 , +uL2 . !

Approach: We find auxillary equationto kill (B),(C). In particular:

find gradient potential kill pressure p get formally biharmonic eq -coupled 2nd order (like 2nd order Elliptic equation) for which Sobolev Wk,p

theory holds.Claim v Wm+2, such that

divv = g in

v = on

(Prove later)If Claim 1 holds, then w = u v where w satisfies

() w + p = f + v = f f has same regularity as f.

divw = 0w = 0


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Case 1: is 1-connected

divw = 0 can find can find extr in 1-form that is closed dW = 0.1 homotopy cohomology is exact translated such that

= x2

,

x1 Condition (B)

We get rid of the divergence-free condition and plug in:

Take = 1 D2 + D1p = f1first component of the vector f

Take derivatives & subtract to eliminatef

:D2(D2) + D212p = D2f1 mixed derivatives cancel

D1(D1) + D221p = D1f2 (D22 + D21) = D2f1 D1f2

2 = D2f D1f

2 Wm1, (1.9)

Note that 1.9, an equation in does not involve p. Further, it is a firstorder differential operator lost 1-order regularity.

dV = 0 tangent derivativenormal derivative = 0

= 0 on

For the second order operatorneed a second condition;this problem as it stands is no tangent gradient = 0

Note that the potential is constant, but because is not unique, pick = 0 on

u = f Wm,2

u = 0 u Wm+2,2

Using finite quotients and integrate by parts, get 2 more derivative estimate.This is possible = 2 For u Wm+2,, Calderon-Zygmund theory o

harmonic analysis does this forhigher order .


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Figure 1.4: Difficulty: cannot shrinkclosed paths to 0. The 1-cohomology

of the domain is like copies H1

() =Zk have k freedoms cannot notspecify constants ci

2u = f u Wm+2,

u

= 0 the needed second condition

Wm1+4,w Wm+2, .

From ppWm+1, =f + Wm,.Case 2: is k-connected

Suppose 2-connected for simplicity.

We cut k (here: 2) times to make 1-connected

is

(k + 1

)-connected, k 1. Let = 0 1 k. R such

that

w = (D2, D1) = = 0 on 0, = c on i, 1 i k

= 0 on


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Figure 1.5: is defined ex-

cept on cut. For to be well-defined there too extend make well-defined trace

For the Multi-Connected case, equation same as in single connectedcase. The biharmonic equation is uniquely solable; but the dirichlet +

neumann boundary conditions are difficult ? Rest of argument is exactlysame.

For x + = +

(x0

)+ d.

lhs,rhs have same limit we can smoothly extend

Proofs of Case 1

Omitted. (see p7,8 handcopy)


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Wednesday, September 14 For the Navier-Stokes Equations we introduction the nonlinear advec-

tion term u u, i.e. the advection operator is

u = u

x+ v

y+ w

z.

Picture advection as transport of salt dumped in a river. If the river isoriginally fresh water and is flowing quickly, the predominant form oftransport of the salt in the water will be advective, as the water flowitself would transport the salt. If the river were not flowing, the salt

would simply disperse outwards from its source in a diffusive manner,which is not advection.

Very few exact solutions of NSE are known in closed form. In general,exact solutions are possible only when the nonlinear terms vanishidentically.

1.17 Navier-Stokes Existence

NS Existence

Triple product B[, , ] Brouwer fixpoint theorem

(N S)

u + u u p = f in divu = 0 in

u = 0 on .

Variation form:

((u, v

))+ B

[u, u, v

]=

(f, v

)v V ((u,v)) like L2 inner product but

with gradients

B[u, vw] = uL

2nn2

L2

v wLn

u, v V, w Vbounded

We to to extend to the

bigger spaceV = V.


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Note that V H10 take closure instead of the space

H10 Ln

which has norm uH10Ln =uH1

0

+uLn for n = 2, H10 Ln = V.Lemma For n 2, B V V V R is bounded:

B[u, vw] ho

uL2vL2wV

replaces with Ln

Sobolev embedding

c(n)uH10 vH10 uVRemember: For 2 n 4 check that

V = V

but in general H10 Ln V 2n

n2 n 2 n 2 n 4

In particular, B V V V R is well defined for 2 n 4; form 5, V V B is not well-defined.

We need to understand trilinear form better; It can induce bilinear form

B

[u, v

] V V R replace with

V

Lemma

1. B[u, v, v] = 0 v Vand w V.2. B[u, v, w] = B[u, w, v] u V, v, w V skew symmetry in last two vars.Proof (1)

B

[u, v, v

]df=

u v v =

uiiv

jvj

=

uiiv

2

2

ibp=

iu

j v22 = 0


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(2) B

[u, v + w

]= 0 by (1).

Lhs =

0B[u, v, v] + B[u, v, w] = B[u, wv] + B[u, w , w]0

= 0.

1.18 Existence - Galerkin

Goal: Solve u V for

((u, v

))+ B

[u, u, v

]=

(fv

)v

V .

Let {wk}k=1 be an orthonormal basis (ONB) of L2() replace with V.The space L2 is infinite-dimensional vector space cut off finite like

Rn have family of solutions if we can prove apriori estimate, we cantake limit function can solve the limiting equation done.

Let Xm = span {w1, . . . , wm} m 1 and let um = mi=1 ami wi coeff tb determined((um, wj)) + B[um, um, wj] = (f, wj) 1 j m choose wj to be basis

N O T E: ONB L2, but H10

amj + Ajila

mi a

m

l = (f, wj)We ask if we can solve the term Ajila

mi a

m

l , being nonlinear. Formally, this

term is not like solving Ax = B, but rather like solving Ax + BxxT = C;does solution exist maybe equal number of equations solutionmaybe not unique though nonlinear terms. Therefore we need to turn tothe following

Theorem 5 (Brouwer)Iff BR BR is continuous x0 BR such that

f(x0) = x0,where BR are closed balls. !


30/83


Figure 1.6: The closed ballBR

Figure 1.7: and P()Proof (contradiction)

If f(x) x x Bn F(x) = f(x)x BR;F BR BR. F(x) = x if x B F is a continuous map and wi is

fixed on the boundary F BR= Id impossible by degree theorems x = 0 on in BR, but boundary is of degree 1. infinite-d versions of Browder

thm very useful for existence ofnonlinear pde.Lemma (Variation of the Brouwer Fixpoint Theorem) Let X

be a finite-dimensional Hilbert space equipt with inner product [, ] andnorm []; P X X is continuous (but not necessarily linear). Suppose[P()] = k > 0 X with []k such that P() = 0.

Proof If p() 0 BR = X [] k, then defineQ() = k p()[p()] BR

and

[Q()] = k.Now Q BR BR continuously and Brouwer 0 BR such thatQ(0) = 0 where automatically Q(x0) is on the boundary and [Q(0), 0] =k.


31/83


= [k p(0)[p(0), 0]] = k [p(0), 0][p(0)]which is impossible because 90degree

Define Pm(u) Xm Xm by[Pm(u), v] = ((u, v)) + B[u, u, v] (f, v).

Need to verify: B[(), ] > 0.((u, u)) (f, u) = u2V fVuV =uv u fV > 0

if uV = k where m > fVV um Xm with umV < k such thatpm(um) = equivalent to amj + Ajilami aml = (f, wj). In fact umV fVV .

Since[pm(um), um]um2V (fum) um2V = (fum) fVumV umV fV

which gives the sought after apriori estimate with uniform bound (so thatwe can now take limits).

ChallangeWrite down Lax-Milgram proof using a Galerkin method approach


32/83


Friday, September 16 Variational Form for Inhomogeneous Navier-Stokes

Uniqueness for Inhomogeneous Navier-Stokes

1.19 Variational Form for Inhomogeneous

Navier-Stokes

We show that the variational form for inhomogeous Navier-Stokes is wellposed.Define the eigenfunction {Xm} of Stokes operator having zero Dirichlet

boundary conditions Xm = span{wi}mi=1.Find u V such that

((u, v)) + B[u, u, v] uuvwell defined for vV

= (f, v) v V , f V

(From last time) For um Xm

((um

, v)) + B[um

, um

, v]skew-symm in last two vars = (f, v) v Xm

where XmV(1.10)

Use Brouwer fpt in finite space

Write v = um. Then 1.10 gives

This is called an apriori

estimate lhs holds for allsequences um, & rhs is

independent of the number m.

((um, um)) = (f, um) umV 1fV

Send m.

By weak compactness properties of V, bounded sequence in V take subsequence so sequence converges weakly:

um u in L2 Rellich compactness um u in L2 um u in V.


33/83


v Xm,

((um, v)) ((u, v)),(f, v) (f, v) vacuouslyB[um, um, v] B[u, u, v] (1.11)

The last of these 1.11 is not obvious. Write

B[um, um, v] =

um um vskew-sym

= B[um, v, um] =

um

tensor

umL1

vL

so that =

u u v = B

[u, v, u

]= B

[u, v, u

]= B

[u, u, v

] ((u, u)) + B[u, u, v] = (f, v) v Xm not for V yetNote: we dont want solution & test function to vary at the same time:

fix solution and send test function ; fix test function & send solution.

we have, then, that

For all v V there exists vm Xm so that

vm v

V .

and so

((u, v)) + B[u, u, v] = (f, v) v V , f V .1.20 Uniqueness for Inhomogeneous

Navier-StokesBrouwer FPT is great for non-

linear you can have many fix-points; LM is only for 1 fixpoint.

Theorem 6 Forn 4 and 1 in the sense that2 c(n)fV !u Vfor

U + p = f u u.

! for small Reynolds number


34/83


Proof n 4 V = V.

(u, v)L2uL2fV

+

0skew-symmB[u, u, v] = (f, v)

the solution is unique in a ball ?Suppose u1, u2 unique solutions. Let w = u1 u2. Then

(ui, vi) + B[ui, ui, v] = (f, v) replace v with w (w, v) + B[u1, u1, v]Note that

(u1u1 u2u2

)=

(u1 u2

)u2 + u2

(u1 u2

) v = wu1v u2wv

(u, v) + B[w, u, v] + B[u2, w, v] = 0.

Test with v = w w22 + B[w, u, w] + B[u2, w, w]0

= 0

w22 w u w =switch

vL4

uL4

wL2

w5u4w2 Sobolev

u4 c(4) w2 .Use

u

4

c(n) fV

c(4)fV


35/83


Curl

For n = 2, = (1

, 2

), curl = D21

D12

. For n = 3, = (1, 2, 3),

curl = ( ) = deti j kx

y

z

1 2 3

For n = 2 the term vanishes not a vector n 4, = (1, . . . , n),

curl =

(R1 , . . . , Rn

)where

Ri =n

j,k=1

aijkDjk

and

j,k

aijk = 0 i.

We want curl H1 Ln.

Theorem 7 at least one u H1

and p D

() solving 1.21, H1

Ln

such that div = 0 and = on !

Proof. Let u = u for

u = 0

u = 0 on In terms of u

(u + ) + (u + ) (u + ) + p = f u + u u

difficult quadratic terms

+

linear first order u + u + p = f + +

f

solve this gives variational

form

For all v V,

(uv) + B[u, u, v] + B[v, u, v] + B[u, w, v] = (f, v). B[, , ] are difficult but cancontrol


36/83


Verify that f H1. Suppose H10 . Then

, well-defined =

Ln

L2 H1

0

Ln L2 H10

H1.


37/83


Monday, September 19 coersivity

For

u p + u u = f

divu = 0

u =

Suppose H2 Ln satisfies div = 0, = 0 on , u = u +u+p = f

wheref = f + H1

.The corresponding weak form is

((u, v)) + B[u, u, v] + B[u, , v] + B[, u, v] = (f, v) v VWe need the coersivity estimate for LM:

For > 0

((u, u)) + B[u, u, u] + B[u, , u] + B[, u, u] u2 (( u, u)) + B[u, , u] u2 .

It suffices to show that

B[u, , u] 2u2 .

B[u, , u] = B[u, u, ] =

u Du uL

2nn2DuL2 L2

Sobolev embedding

c(n)u2Ln


38/83


1.22 Lemmas

How and why we cut off

Why: Chop off to make norm Ln small in the above inequalityB[u, , u] c(n)u2 Ln

How: We introduce the function = curl cannot chop offindirectly: we can cut potential, but cant cut curl:

= curl

not here

cut here.

If we cut , the divergence-free condition breaks down!

Additionally H2 W1,n Lbounded

and = curl ; H2 H12.

curl ()Ln

+Ln , c

Ln(2)

+Ln(2)Lemma 1 Let(x) = dist(x, ) > 0 there exists C2() such that

= 1 if (x) 2 = 0 if (x) , = exp1 (x) if (x)

!


39/83


Proof Define =

(x

)

. I f C2 0 = 0

(

)> 0 from

differential geometry

such that

C

2

()Let =

t t 2

log t 2 t 2

0 t

Then ?(x) = ((x)); = ((x))(x) by chain rule.Why is 1? Since

(x1) = (x2) x1 x2Lip() 1 Lip () 1

we have

((x)) (x) ((x))

(x) .So (?)

f

Ln

+

Ln

curl

(2

)Ln

.

We need to estimate Ln

. Be careful: is small near by use

Poincaire inequality near the boundary (Poincaire inequality with weight)

Lemma 2 (Hardy inequality)

v2

2 c1

v2 v H10

!

This is like a Poincaire inequalitywith weight. Cf:

Bv2 c1

Br

(v2)Proof. Near boundary, of 0

0type. It suffices to show > 0 such that

{}

v2

2 c

v2


40/83


Figure 1.8: Folliate.

For

{>}

1

2

v2

use ordinary poinaire away fromboundary

We want to estimate

{}

v2

.

To do this we use Federers co-area formula for the annular region

Federers Co-area formula (Class: 767)

Rn

fg = 0

{g=t}

fdHn1dtwhere f L1, g Lip .

Like a curvilinear Fubini

For level surfaces

For good Lipschitz surfaces, this is just Fubini; for curvilinearits Federer.

Then


41/83


{}

v2

=

0

1t2 {=t} v2 ddt = 0 {=t} v2 dd1t

ibp=

1

t=t

v2 0 +0

1

t

d

dt

=tv2dt = 1

2=

v2 +

0

1

t

=tv v ddt

= t is parallel surface of boundary

2

0

1

t

=tv2v2 dt throw neg terms away & ho

ho - 2= 2

0

1

t2

=t

v2 ddt

1

2

0 =t

v

2

ddt

1

2

v2

2 2

v2

2

1

2

v21

2

,

= { }.


42/83


Wednesday, September 21 regularity estimate

(Last time) v2

2 c v2 v H10 .

If not comfortable with coarea formula do: = B1, (x) = 1 x B1

v2

1 x2 B1 v2

can reduce to polar coors

lhs = 1

0

1(1 r)2 Br v2 ddr do next steps as in Notes last class

Lemma 3 > 0 h1 Ln such that B(u, , u) u2 need for((u,u))+B[u, , u] u2;choose

=

2

!

Proof.

B[u, , u = B[u, u, ]

u u L2u2u wL2 =u uijL2 .

boundary data has curl extension; cut off interior part: = curl () H2 W1,n L and

Use to cut off

uL2 u + u L2


43/83


Figure 1.9: the bump func-tion

set = x (x) . This is the tubular neightborhood . Hassupport in annular region; has support near boundary.

u L2() +u L2() : we restrict on = exp1; L

2.max est

uL2()

+uL

2nn2Ln()

Use 12

= 1n

+ 12nn2

poincaire Sobolev embedding

u Ln() +Ln()uNow we can make the coefficient as small as possible by absolute conti-nuity:

lim0Ln() = 0.1.23 Regularity for NSE for n = 2, 3, 4, 5, 6, 7

n = 2

n = 3

n = 4 critical - papers

n = 5 supercritical - this dimension is the heart of the matter

n = 6 solved

n = 7 open problem


44/83


n=2,3 Any weak solution u

V tou + p = f u u

is smooth up to provided C and f C().n = 2

We begin with minimum regularity: u H10 . Assume f = 0.

u u =

(u u

)uj u

xj=

xj uju

xjuju

0divu=0.u H10()

sob emb (crit dim)

u Lqfor some q < + as long as finite

u u u2 Lq q < + (Lq) W1,q q = qq 1

Last time, if u + p W1,q add 2nd order regularity u

W1,q, p Lq q.Now go back:

u

Lp u

Lp=

(u u

) product Lq u W2,q p W1,q.

Go back:

(u u) = u 2uLq

+ u u W1,q Apply Stokes operator

u W3,q p W2,q u Wk,q p Wk1,q k, q

Apply Morrey embedding u C(), p C().


45/83


n = 3.

u L2 W1,2embedding in L6 for n=3

u L6.

u6

u2

L32 Not L2 lol u + p L32 u W2,32, p W1,32

in 3d, product = dim u Lq u u = (u u) W1,q u W1,q, p Lq keep going

n=4

Previous argument gets stuck:

W1,2(R4) L4 u u = u uL2

W1,2 u + p W1,2 u W1,2, p L2

u W2,4

3p W1,4

3 .

This is good because we have one more derivative. Its bad because wenow have lower integrability (!):

u W1,4

3 (R4) L2 stop!Our efforts are wasted because nothing becomes better.

Idea for n = 4

Consider instead Burgers equation (ignore pressure p):

u =

L2u u

2u = (u u) cancel 1 derivative u =

(u u

)

u

u

2

u

u

2

2

4

=

u

2

4

u4Sobolev emb

cu24 (1.12)Note that the integrability does not improve, but the power in 1.12 does


46/83


Theorem 8 Any solution u L4

(R4

)of u =

(u u

)is zero

uL4(R4) cu2L4(R4) .Then

1 cuL4(R4)uL4(R4) 1cis not true. !


47/83


Monday, October 3Lemma 4 For all m R if f Hm,2(B) & u B Hm+2 12 ,2(B) then

v + p = f v = 0

v = u on B

has a unique solution v Hm+2,2(B) andvHm+2 cfHm +uHm+2 12 ,2 (B)

[From last time - needs to be proved].For m = 1, m Z Lemma has been known via Solonikhovs theorem.For m = 2, by interpolation, it sufficies to prove the lemma for m Z.

If two integers true, then between two integers also true use standard interpolation.

u = f

2uLp1

fLp12uLp2

fLp2Then for all q [p1, p2],

2uLq

cfLq If k > 0, Hk = Hk0 Holder expontent negative, so

use duality to characterize.

m =

v Hm,2(B) divv = 0 m 0Sobolev space

v Hm

(B

) divv = 0

m < 0.

and

m =

u Hm 12 ,2(B), B u d = 0 m 12u Hm 12 (B) u, (S) = 0 m < 12 whereHm+ 12 (S) = Hm 12 (S) . . . . so

define pair


48/83


For any m, ! v m such that

v + p = 0

v = 0

v B = . (divv)() = v() .nonpositive integer case was

last class bitch!To prove above, m = k, k Z. We want to estimate

vHk(B) Crk .Decompose:

v = v1+

swhere s satisfies

() s = 0s

= on B . In order for this to be solvable,the average on the boundarymust be 0. We know that byLax-Milgram (?)

We know

sHm,2(B) Cm .v1 = u s

v1 + p = 0 in B1

v1 = 0 in B1

From the boundary condition of (*), v1 = v s on B and

v1, (S) = 0 is compatibleIt suffices to estimate

v1

Hk

(B) C

Hk

(B)

That is,

v1Hk(B) = supfHk(B)

v1, f fHk(B)


49/83


Let 0 f Hk0

(B

). Let v k+2 solve

v + q = f

v = 0v = 0

Solenekovs estimate

vHk+2(B) cfHk(B)and

v1, f

=

v1 v

+ q

= v1 v + v1, q ibpdiv v1, q divv, q=

Bv1, v

v1

, v1

= B

v1, v + v1, q0

p, v0divp,vpdivv

where is 0 because div(v1, q B v1, q)n = 0. Then

v1Hk = supfHk0f0B v1, v

fHk

0

= 1.

Continuing,

v

Hm+1

2 (B)

v1Hk 1

2 (B)vHk+2(B) vHk12 +sHk12 (B)

fHk0(B) vHm(B) +Hm12 (B)

Using

s =?

s

=


50/83


and

s Hm12 (B) I can talk about trace if I loose 1

2derivatives.

cfHk0(B)Hm12 (B) cHm 12 (B)

Stokes operator is self-dual, so operator same operator .

Conclusion: for every m R,

S m Zmis an isomorphism V.

v + p = 0

v = 0

v =

un cm(B) . interpolate true for all whole numbers.

duality interpolate negative numbers.

Using conclusion,

v + p = 0

v = 0

v B = u B H1(B).


51/83


Global integrability/ differentiablity

v H3

2 (B) W1, 83 (B.)Claimv C. Because p = 0,, p C and ddp C so v C

and v C.

Need to prove for 0 < < 12

,

v2L2(B) c4v2L2(B)v2L2(B) c4v2L2(B) .This estimate is not obvious, because of p term: Let v = v B u

average on boundary stillsatisfies boundary conditions.

Then

vL83 (B)

embedding=v u

H32 (B)

u uH1(B) Poincaire uL2(B)For q, we need estimate. Assume p = 0.

pL2(B) vL2(B) CuL2(B) . p is harmonic we can control L2 norm, control all high order deriva-

tives of harmonic fnction inside

qHm,2(B) cuL2(B) .By embedding

k+1pL(B)

k+2vL(B)

cuL2(B)

This concludes Streuwes paper reduced to the n = 4 case.


52/83


Sunday, October 28 3d regularity For n = 3, f L(0, T; H), f L1(0, T; H), u0 H2 V.

If 1 or u0 sufficiently small then ! u R3 such that

u L2(0, T; V) L(0, T; H) u L(0, T; H2())and

f L2(0, T; V) u L2 u L(0, T; H) L(0, T; V).Proof (tricky)

ut u + u utrouble term

+ p = f

u = 0

assume smooth, carry out argument

u ut u + p + (u u) = fd

dt

u22

+ 2

u

+ 2B

[u, u , u

]= 2

f, u

2

f

L2

u

L2 ,

B[u4

, u2

, u4] u2L2uL2 Cu2L2uL2

where we have used that, in 3d, u24 u 142 u 342 2poincaire

u22d

dtu22 L

+ 2 cu2if 0 then ok.

u2 u L2 2

f

L2

u

L2

If the term is not 0, we impose condition on the initial data:cu02 > positive. When we turn on evolution the term cu2


53/83


grows, so it could become negative. We estimate:

u (u u + p + u u) = f u,u2L2 = (f, u) 2(u, u)

1

fV + 2u2L2 + (u, u)

u2L2 1f2V 2(u, u).Call

d1 =

f

(0

)L2

+ c0

u0

H2 + c0

u0

2

H2 ,

d2 =fL ,d3 =

d2

+ 2u02L2 + T d2

1

2

d1

d4 =d2

+ 1 + d21u02L2 + T d2 exp

T

0f(s)

L2

3

c2

can make true for either large coefficient or small data.

Clearly d3 d4

u

(0

)2

L2 d3 d4 0.

We show that T can be extended all the way to T.

d

dtu2L2 2fL2uL2add

d

dt 1 +u

2

L2 1 +u

2

L2f

L2 as long as positive we havegronwell inequality, which tellsinitial control1 +u2L2 exp t

0f(s)

L2ds 0


54/83


Then

1 +u(t)2L2 1 + d21 expT0 f(s)L2 dsand

u d4 cu

geq0.

Contradiction cannot stop.

Need to prove that next time L

L2

.


55/83

Chapter 2

Evolution NSE

Wednesday, October 5We study Evolution NSE on

bounded domain

entire space

((u, v)) = Du Dv u, v V, u = ((u, u)) (H, ) and (V,) are Hilbert. V H inclusion map

H compact linear

H bounded linear functional H V

V H H V sort of like divergence-free:= H1 (H10); Equivalent,because its actually(V) (H1

0) = H1.

identify use RieszReprentation Theorem

2.1 Stokes operator

The space u V can be viewed as an element in the dual as follows:v is a bounded linear functional on V, so by RRT another element,

55


56/83

CHAPTER 2. EVOLUTION NSE 56

Au V such that

((u, v))=

Au, vV

,V .

For u Au, V V. Then Stokes operator is clearly a boundedlinear functional.

For u H10 , A is if drop the divergence-free condtion. Then

A u, v = Au vThis is (somehow ) like

Du Dv u, v V. Stokes operator for u

t u + p

Stokes

= f.

When unbounded, need to adjust inner product Poincaire not hold anymore. Then

[[u, v]] = (uv + Du Dv) like inner product in H1 except

u, v V instead of H1.

can define

[[u, v]] = Au, vV,Vbut A not as good as linear case:

A = + I.

2.2 Function spaces

For 1 +,

L(a, b; X ) = f (a, b)X fL(a,b;X)


57/83


C

(a, b; X

)=

f

(a, b

)X

f is continuous

A. Case [a, b]: can define norm:fC(a,b;X) = supatb

f(t)X

.

- cannot discount measure zero set (!)

-Compare with norm having ess sup not defined everywhere, so wecan discount.

Lemma. Let X Banach space with dual X and u, g L1(a, b; X).Following are equiv:

1. u a.e. integrable equal to primative function of g

u(t) = +t0

g(s)ds2. For every test function = D((a, b)),

b

au(t)(t)dt = b

ag(t)(t)dt read D as smooth with

compact support.

3. ddtu, = g, in the sense of distribution.

-If (i) (iii) hold, u is a.e. equal to a continuous function from[a, b]X after mollifying.Proof

(i) (ii) Easy to see

(iii) (ii). For all D((a, b)), true in sense of distributions:

b

au, (t)dt = b

ag, (t) dt

u

(t

), u

(t

)+ g

(t

)

(t

)

dt = 0 or get rid of t var.

ba u(t)(t) + g(t)(t) dt,must =0

ba

b(u(t)(t) + g(t)(t)) dt = 0which proves (iii) (ii).


58/83


(ii) (i). Suppose g 0.

need to prove that u(t) = + g(s)gone

= constant. Then

u = 0.

Let 0 D((a, b)) be such that the mass ba 0(t)dt = 1. Then

b

a( 0

compact support/smooth

)(t)dt = 0. D

((a, b

))such that

( 0

)(t) = ta ((t) 0())d Use (ii).

b

au(t)(t)dt = 0.For g = 0 and =

b

au(t)((t)dt) = b

a(t)dt

ba (t)dt

ba

u(t)0(t)dx

0= constant

b

a u() =0

(t)dt = 0 u(t) = 0 a.e. If g 0, look at v0(t) = ta g()d. Then

b

a(u v0)(t)dt = 0 u v0 =

so v0 = g(t). If f L1(a, b; X) satisfies

b

a f(t

)

(t

)dt =

0 C

(a, b

;R

)Since, actually the spatial

variation is the Banach space

X, he replaces R with X.thenf = 0 a.e.


59/83


Friday, October 7

For u L2(0, t; V) u( 0, Tlike L2

;H1

V ), thenu C([0, T], H) like L2 L2 but preserves

continuity.

d

dtu2 = u(t), u(t)

V,V L1(0, T) on real number function (not

X). u2 (t) W1,1(0, T) C([0, T])

Stokes systemut u + p = f

u = 0u = 0 = u0

has two cond: u0 = 0u0 = 0

match initial data. cannot talk a priori about trace. But with regu-larity above we can.

Question: for f L2(0, T; V) , u0 H compatibility conditionare given, we seek u L2

(0, T; V

)such that

ddt(u, v) + ((u, v)) = f, v v V

u(0) = u0.2.3 Galerkin

solve using limiting process

adjust nonlinearity to look more linear.

Let {wi}

i=1 be complete (but not orthogonal) base ofV & orthonormalbase in H.

(wij)L2 = ij for i j((wi, wj)) = 0 otherwise.


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Vm = span

{wi

}

i=1 m.

Find mi=1 gim(t)w(x) such thatd

dt(um, wj) + ((um, wj)) = f , wj not fm tail disappears.

The boundary u(0) = u0:um(0) = um0project= m

i=1

(u0, wi)wi

(um, wj

)= gim

(t

)wi, wj

orthon.((um, wj))H10

inner product

=

m

j=1

gim(t)((wi, wj)) = mj=1

ijgim(t) where ij = ((wi, wj))Fj(t) = f, wj gim(0) = (u0, wi)

and

gim(t) = mj=1

jigjm(t) = Fi(t)f or l i mgim(

0

)=

(u

0, w

i).

d

dt

g1m

gmm

+

11 m1

1m mm

g1m

gmm

=

f1

fm

X + AX = B(t)X(0) = X0

Note: We do not need to use Brouwer fixpoint here, because we havetime. time is actually nice right?

take limit: Clearly {um} L2(0, T; Vm).


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how to take limit? Use compactness. Compactness introduces some

form of control (trapped in a ball) - use apriori estimate.

d

dtum2 = 2 d

dtum, um ddt(um,uj) = (um, ddtuj)

d

dt1

2um2 = d

dtum, um = d

dtum, um = dum

dt, wj

Using that 2ab a2 + 1

b2,

d

dt 1

2 um2 + ((um, um)) = f, umd

dtum2 + 2um2 schwarz 1

f2V + um2V

um2 (t) + t0um()2 d 1

t

0f2V d m

sup0tT

um2H(t) + T0um() d um0 2 + 1

t

0f2V d Lax-Hopf condition

so (or where)um L

t L

2x L

2t H

1x

Then

um2L(0,T;H) + um2L2(0,T;V) u02H + 1t

0f(t)2

Vdt

pass limit. cf Heat equation.


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Monday, October 10

2.4 Pass to limits

in finite dimension, weak convergence & strong convergence samething

take subsequence , we may Without loss of generality v L(0, T; H)and w L2(0, T; V) such that um v in L ( 0, T ; H ) & um win L2(0, T; V) = (L1);

is Banach sp but not H-sp;is H-sp

B1

L is weak star compact 1. This means:

um, v1, i.e. um, w, L2(0, T; v) um,umL 1 u L such that um u L1, um, u, or

um u L1(0, T; H) Claim: v = w.

Note for T < +, L2(0, T; V) L1(0, T; H).If I pick an element in the smaller test space, then L2 = L1 - they

are the same. Remember that we can cook up a sequence having 2different limits in different spaces, for example

sinxn 0 in L2

but

sinxn

where is the Dirac mass in the Radon measure space M

([0, 1

]).

1The weak star topolgy is by definition, the weakest one that makes all functionals

x x, xcontinuous


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So

v,

=

w,

L2

(0, T; V

)

v = w

.

pass to limit: ddt

um, wi + (( umL2uwi

,

fixedwi) ) = f, wi

f,wi

.

The term is converges:

d

dt

wm, wi

w, wi

.

Distributions of convergent sequences convergeIf function convergent Distribution also convergent. More-over, the distribution converges in the distributional derivativesense

d

dtfi

d

dtf

Shift test function to see it converges:

ddt

f = f ddt

.

Then the sequence um0 is convergent,

um0 =m

i=0

gim(0)wi u0 in H. Then

d

dt(u, wi) + ((u, wi)) = f, wi

d

dt(u, v) + ((u, wi)) = f, wi for all v V.

v satisfies the equation in the distributional sense and the boundaryvalue

u = 0

is kosher because of the yellow boxed explanation above.


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initial-time condition holds:

u t= 0 = u0Show: has well-defined limit in L2-sp. u C([0, 1], H) u inH-norm as function of t has well-defined limit.

pf Usingd

dt= u2 = 2 u, uV,V read in V, V duel

in the sense of distribution in (0, T). Once prove this then prove that L2 norm in X-var is continuous in

t-var.

Notice that2 u, uV,V L1(0, T)

by Holder.

u H, u2 L1(0, T), ddtu2 L1(0, T)

u2 (t) W1,1(0, T) embedding

C1

2 (0, T)

u2 (t) u2 (t0)ftc= 2 tt0 u, u (s) dS 0 as t t0 by absolutecontinuity, because the L2-norm

is continuous.. .u2L2 (t) u2 as t t0 u(t) u(t0)2 0

The last convergence holds because u u(t0) < u(t)2 +u(t0)2 2 u(t), u(t0) 0 2 u(t0)2 2 u(t0), u(t0) = 0. Leray-Hopf type energy inequality:

sup0tT

u(t)2 + T0um2 dt

lower semi-continuity

u02 + 1

T

0f(t)2

Vdt


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Lower semicontinuity

weaker than continuity

A function f is upper (lower) semi-continuous at a point x0 if,roughly speaking, the function values for arguments near x0 areeither close to f(x0) or less than f(x0).

A lower semi-continuous function. The solid red dot indicatesf(x0)

In particular, f is lower semi-continuous at x0 if for every > 0there exists a neighborhood U of x0 such that f

(x

) f

(x0

)

for all x U. Equivalently, this can be expressed as

liminfxx0

f(x) f(x0) A function is lower semi-continuous iff x X f(x) > is an

open set for every R.

A function is continuous at x0 iff it is upper and lower semi-continuous there.

Using Stokes operator

u

= Au + ff H1, Au L2(0, T , V ), f L2(0, T; V) so

u L(0, T; H) L2(0, T; V)Stability implies uniquenessLike Heat equation: insteadof laplace operator its Stokesoperator.


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Prove: ddt

u

2

= 2

u, u

V,V

C(0, T) L2 dense in L2C(0, T; V) L2(0, T; V)dense

C(0, T; V) L2(0, T; V)dense.In fact, mollification in t-var um = u {vm} C such that

vm u in L2(0, T; V)

and

vm u in L2(0, T; V) use 1 sequence to do bothmollificationUse Holder

2 um, vm u, u in L1(0, T)using that d

dtvm2 ddt u2.

2.5 Uniqueness

2.6 Stability Properties


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Wednesday, October 12


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Figure 2.1: jump andextension

Wednesday, October 19Suppose u L2(0, T; X0) u L1(0, T; X1) L2. Then there exists

< 12

such that () Dt u L2(0, T; X1) use Fourier transform

u =

u t [0, T]0 otherwise

Then

uu = u[0,T]g

+u(0)0 u(T)T extension generates Fourier transform can be applied to distribution.

2iu() = g + u(0) 1 u()e2iT Then g L2.0

(T

)=

e2i0

(t

)dt

Condition (*) is equivalent to, in terms of u,


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R 2 u()2X1 d < +for every L1 function, f() = e2itT is bounded, and L1 L1 L H2 by Holder: f2

f f f2L2 fL1fL

.

2 u()2X1

g(t)2X1

bounded

+

just numbers - boundedu(0)2X1

+u()2X1

C

cannot integrate; derivatives in L2

, but this term not L2

so sacrificederivatives

2 2(1 + 2)1 + 2(1)

2 + 2 2(1 + 2) 2 < 2 + 2.We need to estimate

R

2u(t)2X1

d R

1 + 2

1 + (2)1 u()2X1split

11 +

2(1)

1 u

(

)2

X1

parseval: Ru2X1d +

R

2

1 + 2(1)

u

X1

d

R

12

1 + 2(1)1

g2X1 + CL2

R

g()2X1

+R

C1+(1)

d singularity at

but0 12(1) d


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recall Trilinear form B

[u

L103

, v

L2

, w

L103

]=

(u v

) w u, v, w V is

well defined for n 4: after Holder Ln not preserved.

B[u, v] V R. B[u] = B[u, u] NS can be reformulated into

utV

AuH1

+ BuV

= fV

where it follows that ut V (after looking at where other terms are).Note that V is integrable in t.

Then the regularity ofu is improved: u Cfalse, not continuous!!

(0, T; H), byAlbin-Lions above (AL: V H V)

Error (from side note above). u L2

Au L2(0, T; V), f L2(0, T; V), BuV u2V ,BuV =B[u, u]V supuV

CuuVwVso

Bu

V is L

1 integrable.

not continuous. u C([0, T; Hweak]. weak continuous can stilltake weak limit. For all H, u(t), C([0, T]).

nonlinear terms are trouble; take care ofBu , being nonlinear (because

of square): T0 BuV T0 u2V.Proof (Galerkin & Aubin-Lions Lem).

{wi}i=1, Vm = span{w1, , . . . , wm}, um = mi=1 gim(t)wi(x). Solve d

dt

(um, wj

)+

((um, wj

))Dirichlet inner product+ B

[um, um, wj

]=

f

(t

), wj

.

truncate initial data: um(0) = projVm(u0) = u0m, where u0m = mi=1(u0, wi)wi


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Figure 2.2: solution couldblow up if [0, T]

nonlin. ODE:

mi=1(wi, wj) gim(t) + mi=1

((wi, wj))g(t) + mi=1,

Bwi, w, wjgim(t)gm(t)where the term is nondegenerate because it forms a base. non-singular. (wi, wj)1i, jm GL(m).

invert: gim +mj=1 ijgjm(t)+mj,k=1 ijkgjmgkm = mj=1 ij f, wj whereij, ij, ijk R.

nonlinear, cannot solve globally (like X = X2.)

can solve locally; Fixpoint, X = F(t, X). fm > 0 & gim(0, fm) R

If tm is the maximum value then gim(t)

as tm

; however,

ddt

um, umt((um, um)) + Bum, um, um=0

= f, um . here have cancelation, so even though the equation is nonlinear, the

estimate we have is still linear (!)

d

dtum2 + 2um2 fVumV

norm does not blow up, so we continue

um2V + 1f2V


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um(t

)2

bounded+

T

0 um(

)2

Vdt

um(0)2

+1

t

0 f(t

)2

Vdt

orthon. proj

u02

+1

t

0 f(

)2

Vd

so gim(0) = (u0, wi).


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Friday, October 21 Galerkin - finish convergence

uniqueness for m = 2.

Leray-Hopf solution for n = 3

regularity.

Suppose u L(0, T; H) L2(0, T; V) solvesut u + u u + p = 0

u = 0

u t=0= u0. zeroand suppose (mult by u, integrate) you can have turbulance and

still have regularity. . .

R3u(t) + 2T

0R3u2 dxdt

R3u02 .

um:

um

2

+ T

0R3

um

2

u0

+

1

t

0

f

2

V generalized energy inequalty.

This says:

umL(0,T;H) +um(0,T;V) C. pass to limit, possible for the subsequence.

um u in L(0, T; H)

, where is weak star convergence because of the L, and

um u in L2(0, T; V)


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um = fm where fm = f Aum

L2

(0,T;V

)

Bum

B[um,um]V2

In fact, B

[um, um

] V:

BumV supwV=1B[um,um,w]

= supwV=1

R

u um w

Sobolev

um22 ,so then Bum L1(0, t; V).Moreover T0 BumV dt T0 um22 dt C. Upshot:

fm L1(0, T; V).

have better control of derivatives,., so um u in L2(0, T; H). ?? Of course need to apply Albin-Lions; need to control time deriva-

tives, so

umL1(0,T;V) C time derivative bounded (!)where we used

R

um um w um4um2u4 Cum22 = Cum2 . so um L2(0, T; V) & um L1(0, T; V) L2(0, T; H) apply Albin-Lions, withV H V indeed, is L2 compact.

Then um u in L2(0, T; H) strong convergence inH-space.

dumdt

, wj + ((um, wj))((um,wj))

+ B[um, um, wj] = f, wjconstantconverge

where converge because um uumt

ut

in distribution & w is

smooth

ut

, wj

in L1 by lower semicontinuity.

It suffices to show

B[um, um, wj] B[u, u, wj].Note that um um wj where u in L2.


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umwj uwj (without wj strong convergence.)

umwj uwj2 =(um u)wj2 wjum u2 0which proves existence

Need to prove:

u L2(0, T; V), u L1(0, T; V) C(0, T; H). initial data for approx solution pass to limit (will not repeat; see

previous Notes).

dudt

, v + ((u, v)) + B[u, u, v = f, v] v V

du

dt A + Bu = f

u(0) = u0.

The existence argument is due to Lions. E. Hopf

Navier-Stokes Seminar

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