# The Periodic Boundary Value Problem for a Quasilinear Evolution Equation in Besov Spaces

Authors: Yun Wu; Zhengrong Liu; Xiang Zhang
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

This paper is concerned with the periodic boundary value problem for a quasilinear evolution equation of the following type:∂
t
u
+
f
(
u
)
∂
x
u
+
F
(
u
)
=
0, x
∈
T
=
R
/
2
π
Z, t
∈
R
+. Under some conditions, we prove that this equation is locally well-posed in Besov space B
p
,
r
s
(
T
). Furthermore, we study the continuity of the solution map for this equation in B
2
,
r
s
(
T
). Our work improves some earlier results.


---

## Body

## 1. Introduction and Results

In this paper, we consider the following quasilinear evolution equation with periodic boundary condition:(1)
∂
t
u
+
f
u
∂
x
u
+
F
u
=
0
,
x
∈
T
=
R
2
π
Z
,
t
>
0
,
(2)
u
0
,
x
=
u
0
x
,
x
∈
T
.Clearly, (1) can be viewed as a generalized Burgers-type equation:(3)
∂
t
u
+
f
u
∂
x
u
=
0
,with a nonlocal perturbation F
(
u
). Note that (1) contains the following important equations.Whenf
(
u
)
=
u and F
(
u
)
=
∂
x
1
-
∂
x
2
-
1
(
u
2
+
1
/
2
u
x
2
), (1) changes into the CH equation:(4)
u
t
-
u
x
x
t
+
3
u
u
x
=
2
u
x
u
x
x
+
u
u
x
x
x
.Whenf
(
u
)
=
u and F
(
u
)
=
∂
x
1
-
∂
x
2
-
1
(
3
/
2
u
2
), (1) reduces to the DP equation:(5)
u
t
-
u
x
x
t
+
4
u
u
x
=
3
u
x
u
x
x
+
u
u
x
x
x
.Whenf
(
u
)
=
u and F
(
u
)
=
∂
x
(
1
-
∂
x
2
)
-
1
(
b
/
2
u
2
+
3
-
b
/
2
u
x
2
), (1) becomes the b-equation:(6)
u
t
-
u
x
x
t
+
b
+
1
u
u
x
=
b
u
x
u
x
x
+
u
u
x
x
x
.Whenf
(
u
)
=
u
2 and F
(
u
)
=
(
1
-
∂
x
2
)
-
1
∂
x
u
3
+
3
/
2
u
u
x
2
+
(
1
-
∂
x
2
)
-
1
1
/
2
u
x
3, (1) turns out to be the Novikov [1] equation:(7)
u
t
-
u
x
x
t
+
4
u
2
u
x
=
3
u
u
x
u
x
x
+
u
2
u
x
x
x
.The CH equation was derived independently by two groups of researcheres, Fuchssteiner and Fokas [2] and Camassa and Holm [3]. Fuchssteiner and Fokas derived (4) in studying completely integrable generalizations of the KdV equation with bi-Haniltonian structures, while Camassa and Holm proposed (4) to describe the unidirectional propagation of shallow water waves over a flat bottom.Many physicists and mathematicians have paid much attention to the CH equation and a series of achievements had been made. For example, consider the following authors. Constantin [4] and Misiolek [5], investigated the Cauchy problem. Constantin, Escher, and McKean, and so forth [6–8] studied the wave-breaking and so on. Xin and Zhang [9] proved that there are global weak solutions for any data in H
1
(
R
) without any sign conditions on the initial value. Bressan and Constantin [10, 11] and Holden and Raynaud [12, 13] proved the existence of the global conservative and dissipative solutions. The CH equation arises also as an equation of the geodesic flow for the H
1 metrics on the Bott-Virasoro group [14, 15]. Ivanov [16] extended the CH hierarchy and obtained their conserved quantities. Constantin and Ivanov [17] studied an integrable two-component CH shallow water system.The DP equation can be regarded as a model for nonlinear shallow water dynamics derived by Degasperis and Procesi [18] in 1999. In 2003, the DP equation was derived by Dullin et al. [19] as a shallow water approximation to the Euler equation. The Cauchy problem for the DP equation has been studied extensively. The local and global well-posedness for the strong solutions, the global existence of weak solution, the blow-up phenomena, and nonuniformly continuous dependence on the initial data can be seen in [20, 21] and the references therein.In 2003, Holm and Staley [22] studied the exchange of stability in the dynamics of solitary wave solutions for b-equation. The well-posedness, blow-up phenomena, and global solutions for the b-equation can be found in [23, 24] and the additional references therein.It is showed that Novikov equation possesses a bi-Hamilton structure and an infinite sequence of conserved quantities and admits exact peaked solutionsu
(
t
,
x
)
=
±
c
e
-
x
-
c
t with c
>
0 (see [25]), as well as the explicit formulas for multipeakon solutions (see [25, 26]). By using the Littlewood-Paley decomposition and Kato’s theory, the Novikov equation’s well-posedness was studied in Besov spaces B
p
,
r
s
(
R
) and in the Sobolev space H
s
(
R
) (see [27, 28]). Wu and Yin [29] established some results on the existence and uniqueness of global weak solutions to the Novikov equation. Jiang and Ni [30] obtained some results about blow-up phenomena of the strong solution to the Cauchy problem for the Novikov equation. For the periodic case, Tiğlay [31] investigated the Cauchy problem for the periodic Novikov equation and proved that for s
>
5
/
2 the periodic Novikov equation is local well-posedness in H
s
(
T
). The range of local well-posedness for the periodic Novikov equation was extended to the s
>
3
/
2 in [32]. When s
<
3
/
2, Grayshan [33] proved that the properties of the solution map for the Novikov equation are not uniformly continuous in Sobolev spaces H
s.The issue of continuity of the solution map has been the subject of many papers. For the Burgers equation, Kato [34] showed that the solution map u
0
↦
u is not Hölder continuous regardless of the Hölder exponent. However, for certain general quasilinear hyperbolic systems, Kato obtained uniform continuity of the data to solution map for initial data in Sobolev spaces with integer index (measured in a weaker Sobolev norm). Tao [35] obtained Lipschitz continuity of the solution map for the Benjamin-Ono equation for H
1
(
R
) initial data measured in L
2. Herr et al. and so forth [36] have also obtained Lipschitz continuity in a weaker topology for the Benjamin-Ono with generalized dispersion. Davidson [37] studied the continuity of the solution map for the generalized reduced Ostrovsky equation. Karapetyan [38] proved the Hölder continuity of the solution map for the hyperelastic rod equation. For the continuity of solution map for some CH type equation and incompressible Euler equations in Besov spaces, we refer to [39–41]. These works lead to a natural question, whether a result similar to these holds for (1) when f, F satisfy some assumptions.Motivated by the results mentioned above, this paper deals with the problem (1)-(2). The aim of this paper is to prove local well-posedness of strong solutions in Besov spaces, Hölder continuity of the solution map in B
p
,
r
s equipped with a weak topology.We formulate the periodic boundary value problem (1)-(2) in Besov space as(8)
∂
t
u
+
f
u
∂
x
u
+
F
u
=
0
,
x
∈
T
,
t
∈
R
+
,
u
0
,
x
=
u
0
x
∈
B
p
,
r
s
T
.Now we introduce some notations and make some assumptions tof
(
u
) and F
(
u
).Notations. In this paper we adopt the following notations.The notation≲ denotes the estimates that hold up to some “harmless” constant which may change from line to line but whose meaning is clear from the context. D
(
T
) is the space of all infinitely differentiable functions on T and D
′
(
T
) is its dual space (the details on the periodic distributions can be found in, e.g., [42]). Assuming that all function spaces are over T, hence we drop T in our notations of function spaces if there is no ambiguity.ForT
>
0, s
∈
R, and 1
≤
p
≤
∞, we define(9)
E
p
,
r
s
T
=
C
0
,
T
;
B
p
,
r
s
∩
C
1
0
,
T
;
B
p
,
r
s
-
1
,
E
~
p
,
r
s
T
=
C
0
,
T
;
B
p
,
r
s
if
r
<
∞
,
E
p
,
∞
s
T
=
L
∞
0
,
T
;
B
p
,
∞
s
∩
Lip
⁡
0
,
T
;
B
p
,
∞
s
-
1
⁡
,
E
~
p
,
∞
s
T
=
L
∞
0
,
T
;
B
p
,
∞
s
if
r
=
∞
,
E
p
,
r
s
=
⋂
T
>
0
E
p
,
r
s
T
.
Assumptions. Assume that, for 1
≤
p, r
≤
∞, there exists a real number s
0
≥
1
+
1
/
p and an n
∈
N such that f
(
·
), F
(
·
) satisfy the following conditions.(A1)
f
∈
C
1
(
R
) and f
u
B
p
,
r
s
≲
u
B
p
,
r
s
n for s
>
1
/
p. Besides, for s
>
1
/
p and any u, v in B
p
,
r
s, f satisfies(10)
f
u
-
f
v
B
p
,
r
s
≲
u
B
p
,
r
s
+
v
B
p
,
r
s
n
-
1
∂
x
u
-
∂
x
v
B
p
,
r
s
-
1
≲
u
B
p
,
r
s
+
v
B
p
,
r
s
n
-
1
u
-
v
B
p
,
r
s
.
(A2)
Fors
>
s
0, F
:
B
p
,
r
s
↦
B
p
,
r
s is continuous and F
u
B
p
,
r
s
≲
u
B
p
,
r
s
n
+
1. Moreover, for s
>
s
0 and any u, v in B
p
,
r
s, F satisfies(11)
F
u
-
F
v
B
p
,
r
s
≲
u
B
p
,
r
s
+
v
B
p
,
r
s
n
∂
x
u
-
∂
x
v
B
p
,
r
s
-
1
≲
u
B
p
,
r
s
+
v
B
p
,
r
s
n
u
-
v
B
p
,
r
s
.
(A3)
Fors
0
-
1
<
σ
<
s
0 and any u, v in B
p
,
r
σ
+
1, it holds that(12)
F
u
-
F
v
B
p
,
r
σ
≲
u
B
p
,
r
σ
+
1
+
v
B
p
,
r
σ
+
1
n
∂
x
u
-
∂
x
v
B
p
,
r
σ
-
1
≲
u
B
p
,
r
σ
+
1
+
v
B
p
,
r
σ
+
1
n
u
-
v
B
p
,
r
σ
.Our main results are as follows.Theorem 1.
For1
≤
p, r
≤
∞, if there exist s
0
≥
1
+
1
/
p and n such that f
(
u
) and F
(
u
) satisfy the assumptions (A1)–(A3), then we have the following results.(1)
Fors
>
s
0, there is T
=
T
(
u
0
)
>
0 such that problem (8) has a unique solution u in E
p
,
r
s
(
T
). Moreover, the solution u satisfies(13)
u
t
B
p
,
r
s
≤
2
u
0
B
p
,
r
s
,
f
o
r
0
≤
t
≤
T
≤
1
2
n
C
u
0
B
p
,
r
s
n
,
whereC
>
0 depends on p, r, s but is independent of u
0.
(2)
Whens
>
s
0, the solution map u
0
↦
u defined by the problem (8) is continuous from B
p
,
r
s into E
p
,
r
s
′
(
T
) for every s
′
<
s. Moreover, when r
<
∞, the solution map is continuous from B
p
,
r
s into E
p
,
r
s
(
T
)
.
(3)
Whens
>
s
0, the solution map u
0
↦
u is Hölder continuous from B
(
0
,
R
) in B
p
,
r
s, with the B
p
,
r
q
1 topology, to C
(
[
0
,
T
]
,
B
p
,
r
q
2
), where T
=
T
(
R
). More precisely, let u
0
,
v
0
∈
B
(
0
,
R
)
⊂
B
p
,
r
s and u, v be the solutions corresponding to the initial data u
0, v
0, respectively. Then we have(14)
u
-
v
B
p
,
r
q
2
≲
u
0
-
v
0
B
p
,
r
q
1
α
,
where the exponentα satisfies(15)
α
=
1
,
i
f
s
≠
2
+
1
p
,
s
0
-
1
<
q
1
=
q
2
≤
s
-
1
,
q
1
=
q
2
≠
1
+
1
p
,
1
p
2
2
s
-
q
2
,
i
f
s
≠
2
+
1
p
,
s
-
1
=
q
1
<
q
2
<
s
.Remark 2.
As mentioned before, we can view (1) as a generalized Burgers-type equation: (16)
∂
t
u
+
f
u
∂
x
u
=
0
,with a nonlocal perturbation F
(
u
). To obtain the well-posedness, we need to assume some smooth properties on f
(
u
) and F
(
u
). Condition (A1) implies that, for 1
≤
p, r
≤
∞, s
>
1
/
p, f
:
B
p
,
r
s
↦
B
p
,
r
s is continuous and (A2) shows that when 1
≤
p, r
≤
∞, s
>
s
0
≥
1
+
1
/
p, F
:
B
p
,
r
s
↦
B
p
,
r
s is continuous. Therefore, it is reasonable to expect the well-posedness of (1) in B
p
,
r
s with 1
≤
p, r
≤
∞, s
>
s
0. Basically, condition (A3) is used to estimate u
-
v
B
p
,
r
s
′ for two solutions u, v and s
′
≤
s
-
1.We note that the issue of nonuniform dependence on initial data has been the subject of many papers (see, e.g., [32, 43]). In this paper, it is to be regretted that we can not find a feasible method to study the uniform continuity of the solution map u
0
↦
u defined by problem (1)-(2) for general f
(
u
) and F
(
u
). But when f
(
u
)
=
u and F
(
u
)
=
u
2, the solution map u
0
↦
u defined by problem (1)-(2) is not uniformly continuous from any bounded subset of B
2
,
r
s into E
~
2
,
r
s
(
T
) for any T
>
0 with s
>
3
/
2, 1
≤
r
≤
∞. As a special case, we have the following proposition.Proposition 3.
Whenf
(
u
)
=
u and F
(
u
)
=
u
2, problem (1)-(2) satisfies (A1)–(A3) automatically. In this case, Theorem 1 holds for s
0
=
3
/
2. However, the solution map u
0
↦
u defined by problem (1)-(2) is not uniformly continuous from any bounded subset of B
2
,
r
s into E
~
2
,
r
s
(
T
) for any T
>
0 with s
>
3
/
2 and 1
≤
r
≤
∞. More precisely, there exist two sequences of solutions u
k and v
k in E
~
2
,
r
s
(
T
) such that(17)
u
k
t
B
2
,
r
s
+
v
k
t
B
2
,
r
s
≲
1
,
(18)
lim
k
→
∞
⁡
u
k
0
-
v
k
0
B
2
,
r
s
=
0
,
(19)
liminf
k
→
∞
⁡
u
k
t
-
v
k
t
B
2
,
r
s
≳
sin
⁡
t
2
,
0
≤
t
≤
T
.We now conclude this introduction by outlining the rest of the paper. In Section2, we give some preliminaries. In Section 3, we prove Theorem 1. In Section 4, we demonstrate Proposition 3.


## 2. Preliminaries

In this section, we shall recall some basic facts on the Littlewood-Paley theory, Besov spaces, and the transport equations theory that will be used in this paper. We refer to [44–46] for the elementary properties of them.Letχ, ϕ be two functions satisfying χ, ϕ
∈
C
c
∞
(
R
), 0
≤
χ, ϕ
≤
1, χ
(
ξ
)
=
1 for ξ
≤
3
/
4, supp
⁡
χ
=
{
ξ
∈
R
,
ξ
≤
4
/
3
} and ϕ
(
ξ
)
=
χ
(
2
-
1
ξ
)
-
χ
(
ξ
). Then(20)
χ
ξ
+
∑
j
∈
N
ϕ
2
-
j
ξ
=
1
⁡
,
∀
ξ
∈
R
,
supp
⁡
ϕ
2
-
j
·
⋂
supp
⁡
ϕ
2
-
j
′
·
=
⌀
i
f
j
-
j
′
≥
2
,
supp
⁡
χ
·
⋂
supp
⁡
ϕ
2
-
j
·
=
⌀
i
f
j
≥
1
.We decomposeu
∈
D
′
(
T
) on the circle T into Fourier series u
=
1
/
2
π
∑
ξ
∈
Z
F
x
u
(
ξ
)
e
i
x
ξ, where F
x
u
(
ξ
)
=
∫
T
e
-
i
x
ξ
u
(
x
)
d
x is the Fourier transform on the circle. The inverse relation is given by F
x
-
1
u
(
x
)
=
1
/
2
π
∑
ξ
∈
Z
u
(
ξ
)
e
i
x
ξ. Define the periodic dyadic blocks as(21)
Δ
-
1
u
=
F
x
-
1
χ
F
x
u
=
1
2
π
∑
ξ
∈
Z
χ
ξ
F
x
u
ξ
e
i
x
ξ
,
Δ
j
u
=
F
x
-
1
ϕ
2
-
j
ξ
F
x
u
=
1
2
π
∑
ξ
∈
Z
ϕ
2
-
j
ξ
F
x
u
ξ
e
i
x
ξ
,
if
j
≥
0
.Then, we define the the low frequency cut-offS
j as S
j
u
=
∑
i
=
-
1
j
-
1
Δ
i
u. Direct computation implies that, for any 1
≤
p
≤
∞, we have the quasi-orthogonality properties with our choice of ϕ and ∀
u, v
∈
D
′
(
T
): (22)
Δ
i
Δ
j
u
≡
0
,
i
f
i
-
j
≥
2
,
Δ
j
S
i
-
1
u
Δ
i
v
≡
0
,
i
f
i
-
j
≥
5
.Furthermore, for all u
∈
L
p, Δ
i
u
L
p
≤
C
u
L
p, and S
j
u
L
p
≤
C
u
L
p
.Definition 4 (Besov spaces).
Lets
∈
R, 1
≤
p, r
≤
+
∞. The nonhomogeneous Besov space B
p
,
r
s
(
T
) is defined by(23)
B
p
,
r
s
=
f
∈
D
′
T
:
f
B
p
,
r
s
T
<
∞
,where f
B
p
,
r
s
T
=
2
j
s
Δ
j
f
l
r
(
L
p
)
=
2
j
s
Δ
j
f
L
p
j
≥
-
1
l
r. In particular, B
p
,
r
∞
=
⋂
s
′
∈
R
B
p
,
r
s
′
.Remark 5.
Whenp
=
2, 1
≤
r
≤
+
∞, the above definition is equivalent to the following:(24)
u
B
2
,
r
s
T
=
∑
ξ
=
-
1
1
1
+
ξ
2
s
F
x
u
2
r
/
2
+
∑
j
∈
N
∑
ξ
=
2
j
2
j
+
1
1
+
ξ
2
s
F
x
u
2
r
/
2
1
/
r
,
r
<
∞
,
sup
j
∈
N
⁡
∑
ξ
=
-
1
1
1
+
ξ
2
s
F
x
u
2
1
/
2
,
∑
ξ
=
2
j
2
j
+
1
1
+
ξ
2
s
F
x
u
2
1
/
2
,
r
=
∞
.When p
=
r
=
2, then B
2,2
s turn out to be the Sobolev spaces H
s.The following lemma summarizes some useful properties ofB
p
,
r
s.Lemma 6 (see [44–46]).
Lets
∈
R, 1
≤
p
,
r
,
p
j
,
r
j
≤
∞, j
=
1,2; then consider the following.(1)
B
p
,
r
s
(
T
) is a Banach space. When 1
<
p, r
<
∞, then the set of all trigonometric polynomial is dense in B
p
,
r
s
(
T
)
.
(2)
B
p
1
,
r
1
s
1
↪
B
p
2
,
r
2
s
2 if p
1
≤
p
2; r
1
≤
r
2 and s
2
=
s
1
-
(
1
/
p
1
-
1
/
p
2
).
B
p
,
r
1
s
1
↪
B
p
,
r
2
s
2 is locally compact if s
2
<
s
1; r
1
≤
r
2.
(3)
∀
s
>
0, B
p
,
r
s
∩
L
∞ is a Banach algebra. B
p
,
r
s is a Banach algebra ⇔
B
p
,
r
s
↪
L
∞
⇔
s
>
1
/
p  (or s
≥
1
/
p and r
=
1
).
(4)
For allθ
∈
[
0,1
], s
=
θ
s
1
+
(
1
-
θ
)
s
2,(25)
f
B
p
,
r
s
≤
C
f
B
p
,
r
s
1
θ
f
B
p
,
r
s
2
1
-
θ
,
∀
f
∈
B
p
,
r
s
1
∩
B
p
,
r
s
2
.
(5)
For allθ
∈
(
0,1
), s
1
>
s
2, s
=
θ
s
1
+
(
1
-
θ
)
s
2,(26)
u
B
p
,
1
s
≤
C
θ
C
1
-
θ
s
1
-
s
2
u
B
p
,
∞
s
1
θ
u
B
p
,
∞
s
2
1
-
θ
,
∀
u
∈
B
p
,
∞
s
1
.
(6)
Ifu
k
k
∈
N is bounded in B
p
,
r
s and u
k converges to u in D
′
(
T
), then u
∈
B
p
,
r
s and(27)
u
B
p
,
r
s
≤
liminf
k
→
∞
⁡
u
k
B
p
,
r
s
.Now we state some results in Besov spacesB
p
,
r
s of the transport equation. We refer to [45–48] for the details.Lemma 7 (a priori estimates).
Let1
≤
p, r
≤
∞ and s
>
-
m
i
n
1
/
p
,
1
-
1
/
p
. Assume that f
0
∈
B
p
,
r
s, F
∈
L
1
(
0
,
T
;
B
p
,
r
s
) and ∂
x
v
∈
L
1
(
0
,
T
;
B
p
,
r
s
-
1
) if s
>
1
+
1
/
p or ∂
x
v
∈
L
1
(
0
,
T
;
B
p
,
∞
1
/
p
∩
L
∞
) otherwise. If f
∈
L
∞
(
0
,
T
;
B
p
,
r
s
)
∩
C
(
[
0
,
T
]
;
D
′
) solves the following 1
-
D linear transport equation:(28)
f
t
+
v
f
x
=
F
,
x
∈
T
,
t
>
0
,
f
0
,
x
=
f
0
,
x
∈
T
,then there exists a constant C depending only on s, p, r as follows.(1)
For allt
∈
[
0
,
T
]
(29)
f
B
p
,
r
s
≤
f
0
B
p
,
r
s
+
∫
0
t
F
τ
B
p
,
r
s
d
τ
+
C
∫
0
t
V
′
τ
f
τ
B
p
,
r
s
d
τ
,
and hence,(30)
f
B
p
,
r
s
≤
e
C
V
t
f
0
B
p
,
r
s
+
∫
0
t
e
-
C
V
τ
F
τ
B
p
,
r
s
d
τ
,
where(31)
V
t
=
∫
0
t
v
x
τ
B
p
,
∞
1
/
p
∩
L
∞
d
τ
,
i
f
s
<
1
+
1
p
,
∫
0
t
v
x
τ
B
p
,
r
s
-
1
d
τ
,
i
f
s
>
1
+
1
p
o
r
s
=
1
+
1
p
,
r
=
1
.
Moreover, ifr
<
∞, then f
∈
C
(
[
0
,
T
]
;
B
p
,
r
s
). If r
=
∞, then f
∈
C
(
[
0
,
T
]
;
B
p
,
1
s
′
) for all s
′
<
s
.
(2)
Ifv
=
f, then (1) holds true for all s
>
0 with V
(
t
)
=
∫
0
t
v
x
τ
L
∞
d
τ.Lemma 8 (existence and uniqueness).
Letp, r, s, f
0 and F be as in the statement of Lemma 7. Assume that v
∈
L
ρ
(
0
,
T
;
B
∞
,
∞
-
M
) for some ρ
>
1, M
>
0 and v
x
∈
L
1
(
0
,
T
;
B
p
,
r
s
-
1
) if s
>
1
+
1
/
p or s
=
1
+
1
/
p and r
=
1 and v
x
∈
L
1
(
0
,
T
;
B
p
,
∞
1
/
p
∩
L
∞
) if s
<
1
+
1
/
p. Then, problem (28) has a unique solution f
∈
L
∞
(
0
,
T
;
B
p
,
r
s
)
∩
C
(
[
0
,
T
]
;
B
p
,
r
s
′
) for any s
′
<
s and the inequalities of Lemma 7 hold true. Moreover, if r
<
∞, then f
∈
C
(
[
0
,
T
]
;
B
p
,
r
s
)
.Lemma 9 (see [49]).
Letting1
≤
p, r
≤
∞, then we have the following estimates.(1)
Fors
>
0,(32)
f
g
B
p
,
r
s
≤
C
f
B
p
,
r
s
g
L
∞
+
f
L
∞
g
B
p
,
r
s
∀
f
,
g
∈
B
p
,
r
s
∩
L
∞
.
(2)
For alls
1
≤
1
/
p
<
s
2  (
s
2
≥
1
/
p if r
=
1
) and s
1
+
s
2
>
0, we have(33)
f
g
B
p
,
r
s
1
≤
C
f
B
p
,
r
s
1
g
B
p
,
r
s
2
∀
f
∈
B
p
,
r
s
1
,
g
∈
B
p
,
r
s
2
.In several cases, we need to estimateu
-
v
B
p
,
r
s
-
1 for two solutions u, v.Lemma 10.
Let1
≤
p, r
≤
∞. Suppose there exists a real number s
0
≥
1
+
1
/
p and a nature number n  (independent of s
0
) such that f
(
u
), F
(
u
) satisfy the assumptions (A1)–(A3). Let u
,
v
∈
C
(
[
0
,
T
]
;
D
′
(
T
)
)
⋂
L
∞
(
0
,
T
;
B
p
,
r
s
) be two solutions to (1) with initial data u
0
,
v
0
∈
B
p
,
r
s for s
>
s
0, respectively. Then the following statements hold.(1)
Whens
0
-
1
<
s
′
≤
s
-
1, s
′
≠
1
+
1
/
p, s
≠
2
+
1
/
p, for all t
∈
[
0
,
T
],(34)
u
-
v
B
p
,
r
s
′
≲
e
C
T
,
u
0
,
v
0
u
0
-
v
0
B
p
,
r
s
′
.
(2)
For anys
-
1
<
s
′
<
s, s
≠
2
+
1
/
p, there exists a θ
=
s
-
s
′
∈
(
0,1
) such that for all t
∈
[
0
,
T
],(35)
u
-
v
B
p
,
r
s
′
≲
e
θ
C
T
,
u
0
,
v
0
u
0
-
v
0
B
p
,
r
s
-
1
θ
.Proof.
Sinceu, v are two solutions to (1) with initial data u
0
,
v
0
∈
B
p
,
r
s, respectively, we know that(36)
u
t
+
f
u
∂
x
u
+
F
u
=
0
,
v
t
+
f
v
∂
x
v
+
F
v
=
0
.Subtracting the second equation of (36) from the first one, then w
=
u
-
v satisfies(37)
w
t
+
f
v
∂
x
w
=
-
f
u
-
f
v
∂
x
u
-
F
u
-
F
v
,
w
0
=
u
0
-
v
0
.
Whens
>
2
+
1
/
p
>
s
0 or 2
+
1
/
p
>
s
>
s
0 or s
>
s
0
≥
2
+
1
/
p, and s
′
≠
1
+
1
/
p, by using Lemma 7, we have(38)
w
t
B
p
,
r
s
′
≤
exp
⁡
C
W
t
w
0
B
p
,
r
s
′
+
C
∫
0
t
exp
⁡
C
W
t
-
C
W
τ
F
^
B
p
,
r
s
′
d
τ
,where F
^
=
-
f
(
u
)
-
f
(
v
)
∂
x
u
-
F
(
u
)
-
F
(
v
) and(39)
W
t
=
∫
0
t
∂
x
f
v
τ
B
d
τ
≤
C
∫
0
t
v
τ
B
p
,
r
s
n
d
τwith B
=
B
p
,
∞
1
/
p
∩
L
∞ if s
′
<
1
+
1
/
p or B
=
B
p
,
r
s
′
-
1 if s
′
>
1
+
1
/
p. Since u
,
v
∈
L
∞
(
0
,
T
;
B
p
,
r
s
), we know(40)
v
t
B
p
,
r
s
+
u
t
B
p
,
r
s
≲
1
,
0
≤
t
≤
T
.Hence, we obtain W
(
t
)
≲
1
. Since s
-
1
>
s
0
-
1, by (A2) (if s
′
>
s
0) or (A3) (if s
0
-
1
<
s
′
<
s
0) and (40), we have(41)
F
u
-
F
v
B
p
,
r
s
′
≲
u
B
p
,
r
s
+
v
B
p
,
r
s
n
u
-
v
B
p
,
r
s
′
≲
u
-
v
B
p
,
r
s
′
.
Similarly, when2
+
1
/
p
>
s
>
s
0, it follows that s
0
≥
1
+
1
/
p
>
s
′
>
s
0
-
1. Then (A3) and (40) also yield(42)
F
u
-
F
v
B
p
,
r
s
′
≲
u
B
p
,
r
s
+
v
B
p
,
r
s
n
u
-
v
B
p
,
r
s
′
≲
u
-
v
B
p
,
r
s
′
.As s
′
>
s
0
-
1
≥
1
/
p and B
p
,
r
s
′ is an algebra, by (A1) and (40), we obtain (43)
f
u
-
f
v
∂
x
u
B
p
,
r
s
′
≲
f
u
-
f
v
B
p
,
r
s
′
u
B
p
,
r
s
≲
u
-
v
B
p
,
r
s
′
.
Putting all of these results together we arrive atF
^
B
p
,
r
s
′
≲
u
-
v
B
p
,
r
s
′ and therefore (44)
w
t
B
p
,
r
s
′
≲
w
0
B
p
,
r
s
′
+
∫
0
t
u
-
v
B
p
,
r
s
′
d
τ
.By Gronwall’s inequality we obtain (34).
Finally, ifs
-
1
<
s
′
<
s, by using interpolation and (40) again, we have(45)
w
B
p
,
r
s
′
≤
w
B
p
,
r
s
-
1
θ
w
B
p
,
r
s
1
-
θ
≤
w
B
p
,
r
s
-
1
θ
u
B
p
,
r
s
+
v
B
p
,
r
s
1
-
θ
≲
w
B
p
,
r
s
-
1
θ
,where θ
=
s
-
s
′. Using (34), we complete the proof for this lemma.


## 3. Proof of Theorem1

The proof of Theorem1 includes the following several steps.


### 3.1. Approximate Solution

Starting fromu
1
=
0 and by induction, we define a sequence of smooth functions {
u
k
}, k
∈
N by solving the following transport equation iteratively:(46)
∂
t
u
k
+
1
+
f
u
k
∂
x
u
k
+
1
=
-
F
u
k
,
(47)
u
k
+
1
0
,
x
=
S
k
+
1
u
0
,Since f
(
u
), F
(
u
) satisfy the continuous assumption and all the data belong to B
p
,
r
∞, from Lemma 8 and by induction, we see that, for all k
≥
1, the above equation has a global solution u
k
+
1 belonging to C
(
R
+
,
B
p
,
r
∞
)
.


### 3.2. Uniform Bounds and Lifespan of the Approximate Solutions

Fork
∈
N, set U
k
(
t
)
=
∫
0
t
u
k
B
p
,
r
s
n
d
τ with 1
≤
p, r
≤
∞ and s
>
s
0
≥
1
+
1
/
p; then we have(48)
u
k
+
1
t
B
p
,
r
s
≤
e
C
U
k
t
u
0
B
p
,
r
s
+
C
∫
0
t
e
-
C
U
k
τ
u
k
B
p
,
r
s
n
+
1
d
τ
.In fact, since u
k
+
1
0
,
x
B
p
,
r
s
≲
u
0
B
p
,
r
s, from the estimate (30) of Lemma 7, the assumptions (A1), (A2), and the algebra property of B
p
,
r
s, we have(49)
u
k
+
1
t
B
p
,
r
s
≤
u
0
B
p
,
r
s
·
exp
⁡
C
∫
0
t
∂
x
f
u
k
t
′
B
p
,
r
s
-
1
d
t
′
+
∫
0
t
exp
⁡
C
∫
τ
t
∂
x
f
u
k
t
′
B
p
,
r
s
-
1
d
t
′
·
F
u
k
B
p
,
r
s
d
τ
≤
u
0
B
p
,
r
s
·
exp
⁡
C
∫
0
t
u
k
t
′
B
p
,
r
s
n
d
t
′
+
∫
0
t
exp
⁡
C
∫
τ
t
u
k
t
′
B
p
,
r
s
n
d
t
′
u
k
B
p
,
r
s
n
+
1
d
τ
,which gives (48). Let T
>
0 such that 2
n
C
u
0
B
p
,
r
s
n
T
<
1; we claim that, for any k
∈
N,(50)
u
k
B
p
,
r
s
≤
u
0
B
p
,
r
s
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
n
,
∀
t
∈
0
,
T
.Assume that (50) is true for k. We now prove that it also holds true for k
+
1. Since U
k
(
t
)
=
∫
0
t
u
k
B
p
,
r
s
n
d
τ, by using (50), we have(51)
e
C
U
k
t
-
C
U
k
τ
≤
exp
⁡
C
∫
τ
t
u
0
B
p
,
r
s
n
1
-
2
n
C
u
0
B
p
,
r
s
n
t
′
d
t
′
=
exp
⁡
-
1
2
n
∫
τ
t
d
1
-
2
n
C
u
0
B
p
,
r
s
n
t
′
1
-
2
n
C
u
0
B
p
,
r
s
n
t
′
=
1
-
2
n
C
u
0
B
p
,
r
s
n
τ
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
2
n
.From the above equation, we see clearly that when τ
=
0, U
k
(
0
)
=
0. We obtain(52)
e
C
U
k
t
≤
1
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
2
n
.Using (48), (50), (51), and (52) gives rise to(53)
u
k
+
1
t
B
p
,
r
s
≤
e
C
U
k
t
u
0
B
p
,
r
s
+
C
∫
0
t
e
C
U
k
t
-
C
U
k
τ
u
k
τ
B
p
,
r
s
n
+
1
d
τ
≤
1
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
2
n
u
0
B
p
,
r
s
+
∫
0
t
C
u
0
B
p
,
r
s
n
+
1
1
-
2
n
C
u
0
B
p
,
r
s
n
τ
1
+
1
/
2
n
d
τ
=
1
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
2
n
u
0
B
p
,
r
s
-
u
0
B
p
,
r
s
2
n
∫
0
t
d
1
-
2
n
C
u
0
B
p
,
r
s
n
τ
1
-
2
n
C
u
0
B
p
,
r
s
n
τ
1
+
1
/
2
n
≤
u
0
B
p
,
r
s
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
n
.Hence, (50) is true and {
u
k
} is uniformly bounded in C
(
[
0
,
T
]
;
B
p
,
r
s
) for all t
∈
[
0
,
T
]. Setting T
>
0 such that 2
n
C
u
0
B
p
,
r
s
n
T
<
1, for n
∈
N, we can conclude that the solution u
k exists for 0
≤
t
≤
T and satisfies the bound(54)
u
k
t
B
p
,
r
s
≤
2
u
0
B
p
,
r
s
,
0
≤
t
≤
T
.Furthermore, for s
>
1
+
1
/
p and t
∈
[
0
,
T
], using (46) yields(55)
∂
t
u
k
B
p
,
r
s
-
1
≤
f
u
k
∂
x
u
k
+
1
B
p
,
r
s
-
1
+
F
u
k
B
p
,
r
s
-
1
≤
u
k
B
p
,
r
s
-
1
n
∂
x
u
k
+
1
B
p
,
r
s
-
1
+
F
u
k
B
p
,
r
s
≲
u
0
B
p
,
r
s
n
+
1
.Hence, we conclude that {
u
k
}
⊂
E
p
,
r
s
(
T
) is uniformly bounded.


### 3.3. Convergence inC
(
[
0
,
T
]
,
B
p
,
r
s
-
1
)

We are going to show that{
u
k
} is a Cauchy sequence in C
(
[
0
,
T
]
;
B
p
,
r
s
-
1
). It is sufficient to estimate the difference of the iteration. For k
,
l
∈
N, from (46) we have(56)
∂
t
u
k
+
l
+
1
-
u
k
+
1
+
f
u
k
+
l
∂
x
u
k
+
l
+
1
-
u
k
+
1
=
f
u
k
-
f
u
k
+
l
∂
x
u
k
+
1
+
F
u
k
-
F
u
k
+
l
.


#### 3.3.1. Case ofs
>
2
+
1
/
p
>
s
0 or 2
+
1
/
p
>
s
>
s
0 or s
>
s
0
≥
2
+
1
/
p

Using estimates analogous to those in Lemma10, we obtain that (57)
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
-
1
≲
C
T
u
k
+
l
+
1
-
u
k
+
1
0
B
p
,
r
s
-
1
+
∫
0
t
u
k
+
l
τ
-
u
k
τ
B
p
,
r
s
-
1
d
τ
,where C
T is a big constant depending on u
0
B
p
,
r
s and T. Note that (see [50] for detailed computations)(58)
u
k
+
l
+
1
-
u
k
+
1
0
B
p
,
r
s
-
1
=
∑
i
=
k
+
1
k
+
l
Δ
i
u
0
B
p
,
r
s
-
1
≤
C
2
-
k
u
0
B
p
,
r
s
-
1
.It follows that(59)
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
-
1
≲
C
T
2
-
k
+
∫
0
t
u
k
+
l
τ
-
u
k
τ
B
p
,
r
s
-
1
d
τ
.For simplicity, we denote I
=
[
0
,
T
]. Arguing by induction with respect to the index k, we obtain(60)
u
k
+
l
+
1
-
u
k
+
1
C
I
;
B
p
,
r
s
-
1
≤
T
C
T
k
+
1
k
+
1
!
u
l
C
I
;
B
p
,
r
s
-
1
+
2
-
k
C
T
∑
j
=
0
k
2
j
T
C
T
j
j
!
.Since u
l
B
p
,
r
s is uniformly bounded by 2
u
0
B
p
,
r
s, hence (60) implies that {
u
k
} is a Cauchy sequence in C
(
I
;
B
p
,
r
s
-
1
) and therefore {
u
k
} converges to some function u
∈
C
(
I
;
B
p
,
r
s
-
1
).


#### 3.3.2. Case ofs
=
2
+
1
/
p
>
s
0
≥
1
+
1
/
p

A simple interpolation argument like the one in Lemma10 leads to (61)
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
-
1
≲
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
1
θ
u
0
B
p
,
r
s
1
-
θ
,where we choose s
1, s
2 such that s
0
-
1
<
s
1
<
1
+
1
/
p
≤
s
0
<
s
2
<
2
+
1
/
p
=
s and s
-
1
=
1
+
1
/
p
=
θ
s
1
+
(
1
-
θ
)
s
2 with 0
<
θ
<
1. Therefore, {
u
k
} is also a Cauchy sequence in C
(
I
;
B
p
,
r
s
-
1
).Gathering the above results, we know that{
u
k
}
⊂
C
(
I
;
B
p
,
r
s
-
1
) is a Cauchy sequence for s
>
s
0 and converges to some function u
∈
C
(
I
;
B
p
,
r
s
-
1
).


### 3.4. Existence of the Solution

Firstly, asu
k
B
p
,
r
s is uniformly bounded by 2
u
0
B
p
,
r
s, Fatou property guarantees u
B
p
,
r
s
≤
2
u
0
B
p
,
r
s which gives (13). Repeat the process in Lemma 10; for any s
′
<
s
-
1, we obtain(62)
u
k
-
u
B
p
,
r
s
′
≤
u
k
-
u
B
p
,
r
s
-
1
.If s
-
1
<
s
′
<
s, by using interpolation again, we have(63)
u
k
-
u
B
p
,
r
s
′
≤
u
k
-
u
B
p
,
r
s
-
1
θ
u
k
-
u
B
p
,
r
s
1
-
θ
≤
u
k
-
u
B
p
,
r
s
-
1
θ
4
u
0
B
p
,
r
s
1
-
θ
,where θ
=
s
-
s
′. From (62) and (63), we see that {
u
k
} converges to u in C
(
I
;
B
p
,
r
s
′
) for all s
′
<
s; this enables us to deduce that u indeed solves (1) in the sense of distributions.


### 3.5. Regularity and Uniqueness of the Solution

Whenr
<
∞, from Lemma 8, we know that u
∈
C
(
[
0
,
T
]
;
B
p
,
r
s
); thus by the equation itself in (8), we know that u
t
∈
C
(
[
0
,
T
]
;
B
p
,
r
s
-
1
). When r
=
∞, directly from the equation in (8), we know u
t
∈
L
∞
(
0
,
T
;
B
p
,
∞
s
-
1
). Therefore we have u
∈
E
p
,
∞
s
(
T
). The uniqueness of the solution is a corollary of Lemma 10.


### 3.6. Continuity of the Solution Map

The continuity with respect to the initial data in(64)
C
0
,
T
;
B
p
,
r
s
′
⋂
C
0
,
T
;
B
p
,
r
s
′
-
1
,
∀
s
′
<
sor in(65)
C
0
,
T
;
B
p
,
r
s
⋂
C
0
,
T
;
B
p
,
r
s
-
1
,
s
=
2
+
1
pcan be obtained directly by Lemma 10. We now prove that the continuity holds true up to index s
>
s
0
≥
1
+
1
/
p, s
≠
2
+
1
/
p. We state the following lemma first.Lemma 11.
DenoteN
¯
=
N
∪
{
∞
}
. Let n, f satisfy the assumption (A1). Suppose that (
p
,
r
)
∈
1
,
+
∞
2, r
<
∞, and s
>
1
+
1
/
p, s
≠
2
+
1
/
p. Let a
k
k
∈
N
¯ be a sequence of continuous bounded functions on [
0
,
T
]
×
R with s
u
p
k
∈
N
¯
a
k
(
t
)
B
p
,
r
s
n
≤
α
(
t
) for some α
(
t
)
∈
L
∞
0
,
T if n
>
1 or α
(
t
)
∈
L
1
0
,
T if n
=
1. Assume that v
k
k
∈
N
¯ is the solution to(66)
∂
t
v
k
+
f
a
k
∂
x
v
k
=
g
,
v
k
0
=
v
0
,with v
0
∈
B
p
,
r
s
-
1, g
∈
L
1
0
,
T
;
B
p
,
r
s
-
1. If a
k
→
a
∞ in L
1
0
,
T
;
B
p
,
r
s
-
1 then v
k
→
v
∞ in C
(
[
0
,
T
]
;
B
p
,
r
s
-
1
)
.Proof.
Letw
k
=
v
k
-
v
∞. It is obvious that w
k solves the transport equation: (67)
w
t
+
f
a
k
∂
x
w
=
-
f
a
k
-
f
a
∞
∂
x
v
∞
.We first consider v
0
∈
B
p
,
r
s and g
∈
L
1
0
,
T
;
B
p
,
r
s. Note that r
<
∞; by Lemma 7, we obtain that, for k
∈
N
¯, v
k
∈
C
[
0
,
T
]
;
B
p
,
r
s and(68)
v
k
t
B
p
,
r
s
≤
v
0
B
p
,
r
s
exp
⁡
C
∫
0
t
α
τ
d
τ
+
∫
0
t
exp
⁡
C
∫
τ
t
α
τ
′
d
τ
′
g
B
p
,
r
s
d
τ
≲
v
0
B
p
,
r
s
+
∫
0
t
g
B
p
,
r
s
d
τ
.Since s
≠
2
+
1
/
p, for k
∈
N
¯, Lemma 7 tells us that(69)
w
k
t
B
p
,
r
s
-
1
≤
∫
0
t
exp
⁡
C
∫
τ
t
α
τ
′
d
τ
′
×
f
a
k
-
f
a
∞
∂
x
v
∞
B
p
,
r
s
-
1
d
τ
≲
∫
0
t
f
a
k
-
f
a
∞
∂
x
v
∞
B
p
,
r
s
-
1
d
τ
.Since s
>
1
+
1
/
p, using (A1) yields that(70)
f
a
k
-
f
a
∞
∂
x
v
∞
B
p
,
r
s
-
1
≲
a
k
B
p
,
r
s
-
1
+
a
∞
B
p
,
r
s
-
1
n
-
1
a
k
-
a
∞
B
p
,
r
s
-
1
v
∞
B
p
,
r
s
≲
1
+
a
k
B
p
,
r
s
-
1
+
a
∞
B
p
,
r
s
-
1
n
a
k
-
a
∞
B
p
,
r
s
-
1
·
v
∞
B
p
,
r
s
≲
1
+
α
t
L
∞
a
k
-
a
∞
B
p
,
r
s
-
1
v
∞
B
p
,
r
s
.Therefore, we obtain the following estimate by gathering the above inequalities:(71)
w
k
t
B
p
,
r
s
-
1
≤
∫
0
t
exp
⁡
C
∫
τ
t
α
τ
′
d
τ
′
×
f
a
k
-
f
a
∞
∂
x
v
∞
B
p
,
r
s
-
1
d
τ
≲
v
0
B
p
,
r
s
+
∫
0
t
g
B
p
,
r
s
d
τ
∫
0
t
a
k
-
a
∞
B
p
,
r
s
-
1
d
τ
,which yields the desired result of convergence.
For the general casev
0
∈
B
p
,
r
s
-
1 and g
∈
L
1
0
,
T
;
B
p
,
r
s
-
1, we have(72)
w
k
B
p
,
r
s
-
1
≤
v
k
-
v
k
j
B
p
,
r
s
-
1
+
v
k
j
-
v
∞
j
B
p
,
r
s
-
1
+
v
∞
j
-
v
∞
B
p
,
r
s
-
1
,where v
k
j satisfies the “cut-off” equation: (73)
∂
t
v
k
j
+
f
a
k
∂
x
v
k
j
=
S
j
g
=
∑
i
=
-
1
j
-
1
Δ
i
g
,
v
k
j
0
=
S
j
v
0
=
∑
i
=
-
1
j
-
1
Δ
i
v
0
.For v
k
-
v
k
j, we see that v
k
-
v
k
j solves the equation: (74)
∂
t
u
+
f
a
k
∂
x
u
=
g
-
S
j
g
,
u
0
=
v
0
-
S
j
v
0
.By Lemma 7 and (A1) once again, it follows that(75)
v
k
-
v
k
j
B
p
,
r
s
-
1
≤
v
0
-
S
j
v
0
B
p
,
r
s
-
1
e
C
∫
0
t
α
τ
d
τ
+
∫
0
t
e
C
∫
τ
t
α
τ
′
d
τ
′
g
-
S
j
g
B
p
,
r
s
-
1
d
τ
≲
v
0
-
S
j
v
0
B
p
,
r
s
-
1
+
∫
0
t
g
-
S
j
g
B
p
,
r
s
-
1
d
τ
.Note that v
k
j
(
0
)
∈
B
p
,
r
s and S
j
g
∈
L
1
0
,
T
;
B
p
,
r
s. Thanks to (71), we have(76)
v
k
j
-
v
∞
j
t
B
p
,
r
s
-
1
≲
v
k
j
0
B
p
,
r
s
+
∫
0
t
S
j
g
B
p
,
r
s
d
τ
·
∫
0
t
a
k
-
a
∞
B
p
,
r
s
-
1
d
τ
.Similarly, we can obtain(77)
v
∞
-
v
∞
j
B
p
,
r
s
-
1
≲
v
0
-
S
j
v
0
B
p
,
r
s
-
1
+
∫
0
t
g
-
S
j
g
B
p
,
r
s
-
1
d
τ
.Inserting (75), (76), and (77) into (72), we have(78)
w
k
B
p
,
r
s
-
1
≲
v
0
-
S
j
v
0
B
p
,
r
s
-
1
+
∫
0
t
g
-
S
j
g
B
p
,
r
s
-
1
d
τ
+
v
k
j
0
B
p
,
r
s
+
∫
0
t
S
j
g
B
p
,
r
s
d
τ
·
∫
0
t
a
k
-
a
∞
B
p
,
r
s
-
1
d
τ
.
By the definition ofS
j, we have g
τ
-
S
j
g
τ
B
p
,
r
s
-
1
≲
g
τ
B
p
,
r
s
-
1
∈
L
1
(
0
,
T
)
. Using the Lebesgue dominated convergence theorem yields that the first two terms of the right-hand side of the above estimate may be arbitrary small for j large enough. Take j large enough; then fix j. Let k tend to infinity so that the last term of right-hand side tends to zero. We complete the proof of Lemma 11.Sets
>
s
0
≥
1
+
1
/
p, s
≠
2
+
1
/
p. Fix u
0
,
∞
∈
B
p
,
r
s and let u
0
,
k
∈
B
p
,
r
s be a sequence with u
0
,
k
→
u
0
,
∞. For simplicity, we denote I
=
[
0
,
T
] and we let u
∞ (resp., u
k) be the solution to (1) with initial data u
0
,
∞ (resp., u
0
,
k); we need to demonstrate that(79)
lim
k
→
∞
⁡
u
k
=
u
∞
in
C
I
;
B
p
,
r
s
.By Lemma10, we have u
k
→
u
∞ in C
(
I
;
B
p
,
r
s
-
1
). To prove (79), it suffices to prove that(80)
lim
k
→
∞
⁡
∂
x
u
k
=
∂
x
u
∞
in
C
I
;
B
p
,
r
s
-
1
.Set v
k
=
∂
x
u
k, k
∈
N
¯; then v
k satisfies(81)
∂
t
v
k
+
f
u
k
∂
x
v
k
=
-
f
′
u
k
∂
x
u
k
2
-
∂
x
F
u
k
=
g
k
,
v
k
0
,
x
=
∂
x
u
0
,
k
.Motivated by the work in [34], for k
∈
N
¯, we decompose v
k
=
z
k
+
h
k such that(82)
∂
t
z
k
+
f
u
k
∂
x
z
k
=
g
k
-
g
∞
,
z
k
0
,
x
=
∂
x
u
0
,
k
-
∂
x
u
0
,
∞
,
∂
t
h
k
+
f
u
k
∂
x
h
k
=
g
∞
,
h
k
0
,
x
=
∂
x
u
0
,
∞
.Since u
0
,
k
→
u
0
,
∞ in B
p
,
r
s, we have u
0
,
k
B
p
,
r
s
≲
u
0
,
∞
B
p
,
r
s
≲
1. Then (13) shows that(83)
u
k
B
p
,
r
s
+
u
∞
B
p
,
r
s
≲
2
u
0
,
k
B
p
,
r
s
+
2
u
0
,
∞
B
p
,
r
s
≲
1
.By the algebra property of B
p
,
r
s
-
1, the assumptions (A1), (A2), and (83), we have (84)
g
k
-
g
∞
B
p
,
r
s
-
1
=
f
′
u
∞
∂
x
u
∞
2
-
f
′
u
k
∂
x
u
k
2
+
∂
x
F
u
∞
-
∂
x
F
u
k
B
p
,
r
s
-
1
≲
∂
x
u
∞
-
∂
x
u
k
B
p
,
r
s
-
1
.Since s
>
1
+
1
/
p, s
≠
2
+
1
/
p, by Lemma 7, we arrive at (85)
z
k
t
B
p
,
r
s
-
1
≲
z
0
B
p
,
r
s
-
1
e
C
∫
0
t
f
u
k
τ
B
p
,
r
s
d
τ
+
∫
0
t
e
C
∫
τ
t
f
u
k
τ
′
B
p
,
r
s
d
τ
′
g
k
-
g
∞
B
p
,
r
s
-
1
d
τ
≲
∂
x
u
0
,
∞
-
∂
x
u
0
,
k
B
p
,
r
s
-
1
+
∫
0
t
∂
x
u
∞
-
∂
x
u
k
B
p
,
r
s
-
1
d
τ
.By (83) and the convergence u
k
→
u
∞ in C
(
I
;
B
p
,
r
s
-
1
), we can get h
k
→
h
∞
=
v
∞
=
∂
x
u
∞ in C
(
I
;
B
p
,
r
s
-
1
) by applying Lemma 11. That is to say, there is a K
>
0 such that(86)
k
>
K
⟹
h
k
-
h
∞
C
I
;
B
p
,
r
s
-
1
<
ɛ
.Therefore, whenk
>
K, we have(87)
∂
x
u
∞
-
∂
x
u
k
B
p
,
r
s
-
1
=
v
k
-
v
∞
B
p
,
r
s
-
1
≤
z
k
t
B
p
,
r
s
-
1
+
h
k
t
B
p
,
r
s
-
1
≲
∂
x
u
0
,
∞
-
∂
x
u
0
,
k
B
p
,
r
s
-
1
+
∫
0
t
∂
x
u
∞
-
∂
x
u
k
B
p
,
r
s
-
1
d
τ
+
ɛ
.By using the Gronwall’s inequality, we have (88)
∂
x
u
∞
-
∂
x
u
k
C
I
;
B
p
,
r
s
-
1
≲
ɛ
C
T
,
u
0
,
∞
a
s
k
>
K
,where C
(
T
,
u
0
,
∞
) is a enough big constant depending on T and u
0
,
∞. Hence we obtain (80).


### 3.7. Hölder Continuity of the Solution Map with Weak Topology

Basically, direct from Lemma10, (14) is proved. Hereto, we complete the proof of Theorem 1.


## 3.1. Approximate Solution

Starting fromu
1
=
0 and by induction, we define a sequence of smooth functions {
u
k
}, k
∈
N by solving the following transport equation iteratively:(46)
∂
t
u
k
+
1
+
f
u
k
∂
x
u
k
+
1
=
-
F
u
k
,
(47)
u
k
+
1
0
,
x
=
S
k
+
1
u
0
,Since f
(
u
), F
(
u
) satisfy the continuous assumption and all the data belong to B
p
,
r
∞, from Lemma 8 and by induction, we see that, for all k
≥
1, the above equation has a global solution u
k
+
1 belonging to C
(
R
+
,
B
p
,
r
∞
)
.


## 3.2. Uniform Bounds and Lifespan of the Approximate Solutions

Fork
∈
N, set U
k
(
t
)
=
∫
0
t
u
k
B
p
,
r
s
n
d
τ with 1
≤
p, r
≤
∞ and s
>
s
0
≥
1
+
1
/
p; then we have(48)
u
k
+
1
t
B
p
,
r
s
≤
e
C
U
k
t
u
0
B
p
,
r
s
+
C
∫
0
t
e
-
C
U
k
τ
u
k
B
p
,
r
s
n
+
1
d
τ
.In fact, since u
k
+
1
0
,
x
B
p
,
r
s
≲
u
0
B
p
,
r
s, from the estimate (30) of Lemma 7, the assumptions (A1), (A2), and the algebra property of B
p
,
r
s, we have(49)
u
k
+
1
t
B
p
,
r
s
≤
u
0
B
p
,
r
s
·
exp
⁡
C
∫
0
t
∂
x
f
u
k
t
′
B
p
,
r
s
-
1
d
t
′
+
∫
0
t
exp
⁡
C
∫
τ
t
∂
x
f
u
k
t
′
B
p
,
r
s
-
1
d
t
′
·
F
u
k
B
p
,
r
s
d
τ
≤
u
0
B
p
,
r
s
·
exp
⁡
C
∫
0
t
u
k
t
′
B
p
,
r
s
n
d
t
′
+
∫
0
t
exp
⁡
C
∫
τ
t
u
k
t
′
B
p
,
r
s
n
d
t
′
u
k
B
p
,
r
s
n
+
1
d
τ
,which gives (48). Let T
>
0 such that 2
n
C
u
0
B
p
,
r
s
n
T
<
1; we claim that, for any k
∈
N,(50)
u
k
B
p
,
r
s
≤
u
0
B
p
,
r
s
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
n
,
∀
t
∈
0
,
T
.Assume that (50) is true for k. We now prove that it also holds true for k
+
1. Since U
k
(
t
)
=
∫
0
t
u
k
B
p
,
r
s
n
d
τ, by using (50), we have(51)
e
C
U
k
t
-
C
U
k
τ
≤
exp
⁡
C
∫
τ
t
u
0
B
p
,
r
s
n
1
-
2
n
C
u
0
B
p
,
r
s
n
t
′
d
t
′
=
exp
⁡
-
1
2
n
∫
τ
t
d
1
-
2
n
C
u
0
B
p
,
r
s
n
t
′
1
-
2
n
C
u
0
B
p
,
r
s
n
t
′
=
1
-
2
n
C
u
0
B
p
,
r
s
n
τ
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
2
n
.From the above equation, we see clearly that when τ
=
0, U
k
(
0
)
=
0. We obtain(52)
e
C
U
k
t
≤
1
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
2
n
.Using (48), (50), (51), and (52) gives rise to(53)
u
k
+
1
t
B
p
,
r
s
≤
e
C
U
k
t
u
0
B
p
,
r
s
+
C
∫
0
t
e
C
U
k
t
-
C
U
k
τ
u
k
τ
B
p
,
r
s
n
+
1
d
τ
≤
1
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
2
n
u
0
B
p
,
r
s
+
∫
0
t
C
u
0
B
p
,
r
s
n
+
1
1
-
2
n
C
u
0
B
p
,
r
s
n
τ
1
+
1
/
2
n
d
τ
=
1
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
2
n
u
0
B
p
,
r
s
-
u
0
B
p
,
r
s
2
n
∫
0
t
d
1
-
2
n
C
u
0
B
p
,
r
s
n
τ
1
-
2
n
C
u
0
B
p
,
r
s
n
τ
1
+
1
/
2
n
≤
u
0
B
p
,
r
s
1
-
2
n
C
u
0
B
p
,
r
s
n
t
1
/
n
.Hence, (50) is true and {
u
k
} is uniformly bounded in C
(
[
0
,
T
]
;
B
p
,
r
s
) for all t
∈
[
0
,
T
]. Setting T
>
0 such that 2
n
C
u
0
B
p
,
r
s
n
T
<
1, for n
∈
N, we can conclude that the solution u
k exists for 0
≤
t
≤
T and satisfies the bound(54)
u
k
t
B
p
,
r
s
≤
2
u
0
B
p
,
r
s
,
0
≤
t
≤
T
.Furthermore, for s
>
1
+
1
/
p and t
∈
[
0
,
T
], using (46) yields(55)
∂
t
u
k
B
p
,
r
s
-
1
≤
f
u
k
∂
x
u
k
+
1
B
p
,
r
s
-
1
+
F
u
k
B
p
,
r
s
-
1
≤
u
k
B
p
,
r
s
-
1
n
∂
x
u
k
+
1
B
p
,
r
s
-
1
+
F
u
k
B
p
,
r
s
≲
u
0
B
p
,
r
s
n
+
1
.Hence, we conclude that {
u
k
}
⊂
E
p
,
r
s
(
T
) is uniformly bounded.


## 3.3. Convergence inC
(
[
0
,
T
]
,
B
p
,
r
s
-
1
)

We are going to show that{
u
k
} is a Cauchy sequence in C
(
[
0
,
T
]
;
B
p
,
r
s
-
1
). It is sufficient to estimate the difference of the iteration. For k
,
l
∈
N, from (46) we have(56)
∂
t
u
k
+
l
+
1
-
u
k
+
1
+
f
u
k
+
l
∂
x
u
k
+
l
+
1
-
u
k
+
1
=
f
u
k
-
f
u
k
+
l
∂
x
u
k
+
1
+
F
u
k
-
F
u
k
+
l
.


### 3.3.1. Case ofs
>
2
+
1
/
p
>
s
0 or 2
+
1
/
p
>
s
>
s
0 or s
>
s
0
≥
2
+
1
/
p

Using estimates analogous to those in Lemma10, we obtain that (57)
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
-
1
≲
C
T
u
k
+
l
+
1
-
u
k
+
1
0
B
p
,
r
s
-
1
+
∫
0
t
u
k
+
l
τ
-
u
k
τ
B
p
,
r
s
-
1
d
τ
,where C
T is a big constant depending on u
0
B
p
,
r
s and T. Note that (see [50] for detailed computations)(58)
u
k
+
l
+
1
-
u
k
+
1
0
B
p
,
r
s
-
1
=
∑
i
=
k
+
1
k
+
l
Δ
i
u
0
B
p
,
r
s
-
1
≤
C
2
-
k
u
0
B
p
,
r
s
-
1
.It follows that(59)
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
-
1
≲
C
T
2
-
k
+
∫
0
t
u
k
+
l
τ
-
u
k
τ
B
p
,
r
s
-
1
d
τ
.For simplicity, we denote I
=
[
0
,
T
]. Arguing by induction with respect to the index k, we obtain(60)
u
k
+
l
+
1
-
u
k
+
1
C
I
;
B
p
,
r
s
-
1
≤
T
C
T
k
+
1
k
+
1
!
u
l
C
I
;
B
p
,
r
s
-
1
+
2
-
k
C
T
∑
j
=
0
k
2
j
T
C
T
j
j
!
.Since u
l
B
p
,
r
s is uniformly bounded by 2
u
0
B
p
,
r
s, hence (60) implies that {
u
k
} is a Cauchy sequence in C
(
I
;
B
p
,
r
s
-
1
) and therefore {
u
k
} converges to some function u
∈
C
(
I
;
B
p
,
r
s
-
1
).


### 3.3.2. Case ofs
=
2
+
1
/
p
>
s
0
≥
1
+
1
/
p

A simple interpolation argument like the one in Lemma10 leads to (61)
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
-
1
≲
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
1
θ
u
0
B
p
,
r
s
1
-
θ
,where we choose s
1, s
2 such that s
0
-
1
<
s
1
<
1
+
1
/
p
≤
s
0
<
s
2
<
2
+
1
/
p
=
s and s
-
1
=
1
+
1
/
p
=
θ
s
1
+
(
1
-
θ
)
s
2 with 0
<
θ
<
1. Therefore, {
u
k
} is also a Cauchy sequence in C
(
I
;
B
p
,
r
s
-
1
).Gathering the above results, we know that{
u
k
}
⊂
C
(
I
;
B
p
,
r
s
-
1
) is a Cauchy sequence for s
>
s
0 and converges to some function u
∈
C
(
I
;
B
p
,
r
s
-
1
).


## 3.3.1. Case ofs
>
2
+
1
/
p
>
s
0 or 2
+
1
/
p
>
s
>
s
0 or s
>
s
0
≥
2
+
1
/
p

Using estimates analogous to those in Lemma10, we obtain that (57)
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
-
1
≲
C
T
u
k
+
l
+
1
-
u
k
+
1
0
B
p
,
r
s
-
1
+
∫
0
t
u
k
+
l
τ
-
u
k
τ
B
p
,
r
s
-
1
d
τ
,where C
T is a big constant depending on u
0
B
p
,
r
s and T. Note that (see [50] for detailed computations)(58)
u
k
+
l
+
1
-
u
k
+
1
0
B
p
,
r
s
-
1
=
∑
i
=
k
+
1
k
+
l
Δ
i
u
0
B
p
,
r
s
-
1
≤
C
2
-
k
u
0
B
p
,
r
s
-
1
.It follows that(59)
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
-
1
≲
C
T
2
-
k
+
∫
0
t
u
k
+
l
τ
-
u
k
τ
B
p
,
r
s
-
1
d
τ
.For simplicity, we denote I
=
[
0
,
T
]. Arguing by induction with respect to the index k, we obtain(60)
u
k
+
l
+
1
-
u
k
+
1
C
I
;
B
p
,
r
s
-
1
≤
T
C
T
k
+
1
k
+
1
!
u
l
C
I
;
B
p
,
r
s
-
1
+
2
-
k
C
T
∑
j
=
0
k
2
j
T
C
T
j
j
!
.Since u
l
B
p
,
r
s is uniformly bounded by 2
u
0
B
p
,
r
s, hence (60) implies that {
u
k
} is a Cauchy sequence in C
(
I
;
B
p
,
r
s
-
1
) and therefore {
u
k
} converges to some function u
∈
C
(
I
;
B
p
,
r
s
-
1
).


## 3.3.2. Case ofs
=
2
+
1
/
p
>
s
0
≥
1
+
1
/
p

A simple interpolation argument like the one in Lemma10 leads to (61)
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
-
1
≲
u
k
+
l
+
1
-
u
k
+
1
B
p
,
r
s
1
θ
u
0
B
p
,
r
s
1
-
θ
,where we choose s
1, s
2 such that s
0
-
1
<
s
1
<
1
+
1
/
p
≤
s
0
<
s
2
<
2
+
1
/
p
=
s and s
-
1
=
1
+
1
/
p
=
θ
s
1
+
(
1
-
θ
)
s
2 with 0
<
θ
<
1. Therefore, {
u
k
} is also a Cauchy sequence in C
(
I
;
B
p
,
r
s
-
1
).Gathering the above results, we know that{
u
k
}
⊂
C
(
I
;
B
p
,
r
s
-
1
) is a Cauchy sequence for s
>
s
0 and converges to some function u
∈
C
(
I
;
B
p
,
r
s
-
1
).


## 3.4. Existence of the Solution

Firstly, asu
k
B
p
,
r
s is uniformly bounded by 2
u
0
B
p
,
r
s, Fatou property guarantees u
B
p
,
r
s
≤
2
u
0
B
p
,
r
s which gives (13). Repeat the process in Lemma 10; for any s
′
<
s
-
1, we obtain(62)
u
k
-
u
B
p
,
r
s
′
≤
u
k
-
u
B
p
,
r
s
-
1
.If s
-
1
<
s
′
<
s, by using interpolation again, we have(63)
u
k
-
u
B
p
,
r
s
′
≤
u
k
-
u
B
p
,
r
s
-
1
θ
u
k
-
u
B
p
,
r
s
1
-
θ
≤
u
k
-
u
B
p
,
r
s
-
1
θ
4
u
0
B
p
,
r
s
1
-
θ
,where θ
=
s
-
s
′. From (62) and (63), we see that {
u
k
} converges to u in C
(
I
;
B
p
,
r
s
′
) for all s
′
<
s; this enables us to deduce that u indeed solves (1) in the sense of distributions.


## 3.5. Regularity and Uniqueness of the Solution

Whenr
<
∞, from Lemma 8, we know that u
∈
C
(
[
0
,
T
]
;
B
p
,
r
s
); thus by the equation itself in (8), we know that u
t
∈
C
(
[
0
,
T
]
;
B
p
,
r
s
-
1
). When r
=
∞, directly from the equation in (8), we know u
t
∈
L
∞
(
0
,
T
;
B
p
,
∞
s
-
1
). Therefore we have u
∈
E
p
,
∞
s
(
T
). The uniqueness of the solution is a corollary of Lemma 10.


## 3.6. Continuity of the Solution Map

The continuity with respect to the initial data in(64)
C
0
,
T
;
B
p
,
r
s
′
⋂
C
0
,
T
;
B
p
,
r
s
′
-
1
,
∀
s
′
<
sor in(65)
C
0
,
T
;
B
p
,
r
s
⋂
C
0
,
T
;
B
p
,
r
s
-
1
,
s
=
2
+
1
pcan be obtained directly by Lemma 10. We now prove that the continuity holds true up to index s
>
s
0
≥
1
+
1
/
p, s
≠
2
+
1
/
p. We state the following lemma first.Lemma 11.
DenoteN
¯
=
N
∪
{
∞
}
. Let n, f satisfy the assumption (A1). Suppose that (
p
,
r
)
∈
1
,
+
∞
2, r
<
∞, and s
>
1
+
1
/
p, s
≠
2
+
1
/
p. Let a
k
k
∈
N
¯ be a sequence of continuous bounded functions on [
0
,
T
]
×
R with s
u
p
k
∈
N
¯
a
k
(
t
)
B
p
,
r
s
n
≤
α
(
t
) for some α
(
t
)
∈
L
∞
0
,
T if n
>
1 or α
(
t
)
∈
L
1
0
,
T if n
=
1. Assume that v
k
k
∈
N
¯ is the solution to(66)
∂
t
v
k
+
f
a
k
∂
x
v
k
=
g
,
v
k
0
=
v
0
,with v
0
∈
B
p
,
r
s
-
1, g
∈
L
1
0
,
T
;
B
p
,
r
s
-
1. If a
k
→
a
∞ in L
1
0
,
T
;
B
p
,
r
s
-
1 then v
k
→
v
∞ in C
(
[
0
,
T
]
;
B
p
,
r
s
-
1
)
.Proof.
Letw
k
=
v
k
-
v
∞. It is obvious that w
k solves the transport equation: (67)
w
t
+
f
a
k
∂
x
w
=
-
f
a
k
-
f
a
∞
∂
x
v
∞
.We first consider v
0
∈
B
p
,
r
s and g
∈
L
1
0
,
T
;
B
p
,
r
s. Note that r
<
∞; by Lemma 7, we obtain that, for k
∈
N
¯, v
k
∈
C
[
0
,
T
]
;
B
p
,
r
s and(68)
v
k
t
B
p
,
r
s
≤
v
0
B
p
,
r
s
exp
⁡
C
∫
0
t
α
τ
d
τ
+
∫
0
t
exp
⁡
C
∫
τ
t
α
τ
′
d
τ
′
g
B
p
,
r
s
d
τ
≲
v
0
B
p
,
r
s
+
∫
0
t
g
B
p
,
r
s
d
τ
.Since s
≠
2
+
1
/
p, for k
∈
N
¯, Lemma 7 tells us that(69)
w
k
t
B
p
,
r
s
-
1
≤
∫
0
t
exp
⁡
C
∫
τ
t
α
τ
′
d
τ
′
×
f
a
k
-
f
a
∞
∂
x
v
∞
B
p
,
r
s
-
1
d
τ
≲
∫
0
t
f
a
k
-
f
a
∞
∂
x
v
∞
B
p
,
r
s
-
1
d
τ
.Since s
>
1
+
1
/
p, using (A1) yields that(70)
f
a
k
-
f
a
∞
∂
x
v
∞
B
p
,
r
s
-
1
≲
a
k
B
p
,
r
s
-
1
+
a
∞
B
p
,
r
s
-
1
n
-
1
a
k
-
a
∞
B
p
,
r
s
-
1
v
∞
B
p
,
r
s
≲
1
+
a
k
B
p
,
r
s
-
1
+
a
∞
B
p
,
r
s
-
1
n
a
k
-
a
∞
B
p
,
r
s
-
1
·
v
∞
B
p
,
r
s
≲
1
+
α
t
L
∞
a
k
-
a
∞
B
p
,
r
s
-
1
v
∞
B
p
,
r
s
.Therefore, we obtain the following estimate by gathering the above inequalities:(71)
w
k
t
B
p
,
r
s
-
1
≤
∫
0
t
exp
⁡
C
∫
τ
t
α
τ
′
d
τ
′
×
f
a
k
-
f
a
∞
∂
x
v
∞
B
p
,
r
s
-
1
d
τ
≲
v
0
B
p
,
r
s
+
∫
0
t
g
B
p
,
r
s
d
τ
∫
0
t
a
k
-
a
∞
B
p
,
r
s
-
1
d
τ
,which yields the desired result of convergence.
For the general casev
0
∈
B
p
,
r
s
-
1 and g
∈
L
1
0
,
T
;
B
p
,
r
s
-
1, we have(72)
w
k
B
p
,
r
s
-
1
≤
v
k
-
v
k
j
B
p
,
r
s
-
1
+
v
k
j
-
v
∞
j
B
p
,
r
s
-
1
+
v
∞
j
-
v
∞
B
p
,
r
s
-
1
,where v
k
j satisfies the “cut-off” equation: (73)
∂
t
v
k
j
+
f
a
k
∂
x
v
k
j
=
S
j
g
=
∑
i
=
-
1
j
-
1
Δ
i
g
,
v
k
j
0
=
S
j
v
0
=
∑
i
=
-
1
j
-
1
Δ
i
v
0
.For v
k
-
v
k
j, we see that v
k
-
v
k
j solves the equation: (74)
∂
t
u
+
f
a
k
∂
x
u
=
g
-
S
j
g
,
u
0
=
v
0
-
S
j
v
0
.By Lemma 7 and (A1) once again, it follows that(75)
v
k
-
v
k
j
B
p
,
r
s
-
1
≤
v
0
-
S
j
v
0
B
p
,
r
s
-
1
e
C
∫
0
t
α
τ
d
τ
+
∫
0
t
e
C
∫
τ
t
α
τ
′
d
τ
′
g
-
S
j
g
B
p
,
r
s
-
1
d
τ
≲
v
0
-
S
j
v
0
B
p
,
r
s
-
1
+
∫
0
t
g
-
S
j
g
B
p
,
r
s
-
1
d
τ
.Note that v
k
j
(
0
)
∈
B
p
,
r
s and S
j
g
∈
L
1
0
,
T
;
B
p
,
r
s. Thanks to (71), we have(76)
v
k
j
-
v
∞
j
t
B
p
,
r
s
-
1
≲
v
k
j
0
B
p
,
r
s
+
∫
0
t
S
j
g
B
p
,
r
s
d
τ
·
∫
0
t
a
k
-
a
∞
B
p
,
r
s
-
1
d
τ
.Similarly, we can obtain(77)
v
∞
-
v
∞
j
B
p
,
r
s
-
1
≲
v
0
-
S
j
v
0
B
p
,
r
s
-
1
+
∫
0
t
g
-
S
j
g
B
p
,
r
s
-
1
d
τ
.Inserting (75), (76), and (77) into (72), we have(78)
w
k
B
p
,
r
s
-
1
≲
v
0
-
S
j
v
0
B
p
,
r
s
-
1
+
∫
0
t
g
-
S
j
g
B
p
,
r
s
-
1
d
τ
+
v
k
j
0
B
p
,
r
s
+
∫
0
t
S
j
g
B
p
,
r
s
d
τ
·
∫
0
t
a
k
-
a
∞
B
p
,
r
s
-
1
d
τ
.
By the definition ofS
j, we have g
τ
-
S
j
g
τ
B
p
,
r
s
-
1
≲
g
τ
B
p
,
r
s
-
1
∈
L
1
(
0
,
T
)
. Using the Lebesgue dominated convergence theorem yields that the first two terms of the right-hand side of the above estimate may be arbitrary small for j large enough. Take j large enough; then fix j. Let k tend to infinity so that the last term of right-hand side tends to zero. We complete the proof of Lemma 11.Sets
>
s
0
≥
1
+
1
/
p, s
≠
2
+
1
/
p. Fix u
0
,
∞
∈
B
p
,
r
s and let u
0
,
k
∈
B
p
,
r
s be a sequence with u
0
,
k
→
u
0
,
∞. For simplicity, we denote I
=
[
0
,
T
] and we let u
∞ (resp., u
k) be the solution to (1) with initial data u
0
,
∞ (resp., u
0
,
k); we need to demonstrate that(79)
lim
k
→
∞
⁡
u
k
=
u
∞
in
C
I
;
B
p
,
r
s
.By Lemma10, we have u
k
→
u
∞ in C
(
I
;
B
p
,
r
s
-
1
). To prove (79), it suffices to prove that(80)
lim
k
→
∞
⁡
∂
x
u
k
=
∂
x
u
∞
in
C
I
;
B
p
,
r
s
-
1
.Set v
k
=
∂
x
u
k, k
∈
N
¯; then v
k satisfies(81)
∂
t
v
k
+
f
u
k
∂
x
v
k
=
-
f
′
u
k
∂
x
u
k
2
-
∂
x
F
u
k
=
g
k
,
v
k
0
,
x
=
∂
x
u
0
,
k
.Motivated by the work in [34], for k
∈
N
¯, we decompose v
k
=
z
k
+
h
k such that(82)
∂
t
z
k
+
f
u
k
∂
x
z
k
=
g
k
-
g
∞
,
z
k
0
,
x
=
∂
x
u
0
,
k
-
∂
x
u
0
,
∞
,
∂
t
h
k
+
f
u
k
∂
x
h
k
=
g
∞
,
h
k
0
,
x
=
∂
x
u
0
,
∞
.Since u
0
,
k
→
u
0
,
∞ in B
p
,
r
s, we have u
0
,
k
B
p
,
r
s
≲
u
0
,
∞
B
p
,
r
s
≲
1. Then (13) shows that(83)
u
k
B
p
,
r
s
+
u
∞
B
p
,
r
s
≲
2
u
0
,
k
B
p
,
r
s
+
2
u
0
,
∞
B
p
,
r
s
≲
1
.By the algebra property of B
p
,
r
s
-
1, the assumptions (A1), (A2), and (83), we have (84)
g
k
-
g
∞
B
p
,
r
s
-
1
=
f
′
u
∞
∂
x
u
∞
2
-
f
′
u
k
∂
x
u
k
2
+
∂
x
F
u
∞
-
∂
x
F
u
k
B
p
,
r
s
-
1
≲
∂
x
u
∞
-
∂
x
u
k
B
p
,
r
s
-
1
.Since s
>
1
+
1
/
p, s
≠
2
+
1
/
p, by Lemma 7, we arrive at (85)
z
k
t
B
p
,
r
s
-
1
≲
z
0
B
p
,
r
s
-
1
e
C
∫
0
t
f
u
k
τ
B
p
,
r
s
d
τ
+
∫
0
t
e
C
∫
τ
t
f
u
k
τ
′
B
p
,
r
s
d
τ
′
g
k
-
g
∞
B
p
,
r
s
-
1
d
τ
≲
∂
x
u
0
,
∞
-
∂
x
u
0
,
k
B
p
,
r
s
-
1
+
∫
0
t
∂
x
u
∞
-
∂
x
u
k
B
p
,
r
s
-
1
d
τ
.By (83) and the convergence u
k
→
u
∞ in C
(
I
;
B
p
,
r
s
-
1
), we can get h
k
→
h
∞
=
v
∞
=
∂
x
u
∞ in C
(
I
;
B
p
,
r
s
-
1
) by applying Lemma 11. That is to say, there is a K
>
0 such that(86)
k
>
K
⟹
h
k
-
h
∞
C
I
;
B
p
,
r
s
-
1
<
ɛ
.Therefore, whenk
>
K, we have(87)
∂
x
u
∞
-
∂
x
u
k
B
p
,
r
s
-
1
=
v
k
-
v
∞
B
p
,
r
s
-
1
≤
z
k
t
B
p
,
r
s
-
1
+
h
k
t
B
p
,
r
s
-
1
≲
∂
x
u
0
,
∞
-
∂
x
u
0
,
k
B
p
,
r
s
-
1
+
∫
0
t
∂
x
u
∞
-
∂
x
u
k
B
p
,
r
s
-
1
d
τ
+
ɛ
.By using the Gronwall’s inequality, we have (88)
∂
x
u
∞
-
∂
x
u
k
C
I
;
B
p
,
r
s
-
1
≲
ɛ
C
T
,
u
0
,
∞
a
s
k
>
K
,where C
(
T
,
u
0
,
∞
) is a enough big constant depending on T and u
0
,
∞. Hence we obtain (80).


## 3.7. Hölder Continuity of the Solution Map with Weak Topology

Basically, direct from Lemma10, (14) is proved. Hereto, we complete the proof of Theorem 1.


## 4. Proof of Proposition3

In this section, we will show that whenf
(
u
)
=
u, F
(
u
)
=
u
2, the solution map defined by the problem (1)-(2) is not uniformly continuous from any bounded subset of B
2
,
r
s into E
~
2
,
r
s
(
T
) for any T
>
0 with s
>
3
/
2, 1
≤
r
≤
∞. Actually, in this case, s
0
=
3
/
2 and f
(
·
), F
(
u
) satisfy the assumptions (A1)–(A3) with n
=
1.To prove Proposition3, we need to show that, for any given T
>
0, there are two sequences of solutions u
k and v
k which satisfy (17)–(19) when t
∈
[
0
,
T
].We will use the following estimates for trigonometric functions cosine and sine.Lemma 12 (see [43, 51]).
Letσ
,
α
∈
R. If λ
∈
Z
+ and λ
≫
1, then(89)
sin
⁡
λ
x
-
α
B
2
,
r
σ
T
=
cos
⁡
λ
x
-
α
B
2
,
r
σ
T
=
2
1
/
r
π
1
+
λ
2
σ
/
2
≈
λ
σ
,
r
<
∞
,
π
1
+
λ
2
σ
/
2
≈
λ
σ
,
r
=
∞
.From now on, we assume that1
/
2
<
σ
<
min
⁡
{
s
-
1,3
/
2
}, f
(
u
)
=
u, and F
(
u
)
=
u
2.


### 4.1. Approximate Solutions and Actual Solutions

Following the approach in [32], we choose approximate solutions as(90)
u
ω
,
k
t
,
x
=
ω
k
-
1
+
k
-
s
cos
⁡
θ
,
where
θ
=
k
x
-
ω
t
,
k
∈
Z
+
,
ω
=
0,1
.Substituting these functions into (8) gives rise to the error which is defined to be (91)
E
=
∂
t
u
ω
,
k
+
u
ω
,
k
∂
x
u
ω
,
k
+
u
ω
,
k
2
.We begin by estimating the B
2
,
r
σ norm of E.Lemma 13.
Fork
≫
1, 1
≤
r
≤
∞, and 1
/
2
<
σ
<
m
i
n
{
s
-
1,3
/
2
} and σ
+
2
>
s, then (92)
E
B
2
,
r
σ
≲
k
r
s
,where r
s
<
0 and(93)
r
s
=
-
2
s
+
1
+
σ
<
0
,
i
f
3
2
<
s
≤
3
+
σ
2
,
-
2
,
i
f
s
>
3
+
σ
2
.Proof.
Firstly, we have(94)
E
B
2
,
r
σ
≤
∂
t
u
ω
,
k
+
u
ω
,
k
∂
x
u
ω
,
k
B
2
,
r
σ
+
u
ω
,
k
2
B
2
,
r
σ
.As the u
ω
,
k are explicitly given, direct computation shows that(95)
∂
t
u
ω
,
k
+
u
ω
,
k
∂
x
u
ω
,
k
B
2
,
r
σ
≲
-
1
2
k
-
2
s
+
1
sin
⁡
2
k
x
-
2
ω
t
B
2
,
r
σ
≲
k
-
2
s
+
1
+
σUse Lemma 12 to get(96)
u
ω
,
k
2
B
2
,
r
σ
≲
k
-
2
.Putting our estimates for both terms of (94) together, it follows that(97)
E
B
2
,
r
σ
≲
max
⁡
k
-
2
s
+
1
+
σ
,
k
-
2
≲
k
r
s
,which completes the estimate.When1
≤
r
≤
∞, Lemma 12 shows that for k
≫
1 and any t
⩾
0
(98)
u
ω
,
k
t
,
x
B
2
,
r
s
=
ω
k
-
1
+
k
-
s
cos
⁡
k
x
-
ω
t
B
2
,
r
s
≲
1
.Therefore, according to Theorem 1, the following periodic boundary value problem(99)
∂
t
u
w
,
k
+
u
w
,
k
∂
x
u
w
,
k
=
-
F
u
w
,
k
,
x
∈
T
,
t
>
0
,
u
0
,
x
=
u
ω
,
k
0
,
x
=
ω
k
-
1
+
k
-
s
cos
⁡
k
x
∈
B
2
,
r
shas a unique solution u
ω
,
k
(
t
,
x
) in E
~
2
,
r
s
(
T
~
), where T
~
>
0 can be chosen independent of k
≫
1 by using (98) and(100)
T
~
=
1
2
C
s
,
r
u
ω
,
k
0
B
2
,
r
s
≳
1
.Let T
1
=
m
i
n
{
T
~
,
T
}
. Then, u
ω
,
k
∈
E
~
2
,
r
s
(
T
1
) and (13) turn out to be(101)
u
ω
,
k
t
,
x
B
2
,
r
s
≲
u
ω
,
k
0
,
x
B
2
,
r
s
≲
1
,
0
≤
t
≤
T
1
.Letv
=
u
ω
,
k
-
u
ω
,
k. Then, v satisfies the initial value problem:(102)
∂
t
v
+
u
w
,
k
∂
x
v
=
E
-
v
∂
x
u
w
,
k
-
F
u
ω
,
k
-
F
u
ω
,
k
,
x
∈
T
,
t
∈
R
+
,
v
0
,
x
=
0
,
x
∈
T
.


### 4.2. Estimating the Difference between Approximate and True Solutions

We now calculate theB
2
,
r
σ norm and B
2
,
r
2
s
-
σ norm of v for 1
/
2
<
σ
<
m
i
n
{
s
-
1,3
/
2
}.Lemma 14.
Letk
≫
1, 1
/
2
<
σ
<
m
i
n
{
s
-
1,3
/
2
}, σ
+
2
>
s, and r
s be given in Lemma 13; then(103)
v
t
B
2
,
r
σ
≲
k
r
s
,
v
t
B
2
,
r
2
s
-
σ
≲
k
s
-
σ
,
f
o
r
0
≤
t
≤
T
1
.Proof.
Rewrite the equation in (102) as(104)
∂
t
v
+
u
w
,
k
∂
x
v
=
E
-
v
∂
x
u
w
,
k
-
F
u
ω
,
k
-
F
u
ω
,
k
.
By Lemma7 and v
(
0
,
x
)
=
0, we have(105)
v
t
B
2
,
r
σ
≤
C
∫
0
t
exp
⁡
C
∫
τ
t
∂
x
u
ω
,
n
τ
′
B
2
,
r
σ
-
1
d
τ
′
×
F
′
B
2
,
r
σ
d
τ
,where F
′
=
E
-
v
∂
x
u
w
,
k
-
F
u
ω
,
k
-
F
u
ω
,
k. Note that B
2
,
r
σ is a Banach algebra; from (98), (101), we obtain(106)
v
t
B
2
,
r
σ
≲
∫
0
t
v
B
2
,
r
σ
d
τ
+
∫
0
t
E
B
2
,
r
σ
d
τ
.Applying the Gronwall’s inequality and Lemma 13 yields(107)
v
t
B
2
,
r
σ
≲
e
C
s
,
r
T
∫
0
t
k
r
s
d
s
≲
k
r
s
.
To estimate theB
2
,
r
2
s
-
σ norm of v, we first estimate the B
2
,
r
2
s
-
σ norm of u
w
,
k
(
t
). Applying the property (2) in Lemma 7, the embedding B
2
,
r
s
-
1
↪
L
∞, and (101), it follows that(108)
u
w
,
k
B
2
,
r
2
s
-
σ
-
u
w
,
k
0
B
2
,
r
2
s
-
σ
≤
C
∫
0
t
F
u
w
,
k
B
2
,
r
2
s
-
σ
+
∂
u
w
,
k
L
∞
u
w
,
k
B
2
,
r
2
s
-
σ
d
τ
≤
C
∫
0
t
u
w
,
k
L
∞
u
w
,
k
B
2
,
r
2
s
-
σ
+
u
w
,
k
B
2
,
r
s
u
w
,
k
B
2
,
r
2
s
-
σ
d
τ
≤
C
∫
0
t
u
w
,
k
B
2
,
r
2
s
-
σ
+
u
w
,
k
B
2
,
r
s
u
w
,
k
B
2
,
r
2
s
-
σ
d
τ
≲
∫
0
t
u
w
,
k
B
2
,
r
2
s
-
σ
d
τ
.From the above inequality and Lemma 12, we have(109)
u
w
,
k
t
B
2
,
r
2
s
-
σ
≤
u
w
,
k
0
B
2
,
r
2
s
-
σ
C
s
,
r
T
e
C
s
,
r
T
≲
k
s
-
σ
.Thanks to (109), we have(110)
v
t
B
2
,
r
2
s
-
σ
≤
u
ω
,
k
t
B
2
,
r
2
s
-
σ
+
u
ω
,
k
t
B
2
,
r
2
s
-
σ
≲
ω
k
-
1
+
k
-
s
c
o
s
k
x
-
ω
t
B
2
,
r
2
s
-
σ
+
u
ω
,
k
t
B
2
,
r
2
s
-
σ
≲
k
s
-
σ
.Combining these results gives rise to the desired estimates.


### 4.3. Proof for Proposition3

It suffices to show thatu
0
,
k and u
1
,
k are two sequences of solutions satisfying the three conditions (17), (18), and (19) for t
∈
[
0
,
T
1
].By the construction ofu
ω
,
k, we see that (101) implies (17), and(111)
u
0
,
k
0
-
u
1
,
k
0
B
2
,
r
s
=
u
0
,
k
0
-
u
1
,
k
0
B
2
,
r
s
=
k
-
1
⟶
0
,
w
h
e
n
k
⟶
∞gives (18). For (19), first we have that, for t
∈
[
0
,
T
1
],(112)
liminf
k
→
∞
⁡
u
0
,
k
t
-
u
1
,
k
t
B
2
,
r
s
≥
liminf
k
→
∞
⁡
u
0
,
k
t
-
u
1
,
k
t
B
2
,
r
s
-
lim
k
→
∞
⁡
u
0
,
k
t
-
u
0
,
k
t
B
2
,
r
s
-
lim
k
→
∞
⁡
u
1
,
k
t
-
u
1
,
k
t
B
2
,
r
s
.By Lemma 14 and interpolation inequality, we obtain(113)
v
B
2
,
r
s
≤
v
B
2
,
r
σ
1
/
2
v
B
2
,
r
2
s
-
σ
1
/
2
≲
k
r
s
·
1
/
2
k
s
-
σ
·
1
/
2
=
k
r
s
′
,where r
s
′ can be simplified as(114)
r
s
′
=
r
s
·
1
2
+
s
-
σ
·
1
2
=
-
s
+
1
2
,
i
f
3
2
<
s
≤
3
+
σ
2
,
3
+
σ
2
2
2
-
2
+
s
-
σ
2
,
i
f
s
>
3
+
σ
2
.As r
s
′, we can deduce that(115)
lim
k
→
∞
⁡
u
0
,
k
t
-
u
0
,
k
t
B
2
,
r
s
=
lim
k
→
∞
⁡
u
1
,
k
t
-
u
1
,
k
t
B
2
,
r
s
=
0
.Therefore, using Lemma 12, for t
∈
[
0
,
T
1
], we have(116)
liminf
k
→
∞
⁡
u
0
,
k
t
-
u
1
,
k
t
B
2
,
r
s
≳
liminf
k
→
∞
⁡
u
0
,
k
t
-
u
1
,
k
t
B
2
,
r
s
≥
liminf
k
→
∞
⁡
2
k
-
s
sin
⁡
k
x
-
t
2
sin
⁡
t
2
B
2
,
r
s
-
limsup
k
→
∞
⁡
k
-
1
B
2
,
r
s
=
liminf
k
→
∞
⁡
2
k
-
s
sin
⁡
k
x
-
t
2
B
2
,
r
s
sin
⁡
t
2
≳
sin
⁡
t
2
.We complete the proof for Proposition3.


## 4.1. Approximate Solutions and Actual Solutions

Following the approach in [32], we choose approximate solutions as(90)
u
ω
,
k
t
,
x
=
ω
k
-
1
+
k
-
s
cos
⁡
θ
,
where
θ
=
k
x
-
ω
t
,
k
∈
Z
+
,
ω
=
0,1
.Substituting these functions into (8) gives rise to the error which is defined to be (91)
E
=
∂
t
u
ω
,
k
+
u
ω
,
k
∂
x
u
ω
,
k
+
u
ω
,
k
2
.We begin by estimating the B
2
,
r
σ norm of E.Lemma 13.
Fork
≫
1, 1
≤
r
≤
∞, and 1
/
2
<
σ
<
m
i
n
{
s
-
1,3
/
2
} and σ
+
2
>
s, then (92)
E
B
2
,
r
σ
≲
k
r
s
,where r
s
<
0 and(93)
r
s
=
-
2
s
+
1
+
σ
<
0
,
i
f
3
2
<
s
≤
3
+
σ
2
,
-
2
,
i
f
s
>
3
+
σ
2
.Proof.
Firstly, we have(94)
E
B
2
,
r
σ
≤
∂
t
u
ω
,
k
+
u
ω
,
k
∂
x
u
ω
,
k
B
2
,
r
σ
+
u
ω
,
k
2
B
2
,
r
σ
.As the u
ω
,
k are explicitly given, direct computation shows that(95)
∂
t
u
ω
,
k
+
u
ω
,
k
∂
x
u
ω
,
k
B
2
,
r
σ
≲
-
1
2
k
-
2
s
+
1
sin
⁡
2
k
x
-
2
ω
t
B
2
,
r
σ
≲
k
-
2
s
+
1
+
σUse Lemma 12 to get(96)
u
ω
,
k
2
B
2
,
r
σ
≲
k
-
2
.Putting our estimates for both terms of (94) together, it follows that(97)
E
B
2
,
r
σ
≲
max
⁡
k
-
2
s
+
1
+
σ
,
k
-
2
≲
k
r
s
,which completes the estimate.When1
≤
r
≤
∞, Lemma 12 shows that for k
≫
1 and any t
⩾
0
(98)
u
ω
,
k
t
,
x
B
2
,
r
s
=
ω
k
-
1
+
k
-
s
cos
⁡
k
x
-
ω
t
B
2
,
r
s
≲
1
.Therefore, according to Theorem 1, the following periodic boundary value problem(99)
∂
t
u
w
,
k
+
u
w
,
k
∂
x
u
w
,
k
=
-
F
u
w
,
k
,
x
∈
T
,
t
>
0
,
u
0
,
x
=
u
ω
,
k
0
,
x
=
ω
k
-
1
+
k
-
s
cos
⁡
k
x
∈
B
2
,
r
shas a unique solution u
ω
,
k
(
t
,
x
) in E
~
2
,
r
s
(
T
~
), where T
~
>
0 can be chosen independent of k
≫
1 by using (98) and(100)
T
~
=
1
2
C
s
,
r
u
ω
,
k
0
B
2
,
r
s
≳
1
.Let T
1
=
m
i
n
{
T
~
,
T
}
. Then, u
ω
,
k
∈
E
~
2
,
r
s
(
T
1
) and (13) turn out to be(101)
u
ω
,
k
t
,
x
B
2
,
r
s
≲
u
ω
,
k
0
,
x
B
2
,
r
s
≲
1
,
0
≤
t
≤
T
1
.Letv
=
u
ω
,
k
-
u
ω
,
k. Then, v satisfies the initial value problem:(102)
∂
t
v
+
u
w
,
k
∂
x
v
=
E
-
v
∂
x
u
w
,
k
-
F
u
ω
,
k
-
F
u
ω
,
k
,
x
∈
T
,
t
∈
R
+
,
v
0
,
x
=
0
,
x
∈
T
.


## 4.2. Estimating the Difference between Approximate and True Solutions

We now calculate theB
2
,
r
σ norm and B
2
,
r
2
s
-
σ norm of v for 1
/
2
<
σ
<
m
i
n
{
s
-
1,3
/
2
}.Lemma 14.
Letk
≫
1, 1
/
2
<
σ
<
m
i
n
{
s
-
1,3
/
2
}, σ
+
2
>
s, and r
s be given in Lemma 13; then(103)
v
t
B
2
,
r
σ
≲
k
r
s
,
v
t
B
2
,
r
2
s
-
σ
≲
k
s
-
σ
,
f
o
r
0
≤
t
≤
T
1
.Proof.
Rewrite the equation in (102) as(104)
∂
t
v
+
u
w
,
k
∂
x
v
=
E
-
v
∂
x
u
w
,
k
-
F
u
ω
,
k
-
F
u
ω
,
k
.
By Lemma7 and v
(
0
,
x
)
=
0, we have(105)
v
t
B
2
,
r
σ
≤
C
∫
0
t
exp
⁡
C
∫
τ
t
∂
x
u
ω
,
n
τ
′
B
2
,
r
σ
-
1
d
τ
′
×
F
′
B
2
,
r
σ
d
τ
,where F
′
=
E
-
v
∂
x
u
w
,
k
-
F
u
ω
,
k
-
F
u
ω
,
k. Note that B
2
,
r
σ is a Banach algebra; from (98), (101), we obtain(106)
v
t
B
2
,
r
σ
≲
∫
0
t
v
B
2
,
r
σ
d
τ
+
∫
0
t
E
B
2
,
r
σ
d
τ
.Applying the Gronwall’s inequality and Lemma 13 yields(107)
v
t
B
2
,
r
σ
≲
e
C
s
,
r
T
∫
0
t
k
r
s
d
s
≲
k
r
s
.
To estimate theB
2
,
r
2
s
-
σ norm of v, we first estimate the B
2
,
r
2
s
-
σ norm of u
w
,
k
(
t
). Applying the property (2) in Lemma 7, the embedding B
2
,
r
s
-
1
↪
L
∞, and (101), it follows that(108)
u
w
,
k
B
2
,
r
2
s
-
σ
-
u
w
,
k
0
B
2
,
r
2
s
-
σ
≤
C
∫
0
t
F
u
w
,
k
B
2
,
r
2
s
-
σ
+
∂
u
w
,
k
L
∞
u
w
,
k
B
2
,
r
2
s
-
σ
d
τ
≤
C
∫
0
t
u
w
,
k
L
∞
u
w
,
k
B
2
,
r
2
s
-
σ
+
u
w
,
k
B
2
,
r
s
u
w
,
k
B
2
,
r
2
s
-
σ
d
τ
≤
C
∫
0
t
u
w
,
k
B
2
,
r
2
s
-
σ
+
u
w
,
k
B
2
,
r
s
u
w
,
k
B
2
,
r
2
s
-
σ
d
τ
≲
∫
0
t
u
w
,
k
B
2
,
r
2
s
-
σ
d
τ
.From the above inequality and Lemma 12, we have(109)
u
w
,
k
t
B
2
,
r
2
s
-
σ
≤
u
w
,
k
0
B
2
,
r
2
s
-
σ
C
s
,
r
T
e
C
s
,
r
T
≲
k
s
-
σ
.Thanks to (109), we have(110)
v
t
B
2
,
r
2
s
-
σ
≤
u
ω
,
k
t
B
2
,
r
2
s
-
σ
+
u
ω
,
k
t
B
2
,
r
2
s
-
σ
≲
ω
k
-
1
+
k
-
s
c
o
s
k
x
-
ω
t
B
2
,
r
2
s
-
σ
+
u
ω
,
k
t
B
2
,
r
2
s
-
σ
≲
k
s
-
σ
.Combining these results gives rise to the desired estimates.


## 4.3. Proof for Proposition3

It suffices to show thatu
0
,
k and u
1
,
k are two sequences of solutions satisfying the three conditions (17), (18), and (19) for t
∈
[
0
,
T
1
].By the construction ofu
ω
,
k, we see that (101) implies (17), and(111)
u
0
,
k
0
-
u
1
,
k
0
B
2
,
r
s
=
u
0
,
k
0
-
u
1
,
k
0
B
2
,
r
s
=
k
-
1
⟶
0
,
w
h
e
n
k
⟶
∞gives (18). For (19), first we have that, for t
∈
[
0
,
T
1
],(112)
liminf
k
→
∞
⁡
u
0
,
k
t
-
u
1
,
k
t
B
2
,
r
s
≥
liminf
k
→
∞
⁡
u
0
,
k
t
-
u
1
,
k
t
B
2
,
r
s
-
lim
k
→
∞
⁡
u
0
,
k
t
-
u
0
,
k
t
B
2
,
r
s
-
lim
k
→
∞
⁡
u
1
,
k
t
-
u
1
,
k
t
B
2
,
r
s
.By Lemma 14 and interpolation inequality, we obtain(113)
v
B
2
,
r
s
≤
v
B
2
,
r
σ
1
/
2
v
B
2
,
r
2
s
-
σ
1
/
2
≲
k
r
s
·
1
/
2
k
s
-
σ
·
1
/
2
=
k
r
s
′
,where r
s
′ can be simplified as(114)
r
s
′
=
r
s
·
1
2
+
s
-
σ
·
1
2
=
-
s
+
1
2
,
i
f
3
2
<
s
≤
3
+
σ
2
,
3
+
σ
2
2
2
-
2
+
s
-
σ
2
,
i
f
s
>
3
+
σ
2
.As r
s
′, we can deduce that(115)
lim
k
→
∞
⁡
u
0
,
k
t
-
u
0
,
k
t
B
2
,
r
s
=
lim
k
→
∞
⁡
u
1
,
k
t
-
u
1
,
k
t
B
2
,
r
s
=
0
.Therefore, using Lemma 12, for t
∈
[
0
,
T
1
], we have(116)
liminf
k
→
∞
⁡
u
0
,
k
t
-
u
1
,
k
t
B
2
,
r
s
≳
liminf
k
→
∞
⁡
u
0
,
k
t
-
u
1
,
k
t
B
2
,
r
s
≥
liminf
k
→
∞
⁡
2
k
-
s
sin
⁡
k
x
-
t
2
sin
⁡
t
2
B
2
,
r
s
-
limsup
k
→
∞
⁡
k
-
1
B
2
,
r
s
=
liminf
k
→
∞
⁡
2
k
-
s
sin
⁡
k
x
-
t
2
B
2
,
r
s
sin
⁡
t
2
≳
sin
⁡
t
2
.We complete the proof for Proposition3.



---

*Source: 862593-2015-06-22.xml*