Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 93, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
UNIFORM DECAY OF SOLUTIONS TO CAUCHY
VISCOELASTIC PROBLEMS WITH DENSITY
MOHAMMAD KAFINI
Abstract. In this article we consider the decay of solutions to a linear Cauchy
viscoelastic problem with density. This study includes the exponential and
polynomial rates as particular cases. To compensate for the lack of Poincare’s
inequality in the whole space, we consider the solutions in spaces weighted by
the density.
1. Introduction
In this article we are concerned with the initial-value problem
Z t
ρ(x)utt − ∆u(x) +
g(t − s)∆u(x, s)ds = 0, x ∈ Rn , t > 0,
0
u(x, 0) = u0 (x),
ut (x, 0) = u1 (x),
(1.1)
x ∈ Rn ,
where u0 , u1 are initial data chosen in suitable spaces and g is the relaxation function subjected to some conditions to be specified later. The density ρ(x) satisfying
the following conditions
(H1) ρ : Rn → R, n ≥ 2, ρ(x) > 0, ρ(x) ∈ C 0,γ (Rn ) with γ ∈ (0, 1) and
ρ ∈ Ln/2 (Rn ) ∩ L∞ (Rn ).
In the whole space case, Poincare’s inequality and some Lebesgue and Sobolev
embedding inequalities are not valid. To overcome this difficulty in this case, we
exploit the density to introduce weighted spaces for solutions of our problem.
The work with weighted spaces was studied by many authors. Papadopoulos and
Stavarakakis [11] established existence of a global solutions and blow up results for
the non local quasilinear hyperbolic problem of Kirchhoff type
utt − φ(x)k∇u(t)k2 ∆u + δut = |u|a u,
x ∈ Rn , t ≥ 0,
in the case where n ≥ 3, δ ≥ 0 and ρ(x) = (φ(x))−1 is a positive function lying in
Ln/2 (Rn ) ∩ L∞ (Rn ). Brown and Stavarakakis [1] proved the existence of positive
2000 Mathematics Subject Classification. 35B05, 35L05, 35L15, 35L70.
Key words and phrases. Cauchy problem; relaxation function; viscoelastic; weighted spaces.
c
2011
Texas State University - San Marcos.
Submitted May 25, 2011. Published July 18, 2011.
1
2
M. KAFINI
EJDE-2011/93
solutions for the semilinear elliptic equation
−∆u(x) = λg(x)f (u(x)),
lim
|x|→+∞
0 < u < 1, x ∈ R
u(x) = 0
and for g ∈ Ln/2 (Rn ) for n = 1, 2, 3. Karachalios and Stavarakakis [6] proved a local
existence of solutions and global attractor in the energy space D1,2 (Rn ) × L2g (Rn )
for a semilinear hyperbolic problem
utt − φ(x)∆u + δut + λf (u) = η(x),
x ∈ Rn , t > 0,
in the case where δ > 0, n ≥ 3 and ρ(x) = (φ(x))−1 lies in Ln/2 (Rn ).
It is also worth mentioning the work of Zhou [12] and Cavalcanti citec2. In this
work, the following nonlinear wave equation with damping and source terms of the
form
utt − φ(x)∆u + a|ut |m−1 ut = f (x, u),
x ∈ Rn , t > 0,
was considered. Where the author proved, in the linear damping case, that the
solution blows up in finite time even for vanishing initial energy. Criteria to guarantee blow up of solutions with positive initial energy were established for both
linear and nonlinear cases. Global existence and large time behavior also proved.
In [9], a class of abstract viscoelastic systems of the form
utt (t) + Au(t) + βu(t) − (g ∗ Aα u)(t) = 0
u(0) = u0 ,
ut (0) = u1 ,
(1.2)
for 0 ≤ α ≤ 1, β ≥ 0, were investigated. The main focus was on the case when
0 < α < 1 and the main result was that solutions for (1.2) decay polynomially
even if the kernel g decay exponentially. This result is sharp (see [9, Theorem 12].
This result has been improved by Rivera et al. [10], where the authors studied a
more general abstract problem than (1.2) and established a necessary and sufficient
condition to obtain an exponential decay. In the case of lack of exponential decay,
a polynomial decay has been proved.
Kafini and Messaoudi [4] looked into the following Cauchy viscoelastic problem
Z
utt − ∆u +
t
g(t − s)∆u(x, s)ds = 0,
x ∈ Rn , t > 0
0
u(x, 0) = u0 (x),
ut (x, 0) = u1 (x),
x ∈ Rn
and showed that, for compactly supported initial data u0 , u1 and for an exponentially decaying relaxation function g, the decay of the first energy of solution is
polynomial. The finite-speed propagation is used to compensate for the lack of
Poincaré’s inequality in Rn . For nonexistence, the same authors [5] established a
blow-up result to the following Cauchy viscoelastic problem
Z t
utt − ∆u +
g(t − s)∆u(x, s)ds + ut = |u|p−1 u, x ∈ Rn , t > 0
0
u(x, 0) = u0 (x),
ut (x, 0) = u1 (x),
x ∈ Rn ,
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UNIFORM DECAY OF SOLUTIONS
3
under the conditions
Z
t
g(s)ds) <
0
E(0) =
2p − 2
,
2p − 1
g 0 (t) ≤ 0,
1
1
1
ku1 k22 + k∇u0 k22 −
ku0 kp+1
p+1 ≤ 0.
2
2
p+1
This result extends the one of [13], established for the wave equation in bounded
domain.
In this article, we will extend the result in [12] to our viscoelastic problem. We
aim to study the effect of the density to the decay rates. In the case ρ(x) = 1 (as
in [4]), the best decay obtained is polynomial. Here, we establish a general decay
result for solutions where the exponential and polynomial are only special cases.
This result does not contradict the past results in [4, 9, 10]. The choice of spaces
of solutions and the density one make it possible to get an exponential decay rate.
Where we have ρ(x) is a continuous and Ln/2 (Rn ) ∩ L∞ (Rn ) function that make
most of its contribution concentrate at the early time in the contrast of the late
time (t → ∞). Obviously, ρ(x) cannot be a constant here. In our proof, we use
the multiplier method together with the Lyapunov functional method as in [7] with
some necessary modification due to the nature of the problem. The paper organized
as follows. In Section 2, we define our function space and the assumptions on the
kernel g. In section 3, we state and prove our main result.
2. Preliminaries
To achieve our result, we assume the following assumptions on the relaxation
function g:
(G1) g : R+ → R+ is a differentiable function such that
Z ∞
g(0) > 0, 1 −
g(s)ds = l > 0.
0
(G2) There exists a nonincreasing differentiable function ξ : R+ → R+ such that
Z ∞
g 0 (t) ≤ −ξ(t)g(t), t ≥ 0,
ξ(t)dt = +∞.
0
There are many functions satisfying (G1) and (G2), for example
α
g1 (t) =
, ν>1
(1 + t)ν
p
g2 (t) = αe−β(t+1) , 0 < p ≤ 1
α
g3 (t) =
, ν>1
(1 + t)[ln(1 + t)]ν
where α and β are chosen properly.
Remark 2.1. Condition (G1) is necessary to guarantee the hyperbolicity of the
system.
We define the function space of our problem and its norm, as in [1, 11], as follows:
(A) The function space for (1.1) is X = D1,2 (Rn ) × L2ρ (Rn ), with
2n
D1,2 (Rn ) = {f ∈ L n−2 (Rn ) : ∇f ∈ L2 (Rn )}.
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M. KAFINI
EJDE-2011/93
(B) The space L2ρ (Rn ) is defined to be the closure of C0∞ (Rn ) functions with respect
to the inner product
Z
(f, h)L2ρ (Rn ) =
ρf h dx.
Rn
One can easily check that L2ρ (Rn ) is a separable Hilbert space and
kf k2L2ρ (Rn ) = (f, f )L2ρ (Rn ) .
(C) For 1 < p < ∞, if f is a measurable function on Rn , we define
Z
1/p
ρ|f |p dx
kf kLpρ =
Rn
Lpρ (Rn )
and let
consist of all f for which kf kLpρ (Rn ) < ∞.
For the weighted space Lpρ (Rn ), we have the following lemma
Lemma 2.2 ([6]). Let ρ satisfies (H1), then for any u ∈ D1,2 (Rn ),
2n
2n
kukLqρ ≤ kρkLs k∇ukL2 , with s =
, 2≤q≤
.
2n − qn + 2q
n−2
Corollary 2.3. If q = 2, then Lemma 2.2. yields
kukL2ρ ≤ kρkLn/2 k∇ukL2 ,
where we can assume kρkLn/2 = C0 > 0 to get
kukL2ρ (Rn ) ≤ C0 k∇ukL2 .
(2.1)
Theorem 2.4 ([12]). Suppose that (H1) holds and g satisfies (G1). Assume that
n+2
1 ≤ p ≤ n−2
if n ≥ 2 or 1 ≤ p if n = 2. Then for any initial data
u0 ∈ D1,2 (Rn )
and
u1 ∈ L2ρ (Rn ),
and
ut ∈ C([0, T ); L2ρ (Rn )),
problem (1.1) has a unique solution
u ∈ C([0, T ); D1,2 (Rn ))
for T small enough.
We now introduce the ‘modified’ energy functional
Z t
1
1
1
1−
g(s)ds k∇uk22 + (g ◦ ∇u),
E(t) = kut k2L2ρ +
2
2
2
0
where
Z
(2.2)
t
g(t − s)kv(t) − v(s)k22 ds,
(g ◦ v)(t) =
∀v ∈ L2 (Rn ).
0
3. Decay of solutions
In this section, we state and prove our main result. For this purpose, we set
F (t) := E(t) + ε1 Ψ(t) + ε2 χ(t),
where ε1 and ε2 are positive constants and
Z
Ψ(t) :=
ρuut dx,
Rn
Z t
Z
χ(t) := −
ρut
g(t − s) u(t) − u(s) ds dx.
Rn
0
(3.1)
(3.2)
(3.3)
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UNIFORM DECAY OF SOLUTIONS
5
Lemma 3.1. Along the solution of (1.1), the ‘modified’ energy satisfies
E 0 (t) =
1 0
1
1
(g ◦ ∇u) − g(t)k∇uk22 ≤ (g 0 ◦ ∇u) ≤ 0.
2
2
2
(3.4)
Proof. By multiplying (1.1) by ut and integrating over Rn , using integration by
parts, hypotheses (G1) and (G2) and some manipulations as in [2, 3, 8], we reach
the result.
Lemma 3.2. For any ε1 and ε2 small enough,
α1 F (t) ≤ E(t) ≤ α2 F (t),
(3.5)
holds for two positive constants α1 and α2 .
Proof. By applying Young’s inequality to (3.1) and using (3.2) and (3.3), we obtain
1
F (t) ≤ E(t) + ε1 δkut k2L2ρ + kuk2L2ρ
4δ
Z t
1
+ ε2 δkut k2L2ρ + k
g(t − s) u(t) − u(s) dsk2L2ρ
4δ 0
C0
C0
k∇uk22 + ε2 δkut k2L2ρ +
(1 − l)(g ◦ ∇u)
≤ E(t) + ε1 δkut k2L2ρ +
4δ
4δ
ε1 C0
ε2 C0
2
2
≤ E(t) + δ(ε1 + ε2 )kut kL2ρ +
k∇uk2 +
(1 − l)(g ◦ ∇u)
4δ
4δ
≤ E(t) + βE(t) ≤ α2 F (t).
Also, for ε1 and ε2 small enough, we have
F (t) ≥ E(t) − ε1 Ψ(t) − ε2 χ(t)
≥ E(t) − δ(ε1 + ε2 )kut k2L2ρ −
ε2 C0
ε1 C0
k∇uk22 −
(1 − l)(g ◦ ∇u)
4δ
4δ
≥ E(t) − βE(t) ≥ α1 F (t).
Consequently, (3.5) follows.
Lemma 3.3. Assume (H1), (G1), (G2). Along the solution of (1.1), the functional
Z
Ψ(t) :=
ρuut dx,
Rn
satisfies
Ψ0 (t) ≤ kut k2L2ρ − (l − δ)k∇uk22 +
1
(1 − l)(g ◦ ∇u)(t).
4δ
Proof. From the definition of Ψ(t) in (3.2) we have
Z
Z
Ψ0 (t) =
ρu2t dx +
ρuutt dx.
Rn
(3.6)
(3.7)
Rn
To estimate the last term in (3.7), we multiply (1.1) by u and integrate by parts
over Rn . So, we obtain
Z
Z t
Z
Z
g(t − s)∇u(s) ds dx −
|∇u(t)|2 dx.
(3.8)
ρuutt dx =
∇u(t) ·
Rn
Rn
0
Rn
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M. KAFINI
EJDE-2011/93
The first term in (3.8) can be estimated as follows
Z
Z
t
∇u(t) ·
g(t − s)∇u(s) ds dx
Z
Z t
=
∇u(t) ·
g(t − s)(∇u(s) − ∇u(t) + ∇u(t)) ds dx
Rn
0
Z
Z t
Z
Z
t
2
=
g(s)ds
|∇u(t)| dx +
∇u(t) ·
g(t − s)(∇u(s) − ∇u(t)) ds dx
0
Rn
Rn
0
Z
Z
Z
Z
t
1 t
2
2
g(s)ds (g ◦ ∇u)(t).
≤
g(s)ds
|∇u(t)| dx + δ
|∇u(t)| dx +
4δ 0
0
Rn
Rn
(3.9)
Rn
0
Recalling that
Z
t
Z
g(s)ds ≤
0
∞
g(s)ds = 1 − l,
0
we obtain the result.
Lemma 3.4. Assume (G1), (G2). Along the solution of (1.1), for any δ > 0, the
functional
Z t
Z
χ(t) := −
ρut
g(t − s)(u(t) − u(s)) ds dx
Rn
0
satisfies
1
1
χ0 (t) ≤ δ(1 + 2(1 − l)2 )k∇uk22 + (1 − l) (2δ + ) +
(g ◦ ∇u)(t)
2δ
4δ
Z t
Z
g(0)
ρu2t dx.
C0 (−(g 0 ◦ ∇u)(t)) −
g(s)ds − δ
4δ
Rn
0
(3.10)
Proof. From the definition of χ(t) in (3.3), we have
Z t
ρutt
g(t − s)(u(t) − u(s)) ds dx
Rn
0
Z t
Z
Z t
Z
−
ρut
g 0 (t − s)(u(t) − u(s)) ds dx −
g(s)ds
χ0 (t) = −
Z
Rn
0
To simplify the first term in (3.11), we multiply (1.1) by
and integrate by parts over Rn . So we obtain
Z
Rn
0
Z
Rt
0
(3.11)
ρu2t dx.
g(t − s)(u(t) − u(s))ds
t
g(t − s)(u(t) − u(s)) ds dx
Z
Z t
=
∆u(x)
g(t − s)(u(t) − u(s)) ds dx
Rn
0
Z Z t
Z t
−
(
g(t − s)(u(t) − u(s))ds
g(t − s)∆u(s)ds)dx.
ρutt
Rn
0
Rn
0
0
(3.12)
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UNIFORM DECAY OF SOLUTIONS
The first term in the right side of (3.12) is estimated as follows
Z
Z t
∆u(x)
g(t − s)(u(t) − u(s)) ds dx
Rn
0
Z
Z t
≤−
∇u(x) ·
g(t − s)(∇u(t) − ∇u(s)) ds dx
Rn
0
Z
Z t
≤
∇u(x) ·
g(t − s)(∇u(s) − ∇u(t)) ds dx
Rn
0
Z t
Z
1
g(s)ds)(g ◦ ∇u)(t)
≤δ
|∇u(t)|2 dx + (
4δ 0
n
ZR
1
≤δ
|∇u(t)|2 dx + (1 − l)(g ◦ ∇u)(t),
4δ
Rn
7
(3.13)
while the second term becomes, as in (3.9),
Z Z t
Z t
−
(
g(t − s)(u(t) − u(s))ds
g(t − s)∆u(x, s)ds)dx
Rn 0
0
Z Z t
Z t
=
(
g(t − s)∇u(s)ds) · (
g(t − s)(∇u(t) − ∇u(s))ds)dx
Rn 0
0
Z
Z t
Z
Z t
1
≤δ
|
g(t − s)∇u(s)ds|2 dx +
|
g(t − s)(∇u(t) − ∇u(s))ds|2 dx
4δ
n
n
R
0
R
0
Z Z t
1
(
g(t − s)(∇u(t) − ∇u(s))ds)2 dx
≤ (2δ + )
4δ Rn 0
Z
2δ(1 − l)2
|∇u(t)|2 dx
n
R
Z
1
≤ (2δ + )(1 − l)(g ◦ ∇u)(t) + 2δ(1 − l)2
|∇u(t)|2 dx.
4δ
Rn
(3.14)
Back to (3.11), the second term can be estimated as follows
Z
Z t
−
ρut
g 0 (t − s)(u(t) − u(s)) ds dx
n
R
0
Z
Z t
1
2
ρut dx + k
−g 0 (t − s)(u(t) − u(s))dsk2L2ρ
≤δ
4δ 0
Rn
Z
g(0)
≤δ
ρu2t dx +
C0 (−(g 0 ◦ ∇u)(t)).
4δ
n
R
By combining (3.11)-(3.15), the assertion of the lemma is established.
(3.15)
Theorem 3.5. Let u0 ∈ D1,2 (Rn ) and u1 ∈ L2ρ (Rn ) be given. Assume that (H1)
holds and g satisfies (G1) and (G2). Then, for each t0 > 0, there exist strictly
positive constants C1 and C2 such that the energy of solution given by (1.1) satisfies,
for all t ≥ t0 ,
Rt
ξ(s)ds
−C
E(t) ≤ C1 E(t0 )e 2 t0
, ∀t > t0 .
(3.16)
Proof. Since g is positive 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
8
M. KAFINI
EJDE-2011/93
Differentiation of (2.1) using (3.4), (3.6) and (3.10), yields
F 0 (t) = E 0 (t) + ε1 Ψ0 (t) + ε2 χ0 (t)
Z
g(0)
1
C0 ]((g 0 ◦ ∇u)(t))
≤ −(ε2 (g0 − δ) − ε1 )
ρu2t dx + [ − ε2
2
4δ
n
R
(3.17)
− [ε1 (l − δ) − ε2 δ(1 + 2(1 − l)2 )]k∇uk22
ε1
1
1 (1 − l) + ε2 (1 − l)[ 2δ +
+ ] (g ◦ ∇u)(t)
4δ
2δ
4δ
At this point we choose δ so small that
1
max{g0 − δ, l − δ} > g0 ,
2
1
2
δ(1 + 2(1 − l) ) < g0 .
4
Whence δ is fixed, the choice of any two positive constants ε1 and ε2 satisfying
1
1
g0 ε2 < ε1 < g0 ε2 ,
(3.18)
4
2
will make
k1 = ε2 (g0 − δ) − ε1 > 0,
k2 = ε1 (l − δ) − ε2 δ(1 + 2(1 − l)2 ) > 0.
Then we pick ε1 and ε2 so small that (3.5) and (3.18) remain valid and
1
g(0)
− ε2
C0 > 0.
2
4δ
Therefore, for some positive constants β, β1 and β2 , we have
Z
0
F (t) ≤ −β(
ρut 2 dx + k∇uk22 ) + β1 (g ◦ ∇u)(t),
Rn
(3.19)
≤ −βE(t) + β2 (g ◦ ∇u)(t) ∀t ≥ t0 .
Multiplying (3.19) by ξ(t) gives
ξ(t)F 0 (t) ≤ −βξ(t)E(t) + β2 ξ(t)(g ◦ ∇u)(t),
∀t ≥ t0 .
The last term can be estimated, using (H2), as follows
β2 ξ(t)(g ◦ ∇u)(t) ≤ −β2 (g 0 ◦ ∇u)(t) ≤ −2β2 E 0 (t).
Thus, (3.19) becomes
ξ(t)F 0 (t) ≤ −βξ(t)E(t) − 2β2 E 0 (t)
ξ(t)F 0 (t) + 2β2 E 0 (t) ≤ −βξ(t)E(t).
(3.20)
It is clear that
F1 (t) = ξ(t)F (t) + 2β2 E(t) ∼ E(t).
Therefore, using (3.20) and the fact that ξ 0 (t) ≤ 0, we arrive at
F10 (t) = (ξ(t)F (t) + 2β2 E(t))0 ≤ −βξ(t)E(t).
(3.21)
Integration over (t0 , t) leads to, for some constant C2 > 0 such that
−C2
F1 (t) ≤ F1 (t0 )e
Rt
t0
ξ(s)ds
,
∀t > t0 .
Recalling (3.5), estimate (3.11) yields the desired result (3.16).
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UNIFORM DECAY OF SOLUTIONS
9
R∞
Remark 3.6. Our result is established without using the condition 0 ξ(s)ds =
+∞, which is crucial for obtaining uniform stability.
Exponential decay is obtained for ξ(t) ≡ a, and polynomial decay for ξ(t) =
a(1 + t)−1 , where a is a positive constant.
Estimate (3.16) is also true for t ∈ [0, t0 ] by virtue of continuity and boundedness
of E(t) and ξ(t).
3.1. Acknowledgments. The author would like to express his sincere thanks to
King Fahd University of Petroleum and Minerals for its support.
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Mohammad Kafini
Department of Mathematics and Statistics, KFUPM, Dhahran 31261, Saudi Arabia
E-mail address: mkafini@kfupm.edu.sa