Volume 2007, Article ID 42954,51pages 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 construct for every 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}, where*h*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*

*(1,x)*

_{t}*=*

*u*1(x)

*∈*

*H*˙

^{−}^{1}(R

*), where*

^{n}*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) in the formu(t,r)*

_{=}*v(t)ω(r) for which lim*

_{t}*0*

_{→}*u*

_{L}^{2}([0,

*))*

_{∞}*= ∞*. When

*g*(r)

*0, we prove that the above Cauchy problem has a nontrivial solution*

_{=}*u(t,r) in the form*

*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 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
*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 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}, (1.1)

the functions*h*1(r),*h*2(r) satisfy the conditions

*h*1(r),h2(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)h2(τ)dτ ds <*∞*,
_{∞}

0

_{∞}

*r*

*h*2(s)
*h*1(s)

_{∞}

*s*

*h*2(τ)
*h*1(τ)*C*1+*C*2

*h*1(τ)h2(τ)
2

*dτ*
1/2

*ds*
2

*dr <**∞*,
*C*1,C2are 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)h2(τ)

*dτ*

*ds*
2

*dr <**∞*,
*C*1,C2are arbitrary nonnegative constants,

*r**∈*max[0,*∞*)

*h*1(r)h2(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)h2(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 positive constants,

*g*

*∈*Ꮿ(R

^{1}),

*g*(

*|*

*x*

*|*)

*≥*0 for

*|*

*x*

*| ∈*[0,

*∞*). (InSection 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)h2(r)*∂**r*

*h*1(r)
*h*1(r)h2(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.

*Theorem 1.1. Supposen**≥**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*

*|*

*,aandbare positive constants. Then the homoge-*

*neous problem of Cauchy (1), (2) has nontrivial solutionu*

*=*

*u(t,r)*

*∈*Ꮿ((0, 1]L

^{2}([0,

*∞*)))

*for which*

lim*t**→*0 *u* *L*^{2}([0,* _{∞}*))

*= ∞*

*.*(1.5)

*Theorem 1.2. Supposen*

*≥*

*2 is fixed,h*1(r),

*h*2(r) satisfy the conditions (i1). Suppose also

*thataandbare 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 everyr*

*≥*

*0,g*(r)

*≤*

*b/2*

*−*

*f(1) for ev-*

*eryr*

*∈*[0,

*∞*

*). Then the nonhomogeneous 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,* _{∞}*))

*= ∞*

*.*(1.6) When

*g*

*s*is the Minkowski metric and

*u*0,u1

*∈*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}) for the 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 inᏯ

*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 inᏯ* ^{∞}*0(R

^{3}) in [3] (see also [2, Section 6.3]) it is proved that there exists a number0

*>*0 such that for any data (u0,u1)

*∈*Ꮿ

*0(R*

^{∞}^{3}) with

*E(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}*p,p*

^{−}^{1}R

^{3}

, (1.11)

where*m**=*0 is constant and*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*0and

*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 nontrivial 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. InSection 2, we will prove some preliminary results.

InSection 3, we will proveTheorem 1.1. InSection 4, we will proveTheorem 1.2. In the appendix we will prove some results which are used for the proof of Theorems1.1and 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 fixedt**∈**[0, 1] the functionu(t,r)**=**v(t)ω(r), wherev(t)**∈*Ꮿ^{4}*([0, 1]),v(t)**=*0
*for everyt**∈**[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)h2(τ)*f*(u)

*dτ ds* (1* ^{∗}*)

*for every fixedt*

*∈*

*[0, 1].*

*Proof. Suppose thatt**∈*[0, 1] is fixed and the function*u(t,r)**=**v(t)ω(r),v(t)**∈*Ꮿ^{4}([0, 1]),
*v(t)**=*0 for every*t**∈*[0, 1],*ω(r)**∈*Ꮿ^{2}([0,*∞*)),*ω(**∞*)*=**ω** ^{}*(

*∞*)

*=*0, satisfies (1). Then for

every fixed*t**∈*[0, 1] and for*r**∈*[0,*∞*) we have
*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)h2(r)*∂**r*

*h*1(r)

*h*1(r)h2(r)*u**r*(t,*r)*

*=**f*(u),
1

*h*1(r)h2(r)*∂**r*

*h*1(r)

*h*1(r)h2(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)h2(r)*u**r*(t,*r)*

*=*

*h*2(r)
*h*1(r)

*v** ^{}*(t)

*v(t)* *u(t,r)**−*

*h*1(r)h2(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*

*(t,*

_{r}*∞*)

*=*

*v(t)ω*

*(*

^{}*∞*)

*=*0, then we get

*−* *h*1(r)

*h*1(r)h2(r)*u** _{r}*(t,r)

*=*

_{∞}*r*

*h*2(τ)
*h*1(τ)

*v** ^{}*(t)

*v(t)u(t,τ)**−*

*h*1(τ)h2(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)h2(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)h2(τ)*f*(u)

*dτ.*

(2.2)

Now we integrate the last equality from*r*to*∞*; we use 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)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)*

*∈*Ꮿ

^{4}([0, 1]),

*v(t)*

*=*0 for every

*t*

*∈*[0, 1],

*ω(r)*

*∈*Ꮿ

^{2}([0,

*∞*)),

*ω(*

*∞*)

*=*

*ω*

*(*

^{}*∞*)

*=*0.

*Proposition 2.2. Leth*1(r),*h*2(r) satisfy the conditions (i1), *f* *∈*Ꮿ(*−∞*,*∞**),g**≡**0. If for*
*every fixedt**∈**[0, 1] the functionu(t,r)**=**v(t)ω(r), wherev(t)**∈*Ꮿ^{4}*([0, 1]),v(t)**=**0 for*
*everyt**∈**[0, 1],ω(r)**∈*Ꮿ^{2}([0,*∞**)),ω(**∞*)*=**ω** ^{}*(

*∞*)

*=*

*0, satisfies the integral equation (1*

^{∗}*)*

*then the functionu(t,r)*

*=*

*v(t)ω(r) satisfies (1) for every fixedt*

*∈*

*[0, 1].*

*Proof. Lett**∈*[0, 1] be fixed and let 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, satisfy the integral equation (1

*). From here and from*

^{∗}*f*

*∈*Ꮿ(

*−∞*,

*∞*), 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)h2(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)h2(τ)*f*(u)

*dτ,*
*h*1(r)

*h*1(r)h2(r)*u**r*(t,*r)**= −*
_{∞}

*r*

*h*2(r)
*h*1(r)

*v** ^{}*(t)

*v(t)* *u(t,τ)**−*

*h*1(τ)h2(τ)*f*(u)

*dτ,*

*∂*_{r}

*h*1(r)

*h*1(r)h2(r)*u** _{r}*(t,r)

*=*

*h*2(r)
*h*1(r)

*v** ^{}*(t)

*v(t)* *u(t,r)**−*

*h*1(r)h2(r)*f*(u),
*h*2(r)

*h*1(r)
*v** ^{}*(t)

*v(t)* *u(t,r)**−**∂**r*

*h*1(r)

*h*1(r)h2(r)*u**r*(t,*r)*

*=*

*h*1(r)h2(r)*f*(u),
1

*h*1(r)
*v** ^{}*(t)

*v(t)* *u(t,r)**−* 1
*h*1(r)h2(r)*∂**r*

*h*1(r)

*h*1(r)h2(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)h2(r)*∂**r*

*h*1(r)

*h*1(r)h2(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), 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 it*

^{∗}satisfies (1) for every fixed*t**∈*[0, 1].

*Proposition 2.3. Leth*1(r),*h*2(r) satisfy the conditions (i1), *f**∈*Ꮿ(*−∞*,*∞**),g**∈*Ꮿ([0,*∞**)),*
*g*(r)*≥**0 for everyr**≥**0. If for every fixedt**∈**[0, 1] the functionu(t,r)**=**v(t)ω(r), where*
*v(t)**∈*Ꮿ^{4}*([0, 1]),v(t)**=**0 for everyt**∈**[0, 1],ω(r)**∈*Ꮿ^{2}([0,*∞**)),ω(**∞*)*=**ω** ^{}*(

*∞*)

*=*

*0, sat-*

*isfies (1), then the functionu(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)h2(τ)^{}*f*(u) +*g(r)*^{}

*dτ ds*
(1* ^{∗∗}*)

*for every fixedt*

*∈*

*[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)**=*0 for every*t**∈*[0, 1],*ω(r)**∈*Ꮿ^{2}([0,*∞*)),*ω(**∞*)*=**ω** ^{}*(

*∞*)

*=*0, satisfy (1). Then for

every fixed*t**∈*[0, 1] and for*r**∈*[0,*∞*) we have

*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)h2(r)*∂**r*

*h*1(r)

*h*1(r)h2(r)*u**r*(t,r)

*=*

*f*(u) +*g*(r)^{},
1

*h*1(r)h2(r)*∂**r*

*h*1(r)

*h*1(r)h2(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)h2(r)*u** _{r}*(t,r)

*=*
*h*2(r)

*h*1(r)
*v** ^{}*(t)

*v(t)* *u(t,r)**−*

*h*1(r)h2(r)^{}*f*(u) +*g(r)*^{}*.*

(2.7)

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)h2(r)*u** _{r}*(t,r)

*=*

_{∞}*r*

*h*2(τ)
*h*1(τ)

*v** ^{}*(t)

*v(t)u(t,τ)**−*

*h*1(τ)h2(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)h2(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)h2(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)h2(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)*

*=*0 for every

*t*

*∈*[0, 1],

*ω(r)*

*∈*Ꮿ

^{2}([0,

*∞*)),

*ω(*

*∞*)

*=*

*ω*

*(*

^{}*∞*)

*=*0.

*Proposition 2.4. Leth*1(r),*h*2(r) satisfy the conditions (i1), *f**∈*Ꮿ(*−∞*,*∞**),g**∈*Ꮿ([0,*∞**)),*
*g*(r)*≥**0 for everyr**≥**0. If for every fixedt**∈**[0, 1] the functionu(t,r)**=**v(t)ω(r), where*

*v(t)**∈*Ꮿ^{4}*([0, 1]),v(t)**=**0 for everyt**∈**[0, 1],ω(r)**∈*Ꮿ^{2}([0,*∞**)),ω(**∞*)*=**ω** ^{}*(

*∞*)

*=*

*0, sat-*

*isfies the integral equation (1*

^{∗∗}*), then the functionu(t,r)*

*=*

*v(t)ω(r) satisfies (1) for every*

*fixedt*

*∈*

*[0, 1].*

*Proof. Lett**∈*[0, 1] be fixed and let 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, 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)h2(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)h2(τ)^{}*f*(u) +*g*(r)^{}

*dτ,*
*h*1(r)

*h*1(r)h2(r)*u**r*(t,r)*= −*
_{∞}

*r*

*h*2(r)
*h*1(r)

*v** ^{}*(t)

*v(t)u(t,τ)**−*

*h*1(τ)h2(τ)^{}*f*(u) +*g*(r)^{}

*dτ,*

*∂*_{r}

*h*1(r)

*h*1(r)h2(r)*u** _{r}*(t,

*r)*

*=*

*h*2(r)
*h*1(r)

*v** ^{}*(t)

*v(t)* *u(t,r)**−*

*h*1(r)h2(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)h2(r)*u**r*(t,r)

*=*

*h*1(r)h2(r)^{}*f*(u) +*g(r)*^{},
1

*h*1(r)
*v** ^{}*(t)

*v(t)* *u(t,r)**−* 1
*h*1(r)h2(r)*∂**r*

*h*1(r)

*h*1(r)h2(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)h2(r)*∂**r*

*h*1(r)

*h*1(r)h2(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), 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 it*

^{∗∗}satisfies (1) for every fixed*t**∈*[0, 1].

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 constantsc,d,A,B,A*1,*A*2satisfy 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)h2(r)*≥*0 for every*r**∈*[0,*∞*),
_{∞}

*r*

*h*2(s)
*h*1(s)

_{∞}

*s*

*h*2(τ)
*h*1(τ)*A*1*−* *b*

2B

*h*1(τ)h2(τ)

*dτ ds**≥*1 for*r**∈*[c,d],
_{∞}

1

*h*2(s)
*h*1(s)

_{∞}

*s*

*A*1

*h*2(τ)
*h*1(τ)^{−}

*b*
*B*

*h*1(τ)h2(τ)

*dτ ds**≥* *A*1

10^{10}

(H1)

*r**∈*max[0,* _{∞}*)

_{∞}

*r*

*h*2(s)
*h*1(s)

_{∞}

*s*

*h*2(τ)
*h*1(τ)*A*2+

*h*1(τ)h2(τ) *b*
2B

*dτ ds**≤*1,

*r*max*∈*[0,*∞*)

*h*2(r)
*h*1(r)

_{∞}

*r*

*h*2(τ)

*h*1(τ)*A*2+^{}*h*1(τ)h2(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)h2(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)h2(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)h2(τ)*b*
*B*

2

*dτ*
1/2

*ds*
2

*dr <*1. (H4)

*Example 2.5. Let 0<*1/3 be enough small, *n**≥*2 is fixed. We choose*c >*0,*d >*0,
*c**≤**d <**∞*such that for every*r**∈*[c,d] we have

*π*

4 * ^{≤}*arct

*g(d*+ 1

*−*

*r)*

^{3}, arct

*gd*

^{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)

We note that the functions*h*1(r) and*h*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)h2(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)h2(τ)

*dτ ds**≥*1 for*r**∈*[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*2are 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. Leta,b,c,d,A*1,*A*2,*B,A*be the constants from the above example. Then
*a/2**=**A*2and

*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)*≡*0 we put

*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)h2(τ)*f*^{}*v(1)ω(τ)*^{}

*dτ ds,*
*u*1*≡*0.

(1* ^{}*)
InSection 3, we will prove that (1

*) has unique nontrivial solution*

^{}*ω(r)*

*∈*

*L*

^{2}([0,

*∞*)).

When*g*(r)*=*0 we put
*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)h2(τ)^{}*f*^{}*v(1)ω(τ)*^{}+*g*(τ)^{}

*dτ ds,*
*u*1*≡*0.

(1* ^{}*)
InSection 4, we will prove that (1

*) has unique nontrivial solution*

^{}*ω(r)*

*∈*

*L*

^{2}([0, 1)).

**3. Proof ofTheorem 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 (H*5), (H6), and (H7) we con-
sider 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)h2(τ)*f*(u)

*dτ ds.* (1* ^{}*)

*Theorem 3.1. Letn*

*≥*

*2 be fixed, leth*1(r),

*h*2(r) fixed, which satisfy the conditions (i1),

*let the positive constantsa,bbe fixed,a*

*≤*

*b, let the positive constantsc,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*

*|*

*. Let alsov(t) be fixed function which satisfies the conditions (H*5),

*. . ., (H7).*

*Then (1*^{}*) has unique nontrivial solutionu(t,r)**=**v(t)ω(r) for whichu(t,r)**∈*Ꮿ([0, 1]*×*
[0,*∞**)),u(t,r)**≤*1/B *for everyt**∈**[0, 1] and for everyr**∈*[0,*∞**),* *u(t,r)**≥*1/A*for every*
*t**∈**[0, 1] and for everyr**∈*[c,d],*u(t,r)**≥**0 for everyt**∈**[0, 1] and for everyr**∈*[0,*∞**),*
*u(t,**∞*)*=**u** _{r}*(t,

*∞*)

*=*

*0 for everyt*

*∈*

*[0, 1],u(t,r)*

*∈*

*C((0, 1]L*

^{2}([0,

*∞*

*))).*

*Proof. LetM*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)h2(τ)*f*(u)

*dτ ds* (3.2)
for*u**∈**M. First we will see thatL*:*M**→**M*. Let*u**∈**M. Then the following holds.*