J. Math. Anal. Appl. 376 (2011) 741–749

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Journal of Mathematical Analysis andApplications

www.elsevier.com/locate/jmaa

Infinitely many radial solutions for a p-Laplacian problem p-superlinearat the origin ✩

Jorge Cossio ∗, Sigifredo Herrón, Carlos Vélez

Escuela de Matemáticas, Universidad Nacional de Colombia, Apartado Aéreo 3840, Medellín, Colombia

a r t i c l e i n f o a b s t r a c t

Article history:Received 10 April 2010Available online 3 November 2010Submitted by Goong Chen

Keywords:p-Laplace operatorRadial solutionShooting method

We prove the existence of infinitely many radial solutions for a p-Laplacian Dirichlet prob-lem which is p-superlinear at the origin. The main tool that we use is the shooting method.We extend for more general nonlinearities the results of J. Iaia in [J. Iaia, Radial solutionsto a p-Laplacian Dirichlet problem, Appl. Anal. 58 (1995) 335–350]. Previous developmentsrequire a behavior of the nonlinearity at zero and infinity, while our main result only needsa condition of the nonlinearity at zero.

© 2010 Elsevier Inc. All rights reserved.

1. Introduction

In this paper we study the nonlinear Dirichlet problem{�pu + g(u) = 0 in B1(0) ⊂ R

N ,

u = 0 on ∂ B1(0),(1)

where �p denotes the p-Laplacian which is defined as �pu = div(|∇u|p−2∇u) with p > 1 and g is a locally Lipschitzcontinuous function satisfying the following conditions:

(g1) For every u �= 0, ug(u) > 0.

(g2) There exist α < 0, � > 0 and a > 0 such that

|u| � � �⇒ ∣∣g(u)∣∣ � a|u|p−1+α.

We observe that condition (g2) and Lipschitzian behavior of g imply 2 − p � α. Moreover,

limu→0

g(u)

|u|p−2u= ∞.

There are similar results related with the existence of radial solutions to problem (1). Some of them are [3–5,7] and refer-ences therein. In [3], A. Castro and A. Kurepa studied the semilinear case in the superlinear setting. Multiplicity of solutionswas proved by J. Cossio and S. Herrón in [4] by using bifurcation techniques. In [5], A. El Hachimi and F. De Thelin extended,for p �= 2, the result that A. Castro and A. Kurepa showed in [3] by using phase-plane analysis. Finally, in [7], J. Iaia studied

✩ This research was in part supported by Colciencias, under Contract 574-2009, and by DIME, under code 20101007172.

* Corresponding author.E-mail addresses: jcossio@unal.edu.co (J. Cossio), sherron@unal.edu.co (S. Herrón), cauvelez@unal.edu.co (C. Vélez).

0022-247X/$ – see front matter © 2010 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2010.10.075

742 J. Cossio et al. / J. Math. Anal. Appl. 376 (2011) 741–749

problem (1) in case g(u) = |u|q−1u where 1 < q + 1 < p < N , using the shooting method technique. This previous papermotivated our work. In this paper we extend for more general nonlinearities the results that J. Iaia proved in [7]. Devel-opments such as those in [5] and [7] require a behavior of the nonlinearity g at zero and infinity, while our main result(see Theorem B below) only needs a condition on g at zero. Several proofs that we present are inspired in [7]. Other workswhere the reader can find related results are [1,2,6,8].

The radial version of (1) has the form{(rN−1Ψp

(u′))′ + rN−1 g

(u(r)

) = 0, 0 < r < 1,

u(1) = 0, u′(0) = 0,(2)

where Ψp(x) := x|x|p−2 and Ψp(0) := 0. It is clear that Ψ ′p(x) = (p − 1)|x|p−2 and assumption p > N � 2 implies that

Ψp ∈ C1(R). In addition, we observe that Ψp′ is the inverse of Ψp, where 1/p + 1/p′ = 1.

Our main results read as follows.

Theorem A. Assume p > N and g is a locally Lipschitz continuous function satisfying (g1), (g2) and

(g3) There exist β ∈ (1, p), M > 0 and b1,b2 > 0 such that

|u| > M �⇒ b1|u|p−β �∣∣g(u)

∣∣ � b2|u|p−β .

Then for each integer k � 0 problem (1) has a radial solution with exactly k interior zeroes in [0,1]. In particular (1) has infinitelymany radial solutions.

Theorem B. If p > N and g is a locally Lipschitz continuous function satisfying (g1) and (g2), then there exists a positive integer k0such that, for each k � k0 , problem (1) has a radial solution with exactly k interior zeroes in [0,1]. In particular (1) has infinitely manyradial solutions.

This paper is organized as follows. In Section 2, by applying the Contraction Mapping Principle, we prove the existence ofa unique local solution for an initial value problem associated with (2). Then we show that the solution is actually definedon [0,+∞). In Section 3, we study the qualitative properties of the solution which will be used for proving Theorem A inSection 4. In Section 5 we prove Theorem B.

2. Preliminaries

Throughout Sections 2 to 4 we assume g satisfies (g1)–(g3).In order to study problem (2) we consider the initial value problem{(

rN−1Ψp(u′))′ + rN−1 g

(u(r)

) = 0,

u(r0) = d �= 0, u′(r0) = γ ,(3)

where γ ∈ R and r0 � 0. A local solution of (3) is obtained as a fixed point of the operator defined by

(T u)(r) := d +r∫

r0

t1−Np−1 Ψp′

(rN−1

0 Ψp(γ ) −t∫

r0

sN−1 g(u)ds

)dt. (4)

For ε > 0, R > 0 and d �= 0 we define

BεR(d, r0) := {

u ∈ C([r0, r0 + ε],R

): ‖u − d‖∞ � R

}.

Lemma 2.1. Let R < 12 |d|. There exists ε(d) > 0 small enough such that T maps Bε

R(d, r0) into itself. Moreover, T : BεR(d, r0) →

BεR(d, r0) is a contraction mapping and there exists a unique local solution u ∈ C1[r0, r0 + ε].

Proof. The first part follows easily from the continuity of g on [− 32 |d|, 3

2 |d|]. Regarding the second part, let u, v ∈ BεR(d, r0).

We define

ηw(t) := rN−10 Ψp(γ ) −

t∫r0

sN−1 g(w)ds, Θ(λ, t) := Ψp′(ληv(t) + (1 − λ)ηu(t)

),

for 0 � λ � 1.

From (4) it follows that |(T u − T v)(r)| � ∫ r t1−Np−1 |Θ(1, t) − Θ(0, t)|dt.

r0

J. Cossio et al. / J. Math. Anal. Appl. 376 (2011) 741–749 743

We recall that p > N � 2. Because of the Mean Value Theorem there exists λ̄ ∈ (0,1) such that

Θ(1, t) − Θ(0, t) = Θλ(λ̄, t) = (p′ − 1

)∣∣Θ(λ̄, t)∣∣ 2−p

p−1(ηv(t) − ηu(t)

)

= (p′ − 1

)∣∣∣∣∣rN−10 Ψp(γ ) −

t∫r0

sN−1[λ̄g(v) + (1 − λ̄)g(u)]

ds

∣∣∣∣∣2−pp−1 t∫

r0

sN−1[g(v) − g(u)]

ds.

Since g is continuous at d �= 0, hypothesis (g1) implies the existence of δ(d) > 0 such that

|x − d| < δ �⇒ 1

2

∣∣g(d)∣∣ <

∣∣g(x)∣∣ <

3

2

∣∣g(d)∣∣. (5)

On the other hand, by using the continuity of the functions u and v at r0, there exists a positive number ρ(d) > 0 such that

r0 � s � r0 + ρ �⇒ ∣∣w(s) − d∣∣ < δ, (6)

where w denotes either u or v . From (5) and (6) we have |g(w(s))| > 12 |g(d)| for all r0 � s � r0 + ρ . Again, the continuity

of g and u implies that w has the same sign of d = w(r0) and g(w(s)) has also the same sign of g(d) for all r0 � s � r0 +ρ .

Now, by choosing ε < min{1,rN−1

0 |γ |p−1

3|g(d)|(r0+1)N−1 }, from (5), we find that

∣∣∣∣∣rN−10 Ψp(γ ) −

t∫r0

sN−1[λ̄g(v) + (1 − λ̄)g(u)]

ds

∣∣∣∣∣� rN−1

0 |γ |p−1 −∣∣∣∣∣

t∫r0

sN−1[λ̄g(v) + (1 − λ̄)g(u)]

ds

∣∣∣∣∣ � rN−10 |γ |p−1 − 3/2

∣∣g(d)∣∣(r0 + 1)N−1ε

> (1/2)rN−10 |γ |p−1 > 0,

provided r0 > 0 and γ �= 0. The case r0 = 0 = γ is simple. Coming back to the above estimates, and because Lipschitzianbehavior of g , we immediately have that

‖T u − T v‖∞ < K0‖u − v‖∞ for a constant K0 < 1.

Thus, by the Contraction Mapping Principle, there is a unique local solution u ∈ C[r0, r0 + ε] for ε > 0 small enough. It isclear that u ∈ C1[r0, r0 + ε]. �

Associated to the differential equation in (3), we define the energy of a solution u as follows: if u is defined on [0, t0),

E (r) := 1

p′∣∣u′(r)

∣∣p + G(u(r)

), for each r ∈ [0, t0), where G(t) =

t∫0

g(s)ds � 0.

It is easy to check that for a given solution of (3), E ′(r) = − N−1r |u′|p � 0, that is, the energy is a decreasing function.

Remark. Since E (r) � E (0) for all r ∈ [0, t0), u′ is bounded. More precisely,

∣∣u′(r)∣∣ �

(p′G(d)

) 1p .

From now on, unless otherwise is stated, we consider the unique local solution u of problem (3) with r0 = 0 and γ = 0.Also, in order to simplify our arguments, we assume, without loss of generality, that d > 0.

Lemma 2.2. If u is a solution of (3) then

∣∣u(r)∣∣ �

((p − β + 1)

b1

(G(d) − m

))1/(p−β+1)

, ∀r ∈ R,

where m = min[−M,M] G(s) − b1 |s|p−β+1 .

(p−β+1)

744 J. Cossio et al. / J. Math. Anal. Appl. 376 (2011) 741–749

Proof. This is a consequence of the first inequality on the right-hand side of (g3). Integrating this inequality, we get that

G(u) � b1

p − β + 1|u|p−β+1 +

(G(M) − b1

p − β + 1M p−β+1

), ∀u > M.

Similarly,

G(u) � b1

p − β + 1|u|p−β+1 +

(G(−M) − b1

p − β + 1M p−β+1

), ∀u < −M.

Thus,

G(u) � b1

p − β + 1|u|p−β+1 + m.

Now, given the solution u of (3), with initial data u(0) = d and u′(0) = 0, since the energy E is decreasing, we have that

b1

p − β + 1

∣∣u(r)∣∣p−β+1 + m � p

p − 1

∣∣u′(r)∣∣p + G

(u(r)

) = E (r) � E (0) = G(d),

for all r ∈ R. The desired inequality follows. �From the boundedness of u and u′ it follows the following lemma.

Lemma 2.3. There exists a unique solution u of (3) defined on [0,+∞).

3. Qualitative properties of the solution

In this section we prove some qualitative properties of the solution u of (3) which will be used for proving Theorem Ain the next section. We observe that Eq. (3) is equivalent to the integral relation given by

∣∣u′∣∣p−2u′(r) = − 1

rN−1

r∫0

sN−1 g(u)ds. (7)

Lemma 3.1. If u is a solution of (3) in [0,+∞) with u(0) = d then u has at least one zero.

Proof. Since u(0) = d > 0 and u is a continuous function, u > 0 near the origin. Suppose, by contradiction, that the lemmawere not true. Then u would be positive in [0,+∞). From (7) we see that u′ < 0, which means u is decreasing in [0,+∞).

We claim that there exists r̄ > 0 such that u(r̄) < � (see (g2)). Assume on the contrary that u is bounded from below.Hence there exists L � � so that u(r) → L as r → +∞. Dividing (7) by r > 0, keeping in mind that u′ is bounded, and usingL’Hôspital’s rule, we get g(L) = 0, this contradiction proves the claim.

Because of (g2), it follows that for all r > r̄ we get g(u) � aup−1+α . Now, by (7) and the monotonicity of u,

∣∣u′∣∣p−1 = 1

rN−1

r̄∫0

sN−1 g(u)ds + 1

rN−1

r∫r̄

sN−1 g(u)ds

� 1

rN−1

r∫r̄

sN−1 g(u)ds � a

rN−1

r∫r̄

sN−1up−1+α ds

� a

NrN−1

[u(r)

]p−1+α(rN − r̄N)

.

If r � 21/Nr̄ := r0 then

∣∣u′∣∣p−1 � ar̄N

NrN−1

[u(r)

]p−1+α, i.e., −u′[u(r)

](1−p−α)(p′−1) � C1r(1−N)(p′−1),

where C1 = ( ar̄N

N )p′−1 > 0. Integrating the last inequality on (r0, r), with r > r0, we have

− p − 1

α

[u(r)

]−α(p′−1) � −C1(p − 1)

p − Nr(p−N)(p′−1) + C1r(p−N)(p′−1)

0 − p − 1

α

[u(r0)

]−α(p′−1).

Since α < 0 it follows that u < 0 for r > 0 sufficiently large. This contradiction shows the lemma. �

J. Cossio et al. / J. Math. Anal. Appl. 376 (2011) 741–749 745

Lemma 3.2. If u is a solution of (3) then u has an infinite number of zeroes in [0,+∞), all of which are simple.

Proof. By the previous lemma, u has at least one zero. We prove the lemma by induction. Suppose that u has k � 1 zeroes,we prove that u has k + 1 zeroes. Let us denote by z1 < z2 < · · · < zk the zeroes of u. We claim that u has only one localextremum on (z j, z j+1), j = 1,2, . . . ,k − 1. There is no loss of generality in assuming that u < 0 on (z j, z j+1). If m j is thefirst local minimum of u on (z j, z j+1), integrating Eq. (3) on (m j, r), with m j � r � z j+1, we see that

rN−1∣∣u′∣∣p−2

u′(r) = −r∫

m j

sN−1 g(u)ds,

which implies u′ > 0 in (m j, z j+1). Thus m j is the unique critical point in (z j, z j+1), where u has a minimum. We observethat if r = z j+1 in the previous equality it follows that z j+1 is a simple zero.

Next we show that there exists mk > zk so that u has a local extremum in mk: If u′ �= 0 in (zk,+∞), assuming u < 0 in(zk−1, zk) (i.e. u′ > 0 and u is increasing), after integration on (zk, r) we get

rN−1∣∣u′∣∣p−2

u′(r) − zN−1k

∣∣u′(zk)∣∣p−2

u′(zk) = −r∫

zk

sN−1 g(u)ds.

Because u is bounded from above and increasing, there exists L̃ > 0 such that u(r) → L̃ as r → +∞. Arguing as in theproof of Lemma 3.1 we get the contradiction g(L̃) = 0, which proves the assertion. In addition, reasoning as above weconclude u′ < 0 in (mk, r). Also, we know that u′ > 0 in (r,mk) (we are assuming u < 0 in (zk−1, zk)). Thus, u attains a localmaximum at r = mk . Furthermore, as we proved in the previous lemma, it is easy to show that there exists r̂ > mk so thatu(r̂) < �. Now, if r � 21/Nr̂ := r1, arguing as at the end of Lemma 3.1, we have

∣∣u′∣∣p−1 = 1

rN−1

r̂∫mk

sN−1 g(u)ds + 1

rN−1

r∫r̂

sN−1 g(u)ds

� 1

rN−1

r∫r̂

sN−1 g(u)ds � a

rN−1

r∫r̂

sN−1up−1+α ds

� a

NrN−1

[u(r)

]p−1+α(rN − r̂N)

� ar̂N

NrN−1

[u(r)

]p−1+α,

which means −u′[u(r)](1−p−α)(p′−1) � C2 r(1−N)(p′−1), where C2 = ( ar̂N

N )p′−1 > 0. Integrating the last inequality on (r1, r),with r > r1, we obtain

− p − 1

α

[u(r)

]−α(p′−1) � −C2(p − 1)

p − Nr(p−N)(p′−1) + C2 r(p−N)(p′−1)

1 − p − 1

α

[u(r1)

]−α(p′−1).

Since α < 0 we have u < 0 for r > 0 large enough. Therefore, there exists zk+1 > mk such that u(zk+1) = 0, which provesthe lemma. �Lemma 3.3. If u is a solution of (3) with u(0) = d and d > 0 is large enough then u has no zeroes in (0,1).

Proof. Let d > M , where M > 0 is as in (g3). Let rd > 0 be such that u(rd) = M . Therefore u(r) > M for every r ∈ [0, rd)

and g(u) < b2up−β � b2dp−β . From this fact and (7) we have −u′ � (b2/N)p′−1d(p−β)/(p−1)r p′−1. Integrating the previousinequality on (0, rd) we see that

d(β−1)/(p−1) − M

d(p−β)/(p−1)< C3r p′

d , ∀d > M,

where C3 = p−1p ( b2

N )p′−1 > 0. If d → +∞ then rd → +∞, and the lemma is proved. �Lemma 3.4. The solution u of (3) depends continuously on the initial data d.

Proof. Let u0 be a solution of (3) with u0(0) = d0 > 0. Let dn → d0 as n → ∞ and let us prove that the correspondingsequence of solutions with initial condition dn , say un(·,dn) ≡ un , converges uniformly to u0 on compact subsets of [0,+∞).We claim that there exists a subsequence {un} of solutions of (3), which converges uniformly to u0 on compact subsets of

746 J. Cossio et al. / J. Math. Anal. Appl. 376 (2011) 741–749

[0,+∞). In fact, first we can assume, without loss of generality, that dn > 0 and thus there exists a constant C > 0 withdn < C . This implies that G(dn) �

∫ C0 g(s)ds =: C4. From the monotonicity of the energy it follows that |u′

n| � [p′C4]1/p . Then{un} is equicontinuous and uniformly bounded on compact subsets, the Arzelá–Ascoli’s Theorem gives us a subsequence{unk } which is uniformly convergent to a function u on compact subsets of [0,+∞). Now, we show that u = u0. From (7)we observe that

−u′nk

(r) −→ v(r) ≡ Ψp′

( r∫0

(s/r)N−1 g(u)ds

),

uniformly on compact subsets of [0,+∞). To prove this we recall that

−u′n(r) = Ψp′

( r∫0

(s/r)N−1 g(un)ds

),

{unk } converges uniformly on compact subsets of [0,+∞),Ψp′ is continuous and g ◦ un is uniformly convergent. Further-more, the uniform convergence of {un} on compact subsets and the equality un(r) − dn = ∫ r

0 u′n(s)ds imply u(r) − d0 =

− ∫ r0 v(s)ds, in other words −u′(r) = Ψp′ (

∫ r0 (s/r)N−1 g(u)ds). Straightforward calculations show that v(0+) = 0 and so

u′(0) = 0. Thus u solves problem (3) with r0 = 0 and d = d0. By uniqueness of the initial value problem we see thatu = u0. Therefore, we have shown that un → u0 uniformly on compacts subsets of [0,+∞), it means the claim has beenproved. In order to complete the proof, we need to show that this is true for the full sequence {un}. So, suppose the resultis not true. Then there would exist a compact set K ⊂ [0, M], a sequence {r j} ⊂ K , and a subsequence {u j} such that forsome ε > 0,∣∣u j(r j) − u0(r j)

∣∣ � ε > 0. (8)

As above, we could find a subsequence of {u j} that converges uniformly to u0 on [0, M] violating (8). Therefore, the fullsequence {un} must converge uniformly on compact subsets of [0,∞) to u0. This finishes the proof of the lemma. �

Let us denote by u(·,d) the solution of (3) with initial condition u(0) = d. Arguments relying on the continuous depen-dence of the solution u(·,d) on the initial data will be used to show the next lemma.

Lemma 3.5. If u(·,d∗) is a solution of (3) with exactly k interior zeroes in [0,1] and u(1,d∗) = 0 then there exists ε̄ > 0 such that ford ∈ (d∗ − ε̄,d∗ + ε̄) we have that u(·,d) has at most k + 1 interior zeroes in [0,1].

Proof. First, we claim that u′(·,d) → u′(·,d∗) uniformly on [0,1] as d → d∗. In fact, if r ∈ [0,1] we have

−rN−1Ψp(u′(r,d)

) =r∫

0

sN−1 g(u(s,d)

)ds.

A similar equality holds replacing d by d∗. From this, we get

∣∣Ψp(u′(r,d)

) − Ψp(u′(r,d∗))∣∣ �

r∫0

∣∣g(u(s,d)

) − g(u(s,d∗))∣∣ds.

Since u(·,d) → u(·,d∗) uniformly on [0,1] as d → d∗, and g is continuous, then g(u(s,d)) → g(u(s,d∗)) uniformly on [0,1].The previous inequality implies that Ψp(u′(·,d)) → Ψp(u′(·,d∗)) uniformly on [0,1]. Finally, the continuity of Ψp′ = Ψ −1

pimplies our claim. In the sequel we prove the lemma. Suppose u(·,d) has k + 1 zeroes and let us see that u(r,d) does nothave an additional zero on [0,1] when d is sufficiently near d∗ . If we denote by mk(d) the point where u(r,d) attains theextremum in (zk(d), zk+1(d)) then integrating Eq. (3) on [mk(d∗),1] we get

∣∣u′∣∣p−2u′(1,d∗) = −

1∫mk(d∗)

rN−1 g(u(r,d∗))dr := λ.

Without loss of generality we can assume that u′(1,d∗) > 0. Hence u′(1,d∗) = λp′−1 := ε > 0. The continuity of u′(·,d∗)implies the existence of 0 < γ < 1 such that |u′(r,d∗) − ε| < ε/2 provided 1 − r � γ . More precisely, if r0 := 1 − γ � r � 1then u′(r,d∗) > ε/2. Because u′(·,d) → u′(·,d∗) uniformly on [r0,1] as d → d∗ we have for d near d∗ and for r ∈ [r0,1]that |u′(r,d) − u′(r,d∗)| < ε/4. From this it follows that u′(r,d) > ε/4 for d near d∗ and r ∈ [r0,1]. On the other hand,the continuous dependence on the initial data implies that zk+1(d) → 1 as d → d∗ . Therefore, there exists ε̄ � ε such that

J. Cossio et al. / J. Math. Anal. Appl. 376 (2011) 741–749 747

1−zk+1(d) < γ provided d ∈ (d∗− ε̄,d∗+ ε̄), which implies that r0 < zk+1(d). Consequently, u′(r,d) > 0 for all r ∈ [zk+1(d),1]whenever d ∈ (d∗ − ε̄,d∗ + ε̄), which means u is strictly increasing in [zk+1(d),1] and, hence, u(r,d) has no other zero on[0,1] when d ∈ (d∗ − ε̄,d∗ + ε̄). This proves the lemma. �Lemma 3.6. Let zk(d) be the kth zero of u(·,d). Then zk(d) → 0 as d → 0.

Proof. Let z1 < z2 < · · · be the zeroes of u. We recall that there exists a unique mk ∈ (zk, zk+1) where u attains a localextremum. Without loss of generality we assume that d < � (see hypothesis (g2)). Therefore u(r) � d < � and g(u) �aup−1+α for r � 0. First, we will prove the lemma for k = 1. From (7) we observe that

∣∣u′∣∣p−2u′(r) = − 1

rN−1

r∫0

sN−1 g(u)ds � −a

rN−1

r∫0

sN−1up−1+α ds.

Hence |u′|p−1 � arN [u(r)]p−1+α for r ∈ [0, z1(d)], which implies −u′ � Cu

p−1+αp−1 r p′−1. Integrating on (0, z1(d)) we obtain

C

p′ zp′1 � − p − 1

αd

−αp−1 .

Consequently, z1(d) → 0 as d → 0.

Now, we suppose that zk(d) → 0 as d → 0 and we will show the assertion for k + 1. In order to do that we assumeu′ > 0 in (zk,mk). Integrating (3) on (r,mk) we have

rN−1(u′)p−1 � a

mk∫r

sN−1up−1+α ds � a

N

(mN

k − rN)[u(r)

]p−1+α.

Therefore (u′)p−1 � aN (mk − r)[u(r)]p−1+α . Here we use that mN

k /rN−1 > mk. Thus, it is clear that u′[u(r)] 1−p−αp−1 � C (mk −

r)p′−1 for r ∈ [zk,mk]. Integrating this inequality on [zk,mk] we see that

d−αp−1 �

[u(mk)

] −αp−1 � C0(mk − zk)

p′> 0,

where C0 is a positive constant which does not depend on d. In this way, we have proved that mk − zk → 0 as d → 0. Byassumption, zk → 0 as d → 0, thus mk → 0 as d → 0. In a similar way as above, we can prove that

d−αp−1 �

[u(mk)

] −αp−1 � C0(zk+1 − mk)

p′> 0,

where C0 is a positive constant. Combining the previous inequalities and the fact that mk → 0 as d → 0 we see thatzk+1(d) → 0 as d → 0. This completes the proof of Lemma 3.6. �4. Proof of Theorem A

For k ∈ N ∪ {0} we define

Ak := {d

∣∣ u(r,d) has exactly k interior zeroes on [0,1]}and let us denote dk = inf Ak. The theorem will be a consequence of the following facts:

(a) dk > 0,

(b) u(r,dk) has exactly k interior zeroes in [0,1],(c) u(1,dk) = 0,

which will be proved by induction. First, we prove the statements (a), (b), and (c) for k = 0. Lemma 3.3 implies A0 �= ∅.

Assume, by contradiction, that d0 = 0. Hence, there exists a sequence {dn} ⊂ A0 such that dn → 0. By Lemmas 3.2 and 3.6,functions u(·,dn) have an infinite number of zeroes with z1(dn) → 0. By choosing N ∈ N sufficiently large we see thatu(r,dN ) has no interior zeroes on [0,1] which contradicts the fact that z1(dN ) < 1. Thus d0 > 0.

We now prove that u(r,d0) has no interior zeroes on [0,1]. In order to do so, let dn → d0. Continuous dependence onthe initial datum implies that u(r,dn) → u(r,d0). By continuity of u, u(r,d0) � 0 on [0,1]. Now, if 0 < z1(d0) < 1, a simpleanalysis shows that u′(z1(d0)) = 0, which is a contradiction because the zeroes of u are simple (see Lemma 3.2). Therefore,z1(d0) � 1 and u(r,d0) > 0 for 0 � r < 1. Now, by continuity of u(·,d0) at r, we have that u(1,d0) � 0. If u(1,d0) > 0, byusing again the continuous dependence on the initial datum there exists d < d0 such that u(r,d) > 0 on [0,1], which is acontradiction with the definition of d0. Consequently u(1,d0) = 0. The proof of the case k = 0 is completed.

748 J. Cossio et al. / J. Math. Anal. Appl. 376 (2011) 741–749

Suppose that (a), (b) and (c) are true for k � 0. We will prove that the same assertions hold for k + 1. By the inductiveassumption and Lemma 3.5, there exists d̂ < dk such that u(·, d̂) has exactly k + 1 interior zeroes in [0,1], which impliesthat Ak+1 is nonempty and dk+1 � d̂ < dk. Now, we claim that dk+1 > 0. Suppose, by contradiction, that dk+1 = 0. Therefore,there exists a sequence {dn} ⊂ Ak+1 such that dn → 0 and u(·,dn) possesses exactly k + 1 interior zeroes in [0,1]. Sincezi(dn) → 0 as n → ∞, for ε < 1 there exists N large enough such that zi(dN ) < ε < 1 with i > k + 1. This means thatu(·,dN ) does not have exactly k + 1 interior zeroes on [0,1], which is a contradiction. Thus, dk+1 > 0. Now we prove thatu(·,dk+1) has exactly k + 1 interior zeroes on [0,1]. In fact, it has at least k + 1 interior zeros. If not, then dk+1 ∈ An forsome 0 � n � k and so dk+1 � inf An = dn � dk which is impossible. Also, if u(·,dk+1) has more than k + 1 interior zeros,by definition of infimum, there is a sequence {dn} ⊂ Ak+1 with dn → dk+1 and u(·,dn) has exactly k + 1 interior zeros on[0,1]. The continuous dependence of the initial value problems on the initial condition implies u(r,dn) → u(r,dk+1) in theC1[0,1]-norm. Thus, for N large enough, u(·,dk+1) inherits the exactly number of zeroes of u(·,dN ), which is a contradiction.Thus, u(·,dk+1) has exactly k + 1 interior zeroes. Finally, we show that u(1,dk+1) = 0. Without loss of generality, assumethat u(1,dk+1) > 0 and denote by w(r) = u(r,dk+1). Let ν > 0 be sufficiently small such that for some γ1 > 0 and γ2 > 0,

(i) |w ′(s)| � γ1 ∀s ∈ C := ⋃k+1j=1[z j − ν, z j + ν] ∪ [1 − ν,1];

(ii) |w(s)| � γ2 ∀s ∈ D := [0,1] � C.

Let m = min{w(s): s ∈ [1 − ν,1]}, we observe that m > 0. Let us take ε < 12 min{γ1, γ2,m}, d ∈ (dk+1 − ε,dk+1) and v such

that ‖w − v‖E < ε with v(0) = d, where E = C1([0,1]). We will show that v possesses exactly k + 1 interior zeroes on[0,1], which contradicts the definition of dk+1. The following estimate shows that v has no zeroes in D: If s ∈ D

∣∣v(s)∣∣ �

∣∣w(s)∣∣ − ∣∣w(s) − v(s)

∣∣ � γ2 − ∣∣w(s) − v(s)∣∣ > γ2 − ε > γ2 − 1

2γ2 > 0.

Now, it is easy to check that v(z j − ν)v(z j + ν) < 0 for all j = 1,2, . . . ,k + 1. By the Intermediate Value Theorem, v has atleast one zero in [z j − ν, z j + ν]. Actually, since v is strict monotone, v has only one zero there: In fact, there is no loss ofgenerality in assuming that w ′ > 0 on [z j − ν, z j + ν]. Thus, v ′(s) = w ′(s) − (w ′(s) − v ′(s)) � w ′(s) − ε � γ1 − ε > 1

2 γ1 > 0

for s ∈ [z j − ν, z j + ν]. In this way we have shown that v possesses exactly k + 1 simple zeroes in⋃k+1

j=1[z j − ν, z j + ν].Finally, we prove that v has no zeroes on [1 − ν,1]. In fact, v(s) � w(s) − |w(s) − v(s)| � w(s) − ε � m − ε > m/2 > 0 forall s ∈ [1 − ν,1]. This proves Theorem A.

Remark. We observe that dk → 0 as k → ∞: indeed, the sequence {dk}k is decreasing and bounded below by zero. Thusit converges to some d̄ � 0. We now prove that there is a contradiction if we assume d̄ > 0. Consider the initial valueproblem (3) with u(0) = d̄ > 0 and u′(0) = 0. Either the unique solution u to this problem has infinitely many zeroes in[0,1] or it has a finite number of zeroes.

If u had an infinite number of zeroes in [0,1], they would have at least an accumulation point r0 ∈ [0,1]. Then, theunique global solution of problem (3) with u(r0) = 0 and u′(r0) = 0 would be zero, which contradicts u(0) = d̄ > 0. If u hada finite number of zeroes in [0,1], let us say u had n interior zeroes in [0,1], u(·,dk) would have at most n + 1 interiorzeroes in [0,1] for k large enough, in virtue of Lemma 3.5. This is a contradiction.

5. Proof of Theorem B

Let g be a locally Lipschitz continuous function that satisfies (g1)–(g2). Let us take the locally Lipschitz continuousfunction g̃ : R → R defined as

g̃(u) =⎧⎨⎩

g(u), |u| � �,

a|u|p−1+α + C1, u > �,

−a|u|p−1+α + C2, u < −�,

where C1 = g(�) − a|�|p−1+α � 0 and C2 = g(−�) + a|�|p−1+α � 0 (see hypothesis (g2)). In other words, we take anextension g̃ of the function g/[−�,�] which satisfies (g1)–(g3) with p − β = p − 1 + α and 0 < b1 < a < b2. Hence we canapply the results of the previous sections to this nonlinearity.

Let us denote by {dk(g̃)}k the sequence of numbers which satisfy (a), (b) and (c) at the beginning of Section 4. Inparticular, for each k � 0, problem{(

rN−1Ψp(u′))′ + rN−1 g̃

(u(r)

) = 0,

u(0) = dk(g̃), u′(0) = 0,(9)

has a unique (global) solution uk . Observe that, in this case,

|u| > � �⇒ a |u|p−β �∣∣g̃(u)

∣∣.

2

J. Cossio et al. / J. Math. Anal. Appl. 376 (2011) 741–749 749

Thus, because of Lemma 2.2,

∣∣uk(r)∣∣ �

(2(p − β + 1)

a

(G̃(dk(g̃)

) − m))1/(p−β+1)

, ∀r ∈ [0,1],

where m = min[−�,�] G̃(s) − a2(p−β+1)

|s|p−β+1. Given the particular form g̃ has, m = 0. Since dk(g̃) → 0 as k → ∞ (seeRemark at the end of Section 4), the right-hand side of this inequality tends to zero as k → ∞. Consequently, uk → 0uniformly in [0,1]. It follows that there exists k0 such that, for k � k0, |uk(r)| � � for all r ∈ [0,1]. Since g̃(u) = g(u) for|u| � �, we conclude that, for k � k0, uk satisfies the differential equation in (9) changing g̃ for g . Because of (b) and (c)from Section 4, uk is a solution of{(

rN−1Ψp(u′))′ + rN−1 g

(u(r)

) = 0, 0 < r < 1,

u(1) = 0, u′(0) = 0,(10)

which has exactly k interior zeroes in [0,1]. This completes the proof of Theorem B.

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