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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 construc t for ever y fixed n
≥ 2 the metric g
s
= h
1
(r)dt
2
− h
2
(r)dr
2
− k
1
(ω)dω
2
1
−
···−
k
n−1
(ω)dω
2

n
−1
,whereh
1
(r), h
2
(r), k
i
(ω), 1 ≤ i ≤ n −1, are continuous functions,
r
=|x|, for which we consider the Cauchy problem (u
tt
−Δu)
gs
= f (u)+g(|x|), where
x
∈ R
n
, n ≥ 2; u(1,x) = u
◦
(x) ∈ L
2
(R
n
), u
t
(1,x) = u
1
(x) ∈
˙

H
−1
(R
n
), where f ∈ Ꮿ
1
(R
1
),
f (0)
= 0, a|u|≤f

(u) ≤ b|u|, g ∈ Ꮿ(R
+
), g(r) ≥ 0, r =|x|, a and b are 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 lim
t→0
u
L
2
([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 lim
t→0
u

L
2
([0,∞))
=∞.
Copyright © 2007 Svetlin Georgiev Georgiev. This is an open access article distributed
under the Creative Commons Attribution License, which p ermits 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

u
tt
−Δu

g
s
= f (u)+g

|
x|

, x ∈ R
n
, n ≥2, (1)
u(1,x)
= u
◦
(x) ∈ L
2


R
n

, u
t
(1,x) = u
1
(x) ∈
˙
H
−1

R
n

,(2)
where g
s
is the met ric
g
s
= h
1
(r)dt
2
−h
2
(r)dr
2
−k

1
(ω)dω
2
1
−···−k
n−1
(ω)dω
2
n
−1
, (1.1)
2 Boundary Value Problems
the functions h
1
(r), h
2
(r) satisfy the conditions
h
1
(r),h
2
(r) ∈Ꮿ
1

[0,∞)

, h
1
(r) > 0, h
2

(r) ≥0 ∀r ∈[0,∞),

∞
0

h
2
(s)
h
1
(s)

∞
s

h
2
(τ)
h
1
(τ)
dτ <
∞,

∞
0

h
2
(s)

h
1
(s)

∞
s

h
1
(τ)h
2
(τ)dτds < ∞,

∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s



h
2
(τ)
h
1
(τ)
C
1
+ C
2

h
1
(τ)h
2
(τ)

2
dτ

1/2
ds

2
dr < ∞,
C
1
,C

2
are arbitrary nonnegative constants,

∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
C
1

+ C
2

h
1
(τ)h
2
(τ)

dτ

ds

2
dr < ∞,
C
1
,C
2
are arbitrary nonnegative constants,
max
r∈[0,∞)

h
1
(r)h
2
(r) < ∞,

∞

0


∞
r

h
2
(s)
h
1
(s)

∞
s

h
2
(τ)
h
1
(τ)
dτ ds

2
dr < ∞,

∞
0



∞
r

h
2
(s)
h
1
(s)


∞
s
h
1
(τ)h
2
(τ)dτ

1/2
ds

2
dr < ∞,

∞
0



∞
r

h
2
(s)
h
1
(s)


∞
s
h
2
(τ)
h
1
(τ)
dτ

1/2
ds

2
dr < ∞,

∞
0



∞
r

h
2
(s)
h
1
(s)
ds

2
dr < ∞,
(i1)
k
i
(ω) ∈ Ꮿ
1
([0,2π] ×···×[0,2π]), i = 1, ,n −1, f ∈ Ꮿ
1
(᏾
1
), f (0) = 0, a|u|≤f

(u) ≤
b|u|, a and b are p ositive constants, g ∈ Ꮿ(R
1
), g(|x|) ≥ 0for|x|∈[0,∞). (In Section 2
we will give example for such metric g

s
.)
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
h
1
(r)
u
tt
−
1

h
1
(r)h
2
(r)
∂
r

h
1
(r)

h
1
(r)h
2

(r)
u
r

=
f (u)+g(r), (1.2)
u(1,r)
= u
◦
∈ L
2

[0,∞)

, u
t
(1,r) =u
1
∈
˙
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→0
u
L
2
([0,∞))
=∞. (1.4)
Our main results are the following.
Svetlin Georgiev Georgiev 3
Theorem 1.1. Suppose n
≥ 2 is fixed, h
1
(r), h
2
(r) satisfy the conditions (i1), g ≡ 0, f ∈
Ꮿ
1
(R
1
), f (0) =0, a|u|≤f

(u) ≤ b|u|, a and b are positive constants. Then the homoge-
neous problem of Cauchy (1), (2) has nontrivial solution u
= u(t,r) ∈ Ꮿ((0,1]L
2
([0,∞)))
for which
lim
t→0
u
L

2
([0,∞))
=∞. (1.5)
Theorem 1.2. Suppose n
≥ 2 is fixed, h
1
(r), h
2
(r) satisfy the conditions (i1). Suppose also
that a and b are fixed positive constants, a
≤ b, f ∈Ꮿ
1
(R
1
), f (0) = 0, a|u|≤f

(u) ≤ b|u|,
b/2
≥ f (1) ≥ a/2, g = 0, g ∈ Ꮿ([0,∞)), g(r) ≥ 0 for every r ≥ 0, g(r) ≤ b/2 − f (1) for ev-
ery r
∈ [0,∞).ThenthenonhomogeneousproblemofCauchy(1), (2) has nontrivial solut ion
u
= u(t,r) ∈ Ꮿ((0,1]L
2
([0,∞))) for which
lim
t→0
u
L
2

([0,∞))
=∞. (1.6)
When g
s
is the Minkowski metric and u
0
,u
1
∈ Ꮿ
∞
0
(R
3
)in[1] (see also [2, Section 6.3]),
it is proved that there exists T>0 and a unique local solution u
∈ Ꮿ
2
([0,T) ×R
3
)forthe
Cauchy problem

u
tt
−Δu

g
s
= f (u), f ∈ Ꮿ
2

(R), t ∈ [0, T], x ∈ R
3
,
u


t=0
= u
0
, u
t


t=0
= u
1
,
(1.7)
for which
sup
t<T,x∈R
3


u(t,x)


=∞
. (1.8)
When g

s
is the Minkowski metric, 1 ≤ p<5 and initial data are in Ꮿ
∞
0
(R
3
)in[1] (see
also [2, Section 6.3]), it is proved that the initial value problem

u
tt
−Δu

g
s
= u|u|
p−1
, t ∈ [0,T], x ∈ R
3
,
u


t=0
= u
0
, u
t



t=0
= u
1
(1.9)
admits a global smooth solution.
When g
s
is the Minkowski metric and initial data are in Ꮿ
∞
0
(R
3
)in[3](seealso[2,
Section 6.3]) it is proved that there exists a number

0
> 0 such that for any data (u
0
,u
1
) ∈
Ꮿ
∞
0
(R
3
)withE(u(0)) < 
0
, the initial value problem


u
tt
−Δu

g
s
= u
5
, t ∈ [0,T], x ∈ R
3
,
u


t=0
= u
0
, u
t


t=0
= u
1
(1.10)
admits a global smooth solution.
When g
s
is the Reissner-Nordstr
¨

om metric in [4], it is proved that the Cauchy problem

u
tt
−Δu

g
s
+ m
2
u = f (u), t ∈[0,1], x ∈ R
3
,
u(1,x)
= u
0
∈
˙
B
γ
p,p

R
3

, u
t
(1,x) = u
1
∈

˙
B
γ−1
p,p

R
3

,
(1.11)
4 Boundary Value Problems
where m
= 0isconstantand f ∈Ꮿ
2
(R
1
), a|u|≤|f
(l)
(u)|≤b|u|, l =0,1, a and b are pos-
itive constants, has unique nontrivial solution u(t, r)
∈ Ꮿ((0,1]
˙
B
γ
p,p
(R
+
)), r =|x|, p>1,
for which
lim

t→0
u
˙
B
γ
p,p
(R
+
)
=∞. (1.12)
When g
s
is the Minkowski metric in [5], it is proved that the Cauchy problem

u
tt
−Δu

g
s
= f (u), t ∈[0,1], x ∈ R
3
,
u(1,x)
= u
0
, u
t
(1,x) = u
1

(1.13)
has global solution. Here f
∈ Ꮿ
2
(R), f (0) = f

(0) = f

(0) = 0,


f

(u) − f

(v)


≤
B|u −v|
q
1
(1.14)
for
|u|≤1, |v|≤1, B>0,
√
2 −1 <q
1
≤ 1, u
0

∈ Ꮿ
5
◦
(R
3
), u
1
∈ Ꮿ
4
◦
(R
3
), u
0
(x) = u
1
(x) = 0
for
|x −x
0
| >ρ, x
0
and ρ are suitable chosen.
When g
s
is the Reissner-Nordstr
¨
om metric, n = 3, p>1, q ≥ 1, γ ∈ (0,1) are fixed
constants, f
∈ Ꮿ

1
(R
1
), f (0) = 0, a|u|≤f

(u) ≤ b|u|, g ∈ Ꮿ(R
+
), g(|x|) ≥ 0, g(|x|) = 0
for
|x|≥r
1
, a and b are positive constants, r
1
> 0 is suitable chosen, in [6], it is proved
that the initial value problem (1), (2) has nontriv ial solution u
∈ Ꮿ((0,1]
˙
B
γ
p,q
(R
+
)) in the
form
u(t,r)
=
⎧
⎨
⎩
v(t)ω(r), for r ≤ r

1
, t ∈ [0, 1],
0, for r
≥ r
1
, t ∈ [0, 1],
(1.15)
where r
=|x|, for which lim
t→0
u
˙
B
γ
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 h
1
(r), h
2
(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) ∈ Ꮿ
4
([0,1]), v(t) = 0
for every t
∈ [0,1], ω(r) ∈ Ꮿ
2
([0,∞)), ω(∞) =ω

(∞) = 0,satisfies(1 ), then the function
u(t,r)
= v(t)ω(r) satisfies the integral equation
u(t,r)
=

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2

(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(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) ∈ Ꮿ
4
([0,1]),
v(t)
= 0foreveryt ∈[0,1], ω(r) ∈Ꮿ
2
([0,∞)), ω(∞) = ω


(∞) = 0, satisfies (1). Then for
Svetlin Georgiev Georgiev 5
every fixed t
∈ [0,1] and for r ∈[0,∞)wehave
u
tt
(t,r) =
v

(t)
v(t)
u(t,r),
1
h
1
(r)
v

(t)
v(t)
u(t,r)
−
1

h
1
(r)h
2
(r)

∂
r

h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=
f (u),
1

h
1
(r)h
2
(r)
∂
r

h
1

(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=
1
h
1
(r)
v

(t)
v(t)
u(t,r)
− f (u),
∂
r

h
1
(r)

h

1
(r)h
2
(r)
u
r
(t,r)

=

h
2
(r)
h
1
(r)
v

(t)
v(t)
u(t,r)
−

h
1
(r)h
2
(r) f (u).
(2.1)
Now we integrate the last equality from r to

∞ here we suppose that u
r
(t,r) = v(t)ω

(r),
u
r
(t,∞) = v(t)ω

(∞) = 0, then we get
−
h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r) =

∞
r


h
2

(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ) f (u)

dτ,
−

h
1
(r)
h
2
(r)
u
r
(t,r) =


∞
r


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ) f (u)

dτ,
−u
r
(t,r) =

h
2

(r)
h
1
(r)

∞
r


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ) f (u)

dτ.
(2.2)

Now we integrate the last equality from r to
∞;weusethatu(t, ∞) = v(t)ω(∞) = 0, then
we get
u(t,r)
=

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)

u(t,τ)
−

h
1
(τ)h
2
(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) ∈ Ꮿ
4
([0,1]),
v(t)
= 0foreveryt ∈ [0,1], ω(r) ∈Ꮿ
2
([0,∞)), ω(∞) = ω

(∞) = 0. 
Proposition 2.2. Let h
1
(r), h
2
(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) ∈ Ꮿ
4
([0,1]), v(t) = 0 for
every t
∈ [0,1], ω(r) ∈ Ꮿ
2
([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) ∈Ꮿ
4
([0,1]),
v(t)
= 0foreveryt ∈ [0,1], ω(r) ∈ Ꮿ
2
([0,∞)), ω(∞) = ω

(∞) = 0, satisfy the integral
equation (1
∗
). From here and from f ∈ Ꮿ(−∞,∞), for every fixed t ∈ [0,1] we have
6 Boundary Value Problems
u(t,r)
∈ Ꮿ
2

([0,∞)) and
u
r
(t,r) =−

h
2
(r)
h
1
(r)

∞
r


h
2
(r)
h
1
(r)
v

(t)
v(t)
u(t,τ)
−

h

1
(τ)h
2
(τ) f (u)

dτ,

h
1
(r)
h
2
(r)
u
r
(t,r) =−

∞
r


h
2
(r)
h
1
(r)
v

(t)

v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ) f (u)

dτ,
h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r) =−

∞
r


h
2

(r)
h
1
(r)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ) f (u)

dτ,
∂
r

h
1
(r)

h
1
(r)h
2
(r)

u
r
(t,r)

=

h
2
(r)
h
1
(r)
v

(t)
v(t)
u(t,r)
−

h
1
(r)h
2
(r) f (u),

h
2
(r)
h
1

(r)
v

(t)
v(t)
u(t,r)
−∂
r

h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=

h
1
(r)h
2
(r) f (u),
1

h
1
(r)
v

(t)
v(t)
u(t,r)
−
1

h
1
(r)h
2
(r)
∂
r

h
1
(r)

h
1
(r)h
2
(r)
u
r

(t,r)

=
f (u).
(2.4)
Since for every fixed t
∈ [0,1] we have
v

(t)
v(t)
u(t,r)
= u
tt
(t,r), (2.5)
we get
1
h
1
(r)
u
tt
(t,r) −
1

h
1
(r)h
2
(r)

∂
r

h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=
f (u), (2.6)
that is, for every fixed t
∈ [0,1] if the function u(t,r) =v(t)ω(r), where v(t) ∈ Ꮿ
4
([0,1]),
v(t)
= 0foreveryt ∈ [0,1], ω(r) ∈ Ꮿ
2
([0,∞)), ω(∞) = ω

(∞) = 0, satisfies (1
∗
), then it

satisfies (1)foreveryfixedt
∈ [0,1]. 
Proposition 2.3. Let h
1
(r), h
2
(r) satisfy the conditions (i1), f ∈Ꮿ(−∞,∞), g ∈Ꮿ([0,∞)),
g(r)
≥ 0 for every r ≥ 0.Ifforeveryfixedt ∈ [0,1] the function u(t,r) = v(t)ω(r),where
v(t)
∈ Ꮿ
4
([0,1]), v(t) = 0 for every t ∈ [0,1], ω(r) ∈ Ꮿ
2
([0,∞)), ω(∞) = ω

(∞) = 0,sat-
isfies (1), then the function u(t,r)
= v(t)ω(r) satisfies the integral equation
u(t,r)
=

∞
r

h
2
(s)
h
1

(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(r)


dτ ds
(1

∗∗
)
for every fixed t
∈ [0,1].
Proof. Let t
∈ [0,1] be fixed and let the function u(t, r) = v(t)ω(r), v(t) ∈ Ꮿ
4
([0,1]),
v(t)
= 0foreveryt ∈[0,1], ω(r) ∈ Ꮿ
2
([0,∞)), ω(∞) = ω

(∞) = 0, satisfy (1). Then for
Svetlin Georgiev Georgiev 7
every fixed t
∈ [0,1] and for r ∈[0,∞)wehave
u
tt
(t,r) =
v

(t)
v(t)
u(t,r),
1
h
1
(r)
v


(t)
v(t)
u(t,r)
−
1

h
1
(r)h
2
(r)
∂
r

h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=


f (u)+g(r)

,
1

h
1
(r)h
2
(r)
∂
r

h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=
1
h
1

(r)
v

(t)
v(t)
u(t,r)
−

f (u)+g(r)

,
∂
r

h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=

h

2
(r)
h
1
(r)
v

(t)
v(t)
u(t,r)
−

h
1
(r)h
2
(r)

f (u)+g(r)

.
(2.7)
Now we integrate the last e quality from r to
∞;herewesupposethatu
r
(t,r) =v(t)ω

(r),
u
r

(t,∞) = v(t)ω

(∞) = 0, then we get
−
h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r) =

∞
r


h
2
(τ)
h
1
(τ)
v

(t)

v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(r)


dτ,
−

h
1
(r)
h
2
(r)
u
r
(t,r) =

∞
r



h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(r)


dτ,
−u
r
(t,r) =

h
2
(r)

h
1
(r)

∞
r


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(r)



dτ.
(2.8)
Now we integrate the last equality from r to
∞; we suppose that u(t,∞) = v(t)ω(∞) = 0,
then we get
u(t,r)
=

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v


(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(r)


dτ ds,
(2.9)
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) ∈ Ꮿ
4
([0,1]),
v(t)
= 0foreveryt ∈ [0,1], ω(r) ∈Ꮿ
2
([0,∞)), ω(∞) = ω

(∞) = 0. 

Proposition 2.4. Let h
1
(r), h
2
(r) satisfy the conditions (i1), f ∈Ꮿ(−∞,∞), g ∈Ꮿ([0,∞)),
g(r)
≥ 0 for every r ≥ 0.Ifforeveryfixedt ∈ [0,1] the function u(t,r) = v(t)ω(r),where
8 Boundary Value Problems
v(t)
∈ Ꮿ
4
([0,1]), v(t) = 0 for every t ∈ [0,1], ω(r) ∈ Ꮿ
2
([0,∞)), ω(∞) = ω

(∞) = 0,sat-
isfies the integral equation (1
∗∗
), then the function u(t,r) = v(t)ω(r) satisfies (1)forevery
fixed t
∈ [0,1].
Proof. Let t
∈ [0,1] be fixed and let the function u(t,r)=v(t)ω(r), where v(t) ∈Ꮿ
4
([0,1]),
v(t)
= 0foreveryt ∈ [0,1], ω(r) ∈ Ꮿ
2
([0,∞)), ω(∞) = ω


(∞) = 0, satisfy the integral
equation (1
∗∗
). From here and from f ∈ Ꮿ(−∞, ∞), g ∈ Ꮿ([0, ∞)), for every fixed t ∈
[0,1] we have u(t,r) ∈Ꮿ
2
([0,∞)) and
u
r
(t,r) =−

h
2
(r)
h
1
(r)

∞
r


h
2
(r)
h
1
(r)
v


(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(r)


dτ,

h
1
(r)
h
2
(r)
u
r
(t,r) =−

∞
r



h
2
(r)
h
1
(r)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(r)


dτ,
h
1
(r)

h
1
(r)h

2
(r)
u
r
(t,r) =−

∞
r


h
2
(r)
h
1
(r)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(r)



dτ,
∂
r

h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=

h
2
(r)
h
1
(r)
v

(t)

v(t)
u(t,r)
−

h
1
(r)h
2
(r)

f (u)+g(r)

,

h
2
(r)
h
1
(r)
v

(t)
v(t)
u(t,r)
−∂
r

h
1

(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=

h
1
(r)h
2
(r)

f (u)+g(r)

,
1
h
1
(r)
v

(t)
v(t)

u(t,r)
−
1

h
1
(r)h
2
(r)
∂
r

h
1
(r)

h
1
(r)h
2
(r)
u
r
(t,r)

=
f (u)+g(r).
(2.10)
Since for every fixed t
∈ [0,1] we have

v

(t)
v(t)
u(t,r)
= u
tt
(t,r), (2.11)
we get
1
h
1
(r)
u
tt
(t,r) −
1

h
1
(r)h
2
(r)
∂
r

h
1
(r)


h
1
(r)h
2
(r)
u
r
(t,r)

=
f (u)+g(r), (2.12)
that is, for every fixed t
∈ [0,1] if the function u(t,r) =v(t)ω(r), where v(t) ∈ Ꮿ
4
([0,1]),
v(t)
= 0foreveryt ∈ [0, 1], ω(r) ∈Ꮿ
2
([0,∞)), ω(∞) = ω

(∞) = 0, satisfies (1
∗∗
), then it
satisfies (1)foreveryfixedt
∈ [0,1]. 
Svetlin Georgiev Georgiev 9
For fixed n
≥ 2, h
1
(r), h

2
(r) which satisfy the conditions (i1) and fixed positive con-
stants a and b, we suppose that the positive constants c, d, A, B, A
1
, A
2
satisfy the condi-
tions
c
≤ d, A ≥ B, A
1
≤ A
2
, A
1
−
b
2B
> 0,

h
2
(r)
h
1
(r)
A
1
−
b

B

h
1
(r)h
2
(r) ≥0foreveryr ∈[0,∞),

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1

−
b
2B

h
1
(τ)h
2
(τ)

dτ ds ≥ 1forr ∈[c,d],

∞
1

h
2
(s)
h
1
(s)

∞
s

A
1

h
2

(τ)
h
1
(τ)
−
b
B

h
1
(τ)h
2
(τ)

dτ ds ≥
A
1
10
10
(H1)
max
r∈[0,∞)

∞
r

h
2
(s)
h

1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
2B

dτ ds ≤ 1,
max
r∈[0,∞)

h

2
(r)
h
1
(r)

∞
r


h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
2B

dτ ≤ 1,
(H2)


∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2
+


h
1
(τ)h
2
(τ)
b
2B

2
dτ

1/2
ds

2
dr <
1
7
,

∞
0


∞
r

h
2
(s)

h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
2
+
b
B

h
1
(τ)h
2
(τ)

dτ ds

2
dr < ∞,

(H3)

∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2
+


h
1
(τ)h
2
(τ)
b
B

2
dτ

1/2
ds

2
dr < 1. (H4)
Example 2.5. Let 0 <


1/3 be enough small, n ≥ 2isfixed.Wechoosec>0, d>0,
c
≤ d<∞ such that for every r ∈ [c,d]wehave
π
4
≤ arctg(d +1−r)
3
,arctgd
3
<

π
3
. (2.13)
Let also b
= 8
3
, a = 4
3
, A = 60, B =40, A
1
= 
3
, A
2
= 2
3
.Let
h
1
(r) =

B
b

−
1+

1+2
A
1

b
B

2
, h
2
(r) =
144(d +1−r)
4

(d +1−r)
6
+1

2
. (2.14)
10 Boundary Value Problems
We note that the functions h
1
(r)andh
2
(r) satisfy all conditions of (i1)and
A
1

h
1
(r)
−
b

2B

h
1
(r) =1,

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1
−
b

2B

h
1
(τ)h
2
(τ)

dτ ds
≥

d+1
r

h
2
(s)
h
1
(s)

d+1
s


h
2
(τ)
h
1

(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)

dτ ds ≥ 1forr ∈[c,d] .
(2.15)
We note that

h
1
(r) ∼ 1,6.
For fixed n
≥ 2, h
1
(r), h
2
(r), which satisfy the conditions (i1), the constants a, b, c, d,
A, B, A
1
, A
2

are fixed which satisfy the conditions (H1), ,(H4), then we suppose that
the function v(t) is fixed function and satisfies the conditions
v(t)
∈ Ꮿ
4

[0,1]

,
v

(t)
v(t)
> 0, v(t) > 0,
∀t ∈ [0,1], (H5)
A
1
≤
v

(t)
v(t)
≤ A
2
, v

(1) = 0, v

(1) = 0, (H6)
lim

t→0

v

(t)
v(t)
−
a
2

=
+0. (H7)
Example 2.6. Let a, b, c, d, A
1
, A
2
, B, A be the constants from the above example. Then
a/2
= A
2
and
v(t)
= C

e
√
A
2
(t−1)
+ e

−
√
A
2
(t−1)

, (2.16)
where C is arbitrary positive constant, satisfiing the hypotheses (H5), (H6), (H7).
Here and below we suppose that v(t) is fixed function which satisfies the conditions
(H5), ,(H7).
When g(r)
≡ 0weput
u
◦
:=v(1)ω(r)=

∞
r

h
2
(s)
h
1
(s)

∞
s



h
2
(τ)
h
1
(τ)
v

(1)ω(τ)−

h
1
(τ)h
2
(τ) f

v(1)ω(τ)


dτ ds,
u
1
≡ 0.
(1

)
In Section 3,wewillprovethat(1

) has unique nontrivial solution ω(r) ∈L
2

([0,∞)).
Svetlin Georgiev Georgiev 11
When g(r)
= 0weput
u
◦
:= v(1)ω(r)
=

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v


(1)ω(τ) −

h
1
(τ)h
2
(τ)

f

v(1)ω(τ)

+ g(τ)


dτ ds,
u
1
≡ 0.
(1

)
In Section 4,wewillprovethat(1

) has unique nontrivial solution ω(r) ∈L
2
([0,1)).
3. Proof of Theorem 1.1
3.1. Local existence of nontrivial solutions of homogeneous Cauchy problem (1), (2).

In this section, we will prove that the homogeneous Cauchy problem (1), (2) has non-
trivial solution in the form u(t,r)
= v(t)ω(r).
For fixed function v(t), which satisfies the conditions (H5), (H6), and (H7)wecon-
sider the integ ral equation
u(t,r)
=

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v


(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ) f (u)

dτ ds. (1

)
Theorem 3.1. Let n
≥ 2 be fixed, le t h
1
(r), h
2
(r) fixed, which satisfy the conditions (i1),
let the positive constants a, b be fixed, a
≤ b, let the positive constants c, d. A, B, A
1
, A
2
be
fixed which satisfy the conditions (H1), , (H4)and f
∈ Ꮿ
1
((−∞,∞)), f (0) = 0, a|u|≤

f

(u) ≤ b|u|.Letalsov( t) be fixed function which satisfies the conditions (H5), , (H7).
Then (1

) has unique nontrivial solution u(t,r) = v(t)ω(r) for which u(t,r) ∈ Ꮿ([0,1] ×
[0,∞)), u(t,r) ≤ 1/B for every t ∈ [0,1] and for every r ∈ [0,∞), u(t,r) ≥ 1/A for every
t
∈ [0,1] and for every r ∈ [c,d], u(t,r) ≥ 0 for every t ∈ [0,1] and for every r ∈ [0,∞),
u(t,
∞) = u
r
(t,∞) = 0 for every t ∈[0,1], u(t, r) ∈C((0,1]L
2
([0,∞))).
Proof. Let M be the set
M
=

u(t,r):u(t,r) ∈Ꮿ

[0,1] ×[0,∞)

, u(t,∞) = u
r
(t,∞) = 0 ∀t ∈[0,1],
u(t,r)
≥
1
A

for t
∈ [0,1], r ∈[c,d], u(t,r) ≤
1
B
∀t ∈ [0,1], ∀r ∈[0,∞),
u(t,r)
≥ 0 ∀t ∈[0,1], ∀r ∈[0,∞), u(t,r) ∈ L
2

[0,∞)

for every t ∈ (0,1]

.
(3.1)
Let t
∈ [0,1] be fixed. We define the operator L as follows:
L(u)(t,r)
=

∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ) f (u)

dτ ds (3.2)
for u
∈ M. First we will see that L : M →M.Letu ∈ M. Then the fol lowing holds.
12 Boundary Value Problems
(1) Since v(t)
∈ Ꮿ
4

([0,1]), u(t,r) ∈ Ꮿ([0,1] ×[0, ∞)), and f ∈ Ꮿ(−∞,∞)andfrom
(i1)wehavethatL(u)
∈ Ꮿ([0,1] ×[0,∞)). Also we have
L(u)


r=∞
= 0,
∂
∂r
L(u)
=−

h
2
(r)
h
1
(r)

∞
r


h
2
(τ)
h
1
(τ)

v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ) f (u)

dτ,
∂
∂r
L(u)


r=∞
= 0.
(3.3)
(2) Now we will prove that for every fixed t
∈ [0, 1] and for every r ∈ [0,∞)wehave
that L(u)
≥ 0. Really,
L(u)
=

∞

r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u
−

h
1
(τ)h
2

(τ) f (u)

dτ ds
≥

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
min
t∈[0,1]
v

(t)

v(t)
u
−

h
1
(τ)h
2
(τ) f (u)

dτ ds
≥

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)

h
1
(τ)
A
1
u −

h
1
(τ)h
2
(τ) f (u)

dτ ds.
(3.4)
Now we suppose that for every fixed t
∈ [0,1] and for every r ∈ [0,∞)wehaveu(t,r) ≥0;
from here f

(u) ≤ bu, since f (0) = 0andu(t,r) ≤ 1/B for every t ∈ [0, 1] and for every
r
∈ [0,∞)weget
f (u)
≤
b
2
u
2
≤
b

2B
u(t,r)
≥

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1
u −
b
2B


h
1
(τ)h
2
(τ)u

dτ ds
≥

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A

1
−
b
2B

h
1
(τ)h
2
(τ)

udτds
≥

∞
r

h
2
(s)
h
1
(s)

∞
s
min
τ∈[0,∞)



h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)

udτds.
(3.5)
From (H1)wehavethat
min
τ∈[0,∞)


h
2
(τ)
h
1

(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)

≥
0 (3.6)
Svetlin Georgiev Georgiev 13
and since u(t,r)
≥ 0foreveryfixedt ∈ [0,1] and for every r ∈ [0,∞)weget

∞
r

h
2
(s)
h
1
(s)

∞

s
min
τ∈[0,∞)


h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)

udτds≥ 0, (3.7)
that is, for every fixed t
∈ [0,1] and for every r ∈ [0,∞)wehave
L(u)
≥ 0. (3.8)
(3) Now we will see that for every fixed t
∈ [0,1] and for every r ∈ [c,d]wehave

L(u)
≥ 1/A.Really,foreveryr ∈ [c,d]wehave
L

(u) =

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)

−

h
1
(τ)h
2
(τ) f

(u)

dτ ds
≥

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)

h
1
(τ)
A
1
−
b
B

h
1
(τ)h
2
(τ)

dτ ds ≥ 0.
(3.9)
(See (H1).) From here, since for every r
∈ [c, d]wehaveu(t,r) ≥1/A,weget
L(u)
≥ L

1
A

=

∞
r


h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
1
A
−

h
1
(τ)h
2
(τ) f


1
A


dτ ds
≥

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1

A
−
b
2A
2

h
1
(τ)h
2
(τ)

dτ ds
=
1
A

∞
r

h
2
(s)
h
1
(s)

∞
s



h
2
(τ)
h
1
(τ)
A
1
−
b
2A

h
1
(τ)h
2
(τ)

dτ ds
≥
1
A

∞
r

h
2
(s)

h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)

dτ ds ≥
1
A
.

(3.10)
(Hereweuse(H1).) Consequently, for every fixed t
∈ [0,1] and for every r ∈ [c,d]we
have that
L(u)
≥
1
A
. (3.11)
(4) Now we will prove that for every fixed t
∈ [0, 1] and for every r ∈ [0,∞)wehave
that
L(u)
≤
1
B
. (3.12)
14 Boundary Value Problems
Really , for every fixed t
∈ [0,1] and for every r ∈ [0,∞)wehave
L(u)
=

∞
r

h
2
(s)
h

1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u
−

h
1
(τ)h
2
(τ) f (u)

dτ ds
≤

∞

r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u +

h
1
(τ)h
2
(τ)



f (u)



dτ ds
≤

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
max

t∈[0,1]
v

(t)
v(t)
|u|+

h
1
(τ)h
2
(τ)


f (u)



dτ ds.
(3.13)
Now we suppose that


f (u)


≤
b
2
u

2
≤
b
2B
u,max
t∈[0,1]
v

(t)
v(t)
≤ A
2
,

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2

(τ)
h
1
(τ)
A
2
u +

h
1
(τ)h
2
(τ)
b
2B
u

dτ ds
=

∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
2B

udτds;
(3.14)
here we use
u
≤
1
B
≤

1
B

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h

2
(τ)
b
2B

dτ ds
≤
1
B
max
r∈[0,∞)

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h

1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
2B

dτ ds ≤
1
B
.
(3.15)
In the last inequality, we use (H2). Consequently, for every fixed t
∈ [0,1] and for every
r
∈ [0,∞)wehave
L(u)
≤
1
B
. (3.16)
(5) Now we will prove that for every fixed t
∈ [0,1] we have that L(u) ∈ L

2
([0∞)).
Really , for every fixed t
∈ [0,1] after we use the inequality (3.13)weget

∞
0


L(u)


2
dr
≤

∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
2B

|
u|dτ ds

2
dr.
(3.17)
Svetlin Georgiev Georgiev 15
Now we use the H

¨
older inequality

∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2

+

h
1
(τ)h
2
(τ)
b
2B

2
dτ

1/2
×


∞
s
|u|
2
dτ

1/2
ds

2
dr
≤


∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2
+

h

1
(τ)h
2
(τ)
b
2B

2
dτ

1/2
×


∞
0
|u|
2
dτ

1/2
ds

2
dr
=u
2
L
2
([0,∞))


∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2
+


h
1
(τ)h
2
(τ)
b
2B

2
dτ

1/2
ds

2
dr
≤u
2
L
2
([0,∞))
1
7
.
(3.18)
(In the last inequality, we use the condition (H3).) Consequently, for every fixed t
∈ [0,1]
we have L(u)
∈ L
2

([0,∞)).
From (1), (2), (3), (4), and (5) we get that for every fixed t
∈ [0,1] we have
L : M
−→ M. (3.19)
Now we will prove that the operator L : M
→ M is contractive operator. Let u
1
and u
2
be two elements of the set M.Then,foreveryfixedt ∈ [0,1] we have


L

u
1

−
L

u
2



=







∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)

u
1

−u
2

−

h
1
(τ)h
2
(τ)

f

u
1

−
f

u
2


dτ ds





≤


∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)


u
1
−u

2


+

h
1
(τ)h
2
(τ)


f

u
1

−
f

u
2




dτ ds,
(3.20)
16 Boundary Value Problems
then from the middle-point theorem we have

|f (u
1
) − f (u
2
)|=|f

(ξ)u
1
−u
2
|, |ξ|≤
max{|u
1
|,|u
2
|}, |f

(ξ)|≤b|ξ|≤b/B,

∞
r

h
2
(s)
h
1
(s)

∞

s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)


u
1
−u
2


+

h
1
(τ)h
2
(τ)
b
B



u
1
−u
2



dτ ds
=

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h

1
(τ)
v

(t)
v(t)
+

h
1
(τ)h
2
(τ)
b
B



u
1
−u
2


dτ ds
≤

∞
r


h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
max
t∈[0,1]
v

(t)
v(t)
+

h
1
(τ)h
2
(τ)

b
B



u
1
−u
2


dτ ds
≤

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2

(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
B



u
1
−u
2


dτ ds,
(3.21)
that is, for every fixed t
∈ [0,1] and for every r ∈ [0,∞)wehave



L

u
1

−
L

u
2



≤

∞
r

h
2
(s)
h
1
(s)

∞
s


h

2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
B



u
1
−u
2


dτ ds.
(3.22)
From here



L

u
1

−
L

u
2



2
L
2
([0,∞))
≤

∞
0


∞
r

h
2
(s)
h

1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
B



u
1
−u

2


dτ ds

2
dr.
(3.23)
Now we will use the H
¨
older inequality

∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s



h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
B

2
dτ

1/2
×


∞
s



u
1
−u
2


2
dτ

1/2
ds

2
dr
≤

∞
0


∞
r

h
2
(s)
h
1
(s)



∞
s


h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
B

2
dτ

1/2
×



∞
0


u
1
−u
2


2
dτ

1/2
ds

2
dr
=


u
1
−u
2


2
L
2

([0,∞))
×

∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2

+

h
1
(τ)h
2
(τ)
b
B

2
dτ

1/2
ds

2
dr,
(3.24)
Svetlin Georgiev Georgiev 17
that is, for every fixed t
∈ [0,1] we have


L

u
1

−

L

u
2



2
L
2
([0,∞))
≤


u
1
−u
2


2
L
2
([0,∞))

∞
0


∞

r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b

B

2
dτ

1/2
ds

2
dr,
(3.25)
from (H4)wehave

∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s



h
2
(τ)
h
1
(τ)
A
2
+

h
1
(τ)h
2
(τ)
b
B

2
dτ

1/2
ds

2
dr < 1. (3.26)
From here and from (3.25)weget



L

u
1

−
L

u
2



2
L
2
([0,∞))
<


u
1
−u
2


L
2
([0,∞))

. (3.27)
Consequently, the operator L : M
→ M is contractive operator. We note that the set M is
closed subset of the space Ꮿ((0,1]L
2
([0,∞))) (for the proof see Lemma A.1 in the appen-
dix of this paper). Therefore, (1

) has unique nontrivial solution in the set M. 
Let u be the solution from Theorem 3.1, that is, u is a solution to the integral equation
(1

). From Proposition 2.2,wehavethatu satisfies (1). Consequently, u is solution to the
Cauchy problem (1), (2) with initial data
u
◦
= v(1)ω(r) =

∞
r

h
2
(s)
h
1
(s)

∞
s



h
2
(τ)
h
1
(τ)
v

(1)ω(τ) −

h
1
(τ)h
2
(τ) f

v(1)ω(τ)


dτ ds,
u
1
≡ 0.
(3.28)
We have
u ∈ Ꮿ((0,1]L
2
([0,∞))), u

◦
∈ L
2
([0,∞)), u
1
∈
˙
H
−1
([0,∞)).
3.2. Blow up of the solutions of homogeneous Cauchy problem (1), (2). Let v(t)bethe
same function as in Theorem 3.1.
Theorem 3.2. Let n
≥ 2 be fixed, let h
1
(r), h
2
(r) fixed, which satisfy the conditions (i1), be
the positive constants a, b be fixed, let the positive constants c, d, A, B, A
1
, A
2
be fixed which
satisfy the conditions (H1), , (H4)and f
∈ Ꮿ
1
((−∞,∞)), f (0) = 0, a|u|≤f

(u) ≤ b|u|.
Then for the solution

u of the Cauchy problem (1), (2) one has
lim
t→0
u
L
2
([0,∞))
=∞. (3.29)
Proof. For every fixed t
∈ (0,1] and for every r ∈ [0,∞)wehave
u(t,r) =

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h

1
(τ)
v

(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds. (3.30)
18 Boundary Value Problems
Since for every fix ed t
∈ (0, 1] and for every r ∈ [c,d]wehavethatu ≥ 1/A follows that
there exists subinterval Δ in [0,
∞)suchthat
u ≥
1
A
for r
∈ Δ. (3.31)
Let us fix the subinterval Δ.Fromhere
u
2
L
2

([0,∞))
=

∞
0


∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)

v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds

2
dr
=

[0,∞)\Δ


∞
r

h
2
(s)
h
1
(s)

∞
s



h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds

2
dr
+

Δ


∞

r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)


dτ ds

2
dr.
(3.32)
Let
I
1
: =

[0,∞)\Δ


∞
r

h
2
(s)
h
1
(s)

∞
s


h
2

(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds

2
dr,
I
2
: =

Δ


∞
r


h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds


2
dr.
(3.33)
Then
u
2
L
2
([0,∞))
= I
1
+ I
2
. (3.34)
For I
1
we have the following estimate:
I
1
≤

∞
0


∞
r

h

2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)+

h
1
(τ)h
2
(τ)


f (u)




dτ ds

2
dr
≤

∞
0


∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)

h
1
(τ)
max
t∈[0,1]
v

(t)
v(t)
u(t,τ)+

h
1
(τ)h
2
(τ)


f (u)



dτ ds

2
dr
≤

∞
0



∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
2
u(t,τ)+

h
1
(τ)h
2

(τ)


f (u)



dτ ds

2
dr.
(3.35)
Svetlin Georgiev Georgiev 19
Now we suppose that f

(u) ≤ bu, f (0) = 0, u ≤1/B for every fixed t ∈ [0,1] and for every
r
∈ [0, ∞). Therefore |f (u)|≤(b/2)u
2
≤ (b/2B)u for every fixed t ∈ [0,1] and for every
r
∈ [0,∞).

∞
0


∞
r


h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
2
u(t,τ)+
b
2B

h
1
(τ)h
2
(τ)u

dτ ds


2
dr
=

∞
0


∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A

2
+
b
2B

h
1
(τ)h
2
(τ)


udτds

2
dr.
(3.36)
Now we apply the H
¨
older inequality

∞
0


∞
r

h
2

(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2
+
b
2B

h
1
(τ)h
2
(τ)

2
dτ


1/2


∞
s
|u|
2
dτ

1/2
ds

2
dr
≤

∞
0


∞
r

h
2
(s)
h
1
(s)



∞
s


h
2
(τ)
h
1
(τ)
A
2
+
b
2B

h
1
(τ)h
2
(τ)

2
dτ

1/2



∞
0
|u|
2
dτ

1/2
ds

2
dr
=

u
2
L
2
[0,∞)

∞
0


∞
r

h
2
(s)
h

1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
2
+
b
2B

h
1
(τ)h
2
(τ)

2
dτ

1/2

ds

2
dr.
(3.37)
Let
Q :
=

∞
0


∞
r

h
2
(s)
h
1
(s)


∞
s


h
2

(τ)
h
1
(τ)
A
2
+
b
2B

h
1
(τ)h
2
(τ)

2
dτ

1/2
ds

2
dr. (3.38)
Then
I
1
≤ Qu
2
L

2
([0,∞))
. (3.39)
Now we consider I
2
. For it we have
I
2
=

Δ


∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)

h
1
(τ)
v

(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds

2
dr
≤

Δ


∞
r

h
2
(s)

h
1
(s)

∞
0


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds

2
dr

=

Δ


∞
r

h
2
(s)
h
1
(s)

Δ


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ) −


h
1
(τ)h
2
(τ) f (u)

dτ ds
+

∞
r

h
2
(s)
h
1
(s)

[0,∞)\Δ


h
2
(τ)
h
1
(τ)
v


(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds

2
dr.
(3.40)
20 Boundary Value Problems
Let
I
21
=

∞
r

h
2
(s)
h
1
(s)


Δ


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds,
I
22
=

∞
r


h
2
(s)
h
1
(s)

[0,∞)\Δ


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds.
(3.41)

Consequently,
I
2
=

Δ

I
21
+ I
22

2
dr ≤ 2

Δ
I
2
21
dr +2

Δ
I
2
22
dr. (3.42)
Also we have

Δ
I

2
21
dr
=

Δ


∞
r

h
2
(s)
h
1
(s)

Δ


h
2
(τ)
h
1
(τ)
v

(t)

v(t)
u(t,τ) −

h
1
(τ)h
2
(τ) f (u)

dτ ds

2
dr
=

Δ


∞
r

h
2
(s)
h
1
(s)

Δ


h
1
(τ)h
2
(τ)

1
h
1
(τ)
v

(t)
v(t)
u(t,τ) − f (u)

dτ ds

2
dr
≤

Δ


∞
r

h
2

(s)
h
1
(s)

Δ
max
τ∈[0,∞)

h
1
(τ)h
2
(τ)

max
τ∈[0,∞)
1
h
1
(τ)
v

(t)
v(t)
u(t,τ)−f (u)

dτ ds

2

dr.
(3.43)
Case 1. Let
max
r∈[0,∞)
1
h
1
(r)
≤
1
A
. (3.44)
Then, after we suppose that for every fixed t
∈ [0,1] and for every r ∈Δ we have f (u) ≥
(a/2)u
2
≥ (a/2A)u,weget

Δ


∞
r

h
2
(s)
h
1

(s)

Δ
max
τ∈[0,∞)

h
1
(τ)h
2
(τ)

max
τ∈[0,∞)
1
h
1
(τ)
v

(t)
v(t)
u(t,τ) −
a
2A
u

dτ ds

2

dr
≤

Δ


∞
r

h
2
(s)
h
1
(s)

Δ
max
τ∈[0,∞)

h
1
(τ)h
2
(τ)
1
A

v


(t)
v(t)
−
a
2


udτds

2
dr.
(3.45)
Svetlin Georgiev Georgiev 21
Now we suppose that for every fixed t
∈ [0,1] and for every r ∈ Δ we have u ≥ 1/A,
A
u ≥ 1,

max
τ∈[0,∞)

h
1
(τ)h
2
(τ)
1
A

v


(t)
v(t)
−
a
2

2

Δ


∞
r

h
2
(s)
h
1
(s)

Δ
Au
1
A
dτ ds

2
dr

≤

max
τ∈[0,∞)

h
1
(τ)h
2
(τ)
1
A

v

(t)
v(t)
−
a
2

2

Δ


∞
r

h

2
(s)
h
1
(s)

Δ
A
2
u
2
1
A
dτ ds

2
dr
≤ A
2

max
τ∈[0,∞)

h
1
(τ)h
2
(τ)
1
A


v

(t)
v(t)
−
a
2

2

∞
0


∞
r

h
2
(s)
h
1
(s)

∞
0
u
2
dτ ds


2
dr
=

u
4
L
2
([0,∞))
A
2

max
τ∈[0,∞)

h
1
(τ)h
2
(τ)
1
A

v

(t)
v(t)
−
a

2

2

∞
0


∞
r

h
2
(s)
h
1
(s)
ds

2
dr,
(3.46)
that is,

∞
0
I
2
21
dr ≤u

4
L
2
([0,∞))
A
2

max
τ∈[0,∞)

h
1
(τ)h
2
(τ)
1
A

v

(t)
v(t)
−
a
2

2

∞
0



∞
r

h
2
(s)
h
1
(s)
ds

2
dr.
(3.47)
Let
F
= A
2

max
τ∈[0,∞)

h
1
(τ)h
2
(τ)
1

A

v

(t)
v(t)
−
a
2

2

∞
0


∞
r

h
2
(s)
h
1
(s)
ds

2
dr. (3.48)
Then


Δ
I
2
21
dr ≤ Fu
4
L
2
[0,∞)
. (3.49)
Case 2. Let
max
r∈[0,∞)
1
h
1
(r)
≥
1
A
. (3.50)
Let
max
r∈[0,∞)
1
h
1
(r)
= G. (3.51)

Then

Δ
I
2
21
dr ≤2

Δ


∞
r

h
2
(s)
h
1
(s)

Δ
max
τ∈[0,∞)

h
1
(τ)h
2
(τ)


G−
1
A

v

(t)
v(t)
u(t,τ)− f (u)

dτ ds

2
dr
+2

Δ


∞
r

h
2
(s)
h
1
(s)


Δ
max
τ∈[0,∞)

h
1
(τ)h
2
(τ)
1
A

v

(t)
v(t)
u(t,τ) −
a
2
u

dτ ds

2
dr.
(3.52)
22 Boundary Value Problems
As mentioned above
2Q
u

2
L
2
([0,∞))
+2Fu
4
L
2
([0,∞))
, (3.53)
that is,

Δ
I
2
21
dr ≤ 2Qu
2
L
2
([0,∞))
+2Fu
4
L
2
([0,∞))
. (3.54)
From (3.49)and(3.54)wehave

Δ

I
2
21
dr ≤ 2Qu
2
L
2
([0,∞))
+2Fu
4
L
2
([0,∞))
. (3.55)
As in the estimate for I
1
we have

∞
0
I
2
22
dr ≤ Qu
2
L
2
[0,∞)
. (3.56)
From (3.42), (3.55), and (3.56)weget

I
2
≤ 4Fu
4
L
2
([0,∞))
+6Qu
2
L
2
[0,∞)
. (3.57)
From the last inequality and from (3.34), (3.39)wehave
u
2
L
2
([0,∞))
≤ 4Fu
4
L
2
([0,∞))
+7Qu
2
L
2
([0,∞))
,

(1
−7Q)u
2
L
2
([0,∞))
≤ 4Fu
4
L
2
([0,∞))
.
(3.58)
From (H3)wehave
Q<
1
7
, (3.59)
from here
u
2
L
2
([0,∞))
≥
1 −7Q
4F
. (3.60)
From (H7)wehave
lim

t→0
F =+0. (3.61)
Therefore
lim
t→0
u
L
2
([0,∞))
=∞. (3.62)

Svetlin Georgiev Georgiev 23
4. Proof of Theorem 1.2
4.1. Local existence of nontrivial solutions of nonhomogeneous Cauchy problem (1),
(2). In this section we will prove that the nonhomogeneous Cauchy problem (1), (2)has
nontrivial solution in the form u(t,r)
= v(t)ω(r).
Let us consider the integral equation
u(t,r)
=

∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(τ)


dτ ds.
(1


)
Theorem 4.1. Let n
≥ 2 be fixed, let h
1
(r), h
2
(r) be fixed, which satisfy the conditions (i1),
the positive constants a, b, a
≤ b are fixed, and let the positive constants c, d, A, B, A
1
,
A
2
be fixed which satisfy the conditions (H1), , (H4)and f ∈ Ꮿ
1
((−∞,∞)), f (0) = 0,
a
|u|≤f

(u) ≤ b|u|, b/2 ≥ f (1) ≥ a/2, g ∈ Ꮿ([0,∞)), g(r) ≥0 for every r ∈ [0,∞), g(r) ≤
b/2 − f (1) for every r ∈[0,∞).Letalsov(t) be fixed function which satisfies the conditions
(H5), , (H7). Then (1

) has unique nontrivial solution u(t,r) = v(t)ω(r) for which
u(t,r)
∈ Ꮿ([0,1] ×[0,∞)), u(t, r) ≤ 1/B for every t ∈ [0,1] and for every r ∈ [0,∞),for
every t
∈ [0,1] and for every r ∈[c,d], u(t,r) ≥1/A, u(t,r) ≥ 0 for every t ∈ [0,1] and for
every r
∈ [0,∞), u(t,∞) = u

r
(t,∞) = 0 for every t ∈[0,1], u(t, r) ∈C((0,1]L
2
([0,∞))).
Proof. Let M be the set of the proof of Theorem 3.1.Lett
∈ [0,1] be fixed. We define the
operator R as follows:
R(u)(t,r)
=

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)

v

(t)
v(t)
u(t,τ)
−

h
1
(τ)h
2
(τ)

f (u)+g(τ)


dτ ds
(4.1)
for u
∈ M. First we will see that R : M →M.Letu ∈ M. Then the fol lowing holds.
(1) Since v(t)
∈ Ꮿ
4
([0,1]), u(t, r) ∈ Ꮿ([0, 1] ×[0,∞)) and f ∈ Ꮿ(−∞, ∞)andfrom
(i1)wehavethatR(u)
∈ Ꮿ([0,1] ×[0,∞)). Also we have
R(u)
|
r=∞
= 0,

∂
∂r
R(u)
=−

h
2
(r)
h
1
(r)

∞
r


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u(t,τ)
−

h

1
(τ)h
2
(τ)

f (u)+g(τ)


dτ,
∂
∂r
R(u)
|
r=∞
= 0.
(4.2)
24 Boundary Value Problems
(2) Now we will prove that for every fixed t
∈ [0, 1] and for every r ∈ [0,∞)wehave
that R(u)
≥ 0. Really,
R(u)
=

∞
r

h
2
(s)

h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
v

(t)
v(t)
u
−

h
1
(τ)h
2
(τ)

f (u)+g(τ)



dτ ds
≥

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
min
t∈[0,1]
v

(t)
v(t)
u

−

h
1
(τ)h
2
(τ)

f (u)+g(τ)


dτ ds
≥

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2

(τ)
h
1
(τ)
A
1
u −

h
1
(τ)h
2
(τ)

f (u)+g(τ)


dτ ds.
(4.3)
Now we suppose that for every fixed t
∈ [0,1] and for every r ∈ [0,∞)wehaveu(t,r) ≥0,
from here f

(u) ≤ bu after integration from 1 to u then we get that
f (u)
− f (1) ≤
b
2
u
2

−
b
2
,
f (u)
≤
b
2
u
2
−
b
2
+ f (1),
f (u)+g(r)
≤
b
2
u
2
−
b
2
+ f (1) + g(r).
(4.4)
From the conditions of Theorem 4.1 we have that
g(r)
≤
b
2

− f (1). (4.5)
Therefore
f (u)+g(r)
≤
b
2
u
2
. (4.6)
From here and from u(t,r)
≤ 1/B for every fixed t ∈ [0,1] and for every r ∈ [0,∞)weget
f (u)+g(r)
≤
b
2B
u. (4.7)
Consequently, from (4.3)weget
R(u)
≥

∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
1
u −
b
2B

h
1
(τ)h
2
(τ)u

dτ ds
≥

∞
r

h
2

(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)

udτds
≥


∞
r

h
2
(s)
h
1
(s)

∞
s
min
τ∈[0,∞)


h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1

(τ)h
2
(τ)

udτds.
(4.8)
Svetlin Georgiev Georgiev 25
From (H1)wehavethat
min
τ∈[0,∞)


h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)


≥
0 (4.9)
and since u(t,r)
≥ 0foreveryfixedt ∈ [0,1] and for every r ∈ [0,∞)from(4.8)weget

∞
r

h
2
(s)
h
1
(s)

∞
s
min
τ∈[0,∞)


h
2
(τ)
h
1
(τ)
A
1
−

b
2B

h
1
(τ)h
2
(τ)

udτds≥ 0, (4.10)
that is, for every fixed t
∈ [0,1] and for every r ∈ [0,∞)wehave
R(u)
≥ 0. (4.11)
(3) Now we will see that for every fixed t
∈ [0,1] and for every r ∈ [c,d]wehave
R(u)
≥ 1/A. Really, after we use (4.7),
R(u)
≥

∞
r

h
2
(s)
h
1
(s)


∞
s


h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)

udτds. (4.12)
Let
R
1
(u) =

∞

r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)


udτds. (4.13)
From here
R

1
(u) =

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1

−
b
2B

h
1
(τ)h
2
(τ)

dτ ds ≥ 0. (4.14)
Since for every r
∈ [c, d]wehaveu(t, r) ≥1/A,weget
R
1
(u) ≥ R
1

1
A

=

∞
r

h
2
(s)
h

1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1
(τ)h
2
(τ)

1
A
dτ ds
=
1

A

∞
r

h
2
(s)
h
1
(s)

∞
s


h
2
(τ)
h
1
(τ)
A
1
−
b
2B

h
1

(τ)h
2
(τ)

dτ ds ≥
1
A
(4.15)