Introduction to the Theory of the Navier–Stokes Equations for Incompressible...

Introduction to the Theory of the Navier–StokesEquations for Incompressible Fluid

Jirı Neustupa

Mathematical Institute of the Czech Academy of Sciences

Prague, Czech Republic

A mini–course, part I

Tata Institute for Fundamental Research, Bangalore

Centre of Applicable Mathematics

June 4–12, 2014

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Part I – Contents

1. Basic notions, equations and function spaces(a physical background, theNavier–Stokes equations, function spaceL2

σ(Ω), Helmholtz decomposition)

2. Weak solution to the Navier–Stokes equations I(first observations and defini-tion)

3. The Stokes problem (steady and non–steady Stokes’ problem, weak and strongsolutions, the Stokes operator)

4. Weak solution to the Navier–Stokes equations II(other equivalent definitions,subtler properties)

5. Global in time existence of the so called Leray–Hopf weak solution(principles of the proof by Galerkin’s method)

References

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1. Basic notions, equations and function spaces

A physical background

We assume thatthe fluid is a continuum. We denote

v = v(x, t) . . . thevelocity, p = p(x, t) . . . thepressure,

ρ = ρ(x, t) . . . thedensity, θ = θ(x, t) . . . thetemperature

Stokes’ postulatesfor the fluid (19th century) require that

a) the stress tensorT depends on the velocity only through the stretching tensor(= rate of strain tensor, rate of deformation tensor)D := (∇v)sym,

b) the stress tensorT does not explicitly depend on positionx and timet,

c) the continuum is isotropic, i.e. it contains no preferred directions,

d) if the fluid is at rest thenT is a multiple of the identity tensorI by a scalar.

1. Basic notions, equations and function spaces 3 / 50

Nowadays,the postulates are usually formulated in another way:

a) the stress tensorT depends on the velocity only through the stretching tensorD,

b) the stress tensorT does not explicitly depend on positionx and timet,

c’) the way tensorT depends on tensorD is material frame indifferent.

Postulate c’ means that IF

Q = Q(t) is an arbitrary unitary matrix,

x∗ = c(t) +Q(t) · (x− x) is another observer’s frame,

T andD have the representationsT∗ andD∗ in framex∗

THEN T∗ depends onD∗ in the same way asT depends onD.


One can derive from these postulates that

T = −p I+ Td,

whereTd is the “dynamic stress tensor” (it equals the zero tensor if the fluid is atrest), andTd depends on tensorD through the formula

Td = α I+ β D+ γ D2.

The coefficientsα, β andγ may generally depend on the so called state variables(pressure, density, temperature) and on the principal invariants of tensorD.

Ideal fluid (= inviscid fluid): Td = O, i.e. T = −p I

Newtonian fluid: T depends linearly onD. One can deduce from this assumption(also using Kirchhoff’s formulap = −1

3 T + µ′ div v, whereµ′ is the so calledcoefficient of bulk viscosity) that

T =[−p+ (µ′ − 2

3µ) div v]I+ 2µD, (1.1)

whereµ is thedynamic coefficient of viscosity.


Condition of incompressibility: div v = 0 (1.2)

Condition (1.2) expresses the fact that each part of the fluid preserves its volume.

Conservation of mass: ∂tρ+ div (ρv) = 0 . . . equation of continuity (1.3)

Note that condition (1.2) follows from (1.3) in the special case whenρ = const. Dueto this and also other historical reasons, equation (1.2) is often calledequation ofcontinuity for incompressible fluid.

However, note that there exist incompressible fluids with non–constant density (e.g.mixtures of several liquids). In these cases, (1.3) reduces to the transport equationfor densityρ:

∂tρ+ v · ∇ρ = 0.

Conservation of momentum: ρ ∂tv + ρv · ∇v = −DivT+ ρf (1.4)


From now, we use the IMPORTANT ASSUMPTIONS:

• the fluid is incompressible with the constant densityρ = const. = 1,

• the fluid is Newtonian,

• µ = const. > 0.

We denoteν := µ/ρ = µ . . . thekinematic coefficient of viscosity.

Formula (1.1) now reduces toT = −p I+ 2ν D.

Substituting this to (1.4), we obtain the so calledNavier–Stokes equation forviscous incompressible fluid(H. Navier 1824, G. Stokes 1845)

∂tv + v · ∇v = −∇p+ ν∆v + f .

This vectorial equation is equivalent to the system of three scalar equations

∂vi + vj ∂jvi = −∂ip+ ν∆vi + fi (i = 1, 2, 3).


Boundary conditionson a fixed material boundary:

the no–slip condition: v = 0, (1.5)

Navier’s slip condition: [T · n]τ + γvτ = 0. (1.6)

Here,τ denotes the tangential component andn denotes the outer normal vector onthe boundary. Condition (1.6) is used together with

the condition of impermeability: v · n = 0. (1.7)

γ . . . the coefficient of friction between the fluid and the wall

γ →∞ . . . conditions (1.6) and (1.7) lead to (1.5)

γ = 0 . . . the so calledfree slip

Further, we mostly consider the no–slip condition (1.5).


The Navier–Stokes initial–boundary value problem

Ω . . . a domain inR3 . . . the domain where we consider the motion of the fluid

(0, T ) . . . the time interval (0 < T ≤ ∞)

QT := Ω× (0, T )

Thus,the Navier–Stokes initial–boundary value problemwe deal with is:

∂tv + v · ∇v = −∇p+ ν∆v + f in QT , (1.8)

div v = 0 in QT , (1.9)

v = 0 on∂Ω× (0, T ), (1.10)

v = v0 in Ω× 0. (1.11)


Function spaceL2σ(Ω)

Let C∞0,σ(Ω) be the linear space of all infinitely differentiable divergence–free vectorfunctions inΩ with a compact support inΩ. We denote byL2

σ(Ω) the closure ofC∞0,σ(Ω) in L2(Ω).

Remark. Assume thatΩ has a locally Lipschitzian boundary.

Then the spaceL2σ(Ω) can be characterized as a space of functions fromL2(Ω),

whose divergence equals zero inΩ (in the sense of distributions), whose normalcomponent on∂Ω) is zero (as an element of the spaceW−1/2,2(∂Ω)) in the sense oftraces. (See e.g. [1] or [11].)

Helmholtz decomposition(see e.g. [1] or [11] for more details):

Lemma. Let Ω be any domain inR3. Then

L2σ(Ω)⊥ = G2(Ω) :=

w ∈ L2(Ω); w = ∇ϕ for someϕ ∈ W 1,2

loc (Ω).

Consequently,L2(Ω) = L2σ(Ω) ⊕G2(Ω), whereL2

σ(Ω) andG2(Ω) are closed or-thogonal subspaces ofL2(Ω).


A part of the proof: Only the part ⊃ . (See e.g. [1] or [11] for the oppositeinclusion.) Thus, letw = ∇ϕ ∈ G2(Ω). It suffices to show that(u,w)2 = 0 for allu ∈ C∞0,σ(Ω). However,

(u,w)2 =

∫Ω

u · ∇ϕ dx = −∫

Ωdiv u ϕ dx = 0.

The orthogonal projection ofL2(Ω) ontoL2σ(Ω) is called theHelmholtz projection.

It is usually denoted byP2, orP 2σ or only byPσ.

Remark. Letv ∈ L2(Ω). The Helmholtz decomposition ofv is: v = Pσv +∇ϕwhereϕ is a weak solution of the Neumann problem

∆ϕ = div v in Ω,∂ϕ

∂n= v · n on∂Ω.

Remark. An analogous decomposition inLq(Ω) (for 1 < q < ∞) is possible⇐⇒this Neumann problem has a weak solutionϕ in D1,q′(Ω) := w ∈ L1

loc(Ω); ∇w ∈Lq′(Ω), where1/q + 1/q′ = 1.


2. Weak solution to the Navier–Stokes equations I

(First observations and definition)

Further important function spaces:W1,2

0,σ(Ω) := W1,20 (Ω) ∩ L2

σ(Ω) (a closed subspace ofW1,20 (Ω)), dense inL2

σ(Ω))

W−1,20,σ (Ω) . . . the dual toW1,2

0,σ(Ω)

Formal considerations (steps 1 and 2)

Step 1: the integral equation. Let φ be an arbitrary test function from the spaceC∞0([0, T ); W1,2

0,σ(Ω)). If we multiply (1.8) byφ, integrate inΩ, and apply the

integration by parts, we obtain the integral equation

∫ T

0

∫Ω

[v · ∂tφ− ν∇v : ∇φ− v · ∇v · φ

]dx dt

= −∫ T

0〈f ,φ〉 dt−

∫Ω

v0 · φ(0) dx

(2.1)

2. Weak solution to the Navier–Stokes equations I 12 / 50

Step 2: a priori estimates. Let us multiply formally the Navier–Stokes equation(1.8) by functionv, integrate inΩ, and apply the integration by parts. We obtain

d

dt

1

2

∫Ω|v|2 dx + ν

∫Ω

∆v · v dx =

∫Ω

f · v dx,

d

dt

1

2‖v‖2

2 + ν ‖∇v‖22 = (f ,v)2 ≤ ‖f‖−1,2 ‖v‖1,2

≤ C ‖f‖−1,2 ‖∇v‖2 ≤ν

2‖∇v‖2

2 + C(ν) ‖f‖2−1,2 ,

‖v(t)‖22 + ν

∫ t

0‖∇v‖2

2 dτ ≤ ‖v0‖22 +

∫ t

0‖f‖2

−1,2 dτ

≤ ‖v0‖22 + C(ν)

∫ T

0‖f‖2

−1,2 dτ

Omitting at first the second term on the left hand side, we get

‖v(t)‖22 ≤ ‖v0‖2

2 + C(ν)

∫ T

0‖f‖2

−1,2 dτ for all t ∈ (0, T ).

This a priori estimate indicates that solutionv should be inL∞(0, T ; L2σ(Ω)).


Then, omitting the first term on the left hand side and consideringt = T , we obtain

ν

∫ T

0‖∇v‖2

2 dτ ≤ ‖v0‖22 + C(ν)

∫ T

0‖f‖2

−1,2 dτ.

This a priori estimate indicates that solutionv should be inL2(0, T ; W1,20,σ(Ω)).

Both the estimates requirev0 ∈ L2σ(Ω) andf ∈ L2(0, T ; W−1,2

0,σ (Ω)).

Thus, we arrive at the definition:

A weak formulation of the Navier–Stokes problem (1.8)–(1.11).Let v0 ∈ L2σ(Ω)

andf ∈ L2(0, T ; W−1,20,σ (Ω)). A vector functionv ∈ L2(0, T ; W1,2

0,σ(Ω)) ∩ L∞(0, T ;L2σ(Ω)) is said to be aweak solutionof the problem (1.8)–(1.11) if it satisfies integral

equation (2.1) for allφ ∈ C∞0([0, T ); W1,2

0,σ(Ω)).


Remark. In order to show that no important information on the solution waslost when passing from the “classical formulation” (1.8)–(1.11) of the consideredNavier–Stokes initial–boundary value problem to the weak formulation, we assumethatv is a “sufficiently smooth” weak solution. Applying the backward integrationby parts to (2.1), we get∫Ω[v(0)− v0] · φ(0) dx +

∫ T

0

∫Ω

[∂tv + v · ∇v − ν∆v − f

]· φ dx dt = 0 (2.2)

for all φ ∈ C∞0([0, T ); W1,2

0,σ(Ω)). Considering at firstφ ∈ C∞0

((0, T ); W1,2

0,σ(Ω))

(which guarantees thatφ(0) = 0), we obtain∫ T

0

∫Ω

[∂tv + v · ∇v − ν∆v − f

]· φ dx dt = 0.

Hence∂tv + v · ∇v − ν∆v − f ∈ G2(Ω) for a.a.t ∈ (0, T ). Consequently, thereexistsp (also depending ont) such that

∂tv + v · ∇v − ν∆v − f = −∇p,

which is equation (1.8).


Now we return toφ ∈ C∞0([0, T ); W1,2

0,σ(Ω))

and to equation (2.2). It gives

0 =

∫Ω[v(0)− v0] · φ(0) dx +

∫ T

0

∫Ω

[∂tv + v · ∇v − ν∆v − f

]· φ dx dt

=

∫Ω[v(0)− v0] · φ(0) dx−

∫ T

0

∫Ω∇p · φ dx dt

=

∫Ω[v(0)− v0] · φ(0) dx.

As this holds for allφ ∈ C∞0([0, T ); W1,2

0,σ(Ω)), i.e. for allφ(0) ∈ W1,2

0,σ(Ω), weobtainv(0) = v0.

Equation (1.9) (i.e.div v = 0) is satisfied due to the conditionv ∈ L∞(0, T ; L2σ(Ω)).

Boundary condition (1.10) (i.e.v = 0 on ∂Ω × (0, T )) holds due to the conditionv ∈ L2(0, T ; W1,2

0,σ(Ω)).

Conclusion: A “sufficiently smooth” weak solutionv satisfies the system (1.8)–(1.11) in a classical sense.


3. The Stokes problem (see e.g. Galdi [1], Sohr [11] or Temam [12])

The steady Stokes problem

−ν∆v = −∇p+ f in Ω, (3.1)

div v = 0 in Ω, (3.2)

v = 0 on∂Ω. (3.3)

Recall the function spaces

W1,20,σ(Ω) := W1,2

0 (Ω) ∩ L2σ(Ω)

W−1,20,σ (Ω) . . . the dual toW1,2

0,σ(Ω)

Remark. As W1,20,σ(Ω) → L2

σ(Ω), we haveL2σ(Ω) ⊂W−1,2

0,σ (Ω). If g ∈ L2σ(Ω) then

〈g,w〉 := (g,w)2 for all w ∈W1,20,σ(Ω).

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A weak solution formulation of the problem (3.1)–(3.3). Let f ∈ W−1,20,σ (Ω).

Functionv ∈W1,20,σ(Ω) is said to be aweak solutionof the problem (3.1)–(3.3) if

ν (∇v,∇w)2 ≡ ν

∫Ω∇v : ∇w dx = 〈f ,w〉 ∀w ∈W1,2

0,σ(Ω). (3.4)

Remark. (3.4) follows formally from (3.1)–(3.3) if we multiply (3.1) byw andintegrate inΩ. On the other hand, ifv is a “smooth” weak solution, then we can wecan apply the backward integration by parts to (3.4) and get

ν

∫Ω[ν∆v + f ] ·w dx = 0 ∀w ∈W1,2

0,σ(Ω).

Henceν∆v + f ∈ G2(Ω). Thus, there existsp such thatν∆v + f = ∇p, which isequation (3.1).

Operator A: Define a linear operatorA : W1,20,σ(Ω)→W−1,2

0,σ (Ω) by the equation

〈Av,w〉 = (∇v,∇w)2 ∀w ∈W1,20,σ(Ω).

Now, (3.4) is equivalent to the equationνAv = f (in spaceW−1,20,σ (Ω)).


Basic properties of operatorA:

• D(A) = W1,20,σ(Ω) (follows from the definition ofA)

• OperatorA is 1–1.(v ∈ N(A) =⇒ ∀w ∈W1,2

0,σ(Ω) : (∇v,∇w)2 = 0 =⇒ ‖∇v‖2 = 0 =⇒ v = 0)

• OperatorA is bounded (as an operator fromW1,20,σ(Ω) to W−1,2

0,σ (Ω)).(‖Av‖−1,2 = sup

w∈W1,20,σ(Ω), w 6=0

|〈Av,w〉|‖w‖1,2

= supw∈W1,2

0,σ(Ω), w 6=0

|(∇v, ∇w)2|‖w‖1,2

≤ ‖∇v‖2

)

• The range ofA need not be generally the whole spaceW−1,20,σ (Ω).

Proof. OperatorA is closed because its domain is the wholeW1,20,σ(Ω) and it is

bounded. HenceA−1 is also closed.By contradiction: Assume thatR(A) =D(A−1) = W−1,2

0,σ (Ω). Then operatorA−1 is bounded (by the closed graph theorem).


Choosezn ∈W1,20,σ(Ω) so that‖∇zn‖2 → 0 and‖zn‖2 → 1. (This choice is possible

e.g. if Ω is an exterior domain orΩ = R3.) Let fn ∈ W−1,2

0,σ (Ω) be defined by theequation

〈fn,w〉 := (∇zn,∇w)2 + (zn,w)2 ∀w ∈W1,20,σ(Ω). (3.5)

Thenfn is a bounded sequence inW−1,20,σ (Ω). Put un := A−1fn. It means that

fn = Aun. Hence

〈fn,w〉 := (∇un,∇w)2 ∀w ∈W1,20,σ(Ω). (3.6)

Equations (3.5) and (3.6) (withw = zn yield

(∇zn,∇zn)2 + (zn, zn)2 = (∇un,∇zn)2 ≤ ‖∇un‖2 ‖∇zn‖2

The left hand side tends to one, while the right hand side tends to zero (forn→∞).This is the contradiction.


• The range ofA need not generally containL2σ(Ω).

Proof. By contradiction, assume thatL2σ(Ω) ⊂ R(A). Denote byA the restriction

of A toA−1(L2σ(Ω)). OperatorA is closed as an operator fromW1,2

0,σ(Ω) to L2σ(Ω).

(This can be proven similarly as the fact thatA is closed.) HenceA−1 is a boundedoperator fromL2

σ(Ω) to W1,20,σ(Ω) (by the closed graph theorem), becauseD(A−1) =

L2σ(Ω).

f ∈ L2σ(Ω) =⇒ f ∈W−1,2

0,σ (Ω), 〈f ,w〉 = (f ,w)2 ∀w ∈W1,20,σ(Ω)

L2σ(Ω) ⊂ R(A) =⇒ ∃u ∈W1,2

0,σ(Ω) : 〈f ,w〉 = (∇u,∇w)2 ∀w ∈W1,20,σ(Ω)

Hence (f ,w)2 = (∇u,∇w)2 ∀w ∈W1,20,σ(Ω), whereu = A−1f .

Choosefn in W1,20,σ(Ω) so that‖fn‖2 → 1 and‖∇fn‖2 → 0. (A sequence with

these properties exists e.g. ifΩ is an exterior domain orΩ = R3.) Putwn = fn. Then

‖fn‖2 = (∇un,∇fn)2 ≤ ‖∇un‖2 ‖fn‖2,

whereun = A−1fn. The right hand side tends to zero, while the left hand side tendsto one (forn→∞). This is a contradiction.


Domain and range of operatorA:

W1,20,σ(Ω) W1,2

0,σ(Ω)

L2σ(Ω) W−1,2

0,σ (Ω)

R(A)

A

D(A) = W1,20,σ(Ω)


W1,20,σ(Ω) W1,2

0,σ(Ω)

L2σ(Ω)

W−1,20,σ (Ω) = R(A)

A

D(A) = W1,20,σ(Ω)

The special case ofa bounded domainΩ:

• If Ω is bounded thenR(A) = W−1,20,σ (Ω).

Proof. The scalar product(∇v,∇w)2 is equivalent to the scalar product(v,w)1,2

in W1,20,σ(Ω). Hence, givenf ∈ W−1,2

0,σ (Ω), there existsu ∈ W1,20,σ(Ω) such that

〈f ,w〉 = (∇u,∇w)2 for all w ∈ W1,20,σ(Ω) (by the Riesz lemma). It means that

f = Au (the identity inW−1,20,σ (Ω)).

Corollary: If Ω is bounded thenA−1 is bounded fromW−1,20,σ (Ω) to W1,2

0,σ(Ω).


D(A) ⊂W1,20,σ(Ω)

A

R(A)

W1,20,σ(Ω)

W1,20,σ(Ω)

L2σ(Ω) W−1,2

0,σ (Ω)

Denote byA the part of operatorA with the rangeR(A) ∩ L2σ(Ω). Thus,A is the

restriction ofA to

D(A) :=u ∈W1,2

0,σ(Ω); Au ∈ L2σ(Ω)

= A−1[R(A) ∩ L2

σ(Ω)].

OperatorA is an operator inL2σ(Ω). It is often called theStokes operator.


Some properties of operatorA: (see, e.g., [11])

• A is a 1–1 positive and self–adjoint operator inL2σ(Ω). Its domain satisfies the

inclusionsC∞0,σ(Ω) ⊂ D(A) ⊂W1,20,σ(Ω).

• If f ∈ L2σ(Ω) then the steady Stokes problem is equivalent toνAv = f . It

means that there existsp ∈ L2loc(Ω) (unique up to an additive constant), such that

−ν∆v +∇p = f (in the sense of distributions inΩ). (3.7)

Principle of the proof. The equationνAv = f means that

ν (∇v,∇w)2 = (f ,w)2 ∀w ∈W1,20,σ(Ω).

Hence ν 〈−∆v,w〉 = (f ,w)2 ∀w ∈W1,20,σ(Ω),

〈−ν∆v − f ,w〉 = 0 ∀w ∈W1,20,σ(Ω).

Then there existsp ∈ L2loc(Ω) such thatν∆v − f = ∇p (in the sense of distribu-

tions inΩ); see [11].


• If Ω is bounded thenR(A) ≡ D(A−1) = L2σ(Ω) and operatorA−1 is bounded

fromL2σ(Ω) to W1,2

0,σ(Ω).

Proof. 1st possibility: If u ∈ D(A) then (Au,w)2 = (∇u,∇w)2 ∀w ∈W1,2

0,σ(Ω). Hence‖∇u‖22 ≤ ‖Au‖2 ‖u‖2 ≤ c ‖Au‖2 ‖∇u‖2. (Constantc is the

constant from Poincare’s inequality‖u‖2 ≤ c ‖∇u‖2.) This yields

‖∇A−1f‖2 ≤ c ‖f‖2 for f = Au.

2nd possibility: by the closed graph theorem

• If Ω is a boundedC2–domain thenD(A) = W1,20,σ(Ω) ∩W2,2(Ω), A = −Pσ∆,

and‖u‖2,2 + ‖∇p‖2 ≤ c ‖f‖2 = c ‖Au‖2

for all u, p andf satisfying−∆u +∇p = f (i.e.Au = f ).

This is a deep statement. It shows that operatorA has the so calledmaximumregularity property.


The non–steady Stokes problem

∂tv − ν∆v = −∇p+ f in QT , (3.8)

div v = 0 in QT , (3.9)

v = 0 on∂Ω× (0, T ), (3.10)

v = v0 in Ω× 0. (3.11)

A weak formulation of the non–steady Stokes problem (3.8)–(3.11).Let v0 ∈L2σ(Ω) andf ∈ L2(0, T ; W−1,2

0,σ (Ω)). A vector functionv ∈ L2(0, T ; W1,20,σ(Ω)) is

said to be aweak solutionof the problem (3.8)–(3.10) if∫ T

0

∫Ω

[−v · ∂tφ+ ν∇v : ∇φ

]dx dt =

∫ T

0〈f ,φ〉 dt+

∫Ω

v0 · φ(0) dx (3.12)

for all φ ∈ C∞0([0, T ); W1,2

0,σ(Ω)).


Remark. As each functionφ = φ(x, t) ∈ C∞0 ([0, T ); W1,20,σ(Ω)) can be ap-

proximated, with an arbitrary accuracy in the norm ofC10([0, T ); W1,2

0,σ(Ω)), by asum of finitely many functions of the typeϕ(x)ϑ(t), whereϕ ∈ W1,2

0,σ(Ω) andϑ ∈ C∞0 ([0, T )), (3.12) is equivalent to

∀ϕ ∀ϑ :

∫ T

0

∫Ω

[−v ·ϕ ϑ(t) + ν∇v : ∇ϕ ϑ(t)

]dx dt

=

∫ T

0〈f ,ϕ〉 ϑ(t) dt+ ϑ(0)

∫Ω

v0 ·ϕ dx.

This can also be written in the form

∀ϕ ∀ϑ :

∫ T

0(v,ϕ)2 ϑ(t) dt−

∫ T

0

⟨νAv − f , ϕ

⟩ϑ(t) dt = −

(v0,ϕ

)2 ϑ(0),

which is equivalent to

∀ϕ :d

dt(v,ϕ)2 +

⟨νAv − f ,ϕ

⟩= 0 a.e. in(0, T ). (3.13)

The derivative with respect tot is understood in the sense of distributions.


Lemma. (follows e.g. from [12, Lemma III.1.1])Let H be a Hilbert space. Letu, g ∈ L1(0, T ; H). Then the next conditions are equivalent:

• u = g a.e. in(0, T ), whereu is the distributional derivative ofu in (0, T ),

• d

dt(u, ϕ)H = (g, ϕ)H a.e. in(0, T ), for all ϕ ∈ H, where the derivative with

respect tot is the distributional derivative in(0, T ).

Due to this lemma, (3.13) is equivalent to

v + νAv = f a.e. in(0, T ), (3.14)

which is an equation in spaceW−1,20,σ (Ω).

Equivalent definition of the weak solution to the problem (3.8)–(3.11). Letv0 ∈ L2

σ(Ω) andf ∈ L2(0, T ; W−1,20,σ (Ω)). A vector functionv ∈ L2(0, T ; W1,2

0,σ(Ω))is said to be aweak solution of the problem (3.8)–(3.11) if it satisfies differentialequation (3.13) (or, alternatively, differential equation (3.14)), with the initial condi-tion v(0) = v0.


Remark (in which sense the weak solution satisfies the initial condition).It follows from (3.14) thatv ∈ L2(0, T ; W−1,2

0,σ (Ω)). Hencev is continuous as amapping from(0, T ) to W−1,2

0,σ (Ω). The initial conditionv(0) := v0 now means thatv0 should be equal to the limit (fort→ 0+) of v(t) in the norm ofW−1,2

0,σ (Ω). (Thenext theorem shows thatlimt→0+ v(t) = v0 even in the norm ofL2

σ(Ω).)

Theorem 1. Givenv0 ∈ L2σ(Ω) and f ∈ L2(0, T ; W−1,2

0,σ (Ω)), the problem (3.8)–(3.11) has a unique weak solutionv. The solution satisfies the energy equality

‖v(t)‖22 + 2ν

∫ t

0‖∇v(τ)‖2

2 dτ

= ‖v0‖22 + 2

∫ t

0

⟨f(τ), v(τ)

⟩dτ for all t ∈ [0, T ). (3.15)

Moreover,v ∈ C([0, T ); L2σ(Ω)).

Proof. Existence: the proof by means of Galerkin’s method will be shown later ina more complicated non–linear case of the Navier–Stokes equation.


Lemma (see e.g. Lions, Magenes [8]).LetV → H → V ′ be three Hilbert spacessuch thatV ′ is a dual toV . Letu ∈ L2(0, T ; V ) and u ∈ L2(0, T ; V ′). Thenu is(after a possible redefinition on a set of measure zero) continuous as a function from[0, T ] toH and

1

2

d

dt‖u(t)‖2

H = 〈u(t), u(t)〉 (in the sense of distributions in(0, T )). (3.16)

Energy equality: equation (3.14) implies

〈v,v〉+ ν 〈Av,v〉 = 〈f ,v〉,1

2

d

dt‖v‖2

2 + ν (∇v,∇v)2 = 〈f ,v〉.

Integrating this identity from0 to t, we obtain (3.15).

Uniqueness: let v1, v2 be two solutions, corresponding to the same datav0 andf .Thenv := v1 − v2 is a solution of the same problem, with the body forcef − f = 0and with the initial conditionv(0) = v0 − v0 = 0. Due to (3.15),v ≡ 0.


More regular solutions:

Theorem 2. Let Ω be a boundedC2–domain,v0 ∈ W1,20,σ(Ω) and f ∈ L2(0, T ;

L2σ(Ω)). Then solutionv, given by Theorem 1, is inL2(0, T ; W2,2(Ω)). Its derivative

with respect tot is in L2(0, T ; L2σ(Ω)) and an associated pressurep is in L2(0, T ;

W 1,2(Ω)).

Proof – based on the a priori estimate: we multiply formally equation (3.8) byPσ∆vand integrate inΩ. Applying the integration by parts, we get

1

2

d

dt‖∇v‖2

2 + ν‖Pσ∆v‖22 = (f , Pσ∆v)2

1

2

d

dt‖∇v‖2

2 + ν‖Av‖22 = (f , Av)2 ≤

ν

2‖Av‖2

2 +1

2ν‖f‖2

2

Integrating this estimate with respect tot, we get

|||∇v|||∞; 0,2 + |||Av|||2; 0,2 ≤ c |||f |||2; 0,2 + c ‖v0‖1,2

where|||g|||r; k,s := ‖g‖Lr(0,T ; Wk,s(Ω)).


4. Weak solution to the Navier–Stokes equations II

(other equivalent definitions, subtler properties)

Define

B : W1,20,σ(Ω)2 −→W−1,2

0,σ (Ω) . . . 〈B(u,v),w〉 :=

∫Ω

u · ∇v ·w dx

for u, v, w ∈W1,20,σ(Ω). By analogy with (2.33), we get

∀ϕ ∈W1,20,σ(Ω) :

d

dt(v,ϕ)2 +

⟨νAv + B(v,v)− f ,ϕ

⟩= 0. (4.1)

This is a differential equation in(0, T ). The derivative with respect tot is understoodin the sense of distributions.

4. Weak solution to the Navier–Stokes equations II 33 / 50

OperatorB satisfies

‖B(u,v)‖−1,2 = supw∈W1,2

0,σ(Ω), w 6=0

|〈B(u,v),w〉|‖w‖1,2

= supw∈W1,2

0,σ(Ω), w 6=0

|(u · ∇v, w)2|‖w‖1,2

≤ supw∈W1,2

0,σ(Ω), w 6=0

‖∇v‖2 ‖u‖122 ‖u‖

126 ‖w‖

126

‖w‖1,2≤ c ‖∇v‖2 ‖u‖

122 ‖u‖

126

≤ c ‖∇v‖2 ‖u‖122 ‖∇u‖

122 .

Thus, if v ∈ L∞(0, T ; L2σ(Ω)) ∩ L2(0, T ; W1,2

0 (Ω)) (which is the class for weaksolutions) thenAv ∈ L2(0, T ; W−1,2

0,σ (Ω)) andB(v,v) ∈ L4/3(0, T ; W−1,20,σ (Ω)).

By analogy with (3.14), we deduce that (4.1) is equivalent to

v + νAv + B(v,v) = f (an equation inW−1,20,σ (Ω)). (4.2)

Remark (on the initial condition). As v ∈ L4/3(0, T ; W−1,20,σ (Ω)), v (after a

possible redefinition on a set of measure zero) is continuous as a mapping from(0, T ) to W−1,2

0,σ (Ω). The initial conditionv(0) := v0 now means thatv0 equals thelimit (for t→ 0+) of v(t) in the norm ofW−1,2

0,σ (Ω).


Equivalent definition of the weak solution to the Navier–Stokes initial–boundaryvalue problem (1.8)–(1.11).

Let v0 ∈ L2σ(Ω) andf ∈ L2(0, T ; W−1,2

0,σ (Ω)).

A vector functionv ∈ L2(0, T ; W1,20 (Ω)) ∩ L∞(0, T ; L2

σ(Ω)) is said to be aweaksolution of the problem (1.8)–(1.11) if it satisfies

∀ϕ ∈W1,20,σ(Ω) :

d

dt(v,ϕ)2 +

⟨νAv + B(v,v)− f ,ϕ

⟩= 0 (4.1)

or, alternatively,

v + νAv + B(v,v) = f (an equation inW−1,20,σ (Ω)) (4.2)

a.e. in(0, T ), with the initial conditionv(0) = v0.


Lemma (Hopf 1951, Prodi 1959, Serrin 1963).The weak solutionv to problem(1.3)–(1.4) can be redefined on a set of zero Lebesgue measure so thatv( . , t) ∈L2(Ω) for all t ∈ [0, T ) and for allφ ∈ C∞0 ([0, T ); W1,2

0,σ(Ω)):∫ t

0

∫Ω

[v · ∂τφ− ν∇v : ∇φ− v · ∇v · φ

]dx dτ

= −∫ t

0〈f ,φ〉 dτ +

∫Ω

v(t) · φ(t) dx−∫

Ωv0 · φ(0) dx. (4.3)

Corollary. The weak solutionv is weakly continuous as a mapping from[0, T ) toL2σ(Ω).

Principle of the proof of the lemma: We use aC1 function θh as on the nextfigure. We use (2.1) withφ(x, τ) θh(τ) instead ofφ(x, τ), and we consider the limitfor h→ 0, see Galdi [2].

-

6

1

τ

θh(τ)

T0 t t+ h


5. Global in time existence of the so called Leray–Hopfweak solution

Theorem 3 (existence of a weak solution – Leray 1934, Hopf 1951, et al).Let Ωbe a domain inR3, T > 0, v0 ∈ L2

σ(Ω) and f ∈ L2(QT ). Then there exists at leastone weak solutionv to problem (1.3)–(1.4). The solution satisfies

• the energy inequality (EI)

‖v( . , t)‖22 + 2ν

∫ t

0‖∇v( . , τ)‖2

2 dτ

≤ ‖v0‖22 + 2

∫ t

0

⟨f( . , τ), v( . , τ)

⟩dτ, (5.1)

holds for allt ∈ [0, T ),

• limt→0+

‖v( . , t)− v0‖2 = 0.

5. Global in time existence of the so called Leray–Hopf weak solution 37 / 50

Principles of the proof of Theorem 3

Assume, for simplicity, thatΩ is bounded and Lipschitzian. ThenW1,20,σ(Ω) →→

L2σ(Ω). Consequently,A−1 is a compact operator inL2

σ(Ω).

Let λ1 ≤ λ2 ≤ λ3 ≤ . . . be the eigenvalues of operatorA andu1, u2, u3, . . . be thecorresponding orthonormal eigenfunctions. PutVn := Lu1, . . . ,un.

1) Galerkin’s approximations

Let vn have the formvn(t) =n∑i=1

αi(t) ui and let it satisfy

∂t(vn,ϕ)2 + ν〈Avn,ϕ〉+ 〈B(vn,vn),ϕ〉 = 〈f ,ϕ〉 for all ϕ ∈ Vn. (5.2)

This is equivalent to

∂t(vn,ui) + ν〈Avn,ui〉+ 〈B(vn,vn),ui〉 = 〈f ,ui〉 for i = 1, 2, . . . , n.


It means that

αi + νλiαi +n∑

k,l=1

αkαl〈B(uk,ul),ui〉 = 〈f ,ui〉 for i = 1, 2, . . . , n. (5.3)

This is a system ofn ODE’s for the unknown coefficientsα1(t), . . . , αn(t). Thesystem is solved with the initial conditions

αi(0) = (v0,ui)2 i = 1, . . . , n. (5.4)

We denotev0n :=n∑i=1

αi(0) ui (= the orthogonal projection ofv0 into Vn).


2) A priori estimates and existence of Galerkin’s approximationvn

Multiply i–th equation byαi and sum fori = 1, . . . , n:

d

dt

1

2

n∑i=1

α2i + ν

n∑i=1

λiα2i =

n∑i=1

αi〈f ,ui〉 =⟨f ,

n∑i=1

αiui

⟩≤ ‖f‖−1,2

∥∥∥ n∑i=1

αiui

∥∥∥1,2≤ C ‖f‖−1,2

∥∥∥∇ n∑i=1

αiui

∥∥∥2

= C ‖f‖−1,2

[( n∑i=1

αi∇ui,n∑j=1

αj∇uj

)2

] 12

= C ‖f‖−1,2

[ n∑i=1

n∑j=1

αiαj 〈Aui,uj〉] 1

2

= C ‖f‖−1,2

[ n∑i=1

α2iλi

] 12

≤ ν

2

n∑i=1

λiα2i + C ‖f‖2

−1,2,

whereC = C(Ω, ν). Integrating from0 to t, we get


n∑i=1

α2i (t) + ν

∫ t

0

n∑i=1

λiα2i (τ) dτ ≤ C

∫ t

0‖f‖2

−1,2 dτ +n∑i=1

α2i (0),

n∑i=1

α2i (t) + ν

∫ t

0

n∑i=1

λiα2i (τ) dτ ≤ C

∫ t

0‖f‖2

−1,2 dτ + ‖v0‖22, (5.5)

‖vn(t)‖22 + ν

∫ t

0‖∇vn(τ)‖2

2 dτ ≤ C

∫ t

0‖f‖2

−1,2 dτ + ‖v0‖22. (5.6)

One can deduce from these estimates that the initial–value problem (5.3), (5.4) hasa solutionα1, . . . , αn on (0, T ). The solution satisfies inequality (5.5) for allt ∈(0, T ). Hence the approximate solutionvn satisfies inequality (5.6) for allt ∈ (0, T ).

Note that returning to the first line on the previous page, we also obtaind

dt

1

2‖vn‖2

2 + ν ‖∇vn‖22 = 〈f ,vn〉,

‖vn(t)‖22 + 2ν

∫ t

0‖∇vn(τ)‖2

2 dτ ≤ 2

∫ t

0〈f ,vn〉 dτ + ‖v0‖2

2. (5.7)


3) Convergent subsequences ofvn

Inequality (5.6) provides uniform estimates ofvn inL∞(0, T ; L2σ(Ω)) and inL2(0, T ;

W1,20,σ(Ω)). Hence there exists a sub–sequence ofvn (we denote it in the same way)

andv ∈ L∞(0, T ; L2σ(Ω)) ∩ L2(0, T ; W1,2

0,σ(Ω)) such that

vn −→ v weakly–* inL∞(0, T ; L2σ(Ω)), (5.8)

vn −→ v weakly inL2(0, T ; W1,20,σ(Ω)). (5.9)

In order to obtain an information on a strong convergence of the sequencevn, westill need an information on∂tvn. Since∂tvn ∈ Vn, we have

‖∂tvn‖−1,2 = supw∈W1,2

0,σ(Ω), w 6=0

|〈∂tvn,w〉|‖w‖1,2

= supw∈Vn, w 6=0

|(∂tvn,w)2|‖w‖1,2

= supw∈Vn, w 6=0

|〈−Avn − B(vn,vn) + f ,w〉|‖w‖1,2

≤ ‖Avn‖−1,2 + ‖B(vn,vn)‖−1,2 + ‖f‖−1,2

≤ ‖∇vn‖2 + C ‖∇vn‖322 ‖vn‖

122 + ‖f‖−1,2.


From this, we observe that∂tvn is uniformly bounded inL4/3(0, T ; W−1,20,σ (Ω)).

Lemma (Lions, Aubin). (see e.g. Lions [7] or Temam [12])Let X0, X, X1 bethree Banach spaces such thatX0 andX1 are reflexive andX0 →→ X → X1. Let0 < T <∞, 1 < α1 <∞, 1 < α2 <∞. Denote

Y :=z ∈ Lα0(0, T ; X0), z ∈ Lα1(0, T ; X1)

the Banach space with the norm‖z‖Y := ‖z‖Lα0(0,T ;X0) + ‖z‖Lα1(0,T ;X1).

ThenY →→ Lα0(0, T ; X) (i.e. the injection ofY intoLα0(0, T ; X) is compact.

We use the lemma withX0 = W1,20,σ(Ω), X = L2

σ(Ω), X1 = W−1,20,σ (Ω), α0 = 2,

α1 = 43.

As vn is a bounded sequence inY, it is compact inL2(0, T ; L2σ(Ω)). Hence

there exists a sub–sequence (denoted againvn) that, in addition to (5.8) and (5.9),satisfies

vn −→ v strongly inL2(0, T ; L2σ(Ω)). (5.10)


4) Verification that v satisfies equation (4.2)

Equation (5.2) means that∫ T

0

∫Ω

[−vn ·ϕ ϑ+ ν∇vn : ∇ϕϑ+ vn · ∇vn ·ϕϑ

]dx dt

=

∫ T

0〈f ,ϕ〉ϑ dt+ ϑ(0)

∫Ω

v0n ·ϕ dx (5.11)

for all ϕ = ϕ(x) ∈ Vn and allϑ = ϑ(t) ∈ C∞0 ([0, T )). Particularly, (5.11) alsoholds for allϕ ∈ Vm, wherem ≤ n. Assume, for a while, thatϕ ∈ Vm is fixed.Using all the types (5.8), (5.9), (5.10) of convergence ofvn to v, one can pass to thelimit (for n→∞) in (5.11) and show that∫ T

0

∫Ω

[−v ·ϕ ϑ+ ν∇v : ∇ϕϑ+ v · ∇v ·ϕϑ

]dx dt

=

∫ T

0〈f ,ϕ〉ϑ dt+ ϑ(0)

∫Ω

v0 ·ϕ dx (5.12)

for all ϕ = ϕ(x) ∈ Vm and allϑ = ϑ(t) ∈ C∞0 ([0, T )). Passing now to the limit form→∞, we deduce that (5.12) holds for allϕ = W1,2

0,σ(Ω) and all functionsϑ.


5) The energy inequality

Recall inequality (5.7):

‖vn(t)‖22 + 2ν

∫ t

0‖∇vn(τ)‖2

2 dτ ≤ 2

∫ t

0〈f ,vn〉 dτ + ‖v0‖2

2.

The limit of the right hand side (forn→∞) is

= 2

∫ t

0〈f(τ),v〉 dτ + ‖v0‖2

2.

The limit inferior of the left hand side (forn→∞) is

≥ ‖v(t)‖22 + 2ν

∫ t

0‖∇vn(τ)‖2

2 dτ.

This yields the energy inequality

‖v(t)‖22 + 2ν

∫ t

0‖∇v(τ)‖2

2 dτ ≤ ‖v0‖22 + 2

∫ t

0

⟨f(τ), v(τ)

⟩dτ. (5.1)


6) The strong right L2–continuity of v at time t = 0

The energy inequality implies that

lim supt→0+

‖v(t)‖22 ≤ ‖v0‖2

2 .

On the other hand, asv is weakly continuous from[0, T ) to L2σ(Ω), we have

lim inft→0+

‖v(t)‖22 ≥ ‖v0‖2

2 .

These inequalities yield

limt→0+

‖v(t)‖22 = ‖v0‖2

2 .

This identity, together with the weakL2–continuity, enables us to conclude that

limt→0+

‖v(t)− v0‖22 = 0.

It means thatv(t)→ v0 in L2σ(Ω) for t→ 0+.


Natural questions:

• Does every weak solution satisfy (EI), or even the energy equality (EE)?

Note that we cannot apply formula (3.16) from the Lions–Magenes lemma asin the linear case of the Stokes problem because we do not havev ∈ L2(0, T ;W−1,2

0,σ (Ω)) – now, we only havev ∈ L4/3(0, T ; W−1,20,σ (Ω)). Consequently, the

energy equality (or at least the energy inequality) does not automatically followfrom the definition of the weak solution.

• Is the weak solution unique?

• Is the weak solution regular provided thatv0 and f are regular?

and many others


References

[1] G. P. Galdi:An Introduction to the Mathematical Theory of the Navier–Stokes Equations.2ndedition, Springer 2011.

[2] G. P. Galdi: An Introduction to the Navier–Stokes initial–boundary value problem. InFunda-mental Directions in Mathematical Fluid Mechanics,ed. G. P. Galdi, J. Heywood, R. Rannacher,series “Advances in Mathematical Fluid Mechanics”. Birkhauser, Basel 2000, 1–98.

[3] E. Hopf: Uber die Anfganswertaufgabe fur die Hydrodynamischen Grundgleichungen.Math. Nachr.4, 1951, 213–231.

[4] O. A. Ladyzhenskaya:The Mathematical Theory of Viscous Incompressible Flow.Gordon andBreach, New York 1963.

[5] J. Leray: Essai sur les mouvements plans dun liquide visqueux que limitent des parois.J. Math. Pures Appl.13, 1934, 331– .

[6] J. Leray: Sur le Mouvements dun Liquide Visqueux Emplissant lEspace.Acta Math.63, 1934,193–248.

[7] J. L. Lions: Quelques Methodes de Resolution des Problemes aux Limites Non Lineaires.Gauthier–Villars, Paris 1969.

[8] J. L. Lions, E. Magenes:Problemes aux limites non homogenes et applications.Dunod, Paris1968.

References 48 / 50

[9] G. Prodi: Un teorema di unicit‘a per le equazioni di Navier–Stokes.Ann. Mat. Pura Appl.48,1959, 173–182.

[10] J. Serrin: The initial value problem for the Navier–Stokes equations.Nonlinear Problems,ed. R. E. Langer, Madison: University of Wisconsin Press 9, 1963, 69–98.

[11] H. Sohr: The Navier–Stokes Equations. An Elementary Functional Analytic Approach.Birkhauser Advanced Texts, Basel–Boston–Berlin 2001.

[12] R. Temam:Navier–Stokes Equations.North–Holland, Amsterdam–New York–Oxford 1977.

References 49 / 50

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