Global existence of smooth solutions to two-dimensional
compressible isentropic Euler equations for Chaplygin gases
De-Xing Konga
and
Yu-Zhu Wangb
a
Center of Mathematical Sciences, Zhejiang University
Hangzhou 310027, China
b
Department of Mathematics, Shanghai Jiao Tong University
Shanghai 200240, China
Abstract
In this paper we investigate the two-dimensional compressible isentropic Euler
equations for Chaplygin gases. Under the assumption that the initial data is close to
a constant state and the vorticity of the initial velocity vanishes, we prove the global
existence of the smooth solution to the Cauchy problem for two-dimensional flow of
Chaplygin gases.
Key words and phrases: Chaplygin gases, compressible Euler equations, irrotational flow, null condition, Cauchy problem, smooth solution.
2000 Mathematics Subject Classification: 76N10; 35Q35; 35L99.
1
1
Introduction
This paper concerns the following two-dimensional compressible isentropic Euler equations



 ∂t ρ + u · ∇ρ + ρ∇ · u = 0,
(1.1)
1


 ∂t u + u · ∇u + ∇P = 0,
ρ
where ρ(x, t) > 0 is the density, u(x, t) = (u1 (x, t), u2 (x, t))T stands for the velocity, P
denotes the pressure and the state equation reads
P = P0 − Aρ−1 ,
in which P0 , A are two positive constants. We are interested in the global existence on
the smooth solution of the Cauchy problem for the system (1.1) with the initial data
t=0:
ρ = ρ̄ + ερ0 (x), u = εu0 (x),
(1.2)
where ε is a positive small parameter, ρ̄ stands for a positive constant and satisfies
P0 − Aρ̄−1 > 0,
ρ0 (x) is a given smooth function with compact support, and u0 (x) = (u10 (x), u20 (x))T is a
given vector-valued smooth function with compact support, without loss of generality, we
may assume that there exists a positive number R such that
supp{ρ0 }, supp{u10 }, supp{u20 } ⊂ {x ∈ R2 | |x| ≤ R}.
In (1.1) and here after, we use the following notations: x = (x1 , x2 ) ∈ R2 , ∇ = (∂x1 , ∂x2 )
and
∂0 = ∂t =
∂
,
∂t
∂i = ∂xi =
∂
(i = 1, 2).
∂xi
The compressible Euler equations comprise a nonlinear symmetric hyperbolic system
of PDEs that model the ideal fluid flow. The characteristic wave speeds of this system are
given by the fluid velocity and the local sound speed. The two wave families associated to
these speeds may be considered separately. Smooth Eulerian flows of ideal polytropic gases
with initial data close to a constant state do not, in gereral, exist for all positive times (see
[6, 17, 18] ). Godin [6] obtained the precise information on the asymptotic behavior of the
life-span of the smooth solution to three-dimensional spherically symmetric Eulerian flows
of ideal polytropic gases with variable entropy, provided that the initial data is a smooth
2
small perturbation with compact support to a constant state. Sideris [17] showed that
the classical solution to the three-dimensional Euler equations for a polytropic, ideal fluid
must blow up in finite time under some assumption on the initial data. In Sideris [18], the
estimates on life-span of smooth solutions in two space dimensions were obtained under
suitable assumptions. In particular, we would like to mention that Grassin [8] proved
that the suitable initial data, which force particles to spread out, may lead to the global
existence in time for ideal polytropic gases.
When the adiabatic constant γ of the gas is equal to −1 (in this case, the equation
of state is suitably modified to keep the pressure and its derivative with respect to the
density positive), the gas is named Chaplygin gas, or called Karman-Tsien gas (see [20,
21], and in the isentropic case, see [4]). Godin [7] obtained the global existence of a
class of smooth three-dimensional spherically symmetric flows of Chaplygin gases with
the variable entropy, provided that the initial data is a smooth small perturbation with
compact support to a constant state.
By the light-cone gauge, Bordemann and Hoppe [2] simplified the description of relativistic membrane moving in the Minkowski space. By performing variables transformations, they established the relationship between the dynamics of relativistic membrane
and two-dimensional fluid dynamics. The equations for the two-dimensional fluid dynamics derived in [2] are nothing but the system (1.1). This implies that the system (1.1) can
be used to describe the motion of relativistic membrane moving in the Minkowski space
R1+3 . On this research topic, we refer Huang and Kong [12].
In the present paper, we investigate the global existence of the smooth solution of the
two-dimensional flow of Chaplygin gases. More precisely, under the assumption that the
initial data is close to a constant state and the vorticity of the initial velocity vanishes, we
show the global existence of the smooth solution of the Cauchy problem (1.1)-(1.2). The
main result in this paper is the following theorem.
Theorem 1.1 Suppose that the initial velocity is irrotational, i.e., u0 satisfies
∇⊥ · u0 , ∂1 u20 − ∂2 u10 = 0.
Then there exists ε0 > 0 such that the Cauchy problem (1.1)-(1.2) has a unique global
smooth solution for any ε ∈ [0, ε0 ].
The key idea to prove Theorem 1.1 is as follows: (i) write the velocity as the gradient
of a potential function; (ii) reduce the system (1.1) to a quasilinear wave equation for
3
the potential function; (iii) verify that this quasilinear wave equation satisfies the null
condition in two space dimensions; (iv) by null condition, L2 -L∞ estimates and energy
estimates, obtain the global existence on the smooth solution under consideration.
Remark 1.1 Theorem 1.1 still holds if the initial data u(x, 0) = εu0 (x) is replaced by
u(x, 0) = ū + εu0 (x), where ū ∈ R2 is a constant vector. In fact, as is well known,
replacing x by x − tū and u by u − ū leads to a new solution of (1.1)-(1.2) (with the same
pressure law), and so we reduce to the assumptions of Theorem 1.1 immediately.
The paper is organized as follows. In Section 2 we reduce the Cauchy problem (1.1)(1.2) to a Cauchy problem for a quasilinear wave equation (see (2.16)-(2.17) below). In
Section 3, we introduce some notations and prove some lemmas, these lemmas play an
important role in the proof of Theorem 1.1. Sections 4-5 are devoted to establishing the
L2 -L∞ estimates and energy estimates of the smooth solution, respectively. Theorem 1.1
is proved in Section 6 by the bootstrap argument.
2
Two-dimensional irrotational flow
Let
∇⊥ = (−∂2 , ∂1 ),
ω = (ω 1 , ω 2 ),
then
∇⊥ · ω = ∂1 ω 2 − ∂2 ω 1 .
As mentioned before, the key idea of the proof of Theorem 1.1 is to write the velocity u as
the gradient of a potential function. To do so, we first prove the following lemma which
plays an important roles in the present paper.
Lemma 2.1 Let T be a given positive real number and assume that (ρ, u) is a smooth
solution to the Cauchy problem (1.1)-(1.2) in the domain R2 × [0, T ). If the initial velocity
u0 (x) satisfies
∇⊥ · u0 = 0,
(2.1)
then
∇⊥ · u = 0,
∀ (x, t) ∈ R2 × [0, T ).
4
(2.2)
Proof. Noting u = (u1 , u2 ), by a direct calculation we have
∇⊥ · (u · ∇u) = ∂1 u1 ∂1 u2 + u1 ∂12 u2 + ∂1 u2 ∂2 u2 + u2 ∂1 ∂2 u2
−∂2 u1 ∂1 u1
−
u1 ∂2 ∂1 u1
−
∂2 u2 ∂2 u1
−
(2.3)
u2 ∂22 u1 ,
(u · ∇)(∇⊥ · u) = u1 ∂12 u2 − u1 ∂1 ∂2 u1 + u2 ∂1 ∂2 u2 − u2 ∂22 u1
(2.4)
(∇ · u)(∇⊥ · u) = ∂1 u1 ∂1 u2 − ∂1 u1 ∂2 u1 + ∂2 u2 ∂1 u2 − ∂2 u1 ∂2 u2 .
(2.5)
and
(2.3)-(2.5) gives
∇⊥ · (u · ∇u) = (u · ∇)(∇⊥ · u) + (∇ · u)(∇⊥ · u).
(2.6)
Noting (2.6) and taking the curl of the second equation in (1.1) gives
∂t (∇⊥ · u) + u · ∇(∇⊥ · u) + (∇ · u)(∇⊥ · u) = 0.
(2.7)
Combining (2.1) and (2.7) yields (2.2) immediately. Thus, the proof of Lemma 2.1 is
completed.
¥
Remark 2.1 Lemma 2.1 shows that, for two-dimensional compressible isentropic Euler
equations for Chaplygin gases, if the flow is irrotational at the initial time, then so is it at
any time t ∈ [0, T ), where T stands for the maximal existence time of the smooth solution.
Under the assumptions of Theorem 1.1, for system (1.1) there exists a scalar function
v = v(x, t) such that
u = ∇v
(2.8)
where the scalar function v(x, t) is a smooth function and has a compact support in x
(since u has compact support by the finite propagation speed). Substituting u = ∇v into
the second equation in (1.1) and noting Lemma 2.1 leads to
1
∂t ∇v + ∇|∇v|2 + ∇f (ρ) = 0,
2
(2.9)
where the function f is defined by
f 0 (ρ) =
1 dP
ρ dρ
(2.10)
and
f (ρ̄) = 0.
5
(2.11)
It follows from (2.10) and (2.11) that
Z
f (ρ) =
ρ
ρ̄
where
c̄2
A
1 dP
dλ =
− 2,
λ dλ
2
2ρ
½
c̄ =
¾1
2
dP
.
(ρ̄)
dρ
(2.12)
Noting (2.9), we have
1
∂t v + |u|2 + f (ρ) = 0.
2
Thus, by (2.8) and (2.13), we obtain



∇ · u = 4v,




∂t2 v = −u · ∂t u − f 0 (ρ)∂t ρ,





 ∂ ∇v = −u · ∇u − f 0 (ρ)∇ρ.
t
It follows from (2.14) that


∇ · u = 4v,






1
∂t ρ = − 0 (∂t2 + ∇v · ∂t ∇v),
f (ρ)




1


 ∇ρ = − 0 (∇∂t v + ∇v · ∇∇v).
f (ρ)
(2.13)
(2.14)
(2.15)
Substituting (2.15) into the first equation in (1.1) yields
∂t2 v − c̄2 4v + 2
2
X
∂i v∂t ∂i v − 2∂t v4v +
i=1
2
X
∂i v∂j v∂i ∂j v − |∇v|2 4v = 0.
(2.16)
i, j=1
On the other hand, noting (1.2), (2.8) and (2.13), we get
Z x1
1
t=0: v=ε
u10 (y, x2 )dy, vt = − ε2 |u0 (x)|2 − f (ρ̄ + ερ0 (x)).
2
R
(2.17)
Remark 2.2 For quasilinear wave equations in two space dimensions satisfying the null
condition (resp. both null conditions), Alinhac [1] proved the quasi-global existence (resp.
global existence) for their Cauchy problems. His proof relies on the construction of an
approximate solution, combined with an energy integral method which displays the null
condition(s). The method used in the present paper is different from one in Alinhac [1].
By (2.17), it is easy to see that v(x, 0), ∂t (x, 0) are smooth functions of x ∈ R2 and
has compact support, i.e.,
supp{v(x, 0)}, supp{∂t (x, 0)} ⊂ {x ∈ R2 | |x| ≤ R}.
6
Hence by the finite propagation speed, we obtain
∀ x ∈ {x ∈ R2 | |x| ≥ R + c̄t}.
v(x, t) = 0,
(2.18)
Obviously, (2.16) can be rewritten as
∂t2 v
2
− c̄ 4v +
2
X
k
gij
∂k v∂i ∂j v
i,j,k=0
where
2
X
+
2
X
kl
gij
∂k v∂l v∂i ∂j v = 0,
(2.19)
i,j,k,l=0
k
gij
∂k v∂i ∂j v = 2∂1 v∂0 ∂1 v + 2∂2 v∂0 ∂2 v − 2∂0 v∂12 v − 2∂0 v∂22 v
(2.20)
i,j,k=0
and
2
X
kl
gij
∂k v∂l v∂i ∂j v = 2∂1 v∂2 v∂1 ∂2 v − (∂2 v)2 ∂12 v − (∂1 v)2 ∂22 v.
(2.21)
i,j,k,l=0
For any given vector (X0 , X1 , X2 ) ∈ R3 with X02 = c̄2 (X12 + X22 ), we have
2
X
k
gij
Xi Xj Xk = 2X0 X12 + 2X0 X22 − 2X0 X12 − 2X0 X22 = 0
(2.22)
i,j,k=0
and
2
X
kl
gij
Xi Xj Xk Xl = 2X1 X2 X1 X2 − X12 X22 − X12 X22 = 0.
(2.23)
i,j,k,l=0
This shows that the wave equation (2.16) satisfies the null condition in two space dimensions (see[1]). The concept of null condition was introduced by Christodoulou and
Klainerman for nonlinear wave equations in three space dimensions (see [3] and [14]).
Therefore, we can rewrite the terms
2
X
k
gij
∂k v∂i ∂j v
and
i,j,k=0
as
2
X
2
X
kl
gij
∂k v∂l v∂i ∂j v
i,j,k,l=0
k
gij
∂k v∂i ∂j v = 2Q01 (∂1 v, v) + 2Q02 (∂2 v, v)
(2.24)
i,j,k=0
and
2
X
kl
gij
∂k v∂l v∂i ∂j v = −∂2 vQ12 (∂1 v, v) + ∂1 vQ12 (∂2 v, v),
i,j,k,l=0
where Qij (ϕ, ψ) are defined by
Qij (ϕ, ψ) = ∂i ϕ∂j ψ − ∂i ψ∂j ϕ
which are null forms.
7
(0 ≤ i < j ≤ 2),
(2.25)
3
Preliminaries
In order to prove Theorem 1.1, in this section we give some preliminaries.
Following Klainerman [13], we introduce the following vector fields
Γ = (Γ0 , · · · , Γ4 ) = (∂0 , ∂1 , ∂2 , Ω, S),
(3.1)
where
Ω = x1 ∂2 − x2 ∂1 ,
S = t∂t + r∂r .
(3.2)
It is easy to show that



[Γi , ¤] = −2δ4i (i = 0, · · · , 4),




[∂i , ∂j ] = 0 (i, j = 0, 1, 2), [S, ∂i ] = −∂i (i = 0, 1, 2),





 [Ω, ∂ ] = −∂ , [Ω, ∂ ] = ∂ , [Ω, ∂ ] = ∂ , [S, Ω] = 0,
1
2
2
1
0
0
(3.3)
where [·, ·] denotes the usual commutator of linear operators and δij is the Kronecker
symbol.
In what follows, we shall use the following notations
|v(x, t)|κ =
X
|Γa v(x, t)|,
|v(t)|κ = sup |v(x, t)|κ ,
x∈R2
|a|≤κ
[v(x, t)]κ =
X
1
(1 + |x|) 2 (1 + |x| − c̄t)|Γa v(x, t)|,
[v(t)]κ = sup [v(x, t)]κ
x∈R2
|a|≤κ
and
X
hv(x, t)i =
1
(1 + |x| + t) 2 |Γa v(x, t)|,
|a|≤κ
hv(t)i = sup hv(x, t)i,
kv(t)kκ =
x∈R2
X
kΓa v(·, t)kL2 ,
|a|≤κ
where κ is a nonnegative integer, a = (a0 , · · · , a4 ) is a multi-index and |a| = |a0 |+· · ·+|a4 |.
Moreover, we also use the following notations
|v|κ,t = sup |v(s)|κ ,
[v]κ,t = sup [v(s)]κ
0≤s<t
0≤s<t
and
hviκ,t = sup hv(s)iκ ,
0≤s<t
||v||κ,t = sup ||v(s)||κ .
0≤s<t
Let
1
c0 = c̄
2
8
and define
Λ(T ) = {(x, t) ∈ R2 × [0, T ) | ||x| − c̄t| ≤ c0 t}.
(3.4)
Thanks to the null condition, we will be able to show that the null form possesses a good
decay in the domain Λ(T ). In the our argument, the following operators
Z = (Z1 , Z2 ) =
(x1 , x2 )S + (−x2 , x1 )Ω
c̄t − |x|
(∂1 , ∂2 ) +
t
|x|t
(3.5)
will play an important role. It follows from (3.5) that
Zi = c̄∂i +
and
µ
|Zϕ(x, t)| ≤ C
xi
∂t
|x|
(i = 1, 2)
¶
||x| − c̄t|
1
|∂ϕ(x, t)| + |ϕ(x, t)|1 ,
t
t
(3.6)
(3.7)
here and hereafter C stands for a common absolute constant. We have
Lemma 3.1 Assume that hkij are constants and it holds that
2
X
hkij Xi Xj Xk = 0
(3.8)
i,j,k=0
for any vector (X0 , X1 , X2 ) satisfying X02 = c̄2 (X12 + X22 ). If ϕ, ψ ∈ C ∞ (R2 × [0, T )), then
¯
¯
¯ 2
¯
X
¯
¯
£
¤
k
¯
hij ∂k ψ∂i ∂j ϕ(x, t)¯¯ ≤ C |Zψ(x, t)||∂ 2 ϕ(x, t)| + |∂ψ(x, t)||Z∂ϕ(x, t)| .
(3.9)
¯
¯i,j,k=0
¯
Lemma 3.2 Assume that hkl
ij are constants and it holds that
2
X
hkl
ij Xi Xj Xk Xl = 0
(3.10)
i,j,k,l=0
for any vector (X0 , X1 , X2 ) satisfying X02 = c̄2 (X12 + X22 ). If ϕ, ψ, φ ∈ C ∞ (R2 × [0, T )),
then
¯
¯
¯ 2
¯
X
¯
¯
kl
¯
hij ∂k ψ∂l φ∂i ∂j ϕ(x, t)¯¯
¯
¯i,j,k,l=0
¯
£
¤
≤ C |∂ψ(x, t)||∂φ(x, t)||Z∂ϕ(x, t)| + |∂ψ(x, t)||Zφ(x, t)||∂ 2 ϕ(x, t)| + |Zψ(x, t)||∂φ(x, t)||∂ 2 ϕ(x, t)| .
(3.11)
In what follows, we only prove Lemma 3.2, because the proof of Lemma 3.1 is of the
same type and even somewhat simpler.
9
Proof. In fact, it follows from (3.10) that
2
X
hkl
ij ∂k ψ∂l φ∂i ∂j ϕ
i,j,k,l=0
2
X
1 kl
=
h (c̄∂k + ωk ∂t )ψ∂l φ∂i ∂j ϕ
c̄ ij
i,j,k,l=0
2
X
−
hkl
ij
i,j,k.l=0
2
X
−
i,j,k,l=0
hkl
ij
i,j,k,l=0
2
X
=
i,j,k,l=0
2
X
ωk ωl ωi ωj
ωk ωl ωi
∂
ψ∂
φ(c̄∂
+
ω
∂
)∂
ϕ
+
hkl
∂t ψ∂t φ∂t2 ϕ
j
j
t
t
k
l
ij
4
c̄
c̄4
i,j,k,l=0
2
X
1
ωk
hkl
(c̄∂
+
ω
∂
)ψ∂
φ∂
∂
ϕ
−
hkl
i j
k
k t
l
ij
ij 2 ∂k ψ(c̄∂l + ωl ∂t )φ∂i ∂j ϕ
c̄
c̄
2
X
+
2
X
ωk
ωk ωl
∂
ψ(c̄∂
+
ω
∂
)φ∂
∂
ϕ
+
hkl
∂k ψ∂l φ(c̄∂i + ωi ∂t )∂j ϕ
t
i
j
k
l
l
ij
2
c̄
c̄3
i,j,k,l=0
i,j,k,l=0
hkl
ij
2
X
ωk ωl
ωk ωl ωi
∂
ψ∂
φ(c̄∂
+
ω
∂
)∂
ϕ
−
hkl
∂k ψ∂l φ(c̄∂j + ωj ∂t )∂t ϕ,
i
i t j
k
l
ij
3
c̄
c̄4
i,j,k,l=0
(3.12)
where
µ
(ω0 , ω1 , ω2 ) =
−c̄,
x1 x2
,
|x| |x|
¶
.
Noting (3.6) and (3.12), we obtain (3.11) immediately. This completes the proof of Lemma
3.2.
¥
2
X
k such that (i)
Lemma 3.3 There exist constants g̃ij
k
g̃ij
∂k v∂i ∂j v satisfies the null
i,j,k=0
condition, that is, for any vector (X0 , X1 , X2 ) satisfying X02 = c̄2 (X12 + X22 ), it holds that
2
X
k
g̃ij
Xi Xj Xk = 0;
(3.13)
i,j,k=0
(ii) the following equality holds


2
2
2
2
X
X
X
X
k
k
k
k


Γ
gij ∂k v∂i ∂j v =
gij ∂k Γv∂i ∂j v +
gij ∂k v∂i ∂j Γv +
g̃ij
∂k v∂i ∂j v.
i,j,k=0
i,j,k=0
i,j,k=0
i,j,k=0
(3.14)
kl such that (I)
Lemma 3.4 There exist constants g̃ij
2
X
kl
g̃ij
∂k v∂l v∂i ∂j v satisfies the
i,j,k,l=0
null condition, that is, for any vector (X0 , X1 , X2 ) satisfying X02 = c̄2 (X12 + X22 ), it holds
that
2
X
kl
g̃ij
Xi Xj Xk Xj = 0;
i,j,k,l=0
10
(3.15)
(II) the following equality holds


2
X
kl
Γ
gij
∂k v∂l v∂i ∂j v  =
i,j,k,l=0
2
X
i,j,k,l=0
kl
gij
∂k v∂l Γv∂i ∂j v
i,j,k,l=0
2
X
+
2
X
kl
gij
∂k Γv∂l v∂i ∂j v +
kl
gij
∂k v∂l ∂i ∂j Γv
2
X
+
i,j,k,l=0
kl
g̃ij
∂k v∂l v∂i ∂j v.
i,j,k,l=0
(3.16)
As before, in what follows we only prove Lemma 3.4, because the proof of Lemma 3.3
is of the same type and even somewhat simpler.
Proof. It follows from (3.3) that


2
2
X
X
kl
kl
∂m 
gij
∂k v∂l v∂i ∂j v  =
gij
∂k ∂m v∂l v∂i ∂j v+
i,j,k,l=0
i,j,k,l=0
2
X
kl
gij
∂k v∂l ∂m v∂i ∂j v +
S

2
X
kl
gij
∂k v∂l v∂i ∂j v 
kl
gij
∂k v∂l v∂i ∂j ∂m v,
=
2
X
kl
gij
∂k Sv∂l v∂i ∂j v
kl
gij
∂k v∂l ∂i ∂j Sv − 4
i,j,k,l=0
2
X
+
kl
gij
∂k v∂l Sv∂i ∂j v+
i,j,k,l=0
i,j,k,l=0
i,j,k,l=0
2
X
(3.17)
i,j,k,l=0
i,j,k,l=0

(m = 0, 1, 2),
2
X
2
X
kl
gij
∂k v∂l v∂i ∂j v
i,j,k,l=0
(3.18)
and

Ω

2
X
kl
gij
∂k v∂l v∂i ∂j v  = 2∂1 Ωv∂2 v∂1 ∂2 v + 2∂1 v∂2 Ωv∂1 ∂2 v + 2∂1 v∂2 v∂1 ∂2 Ωv−
i,j,k,l=0
2∂1 Ωv∂1 v∂22 v − (∂1 v)2 ∂22 Ωv − 2∂2 v∂2 Ωv∂12 v − (∂2 v)2 ∂12 Ωv
=
2
X
i,j,k,l=0
kl
gij
∂k Ωv∂l v∂i ∂j v +
2
X
kl
gij
∂k v∂l Ωv∂i ∂j v +
i,j,k,l=0
2
X
kl
gij
∂k v∂l ∂i ∂j Ωv.
i,j,k,l=0
(3.19)
The desired (3.15)-(3.16) follows from (3.17)-(3.19) immediately. This proves Lemma 3.4.
¥
By Lemmas 3.1-3.4, we can obtain the following lemma.
Lemma 3.5 Let v = v(x, t) be a smooth solution of the Cauchy problem (2.16)-(2.17) on
the domain R2 × [0, T ) and κ be a positive integer, where T is a given positive real number.
11
Then there exists a positive constant Cκ depending on κ but independent of T such that
T
the following inequalities hold for every (x, t) ∈ Λ(T ) {(y, s) | s ≥ 1}
¯
¯
¯ X
¯
X
¯ 2
¯
k
¯
¯ ≤ Cκ
g
∂
v∂
∂
v(x,
t)
|ZΓa v(x, t)||Γb ∂v(x, t)|,
(3.20)
i j
ij k
¯
¯
¯i,j,k=0
¯
|a+b|≤κ+1
κ
¯
¯ 

¯
¯
2
2
X
X
X¯
¯
a
k
k
a
¯Γ 
gij ∂k v∂i ∂j v  (x, t) −
gij ∂k v∂i ∂j Γ v(x, t)¯¯
¯
¯
i,j,k=0
i,j,k=0
|a|≤κ ¯
X
≤ Cκ
(3.21)
|ZΓb v(x, t)||Γc ∂v(x, t)|
|b+c|≤κ+1,|b|,|c|≤κ
and
¯
¯
¯
¯ 2
X
X
¯
¯
kl
¯
¯
≤
C
|ZΓa v(x, t)||Γb ∂v(x, t)||Γc ∂v(x, t)|, (3.22)
g
∂
v∂
v∂
∂
v(x,
t)
κ
i
j
k
l
ij
¯
¯
¯
¯i,j,k,l=0
|a+b+c|≤κ+1
κ
where the terms
2
X
k
gij
∂k v∂i ∂j v
and
i,j,k=0
2
X
kl
gij
∂k v∂l v∂i ∂j v
i,j,k,l=0
are defined by (2.20) and (2.21), respectively. Moreover, it holds that
¯
¯
¯ X
¯
¯ 2
¯
||x| − c̄t|
k
¯
gij ∂k v∂i ∂j v(x, t)¯¯ ≤ Cκ
|∂v(x, t)|[ κ+1 ] |∂v(x, t)|κ+1 +
¯
2
1
+ |x| + t
¯i,j,k=0
¯
κ
³
´
1
|v(x, t)|[ κ+1 ]+1 |∂v(x, t)|κ+1 + |v(x, t)|[ κ+1 ] |v(x, t)|κ+2 ,
2
2
1 + |x| + t
(3.23)
¯
¯
¯ 2
¯
¯ X kl
¯
||x| − c̄t|
¯
gij ∂k v∂l v∂i ∂j v(x, t)¯¯ ≤ Cκ
|∂v(x, t)|2[κ+1 ] |∂v(x, t)|κ+1 +
¯
1
+
|x|
+
t
2
¯i,j,k,l=0
¯
Cκ
κ
Cκ
³
´
1
|∂v(x, t)|[ κ+1 ] |v(x, t)|[ κ+1 ]+1 |∂v(x, t)|κ+1 + |v(x, t)|[ κ+1 ] |v(x, t)|κ+2
2
2
2
1 + |x| + t
(3.24)
and
¯ 
¯

¯
2
2
X ¯¯
X
X
¯
a
k
k
a
¯Γ 
gij ∂k v∂i ∂j v  (x, t) −
gij ∂k v∂i ∂j Γ v(x, t)¯¯
¯
¯
i,j,k=0
i,j,k=0
|a|≤κ ¯
||x| − c̄t|
(3.25)
|∂v(x, t)|[ κ+1 ] |∂v(x, t)|κ +
2
1 + |x| + t
³
´
1
|v(x, t)|[ κ+1 ]+1 |∂v(x, t)|κ + |v(x, t)|[ κ+1 ] |v(x, t)|κ+1 .
Cκ
2
2
1 + |x| + t
≤ Cκ
12
Proof. We only prove (3.20), (3.21), (3.23) and (3.25), because the proof of (3.22) and
(3.24) is similar.
By Lemma 3.3, we obtain


2
X
X
k
Γa 
gij
∂k v∂i ∂j v  =
i,j,k=0
where
2
X
2
X
ka
gijbc
∂k Γb v∂i ∂j Γc v,
(3.26)
b+c≤a i,j,k=0
ka
gijbc
∂k Γb v∂i ∂j Γc v satisfies the null condition, that is, for any vector (X0 , X1 , X2 )
i,j,k=0
satisfying X02 = c̄2 (X12 + X22 ), it holds that
2
X
ka
gijbc
Xi Xj Xk = 0.
i,j,k=0
Noting (3.3), (3.9) and (3.26), we get (3.20) immediately.
By Lemma 3.3 again, we have


2
2
X
X
k
k
gij
∂k v∂i ∂j Γa v =
gij
∂k v∂i ∂j v  −
Γa 
i,j,k=0
i,j,k=0
X
2
X
ka
∂k Γb v∂i ∂j Γc v,
gijbc
b+c≤a,c<a i,j,k=0
(3.27)
where
2
X
ka
gijbc
∂k Γb v∂i ∂j Γc v satisfies the null condition, that is, for any vector (X0 , X1 , X2 )
i,j,k=0
satisfying X02 = c̄2 (X12 + X22 ), it holds that
2
X
ka
Xi Xj Xk = 0.
gijbc
i,j,k=0
Thus, using (3.3), (3.9) and (3.27), we obtain (3.21).
Finally, noting (3.7), (3.20) and using (3.21), we get (3.23) and (3.25) immediately.
Thus, the proof of Lemma 3.5 is completed.
4
¥
L2 -L∞ estimates
The L2 -L∞ estimates play an important role in the proof of Theorem 1.1. This section is
devoted to establishing these estimates.
Theorem 4.1 Let v = v(x, t) be a smooth solution of the Cauchy problem (2.16)-(2.17)
on the domain R2 × [0, T ), where T is a given positive real number, and let κ be a positive
integer, and δ be a positive small constant. Suppose that
[∂v][ κ+6 ],T + hvi[ κ+6 ],T ≤ δ,
2
2
13
(4.1)
then there exist positive constants C0 , C1 and ν such that it holds that
³
´
[∂v(t)]κ +hv(t)iκ+1 ≤ C0 ε+C1 [∂v][ κ+6 ],t + hvi[ κ+6 ],t sup {(1+τ )−ν k∂v(τ )kκ+8 } (4.2)
2
2
0≤τ <t
for every t ∈ [0, T ).
Proof. For t ∈ [0, 1], we have
|∂t v(x, t)|κ ≤ C|v(x, t)|κ+1 ≤
C
|v(x, t)|κ+1 .
1+t
(4.3)
On the other hand, noting the fact that
∂t =
S x·∇
−
t
t
and using (2.18), we obtain
|∂t v(x, t)|κ ≤ C|∇v(x, t)|κ +
C
|v(x, t)|κ+1
1+t
for t ≥ 1.
(4.4)
Thus, it follows from (4.3) and (4.4) that, for every t ∈ [0, T )
|∂t v(x, t)|κ ≤ C|∇v(x, t)|κ +
C
|v(x, t)|κ+1 .
1+t
(4.5)
Thanks to (4.5), in what follows, we only need to prove (4.2) with ∂ replaced by ∂i (i = 1, 2)
in the left-hand side. To do so, we can rewrite the solution v = v(x, t) of the Cauchy
problem (2.16)-(2.17) as
v = v1 + Φ(F ),
where v1 is the solution of the following Cauchy problem


2
2

 ∂t v1 − c̄ 4v1 = 0,
Z x1
1


u10 (y, x2 )dy, ∂t v1 = − ε2 |u0 (x)|2 − f (ρ̄ + ερ0 (x)),
 t = 0 : v1 = ε
2
M
while Φ(F ) is the solution of the following Cauchy problem


 ∂t2 Φ(F ) − c̄2 4Φ(F ) = F,

 t = 0 : Φ(F ) = 0,
in which
F =−
2
X
k
gij
∂k v∂i ∂j v
−
i,j,k=0
(4.6)
(4.7)
∂t Φ(F ) = 0,
2
X
kl
gij
∂k v∂l v∂i ∂j v.
(4.8)
i,j,k,l=0
On the other hand, it follows from (4.6) that (see [5] and [16])
[∂v1 (t)]κ + hv1 (t)iκ+1 ≤ Cε
14
for t ≥ 0.
(4.9)
In order to prove (4.2), it suffices to show
[∇Φ(F )(t)]κ + hΦ(F )(t)iκ+1 ≤
³
´
©
ª
Cε + C1 [∂v][ κ+6 ], t + hvi[ κ+6 ], t sup (1 + τ )−ν k∂v(τ )kκ+8
2
2
(4.10)
0≤τ <t
for t ∈ [0, T ). Here and hereafter, we interpret [∇Φ(F )(t)]κ and hΦ(F )(t)iκ+1 as
[∇Φ(F )(t)]κ = sup
X
1
(1 + |x|) 2 (1 + |x| − c̄t)|Γa ∇Φ(F )(x, t)|
x∈R2 |a|≤κ
and
X
hΦ(F )(t)iκ = sup
1
(1 + |x| + t) 2 |Γa Φ(F )(x, t)|.
x∈R2 |a|≤κ
To prove (4.10), we need the following lemmas.
Lemma 4.1 Suppose that F ∈ C ∞ (R2 × [0, T )). Then for any given positive real number
µ > 0 there exists a positive constant C = C(µ) such that the following inequalities holds
hΦ(F )(x, t)ik ≤ CLµ, κ (F )(x, t)
(4.11)
[∇Φ(F )(x, t)]k ≤ CLµ, κ+1 (F )(x, t),
(4.12)
and
where
Lµ, κ (F )(x, t) =
sup
(y,τ )∈D(x,t)
T
sup T
(y,τ )∈D(x,t)
Λ(t)
n 1
o
1+µ
2
|y| (1 + |y| + τ )
(1 + ||y| − c̄τ |)|F (y, τ )|κ +
Λ(t)c
n 1
o
|y| 2 (1 + |y| + τ )1+µ (1 + |y|)|F (y, τ )|κ ,
(4.13)
in which
©
ª
D(x, t) = (y, τ ) ∈ R2 × [0, t] | |y − x| ≤ c̄(t − τ ) .
(4.14)
and Λ(t)c denotes the complementary set of Λ(t).
Lemma 4.1 can be found in [10] (see also [11]). The following Lemma comes from [15].
Lemma 4.2 Suppose that h ∈ C 2 (R2 × [0, T )) and satisfies kh(t)k2 < ∞. Then there
exits a positive constant C∗ independent of T such that the following estimate holds:
1
|x| 2 |h(x, t)| ≤ C∗ kh(t)k2 ,
15
∀ t ∈ [0, T ).
(4.15)
In what follows, we prove a more general lemma (see Lemma 4.3 below), whose special
case is nothing but (4.10) (more precisely, the case θ = 0 in Lemma 4.3 gives (4.10)).
After the proof of Theorem 4.1, we shall prove Lemma 4.3.
Lemma 4.3 Let v = v(x, t) be the smooth solution of the Cauchy problem (2.16)-(2.17)
on the domain R2 × [0, T )). Suppose that the following estimate holds
[∂v][ κ+6 ]−θ, T + hvi[ κ+6 ]−θ, T ≤ δ,
2
(4.16)
2
where the positive constants κ and δ are defined in Theorem 4.1. Then there exists positive
constant ν such that for any t ∈ [0, T ), it holds that
1
(1 + |x| + t)(− 2 +µ)θ ([∇Φ(F )(x, t)]κ + hΦ(F )(x, t)iκ+1 )
¶
³
´µ
−ν
≤ C [∂v][ κ+6 ]−θ, t + (1 − δ2θ )hvi[ κ+6 ]−θ, t
ε + sup {(1 + τ ) k∂v(τ )kκ+8−2θ } ,
2
2
0≤τ <t
¡
(4.17)
¢
1
where µ is a positive constant satisfying µ ∈ 0, 6 , θ takes its value in {0, 1, 2} and δ2θ is
the Kronecker symbol.
We will prove Lemma 4.3 after the proof of Theorem 4.1.
Proof of Theorem 4.1. Obviously, taking θ = 0 in (4.17) gives the desired (4.10)
immediately. This proves Theorem 4.1.
¥
We now prove Lemma 4.3.
Proof of Lemma 4.3. We distinguish three cases.
Case 1: θ = 2
In this case, it follows from (4.15) and (4.16) that, for every (y, τ ) ∈ D(x, t)
T
Λ(t) it
holds that
1
|y| 2 (1 + |y| + τ )3µ (1 + ||y| − c̄τ |)|F (y, τ )|κ+1
1
≤ C|y| 2 (1 + |y| + τ )3µ (1 + ||y| − c̄τ |)|∂v(y, τ )|[ κ+2 ] |∂v(y, τ )|κ+2
2
1
2
− 21
≤ C|y| (1 + |y| + τ )3µ (1 + |y|)
(1 + |y|) (1 + ||y| − c̄τ |)|∂v(y, τ )|[ κ+2 ] |∂v(y, τ )|κ+2
− 12 +3µ
≤ C[∂v][ κ+2 ], t sup {(1 + τ )
2
0≤τ <t
1
2
2
k∂v(τ )kκ+4 }.
(4.18)
16
Similarly, we obtain from (2.18), (4.15) and (4.16) that
1
|y| 2 (1 + |y| + τ )3µ (1 + |y|)|F (y, τ )|κ+1
1
≤ C|y| 2 (1 + |y| + τ )3µ (1 + |y|)|∂v(y, τ )|[ κ+2 ] |∂v(y, τ )|κ+2
(4.19)
2
1
≤ C[∂v][ κ+2 ] t sup {(1 + τ )− 2 +3µ k∂v(τ )kκ+4 }
2
for (y, τ ) ∈ D(x, t)
T
Λ(t)c .
0≤τ <t
On the other hand, by (4.11) and (4.12) and the fact that
1 + |y| + τ ≤ C(1 + |x| + t),
∀ (y, τ ) ∈ D(x, t),
(4.20)
we have
(1 + |x| + t)−1+2µ {[∇Φ(F )(x, t)]κ + hΦ(F )(x, t)iκ+1 }
(
o
n 1
≤C
sup T
|y| 2 (1 + |y| + τ )3µ (1 + ||y| − c̄τ |)|F (y, τ )|κ+1
(y,τ )∈D(x,t)
+
sup T
(y,τ )∈D(x,t)
Λ(t)
Λ(t)c
)
n 1
o
3µ
|y| 2 (1 + |y| + τ ) (1 + |y|)|F (y, τ )|κ+1
.
(4.21)
¢
Combining (4.18)-(4.19), (4.21) and choosing ν ∈ 0, 12 − 3µ gives (4.17) in the case θ = 2.
¡
Case 2: θ = 1
In the present situation, it follows from (4.8), (4.11), (4.12) and (4.20) that
1
(1 + |x| + t)− 2 +µ ([∇Φ(F )(x, t)]κ + hΦ(F )(x, t)iκ+1 )
(
)
≤C
sup
(y,τ )∈D(x,t)
T
A(y, τ ) +
Λ(t)
sup T
(y,τ )∈D(x,t)
B(y, τ ) ,
(4.22)
Λ(t)c
where
1
1
A(y, τ ) = |y| 2 (1 + |y| + τ ) 2 +2µ (1 + ||y| − c̄τ |)
¯
¯
¯
¯
¯ X
¯
¯ X
¯
2
2
¯
¯
¯
¯
k
kl
¯
¯
¯

× ¯
gij ∂k v∂i ∂j v(y, τ )¯
+¯
gij ∂k v∂l v∂i ∂j v(y, τ )¯¯
¯i,j,k=0
¯
¯i,j,k,l=0
¯
κ+1


κ+1
and
¯
¯ 
¯ 2
¯
¯ X kl
¯
¯
+¯
gij ∂k v∂l v∂i ∂j v(y, τ )¯¯  .
¯i,j,k,l=0
¯
κ+1
κ+1
T
When τ ∈ [0, 1], it follows from (4.15) and (4.16) that, for every (y, τ ) ∈ D(x, t) Λ(t)
¯ 
¯
¯
¯
¯ 2
¯
¯ 2
¯
X
X
¯
¯
¯
¯
1
1
+2µ
k
kl
¯
¯
¯

(1 + |y| + τ ) 2
+¯
(1 + ||y| − c̄τ |)|y| 2 ¯
gij ∂k v∂i ∂j v(y, τ )¯
gij ∂k v∂l v∂i ∂j v(y, τ )¯¯ 
¯i,j,k=0
¯
¯i,j,k,l=0
¯
¯
¯
¯ 2
¯
X
¯
¯
1
1
+2µ
k
¯

2
2
B(y, τ ) = |y| (1+|y|+τ )
(1+|y|) ¯
gij ∂k v∂i ∂j v(y, τ )¯¯
¯i,j,k=0
¯
κ+1
≤ C[∂v][ κ+2 ],t k∂v(τ )kκ+4 ≤ C[∂v][ κ+2 ],t (1 + τ )−ν k∂v(τ )kκ+4 ,
2
2
κ+1
∀ ν ∈ R.
(4.23)
17
Similarly, when τ ∈ [0, 1], it follows from (2.18), (4.15) and (4.16) that, for every (y, τ ) ∈
T
D(x, t) Λ(t)c
¯
¯
¯
¯
¯ X
¯
¯ X
¯
2
2
¯
¯
¯
¯
1
1
k
kl
(1 + |y|)(1 + |y| + τ ) 2 +2µ |y| 2 ¯¯
gij
∂k v∂i ∂j v(y, τ )¯¯
+ ¯¯
gij
∂k v∂l v∂i ∂j v(y, τ )¯¯
¯i,j,k=0
¯
¯i,j,k,l=0
¯
κ+1
≤ C[∂v][ κ+2 ],t (1 + τ )−ν k∂v(τ )kκ+4 ,
2


κ+1
∀ ν ∈ R.
(4.24)
Thus, in what follows, it suffices to investigate the case τ ≥ 1.
By (3.23), (4.5), (4.9) and (4.17) with θ = 2, for (y, τ ) ∈ D(x, t)
T
T
Λ(t) {(z, s)|s ≥ 1}
we have
¯
¯
¯ X
¯
¯
¯ 2
k
(1 + |y| + τ )
gij ∂k v∂i ∂j v(y, τ )¯¯
|y| (1 + ||y| − c̄τ |) ¯¯
¯
¯i,j,k=0
κ+1
n
1
1
≤ C|y| 2 (1 + |y| + τ )− 2 +2µ (1 + ||y| − c̄τ |)2 |∂v(y, τ )|[ κ+2 ] |∂v(y, τ )|κ+2 +
2
o
(1 + ||y| − c̄τ |)(|∂v(y, τ )|[ κ+2 ] |v(y, τ )|κ+3 + |v(y, τ )|[ κ+4 ] |∂v(y, τ )|κ+2 )
2
2
n
³
´ 1
1
1
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t |y| 2 (1 + |y| + τ )− 2 +2µ (1 + |y|)− 2 |v(y, τ )|κ+3 +
1
+2µ
2
1
2
2
2
− 12
1
(1 + |y|) (1 + ||y| − c̄τ |)|∂v(y, τ )|κ+2 + (1 + ||y| − c̄τ |)(1 + |y| + τ )− 2 |∂v(y, τ )|κ+2
³
´ 1
n
1
1
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t |y| 2 (1 + |y| + τ )− 2 +2µ (1 + |y|)− 2 (1 + ||y| − c̄τ |)×
2
o
2
1
(|∇v(y, τ )|κ+2 + (1 + τ )−1 |v(y, τ )|κ+3 ) + (1 + |y|)− 2 |v(y, τ )|κ+3 +
o
1
(1 + ||y| − c̄τ |)(1 + |y| + τ )− 2 (|∇v(y, τ )|κ+2 + (1 + τ )−1 |v(y, τ )|κ+3 )
³
´n
1
1
1
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t ε + |y| 2 (1 + |y| + τ )− 2 +2µ (1 + |y|)− 2 (1 + ||y| − c̄τ |)×
2
2
o
1
1
1
|∇Φ(F )(y, τ )|κ+2 + |y| 2 (1 + |y| + τ )− 2 +2µ (1 + |y|)− 2 |Φ(F )(y, τ )|κ+3
´©
³
ª
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t ε + (1 + |y| + τ )−1+2µ ([∇Φ(F )(y, τ )]κ+2 + hΦ(F )(y, τ )iκ+3 )
2
2
¶
³
´µ
−ν
ε + [∂v][ κ+4 ], t sup {(1 + τ ) k∂v(τ )kκ+6 } .
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t
2
2
2
0≤τ <t
(4.25)
18
Making use of (2.18), (4.5), (4.9) and (4.17) with θ = 2 gives
¯
¯
¯ 2
¯
X
¯
¯
1
1
+2µ
k
¯
2
2
(1 + |y| + τ )
|y| (1 + |y|) ¯
gij ∂k v∂i ∂j v(y, τ )¯¯
¯i,j,k=0
¯
κ+1
1
2
≤ C|y| (1 + |y| + τ )
1
+2µ
2
(1 + |y|)|∂v(y, τ )|[ κ+2 ] |∂v(y, τ )|κ+2
2
1
2
1
+2µ
2
1
1
≤ C[∂v][ κ+2 ], t |y| (1 + |y| + τ )
2
1
2
(1 + |y|) (1 + ||y| − c̄τ |)−1 |∂v(y, τ )|κ+2
1
≤ C[∂v][ κ+2 ], t |y| 2 (1 + |y| + τ ) 2 +2µ (1 + |y|) 2 (1 + ||y| − c̄τ |)−1 ×
2
(|∇v(y, τ )|κ+2 + (1 + τ )−1 |v(y, τ )|κ+3 )
n
1
1
≤ C[∂v][ κ+2 ], t ε + |y| 2 (1 + |y| + τ ) 2 +2µ (1 + ||y| − c̄τ |)−2 [∇Φ(F )(y, τ )]κ+2 +
2
o
1
1
|y| 2 (1 + |y| + τ )2µ (1 + |y|)− 2 (1 + ||y| − c̄τ |)−1 hΦ(F )(y, τ )iκ+3
©
ª
≤ C[∂v][ κ+2 ], t ε + (1 + |y| + τ )−1+2µ ([∇Φ(F )(y, τ )]κ+2 + hΦ(F )(y, τ )iκ+3 )
2
µ
¶
−ν
≤ C[∂v][ κ+2 ], t ε + [∂v][ κ+4 ], t sup {(1 + τ ) k∂v(τ )kκ+6 }
2
T
2
0≤τ <t
(4.26)
T
Λ(t)c {(z, s)|s ≥ 1}. On the other hand, by (4.15), for (y, τ ) ∈
for (y, τ ) ∈ D(x, t)
T
T
D(x, t) Λ(t) {(z, s)|s ≥ 1} we have
(1 + |y| + τ )
1
+2µ
2
¯
¯
¯ 2
¯
¯ X kl
¯
|y| (1 + ||y| − c̄τ |) ¯¯
gij ∂k v∂l v∂i ∂j v(y, τ )¯¯
¯i,j,k,l=0
¯
1
2
1
2
≤ C[∂v]2[κ+2 ],t |y| (1 + |y| + τ )
1
+2µ
2
2
κ+1
(1 + ||y| − c̄τ |)−1 (1 + |y|)−1 |∂v(y, τ )|κ+2
≤ C[∂v]2[κ+2 ],t sup {(1 + τ )−ν k∂v(τ )kκ+4 }.
0≤τ <t
2
(4.27)
T
T
It follows from (2.18) and (4.15) that, for (y, τ ) ∈ D(x, t) Λ(t)c {(z, s)|s ≥ 1}
¯
¯
¯ 2
¯
X
¯
¯
1
1
+2µ
kl
¯
|y| 2 (1 + |y|) ¯
gij ∂k v∂l v∂i ∂j v(y, τ )¯¯
(1 + |y| + τ ) 2
¯i,j,k,l=0
¯
κ+1
1
2
≤ C[∂v]2[κ+2 ], t |y| (1 + |y| + τ )
1
+2µ
2
2
≤
C[∂v]2[κ+2 ], t
2
(1 + ||y| − c̄τ |)−2 |∂v(y, τ )|κ+2
sup {(1 + τ )−ν k∂v(τ )kκ+4 }.
0≤τ <t
(4.22)-(4.28) yields (4.17) in the case θ = 1.
Case 3: θ = 0
19
(4.28)
T
Λ(t) {(z, s)|s ≤ 1}, we obtain from (4.15) and (4.16) that
¯
¯
¯
¯
¯ 2
¯
¯ 2
¯
X
X
¯
¯
¯
¯
1
1+µ
k
kl
¯
¯
¯

2
(1 + |y| + τ )
|y| (1 + ||y| − c̄τ |) ¯
gij ∂k v∂i ∂j v(y, τ )¯
+¯
gij ∂k v∂l v∂i ∂j v(y, τ )¯¯
¯i,j,k=0
¯
¯i,j,k,l=0
¯
For (y, τ ) ∈ D(x, t)
T
κ+1
1
2
− 12
≤ C[∂v][ κ+2 ], t |y| (1 + |y| + τ )1+µ (1 + |y|)
2
|∂v(y, τ )|κ+2
≤ C[∂v][ κ+2 ], t sup {(1 + τ )−ν k∂v(τ )kκ+4 }.
2
0≤τ <t
Similarly, it follows from (2.18), (4.15) and (4.16) that, for (y, τ ) ∈ D(x, t)
T
(4.29)
T
Λ(t)c {(z, s)|s ≤
1}
¯
¯
¯
¯ 2
X
¯
¯
1
1+µ
k
¯
(1 + |y| + τ )
|y| 2 (1 + |y|) ¯
gij ∂k v∂i ∂j v(y, τ )¯¯
¯i,j,k=0
¯
1
κ+1
¯
¯
¯
¯ 2
X
¯
¯
kl
¯
+¯
gij ∂k v∂l v∂i ∂j v(y, τ )¯¯
¯
¯i,j,k,l=0
1
≤ C[∂v][ κ+2 ], t |y| 2 (1 + |y| + τ )1+µ (1 + |y|) 2 (1 + ||y| − c̄τ |)−1 |∂v(y, τ )|κ+2
2
≤ C[∂v][ κ+2 ], t sup {(1 + τ )−ν k∂v(τ )kκ+4 }.
2
0≤τ <t
(4.30)
We next consider the case τ ≥ 1.
20


κ+1


κ+1
Noting (3.23)-(3.24), (4.5), (4.9), (4.16) and using (4.17) in the case θ = 1, for (y, τ ) ∈
T
T
D(x, t) Λ(t) {(z, s)|s ≥ 1} we have
¯
¯
¯ 
¯
¯ X
¯
¯ X
¯
2
2
¯
¯
¯
¯
1
k
kl
(1 + |y| + τ )1+µ (1 + ||y| − c̄τ |)|y| 2 ¯¯
gij
∂k v∂i ∂j v(y, τ )¯¯
+ ¯¯
gij
∂k v∂l v∂i ∂j v(y, τ )¯¯ 
¯i,j,k=0
¯
¯i,j,k,l=0
¯
κ+1
κ+1
n
1
≤ C|y| 2 (1 + |y| + τ )µ (1 + ||y| − c̄τ |)2 |∂v(y, τ )|[ κ+2 ] |∂v(y, τ )|κ+2 +
2
o
(1 + ||y| − c̄τ |)(|∂v(y, τ )|[ κ+2 ] |v(y, τ )|κ+3 + |v(y, τ )|[ κ+4 ] |∂v(y, τ )|κ+2 )
2
2
³
´ 1
n
1
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t |y| 2 (1 + |y| + τ )µ (1 + |y|)− 2 (1 + ||y| − c̄τ |)|∂v(y, τ )|κ+2 +
2
2
o
1
1
(1 + |y|)− 2 |v(y, τ )|κ+3 + (1 + ||y| − c̄τ |)(1 + |y| + τ )− 2 |∂v(y, τ )|κ+2
³
´ 1
n
1
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t |y| 2 (1 + |y| + τ )µ (1 + |y|)− 2 (1 + ||y| − c̄τ |)×
2
¡
2
¢
1
|∇v(y, τ )|κ+2 + (1 + τ )−1 |v(y, τ )|κ+3 + (1 + |y|)− 2 |v(y, τ )|κ+3 +
o
1
(1 + ||y| − c̄τ |)(1 + |y| + τ )− 2 (|∇v(y, τ )|κ+2 + (1 + τ )−1 |v(y, τ )|κ+3 )
´n
³
1
1
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t ε + |y| 2 (1 + |y| + τ )µ (1 + |y|)− 2 (1 + ||y| − c̄τ |)×
2
2
o
1
1
|∇Φ(F )(y, τ )|κ+2 + |y| 2 (1 + |y| + τ )µ (1 + |y|)− 2 |Φ(F )(y, τ )|κ+3
´n
³
o
1
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t ε + (1 + |y| + τ )− 2 +µ ([∇Φ(F )(y, τ )]κ+2 + hΦ(F )(y, τ )iκ+3 )
2
2
¾
´½
³
´
³
−ν
ε + [∂v][ κ+6 ], t + hvi[ κ+6 ], t sup {(1 + τ ) k∂v(τ )kκ+8 } .
≤ C [∂v][ κ+2 ], t + hvi[ κ+4 ], t
2
2
2
2
0≤τ <t
(4.31)
21
Using (2.18), (4.5), (4.9), (4.16) and noting (4.17) with θ = 1, for (y, τ ) ∈ D(x, t)
T
Λ(t)c
T
{(z, s)|s ≥
1} we obtain
¯
¯
¯ 2
¯
X
¯
¯
1
k
1+µ
¯

gij ∂k v∂i ∂j v(y, τ )¯¯
(1 + |y| + τ )
(1 + |y|)|y| 2 ¯
¯i,j,k=0
¯
1
κ+1
¯
¯
¯ 2
¯
X
¯
¯
kl
¯
+¯
gij ∂k v∂l v∂i ∂j v(y, τ )¯¯
¯i,j,k,l=0
¯


κ+1
1
≤ C[∂v][ κ+2 ], t |y| 2 (1 + |y| + τ )1+µ (1 + |y|) 2 (1 + ||y| − c̄τ |)−1 |∂v(y, τ )|κ+2
2
¡
¢
1
1
≤ C[∂v][ κ+2 ], t |y| 2 (1 + |y| + τ )1+µ (1 + |y|) 2 (1 + ||y| − c̄τ |)−1 |∇v(y, τ )|κ+2 + (1 + τ )−1 |v(y, τ )|κ+3
2
n
1
≤ C[∂v][ κ+2 ], t ε + |y| 2 (1 + |y| + τ )1+µ (1 + ||y| − c̄τ |)−2 [∇Φ(F )(y, τ )]κ+2 +
2
o
1
1
1
|y| 2 (1 + |y| + τ ) 2 +µ (1 + |y|)− 2 (1 + ||y| − c̄τ |)−1 hΦ(F )(y, τ )iκ+3
n
o
1
≤ C[∂v][ κ+2 ], t ε + (1 + |y| + τ )− 2 +µ ([∇Φ(F )(y, τ )]κ+2 + hΦ(F )(y, τ )iκ+3 )
2
½
¶¾
³
´µ
≤ C[∂v][ κ+2 ], t ε + [∂v][ κ+6 ], t + hvi[ κ+6 ], t
ε + sup {(1 + τ )−ν k∂v(τ )kκ+8 }
2
2
2
0≤τ <t
´µ
³
≤ C [∂v][ κ+6 ], t + hvi[ κ+6 ], t
2
−ν
ε + sup {(1 + τ )
2
0≤τ <t
¶
k∂v(τ )kκ+8 } .
(4.32)
Therefore, combining (4.11)-(4.12) and (4.29)-(4.32) gives (4.17) in the case θ = 0 immediately. Thus, the proof of Lemma 4.3 is completed.
5
¥
Energy estimate
This section is devoted to establishing the energy estimates, which play an important role
in the proof of Theorem 1.1.
Theorem 5.1 Let v = v(x, t) be a smooth solution of the Cauchy problem (2.16)-(2.17)
on the domain R2 × [0, T )). If there exists a positive small constant δ such that
[∂v][ κ+1 ],T + hvi[ κ+1 ]+1,T ≤ δ,
2
(5.1)
2
then, for every t ∈ [0, T ), it holds that
«
„
C3 [∂v][ κ+1 ],T +hvi[ κ+1 ]+1, T
k∂v(t)kκ ≤ C2 ε(1 + t)
2
2
,
(5.2)
where C2 and C3 are two positive constants independent of T .
Proof. Define
Z
λ
q(λ) =
−∞
1
d%
(1 + |%|)2
22
(5.3)
and
p(x, t) = q(||x| − c̄t|),
ṗ(x, t) = q 0 (|x| − c̄t).
Obviously,
q ∈ C 1 (R),
p ∈ C 1 (R2 × R+ )
and ṗ ∈ C 0 (R2 × R+ ).
Moreover,
0 ≤ p(x, t) ≤ 2
(5.4)
and
ṗ(x, t) =
1
(1 + ||x| − c̄t|)2
(5.5)
for any (x, t) ∈ R2 × (0, +∞).
On the other hand, we introduce the following energy functions
½Z
E0 (v(t)) =
¡
R2
2
2
2
|∂t v(x, t)| + c̄ |∇v(x, t)|
X
Eκ (v(t)) =
¢
¾1
2
dx
,
(5.6)
E0 (Γa v(t)),
(5.7)
|a|≤κ
½Z
Ẽ0 (v(t)) =
p
R2
¡
2
2
2
e (x, t) |∂t v(x, t)| + c̄ |∇v(x, t)|
and
X
Ẽκ (v(t)) =
¢
¾1
dx
2
Ẽ0 (Γa v(t)).
(5.8)
(5.9)
|a|≤κ
Thus, it follows from (5.4), (5.6)-(5.9) that
1
Ĉ
k∂v(t)kκ ≤
1
Eκ (v(t)) ≤ Ẽκ (v(t)) ≤ C̃Eκ (v(t)) ≤ Ĉk∂v(t)kκ
C̃
(5.10)
for some positive constant Ĉ and C̃ which are independent of t. Therefore, in order to
prove (5.2), it suffices to show
«
„
C [∂v][ κ+1 ],T +hvi[ κ+1 ]+1,T
Ẽκ (v(t)) ≤ Cε(1 + t)
2
for t ∈ [0, T ).
23
2
(5.11)
To do so, noting (2.16), we have
∂t2 Γa v
2
a
− c̄ 4Γ v + 2
2
X
a
a
∂i v∂t ∂i Γ v − 2∂t v4Γ v +
i=1

=
X
Cb Γb −2
2
X
∂i v∂t ∂i v + 2∂t v4v −
i=1
2
X
∂i v∂j v∂i ∂j Γa v − |∇v|2 4Γa v
i,j=1
b≤a
2
2
X
2
X

∂i v∂j v∂i ∂j v + |∇v|2 4v  +
i, j=1
∂i v∂t ∂i Γa v − 2∂t v4Γa v +
i=1
2
X
∂i v∂j v∂i ∂j Γa v − |∇v|2 4Γa v
i, j=1
(5.12)
for any given multi-index a, where Cb is a constant which is determined by (3.3). Notice
that Ca = 1. Multiplying ep ∂t Γa v to (5.12) and then integrating with respect to x on R2
leads to
1d
2 dt
Z
©
R2
¡
¢ª
ep |∂t Γa v(x, t)|2 + c̄2 |∇Γa v(x, t)|2 dx+
Ã
Z
ep
R2
2
X
2
X
c̄
ṗ|ZΓa v|2 + 2
∂i v∂t ∂i Γa v∂t Γa v − 2∂t v4Γa v∂t Γa v+
2
i=1

(5.13)
∂i v∂j v∂i ∂j Γa v∂t Γa v − |∇v|2 4Γa v∂t Γa v  dx = I1 (t),
i, j=1
where
Z
I1 (t) =
R2
2
ep ∂t Γa v
2
X
i=1

X

b≤a

Cb Γb −2
2
X
∂i v∂t ∂i v + 2∂t v4v −
i=1
∂i v∂t ∂i Γa v − 2∂t v4Γa v +
2
X

∂i v∂j v∂i ∂j v + |∇v|2 4v  +
i, j=1
2
X
i, j=1
∂i v∂j v∂i ∂j Γa v − |∇v|2 4Γa v



dx.
(5.14)
24
Notice that


2
2
X
X
ep 2
∂i v∂t ∂i Γa v∂t Γa v − 2∂t v4Γa v∂t Γa v +
∂i v∂j v∂i ∂j Γa v∂t Γa v − |∇v|2 4Γa v∂t Γa v 
i=1
=
i, j=1
2
X
∂i (ep ∂i v∂t Γa v∂t Γa v) − 2
2
X
i=1
∂i (ep ∂t v∂i Γa v∂t Γa v) +
i=1
2 X
2
X
i=1
ṗ p
e
2
Ã
2
2
X
a
1
2
a
i=1 j=1
∂t (ep |∇v|2 ∂i Γa v∂i Γa v)−
i=1
2
X
1 p
e
2
2
X
2
2
X
2
X
−4
2
X
∂i v∂j v∂j Γa v∂i Γa v−
!
a
∂i ∂t v∂i Γ v∂t Γ v + 2
i=1
2 X
2
X
i=1
∂t v∂i Γa v∂i Γa v+
|∇v|2 ∂i Γa v∂i Γa v −
a
∂i (∂i v∂j v)∂j Γa v∂t Γa v − 2
2 X
2
X
i=1 j=1
∂i (|∇v|2 )∂i Γa v∂t Γa v +
2
X
∂t2 v∂i Γa v∂i Γa v+
i=1
i=1 j=1
2
2
X
i=1
∂i2 v∂t Γa v∂t Γa v
2
X
2 X
2
X
i=1 j=1
ωi |∇v|2 ∂i Γa v∂t Γa v − c̄
i=1
2
a
i=1
ωi ∂i v∂j v∂j Γa v∂t Γa v + c̄
i=1
Ã
a
ωi ∂t v∂i Γ v∂t Γ v − 2c̄
i=1 j=1
2
∂t (ep ∂i v∂j v∂j Γa v∂i Γa v)−
i=1
2 X
2
X
2
2
X
ωi ∂i v∂t Γ v∂t Γ v − 4
i=1
2 X
2
X
1
2
∂i (ep ∂i v∂j v∂j Γa v∂t Γa v) −
∂i (ep |∇v|2 ∂i Γa v∂t Γa v) +
∂t (ep ∂i v∂i Γa v∂i Γa v)+
i=1
i=1 j=1
2
X
2
X
2
X
i=1
∂t (∂i v∂j v)∂j Γa v∂i Γa v−
!
∂t (|∇v|2 )∂i Γa v∂i Γa v ,
(5.15)
xi
where ωi =
(i = 1, 2). It follows from (5.13) and (5.15) that
|x|
Z
Z
© p¡
¢ª
1d
c̄ p
a
2
2
a
2
I1 (t) =
e |∂t Γ v(x, t)| + c̄ |∇Γ v(x, t)|
dx +
e ṗ|ZΓa v|2 dx+
2 dt R2
2
2
R

Z
2
2
XX
d
ep ∂i v∂i Γa v∂i Γa v − 1
ep ∂i v∂j v∂j Γa v∂i Γa v
dt R2
2
i=1 j=1
!
1X p
+
e |∇v|2 ∂i Γa v∂i Γa v dx + I2 (t) + I3 (t),
2
i=1
(5.16)
25
where
Z
I2 (t) = −
R2
ṗ p
e
2
Ã
2
2
X
ωi ∂i v∂t Γa v∂t Γa v − 4
i=1
ωi ∂t v∂i Γa v∂t Γa v − 2c̄
2
X
2 X
2
X
ωi ∂i v∂j v∂j Γa v∂t Γa v + c̄
∂i v∂j v∂j Γa v∂i Γa v−
i=1 j=1
2
a
a
ωi |∇v| ∂i Γ v∂t Γ v − c̄
i=1
∂t v∂i Γa v∂i Γa v+
i=1
i=1 j=1
2
2
X
i=1
2 X
2
X
2
2
X
2
X
2
!
a
a
|∇v| ∂i Γ v∂i Γ v dx
i=1
(5.17)
and
Z
I3 (t) = −
R2
1 p
e
2
Ã
2
2
X
∂i2 v∂t Γa v∂t Γa v − 4
i=1
2
2
X
∂i ∂t v∂i Γa v∂t Γa v + 2
i=1
2 X
2
X
∂i (∂i v∂j v)∂j Γa v∂t Γa v − 2
2
2 X
2
X
∂t (∂i v∂j v)∂j Γa v∂i Γa v−
i=1 j=1
2
a
a
∂i (|∇v| )∂i Γ v∂t Γ v +
i=1
∂t2 v∂i Γa v∂i Γa v+
i=1
i=1 j=1
2
X
2
X
2
X
2
!
a
a
∂t (|∇v| )∂i Γ v∂i Γ v dx.
i=1
(5.18)
Combining (5.1) and (5.16) yields
Z t
X Z tZ
2
p
a
2
2
Ẽκ (v(t)) +
e ṗ|ZΓ v(x, τ )| dxdτ ≤ C Ẽκ (v(0)) + C
|I(τ )|dτ
|a|≤κ 0
R2
(5.19)
0
for t ∈ [0, T ), where
I(t) =
X
(|I1 (t)| + |I2 (t)| + |I3 (t)|).
(5.20)
|a|≤κ
Now we claim that
C|I(t)| ≤
Z
1 X
ṗep |ZΓa v(x, t)|2 dx+
2
R2
|a|≤κ
´
C ³
[∂v][ κ+1 ],T + hvi[ κ+1 ]+1,T
2
2
1+t
(5.21)
Z
p
e
R2
|∂v(x, t)|2κ dx
for t ∈ [0, T ).
For the time being, we assume that (5.21) is true. We will prove it later. Thus, it
follows from (5.19) and (5.21) that
Ẽκ (v(t))2 +
Z Z
1 X t
ep ṗ|ZΓa v(x, τ )|2 dxdτ
2
2
0
R
|a|≤κ
≤ C Ẽκ (v(0))2 + C
Z
0
t
´
1 ³
[∂v][ κ+1 ],T + hvi[ κ+1 ]+1,T Ẽκ (v(τ ))2 dτ.
2
2
1+τ
26
(5.22)
Noting (2.17), (5.22) and using Gronwall inequality gives (5.11) immediately.
It suffices to prove (5.21) in the rest of this section. To prove (5.21), we distinguish
two cases.
Case I: t ∈ [0, 1]
In this case, (5.21) comes from (5.14), (5.17) and (5.18) immediately.
Case II: t ∈ [1, T )
We first estimate |I1 (t)|.
Noting (2.18), (2.20), (2.21), (3.7), (3.20), (3.21), (5.1), (5.14) and using Cauchy inequality, we have
¯
¯ 

¯
¯
2
2
X
X
¯
¯ a
k
k
a ¯
¯


gij ∂k v∂i ∂j v −
gij ∂k v∂i ∂j Γ v ¯ |∂t Γa v|ep dx+
|I1 (t)| ≤
¯Γ
(x,t)∈Λ(T ) ¯
¯
i,j,k=0
i,j,k=0
¯

¯
¯
¯
2
XZ
X
¯
¯
b
k
¯Cb Γ 
gij ∂k v∂i ∂j v ¯¯ |∂t Γa v|ep dx+
¯
¯
b<a (x,t)∈Λ(T ) ¯
i,j,k=0
¯
¯


¯
¯
2
2
X
X
XZ
¯
¯
k
a ¯
k
¯Cb Γb 

gij ∂k v∂i ∂j Γ v ¯ |∂t Γa v|ep dx+
gij ∂k v∂i ∂j v −
¯
c
¯
i,j,k=0
i,j,k=0
b≤a (x,t)∈Λ(T ) ¯
¯
¯


¯
¯
2
2
XZ ¯
X
X
¯
kl
kl
a ¯
a
p
¯Cb Γb 

g
∂
v∂
v∂
∂
v
−
g
∂
v∂
v∂
∂
Γ
v
i
j
i
j
k
l
k
l
ij
ij
¯
¯ |∂t Γ v|e dx
2 ¯
R
¯
b≤a
i,j,k,l=0
i,j,k,l=0
µ
¶
Z
||x| − c̄t|
1
|∂v|[ κ+1 ] +
|v|[ κ+1 ]+1 ep |∂v|2κ dx+
≤ C
2
2
1 + |x| + t
(x,t)∈Λ(T ) 1 + |x| + t
Z
Z
X
C
ep |∂v|[ κ+1 ] |ZΓb v||∂v|κ dx + C
ep |∂v|[ κ+1 ] |∂v|2κ dx+
Z
2
|b|≤κ (x,t)∈Λ(T )
(x,t)∈Λ(T )c
2
Z
C
R2
ep |∂v|2[κ+1 ] |∂v|2κ dx
2
Z
≤ C
(
(x,t)∈Λ(T )
C
Z
XZ
1
||x| − c̄t|
|∂v|[ κ+1 ] +
|v| κ+1 )ep |∂v|2κ dx+
2
1 + |x| + t
1 + |x| + t [ 2 ]+1
|b|≤κ (x,t)∈Λ(T )
ep |∂v|[ κ+1 ] |ZΓb v||∂v|κ dx+
2
1
ep (
[∂v][ κ+1 ],T + |∂v|2[κ+1 ] )|∂v|2κ dx
2
1
+
t
2
2
R
´Z
p
XZ
e
C² ³
b 2
ep |∂v(x, t)|2κ dx
≤ ²
|ZΓ v| dx +
[∂v][ κ+1 ],T + hvi[ κ+1 ]+1,T
2
2
2
1+t
2 (1 + ||x| − c̄t|)
2
R
R
|b|≤κ
´Z
XZ
C² ³
p
b 2
ṗe |ZΓ v| dx +
≤ ²
[∂v][ κ+1 ],T + hvi[ κ+1 ]+1,T
ep |∂v(x, t)|2κ dx,
2
2
1+t
R2
R2
|b|≤κ
(5.23)
27
where ² is a positive small constant.
We now estimate |I2 (t)|.
Noting (2.18), (3.7), (5.1), (5.17) and using the null condition and Cauchy inequality,
we obtain
¯
2
2
¯X
X
|I2 (t)| ≤
ṗe ¯
ωi ∂i v∂t Γa v∂t Γa v − 2
ωi ∂t v∂i Γa v∂t Γa v −
¯
(x,t)∈Λ(T )
i=1
i=1
¯
¯
Z
2
2
¯
¯X
X
¯
¯
ωi ∂i v∂t Γa v∂t Γa v−
ṗep ¯
c̄
∂t v∂i Γa v∂i Γa v ¯ dx +
¯
¯
c
(x,t)∈Λ(T )
i=1
i=1
¯
2
2
¯
X
X
¯
2
ωi ∂t v∂i Γa v∂t Γa v − c̄
∂t v∂i Γa v∂i Γa v ¯ dx+
¯
i=1
i=1
¯
¯ 2 2
Z
2 X
2
X
ṗ p ¯¯ X X
a
a
e ¯2
ωi ∂i v∂j v∂j Γ v∂t Γ v + c̄
∂i v∂j v∂j Γa v∂i Γa v−
R2 2
¯ i=1 j=1
i=1 j=1
¯
2
2
¯
X
X
2
a
a
2
a
a ¯
2
ωi |∇v| ∂i Γ v∂t Γ v − c̄
|∇v| ∂i Γ v∂i Γ v ¯ dx
¯
i=1
i=1
¯
Z
2
2
¯X
X
¯
≤
ṗep ¯
ωi ∂i v∂t Γa v∂t Γa v − 2
ωi ∂t v∂i Γa v∂t Γa v−
¯
(x,t)∈Λ(T )
i=1
i=1
¯
ÃZ
!
Z
2
¯
X
a
a ¯
2
2
2
c̄
∂t v∂i Γ v∂i Γ v ¯ dx + C
|∂v||∂v|κ dx +
|∂v| |∂v|κ dx
¯
(x,t)∈Λ(T )c
R2
i=1
Z
¯ω
ω2
¯ 1
≤
ṗep ¯ (c̄∂1 + ω1 ∂t )v(∂t Γa v)2 + (c̄∂2 + ω2 ∂t )v(∂t Γa v)2 −
c̄
c̄
(x,t)∈Λ(T )
Z
p¯
1
1
∂t v(c̄∂1 + ω1 ∂t )Γa v(c̄∂1 + ω1 ∂t )Γa v − ∂t v(c̄∂2 + ω2 ∂t )Γa v(c̄∂2 + ω2 ∂t )Γa v−
c̄
c̄
¯
ω
ω1
¯
2
∂t v(c̄∂1 + ω1 ∂t )Γa v∂t Γa v − ∂t v(c̄∂2 + ω2 ∂t )Γa v∂t Γa v ¯ dx+
c̄
c̄
Z
C
[∂v]0,T
ep |∂v|2κ dx
1+t
R2
Z
Z
¡
¢
C
ṗep |Zv||∂Γa v|2 + |∂v||ZΓa v||∂Γa v| dx +
ep |∂v(x, t)|2κ dx
≤ C
[∂v]0,T
1
+
t
2
(x,t)∈Λ(T )
R
Z
XZ
C²
ṗep |ZΓa v(x, t)|2 dx +
≤ ²
([∂v]0,T + hvi1,T )
ep |∂v(x, t)|2κ dx.
1
+
t
2
2
R
R
|a|≤κ
(5.24)
where ωi =
xi
|x| (i
= 1, 2) and ² is a positive small constant.
Similarly, using (2.18), (3.7), (5.1), (5.18), the null condition and Cauchy inequality,
we have
|I3 (t)| ≤ ²
XZ
|a|≤κ
C²
ṗe |ZΓ v(x, t)| dx +
[∂v]1,T
1+t
R2
p
a
Z
2
28
R2
ep |∂v(x, t)|2κ dx.
(5.25)
Choosing ² suitably small, by (5.23)-(5.25) we see that (5.21) holds for t ∈ [1, T ). Thus
the proof of Theorem 5.1 is completed.
6
¥
Proof of Theorem 1.1
This section is devoted to the proof of Theorem 1.1. In order to prove Theorem 1.1, it
suffices to show the following theorem.
Theorem 6.1 Under the assumptions of Theorem 1.1, there exists a positive constant ε0
such that, for every ε ∈ [0, ε0 ], the smooth solution v = v(x, t) of the Cauchy problem
(2.16)-(2.17) exists globally in time.
Proof. The local existence of classical solutions has been proved by the method of Picard
iteration (see Sogge [19] and Hörmander [9]). It suffices to show that if v = v(x, t) is a
smooth solution of the Cauchy problem (2.16)-(2.17) on the domain R2 × [0, T )), then
X
|∂ α v(x, t)| ∈ L∞ (R2 × [0, T )).
|α|≤4
In what follows, we prove Theorem 6.1 by the method of continuous induction, or say,
by the bootstrap argument.
Taking an integer κ ≥ 8, we have
·
¸
κ+9
≤ κ.
2
According to the bootstrap argument, we first assume that
[∂v]κ, T + hviκ+1, T ≤ M ε,
(6.1)
and then we prove that, by choosing M sufficiently large and ε suitably small we have
1
[∂v]κ, T + hviκ+1, T ≤ M ε.
2
(6.2)
δ
To do so, let δ be a positive and small constant. Choosing ε1 ∈ [0, M
], we have
[∂v][ κ+6 ], T + hvi[ κ+6 ], T
2
2
≤ [∂v][ κ+9 ], T + hvi[ κ+9 ]+1, T
2
2
≤ [∂v]κ, T + hviκ+1, T ≤ M ε ≤ δ
29
(6.3)
δ
for ε ∈ [0, M
]. Thus, it follows from Theorems 4.1 and 5.1 that
[∂v]κ, T + hviκ+1, T
³
´
≤ C0 ε + C1 [∂v][ κ+6 ], T + hvi[ κ+6 ], T sup {(1 + τ )−ν k∂v(τ )kκ+8 }
2
2
0≤τ <T
≤ C0 ε + C1 C2 M ε2 sup {(1 + τ )C3 M ε−ν }
0≤τ <T
≤ 2C0 ε
(6.4)
for ε ∈ [0, ε2 ], provided that
·
½
ε2 ∈ 0, min
C0
ν
,
C1 C2 M C3 M
¾¸
,
where ν is the constant appearing in Theorem 4.1. Therefore, taking ε0 = min{ε1 , ε2 } and
M ≥ 4C0 , we obtain (6.2) from (6.4) immediately. This implies that
X
|∂ α v(x, t)| ∈ L∞ (R2 × [0, T )).
|α|≤4
Thus, the proof of Theorem 6.1 is completed.
Acknowledgements.
¥
This work was supported by the NNSF of China (Grant No.
10671124) and the Qiu-Shi Chair Professor Fellowship from Zhejiang University.
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