Blow up of the Solutions of Nonlinear Wave Equation

Volume 2007, Article ID 42954,51pages doi:10.1155/2007/42954

Research Article

Blow up of the Solutions of Nonlinear Wave Equation

Received 14 March 2007; Accepted 26 May 2007 Recommended by Peter Bates

We construct for every fixed n≥2 the metric g_s=h1(r)dt²−h2(r)dr²−k1(ω)dω²₁−

··· −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−Δu)gs= f(u) +g(|x|), where x∈Rⁿ,n≥2;u(1,x)=u_◦(x)∈L²(Rⁿ),u_t(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|,aandb are positive con- stants. Wheng(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 _L²([0,_∞))= ∞. Wheng(r)₌0, we prove that the above Cauchy problem has a nontrivial solutionu(t,r) in the form u(t,r)=v(t)ω(r) for which limt→0 u _L²([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−Δu_gs=f(u) +g|x|

, x∈Rⁿ,n≥2, (1)

u(1,x)=u_◦(x)∈L²Rⁿ

, ut(1,x)=u1(x)∈H˙⁻¹Rⁿ

, (2)

whereg_sis the metric

g_s=h1(r)dt²−h2(r)dr²−k1(ω)dω²₁− ··· −k_n−1(ω)dω²_n−1, (1.1)

the functionsh1(r),h2(r) satisfy the conditions

h1(r),h2(r)∈Ꮿ¹[0,∞), h1(r)>0, h2(r)≥0 ∀r∈[0,∞), _∞

0

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)dτ <∞,

_∞

0

h2(s) h1(s)

_∞

s

h1(τ)h2(τ)dτ ds <∞, _∞

0

_∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)C1+C2

h1(τ)h2(τ) 2

dτ 1/2

ds 2

dr <∞, C1,C2are arbitrary nonnegative constants,

_∞

0

_∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)C1+C2

h1(τ)h2(τ)

dτ

ds 2

dr <∞, C1,C2are arbitrary nonnegative constants,

r∈max[0,∞)

h1(r)h2(r)<∞, _∞

0

_∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)dτ ds

2

dr <∞, _∞

0

_∞

r

h2(s) h1(s)

_∞

s h1(τ)h2(τ)dτ 1/2

ds 2

dr <∞, _∞

0

_∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)dτ

1/2

ds 2

dr <∞, _∞

0

_∞

r

h2(s) h1(s)ds

2

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|,aandbare positive constants,g∈Ꮿ(R¹),g(|x|)≥0 for|x| ∈[0,∞). (InSection 2 we will give example for such metricgs.)

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)∂r

h1(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

limt→0 u L²([0,∞))= ∞. (1.4) Our main results are the following.

Theorem 1.1. Supposen≥2 is fixed,h1(r),h2(r) satisfy the conditions (i1),g≡0, f ∈ Ꮿ¹(R¹), 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]L²([0,∞))) for which

limt→0 u L²([0,_∞))= ∞. (1.5) Theorem 1.2. Supposen≥2 is fixed,h1(r),h2(r) satisfy the conditions (i1). Suppose also thataandbare 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)≥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]L²([0,∞))) for which

limt→0 u L²([0,_∞))= ∞. (1.6) Whengsis the Minkowski metric andu0,u1∈Ꮿ^∞0(R³) in [1] (see also [2, Section 6.3]), it is proved that there existsT >0 and a unique local solutionu∈Ꮿ²([0,T)×R³) for the Cauchy problem

u_tt−Δu_gs=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)

Wheng_sis the Minkowski metric, 1≤p <5 and initial data are inᏯ^∞0(R³) in [1] (see also [2, Section 6.3]), it is proved that the initial value problem

utt−Δu_gs=u|u|^p−1, t∈[0,T], x∈R³, u_t=0=u0, u_tt=0=u1

(1.9) admits a global smooth solution.

Whengsis the Minkowski metric and initial data are inᏯ^∞0(R³) 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(R³) withE(u(0))<0, the initial value problem

utt−Δu_gs=u⁵, t∈[0,T],x∈R³, u_t=0=u0, u_tt=0=u1

(1.10) admits a global smooth solution.

Whengsis the Reissner-Nordstr¨om metric in [4], it is proved that the Cauchy problem utt−Δu_gs+m²u=f(u), t∈[0, 1],x∈R³,

u(1,x)=u0∈B˙^γp,p R³

, u_t(1,x)=u1∈B˙^γp,p⁻¹ R³

, (1.11)

wherem=0 is constant andf ∈Ꮿ²(R¹),a|u| ≤ |f^(l)(u)| ≤b|u|,l=0, 1,aandbare pos- itive constants, has unique nontrivial solutionu(t,r)∈Ꮿ((0, 1] ˙B^γp,p(R⁺)),r= |x|,p >1, for which

limt→0 u B˙^γp,p(R⁺)= ∞. (1.12) Whengsis the Minkowski metric in [5], it is proved that the Cauchy problem

utt−Δu_gs=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< q1≤1,u0∈Ꮿ⁵_◦(R³),u1∈Ꮿ⁴_◦(R³),u0(x)=u1(x)=0 for|x−x0|> ρ,x0andρare suitable chosen.

Wheng_s is 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,aandb are positive constants,r1>0 is suitable chosen, in [6], it is proved that the initial value problem (1), (2) has nontrivial solutionu∈Ꮿ((0, 1] ˙B^γp,q(R⁺)) in the form

u(t,r)=

⎧⎨

⎩

v(t)ω(r), forr≤r1,t∈[0, 1],

0, forr≥r1,t∈[0, 1], (1.15)

wherer= |x|, for which limt→0 u B˙^γp,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 h1(r), h2(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)∈Ꮿ⁴([0, 1]),v(t)=0 for everyt∈[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)= _∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u)

dτ ds (1^∗) for every fixedt∈[0, 1].

Proof. Suppose thatt∈[0, 1] is fixed and the functionu(t,r)=v(t)ω(r),v(t)∈Ꮿ⁴([0, 1]), v(t)=0 for everyt∈[0, 1],ω(r)∈Ꮿ²([0,∞)),ω(∞)=ω(∞)=0, satisfies (1). Then for

every fixedt∈[0, 1] and forr∈[0,∞) we have u_tt(t,r)=v(t)

v(t) u(t,r), 1

h1(r) v(t)

v(t) u(t,r)− 1 h1(r)h2(r)∂r

h1(r)

h1(r)h2(r)ur(t,r)

=f(u), 1

h1(r)h2(r)∂r

h1(r)

h1(r)h2(r)ur(t,r)

= 1 h1(r)

v(t)

v(t) u(t,r)−f(u),

∂r

h1(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 fromrto∞here we suppose thatur(t,r)=v(t)ω(r), u_r(t,∞)=v(t)ω(∞)=0, then we get

− h1(r)

h1(r)h2(r)u_r(t,r)= _∞

r

h2(τ) h1(τ)

v(t)

v(t)u(t,τ)−

h1(τ)h2(τ)f(u)

dτ,

− h1(r)

h2(r)ur(t,r)= _∞

r

h2(τ) h1(τ)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u)

dτ,

−u_r(t,r)=

h2(r) h1(r)

_∞

r

h2(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 fromrto∞; we use thatu(t,∞)=v(t)ω(∞)=0, then we get

u(t,r)= _∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u)

dτ ds, (2.3)

that is, for every fixedt∈[0, 1] if the functionu(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)=0 for everyt∈[0, 1],ω(r)∈Ꮿ²([0,∞)),ω(∞)=ω(∞)=0.

Proposition 2.2. Leth1(r),h2(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)∈Ꮿ⁴([0, 1]),v(t)=0 for everyt∈[0, 1],ω(r)∈Ꮿ²([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 functionu(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, satisfy the integral equation (1^∗). From here and from f ∈Ꮿ(−∞,∞), for every fixed t∈[0, 1] we have

u(t,r)∈Ꮿ²([0,∞)) and ur(t,r)= −

h2(r) h1(r)

_∞

r

h2(r) h1(r)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u)

dτ, h1(r)

h2(r)u_r(t,r)= − _∞

r

h2(r) h1(r)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u)

dτ, h1(r)

h1(r)h2(r)ur(t,r)= − _∞

r

h2(r) h1(r)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u)

dτ,

∂_r

h1(r)

h1(r)h2(r)u_r(t,r)

=

h2(r) h1(r)

v(t)

v(t) u(t,r)−

h1(r)h2(r)f(u), h2(r)

h1(r) v(t)

v(t) u(t,r)−∂r

h1(r)

h1(r)h2(r)ur(t,r)

=

h1(r)h2(r)f(u), 1

h1(r) v(t)

v(t) u(t,r)− 1 h1(r)h2(r)∂r

h1(r)

h1(r)h2(r)ur(t,r)

= f(u).

(2.4)

Since for every fixedt∈[0, 1] we have v(t)

v(t) u(t,r)=u_tt(t,r), (2.5)

we get

1

h1(r)utt(t,r)− 1 h1(r)h2(r)∂r

h1(r)

h1(r)h2(r)ur(t,r)

=f(u), (2.6) that is, for every fixedt∈[0, 1] if the functionu(t,r)=v(t)ω(r), wherev(t)∈Ꮿ⁴([0, 1]), v(t)=0 for everyt∈[0, 1],ω(r)∈Ꮿ²([0,∞)),ω(∞)=ω(∞)=0, satisfies (1^∗), then it

satisfies (1) for every fixedt∈[0, 1].

Proposition 2.3. Leth1(r),h2(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)∈Ꮿ⁴([0, 1]),v(t)=0 for everyt∈[0, 1],ω(r)∈Ꮿ²([0,∞)),ω(∞)=ω(∞)=0, sat- isfies (1), then the functionu(t,r)=v(t)ω(r) satisfies the integral equation

u(t,r)= _∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)

v(t)

v(t) u(t,τ)−

h1(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 functionu(t,r)=v(t)ω(r), v(t)∈Ꮿ⁴([0, 1]), v(t)=0 for everyt∈[0, 1],ω(r)∈Ꮿ²([0,∞)),ω(∞)=ω(∞)=0, satisfy (1). Then for

every fixedt∈[0, 1] and forr∈[0,∞) we have

u_tt(t,r)=v(t) v(t) u(t,r), 1

h1(r) v(t)

v(t) u(t,r)− 1 h1(r)h2(r)∂r

h1(r)

h1(r)h2(r)ur(t,r)

=

f(u) +g(r), 1

h1(r)h2(r)∂r

h1(r)

h1(r)h2(r)ur(t,r)

= 1 h1(r)

v(t)

v(t) u(t,r)−

f(u) +g(r),

∂_r

h1(r)

h1(r)h2(r)u_r(t,r)

= h2(r)

h1(r) v(t)

v(t) u(t,r)−

h1(r)h2(r)f(u) +g(r).

(2.7)

Now we integrate the last equality fromrto∞; here we suppose thatur(t,r)=v(t)ω(r), ur(t,∞)=v(t)ω(∞)=0, then we get

− h1(r)

h1(r)h2(r)u_r(t,r)= _∞

r

h2(τ) h1(τ)

v(t)

v(t)u(t,τ)−

h1(τ)h2(τ)f(u) +g(r)

dτ,

− h1(r)

h2(r)ur(t,r)= _∞

r

h2(τ) h1(τ)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u) +g(r)

dτ,

−u_r(t,r)=

h2(r) h1(r)

_∞

r

h2(τ) h1(τ)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u) +g(r)

dτ.

(2.8)

Now we integrate the last equality fromrto∞; we suppose thatu(t,∞)=v(t)ω(∞)=0, then we get

u(t,r)= _∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u) +g(r)

dτ ds, (2.9)

that is, for every fixedt∈[0, 1] if the functionu(t,r)=v(t)ω(r) satisfies (1), then the functionu(t,r)=v(t)ω(r) satisfies the integral equation (1^∗∗). Herev(t)∈Ꮿ⁴([0, 1]), v(t)=0 for everyt∈[0, 1],ω(r)∈Ꮿ²([0,∞)),ω(∞)=ω(∞)=0.

Proposition 2.4. Leth1(r),h2(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)∈Ꮿ⁴([0, 1]),v(t)=0 for everyt∈[0, 1],ω(r)∈Ꮿ²([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 functionu(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, satisfy the integral equation (1^∗∗). From here and from f ∈Ꮿ(−∞,∞),g∈Ꮿ([0,∞)), for every fixedt∈ [0, 1] we haveu(t,r)∈Ꮿ²([0,∞)) and

ur(t,r)= − h2(r)

h1(r) _∞

r

h2(r) h1(r)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u) +g(r)

dτ, h1(r)

h2(r)u_r(t,r)= − _∞

r

h2(r) h1(r)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u) +g(r)

dτ, h1(r)

h1(r)h2(r)ur(t,r)= − _∞

r

h2(r) h1(r)

v(t)

v(t)u(t,τ)−

h1(τ)h2(τ)f(u) +g(r)

dτ,

∂_r

h1(r)

h1(r)h2(r)u_r(t,r)

=

h2(r) h1(r)

v(t)

v(t) u(t,r)−

h1(r)h2(r)f(u) +g(r), h2(r)

h1(r) v(t)

v(t) u(t,r)−∂r

h1(r)

h1(r)h2(r)ur(t,r)

=

h1(r)h2(r)f(u) +g(r), 1

h1(r) v(t)

v(t) u(t,r)− 1 h1(r)h2(r)∂r

h1(r)

h1(r)h2(r)ur(t,r)

= f(u) +g(r).

(2.10) Since for every fixedt∈[0, 1] we have

v(t)

v(t) u(t,r)=utt(t,r), (2.11) we get

1

h1(r)utt(t,r)− 1 h1(r)h2(r)∂r

h1(r)

h1(r)h2(r)ur(t,r)

= f(u) +g(r), (2.12)

that is, for every fixedt∈[0, 1] if the functionu(t,r)=v(t)ω(r), wherev(t)∈Ꮿ⁴([0, 1]), v(t)=0 for everyt∈[0, 1],ω(r)∈Ꮿ²([0,∞)),ω(∞)=ω(∞)=0, satisfies (1^∗∗), then it

satisfies (1) for every fixedt∈[0, 1].

For fixedn≥2,h1(r),h2(r) which satisfy the conditions (i1) and fixed positive con- stantsaandb, we suppose that the positive constantsc,d,A,B,A1,A2satisfy the condi- tions

c≤d, A≥B, A1≤A2, A1− b 2B>0, h2(r)

h1(r)A1−b B

h1(r)h2(r)≥0 for everyr∈[0,∞), _∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)A1− b

2B

h1(τ)h2(τ)

dτ ds≥1 forr∈[c,d], _∞

1

h2(s) h1(s)

_∞

s

A1

h2(τ) h1(τ)⁻

b B

h1(τ)h2(τ)

dτ ds≥ A1

10¹⁰

(H1)

r∈max[0,_∞)

_∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)A2+

h1(τ)h2(τ) b 2B

dτ ds≤1,

rmax∈[0,∞)

h2(r) h1(r)

_∞

r

h2(τ)

h1(τ)A2+h1(τ)h2(τ) b 2B

dτ≤1,

(H2)

_∞

0

_∞

r

h2(s) h1(s)

_∞

s

h2(τ)

h1(τ)A2+h1(τ)h2(τ) b 2B

2

dτ 1/2

ds 2

dr <1 7, _∞

0

_∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)A2+b

B

h1(τ)h2(τ)

dτ ds 2

dr <∞,

(H3)

_∞

0

_∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)A2+

h1(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 choosec >0,d >0, c≤d <∞such that for everyr∈[c,d] we have

π

4 ^≤arctg(d+ 1−r)³, arctgd³<π

3. (2.13)

Let alsob=8³,a=4³,A=60,B=40,A1=³,A2=2³. Let

h1(r)= B

b

−1 +

1 + 2A1b B

2

, h2(r)= 144(d+ 1−r)⁴

(d+ 1−r)⁶+ 1². (2.14)

We note that the functionsh1(r) andh2(r) satisfy all conditions of (i1) and A1

h1(r)⁻ b 2B

h1(r)₌1, _∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)A1− b

2B

h1(τ)h2(τ)

dτ ds

≥ _d+1

r

h2(s) h1(s)

_d+1

s

h2(τ) h1(τ)A1− b

2B

h1(τ)h2(τ)

dτ ds≥1 forr∈[c,d].

(2.15) We note that h1(r)∼1, 6.

For fixedn_≥2,h1(r),h2(r), which satisfy the conditions (i1), the constantsa,b,c,d, A,B,A1,A2are fixed which satisfy the conditions (H1),. . ., (H4), then we suppose that the functionv(t) is fixed function and satisfies the conditions

v(t)∈Ꮿ⁴[0, 1], v(t)

v(t) >0, v(t)>0, ∀t∈[0, 1], (H5) A1≤v(t)

v(t) ^≤A2, v(1)=0, v(1)=0, (H6) limt→0

v(t) v(t) ⁻

a 2

=+0. (H7)

Example 2.6. Leta,b,c,d,A1,A2,B,Abe the constants from the above example. Then a/2=A2and

v(t)=Ce^√A2(t−1)+e^{−√A2(t−1)}, (2.16) whereCis arbitrary positive constant, satisfiing the hypotheses (H5), (H6), (H7).

Here and below we suppose thatv(t) is fixed function which satisfies the conditions (H5),. . ., (H7).

Wheng(r)≡0 we put

u_◦:=v(1)ω(r)= _∞

r

h2(s) h1(s)

_∞

s

h2(τ)

h1(τ)v(1)ω(τ)−

h1(τ)h2(τ)fv(1)ω(τ)

dτ ds, u1≡0.

(1) InSection 3, we will prove that (1) has unique nontrivial solutionω(r)∈L²([0,∞)).

Wheng(r)=0 we put u_◦:=v(1)ω(r)

= _∞

r

h2(s) h1(s)

_∞

s

h2(τ)

h1(τ)v(1)ω(τ)−

h1(τ)h2(τ)fv(1)ω(τ)+g(τ)

dτ ds, u1≡0.

(1) InSection 4, we will prove that (1) has unique nontrivial solutionω(r)∈L²([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 formu(t,r)=v(t)ω(r).

For fixed functionv(t), which satisfies the conditions (H5), (H6), and (H7) we con- sider the integral equation

u(t,r)= _∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)

v(t)

v(t)u(t,τ)−

h1(τ)h2(τ)f(u)

dτ ds. (1) Theorem 3.1. Letn≥2 be fixed, leth1(r),h2(r) fixed, which satisfy the conditions (i1), let the positive constantsa,bbe fixed,a≤b, let the positive constantsc,d.A,B,A1,A2be fixed which satisfy the conditions (H1),. . ., (H4) and f ∈Ꮿ¹((−∞,∞)), f(0)=0,a|u| ≤ f(u)≤b|u|. Let alsov(t) be fixed function which satisfies the conditions (H5),. . ., (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/Afor 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]L²([0,∞))).

Proof. LetMbe the set M=

u(t,r) :u(t,r)∈Ꮿ[0, 1]×[0,∞),u(t,∞)=ur(t,∞)=0∀t∈[0, 1], u(t,r)≥ 1

Afort∈[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)∈L²[0,∞)for everyt∈(0, 1]

. (3.1) Lett∈[0, 1] be fixed. We define the operatorLas follows:

L(u)(t,r)= _∞

r

h2(s) h1(s)

_∞

s

h2(τ) h1(τ)

v(t)

v(t) u(t,τ)−

h1(τ)h2(τ)f(u)

dτ ds (3.2) foru∈M. First we will see thatL:M→M. Letu∈M. Then the following holds.