Electronic Journal of Differential Equations, Vol. 2005(2005), No. 125, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
DECAY OF SOLUTIONS TO EQUATIONS MODELLING
INCOMPRESSIBLE BIPOLAR NON-NEWTONIAN FLUIDS
BO-QING DONG
Abstract. This article concerns systems of equations that model incompressible bipolar non-Newtonian fluid motion in the whole space Rn . Using the
improved Fourier splitting method, we prove that a weak solution decays in
the L2 norm at the same rate as (1 + t)−n/4 as the time t approaches infinity.
Also we obtain optimal L2 error-estimates for Newtonian and Non-Newtonian
flows.
1. Introduction
Consider the viscous incompressible fluid motion governed by the momentum
and continuity equations
∂t u + (u · ∇)u − ∇ · τ v + ∇π = 0 in Rn × (0, ∞),
∇·u=0
in Rn × (0, ∞),
(1.1)
(1.2)
with the initial condition
u(x, 0) = u0 in Rn .
(1.3)
Here n ≥ 2, the gradient ∇ = (∂x1 , . . . , ∂xn ), u = (u1 , . . . , un ) and π denote the
v
unknown velocity and pressure of the fluid motion. τ v = (τij
) is the stress tensor
specified in the form
v
τij
= 2 (µ0 + µ1 |e(u)|p−2 ) eij (u) − 2µ2 ∆eij (u)
(1.4)
with the constant viscosities µ0 > 0, µ1 , µ2 ≥ 0 and the symmetric deformation
velocity tensor e(u) = (eij (u)),
∂uj 1 ∂ui
+
, |e(u)| = (eij (u)eij (u))1/2 .
(1.5)
eij (u) =
2 ∂xj
∂xi
When µ1 = µ2 = 0, the Stokes Law
v
τij
= 2µ0 eij (u)
(1.6)
holds. The fluids, such as water and alcohol, satisfying the linear equation expressed
by (1.6) is said to be Newtonian, and (1.1) turns out to be the famous Navier-Stokes
equations (refer to [10, 16]). However the nonlinear constitutive equation expressed
by (1.4) with µ1 , µ2 > 0 is related to other non-Newtonian fluids such as the molten
2000 Mathematics Subject Classification. 35B40, 35Q35, 76A05.
Key words and phrases. Decay; bipolar non-Newtonian fluids; Fourier splitting method.
c
2005
Texas State University - San Marcos.
Submitted March 28, 2005. Published November 7, 2005.
1
2
B.-Q. DONG
EJDE-2005/125
plastics, dyes, adhesives, paints and greases. When µ2 = 0, system (1.1)-(1.4) was
first proposed by Ladyzhenskaya [11] and is known as the Ladyzhenskaya equations.
The fluid is said to be monopolar because the only first order derivative of the
velocity field is involved in the stress tensor (see (1.4) for µ2 = 0), whereas the fluid
is bipolar if the second order derivative arises in τ v (see (1.4) for µ1 , µ2 > 0). The
theory of bipolar fluids is compatible with the basic principles of thermodynamics,
including the Clausius-Duhem inequality and the material frame indifference (See
[1] for a detailed description of multipolar fluids). Moreover, the fluid is shear
thinning if p < 2 and shear thickening if p > 2 (when p = 2, the system turns out
to be Navier-Stokes equations).
There is an extensive literature on the solutions of the incompressible nonNewtonian fluids. Ladyzhenskaya [11] and Lions [12] first discussed the existence
and uniqueness of weak solutions of the sort monopolar model (see (1.4) for µ2 = 0),
and more recently, Du and Gunzguiger [5] studied the somewhat more general existence and uniqueness results in bounded domains. Pokorny [14] investigated the
Cauchy problem for both monopolar and bipolar fluids in whole spaces. As for the
decay properties of solutions, on the one hand, Necăsová and Penel [13] recently
examined the logarithmic decay in R2 and algebraic decay in R3 with respect to
the monopolar shear thickening fluids (p ≥ 3) by the Schonbek’s Fourier splitting
method [15]. With the aid of Wiegner’s method [17], Guo and Zhu [6] improved
the algebraic decay rates. Higher decay rates were recently proved by Dong [3]
based on the arguments of Kajikiya and Miyakawa [8]. In particular, by improving
Schonbek’s Fourier splitting method, the optimal algebraic decay rate in R2 of this
monopolar model was obtained by Dong and Li [2] in the following form
ku(t)kL2 ≤ C(1 + t)−1/2 ,
ku(t) − et∆ u0 kL2 ≤ C(1 + t)−3/4 ,
∀ t > 0.
(1.7)
On the other hand, Guo and Zhu [7] considered also the decay of the weak solution
of the bipolar fluids (see (1.4) for µ1 , µ2 > 0). Based on the fourth order linear
parabolic equation and the Wiegner’s method [17], the decay rate of the L2 norm
they obtained is only one-half of the decay rate to the linear heat equation, that is
n
n
ku(t)kL2 ≤ C(1 + t)− 4r − 8 , ∀ t > 0;
ku(t) − et∆ u0 kL2 → 0,
as t → ∞.
(1.8)
assuming Lr ∩L2 integrability of the initial data. Furthermore, as for the time decay
of Navier-Stokes equations in whole spaces, the sharp decay rates were obtained by
Schonbek[15], Kajikiya and Miyakawa [8], Wiegner[17] and references cited therein.
The aim of this paper is investigate the optimal rate of decay of global solutions to
the Cauchy problem of the bipolar shear thinning fluids (1.1)-(1.4) (p ≥ 3, µ1 , µ2 >
0 in (1.4)). We use the improved Fourier splitting methods developed by Dong et
al [2, 4] and Zhang [18], and the rigorous analysis of the lower frequency effect of
the lower dissipative term ∆u which determines mainly the time decay rates of the
solutions. We obtain the optimal L2 -decay rate, which is the same as that of the
linear heat equation
ku(t)kL2 ≤ C(1 + t)−n/4 ,
1
2
∀ t > 0,
(1.9)
assuming L ∩ L integrability of the initial data. Furthermore, since the bipolar
non-Newtonian flow is modified from the Newtonian flow, we examine the L2 decay estimates of the error u(t) − ũ(t). Here u(t) denotes weak solution of the
non-Newtonian system (1.1)-(1.4) with µ1 , µ2 > 0, whereas ũ denotes the weak
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DECAY OF SOLUTIONS
3
solution of the Newtonian system of (1.1)-(1.4) with µ1 = µ2 = 0. The optimal
error estimates we obtained are the following
(1.10)
ku(t) − ũ(t)k = o (1 + t)−n/4 , as t → ∞.
This paper is organized as follows. In Sections 2 we define weak solutions and
state some preliminary lemmas. Decay estimates of the non-Newtonian flow are
described in Section 3. Decay estimates of the error between the non-Newtonian
and Newtonian flows u(t) − ũ(t) are derived in Section 4.
2. Preliminaries
Let k · kq = k · kLq (k · k = k · k2 ) be the norm of the usual scalar and vector
Lebesgue space Lq (Rn ) and k · km,p = k · kW m,p be the norm of the Sobolev space
∞
W m,p (Rn ). The space H denotes the L2 −closure of C0,σ
(Rn ), which is the set
of smooth divergence-free vector fields with compact supports in Rn . The space
1,q
∞
W0,σ
(Rn ) denotes the closure of C0,σ
(Rn ) in W 1,q (Rn ). The Fourier transformation
of a function f is denoted by fˆ or F [f ]. C > 0, independent of the quantities t, x,
ρ, u and ũ, is a generic constant, which may depend on the initial data u0 .
Without loss of generality, we assume that µ0 = µ1 = µ2 = 1 in (1.4). Substitution of (1.4) into (1.1) produces
ut − 4u + ∆2 u + (u · ∇)u − ∇ · (|∇u|p−2 ∇u) + ∇π = 0,
(2.1)
in Rn × (0, ∞).
By a weak solution of the initial value problem (1.2)-(1.4) and (2.1) for n ≥ 2
2n
and p ≥ 1 + n+2
(see [11, 12, 14]), we mean a vector field
1,p
2,2
u ∈ Lp (0, T ; W0,σ
(Rn )) ∩ L∞ (0, T ; H) ∩ L2 (0, T ; W0,σ
(Rn )),
∀T > 0
satisfying
Z tZ
∂ϕ
dxds
u(t) · ϕ(t) dx −
u·
∂s
Rn
0
Rn
(2.2)
Z tZ
Z tZ
Z
∂ui
+
ϕi dxds +
uj
τij (e(u)) · eij (ϕ)dxds =
u0 · ϕ(0) dx
∂xj
0
Rn
0
Rn
Rn
Z
2,2
1,p
a. e. t ∈ (0, T ) for every ϕ ∈ C 1 ([0, T ), H) ∩ C([0, T ), W0,σ
(Rn ) ∩ W0,σ
(Rn )) and
ϕ(x, T ) = 0. Moreover, we assume that the weak solution also satisfies the following
energy inequality
Z
Z
Z
Z
1 d
|u|2 dx +
|∇u|2 dx +
|∆u|2 dx +
|∇u|p dx ≤ 0.
(2.3)
2 dt Rn
n
n
n
R
R
R
It should be noted that a weak solution can be specified as a limit of a sequence
of smooth approximate solutions in a local L2 -norm due to the standard FaedoGalerkin argument. Thus the decay estimates with respect to the weak solution
become limits of those of the smooth approximate solutions (see, for example,
Kajikiya and Miyakawa [8]). Since we only consider the decay estimates of the
weak solution in L2 -norm, without loss of generality, we may suppose that the
weak solution admit enough regularity so that we can work on the weak solution
directly rather than on the sequence of smooth approximate solutions.
Let us now recall some preliminary lemmas.
4
B.-Q. DONG
EJDE-2005/125
lemma 2.1 (Gronwall Inequality). Let f (t), g(t), h(t) be nonnegative continuous
functions and satisfying the inequality
Z t
g(t) ≤ f (t) +
g(s)h(s)ds, ∀ t > 0,
0
where f 0 (t) ≥ 0. Then
g(t) ≤ f (t) exp
t
Z
h(s)ds ,
∀ t > 0.
(2.4)
0
lemma 2.2. Assume that u0 ∈ H ∩ L1 (Rn ) and u is a weak solution of (1.2)-(1.4)
and (2.1). Then
sup ku(t)k ≤ ku0 k,
(2.5)
0≤t≤∞
and (i) 2 < p < 3, n = 2,
t
Z
|û(ξ, t)| ≤ C + C|ξ|
Z t
4−p
2
2
ku(s)k2 ds + C|ξ|
ku(s)k 4−p ds
,
0
(ii) 1 +
2n
n+2
(2.6)
0
≤ p < 3, n ≥ 3,
Z
|û(ξ, t)| ≤ C + C|ξ|
t
Z
2
ku(s)k ds + C|ξ|
0
t
ku(s)k
2α
2−β
2−β
2
ds
,
(2.7)
0
where α = 2n−(n−2)(p−1)
, β=
4
(iii) p ≥ 3, n ≥ 2,
(n+2)(p−1)−2n
,
4
Z
|û(ξ, t)| ≤ C + C|ξ|
t
ku(s)k2 ds + C|ξ|.
(2.8)
0
Proof. From the energy inequality (2.3), it is easy to get the first inequality (2.5).
We now prove (2.6)-(2.8), first, applying the Fourier transformation of (2.1) we
have
ût + (|ξ|2 + |ξ|4 )û = F [∇ · (|e(u)|p−2 e(u)) − (u · ∇)u − ∇π] =: G(ξ, t).
(2.9)
Now we estimate G(ξ, t). Taking divergence in (2.1) to get,
X ∂2
[−ui uj + |e(u)|p−2 eij (u)].
∆π =
∂x
∂x
i
j
i,j
The Fourier transformation yields
X
|ξ|2 F [π] =
ξi ξj F [−ui uj + |e(u)|p−2 eij (u)],
i,j
and thus
|F [∇π]| = |ξ|F [π] ≤ |F [∇ · (|e(u)|p−2 e(u))]| + |F [(u · ∇)u]|.
(2.10)
Furthermore,
|F [(u · ∇)u]| = |F [div(u ⊗ u)]| ≤
XZ
i,j
|ui uj | |ξj |dx ≤ |ξ| kuk2 ,
(2.11)
Rn
|F [∇ · (|e(u)|p−2 e(u))]| ≤ |ξ||F [|e(u)|p−2 e(u)]| ≤ |ξ|k∇ukp−1
p−1 .
(2.12)
So inserting (2.10)-(2.12) into G(ξ, t), we have
|G(ξ, t)| ≤ C|ξ|kuk2 + C|ξ|k∇ukp−1
p−1 .
(2.13)
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DECAY OF SOLUTIONS
5
From (2.9), it follows easily that,
2
4
d (|ξ|2 +|ξ|4 )t ≤ G(ξ, t)e(|ξ| +|ξ| )t .
ûe
dt
Integrating in time gives,
Z t
−(|ξ|2 +|ξ|4 )t
−(|ξ|2 +|ξ|4 )(t−s) |û(ξ, t)| ≤ e
û0 (ξ) +
G(ξ, t)e
ds
0
t
Z
(2.14)
|G(ξ, t)|ds
≤ |û0 (ξ)| +
0
Z t
Z t
p−1
k∇u(s)kp−1
ds.
ku(s)k2 ds + C|ξ|
≤ C + |ξ|
0
0
Rt
Now we estimate 0 k∇u(s)kp−1
p−1 ds in three cases (note that the case p = 2, n = 2
is the Navier-Stokes equations [10, 16]):
(i) 2 < p < 3, n = 2;
2n
(ii) 1 + n+2
≤ p < 3, n ≥ 3;
(iii) p ≥ 3, n ≥ 2.
Case (i): 2 < p < 3, n = 2. By Gagliardo-Nirenberg inequality (refer to [9]),
Z t
Z t
k∇u(s)kp−1
ds
≤
|ξ|ku(s)kkD2 u(s)kp−2 ds
p−1
0
0
≤
t
Z
2
4−p
Z
2
2
4−p
2
ku(s)k 4−p ds
0
≤
kD2 u(t)k2 dt
p−2
2
0
t
Z
∞
ku(s)k 4−p ds
,
0
R∞
noting that 0 kD2 u(t)k2 dt ≤ C.
2n
≤ p < 3, n ≥ 3.
Case (ii): 1 + n+2
Z
t
k∇u(s)kp−1
p−1 ds ≤
0
Z
t
ku(s)kα kD2 u(s)kβ ds
0
≤
t
Z
2α
ku(s)k 2−β ds
2−β
Z
2
t
Z
kD2 u(t)k2 dt
β2
(2.15)
0
0
≤
∞
2α
ku(s)k 2−β ds
2−β
2
,
0
where α = 2n−(n−2)(p−1)
, β = (n+2)(p−1)−2n
, and 0 < β < 1.
4
4
Case (iii): p ≥ 3, n ≥ 2. With the above definition of the weak solution, we know
also that ∇u ∈ L2 ((0, ∞) × Rn ) ∩ Lp ((0, ∞) × Rn ), so by using the interpolation
technology, ∇u ∈ Lp−1 ((0, ∞) × Rn ), i.e.
Z ∞
k∇u(s)kp−1
(2.16)
p−1 ds ≤ C.
0
Hence (2.14)-(2.16) imply the assertions of lemma 2.2 and the proof is complete. 6
B.-Q. DONG
EJDE-2005/125
3. Decay estimates of the non-Newtonian flows
As is well known that the weak solutions of Navier-Stokes equations have the
optimal decay estimates in the whole space [8, 15, 17]. In this section, we show
that the non-Newtonian flow has also the same optimal L2 time decay estimates.
The results read as follows.
Theorem 3.1. Assume that u0 ∈ H ∩ L1 (Rn ). Let u(t) be a weak solution of
2n
(1.2)-(1.4) and (2.1). Then, for n = 2, p > 2 and for n ≥ 3, p ≥ 1 + n+2
, we have
ku(t)k ≤ C(1 + t)−n/4 ,
∀ t > 0.
(3.1)
Proof. From the energy inequality (2.3), it follows that
Z
Z
1 d
2
|u| dx +
|∇u|2 ≤ 0.
2 dt Rn
Rn
(3.2)
Applying Plancherel’s theorem to (3.2) yields
Z
Z
1 d
|û(ξ, t)|2 dξ +
|ξ|2 |û(ξ, t)|2 dξ ≤ 0.
2 dt Rn
Rn
(3.3)
Let f (t) be a smooth function of t with f (0) = 1, f (t) > 0 and f 0 (t) > 0, then
Z
Z
Z
d
f (t)
|û(ξ, t)|2 dξ + 2f (t)
|ξ|2 |û(ξ, t)|2 dξ ≤ f 0 (t)
|û(ξ, t)|2 dξ.
dt
n
n
n
R
R
R
Set B(t) = {ξ ∈ Rn : 2f (t)|ξ|2 ≤ f 0 (t)}. Then
Z
2f (t)
|ξ|2 |û(ξ, t)|2 dξ
Rn
Z
Z
= 2f (t)
|ξ|2 |û(ξ, t)|2 dξ + 2f (t)
|ξ|2 |û(ξ, t)|2 dξ
c
B(t)
B(t)
Z
2
2
≥ 2f (t)
|ξ| |û(ξ, t)| dξ
B(t)c
Z
Z
0
2
0
≥ f (t)
|û(ξ, t)| dξ − f (t)
|û(ξ, t)|2 dξ.
Rn
B(t)
Therefore,
d
f (t)
dt
Z
2
Z
0
|û(ξ, t)|2 dξ.
|û(ξ, t)| dξ ≤ f (t)
Rn
B(t)
Integrating in time yields
Z
f (t)
Rn
|û(ξ, t)|2 dξ ≤
Z
Rn
|û0 |2 dξ + C
Z
0
t
f 0 (s)
Z
|û(ξ, s)|2 dξds.
(3.4)
B(s)
Now we study three cases: (i) 2 < p < 3, n = 2; (ii) 1 +
p ≥ 3, n ≥ 2.
2n
n+2
≤ p < 3, n ≥ 3; (iii)
EJDE-2005/125
DECAY OF SOLUTIONS
(i) Case 2 < p < 3, n = 2. Let A2 =
According to (2.6),
Z
f 0 (t)
2f (t) ,
7
and ωn be volume of unit ball in Rn .
|û(ξ, t)|2 dξ
f (t)
R2
Z
≤
t
Z
|û0 |2 dξ + Cωn
R2
f 0 (s)
A
n
1
0
0
s
Z
s
4−p
o2
2
2
ku(τ )k2 dτ + ρ
ku(τ )k 4−p dτ
ρ dρ ds
0
0
Z t
n f 0 (s) f 0 (s) 2 Z s
2 o
≤C +C
f 0 (s)
ku(τ )k2 dτ
+
ds
2f (s)
2f (s)
0
0
Z t
n f 0 (s) 2 Z s
4−p o
2
0
f (s)
ku(τ )k 4−p dτ
+C
ds.
2f (s)
0
0
Z
Z
+ρ
(3.5)
On the one hand, using (2.5) on the right hand side of (3.5), we obtain
Z
f (t)
|û(ξ, t)|2 dξ ≤ C + C
R2
Z
0
t
o
n f 0 (s)
f 0 (s) 2 2
+
(s + s4−p ) ds.
f 0 (s)
2f (s)
2f (s)
(3.6)
0
4
(t)
5
Let f (t) = (ln(e + t))5 . Then f 0 (t) = 5(ln(e+t))
, and ff (t)
= (e+t) ln(e+t)
. By (3.6)
e+t
and an elementary calculation based on Plancherel’s theorem, we have
(ln(e + t))5
Z
|u(x, t)|2 dx
Z
5
= (ln(e + t))
|û(ξ, t)|2 dξ
R2
Z tn
(ln(e + s))3
s2 (ln(e + s))2
s4−p (ln(e + s))2 o
≤C +C
+
+
ds
(e + s)2
(e + s)3
(e + s)3
0
Z t
(ln(e + s))2
≤C +C
ds (because 1 < 4 − p < 2)
e+s
0
≤ C(ln(e + t))3 ,
R2
and so
ku(t)k ≤ C(ln(e + t))−1 .
(3.7)
By the inductive argument, we suppose that
ku(t)k ≤ C(ln(e + t))−m
∀ m ∈ N.
(3.8)
Inserting (3.8) into the right hand side of (3.5), letting f (t) = (ln(e + t))2m+3 , and
Rt
using (3.5) and the inequality 0 (ln(e + s))−m ds ≤ C(e + t) ln(e + t))−m , thus from
8
B.-Q. DONG
EJDE-2005/125
(3.5)
(ln(e + t))2m+3
Z
|u(x, t)|2 dx
Z
2m+3
= (ln(e + t))
|û(ξ, t)|2 dξ
R2
R2
Z tn
(ln(e + s))2m+1
1
(ln(e + s))2m−(4−p)m o
≤C +C
+
ds
+
(e + s)2
e+s
(e + s)(e + s)p−2
0
Z t
1
≤C +C
ds
0 e+s
≤ C ln(e + t).
Here we used that (ln(e + s))k ≤ c(k)(e + s), for all k > 0. The above inequality
implies
ku(t)k ≤ C(ln(e + t))−m−1
∀ m ∈ N.
(3.9)
On the other hand, from (3.5) and the Hölder inequality,
Z
f (t)
|û(ξ, t)|2 dξ
R2
Z t
Z t
f 0 (s) 2 Z t
f 0 (s)
ds +
sf 0 (s)
ds
ku(s)k4 ds
≤C +C
f 0 (s)
2f (s)
2f (s)
0
0
0
Z t
4−p
f 0 (s) 2 Z t
4−p
4
2
ds
.
ku(s)k 4−p ds
+C
s 2 f 0 (s)
2f
(s)
0
0
Let f (t) = (1 + t)2 . Then
Z
(1 + t)2
|û(ξ, t)|2 dξ
R2
Z
≤ C(1 + t) + C(1 + t)
t
4−p
2
ku(s)k4 ds + C(1 + t)
0
1
2
Noting that <
(3.10), we have
4−p
2
Z
t
Z
4
ku(s)k 4−p ds
(3.10)
4−p
2
.
0
< 1 and applying the Young inequality to the last term of
t
ku(s)k
4
4−p
ds
4−p
2
Z
t
≤C
4
ku(s)k 4−p ds + C.
(3.11)
0
0
Inserting (3.8) and (3.11) into (3.10), we get the following estimate
Z
Z t
n
2
(1 + t)
|û(ξ, t)| dξ ≤ C + C
ku(s)k2 (1 + s) (1 + s)−1 (ln(e + s))−m
R2
0
o
m(2p−4)
+ (1 + s)−1 (ln(e + s))− 4−p
ds.
Let
Z
2
Z
|û(ξ, t)| dξ = (1 + t)
g(t) = (1 + t)
R2
|u(x, t)|2 dx, f (t) = C,
R2
h(t) = (1 + t)−1 (ln(e + t))−m + (1 + t)−1 (ln(e + t))−
m(2p−4)
4−p
.
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DECAY OF SOLUTIONS
9
When the integer m is suitable large, it is simple to deduce that
Applying Lemma 2.1, we have
Z ∞
g(t) ≤ C exp
h(t)dt ≤ C,
R∞
0
h(t)dt < ∞.
0
and thus
ku(t)k ≤ C(1 + t)−1/2 .
2n
(ii) Case 1 + n+2
≤ p < 3, n ≥ 3. Inserting (2.7) into the right hand of (3.4) and
using Hölder inequality and Young inequality, we have
Z
f (t)
|û(ξ, t)|2 dξ
Rn
Z
t
Z
≤
f 0 (s)
|û0 |dξ + Cωn
Rn
Z
0
A
n
1
0
2−β
o2
Z s
2α
2
ρn−1 dρds
+ρ
ku(τ )k2 dτ + ρ
ku(τ )k 2−β dτ
0
0
Z s
Z t
o
n f 0 (s) n/2 f 0 (s) n+2
2
+
s
≤C +C
ku(τ )k4 dτ ds
f 0 (s)
2f (s)
2f (s)
0
0
Z t
Z t
2−β
f 0 (s) n+2
2−β
4α
2
2
0
2−β
2
s
ds
+C
f (s)
ds
ku(s)k
2f
(s)
0
0
Z t
Z t
n f 0 (s) n/2 f 0 (s) n+2
o
2
0
≤C +C
s
f (s)
+
ku(s)k4 ds ds
2f (s)
2f (s)
0
0
Z t
Z t
f 0 (s) n+2
2−β
4α
2
s 2 ds
+C
ku(s)k 2−β ds + C .
f 0 (s)
2f (s)
0
0
Z
s
Let f (t) = (1 + t)n . Noting that
Z
(1 + t)n
|û(ξ, t)|2 dξ
1
2
<
2−β
2
< 1,
4α
2−β
> 2 and ku(t)k ≤ C, we have
Rn
n/2
≤ C(1 + t)
n/2
Z
t
4
Z
ku(s)k ds + C(1 + t)
+ C(1 + t)
0
≤ C(1 + t)n/2 + C(1 + t)n/2
n/2
Z
t
4α
ku(s)k 2−β ds
0
t
ku(s)k2 ds.
0
This yields
Z
(1 + t)n/2 ku(t)k2 = (1 + t)n/2
|û(ξ, t)|2 dξ
Rn
Z t
n
≤C+
(1 + s)n/2 ku(s)k2 (1 + s)− 2 ds.
0
Letting
g(t) = (1 + t)n/2 ku(t)k2 , h(t) = (1 + t)−n/2 ,
R∞
applying Lemma 2.1 and the bound 0 h(s)ds ≤ C, we deduce readily that
f (t) = C,
ku(t)k ≤ C(1 + t)−n/4 .
10
B.-Q. DONG
EJDE-2005/125
(iii) Case p ≥ 3, n ≥ 2. Inserting (2.8) into the right hand of (3.4),
Z
f (t)
|û(ξ, t)|2 dξ
Rn
Z t
Z A
Z s
2
≤ C + Cωn
f 0 (s)
1+ρ
ku(τ )k2 dτ + ρ ρn−1 dρ ds
0
0
0
0
Z t
n/2 f 0 (s) n+2
Z s
2
f
(s)
2
+
ku(τ )k2 dτ + 1
ds.
≤C +C
f 0 (s)
2f (s)
2f (s)
0
0
(3.12)
First, we discuss the case n = 2. It follows from (3.12) and the bound ku(t)k ≤ C
that
Z
Z t
f 0 (s) f 0 (s) 2
f (t)
+
(s2 + 1) ds.
|û(ξ, t)|2 dξ ≤ C + C
f 0 (s)
2f (s)
2f (s)
R2
0
Let f (t) = (ln(e + t))5 . By the same calculation as that of (3.7), we have
ku(t)k ≤ C(ln(e + t))−1 .
(3.13)
Hence, letting f (t) = (1 + t)2 in (3.10) and using the Hölder inequality,
Z
Z t
Z s
2
(1 + t)2
|û(ξ, t)|2 dξ ≤ C(1 + t) + C
(1 + s)−1
ku(τ )k2 dτ ds
R2
0
0
Z tZ s
≤ C(1 + t) + C
ku(τ )k4 dτ ds
0
0
Z t
ku(s)k4 ds.
≤ C(1 + t) + C(1 + t)
0
By (3.11), we obtain the inequality
Z
Z t
(1 + t)
|û(ξ, t)|2 dξ ≤ C + C
ku(s)k2 (1 + s) (1 + s)−1 (ln(e + s))−2 ds.
R2
0
Let
Z
Z
2
|û(ξ, t)| dξ = (1 + t)
g(t) = (1 + t)
R2
|u(x, t)|2 dx,
R2
h(t) = (1 + t)−1 (ln(e + t))−2 ,
f (t) = C .
Applying Lemma 2.1, we have
g(t) ≤ C exp
Z
∞
h(t)dt ≤ C,
0
and so
ku(t)k ≤ C(1 + t)−1/2 .
Next, we carry out the proof in the case n ≥ 3. Letting f (t) = (1 + t)n in (3.10)
and using the Hölder inequality, we have, similar to the argument in the case of
n = 2,
Z
n
(1 + t)
|û(ξ, t)|2 dξ
Rn
Z t
− n+2
−n
−n
2
2
2
+ C(1 + t)
+ C(1 + t)
ku(s)k4 ds.
≤ C + C(1 + t)
0
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DECAY OF SOLUTIONS
11
Thus
(1 + t)n/2
Z
|û(ξ, t)|2 dξ ≤ C + C
Rn
Since
R∞
0
Z
t
n
ku(s)k2 (1 + s)n/2 (1 + s)− 2 ds.
0
n
(1 + s)− 2 ds ≤ C when n ≥ 3. Thus applying Lemma 2.1, we have
Z
Z ∞
n
(1 + t)n/2
|û(ξ, t)|2 dξ ≤ C exp
(1 + t)− 2 dt ≤ C.
Rn
−n/4
Hence ku(t)k ≤ C(1 + t)
0
. The proof of Theorem 3.1 is complete.
4. Error estimates for Newtonian and non-Newtonian flows
Theorem 4.1. In addition to the assumption of Theorem 3.1, suppose that ũ denotes the weak solution of the Newtonian system (1.1)-(1.4) with µ1 = µ2 = 0.
Then
ku(t) − ũ(t)k = o (1 + t)−n/4 , as t → ∞.
Note that the estimates of Theorem 4.1 with u(t) − ũ(t) replaced by et∆ u0 − ũ(t)
also hold (see Kajikiya and Miyakawa [8]). Thus from the inequality
ku(t) − ũ(t)k ≤ ket∆ u0 − ũ(t)k + ket∆ u0 − u(t)k,
we need to prove the validity of the estimates of Theorem 4.1 with u(t) − ũ(t)
replaced by et∆ u0 − u(t). So we only need the following lemma.
lemma 4.2. In addition to the assumption of Theorem 3.1, let v(t) = et∆ u0 be the
solution of the linear heat equation with the same initial data u0 , then for t ≥ 1,

−p/2

,
2 < p < 3, n = 2
(1 + t)
n
1
2
2n
ku(t) − v(t)k ≤ C (1 + t)− 2 − 2 , 1 + n+2
≤ p < 3, n ≥ 3

n
1

− 2 −2
, p ≥ 3, n ≥ 2.
(1 + t)
We remark that from Lemma 4.2, it is readily seen that when u0 ∈ H ∩ L1 ,
ku(t) − v(t)k = o (1 + t)−n/4 , t → ∞.
Proof of Lemma 4.2. Denote the difference w(t) = u(t) − v(t). Thus w(t) satisfies
wt − 4w + ∆2 w = B(u, v),
w(x, 0) = 0,
p−2
(4.1)
2
where B(u, v) = −(u · ∇)u + ∇ · (|e(u)| e(u)) − ∆ v − ∇π. Since u0 ∈ H, v is
divergence free, and so is w.
Multiplying by w and integrating with respect to Rn , it follows that
d
kwk2 + 2k∇wk2 + 2k∆wk2 =: 2B(u, v, w),
(4.2)
dt
where
B(u, v, w) = −((u · ∇)u, w) − ((|e(u)|p−2 e(u)), ∇w) − (∆2 v, w)
= ((u · ∇)w, w + v) + (|e(u)|p−2 e(u), ∇v) − k∇ukpp + (∆v, ∆w) (4.3)
= ((u · ∇)w, v) + (|e(u)|p−2 e(u), ∇v) + (∆v, ∆w) − k∇ukpp .
Since, for 1 ≤ q ≤ ∞ and k ∈ N ,
n
1
k
kDk v(t)kq ≤ (1 + t)− 2 (1− q )− 2 ku0 k1
∀ t ≥ 1,
(4.4)
12
B.-Q. DONG
EJDE-2005/125
which follows from the properties of the heat kernel (see Kajikiya and Miyakawa
[8]). Thus we estimate the first three terms. Noting that ku(t)k ≤ C(1 + t)−n/4 ,
we have
2|((u · ∇)w, v) + (|e(u)|p−2 e(u), ∇v) + (∆v, ∆w)|
p−1
≤ 2kukk∇wkkvk∞ + 2k∇ukp−1
k∇vk∞ + 2k∆vkk∆wk
1
2
2
≤ k∇wk2 + kuk2 kvk2∞ + 2k∇ukp−1
p−1 k∇vk∞ + 2k∆wk + k∆vk
2
3n
n
1
1
−n
2 −2 .
≤ k∇wk2 + 2k∆wk2 + (1 + t)− 2 + 2(1 + t)− 2 − 2 k∇ukp−1
p−1 + (1 + t)
2
(4.5)
Hence (4.2)-(4.5) yield
n
1
1
d
−n
2 −2 .
kwk2 + k∇wk2 ≤ 2(1 + t)− 2 − 2 k∇ukp−1
(4.6)
p−1 + (1 + t)
dt
2
Similar to (3.4) with f (t) = (1 + t)2n , the derivation of (4.6) implies
Z
(1 + t)2n
|ŵ(ξ, t)|2 dξ
Rn
Z t
Z
3n
1
3n
p−1
k∇ukp−1
ds.
≤ C(1 + t)2n
|ŵ(ξ, s)|2 dξ + C(1 + t) 2 −1 + C(1 + t) 2 − 2
0
B(t)
Similar to the proof of the Lemma 2.2, we use (4.1) and kv(t)k1 ≤ ku0 k1 ≤ C to
obtain
Z t
Z t
Z t
p−1
2
4
|ŵ(ξ, t)| ≤ C|ξ|
ku(s)k ds + C|ξ|
k∇ukp−1 ds + C|ξ|
kv(s)k1 ds
0
0
0
(4.7)
Z t
Z t
p−1
2
4
≤ C|ξ|
ku(s)k ds + C|ξ|
k∇ukp−1 ds + C|ξ| t.
0
0
Therefore, (4.6) and (4.7) yield
Z
Z t
2
n
n
ku(s)k2 ds + C(1 + t)− 2 −1
kw(t)k2 =
|ŵ(ξ, t)|2 dξ ≤ C(1 + t)− 2 −1
0
Rn
Z t
Z t
2
n
1
n
p−1
p−1
+ C(1 + t)− 2 − 2
k∇ukp−1
k∇ukp−1
ds + C(1 + t)− 2 −1
ds .
0
0
When 2 < p < 3, n = 2, it follows from (2.15) and ku(t)k ≤ C(1 + t)1/2 that
p
kw(t)k2 ≤ C(1 + t)−2 (ln(1 + t))2 + C(1 + t)−(p−1) + C(1 + t)− 2
p
≤ C(1 + t)− 2 .
When 1 + n+2
2n ≤ p < 3 and n ≥ 3, equation (2.15) and the inequalities ku(t)k ≤
nα
C(1 + t)−n/4 and 4−2β
> 1 imply
Z t
Z t
nα
ds
=
(1 + s)− 4−2β ds ≤ C,
k∇ukp−1
p−1
0
0
which yields
n
n
1
n
1
kw(t)k2 ≤ C(1 + t)− 2 −1 + C(1 + t)− 2 − 2 ≤ C(1 + t)− 2 − 2 .
Similarly, for the case of p ≥ 3 and n ≥ 2, we derive from (2.16) that
n
1
kw(t)k2 ≤ C(1 + t)− 2 − 2 .
EJDE-2005/125
DECAY OF SOLUTIONS
Hence the proof of Lemma 4.2 is complete.
13
Acknowledgements. The author would like to express his gratitude to the referees for his/her valuable comments and suggestions. He is also grateful to Zhi-Min
Chen and Yongsheng Li for many helpful discussions.
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Bo-Qing Dong
School of Mathematical Sciences, Nankai University, Tianjin 300071, China
E-mail address: bqdong@mail.nankai.edu.cn