More precisely we prove that the Cauchy problem has unique nontrivial solution in C((0,1] ˙Bγp,p(R

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-NORDSTR ¨OM METRIC

SVETLIN G. GEORGIEV

Abstract. In this paper, we study the solutions to the Cauchy problem (utt−∆u)gs+m²u=f(u), t∈(0,1], x∈R³,

u(1, x) =u0∈B˙_p,p^γ (R³), ut(1, x) =u1∈B˙^γ−1_p,p (R³),

where gs is the Reissner-Nordstr¨o m metric; p >1, γ ∈ (0,1),m 6= 0 are constants,f ∈ C²(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 C((0,1] ˙B^γp,p(R⁺)),

u(t, r) =

(v(t)ω(r) fort∈(0,1], r≤r1

0 fort∈(0,1], r≥r1, wherer=|x|, and limt→0kuk_B˙γ

p,p(R⁺)=∞.

1. Introduction

In this paper, we study properties of the solutions to the Cauchy problem (utt−∆u)gs+m²u=f(u), t∈(0,1], x∈R³, (1.1) u(1, x) =u₀∈B˙_p,p^γ (R³), u_t(1, x) =u₁∈B˙_p,p^γ−1(R³), (1.2) wheregs is the Reissner-Nordstr¨om [2],

g_s= r²−Kr+Q²

r² dt²− r²

r²−Kr+Q²dr²−r²dφ²−r²sin²φdθ²,

the constants K and Qare positive, m 6= 0, p∈ (1,∞) and γ ∈ (0,1) are fixed, f ∈ C²(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

2000Mathematics 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

C((0,1] ˙B_p,p^γ (R⁺)) such that limt→0kukB˙^γ_p,p(R⁺) =∞. The Cauchy problem (1.1)- (1.2) may rewrite in the form

r²

r²−Kr+Q²u_tt− 1

r²∂_r((r²−Kr+Q²)u_r)

− 1

r²sinφ∂_φ(sinφu_φ)− 1

r²sin²φu_θθ+m²u=f(u),

(1.3)

u(1, r, φ, θ) =u₀∈B˙^γ_p,p(R⁺×[0,2π]×[0, π]),

ut(1, r, φ, θ) =u1∈B˙_p,p^γ−1(R⁺×[0,2π]×[0, π]), (1.4) wherex=rcosφcosθ,y=rsinφcosθ,z=rsinθ, 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∈ C²([0, T)×R³) of (1.1)-(1.2) such that

sup

t<T , x∈R³

|u(t, x)|=∞.

Wheng_s 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.

Whenφ6= 0, π,2π,θ6= 0 are fixed constants we obtain the Cauchy problem r²

r²−Kr+Q²utt− 1

r²∂r((r²−Kr+Q²)ur) +m²u=f(u), (1.5) u(1, r) =u0∈B˙_p,p^γ (R⁺), ut(1, r) =u1∈B˙^γ−1_p,p (R⁺). (1.6) Our main result is as follows.

Theorem 1.1. Let mbe a non-zero constant, p∈(1,∞),γ∈(0,1) and K, Qbe positive constants for which

K²>4Q², 1

1−K+Q² >1, 1−K+Q²>0, with1−K+Q² is small enough such that

K−p

K²−4Q²

2 −2p

1−K+Q²>0.

Also let f ∈ C²(R¹), 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) ∈ C((0,1] ˙B_p,p^γ (R⁺))for which

t→0limkukB˙^γ_p,p(R⁺)=∞.

This paper is organized as follows: In section 2 we prove that the Cauchy problem (1.1)-(1.2) has unique nontrivial solution ˜u = v(t)ω(r) ∈ C((0,1] ˙B^γ_p,p(R⁺)). In section 3 we prove that

t→0limk˜ukB˙^γ_p,p(R⁺)=∞, where ˜uis the solution, which is received in section 2.

Let

C= pγ.2^pγ 2^pγ−1

1/p

.

LetA >0,Q >0,B >0,K >0, 1< β < αbe constants for which

(H1) _1−K+Q⁸ 2

8 1−K+Q²

2m²

α²A + 4m²

≤1, ^αA_m >1 (H2)

1 1−αK+α²Q²

1 1−αK+α²Q²

m²

α⁴A² −2m²r²₁ r₁−1

β 2

≥1 and _{α4(1−αK+α}^m2 _2Q2)A2 −2m²r²₁≥0,

(H3)

C2^p(1−γ) p(1−γ)

^1/p 8 1−K+Q²

8 1−K+Q²

2m²

α²A² +m²+6m² AB

<1 (H4) _α¹2

1 1−αK+α²Q²

m² α²A⁴ −_β¹2

m² A² >0 (H5)

2^p(1−γ) p(1−γ)

^1/p

16C 1−K+Q²

8 1−K+Q²

2m²

α²A² + 4m²

<1 (H6) K²>4Q²,A≥ _1−K+Q⁸ ₂ >1, _AB⁶ <1, 1> ^2Q_K² > ^K−

√

K²−4Q²

2 , 1−K+

Q²>0 is small enough such that 1> K−p

K²−4Q²

2 −3p

1−K+Q²>0, 2

K−p

K²−4Q²−2p

1−K+Q² ≤β < α≤3, where

r₁=K−p

K²−4Q²

2 −

√2 4

p1−K+Q². Example. Let

A= 1

⁴, B= 1

, p= 3

2, γ= 1

3, α= 3, 1

β = K−p

K²−4Q²

2 −3

2

p1−K+Q², K= 4 3 +1

6²⁰−3 2², Q²=1

3 +1 6²⁰−1

2², m²=⁴,

where 0< <<1 is enough small such that (H1)-(H6) hold. Then 1−αK+α²Q²= 1−3K+ 9Q²=²⁰,

1−K+Q²=².

Remark 1.2. Let²= 1−K+Q². Note that from (H6) we haveg(r) =r²−Kr+

Q²>0 forr∈[0, r1],g(r) is decrease function forr∈[0, r1]. Also (forr∈[0, r1]) we have

r²

r²−Kr+Q² ≤ 1 1−K+Q². In deed, letting ˜r= ^K−

√

K²−4Q²

2 , we haver1= ˜r−

√2

4 . Note that function r²

r²−Kr+Q² is increasing forr∈[0, r₁]. Therefore,

r²

r²−Kr+Q² ≤ r₁²

r₁²−Kr₁+Q² ≤ 8

² = 8

1−K+Q².

Note that the function

1 r²−Kr+Q² is increasing forr∈[0, r1]. Therefore, forr∈[0, r1],

1

r²−Kr+Q² ≤ 8 1−K+Q².

Here we will use the following definition of the ˙B^γ_p,p(M)-norm (γ∈(0,1),p >1) (see [3, p.94, def. 2], [1])

kukB˙_p,p^γ (M)=Z 2 0

h^−1−pγk∆huk^p_Lp(M)dh^1/p , where ∆_hu=u(x+h)−u(x).

Lemma 1.3. Let u(x)∈ C²([0, r1]), u(x) = 0 forx ≥r1, 0 < r1 <1. Then for γ∈(0,1),p >1 we have

CkukB˙^γp,p([0,r1])≥ kuk_Lp([0,r1]). Proof. We have

kuk^p_˙

B_p,p^γ ([0,r₁])= Z 2

0

h^−1−pγk∆huk^p_Lp([0,r₁])dh

= Z 2

0

h^−1−pγku(x+h)−u(x)k^p_Lp([0,r₁])dh

≥ Z 2

1

h^−1−pγku(x+h)−u(x)k^p_Lp([0,r₁])dh

= Z 2

1

h^−1−pγku(x)k^p_Lp([0,r₁])dh

=ku(x)k^p_Lp([0,r₁])

Z 2 1

h^−1−pγdh

=ku(x)k^p_Lp([0,r1])

2^pγ−1 pγ2^pγ ; i.e.,

kuk^p_˙

B_p,p^γ ([0,r₁])≥2^pγ−1

pγ2^pγ ku(x)k^p_Lp([0,r₁]).

From this estimate, we haveCkukB˙_p,p^γ ([0,r1])≥ ku(x)k_Lp([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 constantsA, K,Q,B, 1< β < α satisfy (H1)-(H6). Lett∈(0,1]. Letv(t) be function which satisfies the hypotheses:

(H7) v(t)∈ C³[0,∞),v(t)>0 for allt∈[0,1]

(H8) v⁰⁰(t)>0 for allt∈[0,1],v⁰(1) =v⁰⁰⁰(1) = 0,v(1)6= 0

(H9)

min

t∈[0,1]v(t)≥ 1

A, max

t∈[0,1]v(t)≤ 2 A, min

t∈[0,1]

v⁰⁰(t) v(t) ≥ m²

α²A², max

t∈[0,1]

v⁰⁰(t)

v(t) ≤ 2m² α²A²; limt→0[v⁰⁰(t)− m²

α²A²v(t)] = +0, v⁰⁰(t)− m²

α²A²v(t)≥0 fort∈[0,1].

Note that there exist a functions v(t) for which (H7)-(H9) hold. For example consider the function

v(t) = (t−1)²+^2α_m^2A2² −1 A³_m^α22

. (2.1)

Thenv(t)∈ C³[0,∞);v(t)>0 for allt∈[0,1] because (H1), we have ^αA_m >1; i.e., (H7) holds. Since

v⁰(t) =2(t−1)

A³_m^α2₂ , v⁰(1) = 0, v⁰⁰(t) = 2

A³_m^α2₂ ≥0 ∀t∈[0,1], v⁰⁰⁰(t) = 0, v⁰⁰⁰(1) = 0, it follows (H8). On the other hand

min

t∈[0,1]v(t)≥ 1

A, max

t∈[0,1]v(t)≤ 2

A, v⁰⁰(t)

v(t) = 2

(t−1)²+^2α_m^2A₂² −1, which implies

min

t∈[0,1]

v⁰⁰(t) v(t) ≥ m²

α²A², max

t∈[0,1]

v⁰⁰(t)

v(t) ≤ 2m² α²A², v⁰⁰(t)− m²

α²A²v(t) = m⁴

α⁴A⁵(2−t)t, lim

t→0[v⁰⁰(t)− m²

α²A²v(t)] = +0;

i.e., (H9) holds.

Here and below we suppose thatv(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

u(t, r) =

(v(t)ω(r) forr≤r1, 0 forr≥r₁, withtin (0,1] andu∈ C((0,1] ˙B^γ_p,p(R⁺)).

Let us consider the integral equation

u(t, r) =









 Rr1

r 1 τ²−Kτ+Q²

Rr1

τ

s⁴ s²−Ks+Q²

v⁰⁰(t) v(t)u(t, s) +s²m²u(t, s)−f(u(t, s))s²

ds dτ, for 0≤r≤r1,

0 forr≥r1,

(2.2)

whereu(t, r) =v(t)ω(r) andt∈(0,1].

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 ∈ C²(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∈ C²[0, r1], u(t, r1) = ur(t, r1) =urr(t, r1) = 0for t∈(0,1], u(t, r)∈ C((0,1] ˙B_p,p^γ [0, r1]), forr∈[_α¹,¹_β]andt∈(0,1]u(t, r)≥_A¹2, forr∈[_α¹, r1] andt∈(0,1]u(t, r)≥0, forr∈[0, r₁]and t∈(0,1]|u(t, r)| ≤ _AB² ,u(t, r) = 0for r≥r₁,t∈(0,1].

Proof. LetN ={u(t, r)∈ C([0, r1]) :t∈(0,1]}withu(t, r) =ur(t, r) =urr(t, r) = 0 for t ∈(0,1], r ≥r1, u(t, r)∈ C((0,1] ˙B_p,p^γ [0, r1]). For r ∈[_α¹,_β¹] and t ∈(0,1], we have u(t, r) ≥ _A¹2. Forr ∈ [0, r₁] and t ∈ (0,1], we have |u(t, r)| ≤ _AB² . For r∈[_α¹, r1] andt∈(0,1], we haveu(t, r)≥0}.

We remark that ifu ∈N is a solution of (2.2),u ∈ C²([0, r₁]). We define the operatorR as follows

R(u) = Z r1

r

1 τ²−Kτ +Q²

Z r1

τ

s⁴ s²−Ks+Q²

v⁰⁰(t) v(t) u(t, s) +s²m²u(t, s)−s²f(u)

ds dτ, for 0≤r≤r1 andt∈(0,1].

First we show that R : N → N. For each u ∈ N, we have the following five statements:

(1) Sinceu∈ C([0, r1]) andf ∈ C²(R¹), from the definition of the operator R we haveR(u)∈ C²([0, r1]),R(u)|r=r1 = 0,

∂

∂rR(u) = 1 r²−Kr+Q²

Z r r1

[ s⁴

s²−Ks+Q² v⁰⁰(t)

v(t)u+s²(m²u−f(u))]ds,

∂

∂rR(u) _r=r

1 = 0,

∂²

∂r²R(u) = K−2r (r²−Kr+Q²)²

Z r r1

[ s⁴

s²−Ks+Q² v⁰⁰(t)

v(t) u+s²(m²u−f(u))]ds

+ r⁴

(r²−Kr+Q²)² v⁰⁰(t)

v(t) u(t, r) + r²

r²−Kr+Q²(m²u(t, r)−f(u)).

Sinceu(t, r1) = 0,f(u(t, r1)) =f(0) = 0 we obtain

∂²

∂r²R(u) _r=r

1= 0.

Note that R(u) = 0 forr≥r1, t ∈(0,1] becauseu(t, r) = 0 forr ≥r1, t∈(0,1]

andf(u(t, r)) =f(0) = 0 forr≥r1,t∈(0,1].

(2) Forr∈[0, r1],t∈(0,1] we have|u(t, r)| ≤ _AB² . Then

|R(u)|

=

Z r r1

1 τ²−Kτ+Q²

Z τ r1

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) u+s²(m²u−f(u)) dsdτ

≤ Z r

r₁

1 τ²−Kτ+Q²

Z τ r₁

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) |u|+s²(m²|u|+|f(u)|) dsdτ .

Since|f(u)| ≤3m²|u|), the above quantity is lees than or equal to Z r

r1

1 τ²−Kτ+Q²

Z τ r1

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) |u|+ 4s²m²|u|

dsdτ

= Z r

r₁

1 τ²−Kτ+Q²

Z τ r₁

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) + 4s²m²

|u|dsdτ

≤ 2 AB

Z r r₁

1 τ²−Kτ+Q²

Z τ r₁

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) + 4s²m² dsdτ

where we use _r2−Kr+Q^{r2 2} ≤ _1−K+Q⁸ 2, _r2−Kr+Q^{1 2} ≤ _1−K+Q⁸ 2 for r ∈ [0, r1]. The above estimate is also less than or equal to

2 AB

8 1−K+Q²

8

1−K+Q² max

t∈[0,1]

v⁰⁰(t)

v(t) + 4m²

= 2 AB

8 1−K+Q²

8 1−K+Q²

2m²

α²A² + 4m²

≤ 2 AB.

In the above inequality we use (H1). Consequently,

|R(u)| ≤ 2

AB forr∈[0, r1], t∈(0,1].

(3)Forr∈[_α¹, r1] andt∈(0,1] we haveu(t, r)≥0. Then R(u) =

Z r₁ r

1 τ²−Kτ +Q²

Z r₁ τ

s⁴ s²−Ks+Q²

v⁰⁰(t) v(t) u(t, s) +s²m²u(t, s)−s²f(u)

ds dτ

(where we usef(u)≤3m²uforr∈[_α¹, r₁],t∈(0,1]. The above quantity is greater than or equal to

Z r₁ r

1 τ²−Kτ +Q²

Z r₁ τ

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) u(t, s) +s²(m²u(t, s)−3m²u) ds dτ

≥ Z r1

r

1 τ²−Kτ+Q²

Z r1

τ

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) −2m²s²

u(t, s)ds dτ

≥ Z r1

r

1 τ²−Kτ+Q²

Z r1

τ

1

α²(1−αK+α²Q²) min

t∈[0,1]

v⁰⁰(t)

v(t) −2m²r²₁

u(t, s)dsdτ

= Z r1

r

1 τ²−Kτ+Q²

Z r1

τ

m²

α⁴(1−αK+α²Q²)A² −2m²r²₁

u(t, s)ds dτ . From (H2), we have

m²

α⁴(1−αK+α²Q²)A² −2m²r²₁≥0.

From this inequality and fromu(t, r)≥0 forr∈[_α¹, r1],t∈(0,1],r²−Kr+Q²>0, forr∈[0, r₁], we get

R(u)≥0 f or r∈[1

α, r₁], t∈(0,1].

(4) Forr∈[_α¹,_β¹] andt∈(0,1] we have thatu(t, r)≥ _A¹2. Using f(u)≤3m²ufor r∈[_α¹,_β¹], t∈(0,1], we have

R(u)≥ Z r

r₁

1 τ²−Kτ +Q²

Z τ r₁

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) u−2s²m²u dsdτ

≥ Z r

r1

1 τ²−Kτ +Q²

Z τ r1

s⁴

s²−Ks+Q² min

t∈[0,1]

v⁰⁰(t)

v(t) u−2s²m²u ds dτ

= Z r

r₁

1 τ²−Kτ +Q²

Z τ r₁

s² s²

s²−Ks+Q² min

t∈[0,1]

v⁰⁰(t)

v(t) u−2m²u ds dτ

≥ Z r₁

r

1 τ²−Kτ+Q²

Z ¹_β

1 α

s² s²

s²−Ks+Q² min

t∈[0,1]

v⁰⁰(t)

v(t) u−2m²u ds dτ

≥ Z r₁

1 β

1 τ²−Kτ+Q²

Z ¹_β

1 α

s² s²

s²−Ks+Q² min

t∈[0,1]

v⁰⁰(t)

v(t) u−2m²u ds dτ

≥ 1 A²

1 1−αK+α²Q²

m²

α⁴A² −2m²r²₁ r1− 1

β

² 1

1−αK+α²Q² ≥ 1 A², (see (H2)); i.e., forr∈[_α¹,¹_β] andt∈(0,1] we haveR(u)≥ _A¹2.

(5) We have the estimate k∆hR(u)k^p_Lp=

Z r₁ 0

Z r+h r

1 τ²−Kτ +Q²

Z r₁ τ

s⁴ s²−Ks+Q²

v⁰⁰(t) v(t) u(t, s) +s²(m²u−f(u))

ds dτ

^p dr

≤ Z r1

0

Z r+h r

1 τ²−Kτ +Q²

Z r1

τ

s⁴ s²−Ks+Q²

v⁰⁰(t) v(t) |u|

+ 4m²|u(t, s)|s²

ds, dτp

dr

where we use that for s∈[0, r₁], _s2−Ks+Q^s4 ₂ ≤ _1−K+Q⁸ 2, _s2−Ks+Q¹ ₂ ≤ _1−K+Q⁸ 2 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

0

Z r+h r

8 1−K+Q²

Z r1

τ

8

1−K+Q² max

t∈[0,1]

v⁰⁰(t)

v(t) |u|+ 4m²|u|

dsdτ^p dr

≤ Z r₁

0

Z r+h r

8 1−K+Q²

Z r₁ τ

8 1−K+Q²

2m²

α²A²|u|+ 4m²|u|

dsdτ^p dr

≤ Z r1

0

Z r+h r

64 (1−K+Q²)²

2m² α²A²

Z r1

0

|u|ds+ 8

1−K+Q²4m² Z r1

0

|u|ds dτp

dr

≤h^p 64 (1−K+Q²)²

2m²

α²A²kukL^p[0,r₁]+ 8

1−K+Q²4m²kukL^p[0,r₁]

^p

; i.e,,

k∆hR(u)k^p_Lp[0,r1]

≤h^p 64 (1−K+Q²)²

2m²

α²A²kukL^p[0,r₁]+ 8

1−K+Q²4m²kukL^p[0,r₁]

^p .

Consequently, kR(u)k^p_˙

B^γ_p,p[0,r1]

= Z 2

0

h^−1−pγk∆_hR(u)k^p_Lp[0,r

1]dh

≤ 64

(1−K+Q²)² 2m²

α²A²kuk_Lp[0,r₁]+ 8.4m²

(1−K+Q²)kuk_Lp[0,r₁]

pZ 2 0

h^{−1+p(1−γ)}dh

= 64

(1−K+Q²)² 2m²

α²A²kukL^p[0,r₁]+ 8.4m²

(1−K+Q²)kukL^p[0,r₁]

^p 2^p(1−γ) p(1−γ). Therefore,

kR(u)kB˙^γp,p[0,r1]

≤ 64

(1−K+Q²)² 2m²

α²A²kuk_Lp[0,r1]+ 8.4m²

(1−K+Q²)kuk_Lp[0,r1]

2^p(1−γ) p(1−γ)

^1/p . From Lemma 1.3, we have

kR(u)kB˙^γ_p,p[0,r₁]≤C 64 (1−K+Q²)²

2m²

α²A²kukB˙_p,p^γ [0,r₁]

+ 8.4m²

(1−K+Q²)kukB˙_p,p^γ [0,r₁]

2^p(1−γ) p(1−γ)

^1/p .

From the above inequality, ifu∈B˙_p,p^γ [0, r1] we get R(u)∈B˙_p,p^γ [0, r1] fort∈(0,1].

From statements (1)–(5) above,R:N→N.

Now, letu, u1∈N. Then k∆_h(R(u)−R(u₁))k^p_Lp

= Z r1

0

Z r+h r

1 τ²−Kτ+Q²

Z r1

τ

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t)(u(t, s)−u₁) +s²(m²(u−u₁)−(f(u)−f(u₁)))

ds dτ

p

dr.

From the mean value theorem,|f(u)−f(u1)|=|u−u1kf⁰(ξ)|whereξ∈(u, u1) or ξ∈(u1, u). Then

|f(u)−f(u₁)| ≤3m²|ξ||u−u₁| ≤3m²|u−u₁kq|,

where|q|= max{|u|,|u1|}. Since|u| ≤ _AB² forr∈[0, r1],t∈(0,1] we have

|f(u)−f(u1)| ≤ 6m²

AB|u−u1|.

Now, we use thatu(t, r) = 0 forr≥r1 andt∈(0,1], to obtain

k∆h(R(u)−R(u1))k^p_Lp

≤ Z r1

0

Z r+h r

1 τ²−Kτ+Q²

Z r1

τ

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t)|u(t, s)−u1| +s²m²|u−u1|+|f(u)−f(u1)|

ds dτp

dr

≤ Z r₁

0

Z r+h r

1 τ²−Kτ+Q²

Z r₁ τ

s⁴

s²−Ks+Q² max

t∈[0,1]

v⁰⁰(t)

v(t)|u(t, s)−u1| +s²m²|u−u1|+|f(u)−f(u1)|

ds dτp

dr

≤ Z r₁

0

Z r+h r

1 τ²−Kτ+Q²

Z r₁ τ

8

1−K+Q² max

t∈[0,1]

v⁰⁰(t)

v(t) |u(t, s)−u1| +m²|u−u₁|+6m²

AB|u−u₁| ds dτ^p

dr

≤ Z r₁

0

Z r+h r

8 1−K+Q²

Z r₁ τ

8 1−K+Q²

2m²

α²A²+m²+6m² AB

× |u−u1|dsdτp

dr

≤h^p 8 1−K+Q²

p 8 (1−K+Q²)

2m²

α²A² +m²+6m² AB

p

ku−u1k^p_Lp; i.e.,

k∆h(R(u)−R(u₁))k^p_Lp[0,r1]

≤h^p 8 1−K+Q²

^p 8 (1−K+Q²)

2m²

α²A² +m²+6m² AB

^p

ku−u₁k^p_Lp. From the last inequality we get

kR(u)−R(u1)k^p_˙

B_p,p^γ [0,r1]≤ 8 1−K+Q²

^p 8 (1−K+Q²)

2m²

α²A² +m²+6m² AB

^p

× ku−u1k^p_Lp

Z 2 0

h^{−1+p(1−γ)}dh.

From the above inequality and Lemma 1.3, kR(u)−R(u₁)kB˙^γ_p,p[0,r₁]≤ 2^p(1−γ)

p(1−γ)

^1/p 8 1−K+Q²

8 (1−K+Q²)

2m² α²A² +m²+6m²

AB

ku−u1kL^p[0,r₁]

≤C 2^p(1−γ) (p(1−γ)

1/p 8 1−K+Q²

8 (1−K+Q²)

2m² α²A² +m²+6m²

AB

ku−u₁kB˙_p,p^γ [0,r₁]

<ku−u1kB˙_p,p^γ [0,r₁]

(see i3)). i.e.,

kR(u)−R(u1)kB˙_p,p^γ [0,r₁]<ku−u1kB˙_p,p^γ [0,r₁].

Consequently, the operatorR:N →N is contractive operator.

Lemma 2.2. The set N is closed subset ofC((0,1] ˙B_p,p^γ (R⁺)).

Proof. Lett∈(0,1] be fixed. Let{un}be a sequence of elements of the setN for which

n→∞lim ku_n−uk˜˜ B˙_p,p^γ (R⁺)= 0, where ˜u˜∈B˙_p,p^γ (R⁺). We have

n→∞lim ku_n−uk˜˜ B˙_p,p^γ ([0,r₁])= 0.

We define

˜ u=

(u˜˜ forr∈[0, r1], 0 forr > r1. We have

n→∞lim kun−uk˜ B˙_p,p^γ ([0,r₁])= 0.

First we note that for u ∈ N, R(u) is continuous function of u and there exists R⁰(u) becausef(u)∈ C²(R¹). In fact,

R⁰(u) = Z r₁

r

1 τ²−Kτ+Q²

Z r₁ τ

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) +s²m²−s²f⁰(u) ds dτ.

From which,

|R⁰(u)|

≥ Z r₁

r

1 τ²−Kτ+Q²

Z r₁ τ

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) +s²m²−s²|f⁰(u)|

ds dτ

≥ Z r₁

r

1 τ²−Kτ+Q²

Z r₁ τ

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) +s²m²−3m²s²|u|

ds, dτ

≥ Z r1

r

1 τ²−Kτ+Q²

Z r1

τ

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) +s²m²−6m² ABs²

ds dτ

= Z r1

r

1 τ²−Kτ+Q²

Z r1

τ

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) +s²m² 1− 6

AB

ds dτ . From (H6), 1>6/(AB). Therefore, for s∈[0, r₁] we have

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) +s²m² 1− 6

AB ≥0.

Then forr∈[0, r₁] we have

|R⁰(u)|

≥ Z r₁

1 α

1 τ²−Kτ+Q²

Z r₁

1 α

s⁴ s²−Ks+Q²

v⁰⁰(t)

v(t) +s²m² 1− 6

AB

s² ds dτ

≥ r₁− 1

α

² m²

α²A²(1−αK+α²Q²)² >0.

From this, foru∈N, there exists M := min

x∈[0,r1]|R⁰(u)(x)|>0

becauseR⁰(u)(x) is continuous function of x∈[0, r1]. Let M1= max

r∈[0,r1]

∂

∂r(R⁰(u))(r) .

Now we prove that for each >0 there existsδ=δ()>0 such that

|x−y|< δ implies |um(x)−u_m(y)|< ∀m.

We suppose that there exists ˜ > 0 such that for every δ > 0 there exist natural m andx, 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](Foru∈N the function R(u)(r) is uniformly continuous function ofr∈ [0, r1] becauseR(u)(r)∈ C²([0, r1]) and as in point (2) we have

∂

∂rR(u)(r)

≤ _AB² ).

Then there existsδ1=δ1(˜˜)>0 such that for everyu∈N

|R(u)(x)−R(u)(y)|<˜˜ for for allx, y∈[0, r1]: |x−y|< δ1. Then we may choose 0< δ <min

δ1,^(M˜−_2M^˜˜)AB

1 such that there exist naturalm andx1∈[0, r1],x2∈[0, r1] for which|x1−x2|< δand|um(x1)−um(x2)| ≥˜. In particular

|R(um)(x₁)−R(u_m)(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(u_m)(x₁)−R(u_m)(x₂)|

=|R⁰(ξ)(x₁)u_m(x₁)−R⁰(ξ)(x₂)u_m(x₂)|

=|R⁰(ξ)(x₁)u_m(x₁)−R⁰(ξ)(x₁)u_m(x₂) +R⁰(ξ)(x₁)u_m(x₂)−R⁰(ξ)(x₂)u_m(x₂)|

≥ |R⁰(ξ)(x1)um(x1)−R⁰(ξ)(x1)um(x2)| − |R⁰(ξ)(x1)−R⁰(ξ)(x2)kum(x2)|

=|R⁰(ξ)(x₁)ku_m(x₁)−u_m(x₂)|−

∂

∂r(R⁰(ξ))(η)

|x₁−x₂ku_m(x₂)|

≥M˜−M1δ 2 AB ≥˜˜,

which is a contradiction to (2.3).

Consequently, for each >0 there existsδ=δ()>0 such that

|x−y|< δ implies|u_m(x)−u_m(y)|< ∀m. (2.4) On the other hand, from the definition of the setN we have

|um| ≤ 2

AB ∀m. (2.5)

From (2.4) and (2.5), we conclude that the set{u_n} is compact subset ofC([0, r₁]).

Then there exists subsequence {un_k} and function u ∈ C([0, r1]) for which: for every > 0 there exists M = M() > 0 such that for every nk > M we have

|un_k(x)−u(x)|< for everyx∈[0, r1]; u(x) = 0 forx > r1. From this and from lim_k→∞kun_k−uk˜ B˙^γp,p([0,r1]) = 0 we have: For every >0 ∃M =M()>0 such that for everyn_k> M we have

max

x∈[0,r1]|un_k−u|< , kun_k−uk˜ B˙^γp,p([0,r1])< .

Then for everynk > M we have

|u−u| ≤ |u˜ −u_nk|+|u_nk−u|˜ < +|˜u−u_nk|, Z r₁

0

|u−u|dx < r˜ 1+ Z r₁

0

|˜u−un_k|dx, Using the H¨older’s inequality,

ku−uk˜ _L1[0,r1]< r1+r₁^1/qZ r1

0

|˜u−un_k|^pdx1/p

, 1 p+1

q = 1, forh >0, we have

h^−1−pγku−uk˜ L¹[0,r₁]< h^−1−pγr1+r^1/q₁ h^−1−pγZ r₁ 0

|˜u−un_k|^pdx^1/p ,

Z 2 1

h^−1−pγdhku−uk˜ L¹[0,r₁]

<

Z 2 1

h^−1−pγdhr₁+r₁^1/q Z 2

1

h^−1−pγZ r₁ 0

|˜u−u_nk|^pdx^1/p dh, Using H¨older’s inequality and that forh >1 we haveh^{(−1−pγ)p}≤h^−1−pγ,

1−2^−pγ

pγ ku−uk˜ L¹[0,r₁]

<1−2^−pγ

pγ r1+r^1/q₁ Z 2

1

h^−1−pγZ r1

0

|˜u−un_k|^pdx1/p

dh

≤1−2^−pγ

pγ r1+r^1/q₁ Z 2 1

h^{(−1−pγ)p} Z r1

0

|˜u−un_k|^pdxdh1/p

≤1−2^−pγ

pγ r1+r^1/q₁ Z 2 1

h^−1−pγ Z r1

0

|˜u−un_k|^pdxdh1/p

. Using that forx > r1,un_k(x) = ˜u(x) = 0, the above expression equals

1−2^−pγ

pγ r₁+r^1/q₁ Z 2 1

h^−1−pγ Z r1

0

|∆h(˜u−u_nk)|^pdxdh^1/p

≤ 1−2^−pγ

pγ r₁+r^1/q₁ Z 2 0

h^−1−pγ Z r1

0

|∆h(˜u−u_nk)|^pdxdh^1/p

= 1−2^−pγ

pγ r1+r^1/q₁ k˜u−un_kkB˙^γp,p([0,r1])

< (1−2^−pγ

pγ r₁+r^1/q₁ ) i.e., for every >0,

1−2^−pγ

pγ ku−uk˜ _L1[0,r1] < 1−2^−pγ

pγ r1+r^1/q₁ .

Consequently u = ˜u a.e.(almost everywhere) in [0, r₁], |u|^p = |˜u|^p a.e. in [0, r₁].

¿From here |un−u| = |un −u|˜ a.e., |un−u|^p = |un −u|˜^p a.e. in [0, r₁]. Since

un(x) = u(x) = 0 for x > r1 we have |∆h(un−u)|^p =|∆h(un−u)|˜ ^p, |∆hu|^p =

|∆hu|˜^p a.e. in [0, r1], forh >0. Therefore,u∈B˙_p,p^γ ([0, r1]) and Z r1

0

|un−u|^pdx= Z r1

0

|un−u|˜^pdx.

Now, we show that lim_n→∞kun−ukB˙p,p^γ ([0,r1])= 0. Note that kun−ukB˙^γ_p,p([0,r₁])

=Z 2 0

h^−1−pγ Z r₁

0

|∆h(un−u)|^pdx dh^1/p

=Z 2 0

h^−1−pγ Z r₁

0

|∆h(un−u)|˜ ^pdx dh^1/p

=kun−uk˜ B˙^γ_p,p([0,r₁])→n→∞0.

Consequently, for every sequence {u_n} with elements from N, which converges in ˙B_p,p^γ ([0, r₁]) there exists function u ∈ C([0, r₁]), u ∈ B˙_p,p^γ ([0, r₁]), for which kun−ukB˙_p,p^γ ([0,r₁])→n→∞0.

Below we suppose that the sequence {un} is a sequence of elements of the set N which converges in ˙B^γ_p,p([0, r1]). Then there existsu∈ C([0, r1]), u(x) = 0 for x > r1,u∈B˙_p,p^γ ([0, r1]),kun−ukB˙^γp,p([0,r1])→_n→∞0.

Now we suppose thatu(r₁)6= 0. Sinceu∈ C([0, r₁]), u_n ∈ C([0, r₁]),u_n(r₁) = 0 for every naturaln, there exist₂>0 and ∆₁⊂[0, r₁], r₁∈∆₁, such that

|u_n|< ₂

2, |u|> ₂

for every natural n and every x ∈ ∆1. Then for every natural n and for every x∈∆1,

|un(x)−u(x)|> 2

2. Let3>0 be such that

3<2

2

1−2^−pγ

pγ µ(∆1)r⁻

1 q

1 , 1

p+1

q = 1, (2.6)

whereµ(∆₁) is the measure of the set ∆₁. There existsM >0 such that for every n > M we have ku_n−ukB˙_p,p^γ ([0,r1]) < ₃. Consequently for every n > M and for everyx∈∆1we have

|un(x)−u(x)|>2

2, kun−ukB˙^γp,p([0,r1])< 3. Also using the H¨older’s inequality, we have

2

2µ(∆1)<

Z

∆₁

|un(x)−u(x)|dx

≤ Z r₁

0

|un(x)−u(x)|dx

≤Z r₁ 0

|un(x)−u(x)|^pdx^1/p r^1/q₁ .