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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 5
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 6
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 7
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 8
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 9
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 10
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 11
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 12
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 13
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 14
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 15
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 16
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 17
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 18
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 19
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 20
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 21
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 22
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 23
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 24
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 25
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 26
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 27
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 28
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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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 29
(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. !
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 30
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.
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 31
= [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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 32
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.
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 33
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 34
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 35
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 36
Verify that f H1. Suppose H10 . Then
, well-defined =
Ln
L2 H1
0
Ln L2 H10
H1.
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 37
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 38
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)
!
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 39
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 40
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 41
{}
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
,
= { }.
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 42
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 43
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 44
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().
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 45
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 46
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. !
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 47
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 48
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)
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 49
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
=
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 50
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).
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 51
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.
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 52
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 53
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
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CHAPTER 1. LINEAR & NONLINEAR STATIC NSE 54
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
.
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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
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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)
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CHAPTER 2. EVOLUTION NSE 57
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).
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CHAPTER 2. EVOLUTION NSE 58
(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.
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CHAPTER 2. EVOLUTION NSE 59
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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CHAPTER 2. EVOLUTION NSE 60
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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CHAPTER 2. EVOLUTION NSE 61
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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CHAPTER 2. EVOLUTION NSE 62
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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CHAPTER 2. EVOLUTION NSE 63
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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CHAPTER 2. EVOLUTION NSE 64
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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CHAPTER 2. EVOLUTION NSE 65
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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CHAPTER 2. EVOLUTION NSE 66
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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CHAPTER 2. EVOLUTION NSE 67
Wednesday, October 12
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CHAPTER 2. EVOLUTION NSE 68
Wednesday, October 14
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CHAPTER 2. EVOLUTION NSE 69
Wednesday, October 17
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CHAPTER 2. EVOLUTION NSE 70
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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CHAPTER 2. EVOLUTION NSE 71
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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CHAPTER 2. EVOLUTION NSE 72
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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CHAPTER 2. EVOLUTION NSE 73
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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CHAPTER 2. EVOLUTION NSE 74
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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CHAPTER 2. EVOLUTION NSE 75
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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CHAPTER 2. EVOLUTION NSE 76
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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CHAPTER 2. EVOLUTION NSE 77
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
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