Existence results of sign-changing solutions for singular one-dimensional p -Laplacian problems

Nonlinear Analysis 68 (2008) 1195–1209 www.elsevier.com/locate/na

Existence results of sign-changing solutions for singular one-dimensional p-Laplacian problems Yong-Hoon Lee ∗ , Inbo Sim Department of Mathematics, Pusan National University, Pusan 609-735, Republic of Korea Received 3 August 2006; accepted 7 December 2006

Abstract Consider singular one-dimensional p-Laplacian problems with Dirichlet boundary condition  (ϕ p (u 0 (t)))0 + h(t) f (u(t)) = 0, t ∈ (0, 1), (P) u(0) = 0 = u(1), (D) where ϕ p : R → R is defined by ϕ p (x) = |x| p−2 x, p > 1, h a nonnegative measurable function on (0, 1) which may be singular at t = 0 and/or t = 1 and f ∈ C(R, R). By applying the global bifurcation theorem and deriving the shape of the unbounded subcontinua of solutions, we obtain the existence and multiplicity results of sign-changing solutions for (P) + (D). c 2007 Elsevier Ltd. All rights reserved.

MSC: 34B15; 34A23 Keywords: Singular boundary value problem; p-Laplacian; Global bifurcation; Existence; Multiplicity

1. Introduction In this paper, we show the existence and multiplicity of sign-changing solutions for the following problem under Dirichlet boundary conditions  (ϕ p (u 0 (t)))0 + h(t) f (u(t)) = 0, t ∈ (0, 1), (P) u(0) = 0 = u(1), (D) where ϕ p : R → R is defined by ϕ p (x) = |x| p−2 x, p > 1, h a nonnegative measurable function on (0, 1) which may be singular at t = 0 and/or t = 1 and f ∈ C(R, R). Recently, p-Laplacian problems with a singular indefinite weight was studied by Lee and Sim [5] who gave global existence results with respect to given parameter λ for the following problem  (ϕ p (u 0 (t)))0 + λh(t) f (u(t)) = 0, a.e. (0, 1), (Pλ ) u(0) = 0 = u(1). (D) ∗ Corresponding author. Tel.: +82 51 510 2295; fax: +82 51 581 1458.

E-mail address: [email protected] (Y.-H. Lee). c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.12.015

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1 They mainly considered the case 0 < f 0 , lim|u|→0 ϕfp(u) (u) < ∞ with indefinite weight h ∈ L (0, 1), h ≥ 0 a.e. and R I h(s)ds > 0 for any compact subinterval I in (0, 1) and they gave global analyses not only for positive solutions but also for sign-changing solutions employing a bifurcation argument of Rabinowitz [13]. As for results of sign-changing ∗ ∗ solutions, assuming f ∞ , lim|u|→∞ ϕfp(u) (u) = 0, they proved that for each k ∈ N, there exists λ with 0 < λ ≤ µk ( p) such that the problem (Pλ ) + (D) has at least one sign-changing solution with k − 1 interior zeros for all λ > λ∗ , where µk ( p) is the k-th eigenvalue of the problem  (ϕ p (u 0 (t)))0 + µf 0 h(t)ϕ p (u(t)) = 0, a.e. (0, 1), u(0) = u(1) = 0.

Moreover, assuming f ∞ = ∞, they proved that for each k ∈ N, there exist λ∗ ≥ λ∗ > 0 such that (Pλ ) + (D) has at least one sign-changing solution with k − 1 interior zeros for λ < λ∗ and no such sign-changing solutions for λ > λ∗ . They also gave two more corollaries in detail regarding this case. In the sequel, it is natural to ask about the existence of sign-changing solutions for the problem (Pλ ) + (D) when f satisfies f 0 = 0 or f 0 = ∞. Although we can not expect bifurcation phenomena with respect to the parameter λ for both cases, we believe that the bifurcation argument is still powerful in answering these questions. Therefore, we better consider the problem (P) + (D) and make use of global bifurcation arguments in the frame of techniques nicely employed by Ma and Thompson [7]. More precisely, for the case f 0 = 0, we consider the auxiliary problem for (P) + (D) as follows  (ϕ p (u 0 (t)))0 + µh(t)ϕ p (u(t)) + h(t) f (u(t)) = 0, a.e. (0, 1), (A P1 ) u(0) = u(1) = 0. (D) Using Rabinowitz’s bifurcation argument, we establish the unbounded continua Ck in R × C 1 [0, 1] for k ∈ N. Adding the condition f ∞ = ∞ and using Picone’s type identity, we derive the shape of Ck which crosses the axis of C 1 [0, 1] and provides us with a sign-changing solution with k − 1 interior zeros. On the other hand, for the case f 0 = ∞, we cannot apply the above process directly. Instead, we first change the variable u into z which satisfies   z(t)  p = 0. (ϕ p (z 0 (t)))0 + kzk∞ h(t) f  p p−1 kzk∞ And then apply the above process to the auxiliary problem      z(t)  p  = 0, (ϕ p (z 0 (t)))0 + µϕ p (z(t)) + kzk∞ h(t) f  p p−1  kzk ∞   z(0) = z(1) = 0.

t ∈ (0, 1),

(A P2 ) (D)

With an additional condition f ∞ = 0, the bifurcation argument guarantees the existence of unbounded continua Dk in R × C 1 [0, 1] for k ∈ N. Moreover, using Picone’s type identity, we derive the shape of Dk which meets the axis of C 1 [0, 1]. The existence of positive solutions for singular one-dimensional p-Laplacian problems with Dirichlet boundary condition has been studied extensively in [1,3,14,16,17] and references therein. Proofs of results mainly make use of upper and lower solution methods, fixed point theorems and fixed point index theory on cones under more general conditions on h as follows ! ! Z 1 Z 1 Z 1 Z s 2 2 −1 −1 ϕp h(τ )dτ ds + ϕp h(τ )dτ ds < ∞. 0

s

1 2

1 2

However, the methods do not seem to be effective in managing sign-changing solutions. Up to now, the literature devoted to the existence of sign-changing solutions is not that extensive. E.g. in [10], Naito and Tanaka established the precise condition on h concerning the behavior of f 0 and f ∞ for the existence of sign-changing solutions to the problem for the case p = 2:  00 u + h(t) f (u(t)) = 0, t ∈ (0, 1), (S P) u(0) = 0 = u(1). (D)

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They mainly employed the shooting method and Sturm’s comparison theorem. Recently, in [11], they extended the above results to the p-Laplacian problem (P) + (D), using similar arguments based on the shooting method together with the qualitative theory for half-linear differential equations. They considered an indefinite weight h satisfying the following conditions h ∈ C 1 [0, 1]

h > 0.

and

Ma and Thompson [7] and Ma [6] showed the existence and multiplicity results of sign-changing solutions for (S P) + (D) by using the global continuation technique. They considered conditions on h up to: (h) h ∈ C 1 ([0, 1], [0, ∞)) and h(t0 ) > 0, for some t0 ∈ [0, 1]. Our concern in this paper is to deal with indefinite weight h singular at the boundary so that we relax Ma and Thompson’s condition on h without losing the existence and multiplicity results. For this purpose, we assume the following condition on h for the case f 0 = 0 R (H1 ) h ∈ L 1 (0, 1), h ≥ 0 a.e. with I h(s)ds > 0 for any compact subinterval I in (0, 1), and the following for the case f 0 = ∞ (H2 ) h ∈ C 1 (0, 1) ∩ L 1 (0, 1), h ≥ 0 and limt→0+ th(t) and limt→1− (1 − t)h(t) exist. For nonlinear term f , we assume the same condition as Ma and Thompson given as follows (F) f ∈ C(R, R) with s f (s) > 0 for s 6= 0. This paper is organized as follows: In Section 2, we introduce the operator equation which is equivalent to problem (P) + (D) and get the comparison theorem by Picone’s identity and obtain an easy consequence for the nonresonance case. In Section 3, we consider the case of f 0 = 0. Using the global bifurcation theorem and deriving the shape of the unbounded subcontinuum, we establish existence and multiplicity results of sign-changing solutions for (P) + (D). In Section 4, we consider the case of f 0 = ∞. Using the change of variable and the similar arguments as in Section 3, we show existence and multiplicity results of sign-changing solutions for (P) + (D). 2. Preliminaries In this section, we introduce the operator equation which is equivalent to problem (P) + (D) and get a comparison theorem by Picone’s identity, and give a simple consequence about nonexistence of solutions for the nonresonance case. Consider the auxiliary problem  (ϕ p (u 0 (t)))0 + h = 0, a.e. (0, 1), (A P) u(0) = u(1) = 0, (D) where h ∈ L 1 (0, 1). We introduce the equivalent operator form known by Man´asevich and Mawhin [8,9]. Problem (A P) + (D) is equivalently written as   Z t Z s −1 u(t) = G p (h)(t) , ϕ p a(h) + h(τ )dτ ds, 0

0

L 1 (0, 1)

where a : → R is a continuous function which sends bounded sets of L 1 into bounded sets of R and satisfying   Z 1 Z s h(τ )dτ dt = 0. (2.1) ϕ −1 a(h) + p 0

0

It is known that G p : L 1 (0, 1) → C01 [0, 1] is continuous and maps equi-integrable sets of L 1 (0, 1) into relatively compact sets of C01 [0, 1]. One may refer to Man´asevich and Mawhin [8,9] and Garc´ıaHuidobro–Man´asevich–Ward [2] for more details. Next, consider the eigenvalue problem  (ϕ p (u 0 (t)))0 + λh(t)ϕ p (u(t)) = 0, a.e. (0, 1), (E) u(0) = 0 = u(1). (D)

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The properties of eigenvalues and corresponding eigenfunctions for (E) + (D) are given in Proposition 2.6 [5] as follows: (i) the set of all eigenvalues of (E) + (D) is a countable set {λk ( p)|k ∈ N} satisfying 0 < λ1 ( p) < λ2 ( p) < · · · < λk ( p) < · · · → ∞, p p (ii) for each k, Ker(I − Tλk ( p) ), where Tλ (u) = (G p ◦ H1 )(λ, u), with the Nemitskii operator H1 : R × C01 [0, 1] → L 1 (0, 1) given by H1 (λ, u)(t) = λh(t)ϕ p (u(t)) is a subspace of C 1 [0, 1] and its dimension is 1, (iii) let u k be a corresponding eigenfunction to λk ( p), then the number of interior zeros of u k is k − 1. p

Obviously, Tλ is completely continuous and thus, we have Lemma 2.1 (Lemma 2.8, [5]). For fixed p > 1, all r > 0 and each k ∈ N, we have  1, if 0 < µ < λ1 ( p) d L S (I − Tµp , Br (0), 0) = (−1)k , if µ ∈ (λk ( p), λk+1 ( p)). The following lemma is known as the generalized Picone identity. Let us consider the following two operators: l p [y] = (ϕ p (y 0 ))0 + b1 (t)ϕ p (y),

(2.2)

L p [z] = (ϕ p (z 0 ))0 + b2 (t)ϕ p (z).

(2.3)

Lemma 2.2 (pp. 382, [4]). Let b1 , b2 ∈ L 1 (I ), I an interval and if y and z are functions such that y, z, ϕ p (y 0 ), and ϕ p (z 0 ) are differentiable on I and z(t) 6= 0 for t ∈ I . Then we have the following identity: d dt



 |y| p ϕ p (z 0 ) − yϕ p (y 0 ) ϕ p (z)

= (b1 − b2 )|y| p 0 p   0  yz z 0 p 0 − |y | + ( p − 1) − pϕ p (y)y ϕ p z z − yl p (y) +

|y| p L p (z). ϕ p (z)

(2.4) (2.5) (2.6) (2.7)

Remark 2.3. By Young’s inequality, we get 0 p  0 yz z 0 p 0 − pϕ p (y)y ϕ p |y | + ( p − 1) ≥ 0, z z 0 p 0 p and the equality holds if and only if sgn y 0 = sgn z 0 and yy = zz . With Lemma 2.2, Remark 2.3 and a slight modification of the proof of Lemma 4 [17], we have the following. Lemma 2.4. Let b2 (t) ≥ b1 (t) > 0 a.e. and bi (t) ∈ L 1 (0, 1), i = 1, 2. Also let u 1 , u 2 be nontrivial solutions of l p (u 1 ) = 0, L p (u 2 ) = 0, respectively. If (c, d) ⊂ (0, 1), u 1 (c) = u 1 (d) = 0 and u 1 (t) 6= 0 on (c, d), then either there exists τ ∈ (c, d) such that u 2 (τ ) = 0 or b2 = b1 and u 2 (t) = cu 1 (t) for some constant c 6= 0. As an immediate consequence, we obtain the following comparison. Lemma 2.5. Let b2 (t) > b1 (t) > 0 a.e. and bi (t) ∈ L 1 (0, 1), i = 1, 2. Also let u 1 , u 2 be nontrivial solutions of l p (u 1 ) = 0, L p (u 2 ) = 0, respectively. If u 1 has k-many interior zeros, then u 2 has at least k + 1-many interior zeros. Using this comparison theorem, we have a non-existence result when f satisfies a nonresonance condition.

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Theorem 2.6. Assume that there exists an integer k ∈ N such that λk ( p) <

f (s) < λk+1 ( p) ϕ p (s)

for s 6= 0.

(2.8)

Then problem (P) + (D) has no non-trivial solution. Proof. Assume to the contrary that (P) + (D) has a solution u of C 1 class. We see that u satisfies (ϕ p (u 0 ))0 + b(t)ϕ p (u) = 0,

a.e. (0, 1),

where b(t) = h(t)

f (u(t)) . ϕ p (u(t))

Note that f 0 ≤ λk+1 ( p) < ∞ by (2.8) and hence

f (s) ϕ p (s)

can be regarded as a continuous function on R. Thus we

L 1 (0, 1).

get b ∈ Also, notice that a nontrivial solution of (P) + (D) has a finite number of zeros (see, Lemma 2.11 in [5]). From (2.8) and the above fact, λk ( p)h(t) < b(t) < λk+1 ( p)h(t)

a.e. (0, 1).

We know that the eigenfunction φk corresponding to λk has exactly k −1 zeros in (0, 1). By applying Lemma 2.5 to φk and u, we see that u has at least k zeros in (0, 1). Now, by applying Lemma 2.5 again to u and φk+1 , we find that φk+1 has at least k + 1 zeros in (0, 1). This contradiction implies that problem (P) + (D) has no non-trivial solution.  3. The case of f0 = 0 In this section, we show the existence and multiplicity results of sign-changing solutions for (P) + (D) when f 0 = 0 with the help of the global bifurcation theorem [15] for nonlinear operators. The main conditions we consider in this section are (H1 ), (F) and f 0 = 0. First, let us consider the auxiliary problem  (ϕ p (u 0 (t)))0 + µh(t)ϕ p (u(t)) + h(t) f (u(t)) = 0, a.e. (0, 1), (A P1 ) u(0) = u(1) = 0. (D) Note that a solution of (A P1 )+(D) of the form (0, u) corresponds to a solution u of (P)+(D). To obtain the existence of bifurcation branches for problem (A P1 ) + (D), we will make use of the following well-known theorem. Proposition 3.1 ([15]). Let E be a real Banach space and F : R × E → E completely continuous such that F(µ, 0) = 0 for all µ ∈ R. Suppose that there exist constants ρ, η ∈ R with ρ < η such that (ρ, 0) and (η, 0) are not bifurcation points for the equation u − F(µ, u) = 0.

(3.1)

Furthermore, assume that d L S (I − F(ρ, ·), Br (0), 0) 6= d L S (I − F(η, ·), Br (0), 0), where Br (0) = {u ∈ E : kuk E < r } is an isolating neighborhood of the trivial solution for both constants ρ and η. Let S = {(λ, u) : (λ, u) is a solution of (3.1) with u 6= 0} ∪ ([ρ, η] × {0}), and let C be the component of S containing [ρ, η] × {0}. Then either (i) C is unbounded in R × E, or (ii) C ∩ [(R \ [ρ, η]) × {0}] 6= ∅.

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Equation (A P1 ) + (D) can be rewritten in the operator form   u = G p ◦ (H1 + H2 ) (µ, u) , F(µ, u), where Nemitskii operators Hi : R × C01 [0, 1] → L 1 (0, 1) are defined by H1 (µ, u)(t) , µh(t)ϕ p (u(t))

and

H2 (µ, u)(t) , h(t) f (u(t)),

respectively. Then Hi , i = 1, 2 are continuous operators which send bounded sets of R × C01 [0, 1] into equiintegrable sets of L 1 (0, 1). Combining the properties of G p , we easily see that F is completely continuous in R × C01 [0, 1] → C01 [0, 1] and obviously, F(µ, 0) = 0 for all µ ∈ R. For the proof of the following theorem, one may refer to Theorem 2.10 in [5]. It is worth noticing that the conclusion of Theorem 2.10 [5] is still true without giving the oddness of f . Theorem 3.2. Assume (H1 ) and (F). Also assume f 0 = 0. Then (λk ( p), 0) is a bifurcation point for (A P1 ) + (D) and the associated global bifurcation branch Ck satisfies the alternatives in Proposition 3.1. To get unbounded continua from the alternatives, we shall follow the argument of Rabinowitz [12]. We first need the following lemma. Lemma 3.3. If (µ, u) is a solution of (A P1 ) + (D) and u has a double zero (i.e., u(t) = 0 = u 0 (t) for some t ∈ [0, 1]), then u ≡ 0. Proof. Let u be a solution of (A P1 ) + (D) and t ∗ ∈ [0, 1] be a double zero. Then ! Z t∗ Z t∗ −1 u(t) = ϕp −µh(τ )ϕ p (u(τ )) − h(τ ) f (u(τ ))dτ ds. s

t

By f 0 = 0, we may choose Cu > 0 such that | f (v)| ≤ Cu ϕ p (|v|), for all |v| ∈ [0, kuk∞ + 1]. For t ≤ t ∗ , we get ! Z t∗ Z t∗ |u(t)| ≤ ϕ −1 |µ|h(τ )ϕ p (|u(τ )|) + |h(τ ) f (u(τ ))|dτ ds p s

t

t∗

Z ≤ t

≤

ϕ −1 p

! |µ|h(τ )ϕ p (|u(τ )|) + Cu h(τ )ϕ p (|u(τ )|)dτ ds

s t∗

Z

ϕ −1 p

t∗

Z

! |µ|h(τ )ϕ p (|u(τ )|) + Cu h(τ )ϕ p (|u(τ )|)dτ .

t

This implies ϕ p (|u(t)|) ≤

t∗

Z

[|µ|h(τ ) + Cu h(τ )]ϕ p (|u(τ )|)dτ.

t

By Gronwall’s inequality, we get u ≡ 0 on [0, t ∗ ]. Similarly, we can get u ≡ 0 on [t ∗ , 1] and the proof is complete.  Let us denote Nk+ = {u ∈ C01 [0, 1]|u has exactly k − 1 simple zeros in (0, 1), u > 0 near 0 and all zeros of u in [0, 1] are simple} and let Nk− = −Nk+ and Nk = Nk− ∪ Nk+ . They are disjoint and open in C01 [0, 1]. Let φk be the corresponding normalized eigenfunction to eigenvalue λk for (E) + (D) and Tk+ = R × Nk+ , Tk− = R × Nk− and Tk = R × Nk . From Lemma 3.3, any solutions (λ, u) of (A P1 ) + (D) with u ∈ ∂ Nk+ has u ≡ 0 and thus, Ckν preserves the following property; if (λn , u n ) → (λ, u) 6≡ (0, 0), (λn , u n ) a solution of (A P1 ) + (D) and u n ∈ Nk+ , then u ∈ Nk+ . Thus, we have Ck+ ∩ Ck− = ∅. Using the same argument as in [12] together with the simplicity of λk , (G p ◦ H2 )(λ, u) = o(kuk) and the oddness of (G p ◦ H1 )(λ, u) in u, it is possible to show that there are two maximal

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connected subsets of Ck bifurcating from (λk , 0) in the directions φk and −φk (see; Lemma 1.24 and Theorem 1.25 in [12]). This implies that Ck can be decomposed into Ck+ ∪ Ck− and it does not satisfy the latter possibility given in Theorem 3.2. Since both Ck+ and Ck− preserve the nodal properties of φk and −φk , respectively, neither Ck+ nor Ck− contains a pair of (λ, u), (λ, −u). Following the proof of Theorem 1.27 in [12], Ck+ and Ck− are unbounded. Therefore, we have the following theorem. Theorem 3.4. For each k ≥ 1 and ν ∈ {+, −}, there exist two unbounded continua Ckν of nontrivial solutions for (A P1 ) + (D) in Tkν ∪ {(λk , 0)}. Now, let us derive the shape of Ckν . First, we have a block in the µ-direction. One may refer to Section 3 in [5] for more detailed computations. Lemma 3.5. For each k ≥ 1, let u νk be a solution in Ckν of (A P1 ) + (D) at µ. Then µ ≤ λk ( p). + Proof. Case I: Let u + 1 be a positive solution in C1 of (A P1 ) + (D). Then 0

0

+ + + + 0 0 0 = ϕ p (u + 1 (t)) + µh(t)ϕ p (u 1 (t)) + h(t) f (u 1 (t)) ≥ ϕ p (u 1 (t)) + µh(t)ϕ p (u 1 (t)).

Let φ1+ be a positive eigenfunction corresponding to the first eigenvalue λ1 ( p) of (E) + (D). Taking z = u + 1 , b2 (t) = µh(t) and y = φ1+ , b1 (t) = λ1 ( p)h(t) in Lemma 2.2 and integrating (2.4)–(2.7), we have Z 1 p (λ1 ( p)h(t) − µh(t))|u + 1 (t)| dt ≥ 0. 0

p

|y| − Hence λ1 ( p) − µ ≥ 0. Similarly, taking z = u − 1 and y = φ1 and using the negativity in ϕ p (z) L p (z), we have the same result. + + Case II: Let u + k be a solution in Ck of (A P1 ) + (D) and φk be a positive eigenfunction corresponding to the k-th + + ∗ ∗ eigenvalue λk ( p) of (E) + (D). Let t1 and t1 be the first interior zero of φk+ and u + k with φk > 0 in (0, t1 ) and u k > 0 + + ∗ in (0, t1 ) and let tk−1 and tk−1 be the last interior zero of φk and u k , respectively. In the case k = 2, let us suppose ∗ t1 ≤ t1 first. Then we get   Z t1  + p + 0 ( p−1) 0 |u 2 | φ2 dt = 0, ( p−1)  0  φ+ 2

and t1

Z 0

n o0 + 0 ( p−1) − u+ u dt = 0, 2 2

where x ( p−1) = ϕ p (x). Using the same argument as in Case I, we get µ ≤ λ2 ( p). Second, suppose t1∗ ≤ t1 . Then we get   Z 1  + p + 0 ( p−1) 0 |u 2 | φ2 dt = 0, ( p−1)  t1  φ+ 2

and Z

1 t1

n o0 + 0 ( p−1) − u+ u dt = 0, 2 2

so we similarly get µ ≤ λ2 ( p).

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Using the same argument as in the case k = 1 for u − 2 , we obtain µ ≤ λ2 ( p). ∗ In case of k ≥ 3, if t1 ≤ t1∗ or tk−1 ≤ tk−1 , then we obtain µ ≤ µk ( p) by the same process as in case of k = 2. If ∗ ∗ ∗ ) for some i, 1 < i < k, and we have t1 < t1 and tk−1 < tk−1 , then there exists an interval (ti , ti+1 ) ⊂ (ti∗ , ti+1 )0 Z ti+1 ( ν p ν 0 ( p−1) |u k | φk ( p−1) − u k u νk 0 dt = 0. ν( p−1) ti φk

Either for u k > 0 in (ti , ti+1 ) or for u k < 0 in (ti , ti+1 ), we have −u k [(ϕ p (u νk 0 (t)))0 + µh(t)ϕ p (u νk (t))] ≥ 0. 

Thus following the argument in the case k = 2, we get the conclusion.

To get an a priori estimate for parameter µ, we need to add conditions on f . Lemma 3.6. Assume (H1 ) and (F). Moreover, assume f ∞ = ∞. Let J be a compact interval in (−∞, λk ( p)]. Then there exists b J > 0 such that for all µ ∈ J and all possible solutions u of (A P1 ) + (D) at µ, one has kuk ≤ b J . Proof. Suppose on the contrary that there exists a sequence {(µn , u n )} of solutions for (A P1 ) + (D) with µn ∈ J, u n ∈ Ck , and ku n k → ∞ as n → ∞. Let 0 = z 0(n) , z 1(n) , . . . , z (k−1)(n) , z k(n) = 1 denote the zeros of u n in [0, 1]. Then at least one subinterval (z j (n) , z ( j+1)(n) ) , I j (n) is of length at least k1 , for some j ∈ {0, 1, 2, . . . , k − 1}. In fact, {max I j (n) |u n |}∞ n=1 is an unbounded sequence (see, the proof of Lemma 4.5 in [5]). As subsequences if necessary, put limn→∞ z j (n) = z j (∞) , limn→∞ z ( j+1)(n) = z ( j+1)(∞) and limn→∞ δn = δ, where u n (δn ) = max I j (n) u n (t), for u n (t) > 0 or u n (δn ) = min I j (n) u n (t), for u n (t) < 0. Then there exists j ∗ ∈ {0, 1, 2, . . . , k − 1} such that z j ∗ (∞) < z ( j ∗ +1)(∞) . Without loss of generality, we may assume z j ∗ (∞) < δ < z ( j ∗ +1)(∞) (cases of z j ∗ (∞) = δ or z ( j ∗ +1)(∞) = δ can be considered similarly as in the proof of Lemma 3.8 in [5]). In the rest of the proof, ku n k∞ denotes ku n k∞ on I j ∗ (n) . By the well-known fact (Lemma 1 in [14]), for any 0 <  < |u n (t)| ≥ m 2 ku n k∞ n where m = min k(δ−z2 ∗

j (∞)

δ−z j ∗ (∞) , 4

we get

for all t ∈ [z j ∗ (n) + , z ( j ∗ +1)(n) − ] , J j ∗ (n) , o 2 , ) k(z ∗ −δ) . By the condition f ∞ = ∞, we may choose R1 > 0 such that ( j +1)(∞)

f (u) ≥ ηu p−1 for u ≥ R1 or f (u) ≤ ηu p−1 for u ≤ −R1 , where η > 0 can be given as ! Z z j ∗ (∞) +2 Z z j ∗ (∞) +2 −1 ( p−1) 2( p−1) −1 ϕ p ((µn + η)m  ) ϕp h(τ )dτ > 1. z j ∗ (∞) +

Since ku n k∞ → ∞, ku n k∞ >

R1 m 2

s

for sufficiently large n. Thus

ν(−1) j∗ u n (t) ≥ m 2 ku n k∞ > R1 , for t ∈ J j ∗ (n) and we get, for u n (t) > 0, Z δn  Z δn −1 ϕp µn h(τ )ϕ p (u n (τ )) + h(τ ) f (u n (τ ))dτ ds u n (δn ) = z j ∗ (n)

Z

s

z j ∗ (∞) +2

≥ z j ∗ (∞) +

≥

≥

ϕ −1 p (µn

ϕ −1 p

+ η)

ϕ −1 p ((µn

(µn + η)

z j ∗ (∞) +

+ η)m

! h(τ )ϕ p (u n (τ ))dτ ds

s z j ∗ (∞) +2

Z

z j ∗ (∞) +2

Z

( p−1) 2( p−1)



)

z j ∗ (∞) +2

Z

ϕ −1 p

! p−1 u n h(τ )dτ

ds

s

Z

z j ∗ (∞) +2 z j ∗ (∞) +

ϕ −1 p

z j ∗ (∞) +2

Z s

! h(τ )dτ ds · ku n k∞ .

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Y.-H. Lee, I. Sim / Nonlinear Analysis 68 (2008) 1195–1209

This is a contradiction to the choice of η. Similarly, for u n (t) < 0, we get u n (δn ) =

δn

Z

z j ∗ (n)

ϕ −1 p

z j ∗ (∞) +2

Z ≤

z j ∗ (∞) +

≤

≤

ϕ −1 p (µn

 µn h(τ )ϕ p (u n (τ )) + h(τ ) f (u n (τ ))dτ ds

s

ϕ −1 p

+ η)

ϕ −1 p ((µn

δn

Z

(µn + η)

z j ∗ (∞) +

+ η)m

! h(τ )ϕ p (u n (τ ))dτ ds

s z j ∗ (∞) +2

Z

z j ∗ (∞) +2

Z

( p−1) 2( p−1)



)

z j ∗ (∞) +2

Z

ϕ −1 p

! p−1 u n h(τ )dτ

ds

s

Z

z j ∗ (∞) +2 z j ∗ (∞) +

ϕ −1 p

z j ∗ (∞) +2

Z

! h(τ )dτ ds · u n (δn ).

s

This is also a contradiction to the choice of η and completes the proof.



Now we have the main theorem in this section as follows. Theorem 3.7. Assume (H1 ) and (F). Also assume f 0 = 0 and f ∞ = ∞. Then for each k ∈ N, problem (P) + (D) − + − has two solutions u + k and u k such that u k has exactly k − 1 zeros and is positive near t = 0, and u k has exactly k − 1 zeros and is negative near t = 0. Proof. Using Theorem 3.4 and Lemmas 3.5 and 3.6, we get the conclusion.



4. The case of f0 = ∞ In this section, we establish an existence and multiplicity result of sign-changing solutions for (P) + (D) when f 0 = ∞ with the help of the global bifurcation theorem and the change of variable. To make a connection with the global bifurcation theorem, we need to assume f ∞ = 0 here. Thus, without further mention, we will assume conditions (H2 ), (F), f 0 = ∞ and f ∞ = 0 throughout this section. For u ∈ C[0, 1] and u 6≡ 0, let us change the variable by z :=

u p . kuk∞

(4.1)

Then problem (P) + (D) is transformed to problem    





z(t)  p = 0, t ∈ (0, 1), (ϕ p (z 0 (t)))0 + kzk∞ h(t) f  p p−1  kzk∞   z(0) = z(1) = 0.

(4.2)

It is clear that z is a nontrivial solution of (4.2) if and only if u :=

z p p−1 kzk∞

is a nontrivial solution of (P) + (D). As in Section 3, let us consider the auxiliary problem      z(t)   p (ϕ p (z 0 (t)))0 + µϕ p (z(t)) + kzk∞ h(t) f  = 0, t ∈ (0, 1), (A P2 ) p p−1  kzk∞   z(0) = z(1) = 0. (D)

(4.3)

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Y.-H. Lee, I. Sim / Nonlinear Analysis 68 (2008) 1195–1209

Notice that we add µϕ p (z(t)) not µh(t)ϕ p (z(t)) in (4.2). Define F : C01 [0, 1] → L 1 (0, 1) by  0, z = 0,      z(t)  F(z)(t) := p , z 6= 0. kzk h(t) f   p   ∞ p−1 kzk∞

(4.4)

Then (A P2 ) + (D) can be rewritten in the operator form   z = G p ◦ (H3 + H4 ) (µ, z) , G(µ, z), where the Nemitskii operators Hi : R × C01 [0, 1] → L 1 (0, 1) are H3 (µ, z)(t) , µϕ p (z(t))

and

H4 (µ, z)(t) , F(z)(t),

respectively. Lemma 4.1. Assume f ∞ = 0. Then we have kF(z)k L 1 (0,1) = o(kzk p−1 ),

as kzk → 0.

Proof. First, note that kzk → 0 implies kzk∞ → 0. Define the nondecreasing function f ∗ : [0, ∞) → [0, ∞) by f ∗ (s) = max | f (r )|. −s≤r ≤s

(4.5)

Then it follows from condition f ∞ = 0 that for given  > 0, there exists N1 > 0 such that for |r | > N1 , we have f (r )  ϕ (r ) ≤ 2 . p Thus for s ≥ |r | > N1 , we get f (r ) f (r )  ≤ ϕ (s) ϕ (r ) ≤ 2 p p and this implies f (r )  ≤ . max s≥|r |>N1 ϕ p (s) 2 On other hand, there exists N2 > N1 > 0 such that for s ≥ N2 , f (r )  ≤ . max −N1 ≤r ≤N1 ϕ p (s) 2 Therefore, we have f (r )  ≤ max −N1 ≤r ≤N1 ϕ p (s) 2 and this implies that for s ≥ N2 , we have ∗ f (s) = max f (r ) + max f (r ) ≤ . ϕ (s) |r |≤N1 ϕ (s) |r |>N1 ϕ (s) p p p Thus, we obtain lim

s→∞

f ∗ (s) = 0. s p−1

(4.6)

Y.-H. Lee, I. Sim / Nonlinear Analysis 68 (2008) 1195–1209

1205

This together with (F) implies ! ! p p ∗ |z(t)| z(t) h(t)kzk∞ f h(t)kzk∞ f p p p−1 p−1 |F(z)(t)| kzk∞ kzk∞ = ≤ p−1 p−1 p−1 kzk∞ kzk∞ kzk∞ ! ! p

≤

p

kzk∞

h(t)kzk∞ f ∗

p p−1 kzk∞

p−1

h(t)kzk∞ f ∗

1 1 p−1

kzk∞ p−1 kzk∞

=

kzk∞

! 1

f∗ = h(t)

1 p−1

kzk∞

! p−1 .

(4.7)

1 1 p−1

kzk∞

Combining (4.6) with (4.7), we get |F(z)(t)| p−1

→ 0,

as kzk → 0.

kzk∞

Therefore, we conclude kF(z)k L 1 (0,1) = o(kzk p−1 ),

as kzk → 0.



Lemma 4.2. Each Hi , i = 3, 4 is a continuous operator which sends bounded sets of R×C01 [0, 1] into equi-integrable sets of L 1 (0, 1). Proof. Continuity of H3 is obvious. If z n → z in C01 [0, 1], then it is easy to check H4 (z n ) → H4 (z), when z 6≡ 0. Let z ≡ 0. Then by (4.7), we have ! 1

f∗ p−1

|F(z)(t)| ≤ kzk∞ h(t)

1 p−1

kzk∞

! p−1

(4.8)

1 1 p−1

kzk∞

and this inequality guarantees the conclusion. Once again, (4.8) shows that H4 sends bounded sets of R × C01 [0, 1] into equi-integrable sets of L 1 (0, 1).  We conclude by standard arguments that G is completely continuous in R × C01 [0, 1] → C01 [0, 1] and obviously G(µ, 0) = 0, for all µ ∈ R. Theorem 4.3. (λk ( p), 0), where λk ( p) is an eigenvalue for (E) + (D) with h(t) ≡ 1, is a bifurcation point for (A P2 ) + (D) and the associated global bifurcation branch Dk satisfies the alternatives in Proposition 3.1. We notice that due to the form of auxiliary problem (A P2 ), the indefinite weight h in problem (E) + (D), in what follows, is identically 1 without further mention. Proof of Theorem 4.3. Let p > 1 be given. Take ρ = λk ( p) − δk and η = λk ( p) + δk with sufficiently small δk > 0 so that ρ and η are not eigenvalues of (E) + (D). We shall prove that 0 is an isolated solution of (A P2 ) + (D) with µ = ρ, η and for sufficiently small r > 0, d L S (I − G(ρ, ·), Br (0), 0) = d L S (I − (G p ◦ H3 )(ρ, ·), Br (0), 0),

(4.9)

d L S (I − G(η, ·), Br (0), 0) = d L S (I − (G p ◦ H3 )(η, ·), Br (0), 0).

(4.10)

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Y.-H. Lee, I. Sim / Nonlinear Analysis 68 (2008) 1195–1209

To show this, especially (4.9) here, we only need to show that there exists r > 0, such that, for all τ ∈ [0, 1], the following equation z = J (τ, z) , τ (G p ◦ H3 )(ρ, z) + (1 − τ )G(ρ, z)

(4.11)

has no nontrivial solution in Br (0). Indeed, suppose on the contrary that there exist sequences {z n } ⊂ C01 [0, 1] and {τn } ⊂ [0, 1] such that kz n k → 0 with z n = J (τn , z n ). Then   Z s Z t ρϕ (z (τ ))dτ ds z n (t) = τn ϕ −1 a + p n n p 0 0      Z s Z t z (τ ) n p ρϕ p (z n (τ )) + kz n k∞ h(τ ) f  bn +  dτ  ds, ϕ −1 + (1 − τn ) p p p−1 0 0 kz n k∞ !! p

where an = a(ρϕ p (z n )), bn = a ρϕ p (z n ) + kz n k∞ h f

and function a : L 1 (0, 1) → R is given in

zn

p p−1 kz n k∞

Section 2. Assume τn → τ0 ∈ [0, 1] and let vn , kzz nn k , then we get   Z s Z t a ˆ + vn (t) = τn ϕ −1 ρϕ (v (τ ))dτ ds n p n p 0

0

 + (1 − τn )

t

Z 0

where aˆ n = aˆ n =

an kz n k p−1

and bˆn =

 

ˆ ϕ −1 p bn  bn . kz n k p−1

 s

Z + 0

z n (τ )

f p  p−1  kz k p n ∞ ρϕ p (vn (τ )) + kz n k∞ h(τ )  kz n k p−1 

!



     dτ  ds,    

Since function a is homogeneous,

 
a(ρϕ p (z n )) ϕ p (z n ) an = = a ρ = a ρϕ p (vn ) p−1 p−1 p−1 kz n k kz n k kz n k

(4.12)

as well as 

zn

!

f p  p−1  kz k p n ∞ bˆn = a  ρϕ p (vn ) + kz n k∞ h kz k p−1 n 

  .  

(4.13)

Therefore, we have vn0 (t)

=

τn ϕ −1 p

  Z t aˆ n + ρϕ p (vn (s))ds 0

  

ˆ + (1 − τn )ϕ −1 p bn 

 Z t  p ρϕ p (vn (s)) + kz n k∞ + h(s)  0 

f

z n (s)

p p−1

kz n k∞ kz n k p−1

!



     ds  .    

Since all {τn }, {aˆ n }, {bˆn } are bounded, {vn }, {z n } are uniformly bounded and we conclude, from Lemmas 4.1 and 4.2, that {vn0 } is also uniformly bounded. Thus by the Arzela–Ascoli Theorem, {vn } has a uniformly convergent subsequence in C[0, 1]. Let vn → v, then (4.12) and (4.13) imply
lim aˆ n = a ρϕ p (v) = lim bˆn . n→∞

n→∞

Y.-H. Lee, I. Sim / Nonlinear Analysis 68 (2008) 1195–1209

1207

By using the Lebesgue Dominated Convergence Theorem,   Z s Z t
−1 ρϕ p (v(τ ))dτ ds. ϕ p a ρϕ p (v) + v(t) = 0

0

This implies that ρ is an eigenvalue of (E) + (D) and this contradiction shows that there exists r > 0 such that (4.11) has only the trivial solution in Br (0), for all τ ∈ [0, 1]. Thus d L S (I − J (τ, ·), Br (0), 0) is well-defined for all τ ∈ [0, 1] and by the property of homotopy invariance, we get d L S (I − G p ◦ H3 (ρ, ·), Br (0), 0) = d L S (I − J (1, ·), Br (0), 0) = d L S (I − J (0, ·), Br (0), 0) = d L S (I − G(ρ, ·), Br (0), 0).

(4.14)

Furthermore, we know by Lemma 2.1 that d L S (I − G p ◦ H3 (ρ, ·), Br (0), 0) = (−1)k−1 ,

for ρ = λk ( p) − δk .

A similar equality to (4.10) holds for µ = η with the same ball Br (0). Since η = λk ( p) + δk , again by Lemma 2.1, we get d L S (I − G p ◦ H3 (η, ·), Br (0), 0) = (−1)k . Therefore d L S (I − G(ρ, ·), Br (0), 0) 6= d L S (I − G(η, ·), Br (0), 0) 

and the theorem is a consequence of Proposition 3.1.

To get the unboundedness of Dk , we need the following lemma. Lemma 4.4. If (µ, z) is a solution of (A P2 ) + (D) and z has a double zero, then z ≡ 0. Proof. Suppose that z ∈ C01 [0, 1] is a non-trivial solution of (A P2 ) + (D) and has a double zero at t0 ∈ [0, 1]. Define an energy function E : C01 [0, 1] → C[0, 1] given by   Z z(t) p2 µ σ  p−1 0 p−1 |z (t)| p + |z(t)| p + kzk∞ h(t) f dσ. E[z](t) , p p p p−1 0 kzk∞ From condition (H2 ) and z ∈ C01 [0, 1], z(t)

Z h(t) 0





σ

f

R z(t)

p

0

σ

f

!

p p−1

dσ

kzk∞

 dσ = th(t)

t

p−1 kzk∞

R z(t) 0

f

σ

p p−1

! dσ

kzk∞

= (1 − t)h(t)

1−t

is well-defined !and continuous at t = 0 and t = 1. Thus E is well-defined. By (F), it is easy to see R z(t) σ f dσ > 0 for all z(t) 6= 0. Note that E[z](t) = 0 in a subinterval I in [0, 1] implies z(t) = 0 in p 0 p−1

kzk∞

I . For t0 < t, we have p2 p−1

d E[z](t) = kzk∞ h 0 (t) dt

Z 0

z(t)

 f

σ p p−1

kzk∞



0  dσ ≤ [h (t)]+ E[z](t), h(t)

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Y.-H. Lee, I. Sim / Nonlinear Analysis 68 (2008) 1195–1209

where [s]+ = max{s, 0}. Using Gronwall’s inequality in differential form, we obtain Z t 0  [h (t)]+ E[z](t) ≤ E[z](t0 ) exp dt , for t0 < t. h(t) t0 In a similar fashion, for t < t0 , we get Z t0 0  [h (t)]− E[z](t) ≤ E[z](t0 ) exp dt , h(t) t

for t < t0 ,

(4.15)

(4.16)

where [s]− = max{−s, 0}. From (4.15) and (4.16), we conclude z ≡ 0 and this contradiction completes the proof.  Using the simplicity of eigenvalues λk and Lemma 4.1, we have the following multiplicity result. Theorem 4.5. For each integer k ≥ 1 and ν ∈ {+, −}, there exist two unbounded continua Dkν of nontrivial solutions for (A P2 ) + (D) in Tkν ∪ {(λk , 0)}. We now derive the shape of Dkν . First, using a similar argument as in the proof of Lemma 3.5, we have the following lemma. Lemma 4.6. For each integer k ≥ 1, let z kν be a solution in Dkν of (A P2 ) + (D) at µ. Then µ ≤ λk ( p). An a priori estimate of solutions is essential and guaranteed as follows. Lemma 4.7. Let K be a compact interval in (−∞, λk ( p)]. Then there exists c K > 0 such that for all µ ∈ K and all possible solutions z of (A P2 ) + (D) at µ, on has kzk ≤ c K . Proof. Suppose on the contrary that there exists a sequence {(µn , z n )} of solutions for (A P2 )+(D) with µn ∈ K , z n ∈ Dkν and kz n k → ∞ as n → ∞. Let 0 = w0(n) , w1(n) , . . . , w(k−1)(n) , wk(n) = 1 denote the zeros of z n in [0, 1]. Let limn→∞ w j (n) = w j (∞) , then there exists j ? ∈ {0, 1, 2, . . . , k − 1} such that w j ? (∞) < w( j ? +1)(∞) .

(4.17)

The concavity of ν(−1)l z n implies the property (b) of Lemma 1 in [14], thus we obtain for j ? , o n  ? min ν(−1) j z n (t)|t ∈ L j ∗ (n) ≥ γ max |z n (t)||t ∈ [w j ? (∞) , w( j ? +1)(∞) ] ,

(4.18)

where 

L

j ? (n)

:= w

j ? (n)

 w( j ? +1)(n) − w j ? (n) w( j ? +1)(n) − w j ? (n) ? + , w( j +1)(n) − , 4 4

(4.19)

for some γ > 0. Combining (4.17) with (4.18), it follows that there exists a closed interval K j ? (∞) ⊂ (w j ? (∞) , w( j ? +1)(∞) ) with positive length such that ?

ν(−1) j z n (t) → ∞,

as n → ∞

(4.20)

uniformly for t ∈ K j ? (∞) . Let us rewrite (A P2 ) as (ϕ p (z n0 (t)))0 + g(µn , z n )(t)ϕ p (z n (t)) = 0,

t ∈ K j ? (∞) ,

(4.21)

where p kz n k∞ h(t) f

g(µn , z n )(t) = µn +

z n (t)

p p−1

kz n k∞

ϕ p (z n (t))

! .

(4.22)

Y.-H. Lee, I. Sim / Nonlinear Analysis 68 (2008) 1195–1209

1209

From (H2 ), (F), (4.20) and (4.22) and f 0 = ∞ with the fact ! ! p

F(z n (t)) = ϕ p (z n (t))

kz n k∞ h(t) f

z n (t)

p p−1 kz n k∞

ϕ p (z n (t))

h(t) f

z n (t)

p p−1

kz n k∞

= ϕp

z n (t)

! ,

p p−1

kz n k∞

we have lim g(µn , z n )(t) = ∞,

n→∞

uniformly for t ∈ K j ? (∞) .

(4.23)

Let us consider  (ϕ p (v 0 (t)))0 + β1 ϕ p (v(t)) = 0, t ∈ K j ? (∞) , v|∂ K j ? (∞) = 0,

(4.24)

where β1 and v are the first eigenvalue and corresponding eigenfunction of (4.24), respectively. By the comparison theorem (Lemma 2.5) to v and z n , z n must change sign on K j ? (∞) , for all n sufficiently large. This contradicts (4.20) and the proof is done.  Now we have the main theorem in this section as follows. Theorem 4.8. Assume (H2 ) and (F). Also assume f 0 = ∞ and f ∞ = 0. Then for each k ∈ N, problem (P) + (D) − + − has two solutions u + k and u k such that u k has exactly k − 1 zeros and is positive near t = 0, and u k has exactly k − 1 zeros and is negative near t = 0. Proof. Using Theorem 4.5, Lemmas 4.6 and 4.7, we get the conclusion.



Acknowledgement This work was supported by the Korea Research Foundation Grant funded by the Korean Government (MOEHRD)” (KRF-2005-070-C00010). References [1] R.P. Agarwal, H. L¨u, D. O’Regan, Eigenvalues and the one-dimensional p-Laplacian, J. Math. Anal. Appl. 266 (2002) 383–400. [2] M. Garc´ıa-Huidobro, R. Man´asevich, J.R. Ward, A homotopy along p for systems with a vector p-Laplace operator, Adv. Differential Equations 8 (2003) 337–356. [3] L. Kong, J. Wang, Multiple positive solutions for the one-dimensional p-Laplacian, Nonlinear Anal. 42 (2000) 1327–1333. [4] T. Kusano, T. Jaros, N. Yoshida, A Picone-type identity and Sturmian comparison and oscillation theorems for a class of half-linear partial differential equations of second order, Nonlinear Anal. 40 (2000) 381–395. [5] Y.H. Lee, I. Sim, Global bifurcation phenomena for singular one-dimensional p-Laplacian, J. Differential Equations 229 (2006) 229–256. [6] R. Ma, Nodal solutions second-order two-point boundary value problems with superlinear or sublinear nonlinearities, J. Math. Anal. Appl. (2006) (online available). [7] R. Ma, B. Thompson, Multiplicity results for second-order two-point boundary value problems with superlinear or sublinear nonlinearities, J. Math. Anal. Appl. 303 (2005) 726–735. [8] R. Man´asevich, J. Mawhin, Periodic solutions of nonlinear systems with p-Laplacian-like operators, J. Differential Equations 145 (1998) 367–393. [9] R. Man´asevich, J. Mawhin, Boundary value problems for nonlinear perturbations of vector p-Laplacian-like operators, J. Korean Math. Soc. 37 (2000) 665–685. [10] Y. Naito, S. Tanaka, On the existence of multiple solutions of the boundary value problem for nonlinear second-order differential equations, Nonlinear Anal. 56 (2004) 919–935. [11] Y. Naito, S. Tanaka, Multiplicity of solutions for a class of two-point boundary value problems involving one-dimensional p-Laplacian (preprint). [12] P.H. Rabinowitz, Some global results for nonlinear eigenvalue problems, J. Funct. Anal. 7 (1971) 487–513. [13] P.H. Rabinowitz, Some aspects of nonlinear eigenvalue problems, Rocky Mountain J. Math. 3 (1973) 161–202. [14] J. S´anchez, Multiple positive solutions of singular eigenvalue type problems involving the one-dimensional p-Laplacian, J. Math. Anal. Appl. 292 (2004) 401–414. [15] K. Schmitt, R. Thompson, Nonlinear analysis and differential equations: An introduction, University of Utah Lecture Note, Salt Lake City, 2004. [16] J. Wang, The existence of positive solutions for the one-dimensional p-Laplacian, Proc. Amer. Math. Soc. 125 (1997) 2275–2283. [17] X. Yang, Sturm type problems for singular p-Laplacian boundary value problems, Appl. Math. Comput. 136 (2003) 181–193.