Nonuniform nonresonant singular Dirichlet boundary value problems for the one-dimensional p -Laplacian with sign changing nonlinearity

Nonlinear Analysis 68 (2008) 1155–1168 www.elsevier.com/locate/na

Nonuniform nonresonant singular Dirichlet boundary value problems for the one-dimensional p-Laplacian with sign changing nonlinearity Daqing Jiang ∗ , Huina Zhang School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, PR China Received 5 March 2006; accepted 7 December 2006

Abstract Existence results are presented for the nonuniform nonresonant singular boundary value problem  [φ(u 0 )]0 + f (t, u, u 0 ) = 0, for t ∈ (0, 1), u(0) = u(1) = 0, where φ(s) = |s| p−2 s, p > 1. The singularity may appear at u = 0, t = 0 or t = 1, and the function f may change sign. The existence of solutions is obtained via an upper and lower solutions method. c 2007 Elsevier Ltd. All rights reserved.

MSC: primary 34B16; secondary 34B15 Keywords: Singular boundary value problem; Nonuniform nonresonance; Upper and lower solutions approach

1. Introduction In this paper, we discuss the singular Dirichlet boundary value problem  [φ(u 0 )]0 + f (t, u, u 0 ) = 0, for t ∈ (0, 1), u(0) = u(1) = 0,

(1.1)

where φ(s) = |s| p−2 s, p > 1. The singularity may appear at u = 0, t = 0 or t = 1, and the function f may change sign. Recently, problem (1.1) has been studied extensively. Readers may refer to [8,9,11,12,14,18–22] for the details. In [14,18,20,22], the problem is not able to possess singularity. In [14,18,20], the authors present an upper and lower solutions approach. In [22], the author studied uniform and nonuniform nonresonance at the first eigenvalue of the p-Laplacian by a degree method. Some basic results on the singular boundary value problems were obtained in many

∗ Corresponding author.

E-mail address: [email protected] (D. Jiang). c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.12.012

1156

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

papers; for example, see [1,2,8,19]. In all these papers the arguments rely on the assumption that f is positive. This implies that the solutions are concave. Very recently, many authors have paid much attention to the singular problem when f is allowed to change sign, and f (t, u, v) may be singular at u = 0, t = 0 and t = 1. We refer the reader to, for example, [3–7,10,17] ( p = 2) and [9,11,12,21]. In all these papers, the singular problem is well explained by applying the modified upper and lower solutions method. As regards the singular problem, most of the literature is on the cases in which f is allowed to be superlinear at ∞. In [12], the authors studied singular problem (1.1) and uniform nonresonance at the first eigenvalue. However, f (t, u, v) is only suitable for linear growth of (u, v); for example, f (t, u, v) = t −m (1 − t)−n u −a + a0 u p−1 + b0 |v| p−1 + sin(8t) −1/ p

with 0 ≤ m, n < 1, a > 0, a0 ≥ 0, b0 ≥ 0, and a0 , b0 satisfying a0 λ−1 < 1. Here λ1 is the first eigenvalue 1 + b0 λ1 of the problem [φ(u 0 )]0 + λφ(u) = 0, u(0) = u(1) = 0. When p = 2 and f is singular at t = 0 or t = 1, for the nonresonant Dirichlet and mixed boundary data problem, we refer the reader to the works of O’Regan [13,15,16] using degree methods. The present work is a direct extension of some results in [12,22] for the singular problem, i.e., f is singular at u = 0 and there is nonuniform nonresonance at the first eigenvalue. Our technique relies essentially on a method of upper and lower solutions in [18,20] which we believe to be well adapted to this type of problem. The nonlinear eigenvalue to the problem (1.1) is  [φ(u 0 )]0 + λφ(u) = 0, for t ∈ (0, 1), (1.2) u(0) = u(1) = 0. It is well known (see [22]) that (1.2) has eigenvalues 0 < λ1 < λ2 < · · · < λn < · · · ,

λn → ∞ as n → ∞.

In what follows, let the L p -norm in L p [0, 1] be defined by !1/ p Z 1

kuk p =

,

|u(t)| p dt

0

and the L ∞ -norm in L ∞ [0, 1] be kuk∞ = ess sup{|u(t)| : t ∈ [0, 1]}. Let u ∈ C 1 ([0, 1], R) with u(0) = u(1) = 0; we have the following two lemmas (see [22]). Lemma 1.1. We have −1/ p

kuk p ≤ λ1

ku 0 k p ,

kuk∞ ≤ (1/2)1−1/ p ku 0 k p .

Lemma 1.2. Suppose that c ∈ L ∞ [0, 1] ∩ C((0, 1), R + )(R + = [0, ∞)) satisfies that c(t) < kck∞ holds on a subset of (0, 1) of positive measure. Then there exists an η > 0 such that Z 1 0 p c(t)|u(t)| p dt ≤ (kck∞ λ−1 1 − η)ku k p . 0

We state the following existence principle (see [12]). Lemma 1.3. Let f (t, x, z) : (0, 1) × R 2 → R be continuous, and | f (t, x, z)| ≤ q(t)F(t, x, z),

(t, x, z) ∈ (0, 1) × R 2 ,

1157

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

where F(t, x, z) : [0, 1] × R 2 → R + is continuous, and q ∈ L 1 [0, 1] ∩ C((0, 1), R + ). Then the following boundary value problem:  [φ(u 0 )]0 + f (t, u, u 0 ) = 0, for t ∈ (0, 1), u(0) = A, u(1) = B has at least one solution in C 1 ([0, 1], R) if there is a real number M, independent of λ, such that |u 0 (t)| ≤ M on [0, 1] for each λ ∈ (0, 1) and for any solution u(t) to the boundary value problems  [φ(u 0 )]0 + λ f (t, u, u 0 ) = 0, for t ∈ (0, 1), u(0) = A, u(1) = B. 2. Main results In this section we use the former lemmas to establish new existence results for singular boundary value problem (1.1). Motivated by the example f (t, u, v) = t −m (1 − t)−n u −a + a0 u p−1 + b0 |v| p−1 − c0 u c + sin(8t) −1/ p

with 0 ≤ m, n < 1, a > 0, 0 ≤ c < p, a0 ≥ 0, b0 ≥ 0, c0 ≥ 0, and a0 , b0 satisfying a0 λ−1 1 + b0 λ1 the following result.

< 1, we have

Theorem 2.1. Let n 0 ∈ {1, 2 . . .} be fixed and suppose the following conditions are satisfied: f : (0, 1) × (0, ∞) × R → R is continuous;  let n ∈ {n 0 , n 0 + 1 . . .}, and associated with each n we have a constant ρn such that   1 {ρn } is a nonincreasing sequence with lim ρn = 0 and such that for n+1 ≤ t ≤ 1,  n→∞ 2  we have f (t, ρn , 0) ≥ 0;  if h : (0, 1) × (0, ∞) × R → R is any continuous function with h(t, u, z) ≥ f (t, u, z) for all    (t, u, z) ∈ (0, 1) × (0, ∞) × R and if u ∈ C 1 [0, 1], u(t) > 0 for t ∈ [0, 1], is any solution of [φ(u 0 )]0 + h(t, u, u 0 ) = 0, 0 < t < 1 then there exists a α ∈ C 1 [0, 1], α(0) = α(1) = 0, α(t) > 0   for t ∈ (0, 1), with u(t) ≥ α(t) for t ∈ (0, 1);  for any ε > 0, ∃a, b, with a ≥ 0, b ≥ 0 a.e. on [0, 1], and ∃α, β with 1 ≤ α < p, 1 ≤ β < p, α + β = p such that a, b ∈ C[0, 1] and if b 6= 0 with u f (t, u, z) ≤ a(t)u p + b(t)u α |z|β for  t ∈ (0, 1), u ≥ ε and z ∈ R; kbk∞ kak∞ + α/ p ≤ 1; λ1 λ1   1 < kak∞ either (a) a 2n 0 +1

(2.1) (2.2)

(2.3)

(2.4)

(2.5)  or

(b) b



1 2n 0 +1

< kbk∞ ;

(2.6)

and  on [0, 1], and ∃δ, γ with for any ε > 0, ∃a0 , b0 , ηε with a0 ≥ 0, b0 ≥ 0, ηε ≥ 0 a.e. p 1 p−γ 0 ≤ δ < p, 0 ≤ γ < p such that a0 ∈ L [0, 1], b0 ∈ L [0, 1] and ηε ∈ L 1 [0, 1] with | f (t, u, z)|  ≤ a0 (t)u δ + b0 (t)|z|γ + ηε (t) for t ∈ (0, 1), u ≥ ε and z ∈ R. Then (1.1) has a solution u ∈ C[0, 1] with u(t) ≥ α(t) for t ∈ [0, 1] (here α is given in (2.3)). Proof. For n = n 0 , n 0 + 1, . . . , let   1 en = n+1 , 1 , θn (t) = max{1/2n+1 , t}, 2

0≤t ≤1

(2.7)

1158

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

and f n (t, x, z) = max{ f (θn (t), x, z), f (t, x, z)}. Next we define inductively gn 0 (t, x, z) = f n 0 (t, x, z) and gn (t, x, z) = min{ f n 0 (t, x, z), . . . , f n (t, x, z)},

n = n 0 + 1, n 0 + 2 . . . .

Notice that f (t, x, z) ≤ · · · ≤ gn+1 (t, x, z) ≤ gn (t, x, z) ≤ · · · ≤ gn 0 (t, x, z) for (t, x, z) ∈ (0, 1) × (0, ∞) × R and gn (t, x, z) = f (t, x, z) for (t, x, z) ∈ en × (0, ∞) × R. We begin with the boundary value problems  [φ(u 0 )]0 + gn (t, u, u 0 ) = 0, for t ∈ (0, 1), u(0) = u(1) = ρn .

(2.8n )

First, we consider the boundary value problem  [φ(u 0 )]0 + gn∗0 (t, u, u 0 ) = 0, for t ∈ (0, 1), u(0) = u(1) = ρn 0 ,

(2.9)

where gn∗0 (t, u, z)

 =

gn 0 (t, ρn 0 , z) + r (ρn 0 − u), gn 0 (t, u, z),

u < ρn 0 ; u ≥ ρn 0

with r : R → [−1, 1] the radial retraction defined by ( u, |u| ≤ 1; r (u) = u , |u| > 1. |u| To show that (1.1) has a solution, we consider the family of problems  [φ(u 0 )]0 + λgn∗0 (t, u, u 0 ) = 0, for t ∈ (0, 1) u(0) = u(1) = ρn 0

(2.10λ )

where λ ∈ (0, 1). Let u be any solution of (2.10λ ) for some λ ∈ (0, 1). We first show that u(t) ≥ ρn 0 ,

t ∈ [0, 1].

(2.11)

Suppose (2.11) is not true; then there exists a t0 ∈ (0, 1) with u(t0 ) < However, note that

ρn 0 , u 0 (t0 )

= 0, and

[φ(u 0 )]0 (t0 ) = −λgn∗0 (t0 , u(t0 ), u 0 (t0 )) = −λ[gn 0 (t0 , ρn 0 , 0) + r (ρn 0 − u(t0 ))]. We need to discuss two cases, namely t0 ∈ [

1 , 1) 2n 0 +1

and t0 ∈ (0,

1 ). 2n 0 +1

Case (A): t0 ∈ [ n01+1 , 1). 2 Since gn 0 (t0 , u, z) = f (t0 , u, z) for (u, z) ∈ (0, ∞) × R (note that t0 ∈ en 0 ), we have [φ(u 0 )]0 (t0 ) = −λ[ f (t0 , ρn 0 , 0) + r (ρn 0 − u(t0 ))] < 0, which is a contradiction.

[φ(u 0 )]0 (t0 )

≥ 0.

1159

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

Case (B): t0 ∈ (0,

1 ). 2n 0 +1

Since gn 0 (t0 , u, z) = max{ f (

1 , u, z), 2n 0 +1

f (t0 , u, z)}, we have gn 0 (t0 , u, z) ≥ f (t0 , u, z) and gn 0 (t0 , u, z) ≥

f ( n0 +1 , u, z) for u ∈ (0, ∞) and z ∈ R. 2 Thus 1

[φ(u 0 )]0 (t0 ) = −λ[gn 0 (t0 , ρn 0 , 0) + r (ρn 0 − u(t0 ))]     1 ≤ −λ f , ρn 0 , 0 + r (ρn 0 − u(t0 )) < 0, 2n 0 +1 which is a contradiction again. Consequently (2.11) is true. Next we show that for t ∈ [0, 1].

u(t) ≤ Mn 0

(2.12)

Here Mn 0 (≥ρn 0 ) is a predetermined constant (see (2.25)). Notice that (2.7) (with ε = ρn 0 ) guarantees the existence of a0 , b0 , ηε , δ and γ (as described in (2.7)). p

By the definition of gn 0 (t, u, z), there exists φ1 (t) ∈ L 1 [0, 1], φ2 (t) ∈ L p−γ [0, 1], φ3 (t) ∈ L 1 [0, 1] such that |gn∗0 (t, u, u 0 )| = |gn 0 (t, u, u 0 )| = | f n 0 (t, u, u 0 )| ≤ φ1 (t)u δ + φ2 (t)|u 0 |γ + φ3 (t)

(2.13)

for t ∈ (0, 1), u ≥ ρn 0 and u 0 ∈ R, where φ1 (t) = max{a0 (t), a0 (θn 0 (t))} ≥ 0, φ2 (t) = max{b0 (t), b0 (θn 0 (t))} ≥ 0, φ3 (t) = max{ηε (t), ηε (θn 0 (t))} ≥ 0. Next notice that (2.4) (with ε = ρn 0 ) guarantees the existence of a, b, α, β (as described in (2.4)), with ugn∗0 (t, u, u 0 ) = u f n 0 (t, u, u 0 ) ≤ φ4 (t)u p + φ5 (t)u α |u 0 |β ,

(2.14)

where φ4 (t) = max{a(t), a(θn 0 (t))} ≥ 0, φ5 (t) = max{b(t), b(θn 0 (t))} ≥ 0,

φ4 (t) ∈ C[0, 1], φ5 (t) ∈ C[0, 1].

Let v(t) := u(t) − ρn 0 ; then v(t) ≥ 0, v(0) = v(1) = 0. Multiplying the Eq. (2.10λ ) by v(t) and then integrating from 0 to 1, we obtain Z 1 Z 1 [φ(u 0 )]0 v(t)dt = λ gn∗0 (t, u, u 0 )v(t)dt, − 0

0

namely p

kv 0 k p ≤

Z

1

Z0

1

gn 0 (t, u(t), v 0 (t))v(t)dt Z

1

ρn 0 |gn 0 (t, u(t), v 0 (t))|dt gn 0 (t, u(t), v (t))u(t)dt + 0 0 Z 1 Z 1 Z 1 φ4 (t)u p (t)dt + φ5 (t)u α (t)|v 0 (t)|β dt + ρn 0 φ1 (t)u δ (t)dt ≤ 0 0 0 Z 1 Z 1 0 γ + ρn 0 φ2 (t)|v (t)| dt + ρn 0 φ3 (t)dt, 0

≤

0

ρn 0

1

Z 0

(2.15)

0

φ1 (t)u δ (t)dt ≤ ρn 0 kukδ∞

Z

1

φ1 (t)dt

0 δ−δ/ p

≤ ρn 0 (1/2)

kv 0 kδp kφ1 k1 ,

(2.16)

1160

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

ρn 0

1

Z 0

0 γ

φ2 (t)|v | (t)dt ≤ ρn 0

"Z

1

0

= ρn 0 kφ2 k ρn 0

1

Z 0

|φ2 (t)|

p p−γ

# p−γ "Z p dt

#γ

p

γ · γp

|v (t)| 0

dt

0

γ kv 0 k p ,

p p−γ

1

(2.17)

φ3 (t)dt ≤ ρn 0 kφ3 k1 .

(2.18)

Case (A): Suppose a(t) < kak∞ on a subset of [0, 1] of positive measure and a( n01+1 ) < kak∞ . 2 This implies φ4 (t) < kφ4 k∞ = kak∞ on a subset of [0, 1] of positive measure. From Lemma 1.2, there exists an η1 > 0 with   Z 1 kak∞ p φ4 (t)u p (t)dt ≤ − η1 kv 0 k p , λ1 0 "Z # α "Z #β Z 1

1

φ5 (t)u α (t)|v 0 (t)|β dt ≤ kφ5 k∞

0

0

p

|(v(t) + ρn 0 )α | α dt

p

1

p

(2.19)

p

·β

|v 0 (t)| β dt

0

= kbk∞ kv + ρn 0 kαp kv 0 kβp ≤ kbk∞ (kvk p + ρn 0 )α kv 0 kβp α  kv 0 k p ≤ kbk∞  1 + ρn 0  kv 0 kβp . λ1p

(2.20)

From (2.15)–(2.20) and (2.5) we obtain p kv 0 k p

 ≤

 α  0k kv kak∞ p p − η1 kv 0 k p + kbk∞  1 + ρn 0  kv 0 kβp λ1 λ1p

+ ρn 0 (1/2)δ−δ/ p kv 0 kδp kφ1 k1 + ρn 0 kφ2 k

p p−γ

γ

kv 0 k p + ρn 0 kφ3 k1 .

(2.21)

Thus there exists K n 0 (independent of λ) with K n 0 ≥ ρn 0 ,

ku 0 k p = kv 0 k p ≤ K n 0

(2.22)

holding. If (2.22) is not true, then kv 0 k p = ∞; from Eq. (2.21) we can obtain 1≤

kak∞ kbk∞ − η1 + α/ p , λ1 λ1

a contradiction. Case (B): Suppose b(t) < kbk∞ on a subset of [0, 1] of positive measure and b( n01+1 ) < kbk∞ . This implies that 2 φ5 (t) < kφ5 k∞ = kbk∞ on a subset of [0, 1] of positive measure, From Lemma 1.2, there exists an η2 > 0 such that   p Z 1 1 α kbk · p ∞ p φ5α (t)|u(t)| p dt ≤  − η2  kv 0 k p , λ1 0 and there exists an η3 > 0 such that p

− αp

α

α −1 λ1 − η2 ) p ≤ kbk∞ λ1 (kbk∞

− η3 ,

and then Z 0

1

α

β

φ5 (t)u (t)|v (t)| dt ≤ 0

"Z 0

1

1 α ·α

|φ5

# α "Z p u (t)| · α

1

p α

0

β· βp

|v (t)| 0

#β

p

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168 p

α

1161

p

α −1 ≤ (kbk∞ λ1 − η2 ) p kv 0 k p

− αp

≤ (kbk∞ λ1 Also Lemma 1.1 gives Z 1 Z φ4 (t)u p (t)dt ≤ kφ4 k∞ 0

1

p

− η3 )kv 0 k p .

(2.23)

|u(t)| p dt

0 p

≤ kak∞ kuk p ≤

kak∞ 0 p kv k p . λ1

(2.24)

Form (2.15)–(2.18), (2.23) and (2.24) we obtain p

δ

η3 kv 0 k p ≤ 2 p

−δ

ρn 0 ka0 k1 kv 0 kδp + ρn 0 kb0 k

p p−γ

γ

kv 0 k p + ρn 0 kηε k1 .

Thus there exists a K n 0 , K n 0 ≥ ρn 0 such that (2.22) holds. In both cases (2.22) holds, and now since kvk∞ ≤ (1/2)1−1/ p kv 0 k p , we have kvk∞ ≤ (1/2)1−1/ p K n 0 , and as a result we have kuk∞ ≤ (1/2)1−1/ p K n 0 + ρn 0 ≡ Mn 0

and

ku 0 k p ≤ K n 0 .

(2.25)

For any solution u to (2.10λ ), (2.7) (with ε = ρn 0 ) and (2.25) imply Z 1 Z 1 0 0 |[φ(u (t))] |dt = λ |gn∗0 (t, u(t), u 0 (t))|dt 0

0 1

Z

[φ1 (t)u δ (t) + φ2 (t)|u 0 (t)|γ + φ3 (t)]dt

≤ 0

γ

≤ kφ1 k1 Mnδ0 + K n 0 kφ2 k

p p−γ

+ kφ3 k1 ≡ Q n 0 ,

R1 and since u(0) = u(1) = ρn 0 we have |φ(u 0 (t))| ≤ 0 |[φ(u 0 (t))]0 |dt ≡ Q n 0 , so there exists an Rn 0 such that ku 0 k∞ ≤ Rn 0 . From Lemma 1.3, (2.9) has a positive solution β ∈ C 1 ([0, 1], R), β(t) ≥ ρn 0 . We now see that β(t) and ρn 0 are, respectively, an upper and a lower solution for the problem (2.8n0 ); by the method of upper and a lower solutions, there is a solution u n 0 (t) ∈ C 1 [0, 1] of (2.8n0 ) with ρn 0 ≤ u n 0 (t) ≤ Mn 0 ,

t ∈ [0, 1],

ku 0n 0 k∞ ≤ Rn 0 .

Also notice that if we take h(t, u, z) = gn 0 (t, u, z) in (2.3), then since gn 0 ≥ f and u n 0 satisfies [φ(u 0n 0 )]0 + gn 0 (t, u n 0 , u 0n 0 ) = 0, t ∈ (0, 1), with ρn 0 ≤ u n 0 (t) for t ∈ (0, 1), then u n 0 (t) ≥ α(t)

for t ∈ [0, 1].

Next we consider the boundary value problem  [φ(u 0 )]0 + gn∗0 +1 (t, u, u 0 ) = 0, for t ∈ (0, 1) u(0) = u(1) = ρn 0 +1 where  gn 0 +1 (t, ρn 0 +1 , z ∗ ) + r (ρn 0 +1 − u), u ≤ ρn 0 +1 ; ∗ ρn 0 +1 ≤ u ≤ u n 0 (t); gn 0 +1 (t, u, z) = gn 0 +1 (t, u, z ∗ ),  gn 0 +1 (t, u n 0 (t), z ∗ ) + r (u n 0 (t) − u), u ≥ u n 0 (t) with   Rn 0 +1 , z ∗ = z,  −Rn 0 +1 ,

z > Rn 0 +1 ; −Rn 0 +1 ≤ z ≤ Rn 0 +1 ; z ≤ −Rn 0 +1

where Rn 0 +1 ≥ Rn 0 is a predetermined constant (see (2.30)).

(2.26)

(2.27)

1162

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

Now Schauder’s fixed point theorem guarantees that there exists a solution u n 0 +1 ∈ C 1 [0, 1] to (2.27). We first show that u n 0 +1 (t) ≥ ρn 0 +1 ,

t ∈ [0, 1].

(2.28)

Suppose (2.28) is not true; then there exists a t1 ∈ (0, 1) with u n 0 +1 (t1 ) < ρn 0 +1 ,

u 0n 0 +1 (t1 ) = 0

We need to discuss two cases, namely t1 ∈ [

and

[φ(u 0n 0 +1 )]0 (t1 ) ≥ 0.

1 , 1) 2n 0 +2

and t1 ∈ (0,

1 ). 2n 0 +2

Case (A): t1 ∈ [ n01+2 , 1). 2 Then since gn 0 +1 (t1 , u, z) = f (t1 , u, z) for (u, z) ∈ (0, ∞) × R (note that t1 ∈ en 0 +1 ) we have [φ(u 0n 0 +1 )]0 (t1 ) = − f (t1 , ρn 0 +1 , 0) − r (ρn 0 +1 − u n 0 +1 (t1 )) < 0 which is a contradiction. Case (B): t1 ∈ (0, n01+2 ). 2 Then since    gn 0 +1 (t1 , u, z) = min max f

    , u, z , f (t , u, z) , max f 1 +1

  u, z , f (t , u, z) , 1 +2

1

1

2n 0

2n 0

we have gn 0 +1 (t1 , u, z) ≥ f (t1 , u, z) and gn 0 +1 (t1 , u, z) ≥ min

  f

  , u, z , f +1

1

1

2n 0

2n 0

, u, z +2



for u ∈ (0, ∞) and z ∈ R. Thus we have [φ(u 0n 0 +1 )]0 (t1 ) = −gn∗0 +1 (t1 , u n 0 +1 (t1 ), u 0n 0 +1 (t1 )) = −gn 0 +1 (t1 , ρn 0 +1 , 0) − r (ρn 0 +1 − u n 0 +1 (t1 ))      1 1 ≤ − min f , ρ , 0 , f , ρ , 0 − r (ρn 0 +1 − u n 0 +1 (t1 )) < 0, n +1 n +1 0 0 2n 0 +1 2n 0 +2 since f ( n01+1 , ρn 0 +1 , 0) ≥ 0, f (t, ρn 0 +1 , 0) ≥ 0 for t ∈ [ n01+2 , 1] and 2 2 true. Next we show that u n 0 +1 (t) ≤ u n 0 (t)

1 2n 0 +1

∈(

1 , 1), 2n 0 +2

and consequently (2.28) is

for t ∈ [0, 1].

(2.29)

If (2.29) is not true, then u n 0 +1 − u n 0 would have a positive absolute maximum at τ0 ∈ (0, 1) in which case u 0n 0 +1 (τ0 ) = u 0n 0 (τ0 ) and [φ(u 0n 0 +1 )]0 (τ0 ) ≤ [φ(u 0n 0 )]0 (τ0 ). Then u n 0 +1 (τ0 ) > u n 0 (τ0 ) together with gn 0 (τ0 , u, z) ≥ gn 0 +1 (τ0 , u, z), for (u, z) ∈ (0, ∞) × R gives (note that (u 0n 0 +1 (τ0 ))∗ = (u 0n 0 (τ0 ))∗ = u 0n 0 (τ0 ), since Rn 0 +1 ≥ Rn 0 and ku 0n 0 k ≤ Rn 0 ) {[φ(u 0n 0 +1 )]0 − [φ(u 0n 0 )]0 }(τ0 ) = −gn∗0 +1 (τ0 , u n 0 +1 (τ0 ), u 0n 0 +1 (τ0 )) + gn∗0 (τ0 , u n 0 (τ0 ), u 0n 0 (τ0 )) = −gn 0 +1 (τ0 , u n 0 (τ0 ), (u 0n 0 +1 (τ0 ))∗ ) − r (u n 0 (τ0 ) − u n 0 +1 (τ0 )) + gn 0 (τ0 , u n 0 (τ0 ), u 0n 0 (τ0 )) = (gn 0 − gn 0 +1 )(τ0 , u n 0 (τ0 ), u 0n 0 (τ0 )) + r (u n 0 +1 − u n 0 )(τ0 ) > 0, a contradiction. Thus (2.29) holds. In addition, since ku n 0 +1 k∞ ≤ ku n 0 k∞ ≤ Mn 0 , then (2.7) (with ε = ρn 0 +1 ) guarantees the existence of a0 , b0 , ηε , δ and γ (as described in (2.7)) with (note that gn∗0 +1 (t, u n 0 +1 (t), u 0n 0 +1 (t)) =

1163

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

gn 0 +1 (t, u n 0 +1 (t), (u 0n 0 +1 (t))∗ )) |gn∗0 +1 (t, u n 0 +1 (t), u 0n 0 +1 (t))| ≤ φ6 (t)[u n 0 +1 (t)]δ + φ7 (t)|(u 0n 0 +1 (t))∗ |γ + φ8 (t) ≤ φ6 (t)Mnδ0 + φ7 (t)|u 0n 0 +1 (t)|γ + φ8 (t) and for t ∈ (0, 1) (note that |z ∗ | ≤ |z|) here, φ6 (t) = max{a0 (t), a0 (θn 0 (t)), a0 (θn 0 +1 (t))}, φ7 (t) = max{b0 (t), b0 (θn 0 (t)), b0 (θn 0 +1 (t))}, φ8 (t) = max{ηε (t), ηε (θn 0 (t)), ηε (θn 0 +1 (t))}. As a result p ku 0n 0 +1 k p

1

Z = 0

≤

(u n 0 +1 (t) − ρn 0 +1 )gn∗0 +1 (t, u n 0 +1 (t), u 0n 0 +1 (t))dt

Mnδ0 (Mn 0

+ ρn 0 +1 )

+ (Mn 0 + ρn 0 +1 ) ≤

Mnδ0 (Mn 0

1

Z 0

"Z

1

φ6 (t)dt + (Mn 0 + ρn 0 +1 ) |φ7 (t)|

p p−γ

# p−γ "Z p dt 0

0

1

1

Z

φ8 (t)dt

0

γ· p |u 0n 0 +1 (t)| γ dt

+ ρn 0 +1 )kφ6 k1 + (Mn 0 + ρn 0 +1 )kφ7 k

p p−γ

#γ

p

γ

ku 0n 0 +1 k p + (Mn 0 + ρn 0 +1 )kφ8 k1 ,

so there exists a constant K n 0 +1 ≥ ρn 0 +1 with p

ku 0n 0 +1 k p ≤ K n 0 +1 . Also since u n 0 +1 (0) = u n 0 +1 (1) = ρn 0 +1 , we have Z 1 0 |φ(u n 0 +1 (t))| ≤ |[φ(u 0n 0 +1 (s))]0 |ds 0

1

Z = 0

|gn∗0 +1 (t, u n 0 +1 (t), u 0n 0 +1 (t))|dt γ

≤ kφ6 k1 Mnδ0 + K n 0 +1 kφ7 k

p p−γ

+ kφ8 k1 ≡ Q n 0 +1 ,

so there exists an Rn 0 +1 ≥ Rn 0 such that ku 0n 0 +1 k∞ ≤ Rn 0 +1 .

(2.30)

Also notice that if we take h(t, u, z) = gn 0 +1 (t, u, z) in (2.3), then since gn 0 +1 ≥ f and u n 0 +1 satisfies [φ(u 0n 0 +1 )]0 + gn 0 +1 (t, u n 0 +1 (t), u 0n 0 +1 (t)) = 0, t ∈ (0, 1), we then have u n 0 (t) ≥ α(t)

for t ∈ [0, 1].

(2.31)

Now we proceed inductively to a construction; we consider u n 0 +2 , u n 0 +3 , . . . as follows. Suppose we have u k for some k ∈ {n 0 + 1, n 0 + 2, . . .} with α(t) ≤ u k (t) ≤ u k−1 (t) for t ∈ [0, 1]. Then we consider the boundary value problem  ∗ [φ(u 0 )]0 + gk+1 (t, u, u 0 ) = 0, for t ∈ (0, 1), (2.32) u(0) = u(1) = ρk+1 , where  gk+1 (t, ρk+1 , z ∗ ) + r (ρk+1 − u), ∗ gk+1 (t, u, z) = gk+1 (t, u, z ∗ ),  gk+1 (t, u k (t), z ∗ ) + r (u k (t) − u),

u ≤ ρk+1 ; ρk+1 ≤ u ≤ u k (t); u ≥ u k (t)

1164

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

with   Rk+1 , z ∗ = z,  −Rk+1 ,

z > Rk+1 ; −Rk+1 ≤ z ≤ Rk+1 ; z ≤ −Rk+1

where Rk+1 ≥ Rk is a predetermined constant. Now Schauder’s fixed point theorem guarantees that (2.32) has a solution u k+1 ∈ C 1 [0, 1], and as above yields ρk+1 ≤ u k+1 (t) ≤ u k (t) for t ∈ [0, 1], ku 0k+1 k∞ ≤ Rk+1 , with u k+1 (t) ≥ α(t)

for t ∈ [0, 1]

[φ(u 0k+1 )]0 + gk+1 (t, u k+1 (t), u 0k+1 (t)) = 0

and

on (0, 1).

We obtain a sequence {u n (t)} of solutions to (2.8n ). Firstly, we claim  { j} ∞ {u n }n=n 0 +1 equicontinuous   j = 0, 1 is a bounded, 1 1 family on , 1 − n +1 . 2n 0 +1 2 0

(2.33)

Note that  e = [a, b] =

1 2n 0



1

,1 − +1

2n 0 +1

,

 = min α(t),

ku n k∞ ≤ ku n 0 k∞ ≤ Mn 0 ,

t∈e

n ≥ n 0 + 1,

and then (2.7) guarantees the existence of a0 , b0 , ηε , δ and γ (as described in (2.7)) with |gn (t, u n (t), u 0n (t))| = | f (t, u n (t), u 0n (t))| ≤ a0 (t)Mnδ0 + b0 (t)|u 0n (t)|γ + ηε (t).

(2.34)

For t ∈ e ⊂ (0, 1), and n ≥ n 0 + 1, let   u n (b) − u n (a) (t − a) . rn (t) = u n (t) − u n (a) + b−a Then rn (a) = rn (b) = 0. So for n ≥ n 0 + 1, we have Z b Z b Z u n (b) − u n (a) b 0 p 0 [φ(u 0 (t))]0rn (t)dt = − |u n (t)| dt + φ(u n (t))dt . n b−a a a a Next, since rn (t) ≤ 4Mn 0 for t ∈ e, we have for any n ≥ n 0 + 1 that Z b Z Z b 2Mn 0 b 0 p 0 |φ(u n (t))|dt + 4Mn 0 [φ(u 0n (t))]0 dt |u n (t)| dt ≤ b−a a a a Z b  p−1 Z b p 2Mn 0 0 p δ+1 ≤ |u (t)| dt a0 (t)dt + 4M n n0 p−1 a a (b − a) p Z b Z b + 4Mn 0 b0 (t)|u 0n (t)|γ dt + 4Mn 0 ηε (t)dt a

2Mn 0

≤

(b − a)

a

Z

p−1 p

a

b

|u 0n (t)| p dt Z

+ 4Mn 0 kb0 k

p p−γ

b

a

 p−1 p

|u 0n (t)| p dt

+ 4Mnδ+1 ka0 k1 0  γp

+ 4Mn 0 kηε k1 .

So there exists a Cn 0 with Z a

b

|u 0n (t)| p dt ≤ Cn 0

for n ≥ n 0 + 1.

(2.35)

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

1165

n (a) Also there exists tn ∈ e such that u 0n (tn ) = u n (b)−u , so for n ≥ n 0 + 1 we have (using (2.35)) b−a Z b Z b |gn (t, u n (t), u 0n (t))|dt |[φ(u 0n (t))]0 |dt =

a

a

Z

b

≤ a

≤

Mnδ0 ka0 k1

b

Z

a0 (t)Mnδ0 dt

+ a

Z + kb0 k

p p−γ

b

a

|u 0n (t)| p dt

b

Z

b0 (t)|u 0n (t)|γ dt

+ a

 γp

ηε (t)dt

+ kηε k1 < C

and φ(u 0n (t))

=φ



u n (b) − u n (a) b−a



Z

t

+ tn

[φ(u 0n (s))]0 ds,

t ∈ e.

Then |u 0n (t)| p−1

≤C +φ



u n 0 (b) + u n 0 (a) b−a



=: C1 (a, b),

t ∈ e,

which means that u 0n is bounded on e. Rb Since a |[φ(u 0n (t))]0 |dt < C, thus φ(u 0n ) is equicontinuous on e. The equicontinuity of u 0n follows from the 1/( p−1) 1/( p−1) uniform continuity of φ −1 on [−C1 (a, b), C1 (a, b)], and |u 0n (t) − u 0n (s)| = |φ −1 (φ(u 0n (t))) − φ −1 (φ(u 0n (s)))|,

t, s ∈ e.

Thus the Arzela–Ascoli theorem guarantees the existence of a subsequence Nn 0 of integers and a function { j} { j} z n 0 ∈ C 1 [ n01+1 , 1 − n01+1 ] with u n , j = 0, 1, converging uniformly to z n 0 on [ n01+1 , 1 − n01+1 ] as n → ∞ 2 2 2 2 through Nn 0 . Similarly  { j} ∞ {u n }n=n 0 +2 equicontinuous   j = 0, 1 is a bounded, 1 1 family on , 1 − n +2 , 2n 0 +2 2 0 so there is a subsequence Nn 0 +1 of Nn 0 and a function   1 1 z n 0 +1 ∈ C 1 n +2 , 1 − n +2 2 0 2 0 { j}

{ j}

with u n , j = 0, 1, converging uniformly to z n 0 +1 on [

1 ,1 2n 0 +2

−

z n 0 +1 = z n 0 on [ n0 +1 , 1 − n0 +1 ] since Nn 0 +1 ⊆ Nn 0 . 2 2 We proceed inductively to obtain subsequences of integers 1

1

1 ] 2n 0 +2

Nn 0 ⊇ Nn 0 +1 ⊇ · · · ⊇ Nk ⊇ · · · and functions zk ∈ C 1



1

2

,1 − k+1

1



2k+1

with { j} un ,

j = 0, 1, converging uniformly to

as n → ∞ through Nk , and   1 1 ,1 − k . z k = z k−1 on 2k 2

{ j} zk

 on

1 2k+1

,1 −

1 2k+1



as n → ∞ through Nn 0 +1 . Note that

1166

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

1 1 , 1 − 2k+1 ] and u(0) = u(1) = 0. Notice that u is Define a function u : [0, 1] → [0, ∞] by u(t) = z k (t) on [ 2k+1 well defined and α(t) ≤ u(t) ≤ u n 0 (t) for t ∈ (0, 1). Next fix t ∈ (0, 1) (without loss of generality, assume t 6= 12 ) 1 1 and let m ∈ {n 0 , n 0 + 1, . . .}, such that t ∈ ( 2m+1 , 1 − 2m+1 ). Let Nm∗ = {n ∈ Nm : n ≥ m}. Now u n , n ∈ Nm∗ satisfies the integral equation    Z t 1 0 + gn (s, u n (s), u 0n (s))ds φ(u n (t)) = φ u 0n 1 2 2 1 ,1 − for t ∈ [ 2m+1

1 ]. 2m+1

Let n → ∞ through Nm∗ to obtain    Z t 1 0 0 0 φ(z m (t)) = φ z m + gn (s, z m (s), z m (s))ds 1 2 2    Z t 1 0 0 = φ zm + f (s, z m (s), z m (s))ds, 1 2 2

i.e.,    Z t 1 φ(u (t)) = φ u 0 f (s, u(s), u 0 (s))ds. + 1 2 2 0

So we have [φ(u 0 )]0 + f (t, u, u 0 ) = 0,

t ∈ (0, 1).

It remains to show that u is continuous at 0 and 1. Let  > 0 be given. Now since limn→+∞ u n (0) = 0, there exists n 1 ∈ {n 0 , n 0 + 1 . . .} with u n 1 (0) < u n 1 ∈ C[0, 1], there exists δn 1 > 0 with ε u n 1 (t) < for t ∈ [0, δn 1 ]. 2 Now for n ≥ n 1 , since {u n (t)} is nonincreasing for each t ∈ [0, 1], then we have ε α(t) ≤ u n (t) ≤ u n 1 (t) < for t ∈ [0, δn 1 ]. 2 Consequently ε α(t) ≤ u(t) ≤ < ε for t ∈ (0, δn 1 ], 2 and so u is continuous at 0. Similarly u is continuous at 1. As a result u ∈ C[0, 1]. The proof of Theorem 2.1 is complete.  Remark 1. From the proof we see that u f (t, u, z) ≤ a(t)u p + b(t)u α |z|β in (2.4) could be replaced by u f (t, u, z) ≤ a(t)u p + b(t)u α |z|β + c(t)u γ + d(t)u|z|τ + uh ε (t), p

where c ∈ L 1 [0, 1], d ∈ L p−τ [0, 1] and h ε ∈ L 1 [0, 1], 1 ≤ γ < p, 0 ≤ τ < p − 1. Remark 2. In (2.2) it is possible to replace

1 2n+1

≤ t ≤ 1 with either

1 (i) 0 ≤ t ≤ 1 − 2n+1 , 1 1 (ii) 2n+1 ≤ t ≤ 1 − 2n+1 or (iii) 0 ≤ t ≤ 1.

This is clear once one changes the definition of en , and θn has also to be adjusted correspondingly. Remark 3. For any n ∈ {1, 2, . . .}, n ≥ 1, set   1 1 en := n+1 , 1 − n+1 . 2 2

ε 2.

Since

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

1167

There exists a λ ∈ C 1 [0, 1] such that (see [21], Lemma 2) (1) φ(λ0 (t)) ∈ C 1 [0, 1] and max0≤t≤1 |[φ(λ0 (t))]0 | > 0, (2) λ(0) = λ(1) = 0, λ(t) > 0, ∀t ∈ (0, 1), (3) λ(t) ≤ ε1 , t ∈ e1 , λ(t) ≤ εn , ∀t ∈ en /en−1 , n ≥ 2. Remark 4. In [12], the authors use the following condition instead of (2.2) in Theorem 2.1: (H1 ): For each fixed compact set e ⊂ (0, 1), there exists a constant εe > 0 such that for any M > 0, there is L M > 0 such that f (t, x, z) > L M ,

for all (t, x, z) ∈ e × (0, εe ] × [−M, M].

By (H1 ), then for each fixed en ⊂ (0, 1), there exists a constant εn > 0 (εn ↓ 0) such that for any M > 0, there is an L M > 0 such that f (t, x, z) > L M ,

∀(t, x, z) ∈ en × (0, εn ] × [−M, M].

Having chosen a function λ with the above described properties as in Remark 3, we note that f (t, x, z) > L M ,

∀t ∈ (0, 1), 0 < x ≤ λ(t), z ∈ [−M, M].

Set Mλ = kλ0 k∞ = supt∈[0,1] |λ0 (t)|, and set (   1 ) p−1 L Mλ m := min 1, . (k[φ(λ0 )]0 k∞ + 1) Lemma A (See [12]). Let h(t, x, z) ≥ f (t, x, z), for all (t, x, z) ∈ (0, 1)×(0, ∞)×R, with h : (0, 1)×(0, ∞)×R → R a continuous function, and let v ∈ C([0, 1], [0, ∞)) ∩ C 1 ([0, 1], R) be any solution of [φ(v 0 (t))]0 + h(t, v, v 0 ) = 0 on (0, 1) with v(0) ≥ 0, v(1) ≥ 0. Then, v(t) ≥ mλ(t),

0 ≤ t ≤ 1.

We can see that if (H1 ) is satisfied, then (2.3) holds in Theorem 2.1 with α(t) = mλ(t). Acknowledgements This work was supported by NNSF of China (No: 10571021) and the Key Laboratory for Applied Statistics of MOE (KLAS). References [1] R.P. Agarwal, D. O’Regan, Singular boundary value problems for superlinear second ordinary and delay differential equations, J. Differential Equations 130 (1996) 335–355. [2] R.P. Agarwal, D. O’Regan, Nonlinear superlinear singular and nonsingular second order boundary value problems, J. Differential Equations 143 (1998) 60–95. [3] R.P. Agarwal, D. O’Regan, Singular initial and boundary value problems with sign changing nonlinearities, IMA J. Appl. Math. 65 (2000) 173–198. [4] R.P. Agarwal, D. O’Regan, Some new existence results for singular problems with sign changing nonlinearities, J. Comput. Appl. Math. 113 (2000) 1–15. [5] L. Bobisud, D. O’Regan, W.D. Royalty, Solvability of some nonlinear boundary value problems, Nonlinear Anal. 12 (1988) 855–869. [6] P. Habets, F. Zanolin, Upper and lower solutions for a generalized Emden–Fowler equation, J. Math. Anal. Appl. 181 (1994) 684–700. [7] J. Janus, J. Myjak, A generalized Emden–Fowler equation with a negative exponent, Nonlinear Anal. 23 (1994) 953–970. [8] D. Jiang, H. Liu, On the existence of nonnegative radial solutions for the one-dimensional p-Laplacian elliptic systems, Ann. Polon. Math. 71 (1999) 19–29. [9] D. Jiang, W. Gao, Upper and lower solution method and a singular boundary value problem for the one-dimension p-Laplacian, J. Math. Anal. Appl. 252 (2000) 631–648. [10] D.Q. Jiang, Upper and lower solutions for a superlinear singular boundary value problem, Comput. Math. Appl. 41 (2001) 563–569. [11] D.Q. Jiang, Upper and lower solutions method and a superlinear singular boundary value problem for the one-dimension p-Laplacian, Comput. Math. Appl. 42 (2001) 927–940.

1168

D. Jiang, H. Zhang / Nonlinear Analysis 68 (2008) 1155–1168

[12] D.Q. Jiang, P.Y.H. Pang, R.P. Agarwal, Nonresonant singular boundary value problems for the one-dimension p-Laplacian, Dynam. Systems Appl. 11 (2002) 449–458. [13] D. O’Regan, Nonresonant nonlinear singular problems in the limit circle case, J. Math. Anal. Appl. 197 (1996) 708–725. [14] D. O’Regan, Some general existence principles and results for [φ(y 0 )]0 = q(t) f (t, y, y 0 ), (0 < t < 1), SIAM J. Math. Anal. 24 (1993) 648–668. [15] D. O’Regan, Singular Dirichlet boundary value problems—I, superlinear and nonresonant case, Nonlinear Anal. 29 (1997) 221–245. [16] D. O’Regan, Existence theory for nonresonant singular boundary value problems, Proc. Edinb. Math. Soc. 38 (1995) 431–437. [17] D. O’Regan, R.P. Agarwal, Singular problems: An upper and lower solution approach, J. Math. Anal. Appl. 251 (2000) 230–250. [18] J. Wang, W. Gao, Z. Lin, Boundary value problems for general second order equations and similarity solutions to the Rayleigh problem, Tohoku Math. J. 47 (1995) 327–344. [19] J. Wang, W. Gao, A singular boundary value problem for the one-dimensional p-Laplacian, J. Math. Anal. Appl. 201 (1996) 851–866. [20] J. Wang, D.Q. Jiang, A unified approach to some two-point, three-point and four-point boundary value problems with Caratheodory functions, J. Math. Anal. Appl. 211 (1997) 223–232. [21] Q. Yao, H. Lu, Positive solution of one-dimensional singular p-Laplacian equations, Acta Math. Sinica 41 (1998) 1255–1264 (in Chinese). [22] M.R. Zhang, Nonuniform nonresonance at the first eigenvalue of the p-Laplacian, Nonlinear Anal. 29 (1997) 41–51.