Introduction to the Theory of the Navier–Stokes

Equations for Incompressible Fluid

Jiřı́ 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

1 / 50
 Part I – Contents

1. Basic notions, equations and function spaces (a physical background, the

Navier–Stokes equations, function space L2σ (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 strong
solutions, 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

2 / 50
 1. Basic notions, equations and function spaces

A physical background

We assume that the fluid is a continuum. We denote

v = v(x, t) . . . the velocity, p = p(x, t) . . . the pressure,
ρ = ρ(x, t) . . . the density, θ = θ(x, t) . . . the temperature

Stokes’ postulates for the fluid (19th century) require that

a) the stress tensor T depends on the velocity only through the stretching tensor
(= rate of strain tensor, rate of deformation tensor) D := (∇v)sym ,
b) the stress tensor T does not explicitly depend on position x and time t,
c) the continuum is isotropic, i.e. it contains no preferred directions,
d) if the fluid is at rest then T is a multiple of the identity tensor I by a scalar.

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

Nowadays, the postulates are usually formulated in another way:
a) the stress tensor T depends on the velocity only through the stretching tensor D,
b) the stress tensor T does not explicitly depend on position x and time t,
c’) the way tensor T depends on tensor D 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 and D have the representations T∗ and D∗ in frame x∗
THEN T∗ depends on D∗ in the same way as T depends on D.

1. Basic notions, equations and function spaces 4 / 50

One can derive from these postulates that
T = −p I + Td ,
where Td is the “dynamic stress tensor” (it equals the zero tensor if the fluid is at
rest), and Td depends on tensor D through the formula
Td = α I + β D + γ D2 .
The coefficients α, β and γ may generally depend on the so called state variables
(pressure, density, temperature) and on the principal invariants of tensor D.

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

Newtonian fluid: T depends linearly on D. One can deduce from this assumption
(also using Kirchhoff’s formula p = − 13 T + µ0 div v, where µ0 is the so called
coefficient of bulk viscosity) that
T = −p + (µ0 − 32 µ) div v I + 2µ D,
 
(1.1)
where µ is the dynamic coefficient of viscosity.

1. Basic notions, equations and function spaces 5 / 50

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: ∂t ρ + div (ρv) = 0 . . . equation of continuity (1.3)

Note that condition (1.2) follows from (1.3) in the special case when ρ = const. Due
to this and also other historical reasons, equation (1.2) is often called equation of
continuity 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 equation
for density ρ:
∂t ρ + v · ∇ρ = 0.

Conservation of momentum: ρ ∂t v + ρv · ∇v = −Div T + ρf (1.4)

1. Basic notions, equations and function spaces 6 / 50

From now, we use the IMPORTANT ASSUMPTIONS:

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

• the fluid is Newtonian,
• µ = const. > 0.

We denote ν := µ/ρ = µ . . . the kinematic coefficient of viscosity.

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

Substituting this to (1.4), we obtain the so called Navier–Stokes equation for
viscous incompressible fluid (H. Navier 1824, G. Stokes 1845)
∂t v + v · ∇v = −∇p + ν∆v + f .

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

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

1. Basic notions, equations and function spaces 7 / 50

Boundary conditions on 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 and n denotes the outer normal vector on
the 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 called free slip

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

1. Basic notions, equations and function spaces 8 / 50

 The Navier–Stokes initial–boundary value problem

Ω . . . a domain in R3 . . . 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 problem we deal with is:

∂t v + 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)

1. Basic notions, equations and function spaces 9 / 50

 Function space L2σ (Ω)

Let C∞ 0,σ (Ω) be the linear space of all infinitely differentiable divergence–free vector
functions in Ω with a compact support in Ω. We denote by L2σ (Ω) the closure of
C∞ 2
0,σ (Ω) in L (Ω).

Remark. Assume that Ω has a locally Lipschitzian boundary.

Then the space L2σ (Ω) can be characterized as a space of functions from L2 (Ω),
whose divergence equals zero in Ω (in the sense of distributions), whose normal
component on ∂Ω) is zero (as an element of the space W −1/2,2 (∂Ω)) in the sense of
traces. (See e.g. [1] or [11].)

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

Lemma. Let Ω be any domain in R3 . Then
1,2
L2σ (Ω)⊥ = G2 (Ω) := w ∈ L2 (Ω); w = ∇ϕ for some ϕ ∈ Wloc

(Ω)}.
Consequently, L2 (Ω) = L2σ (Ω) ⊕ G2 (Ω), where L2σ (Ω) and G2 (Ω) are closed or-
thogonal subspaces of L2 (Ω).

1. Basic notions, equations and function spaces 10 / 50

A part of the proof: Only the part ⊃ . (See e.g. [1] or [11] for the opposite
inclusion.) Thus, let w = ∇ϕ ∈ G2 (Ω). It suffices to show that (u, w)2 = 0 for all
u ∈ C∞ 0,σ (Ω). However,
Z Z
(u, w)2 = u · ∇ϕ dx = − div u ϕ dx = 0. 
Ω Ω

The orthogonal projection of L2 (Ω) onto L2σ (Ω) is called the Helmholtz projection.
It is usually denoted by P2 , or Pσ2 or only by Pσ .

Remark. Let v ∈ L2 (Ω). The Helmholtz decomposition of v is: v = Pσ v + ∇ϕ

where ϕ is a weak solution of the Neumann problem
∂ϕ
∆ϕ = div v in Ω, = v · n on ∂Ω.
∂n

Remark. An analogous decomposition in Lq (Ω) (for 1 < q < ∞) is possible ⇐⇒

0
this Neumann problem has a weak solution ϕ in D1,q (Ω) := {w ∈ L1loc (Ω); ∇w ∈
0
Lq (Ω)}, where 1/q + 1/q 0 = 1.

1. Basic notions, equations and function spaces 11 / 50

 2. Weak solution to the Navier–Stokes equations I
(First observations and definition)

Further important function spaces:

1,2
W0,σ (Ω) := W01,2 (Ω) ∩ L2σ (Ω) (a closed subspace of W01,2 (Ω)), dense in L2σ (Ω))
−1,2 1,2
W0,σ (Ω) . . . the dual to W0,σ (Ω)

Formal considerations (steps 1 and 2)

Step 1: the integral
equation. Let φ be an arbitrary test function from the space
1,2
C0∞ [0, T ); W0,σ (Ω) . If we multiply (1.8) by φ, integrate in Ω, and apply the
integration by parts, we obtain the integral equation

Z T Z
 
v · ∂t φ − ν ∇v : ∇φ − v · ∇v · φ dx dt
0 Ω (2.1)
Z T Z
= − hf , φi dt − v0 · φ(0) dx
0 Ω

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 function v, integrate in Ω, and apply the integration by parts. We obtain
Z Z Z
d 1 2
|v| dx + ν ∆v · v dx = f · v dx,
dt 2 Ω Ω Ω
d 1
kvk22 + ν k∇vk22 = (f , v)2 ≤ kf k−1,2 kvk1,2
dt 2
ν
≤ C kf k−1,2 k∇vk2 ≤ k∇vk22 + C(ν) kf k2−1,2 ,
2
Z t Z t
2 2 2
kv(t)k2 + ν k∇vk2 dτ ≤ kv0 k2 + kf k2−1,2 dτ
0 0
Z T
≤ kv0 k22 + C(ν) kf k2−1,2 dτ
0
Omitting at first the second term on the left hand side, we get
Z T
kv(t)k22 ≤ kv0 k22 + C(ν) kf k2−1,2 dτ for all t ∈ (0, T ).
0
This a priori estimate indicates that solution v should be in L∞ (0, T ; L2σ (Ω)).

2. Weak solution to the Navier–Stokes equations I 13 / 50

Then, omitting the first term on the left hand side and considering t = T , we obtain
Z T Z T
2 2
ν k∇vk2 dτ ≤ kv0 k2 + C(ν) kf k2−1,2 dτ.
0 0
1,2
This a priori estimate indicates that solution v should be in L2 (0, T ; W0,σ (Ω)).
−1,2
Both the estimates require v0 ∈ L2σ (Ω) and f ∈ L2 (0, T ; W0,σ (Ω)).

Thus, we arrive at the definition:

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

−1,2 1,2
and f ∈ L2 (0, T ; W0,σ (Ω)). A vector function v ∈ L2 (0, T ; W0,σ (Ω)) ∩ L∞ (0, T ;
L2σ (Ω)) is said to be a weak solution of the problem

(1.8)–(1.11) if it satisfies integral
∞ 1,2
equation (2.1) for all φ ∈ C0 [0, T ); W0,σ (Ω) .

2. Weak solution to the Navier–Stokes equations I 14 / 50

 Remark. In order to show that no important information on the solution was
lost when passing from the “classical formulation” (1.8)–(1.11) of the considered
Navier–Stokes initial–boundary value problem to the weak formulation, we assume
that v is a “sufficiently smooth” weak solution. Applying the backward integration
by parts to (2.1), we get
Z Z TZ
 
[v(0) − v0 ] · φ(0) dx + ∂t v + v · ∇v − ν∆v − f · φ dx dt = 0 (2.2)
Ω 0 Ω
1,2 1,2
for all φ ∈ C0∞ [0, T ); W0,σ (Ω) . Considering at first φ ∈ C0∞ (0, T ); W0,σ



(Ω)
(which guarantees that φ(0) = 0), we obtain
Z TZ
 
∂t v + v · ∇v − ν∆v − f · φ dx dt = 0.
0 Ω
Hence ∂t v + v · ∇v − ν∆v − f ∈ G2 (Ω) for a.a. t ∈ (0, T ). Consequently, there
exists p (also depending on t) such that
∂t v + v · ∇v − ν∆v − f = −∇p,
which is equation (1.8).

2. Weak solution to the Navier–Stokes equations I 15 / 50

 1,2
Now we return to φ ∈ C0∞ [0, T ); W0,σ


(Ω) and to equation (2.2). It gives
Z Z TZ
 
0 = [v(0) − v0 ] · φ(0) dx + ∂t v + v · ∇v − ν∆v − f · φ dx dt
Ω 0 Ω
Z Z T Z
= [v(0) − v0 ] · φ(0) dx − ∇p · φ dx dt
Ω 0 Ω
Z
= [v(0) − v0 ] · φ(0) dx.
Ω
1,2 1,2
As this holds for all φ ∈ C0∞ [0, T ); W0,σ


(Ω) , i.e. for all φ(0) ∈ W0,σ (Ω), we
obtain v(0) = v0 .
Equation (1.9) (i.e. div v = 0) is satisfied due to the condition v ∈ L∞ (0, T ; L2σ (Ω)).
Boundary condition (1.10) (i.e. v = 0 on ∂Ω × (0, T )) holds due to the condition
1,2
v ∈ L2 (0, T ; W0,σ (Ω)).

Conclusion: A “sufficiently smooth” weak solution v satisfies the system (1.8)–

(1.11) in a classical sense.

2. Weak solution to the Navier–Stokes equations I 16 / 50

 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

1,2
W0,σ (Ω) := W01,2 (Ω) ∩ L2σ (Ω)
−1,2 1,2
W0,σ (Ω) . . . the dual to W0,σ (Ω)
1,2 −1,2
Remark. As W0,σ (Ω) ,→ L2σ (Ω), we have L2σ (Ω) ⊂ W0,σ (Ω). If g ∈ L2σ (Ω) then
1,2
hg, wi := (g, w)2 for all w ∈ W0,σ (Ω).

3. The Stokes problem 17 / 50

 −1,2
A weak solution formulation of the problem (3.1)–(3.3). Let f ∈ W0,σ (Ω).
1,2
Function v ∈ W0,σ (Ω) is said to be a weak solution of the problem (3.1)–(3.3) if
Z
1,2
ν (∇v, ∇w)2 ≡ ν ∇v : ∇w dx = hf , wi ∀w ∈ W0,σ (Ω). (3.4)
Ω

Remark. (3.4) follows formally from (3.1)–(3.3) if we multiply (3.1) by w and

integrate in Ω. On the other hand, if v is a “smooth” weak solution, then we can we
can apply the backward integration by parts to (3.4) and get
Z
1,2
ν [ν∆v + f ] · w dx = 0 ∀w ∈ W0,σ (Ω).
Ω
Hence ν∆v + f ∈ G2 (Ω). Thus, there exists p such that ν∆v + f = ∇p, which is
equation (3.1).
1,2 −1,2
Operator A: Define a linear operator A : W0,σ (Ω) → W0,σ (Ω) by the equation
1,2
hAv, wi = (∇v, ∇w)2 ∀w ∈ W0,σ (Ω).
−1,2
Now, (3.4) is equivalent to the equation νAv = f (in space W0,σ (Ω)).

3. The Stokes problem 18 / 50

Basic properties of operator A:
1,2
• D(A) = W0,σ (Ω) (follows from the definition of A)

• Operator A is 1–1.
 
1,2
v ∈ N (A) =⇒ ∀w ∈ W0,σ (Ω) : (∇v, ∇w)2 = 0 =⇒ k∇vk2 = 0 =⇒ v = 0

1,2 −1,2
• Operator A is bounded (as an operator from W0,σ (Ω) to W0,σ (Ω)).
 |hAv, wi| |(∇v, ∇w)2 | 
kAvk−1,2 = sup = sup ≤ k∇vk2
w∈W1,2 (Ω), w6=0 kwk1,2
0,σ w∈W1,2 (Ω), w6=0
0,σ
kwk1,2

−1,2
• The range of A need not be generally the whole space W0,σ (Ω).
1,2
Proof. Operator A is closed because its domain is the whole W0,σ (Ω) and it is
−1
bounded. Hence A is also closed. By contradiction: Assume that R(A) =
−1,2
D(A−1 ) = W0,σ (Ω). Then operator A−1 is bounded (by the closed graph theorem).

3. The Stokes problem 19 / 50

 1,2
Choose zn ∈ W0,σ (Ω) so that k∇zn k2 → 0 and kzn k2 → 1. (This choice is possible
−1,2
e.g. if Ω is an exterior domain or Ω = R3 .) Let fn ∈ W0,σ (Ω) be defined by the
equation
1,2
hfn , wi := (∇zn , ∇w)2 + (zn , w)2 ∀w ∈ W0,σ (Ω). (3.5)
−1,2
Then {fn } is a bounded sequence in W0,σ (Ω). Put un := A−1 fn . It means that
fn = Aun . Hence
1,2
hfn , wi := (∇un , ∇w)2 ∀w ∈ W0,σ (Ω). (3.6)

Equations (3.5) and (3.6) (with w = zn yield

(∇zn , ∇zn )2 + (zn , zn )2 = (∇un , ∇zn )2 ≤ k∇un k2 k∇zn k2

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

3. The Stokes problem 20 / 50

• The range of A need not generally contain L2σ (Ω).
Proof. By contradiction, assume that L2σ (Ω) ⊂ R(A). Denote by A the restriction
1,2
of A to A−1 (L2σ (Ω)). Operator A is closed as an operator from W0,σ (Ω) to L2σ (Ω).
(This can be proven similarly as the fact that A is closed.) Hence A−1 is a bounded
1,2
operator from L2σ (Ω) to W0,σ (Ω) (by the closed graph theorem), because D(A−1 ) =
L2σ (Ω).
−1,2 1,2
f ∈ L2σ (Ω) =⇒ f ∈ W0,σ (Ω), hf , wi = (f , w)2 ∀w ∈ W0,σ (Ω)
1,2 1,2
L2σ (Ω) ⊂ R(A) =⇒ ∃u ∈ W0,σ (Ω) : hf , wi = (∇u, ∇w)2 ∀w ∈ W0,σ (Ω)
1,2
Hence (f , w)2 = (∇u, ∇w)2 ∀w ∈ W0,σ (Ω), where u = A−1 f .
1,2
Choose {fn } in W0,σ (Ω) so that kfn k2 → 1 and k∇fn k2 → 0. (A sequence with
these properties exists e.g. if Ω is an exterior domain or Ω = R3 .) Put wn = fn . Then
kfn k2 = (∇un , ∇fn )2 ≤ k∇un k2 kfn k2 ,
where un = A−1 fn . The right hand side tends to zero, while the left hand side tends
to one (for n → ∞). This is a contradiction. 

3. The Stokes problem 21 / 50

Domain and range of operator A:

−1,2
L2σ (Ω) W0,σ (Ω)

A
1,2 1,2
W0,σ (Ω) W0,σ (Ω)

1,2
D(A) = W0,σ (Ω) R(A)

3. The Stokes problem 22 / 50

 −1,2
The special case of W0,σ (Ω) = R(A)
L2σ (Ω)
a bounded domain Ω:

A
1,2 1,2
W0,σ (Ω) W0,σ (Ω)

1,2
D(A) = W0,σ (Ω)

−1,2
• If Ω is bounded then R(A) = W0,σ (Ω).
Proof. The scalar product (∇v, ∇w)2 is equivalent to the scalar product (v, w)1,2
1,2 −1,2 1,2
in W0,σ (Ω). Hence, given f ∈ W0,σ (Ω), there exists u ∈ W0,σ (Ω) such that
1,2
hf , wi = (∇u, ∇w)2 for all w ∈ W0,σ (Ω) (by the Riesz lemma). It means that
−1,2
f = Au (the identity in W0,σ (Ω)). 
−1,2 1,2
Corollary: If Ω is bounded then A−1 is bounded from W0,σ (Ω) to W0,σ (Ω).

3. The Stokes problem 23 / 50

 −1,2
L2σ (Ω) W0,σ (Ω)
1,2
W0,σ (Ω)
A

1,2
D(A) ⊂ W0,σ (Ω)
1,2
R(A) W0,σ (Ω)

Denote by A the part of operator A with the range R(A) ∩ L2σ (Ω). Thus, A is the
restriction of A to
1,2
(Ω); Au ∈ L2σ (Ω) = A−1 [R(A) ∩ L2σ (Ω)].

D(A) := u ∈ W0,σ

Operator A is an operator in L2σ (Ω). It is often called the Stokes operator.

3. The Stokes problem 24 / 50

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

• A is a 1–1 positive and self–adjoint operator in L2σ (Ω). Its domain satisfies the
1,2
inclusions C∞
0,σ (Ω) ⊂ D(A) ⊂ W0,σ (Ω).

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

means that there exists p ∈ L2loc (Ω) (unique up to an additive constant), such that
−ν∆v + ∇p = f (in the sense of distributions in Ω). (3.7)

Principle of the proof. The equation νAv = f means that

1,2
ν (∇v, ∇w)2 = (f , w)2 ∀w ∈ W0,σ (Ω).
1,2
Hence ν h−∆v, wi = (f , w)2 ∀w ∈ W0,σ (Ω),
1,2
h−ν∆v − f , wi = 0 ∀w ∈ W0,σ (Ω).
Then there exists p ∈ L2loc (Ω) such that ν∆v − f = ∇p (in the sense of distribu-
tions in Ω); see [11]. 

3. The Stokes problem 25 / 50

• If Ω is bounded then R(A) ≡ D(A−1 ) = L2σ (Ω) and operator A−1 is bounded
1,2
from L2σ (Ω) to W0,σ (Ω).
Proof. 1st possibility: If u ∈ D(A) then (Au, w)2 = (∇u, ∇w)2 ∀w ∈
1,2
W0,σ (Ω). Hence k∇uk22 ≤ kAuk2 kuk2 ≤ c kAuk2 k∇uk2 . (Constant c is the
constant from Poincaré’s inequality kuk2 ≤ c k∇uk2 .) This yields
k∇A−1 f k2 ≤ c kf k2 for f = Au.

2nd possibility: by the closed graph theorem 

1,2
• If Ω is a bounded C 2 –domain then D(A) = W0,σ (Ω) ∩ W2,2 (Ω), A = −Pσ ∆,
and
kuk2,2 + k∇pk2 ≤ c kf k2 = c kAuk2
for all u, p and f satisfying −∆u + ∇p = f (i.e. Au = f ).
This is a deep statement. It shows that operator A has the so called maximum
regularity property.

3. The Stokes problem 26 / 50

 The non–steady Stokes problem

∂t v − 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 ∈

−1,2 1,2
L2σ (Ω) and f ∈ L2 (0, T ; W0,σ (Ω)). A vector function v ∈ L2 (0, T ; W0,σ (Ω)) is
said to be a weak solution of the problem (3.8)–(3.10) if
Z TZ Z T Z
 
−v · ∂t φ + ν ∇v : ∇φ dx dt = hf , φi dt + v0 · φ(0) dx (3.12)
0 Ω 0 Ω

1,2
for all φ ∈ C0∞ [0, T ); W0,σ


(Ω) .

3. The Stokes problem 27 / 50

 1,2
Remark. As each function φ = φ(x, t) ∈ C0∞ ([0, T ); W0,σ (Ω)) can be ap-
1 1,2
proximated, with an arbitrary accuracy in the norm of C0 ([0, T ); W0,σ (Ω)), by a
1,2
sum of finitely many functions of the type ϕ(x)ϑ(t), where ϕ ∈ W0,σ (Ω) and
ϑ ∈ C0∞ ([0, T )), (3.12) is equivalent to
Z TZ
 
∀ϕ ∀ϑ : −v · ϕ ϑ̇(t) + ν ∇v : ∇ϕ ϑ(t) dx dt
0 Ω
Z T Z
= hf , ϕi ϑ(t) dt + ϑ(0) v0 · ϕ dx.
0 Ω
This can also be written in the form
Z T Z T

∀ϕ ∀ϑ : (v, ϕ)2 ϑ̇(t) dt − νAv − f , ϕ ϑ(t) dt = − v0 , ϕ 2
ϑ(0),
0 0
which is equivalent to
d
∀ϕ : (v, ϕ)2 + νAv − f , ϕ = 0 a.e. in (0, T ). (3.13)
dt
The derivative with respect to t is understood in the sense of distributions.

3. The Stokes problem 28 / 50

Lemma. (follows e.g. from [12, Lemma III.1.1]) Let H be a Hilbert space. Let
u, g ∈ L1 (0, T ; H). Then the next conditions are equivalent:
• u̇ = g a.e. in (0, T ), where u̇ is the distributional derivative of u in (0, T ),
d
• (u, ϕ)H = (g, ϕ)H a.e. in (0, T ), for all ϕ ∈ H, where the derivative with
dt
respect to t is the distributional derivative in (0, T ).

Due to this lemma, (3.13) is equivalent to

v̇ + νAv = f a.e. in (0, T ), (3.14)
−1,2
which is an equation in space W0,σ (Ω).

Equivalent definition of the weak solution to the problem (3.8)–(3.11). Let

−1,2 1,2
v0 ∈ L2σ (Ω) and f ∈ L2 (0, T ; W0,σ (Ω)). A vector function v ∈ L2 (0, T ; W0,σ (Ω))
is said to be a weak solution of the problem (3.8)–(3.11) if it satisfies differential
equation (3.13) (or, alternatively, differential equation (3.14)), with the initial condi-
tion v(0) = v0 .

3. The Stokes problem 29 / 50

Remark (in which sense the weak solution satisfies the initial condition).
−1,2
It follows from (3.14) that v̇ ∈ L2 (0, T ; W0,σ (Ω)). Hence v is continuous as a
−1,2
mapping from (0, T ) to W0,σ (Ω). The initial condition v(0) := v0 now means that
−1,2
v0 should be equal to the limit (for t → 0+) of v(t) in the norm of W0,σ (Ω). (The
2
next theorem shows that limt→0+ v(t) = v0 even in the norm of Lσ (Ω).)

−1,2
Theorem 1. Given v0 ∈ L2σ (Ω) and f ∈ L2 (0, T ; W0,σ (Ω)), the problem (3.8)–
(3.11) has a unique weak solution v. The solution satisfies the energy equality
Z t
2
kv(t)k2 + 2ν k∇v(τ )k22 dτ
0
Z t
= kv0 k22 + 2 f (τ ), v(τ ) dτ for all t ∈ [0, T ). (3.15)
0

Moreover, v ∈ C([0, T ); L2σ (Ω)).

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

3. The Stokes problem 30 / 50

Lemma (see e.g. Lions, Magenes [8]). Let V ,→ H ,→ V 0 be three Hilbert spaces
such that V 0 is a dual to V . Let u ∈ L2 (0, T ; V ) and u̇ ∈ L2 (0, T ; V 0 ). Then u is
(after a possible redefinition on a set of measure zero) continuous as a function from
[0, T ] to H and
1 d
ku(t)k2H = hu̇(t), u(t)i (in the sense of distributions in (0, T )). (3.16)
2 dt

Energy equality: equation (3.14) implies

hv̇, vi + ν hAv, vi = hf , vi,

1 d
kvk22 + ν (∇v, ∇v)2 = hf , vi.
2 dt
Integrating this identity from 0 to t, we obtain (3.15).
Uniqueness: let v1 , v2 be two solutions, corresponding to the same data v0 and f .
Then v := v1 − v2 is a solution of the same problem, with the body force f − f = 0
and with the initial condition v(0) = v0 − v0 = 0. Due to (3.15), v ≡ 0. 

3. The Stokes problem 31 / 50

More regular solutions:
1,2
Theorem 2. Let Ω be a bounded C 2 –domain, v0 ∈ W0,σ (Ω) and f ∈ L2 (0, T ;
L2σ (Ω)). Then solution v, given by Theorem 1, is in L2 (0, T ; W2,2 (Ω)). Its derivative
with respect to t is in L2 (0, T ; L2σ (Ω)) and an associated pressure p is in L2 (0, T ;
W 1,2 (Ω)).

Proof – based on the a priori estimate: we multiply formally equation (3.8) by Pσ ∆v

and integrate in Ω. Applying the integration by parts, we get
1 d
k∇vk22 + νkPσ ∆vk22 = (f , Pσ ∆v)2
2 dt
1 d ν 1
k∇vk22 + νkAvk22 = (f , Av)2 ≤ kAvk22 + kf k22
2 dt 2 2ν
Integrating this estimate with respect to t, we get
|||∇v|||∞; 0,2 + |||Av|||2; 0,2 ≤ c |||f |||2; 0,2 + c kv0 k1,2
where |||g|||r; k,s := kgkLr (0,T ; Wk,s (Ω)) . 

3. The Stokes problem 32 / 50

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

Define
Z
1,2 −1,2
B : W0,σ (Ω)2 −→ W0,σ (Ω) ... hB(u, v), wi := u · ∇v · w dx
Ω
1,2
for u, v, w ∈ W0,σ (Ω). By analogy with (2.33), we get

1,2 d
∀ ϕ ∈ W0,σ (Ω) : (v, ϕ)2 + νAv + B(v, v) − f , ϕ = 0. (4.1)
dt

This is a differential equation in (0, T ). The derivative with respect to t is understood

in the sense of distributions.

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

Operator B satisfies
|hB(u, v), wi| |(u · ∇v, w)2 |
kB(u, v)k−1,2 = sup = sup
w∈W1,2 (Ω), w6=0 kwk1,2 w∈W1,2 (Ω), w6=0 kwk1,2
0,σ 0,σ
1 1 1
k∇vk2 kuk2 kuk6 kwk6
2 2 2 1 1
≤ sup ≤ c k∇vk2 kuk22 kuk62
w∈W1,2 (Ω), w6=0 kwk1,2
0,σ
1 1
≤ c k∇vk2 kuk2 k∇uk2 .
2 2

Thus, if v ∈ L∞ (0, T ; L2σ (Ω)) ∩ L2 (0, T ; W01,2 (Ω)) (which is the class for weak
−1,2 −1,2
solutions) then Av ∈ L2 (0, T ; W0,σ (Ω)) and B(v, v) ∈ L4/3 (0, T ; W0,σ (Ω)).
By analogy with (3.14), we deduce that (4.1) is equivalent to
−1,2
v̇ + νAv + B(v, v) = f (an equation in W0,σ (Ω)). (4.2)

−1,2
Remark (on the initial condition). As v̇ ∈ L4/3 (0, T ; W0,σ (Ω)), v (after a
possible redefinition on a set of measure zero) is continuous as a mapping from
−1,2
(0, T ) to W0,σ (Ω). The initial condition v(0) := v0 now means that v0 equals the
−1,2
limit (for t → 0+) of v(t) in the norm of W0,σ (Ω).

4. Weak solution to the Navier–Stokes equations II 34 / 50

Equivalent definition of the weak solution to the Navier–Stokes initial–boundary
value problem (1.8)–(1.11).
−1,2
Let v0 ∈ L2σ (Ω) and f ∈ L2 (0, T ; W0,σ (Ω)).
A vector function v ∈ L2 (0, T ; W01,2 (Ω)) ∩ L∞ (0, T ; L2σ (Ω)) is said to be a weak
solution of the problem (1.8)–(1.11) if it satisfies

1,2 d
∀ ϕ ∈ W0,σ (Ω) : (v, ϕ)2 + νAv + B(v, v) − f , ϕ = 0 (4.1)
dt

or, alternatively,
−1,2
v̇ + νAv + B(v, v) = f (an equation in W0,σ (Ω)) (4.2)

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

4. Weak solution to the Navier–Stokes equations II 35 / 50

Lemma (Hopf 1951, Prodi 1959, Serrin 1963). The weak solution v to problem
(1.3)–(1.4) can be redefined on a set of zero Lebesgue measure so that v( . , t) ∈
1,2
L2 (Ω) for all t ∈ [0, T ) and for all φ ∈ C0∞ ([0, T ); W0,σ (Ω)):
Z tZ
 
v · ∂τ φ − ν ∇v : ∇φ − v · ∇v · φ dx dτ
0 Ω
Z t Z Z
= − hf , φi dτ + v(t) · φ(t) dx − v0 · φ(0) dx. (4.3)
0 Ω Ω

Corollary. The weak solution v is weakly continuous as a mapping from [0, T ) to

L2σ (Ω).

Principle of the proof of the lemma: We use a C 1 function θh as on the next

figure. We use (2.1) with φ(x, τ ) θh (τ ) instead of φ(x, τ ), and we consider the limit
for h → 0, see Galdi [2].
6
θh (τ )
1
τ -
0 t t+h T 

4. Weak solution to the Navier–Stokes equations II 36 / 50

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

Theorem 3 (existence of a weak solution – Leray 1934, Hopf 1951, et al). Let Ω
be a domain in R3 , T > 0, v0 ∈ L2σ (Ω) and f ∈ L2 (QT ). Then there exists at least
one weak solution v to problem (1.3)–(1.4). The solution satisfies
• the energy inequality (EI)
Z t
kv( . , t)k22 + 2ν k∇v( . , τ )k22 dτ
0
Z t
2
≤ kv0 k2 + 2 f ( . , τ ), v( . , τ ) dτ, (5.1)
0

holds for all t ∈ [0, T ),

• lim kv( . , t) − v0 k2 = 0.
t→0+

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

 Principles of the proof of Theorem 3
1,2
Assume, for simplicity, that Ω is bounded and Lipschitzian. Then W0,σ (Ω) ,→,→
L2σ (Ω). Consequently, A−1 is a compact operator in L2σ (Ω).

Let λ1 ≤ λ2 ≤ λ3 ≤ . . . be the eigenvalues of operator A and u1 , u2 , u3 , . . . be the

corresponding orthonormal eigenfunctions. Put Vn := L{u1 , . . . , un }.

1) Galerkin’s approximations
n
X
Let vn have the form vn (t) = αi (t) ui and let it satisfy
i=1

∂t (vn , ϕ)2 + νhAvn , ϕi + hB(vn , vn ), ϕi = hf , ϕi for all ϕ ∈ Vn . (5.2)

This is equivalent to

∂t (vn , ui ) + νhAvn , ui i + hB(vn , vn ), ui i = hf , ui i for i = 1, 2, . . . , n.

5. Global in time existence of the so called Leray–Hopf weak solution 38 / 50

It means that
n
X
α̇i + νλi αi + αk αl hB(uk , ul ), ui i = hf , ui i for i = 1, 2, . . . , n. (5.3)
k,l=1

This is a system of n ODE’s for the unknown coefficients α1 (t), . . . , αn (t). The
system is solved with the initial conditions

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

n
X
We denote v0n := αi (0) ui (= the orthogonal projection of v0 into Vn ).
i=1

5. Global in time existence of the so called Leray–Hopf weak solution 39 / 50

2) A priori estimates and existence of Galerkin’s approximation vn
Multiply i–th equation by αi and sum for i = 1, . . . , n:
n n n n
d 1 X 2 X
2
X D X E
α +ν λi αi = αi hf , ui i = f , αi ui
dt 2 i=1 i i=1 i=1 i=1
n
X n
X
≤ kf k−1,2 αi ui ≤ C kf k−1,2 ∇ αi ui
1,2 2
i=1 i=1
n
X n
X   12
= C kf k−1,2 αi ∇ui , αj ∇uj
2
i=1 j=1
n X
X n  12 n
X  12
= C kf k−1,2 αi αj hAui , uj i = C kf k−1,2 αi2 λi
i=1 j=1 i=1
n
ν X
≤ λi αi2 + C kf k2−1,2 ,
2 i=1

where C = C(Ω, ν). Integrating from 0 to t, we get

5. Global in time existence of the so called Leray–Hopf weak solution 40 / 50

 n
X n
Z tX Z t n
X
αi2 (t) +ν λi αi2 (τ ) dτ ≤ C kf k2−1,2 dτ + αi2 (0),
i=1 0 i=1 0 i=1
Xn n
Z tX Z t
αi2 (t) + ν λi αi2 (τ ) dτ ≤ C kf k2−1,2 dτ + kv0 k22 , (5.5)
i=1 0 i=1 0
Z t Z t
kvn (t)k22 + ν k∇vn (τ )k22 dτ ≤ C kf k2−1,2 dτ + kv0 k22 . (5.6)
0 0
One can deduce from these estimates that the initial–value problem (5.3), (5.4) has
a solution α1 , . . . , αn on (0, T ). The solution satisfies inequality (5.5) for all t ∈
(0, T ). Hence the approximate solution vn satisfies inequality (5.6) for all t ∈ (0, T ).
Note that returning to the first line on the previous page, we also obtain
d 1
kvn k22 + ν k∇vn k22 = hf , vn i,
dt 2
Z t Z t
2
kvn (t)k2 + 2ν k∇vn (τ )k2 dτ ≤ 2 hf , vn i dτ + kv0 k22 .
2
(5.7)
0 0

5. Global in time existence of the so called Leray–Hopf weak solution 41 / 50

3) Convergent subsequences of {vn }
Inequality (5.6) provides uniform estimates of vn in L∞ (0, T ; L2σ (Ω)) and in L2 (0, T ;
1,2
W0,σ (Ω)). Hence there exists a sub–sequence of {vn } (we denote it in the same way)
1,2
and v ∈ L∞ (0, T ; L2σ (Ω)) ∩ L2 (0, T ; W0,σ (Ω)) such that
vn −→ v weakly–* in L∞ (0, T ; L2σ (Ω)), (5.8)
1,2
vn −→ v weakly in L2 (0, T ; W0,σ (Ω)). (5.9)
In order to obtain an information on a strong convergence of the sequence {vn }, we
still need an information on ∂t vn . Since ∂t vn ∈ Vn , we have
|h∂t vn , wi| |(∂t vn , w)2 |
k∂t vn k−1,2 = sup = sup
w∈W1,2 (Ω), w6=0 kwk1,2 w∈Vn , w6=0 kwk1,2
0,σ

|h−Avn − B(vn , vn ) + f , wi|

= sup
w∈Vn , w6=0 kwk1,2
≤ kAvn k−1,2 + kB(vn , vn )k−1,2 + kf k−1,2
3 1
≤ k∇vn k2 + C k∇vn k22 kvn k22 + kf k−1,2 .

5. Global in time existence of the so called Leray–Hopf weak solution 42 / 50

 −1,2
From this, we observe that {∂t vn } is uniformly bounded in L4/3 (0, T ; W0,σ (Ω)).

Lemma (Lions, Aubin). (see e.g. Lions [7] or Temam [12]) Let X0 , X, X1 be
three Banach spaces such that X0 and X1 are reflexive and X0 ,→,→ X ,→ X1 . Let
0 < T < ∞, 1 < α1 < ∞, 1 < α2 < ∞. Denote
Y := z ∈ Lα0 (0, T ; X0 ), ż ∈ Lα1 (0, T ; X1 )


the Banach space with the norm kzkY := kzkLα0 (0,T ; X0 ) + kżkLα1 (0,T ; X1 ) .
Then Y ,→,→ Lα0 (0, T ; X) (i.e. the injection of Y into Lα0 (0, T ; X) is compact.

1,2 −1,2
We use the lemma with X0 = W0,σ (Ω), X = L2σ (Ω), X1 = W0,σ (Ω), α0 = 2,
4
α1 = 3 .

As {vn } is a bounded sequence in Y, it is compact in L2 (0, T ; L2σ (Ω)). Hence

there exists a sub–sequence (denoted again {vn }) that, in addition to (5.8) and (5.9),
satisfies
vn −→ v strongly in L2 (0, T ; L2σ (Ω)). (5.10)

5. Global in time existence of the so called Leray–Hopf weak solution 43 / 50

4) Verification that v satisfies equation (4.2)
Equation (5.2) means that
Z TZ
 
−vn · ϕ ϑ̇ + ν∇vn : ∇ϕ ϑ + vn · ∇vn · ϕ ϑ dx dt
0 Ω
Z T Z
= hf , ϕi ϑ dt + ϑ(0) v0n · ϕ dx (5.11)
0 Ω
for all ϕ = ϕ(x) ∈ Vn and all ϑ = ϑ(t) ∈ C0∞ ([0, T )). Particularly, (5.11) also
holds for all ϕ ∈ Vm , where m ≤ n. Assume, for a while, that ϕ ∈ Vm is fixed.
Using all the types (5.8), (5.9), (5.10) of convergence of vn to v, one can pass to the
limit (for n → ∞) in (5.11) and show that
Z TZ
 
−v · ϕ ϑ̇ + ν∇v : ∇ϕ ϑ + v · ∇v · ϕ ϑ dx dt
0 Ω
Z T Z
= hf , ϕi ϑ dt + ϑ(0) v0 · ϕ dx (5.12)
0 Ω
∞
for all ϕ = ϕ(x) ∈ Vm and all ϑ = ϑ(t) ∈ C0 ([0, T )). Passing now to the limit for
1,2
m → ∞, we deduce that (5.12) holds for all ϕ = W0,σ (Ω) and all functions ϑ.

5. Global in time existence of the so called Leray–Hopf weak solution 44 / 50

5) The energy inequality
Recall inequality (5.7):
Z t Z t
kvn (t)k22 + 2ν k∇vn (τ )k22 dτ ≤ 2 hf , vn i dτ + kv0 k22 .
0 0

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

Z t
= 2 hf (τ ), vi dτ + kv0 k22 .
0

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

Z t
2
≥ kv(t)k2 + 2ν k∇vn (τ )k22 dτ.
0

This yields the energy inequality

Z t Z t
kv(t)k22 + 2ν k∇v(τ )k22 dτ ≤ kv0 k22 + 2 f (τ ), v(τ ) dτ. (5.1)
0 0

5. Global in time existence of the so called Leray–Hopf weak solution 45 / 50

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

The energy inequality implies that

lim sup kv(t)k22 ≤ kv0 k22 .

t→0+

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

lim inf kv(t)k22 ≥ kv0 k22 .

t→0+

These inequalities yield

lim kv(t)k22 = kv0 k22 .

t→0+

This identity, together with the weak L2 –continuity, enables us to conclude that

lim kv(t) − v0 k22 = 0.

t→0+

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

5. Global in time existence of the so called Leray–Hopf weak solution 46 / 50

 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 as
in the linear case of the Stokes problem because we do not have v̇ ∈ L2 (0, T ;
−1,2 −1,2
W0,σ (Ω)) – now, we only have v̇ ∈ L4/3 (0, T ; W0,σ (Ω)). Consequently, the
energy equality (or at least the energy inequality) does not automatically follow
from the definition of the weak solution.

• Is the weak solution unique?

• Is the weak solution regular provided that v0 and f are regular?

and many others

5. Global in time existence of the so called Leray–Hopf weak solution 47 / 50

References

[1] G. P. Galdi: An Introduction to the Mathematical Theory of the Navier–Stokes Equations. 2nd
edition, Springer 2011.
[2] G. P. Galdi: An Introduction to the Navier–Stokes initial–boundary value problem. In Funda-
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 für die Hydrodynamischen Grundgleichungen.
Math. Nachr. 4, 1951, 213–231.
[4] O. A. Ladyzhenskaya: The Mathematical Theory of Viscous Incompressible Flow. Gordon and
Breach, 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: Problèmes aux limites non homogènes et applications. Dunod, Paris
1968.

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.
Birkhäuser Advanced Texts, Basel–Boston–Berlin 2001.
[12] R. Temam: Navier–Stokes Equations. North–Holland, Amsterdam–New York–Oxford 1977.

References 49 / 50