Hindawi Publishing Corporation
Boundary Value Problems
Volume 2007, Article ID 42954, 51 pages
doi:10.1155/2007/42954
Research Article
Blow up of the Solutions of Nonlinear Wave Equation
Svetlin Georgiev Georgiev
Received 14 March 2007; Accepted 26 May 2007
Recommended by Peter Bates
We con st ruc t for e ver y fixed n≥2 the metric gs=h1(r)dt2−h2(r)dr2−k1(ω)dω2
1−
···−kn−1(ω)dω2
n−1,whereh1(r), h2(r), ki(ω), 1 ≤i≤n−1, are continuous functions,
r=|x|, for which we consider the Cauchy problem (utt −∆u)gs =f(u)+g(|x|), where
x∈Rn,n≥2; u(1,x)=u◦(x)∈L2(Rn), ut(1,x)=u1(x)∈˙
H−1(Rn), where f∈Ꮿ1(R1),
f(0) =0, a|u|≤f′(u)≤b|u|,g∈Ꮿ(R+), g(r)≥0, r=|x|,aand bare positive con-
stants. When g(r)≡0, we prove that the above Cauchy problem has a nontrivial solution
u(t,r)intheformu(t,r)=v(t)ω(r) for which limt→0uL2([0,∞)) =∞.Wheng(r)= 0,
we prove that the above Cauchy problem has a nontrivial solution u(t,r)intheform
u(t,r)=v(t)ω(r) for which limt→0uL2([0,∞)) =∞.
Copyright © 2007 Svetlin Georgiev Georgiev. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, dis-
tribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
In this paper, we study the properties of the solutions of the Cauchy problem
utt −∆ugs=f(u)+g|x|,x∈Rn,n≥2, (1)
u(1,x)=u◦(x)∈L2Rn,ut(1,x)=u1(x)∈˙
H−1Rn,(2)
where gsis the metric
gs=h1(r)dt2−h2(r)dr2−k1(ω)dω2
1−···−kn−1(ω)dω2
n−1, (1.1)
2 Boundary Value Problems
the functions h1(r), h2(r) satisfy the conditions
h1(r),h2(r)∈Ꮿ1[0,∞),h1(r)>0, h2(r)≥0∀r∈[0,∞),
∞
0h2(s)
h1(s)∞
sh2(τ)
h1(τ)dτ < ∞,∞
0h2(s)
h1(s)∞
sh1(τ)h2(τ)dτ ds < ∞,
∞
0∞
rh2(s)
h1(s)∞
sh2(τ)
h1(τ)C1+C2h1(τ)h2(τ)2
dτ1/2
ds2
dr < ∞,
C1,C2are arbitrary nonnegative constants,
∞
0∞
rh2(s)
h1(s)∞
sh2(τ)
h1(τ)C1+C2h1(τ)h2(τ)dτds2
dr < ∞,
C1,C2are arbitrary nonnegative constants,
max
r∈[0,∞)h1(r)h2(r)<∞,
∞
0∞
rh2(s)
h1(s)∞
sh2(τ)
h1(τ)dτ ds2
dr < ∞,
∞
0∞
rh2(s)
h1(s)∞
sh1(τ)h2(τ)dτ1/2
ds2
dr < ∞,
∞
0∞
rh2(s)
h1(s)∞
s
h2(τ)
h1(τ)dτ1/2
ds2
dr < ∞,
∞
0∞
rh2(s)
h1(s)ds2
dr < ∞,
(i1)
ki(ω)∈Ꮿ1([0,2π]×···×[0,2π]), i=1,...,n−1, f∈Ꮿ1(᏾1), f(0) =0, a|u|≤f′(u)≤
b|u|,aand bare positive constants, g∈Ꮿ(R1), g(|x|)≥0for|x|∈[0,∞). (In Section 2
we will give example for such metric gs.)
We search a solution u=u(t,r) to the Cauchy problem (1), (2). Therefore, if the
Cauchy problem (1), (2) has such solution, it will satisfy the Cauchy problem
1
h1(r)utt −1
h1(r)h2(r)∂rh1(r)
h1(r)h2(r)ur=f(u)+g(r), (1.2)
u(1,r)=u◦∈L2[0,∞),ut(1,r)=u1∈˙
H−1[0,∞).(1.3)
In this paper, we will prove that the Cauchy problem (1), (2) has nontrivial solution
u=u(t,r) for which
lim
t→0uL2([0,∞)) =∞.(1.4)
Our main results are the following.
Svetlin Georgiev Georgiev 3
Theorem 1.1. Suppose n≥2is fixed, h1(r),h2(r)satisfy the conditions (i1), g≡0,f∈
Ꮿ1(R1),f(0) =0,a|u|≤f′(u)≤b|u|,aand bare positive constants. Then the homoge-
neous problem of Cauchy (1), (2) has nontrivial solution u=u(t,r)∈Ꮿ((0,1]L2([0,∞)))
for which
lim
t→0uL2([0,∞)) =∞.(1.5)
Theorem 1.2. Suppose n≥2is fixed, h1(r),h2(r)satisfy the conditions (i1). Suppose also
that aand bare fixed positive constants, a≤b,f∈Ꮿ1(R1),f(0) =0,a|u|≤f′(u)≤b|u|,
b/2≥f(1) ≥a/2,g= 0,g∈Ꮿ([0,∞)),g(r)≥0for every r≥0,g(r)≤b/2−f(1) for ev-
ery r∈[0,∞).ThenthenonhomogeneousproblemofCauchy(
1), (2) has nontrivial solution
u=u(t,r)∈Ꮿ((0,1]L2([0,∞))) for which
lim
t→0uL2([0,∞)) =∞.(1.6)
When gsis the Minkowski metric and u0,u1∈Ꮿ∞
0(R3)in[
1] (see also [2, Section 6.3]),
it is proved that there exists T>0 and a unique local solution u∈Ꮿ2([0,T)×R3)forthe
Cauchy problem
utt −∆ugs=f(u), f∈Ꮿ2(R), t∈[0,T], x∈R3,
u
t=0=u0,ut
t=0=u1,(1.7)
for which
sup
t<T,x∈R3
u(t,x)
=∞.(1.8)
When gsis the Minkowski metric, 1 ≤p<5 and initial data are in Ꮿ∞
0(R3)in[
1] (see
also [2, Section 6.3]), it is proved that the initial value problem
utt −∆ugs=u|u|p−1,t∈[0,T], x∈R3,
u
t=0=u0,ut
t=0=u1
(1.9)
admits a global smooth solution.
When gsis the Minkowski metric and initial data are in Ꮿ∞
0(R3)in[
3](seealso[2,
Section 6.3]) it is proved that there exists a number ǫ0>0 such that for any data (u0,u1)∈
Ꮿ∞
0(R3)withE(u(0)) <ǫ0, the initial value problem
utt −∆ugs=u5,t∈[0,T], x∈R3,
u
t=0=u0,ut
t=0=u1
(1.10)
admits a global smooth solution.
When gsis the Reissner-Nordstr¨
om metric in [4], it is proved that the Cauchy problem
utt −∆ugs+m2u=f(u), t∈[0,1], x∈R3,
u(1,x)=u0∈˙
Bγ
p,pR3,ut(1,x)=u1∈˙
Bγ−1
p,pR3,(1.11)
4 Boundary Value Problems
where m= 0isconstantand f∈Ꮿ2(R1), a|u|≤|f(l)(u)|≤b|u|,l=0,1, aand bare pos-
itive constants, has unique nontrivial solution u(t,r)∈Ꮿ((0,1] ˙
Bγ
p,p(R+)), r=|x|,p>1,
for which
lim
t→0u˙
Bγ
p,p(R+)=∞.(1.12)
When gsis the Minkowski metric in [5], it is proved that the Cauchy problem
utt −∆ugs=f(u), t∈[0,1], x∈R3,
u(1,x)=u0,ut(1,x)=u1
(1.13)
has global solution. Here f∈Ꮿ2(R), f(0) =f′(0) =f′′(0) =0,
f′′(u)−f′′(v)
≤B|u−v|q1(1.14)
for |u|≤1, |v|≤1, B>0, √2−1<q
1≤1, u0∈Ꮿ5
◦(R3), u1∈Ꮿ4
◦(R3), u0(x)=u1(x)=0
for |x−x0|>ρ,x0and ρare suitable chosen.
When gsis the Reissner-Nordstr¨
om metric, n=3, p>1, q≥1, γ∈(0,1) are fixed
constants, f∈Ꮿ1(R1), f(0) =0, a|u|≤f′(u)≤b|u|,g∈Ꮿ(R+), g(|x|)≥0, g(|x|)=0
for |x|≥r1,aand bare positive constants, r1>0 is suitable chosen, in [6], it is proved
that the initial value problem (1), (2) has nontrivial solution u∈Ꮿ((0,1] ˙
Bγ
p,q(R+)) in the
form
u(t,r)=⎧
⎨
⎩
v(t)ω(r), for r≤r1,t∈[0,1],
0, for r≥r1,t∈[0,1], (1.15)
where r=|x|, for which limt→0u˙
Bγ
p,q(R+)=∞.
The paper is organized as follows. In Section 2, we will prove some preliminary results.
In Section 3,wewillproveTheorem 1.1.InSection 4,wewillproveTheorem 1.2.Inthe
appendix we will prove some results which are used for the proof of Theorems 1.1 and
1.2.
2. Preliminary results
Proposition 2.1. Let h1(r),h2(r)satisfy the conditions (i1), f∈Ꮿ(−∞,∞),g≡0.If
for every fixed t∈[0,1] the function u(t,r)=v(t)ω(r),wherev(t)∈Ꮿ4([0,1]),v(t)= 0
for every t∈[0,1],ω(r)∈Ꮿ2([0,∞)),ω(∞)=ω′(∞)=0,satisfies(
1), then the function
u(t,r)=v(t)ω(r)satisfies the integral equation
u(t,r)=∞
rh2(s)
h1(s)∞
sh2(τ)
h1(τ)
v′′(t)
v(t)u(t,τ)−h1(τ)h2(τ)f(u)dτ ds (1∗)
for every fixed t∈[0,1].
Proof. Suppose that t∈[0,1]isfixedandthefunctionu(t,r)=v(t)ω(r), v(t)∈Ꮿ4([0,1]),
v(t)= 0foreveryt∈[0,1], ω(r)∈Ꮿ2([0,∞)), ω(∞)=ω′(∞)=0, satisfies (1). Then for
Svetlin Georgiev Georgiev 5
every fixed t∈[0,1] and for r∈[0,∞)wehave
utt(t,r)=v′′(t)
v(t)u(t,r),
1
h1(r)
v′′(t)
v(t)u(t,r)−1
h1(r)h2(r)∂rh1(r)
h1(r)h2(r)ur(t,r)=f(u),
1
h1(r)h2(r)∂rh1(r)
h1(r)h2(r)ur(t,r)=1
h1(r)
v′′(t)
v(t)u(t,r)−f(u),
∂rh1(r)
h1(r)h2(r)ur(t,r)=h2(r)
h1(r)
v′′(t)
v(t)u(t,r)−h1(r)h2(r)f(u).
(2.1)
Now we integrate the last equality from rto ∞here we suppose that ur(t,r)=v(t)ω′(r),
ur(t,∞)=v(t)ω′(∞)=0, then we get
−h1(r)
h1(r)h2(r)ur(t,r)=∞
rh2(τ)
h1(τ)
v′′(t)
v(t)u(t,τ)−h1(τ)h2(τ)f(u)dτ,
−h1(r)
h2(r)ur(t,r)=∞
rh2(τ)
h1(τ)
v′′(t)
v(t)u(t,τ)−h1(τ)h2(τ)f(u)dτ,
−ur(t,r)=h2(r)
h1(r)∞
rh2(τ)
h1(τ)
v′′(t)
v(t)u(t,τ)−h1(τ)h2(τ)f(u)dτ.
(2.2)
Now we integrate the last equality from rto ∞;weusethatu(t,∞)=v(t)ω(∞)=0, then
we get
u(t,r)=∞
rh2(s)
h1(s)∞
sh2(τ)
h1(τ)
v′′(t)
v(t)u(t,τ)−h1(τ)h2(τ)f(u)dτ ds, (2.3)
that is, for every fixed t∈[0,1] if the function u(t,r)=v(t)ω(r) satisfies (1), then the
function u(t,r)=v(t)ω(r) satisfies the integral equation (1∗). Here v(t)∈Ꮿ4([0,1]),
v(t)= 0foreveryt∈[0,1], ω(r)∈Ꮿ2([0,∞)), ω(∞)=ω′(∞)=0.
Proposition 2.2. Let h1(r),h2(r)satisfy the conditions (i1), f∈Ꮿ(−∞,∞),g≡0.Iffor
every fixed t∈[0,1] the function u(t,r)=v(t)ω(r),wherev(t)∈Ꮿ4([0,1]),v(t)= 0for
every t∈[0,1],ω(r)∈Ꮿ2([0,∞)),ω(∞)=ω′(∞)=0, satisfies the integral equation (1∗)
then the function u(t,r)=v(t)ω(r)satisfies (1)foreveryfixedt∈[0,1].
Proof. Let t∈[0,1] be fixed and let the function u(t,r)=v(t)ω(r), where v(t)∈Ꮿ4([0,1]),
v(t)= 0foreveryt∈[0,1], ω(r)∈Ꮿ2([0,∞)), ω(∞)=ω′(∞)=0, satisfy the integral
equation (1∗). From here and from f∈Ꮿ(−∞,∞), for every fixed t∈[0,1] we have