Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 250, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
NAVIER-STOKES PROBLEM IN VELOCITY-PRESSURE
FORMULATION: NEWTON LINEARIZATION AND
CONVERGENCE
ANIS YOUNES, ABDENNACEUR JARRAY, MOHAMED BOUCHIBA
Abstract. In this article we study the nonlinear Navier-Stokes problem in
velocity-pressure formulation. We construct a sequence of a Newton-linearized
problems and we show that the sequence of weak solutions converges towards
the solution of the nonlinear one in a quadratic way.
1. Introduction
The stationary Navier-Stokes problem may be written in the form
−ν∆u + (u · ∇)u + ∇p = f
div u = 0
u=0
in Ω
in Ω
(1.1)
on Γ = ∂Ω
This equation describes the motion of an incompressible fluid contained in Ω and
subjected to an outside forces f , u is the velocity of fluid flow, p is the pressure and
ν its viscosity.
The variational formulation of the Navier Stokes equations in the classic form
is well studied in [8, 9, 16]. In most publications they uses a trilinear form in the
variational formulation for studying the nonlinear term presented in the equation
of momentum.
This paper is devoted to give another idea: we construct a sequence of a Newtonlinearized problems and we show, using Lax-Milgramm theorem, that the variational formulation of each one has an unique solution. We show then that the
sequence of weak solutions converges towards the solution of the nonlinear one in
a quadratic way.
The outline of the paper is as follows: In Section 2 we start by a Newtonlinearisation of the Navier Stokes equations. We obtain a sequence of linear problems and we show the existence of a weak solution. In Section 3 we show the
quadratic convergence of the sequence of the solutions in Theorem 3.3. In section
4 the nonhomogeneous problem is treated.
2000 Mathematics Subject Classification. 35J20, 49J96.
Key words and phrases. Navier-Stokes equations; Newton’s algorithm; convergence.
c
2014
Texas State University - San Marcos.
Submitted January 3, 2013. Published December 1, 2014.
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A. YOUNES, A. JARRAY, M. BOUCHIBA
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2. Linearized problems
Linearization. Let Ω a bounded domain of R2 with Lipschitz-continuous boundary Γ, and let
V = {v ∈ (H01 (Ω))2 , div v = 0}
with norm kukV = max{ku1 kH01 , ku2 kH01 }. We set L20 = (L20 (Ω))2 , and H01 (Ω) =
(H01 (Ω))2 with norm kukH01 = max{ku1 kH01 , ku2 kH01 } and W = H01 (Ω) × L20 .
The nonlinear term
!
∂u1
1
u1 ∂u
∂x + u2 ∂y
(u · ∇)u =
∂u2
2
u1 ∂u
∂x + u2 ∂y
can be written as
1
∇|u|2 + rot u ∧ u.
2
To solve (1.1) we construct a sequence of Newton-linearized problems. Starting
from an arbitrary u0 ∈ H01 (Ω) and p0 ∈ L20 we consider the iterative scheme:
(u · ∇)u =
−ν∆un+1 + (un+1 · ∇)un + (un · ∇)un+1 + ∇pn+1 = fn
div un+1 = 0
un+1 = 0
in Ω
in Ω
(2.1)
on Γ = ∂Ω
where fn = f + (un · ∇)un . Problem (2.1) is linear.
Variational formulation. The variational formulation of (2.1) is
Find (un+1 , pn+1 ) ∈ W such that
a0 (un+1 , v) + an (un+1 , v) + an (un+1 , v) + b(pn+1 , v) = Ln (v) ∀v ∈ H01 (Ω) (2.2)
b(q, un+1 ) = 0 ∀q ∈ L20
where the bilinear forms a0 , an , an are given for v, u ∈ H01 (Ω) and p ∈ L20 by
Z
Z
n
a0 (u, v) = ν
∇u∇v dx, a (u, v) = (u · ∇un )v dx,
Ω
Ω
Z
Z
Z
an (u, v) = (un · ∇u) v dx, b(p, v) =
∇pv dx = −
p div v dx
Ω
Ω
Ω
and Ln (v) = hfn , vi Using Green formula and div v = 0 we have b(p, v) = 0.
Then we associate to (2.2) the problem
Find un+1 ∈ V such that
a0 (un+1 , v) + an (un+1 , v) + an (un+1 , v) = Ln (v) ∀v ∈ V .
(2.3)
Lemma 2.1. Problem (2.2) is equivalent to problem (2.3).
Proof. Indeed, if (un+1 , pn+1 ) is a solution of problem (2.2) then un+1 is a solution
of (2.3). Reciprocally, if un+1 is a solution of the problem (2.3) then we apply de
Rham’s theorem: Let Ω a bounded regular domain of R2 and K a continuous linear
form on (H01 (Ω))2 . Then the linear form K vanishes on V if and only if there exists
a unique function pn+1 ∈ L2 (Ω)/R such that for all v ∈ H01 (Ω),
Z
K(v) =
pn+1 div v dx .
Ω
Let the linear form satisfies
K(v) = a0 (un+1 , v) + an (un+1 , v) + an (un+1 , v) − Ln (v).
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NAVIER-STOKES PROBLEM
3
Therefore we have K(v) = 0 for all v ∈ V , then de Rham’s theorem implies that
there exists a unique function pn+1 ∈ L2 (Ω)/R such that
Z
a0 (un+1 , v) + an (un+1 , v) + an (un+1 , v) − Ln (v) =
pn+1 div v dx ∀v ∈ H01 (Ω);
Ω
therefore,
a0 (un+1 , v) + an (un+1 , v) + an (un+1 , v) −
Z
pn+1 divvdx = Ln (v) ∀v ∈ H01 (Ω) .
Ω
Which gives the desired result.
Let us now show that problem (2.3) has an unique solution for each n. For this,
we need the following lemma.
Lemma 2.2. For fixed un ∈ V the form (u, v) → an (u, v) and (u, v) → an (u, v)
are continuous on H01 (Ω).
Proof. We have
2 Z
X
an (u, v) =
i,j=1
an (u, v) =
un,j
Ω
2 Z
X
i,j=1
uj
Ω
∂ui
vi dx,
∂xj
∂un,i
vi dx
∂xj
by Holder’s inequality we have
Z
∂ui
∂ui
un,j
vi dx ≤ kun,j kL4 kvi kL4 k
kL2
∂x
∂x
j
j
Ω
(2.4)
According to the Sobolev Imbedding Theorem, the space H 1 (Ω) is continuously
embedded in L4 (Ω). Then there exists C1 > 0 such that
|an (u, v)| ≤ C1 kukH01 (Ω) kvkH01 (Ω) kun kH01 (Ω) .
(2.5)
n
The same result holds with the term a ,
|an (u, v)| ≤ C2 kukH01 (Ω) kvkH01 (Ω) kun kH01 (Ω) .
(2.6)
To show the coercivity of the form a = a0 +an +an we have the following lemma.
Lemma 2.3. We have an (u, u) = 0 for all u ∈ V .
Proof. Note that
Z
an (u, u) =
(un .∇)u u dx =
Ω
where
2
∇(|u| ) =
1
2
Z
∂(u21 +u22 )
∂x
∂(u21 +u22 )
∂y
un ∇(|u|2 ) dx
!
Using Green’s formula and div un = 0 and boundary conditions we have
Z
Z
2
2an (u, u) =
∇|u| un dx = −
div un |u|2 dx = 0.
Ω
(2.7)
Ω
(2.8)
Ω
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A. YOUNES, A. JARRAY, M. BOUCHIBA
EJDE-2014/250
For α > 0, let Bα = {v ∈ V : kvkH01 (Ω) ≤ α}.
Lemma 2.4. We have
a(u, u) ≥ (νC3 − αC2 )kuk2H 1 (Ω)
0
∀u ∈ V, ∀un ∈ Bα
(2.9)
with C3 = min[ 2(Cp1(Ω))2 , 12 ].
Proof. Using (2.8) we obtain a(u, u) = a0 (u, u)+an (u, u). By the Poincare inequality,
kukL2 (Ω) ≤ Cp (Ω)k∇ukL2 (Ω) ,
(2.10)
we obtain
a0 (u, u) = νk∇uk2L2 (Ω) ≥ ν min[
1
1
, ]kuk2H 1 (Ω) .
0
2(Cp (Ω))2 2
Then
a0 (u, u) ≥ νC3 kuk2H 1 (Ω) .
(2.11)
0
Using (2.6) we have
an (u, u) ≤ C2 αkuk2H 1 (Ω)
0
∀u ∈ V, un ∈ Bα
(2.12)
which gives
an (u, u) ≥ −C2 αkuk2H 1 (Ω)
0
∀u ∈ V, un ∈ Bα
(2.13)
with (2.11) we have the result.
Lemma 2.5. For kf k(L2 (Ω))2 small enough or ν large enough there is α∗ > 0
independent of n such that kun kH01 (Ω) ≤ α∗ for all n ∈ N where un is solution of
(2.3) with n instead of n + 1.
3
Proof. We must have (νC3 − C2 α∗ ) > 0. So we choose α∗ < νC
C2 .
Remains to show by induction that if un is solution of (2.3) with n instead of
n + 1, then kun kH01 (Ω) ≤ α∗ for all n ∈ N. Let u0 ∈ Bα∗ and assume that un ∈ Bα∗ .
We note u = un+1 is a solution of (2.3) and kf k2 = kf k(L2 (Ω))2 . We have
Z
a(u, u) = Ln (u) = (f + (un ∇)un )u dx .
(2.14)
Ω
∗2
Then a(u, u) ≤ (kf k2 + Cα )kukH01 (Ω) .
From (2.9) we obtain (νC3 − C2 α∗ )kukH01 (Ω) ≤ (kf k2 + Cα∗ 2 ) which gives
kukH 1 (Ω) ≤
0
kf k2 + Cα∗ 2
.
(νC3 − C2 α∗ )
So to deduce the result we must have
kf k2 + Cα∗ 2
≤ α∗ .
(νC3 − C2 α∗ )
We put
P (α∗ ) = (C + C2 )α∗ 2 − νC3 α∗ + kf k2 ≤ 0,
α∗ <
νC3
C2
Therefore, the discriminant of the polynomial P (α∗ ) must verify
∆ = ν 2 C3 2 − 4(C + C2 )kf k2 > 0 .
(2.15)
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NAVIER-STOKES PROBLEM
Then
kf k2 <
and hence P (α∗ ) has two roots
ν 2 C3 2
4(C + C2 )
√
νC3 − ∆
α1 =
,
2(C + C2 )
5
(2.16)
√
νC3 + ∆
α2 =
2(C + C2 )
3
Since α2 > 0 we can choose 0 < α∗ < min( νC
C2 , α2 ).
Theorem 2.6. (1) For f ∈ (L2 (Ω))2 satisfying (2.14), problem (2.3) has a unique
solution un+1 ∈ V ∩ Bα∗ .
(2) If u0 ∈ Bα∗ ∩ H 2 (Ω), then un+1 ∈ H 2 (Ω).
Proof. (1) Since un ∈ Bα∗ , we have
|Ln (v)| ≤ (kf k2 + Cα∗ 2 )kvkH01 (Ω)
which gives the continuity of Ln and using Lemma 2.1, Lemma 2.3 and Lemma 2.4
with Lax-Milgram Theorem we obtain the result.
(2) We assume that un ∈ H 2 (Ω) then (un ∇)un ∈ (L2 (Ω)2 , which implies that
fn = f + (un ∇)un ∈ (L2 (Ω)2 for f ∈ (L2 (Ω)2 , and by classical regularity Theorem
we have un+1 ∈ H 2 (Ω).
3. Convergence
The sequence (un )n∈N , solutions of (2.3) with n instead of n + 1, satisfy
kun kH01 (Ω) ≤ α∗
∀n ≥ 0,
(3.1)
H01 (Ω).
which implies that the sequence (un )n∈N is bounded in
Then there exist
a subsequence that converges weakly to φ in H01 (Ω). Since the injection of H01 (Ω)
in (L2 (Ω))2 is compact, there exists a subsequence still noted un which converges
strongly to φ in (L2 (Ω))2 .
We need the following result.
Lemma 3.1. For v ∈ V , we have:
(1) limn→∞ a0 (un+1 , v) = a0 (φ, v);
R
˙
(2) limn→∞ an (un+1 , v) = a∞ (φ, v) = Ω (φ∇)φvdx;
R
n
∞
˙
(3) limn→∞ a (un+1 , v) = a (φ, v) = Ω (φ∇)φvdx;
R
˙
(4) We have limn→∞ Ln (v) = L∞ (v) = Ω [f + (φ∇)φ]vdx.
Proof. (1) Since un * φ, and by linearity of u → a0 (u, v) we have a0 (un+1 , v) →
a0 (φ, v) for all v ∈ V .
(2) Let
Z
n
∞
˙ n − (φ∇)φ}v
˙
E = |a (un+1 , v) − a (φ, v)| =
{(un+1 ∇)u
dx
(3.2)
Ω
We can write
(un+1 · ∇)un − (φ.∇)φ = ((un+1 − φ) · ∇)un + (φ · ∇)(un − φ)
(3.3)
2
which gives with un ∈ H (Ω) and using Green’s theorem,
E ≤ C[kun+1 − φk2 kun kH 1 kvkH 1 + kun − φk2 (k∇vkH 1 kφkH 1 + k∇φkH 1 kvkH 1 )].
(3.4)
Since un converges strongly to φ in (L2 (Ω))2 , it follows that E → 0.
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A. YOUNES, A. JARRAY, M. BOUCHIBA
EJDE-2014/250
(3) Let
Z
F = |an (un+1 , v) − a∞ (φ, v)| =
{(un · ∇)un+1 − (φ · ∇)φ}v dx .
Ω
Then
F ≤ C[kun − φk2 kun+1 kH 1 kvkH 1 + kun+1 − φk2 (kvkH 1 k∇φkH 1 + kφkH 1 k∇vkH 1 )];
(3.5)
thus F → 0.
(4) Let
Z
G = |Ln (v) − L∞ (v)| ≤
|(un ∇un ) − (φ∇φ)||∇v| dx.
Ω
Then
G ≤ C[kun − φk2 (kun kH 1 kvkH 1 + kφkH 1 k∇vkH 1 + k∇φkH 1 kvkH 1 )].
(3.6)
Then Lemma 3.1 gives the desired result.
For using de Rham’s Theorem, let L a continuous linear form on (H01 (Ω))2 which
vanishes on V if and only if there exists a unique function ϕ ∈ L2 (Ω)/R such that
for all v ∈ H01 (Ω),
Z
L(v) =
ϕ div v dx.
Ω
Theorem 3.2. We have limn→∞ un = φ in V then φ is a solution of (1.1).
Proof. It follows from Lemma 3.1 that
lim a0 (un+1 , v) + an (un+1 , v) + an (un+1 , v) = a0 (φ, v) + 2a∞ (φ, v) = L∞ (v) .
n→∞
Let the linear form L(v) = a0 (φ, v) + a∞ (φ, v) + a∞ (φ, v) − L∞ (v). Therefore
L(v) = 0 for all v ∈ V , then de Rham’s theorem implies that there exists a unique
function p ∈ L2 (Ω)/R such that
Z
a0 (φ, v) + 2a∞ (φ, v) − L∞ (v) =
p div v dx ∀v ∈ H01 (Ω)
(3.7)
Ω
which gives
Z
Z
Z
Z
˙
ν
∇φ ∇v dx + (φ∇)φv dx −
p div v dx =
f v dx ∀v ∈ H01 (Ω),
Ω
Ω
Ω
Ω
Z
˙ + ∇p − f )v dx = 0 ∀v ∈ H01 (Ω) .
(−ν∆φ + (φ∇)φ
(3.8)
(3.9)
Ω
Then in D0 (Ω),
˙ + ∇p − f = 0 .
− ν∆φ + (φ∇)φ
Since φ ∈ V we conclude that φ is the solution of (1.1).
(3.10)
Theorem 3.3. Let un+1 be the solution of (2.2), and φ be the solution of (1.1).
Then convergence of the sequence (un+1 )n∈N towards φ is quadratic; i.e.,
kun+1 − φkH01 (Ω) ≤ C2 kun − φk2H 1 (Ω)
0
(3.11)
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NAVIER-STOKES PROBLEM
7
Proof. Let ωn = un − φ and χn = pn − p. Subtracting problem (2.1) from (1.1) we
obtain
−ν∆ωn+1 + (ωn+1 ∇)un + (un ∇)ωn+1 + ∇χn+1 = (ωn ∇)ωn in Ω
div ωn+1 = 0
ωn+1 = 0
in Ω
(3.12)
on Γ
The variational formulation of (3.12) is
Find (ωn+1 , χn+1 ) ∈ W such that
a(ωn+1 , v) + b(χn+1 , v) = Fn (v) ∀v ∈ H01 (Ω)
b(q, ωn+1 ) = 0 ∀q ∈
(3.13)
L20 ,
n
where a = a0 + a + an and
Z
b(q, ωn+1 ) = −
q div ωn+1 dx,
Z
(ωn ∇)ωn v dx.
Fn (v) =
Ω
Ω
Since div ωn+1 = 0, using Lemma2.1 and Lemma 2.5, for un ∈ Bα∗ and v = ωn+1 ,
we obtain
(νC1 − Cα∗ )kωn+1 k2H 1 (Ω) ≤ a(ωn+1 , ωn+1 ) = F( ωn+1 )
0
(3.14)
≤ Ckωn k2H 1 (Ω) kωn+1 kH01 (Ω) .
0
This gives 3.10, with C2 =
C
(νC1 −Cα∗ )
and the convergence is quadratic.
4. Nonhomogeneous problem
We are concerned now with the nonhomogeneous problem
−ν∆u + (u · ∇)u + ∇p = f in Ω
div u = 0
u=g
in Ω
(4.1)
on Γ
Where the state u is sought in the space (H 1 (Ω))2 ∩ V .
Throughout this section Ω denotes a bounded domain in R2 , with Lipschitzcontinuous boundary Γ = ∩Γi i = 1, . . . , 4. We assume in this section that
Z
g.ni dσ = 0 with g ∈ H = (H 1/2 (Γ))2 and f ∈ K = (H −1 (Ω))2 .
(4.2)
Γi
We assume also that for a given g ∈ H satisfying 3.11, for any c > 0 there exists a
function w0 ∈ (H 1 (Ω))2 such that
div w0 = 0,
|an (w0 , un )| ≤
w0 |Γ = g,
ckun k2H 1 (Ω)
0
∀un ∈ V.
(4.3)
(4.4)
The existence of w0 satisfying 3.11, 3.14 is a technical result due to Hopf [11].
Theorem 4.1. Given (g, f ) ∈ K × H satisfying 3.14, there exists a pair (u, p) ∈
(H 1 (Ω))2 × L20 (Ω) which is a solution of (4.1).
Proof. Let ξ0 = u0 − w0 where w0 verify 3.11, 3.14 and an arbitrary u0 ∈ V . We
consider the sequence of linear problems
−ν∆ξn+1 + (ξn+1 .∇)ξn + (ξn · ∇)ξn+1 + ∇pn+1 = fn in Ω
div ξn+1 = 0
ξn+1 = 0
in Ω
on Γ
(4.5)
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A. YOUNES, A. JARRAY, M. BOUCHIBA
EJDE-2014/250
with ξn+1 = un+1 − w0 , un+1 ∈ H01 (Ω) and fn = f + (ξn ∇)ξn + ν∆w0 − (w0 · ∇)w0 .
Then ξn+1 is a solution of the variational problem
Find ξn+1 ∈ V such that
a(ξn+1 , v) = Ln (v) ∀v ∈ V ,
(4.6)
where a(ξ, v) = a0 (ξ, v) + an (ξ, v) + an (ξ, v) + a? (ξ, v) with
Z
Z
˙ 0 v dx + (w0 ∇)ξv
˙
a? (ξ, v) = (ξ ∇)w
dx
Ω
Ω
and Ln (v) = hfn , vi.
Taking c > ν and using 2.9 we obtain
|a(ξn+1 , ξn+1 )| ≥ (ν − c)kξn+1 k2H 1 (Ω)
0
(4.7)
Thus we have the coercivity and L is obviously continuous on V . We observe that
problem (4.6) fits into the framework of section 1 and therefore the sequence ξn
converges towards a solution of (4.1).
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EJDE-2014/250
NAVIER-STOKES PROBLEM
9
Anis Younes
Tunis El Manar University, Faculty of Sciences of Tunis, Tunisia
E-mail address: younesanis@yahoo.fr
Abdennaceur Jarray
Tunis El Manar University, Faculty of Sciences of Tunis, Tunisia
E-mail address: abdennaceur.jarray@gmail.com
Mohamed Bouchiba
Carthage University, National Institute of Applied Sciences and Technology, Tunisia
E-mail address: mohamed.bouchiba@yahoo.fr