Electronic Journal of Differential Equations, Vol. 2005(2005), No. 67, pp. 1–22.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
IN THE REISSNER-NORDSTRÖM METRIC
SVETLIN G. GEORGIEV
Abstract. In this paper, we study the solutions to the Cauchy problem
(utt − ∆u)gs + m2 u = f (u),
u(1, x) = u0 ∈
γ
Ḃp,p
(R3 ),
t ∈ (0, 1], x ∈ R3 ,
γ−1
ut (1, x) = u1 ∈ Ḃp,p
(R3 ),
where gs is the Reissner-Nordströ m metric; p > 1, γ ∈ (0, 1), m 6= 0 are
constants, f ∈ C 2 (R1 ), f (0) = 0, 2m2 |u| ≤ f (l) (u) ≤ 3m2 |u|, l = 0, 1. More
precisely we prove that the Cauchy problem has unique nontrivial solution in
γ
C((0, 1]Ḃp,p
(R+ )),
(
v(t)ω(r) for t ∈ (0, 1], r ≤ r1
u(t, r) =
0
for t ∈ (0, 1], r ≥ r1 ,
where r = |x|, and limt→0 kukḂ γ
p,p (R
= ∞.
+)
1. Introduction
In this paper, we study properties of the solutions to the Cauchy problem
(utt − ∆u)gs + m2 u = f (u),
u(1, x) = u0 ∈
γ
Ḃp,p
(R3 ),
t ∈ (0, 1], x ∈ R3 ,
ut (1, x) = u1 ∈
γ−1
Ḃp,p
(R3 ),
(1.1)
(1.2)
where gs is the Reissner-Nordström [2],
gs =
r2 − Kr + Q2 2
r2
dt − 2
dr2 − r2 dφ2 − r2 sin2 φdθ2 ,
2
r
r − Kr + Q2
the constants K and Q are positive, m 6= 0, p ∈ (1, ∞) and γ ∈ (0, 1) are fixed,
f ∈ C 2 (R1 ), f (0) = 0, 2m2 |u| ≤ f (l) (u) ≤ 3m2 |u|, l = 0, 1. More precisely we
prove that the Cauchy problem (1.1)-(1.2) has a unique nontrivial solution u in
2000 Mathematics Subject Classification. 35L05, 35L15.
Key words and phrases. Partial differential equation; Klein-Gordon; blow up.
c 2005 Texas State University - San Marcos.
Submitted March 14, 2005. Published June 27, 2005.
1
2
S. G. GEORGIEV
EJDE-2005/67
γ
γ
C((0, 1]Ḃp,p
(R+ )) such that limt→0 kukḂp,p
(R+ ) = ∞. The Cauchy problem (1.1)(1.2) may rewrite in the form
r2
1
utt − 2 ∂r ((r2 − Kr + Q2 )ur )
2
− Kr + Q
r
1
1
− 2
∂φ (sin φuφ ) − 2 2 uθθ + m2 u = f (u),
r sin φ
r sin φ
r2
γ
u(1, r, φ, θ) = u0 ∈ Ḃp,p
(R+ × [0, 2π] × [0, π]),
γ−1
ut (1, r, φ, θ) = u1 ∈ Ḃp,p
(R+ × [0, 2π] × [0, π]),
(1.3)
(1.4)
where x = r cos φ cos θ, y = r sin φ cos θ, z = r sin θ, φ ∈ [0, 2π], θ ∈ [0, π].
When gs is the Riemann metric, m = 0, f (u) = |u|p ; u0 , u1 ∈ C0∞ (R3 ) in
[1, Section 6.3] is proved that there exists T > 0 and a unique local solution
u ∈ C 2 ([0, T ) × R3 ) of (1.1)-(1.2) such that
sup
|u(t, x)| = ∞.
t<T, x∈R3
When gs is the Riemann metric, m = 0, f (u) = |u|p , 1 ≤ p < 5 and initial data
are in C0∞ (R3 ), in [1] is proved that the initial value problem (1.1)-(1.2) admits a
global smooth solution.
When φ 6= 0, π, 2π, θ 6= 0 are fixed constants we obtain the Cauchy problem
1
r2
utt − 2 ∂r ((r2 − Kr + Q2 )ur ) + m2 u = f (u),
r2 − Kr + Q2
r
(1.5)
γ
γ−1
u(1, r) = u0 ∈ Ḃp,p
(R+ ), ut (1, r) = u1 ∈ Ḃp,p
(R+ ).
(1.6)
Our main result is as follows.
Theorem 1.1. Let m be a non-zero constant, p ∈ (1, ∞), γ ∈ (0, 1) and K, Q be
positive constants for which
1
K 2 > 4Q2 ,
> 1, 1 − K + Q2 > 0,
1 − K + Q2
with 1 − K + Q2 is small enough such that
p
p
K − K 2 − 4Q2
− 2 1 − K + Q2 > 0.
2
2
1
Also let f ∈ C (R ), f (0) = 0, 2m2 |u| ≤ f (l) (u) ≤ 3m2 |u|, l = 0, 1. Then the
Cauchy problem (1.1)-(1.2) has a unique nontrivial solution u(t, r) = v(t)ω(r) ∈
γ
C((0, 1]Ḃp,p
(R+ )) for which
γ
lim kukḂp,p
(R+ ) = ∞.
t→0
This paper is organized as follows: In section 2 we prove that the Cauchy problem
γ
(1.1)-(1.2) has unique nontrivial solution ũ = v(t)ω(r) ∈ C((0, 1]Ḃp,p
(R+ )). In
section 3 we prove that
γ
lim kũkḂp,p
(R+ ) = ∞,
t→0
where ũ is the solution, which is received in section 2.
Let
pγ.2pγ 1/p
C = pγ
.
2 −1
Let A > 0, Q > 0, B > 0, K > 0, 1 < β < α be constants for which
EJDE-2005/67
(H1)
(H2)
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
8
1−K+Q2
8
2m2
1−K+Q2 α2 A
+ 4m2 ≤ 1,
αA
m
3
>1
1
m2
1
1 2
2 2
−
2m
≥1
r
r
−
1
1
1 − αK + α2 Q2 1 − αK + α2 Q2 α4 A2
β
and
m2
α4 (1−αK+α2 Q2 )A2
− 2m2 r12 ≥ 0,
(H3)
C
(H4)
(H5)
2p(1−γ) 1/p
8
2m2
8
6m2 2
+
m
<1
+
p(1 − γ)
1 − K + Q2 1 − K + Q2 α2 A2
AB
1
1
m2
1 m2
α2 1−αK+α2 Q2 α2 A4 − β 2 A2 > 0
p(1−γ) 1/p
16C
8
2m2
2
p(1−γ)
1−K+Q2 1−K+Q2 α2 A2
+ 4m2 < 1
√
K− K 2 −4Q2
2Q2
8
6
(H6) K 2 > 4Q2 , A ≥ 1−K+Q
>
1,
<
1,
1
>
>
, 1−K +
2
AB
K
2
2
Q > 0 is small enough such that
p
p
K − K 2 − 4Q2
1>
− 3 1 − K + Q2 > 0,
2
2
p
p
≤ β < α ≤ 3,
2
2
K − K − 4Q − 2 1 − K + Q2
where
r1 =
K−
p
√
K 2 − 4Q2
2p
−
1 − K + Q2 .
2
4
Example. Let
1
1
3
1
A = 4 , B = , p = , γ = , α = 3,
2
3
p
K − K 2 − 4Q2
3p
3
1
4 1
=
−
1 − K + Q2 , K = + 20 − 2 ,
β
2
2
3 6
2
1 1 20 1 2
2
2
4
Q = + − , m = ,
3 6
2
where 0 < << 1 is enough small such that (H1)-(H6) hold. Then
1 − αK + α2 Q2 = 1 − 3K + 9Q2 = 20 ,
1 − K + Q2 = 2 .
Remark 1.2. Let 2 = 1 − K + Q2 . Note that from (H6) we have g(r) = r2 − Kr +
Q2 > 0 for r ∈ [0, r1 ], g(r) is decrease function for r ∈ [0, r1 ]. Also (for r ∈ [0, r1 ])
we have
r2
1
≤
.
2
r − Kr + Q2
1 − K + Q2
√
√
K− K 2 −4Q2
2
In deed, letting r̃ =
,
we
have
r
=
r̃
−
1
2
4 . Note that function
r2
r2 − Kr + Q2
is increasing for r ∈ [0, r1 ]. Therefore,
r2
r12
8
8
≤
≤ 2 =
.
2
r2 − Kr + Q2
r1 − Kr1 + Q2
1 − K + Q2
4
S. G. GEORGIEV
EJDE-2005/67
Note that the function
1
r2 − Kr + Q2
is increasing for r ∈ [0, r1 ]. Therefore, for r ∈ [0, r1 ],
8
1
≤
.
r2 − Kr + Q2
1 − K + Q2
γ
Here we will use the following definition of the Ḃp,p
(M )-norm (γ ∈ (0, 1), p > 1)
(see [3, p.94, def. 2], [1])
Z 2
1/p
γ
kukḂp,p
=
h−1−pγ k∆h ukpLp (M ) dh
,
(M )
0
where ∆h u = u(x + h) − u(x).
Lemma 1.3. Let u(x) ∈ C 2 ([0, r1 ]), u(x) = 0 for x ≥ r1 , 0 < r1 < 1. Then for
γ ∈ (0, 1), p > 1 we have
γ
CkukḂp,p
([0,r1 ]) ≥ kukLp ([0,r1 ]) .
Proof. We have
kukpḂ γ
Z
p,p ([0,r1 ])
2
h−1−pγ k∆h ukpLp ([0,r1 ]) dh
=
0
Z
2
h−1−pγ ku(x + h) − u(x)kpLp ([0,r1 ]) dh
=
0
Z
2
h−1−pγ ku(x + h) − u(x)kpLp ([0,r1 ]) dh
≥
1
Z
=
1
=
2
h−1−pγ ku(x)kpLp ([0,r1 ]) dh
ku(x)kpLp ([0,r1 ])
= ku(x)kpLp ([0,r1 ])
Z
2
h−1−pγ dh
1
pγ
2 −1
;
pγ2pγ
i.e.,
kukpḂ γ
p,p ([0,r1 ])
≥
2pγ − 1
ku(x)kpLp ([0,r1 ]) .
pγ2pγ
γ
From this estimate, we have CkukḂp,p
([0,r1 ]) ≥ ku(x)kLp ([0,r1 ]) which completes the
proof.
2. Existence of local solutions to the Cauchy problem (1.1)-(1.2)
Here and below we suppose that the positive constants A, K, Q, B, 1 < β < α
satisfy (H1)-(H6). Let t ∈ (0, 1]. Let v(t) be function which satisfies the hypotheses:
(H7) v(t) ∈ C 3 [0, ∞), v(t) > 0 for all t ∈ [0, 1]
(H8) v 00 (t) > 0 for all t ∈ [0, 1], v 0 (1) = v 000 (1) = 0, v(1) 6= 0
EJDE-2005/67
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
5
(H9)
1
,
A
2
,
A
t∈[0,1]
t∈[0,1]
v 00 (t)
m2
2m2
v 00 (t)
≥ 2 2 , max
≤ 2 2;
min
α A
α A
t∈[0,1] v(t)
t∈[0,1] v(t)
2
2
m
m
lim [v 00 (t) − 2 2 v(t)] = +0, v 00 (t) − 2 2 v(t) ≥ 0 for t ∈ [0, 1].
t→0
α A
α A
min v(t) ≥
max v(t) ≤
Note that there exist a functions v(t) for which (H7)-(H9) hold. For example
consider the function
2 2
(t − 1)2 + 2αmA
−1
2
v(t) =
.
(2.1)
2
α
A3 m2
Then v(t) ∈ C 3 [0, ∞); v(t) > 0 for all t ∈ [0, 1] because (H1), we have
(H7) holds. Since
αA
m
> 1; i.e.,
2(t − 1)
, v 0 (1) = 0,
α2
A3 m
2
2
v 00 (t) =
≥ 0 ∀ t ∈ [0, 1],
α2
3
A m
2
v 0 (t) =
v 000 (t) = 0,
v 000 (1) = 0,
it follows (H8). On the other hand
min v(t) ≥
t∈[0,1]
1
,
A
max v(t) ≤
t∈[0,1]
2
,
A
v 00 (t)
2
=
,
2 2
2
v(t)
(t − 1) + 2αmA
−1
2
which implies
m2
2m2
v 00 (t)
v 00 (t)
≥ 2 2 , max
≤ 2 2,
α A
α A
t∈[0,1] v(t)
t∈[0,1] v(t)
2
4
m
m
m2
v 00 (t) − 2 2 v(t) = 4 5 (2 − t)t, lim [v 00 (t) − 2 2 v(t)] = +0;
t→0
α A
α A
α A
min
i.e., (H9) holds.
Here and below we suppose that v(t) is a fixed function satisfying (H7)-(H9). In
this section we will prove that the Cauchy problem (1.1)-(1.2) has unique nontrivial
solution of the form
(
v(t)ω(r) for r ≤ r1 ,
u(t, r) =
0
for r ≥ r1 ,
γ
with t in (0, 1] and u ∈ C((0, 1]Ḃp,p
(R+ )).
Let us consider the integral equation
R r
R r1 v 00 (t)
1
1
s4

2 −Kτ +Q2
2 −Ks+Q2 v(t) u(t, s)

τ
s
r
τ

u(t, r) = +s2 m2 u(t, s) − f (u(t, s))s2 ds dτ,
for 0 ≤ r ≤ r1 ,



0
forr ≥ r1 ,
where u(t, r) = v(t)ω(r) and t ∈ (0, 1].
(2.2)
6
S. G. GEORGIEV
EJDE-2005/67
Theorem 2.1. Let p ∈ (1, ∞), m 6= 0 and γ ∈ (0, 1) be fixed constants and the
positive constants A, B, Q,K, α > β > 1 satisfy (H1)–(H6) and f ∈ C 2 (R1 ),
f (0) = 0, 2m2 |u| ≤ f (l) (u) ≤ 3m2 |u|, l = 0, 1. Let also v(t) is function for which
(H7)–(H9) hold. Then the equation (2.2) has unique nontrivial solution u(t, r) =
v(t)ω(r) for which w ∈ C 2 [0, r1 ], u(t, r1 ) = ur (t, r1 ) = urr (t, r1 ) = 0 for t ∈ (0, 1],
γ
u(t, r) ∈ C((0, 1]Ḃp,p
[0, r1 ]), for r ∈ [ α1 , β1 ] and t ∈ (0, 1] u(t, r) ≥ A12 , for r ∈ [ α1 , r1 ]
2
and t ∈ (0, 1] u(t, r) ≥ 0, for r ∈ [0, r1 ] and t ∈ (0, 1] |u(t, r)| ≤ AB
, u(t, r) = 0 for
r ≥ r1 , t ∈ (0, 1].
Proof. Let N = {u(t, r) ∈ C([0, r1 ]) : t ∈ (0, 1]} with u(t, r) = ur (t, r) = urr (t, r) =
γ
0 for t ∈ (0, 1], r ≥ r1 , u(t, r) ∈ C((0, 1]Ḃp,p
[0, r1 ]). For r ∈ [ α1 , β1 ] and t ∈ (0, 1],
1
2
we have u(t, r) ≥ A2 . For r ∈ [0, r1 ] and t ∈ (0, 1], we have |u(t, r)| ≤ AB
. For
1
r ∈ [ α , r1 ] and t ∈ (0, 1], we have u(t, r) ≥ 0}.
We remark that if u ∈ N is a solution of (2.2), u ∈ C 2 ([0, r1 ]). We define the
operator R as follows
Z r1 Z r1
v 00 (t)
s4
1
u(t, s)
R(u) =
2
2
2
2
τ − Kτ + Q τ
s − Ks + Q v(t)
r
+ s2 m2 u(t, s) − s2 f (u) ds dτ,
for 0 ≤ r ≤ r1 and t ∈ (0, 1].
First we show that R : N → N . For each u ∈ N , we have the following five
statements:
(1) Since u ∈ C([0, r1 ]) and f ∈ C 2 (R1 ), from the definition of the operator R we
have R(u) ∈ C 2 ([0, r1 ]), R(u)|r=r1 = 0,
Z r
∂
1
s4
v 00 (t)
R(u) = 2
[ 2
u + s2 (m2 u − f (u))]ds,
2
2
∂r
r − Kr + Q r1 s − Ks + Q v(t)
∂
R(u)
∂r
∂2
K − 2r
R(u) = 2
2
∂r
(r − Kr + Q2 )2
Z
r
[
r1
r=r1
= 0,
s4
v 00 (t)
u + s2 (m2 u − f (u))]ds
s2 − Ks + Q2 v(t)
r
v 00 (t)
r2
u(t, r) + 2
(m2 u(t, r) − f (u)).
2
2
− Kr + Q ) v(t)
r − Kr + Q2
4
+
(r2
Since u(t, r1 ) = 0, f (u(t, r1 )) = f (0) = 0 we obtain
∂2
R(u) r=r1 = 0.
∂r2
Note that R(u) = 0 for r ≥ r1 , t ∈ (0, 1] because u(t, r) = 0 for r ≥ r1 , t ∈ (0, 1]
and f (u(t, r)) = f (0) = 0 for r ≥ r1 , t ∈ (0, 1].
2
. Then
(2) For r ∈ [0, r1 ], t ∈ (0, 1] we have |u(t, r)| ≤ AB
|R(u)|
Z r
=
Z τ
s4
v 00 (t)
2
2
u
+
s
(m
u
−
f
(u))
dsdτ
2
2
r1
r1 s − Ks + Q v(t)
Z r
Z τ
1
s4
v 00 (t)
≤
|u| + s2 (m2 |u| + |f (u)|) dsdτ .
2
2
2
2
r1 τ − Kτ + Q
r1 s − Ks + Q v(t)
1
τ 2 − Kτ + Q2
EJDE-2005/67
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
7
Since |f (u)| ≤ 3m2 |u|), the above quantity is lees than or equal to
Z r
Z τ
1
s4
v 00 (t)
2 2
|u|
+
4s
m
|u|
dsdτ
2
2
2
2
r1 τ − Kτ + Q
r1 s − Ks + Q v(t)
Z r
Z τ
1
s4
v 00 (t)
2 2
|u|dsdτ
=
+
4s
m
2
2
2
2
r1 τ − Kτ + Q
r1 s − Ks + Q v(t)
Z r
Z τ
2
1
s4
v 00 (t)
≤
+ 4s2 m2 dsdτ
2
2
2
2
AB r1 τ − Kτ + Q r1 s − Ks + Q v(t)
2
8
1
8
r
The
where we use r2 −Kr+Q
2 ≤ 1−K+Q2 , r 2 −Kr+Q2 ≤ 1−K+Q2 for r ∈ [0, r1 ].
above estimate is also less than or equal to
2
8
8
v 00 (t)
max
+ 4m2
2
2
AB 1 − K + Q 1 − K + Q t∈[0,1] v(t)
2
8
8
2m2
2
=
+
4m
AB 1 − K + Q2 1 − K + Q2 α2 A2
2
.
≤
AB
In the above inequality we use (H1). Consequently,
2
for r ∈ [0, r1 ], t ∈ (0, 1].
AB
(3)For r ∈ [ α1 , r1 ] and t ∈ (0, 1] we have u(t, r) ≥ 0. Then
Z r1 Z r1
1
s4
v 00 (t)
u(t, s)
R(u) =
τ 2 − Kτ + Q2 τ
s2 − Ks + Q2 v(t)
r
+ s2 m2 u(t, s) − s2 f (u) ds dτ
|R(u)| ≤
(where we use f (u) ≤ 3m2 u for r ∈ [ α1 , r1 ], t ∈ (0, 1]. The above quantity is greater
than or equal to
Z r1
Z r1 1
s4
v 00 (t)
2
2
2
u(t,
s)
+
s
(m
u(t,
s)
−
3m
u)
ds dτ
τ 2 − Kτ + Q2 τ
s2 − Ks + Q2 v(t)
r
Z r1
Z r1 1
s4
v 00 (t)
2 2
≥
−
2m
s
u(t, s)ds dτ
τ 2 − Kτ + Q2 τ
s2 − Ks + Q2 v(t)
r
Z r1 Z r1
1
v 00 (t)
1
2 2
min
−
2m
r
≥
1 u(t, s)dsdτ
τ 2 − Kτ + Q2 τ
α2 (1 − αK + α2 Q2 ) t∈[0,1] v(t)
r
Z r1
Z r1 1
m2
2 2
=
−
2m
r
1 u(t, s)ds dτ .
τ 2 − Kτ + Q2 τ
α4 (1 − αK + α2 Q2 )A2
r
From (H2), we have
m2
− 2m2 r12 ≥ 0.
α4 (1 − αK + α2 Q2 )A2
From this inequality and from u(t, r) ≥ 0 for r ∈ [ α1 , r1 ], t ∈ (0, 1], r2 −Kr+Q2 > 0,
for r ∈ [0, r1 ], we get
R(u) ≥ 0 f or
1
r ∈ [ , r1 ],
α
t ∈ (0, 1].
8
S. G. GEORGIEV
EJDE-2005/67
(4) For r ∈ [ α1 , β1 ] and t ∈ (0, 1] we have that u(t, r) ≥ A12 . Using f (u) ≤ 3m2 u for
r ∈ [ α1 , β1 ], t ∈ (0, 1], we have
Z τ
Z r
s4
v 00 (t)
1
R(u) ≥
u − 2s2 m2 u dsdτ
2
2
2
2
r1 s − Ks + Q v(t)
r1 τ − Kτ + Q
Z r
Z τ
1
s4
v 00 (t)
2 2
≥
min
u
−
2s
m
u
ds dτ
2
2
2
2
r1 τ − Kτ + Q
r1 s − Ks + Q t∈[0,1] v(t)
Z τ Z r
s2
v 00 (t)
1
2
2
s
min
u
−
2m
u
ds dτ
=
2
2
s2 − Ks + Q2 t∈[0,1] v(t)
r1
r1 τ − Kτ + Q
Z β1 Z r1
v 00 (t)
1
s2
2
2
min
≥
s
u
−
2m
u
ds dτ
τ 2 − Kτ + Q2 α1
s2 − Ks + Q2 t∈[0,1] v(t)
r
Z β1 Z r1
1
s2
v 00 (t)
2
2
s
min
u
−
2m
u
ds dτ
≥
1
τ 2 − Kτ + Q2 α1
s2 − Ks + Q2 t∈[0,1] v(t)
β
m2
1
1
1 1
1 2
− 2m2 r12 r1 −
≥ 2,
≥ 2
2
2
4
2
A 1 − αK + α Q α A
β 1 − αK + α2 Q2
A
(see (H2)); i.e., for r ∈ [ α1 , β1 ] and t ∈ (0, 1] we have R(u) ≥ A12 .
(5) We have the estimate
Z r1 Z r+h
Z r1 1
s4
v 00 (t)
k∆h R(u)kpLp =
u(t, s)
2
2
2
2
τ − Kτ + Q τ
s − Ks + Q v(t)
r
0
p
+ s2 (m2 u − f (u)) ds dτ
dr
Z r1 Z r+h
Z r1 1
s4
v 00 (t)
≤
|u|
2
2
2
2
τ − Kτ + Q τ
s − Ks + Q v(t)
0
r
p
+ 4m2 |u(t, s)|s2 ds, dτ dr
4
s
8
1
8
where we use that for s ∈ [0, r1 ], s2 −Ks+Q
2 ≤ 1−K+Q2 , s2 −Ks+Q2 ≤ 1−K+Q2 and
u(t, r) = 0 for r ≥ r1 and t ∈ (0, 1]. By (H9) the above estimate is less than or
equal to
Z r1 Z r+h
Z r1 p
8
v 00 (t)
8
max
|u| + 4m2 |u| dsdτ dr
2
2
1−K +Q τ
1 − K + Q t∈[0,1] v(t)
0
r
Z r1 Z r+h
Z r1 p
8
8
2m2
2
|u|
+
4m
|u|
dsdτ
dr
≤
1 − K + Q2 τ
1 − K + Q2 α2 A2
0
r
Z r1 Z r+h
Z r1
Z r1
p
64
2m2
8
2
≤
|u|ds
+
4m
|u|ds
dτ dr
(1 − K + Q2 )2 α2 A2 0
1 − K + Q2
0
r
0
p
64
2m2
8
2
p
p
≤ hp
kuk
+
4m
kuk
;
L [0,r1 ]
L [0,r1 ]
(1 − K + Q2 )2 α2 A2
1 − K + Q2
i.e,,
k∆h R(u)kpLp [0,r1 ]
p
2m2
8
64
2
p [0,r ] +
p [0,r ]
≤ hp
kuk
4m
kuk
.
L
L
1
1
(1 − K + Q2 )2 α2 A2
1 − K + Q2
EJDE-2005/67
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
9
Consequently,
kR(u)kpḂ γ [0,r ]
1
p,p
Z 2
h−1−pγ k∆h R(u)kpLp [0,r1 ] dh
=
0
p Z 2
64
2m2
8.4m2
≤
kukLp [0,r1 ] +
h−1+p(1−γ) dh
kukLp [0,r1 ]
(1 − K + Q2 )2 α2 A2
(1 − K + Q2 )
0
p 2p(1−γ)
64
2m2
8.4m2
p [0,r ] +
p [0,r ]
=
kuk
kuk
.
L
L
1
1
(1 − K + Q2 )2 α2 A2
(1 − K + Q2 )
p(1 − γ)
Therefore,
γ
kR(u)kḂp,p
[0,r1 ]
2p(1−γ) 1/p
2m2
8.4m2
64
p
p
kuk
+
kuk
.
L [0,r1 ]
L [0,r1 ]
(1 − K + Q2 )2 α2 A2
(1 − K + Q2 )
p(1 − γ)
From Lemma 1.3, we have
2m2
64
γ
γ
kR(u)kḂp,p
kukḂp,p
≤
C
[0,r1 ]
[0,r1 ]
2
2
(1 − K + Q ) α2 A2
2p(1−γ) 1/p
8.4m2
γ
kuk
+
.
Ḃp,p [0,r1 ]
(1 − K + Q2 )
p(1 − γ)
≤
γ
γ
[0, r1 ] for t ∈ (0, 1].
[0, r1 ] we get R(u) ∈ Ḃp,p
From the above inequality, if u ∈ Ḃp,p
From statements (1)–(5) above, R : N → N .
Now, let u, u1 ∈ N . Then
k∆h (R(u) − R(u1 ))kpLp
Z r1 Z r+h
Z r1 1
s4
v 00 (t)
=
(u(t, s) − u1 )
τ 2 − Kτ + Q2 τ
s2 − Ks + Q2 v(t)
0
r
p
+ s2 (m2 (u − u1 ) − (f (u) − f (u1 ))) ds dτ
dr.
From the mean value theorem, |f (u) − f (u1 )| = |u − u1 kf 0 (ξ)| where ξ ∈ (u, u1 ) or
ξ ∈ (u1 , u). Then
|f (u) − f (u1 )| ≤ 3m2 |ξ||u − u1 | ≤ 3m2 |u − u1 kq|,
where |q| = max{|u|, |u1 |}. Since |u| ≤
2
AB
for r ∈ [0, r1 ], t ∈ (0, 1] we have
6m2
|u − u1 |.
AB
Now, we use that u(t, r) = 0 for r ≥ r1 and t ∈ (0, 1], to obtain
|f (u) − f (u1 )| ≤
10
S. G. GEORGIEV
EJDE-2005/67
k∆h (R(u) − R(u1 ))kpLp
Z r1 Z r+h
Z r1 1
s4
v 00 (t)
≤
|u(t, s) − u1 |
2
2
2
2
τ − Kτ + Q τ
s − Ks + Q v(t)
0
r
p
+ s2 m2 |u − u1 | + |f (u) − f (u1 )| ds dτ dr
Z r1 Z r+h
Z r1 1
s4
v 00 (t)
≤
max
|u(t, s) − u1 |
2
2
2
2
τ − Kτ + Q τ
s − Ks + Q t∈[0,1] v(t)
0
r
p
+ s2 m2 |u − u1 | + |f (u) − f (u1 )| ds dτ dr
Z r1 Z r1 Z r+h
8
v 00 (t)
1
max
|u(t, s) − u1 |
≤
2
2
2
τ − Kτ + Q τ
1 − K + Q t∈[0,1] v(t)
0
r
p
6m2
|u − u1 | ds dτ dr
+ m2 |u − u1 | +
AB
Z r1 Z r+h
Z r1 8
6m2 8
2m2
2
≤
+
m
+
1 − K + Q2 τ
1 − K + Q2 α2 A2
AB
0
r
p
× |u − u1 |dsdτ dr
p 8
2m2
6m2 p
8
2
≤ hp
+
m
+
ku − u1 kpLp ;
1 − K + Q2
(1 − K + Q2 ) α2 A2
AB
i.e.,
k∆h (R(u) − R(u1 ))kpLp [0,r1 ]
p 8
8
2m2
6m2 p
2
≤ hp
+
m
+
ku − u1 kpLp .
1 − K + Q2
(1 − K + Q2 ) α2 A2
AB
From the last inequality we get
p 8
8
2m2
6m2 p
kR(u) − R(u1 )kpḂ γ [0,r ] ≤
+ m2 +
2
2
2
2
1
p,p
1−K +Q
(1 − K + Q ) α A
AB
Z 2
× ku − u1 kpLp
h−1+p(1−γ) dh.
0
From the above inequality and Lemma 1.3,
2p(1−γ) 1/p
8
8
2m2
γ
kR(u) − R(u1 )kḂp,p
[0,r1 ] ≤
2
2
p(1 − γ)
1 − K + Q (1 − K + Q ) α2 A2
6m2 ku − u1 kLp [0,r1 ]
+ m2 +
AB
2p(1−γ) 1/p
8
8
2m2
≤C
(p(1 − γ)
1 − K + Q2 (1 − K + Q2 ) α2 A2
2
6m
γ
ku − u1 kḂp,p
+ m2 +
[0,r1 ]
AB
γ
< ku − u1 kḂp,p [0,r1 ]
(see i3)). i.e.,
γ
γ
kR(u) − R(u1 )kḂp,p
[0,r1 ] < ku − u1 kḂp,p
[0,r1 ] .
EJDE-2005/67
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
Consequently, the operator R : N → N is contractive operator.
11
γ
Lemma 2.2. The set N is closed subset of C((0, 1]Ḃp,p
(R+ )).
Proof. Let t ∈ (0, 1] be fixed. Let {un } be a sequence of elements of the set N for
which
˜ γ + = 0,
lim kun − ũk
Ḃp,p (R )
n→∞
γ
˜ ∈ Ḃp,p
where ũ
(R+ ). We have
˜ γ
lim kun − ũk
Ḃp,p ([0,r1 ]) = 0.
n→∞
We define
ũ =
(
˜
ũ
0
for r ∈ [0, r1 ],
for r > r1 .
We have
γ
lim kun − ũkḂp,p
([0,r1 ]) = 0.
n→∞
First we note that for u ∈ N , R(u) is continuous function of u and there exists
R0 (u) because f (u) ∈ C 2 (R1 ). In fact,
Z r1 Z r1
1
s4
v 00 (t)
2 2
2 0
+
s
m
−
s
f
(u)
ds dτ.
R0 (u) =
τ 2 − Kτ + Q2 τ
s2 − Ks + Q2 v(t)
r
From which,
|R0 (u)|
Z r1
1
≥
2
τ − Kτ + Q2
r
Z r1
1
≥
2
τ − Kτ + Q2
r
Z r1
1
≥
2
τ − Kτ + Q2
r
Z r1
1
=
2
τ − Kτ + Q2
r
r1 s4
v 00 (t)
s2 − Ks + Q2 v(t)
τ
Z r1 s4
v 00 (t)
s2 − Ks + Q2 v(t)
τ
Z r1 s4
v 00 (t)
s2 − Ks + Q2 v(t)
τ
Z r1 s4
v 00 (t)
2
2
s − Ks + Q v(t)
τ
Z
+ s2 m2 − s2 |f 0 (u)| ds dτ
+ s2 m2 − 3m2 s2 |u| ds, dτ
6m2 2 s ds dτ
AB
6 + s2 m2 1 −
ds dτ .
AB
+ s2 m2 −
From (H6), 1 > 6/(AB). Therefore, for s ∈ [0, r1 ] we have
v 00 (t)
6 s4
2 2
+
s
m
1
−
≥ 0.
s2 − Ks + Q2 v(t)
AB
Then for r ∈ [0, r1 ] we have
|R0 (u)|
Z r1
≥
Z r1 1
s4
v 00 (t)
6 2
2 2
+
s
m
1
−
s ds dτ
1
τ 2 − Kτ + Q2 α1 s2 − Ks + Q2 v(t)
AB
α
1 2
m2
≥ r1 −
> 0.
2
2
α α A (1 − αK + α2 Q2 )2
From this, for u ∈ N , there exists
M := min |R0 (u)(x)| > 0
x∈[0,r1 ]
12
S. G. GEORGIEV
EJDE-2005/67
because R0 (u)(x) is continuous function of x ∈ [0, r1 ]. Let
M1 = max
r∈[0,r1 ]
∂
(R0 (u))(r) .
∂r
Now we prove that for each > 0 there exists δ = δ() > 0 such that
|x − y| < δ
implies
|um (x) − um (y)| < ∀m.
We suppose that there exists ˜ > 0 such that for every δ > 0 there exist natural
m and x, y ∈ [0, r1 ], |x − y| < δ for which |um (x) − um (y)| ≥ ˜. We choose ˜˜ > 0
such that ˜
˜ < M ˜. We note that R(um )(x) is uniformly continuous function of
x ∈ [0, r1 ](For u ∈ N the function R(u)(r) is uniformly continuous function of r ∈
∂
2
[0, r1 ] because R(u)(r) ∈ C 2 ([0, r1 ]) and as in point (2) we have ∂r
R(u)(r) ≤ AB
).
˜) > 0 such that for every u ∈ N
Then there exists δ1 = δ1 (˜
|R(u)(x) − R(u)(y)| < ˜˜ for for all x, y ∈ [0, r1 ]: |x − y| < δ1 .
−˜
˜)AB
such that there exist natural m
Then we may choose 0 < δ < min δ1 , (M ˜2M
1
and x1 ∈ [0, r1 ], x2 ∈ [0, r1 ] for which |x1 − x2 | < δ and |um (x1 ) − um (x2 )| ≥ ˜. In
particular
|R(um )(x1 ) − R(um )(x2 )| < ˜˜.
(2.3)
Then by the mean value theorem, R(0) = 0, R(um )(x1 ) = R0 (ξ)(x1 )um (x1 ),
R(um )(x2 ) = R0 (ξ)(x2 )um (x2 ),
|R(um )(x1 ) − R(um )(x2 )|
= |R0 (ξ)(x1 )um (x1 ) − R0 (ξ)(x2 )um (x2 )|
= |R0 (ξ)(x1 )um (x1 ) − R0 (ξ)(x1 )um (x2 ) + R0 (ξ)(x1 )um (x2 ) − R0 (ξ)(x2 )um (x2 )|
≥ |R0 (ξ)(x1 )um (x1 ) − R0 (ξ)(x1 )um (x2 )| − |R0 (ξ)(x1 ) − R0 (ξ)(x2 )kum (x2 )|
= |R0 (ξ)(x1 )kum (x1 ) − um (x2 )|−
∂
(R0 (ξ))(η) |x1 − x2 kum (x2 )|
∂r
2
≥ ˜
˜,
AB
which is a contradiction to (2.3).
Consequently, for each > 0 there exists δ = δ() > 0 such that
≥ M ˜ − M1 δ
|x − y| < δ
implies|um (x) − um (y)| < ∀m.
(2.4)
On the other hand, from the definition of the set N we have
2
∀m.
(2.5)
AB
From (2.4) and (2.5), we conclude that the set {un } is compact subset of C([0, r1 ]).
Then there exists subsequence {unk } and function u ∈ C([0, r1 ]) for which: for
every > 0 there exists M = M () > 0 such that for every nk > M we have
|unk (x) − u(x)| < for every x ∈ [0, r1 ]; u(x) = 0 for x > r1 . From this and from
γ
limk→∞ kunk − ũkḂp,p
([0,r1 ]) = 0 we have: For every > 0 ∃M = M () > 0 such
that for every nk > M we have
|um | ≤
max |unk − u| < ,
x∈[0,r1 ]
γ
kunk − ũkḂp,p
([0,r1 ]) < .
EJDE-2005/67
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
13
Then for every nk > M we have
|u − ũ| ≤ |u − unk | + |unk − ũ| < + |ũ − unk |,
Z r1
Z r1
|u − ũ|dx < r1 +
|ũ − unk |dx,
0
0
Using the Hölder’s inequality,
1/q
ku − ũkL1 [0,r1 ] < r1 + r1
Z
r1
|ũ − unk |p dx
1/p
,
0
1 1
+ = 1,
p q
for h > 0, we have
1/q
h−1−pγ ku − ũkL1 [0,r1 ] < h−1−pγ r1 + r1 h−1−pγ
Z
r1
1/p
|ũ − unk |p dx
,
0
Z
2
h−1−pγ dhku − ũkL1 [0,r1 ]
1
Z
<
2
h
−1−pγ
1
dhr1 +
1/q
r1
Z
2
h
1
−1−pγ
Z
r1
|ũ − unk |p dx
1/p
dh,
0
Using Hölder’s inequality and that for h > 1 we have h(−1−pγ)p ≤ h−1−pγ ,
1 − 2−pγ
ku − ũkL1 [0,r1 ]
pγ
Z 2
Z r 1
1/p
1 − 2−pγ
1/q
<
r1 + r1
h−1−pγ
|ũ − unk |p dx
dh
pγ
1
0
Z r1
Z 2
1/p
1 − 2−pγ
1/q
≤
r1 + r1
h(−1−pγ)p
|ũ − unk |p dxdh
pγ
1
0
Z 2
Z r1
1/p
−pγ
1−2
1/q
≤
r1 + r1
h−1−pγ
|ũ − unk |p dxdh
.
pγ
1
0
Using that for x > r1 , unk (x) = ũ(x) = 0, the above expression equals
Z r1
Z 2
1/p
1 − 2−pγ
1/q
r1 + r1
h−1−pγ
|∆h (ũ − unk )|p dxdh
pγ
1
0
Z r1
Z 2
1/p
1 − 2−pγ
1/q
−1−pγ
≤
r1 + r1
h
|∆h (ũ − unk )|p dxdh
pγ
0
0
1 − 2−pγ
1/q
γ
=
r1 + r1 kũ − unk kḂp,p
([0,r1 ])
pγ
1 − 2−pγ
1/q
< (
r1 + r1 )
pγ
i.e., for every > 0,
1 − 2−pγ
1 − 2−pγ
1/q
.
ku − ũkL1 [0,r1 ] < r1 + r1
pγ
pγ
Consequently u = ũ a.e.(almost everywhere) in [0, r1 ], |u|p = |ũ|p a.e. in [0, r1 ].
¿From here |un − u| = |un − ũ| a.e., |un − u|p = |un − ũ|p a.e. in [0, r1 ]. Since
14
S. G. GEORGIEV
EJDE-2005/67
un (x) = u(x) = 0 for x > r1 we have |∆h (un − u)|p = |∆h (un − ũ)|p , |∆h u|p =
γ
|∆h ũ|p a.e. in [0, r1 ], for h > 0. Therefore, u ∈ Ḃp,p
([0, r1 ]) and
Z r1
Z r1
|un − u|p dx =
|un − ũ|p dx.
0
0
γ
Now, we show that limn→∞ kun − ukḂp,p
([0,r1 ]) = 0. Note that
γ
kun − ukḂp,p
([0,r1 ])
Z
Z r1
2
1/p
−1−pγ
=
h
|∆h (un − u)|p dx dh
0
=
Z
0
2
h−1−pγ
Z
r1
|∆h (un − ũ)|p dx dh
1/p
0
0
γ
= kun − ũkḂp,p
([0,r1 ]) →n→∞ 0.
Consequently, for every sequence {un } with elements from N , which converges
γ
γ
([0, r1 ]), for which
([0, r1 ]) there exists function u ∈ C([0, r1 ]), u ∈ Ḃp,p
in Ḃp,p
γ
kun − ukḂp,p
→
0.
n→∞
([0,r1 ])
Below we suppose that the sequence {un } is a sequence of elements of the set
γ
([0, r1 ]). Then there exists u ∈ C([0, r1 ]), u(x) = 0 for
N which converges in Ḃp,p
γ
γ
x > r1 , u ∈ Ḃp,p ([0, r1 ]), kun − ukḂp,p
([0,r1 ]) →n→∞ 0.
Now we suppose that u(r1 ) 6= 0. Since u ∈ C([0, r1 ]), un ∈ C([0, r1 ]), un (r1 ) = 0
for every natural n, there exist 2 > 0 and ∆1 ⊂ [0, r1 ], r1 ∈ ∆1 , such that
|un | <
2
,
2
|u| > 2
for every natural n and every x ∈ ∆1 . Then for every natural n and for every
x ∈ ∆1 ,
2
|un (x) − u(x)| > .
2
Let 3 > 0 be such that
3 <
−1
2 1 − 2−pγ
µ(∆1 )r1 q ,
2
pγ
1 1
+ = 1,
p q
(2.6)
where µ(∆1 ) is the measure of the set ∆1 . There exists M > 0 such that for every
γ
n > M we have kun − ukḂp,p
([0,r1 ]) < 3 . Consequently for every n > M and for
every x ∈ ∆1 we have
2
γ
|un (x) − u(x)| > , kun − ukḂp,p
([0,r1 ]) < 3 .
2
Also using the Hölder’s inequality, we have
Z
2
µ(∆1 ) <
|un (x) − u(x)|dx
2
∆
Z r11
≤
|un (x) − u(x)|dx
0
Z
r1
1/p
1/q
≤
|un (x) − u(x)|p dx
r1 .
0
EJDE-2005/67
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
15
For h > 0, we have
h
−1−pγ 2
2
µ(∆1 ) ≤ h
−1−pγ
Z
r1
1/p
1/q
|un (x) − u(x)|p dx
r1 ,
0
Z 2
Z r 1
1/p
2
1/q
|un (x) − u(x)|p dx
h−1−pγ
h−1−pγ µ(∆1 )dh ≤
r1 dh,
2
0
1
1
Z r1
1/p
Z 2
1 − 2−pγ 2
1/q
|un (x) − u(x)|p dxdh
µ(∆1 ) ≤
h(−1−pγ)p
r1 .
pγ
2
0
1
Z
2
Since h > 1, h(−1−pγ)p ≤ h−1−pγ and the above expression is less than or equal to
Z r1
Z 2
1/p
1/q
|un (x) − u(x)|p dxdh
h−1−pγ
r1 ≤
0
1
Now using that un = u = 0 for x > r1 , the above expression is less than or equal to
Z r1
Z 2
1/p
1/q
h−1−pγ
|∆h (un (x) − u(x))|p dxdh
r1
1
≤
0
Z
2
h−1−pγ
0
Z
r1
|∆h (un (x) − u(x))|p dxdh
1/p
1/q
r1
0
1/q
γ
= kun − ukḂp,p
([0,r1 ]) r1
1/q
< 3 r1 .
Therefore,
1 − 2−pγ 2
1/q
µ(∆1 ) < 3 r1 ,
pγ
2
which is a contradiction with (2.6). Consequently, u(r1 ) = 0. From this, u(t, r) = 0
for r ≥ r1 . Then ur (t, r) = urr (t, r) = 0 for every r ≥ r1 .
Now we suppose that the inequality
2
AB
is not hold for every r ∈ [0, r1 ]. Since u ∈ C([0, r1 ]) we may take 4 > 0 and
∆2 ⊂ [0, r1 ] such that
2
+ 4 for r ∈ ∆2 .
|u| ≥
AB
Then for every natural n and for every r ∈ ∆2 , we have
|u(t, r)| ≤
|un − u| ≥ |u| − |un | ≥
2
2
+ 4 −
= 4 .
AB
AB
Let 5 > 0 be such that
1 − 2−pγ
1/q
4 µ(∆2 ) > 5 r1 .
pγ
(2.7)
γ
There exist M > 0 such that for every n > M we have kun − ukḂp,p
([0,r1 ]) < 5 .
Consequently for every n > M and for every x ∈ ∆2 we have
|un (x) − u(x)| ≥ 4 ,
γ
kun − ukḂp,p
([0,r1 ]) < 5 .
Also using the Hölder’s inequality, we have
Z
Z
4 µ(∆2 ) <
|un (x) − u(x)|dx ≤
∆2
0
r1
1/p
1/q
|un (x) − u(x)|p dx
r1 .
16
S. G. GEORGIEV
EJDE-2005/67
For h > 0 we have
h−1−pγ 4 µ(∆2 ) ≤ h−1−pγ
r1
Z
1/p
1/q
|un (x) − u(x)|p dx
r1 ,
0
Z
2
h−1−pγ 4 µ(∆2 )dh ≤
2
Z
1
h−1−pγ
r1
Z
1/p
1/q
|un (x) − u(x)|p dx
r1 dh,
0
1
Using Hölder’s inequality and that for h > 1, h(−1−pγ)p ≤ h−1−pγ , we have
1 − 2−pγ
4 µ(∆2 )
pγ
Z
Z 2
≤
h(−1−pγ)p
Z
|un (x) − u(x)|p dxdh
1/p
1/q
r1
0
1
≤
r1
2
h−1−pγ
Z
r1
|un (x) − u(x)|p dxdh
1/p
1/q
r1
.
0
1
Using that un = u = 0 for x > r1 , the above expression is less than or equal to
Z r1
Z 2
1/p
1/q
h−1−pγ
|∆h (un (x) − u(x))|p dxdh
r1
1
≤
0
Z
2
0
h
−1−pγ
Z
r1
|∆h (un (x) − u(x))|p dxdh
1/p
1/q
r1
0
1/q
γ
= kun − ukḂp,p
([0,r1 ]) r1
1/q
< 5 r1 .
Therefore,
1 − 2−pγ
1/q
4 µ(∆2 ) < 5 r1 ,
pγ
2
for every r ∈ [0, r1 ].
which is a contradiction with (2.7). Therefore, |u| ≤ AB
Now suppose that the inequality
1
|u(t, r)| ≥ 2
A
is not true for every r ∈ [ α1 , β1 ]. Since u ∈ C([0, r1 ]) we may take 6 > 0 and
∆3 ⊂ [ α1 , β1 ] such that
1
|u| ≤ 2 − 6 f or r ∈ ∆3 .
A
Then for every natural n and for every r ∈ ∆3 we have
1
1
|un − u| ≥ |un | − |u| ≥ 2 + 6 − 2 = 6 .
A
A
Let 7 > 0 be such that
1 − 2−pγ
1/q
6 µ(∆3 ) > 7 r1 .
pγ
(2.8)
γ
There exist M > 0 such that for every n > M we have kun − ukḂp,p
([0,r1 ]) < 7 .
Consequently, for every n > M and for every x ∈ ∆3 , we have
|un (x) − u(x)| > 6 ,
γ
kun − ukḂp,p
([0,r1 ]) < 7 .
Also using the Hölder’s inequality, we have
Z
Z
6 µ(∆3 ) <
|un (x) − u(x)|dx ≤
∆3
0
r1
1/p
1/q
|un (x) − u(x)|p dx
r1 .
EJDE-2005/67
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
17
For h > 0 we have
h−1−pγ 6 µ(∆3 ) ≤ h−1−pγ
r1
Z
1/p
1/q
|un (x) − u(x)|p dx
r1 ,
0
2
Z
2
Z
h−1−pγ 6 µ(∆3 )dh ≤
1
r1
Z
h−1−pγ
1/p
1/q
|un (x) − u(x)|p dx
r1 dh,
0
1
Using the Hölder’s inequality and that for h > 1, h(−1−pγ)p ≤ h−1−pγ , we have
Z r1
Z 2
1/p
1 − 2−pγ
1/q
(−1−pγ)p
|un (x) − u(x)|p dxdh
6 µ(∆3 ) ≤
h
r1
pγ
0
1
Z r1
Z 2
1/p
1/q
|un (x) − u(x)|p dxdh
≤
h−1−pγ
r1 .
0
1
Using that un = u = 0 for x > r1 , the above expression is less than or equal to
Z r1
Z 2
1/p
1/q
h−1−pγ
|∆h (un (x) − u(x))|p dxdh
r1
1
≤
0
Z
2
h−1−pγ
r1
Z
0
|∆h (un (x) − u(x))|p dxdh
1/p
1/q
r1
0
1/q
1/q
γ
= kun − ukḂp,p
([0,r1 ]) r1
< 7 r1 .
Therefore,
1 − 2−pγ
1/q
6 µ(∆2 ) < 7 r1 ,
pγ
which is a contradiction with (2.8). Therefore, |u| ≥ A12 for every r ∈ [ α1 , β1 ].
Now suppose that the inequality
u(t, r) ≥ 0
is not true for every r ∈
natural n and for every
for every natural n and
[ α1 , r1 ]. Then from u ∈ C([0, r1 ]) and from un ≥ 0 for every
r ∈ [ α1 , r1 ], we may take 8 > 0 and ∆4 ⊂ [ α1 , r1 ] such that
for every r ∈ ∆4 we have
|un − u| ≥ 8 .
Let 9 > 0 be such that
1 − 2−pγ
1/q
8 µ(∆3 ) > 9 r1 .
(2.9)
pγ
γ
There exist M > 0 such that for every n > M we have kun − ukḂp,p
([0,r1 ]) < 9 .
Consequently, for every n > M and for every x ∈ ∆4 we have
|un (x) − u(x)| > 8 ,
γ
kun − ukḂp,p
([0,r1 ]) < 9 .
Also using the Hölder’s inequality,
Z
Z
8 µ(∆4 ) <
|un (x) − u(x)|dx ≤
∆4
r1
1/p
1/q
|un (x) − u(x)|p dx
r1 .
0
For h > 0 ,
h
−1−pγ
8 µ(∆4 ) ≤ h
−1−pγ
Z
r1
1/p
1/q
|un (x) − u(x)|p dx
r1 ,
0
Z
2
h
1
−1−pγ
Z
8 µ(∆4 )dh ≤
2
h
1
−1−pγ
Z
0
r1
1/p
1/q
|un (x) − u(x)|p dx
r1 dh
18
S. G. GEORGIEV
EJDE-2005/67
Using Hölder’s inequality and that for h > 1, h(−1−pγ)p ≤ h−1−pγ we have
Z r1
1/p
Z 2
1 − 2−pγ
1/q
|un (x) − u(x)|p dxdh
8 µ(∆4 ) ≤
h(−1−pγ)p
r1
pγ
0
1
Z r1
1/p
Z 2
1/q
−1−pγ
|un (x) − u(x)|p dxdh
≤
h
r1 .
0
1
Using that un = u = 0 for x > r1 , the above expression is less than or equal to
Z r1
Z 2
1/p
1/q
−1−pγ
h
|∆h (un (x) − u(x))|p dxdh
r1
1
≤
0
Z
0
2
h−1−pγ
Z
r1
|∆h (un (x) − u(x))|p dxdh
1/p
1/q
r1
0
1/q
γ
= kun − ukḂp,p
([0,r1 ]) r1
1/q
< 9 r1 .
Therefore,
1 − 2−pγ
1/q
8 µ(∆4 ) < 9 r1 ,
pγ
which is a contradiction with (2.9). Therefore, |u| ≥ 0 for every r ∈ [ α1 , r1 ].
Consequently u ∈ N . Then for every sequence {un } ⊂ N , which converges in
γ
([0, r1 ]) there exists u ∈ N for which
Ḃp,p
γ
lim kun − ukḂp,p
([0,r1 ]) = 0.
n→∞
γ
([0, r1 ])).
From lemma 2.2 we have that the set N is closed subset of C((0, 1]Ḃp,p
γ
Since C((0, 1]Ḃp,p ([0, r1 ])) is complete metric space and R : N → N is contractive
operator the equation (2.2) has unique nontrivial solution ũ ∈ N . From (2.2) we
have that ũ(r) ∈ C 2 [0, r1 ] and ũ(t, r1 ) = ũr (t, r1 ) = ũrr (t, r1 ) = 0.
Let ũ is the solution from the Theorem 2.1, i.e ũ is the solution to the equation
(2.2). Then ũ is solution to the Cauchy problem (1.1)-(1.2) with initial data
R r
R r1 1
1
s4
00

2 −Kτ +Q2

τ
s2 −Ks+Q2 v (1)ω(s)
r
τ

u0 = +s2 (m2 v(1)ω(s) − f (v(1)ω(s)) ds dτ
for r ≤ r1 ,



0
for r ≥ r1 ,
and
R r
R r1 1
1
s4
000

2 −Kτ +Q2

τ
s2 −Ks+Q2 v (1)ω(s)
r
τ

u1 = +s2 (m2 v 0 (1)ω(s) − f 0 (u)v 0 (1)ω(s) dsdτ = 0



0for r ≥ r1 .
for r ≤ r1 ,
γ
γ−1
From the proof of the Theorem 2.1, we have u0 ∈ Ḃp,p
(R+ ), u1 ∈ Ḃp,p
(R+ ),
γ
ũ ∈ C((0, 1]Ḃp,p
[0, r1 ]).
3. Blow-up of solutions to the Cauchy problem (1.1)-(1.2)
Let v(t) be the same function as in Theorem 2.1.
EJDE-2005/67
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
19
Theorem 3.1. Let m2 6= 0, γ ∈ (0, 1), p > 1 be fixed and the constants A, B,
Q, K, 1 < β < α satisfy the conditions (H1)-(H6). Let f ∈ C 2 (R1 ), f (0) = 0,
2m2 |u| ≤ f (l) (u) ≤ 3m2 |u|, l = 0, 1. Then the solution ũ of the Cauchy problem
(1.1)-(1.2) satisfies
γ
lim kũkḂp,p
[0,r1 ] = ∞.
t→0
Proof. For t ∈ (0, 1], we have
k∆h R(ũ)kpLp
Z r1 Z r+h
=
Z
1
− Kτ + Q2
Z
1
τ 2 − Kτ + Q2
Z
r
0
1
α
Z
1
2
τ − Kτ + Q2
Z
r+h
=
0
τ2
r
Z
r1
Z
+
1
α
r
r+h
r1
[
τ
r1
[
τ
r1
[
τ
p
s4
v 00 (t)
ũ + s2 (m2 ũ − f (ũ))]dsdτ dr
2
2
s − Ks + Q v(t)
s2
p
s4
v 00 (t)
ũ + s2 (m2 ũ − f (ũ))]dsdτ dr
2
− Ks + Q v(t)
p
v 00 (t)
s4
ũ + s2 (m2 ũ − f (ũ))]dsdτ dr.
s2 − Ks + Q2 v(t)
Let
Z
1
α
Z
r+h
I1 =
0
r
2
2
1
τ 2 − Kτ + Q2
Z
r1
[
τ
s4
v 00 (t)
ũ
s2 − Ks + Q2 v(t)
p
+ s (m ũ − f (ũ))]dsdτ dr,
Z r1
Z r1 Z r+h
s4
v 00 (t)
1
I2 =
[
ũ
1
τ 2 − Kτ + Q2 τ s2 − Ks + Q2 v(t)
r
α
p
+ s2 (m2 ũ − f (ũ))]dsdτ dr.
As in proof of Theorem 2.1, for I1 we have the estimate
I1 ≤ C p
p
2p(1−γ)
8p
8
2m2
2
+
4m
kũkpḂ γ hp .
p,p
p(1 − γ) (1 − K + Q2 )p 1 − K + Q2 α2 A2
Using that u(t, r) = 0, f (u(t, r)) = 0 for r ≥ r1 ), for I2 , we have
Z β1
1
s4
v 00 (t)
[ 2
ũ + s2 (m2 ũ − f (ũ))]dsdτ
2
2
2
1
1
τ
−
Kτ
+
Q
s
−
Ks
+
Q
v(t)
r
α
α
Z r+h
Z r1
p
1
s4
v 00 (t)
+
[
ũ + s2 (m2 ũ − f (ũ))]dsdτ dr
2
2
2
2
τ − Kτ + Q β1 s − Ks + Q v(t)
r
Z r1 Z r+h
Z β1
1
s4
v 00 (t)
[ 2
≤
ũ + s2 (m2 ũ − f (ũ))]dsdτ
2
2
2
1
1
τ
−
Kτ
+
Q
s
−
Ks
+
Q
v(t)
r
α
α
Z r+h
Z r1
s4
v 00 (t)
1
+
[ 2
ũ
2
2
2
τ − Kτ + Q β1 s − Ks + Q v(t)
r
p
+ s2 (m2 ũ − f (ũ))]dsdτ
dr.
Z
I2 =
r1
Z
r+h
20
S. G. GEORGIEV
EJDE-2005/67
Let
r+h
Z
I21 =
r
Z
r+h
I22 =
r
1
τ 2 − Kτ + Q2
Z
1
τ 2 − Kτ + Q2
Z
1
β
[
s4
v 00 (t)
ũ + s2 (m2 ũ − f (ũ))]dsdτ,
s2 − Ks + Q2 v(t)
[
v 00 (t)
s4
ũ + s2 (m2 ũ − f (ũ))]dsdτ .
s2 − Ks + Q2 v(t)
1
α
r1
1
β
Then
Z
I2 ≤
r1 I21 + I22
p
dr.
1
α
For I21 we have the following estimate, we use that for r ∈ [ α1 , β1 ] u ≥ 0, f (u) ≥
2m2 u, therefore −f (u) ≤ −2m2 u,
Z β1 Z r+h
s4
v 00 (t)
1
2 2
ũ
−
s
m
ũ
dsdτ
I21 ≤
τ 2 − Kτ + Q2 α1 s2 − Ks + Q2 v(t)
r
Z r+h
Z β1 2 1
s4
v 00 (t) A2 ũ
2 2 A ũ
=
−
s
m
dsdτ
τ 2 − Kτ + Q2 α1 s2 − Ks + Q2 v(t) A2
A2
r
Z r+h
Z β1 1
v 00 (t) 1
s4
2 2 1
=
−
s
m
A2 ũdsdτ.
τ 2 − Kτ + Q2 α1 s2 − Ks + Q2 v(t) A2
A2
r
From (H4) we have that for s ∈ [ α1 , β1 ],
s4
v 00 (t) 1
s2 m2
1
1
m2
1 m2
−
≥
−
> 0.
s2 − Ks + Q2 v(t) A2
A2
α2 1 − αK + α2 Q2 α2 A4
β 2 A2
On the other hand we have ũ ≥ A12 for every t ∈ (0, 1] and every r ∈ [ α1 , β1 ];
therefore, A2 ũ ≥ 1 and A2 ũ ≤ A2p ũp . Consequently
Z r+h
Z β1 1
s4
v 00 (t) 1
2 2 1
I21 ≤
−
s
m
A2p ũp dsdτ.
τ 2 − Kτ + Q2 α1 s2 − Ks + Q2 v(t) A2
A2
r
From (H6),
8
≤ A ≤ A2 .
1 − K + Q2
From this inequality, we have
Z β1 00
Z r+h
v (t)
m2
8
[
− 2 2 ]A2p ũp dsdτ
I21 ≤
2
1 − K + Q α1 v(t)
α A
r
2
Z
1
v 00 (t) − αm
8
2 A2 v
2p
A
ũp dsh
≤
1 − K + Q2
v
0
2
=
v 00 (t) − αm
8
2 A2 v
A2p kũkpLp h
2
1−K +Q
v
2
v 00 (t) − αm
8
2 A2 v
≤
A2p kũkpLp h.
(1 − K + Q2 )2
v
Now we use Lemma 1.3 to get
2
I21
v 00 (t) − αm
82
2 A2 v
≤C
A2p kũkpḂ γ h.
p,p
(1 − K + Q2 )2
v
p
EJDE-2005/67
BLOW UP OF SOLUTIONS FOR KLEIN-GORDON EQUATIONS
As in proof of Theorem 2.1 for I22 we have
8
8
2m2
2
γ h.
I22 ≤ C
kũkḂp,p
+
4m
(1 − K + Q2 ) 1 − K + Q2 α2 A2
Consequently,
I2 ≤ [C
8
8
2m2
2
γ h
kũkḂp,p
+
4m
(1 − K + Q2 ) 1 − K + Q2 α2 A2
2
+ Cp
k∆h R(ũ)kpLp ≤ [C
v 00 (t) − αm
82
2 A2 v
A2p kũkpḂ γ h]p ,
2
2
p,p
(1 − K + Q )
v
8
2m2
8
γ h
+ 4m2 kũkḂp,p
2
2
2
2
(1 − K + Q ) 1 − K + Q α A
2
v 00 (t) − αm
82
2 A2 v
A2p kũkpḂ γ h]p
2
2
p,p
(1 − K + Q )
v
p
2
p
2m
8
8
2
+ Cp
+
4m
kũkpḂ γ hp .
p,p
(1 − K + Q2 )p 1 − K + Q2 α2 A2
+ Cp
Then
kũkpḂ γ
p,p
hh
≤ C
8
2m2
8
2
γ
+
4m
kũkḂp,p
(1 − K + Q2 ) 1 − K + Q2 α2 A2
2
ip
v 00 (t) − αm
82
2 A2 v
p
p
2p
+C
A
kũk
γ
Ḃp,p
(1 − K + Q2 )2
v
p
iZ 2
2
p
8
2m
8
p
p
2
+C
+ 4m
kũkḂ γ
h−1+p(1−γ) dh
p,p
(1 − K + Q2 )p 1 − K + Q2 α2 A2
0
8
8
2m2
2p(1−γ) hh
2
γ
C
+
4m
kũkḂp,p
=
p(1 − γ)
(1 − K + Q2 ) 1 − K + Q2 α2 A2
2
ip
v 00 (t) − αm
82
2 A2 v
p
2p
p
A
kũk
+C
γ
Ḃp,p
(1 − K + Q2 )2
v
p
i
2
p
8
2m
8
p
2
+
4m
kũk
,
+ Cp
γ
Ḃp,p
(1 − K + Q2 )p 1 − K + Q2 α2 A2
and
γ
kũkḂp,p
≤ 2C
8.21−γ
2m2
8
2
γ
+
4m
kũkḂp,p
(1 − K + Q2 )(p(1 − γ))1/p 1 − K + Q2 α2 A2
2
v 00 (t) − αm
82 .21−γ
2 A2 v
+C
A2p kũkpḂ γ .
p,p
v
(1 − K + Q2 )2 (p(1 − γ))1/p
p
Let
D = 2C
8.21−γ
8
2m2
2
+
4m
,
(1 − K + Q2 )(p(1 − γ))1/p 1 − K + Q2 α2 A2
2
v 00 (t) − αm
82 .21−γ
2 A2 v
F =C
A2p .
v
(1 − K + Q2 )2 (p(1 − γ))1/p
p
21
22
S. G. GEORGIEV
EJDE-2005/67
From (H5) we have that D < 1. Then
γ
γ
kũkḂp,p
≤ DkũkḂp,p
+ F kũkpḂ γ .
p,p
from this inequality,
γ
(1 − D)kũkḂp,p
≤ kũkpḂ γ ,
p,p
1−D
p−1
.
≤ kũkḂ
γ
p,p
F
Since limt→0 F = +0, we have
γ
lim kũkḂp,p
= ∞.
t→0
References
[1] Shatah, J., M. Struwe; Geometric wave equation. Courant lecture notes in mathematics
2(1998).
[2] Taylor, M.; Partial differential equations III. Springer-Verlag, 1996.
[3] Runst, T., W. Sickel.; Sobolev spaces of fractional order, Nemytskij operators and nonlinear
partial differential equations, 1996, New York.
Svetlin Georgiev Georgiev
University of Sofia, Faculty of Mathematics and Informatics, Department of Differential Equations, Bulgaria
E-mail address: sgg2000bg@yahoo.com