Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 134, pp. 1–17.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
RATE OF DECAY FOR SOLUTIONS OF VISCOELASTIC
EVOLUTION EQUATIONS
MOHAMMAD KAFINI
Abstract. In this article we consider a Cauchy problem of a nonlinear viscoelastic equation of order four. Under suitable conditions on the initial data
and the relaxation function, we prove polynomial and logarithmic decay of
solutions.
1. Introduction
In this work, we are concerned with the Cauchy problem
Z t
utt − ∆u +
g(t − s)∆u(s)ds − ∆ut − ∆utt = div ϕ(∇u),
x ∈ Rn , t > 0,
0
u(x, 0) = u0 (x),
ut (x, 0) = u1 (x),
x ∈ Rn ,
(1.1)
where u0 , u1 are initial data and g is the relaxation function subjected to some
conditions to be specified later. The nonlinear function ϕ is a conservative vector
field on Rn . It is the gradient of some scalar function (potential) G. This type of a
nonlinear evolution equations of fourth order arises in the study of strain solitary
waves.
In a nonlinear elastic rods the longitudinal wave equation takes the form
utt − [b0 + b1 n(ux )n−1 ]uxx − b2 uxxtt = 0,
(1.2)
where b0 , b2 > 0 are constants, b1 is arbitrary real, n is a natural number (see
[24, 25]). In one-dimension (n = 1) and for a nonlinear function ϕ, equation (1.2)
takes the form
utt − αuxx − βuxxtt = ϕ(ux )x .
(1.3)
Considering an additional damping of the form a2 uxxt , equation (1.3) takes the
form
utt − a1 uxx − a2 uxxt − a3 uxxtt = ϕ(ux )x .
(1.4)
In 1872, Boussinesq described shallow-water waves and derived the equation
utt = uxx + uxxxx + (u2 )xx .
2000 Mathematics Subject Classification. 35B05, 35L05, 35L15, 35L70.
Key words and phrases. Decay; Cauchy problem; relaxation function; viscoelastic.
c
2013
Texas State University - San Marcos.
Submitted October 14, 2012. Published May 29, 2013.
1
(1.5)
2
M. KAFINI
EJDE-2013/134
This equation was improved and took many different forms. While the propagation
of longitudinal deformation waves in an elastic rod is modelled by the nonlinear
partial differential equation
1
utt − uxxtt − uxx − (up )xx = 0,
p
with p = 3 or 5, this equation is called the nonlinear Pochhammer-Chree equation
(see [6, 16]). The general class (Cauchy problem) for this type of problems takes
the form
utt − ∇2 utt − ∇2 u = ∇2 f (u)
in W s,p (Rn ).
(1.6)
Chen [4] studied equation (1.4) with the initial and boundary conditions
u(0, t) = u(1, t) = 0,
u(x, 0) = u0 (x),
t ≥ 0,
ut (x, 0) = u1 (x),
x ∈ [0, t].
(1.7)
He proved the existence of a unique classical solution and gave sufficient conditions
for the blow-up of solutions using the concavity method. Moreover, the same author
proved that problem in (1.3) with conditions given in (1.7) admits a unique global
classical solution. Later on, Chen al [5] considered problem (1.4) with the same
conditions and studied the asymptotic behavior of the solution. Sufficient conditions
for a blow-up result are given. They used the integral inequality given in [10,
Theorem 8.1].
The Cauchy problem for this type of equations have been extensively studied by
many authors. De Godefrroy [8] considered the Cauchy problem of equation (1.5)
and proved the existence of a unique local solution and gave sufficient conditions
for a blow-up result using the concavity method. Constantin et al [7] studied the
local well-posedness to the Cauchy problem
utt = uxxtt + [F (u)]xx
and obtained the global existence of the solution by the extension theorem. Liu
[16] proved by the contraction mapping principle that the Cauchy problem
utt = uxxtt + f (u)xx ,
admits a unique global solution in addition to a blow up result. Wang et al [22]
proved that the multidimensional generalized equation of (1.4) has a unique global
small amplitude solution. In [23], the same authors have proved that the Cauchy
problem of (1.4) has a unique global generalized solution and a unique global classical solution. Later on, they discussed the blow-up of the solution using the concavity
method. However, the asymptotic behavior of the solution has not been studied.
On the viscoelastic problems, we will mention here some results from the literature related to this type of problems. Cavalcanti et al [3] considered the problem
Z t
|ut |ρ utt − ∆u − ∆utt +
g(t − s)∆u(s)ds − γ∆ut = 0,
0
for ρ > 0, and proved a global existence result for γ ≥ 0 and an exponential decay
for γ > 0. This result has been improved by Messaoudi and Tatar [17], for γ = 0,
they established exponential and polynomial decay results in the absence, as well
as in the presence, of a source term.
EJDE-2013/134
RATE OF DECAY FOR SOLUTIONS
3
Kafini and Messaoudi [11] studied the Cauchy problem
Z t
g(t − s)∆u(x, s)ds = 0, x ∈ Rn , t > 0,
utt − ∆u +
0
u(x, 0) = u0 (x),
ut (x, 0) = u1 (x),
x ∈ Rn ,
where u0 , u1 are two compactly supported functions and g is a positive nonincreasing function defined on R+ . They proved that the decay of the solution is
polynomial (respectively logarithmic) if the rate of decay of the relaxation function
is exponential (respectively polynomial). They used the multiplier method and a
lemma by Martinez [17].
For more results related to stability and asymptotic behavior of viscoelastic
equations, we refer the reader to the work of Renardy et al [21], Munoz and Oquendo
[19], Fabrizio and Morro [9], Baretto et al [1], Kafini [12, 13, 14, 15].
In the present article, we study the asymptotic behavior of solutions to (1.1). To
achieve this goal some conditions have to be imposed on the relaxation function g.
Since Poincaré and some embedding inequalities are no longer valid, We will use
the nature of the wave propagation to overcome this difficulties. The proof is based
on the multiplier method and makes use of a lemma by Martinez [17]. This paper
is organized as follows. In Section 2, we present conditions and materials needed
for our work. In section 3, we state and proof our main result.
2. Preliminaries
In this section we present some material needed for the proof of our results. We
will use the following assumptions:
(G1) g : R+ → R+ is a differentiable function such that
Z ∞
1−
g(s)ds = l > 0.
0
(G2) There exists a > 0 such that
g 0 (t) ≤ −ag p (t),
2
1 ≤ p < 3/2,
t ≥ 0.
n
(G3) ϕ ∈ C (R ), and for λ > 0, β ≥ 1, (n − 2)β ≤ n, it satisfies
|ϕ(s)| ≤ λ|s|β , ∀s ∈ Rn .
(G4) There exists a function G : Rn → R such that G(w) ≥ 0, ∇G(w) = ϕ(w),
and 2G(w) ≤ w.ϕ(w) for all w ∈ Rn .
Proposition 2.1 ([4, 5]). Assume that (G1)–(G2) hold and u0 , u1 ∈ H 1 (Rn ), with
compact support, then problem (1.1) has a unique global solution such that
u ∈ C 1 ((0, ∞); H 1 (Rn )),
utt ∈ C((0, ∞); L2 (Rn )).
Proposition 2.2 ([17]). Let E : R+ → R+ be non-increasing function, and φ :
R+ → R an increasing function in C 2 such that φ(0) = 0 and φ(t) → +∞ as
t → +∞. Assume that there exist q ≥ 0 and A > 0 such that
Z +∞
E q+1 (t)φ0 (t)dt ≤ AE(S), 0 ≤ S < +∞.
S
Then
E(t) ≤ CE(0)(1 + φ(t))−1/q
∀t ≥ 0,
if q > 0,
4
M. KAFINI
EJDE-2013/134
and
E(t) ≤ CE(0)e−ωφ(t)
∀t ≥ 0,
if q = 0,
where C and ω are positive constants independent of the initial energy E(0).
Lemma 2.3 (Sobolev, Gagliardo, Nirenberg [2, Thm. 9.9]). Suppose that 1 ≤ p <
n. If u ∈ W 1,p (Rn ), then u ∈ Lp∗ (Rn ), with
1
1
1
= − .
p∗
p n
Moreover there exists a constant C = C(n, p) such that
∀u ∈ W 1,p (Rn ).
kukp∗ ≤ Ck∇ukp ,
Lemma 2.4. If u is the solution of (1.1), then
ku(t)k2 ≤ C(L + t)k∇u(t)k2 .
where L > 0 is such that
supp{u0 (x), u1 (x)} ⊂ B(L) = {x ∈ Rn | |x| < L}.
Proof. Using Lemma 2.3, for p = 2, we have
2n
, if n ≥ 3.
n−2
By using the finite-speed propagation property and Hölder inequality, we have
Z
Z
|u|2 dx =
|u|2 dx
kukp∗ ≤ Ck∇uk2 ,
Rn
p∗ =
B(L+t)
≤
2 Z
1− p∗
1dx
Z
B(L+t)
(|u|2 )
p∗
2
2/p∗
dx
B(L+t)
≤ C(L + t)2 ku(t)k2p∗ .
Hence,
ku(t)k2 ≤ C(L + t)ku(t)kp∗ ≤ C(L + t)k∇u(t)k2 .
Now, we introduce the “modified” energy functional
Z
Z t
Z
1h
E(t) =
|ut |2 dx + 1 −
g(s)ds
|∇u|2 dx
2 Rn
n
0
R
Z
i Z
2
+
|∇ut | dx + (g ◦ ∇u) +
G(∇u)dx,
Rn
Rn
where
Z
(g ◦ u)(t) =
t
Z
g(t − s)
0
|u(s) − u(t)|2 dx ds.
Rn
Lemma 2.5. If u is a solution of (1.1), then the modified energy satisfies
E 0 (t) =
1
1
1
1 0
(g ◦ ∇u) − g(t)k∇uk22 − k∇ut k22 ≤ (g 0 ◦ ∇u) ≤ 0.
2
2
2
2
(2.1)
The proof of the above lemma follows by multiplying equation (1.1) by ut and integrating over Rn , using integration by parts, and repeating the same computations
as in [20].
EJDE-2013/134
RATE OF DECAY FOR SOLUTIONS
5
Corollary 2.6. Under the assumptions (G1)–(G2), we have
(g p ◦ ∇u)(t) ≤ (−g 0 ◦ ∇u)(t) ≤ −2E 0 (t).
(2.2)
The proof of the above corollary follows by using (G2) and (2.1).
Lemma 2.7. Let 1 < p < 2 and u be the solution of (1.1), then for any 0 < θ <
2 − p, there exists C(θ, p) such that
(g ◦ ∇u)
p−1+θ
θ
(t) ≤ C(θ)(g p ◦ ∇u)(t).
(2.3)
Proof. Using Hölder’s inequality, it is easy to see that
Z t
2(p−1)
(p−1)(1−θ)
(g ◦ ∇u)(t) =
g p−1+θ (t − s)k∇u(t) − ∇u(s)k2p−1+θ
0
2θ
pθ
(2.4)
× g p−1+θ (t − s)k∇u(t) − ∇u(s)k2p−1+θ ds
p−1
nZ t
o p−1+θ
θ
(g p ◦ ∇u)(t) p−1+θ .
≤
g 1−θ (s)dsk∇u(t) − ∇u(s)k22
0
Using (G1) and (G2), we easily arrive at
Z t
Z ∞
Z
g 1−θ (s)ds ≤
g 1−θ (s)ds ≤ −
0
0
∞
g 1−θ−p (s)g 0 (s)ds
0
g 2−θ−p (0)
−1
[g 2−θ−p (s)]∞
0 =
(2 − p − θ)
(2 − p − θ)
= C0 < ∞,
=
where 2 − p − θ > 0. Therefore, (2.4) becomes
p−1
n
o p−1+θ
θ
(g p ◦ ∇u)(t) p−1+θ
(g ◦ ∇u)(t) ≤ 2C0 sup k∇uk22
0≤t<∞
p−1
n
o p−1+θ
θ
≤ 2C0 sup E(t)
(g p ◦ ∇u)(t) p−1+θ
(2.5)
0≤t<∞
θ
p−1
≤ {2C0 E(0)} p−1+θ (g p ◦ ∇u)(t) p−1+θ .
Thus, (2.3) is established, with C = {2C0 E(0)}
p−1
θ
.
Lemma 2.8. Let 1 < p < 3/2, then
Z Z t
Z t
1/2
1/2
g(t − s)|∇u(t) − ∇u(s)|ds dx ≤
g 2−p (s)ds
(g p ◦ ∇u)(t)
.
Rn
0
0
Proof. Using Hölder’s inequality, direct calculations lead to
Z Z t
g(t − s)|∇u(t) − ∇u(s)|ds dx
Rn
0
Z Z t
=
g 1−p/2 (t − s)g p/2 (t − s)|∇u(t) − ∇u(s)|ds dx
Rn
0
Z t
1/2
1/2
≤
g 2−p (s)ds
(g p ◦ ∇u)(t)
.
0
6
M. KAFINI
EJDE-2013/134
3. Decay of solutions
In this section, we establish two lemmas then we state and prove our main result.
We take φ(t) = ln(1 + t), in order to apply Proposition 2.2.
Lemma 3.1. Under assumptions (G1)–(G4), for δ, δ1 , δ2 > 0, the solution of (1.1)
satisfies
Z t
Z h
−1 q
g(s)ds |∇u|2
1 − (δ + Cδ1 + δ2 ) −
(1 + t) E (t)
0
Rn
Z
i
C 2
1 2
− 1+
|ut | − 1 +
|∇ut | + (1 + t)−1 E q (t)
2G(∇u)dx
4δ1
4δ2
Rn
Z
+ (1 + t)−1 E q (t)
2G(∇u)dx
Rn
Z
Z
(3.1)
i 1 dh
i
dh
2
−1 q
−1 q
uut dx −
|∇u| dx
≤−
(1 + t) E (t)
(1 + t) E (t)
dt
2 dt
Rn
Rn
Z
i
dh
−
(1 + t)−1 E q (t)
∇ut .∇u dx
dt
Rn
i
h 1
+ 1 (1 + t)−1 + 1 E q−1 (t) − E 0 (t) ,
+C
4δ
where C is a generic positive constant independent of δ, δ1 and δ2 .
Proof. Multiplying equation (1.1) by (1 + t)−1 E q (t)u(t) and integrating by parts
over Rn , we obtain
Z
Z
Z t
hZ
uutt dx −
u∆u dx +
u(t)
(1 + t)−1 E q (t)
g(t − s)∆u(s) ds dx
Rn
Rn
Rn
0
Z
Z
i
(3.2)
−
u∆ut dx −
u∆utt dx
Rn
Rn
Z
= (1 + t)−1 E q (t)
u div ϕ(∇u) dx.
Rn
A direct integration by parts yields
Z
(1 + t)−1 E q (t)
|∇u|2 − |ut |2 − |∇ut |2 dx
Rn
Z
Z t
+ (1 + t)−1 E q (t)
g(t − s)
u(t)∆u(s) dx ds
0
Rn
Z
Z
Z
− q(1 + t)−1 E q−1 (t)E 0 (t)
uut dx +
|∇u|2 dx +
∇ut .∇u dx
Rn
Rn
Rn
Z
Z
Z
(3.3)
+ (1 + t)−2 E q (t)
uut dx +
|∇u|2 dx +
∇ut .∇u dx
Rn
Rn
Rn
Z
Z
i 1 dh
i
dh
=−
(1 + t)−1 E q (t)
uut dx −
(1 + t)−1 E q (t)
|∇u|2 dx
dt
2 dt
n
Rn
ZR
i
dh
−
(1 + t)−1 E q (t)
∇ut .∇u dx
dt
Rn
Z
− (1 + t)−1 E q (t)
∇u.ϕ(∇u)dx,
Rn
EJDE-2013/134
RATE OF DECAY FOR SOLUTIONS
7
the second term in the left side of (3.3) can be estimated as
Z t
Z
(1 + t)−1 E q (t)
g(t − s)
u(t)∆u(s) dx ds
0
Rn
Z t
Z
= −(1 + t)−1 E q (t)
g(t − s)
∇u(t).∇u(s) dx ds
0
Rn
Z
Z t
−1 q
= −(1 + t) E (t)
∇u(t).
g(t − s) (∇u(s) − ∇u(t)) dx ds
0
Rn
Z
Z t
g(t − s)
|∇u(t)|2 dx ds
− (1 + t)−1 E q (t)
0
Rn
Z
Z t
1
−1 q
2
2−p
p
≥ −(1 + t) E (t) δ
|∇u(t)| dx +
g
(s)ds (g ◦ ∇u)(t)
4δ
Rn
0
Z t
Z
− (1 + t)−1 E q (t)
g(s)ds
|∇u(t)|2 dx.
0
Rn
Thus (3.3) becomes
Z h
Z t
i
(1 + t)−1 E q (t)
1−δ−
g(s)ds |∇u|2 − |ut |2 − |∇ut |2 dx
Rn
0
Z t
1
2−p
g
(s)ds (1 + t)−1 E q (t)(g p ◦ ∇u)
−
4δ 0
Z
Z
Z
− q(1 + t)−1 E q−1 E 0 (t)
uut dx +
|∇u|2 dx +
∇ut .∇u dx dt
Rn
Rn
Rn
Z
Z
Z
(3.4)
+ (1 + t)−2 E q (t)
uut dx +
|∇u|2 dx +
∇ut .∇u dx dt
Rn
Rn
Rn
Z
Z
i 1 dh
i
dh
≤−
(1 + t)−1 E q (t)
uut dx −
(1 + t)−1 E q (t)
|∇u|2 dx
dt
2 dt
Rn
Rn
Z
i
dh
−
(1 + t)−1 E q (t)∇ut .∇u dx
dt Rn
Z
−1 q
− (1 + t) E (t)
∇u.ϕ(∇u)dx.
Rn
Adding (1 + t)
−1
q
R
E (t) Rn 2G(∇u)dx to both sides of (3.4), we obtain
Z h
Z t
i
−1 q
(1 + t) E (t)
1−δ−
g(s)ds |∇u|2 − |ut |2 − |∇ut |2 dx
Rn
0
Z
−1 q
+ (1 + t) E (t)
2G(∇u)dx
Rn
Z
1 t 2−p
≤
g
(s)ds (1 + t)−1 E q (t)(g p ◦ ∇u)(t)
4δ 0
Z
Z
Z
+ q(1 + t)−1 E q−1 E 0 (t)
uut dx +
|∇u|2 dx +
∇ut .∇u dx
Rn
Rn
Rn
Z
Z
Z
− (1 + t)−2 E q (t)
uut dx +
|∇u|2 dx +
∇ut .∇u dx
Rn
Rn
Rn
Z
Z
i 1 dh
i
dh
−1 q
−1 q
(1 + t) E (t)
uut dx −
(1 + t) E (t)
|∇u|2 dx
−
dt
2 dt
Rn
Rn
8
M. KAFINI
dh
dt
EJDE-2013/134
Z
i
(1 + t)−1 E q (t)∇ut .∇u dx
Rn
Z
−1 q
+ (1 + t) E (t)
(2G(∇u) − ∇u.ϕ(∇u))dx.
−
Rn
By assumption (G4), the above inequality yields
Z h
Z t
i
(1 + t)−1 E q (t)
1−δ−
g(s)ds |∇u|2 − |ut |2 − |∇ut |2 dx
Rn
0
Z
+ (1 + t)−1 E q (t)
2G(∇u)dx
Rn
Z
Z
Z
≤ −(1 + t)−2 E q (t)
uut dx +
|∇u|2 dx +
∇ut .∇u dx
Rn
Rn
Rn
Z
Z
Z
+ q(1 + t)−1 E q−1 E 0 (t)
uut dx +
|∇u|2 dx +
∇ut .∇u dx
Rn
Rn
Rn
Z t
1
g 2−p (s)ds (1 + t)−1 E q (t)(g p ◦ ∇u)(t)
+
4δ 0
Z
i
i 1 d hZ
dh
−1 q
−
(1 + t)−1 E q (t)|∇u|2 dx
(1 + t) E (t)
uut dx −
dt
2 dt IRn
Rn
Z
i
dh
−
(1 + t)−1 E q (t)∇ut .∇u dx .
dt Rn
(3.5)
Terms in the right side of (3.5) can be estimated using the non-increasing property
of E(t), Cauchy’s inequality, assumption (G2), Lemma 2.4 and Lemma 2.8, the first
term is handled as follows
Z
Z
1/2 Z
1/2
−(1 + t)−2 E q (t)
uut dx ≤ (1 + t)−2 E q (t)
|u|2 dx
|ut |2 dx
Rn
Rn
−2
Rn
q
≤ (1 + t)
E (t)C(1 + t)k∇uk2 kut k2
h
i
1
≤ C(1 + t)−1 E q (t) δ1 k∇uk22 +
kut k22 .
4δ1
The second term satisfies
− (1 + t)−2 E q (t)
Z
|∇u|2 dx ≤ 0,
(3.6)
Rn
The third term satisfies
Z
1
k∇ut k22
−(1 + t)−2 E q (t)
∇ut .∇u dx ≤ (1 + t)−2 E q (t) δ2 k∇uk22 +
4δ2
Rn
(3.7)
1
≤ (1 + t)−1 E q (t) δ2 k∇uk22 +
k∇ut k22 .
4δ2
The fourth term satisfies
Z
1 t 2−p
g
(s)ds (1 + t)−1 E q (t)(g p ◦ ∇u)
4δ 0
C
−C
≤
(1 + t)−1 E q (t)(g p ◦ ∇u) ≤
(1 + t)−1 E q (t)(g 0 ◦ ∇u)
4δ
4δ
−C
≤
(1 + t)−1 E q (t)E 0 (t),
4δ
(3.8)
EJDE-2013/134
RATE OF DECAY FOR SOLUTIONS
where, by the assumptions on g,
Z
Z t
2−p
g
(s)ds <
9
∞
g 2−p (s)ds < ∞.
0
0
The fifth term satisfies
q(1 + t)−1 E q−1 E 0 (t)
Z
uut dx
Z
−1 q−1 0
≤ −C(1 + t) E
E (t)
|u|2 + |ut |2 dx
Rn
Z
q−1
0
≤ −CE
(t)E (t)
|∇u|2 + |∇ut |2 dx
Rn
Rn
≤ −CE
q−1
0
(t)E (t)E(t) ≤ −CE q (t)E 0 (t).
The sixth term satisfies
q(1 + t)−1 E q−1 E 0 (t)
Z
|∇u|2 dx ≤ −C(1 + t)−1 E q E 0 (t).
(3.9)
Rn
The seventh term satisfies
q(1 + t)−1 E q−1 E 0 (t)
Z
∇ut .∇u dx
Z ≤ −C(1 + t)−1 E q−1 E 0 (t)
|∇ut |2 + |∇u|2 dx
Rn
(3.10)
Rn
≤ −C(1 + t)−1 E q−1 E 0 (t)E(t)dt ≤ −C(1 + t)−1 E q (t)E 0 (t).
Combining (3.5)–(3.10) and the Lemma is proved.
Lemma 3.2. Under assumptions (G1), (G4), for δ > 0, the solution of (1.1)
satisfies
Z t
−1 q
(1 + t) E (t)
g(s)ds − δ(1 + C) kut k22 − (1 + t)−1 E q (t)δ(2 − l + k)k∇uk22
0
Z t
− C(1 + t)−1 E q (t) δ −
g(s)ds k∇ut k22
0
Z
Z t
i
dh
(1 + t)−1 E q (t)
ut
g(t − s) (u(t) − u(s)) ds dx
≤
dt
Rn
0
Z
Z t
h
i
d
+
(1 + t)−1 E q (t)
∇ut .
g(t − s) (∇u(t) − ∇u(s)) ds dx
dt
Rn
0
h1
i
1
+ C (1 + t)−1 + 1 +
E q (t) (−E 0 (t)) ,
δ
4δ
(3.11)
where C is a generic positive constant independent of δ.
Proof. Multiplying equation (1.1) by
Z t
−1 q
(1 + t) E (t)
g(t − s) (u(t) − u(s)) ds,
0
10
M. KAFINI
EJDE-2013/134
and integrating over Rn we obtain
Z t
(1 + t)−1 E q (t)
g(s)ds kut k22
0
Z
Z t
i
dh
−1 q
(1 + t) E (t)
ut
g(t − s) (u(t) − u(s)) ds dx
=
dt
0
Rn
Z
Z t
−2 q
+ (1 + t) E (t)
ut
g(t − s) (u(t) − u(s)) ds dx
Rn
0
Z
Z t
−1 q−1
0
− q(1 + t) E
(t)E (t)
ut
g(t − s) (u(t) − u(s)) ds dx
Rn
0
Z
Z t
− (1 + t)−1 E q (t)
ut
g 0 (t − s) (u(t) − u(s)) ds dx
Rn
0
Z
Z t
−1 q
+ (1 + t) E (t)
∇u
g(t − s) (∇u(t) − ∇u(s)) ds dx
Rn
0
Z Z t
− (1 + t)−1 E q (t)
g(t − s)∇u(s)ds
Rn
0
Z t
×
g(t − s) (∇u(t) − ∇u(s)) ds dx
0
Z
Z t
+ (1 + t)−1 E q (t)
∇ut
g(t − s) (∇u(t) − ∇u(s)) ds dx
Rn
0
Z
Z t
+ (1 + t)−1 E q (t)
∇utt
g(t − s) (∇u(t) − ∇u(s)) ds dx
Rn
0
Z
Z t
+ (1 + t)−1 E q (t)
div ϕ(∇u)
g(t − s) (u(t) − u(s)) ds dx.
Rn
(3.12)
0
Terms in the right hand side of the above expression can be treated in a similar
way as in (3.5).
The second term satisfies
Z
Z t
(1 + t)−2 E q (t)
ut
g(t − s) (u(t) − u(s)) ds dx
Rn
0
Z
1/2 Z Z t
2 1/2
≤ (1 + t)−2 E q (t)
|ut |2 dx
g(t − s) (u(t) − u(s)) ds dx
Rn
Rn
0
Z t
Z
1/2
≤ (1 + t)−2 E q (t)kut k2 C(1 + t)
|
g(t − s) ∇u(t) − ∇u(s) ds|2 dx
Rn
0
h Z t
i
1/2
−1 q
2−p
≤ C(1 + t) E (t)kut k2
g
(s)ds (g p ◦ ∇u)
0
h
i
C
≤ C(1 + t) E (t) δkut k22 + (g p ◦ ∇u)
4δ
h
i
C
−1 q
2
≤ C(1 + t) E (t) δkut k2 + (−E 0 (t)) .
4δ
−1
q
The third term satisfies
− q(1 + t)
−1
E
q−1
0
Z
(t)E (t)
Z
Rn
t
g(t − s)(u(t) − u(s)) ds dx
ut
0
(3.13)
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RATE OF DECAY FOR SOLUTIONS
11
Z
1/2
≤ −q(1 + t)−1 E q−1 (t)E 0 (t)
u2t dx
Rn
Z Z t
2 1/2
×
g(t − s)(u(t) − u(s))ds dx
Rn
0
−1
≤ −C(1 + t) E q−1 (t)E 0 (t)kut k2 (1 + t)
Z Z t
2 1/2
×
g(t − s)|∇u(t) − ∇u(s)|ds dx
Rn
0
Z Z t
1/2
q−1
0
≤ −CE
(t)E (t)kut k2 (1 − l)
g(t − s)|∇u(t) − ∇u(s)|2 ds dx
Rn
≤ −CE
q−1
0
(t)E (t)E
1/2
(t)E
1/2
0
(t)
≤ −CE q (t)E 0 (t).
The fourth term satisfies
− (1 + t)−1 E q (t)
Z
Z
ut
Rn
t
g 0 (t − s)(u(t) − u(s)) ds dx
0
Cg(0)
(3.14)
(1 + t)(−g 0 ◦ ∇u)
≤ (1 + t) E (t) δkut k22 +
4δ
C
≤ δ(1 + t)−1 E q (t)kut k22 − E q (t)E 0 (t).
4δ
Using (2.2), the fifth term satisfies
Z
Z t
(1 + t)−1 E q (t)
∇u.
g(t − s) (∇u(t) − ∇u(s)) ds dx
Rn
0
Z
h
i
1 t 2−p
(3.15)
≤ (1 + t)−1 E q (t) δk∇uk22 +
g
(s)ds (g p ◦ ∇u)
4δ 0
C
≤ (1 + t)−1 E q (t) δk∇uk22 + (−E 0 (t)) .
4δ
Using Young’s inequality and lemma 2.8, he sixth term satisfies
Z Z t
Z t
− (1 + t)−1 E q (t)
g(t − s)∇u(s)ds.
g(t − s)(∇u(t) − ∇u(s))ds dx
Rn
0
0
Z t
h
i
1
≤ (1 + t)−1 E q (t) δ(1 − l)2 k∇uk22 +
g 2−p (s)ds (g p ◦ ∇u)
4δ 0
C
≤ (1 + t)−1 E q (t) δ(1 − l)2 k∇uk22 + (g p ◦ ∇u)
4δ
C
−1 q
2
≤ (1 + t) E (t) δ(1 − l)k∇uk2 + (−E 0 (t)) .
4δ
(3.16)
The seventh term satisfies
Z
Z t
(1 + t)−1 E q (t)
∇ut .
g(t − s) (∇u(t) − ∇u(s)) ds dx
Rn
0
Z
h
i
1 t 2−p
−1 q
2
(3.17)
≤ (1 + t) E (t) δk∇ut k2 +
g
(s)ds (g p ◦ ∇u)
4δ 0
C
≤ (1 + t)−1 E q (t) δk∇ut k22 + (−E 0 (t)) .
4δ
−1
q
12
M. KAFINI
EJDE-2013/134
The eighth term satisfies
Z
Z t
(1 + t)−1 E q (t)
∇utt .
g(t − s) (∇u(t) − ∇u(s)) ds dx
Rn
0
Z
Z t
i
dh
=
(1 + t)−1 E q (t)
∇ut .
g(t − s) (∇u(t) − ∇u(s)) ds dx
dt
Rn
0
Z
Z t
+ (1 + t)−2 E q (t)
∇ut .
g(t − s) (∇u(t) − ∇u(s)) ds dx
Rn
0
Z
Z t
− q(1 + t)−1 E q−1 (t)E 0 (t)
∇ut .
g(t − s) (∇u(t) − ∇u(s)) ds dx
Rn
0
Z
Z t
− (1 + t)−1 E q (t)
∇ut .
g 0 (t − s) (∇u(t) − ∇u(s)) ds dx
n
0
R
Z t
− (1 + t)−1 E q (t)
g(s)ds k∇ut k22
0
Z
Z t
i
dh
−1 q
(1 + t) E (t)
∇ut .
g(t − s) (∇u(t) − ∇u(s)) ds dx
≤
dt
Rn
0
C
−2 q
2
+ C(1 + t) E (t) δk∇ut k2 + (−E 0 (t))
4δ
− C(1 + t)−1 E q (t)E 0 (t) − C(1 + t)−1 E q (t)E 0 (t)
Z t
− C(1 + t)−1 E q (t)
g(s)ds k∇ut k22
0
Z
Z t
i
dh
≤
(1 + t)−1 E q (t)
∇ut .
g(t − s) (∇u(t) − ∇u(s)) ds dx
dt
Rn
0
C
+ C(1 + t)−1 E q (t) δk∇ut k22 + (−E 0 (t)) − C(1 + t)−1 E q (t)E 0 (t)
4δ
Z t
−1 q
− C(1 + t) E (t)
g(s)ds k∇ut k22 .
0
The ninth term satisfies
Z
Z t
−1 q
(1 + t) E (t)
div ϕ(∇u)
g(t − s) (u(t) − u(s)) ds dx
Rn
0
Z
Z t
= −(1 + t)−1 E q (t)
ϕ(∇u).
g(t − s) (∇u(t) − ∇u(s)) dx ds
Rn
0
Z
Z
o
n
1 t 2−p
−1 q
2
g
(s)ds (g p ◦ ∇u)
≤ (1 + t) E (t) δ
|ϕ(∇u(t))| dx +
4δ 0
Rn
Z
Z
n
o
1 t 2−p
≤ (1 + t)−1 E q (t) δλ2
|∇u(t)|2β dx +
g
(s)ds (g p ◦ ∇u)
4δ 0
IRn
β−1
Z
n
o
2E(0)
1 t 2−p
≤ (1 + t)−1 E q (t) δλ2
k∇u(t)k22 +
g
(s)ds (g p ◦ ∇u)
l
4δ 0
n
o
C
≤ (1 + t)−1 E q (t) δCk∇u(t)k22 +
(−E 0 (t)) .
4δ
(3.18)
Combining (3.12)–(3.18), the assertion of the lemma is proved.
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RATE OF DECAY FOR SOLUTIONS
13
Now, we multiply (3.11) by N and add it to (3.1) to obtain
h Z t
C i
−1 q
(1 + t) E (t) N
g(s)ds − δ(1 + C) − (1 +
) kut (t)k22
4δ1
0
Z t
h
i
+ (1 + t)−1 E q (t) 1 −
g(s)ds − (Cδ1 + δ2 ) − δ (1 + N (2 − l + k)) k∇u(t)k22
0
Z t
1 k∇ut (t)k22
+ (1 + t)−1 E q (t) N
g(s)ds − δ − 1 +
4δ2
0
Z
−1 q
+ (1 + t) E (t)
2G(∇u)dx
Rn
Z
Z
i 1 dh
i
dh
−1 q
−1 q
≤−
(1 + t) E (t)
(1 + t) E (t)
uut dx −
|∇u|2 dx
dt
2 dt
n
Rn
ZR
h
i
d
(1 + t)−1 E q (t)
∇ut .∇u dx
−
dt
Rn
Z
Z t
i
dh
+N
(1 + t)−1 E q (t)
ut
g(t − s) (u(t) − u(s)) ds dx
dt
Rn
0
Z
Z t
h
i
d
−1 q
+N
(1 + t) E (t)
∇ut .
g(t − s) (∇u(t) − ∇u(s)) ds dx
dt
Rn
0
i
h
1
1
3
−1
(1 + t) + 1 +
+1+
+ 1 (1 + t)−1 E q (t) (−E 0 (t)) .
+ NC
4δ
4δ
4δ
(3.19)
Since g is positive, continuous and g(0) > 0, then for any t0 > 0, we have
Z t
Z t0
g(s)ds ≥
g(s)ds = g0 > 0, ∀t ≥ t0 .
0
0
At this point we choose δ1 , δ2 small such that
Z t
1−
g(s)ds − (Cδ1 + δ2 ) > l − (Cδ1 + δ2 ) > 0,
0
and we choose N large enough so that
C
1
N g0 − (1 +
) > 0, N g0 − (1 +
) > 0.
4δ1
4δ2
Whence δ1 , δ2 and N are fixed, we pick δ small enough such that
1
) − δ(1 + C) > 0,
N g0 − (1 +
4δ1
Z t
1−
g(s)ds − (Cδ1 + δ2 ) − δ(1 + N (2 − l + k)) > 0,
0
Z t
1 N
g(s)ds − δ − 1 +
> 0.
4δ2
0
Consequently, we obtain from (3.19), for some constants α, C > 0,
(1 + t)−1 E q+1 (t)
Z
Z
i α dh
i
dh
(1 + t)−1 E q (t)
uut dx −
(1 + t)−1 E q (t)
|∇u|2 dx
≤ −α
n
dt
2 dt
Rn
ZIR
i
dh
(1 + t)−1 E q (t)
∇ut .∇u dx
−α
dt
Rn
14
M. KAFINI
EJDE-2013/134
Z
Z t
i
dh
(1 + t)−1 E q (t)
ut
g(t − s) (u(t) − u(s)) ds dx
dt
Rn
0
Z
Z t
i
h
d
−1 q
g(t − s) (∇u(t) − ∇u(s)) ds dx
(1 + t) E (t)
∇ut .
+ αN
dt
0
Rn
h
3
1
i
1
−1
+ α NC
+1+
(1 + t) + 1 +
+ 1 (1 + t)−1 E q (t) (−E 0 (t))
4δ
4δ
4δ
+ C(1 + t)−1 E q (t)(g ◦ ∇u).
+ αN
Integrating over (S, T ), where T > S ≥ t0 , gives
Z T
(1 + t)−1 E q+1 (t)dt
S
Z
Z
−1 q
−1 q
≤ −α(1 + T ) E (T )
uut (T )dx + α(1 + S) E (S)
uut (S)dx
n
Rn
ZR
Z
− α(1 + T )−1 E q (T )
|∇u(T )|2 dx + α(1 + S)−1 E q (S)
|∇u(S)|2 dx
n
n
IR
ZR
Z
−1 q
−1 q
− α(1 + T ) E (T )
∇ut .∇u(T )dx + α(1 + S) E (S)
∇ut .∇u(S)dx
Rn
Rn
Z
Z t
+ αN (1 + T )−1 E q (T )
ut (x, T )
g(T − s) (u(T ) − u(s)) ds dx
Rn
0
Z
Z t
− αN (1 + S)−1 E q (S)
ut (x, S)
g(S − s) (u(S) − u(s)) ds dx
Rn
0
Z
Z t
+ αN (1 + T )−1 E q (T )
∇ut (x, T )
g(T − s) (∇u(T ) − ∇u(s)) ds dx
Rn
0
Z
Z t
− αN (1 + S)−1 E q (S)
∇ut (x, S)
g(S − s) (∇u(S) − ∇u(s)) ds dx
Rn
0
− αC E q+1 (T ) − E q+1 (S)
Z T
+C
(1 + t)−1 E q (t)(g ◦ ∇u) dt.
S
Using the estimates
Z
(1 + t)−1
uut dx ≤ Ck∇uk2 kut k2 ≤ CE(t)1/2 E(t)1/2 ≤ CE(t),
n
R
Z
|∇u|2 dx ≤ CE(t),
Rn
Z
∇ut .∇u dx ≤ C k∇ut (t)k22 + k∇u(t)k22 ≤ CE(t),
n
R
Z
Z t
−1
(1 + t)
ut (x, t)
g(t − s)(u(t) − u(s)) ds dx
Rn
−1
≤ C(1 + t)
0
kut (t)k2 (1 + t)(g ◦ ∇u)1/2
≤ CE(t)1/2 E(t)1/2 ≤ CE(t),
Z
Z t
∇ut .
g(t − s)(∇u(t) − ∇u(s)) ds dx
Rn
0
≤ C k∇ut (t)k22 + (g ◦ ∇u) ≤ CE(t),
EJDE-2013/134
RATE OF DECAY FOR SOLUTIONS
we obtain, for all S ≥ t0 ,
Z T
Z
(1 + t)−1 E q+1 (t)dt ≤ CE q+1 (S) + C
S
15
T
(1 + t)−1 E q (t)(g ◦ ∇u)dt.
(3.20)
S
At this point we have two possible cases:
Theorem 3.3. (case p = 1) Let u0, u1 ∈ H01 (Rn ) and assume that (G1), (G2) hold.
Then, there exist positive constants C and ω such that, for any t0 > 0,
C
E(t) ≤
, ∀t ≥ t0 .
(1 + t)ω
Proof. Estimates (3.20) and (2.2) yield
Z T
(1 + t)−1 E q+1 (t)dt ≤ AE q+1 (S),
∀S ≥ t0 ,
S
taking q = 0, we obtain
Z T
(1 + t)−1 E(t)dt ≤ AE(S),
∀t ≥ t0 .
S
Then let T → +∞ to obtain
Z +∞
(1 + t)−1 E(t)dt ≤ AE(S),
∀t ≥ t0 .
S
Thus Proposition 2.2. yields
E(t) ≤ CE(0)e−ωφ(t) ≤ CE(0)e−ω ln(1+t) ≤
C
,
(1 + t)ω
∀t ≥ t0 .
Theorem 3.4. (case p > 1) Let u0, u1 ∈ H01 (Rn ) be given, and assume that (G1),
(G2) hold. Then there exists positive constant C such that, for any 0 < θ < 2 − p
and t0 > 0,
E(t) ≤ C(1 + ln(1 + t))−θ/(p−1) .
Proof. By Lemma 2.7 and using Young’s inequality, for 0 < θ < 1, we have
θ
(1 + t)−1 E q (t)(g ◦ ∇u) ≤ C(1 + t)−1 E q (t)(g p ◦ ∇u) p−1+θ
h q(p−1+θ)
i
≤ C(1 + t)−1 εE p−1 (t) + C(ε)(g p ◦ ∇u) ,
we choose q = (p − 1)/θ so that q(p−1+θ)
= q + 1. Consequently,
p−1
Z T
(1 + t)−1 E q+1 (t)dt ≤ AE(S), ∀S ≥ t0 .
S
Then letting T → +∞, we obtain
Z +∞
(1 + t)−1 E q+1 (t)dt ≤ AE(S).
S
Thus Proposition 2.2. yields
E(t) ≤ CE(0)(1 + φ(t))
This completes the proof.
−1
q
−θ
≤ C(1 + ln(1 + t)) p−1 ,
∀t ≥ t0 .
16
M. KAFINI
EJDE-2013/134
Acknowledgments. The author would like to express his sincere gratitude to
King Fahd University of Petroleum and Minerals for its support.
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EJDE-2013/134
RATE OF DECAY FOR SOLUTIONS
Mohammad Kafini
Department of Mathematics and Statistics, KFUPM, Dhahran 31261, Saudi Arabia
E-mail address: mkafini@kfupm.edu.sa
17