Dynamical Systems, Differential
Equations and Applications
AIMS Proceedings, 2015
doi:10.3934/proc.2015.1105
pp. 1105–1114
BLOW-UP OF SOLUTIONS TO SEMILINEAR WAVE EQUATIONS
WITH NON-ZERO INITIAL DATA
Kyouhei Wakasa
Department of Mathematics, Hokkaido University
Sapporo, 060-0810, Japan
Abstract. In this paper, we are concerned with the initial value problem for utt −∆u =
|ut |p in Rn × [0, ∞) with the initial data u(x, 0) = f (x), ut (x, 0) = g(x), where (f, g)
are slowly decaying.
H. Takamura [13] obtained the blow-up result for the case where f ≡ 0 and g 6≡ 0.
Our purpose in this paper is to show the blow-up result for the case where the both
initial data do not vanish identically.
1. Introduction. In this paper, we consider the following initial-value problem for semilinear wave equation:
utt − ∆u = F (ut )
in Rn × [0, ∞),
(1)
u(x, 0) = f (x), ut (x, 0) = g(x), for x ∈ Rn ,
where n ≥ 2 and u = u(x, t) is a scalar unknown function of space-time variables, and
F (ut ) = |ut |p with p > 1.
In the case where the initial data are compactly supported, we have the following conn+1
. Then, (1) has a global in time solution
jecture by R. T. Glassey [2]. Let pc (n) =
n−1
for “small” data if p > pc (n), and has no global in time solution for some “positive” data if
1 < p ≤ pc (n).
The blow-up part of this conjecture was first obtained by F. John [6] in the case of n = 3
and 1 < p ≤ pc (3). This result was extended to the case, n = 1 and n = 2 by K. Masuda
[8] for p = 2. Later, J. Schaeffer [11] obtained the blow-up result in the case of n = 2
and p = pc (2). R. Agemi [1] also proved in the case of n = 2 and 1 < p ≤ pc (2). For
higher dimensional case, n ≥ 4, M. A. Rammaha [9] proved the blow-up result in the case
of p = pc (n) for odd n and 1 < p < pc (n) for even n. Finally, Y. Zhou [15] obtained the
blow-up result in the case of 1 < p ≤ pc (n). On the other hand, the global existence part was
first proved by T. C. Sideris [12] in the case of n = 3 and p > pc (3) with radially symmetric
solution. Later, K. Hidano & K. Tsutaya [4] and N. Tzvetkov [14] independently proved the
global existence result without the assumption that the solution is radially symmetric in the
case of n = 2 and n = 3 for p > pc (n). Finally, K. Hidano & C. Wang & K. Yokoyama [5]
proved global existence result in the case of p > pc (n) and n ≥ 4 with radially symmetric
solution.
Next, we consider the case where the initial data are slowly decaying. In this case, one
cannot expect to get the global in time solution even for the case p > pc (n). When n = 2
and n = 3, H. Kubo [7] obtained the blow-up result for the initial data (f, g) satisfying
C
(2)
f ≡ 0, g(x) ≥
(1 + |x|)κ
2010 Mathematics Subject Classification. Primary: 35L70; Secondary: 35B44, 35E15.
Key words and phrases. Initial value problem, blow-up, semilinear wave equations, high dimensions,
point-wise estimates.
1105
1106
KYOUHEI WAKASA
with C > 0 and 0 < κ < κ0 , where we set
κ0 =
1
.
p−1
(3)
Later, H. Takamura [13] obtained the blow-up result under the condition (2) for n ≥ 2. On
the other hand, when n = 3, K. Hidano [3] showed the global existence result for radially
symmetric solution and “small” (f, g) satisfying

1

0

,
 |f (r)|, |g(r)| = O
rκ (4)
1

(j+1)
(j)

(r)|, |g (r)| = O
(j
=
1,
2)
as
r
→
∞,
 |f
r1+κ
with κ ≥ κ0 and p > pc (3), where r = |x|.
In view of the condition (2), we may state the question as follows: what about the blowup results in the case where both initial data do not vanish identically. In this paper, we
show the blow-up result under such conditions. Our method of the proof is based on an
iteration argument of point-wise estimates by Takamura[13]. However, due to the fact that
f does not vanish identically, we need to make a modification for the iteration scheme. The
main difficulty is to get the estimate of the first step of the iteration argument in even space
dimensions. Because, the structure of representation formula is more complex than in odd
dimensions. To overcome this difficulty, we use some technique which was introduced in
M. A. Rammaha & H. Takamura & H. Uesaka & K. Wakasa [10] who showed the blow-up
result for utt − ∆u = |u|p with non-zero initial data in high space dimensions. See Lemma
4.2 below.
2. Main theorem. For an unknown function u = u(r, t), r ∈ (0, ∞), t ∈ [0, ∞), we
consider the following radially symmetric version of (1):
utt −
n−1
ur − urr = F (ut )
r
in (0, ∞) × [0, ∞),
(5)
u(r, 0) = f (r), ut (r, 0) = g(r), for r ∈ (0, ∞).
In order to state our main results, we assume that the following assumptions on the
nonlinear term and initial data:
• Let F ∈ C 1 (R) satisfies
F (s) ≥ A|s|p for s ∈ R,
(6)
where p > 1 and A > 0.
• There exists a constant R > 0 such that f ∈ C 2 ((0, ∞)) and g ∈ C 1 ((0, ∞)) satisfy
f (r), g(r) > 0
for r ≥ R,
f is a monotone decreasing function,
C1
f 0 (r) ≥ −
for r ≥ R,
(1 + r)κ
f (r)
C2
−m(m − 1)
+ g(r) ≥
for r ≥ R,
r
(1 + r)κ
(7)
if n = 2m + 1 (m = 1, 2, 3, · · · ), where κ > 0, C1 and C2 are positive constants
satisfying
C2 > 23+κ C1 ,
(8)
BLOW-UP OF SOLUTIONS TO SEMILINEAR WAVE EQUATIONS
1107
and
f (r), g(r) > 0
for r ≥ R,
f (r)
C3
0
for r ≥ R,
Em
+ |f (r)| ≤
(9)
r
(1 + r)κ
f (r)
g(r)
C4
0
−Em
for r ≥ R,
− |f (r)| +
≥
r
4
(1 + r)κ
if n = 2m (m = 1, 2, 3, · · · ), where κ > 0, C3 and C4 are positive constants satisfying
C4 > π23+2κ (3/2)m−1/2 C3
(10)
and Em is defined by

1


if m = 1,
2
Em =
2

 m + 1 + 5ζm (m − 1)
if m = 2, 3, 4, · · · ,
8
3
where ζm > 0 is determined in Lemma 3.1 below.
(11)
Theorem 2.1. Assume (6), and (7), (8) if n = 2m + 1 (m = 1, 2, 3, · · · ), or (9), (10) if
n = 2m (m = 1, 2, 3, · · · ). Let u be a solution of (5). Then u cannot exist globally in time
provided,
0 < κ < κ0 ,
where κ0 is the one in (3).
Remark 1. In view of the condition (4) for the global existence result, our assumption (7)
and (9) are natural with the decay order of the initial data.
3. Preliminaries. First, we recall the representation formula for (5) which was obtained
by (6a) and (6b) in Rammaha [9].
Let u0 = u0 (r, t) be a solution of
(
n−1 0
u0tt −
ur − u0rr = 0
in (0, ∞) × [0, ∞)
(12)
r
u0 (r, 0) = f (r), u0t (r, 0) = g(r), for r ∈ (0, ∞).
When n = 2m + 1 (m = 1, 2, 3, · · · ), then u0 is represented by
u0 (r, t) =
∂ 1
1
I(r, t, u0 (·, 0)) + m I(r, t, u0t (·, 0)),
∂t 2rm
2r
where
r+t
λ2 + r2 − t2
2rλ
|r−t|
where Pk denotes Legendre polynomials of degree k defined by:
Z
I(r, t, w(·)) =
λm w(λ)Pm−1
(13)
dλ,
(14)
dk 2
(z − 1)k .
dz k
When n = 2m (m = 1, 2, 3, · · · ), then u0 is represented by
Pk (z) =
u0 (r, t) =
1
2k k!
∂
2
2
J(r, t, u(·, 0)) + m−1 J(r, t, ut (·, 0)),
m−1
∂t πr
πr
(15)
where
J(r, t, w(·))
Z t
Z r+ρ m
λ w(λ)Tm−1 (λ2 + r2 − ρ2 )/(2rλ) dλ
ρdρ
p
p
p
=
,
λ2 − (r − ρ)2 (r + ρ)2 − λ2
t2 − ρ2 |r−ρ|
0
where Tk denotes Tschebyscheff polynomials of degree k defined by:
Tk (z) =
(−1)k
dk
(1 − z 2 )1/2 k (1 − z 2 )k−(1/2) .
(2k − 1)!!
dz
(16)
1108
KYOUHEI WAKASA
Making use of Duhamel’s principle, the solution u = u(r, t) of (5) must satisfy
u(r, t) = u0 (r, t) + L(F (ut ))(r, t),
where
L(F (ut ))(r, t) =
1
2rm
Z
(17)
t
I(r, t − τ, F (ut )(·, τ )))dτ
(18)
0
if n = 2m + 1 (m = 1, 2, 3, · · · ),
L(F (ut ))(r, t) =
2
πrm−1
Z
t
J(r, t − τ, F (ut )(·, τ )))dτ
(19)
0
if n = 2m (m = 1, 2, 3, · · · ).
The following lemma gives us the positivity for Pm−1 (z), Tm−1 (z) near z = 1.
Lemma 3.1. For m = 2, 3, 4, · · · , there exist positive constant ηm and ζm depending only
on m such that
1
m(m − 1)
1
0
Pm−1 (z) ≥ , 0 < Pm−1
(z) ≤
for
≤ z ≤ 1.
(20)
2
2
1 + ηm
1
1
0
≤ Tm−1 (z) ≤ 1, 0 < Tm−1
(z) ≤ (m − 1)2 for
≤ z ≤ 1.
(21)
2
1 + ζm
For the proof, we refer the reader to Lemma 2.5 in [13] or Lemma 4.1 and Lemma 5.1 in
[10]. Finally, we define the following domain:
Σ = (r, t) ∈ (0, ∞)2 : r − t ≥ max {R, δt} > 0 ,
where
δ=

 0
n 2
2 o
,
 min
η m ζm
if m = 1
if m = 2, 3, 4 · · · .
4. Iteration Scheme. In this section, we introduce our iteration scheme that allows us to
prove Theorem 2.1.
Proposition 1. Let u be a solution of (5). Suppose that the assumptions in Theorem 2.1
are fulfilled. Then, there exists a positive constant C0 such that u satisfies
C0 t
u(r, t) − f (r) ≥
(1 +Zr + t)κ
(22)
r+t
A
+ m
λm (r + t − λ)1−p |u(λ, r + t − λ) − f (λ)|p dλ
8r
r
in Σ.
Lemma 4.1 (Takamura [13]). Let L be the integral in (18) and (19). Suppose that the
assumptions in Theorem 2.1 are fulfilled. Then,
L(F (ut ))(r, t)
Z r+t
A
≥ m
λm (r + t − λ)1−p |u(λ, r + t − λ) − f (λ)|p dλ
8r
r
(23)
holds in Σ.
Noticing that f (6≡ 0), we have inequality (23) by using (4.7) in [13].
Proposition 2. Let u0 be a solution of (12). Suppose that the assumptions in Theorem 2.1
are fulfilled. Then, u0 satisfies
C0 t
(24)
u0 (r, t) − f (r) ≥
(1 + r + t)κ
in Σ, where C0 is the one in Proposition 1.
BLOW-UP OF SOLUTIONS TO SEMILINEAR WAVE EQUATIONS
1109
Proof of Proposition 1. (17) yields
u(r, t) − f (r) = u0 (r, t) − f (r) + L(F (ut ))(r, t).
Making use of Lemma 4.1 and Proposition 2, we have (22) in Σ.
Proof of Theorem 2.1. One can find that the first term of the right-hand side of (22) is
similar to (3.4) in [13]. Hence, in order to show the proof of Theorem 2.1, it is sufficient to
show the estimates of the first step of the iteration argument, that is Proposition 2.
Proof of Proposition 2.
Case n = 2m + 1 (m = 1, 2, 3, · · · ).
First of all, we consider the case of m = 2, 3, 4, · · · . Then, (13) gives us
u0 (r, t) − f (r)
1
{f (r + t)(r + t)m + f (r − t)(r − t)m }
2rm Z r+t
2
λ + r 2 − t2
t
1
m
0
λ f (λ)Pm−1
−
dλ
+ m
2r Zr−t
rλ
2 2rλ
r+t
2
2
1
λ +r −t
+ m
dλ − f (r)
λm g(λ)Pm−1
2r
2rλ
r−t
=
in Σ. Note that the variable of Pm−1 is estimated by
(r − t)2 + r2 − t2
r−t
λ2 + r2 − t2
≥
=
2rλ
2r(r + t)
r+t
for r − t ≤ λ ≤ r + t
and the condition r − t ≥ (2/ηm )t is equivalent to (r − t)/(r + t) ≥ 1/(1 + ηm ). By the
positivity of (f, g) in (7) and Lemma 3.1, we obtain
1
{f (r + t)(r + t)m + f (r − t)(r − t)m }
2rm Z r+t
1
f (λ)
m
λ −m(m − 1)
+ g(λ) dλ − f (r)
+ m
4r
λ
r−t
u0 (r, t) − f (r) ≥
in Σ. Making use of monotonicity of f and decay condition in (7), we have
1
{f (r + t)(r + t)m + f (r − t)(r − t)m }
2rm
Z r+t
C2
λm
−f (r) + m
dλ
κ
4r
r−t (1 + λ)
m
f (r + t){(r + t) + (r − t)m }
≥
− f (r)
2rm
C2 t
+
4(1 + r + t)κ
u0 (r, t) − f (r) ≥
in Σ. It follows from the fact
(r + t)m + (r − t)m ≥ 2rm
for r, t ≥ 0.
that
u0 (r, t) − f (r) ≥ f (r + t) − f (r) +
C2 t
4(1 + r + t)κ
in Σ. By the mean value theorem, there exists ξ ∈ (r, r + t) such that
u0 (r, t) − f (r) ≥ tf 0 (ξ) +
C2 t
4(1 + r + t)κ
in Σ. Noticing that
r=
r
r
r+t
+ ≥
2 2
2
for r − t > 0
1110
KYOUHEI WAKASA
and the decay condition for f 0 in (7), we obtain
u0 (r, t) − f (r) ≥
≥
C1 t
C2 t
+
κ
(1 + r)
4(1 + r + t)κ
C2 t
C1 2κ t
+
−
κ
(1 + r + t)
4(1 + r + t)κ
−
in Σ. Therefore, (24) follows in the case of n = 2m + 1 (m = 2, 3, · · · ) by the assumption
for C1 and C2 in (8).
Next, we consider the case of m = 1 (n = 3). Then, (13) yields
u0 (r, t) − f (r)
1
{f (r + t)(r + t) + f (r − t)(r − t)}
2r Z r+t
1
λg(λ)dλ − f (r)
+
2r r−t
=
in Σ. Making use of the same argument above, we have
C2 t
2(1 + r + t)κ
C1 2κ t
C2 t
≥−
+
κ
(1 + r + t)
2(1 + r + t)κ
u0 (r, t) − f (r) ≥ f (r + t) − f (r) +
in Σ. By the assumption for C1 and C2 in (8), (24) follows in the case of n = 3.
Case n = 2m (m = 1, 2, 3, · · · ).
First of all, we consider the case of m = 2, 3, 4, · · · . Changing the variables ξ =
r+ρ−λ
2ρ
in the λ-integral of (16), we have
J(r, t, w(·))
Z
Z 1
1 t
ρdρ
(r + ρ − 2ρξ)m w(r + ρ − 2ρξ)
p
√
√
√ √
=
×
dρ
2 0
r + ρ − ρξ r − ξρ ξ 1 − ξ
t 2 − ρ2
0
×Tm−1 (Θ(r, ρ, ξ)) dξ
in Σ, where we set
Θ(r, ρ, ξ) =
(r + ρ − 2ρξ)2 + r2 − ρ2
.
2r(r + ρ − 2ρξ)
Making use of integration by parts in the ρ-integral of above and Tm−1 (1) = 1, one has
J(r, t, w(·))
Z
Z 1
1 t
∂ p2
(r + ρ − 2ρξ)m w(r + ρ − 2ρξ)
√
√
√ √
=
−
t − ρ2 dρ
2 0
∂ρ
r + ρ − ρξ r − ξρ ξ 1 − ξ
0
×Tm−1 (Θ(r, ρ, ξ)) dξ
Z
Z
tw(r)rm−1 1
dξ
1 tp 2
√ √
=
+
t − ρ2 dρ
2
2
ξ
1
−
ξ
0
0
Z 1
∂
dξ
×
{K(r, ρ, ξ)w(r + ρ − 2ρξ)Tm−1 (Θ(r, ρ, ξ))} √ √
∂ρ
ξ
1−ξ
0
in Σ, where we set
K(r, ρ, ξ) = √
(r + ρ − 2ρξ)m
√
.
r + ρ − ρξ r − ξρ
BLOW-UP OF SOLUTIONS TO SEMILINEAR WAVE EQUATIONS
1111
Hence,
∂
2
J(r, t, u(·, 0))
∂t πrm−1
Z t
1
tdρ
p
= f (r) + m−1
2 − ρ2
πr
t
0
Z 1
∂
dξ
×
{K(r, ρ, ξ)f (r + ρ − 2ρξ)Tm−1 (Θ(r, ρ, ξ))} √ √
ξ 1−ξ
0 ∂ρ
in Σ.
Making use of Proposition 5.2 in [10] by putting ρ = tη, one can obtain the following:
Lemma 4.2 (M.A.Rammaha & H.Takamura & H.Uesaka & K.Wakasa [10]). Let m =
2, 3, 4 · · · . Assume that w ∈ C 1 ((0, ∞)) and w(y) > 0 for y ≥ R, where R is as in (9).
Then
∂
{K(r, ρ, ξ)w(r + ρ − 2ρξ)Tm−1 (Θ(r, ρ, ξ))}
∂ρ
w(r + ρ − 2ρξ)
≥ − Em
+ |w0 (r + ρ − 2ρξ)| K(r, ρ, ξ)
(25)
r + ρ − 2ρξ
holds for 0 ≤ ρ ≤ t, 0 ≤ ξ ≤ 1 and (r, t) ∈ Σ, where Em is as in (11).
It follows from (15), Lemma 4.2, (21) and the assumption of positivity for g in (9) that
Z 1
Z t
1
tdρ
f (r + ρ − 2ρξ)
p
u0 (r, t) ≥ − m−1
Em
+ |f 0 (r + ρ − 2ρξ)|
πr
r + ρ − 2ρξ
t 2 − ρ2 0
0
K(r, ρ, ξ)dξ
× √ √
ξ 1 −Zξ
Z 1
t
1
ρdρ
K(r, ρ, ξ)g(r + ρ − 2ρξ)dξ
p
√ √
+
+ f (r)
2
2
2πrm−1 0
ξ 1−ξ
t −ρ 0
in Σ. Note that the second term of right-hand side above is estimated from below by
Z t
Z 1
1
tdρ
K(r, ρ, ξ)g(r + ρ − 2ρξ)dξ
p
√ √
≥
.
m−1
2
2
4πr
ξ 1−ξ
t −ρ 0
t/2
in Σ. Then we have
u0 (r, t) − f (r) ≥ J1 (r, t) + J2 (r, t)
in Σ,
where we set
J1 (r, t) = −
1
Z
t/2
πrm−1 0
K(r, ρ, ξ)dξ
× √ √
ξ 1−ξ
tdρ
p
t2 − ρ2
Z
1
Em
0
f (r + ρ − 2ρξ)
+ |f 0 (r + ρ − 2ρξ)|
r + ρ − 2ρξ
and
Z
t
tdρ
p
πrm−1 t/2 t2 − ρ2
Z 1
f (r + ρ − 2ρξ)
g(r + ρ − 2ρξ)
×
−Em
− |f 0 (r + ρ − 2ρξ)| +
r + ρ − 2ρξ
4
0
K(r, ρ, ξ)dξ
× √ √
.
ξ 1−ξ
J2 (r, t) =
1
First, we shall estimate J1 (r, t). By using the simple estimate
1 + r + ρ − 2ξρ ≥ 1 + r − ρ ≥ 1 + r −
t
1+r+t
≥
2
4
for 0 ≤ ξ ≤ 1, 0 ≤ ρ ≤ t/2, r > t,
1112
KYOUHEI WAKASA
and (9), we obtain
C3
J1 (r, t) ≥ − m−1
πr
t/2
Z
Z
tdρ
1
K(r, ρ, ξ)dξ
√
1
−
ξ
ξ(1 + r + ρ − 2ρξ)κ
−
0
Z0 t/2 Z 1
1+2κ
C3 2
K(r, ρ, ξ)dξ
√
√
≥ − 1/2 m−1
dρ
π3 r
(1 + r + t)κ 0
1−ξ ξ
0
p
t2
ρ2
√
in Σ. By noting:
t
3
≤ r
2
2
r + ρ − 2ρξ ≤ r +
for ξ ≥ 0, 0 ≤ ρ ≤
t
, r > t,
2
and
r − ξρ ≥ r − ρ ≥ r −
we obtain
3m
K(r, ρ, ξ) ≤
2m−1/2
t
r
≥
2
2
for 0 ≤ ξ ≤ 1, 0 ≤ ρ ≤ t/2, r > t,
rm−1
for 0 ≤ ξ ≤ 1, 0 ≤ ρ ≤
t
, r > t.
2
Hence,
m− 21
3
t
J1 (r, t) ≥ −C3 2
in Σ.
·
2
(1 + r + t)κ
Next, we shall estimate J2 (r, t). Making use of (9), we obtain
Z t
Z 1
C4
tdρ
K(r, ρ, ξ)dξ
√
p
√
J2 (r, t) ≥ m−1
2
2
πr
1
−
ξ
ξ(1 + r + ρ − 2ρξ)κ
t −ρ 0
t/2
Z t
Z 1/2
C4
1
K(r, ρ, ξ)dξ
√
√
≥ m−1 ·
dρ
πr
(1 + r + t)κ t/2
1−ξ ξ
0
2κ
(26)
in Σ. By making use of:
r + ρ − 2ρξ ≥ r
for 0 ≤ ξ ≤
1
, ρ ≥ 0,
2
and
r − ξρ ≤ r
for ρ, ξ ≥ 0 and r + ρ − ρξ ≤ r + t ≤ 2r for ξ ≥ 0, 0 ≤ ρ ≤ t, r > t,
we conclude
K(r, ρ, ξ) ≥ 2−1/2 rm−1
for 0 ≤ ξ ≤
1
, 0 ≤ ρ ≤ t, r > t.
2
Since,
1/2
Z
0
√
dξ
√
√ ≥ 2
1−ξ ξ
Z
√
1/2
dξ =
0
2
,
2
we have
J2 (r, t) ≥
C4
t
·
4π (1 + r + t)κ
in Σ.
(27)
Combining (26) and (27), yields
0
2κ
u (r, t) − f (r) ≥ −C3 2
m− 21
3
t
C4
t
·
+
·
2
(1 + r + t)κ
4π (1 + r + t)κ
(28)
in Σ. Therefore, (24) follows in the case of n = 2m (m = 2, 3, 4, · · · ) by the assumption for
C3 and C4 in (10).
Next, we consider the case of m = 1 (n = 2). Then, (16) implies
Z t
Z r+ρ
ρdρ
λw(λ)dλ
p
p
p
J(r, t, w(·)) =
2
2
2
λ − (r − ρ)2 (r + ρ)2 − λ2
t − ρ r−ρ
0
BLOW-UP OF SOLUTIONS TO SEMILINEAR WAVE EQUATIONS
1113
r+ρ−λ
in the λ-integral of above, we obtain
2ρ
Z 1
Z
H(r, ρ, ξ)w(r + ρ − 2ρξ)dξ
1 t
ρdρ
p
√ √
J(r, t, w(·)) =
2
2
2 0
ξ 1−ξ
t −ρ 0
in Σ. Changing the variables ξ =
in Σ, where we set
r + ρ − 2ρξ
√
.
r + ρ − ρξ r − ξρ
Making use of integration by parts, we have
H(r, ρ, ξ) = √
u0 (r, t)
Z 1
Z
∂
tdρ
1 t
dξ
p
= f (r) +
{H(r, ρ, ξ)f (r + ρ − 2ρξ)} √ √
2 − ρ2
π 0
ξ 1−ξ
t
0 ∂ρ
Z
Z 1
ρdρ
1 t
H(r, ρ, ξ)g(r + ρ − 2ρξ)dξ
p
√ √
+
2
2
π 0
ξ 1−ξ
t −ρ 0
in Σ. Analogously to the case of m = 2, 3, 4, · · · , we have
Z
Z 1
1 t
tdρ
f (r + ρ − 2ρξ)
p
u0 (r, t) ≥ −
+ |f 0 (r + ρ − 2ρξ)|
π 0
2(r + ρ − 2ρξ)
t 2 − ρ2 0
H(r, ρ, ξ)dξ
× √ √
Zξ 1 − ξ
Z 1
1 t
ρdρ
H(r, ρ, ξ)g(r + ρ − 2ρξ)dξ
p
√ √
+
+ f (r).
2
2
π 0
ξ 1−ξ
t −ρ 0
in Σ. Making use of the same argument above and assumption (9), we have the same
estimate as (28) by putting m = 1. Therefore, we obtain (24) in the case of n = 2. This
completes the proof of Proposition 2.
Acknowledgments. The author would like to thank the referees for his/her valuable comments which helped to improve the manuscript. The author is supported by Grant-in-Aid
for Science Research of JSPS Fellow No.26-2330.
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KYOUHEI WAKASA
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Received September 2014; revised December 2014.
E-mail address: [email protected]