1. Introduction
Let
(
t
,
x
)
∈
[
0
,
∞
)
×
ℝ
n
,
α
>
0
,
a
,
b
∈
ℂ
, and let
Δ
be the Laplace operator. We consider the following Cauchy problem:
(1)
a
u
t
t
(
t
)
(
x
)
+
2
b
u
t
(
t
,
x
)
+
(
-
Δ
)
α
u
(
t
,
x
)
=
0
,
with initial conditions
(2)
u
(
0
,
x
)
=
f
(
x
)
,
u
t
(
0
,
x
)
=
g
(
x
)
.
Here, as usual, the fractional Laplacian
(
-
Δ
)
α
is defined through the Fourier transform:
(3)
(
-
Δ
)
α
f
^
(
ξ
)
=
|
ξ
|
2
α
f
^
(
ξ
)
for all test functions
f
. The partial differential equation in (1) is significantly interesting in mathematics, physics, biology, and many scientific fields. It is the wave equation when
a
=
1
,
b
=
0
, and
α
=
1
and it is the half wave equation when
a
=
0
,
2
b
=
i
, and
α
=
1
/
2
. As known, the wave equation is one of the most fundamental equations in physics. Another fundamental equation in physics is the Schrödinger equation which can be deduced from (1) by letting
a
=
0
,
2
b
=
i
, and
α
=
1
. The Schrödinger equation plays a remarkable role in the study of quantum mechanics and many other fields in physics. Also, (1) is the heat equation when
a
=
0
,
b
=
1
/
2
, and
α
=
1
.
As we all know, wave equation, Schrödinger equation, heat equation, and Laplace equations are most important and fundamental types of partial differential equations. The researches on these equations and their related topics are well-mature and very rich and they are still quite active and robust research fields in modern mathematics. The reader is readily to find hundreds and thousands of interesting papers by searching the Google Scholar or checking the MathSciNet in AMS. Here we list only a few of them that are related to this research paper [1–23].
With an extra damping term
2
b
u
t
(
t
,
x
)
in the wave equation, one obtains the damped wave equation
(4)
u
t
t
(
t
,
x
)
+
2
b
u
t
(
t
,
x
)
-
Δ
u
(
t
,
x
)
=
0
,
b
>
0
.
We observe that there are also a lot of research articles in the literature addressing the above damped wave equation. Among numerous research papers we refer to [24–35] and the references therein. From the reference papers, we find that the damped wave equation (4) is well studied in many interesting topics such as the local and global well-posedness of some linear, semilinear, and nonlinear Cauchy problems and asymptotic and regularity estimates of the solution. We observe that the space frames of these studies focus on the Lebesgue spaces and the Lebesgue Sobolev spaces.
These observations motivate us to consider the Cauchy problem of a more general fractional damped wave equation:
(5)
u
t
t
(
t
,
x
)
+
2
b
u
t
(
t
,
x
)
+
(
-
Δ
)
α
u
(
t
,
x
)
=
0
,
u
(
0
,
x
)
=
f
(
x
)
,
u
t
(
0
,
x
)
=
g
(
x
)
,
where
α
,
b
>
0
are fixed constants. According to our best knowledge, the fractional damped wave equation was not studied in the literature, except the wave case
α
=
1
. So our plan is to first study the linear equation (5) and to prove some
L
p
→
L
q
estimates. In our later works, we will use those estimates to study the well-posedness of certain nonlinear equations. We can easily check that the solution of (5) is formally given by
(6)
u
b
(
f
,
g
)
(
t
,
x
)
=
{
e
-
b
t
cosh
(
t
L
)
f
+
e
-
b
t
sinh
(
t
L
)
L
(
b
f
+
g
)
}
,
where
L
is the Fourier multiplier with symbol
b
2
-
|
ξ
|
2
α
(see Appendix). Thus our interest will focus on the operators
(7)
T
α
,
b
(
t
)
:
=
e
-
b
t
cosh
(
t
L
)
,
S
α
,
b
(
t
)
:
=
e
-
b
t
sinh
(
t
L
)
L
.
Using dilation, we will restrict ourselves to the case
b
=
1
so the theorems are all stated for
u
(
f
,
g
)
=
u
1
(
f
,
g
)
(see Remark 6). We now denote
(8)
S
α
(
t
)
=
S
α
,
1
(
t
)
=
e
-
t
sinh
(
t
1
-
(
-
Δ
)
α
)
1
-
(
-
Δ
)
α
,
T
α
(
t
)
=
T
α
,
1
(
t
)
=
e
-
t
cosh
(
t
1
-
(
-
Δ
)
α
)
.
These two operators are both convolution. We denote their kernels by
Ω
α
(
t
)
and
K
α
(
t
)
. Thus, we may write
(9)
T
α
(
t
)
f
=
K
α
(
t
)
*
f
,
S
α
(
t
)
f
=
Ω
α
(
t
)
*
f
.
To state our main results, we need the following definition of admissible triplet.
Definition 1.
A triplet
(
p
,
q
,
r
)
is called
σ
-admissible if
(10)
1
q
≤
σ
(
1
r
-
1
p
)
,
where
0
<
r
≤
p
≤
+
∞
,
r
<
q
<
∞
, and
σ
>
0
.
The following theorems are part of the main results in the paper.
Theorem 2.
Let
α
>
0
and let
(
p
,
q
,
r
)
be
n
/
2
α
-admissible and
1
≤
p
≤
+
∞
. Then for any
β
>
n
α
|
1
/
p
-
1
/
2
|
, one has
(11)
(
∫
0
∞
∥
K
α
(
t
)
*
f
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
+
∥
f
∥
L
˙
β
p
(
ℝ
n
)
,
(
∫
0
∞
∥
Ω
α
(
t
)
*
f
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
+
∥
f
∥
L
˙
β
-
α
p
(
ℝ
n
)
.
Here,
L
˙
γ
p
(
ℝ
n
)
denotes the homogeneous Sobolev
L
p
space with order
γ
, and
H
r
denotes the real Hardy space.
Theorem 3.
Let
α
=
1
,
(
p
,
q
,
r
)
be
n
/
2
-admissible and
1
≤
p
≤
+
∞
. Then the damped wave operators satisfy
(12)
(
∫
0
∞
∥
K
1
(
t
)
*
f
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
+
∥
f
∥
L
˙
β
p
(
ℝ
n
)
,
(
∫
0
∞
∥
Ω
1
(
t
)
*
f
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
+
∥
f
∥
L
˙
β
-
1
p
(
ℝ
n
)
,
for any
β
>
(
n
-
1
)
|
1
/
p
-
1
/
2
|
.
By the above theorems, we easily obtain the following space-time estimates on the solution
u
(
t
,
x
)
.
Theorem 4.
Let
α
>
0
and let
(
p
,
q
,
r
)
be
n
/
2
α
-admissible and
1
≤
p
≤
+
∞
. For the solution
u
(
t
,
x
)
of (5), one has
(13)
(
∫
0
∞
∥
u
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
+
∥
f
∥
L
˙
β
p
(
ℝ
n
)
+
∥
f
∥
L
˙
β
-
α
p
+
∥
g
∥
H
r
(
ℝ
n
)
+
∥
g
∥
L
˙
β
-
α
p
(
ℝ
n
)
.
Theorem 5.
Let
α
=
1
,
(
p
,
q
,
r
)
be
n
/
2
-admissible and
1
≤
p
≤
+
∞
. The solution
u
(
t
,
x
)
of the damped wave equation satisfies
(14)
(
∫
0
∞
∥
u
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
+
∥
f
∥
L
˙
β
p
(
ℝ
n
)
+
∥
f
∥
L
˙
β
-
1
p
+
∥
g
∥
H
r
(
ℝ
n
)
+
∥
g
∥
L
˙
β
-
1
p
(
ℝ
n
)
,
for any
β
>
(
n
-
1
)
|
1
/
p
-
1
/
2
|
.
Remark 6.
For (5) with general
b
>
0
, it is not hard to see that
(15)
T
a
,
b
(
t
)
f
(
x
)
=
T
a
,
1
(
s
)
f
b
(
b
1
/
α
x
)
,
S
a
,
b
(
t
)
f
(
x
)
=
S
a
,
1
(
s
)
(
1
b
f
b
)
(
b
1
/
α
x
)
,
where
s
=
b
t
and
f
b
(
x
)
=
f
(
b
-
1
/
α
x
)
. Therefore,
(16)
u
b
(
f
,
g
)
(
t
,
x
)
=
u
(
f
b
,
1
b
g
b
)
(
b
t
,
b
1
/
α
x
)
and by applying Theorem 4, we have
(17)
(
∫
0
∞
∥
u
b
∥
L
p
q
d
t
)
1
/
q
⪯
b
(
n
/
α
)
(
1
/
r
-
1
/
p
)
-
1
/
q
(
∥
f
∥
H
r
+
b
-
1
∥
g
∥
H
r
)
+
b
-
β
/
α
-
1
/
q
(
∥
f
∥
L
˙
β
p
+
b
∥
f
∥
L
˙
β
-
α
p
+
∥
g
∥
L
˙
β
-
α
p
)
.
For
α
=
1
, we have a similar result using Theorem 5.
In the statement of these theorems, the notation
A
⪯
B
means that there is a constant
C
>
0
independent of all essential variables such that
A
≤
C
B
. Also, throughout this paper, we use the notation
A
≃
B
to mean that there exist positive constants
C
and
c
, independent of all essential variables such that
(18)
c
B
≤
A
≤
C
B
.
It is easy to see that, by the linearity, we only need to prove Theorems 2 and 3. To this end, we will carefully study the kernels
(19)
K
α
(
t
)
(
x
)
=
e
-
t
∫
ℝ
n
cosh
(
t
1
-
|
ξ
|
2
α
)
e
i
〈
x
,
ξ
〉
d
ξ
,
Ω
α
(
t
)
(
x
)
=
e
-
t
∫
ℝ
n
sinh
(
t
1
-
|
ξ
|
2
α
)
1
-
|
ξ
|
2
α
e
i
〈
x
,
ξ
〉
d
ξ
.
Using the linearization
(20)
(
1
+
s
)
1
/
2
≃
1
+
s
2
,
for small
|
ξ
|
, we have
(21)
cosh
(
t
1
-
|
ξ
|
2
α
)
≃
cosh
(
t
(
1
-
|
ξ
|
2
α
2
)
)
=
e
t
(
1
-
|
ξ
|
2
α
/
2
)
+
e
-
t
(
1
-
|
ξ
|
2
α
/
2
)
2
.
Thus for small
|
ξ
|
,
(22)
e
-
t
cosh
(
t
1
-
|
ξ
|
2
α
)
≃
e
-
t
e
t
(
1
-
|
ξ
|
2
α
/
2
)
+
e
-
t
e
-
t
(
1
-
|
ξ
|
2
α
/
2
)
2
≃
e
-
t
|
ξ
|
2
α
/
2
.
This indicates that, for
|
ξ
|
near zero,
T
α
behaves like the fractional heat operator (see [11, 29, 30, 36, 37]).
For large
|
ξ
|
, we similarly have
(23)
cosh
(
t
1
-
|
ξ
|
2
α
)
=
cosh
(
i
t
|
ξ
|
α
(
1
-
|
ξ
|
-
2
α
)
)
≃
e
i
t
|
ξ
|
α
-
e
-
i
t
|
ξ
|
α
2
.
This indicates that as
|
ξ
|
near
∞
,
e
t
T
α
behaves like the wave operator if
α
=
1
and like the Schrödinger operator if
α
=
2
; see [12, 16, 38, 39].
In the same manner, the operator
S
α
(
t
)
behaves the same as the operator
T
α
. Based on these facts, we will estimate the kernels in their low frequencies, median frequencies, and high frequencies, separately, by using different methods. We will estimate the kernels in Section 2 and complete the proofs of main theorems in Section 3. Finally, in Section 4, we will study the almost everywhere convergence for the solution
u
(
t
,
x
)
as
t
→
0
+
. The similar convergence theorem for Schrödinger operator
e
i
t
Δ
f
(
x
)
has been widely studied; see [3, 40–44].
3. Proof of Theorems 2 and 3
Proof of Theorem 2.
Recalling the definition of
(113)
K
α
,
0
(
t
)
,
K
α
,
m
(
t
)
,
K
α
,
∞
(
t
)
,
Ω
α
,
0
(
t
)
,
Ω
α
,
m
(
t
)
,
Ω
α
,
∞
(
t
)
in Section 2, we have
(114)
|
K
α
(
t
)
*
f
|
⪯
|
K
α
,
0
(
t
)
*
f
|
+
|
K
α
,
m
(
t
)
*
f
|
+
|
K
α
,
∞
(
t
)
*
f
|
,
|
Ω
α
(
t
)
*
f
|
⪯
|
Ω
α
,
0
(
t
)
*
f
|
+
|
K
α
,
m
(
t
)
*
f
|
+
|
Ω
α
,
∞
(
t
)
*
f
|
.
By the triangle inequality and Propositions 8, 10, and 11, we only have to verify that, for any
n
/
2
α
-admissible triplet
(
p
,
q
,
r
)
,
(115)
(
∫
0
∞
∥
K
α
,
0
(
t
)
*
f
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
,
(
∫
0
∞
∥
Ω
α
,
0
(
t
)
*
f
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
.
These two inequalities are obviously true if
(116)
1
q
<
n
2
α
(
1
r
-
1
p
)
.
For
1
/
q
=
(
n
/
2
α
)
(
1
/
r
-
1
/
p
)
, denote
(117)
F
(
t
)
f
=
∥
K
α
,
0
(
t
)
*
f
∥
L
p
(
ℝ
n
)
.
By Proposition 8, we have
(118)
F
(
t
)
f
⪯
(
1
+
t
)
-
1
/
q
∥
f
∥
H
r
.
This indicates that, for any
λ
>
0
, there exists a positive constant
C
independent of
λ
and
f
such that
(119)
|
{
t
:
|
F
(
t
)
f
|
>
λ
}
|
≤
|
{
t
:
C
t
-
1
/
q
∥
f
∥
H
r
>
λ
}
|
≤
(
C
∥
f
∥
H
r
λ
)
q
.
This shows that
K
α
,
0
(
t
)
is a bounded mapping from
H
r
(
ℝ
n
)
to the mixed norm space
L
q
,
∞
(
[
0
,
∞
]
,
L
p
(
ℝ
n
)
)
for any admissible triplet
(
p
,
q
,
r
)
. Now we choose admissible triplets
(
p
,
q
1
,
r
1
)
and
(
p
,
q
2
,
r
2
)
satisfying
(120)
r
1
<
r
<
r
2
<
∞
,
q
1
<
q
<
q
2
<
∞
,
1
q
=
θ
q
1
+
1
-
θ
q
2
,
1
r
=
θ
r
1
+
1
-
θ
r
2
.
Then by the Marcinkiewicz interpolation, we easily obtain
(121)
(
∫
0
∞
∥
K
α
,
0
(
t
)
*
f
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
.
Similarly we can show that, for any
n
/
2
α
-admissible triplet
(
p
,
q
,
r
)
,
(122)
(
∫
0
∞
∥
Ω
α
,
0
(
t
)
*
f
∥
L
p
(
ℝ
n
)
q
d
t
)
1
/
q
⪯
∥
f
∥
H
r
(
ℝ
n
)
.
Proof of Theorem 3.
By checking the above proof, we only need to show the following proposition.
Proposition 12.
There is a
δ
p
>
0
for which if
β
>
(
n
-
1
)
|
1
/
p
-
1
/
2
|
, then
(123)
∥
K
1
,
∞
*
f
∥
L
p
(
ℝ
n
)
⪯
(
1
+
t
)
δ
p
e
-
t
∥
f
∥
L
˙
β
p
(
ℝ
n
)
,
∥
Ω
1
,
∞
*
f
∥
L
p
(
ℝ
n
)
⪯
(
1
+
t
)
δ
p
e
-
t
∥
f
∥
L
˙
β
-
1
p
(
ℝ
n
)
hold for all
1
≤
p
≤
∞
.
Proof.
Let
(124)
W
β
(
t
)
f
(
x
)
=
e
-
t
∫
ℝ
n
e
i
t
|
ξ
|
2
-
1
ϕ
3
(
ξ
)
|
ξ
|
β
f
^
(
ξ
)
e
i
〈
x
,
ξ
〉
d
ξ
,
where
ϕ
3
is defined in Section 2.3 (corresponding to
α
=
1
). We will prove, for any
β
>
(
n
-
1
)
/
2
, that
(125)
∥
W
β
(
t
)
f
(
x
)
∥
L
1
⪯
e
-
t
(
1
+
t
)
λ
∥
f
∥
L
1
,
with some
λ
>
0
. Then by repeating the complex interpolation argument in the proof of Proposition 11, with (81) replaced by (125), we finish the proof of the proposition.
Next we turn to the proof of (125). Denote the kernel of
W
β
(
t
)
by
(126)
Ⓢ
β
(
x
,
t
)
=
e
-
t
∫
ℝ
n
e
i
t
|
ξ
|
2
-
1
ϕ
3
(
ξ
)
|
ξ
|
β
e
i
〈
x
,
ξ
〉
d
ξ
.
By Young's inequality, it suffices to show that if
β
>
(
n
-
1
)
/
2
, then
(127)
∥
Ⓢ
β
(
x
,
t
)
∥
L
1
⪯
(
1
+
t
)
(
n
+
1
)
/
2
e
-
t
.
Let
Φ
be the cutoff function defined in Section 2.3. Then we have
(128)
Ⓢ
β
(
x
,
t
)
=
e
-
t
∑
k
=
6
∞
∫
ℝ
n
e
i
t
|
ξ
|
2
-
1
|
ξ
|
β
Φ
(
2
-
k
|
ξ
|
)
ϕ
3
(
|
ξ
|
)
e
i
〈
ξ
,
x
〉
d
ξ
=
e
-
t
∑
k
=
6
∞
Y
k
(
x
)
,
where, by [49, Ch. 4],
(129)
Y
k
(
x
)
=
∫
ℝ
n
e
i
t
|
ξ
|
2
-
1
|
ξ
|
β
Φ
(
2
-
k
|
ξ
|
)
ϕ
3
(
|
ξ
|
)
e
i
〈
ξ
,
x
〉
d
ξ
≃
∫
0
∞
e
i
t
r
2
-
1
r
β
ϕ
3
(
r
)
Φ
(
2
-
k
r
)
V
(
n
-
2
)
/
2
(
r
|
x
|
)
r
n
-
1
d
r
.
In the last integral,
(130)
V
ν
(
s
)
=
J
ν
(
s
)
s
ν
,
and
J
ν
(
s
)
is the Bessel function of order
ν
.
So, by the Minkowski inequality,
(131)
∥
Ⓢ
β
(
·
,
t
)
∥
L
1
≤
e
-
t
∑
k
=
6
∞
∥
Y
k
∥
L
1
.
First, we assume
t
≥
1
. Changing variables, we have
(132)
Y
k
(
x
)
=
|
x
|
(
2
-
n
)
/
2
2
k
(
β
-
n
/
2
-
1
)
∫
0
∞
e
i
t
2
2
k
r
2
-
1
Φ
(
r
)
ϕ
3
(
2
k
r
)
J
(
n
-
2
)
/
2
x
x
x
x
x
x
x
x
x
x
x
x
x
×
(
2
k
r
|
x
|
)
r
n
/
2
-
β
-
1
/
2
d
r
.
Using the Taylor expansion with integral remainder, for
r
∈
supp
(
ϕ
3
)
, we write
(133)
r
2
-
1
=
r
+
g
(
r
)
,
where
(134)
g
(
1
r
)
=
-
1
2
r
∫
0
1
(
1
-
s
r
2
)
-
1
/
2
d
s
.
This gives
(135)
e
i
t
2
2
k
r
2
-
1
=
e
i
t
2
k
r
e
i
t
g
(
1
/
2
k
r
)
,
for
k
≥
6
and
1
/
2
≤
r
≤
2
. By the definition of
g
it is easy to see that if we denote
h
(
r
)
=
g
(
1
/
2
k
r
)
, then
(136)
|
h
(
m
)
(
r
)
|
⪯
2
-
k
.
Also, for any integer
m
≥
0
,
(137)
|
d
m
d
r
m
e
i
t
g
(
2
k
r
)
|
⪯
1
if
t
2
-
k
≤
1
,
|
d
m
d
r
m
e
i
t
g
(
2
k
r
)
|
⪯
(
t
2
-
k
)
m
if
t
2
-
k
>
1
uniformly for
k
≥
10
and
1
/
2
≤
r
≤
2
.
When
(138)
2
k
|
x
|
≤
2
4
,
using the known estimate
(139)
J
(
n
-
2
)
/
2
(
r
)
=
O
(
r
(
n
-
2
)
/
2
)
,
as
r
⟶
0
,
it is easy to see
(140)
|
Y
k
(
x
)
|
⪯
|
x
|
(
2
-
n
)
/
2
2
-
k
(
β
-
n
/
2
-
1
)
(
2
k
|
x
|
)
(
n
-
2
)
/
2
=
2
-
k
(
β
-
n
)
.
Thus,
(141)
e
-
t
∑
k
=
6
∞
∥
Y
k
χ
{
|
x
|
≤
2
-
k
+
4
}
∥
L
1
⪯
e
-
t
∑
k
=
6
∞
2
-
k
(
β
-
n
)
∫
|
x
|
≤
2
-
k
+
4
d
x
⪯
e
-
t
.
When
(142)
2
k
|
x
|
>
2
4
,
we use the asymptotic expansion of
J
(
n
-
2
)
/
2
(
r
)
: for any integer
N
≥
0
,
(143)
J
ν
(
r
)
≃
e
±
i
r
(
∑
j
=
0
N
c
j
r
j
+
1
/
2
)
+
O
(
r
-
(
N
+
1
)
-
1
/
2
)
,
where
c
1
,
c
2
,
…
,
c
N
are constants.
In this case,
(144)
Y
k
(
x
)
≃
∑
j
=
0
N
c
j
|
x
|
(
1
-
n
)
/
2
-
j
2
k
(
β
-
n
/
2
-
1
/
2
+
j
)
×
∫
0
∞
e
i
2
k
r
(
t
±
|
x
|
)
e
i
t
g
(
2
k
r
)
Φ
(
r
)
ϕ
3
(
2
k
r
)
r
n
/
2
-
β
-
j
d
r
+
O
(
|
x
|
(
-
n
+
1
)
/
2
-
N
-
1
2
-
k
(
β
-
n
/
2
+
1
/
2
+
N
)
)
=
∑
j
=
0
N
c
j
Y
k
,
j
(
x
)
+
O
(
|
x
|
(
-
n
+
1
)
/
2
-
N
-
1
2
-
k
(
β
-
n
/
2
+
1
/
2
+
N
)
)
,
where, without loss of generality, we denote
(145)
Y
k
,
j
(
x
)
=
|
x
|
(
1
-
n
)
/
2
-
j
2
k
(
β
-
n
/
2
-
1
/
2
+
j
)
×
∫
0
∞
e
i
2
k
r
(
t
-
|
x
|
)
e
i
t
g
(
2
k
r
)
Φ
(
r
)
ϕ
3
(
2
k
r
)
r
n
/
2
-
β
-
j
d
r
.
It is easy to see that, for a suitable integer
N
,
(146)
e
-
t
∑
k
=
6
∞
2
-
k
(
β
-
n
/
2
+
1
/
2
+
N
)
∫
|
x
|
≥
2
-
k
+
4
|
x
|
(
-
n
+
1
)
/
2
-
N
-
1
d
x
⪯
e
-
t
.
Thus it remains to show that, for each
j
,
(147)
e
-
t
∑
k
=
6
∞
∥
Y
k
,
j
χ
{
|
x
|
>
2
-
k
+
4
}
∥
L
1
⪯
t
(
n
+
1
)
/
2
e
-
t
.
Since the estimates of all
Y
k
,
j
are similar, we will only show
(148)
e
-
t
∑
k
=
6
∞
∥
Y
k
,
0
χ
{
|
x
|
>
2
-
k
+
4
}
∥
L
1
⪯
t
(
n
+
1
)
/
2
e
-
t
.
Using integration by parts and noting
ϕ
3
(
2
k
r
)
≡
1
if
k
>
2
8
and
r
∈
supp
(
Φ
)
, it is easy to check that one has
(149)
|
∫
0
∞
e
i
2
k
r
(
t
-
|
x
|
)
e
i
t
g
(
2
k
r
)
Φ
(
r
)
Ψ
(
2
k
r
)
r
n
/
2
-
β
-
j
d
r
|
⪯
min
{
1
,
2
-
m
k
|
t
-
|
x
|
|
-
m
(
2
-
k
t
)
m
}
if
2
-
k
t
>
1
, for any positive integer
m
, and
(150)
|
∫
0
∞
e
i
2
k
r
(
t
±
|
x
|
)
e
i
t
g
(
2
k
r
)
Φ
(
r
)
ϕ
3
(
2
k
r
)
r
n
/
2
-
β
-
j
d
t
|
⪯
min
{
1
,
2
-
μ
k
|
t
-
|
x
|
|
-
μ
}
if
2
-
k
t
≤
1
, for any positive integer
μ
. Thus, we have the following lemma.
Lemma 13.
Let
2
k
|
x
|
≥
10
. For any
m
≥
0
, one has
(151)
|
Y
k
,
0
(
x
)
|
⪯
|
x
|
(
1
-
n
)
/
2
2
-
k
(
β
-
n
/
2
-
1
/
2
)
2
-
m
k
|
t
-
|
x
|
|
-
m
if
2
-
k
t
≤
1
.
Also, for any
μ
≥
0
,
(152)
|
Y
k
,
0
(
x
)
|
⪯
|
x
|
(
1
-
n
)
/
2
2
-
k
(
β
-
n
/
2
-
1
/
2
)
2
-
μ
k
|
t
-
|
x
|
|
-
μ
(
2
-
k
t
)
μ
if
2
-
k
t
>
1
.
Now we continue the proof of the proposition. Write
(153)
∑
k
=
6
∞
∥
Y
k
,
0
χ
{
|
x
|
>
2
-
k
+
4
}
∥
L
1
(
ℝ
n
)
⪯
∑
k
=
6
∞
∫
10
/
2
k
<
|
x
|
≤
t
/
2
|
Y
k
,
0
(
x
)
|
d
x
+
∑
k
=
6
∞
∫
|
x
|
>
100
t
|
Y
k
,
0
(
x
)
|
d
x
+
∑
k
=
6
∞
∫
t
/
2
<
|
x
|
≤
100
t
|
Y
k
,
0
(
x
)
|
d
x
=
A
1
+
A
2
+
A
3
.
In
A
1
, noting
β
-
(
n
-
1
)
/
2
>
0
, we use the lemma with
μ
=
1
/
2
and
m
=
1
:
(154)
A
1
⪯
∑
k
=
6
log
t
2
-
k
(
β
-
n
/
2
+
1
/
2
)
∫
10
/
2
k
<
|
x
|
≤
t
/
2
|
x
|
(
-
n
+
1
)
/
2
d
x
+
t
-
1
∑
k
=
log
t
+
1
∞
2
-
k
(
β
-
n
/
2
+
1
/
2
)
∫
10
/
2
k
<
|
x
|
≤
t
/
2
|
x
|
(
-
n
+
1
)
/
2
d
x
⪯
t
(
n
+
1
)
/
2
.
Similarly, in Lemma 13 we let
μ
=
m
=
n
:
(155)
A
2
⪯
∑
k
=
6
log
t
2
-
k
(
β
-
n
/
2
-
1
/
2
)
t
n
∫
100
t
<
|
x
|
|
x
|
(
1
-
n
)
/
2
-
n
2
-
2
n
k
d
x
+
∑
k
=
log
t
+
1
∞
2
-
k
(
β
-
n
/
2
-
1
/
2
)
∫
100
t
<
|
x
|
2
-
n
k
|
x
|
(
1
-
n
)
/
2
-
n
d
x
⪯
t
(
n
+
1
)
/
2
.
Let
(156)
E
=
{
x
∈
ℝ
n
:
t
2
<
|
x
|
≤
100
t
}
,
E
k
=
{
x
∈
ℝ
n
:
|
t
-
|
x
|
|
<
2
-
k
}
.
Using Lemma 13, we write
(157)
A
3
⪯
∑
k
=
6
log
t
2
-
k
(
β
-
n
/
2
-
1
/
2
)
×
∫
E
∖
E
k
|
x
|
(
1
-
n
)
/
2
2
-
μ
k
|
t
-
|
x
|
|
-
μ
(
2
-
k
t
)
μ
d
x
+
∑
k
=
log
t
+
1
∞
2
-
k
(
β
-
n
/
2
-
1
/
2
)
×
∫
E
∖
E
k
|
x
|
(
1
-
n
)
/
2
2
-
m
k
|
t
-
|
x
|
|
-
m
d
x
+
∑
k
=
6
∞
2
-
k
(
β
-
n
/
2
-
1
/
2
)
∫
E
k
|
x
|
(
1
-
n
)
/
2
d
x
=
B
1
+
B
2
+
B
3
.
Here, the last term
(158)
B
3
=
∑
k
=
6
∞
2
-
k
(
β
-
n
/
2
-
1
/
2
)
∫
E
k
|
x
|
1
-
n
2
d
x
⪯
∑
k
=
6
∞
2
-
k
(
β
-
n
/
2
-
1
/
2
)
∫
t
-
2
-
k
t
+
2
-
k
r
(
1
-
n
)
/
2
+
n
-
1
⪯
t
(
n
+
1
)
/
2
.
Use the polar coordinate and Lemma 13 for
μ
=
1
/
2
:
(159)
B
1
=
∑
k
=
6
log
t
2
-
k
(
β
-
n
/
2
+
1
/
2
)
t
1
/
2
∫
E
∖
E
k
|
x
|
(
1
-
n
)
/
2
|
t
-
|
x
|
|
-
1
/
2
d
x
⪯
∑
k
=
6
log
t
2
-
k
(
β
-
n
/
2
+
1
/
2
)
t
1
/
2
(
∫
t
/
2
t
-
2
-
k
∫
t
+
2
-
k
100
t
r
(
n
-
1
)
/
2
(
r
-
t
)
-
1
/
2
d
r
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
+
∫
t
/
2
t
-
2
-
k
r
(
n
-
1
)
/
2
(
r
-
t
)
-
1
/
2
d
r
)
⪯
t
n
/
2
∑
k
=
6
log
t
2
-
k
(
β
-
n
/
2
+
1
/
2
)
(
∫
t
/
2
t
-
2
-
k
∫
t
+
2
-
k
100
t
(
r
-
t
)
-
1
/
2
d
r
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
+
∫
t
/
2
t
-
2
-
k
(
r
-
t
)
-
1
/
2
d
r
)
⪯
t
(
n
+
1
)
/
2
.
Similarly, we can show
(160)
∑
k
=
log
t
+
1
∞
2
-
k
(
β
-
n
/
2
-
1
/
2
)
∫
E
∖
E
k
|
x
|
(
1
-
n
)
/
2
2
-
m
k
|
t
-
|
x
|
|
-
m
d
x
⪯
t
(
n
+
1
)
/
2
.
When
0
<
t
≤
1
, the proof is the same with only minor modifications.
4. Almost Everywhere Convergence
Next we will study the pointwise convergence of the solution
u
(
t
,
x
)
of (5) to the initial data. We will prove the following.
Theorem 14.
Let
s
>
1
/
2
. If
f
belongs to the inhomogeneous Sobolev space
L
s
2
(
ℝ
n
)
and
g
∈
L
s
-
α
2
(
ℝ
n
)
, then the solution
u
(
t
,
x
)
of (5) converges to
f
(
x
)
a.e.
x
∈
ℝ
n
as
t
→
0
+
.
To prove this theorem, we need Lemma 15 and Proposition 16.
Lemma 15 (see [50]).
Let
n
≥
2
and
1
<
d
<
n
. Then
(161)
∫
ℝ
n
|
∫
S
n
-
1
g
(
u
)
e
i
x
·
u
d
σ
(
u
)
|
2
d
x
|
x
|
d
⪯
∫
S
n
-
1
|
g
(
u
)
|
2
d
σ
(
u
)
.
Proposition 16.
Let
n
≥
2
and let
m
(
t
,
|
ξ
|
)
be defined on
ℝ
+
×
ℝ
n
and satisfy
(162)
|
(
1
+
|
ξ
|
)
γ
m
(
t
,
|
ξ
|
)
|
⪯
1
.
Denote the maximal function
(163)
m
*
f
(
x
)
=
sup
t
>
0
|
m
(
t
,
D
)
f
(
x
)
|
.
Then if
γ
>
0
, we have
(164)
∥
m
*
f
(
x
)
∥
L
2
(
|
x
|
-
d
d
x
)
⪯
∥
f
∥
L
˙
l
2
,
d
-
2
γ
2
<
l
<
d
2
,
d
>
1
.
If
γ
≤
0
, then
(165)
∥
m
*
f
(
x
)
∥
L
2
(
|
x
|
-
d
d
x
)
⪯
∥
f
∥
L
l
2
,
l
>
d
-
γ
-
1
2
,
d
>
1
.
Proof.
Making
t
into a function of
x
, we only have to bound
(166)
m
(
t
(
x
)
,
D
)
f
(
x
)
=
∫
ℝ
n
e
i
〈
x
,
ξ
〉
m
(
t
(
x
)
,
|
ξ
|
)
f
^
(
ξ
)
d
ξ
,
where
t
(
x
)
:
ℝ
n
→
ℝ
+
is any measurable function.
By the polar decomposition,
(167)
|
m
(
t
(
x
)
,
D
)
f
(
x
)
|
=
|
∫
0
∞
r
n
-
1
m
(
t
(
x
)
,
r
)
∫
S
n
-
1
f
^
(
r
ξ
′
)
e
i
〈
r
x
,
ξ
′
〉
d
σ
(
ξ
′
)
d
r
|
⪯
∫
0
∞
r
n
-
1
(
1
+
r
)
-
γ
|
∫
S
n
-
1
f
^
(
r
ξ
′
)
e
i
〈
r
x
,
ξ
′
〉
d
σ
(
ξ
′
)
|
d
r
.
By Minkowski's inequality, change of variables, and Lemma 15, we have
(168)
∥
m
(
t
(
x
)
,
D
)
f
(
x
)
∥
L
p
(
|
x
|
-
d
d
x
)
⪯
∫
0
∞
r
n
-
1
(
1
+
r
)
-
γ
×
∥
∫
S
n
-
1
f
^
(
r
ξ
′
)
e
i
〈
r
x
,
ξ
′
〉
d
σ
(
ξ
′
)
∥
L
2
(
d
x
/
|
x
|
d
)
d
r
=
∫
0
∞
r
n
/
2
+
d
/
2
-
1
(
1
+
r
)
-
γ
×
∥
∫
S
n
-
1
f
^
(
r
ξ
′
)
e
i
〈
x
,
ξ
′
〉
d
σ
(
ξ
′
)
∥
L
2
(
d
x
/
|
x
|
d
)
d
r
⪯
∫
0
∞
r
n
/
2
+
d
/
2
-
1
(
1
+
r
)
-
γ
(
∫
S
n
-
1
|
f
^
(
r
ξ
′
)
|
2
d
σ
(
ξ
′
)
)
1
/
2
d
r
=
(
∫
0
2
+
∫
2
∞
)
r
n
/
2
+
d
/
2
-
1
(
1
+
r
)
-
γ
×
(
∫
S
n
-
1
|
f
^
(
r
ξ
′
)
|
2
d
σ
(
ξ
′
)
)
1
/
2
d
r
=
L
1
+
L
2
.
When
γ
>
0
, we have
(169)
L
1
⪯
∫
0
2
r
n
/
2
+
d
/
2
-
1
(
∫
S
n
-
1
|
f
^
(
r
ξ
′
)
|
2
d
σ
(
ξ
′
)
)
1
/
2
d
r
⪯
(
∫
0
2
r
d
-
1
-
2
l
d
r
)
1
/
2
×
(
∫
0
2
r
2
l
r
n
-
1
∫
S
n
-
1
|
f
^
(
r
ξ
′
)
|
2
d
σ
(
ξ
′
)
d
r
)
1
/
2
⪯
(
∫
0
2
r
d
-
1
-
2
l
d
r
)
1
/
2
(
∫
|
ξ
|
≤
2
|
ξ
|
2
l
|
f
^
(
ξ
)
|
2
d
ξ
)
1
/
2
⪯
∥
f
∥
L
˙
l
2
(
ℝ
n
)
.
Here we have to let
(170)
d
-
1
-
2
l
<
-
1
.
On the other hand,
(171)
L
2
⪯
∫
2
∞
r
n
/
2
+
d
/
2
-
1
-
γ
(
∫
S
n
-
1
|
f
^
(
r
ξ
′
)
|
2
d
σ
(
ξ
′
)
)
1
/
2
d
r
⪯
(
∫
2
∞
r
d
-
1
-
2
l
-
2
γ
d
r
)
1
/
2
×
(
∫
2
∞
r
2
l
r
n
-
1
∫
S
n
-
1
|
f
^
(
r
ξ
′
)
|
2
d
σ
(
ξ
′
)
d
r
)
1
/
2
⪯
(
∫
2
∞
r
d
-
1
-
2
l
-
2
γ
d
r
)
1
/
2
(
∫
|
ξ
|
≥
2
|
ξ
|
2
l
|
f
^
(
ξ
)
|
2
d
ξ
)
1
/
2
⪯
∥
f
∥
L
˙
l
2
(
ℝ
n
)
.
Obviously we have to let
(172)
d
-
1
-
2
l
-
2
γ
>
-
1
,
which, together with (170), implies
(173)
d
-
2
α
2
<
l
<
d
2
.
If
γ
≤
0
, then
(174)
L
1
⪯
(
∫
0
2
r
d
-
1
d
r
)
1
/
2
×
(
∫
0
2
r
n
-
1
∫
S
n
-
1
|
f
^
(
r
ξ
′
)
|
2
d
σ
(
ξ
′
)
d
r
)
1
/
2
⪯
∥
f
∥
L
2
,
L
2
⪯
(
∫
2
∞
r
-
d
d
r
)
1
/
2
×
(
∫
2
∞
r
2
d
-
1
-
2
γ
r
n
-
1
∫
S
n
-
1
|
f
^
(
r
ξ
′
)
|
2
d
σ
(
ξ
′
)
d
r
)
1
/
2
⪯
∥
f
∥
L
d
-
γ
-
1
/
2
2
.
Proposition 16 is proved.
Proof of Theorem 14.
Denote
(175)
m
1
(
t
,
|
ξ
|
)
=
e
-
t
cosh
(
t
1
-
|
ξ
|
2
α
)
,
m
2
(
t
,
|
ξ
|
)
=
e
-
t
sinh
(
t
1
-
|
ξ
|
2
α
)
1
-
|
ξ
|
2
α
.
It is not hard to verify that
(176)
|
m
1
(
t
,
|
ξ
|
)
|
⪯
1
,
|
m
2
(
t
,
|
ξ
|
)
|
⪯
(
1
+
|
ξ
|
)
-
α
,
DDDDDDDDDDDDDDDD
∀
(
t
,
ξ
)
∈
ℝ
+
×
ℝ
n
.
Since
(177)
u
(
t
,
x
)
=
m
1
(
t
,
D
)
f
(
x
)
+
m
2
(
t
,
D
)
f
(
x
)
+
m
2
(
t
,
D
)
g
(
x
)
=
w
(
t
,
D
)
f
(
x
)
+
m
2
(
t
,
D
)
g
(
x
)
,
Theorem 14 will be proved if we can show, as
t
→
0
+
,
(178)
m
2
(
t
,
D
)
g
(
x
)
⟶
0
,
a
.
e
.
x
∈
ℝ
n
for
g
∈
L
s
-
α
2
(
ℝ
n
)
and
(179)
w
(
t
,
D
)
f
(
x
)
⟶
f
(
x
)
,
a
.
e
.
x
∈
ℝ
n
for
f
∈
L
s
2
(
ℝ
n
)
. The proof of the two limits is similar and we will only show the second convergence. Note that the above convergence always holds for Schwarz function
f
. So a further boundedness on the maximal function
(180)
w
*
f
(
x
)
=
sup
t
>
0
|
w
(
t
,
D
)
f
(
x
)
|
that
(181)
∥
w
*
f
(
x
)
∥
L
2
(
d
x
/
|
x
|
d
)
⪯
∥
f
∥
L
s
2
(
ℝ
n
)
,
s
>
1
2
,
is enough to imply Theorem 14.
Next we will prove (181). By (176) and Proposition 16,
(182)
∥
m
1
*
f
(
x
)
∥
L
2
(
d
x
/
|
x
|
d
)
⪯
∥
f
∥
L
l
2
(
ℝ
n
)
,
∀
l
>
d
-
1
2
.
Fix
s
>
1
/
2
. Taking
1
<
d
<
s
+
1
/
2
and
l
close to
d
-
1
/
2
, we have
l
<
s
and thus
(183)
∥
m
1
*
f
(
x
)
∥
L
2
(
d
x
/
|
x
|
d
)
⪯
∥
f
∥
L
s
2
(
ℝ
n
)
.
Applying Proposition 16 with
γ
=
α
and
1
<
d
<
1
+
2
α
, we have
(184)
∥
m
2
*
f
(
x
)
∥
L
2
(
d
x
/
|
x
|
d
)
⪯
∥
f
∥
L
˙
1
/
2
2
(
ℝ
n
)
≤
∥
f
∥
L
s
2
(
ℝ
n
)
.
Since
(185)
w
*
f
(
x
)
≤
m
1
*
f
(
x
)
+
m
2
*
f
(
x
)
,
we proved (181) when
n
≥
2
(note that Proposition 16 was proved only when
n
≥
2
).
For
n
=
1
, instead of (181), we will show
(186)
∥
w
*
f
(
x
)
∥
L
2
(
[
-
N
,
N
]
)
⪯
N
1
/
2
∥
f
∥
L
s
2
(
ℝ
)
,
s
>
1
2
,
which is also enough to obtain the pointwise convergence. Taking
h
(
x
)
∈
L
2
(
[
-
N
,
N
]
)
, we have
(187)
∫
-
N
N
w
(
t
(
x
)
,
D
)
f
(
x
)
h
(
x
)
d
x
=
∫
-
N
N
∫
R
w
(
t
(
x
)
,
ξ
)
f
^
(
ξ
)
e
i
〈
x
,
ξ
〉
d
ξ
h
(
x
)
d
x
=
∫
ℝ
f
^
(
ξ
)
∫
-
N
N
w
(
t
(
x
)
,
ξ
)
e
i
〈
x
,
ξ
〉
h
(
x
)
d
x
d
ξ
⪯
(
∫
ℝ
|
f
^
(
ξ
)
|
2
(
1
+
|
ξ
|
2
)
s
d
ξ
)
1
/
2
×
(
∫
ℝ
|
∫
-
N
N
w
(
t
(
x
)
,
ξ
)
e
i
〈
x
,
ξ
〉
h
(
x
)
d
x
|
2
(
1
+
|
ξ
|
2
)
s
d
ξ
)
1
/
2
.
Noting that
(188)
w
(
t
(
x
)
,
ξ
)
=
m
1
(
t
(
x
)
,
ξ
)
+
m
2
(
t
(
x
)
,
ξ
)
⪯
1
,
∀
(
x
,
ξ
)
,
we have
(189)
|
∫
-
N
N
w
(
t
(
x
)
,
ξ
)
e
i
〈
x
,
ξ
〉
h
(
x
)
d
x
|
2
⪯
N
∥
h
∥
L
2
(
[
-
N
,
N
]
)
2
.
Therefore
(190)
∫
-
N
N
w
(
t
(
x
)
,
D
)
f
(
x
)
h
(
x
)
d
x
⪯
N
1
/
2
∥
f
∥
L
s
2
·
∥
h
∥
L
2
(
[
-
N
,
N
]
)
and by duality
(191)
∥
w
(
t
(
x
)
,
D
)
f
(
x
)
∥
L
2
(
[
-
N
,
N
]
)
⪯
N
1
/
2
∥
f
∥
L
s
2
(
ℝ
)
,
s
>
1
2
,
from which (186) follows.