Existence of positive solutions for the one-dimensional singular p -Laplacian

Nonlinear Analysis 68 (2008) 2309–2318 www.elsevier.com/locate/na

Existence of positive solutions for the one-dimensional singular p-LaplacianI Zheng-an Yao a , Wenshu Zhou b,∗ a Department of Mathematics, Zhongshan University, Guangzhou 510275, PR China b Department of Mathematics, Jilin University, Changchun 130012, PR China

Received 14 September 2006; accepted 25 January 2007

Abstract This paper concerns the positive solutions of boundary value problems for the one-dimensional singular p-Laplacian. By the classical method of elliptic regularization, we obtain some existence results which generalize some results of [W. Zhou, X. Wei, Positive solutions to BVPs for a singular differential equation, Nonlinear Anal. (2006), doi:10.1016/j.na.2006.06.015]. c 2007 Elsevier Ltd. All rights reserved.

MSC: 34B15 Keywords: p-Laplacian; Positive solution; Existence; Regularization

1. Introduction This paper concerns the existence of positive solutions for the one-dimensional singular p-Laplacian 

0 |ϕ 0 | p + f (t) = 0, Φ p (ϕ 0 ) − λ ϕ

0 < t < 1,

(1)

with either the Dirichlet boundary conditions ϕ(1) = ϕ(0) = 0,

(2)

or the periodic boundary conditions ϕ(1) = ϕ(0) = ϕ 0 (1) = ϕ 0 (0) = 0, where Φ p (s) = |s| p−2 s, p > 2, λ > 0, f (t) ∈ C[0, 1] and f (t) > 0 on [0, 1].

I Supported in part by NNSFC (10531040, 10471156), Tianyuan Youth Foundation (10626056) and 985 Program of Jilin University. ∗ Corresponding author.

E-mail addresses: [email protected] (Z.-a. Yao), [email protected] (W. Zhou). c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2007.01.049

(3)

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(1) arises in the model, which appears in the studies of a degenerate parabolic equation (see [2–4]), considered by Bertsch and Ughi. In [3], they studied the following singular Laplacian: ϕ 00 +

N −1 0 |ϕ 0 |2 ϕ −λ + 1 = 0, t ϕ

0 < t < 1,

(4)

with the boundary conditions: ϕ(1) = ϕ 0 (0) = 0, where N is a positive integer, and, by theories of ordinary differential equations, obtained a decreasing positive solution. Note that the solutions to BVP (1) and (2) or BVP (1) and (3) must not be decreasing. In [15], the authors studied (1) with p = 2, and proved that (i) if λ > 0, f ∈ C 1 [0, 1] and f > 0 on [0, 1], then BVP (1) and (2) with p = 2 admits at least a positive solution; (ii) if λ > 1/2, f ∈ C 1 [0, 1] and f > 0 on [0, 1], then BVP (1) and (3) with p = 2 admits at least a positive solution. Motivated by [3,15], we consider (1), and the main purpose is to search for positive solutions. By the classical method of elliptic regularization, some existence results are obtained, which generalize some results of [15]. In recent years, the following one-dimensional p-Laplacian differential equations have been studied extensively for when F(t, u, v) does not depend on the first-order derivative:  0 Φ p (ϕ 0 ) + F(t, ϕ, ϕ 0 ) = 0, 0 < t < 1, (5) where F(t, u, v) may be singular; see [1,7,9,12,14]. However, when F depends on the first-order derivative it has not received much attention; see [8,10,11,13] and references therein. In [8], by the upper and lower solution method, Jiang and Gao studied problem (5) and (2), where F may be singular at u = 0, t = 0 and t = 1, and obtained a positive solution. In [10], L¨u, O’Regan and Agarwal also discussed the problem (5) and (2), where F may be singular at u = 0, t = 0 and t = 1, and proved a general existence result in the continuity functions class. We point out that the p case considered here, namely, F(t, u, v) = −λ |v|u + f (t), is not contained in those papers since it does not satisfy some sufficient conditions imposed on F. Another point of interest of the present paper is that the nonlinear term depends on the first-order derivative explicitly. We say ϕ ∈ C 1 [0, 1] is a solution to BVP (1) and (2) if ϕ > 0 in (0, 1), |ϕ 0 | p−2 ϕ 0 ∈ C 1 (0, 1), and it satisfies (1) and (2). Similarly, we say ϕ ∈ C 1 [0, 1] is a solution to BVP (1) and (3) if ϕ > 0 in (0, 1), |ϕ 0 | p−2 ϕ 0 ∈ C 1 (0, 1), and it satisfies (1) and (3). Theorem 1. Let p > 2, λ > 0, f ∈ C[0, 1] and f > 0 on [0, 1]. Then BVP (1) and (2) admits at least one solution. Theorem 2. Let p > 2, λ > ( p − 1)/ p, f (t) ∈ C[0, 1] and f (t) > 0 on [0, 1]. Then BVP (1) and (3) admits at least one solution. The paper is organized as follows. In Section 2 we prove Theorem 1. The proof of Theorem 2 is given in Section 3. In the sequel, the following lemma will be used repeatedly. Lemma 3. Let p > 2. Then we have (|η| p−2 η − |η0 | p−2 η0 ) · (η − η0 ) > C p |η − η0 | p ,

∀η, η0 ∈ R,

where C p is a positive constant depending only on p. We refer the reader to [6] for the proof. 2. Proof of Theorem 1 To study the existence of solutions to problem (1) and (2), we may use the classical method of elliptic regularization. For simplicity, throughout the section we assume that f ∈ C 1 [0, 1]. Otherwise, we can replace f by a mollification function Jε f 0 which, as usual, is defined by (see [16, pp. 21–22]) Z +∞ (Jε f 0 )(t) = jε (t − s) f 0 (s)ds, −∞

Z.-a. Yao, W. Zhou / Nonlinear Analysis 68 (2008) 2309–2318

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where ε ∈ (0, 1), jε is the mollifier, and   f (1), t > 1, f 0 (t) = f (t), 0 6 t 6 1,  f (0), t < 0. Remark. Since f 0 ∈ C(R), we have Jε f 0 ∈ C ∞ (R), min[0,1] f 6 Jε f 0 6 max[0,1] f , and Jε f 0 → f,

uniformly on [0, 1], as ε → 0.

Instead of problem (1) and (2), we consider the following regularized problem:   |ϕ 0 |2 (|ϕ 0 |2 + ε)( p−2)/2 |ϕ 0 |2 (|ϕ 0 |2 + ε)( p−2)/2 1 + ( p − 2) 0 2 sgnε (ϕ) + f (t) = 0, ϕ 00 − λ Iε (ϕ) |ϕ | + ε ϕ(1) = ϕ(0) = ε, where ε ∈ (0, 1), Iε (s) and sgnε (s) are defined as follows:  1,     2s s, s > ε,      2  − s + ε2 Iε (s) = sgnε (s) = ε , |s| < ε, 2s     −s,2ε  +  s 6 −ε,  ε −1,

0 < t < 1,

s > ε, s2 , ε2 2 s , ε2

0 6 s < ε, −ε 6 s < 0, s < −ε.

Clearly, Iε (s), sgnε (s) ∈ C 1 (R), and Iε (s) > ε/2, 1 > |sgnε (s)|, sgnε (s)sgn(s) > 0, |Iε0 (s)| 6 1, 0 6 sgn0ε (s) 6 2/ε in R. For λ > 0, it follows from Theorem 4.1 of Chapter 7 in [5] (see Appendix of the present paper) that for any fixed ε ∈ (0, 1), the above regularized problem has a classical solution ϕε ∈ C 2 [0, 1]. By the maximal principle, it is easy to see that ϕε (t) > ε on [0, 1]. Thus ϕε satisfies   (|ϕ 0 |2 + ε)( p−2)/2 |ϕε0 |2 |ϕε0 |2 0 2 ( p−2)/2 ϕε00 − λ ε (|ϕε | + ε) + f (t) = 0, t ∈ [0, 1], (6) 1 + ( p − 2) 0 2 ϕε |ϕε | + ε ϕε (0) = ϕε (1) = ε. Note that (6) is equivalent to 

(|ϕε0 |2 + ε)( p−2)/2 ϕε0

0

−λ

(|ϕε0 |2 + ε)( p−2)/2 |ϕε0 |2 + f (t) = 0, ϕε

t ∈ [0, 1].

(7)

Lemma 4. Under the assumptions of Theorem 1, for all ε ∈ (0, 1) there exists a positive constant L independent of ε such that |ϕε0 (t)| 6 L ,

t ∈ [0, 1].

Proof. Noticing ϕε (1) = ϕε (0) = ε and ϕε (t) > ε for all t ∈ [0, 1], we have ϕε0 (0) = lim

t→0+

ϕε (t) − ε > 0, t

By (7), we obtain  0 (|ϕε0 |2 + ε)( p−2)/2 ϕε0 + Λ > 0,

ϕε0 (1) = lim

t→1−

t ∈ [0, 1],

1

where Λ = max[0,1] f , i.e.  0 (|ϕε0 |2 + ε)( p−2)/2 ϕε0 + Λt > 0,

t ∈ [0, 1].

ϕε (t) − ε 6 0. t −1

(8)

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Thus the function (|ϕε0 |2 + ε)( p−2)/2 ϕε0 + Λt is nondecreasing on [0, 1]. Combining this with (8), we obtain Λ > (|ϕε0 (1)|2 + ε)( p−2)/2 ϕε0 (1) + Λ

> (|ϕε0 (t)|2 + ε)( p−2)/2 ϕε0 (t) + Λt

> (|ϕε0 (0)|2 + ε)( p−2)/2 ϕε0 (0) > 0, and therefore 0 (|ϕε (t)|2 + ε)( p−2)/2 ϕε0 (t) 6 Λ,

t ∈ [0, 1],

t ∈ [0, 1].

(9)

Noticing p > 2, we have (|ϕε0 |2 + ε)( p−2)/2 > |ϕε0 | p−2 ,

t ∈ [0, 1],

which with (9) implies that |ϕε0 (t)| 6 Λ1/( p−1) ,

t ∈ [0, 1].

This completes the proof of Lemma 4.



It follows from Lemma 4 that |ϕε0 | p−2 6 (|ϕε0 |2 + ε)( p−2)/2 6 (1 + L 2 )( p−2)/2 ,

t ∈ [0, 1].

Obviously, we have 1 6 1 + ( p − 2)

|ϕε0 |2 6 p − 1, |ϕε0 |2 + 1

t ∈ [0, 1].

(10)

Combining the above results with (6), we obtain −ϕε00

min f |ϕε0 |2 [0,1] > 0, +λ − ϕε ( p − 1)(1 + L 2 )( p−2)/2

−( p − 1)|ϕε0 | p−2 ϕε00 + λ

t ∈ [0, 1],

|ϕε0 | p − ( p − 1) max f 6 0, [0,1] ϕε

(11)

t ∈ [0, 1].

(12)

Since p > 2 and ϕε ∈ C 2 [0, 1], it is easy to show that  0 ( p − 1)|ϕε0 | p−2 ϕε00 = |ϕε0 | p−2 ϕε0 , t ∈ [0, 1].

(13)

Therefore (12) can be written as  0 |ϕ 0 | p − |ϕε0 | p−2 ϕε0 + λ ε − ( p − 1) max f 6 0, [0,1] ϕε

t ∈ [0, 1].

(14)

To obtain some uniform bounds of ϕε , we need to establish the following comparison theorem. Proposition 5. Let φi ∈ C 1 [0, 1] satisfy |φi0 |q−2 φi0 ∈ C 1 [0, 1] with q > 2 and φi > 0 on [0, 1] (i = 1, 2). If φ2 > φ1 for t = 0, 1, and 0  |φ 0 |q − |φ20 |q−2 φ20 + % 2 − θ > 0, φ2 where % and θ are positive constants, then φ2 (t) > φ1 (t),

t ∈ [0, 1].

t ∈ [0, 1],

 0 |φ 0 |q − |φ10 |q−2 φ10 + % 1 − θ 6 0, φ1

t ∈ [0, 1],

(15)

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Proof. From (15), we have 0  θ |[g(φ2 )]0 |q−2 [g(φ2 )]0 + % 6 0, φ2  0 θ |[g(φ1 )]0 |q−2 [g(φ1 )]0 + % > 0, φ1

t ∈ [0, 1], t ∈ [0, 1],

where g : (0, +∞) → R is defined as follows:  1−%/(q−1)  s , (% 6= q − 1) g(s) = 1 − %/(q − 1)  ln(s) (% = q − 1). Therefore 0  − |[g(φ1 )]0 |q−2 [g(φ1 )]0 − |[g(φ2 )]0 |q−2 [g(φ2 )]0 + θ

1 1 % − % φ2 φ1

! 6 0,

t ∈ [0, 1],

and then, multiplying the above equality by (g(φ1 ) − g(φ2 ))+ and integrating by parts over (0, 1) and noticing (g(φ1 ) − g(φ2 ))+ = 0 for t = 0, 1, where s+ = max{0, s}, we obtain Z 1  |[g(φ1 )]0 |q−2 [g(φ1 )]0 − |[g(φ2 )]0 |q−2 [g(φ2 )]0 (g(φ1 ) − g(φ2 ))0+ dt 0 ! Z 1 1 1 +θ % − % (g(φ1 ) − g(φ2 ))+ dt 6 0. φ2 φ1 0 By Lemma 3, the first integral of left hand side of the above inequality is nonnegative, and hence ! Z 1 1 1 % − % (g(φ1 ) − g(φ2 ))+ dt 6 0. φ2 φ1 0 Combining this with the fact that g(s) is nondecreasing in (0, ∞), we have ! 1 1 t ∈ (0, 1), % − % (g(φ1 ) − g(φ2 ))+ = 0, φ2 φ1 and therefore φ2 > φ1 on [0, 1]. Thus the proof of Proposition 5 is completed.



Lemma 6. Under the assumptions of Theorem 1, for all ε ∈ (0, 1) there exists a positive constant C independent of ε such that h i2 ϕε (t) > C t (1 − t) + ε 1/2 , t ∈ [0, 1].  2 Proof. Let wε = C t (1 − t) + ε 1/2 , where C ∈ (0, 1] will be determined later. To prove the lemma, by Proposition 5 and noticing (11), it suffices to show that −wε00 + λ

|wε0 |2 − θ L 6 0, wε

t ∈ [0, 1],

(16)

for some sufficiently small positive constant C independent of ε, where θ L = shows that −wε00 + λ

min[0,1] f . ( p−1)(1+L 2 )( p−2)/2

|wε0 |2 − θ L = −2C(1 − 2t)2 + 4C[t (1 − t) + ε 1/2 ] + 4λC(1 − 2t)2 − θ L wε 6 4C[t (1 − t) + ε 1/2 ] + 4λC(1 − 2t)2 − θ L 6 4C(2 + λ) − θ L ,

t ∈ [0, 1].

Simple calculation

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Choosing a positive constant C such that   θL C 6 min 1, , 2+λ we find that (16) holds. Thus the proof of Lemma 6 is completed.



Lemma 7. For any δ ∈ (0, 1/2), there exists a positive constant Cδ independent of ε such that |ϕε0 (t2 ) − ϕε0 (t1 )| 6 Cδ |t2 − t1 |1/( p−1) ,

∀t2 , t1 ∈ [δ, 1 − δ].

Proof. From (6), Lemmas 4 and 6, it is easy to derive that for any δ ∈ (0, 1/2) there exists a positive constant Cδ independent of ε such that  0 2  0 2 00 (|ϕ | + ε)( p−2)/2 1 + ( p − 2) |ϕε | δ 6 t 6 1 − δ. ϕ ε 6 Cδ , ε 0 2 |ϕε | + ε From this and (10) and noticing p > 2, we obtain 0 p−2 00 ϕε 6 Cδ , δ 6 t 6 1 − δ, |ϕε | and then, by (13), we have 0 p−2 0 0 ϕε ) 6 C δ , (|ϕε |

δ 6 t 6 1 − δ.

From this estimate and Lemma 3, we obtain 0 0 0 p−2 0 |ϕε0 (t2 ) − ϕε0 (t1 )| p 6 C −1 ϕε (t2 ) − |ϕε0 (t1 )| p−2 ϕε0 (t1 )] p [ϕε (t2 ) − ϕε (t1 )] · [|ϕε (t2 )|

6 Cδ |ϕε0 (t2 ) − ϕε0 (t1 )||t2 − t1 |,

∀t2 , t1 ∈ [δ, 1 − δ],

and hence |ϕε0 (t2 ) − ϕε0 (t1 )| 6 Cδ |t2 − t1 |1/( p−1) , This completes the proof of Lemma 7.

∀t2 , t1 ∈ [δ, 1 − δ].



From Lemmas 4 and 7 and using the Arzel´a–Ascoli theorem and diagonal sequential process, we see that there exist a subsequence {ϕε j } of {ϕε } and a function ϕ ∈ C 1 (0, 1) ∩ C[0, 1] such that, as ε j → 0, ϕε j → ϕ,

uniformly in C[0, 1],

ϕε j → ϕ,

uniformly in C 1 [δ, 1 − δ],

(17)

where δ ∈ (0, 1/2), and hence from ϕε j (1) = ϕε j (0) = ε j and Lemma 4 it is easy to see that ϕ satisfies (2), and ϕ(t) > Ct 2 (1 − t)2 ,

t ∈ [0, 1],

(18)

and therefore ϕ > 0 in (0, 1). Next we show that ϕ satisfies (1). For 0 < δ 6 t < 1, integrating (7) with ε = ε j over [δ, t] to derive # Z t " (|ϕ 0 |2 + ε )( p−2)/2 |ϕ 0 |2  ( p−2)/2 j εj εj 0 2 0 − f (s) ds |ϕε j (t)| + ε j ϕε j (t) = λ ϕε j δ  ( p−2)/2 + |ϕε0 j (δ)|2 + ε j ϕε0 j (δ), and then letting ε j → 0 and using the Lebesgue dominated convergence theorem and (17), we obtain  Z t |ϕ 0 | p 0 p−2 0 |ϕ (t)| ϕ (t) = λ − f (s) ds + |ϕ 0 (δ)| p−2 ϕ 0 (δ). ϕ δ It follows from this and ϕ ∈ C 1 (0, 1) and f ∈ C 1 [0, 1] that |ϕ 0 (t)| p−2 ϕ 0 (t) ∈ C 1 (0, 1) and  0 |ϕ 0 | p |ϕ 0 | p−2 ϕ 0 − λ + f (t) = 0, 0 < t < 1. ϕ

(19)

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It remains to show that ϕ ∈ C 1 [0, 1]. For this, integrating (7) with ε = ε j over [0, 1] and using (8), we have 1

Z

(|ϕε0 j |2 + ε j )( p−2)/2 |ϕε0 j |2

0

ϕε j

dt 6

1 λ

1

Z

f (t)dt,

0

and then letting ε j → 0 and using Fatou’s lemma and (17), we obtain Z Z 1 0 p |ϕ | 1 1 dt 6 f (t)dt. ϕ λ 0 0 0 p

Fix δ in (19). From the above inequality we deduce that |ϕϕ| ∈ L 1 [0, 1] and therefore, by (19), the function χ (t) := |ϕ 0 (t)| p−2 ϕ 0 (t) = Φ p (ϕ 0 (t)) is absolutely continuous on [0, 1]. Since ϕ 0 (t) = Φq (χ (t)), where 1/ p + 1/q = 1, we see that ϕ ∈ C 1 [0, 1]. This completes the proof of Theorem 1. 3. Proof of Theorem 2 From Theorem 1, we see that for any λ > 0, BVP (1) and (2) admits a solution ϕ which can be approximated by ϕε j satisfying (7) with ε = ε j . It remains to show that ϕ satisfies ϕ 0 (1) = ϕ 0 (0) = 0 for λ > ( p − 1)/ p. We claim that if λ > ( p − 1)/ p, then there exist positive constants C independent of ε j such that  p/( p−1)  ( p−1)/ p −t ϕε j (t) 6 C 1 + ε j   ( p−1)/ p p/( p−1) ϕε j (t) 6 C t + ε j

on [0, 1],

(20)

on [0, 1]. ( p−1)/ p

We first prove (20). Let wε j = C p−1 p (1 + ε j

(21) − t) p/( p−1) , where C >

p p−1

will be determined later. By

Proposition 5 and noticing (14), it suffices to show that 0  |wε0 j | p − ( p − 1) max f > 0, − |wε0 j | p−2 wε0 j + λ [0,1] wε j

t ∈ [0, 1],

(22)

for some sufficiently large positive constant C independent of ε. By a simple calculation, we obtain   0  |wε0 j | p pλ p−1 0 p−2 0 − ( p − 1) max f = C − |wε j | wε j + λ − 1 − ( p − 1) max f, t ∈ [0, 1]. [0,1] [0,1] wε j p−1 Choosing a positive constant C such that   1/( p−1)    ( p − 1)2 max f  p  [0,1]  C > max ,   pλ − p + 1 p−1  and noticing λ > ( p − 1)/ p, we find that (22) holds, and thus the claim (20) is proved. Similarly, we can show (21). Letting ε j → 0 in (20) and (21) we obtain ϕ(t) 6 C min{t p/( p−1) , (1 − t) p/( p−1) },

t ∈ [0, 1].

Combining this with (18) we immediately obtain ϕ 0 (1) = ϕ 0 (0) = 0. This completes the proof of Theorem 2. Acknowledgement The authors would like to thank the referee for his/her important comments which improved this paper.

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Appendix In the appendix we recall Theorem 4.1 of Chapter 7 in [5] and show the solvability of the regularized problem considered in Section 2. We consider the Dirichlet boundary value problem for a second-order quasilinear elliptic equation: −a i j (x, ϕ, Dϕ)Di j ϕ + b(x, ϕ, Dϕ) = 0,

x ∈ Ω , ϕ = ψ, x ∈ ∂Ω ,

(23)

where a i j = a ji , Ω is a bounded domain in R N . By convention, repeated indexes are summed from 1 to N . We make the following assumptions: (F1 ) There exist γ (x, z, r ) > 0 and Γ (x, z, r ) > 0 such that γ |ξ |2 6 ξi a i j (x, z, r )ξ j 6 Γ |ξ |2 , ∀ξ ∈ R N , (x, z, r ) ∈ Ω × R × R N . Γ 6 µ1 (|z|), ∀(x, z, r ) ∈ Ω × R × R N . γ (F2 ) |b(x, (|z|)(1 + r 2 ), ∀(x, z, r ) ∈ Ω × R × RN . P N z, r )| 6 γ µ2−1 (F3 ) i, j=1 (1 + |r |) |Dx a i j | + |Dz a i j | + (1 + |r |)|Dr a i j | 6 γ µ3 (|z|), (1 + |r |)−1 |Dx b| + |Dz b| + (1 + |r |)|Dr b| 6 γ µ3 (|z|)(1 + r 2 ), ∀(x, z, r ) ∈ Ω × R × R N . (F4 ) −b(x, z, r )sgn(z) 6 γ µ(1 + |r |), ∀(x, z, r ) ∈ Ω × R × R N . Here µ is a positive constant, and µi (s) (i = 1, 2, 3) are nonnegative nondecreasing functions defined on [0, ∞). Theorem 4.1 of Chapter 7 in [5] can be stated as follows: Theorem. Let ∂Ω ∈ C 2,α for some α ∈ (0, 1), and let a i j , b ∈ C 1 (Ω ×R×R N ) satisfy (F1 )–(F4 ), and ψ ∈ C 2,α (Ω ). Then the Dirichlet problem (23) has a solution ϕ ∈ C 2,α (Ω ). Now we apply the theorem to show the solvability of the regularized problem. For this, we rewrite the regularized equation as follows: −a(ϕ 0 )ϕ 00 + b(t, ϕ, ϕ 0 ) = 0,

0 < t < 1,

where ( p−2)/2

a(t, z, r ) ≡ a(r ) = (r + ε) 2

b(t, z, r ) = λ



 r2 1 + ( p − 2) 2 , r +ε

(r 2 + ε)( p−2)/2r 2 sgnε (z) − f (t). Iε (z)

Clearly, a ∈ C 1 (R). Since Iε (z), sgnε (z) ∈ C 1 (R) and f ∈ C 1 [0, 1], b ∈ C 1 ([0, 1] × R × R). Let γ (t, z, r ) ≡ γ (r ) = (r 2 + ε)( p−2)/2 , Γ (t, z, r ) ≡ Γ (r ) = ( p − 1)(r 2 + ε)( p−2)/2 . For any fixed ε ∈ (0, 1), we must verify the conditions (F1 )0 − (F4 )0 corresponding to (F1 ) − (F4 ) for N = 1 and Ω = (0, 1). Noticing p > 2, we have 1 6 1 + ( p − 2)

r2 6 p − 1, +ε

r2

and hence γ (r ) 6 a(r ) 6 Γ (r ). and thus (F1 )0 is verified. Next we verify (F2 )0 . Since λ > 0, Iε (z) > ε/2 and |sgnε (z)| 6 1, we have |b(t, z, r )| 6 λ

(r 2 + ε)( p−2)/2r 2 |sgnε (z)| + max f [0,1] Iε (z)

Z.-a. Yao, W. Zhou / Nonlinear Analysis 68 (2008) 2309–2318

2λ 2 γ γ r + ( p−2)/2 max f [0,1] ε ε   max f 2λ [0,1] 6 + ( p−2)/2  γ (1 + r 2 ). ε ε 6

Thus (F2 )0 is verified. Since λ > 0 and sgnε (z)sgn(z) > 0, we have max f −b(t, z, r )sgn(z) 6 max f 6 [0,1]

[0,1]

ε ( p−2)/2

γ (1 + |r |),

and thus (F4 )0 is verified. Finally, we verify (F3 )0 . Simple calculations show that   r r2 2ε Dr a = ( p − 2)γ 2 1 + ( p − 2) 2 + , r +ε r + ε r2 + ε Dt b = − f 0 ,   sgn0ε (z) sgnε (z)Iε0 (z) − Dz b = λγ r 2 , Iε (z) Iε2 (z)   r2 λsgnε (z) γ r 2 + ( p − 2) 2 Dr b = . Iε (z) r +ε We first estimate (1 + |r |)|Dr a|. We have |Dr a| 6 ( p + 1)( p − 2)γ

|r | , +ε

r2

and therefore (1 + |r |)|Dr a| 6 ( p + 1)( p − 2)γ

(1 + |r |)2 . r2 + ε

By the inequality (1 + α)2 6 2(1 + α 2 ) for any α ∈ R and noticing r 2 + ε > ε(1 + r 2 ), we have (1 + |r |)|Dr a| 6

2( p + 1)( p − 2) γ. ε

Next we estimate (1 + |r |)−1 |Dt b|, (1 + |r |)|Dr b| and |Dz b|, respectively. We first obtain max | f 0 | (1 + |r |)

−1

0

|Dt b| 6 |Dt b| 6 max | f | 6 [0,1]

[0,1]

ε ( p−2)/2

γ.

Since Iε (z) > ε/2, |sgnε (z)| 6 1, we have |Dr b| 6

2 pλ γ |r |, ε

and hence (1 + |r |)|Dr b| 6

2 pλ 4 pλ γ (1 + |r |)2 6 γ (1 + r 2 ). ε ε

Since Iε (z) > ε/2, |sgnε (z)| 6 1, |Iε0 (z)| 6 1, 0 6 sgn0ε (z) 6 2/ε, we obtain   0 |sgnε (z)Iε0 (z)| 8λ 2 |sgnε (z)| |Dz b| 6 λγ r + 6 2 γ r 2. 2 Iε (z) Iε (z) ε

2317

2318

Z.-a. Yao, W. Zhou / Nonlinear Analysis 68 (2008) 2309–2318

Combining the above results, we have  (1 + |r |)−1 |Dt b| + |Dz b| + (1 + |r |)|Dr b| 6 

max | f 0 | [0,1]

ε ( p−2)/2

 +

4 pλ 8λ  + 2 γ (1 + r 2 ). ε ε

Thus all conditions (F1 )0 − (F4 )0 are verified, and therefore it follows from the above theorem that for any fixed ε ∈ (0, 1), the regularized problem has a solution ϕε ∈ C 2 [0, 1]. References [1] R.P. Agarwal, H. L¨u, D. O’Regan, Existence theorems for the one-dimensional singular p-Laplacian equation with sign changing nonlinearities, Appl. Math. Comput. 1 (143) (2003) 15–38. [2] G.I. Barenblatt, M. Bertsch, A.E. Chertock, V.M. Prostokishin, Self-similar intermediate asymptotic for a degenerate parabolic filtration–absorption equation, Proc. Natl. Acad. Sci. USA 18 (2000) 9844–9848. [3] M. Bertsch, M. Ughi, Positivity properties of viscosity solutions of a degenerate parabolic equation, Nonlinear Anal. TMA 14 (1990) 571–592. [4] M. Bertsch, R. Dal Passo, M. Ughi, Discontinuous viscosity solutions of a degenerate parabolic equation, Trans. Amer. Math. Soc. 320 (1990) 779–798. [5] Y. Chen, L. Wu, Second Order Elliptic Equations and Elliptic Systems (B. Hu. Trans.), Science Press, Beijing, 1997 (Original work published 1991) (in Chinese). Translations of Mathematical Monographs, vol. 174, American Mathematical Society, Providence, RI, 1998. [6] L. Damascelli, Comparison theorems for some quasilinear degenerate elliptic operations and applications to symmetry and monotonicity results, Ann. Inst. H. Poincar´e Anal. Non Lin`eaire 4 (15) (1998) 493–516. [7] Y. Guo, W. Ge, Three positive solutions for the one-dimension p-Laplacian, J. Math. Anal. Appl. 286 (2003) 491–508. [8] 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. [9] H. L¨u, D. O’Regan, C. Zhang, Multiple positive solutions for the one-dimensional singular p-Laplacian, Appl. Math. Comput. 133 (2002) 407–422. [10] H. L¨u, D. O’Regan, R.P. Agarwal, Nonuniform nonresonance at the first eigenvalue of the one-dimensional singular p-Laplacian, Mem. Differential Equations Math. Phys. 34 (2005) 97–114. [11] D. O’Regan, Some general existence principles and results for (φ(y 0 ))0 = q f (t, y, y 0 ), 0 < t < 1, SIAM, J. Math. Anal. 24 (3) (1993) 648–668. [12] J. Wang, The existence of positive solutions for the one-dimensional p-Laplacian, Proc. Amer. Math. Soc. 125 (1997) 2275–2283. [13] Z. Wang, J. Zhang, Positive solutions for one-dimensional p-Laplacian boundary value problems with dependence on the first order derivative, J. Math. Anal. Appl. 314 (2006) 618–630. [14] D. Hai, R. Shivaji, Existence and uniqueness for a class of quasilinear elliptic boundary value problems, J. Differential Equations 193 (2003) 500–510. [15] W. Zhou, X. Wei, Positive solutions to BVPs for a singular differential equation, Nonlinear Anal. (2006), doi:10.1016/j.na.2006.06.015. [16] W.P. Ziemer, Weakly Differentiable Functions, GTM120, Springer-Verlag, New York, 1989.