Existence of solutions and boundary asymptotic behavior of p(r) -Laplacian equation multi-point boundary value problems

Nonlinear Analysis 72 (2010) 2950–2973

Contents lists available at ScienceDirect

Nonlinear Analysis journal homepage: www.elsevier.com/locate/na

Existence of solutions and boundary asymptotic behavior of p(r )-Laplacian equation multi-point boundary value problemsI Qihu Zhang a,b,c,∗ , Yan Wang a , Zhimei Qiu a a

School of Mathematics and Statistics, Huazhong Normal University, Wuhan, Hubei 430079, China

b

Academic administration, Henan Institute of Science and Technology, Xinxiang, Henan 453003, China

c

Department of Mathematics and Information Science, Zhengzhou University of Light Industry, Zhengzhou, Henan 450002, China

article

info

Article history: Received 11 July 2009 Accepted 20 November 2009 MSC: 34B10

abstract This paper investigates the problem

 ρ(r )f (r , u) = 0, r ∈ (0, R), −∆p(r ) u + m X αi u(ηi ) + e0 , u(r ) → +∞ (as r → R− ), u(0) = i=1

p(r )−2

where −∆p(r ) u = −( u0 u0 )0 is called the p(r )-Laplacian, ρ(r ) is a singular coefficient. The existence and nonexistence of solutions are discussed, and the exact boundary blow-up rate of solutions is given. © 2009 Elsevier Ltd. All rights reserved.

Keywords: p(r )-Laplacian Subsolution Supersolution Boundary blow-up rate

1. Introduction The study of differential equations and variational problems with variable exponent is a new and interesting topic. These problems are interesting in applications (see, e.g., [1–4]) and raise many difficult mathematical problems. Many results have been obtained on these kinds of problems, for example [5–21]. In [8], Fan gave the regularity of weak solutions for differential equations with variable exponent. On the existence of solutions for variable exponent problems, we refer to [6,9,20,21]. In this paper, we consider the following multi-point boundary value problem (P)

 ρ(r )f (r , u) = 0, r ∈ (0, R), −∆p(r ) u + m X αi u(ηi ) + e0 , u(r ) → +∞ (as r → R− ), u(0) = i =1

p(r )−2

where −∆p(r ) u = −( u0 u0 )0 ; 0 < η1 < · · · < ηm < R; αi ≥ 0, (i = 1, . . . , m) and 0 < i=1 αi < 1; e0 is a constant; ρ(r ) is a singular coefficient, 0 < ρ(r ) ∈ C [0, R). On the p-Laplacian multi-point boundary value problems, there are many papers (see [22–28]), but results on p(r )-Laplacian multi-point boundary value problems are rare (see [19,20]). Our aim is to give the existence and nonexistence of boundary blow-up solutions for p(r )-Laplacian multi-point boundary value problems, and give the exact boundary blow-up rate of solutions.

Pm

I Foundation item: Partly supported by the National Science Foundation of China (10701066 & 10826042 & 10971087) and China Postdoctoral Science Foundation funded project (20090460969) and the Natural Science Foundation of Henan Education Committee (2008-755-65) and the Natural Science Foundation of Jiangsu Education Committee (08KJD110007) and The Doctor Grant of Henan Polytechnic University (B2009-41). ∗ Corresponding address: Department of Information and Computation Science, Zhengzhou University of Light Industry, Zhengzhou, Henan, 450002, China. E-mail addresses: [email protected] (Q. Zhang), [email protected] (Y. Wang), [email protected] (Z. Qiu).

0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.11.038

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2951

We denote I = [0, R], C ([a, b]) will be equipped the norm kuk0,[a,b] = maxt ∈[a,b] |u(t )|, C 1 ([a, b]) will be equipped the norm kuk1,[a,b] = kuk0,[a,b] + ku0 k0,[a,b] . We set C = C ([0, R), R), C 1 = {u ∈ C | u0 ∈ C ([0, R), R)}. Denote L1 = L1 (I , R),

RR

with the norm kukL1 = 0 |u(r )| dr. For any h ∈ C [0, R], denote h− = minr ∈[0,R] h(r ), h+ = maxr ∈[0,R] h(r ). Throughout the paper, we assume that p(r ), ρ(r ) and f (r , u) satisfy

(H1 ) ρ ∈ C [0, R), p ∈ C 1 [0, R] and satisfies 1 < p− ≤ p+ ; (H2 ) f (r , ·) is increasing and f (r , 0) = 0 for any r ∈ [0, R]; (H3 ) f : [0, R] × R → R is a continuous function and satisfies |f (r , t )| ≤ C1 + C2 |t |α(r )−1 ,

∀(r , t ) ∈ [0, R] × R,

where C1 , C2 are positive constants, α ∈ C 1 [0, R] and 1 ≤ α(r ), ∀r ∈ [0, R]. The operator −∆p(r ) u is called the one-dimensional p(r )-Laplacian. Especially, if p(r ) ≡ p (a constant), −∆p(r ) is the wellknown p-Laplacian. Because of the non-homogeneity of p(r )-Laplacian, p(r )-Laplacian problems are more complicated than those of p-Laplacian. For example, if Ω ⊂ RN is bounded, then the Rayleigh quotient

R λp(x) =

inf

1,p(x)

u∈W0

(Ω )\{0}

1

Ω p(x)

R

1

|∇ u|p(x) dx

Ω p(x)

|u|p(x) dx

is zero in general, and only under some special conditions λp(x) > 0 (see [7]), and maybe the first eigenvalue and the first eigenfunction of p(x)-Laplacian do not exist, but the fact that the first eigenvalue λp > 0 and the existence of the first eigenfunction are very important in the study of p-Laplacian problems. For examples: in [29], the author considers the existence of positive weak solutions for p-Laplacian systems, the first eigenfunction is used to constructing the subsolution of p-Laplacian problems successfully; in [30], the authors consider the existence of positive weak solutions for p-Laplacian problems, the first eigenvalue is used to deal with the multi-existence of solutions. P2 If f (x, u) can be represented as i=1 hi (x)fi (u), there are many papers on the boundary blow-up solutions of p-Laplacian problems; see for example [31–40]. In the investigation of the existence of boundary blow-up solutions for the following p-Laplacian problems (p is a constant)

− ∆p u + γ (x)f (u) = 0 in Ω ⊂ RN ,

(1)

if γ (x) is positive and bounded on Ω , the following generalized Keller–Osserman condition is crucial ∞

Z 1

1

(F (t ))

1 p

dt < +∞,

where F (t ) =

t

Z

f (s)ds. 0

Throughout the paper, assume that there exists a constant σ ∈ [ηm , R) such that

(A1 ) ϑ0 uq(r )−1 ≤ f (r , u) ≤ ϑ1 uq1 (r )−1 (as u → +∞) for r ∈ [σ , R) uniformly; (A2 ) ρ0 (R − r )−β(r ) ≤ ρ(r ) ≤ ρ1 (R − r )−β1 (r ) for r ∈ [σ , R) uniformly; where ϑ0 , ϑ1 , ρ0 and ρ1 are positive constants. Unless special claim, we always assume that q(r ), q1 (r ), β(r ) and β1 (r ) are Lipschitz continuous on [σ , R], which satisfy 1 ≤ q(r ) ≤ q1 (r ) and β(r ) ≤ β1 (r ) for any r ∈ [σ , R]. When p(r ) is a general function, the typical form of (P) is

− ∆p(r ) u + (R − r )−β(r ) |u|q(r )−2 u = 0. Our main results mean that p(R)−β(R) (i) If q(R)−p(R) > 0, then (P) has boundary blow-up solutions; p(R)−β(R)

(2)

(ii) If q(R)−p(R) < 0, then (P) does not have boundary blow-up solution. Eq. (P) is called positive definite, if there exists a σ ∈ [ηm , R) such that 0 < q(r ) − p(r ) for any r ∈ [σ , R]; Eq. (P) is called negative definite, if there exists a σ ∈ [ηm , R) such that q1 (r ) − p(r ) < 0 for any r ∈ [σ , R]; Eq. (P) is called oscillatory, if Eq. (P) is neither positive definite nor negative definite. In [38], the authors consider the existence and nonexistence of boundary blow-up solutions for (1), when the equation is positive definite or negative definite. In [21], the present author discussed the existence and nonexistence of boundary blow-up solutions of (P), when ρ is bounded and (P) is positive definite or negative definite. The main difficulties of this paper are as follows (i) The non-homogeneity of p(r )-Laplacian. P2 (ii) f (r , u) cannot be represented as the form of i=1 hi (r )fi (u). (iii) The singularity of ρ(r ) (maybe, it will leading to the non-integrability of ρ(r )f (r , u) on [0, R)). We say f satisfies sub-(p− − 1) growth condition, if f satisfies lim

|u|→+∞

f (t , u) = 0, |u|q1 (t )−1

for t ∈ I uniformly,

+ − where q1 (t ) ∈ C (I , R), and 1 < q− 1 ≤ q1 < p .

2952

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

We say f satisfies super-(p+ − 1) growth condition, if f satisfies lim

|u|→+∞

f (t , u) = ∞, |u|q(t )−1

for t ∈ I uniformly,

where q(t ) ∈ C (I , R), and 1 < p− ≤ p+ < q.

p(r )−2

We say u is a solution of (P) if u ∈ C 1 ([0, R), R) with u0 u0 (r ) absolutely continuous on [0, R), and satisfies the boundary value condition and −∆p(r ) u + ρ(r )f (r , u) = 0, a.e. (0, R). This paper is divided into five sections. In the second section, we will do some preparation. In the third section, the fourth section and the fifth section, we will discuss the existence and nonexistence of boundary blow-up solutions of (P), when (P) is negative definite, positive definite and oscillatory, respectively. 2. Preliminaries For any (t , x) ∈ I × R, denote ϕ(t , x) = |x|p(t )−2 x. Obviously, ϕ has the following properties Lemma 2.1 (See [18]). ϕ is a continuous function and satisfies (i) For any t ∈ [0, R], ϕ(t , ·) is strictly monotone; (ii) There exists a function κ : [0, +∞) → [0, +∞), κ(s) → +∞ as s → +∞, such that

ϕ(t , x)x ≥ κ(|x|)|x|,

for all x ∈ R.

It is well known that ϕ(t , ·) is a homeomorphism from R to R for any fixed t ∈ [0, R]. For any t ∈ I, denote by ϕ −1 (t , ·) the inverse operator of ϕ(t , ·), then 2−p(t )

ϕ −1 (t , x) = |x| p(t )−1 x,

ϕ −1 ( t , 0 ) = 0 .

for x ∈ R \ {0},

It is clear that ϕ −1 (t , ·) is continuous and sends bounded sets to bounded sets. Let us now consider the existence of solutions for the following boundary value problems

 ρ(r )f (r , u) = 0, r ∈ (0, η# ), where η# ∈ [ηm , R), −∆p(r ) u + m X αi u(ηi ) + e0 = a0 , u(0) =

(P1 )

i =1

where a0 is a constant. If u is a solution of (P1 ), by integrating the equation from 0 to t, we have

0 p(t )−2 0 p(0)−2 0 u u (t ) = u0 (0) u (0) +

t

Z

ρ(r )f (r , u)dr ,

∀t ∈ [0, η# ].

0 p(0)−2 0 Denote a = u0 (0) u (0). It is easy to see that a is dependent on f (r , u). Define operator F : L1 −→ C as



F (g )(t ) =



t

Z

g (s)ds,

∀g ∈ L1 , ∀t ∈ [0, R].

0

By solving for u0 in (3) and integrating, we find u(t ) = a0 + F {ϕ −1 [t , (a + F (ρ f ))]}(t ), Since

Pm

m X

αi u(ηi ) + e0 = u(0) = a0 , we have  Z ηi −1 a0 + ϕ [t , (a + F (ρ f ))]dt + e0 = a0 ,

i =1

αi



0

i=1 m X

∀t ∈ [0, η# ].

αi

i=1

ηi

Z 0 m P

a0 =

ϕ [t , (a + F (ρ f ))]dt + e0 − a0 1 − −1

m X

! αi

= 0,

i =1

αi

R ηi

i =1

0

ϕ −1 [t , (a + F (ρ f ))]dt + e0 1−

m P

. αi

i =1

For fixed h ∈ C [0, η# ], we denote

Λh (a) =

m X i =1

αi

ηi

Z 0

ϕ [t , (a + h)]dt + e0 − a0 1 − −1

m X i =1

! αi .

(3)

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2953

Lemma 2.2. The function Λh (·) has the following properties (i) For any fixed h ∈ C [0, η# ], the equation

Λ h ( a) = 0

(4)

has a unique solution e a( h ) ∈ R . (ii) The function e a : C [0, η# ] → R, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover

  p+ −1  m P    |e0 | + |a0 | 1 − αi    i=1   |e a(h)| ≤ 3 khk0,[0,η# ] + 1 + . m  P       α η i i       



i=1

Proof. (i) From Lemma 2.1, it is easy to see that if (4) has a solution, then it is unique. Denote

   p+ −1   m P       | | 1 − α e + | a | i 0 0     i = 1   t0 = 3 khk0,[0,η# ] + 1 + . m  P       αi ηi   i=1

If a > t0 , since h ∈ C [0, η# ], it is easy to see that (a + h(t )) > 0 on [0, η# ] and

a + h(t ) ≥ a − khk0,[0,η# ]

  p+ −1  m P    |e0 | + |a0 | 1 − αi    i=1   ≥ 2 khk0,[0,η# ] + 1 + > 0, m  P       α η i i       



∀t ∈ [0, η# ],

i=1

then

1

1

|a + h(t )| p(t )−1 ≥ 2 p+ −1

 p+ −1  p+1−1    |e0 | + |a0 | 1 − αi    i = 1   khk0,[0,η# ] + 1 + , m  P       αi ηi       





m P

∀t ∈ [0, η# ].

i=1

Thus ϕ

−1

[t , (a + F (ρ f ))] is nonzero and keeps the same sign of a, then we can see that

Λ h ( a) > 0 . Similarly, we can see that Λh (a) < 0 when a < −t0 . It means that the existence of solutions of Λh (a) = 0. In this way, we define a function e a(h) : C [0, η# ] → RN , which satisfies

Λh (e a(h)) = 0. (ii) By the proof of (i), we also obtain e a sends bounded sets to bounded sets, and

 p+ −1     |e0 | + |a0 | 1 − αi    i = 1  |e a(h)| ≤ 3 khk0,[0,η# ] +  1 + . m   P       α η i i       





m P

i=1

It only remains to prove the continuity of e a. Let {un } be a convergent sequence in C and un → u as n → +∞. Since {e a(un )} is a bounded sequence, it contains a convergent subsequence {e a(unj )}. Let e a(unj ) → a∗ as j → +∞. Since Λunj (e a(unj )) = 0, letting j → +∞, we have Λu (a∗ ) = 0. From (i), we get a∗ = e a(u), it means that e a is continuous. This completes the proof.  Now, we define a : L1 [0, η# ] → R as a( u) = e a(F (u)).

(5)

It is clear that a(·) is continuous and sends bounded sets of L1 [0, η# ] to bounded sets of R, and hence it is a compact continuous mapping.

2954

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

If u is a solution of (P1 ), then m P

u(t ) =

αi

R ηi 0

i =1

ϕ −1 [t , (a(ρ f ) + F (ρ f ))] + e0 1−

m P

+ F {ϕ −1 [t , (a(ρ f ) + F (ρ f ))]}(t ),

∀t ∈ [0, η# ].

αi

i=1

We denote K (h)(t ) = F {ϕ −1 [t , (a(h) + F (h))]}(t ),

∀t ∈ [0, R].

Lemma 2.3. The operator K is continuous and sends equi-integrable sets of L1 [0, η# ] into relatively compact sets in C 1 [0, η# ]. Proof. Similar to the proof of Lemma 2.3 of [18], we omit it here. 1



1

Let us define P : L → C as m P

P (h) =

αi K (h)(ηi ) + e0

i =1

1−

m P

. αi

i =1

It is easy to see that P is compact continuous. We denote Nf : C 1 → L1 the Nemytskii operator associated to f defined by Nf (u)(t ) = f (t , u(t )),

a.e. on (0, R).

(6)

Now we consider the problem u = P (λρ Nf (u)) + K (λρ Nf (u)).

(7)

Note. It is easy to check that u is a solution of (P1 ) if and only if u is a solution of the abstract equation (7) when λ = 1. Theorem 2.4. If f satisfies sub-(p− − 1) growth condition, then problem (P1 ) has at least a solution. Proof. Denote Ψf (u, λ) := P (λρ Nf (u)) + K (λρ Nf (u)), where Nf (u) is defined in (6). We know that (P1 ) has the same solution of u = Ψf (u, λ),

(8)

when λ = 1. It is easy to see that the operator P is compact continuous from C 1 [0, η# ] to C 1 [0, η# ]. According to Lemmas 2.2 and 2.3, then we can see that Ψf (·, λ) is compact continuous from C 1 [0, η# ] to C 1 [0, η# ] for any λ ∈ [0, 1]. We claim that all the solutions of (8) are uniformly bounded for λ ∈ [0, 1]. In fact, if it is false, we can find a sequence of solutions {(un , λn )} for (8) such that kun k1,[0,η# ] → +∞ as n → +∞, and kun k1,[0,η# ] > 1 for any n = 1, 2, . . . . Since (un , λn ) are solutions of (8), we have

0 p(t )−2 0 p(0)−2 0 u u (t ) = u0 (0) u (0) + n

n

n

t

Z

n

0

)−1 kun k1q1,[(0r,η #]

λn ρ(r )f (r , un ) )−1 kun k1q1,[(0r,η #]

dr ,

∀t ∈ [0, η# ].

(9)

Since f satisfies sub-(p− − 1) growth condition, from (9) and Lemma 2.2, there exist positive constants C1 such that + 0 p(t )−1 q −1 u ≤ C1 kun k11,[0,η# ] , n

Denote α =

0

u (t ) n

∀t ∈ [0, η# ].

+

q 1 −1 , p− −1

0,[0,η# ]

we have

≤ C2 kun kα1,[0,η# ] .

Obviously

Z t |un (t )| = un (0) + u0n (r )dr 0 Z t max |u0n (t )|dr ≤ |un (0)| + 0 t ∈[0,η# ]

≤ C3 kun kα1,[0,η# ] .

(10)

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2955

Thus

kun k0,[0,η# ] ≤ C4 kun kα1,[0,η# ] ,

n = 1, 2, . . . .

(11)

Since α ∈ (0, 1), combining (10) and (11), we get a contradiction. Thus, we obtain that {kun k1,[0,η# ] } is uniformly bounded.

Thus, there exists a large enough R0 > 0 such that all the solutions of (8) belong to B(R0 ) = {u ∈ C 1 [0, η# ] | kuk1,[0,η# ] < R0 }, then the Leray–Schauder degree dLS [I − Ψ (·, λ), B(R0 ), 0] is well defined for λ ∈ [0, 1], and dLS [I − Ψ (·, 1), B(R0 ), 0] = dLS [I − Ψ (·, 0), B(R0 ), 0]. It is easy to see that u = Ψ (u, 0) is equal to

(I0 )

 0, t ∈ (0, η# ), −∆p(t ) u = m X αi u(ηi ) + e0 = a0 . u(0) = i =1

Obviously, system (I0 ) possesses only one solution u0 . Since u0 ∈ B(R0 ), the Leray–Schauder degree dLS [I − Ψf (·, 1), B(R0 ), 0] = dLS [I − Ψf (·, 0), B(R0 ), 0] 6= 0, therefore we obtain that (P1 ) has at least one solution. This completes the proof.



Let us consider

( P2 )

 ρ(r )(f (r , u) + δ h(r , u)) = 0, −∆p(r ) u + m X αi u(ηi ) + e0 = a0 . u(0) =

r ∈ (0, η# ), where η# ∈ [ηm , R),

i=1

Theorem 2.5. If f satisfies sub-(p− − 1) growth condition, and h is continuous, then (P2 ) has a solution when the parameter δ is small enough. Proof. Denote fδ (r , u) = f (r , u) + δ h(r , u). of

Denote Φδ (u, λ) := P (ρ Nfλδ (u)) + K (ρ Nfλδ (u)), where Nfλδ (u) is defined in (6). We know that (P2 ) has the same solution u = Φδ (u, λ),

when λ = 1. Obviously, f0 = f . Thus Φδ (u, 0) = Ψf (u, 1). From the proof of Theorem 2.4, we can see that all the solutions of u = Φδ (u, 0) are uniformly bounded, then there exists  a large enough R0 > 0 such that all the solutions of u = Φδ (u, 0) belong to B(R0 ) = u ∈ C 1 [0, η# ] | kuk1,[0,η# ] < R0 . Since Φδ (·, 0) is compact continuous from C 1 [0, η# ] to C 1 [0, η# ], we have inf

u∈∂ B(R0 )

ku − Φδ (u, 0)k1,[0,η# ] > 0.

Denote aλ (u, δ) = a(ρ Nfλδ (u)), Kλ# (u, δ) = K (Nfλδ (u)), Pλ# (u, δ) = P (Nfλδ (u)). Since f , h are continuous, we have



F (ρ Nf (u)) − F (ρ Nf (u)) → 0 for (u, λ) ∈ B(R0 ) × [0, 1] uniformly, as δ → 0, λδ 0 0,[0,η# ] a (u, δ) − a (u, δ) → 0 for (u, λ) ∈ B(R0 ) × [0, 1] uniformly, as δ → 0, λ 0

#

K (u, δ) − K # (u, δ) → 0 for (u, λ) ∈ B(R0 ) × [0, 1] uniformly, as δ → 0, λ 0 1,[0,η# ] # P (u, δ) − P # (u, δ) → 0 for (u, λ) ∈ B(R0 ) × [0, 1] uniformly, as δ → 0. λ 0 Thus

kΦδ (u, λ) − Φ0 (u, λ)k1,[0,η# ] → 0 for (u, λ) ∈ B(R0 ) × [0, 1] uniformly, as δ → 0. Obviously, Φ0 (u, λ) = Φδ (u, 0) = Φ0 (u, 0). Thus

kΦδ (u, λ) − Φδ (u, 0)k1,[0,η# ] → 0 for (u, λ) ∈ B(R0 ) × [0, 1] uniformly, as δ → 0.

2956

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

Thus, when δ is small enough, we can conclude that inf

(u,λ)∈∂ B(R0 )×[0,1]

ku − Φδ (u, λ)k1,[0,η# ] ≥

inf

u∈∂ B(R0 )

−

ku − Φδ (u, 0)k1,[0,η# ] sup

(u,λ)∈B(R0 )×[0,1]

kΦδ (u, 0) − Φδ (u, λ)k1,[0,η# ] > 0.

Thus u = Φδ (u, λ) has no solution on ∂ B(R0 ) for any λ ∈ [0, 1], when δ is small enough. It means that the Leray–Schauder degree dLS [I − Φδ (u, λ), B(R0 ), 0] is well defined for any λ ∈ [0, 1], and dLS [I − Φδ (u, λ), B(R0 ), 0] = dLS [I − Φδ (u, 0), B(R0 ), 0]. Since Φδ (u, 0) = Ψf (u, 1), from the proof of Theorem 2.4, we can see that the right-hand side is nonzero. Thus (P2 ) has at least a solution, when parameter δ is small enough.  Lemma 2.6 (Comparison Principle, See [17]). If u(0) ≥ v(0), u(R) ≥ v(R) and −∆p(r ) u + ρ(r )f (r , u) ≥ 0, −∆p(r ) v + ρ(r )f (r , v) ≤ 0, then u ≥ v a.e. on [0, R]. Lemma 2.7. If u is a solution of (P1 ), u(0) > max{0, 1−P0m α }, then we have u(η# ) > 0, u0 (η# ) > 0. i=1 i e

Proof. We claim that u(η# ) is the maximum of u. If it is false, then there are two cases. (a) u(0) is the maximum. We have u(0) =

m X

αi u(ηi ) + e0 ≤

i =1

m X

αi u(0) + e0 ,

i =1

thus u(0) ≤ 1−P0m α . It is a contradiction. i=1 i (b) There exists a ξ ∈ (0, η# ) such that u(ξ ) is the maximum of u(r ) on [0, η# ]. Thus u0 (ξ ) = 0 and u(ξ ) > u(0) > 0, so we have e

0 p(r )−2 0 p(ξ )−2 0 u u (r ) = u0 u (ξ ) +

r

Z

ρ(r )f (r , u)dr =

ξ

r

Z ξ

ρ(r )f (r , u)dr > 0,

∀r ∈ (ξ , η# ),

thus when r > ξ , we have u0 (r ) > 0, so u(ξ ) is not the maximum. According to (a) and (b), we can conclude that u(η# ) is the maximum and u(η# ) > u(0), thus there exists a ξ1 ∈ [0, η# ) such that u(ξ1 ) = u(0),

and

u(r ) > u(ξ1 ),

forany r ∈ (ξ1 , η# ].

From mean value theorem, there exists a η ∈ (ξ1 , η# ) such that u0 (η) =

u(η# ) − u(ξ1 )

η# − ξ1

> 0.

Since η ∈ (ξ1 , ηm ), by the definition of ξ1 , we have u(η) > u(ξ1 ) = u(0) > 0. Since

0 p(r )−2 0 p(r )−2 0 u u (r ) = u0 u (η) +

r

Z η

ρ(r )f (r , u)dr ,

∀r ∈ [η, η# ],

we have u0 (r ) > 0 and u(r ) > u(0) > 0, ∀r ∈ (ξ1 , η# ). So u(η# ) > 0 and u0 (η# ) > 0.



Lemma 2.8 (See [21, Theorem 3.5]). If ρ is bounded on [0, R − ε], |f (x, u)| ≤ C |u|q−1 and q < p(x), ∀x ∈ [0, R − ε], then (P) does not have boundary blow-up solution on (0, R − ε) for ε is positive and small enough. Note. The coefficient ρ ∈ C [0, R), therefore ρ is bounded on [0, R − ε].

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2957

3. Equation is negative definite Theorem 3.1. If f satisfies sub-(p− − 1) growth condition, and β(r ) > p(r ), ∀r ∈ [σ , R], then for any a0 > max{0, 1−P0m α }, i=1 i (P) has a blow-up solution u which satisfies u(0) = a0 . e

Proof. Let η# ∈ [σ , R). From Theorem 2.4 and Lemma 2.7, we can see that (P1 ) has a solution u which satisfies u(η# ) > 0,

u0 (η# ) > 0.

Since q1 (r ) < p(r ) on [0, R], there exists a strictly increasing sequence {rn } ⊂ [η# , R), such that r1 = η# and max q1 (r ) <

r ∈[ri ,ri+1 ]

min

r ∈[ri ,ri+1 ]

p(r ),

for any i = 1, 2, . . . .

(12)

Let us now consider the existence of solutions of the following problems

(Ii )



−∆p(r ) u + ρ(r )f (r , u) = 0, u(ri ) = ai , u0 (ri ) = bi ,

r ∈ (ri , ri+1 ),

where ai > 0, bi > 0. Obviously, problem (Ii ) is equivalent to the integral equation u( r ) = ai +

Z r 

t

Z

∗

ρ(s)f (s, u)ds

bi + ri

 p(t1)−1 −1 

t

Z

∗

ρ(s)f (s, u)ds dt ,

bi +

ri



∀r ∈ [ri , ri+1 ],

(13)

ri

where b∗i = |bi |p(ri )−2 bi . Denote Gi (u) = ai ,

zi (u) =

Z r 

∗

ρ(s)f (s, u)ds

bi + ri

 p(t1)−1 −1 

t

Z

b∗i +

ri

t

Z

 ρ(s)f (s, u)ds dt .

ri

It is easy to see that Gi and zi are compact operators from C ([ri , ri+1 ]) to C ([ri , ri+1 ]) (see [18, Lemma 2.3]). If u is a solution of (Ii ), according to (12) and (13), it is easy to see that 1

u0 (r ) =



b∗i +

Z

r

ρ(s)f (s, u)ds



1 −1 p(t )−1



b∗i +

r

Z

ri

1

 ρ(s)f (s, u)ds ,

∀r ∈ [ri , ri+1 ],

ri

then q1 (r )−1 0 u (r ) ≤ c1 + c2 kuk p(r )−1

0,[ri ,ri+1 ]

θ

≤ c3 + c2 kuk1i,[ri ,ri+1 ] ,

i = 1, 2, . . . ,

and then

|u(r )| ≤ |u(0)| +

r

Z ri

0 u (t ) dt ≤ c4 + c5 kukθi 1,[ri ,ri+1 ] ,

where θi , ci (i = 1, . . . , 5) are positive constants, and θi =

i = 1, 2, . . . ,

maxr ∈[r ,r q (r )−1 i i+1 ] 1 minr ∈[r ,r i i+1 ]p(r )−1

∈ (0, 1).

Thus, we have θ

kuk1,[ri ,ri+1 ] ≤ C1 + C2 kuk1i,[ri ,ri+1 ] ,

i = 1, 2, . . . ,

(14)

where C1 and C2 are positive constants. Let us consider the problem u = Φi (u, λ) = Gi (u) + λ zi (u).

(15)

It is easy to see that u is a solution of (Ii ) if and only if u is a solution of the abstract equation (15) when λ = 1. According to (14), it is easy to see that all the solutions of (15) for λ ∈ [0, 1] are uniformly bounded, then, there exists a positive constant R1 > R0 such that all the solutions of u = Φ1 (u, λ) belong to B∗ (R1 ) = {u ∈ C 1 ([r1 , r2 ]) | kuk1,[r1 ,r2 ] < R1 }. Thus, the Leray–Schauder degree dLS [I − Φ1 (·, λ), B∗ (R1 ), 0] is well defined for λ ∈ [0, 1], and dLS [I − Φ1 (·, 1), B∗ (R1 ), 0] = dLS [I − Φ1 (·, 0), B∗ (R1 ), 0]. Since dLS [I − Φ1 (·, 0), B∗ (R1 ), 0] = dB [I − G1 , B∗ (R1 ) ∩ R, 0] 6= 0, we can conclude that u = Φ1 (u, λ) has a solution when λ = 1.

2958

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

It is easy to see that u(r2 ) > 0 and u0 (r2 ) > 0. Thus, the solution of (P1 ) can be extended onto [0, r2 ]. Similarly, the solution of (P1 ) can be extended onto [0, rn ], then it can be extended onto [0, R). It only remains to prove that u is a boundary blow-up solution. Since u(η# ) > 0 and u0 (η# ) > 0, from (H2 ), we can see that u is increasing on [η# , R). Since β(r ) > p(r ) on [σ , R] and ri → R− , let i is big enough such that

β∗− := inf β(r ) > p+ ∗ := sup p(r ), r ∈[ri ,R)

r ∈[ri ,R)

and f (r , u(r )) ≥ C3 ,

∀r ∈ [ri , R), where C3 is a positive constant. p(ri )−2 0

Denote bi = |u (ri )| ∗

0

u (ri ). Obviously

 p(t1)−1 Z t Z r ∗ ρ(s)f (s, u)ds) (bi + u(r ) = u(ri ) + dt ri

ri

≥ u(ri ) +

Z r Z ri

≥ u(ri ) +

ρ(s)C3 ds

Z r

Z

 p(t1)−1 dt

ri

1

−β∗− +1

C3 ρ0 − [(R − t ) β∗ − 1

ri

≥ u(ri ) +

t

r



R+ri 2

The proof is completed.

C3 ρ0

1 2(β∗− − 1)

−β∗− +1

− (R − ri )

−β∗− +1

(R − t )



1 + p∗ −1

 ]

1 + p∗ −1

dt

dt → +∞ (as r → R− ).



Theorem 3.2. If (P) is negative definite, and β(r ) > p(r ), ∀r ∈ [σ , R], then for any a0 > max{0, 1−P0m α }, the following i=1 i problem e

 δρ(r )f (r , u) = 0, r ∈ (0, R), −∆p(r ) u + m X αi u(ηi ) + e0 , u(r ) → +∞, u(0) =

as r → R− ,

i =1

has a solution u which satisfies u(0) = a0 when the positive parameter δ is small enough. Proof. Since q(R) < p(R), there exists a ξ ∈ [σ , R) such that maxr ∈[ξ ,R] max q(r ) < minr ∈[ξ ,R] p(x). We consider the problem (II)

 δρ(r )f (r , u) = 0, r ∈ (0, ξ ), −∆p(r ) u + m X αi u(ηi ) + e0 = a. u(0) = i=1

According to Theorem 2.5, we know that (II) has a solution when positive parameter δ is small enough. From Lemma 2.7, we have u(ξ ) > 0, u0 (ξ ) > 0. Similar to the proof of Theorem 3.1, we know that u can be extended to a solution on [0, R), and u is a boundary blow-up solution (i.e. u(r ) → +∞, as r → R− ).  Theorem 3.3. If (P) is negative definite and β1 (r ) < p(r ), ∀r ∈ [σ , R], then (P) does not have boundary blow-up solution. Proof. Without loss of generality, we may assume that f (r , u) ≤ buq−1 ,

∀r ∈ [σ , R), u ≥ 1,

(16)

where q < p(R) is a positive constant. Suppose, for the contradiction, that (P) has a boundary blow-up solution u. It is easy to see that there exists a point t0 ∈ [σ , R) such that u(t0 ) > 1, u0 (t0 ) > 0, u is increasing on [t0 , R) and u(t0 ) = maxt ∈[0,t0 ] u(t ). According to (16) and the continuity of p(r ), there exists a positive σ1 ∈ [t0 , R) such that

θ := sup

q−1

r ∈[σ1 ,R) p(r ) − 1

1 ≤ u(r ),

< 1,

(17)

∀r ∈ [σ1 , R),

f (r , u(r )) ≤ buq−1 (r ),

∀r ∈ [σ1 , R),

(18)

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2959

and 1 ≤ bu

q −1

(σ1 )

σ1

Z

ρ(s)ds.

(19)

0

Since f (r , u) is increasing with respect to u, we have f (r , u(r )) ≤ f (r , u(σ1 )),

∀r ∈ [0, σ1 ].

(20)

Combining (20) and the continuity of f , we get σ1

Z

ρ(s)f (s, u(s))ds ≤

σ1

Z

ρ(s)f (s, u(σ1 ))ds.

(21)

0

0

Since u(r ) is increasing on [σ1 , R), from (18), (19) and (21), we have σ1

Z

|u0 |p(r )−2 u0 (r ) = |u0 |p(0)−2 u0 (0) +

ρ(s)f (s, u(s))ds +

σ1

0

Z ≤ C4 +

r

Z

ρ(s)f (s, u(s))ds

r

σ1

bρ(s)uq−1 (s)ds

≤ C4 + buq−1 (r ) Z q −1 ≤ C4 bu (σ1 )

r

Z

ρ(s)ds

0

σ1

Z

r

ρ(s)ds + bu (r ) ρ(s)ds 0 0 Z r ≤ (C4 + 1)buq−1 (r ) ρ(s)ds, ∀r ∈ [σ1 , R), q −1

0

where p(0)−1

C4 = |u (0)| 0

Z +

σ1 0

ρ(s)f (s, u(σ1 ))ds .

Thus

|u0 |p(r )−2 u0 ≤ (C4 + 1)buq−1 (r )

r

Z

ρ(s)ds,

∀r ∈ [σ1 , R).

(22)

0

From (17) and (22), we have

  1   1 Z r Z r p(r )−1 p(r )−1 θ q−1 ≤ u (r ) (C4 + 1)b ρ(s)ds , u ≤ (C4 + 1)bu (r ) ρ(s)ds 0

∀r ∈ [σ1 , R),

0

0

and u0 uθ (r )

  1 Z r p(r )−1 ρ(s)ds ≤ (C4 + 1)b ,

∀r ∈ [σ1 , R).

(23)

0

Integrating (23) from σ1 to r, yields

Z

r

σ1

u0 uθ (s)

Z

R

ds ≤ σ1

  1 Z r p(r )−1 (C4 + 1)b ρ(s)ds dr .

(24)

0

Since β1 (R) < p(R), when R − σ1 is small enough, then the right-hand side of (24) is bounded, but the left-hand side of (24) tends to +∞ as r → R− . It is a contradiction. This completes the proof.  4. Equation is positive definite p(r )−2 0 We say u is a supersolution of (P) if u ∈ C 1 ([0, R), R) with u0 u (r ) absolutely continuous on [0, R − ε) for any ε ∈ (0, R), and satisfies



 ρ(r )f (r , u) ≥ 0, x ∈ (0, R), −∆p(r ) u + m X αi u(ηi ) + e0 , u(r ) → +∞ (as r → R− ), u(0) ≥ i=1

2960

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

p(r )−2 0 and we say v is a subsolution of (P) if v ∈ C 1 ([0, R), R) with v 0 v (r ) absolutely continuous on [0, R − ε) for any ε ∈ (0, R), and satisfies



 ρ(r )f (r , v) ≤ 0, r ∈ (0, R), −∆p(r ) v + m X αi v(ηi ) + e0 , v(r ) → +∞ (as r → R− ). v(0) ≤ i=1

Since q(r ) − p(r ) > 0 on [σ , R], there exists an integer n0 ≥ 3 such that q(r ) − p(r ) > g (r , s, ) on [0, R) as

1 n0

 C (R − r )−s + k, R0 ≤ r < R,   Z R0  p(R0 )−1  1  [Cs(R − R0 )−s−1 ] p(t )−1 [sin ε(t − τ )] p(t )−1 dt , C (R − R0 )−s + k − g (r , s, ) = Z r R0   p(R0 )−1 1  −s   C (R − R0 ) + k − [Cs(R − R0 )−s−1 ] p(t )−1 [sin ε(t − τ )] p(t )−1 dt , τ

where s is a positive constant, σ < τ < R0 < R and R − R0 is small enough, ε = constant and C = C (s) = (1 + )



1

ϑ0 ρ0

s

p(R)−1

k=

C



R − R0

ρ0

−s−1−β ∗ # p−p −1

R0

Z

"

+

2

τ < r < R0 , r ≤ τ, ,  ∈ (0, 1) is a small positive

 1 q(R)−p(R) , (s + 1)(p(R) − 1)

+

"

π

2(R0 −τ )

on [σ , R]. Define the function



Cs

R − R0

−s−1 # pp((Rt0)−)−11

2

τ

1

[sin ε(t − τ )] p(t )−1 dt +

|e0 | , m P 1− αi i =1

where β ∗ = maxr ∈[0,R] |β(r )|. Similarly, define the function v(r , s, ) on [0, R) as

 ∗ C (R − r )−s − k∗ , R0 ≤ r < R,   Z R0  p(R0 )−1  1  ∗ C (R − R0 )−s − k∗ − [C ∗ s(R − R0 )−s−1 ] p(t )−1 [sin ε(t − τ )] p(t )−1 dt , v(r , s, ) = Zr R0   p(R0 )−1 1  ∗ −s ∗   C (R − R0 ) − k − [C ∗ s(R − R0 )−s−1 ] p(t )−1 [sin ε(t − τ )] p(t )−1 dt ,

τ < r < R0 , r ≤ τ,

τ

where ∗

C =

C∗ (s)

= (1 − )



1

ϑ1 ρ1

s

p(R)−1

(s + 1)(p(R) − 1)



1 q1 (R)−p(R)

+

" k∗ =

C∗

ρ1



R − R0

−s−1−β1∗ # p−p −1

2

R0

Z

" C ∗s

+



R − R0

τ

,

−s−1 # pp((Rt0)−)−11

2

1

[sin ε(t − τ )] p(t )−1 dt +

|e0 | , m P 1− αi i=1

where β1∗ = maxr ∈[0,R] |β1 (r )|. Obviously, for any positive constant s, we have g (·, s, ), v(·, s, ) ∈ C 1 [0, R). Lemma 4.1. Under the conditions of β1 (R) < p(R) < q(R), we have p(R)−β(R)

(i) g (r , s1 , ) is a supersolution of (P), where s1 = q(R)−p(R) , C = C (s1 ); p(R)−β (R) (ii) v(r , s2 , ) is a subsolution of (P), where s2 = q (R)−p1(R) , C ∗ = C∗ (s2 ). 1

Proof. (i) When r ∈ (R0 , R), we have g 0 = Cs1 (R − r )−s1 −1 , then

0 p(r )−2 0 g g = (Cs1 )p(r )−1 (R − r )−(s1 +1)(p(r )−1) , and

p(r )−2 0 0 ( g 0 g ) = (Cs1 )p(r )−1 (s1 + 1)(p(r ) − 1)(R − r )−(s1 +1)(p(r )−1)−1 + (Cs1 )p(r )−1 (R − r )−(s1 +1)(p(r )−1) (−(s1 + 1)p(r ))0 ln(R − r ) + ((Cs1 )p(r )−1 )0 (R − r )−(s1 +1)(p(r )−1) = (Cs1 )p(r )−1 (s1 + 1)(p(r ) − 1)(R − r )−(s1 +1)(p(r )−1)−1 (1 + h(r )),

(25)

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2961

where h( r ) =

(−(s1 + 1)p(r ))0 ln(R − r ) ((Cs1 )p(r )−1 )0 (R − r ) + (R − r ) (s1 + 1)(p(r ) − 1) (Cs1 )p(r )−1 (s1 + 1)(p(r ) − 1) 1

=

1

n0 (−(s1 + 1)p(r ))0 (R − r ) n0 ln(R − r ) n0

(s1 + 1)(p(r ) − 1)

1

(R − r )

1− n1

0

+

((Cs1 )p(r )−1 )0 (R − r ) n0 1− 1 (R − r ) n0 . p ( r )− 1 (Cs1 ) (s1 + 1)(p(r ) − 1)

It is easy to see that there exist positive constants A, B ≥ 1 (A, B depend on R, p, q, n0 , s1 ) such that

1 1 n0 (−(s1 + 1)p(r ))0 (R − r ) n0 ln(R − r ) n0 ≤ A, ∀r ∈ (R0 , R), (s1 + 1)(p(r ) − 1) 1 ((Cs1 )p(r )−1 )0 n (Cs )p(r )−1 (s + 1)(p(r ) − 1) (R − r ) 0 ≤ B, ∀r ∈ (R0 , R), 1 1 then we have

|h(r )| ≤ (A + B)(R − r )

1

1− n1

0

≤ [(A + B + 1)(R − R0 ) n0 ]n0 −1 ,

∀r ∈ (R0 , R).

(26)

If 0 < R − R0 is small enough, we have 1

(A + B + 1)(R − R0 ) n0 ≤

 4n0

.

(27)

Combining (26) and (27), if 0 < R − R0 is small enough, for any r ∈ [R0 , R), we have p(r )−1

(Cs1 )

(s1 + 1)(p(r ) − 1)(1 + h(r )) ≤ (Cs1 )

p(r )−1

≤ ϑ0 ρ 0 C

(s1 + 1)(p(r ) − 1) 1 +

q(r )−1



 4n1

1

0

1+





n0 −1 !

4n0

.

(28)

Since p(r ), q(r ) and β(r ) are Lipschitz continuous, if 0 < R − R0 is small enough, we have 1

(R − r )−(s1 +1)(p(r )−1)−1 ≤ (R − r )−s1 (q(r )−1)−β(r ) (1 + ) 4n0 ,

∀r ∈ [R0 , R).

(29)

From (25), (28) and (29), we have

p(r )−2 0 0 ( g 0 g ) ≤ (Cs1 )p(r )−1 (s1 + 1)(p(r ) − 1)(R − r )−(s1 +1)(p(r )−1)−1 1 + ≤ ϑ0 ρ0 C q(r )−1



1 1+

 4n1

0





n−1 !

4n0

1

(R − r )−s1 (q(r )−1)−β(r ) (1 + ) 4n0

−β(r )

= ρ0 (R − r ) ϑ0 (C (R − r )−s1 )q(r )−1 q(r )−1 ≤ ρ(r )ϑ0 g ≤ ρ(r )f (r , g ), ∀r ∈ (R0 , R), when 0 < R − R0 is small enough. Thus, when 0 < R − R0 is small enough, we have

p(r )−2 0 0 ( g 0 g ) ≤ ρ(r )f (r , g ) , ∀r ∈ (R0 , R). Obviously, when 0 < R − R0 is small enough, we have

p(r )−2 0 0 ( g 0 g ) = ε(Cs1 (R − R0 )−s1 −1 )(p(R0 )−1) cos(ε(r − τ ))   " # p+ q(r )−1   C  R − R −s1 −1−β ∗ p− −1   0 ≤ ρ(r )ϑ0   2  ρ0  ≤ ρ(r )ϑ0 g q(r )−1 ≤ ρ(r )f (r , g ), ∀r ∈ (τ , R0 ),

(30)

2962

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

and then

p(r )−2 0 0 ( g 0 g ) ≤ ρ(r )f (r , g ),

∀r ∈ (τ , R0 ).

(31)

Obviously

p(r )−2 0 0 ( g 0 g ) = 0 ≤ ρ(r )f (r , g ),

0 ≤ r < τ.

(32)

Notice that g (r , s1 , ) is a C function on B(0, R). From (30), (31) and (32), we can see that 1

−∆p(r ) g + ρ(r )f (r , g ) ≥ 0,

r ∈ (0, R).

It is easy to see that g (0) ≥

m X

αi g (ηi ) + e0 .

i =1

Thus g (r , s1 , ) is a supersolution of (P), when 0 < R − R0 is small enough (R0 depends on R, p, q, β, n0 , s1 ). (ii) Notice that v(·, s2 , ) ∈ C 1 [0, R). Similar to the proof of (i), we can obtain that

p(r )−2 0 0 ( v 0 v ) ≥ ρ(r )ϑ1 v q(r )−1 ≥ ρ(r )f (r , v),

∀r ∈ (R0 , R).

It is easy to see that

p(r )−2 0 0 ( v 0 v ) ≥ 0 ≥ ρ(r )f (r , v),

∀r ∈ (0, R0 ).

It is easy to see that

v(0) ≤

m X

αi v(ηi ) + e0 .

i=1

It means that v(r , s2 , ) is a subsolution of (P). This completes the proof.  Note. It is easy to see that g (r , s1 , ) ≥ v(r , s2 , ), ∀r ∈ [0, R), and g (r , s, ) is a supersolution of (P) for any s ≥ s1 . Let {Rn > σ } be a strictly increasing sequence and Rn → R− as n → +∞. Let us consider the following problems

(P∗1 ) (P01 )

 −∆p(r ) u∗ + ρ(r )f (r , u∗ ) = 0, r ∈ (0, Rn ), u∗ (0) = b, u∗ (Rn ) = v(Rn , s2 , ), where b ∈ [v(0, s2 , ), g (0, s1 , )].  ρ(r )f (r , u) = 0, r ∈ (0, Rn ), −∆p(r ) u + m X αi u(ηi ) + e0 , u(Rn ) = v(Rn , s2 , ). u(0) = i =1

Lemma 4.2. (i) (P∗1 ) has a solution u∗ (r ), which satisfies

v(r , s2 , ) ≤ u∗ (r ) ≤ g (r , s1 , ),

∀r ∈ [0, Rn ];

(ii) (P1 ) has a solution u(r ), which satisfies 0

v(r , s2 , ) ≤ u(r ) ≤ g (r , s1 , ),

∀r ∈ [0, Rn ].

Proof. Denote

v(r ) = v(r , s2 , ),

g (r ) = g (r , s1 , ),

∀r ∈ [0, R).

(i) We denote

 u − g (r )    f (r , g (r )) + 1 + u2 , u ≥ g (r ), e f (r , u) = f (r , u), v(r ) ≤ u ≤ g (r ),   u − v(r )  f (x, v(r )) + , u ≤ v(r ). 1 + u2 We consider the problem

(P2 ) ∗

 −∆p(r ) u + ρ(r )e f (r , u) = 0, r ∈ (0, Rn ), u(0) = b, u(Rn ) = v(Rn ),

where b ∈ [v(0), g (0)]. Obviously u(Rn ) ∈ [v(Rn ), g (Rn )].

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2963

Since f , g and v are continuous, it is easy to see that e f (x, u) is bounded. Similar to the proof of Theorem 3.1, we know that (P∗2 ) has a solution u(r ). If we can prove that v(r ) ≤ u(r ) ≤ g (r ), ∀r ∈ [0, Rn ], then we can see that u is a solution of (P∗1 ). We shall only prove that u(r ) ≤ g (r ) for any r ∈ [0, Rn ]. The argument of v(r ) ≤ u(r ) is similar. Obviously, v(r ) ≤ g (r ), ∀r ∈ [0, R). Assume that u(r ) > g (r ) for some r ∈ (0, Rn ), then there exists an r0 ∈ (0, Rn ) and a positive number δ such that u(r0 ) = g (r0 ) + δ , u(r ) ≤ g (r ) + δ for any r ∈ [0, Rn ]. Hence u0 (r0 ) = g 0 (r0 ).

(33)

There exists a positive number ξ such that u(r ) > g (r ) for any r ∈ J := (r0 − ξ , r0 + ξ ) ⊂ [0, Rn ]. From the definition of g , u and e f , we conclude that

 0 g 0 p(r )−2 g 0 ≤ ρ(r )f (r , g ) = ρ(r )e f (r , g ) < ρ(r )e f (r , u) on [r0 − ξ1 , r0 + ξ1 ], where ξ1 ∈ (0, ξ ) is small enough. For any r ∈ (r0 , r0 + ξ1 ], we have

Z r Z r Z r 0 p(r )−2 0 0 0 p(r )−2 0 0 u g e ρ(r )f (r , u)dr = u dr . g dr <

(34)

r0

r0

r0

From (33) and (34), we have

0 p(r )−2 0 0 p(r )−2 0 g g < u u on (r0 , r0 + ξ1 ], it means that

(g + δ)0 < u0 on (r0 , r0 + ξ1 ]. It is a contradiction to the definition of r0 , so u(r ) ≤ g (r ) for any r ∈ [0, Rn ]. (ii) Let b = v(0), then (P∗1 ) has a solution h0 (r ), which satisfies v(r ) ≤ h0 (r ) ≤ g (r ), ∀r ∈ [0, Rn ], then h0 (0) = v(0) ≤

m X

αi v(ηi ) + e0 ≤

i =1

m X

αi h0 (ηi ) + e0 .

i =1

Hence, h0 (r ) is a subsolution of (P01 ). Pm 0 We may assume that h0 (0) < i=1 αi h0 (ηi ) + e0 , or we get h0 (r ) is a solution of (P1 ). We denote

v1 (r ) = h0 (r ).

(35)

Similarly, let b = g (0), we get a solution h (r ) of (P1 ), which satisfies v1 (r ) ≤ h (r ) ≤ g (r ), ∀r ∈ [0, Rn ], thus ∗

0

h0 (0) = g (0) ≥

m X

αi g (ηi ) + e0 ≥

i =1

m X

0

αi h0 (ηi ) + e0 ,

i=1

hence, h0 (r ) is a supersolution of (P01 ). Pm 0 0 0 We may assume that h0 (0) > i=1 αi h (ηi ) + e0 , or we get that h (r ) is a solution of (P1 ). We denote g1 (r ) = h0 (r ).

(36)

From (35) and (36), we obtain

v(r ) ≤ v1 (r ) ≤ g1 (r ) ≤ g (r ),

∀r ∈ [0, Rn ],

and satisfy

v1 (Rn ) = g1 (Rn ) = v(Rn ), m X v1 (0) ≤ αi v1 (ηi ) + e0 , i =1

g1 (0) ≥

m X

αi g1 (ηi ) + e0 .

i =1

Obviously, v1 (r ) and g1 (r ) are subsolution and supersolution of (P01 ), respectively. Now we assume that h1 (r ) is a solution v (0)+g (0)

of (P∗1 ), when b = 1 2 1 . It is easy to see that h1 (r ) satisfies v1 (r ) ≤ h1 (r ) ≤ g1 (r ), ∀r ∈ [0, Rn ], h1 (Rn ) = v(Rn ). We Pm 0 may assume that h1 (0) 6= i=1 αi h1 (ηi ) + e0 , or we get h1 (r ) is a solution of (P1 ). There are two cases: (a) h1 (0) >

Pm

i =1

αi h1 (ηi ) + e0 , then h1 (r ) is a supersolution of (P01 ). We denote g2 (r ) = h1 (r ), v2 (r ) = v1 (r ).

2964

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

(b) h1 (0) < Let b =

(1◦ ) (2◦ ) (3◦ ) (4◦ )

Pm

i=1

αi h1 (ηi ) + e0 , then h1 (r ) is a subsolution of (P01 ). We denote g2 (r ) = g1 (r ), v2 (r ) = h1 (r ).

v2 (0)+g2 (0) 2

, and repeat the step. Then we get sequences {vk (r )} and {gk (r )} satisfy

v1 (r ) ≤ v2 (r ) ≤ · · · ≤ vk (r ) ≤ · · · ≤ gk (r ) ≤ · · · ≤ g2 (r ) ≤ g1 (r ), ∀r ∈ [0, Rn ], vk (Rn ) =P gk (Rn ) = v(Rn ), k = 1, 2, . .P ., m vk (0) ≤ m i=1 αi vk (ηi ) + e0 , gk (0) ≥ i=1 αi gk (ηi ) + e0 , k = 1, 2, . . ., gk−1 (0)−vk−1 (0) gk (0) − vk (0) = , k = 1 , 2 , . . .. 2

Obviously, when k → +∞, we have lim vk (0) = lim gk (0),

k→+∞

k→+∞

and

[vk+1 (0), vk+1 (0)] ⊆ [vk (0), vk (0)], k = 1, 2, . . . , gk (ηi ) ≥ vk (ηi ), k = 1, 2, . . . , i = 1, . . . , m, therefore, we have

vk (0) ≤

m X

αi vk (ηi ) + e0 ≤

i=1

m X

αi gk (ηi ) + e0 ≤ gk (0),

k = 1, 2, . . . .

i=1

Thus when k → +∞, we have lim

k→+∞

m X

αi vk (ηi ) = lim

k→+∞

i =1

m X

αi gk (ηi ).

i =1

It is easy to know that {vk (r )} and {gk (r )} possess convergent subsequence {vki (r )} and {gkj (r )} in C 1 [0, Rn ], respectively. We may assume that



vki (r ) → v∗ (r ), in C 1 [0, Rn ], gkj (r ) → g∗ (r ), in C 1 [0, Rn ].

Obviously, v∗ (r ), g∗ (r ) are solutions of (P01 ). The proof is completed.  Theorem 4.3. If (P) is positive definite, and β1 (R) < p(R), then (P) has a solution u which satisfies

v(r , s2 , ) ≤ u(r ) ≤ g (r , s1 , ),

∀r ∈ [0, R).

Proof. Denote

v(r ) = v(r , s2 , ),

g (r ) = g (r , s1 , ).

Let us consider the following problem

(Sn )

 ρ(r )f (r , u) = 0, r ∈ (0, Rn ), −∆p(r ) u + m X αi u(ηi ) + e0 , u(Rn ) = v(Rn ), u(0) = i =1

where {Rn > σ } is a strictly increasing sequence and Rn → R− as n → +∞. From (ii) of Lemma 4.2, we can see that (S1 ) has a solution u1 which satisfies

v(r ) ≤ u1 (r ) ≤ g (r ),

∀r ∈ [0, R1 ].

Let us consider

(S ) ∗

 −∆p(r ) u + ρ(r )f (r , u) = 0, r ∈ (0, R2 ), u(0) = u1 (0), u(R2 ) = v(R2 ).

From (i) of Lemma 4.2, we can see that (S∗ ) has a solution u∗1 which satisfies

v(r ) ≤ u∗1 (r ) ≤ g (r ),

∀r ∈ [0, R2 ].

Thus u1 (0) = u1 (0) and u1 (R1 ) ≥ v(R1 ) = u1 (R1 ). From the comparison principle, we get ∗

u1 (r ) ≤ u∗1 (r ),

∗

∀r ∈ [0, R1 ].

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2965

Obviously u∗1 (0) = u1 (0) =

m X

αi u1 (ηi ) + e0 ≤

i=1

m X

αi u∗1 (ηi ) + e0 .

i =1

Thus u1 (r ) is a subsolution of (S2 ), and u1 (r ) ≤ g (r ), ∀r ∈ [0, R2 ]. From (ii) of Lemma 4.2, we get that (S2 ) has a solution u2 (r ) which satisfies ∗

∗

u∗1 (r ) ≤ u2 (r ) ≤ g (r ),

∀r ∈ [0, R2 ].

Thus u1 (r ) ≤ u2 (r ),

∀r ∈ [0, R1 ].

Repeating the step, we get a sequence {un (r )}, where un (r ) is a solution of (Sn ), which satisfies

v(r ) ≤ un (r ) ≤ un+1 (r ) ≤ g (r ),

∀r ∈ [0, Rn ].

Thus {un (r )} is local uniformly bounded on [0, R), and {un (r )} is local uniformly C 1,α regular. Thus there exists a subsequence {unj } of {un } such that unj (r ) → u0 (r ),

∀r ∈ [0, R),

unj (r ) → u0 (r ),

∀r ∈ [0, R).

0

0

Thus

−∆p(r ) u0 + ρ(r )f (r , u0 ) = 0,

r ∈ (0, R).

Obviously u0 (0) =

m X

αi u0 (ηi ) + e0 ,

v(r ) ≤ u0 (r ) ≤ g (r ),

∀r ∈ [0, R).

i =1

Thus u is a solution of (P). This completes the proof.

 q(R)

Theorem 4.4. If β(r ) ≥ 0 and (P) is positive definite, then for every solution u of (P), we have limr →R− u(r )(R − r ) q(R)−p(R) < +∞. In order to prove Theorem 4.4, we need to do some preparation. Let us consider

 ρ(r )f (r , u) = 0, in (0, R∗ ), −∆p(r ) u + m X αi u(ηi ) + e0 . u(0) =

where R∗ ∈ (σ , R), (37)

i =1

p(r )−2

We say u is a supersolution of (37), if u ∈ C 1 [0, R∗ ], with u0

 ρ(r )f (r , u) ≥ 0, −∆p(r ) u + m X αi u(ηi ) + e0 . u(0) ≥

u0 (r ) absolutely continuous on [0, R∗ ], and satisfies

r ∈ (0, R∗ ),

i=1

p(r )−2

We say u is a subsolution of (37), if u ∈ C 1 [0, R∗ ], with u0

 ρ(r )f (r , u) ≤ 0, −∆p(r ) u + m X αi u(ηi ) + e0 . u(0) ≤

u0 (r ) absolutely continuous on [0, R∗ ], and satisfies

x ∈ (0, R∗ ),

i=1

Lemma 4.5. If u is a supersolution of (37), v is a subsolution of (37), and u(R∗ ) ≥ v(R∗ ), then we have (i) If u(0) < v(0), r∗ = min{r |u(r ) ≥ v(r ) }, then r∗ > η1 , e (ii) If v(0) ≤ min{0, 1−P0m α } and u(R∗ ) ≥ v(0), then u(r ) ≥ v(r ), ∀r ∈ [0, R). i=1

i

2966

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

Proof. (i) Obviously, u(r∗ ) = v(r∗ ). Since u(R∗ ) ≥ v(R∗ ), from the comparison principle, we can easily see that u(r ) ≥ v(r ) for any r ∈ [r∗ , R∗ ]. If r∗ ≤ η1 , then we have u(0) ≥

m X

αi u(ηi ) + e0 ≥

i=1

m X

αi v(ηi ) + e0 ≥ v(0).

i=1

It is a contradiction to u(0) < v(0). Thus r∗ > η1 . (ii) If u(0) ≥ v(0), according to the comparison principle, we can see that u(r ) ≥ v(r ), ∀r ∈ [0, R). It only remains to prove that u(0) ≥ v(0). If it is false, i.e. u(0) < v(0), there are two cases. (a) If u(ηi ) ≥ u(0) for any i = 1, . . . , m, then we have u(0) ≥

m X

αi u(ηi ) + e0 ≥

i=1

m X

αi u(0) + e0 ,

i=1

thus u(0) ≥

e0 1−

≥ v(0).

m P

αi

i=1

It is a contradiction. (b) There are some i0 ∈ {1, . . . , m} such that u(ηi0 ) < u(0). Since u(ηi0 ) < u(0) < v(0) ≤ 0, there exists r# ∈ (0, ηi0 ) such that u(r# ) < v(0) and u0 (r# ) < 0. Since u is a supersolution of (37), from (H2 ) we have r

Z

0 p(r )−2 0 p(r )−2 0 u u (r ) ≤ u0 u (r# ) +

ρ(s)f (s, u(s))ds ≤ 0,

∀r ∈ [r# , R∗ ],

r#

then u(r ) ≤ u(r# ) +

r

Z

ϕ

−1



p(r )−2 0 t , u0 u (r# ) +

r#

t

Z



ρ(s)f (s, u(s))ds dt < v(0) ≤ u(R∗ ),

∀r ∈ [r# , R∗ ].

r#

It is a contradiction. Therefore, u(0) ≥ v(0).



Lemma 4.6. If u and v are solutions of (37), u(R∗ ) ≥ v(R∗ ), u(0) < v(0), r∗ = min{r |u(r ) ≥ v(r ) }, then we have (i) r∗ > η1 , (ii) r∗ < ηm . Proof. (i) From Lemma 4.5, we can see that r∗ > η1 . (ii) If it is false, then we have r∗ ≥ ηm . Obviously, u(r∗ ) = v(r∗ ). Since u and v are solutions of (37), we have

0 p(x)−2 0 u u (r ) = a +

r

Z

ρ(s)f (s, u(s))ds,

p(0)−2 0 ∀r ∈ [0, r∗ ], where a = u0 u (0),

0

u(r ) = u(0) +

r

Z

ϕ



t, a +

−1

0

t

Z



ρ(s)f (s, u(s))ds dt ,

∀r ∈ [0, r∗ ],

(38)

0

0 p(x)−2 0 v v (r ) = b +

r

Z

ρ(s)f (s, v(s))ds,

p(0)−2 0 ∀r ∈ [0, r∗ ], where b = v 0 v (0),

0

v(r ) = v(0) +

r

Z

ϕ

−1



t, b +

0

t

Z



ρ(s)f (s, v(s))ds dt ,

∀r ∈ [0, r∗ ].

(39)

0

We claim that u0 (r ) ≥ v 0 (r ), ∀r ∈ [0, r∗ ). If it is false, then there exists a t0 ∈ [0, r∗ ) such that u0 (t0 ) < v 0 (t0 ). Obviously

0 p(r )−2 0 p(t0 )−2 0 u u (r ) = u0 u (t0 ) +

r

Z

ρ(s)f (s, u(s))ds,

∀r ∈ [t0 , r∗ ],

t0

u(r ) = u(t0 ) +

r

Z

ϕ

−1



p(t0 )−2 0 t , u0 u (t0 ) +

t0

Z

t



ρ(s)f (s, u(s))ds dt ,

∀r ∈ [t0 , r∗ ],

t0

0 p(r )−2 0 p(t )−2 v v (r ) = v 0 0 v 0 (t0 ) +

Z

r

ρ(s)f (s, v(s))ds,

∀r ∈ [t0 , r∗ ],

t0

v(r ) = v(t0 ) +

Z

r

t0

ϕ

−1



p(t0 )−2 0 t , v 0 v (t0 ) +

Z

t



ρ(s)f (s, v(s))ds dt , t0

∀r ∈ [t0 , r∗ ].

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2967

Since f (r , ·) is increasing, and u(t ) < v(t ) for any t ∈ [t0 , r∗ ), we have u(r∗ ) < v(r∗ ). It is a contradiction. Thus u0 (r ) ≥ v 0 (r ),

∀r ∈ [0, r∗ ).

(40)

From the boundary value condition, we have m X

αi

u0 (r )dt = u(0) 1 −

0

i =1 m X

ηi

Z

αi

v (r )dt = v(0) 1 − 0

0

i =1

! αi − e0 ,

i=1

ηi

Z

m X

m X

! αi − e0 .

i =1

From (40), we have m X

ηi

Z

αi

u (r )dr ≥ 0

m X

0

i =1

αi

Z

ηi

v 0 (r )dr .

0

i=1

Therefore u(0) ≥ v(0). It is a contradiction. Thus r∗ < ηm .



Proof of Theorem 4.4. Consider the following problem

 ρ(r )f (r , u) = 0, r ∈ (0, Rn ), −∆p(r ) u + m X αi u(ηi ) + e0 , u(r ) → +∞ (as r → R− u(0) = n ),

(In )

i =1

where {Rn > σ } is a strictly increasing sequence and Rn → R− as n → +∞. Obviously, ρ is bounded on [0, Rn ]. From Theorem 4.3, we can see that (In ) has a blow-up solution un , and p(Rn )

lim un (r )(Rn − r ) q(Rn )−p(Rn ) < +∞. −

r →Rn

Let u be a solution of (P). From Lemma 4.6, we can see that {un |[ηm ,Rn ) } is a decreasing sequence, u(r ) ≤ un (r ), ∀r ∈ [ηm , Rn ). Since p and q are Lipschitz continuous, if n → +∞, we have u(r ) ≤ limn→+∞ un (r ), thus limr →R− u(r )(R − q(R)

r ) q(R)−p(R) < +∞.



Theorem 4.7. If (P) is positive definite, and β(R) > p(R), then (P) does not have solution. Proof. Without loss of generality, we may assume that f (r , u) ≥ auq−1

and

ρ(r ) ≥ ρ0 (R − r )−β ,

∀r ∈ [σ , R),

(41)

where q, β > p(R) are positive constants. Let (P) have a boundary blow-up solution u. According to the continuity of p(r ) and (41), there exists σ1 ∈ [σ , R) and a positive constant p such that q, β > p ≥ sup p(r ) > 1, r ∈[σ1 ,R)

f (r , u(r )) ≥ auq−1 (r ), 0 ≤ u (σ1 ),

∀r ∈ [σ1 , R),

1 ≤ u(r ),

0

(42)

∀r ∈ [σ1 , R).

(43)

By computation, we have

|u0 (r )|p(r )−2 u0 (r ) = |u0 (σ1 )|p(σ1 )−2 u0 (σ1 ) +

Z

r

σ1

ρ(s)f (s, u(s))ds.

Combining (42), (43) and (44), we have

|u0 |p(r )−2 u0 (r ) ≥

Z

r

σ1

Z

r

≥ σ1

≥ ρ0 a ≥

ρ(s)f (s, u(s))ds ρ0 (R − s)−β auq−1 (s)ds Z

r

σ1

(R − s)−β ds

ρ0 a [(R − r )−(β−1) − (R − σ1 )−(β−1) ], β −1

∀r ∈ [σ1 , R).

(44)

2968

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

Let σ2 ∈ [σ1 , R) and R − σ2 be small enough, such that

(R − σ1 )−(β−1) ≤

1 2

(R − σ2 )−(β−1) ,

ρ0 a 1 (R − σ2 )−(β−1) ≥ 1, β −12 then we have 0

ρ0 a 1 (R − r )−(β−1) β −12





ρ0 a 1 (R − r )−(β−1) β −12

 p−1 1

u ≥

≥

1 p(r )−1



Let σ3 ∈ [σ2 , R) such that R − σ3 ≤ u(r ) − u(σ2 ) ≥

Z r σ2

,

1 2

∀r ∈ [σ2 , R).

(R − σ2 ), then we have

ρ0 a 1 (R − s)−(β−1) β −12

 p−1 1 ds

h β−p β−p i ≥ C4 (R − r )− p−1 − (R − σ2 )− p−1   p   β− β−p 1 p−1  (R − r )− p−1 , ∀r ∈ [σ3 , R). ≥ C4 1 − 2

Then u(r ) ≥ C5 (R − r )−θ1 ,

∀r ∈ [σ3 , R),

(45)

β−p

where θ1 = p−1 > 0, Ci (i = 4, 5) are positive constants. Replace (43) with (45), repeating the above steps, we will get u(r ) ≥ C6 (R − r )−θ2 , where θ2 =

β+θ1 (q−1)−p p−1

∀r ∈ [σ4 , R),

≥ 2θ1 , and then

u(r ) ≥ Cn+4 (R − r )−θn ,

∀r ∈ [σn+2 , R),

where θn and Cn+4 (n = 1, 2, . . .) are positive constants, θn satisfy

θn =

β + θn−1 (q − 1) − p ≥ θ1 + θn−1 ≥ nθ1 . p−1

When n is sufficiently large, it is a contradiction to Theorem 4.4. The proof is completed.  5. Equation is oscillatory In this section, when (P) is oscillatory, we will discuss the existence and nonexistence of boundary blow-up solutions of (P) in the following three cases: Case (I) There exists a strictly increasing sequence {rk } such that, σ < rk → R− and q(rk ) > p(rk ); Case (II) q1 (r ) < p(r ), ∀r ∈ [σ , R) and β(r ) > p(r ) on [σ , R]; Case (III) q1 (r ) < p(r ) and q1 (r ) is increasing on [σ , R), and β1 (r ) < p(r ) on [σ , R]. 5.1. Case (I) Theorem 5.1. Under the conditions of case (I), if β1 (r ) < p(r ), ∀r ∈ [σ , R], then (P) has at least a boundary blow-up solution. Moreover, there exists a boundary blow-up solution Φ (r ) of (P), such that, for every solution u of problem (P), we have u(r ) ≤ Φ (r ), for any r ∈ [ηm , R). Proof. According to (H3 ), we have f (r , t ) ≤ C1 + C2 |t |α(r )−2 t ,

∀t ≥ 0 .

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2969

p(R)−β (R)

Consider v(r , s5 , ), where s5 = α(R)−p1(R) . Obviously, v(·, s5 , ) ∈ C 1 [0, R). Similar to the proof of Lemma 4.1, by computation, we can see that v(r , s5 , ) is a subsolution of (P). Since q(rk ) > p(rk ) and ρ is bounded on [0, rk ], according to Theorem 4.3, we can conclude that (P) has a boundary blow-up solution gk (r ) on [0, rk ), i.e., gk (r ) is a solution of the following

(P#k )

 ρ(r )f (r , u) = 0 in (0, rk ), −∆p(r ) u + m X αi u(ηi ) + e0 , u(r ) → +∞ as x → rk− , u(0) = i=1

and they are also supersolutions and subsolutions of (P#k ). According to Lemma 4.5, we get that gk (r ) ≥ v(r , s5 , ) for any r ∈ [0, rk ). According to Lemma 4.6, we have gk (r ) ≥ gk+1 (r ) for any r ∈ [ηm , rk ). Since gk (r ) ≥ gk+1 (r ) for any r ∈ [ηm , rk ), we have

v(r , s5 , ) ≤ lim gk (r ), k→+∞

for any r ∈ [ηm , R). 1,α

Since v(r , s5 , ) is locally bounded, gk |[ηm ,rk ) are Cloc functions, according to Ascoli–Arzela Theorem, we can see that {gk |[ηm ,rk ) } possesses a subsequence of {gk |[ηm ,rk ) } (which we denote the same) such that

Φ1 (r ) = lim gk |[ηm ,rk ) (r ),

Φ10 (r ) = lim gk0 |[ηm ,rk ) (r ),

i→+∞

i→+∞

∀r ∈ [ηm , R).

Obviously, g ∗ (r ) ≡ max{g1 (ηm ), 1−P0m α } is a supersolution of (P) on [0, ηm ], and i=1 i |e |

g ∗ (ηm ) ≥ g1 (ηm ) ≥ g2 (ηm ) ≥ · · · ≥ gk (ηm ) ≥ · · · . We claim that g ∗ (0) ≥ gk (0), for any k = 1, 2, . . .. If it is false, there are some k0 such that gk0 (0) > g ∗ (0) ≥ 0. We claim that gk0 (r ) ≥ 0 for any r ∈ [0, rk0 ). If it is false, then there exists some ζ ∈ (0, rk0 ) such that gk0 (ζ ) = infr ∈[0,rk ) gk0 (r ) < 0, combining (H2 ) then we have 0

0 p(r )−2 0 g gk0 (r ) = k0 gk0 (r ) = gk0 (ζ ) +

r

Z

ρ(s)f (s, gk0 (s))ds < 0,

ζ r

Z ζ

∀r ∈ (ζ , rk0 ),

 Z t  ϕ −1 t , ρ(s)f (s, gk0 (s))ds dt < gk0 (ζ ) < 0, ζ

∀r ∈ (ζ , rk0 ).

It is a contradiction. Thus gk0 (r ) ≥ 0 for any r ∈ [0, rk0 ). We claim that gk0 (r ) does not have positive local strictly maximum in (0, rk0 ). If it is false, then there exists ξ ∈ (0, rk0 ) and some small enough δ > 0 such that [ξ − δ, ξ + δ] ⊂ (0, rk0 ) and gk0 (ξ ) > gk0 (r ) > 0,

∀r ∈ (ξ − δ, ξ ) ∪ (ξ , ξ + δ).

Since gk0 is a solution of (P#k ), combining (H2 ), we have

0 p(r )−2 0 g gk0 (r ) = k0 gk0 (r ) = gk0 (ξ ) +

r

Z ξ r

Z ξ

ρ(s)f (s, gk0 (s))ds ≥ 0,

∀r ∈ (ξ , ξ + δ),

 Z t  ϕ −1 t , ρ(s)f (s, gk0 (s))ds dt ≥ gk0 (ξ ), ξ

∀r ∈ (ξ , ξ + δ).

It is a contradiction. Thus gk0 (0) does not have positive local strictly maximum in (0, rk0 ). Therefore, there exists an r0 ∈ (0, ηm ] such that gk0 (r ) is decreasing on [0, r0 ] and increasing on [r0 , ηm ]. Thus gk0 (0) > g ∗ (0) ≥ gk0 (ηm ), then gk0 (0) = maxr ∈[0,ηm ] gk0 (r ). The boundary value condition holds gk0 (0) =

m X

αi gk0 (ηi ) + e0 ≤

i =1

m X

αi gk0 (0) + e0 .

i=1

Thus gk0 (0) ≤ 1−P0m α . It is a contradiction. Thus g ∗ (0) ≥ gk (0), for any k = 1, 2, . . .. i=1 i From the comparison principle, we have g ∗ (r ) ≥ gk (r ), ∀r ∈ [0, ηm ], for any k = 1, 2, . . .. Notice that gk (r ) ≥ v(r , s5 , ) for any x ∈ [0, rk ). We can conclude that {gk |[0,ηm ] } are uniformly bounded on [0, ηm ]. According to mean value theorem, for any k = 1, 2, . . ., there exist tk ∈ (0, ηm ) such that e

gk0 (tk ) =

gk (ηm ) − gk (0)

ηm

.

2970

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

Obviously,

v(ηm , s5 , ) − g ∗ (0) gk (ηm ) − gk (0) g ∗ (ηm ) − v(0, s5 , ) ≤ ≤ . ηm ηm ηm Thus {gk0 (tk )} are uniformly bounded. Since gk are solutions of (P#k ), we can see that

0 p(r )−2 0 p(tk )−2 0 g g (r ) = g 0 g (tk ) + k

k

k

r

Z

ρ(s)f (s, gk (s))ds,

k

∀r ∈ [0, ηm ],

tk

gk (r ) = gk (tk ) +

r

Z

  Z t p(t )−2 ρ(s)f (s, gk (s))ds dt , ϕ −1 t , gk0 k gk0 (tk ) + ξ

tk

∀r ∈ [0, ηm ].

Notice that {gk |[0,ηm ] } and {gk0 (tk )} are uniformly bounded. We can conclude that {gk |[0,ηm ] } and {gk0 |[0,ηm ] } are uniformly bounded and equi-continuous. According to Ascoli–Arzela Theorem, there exists a function Φ2 (r ) such that

Φ2 (r ) = lim gk |[0,ηm ] (r ),

Φ20 (r ) = lim gk0 |[0,ηm ] (r ),

i→+∞

Define Φ (r ) =

n

∀r ∈ [0, ηm ].

i→+∞

Φ1 (r ), r ∈ (ηm , R) Φ2 (r ), r ∈ [0, ηm ] .

Then Φ (r ) is a solution of (P). For any boundary blow-up solution u of (P), according to

the comparison principle, we get that gk (r ) ≥ u(r ) for any r ∈ [ηm , rk ). Thus, we have u(r ) ≤ lim gk (r ) = Φ2 (r ),

for any r ∈ [ηm , R).

i→+∞

This completes the proof.



5.2. Case (II) Theorem 5.2. Under the conditions of q1 (r ) < p(r ) on [σ , R), and β(r ) > p(r ) on [σ , R], then (P) possesses a blow-up solution. Proof. Denote Rj0 = R −

1 , j0

where j0 is an integer such that Rj0 > σ . Let us consider

 −∆p(r ) u + ρ(r )f (r , u) = 0, in (0, Rj0 ),   m  X |e0 | u(0) = αi u(ηi ) + e0 , u(Rj0 ) = b > . m P   i =1  1− αi

(46)

i=1

Since ρ ∈ C ([0, R)), we can see that ρ is bounded on [0, Rj0 ]. Similar to the proof of Lemma 4.2, we can see that (46) has

a solution u0 . If u0 (0) > 1−P0m α , from Lemma 2.7, we can see that i=1 i |e |

u00 (Rj0 ) > 0.

(47)

, since u(Rj0 ) = b > 1− , similar to the proof of Lemma 2.7, we can see that (47) is valid. Thus If u0 (0) ≤ 1− i=1 αi i=1 αi u0 (Rj0 ) > 0. − Since q1 (r ) < p(r ) on [σ , R), there exists an increasing sequence {ri }∞ i=0 such that Rj0 = r0 < ri → R and |e0 | P m

|e0 | P m

0

inf

r ∈[ri ,ri+1 ]

p(r ) >

sup

r ∈[ri ,ri+1 ]

q1 (r ),

for any i = 0, 1, 2, . . . .

(48)

At first, we consider the existence of solution of the following problem (III)

 0 p(r )−2 0 (|u1 | u1 (r ))0 = ρ(r )f (r , u1 (r )), r ∈ (r0 , r1 ), u1 (r0 ) = u0 (r0 ), u01 (r0 ) = u00 (r0 ),

where u0 (r0 ) and u00 (r0 ) are positive. Obviously, problem (III) is equivalent to the integral equation u1 (r ) = u0 (r0 ) +

Z r

Z

t

ρ(s)f (s, u1 (s))ds

b+ r0

 p(t1)−1

dt ,

(49)

r0

where b = |u00 (r0 )|p(r0 )−2 u00 (r0 ). Denote G(u) = u0 (r0 ) and z(u) =

Rr r0

[b +

Rt r0

1

ρ(s)f (s, u(s))ds] p(t )−1 dt. It is easy to see that G and z are compact operators

from C 1 ([r0 , r1 ]) to C 1 ([r0 , r1 ]) (see [18, Lemma 2.3]).

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2971

If u1 is a solution of (III), from (48) and (49), it is easy to see that

ku1 k1,[r0 ,r1 ] ≤ C5 + C6 ku1 kθ1,[r0 ,r1 ] ,

(50)

where θ , C5 and C6 are positive constants, and θ ∈ (0, 1). Lets consider the problem u = Φ (u, λ) = G(u) + λz(u).

(51)

It is easy to see that u is a solution of (III) if and only if u is a solution of the abstract equation (51) when λ = 1. According to (50), it is easy to see that all the solutions of (51) are uniformly bounded for λ ∈ [0, 1], therefore, there exists a positive constant Λ such that all the solutions of (51) belong to B∗ (Λ) = {u ∈ C 1 ([r0 , r1 ]) | kuk1,[r0 ,r1 ] < Λ}. Thus, the Leray–Schauder degree dLS [I − Φ (·, 1), B∗ (Λ), 0] = dLS [I − Φ (·, 0), B∗ (Λ), 0]. Since dLS [I − Φ (·, 0), B∗ (Λ), 0] = dB [I − G, B∗ (R0 ) ∩ R, 0] 6= 0, we can conclude that (51) has a solution. Thus, the solution of (46) can be extended onto [0, r1 ]. Similarly, the solution of (46) can be extended onto [0, rn ], then it can be extended onto [0, R), and we denote it by u. Similarly to the proof of Lemma 2.7, we can see that u(r ) ≥ 0 and

u0 (r ) ≥ 0,

∀r ∈ [r1 , R).

It only remains to prove that u(r ) → +∞ as r → R− . Since β(r ) > p(r ) on [σ , R] and ri → R− , let i is large enough such that

β∗− := inf β(r ) > p+ ∗ := sup p(r ), r ∈[ri ,R)

r ∈[ri ,R)

and f (r , u(r )) ≥ C7 ,

∀r ∈ [ri , R), where C7 is a positive constant. p(ri )−2 0

Denote bi = |u (ri )| 0

u(r ) = u(ri ) +

u (ri ). By computation, we have

Z r

dt

ρ(s)C7 ds

 p(t1)−1 dt

ri

Z r ri

≥ u(ri ) +

t

Z r Z

Z

 p(t1)−1

ri

ri

≥ u(ri ) +

ρ(s)f (s, u(s))ds

bi + ri

≥ u(ri ) +

t

Z

r

1

−β∗− +1

C7 ρ0 − [(R − t ) β∗ − 1



C7 ρ 0

R+ri 2

The proof is completed.

1 2(β∗− − 1)

(R − t )

−β∗− +1

− ( R − ri )

−β∗− +1



1 + p∗ −1

 +1 p∗ −1 dt ]

dt → +∞ (as r → R− ).



5.3. Case (III) Theorem 5.3. Under the conditions: (i) q1 (r ) < p(r ) and q1 (r ) is increasing on [σ , R), (ii) β1 (r ) < p(r ) on [σ , R], then (P) does not have a boundary blow-up solution. Proof. Let (P) have a boundary blow-up solution u. It is easy to see that there exists a constant C8 > 1 such that f (r , u(r )) ≤ C8 (M + u(r ))q1 (r )−1 ,

∀r ∈ [0, R),

where M is a positive constant such that M + u(r ) ≥ 1, ∀r ∈ [0, R). Since β1 (r ) < p(r ) for any r ∈ [σ , R], we have R

Z 0



0 p(0)−1 u (0) +

t

Z

 p(t1)−1

C8 ρ(s)ds 0

dt < +∞,

2972

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

then there exists an r∗ ∈ [σ , R) such that R

Z



0 p(0)−1 u (0) +

t

Z

r∗

C8 ρ(s)ds

 p(t1)−1 dt ≤

0

1 2

.

Without loss of generality, we may assume infr ∈[σ ,R) q1 (r ) > 1. Since u(r ) → +∞ as r → R− , there exists an r1 ∈ (r∗ , R), such that u is increasing on [r1 , R) and

(M + u(t ))q1 (t )−1 ≥ max (M + u(s))q1 (s)−1 , s∈[0,r1 ]

∀t ∈ [r1 , R).

(52)

Since q1 (r ) is increasing on [σ , R), we have

Z

t

ρ(s)C8 (M + u(s))q1 (s)−1 ds ≤ (M + u(t ))q1 (t )−1

Z

t

ρ(s)C8 ds,

∀t ∈ [r1 , R).

(53)

r1

r1

Since u is a solution of (P), combining (52) and (53), we have

 p(t1)−1 Z r Z t 0 p(0)−2 0 u (0) u (0) + ρ(s)f (s, u)ds u(r ) = u(0) + dt , 0

0

 p(t1)−1 Z r Z t 0 p(0)−1 q1 (s)−1 ≤ u(0) + u (0) + ρ(s)C8 (M + u(s)) ds dt 0

≤ u(0) +

0 r1

Z



0 p(0)−1 u (0) +

0

Z r +

t

Z

ρ(s)C8 (M + u(s))q1 (s)−1 ds

0 p(0)−1 u (0) +

r1

Z

ρ(s)C8 (M + u(s))q1 (s)−1 ds +

0 r1

Z



+

t

Z

ρ(s)C8 (M + u(s))q1 (s)−1 ds

 p(t1)−1 dt

r1

0 p(0)−1 u (0) +

0

Z r

dt

0

r1

≤ u(0) +

 p(t1)−1

t

Z

ρ(s)C8 (M + u(s))q1 (s)−1 ds

 p(t1)−1 dt

0

0 p(0)−1 u (0) + (M + u(t ))q1 (t )−1

r1

t

Z

ρ(s)C8 ds

 p(t1)−1

dt .

0

Denote C9 = u(0) +

r1

Z

  p(t1)−1 Z t 0 p(0)−1 u (0) + ρ(s)C8 (M + u(s))q1 (s)−1 ds dt ,

0

0

then we have u(r ) ≤ C9 +

1 2

(M + u(r )),

∀r ∈ [r1 , R).

It means that u(r ) ≤ 2C9 + M. It is a contradiction. The proof is completed.  Acknowledgements The author thanks the referees for his/her careful reading of the manuscript and useful suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]

E. Acerbi, G. Mingione, Regularity results for stationary electro-rheological fluids, Arch. Ration. Mech. Anal. 164 (2002) 213–259. Y. Chen, S. Levine, M. Rao, Variable exponent, linear growth functionals in image restoration, SIAM J. Appl. Math. 66 (4) (2006) 1383–1406. M. Růžička, Electrorheological Fluids: Modeling and Mathematical Theory, in: Lecture Notes in Math, vol. 1784, Springer-Verlag, Berlin, 2000. V.V. Zhikov, Averaging of functionals of the calculus of variations and elasticity theory, Math. USSR. Izv. 29 (1987) 33–36. X.L. Fan, D. Zhao, On the spaces Lp(x) (Ω ) and W m,p(x) (Ω ), J. Math. Anal. Appl. 263 (2001) 424–446. X.L. Fan, Q.H. Zhang, Existence of solutions for p(x)-Laplacian Dirichlet problem, Nonlinear Anal. 52 (2003) 1843–1852. X.L. Fan, Q.H. Zhang, D. Zhao, Eigenvalues of p(x)-Laplacian Dirichlet problem, J. Math. Anal. Appl. 302 (2005) 306–317. X.L. Fan, Global C 1,α regularity for variable exponent elliptic equations in divergence form, J. Differential Equations 235 (2007) 397–417. A. El Hamidi, Existence results to elliptic systems with nonstandard growth conditions, J. Math. Anal. Appl. 300 (2004) 30–42. P. Harjulehto, P. Hästö, M. Koskenoja, S. Varonen, The Dirichlet energy integral and variable exponent Sobolev spaces with zero boundary values, Potential Anal. 25 (2006) 205–222.

Q. Zhang et al. / Nonlinear Analysis 72 (2010) 2950–2973

2973

[11] P. Harjulehto, P. Hästö, V. Latvala, Harnack’s inequality for p(·)-harmonic functions with unbounded exponent p, J. Math. Anal. Appl. 352 (1) (2009) 345–359. [12] O. Kováčik, J. Rákosník, On spaces Lp(x) (Ω ) and W k,p(x) (Ω ), Czechoslovak Math. J. 41 (1991) 592–618. [13] T. Lukkari, Singular solutions of elliptic equations with nonstandard growth, Math. Nachr. 282 (12) (2009) 1770–1787. [14] P. Marcellini, Regularity and existence of solutions of elliptic equations with (p.q)-growth conditions, J. Differential Equations 90 (1991) 1–30. [15] M. Mihˇailescu, V. Rˇadulescu, Continuous spectrum for a class of nonhomogeneous differential operators, Manuscripta Math. 125 (2008) 157–167. [16] S.G. Samko, Densness of C0∞ (RN ) in the generalized Sobolev spaces W m,p(x) (RN ), Dokl. Ross. Akad. Nauk. 369 (4) (1999) 451–454. [17] Q.H. Zhang, A strong maximum principle for differential equations with nonstandard p(x)-growth conditions, J. Math. Anal. Appl. 312 (1) (2005) 24–32. [18] Q.H. Zhang, Existence of solutions for weighted p(r )-Laplacian system boundary value problems, J. Math. Anal. Appl. 327 (1) (2007) 127–141. [19] Q.H. Zhang, Z.M. Qiu, X.P. Liu, Existence of solutions for a class of weighted p(t )-Laplacian system multi-point boundary value problems, J. Inequal. Appl. 2008 (2008) 18 pages. Article ID 791762. [20] Q.H. Zhang, X.P. Liu, Z.M. Qiu, Existence of solutions for weighted p(t )-Laplacian system multi-point boundary value problems, Nonlinear Anal. 71 (2009) 3715–3727. [21] Q.H. Zhang, Existence of asymptotic behavior of blow-up solutions to a class of p(x)-Laplacian equations, J. Math. Anal. Appl. 329 (2007) 472–482. [22] J. Ehrke, J. Henderson, C. Kunkel, Q. Sheng, Boundary data smoothness for solutions of nonlocal boundary value problems for second order differential equations, J. Math. Anal. Appl. 333 (1) (2007) 191–203. [23] H.Y. Feng, H.R. Lian, W.G. Ge, A symmetric solution of a multipoint boundary value problem with one-dimensional p-Laplacian at resonance, Nonlinear Anal. 69 (11) (2008) 3964–3972. [24] P. Kang, Z.L. Wei, Multiple positive solutions of multi-point boundary value problems on the half-line, Appl. Math. Comput. 196 (2008) 402–415. [25] B. Liu, Z.L. Zhao, A note on multi-point boundary value problems, Nonlinear Anal. 67 (9) (2007) 2680–2689. [26] R.Y. Ma, Existence results of a m-point boundary value problem at resonance, J. Math. Anal. Appl. 294 (2004) 147–157. [27] J.X. Sun, X.A. Xu, D. O’Regan, Nodal solutions for m-point boundary value problems using bifurcation methods, Nonlinear Anal. 68 (10) (2008) 3034–3046. [28] F.Y. Xu, Positive solutions for multipoint boundary value problems with one-dimensional p-Laplacian operator, Appl. Math. Comput. 194 (2007) 366–380. [29] C.H. Chen, On positive weak solutions for a class of quasilinear elliptic systems, Nonlinear Anal. 62 (2005) 751–756. [30] E.H. Papageorgiou, N.S. Papageorgiou, A multiplicity theorem for problems with the p-Laplacian, J. Funct. Anal. 244 (2007) 63–77. [31] F. Cîrstea, Y. Du, General uniqueness results and variation speed for blow-up solutions of elliptic equations, Proc. London Math. Soc. 91 (2005) 459–482. [32] J. García-Melián, R. Letelier-Albornoz, J.C. Sabina de Lis, Uniqueness and asymptotic behaviour for solutions of semilinear problems with boundary blow-up, Proc. Amer. Math. Soc. 129 (2001) 3593–3602. [33] Z.M. Guo, F. Zhou, Exact multiplicity for boundary blow-up solutions, J. Differential Equations 228 (2006) 486–506. [34] S. Kichenassamy, Boundary blow-up and degenerate equations, J. Funct. Anal. 215 (2004) 271–289. [35] A.V. Lair, A necessary and sufficient condition for existence of large solutions to semilinear elliptic equations, J. Math. Anal. Appl. 240 (1999) 205–218. [36] J. López-Gómez, The boundary blow-up rate of large solutions, J. Differential Equations 195 (2003) 25–45. [37] H.L. Li, M.X. Wang, Existence and uniqueness of positive solutions to the boundary blow-up problem for an elliptic system, J. Differential Equations 234 (2007) 246–266. [38] A. Mohammed, G. Porcu, G. Porru, Large solutions to some non-linear O.D.E. with singular coefficients, Nonlinear Anal. 47 (2001) 513–524. [39] A. Porretta, L. Véron, Symmetry of large solutions of nonlinear elliptic equations in a ball, J. Funct. Anal. 236 (2006) 581–591. [40] Z.J. Zhang, Boundary blow-up elliptic problems of Bieberbach and Rademacher type with nonlinear gradient terms, Nonlinear Anal. 67 (3) (2007) 727–734.