Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 123, pp. 1–19.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
CAUCHY PROBLEM FOR A GENERALIZED WEAKLY
DISSIPATIVE PERIODIC TWO-COMPONENT
CAMASSA-HOLM SYSTEM
WENXIA CHEN, LIXIN TIAN, XIAOYAN DENG
Abstract. In this article, we study a generalized weakly dissipative periodic
two-component Camassa-Holm system. We show that this system can exhibit
the wave-breaking phenomenon and determine the exact blow-up rate of strong
solution to the system. In addition, we establish a sufficient condition for
having a global solution.
1. Introduction
In recent years, the Camassa-Holm equation [4],
ut − utxx + 3uux = 2ux uxx + uuxxx ,
t > 0, x ∈ R
(1.1)
which models the propagation of shallow water waves has attracted considerable
attention from a large number of researchers, and two remarkable properties of (1.1)
were found. The first one is that the equation possesses the solutions in the form of
peaked solitons or ‘peakons’ [4, 8]. The peakon u(t, x) = ce−|x−ct| , c 6= 0 is smooth
except at its crest and the tallest among all waves of the fixed energy. It is a feature
observed for the traveling waves of largest amplitude which solves the governing
equations for water waves [9, 10, 29, 33]. The other remarkable property is that the
equation has breaking waves [4, 11]; that is, the solution remains bounded while its
slope becomes unbounded in finite time. After wave breaking the solutions can be
continued uniquely as either global conservative [2] or global dissipative solutions
[3].
The Camassa-Holm equation also admits many integrable multicomponent generalizations. The most popular one is
mt − Aux + umx + 2ux m + ρρx = 0
ρt + (ρu)x = 0
(1.2)
m = u − uxx
Notice that the C-H equation can be obtained via the obvious reduction ρ ≡ 0
and A = 0. System (1.2) was derived in [27], where ρ(t, x) is related to the free
surface elevation from the equilibrium (or scalar density), and A ≥ 0 characterizes
2000 Mathematics Subject Classification. 35B44, 35B30, 35G25.
Key words and phrases. Wave-breaking; weakly dissipative; blow-up rate;
periodic two-component Camassa-Holm equation; global solution.
c
2014
Texas State University - San Marcos.
Submitted September 27, 2013. Published May 14, 2014.
1
2
W. CHEN, L. TIAN, X. DENG
EJDE-2014/123
a linear underlying shear flow. Recently, Constantin-Ivanov [12] and Ivanov [23]
established a rigorous justification of the derivation of system (1.2). Mathematical
properties of the system have been also studied further in many works, for example
[1, 6, 7, 14, 15, 19, 22, 26, 28]. Chen, Liu and Zhang [6] established a reciprocal
transformation between the two-component Camassa-Holm system and the first
negative flow of the AKNS hierarchy. Escher, Lechtenfeld, and Yin [14] investigated
local well-posedness for the two-component Camassa-Holm system with initial data
(u0 , ρ0 − 1) ∈ H s × H s−1 with s ≥ 2 by applying Kato’s theory [24] and provided
some precise blow-up scenarios for strong solutions to the system. The local wellposedness is improved by Gui and Liu [20] to the Besov Spaces (especially in the
Sobolev space H s × H s−1 with s > 3/2), and they showed that the finite time
blow-up is determined by either the slope of the first component u or the slope of
the second component ρ [8, 14]. The blow-up criterion is made more precise in [25]
where Liu and Zhang showed that the wave breaking in finite time only depends
on the slope of u. This blow-up criterion is improved to the lowest Sobolev spaces
H s × H s−1 with s > 3/2 [19].
In general, it is difficult to avoid energy dissipation mechanisms in a real world.
We are interested in the effect of the weakly dissipative term on the two-component
Camassa-Holm equation. Wu, Escher and Yin have investigated the blow-up phenomena, the blow-up rate of the strong solutions of the weakly dissipative CH
equation [31] and DP equation [30]. Inspired by the above results, in this paper, we
investigate the following generalized weakly dissipative two-component CamassaHolm system
ut − utxx − Aux + 3uux − σ(2ux uxx + uuxxx ) + λ(u − uxx ) + ρρx = 0,
t > 0, x ∈ R,
ρt + (ρu)x = 0,
t > 0, x ∈ R,
u(0, x) = u0 (x), ρ(0, x) = ρ0 (x),
u(t, x) = u(t, x + 1), ρ(t, x) = ρ(t, x + 1),
(1.3)
x ∈ R,
t ≥ 0, x ∈ R,
or equivalently,
mt − Aux + σ(umx + 2ux m) + 3(1 − σ)uux + λm + ρρx = 0,
ρt + (ρu)x = 0,
(1.4)
m = u − uxx ,
where λm = λ(I − ∂xx )u is the weakly dissipative term, λ ≥ 0 and A are constants,
and σ is a new free parameter. When A = 0, λ = 0 and ρ = 1, Guan and Yin
have obtained a new result of the existence of the strong solution and some new
blow-up results [16]. Meanwhile, they have proved the global existence of the weak
solution about the two-component CH equation [17]. Henry investigates the infinite
propagation speed of the solution for a two-component CH equation [21].
Similar to [12, 14], we can use the method of Besov spaces together with the
transport equation theory to show that system (1.4) is locally well-posedness in
H s ×H s−1 with s > 3/2. The two equations for u and ρ are of a transport structure
∂t f + v∂x f = g. It is well known that most of the available estimates require v to
have some level of regularity. Roughly speaking, the regularity of the initial data
is expected to be preserved as soon as v belongs to L1 (0, T ; Lip). More specially,
u and ρ are “transported” along directions of σu and u respectively. Then, the
EJDE-2014/123
TWO-COMPONENT CAMASSA-HOLM SYSTEM
3
solution can be estimated in a Gronwall way involving kux kL∞ . Hence, one can
RT
use these estimates to derive a criterion which says if 0 kux (τ )kL∞ dτ < ∞, then
solutions can be extended further in time. Compared with the result in [5], we find
that the equation (1.4) has the same blow-up rate when the blow-up occurs. This
fact shows that the blow-up rate of equation (1.4) is not affected by the weakly
dissipative term. But the occurrence of blow-up of equation (1.4) is affected by the
dissipative parameter λ.
The basic elementary framework is as follows. Section 2 gives the local wellposedness of system (1.4) and a wave-breaking criterion, which implies that the wave
breaking only depends on the slope of u, not the slope of ρ. Section 3 improves
the blow-up criterion with a more precise conditions. Section 4 determine the
exact blow-up rate of strong solutions of system (1.4). Finally, section 5 provides
a sufficient condition for global solutions.
Notation. Throughout this paper, we identity periodic function spaces over the
unit S in R2 , i.e. S = R/Z.
2. Formation of singularities for σ 6= 0
We consider the following generalized weakly dissipative two - component Camassa - Holm system:
ut − utxx − Aux + 3uux − σ(2ux uxx + uuxxx ) + λ(u − uxx ) + ρρx = 0,
t > 0, x ∈ R,
ρt + (ρu)x = 0,
u(0, x) = u0 (x),
u(t, x) = u(t, x + 1),
t > 0, x ∈ R,
(2.1)
ρ(0, x) = ρ0 (x),
ρ(t, x) = ρ(t, x + 1),
where λ ≥ 0 and A are constants, and σ is a new free parameter.
System (2.1) can be written in the “transport” form
3−σ 2 σ 2 1 2
u + ux + ρ ) − λu
2
2
2
ρt + (ρu)x = 0 t > 0, x ∈ R
ut + σuux = −∂x G ∗ (−Au +
u(0, x) = u0 (x), ρ(0, x) = ρ0 (x),
u(t, x) = u(t, x + 1), ρ(t, x) = ρ(t, x + 1),
t > 0, x ∈ R
(2.2)
x∈R
t ≥ 0, x ∈ R
cosh(x−[x]− 1 )
where G(x) := 2 sinh(1/2)2 , x ∈ S, and (1 − ∂x2 )−1 f = G ∗ f for all f ∈ L2 (S).
Applying the transport equation theory combined with the method of Besov
spaces, one may follow the similar argument as in [20] to obtain the following local
well-posedness result for the system (2.1). The proof is very similar to that of [20,
Theorem 1.1] and is omitted.
Theorem 2.1. Assume (u0 , ρ0 − 1) ∈ H s (S) × H s−1 (S) with s > 3/2, then there
exist a maximal time T = T (k(u0 , ρ0 − 1)kH s ×H s−1 ) > 0 and a unique solution
(u, ρ − 1) of equation (2.1) in C([0, T ); H s × H s−1 ) ∩ C 1 ([0, T ); H s−1 × H s−2 ) with
initial data (u0 , ρ0 ). Moreover, the solution depends continuously on the initial
data, and T is independent of s.
4
W. CHEN, L. TIAN, X. DENG
EJDE-2014/123
Lemma 2.2 ([26]). Let 0 < s < 1. Suppose that f0 ∈ H s , g ∈ L1 ([0, T ]; H s ),
v, vx ∈ L1 ([0, T ]; L∞ ), and that f ∈ L∞ ([0, T ]; H s ) ∩ C([0, T ); S 0 ) solves the onedimensional linear transport equation
∂t f + v∂x f = g
f (0, x) = f0 (x)
then f ∈ C([0, T ]; H s ). More precisely, there exists a constant C depending only on
s such that
Z t
Z t
kf (τ )kH s V 0 (τ )dτ ,
kf (t)kH s ≤ kf0 kH s + C
kg(τ )kH s dτ +
0
0
then
kf (t)kH s ≤ e
CV (t)
Z
t
kg(τ )kH s dτ ),
(kf0 kH s + C
0
where V (t) =
Rt
0
(kv(τ )kL∞ + kvx (τ )kL∞ )dτ .
We may use [19, Lemma 2.1] to handle the regularity propagation of solutions to
(2.1). In addition, Lemma 2.2 was proved using the Littlewood-Paley analysis for
the transport equation and Moser-type estimates. Using this result and performing
the same argument as in [19], we can obtain the following blow-up criterion.
Theorem 2.3. Let σ 6= 0, (u, ρ) be the solution of (2.1) with initial data (u0 , ρ0 −
1) ∈ H s (S) × H s−1 (S) with s > 3/2, and T be the maximal time of existence. Then
Z t
T <∞⇒
kux (τ )kL∞ dτ = ∞.
(2.3)
0
Regarding the finite time blow-up, we consider the trajectory equation of the
system (2.1),
dq(t, x)
= u(t, q(t, x)), t ∈ [0, T )
(2.4)
dt
q(0, x) = x, x ∈ S,
where u ∈ C 1 ([0, T ); H s−1 ) is the first component of the solution (u, ρ) to (2.1) with
initial data (u0 , ρ0 ) ∈ H s (S) × H s−1 (S) with s > 3/2, and T > 0 is the maximal
time of the existence. Applying Theorem 2.1, we know that q(t, ·) : S → S is the
diffeomorphism for every t ∈ [0, T ), and
Z t
qx (t, x) = exp
ux (τ, q(τ, x))dτ > 0, ∀(t, x) ∈ [0, T ) × S.
(2.5)
0
Hence, the L∞ -norm of any function v(t, ·) ∈ L∞ , t ∈ [0, T ) is preserved under the
diffeomorphism q(t, ·) with t ∈ [0, T ); that is, kv(t, ·)kL∞ = kv(t, q(t, ·))kL∞ .
Lemma 2.4 ([11]). Let T > 0 and v ∈ C 1 ([0, T ); H 1 (R)), then for every t ∈ [0, T ),
there exists at least one point ξ(t) ∈ R with m(t) := inf x∈R [vx (t, x)] = vx (t, ξ(t)).
The function m(t) is absolutely continuous on (0, T ) with
dm(t)
= vtx (t, ξ(t))
dt
a.e. on (0, T ).
Lemma 2.5. Assume (u0 , ρ0 − 1) ∈ H s (S) × H s−1 (S) with s > 3/2, and (u, ρ) is
the solution of system (2.1), then k(u, ρ − 1)k2H 1 ×L2 ≤ k(u0 , ρ0 − 1)k2H 1 ×L2 .
EJDE-2014/123
TWO-COMPONENT CAMASSA-HOLM SYSTEM
5
Proof. Multiplying the first equation in (2.1) by u and using integration by parts
gives
Z
Z
Z
d
(u2 + u2x )dx + 2λ (u2 + u2x )dx + 2 uρρx dx = 0
dt S
S
S
Rewriting the second equation in (2.1) in the form (ρ − 1)t + ρx u + ρux = 0, and
multiplying by (ρ − 1) and using integration by parts, we have
Z
Z
Z
Z
Z
d
(ρ − 1)2 dx + 2 uρρx dx − 2 uρx dx + 2 ux ρ2 dx − 2 ux ρdx = 0.
dt S
S
S
S
S
Combining the above equalities, we have
Z
Z
d
2
2
2
(u + ux + (ρ − 1) )dx + 2λ (u2 + u2x )dx = 0,
dt S
S
Z
Z t
d
(u2 + u2x + (ρ − 1)2 + 2λ
(u2 + u2x )dτ )dx = 0.
dt S
0
So we have
Z
2
u2x
Z
2
t
(u +
+ (ρ − 1) + 2λ
(u2 + u2x )dτ )dx
S
0
Z
=
(u20 + u20x + (ρ0 − 1)2 )dx = k(u0 , ρ0 − 1)k2H 1 ×L2 .
S
Rt
(u + u2x )dτ ≥ 0, we obtain
Z
2
k(u, ρ − 1)kH 1 ×L2 =
(u2 + u2x + (ρ − 1)2 )dx ≤ k(u0 , ρ0 − 1)k2H 1 ×L2 .
Since 2λ
2
0
S
The proof is complete.
Lemma 2.6 ([32]). (1) For all f ∈ H 1 (S), we have
max f 2 (x) ≤
x∈[0,1]
e+1
kf k21 ,
2(e − 1)
e+1
2(e−1)
where
is the best constant.
(2) For all f ∈ H 3 (S), we have
max f 2 (x) ≤ ckf k21 ,
x∈[0,1]
where the possible best constant c ∈ (1, 13
12 ], and the best constant is
e+1
2(e−1) .
Lemma 2.7. If f ∈ H 3 (S), then
max fx2 (x) ≤
x∈[0,1]
1
kf k2H 2 (S) .
12
Proof. From [32, Theorem 2.1], the Fourier expansion of f (x) can be written as
∞
a0 X
f (x) =
+
an cos(2πnx).
2
n=1
Then
fx (x) = −
∞
X
n=1
(2nπan sin(2πnx)).
6
W. CHEN, L. TIAN, X. DENG
Using that
P∞
n=1
EJDE-2014/123
1/n2 = π 2 /6, we have
max fx2 (x)
x∈S
≤
∞
X
2
|2nπan |
n=1
=
∞
X
(2nπ)2 |an |
n=1
≤
∞
X
1 2
2nπ
((2nπ)2 |an |)2
n=1
∞
X
n=1
(
1 2
)
2nπ
∞
1 X
(16n4 π 4 a2n )
≤
24 n=1
∞
1 X
(8n4 π 4 a2n )
12 n=1
Z
1
1
=
f 2 dx ≤
kf k2H 2 (S) .
12 S xx
12
=
The proof is complete.
Applying the above lemmas and the method of characteristics, we may carry out
the estimates along the characteristics q(t, x) which captures supx∈S ux (t, x) and
inf x∈S ux (t, x).
Lemma 2.8. Let σ 6= 0 and (u, ρ) be the solution of (2.1) with initial data (u0 , ρ0 −
1) ∈ H s (S) × H s−1 (S), s > 3/2, and T be the maximal time of existence.
(1) When σ > 0, we have
r
kρ0 k2L∞ + C12
λ2
sup ux (t, x) ≤ ku0x kL∞ +
+
;
(2.6)
2
σ
σ
x∈S
(2) When σ < 0, we have
r
inf ux (t, x) ≥ −ku0x kL∞ −
x∈S
λ2
C2
− 2;
2
σ
σ
where the constants are defined as follows:
s
5(e + 1)
1 + A2
(e + 1)|3 − σ|
C1 =
+(
+
)k(u0 , ρ0 − 1)k2H 1 ×L2 ,
2(e − 1)
2
e−1
s
5(e + 1)
A2
(5 − σ)e + 3 − σ
C2 ==
+(
+
)k(u0 , ρ0 − 1)k2H 1 ×L2 .
2(e − 1)
2
2(e − 1)
(2.7)
(2.8)
(2.9)
Proof. The local well-posedness theorem and a density argument imply that it
suffices to prove the desired estimates for s ≥ 3. Thus, we take s = 3 in the proof.
Here we may assume that u0 6= 0. Otherwise, the results become trivial.
Differentiating the first equation in (2.2) with respect to x and using the identity
−∂x2 G ∗ f = f − G ∗ f , we have
σ
1
3−σ 2
σ
3−σ 2 1 2
utx + σuuxx + u2x = ρ2 +
u + A∂x2 G ∗ u − G ∗ ( u2x +
u + ρ ) − λux .
2
2
2
2
2
2
(2.10)
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TWO-COMPONENT CAMASSA-HOLM SYSTEM
7
(1) When σ > 0, using Lemma 2.4 and the fact that
sup[vx (t, x)] = − inf [−vx (t, x)],
x∈S
x∈S
we can consider m̄(t) and η(t) as
m̄(t) := ux (t, η(t)) = sup(ux (t, x)),
t ∈ [0, T ).
(2.11)
x∈S
This gives
uxx (t, η(t)) = 0
a.e. on t ∈ [0, T )
(2.12)
Take the trajectory q(t, x) defined in (2.4). We know that q(t, ·) : S → S is a
diffeomorphism for every t ∈ [0, T ),then there exists x1 (t) ∈ S such that
q(t, x1 (t)) = η(t),
t ∈ [0, T ).
(2.13)
Let
ζ̄(t) = ρ(t, q(t, x1 )),
t ∈ [0, T ).
(2.14)
Then along the trajectory q(t, x1 (t)), equation (2.10) and the second equation of
(2.1) become
σ
1
m̄0 (t) = − m̄2 (t) − λm̄(t) + ζ̄ 2 (t) + f (t, q(t, x1 ))
2
2
ζ̄ 0 (t) = −ζ̄(t)m̄(t),
(2.15)
where
σ
3−σ 2
3 − σ 2 1 2
u + A∂x2 G ∗ u − G ∗
u2x +
u + ρ .
2
2
2
2
2
Since ∂x G ∗ u = ∂x G ∗ ∂x u, we have
f=
(2.16)
3−σ 2
σ
3−σ 2
1
u + A∂x G ∗ ∂x u − G ∗ ( u2x +
u ) − G ∗ 1 − G ∗ (ρ − 1)
2
2
2
2
1
− G ∗ (ρ − 1)2
2
3−σ 2
3−σ 2
1
≤
u + A∂x G ∗ ∂x u − G ∗ (
u ) − G ∗ 1 − G ∗ (ρ − 1)
2
2
2
|3 − σ| 2
3−σ 2
1
≤
u + A|∂x G ∗ ∂x u| + |G ∗ (
u )| + |G ∗ 1| + |G ∗ (ρ − 1)|.
2
2
2
Based on the following formulas:
f=
|3 − σ|
e+1
|3 − σ| 2
u ≤
·
kuk2H 1 ,
2
2
2(e − 1)
e+1
1
A|∂x G ∗ ∂x u| ≤ AkGx kL2 kux kL2 ≤
+ A2 kux k2L2 ,
2(e − 1) 4
σ 2
σ 2
e+1
σ
|G ∗ ( ux )| ≤ kGx kL∞ k ux kL1 ≤
· kux k2L2 ,
2
2
2(e − 1) 2
3−σ 2
3−σ 2
e+1
|3 − σ|
|G ∗ (
u )| ≤ kGx kL∞ k
u kL1 ≤
·
kuk2L2 ,
2
2
2(e − 1)
2
1
1
e+1
|G ∗ 1| ≤ kGkL∞ ≤
,
2
2
4(e − 1)
e+1
1
+ kρ − 1k2L2 ,
|G ∗ (ρ − 1)| ≤ kGkL2 kρ − 1kL2 ≤
2(e − 1) 4
8
W. CHEN, L. TIAN, X. DENG
EJDE-2014/123
1
1
e+1
|G ∗ (ρ − 1)2 | ≤ kGkL∞ k(ρ − 1)2 kL1 ≤
kρ − 1k2L2 ,
2
2
4(e − 1)
from the above inequalities and Lemma 2.5 we obtain an upper bound of f ,
A2
(e + 1)|3 − σ|
5(e + 1) 1
+ kρ − 1k2L2 + (
+
)kuk2H 1
4(e − 1) 4
4
2(e − 1)
5(e + 1)
A2 + 1 (e + 1)|3 − σ|
(2.17)
≤
+(
+
)k(u0 , ρ0 − 1)k2H 1 ×L2
4(e − 1)
4
2(e − 1)
1
= C12 .
2
Similarly, we obtain a lower bound of f ,
σ
3−σ 2
1
σ−3 2
u + A|∂x G ∗ ∂x u| + |G ∗ ( u2x +
u )| + |G ∗ 1|
−f ≤
2
2
2
2
1
+ |G ∗ (ρ − 1)| + G ∗ (ρ − 1)2
2
e
(e + 1)(|σ| + 2|3 − σ|)
5(e + 1)
A2
+
kρ − 1k2L2 + (
+
)kuk2H 1
≤
4(e − 1) 2(e − 1)
4
4(e − 1)
5(e + 1)
A2
2e + (e + 1)(|σ| + 2|3 − σ|)
≤
+(
+
)k(u0 , ρ0 − 1)k2H 1 ×L2 .
4(e − 1)
4
4(e − 1)
(2.18)
Combining (2.17) and (2.18), we obtain
f≤
|f | ≤
A2
2e + (e + 1)(|σ| + 2|3 − σ|)
5(e + 1)
+(
+
)k(u0 , ρ0 − 1)k2H 1 ×L2 . (2.19)
4(e − 1)
4
4(e − 1)
Since s ≥ 3, it follows that u ∈ C01 (S) and
inf ux (t, x) ≤ 0,
x∈S
sup ux (t, x) ≥ 0,
t ∈ [0, T ).
(2.20)
x∈S
Hence, we obtain
m̄(t) > 0 for t ∈ [0, T ).
From the second equation in (2.15), we have
ζ̄(t) = ζ̄(0)e−
Rt
0
m̄(τ )dτ
,
(2.21)
(2.22)
|ρ(t, q(t, x1 ))| = |ζ̄(t)| ≤ |ζ̄(0)| ≤ kρ0 kL∞ .
For any given x ∈ S, we define
r
kρ0 k2L∞ + C12
λ2
.
+
σ2
σ
Notice that P1 (t) is a C 1 -function in [0, T ) and satisfies
r
kρ0 k2L∞ + C12
λ2
P1 (0) = m̄(0) − ku0x kL∞ −
≤ m̄(0) − ku0x kL∞ ≤ 0.
+
2
σ
σ
Next, we claim that
P1 (t) ≤ 0 for t ∈ [0, T ).
(2.23)
If not, then suppose that there is a t0 ∈ [0, T ) such that P1 (t0 ) > 0. Define
t1 = max{t < t0 : P1 (t) = 0}, then P1 (t1 ) = 0, P10 (t1 ) ≥ 0. That is,
r
kρ0 k2L∞ + C12
λ2
m̄(t1 ) = ku0x kL∞ +
+
, m̄0 (t1 ) = P10 (t1 ) ≥ 0.
2
σ
σ
P1 (t) = m̄(t) − ku0x kL∞ −
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TWO-COMPONENT CAMASSA-HOLM SYSTEM
9
On the other hand, we have
σ
1
m̄0 (t1 ) = − m̄2 (t1 ) − λm̄(t1 ) + ζ̄ 2 (t1 ) + f (t1 , q(t1 , x1 ))
2
2
r
kρ0 k2L∞ + C12
σ
λ2
λ 2
≤ − ku0x kL∞ +
+
+
2
2
σ
σ
σ
λ2
1
1
+
+ kρ0 k2L∞ + C12 < 0.
2σ 2
2
This yields a contraction. Thus, P1 (t) ≤ 0 for t ∈ [0, T ). Since x is chosen
arbitrarily, we obtain (2.6)).
(2) When σ < 0 , we have a finer estimate
1
3−σ 2 1
u + (G ∗ 1) + G ∗ (ρ − 1) + G ∗ (ρ − 1)2
−f ≤ −A(∂x G ∗ ∂x u) + G ∗
2
2
2
1
1
3−σ 2
u | + |G ∗ 1| + |G ∗ (ρ − 1)| + |G ∗ (ρ − 1)2 |
≤ A|∂x G ∗ ∂x u| + |G ∗
2
2
2
5(e + 1)
A2
(5 − σ)e + 3 − σ
1
≤
+(
+
)k(u0 , ρ0 − 1)k2H 1 ×L2 = C22 .
4(e − 1)
4
4(e − 1)
2
(2.24)
We consider the functions m(t) and ξ(t) in Lemma 2.4,
m(t) := inf [ux (t, x)],
x∈S
t ∈ [0, T )
(2.25)
Then uxx (t, ξ(t)) = 0 a.e. on t ∈ [0, T ). Choose x2 (t) ∈ S, such that q(t, x2 (t)) =
ξ(t), t ∈ [0, T ). Let ζ(t) = ρ(t, q(t, x2 )), t ∈ [0, T ). Along the trajectory q(t, x2 ),
equation (2.10) and the second equation of (2.1) become
1
σ
m0 (t) = − m2 (t) − λm(t) + ζ 2 (t) + f (t, q(t, x2 ))
2
2
ζ 0 (t) = −ζ(t)m(t).
q
2
C2
∞
Let P2 (t) = m(t) + ku0x kL + σλ2 − σ2 , ∀x ∈ R. Then P2 (t) is a C 1 -function
in [0, T ) and satisfies
r
C2
λ2
P2 (0) = m(0) + ku0x kL∞ +
− 2 ≥ m(0) + ku0x kL∞ ≥ 0.
2
σ
σ
Now we claim that
P2 (t) ≥ 0 for t ∈ [0, T ).
(2.26)
Assume that there is a t̄0 ∈ [0, T ) such that P2 (t̄0 ) < 0. Define t2 = max{t < t̄0 :
P2 (t) = 0}, then P2 (t2 ) = 0, P20 (t2 ) ≤ 0. That is,
r
C22
λ2
, m0 (t2 ) = P20 (t2 ) ≤ 0.
m(t2 ) = −ku0x kL∞ −
−
σ2
σ
In addition, we have
σ
1
m0 (t2 ) = − m2 (t2 ) − λm(t2 ) + ζ 2 (t2 ) + f (t2 , q(t2 , x2 ))
2
r 2
σ
λ2
C22
λ 2 λ2
1
≥ − (−ku0x kL∞ −
−
+
) +
− C22 > 0.
2
2
σ
σ
σ
2σ 2
This is a contradiction. Then we have P2 (t) ≥ 0 for t ∈ [0, T ), since x is chosen
arbitrarily.
10
W. CHEN, L. TIAN, X. DENG
EJDE-2014/123
Now, we present the following estimates for kρkL∞ (S) , if σux is bounded from
below.
Lemma 2.9 ([5]). Let σ 6= 0 and (u, ρ) be the solution of (2.1) with initial data
(u0 , ρ0 −1) ∈ H s (S)×H s−1 (S), s > 3/2, and T be the maximal time of the existence.
If there is a M ≥ 0 such that inf (t,x)∈[0,T )×S σux ≥ −M , Then we have following
two statements.
(1) If σ > 0, then kρ(t, ·)kL∞ (S) ≤ kρ0 kL∞ (S) eM t/σ .
(2) If σ < 0, then kρ(t, ·)kL∞ (S) ≤ kρ0 kL∞ (S) eN t ,
√
where N = ku0x kL∞ + (C2 / −σ) and C2 is given in (2.24).
Proof. The proof of Lemma 2.9 is similar to that of [5, Proposition 3.8], so we omit
it here.
From the above results, we can get the necessary and sufficient conditions for
the blow-up of solutions.
Theorem 2.10 (Wave-breaking criterion for σ 6= 0). Let σ 6= 0 and (u, ρ) be the
solution of (2.1) with initial data (u0 , ρ0 − 1) ∈ H s (S) × H s−1 (S), s > 3/2, and T
be the maximal time of existence. Then the solution blows up in finite time if and
only if
(2.27)
lim inf σux (t, x) = −∞.
t→T − x∈S
Proof. Assume that T < ∞ and (2.27) is not valid, then there is some positive
number M > 0, such that σux (t, x) ≥ −M , ∀(t, x) ∈ [0, T ) × S. From the above
lemmas, we have |ux (t, x)| ≤ C, where C = C(A, M, σ, λ, k(u0 , ρ0 − 1)kH s ×H s−1 ).
Thus, Theorem 2.3 implies that the maximal existence time T = ∞, which contradicts the assumption T < ∞.
On the other hand, the Sobolev embedding theorem H s ,→ L∞ with s > 1/2
implies that if (2.27) holds, the corresponding solution blows up in finite time. The
proof is complete.
3. Blow-up scenarios
Theorem 3.1. Let σ > 0 and (u, ρ) be the solution of (2.1) with initial data
(u0 , ρ0 − 1) ∈ H s (S) × H s−1 (S), s > 3/2, and T be the maximal time of existence.
Assume that there is some x0 ∈ S such that ρ0 (x0 ) = 0, u0x (x0 ) = inf x∈S u0x (x)
and
k(u0 , ρ0 − 1)k2H 1 ×L2
8e − 10
(3.1)
λ2 4(e − 1)
<
−
,
2
2
18(e − 1) 2σ (18A + 19)e − (18A + 17) + (2|3 − σ| + σ)(e + 1)
then the corresponding solution to system (2.1) blows up in finite time in the following sense: there exists a T such that
2
0<T ≤
+ 72σ(e − 1)(1 + |u0x (x0 )|)
σ−λ
÷ σ(32e − 40 − 324e − 324A2 e + 324A2 + 306) − 36λ2 (e − 1) (3.2)
+ (2|3 − σ| + σ)(e − 1)k(u0 , ρ0 − 1)k2H 1 ×L2
and that lim inf t→T − (inf x∈S ux (t, x)) = −∞.
EJDE-2014/123
TWO-COMPONENT CAMASSA-HOLM SYSTEM
11
Proof. Here we also consider s ≥ 3. We still consider along the trajectory q(t, x2 )
defined as before. In this way, we can write the transport equation of ρ in (2.1)
along the trajectory of q(t, x2 ) as
dρ(t, ξ(t))
= −ρ(t, ξ(t))ux (t, ξ(t)).
dt
By the assumption, we have
(3.3)
m(0) = ux (0, ξ(0)) = inf u0x (x) = u0x (x0 ).
x∈S
Choose ξ(0) = x0 and then ρ0 (ξ(0)) = ρ0 (x0 ) = 0. Then by (3.3), we derive
ρ(t, ξ(t)) = 0,
∀t ∈ [0, T ).
(3.4)
Evaluating the result at x = ξ(t) and combining (3.4) with uxx (t, ξ(t)) = 0, we have
σ
3−σ 2
m0 (t) = − m2 (t) − λm(t) +
u (t, ξ(t)) + A(Gx ∗ ux )(t, ξ(t))
2
2
σ
3−σ 2 1 2
− G ∗ ( u2x +
u + ρ )(t, ξ(t))
2
2
2
(3.5)
σ 2
= − m (t) − λm(t) + f (t, q(t, x2 ))
2
λ
λ2
σ
+ f (t, q(t, x2 )).
= − (m(t) + )2 +
2
σ
2σ
We modify the estimates:
1
e+1
9
A|Gx ∗ ux | ≤ AkGx kL2 kux kL2 ≤
·
+ A2 kux k2L2 ,
18 2(e − 1) 2
1
e+1
9
|G ∗ (ρ − 1)| ≤ kGkL2 kρ − 1kL2 ≤
·
+ kρ − 1k2L2 .
18 2(e − 1) 2
Similarly, we obtain the upper bound of f as
f≤
(18A2 + 19)e − (18A2 + 17) + (2|3 − σ| + σ)(e + 1)
10 − 8e
+
18(e − 1)
4(e − 1)
× k(u0 , ρ0 − 1)k2H 1 ×L2 := −C3 .
By assumption (3.1), we obtain
λ2
2σ
− C3 < 0 and
σ
λ 2 λ 2
λ2
m(t) +
+
− C3 ≤
− C3 < 0, t ∈ [0, T ).
(3.6)
2
σ
2σ
2σ
So m(t) is strictly decreasing in [0, T ). If the solution (u, ρ) of (2.1) exists globally
in time, that is, T = ∞, we will show that it leads to a contradiction.
0x (x0 )|)
. Integrating (3.6) over [0, t1 ] gives
Let t1 = 2σ(1+|u
2σC3 −λ2
Z t1
λ2
m(t1 ) = m(0) +
m0 (t)dt ≤ |u0x (x0 )| + (
− C3 )t1 = −1.
(3.7)
2σ
0
m0 (t) ≤ −
For t ∈ [t1 , T ), we have m(t) ≤ m(t1 ) ≤ −1. From (3.6), we have
σ
λ 2
m0 (t) ≤ − m(t) +
.
2
σ
Integrating over [t1 , T ), by (3.7), yields
1
1
1
1
σ
−
+ λ
≤−
+
≤ − (t − t1 ),
λ
λ
2
m(t) + σλ
−
1
m(t)
+
m(t
)
+
1
σ
σ
σ
(3.8)
t ∈ [t1 , T ),
12
W. CHEN, L. TIAN, X. DENG
m(t) ≤
So, T ≤ t1 +
complete.
2
σ−λ ,
σ
2 (t
1
− t1 ) +
σ
λ−σ
λ
→ −∞,
σ
−
EJDE-2014/123
as t → t1 +
2
.
σ−λ
which is a contradiction to T = ∞. Consequently, the proofis
Theorem 3.2. Let σ 6= 0 and (u, ρ) be the solution of (2.1) with initial data
(u0 , ρ0 − 1) ∈ H s (S) × H s−1 (S), s > 3/2, and T be the maximal time of the
existence.
(1) When σ > 0, assume thatqthere is an x0 ∈ S such that ρ0 (x0 ) = 0, u0x (x0 ) =
C2
2
inf x∈S u0x (x) and u0x (x0 ) < − σλ2 + σ1 − σλ , where C1 is defined in (2.8). Then
the corresponding solution to system (2.1) blows up in finite time in the following
sense: there exists a T1 such that
0 < T1 ≤ −
2(λ + σu0x (x0 ))
,
(λ + σu0x (x0 ))2 − (λ2 + σC12 )
and
lim inf { inf ux (t, x)} = −∞.
t→T1−
x∈S
q (2) When σ < 0, assume that there is some x0 ∈ S such that u0x (x0 ) >
C22
λ
λ2
σ 2 − σ − σ , where C2 is defined in (2.9). Then the corresponding solution
to system (2.1) blows up in finite time in the following sense: there exists a T2 such
that
2(λ + σu0x (x0 ))
0 < T2 ≤ −
,
(λ + σu0x (x0 ))2 − (λ2 − σC22 )
and
lim inf
{sup ux (t, x)} = ∞.
−
t→T2
x∈S
Proof. (1) When σ > 0, using the upper bound of f in (2.17) and (3.4), we have
σ
λ 2 λ 2
1
m0 (t) ≤ − m(t) +
+
+ C12 , t ∈ [0, T ).
2
σ
2σ 2
q
2
C2
By the assumption m(0) = u0x (x0 ) < − σλ2 + σ1 − σλ , we have that m0 (0) < 0
and m(t) is strictly decreasing over [0, T ). Set
λ2
1
1
1 2
1
δ= −
+
C ∈ (0, ).
λ
2 σ(u0x (x0 ) + σ )2 2σ 2 1
2
Since m(t) < m(0) = u0x (x0 ) < − σλ , it holds
σ
λ 2 λ2
1
λ 2
m(t) +
+
+ C12 ≤ −δσ m(t) +
.
2
σ
2σ 2
σ
By a similar argument as in the proof of Theorem 3.1, we obtain
m0 (t) ≤ −
m(t) ≤
λ + σu0x (x0 )
λ
1
− → −∞ as t → −
.
σ + (δσ 2 u0x (x0 ) + λδσ)t σ
λδ + δσu0x (x0 )
Thus, we have 0 < T1 ≤ − λδ+δσu1 0x (x0 ) .
EJDE-2014/123
TWO-COMPONENT CAMASSA-HOLM SYSTEM
13
(2) when σ < 0, we consider the functions m̄(t) and η(t) as defined in (2.11) and
take the trajectory q(t, x1 ) with x1 defined in (2.13), then
σ
1
m̄0 (t) = − m̄2 (t) − λm̄(t) + ρ2 (t, η(t)) + f (t, q(t, x1 ))
2
2
(3.9)
λ 2 λ 2
σ
+
≥ − m̄(t) +
+ f (t, q(t, x1 )).
2
σ
2σ
From the lower bound of f in (2.24), we obtain
λ 2 λ 2
σ
1
m̄0 (t) ≥ − m̄(t) +
+
− C22 , t ∈ [0, T ).
2
σ
2σ 2
q
2
C2
By the assumption m̄(0) ≥ u0x (x0 ) > σλ2 − σ2 − σλ , we have that m̄0 (0) > 0 and
m̄(t) is strictly increasing over [0, T ).
Set
1
(σu0x (x0 ) + λ)2 − (λ2 − σC22 )
∈ (0, ).
θ=
2
2(σu0x (x0 ) + λ)
2
Since m̄(t) > m̄(0) ≥ u0x (x0 ) > − σλ , we obtain
λ 2 λ 2
1
λ 2
σ
− C22 ≥ −θσ m̄(t) +
m̄0 (t) ≥ − m̄(t) +
+
.
2
σ
2σ 2
σ
Similarly, we obtain
λ
1
λ + σu0x (x0 )
− → ∞ as t → −
.
m̄(t) ≥
2
σ + (θσ u0x (x0 ) + λθσ)t σ
λθ + θσu0x (x0 )
Therefore, 0 < T2 ≤ − λθ+θσu1 0x (x0 ) . The proof is complete.
Remark. If σ = 3 and A = 0, then all solutions of system (2.1) with initial data
(u0 , ρ0 − 1) ∈ H s (S) × H s−1 (S) with s > 3/2 satisfying u0 6= 0 and ρ0 (x0 ) = 0 for
some x0 ∈ S, blow up in finite time.
4. Blow-up rate
Theorem 4.1. Let σ 6= 0. If T < ∞ is the blow-up time of the solution (u, ρ)
to (2.1) with initial data (u0 , ρ0 − 1) ∈ H s (S) × H s−1 (S), s > 3/2 satisfying the
assumptions of Theorem 3.2. Then
2
lim { inf ux (t, x)(T − t)} = − , σ > 0,
(4.1)
σ
t→T − x∈S
2
lim {sup ux (t, x)(T − t)} = − , σ < 0.
(4.2)
σ
t→T − x∈S
Proof. We assume that s = 3 to prove the theorem.
(1) when σ > 0, from (3.5) we have
σ
λ 2 λ2
m0 (t) = − m(t) +
+
+ f (t, q(t, x)).
2
σ
2σ
From (2.19), note that
M=
(4.3)
5(e + 1)
A2
2e + (e + 1)(|σ| + 2|3 − σ|)
+(
+
)k(u0 , ρ0 − 1)k2H 1 ×L2 ,
4(e − 1)
4
4(e − 1)
(4.4)
σ
λ
λ2
σ
λ
λ2
(m(t) + )2 −
− M ≤ m0 (t) ≤ − (m(t) + )2 +
+ M.
2
σ
2σ
2
σ
2σ
(4.5)
Then
−
14
W. CHEN, L. TIAN, X. DENG
EJDE-2014/123
Choose ε ∈ (0, σ2 ), since limt→T − m(t) + σλ = −∞, there is some t0 ∈ (0, T ),
2
λ2
such that m(t0 )+ σλ < 0 and m(t0 )+ σλ > 1ε 2σ
+M . Since m is locally Lipschitz,
it follows that m is absolutely continuous. We deduce that m is decreasing on [t0 , T )
and
λ 2
1 λ2
m(t) +
>
+ M , t ∈ [t0 , T ).
(4.6)
σ
ε 2σ
Combining (4.5) with (4.6), we have
σ
1
σ
d
≤ + ε, t ∈ [t0 , T ).
−ε≤
(4.7)
2
dt m(t) + σλ
2
Integrating over (t, T ) with t ∈ [t0 , T ) and noticing that limt→T − m(t)+ σλ = −∞,
we obtain
σ
1
σ
( − ε)(T − t) ≤ −
≤ ( + ε)(T − t).
2
2
m(t) + σλ
Since ε ∈ (0, σ2 ) is arbitrary, in view of the definition of m(t), we have
lim {m(t)(T − t) +
t→T −
λ
2
(T − t)} = − ;
σ
σ
that is, limt→T − {inf x∈S ux (t, x)(T − t)} = − σ2 .
(2) When σ < 0, we consider the functions m̄(t) and η(t) as defined in (2.11).
2
λ2
From (3.9) and (4.4), we have m̄0 (t) ≥ − σ2 m̄(t) + σλ + 2σ
− M.
−
q Because m̄(t) → ∞ as t → T , there is a t1 ∈ (0, T ), such that m̄(t1 ) >
2
λ
2M
λ
0
σ 2 − σ − σ > 0. Thus, we have that m̄ (t) > 0 and m̄(t) is strictly increasing
on [t1 , T ), and
m̄(t) > m̄(t1 ) > 0.
(4.8)
By the transport equation for ρ, we have
dρ(t, η(t))
= −m̄(t)ρ(t, η(t)).
dt
Then
ρ(t, η(t)) = ρ(t1 , η(t1 ))e
−
Rt
t1
m̄(τ )dτ
,
t ∈ [t1 , T ).
(4.9)
Combining (4.8) with (4.9) yields
ρ2 (t, η(t)) ≤ ρ2 (t1 , η(t1 )),
t ∈ [t1 , T )
From (3.9) and (4.10), we have
λ 2 λ2
1
σ
m̄ +
+
− ρ2 (t1 , η(t1 )) − M
−
2
σ
2σ 2
σ
λ 2 λ2
1
−
+ ρ2 (t1 , η(t1 )) + M.
≤ m̄0 ≤ − m̄ +
2
σ
2σ 2
Choose ε ∈ (0, − σ2 ), and pick a t2 ∈ [t1 , T ), such that
λ 2
11 2
λ2 m̄(t2 ) +
>
ρ (t1 , η(t1 )) + M −
.
σ
ε 2
2σ
From (4.11) and (4.12), we have
σ
σ
d
1
−ε≤
≤ + ε, t ∈ [t2 , T ).
2
dt m̄(t) + σλ
2
(4.10)
(4.11)
(4.12)
(4.13)
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TWO-COMPONENT CAMASSA-HOLM SYSTEM
15
Integrating (4.13) over [t, T ) with t ∈ [t2 , T ) and limt→T − m̄(t) = ∞ gives
(
σ
1
− ε)(T − t) ≤ −
2
m̄(t) +
λ
σ
≤(
σ
+ ε)(T − t).
2
Since ε ∈ (0, − σ2 ) is arbitrary, in view of the definition of m̄(t), we have
2
lim {sup ux (t, x)(T − t)} = − .
σ
t→T − x∈S
This completes the proof of Theorem 4.1.
5. Existence of a global solution
In this section, we provide a sufficient condition for the global solution of system
(2.1) in the case when 0 < σ < 2.
Lemma 5.1. Let 0 < σ < 2 and (u, ρ) be the solution of (2.1) with initial data
(u0 , ρ0 − 1) ∈ H s (S) × H s−1 (S), s > 3/2, and T be the maximal time of existence.
Assume that inf x∈S ρ0 (x) > 0.
(1) When 0 < σ ≤ 1, it holds
1
C4 eC3 t ,
inf x∈S ρ0 (x)
1
C3 t
1
| sup ux (t, x)| ≤
C42−σ e 2−σ .
σ
2−σ
x∈S
inf x∈S ρ0 (x)
| inf ux (t, x)| ≤
x∈S
(2) When 1 < σ < 2, it holds
1
| inf ux (t, x)| ≤
1
σ
2−σ
x∈S
C3 t
C42−σ e 2−σ ,
inf x∈S ρ0 (x)
1
| sup ux (t, x)| ≤
C4 eC3 t ,
inf
x∈S ρ0 (x)
x∈S
where constants C3 and C4 are defined as follows:
C3 = 1 +
5(e + 1)
A2
2e + (e + 1)(|σ| + 2|3 − σ|)
+(
+
)k(u0 , ρ0 − 1)k2H 1 ×L2 ,
4(e − 1)
4
4(e − 1)
C4 = 1 + ku0x k2L∞ + kρ0 k2L∞ .
Proof. A density argument indicates that it suffices to prove the desired results for
s ≥ 3. Since s ≥ 3, we have u ∈ C01 (S) and
inf ux (t, x) < 0,
x∈S
sup ux (t, x) > 0,
t ∈ [0, T ).
x∈S
(1) First we will derive the estimate for | inf x∈S ux (t, x)|. Define m(t) and ξ(t)
as in (2.25), and consider along the characteristics q(t, x2 (t)). Then
m(t) ≤ 0
for t ∈ [0, T ).
(5.1)
Let ζ(t) = ρ(t, ξ(t)) and evaluating (2.10) and the second equation of system (2.1)
at (t, ξ(t)), we have
σ
1
m0 (t) = − m2 (t) − λm(t) + ζ 2 (t) + f (t, q(t, x2 ))
2
2
ζ 0 (t) = −ζ(t)m(t),
(5.2)
16
W. CHEN, L. TIAN, X. DENG
EJDE-2014/123
where f is defined in (2.16). The second equation above implies that ζ(t) and ζ(0)
are of the same sign.
Next we construct a Lyapunov function for our system as in [13]. Since here we
have a free parameter σ, we could not find a uniform Lyapunov function. Instead,
we split the case 0 < σ ≤ 1 and the case 1 < σ < 2. From the assumption of the
theorem, we know that ζ(0) = ρ(0, ξ(0)) > 0.
When 0 < σ ≤ 1, we define the Lyapunov function
ω1 (t) = ζ(0)ζ(t) +
ζ(0)
(1 + m2 (t)),
ζ(t)
which is always positive for t ∈ [0, T ). Differentiating ω1 (t) and using (5.2) gives
ζ(0)
2ζ(0)
(1 + m2 (t))ζ 0 (t) +
m(t)m0 (t)
2
ζ (t)
ζ(t)
ζ(0)
= −ζ(0)ζ(t)m(t) − 2 (1 + m2 (t))(−ζ(t)m(t))
ζ (t)
2ζ(0)
σ
1
+
m(t)(− m2 (t) − λm(t) + ζ 2 (t) + f )
ζ(t)
2
2
ζ(0) 3
ζ(0)
2λζ(0) 2
2ζ(0)
= (1 − σ)
m (t) +
m(t) −
m (t) +
m(t)f
ζ(t)
ζ(t)
ζ(t)
ζ(t)
ζ(0)
2ζ(0)
≤
m(t) +
m(t)f
ζ(t)
ζ(t)
ζ(0)
(1 + m2 (t))(1 + |f |) ≤ C3 ω1 (t),
≤
ζ(t)
ω10 (t) = ζ(0)ζ 0 (t) −
(5.3)
where
C3 = 1 +
5(e + 1)
A2
2e + (e + 1)(|σ| + 2|3 − σ|)
+(
+
)k(u0 , ρ0 − 1)k2H 1 ×L2 .
4(e − 1)
4
4(e − 1)
This gives
ω1 (t) ≤ ω1 (0)eC3 t = (ζ 2 (0) + 1 + m2 (0))eC3 t
≤ (1 + ku0x k2L∞ + kρ0 k2L∞ )eC3 t =: C4 eC3 t ,
(5.4)
where C4 = 1 + ku0x k2L∞ + kρ0 k2L∞ .
Recalling that ζ(t) and ζ(0) are of the same sign, the definition of ω1 (t) implies
ζ(t)ζ(0) ≤ ω1 (t) and |ζ(0)||m(t)| ≤ ω1 (t). By (5.4), we obtain
| inf ux (t, x)| = |m(t)| ≤
x∈S
1
ω1 (t)
≤
C4 eC3 t ,
|ζ(0)|
inf x∈S ρ0 (x)
for t ∈ [0, T ).
When 1 < σ < 2, we define the Lyapunov function
ω2 (t) = ζ σ (0)
ζ 2 (t) + 1 + m2 (t)
.
ζ σ (t)
(5.5)
Then
2ζ σ (0)
σ−1 2
σ
m(t)(
ζ (t) − λm(t) + f + )
ζ σ (t)
2
2
(5.6)
σ
σ
ζ (0)
σ
ζ (0)
2
2
≤ σ (1 + m (t))(|f | + ) ≤ σ (1 + m (t))(|f | + 1) ≤ C3 ω2 (t).
ζ (t)
2
ζ (t)
ω20 (t) =
EJDE-2014/123
TWO-COMPONENT CAMASSA-HOLM SYSTEM
17
Thus, we obtain
ω2 (t) ≤ ω2 (0)eC3 t = (ζ 2 (0) + 1 + m2 (0))eC3 t
≤ (1 + ku0x k2L∞ + kρ0 k2L∞ )eC3 t = C4 eC3 t .
Applying Young’s inequality ab ≤
ap
p
+
bq
q
to (5.5) with p =
2
σ
and q =
2
2−σ
yields
2
σ(2−σ) σ2 (1 + m2 ) 2−σ
2−σ
2
ω2 (t)
2
=
ζ
+
σ(2−σ)
ζ σ (0)
ζ 2
2−σ
2
σ σ(2−σ) σ2
2 − σ (1 + m2 ) 2 2−σ
ζ 2
≥
+
σ(2−σ)
2
2
ζ 2
≥ (1 + m2 )
2−σ
2
≥ |m(t)|
2−σ
.
So we have
| inf ux (t, x)| ≤
x∈S
1
ω (t) 2−σ
1
C3 t
1
2
≤
C42−σ e 2−σ .
σ
σ
ζ (0)
inf x∈S ρ02−σ (x)
(2) Now, we estimate | supx∈S ux (t, x)|. Consider m̄(t), η(t), q(t, x1 ) as in (2.11)
and (2.13), and
σ
1
m̄0 (t) = − m̄2 (t) − λm̄(t) + ζ̄ 2 (t) + f (t, q(t, x1 ))
2
2
ζ̄ 0 (t) = −ζ̄(t)m̄(t)
(5.7)
for t ∈ [0, T ), where ζ̄(t) = ρ(t, η(t)). We know that
m̄(t) ≥ 0
for t ∈ [0, T ).
(5.8)
When 0 < σ ≤ 1, we define the Lyapunov function
ω̄1 (t) = ζ̄ σ (0)
ζ̄ 2 (t) + 1 + m̄2 (t)
.
ζ̄ σ (t)
(5.9)
Then from (5.6) and (5.8), we have ω̄10 (t) ≤ C3 ω̄1 (t), then ω̄1 (t) ≤ C4 eC3 t . Hence,
by a similar argument as before, we obtain
ω̄1 (t)
2−σ
≥ |m̄(t)|
.
ζ̄ σ (0)
Then
| sup ux (t, x)| ≤ (
x∈S
1
C3 t
1
ω̄1 (t) 2−σ
1
2−σ
)
≤
C
e 2−σ ,
σ
4
2−σ
ζ̄ σ (0)
inf x∈S ρ0 (x)
t ∈ [0, T ).
When 1 < σ < 2, consider the Lyapunov function
ω̄2 (t) = ζ̄(0)ζ̄(t) +
ζ̄(0)
(1 + m̄2 (t)).
ζ̄(t)
(5.10)
From (5.3) and (5.8), we have ω̄20 (t) ≤ C3 ω̄2 (t) and ω̄2 (t) ≤ C4 eC3 t . Therefore,
| sup ux (t, x)| = |m̄(t)| ≤
x∈S
The proof is complete.
ω̄2 (t)
1
≤
C4 eC3 t ,
inf x∈S ρ0 (x)
ζ̄(0)
t ∈ [0, T ).
18
W. CHEN, L. TIAN, X. DENG
EJDE-2014/123
Theorem 5.2. Let 0 < σ < 2 and (u, ρ) be the solution of (2.1) with initial data
(u0 , ρ0 − 1) ∈ H s (S) × H s−1 (S), s > 3/2, and T be the maximal time of existence.
If inf x∈S ρ0 (x) > 0, then T = +∞ and the solution (u, ρ) is global.
Proof. Assume on the contrary that T < +∞ and the solution blows up in finite
time. It then follows from Theorem 2.3, that
Z T
kux (t)kL∞ dt = ∞.
(5.11)
0
However, from the assumptions of the theorem and Lemma 5.1, we have |ux (t, x)| <
∞ for all (t, x) ∈ [0, T ) × S. This is a contradiction to (5.11). So T = +∞, and it
means that the solution (u, ρ) is global.
Acknowledgements. We like to thank the anonymous referees for their valuable
comments and suggestions. This work was supported by the National Natural Science Foundation of China [11101190, 11171135, 11201186], and Natural Science
Foundation of Jiangsu Province [BK2012282], China Postdoctoral Science Foundation funded project (Nos: 2012M511199, 2013T60499) and Jiangsu University
foundation grant [11JDG117].
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Wenxia Chen
Nonlinear Scientific Research Center, Faculty of Science, Jiangsu University, Zhenjiang, Jiangsu 212013, China
E-mail address: swp@ujs.edu.cn
Lixin Tian
Nonlinear Scientific Research Center, Faculty of Science, Jiangsu University, Zhenjiang, Jiangsu 212013, China
Xiaoyan Deng
Nonlinear Scientific Research Center, Faculty of Science, Jiangsu University, Zhenjiang, Jiangsu 212013, China