Bifurcation of nodal radial solutions for a prescribed mean curvature problem on an exterior domain

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Bifurcation of nodal radial solutions for a prescribed mean curvature problem on an exterior domain Rui Yang a , Yong-Hoon Lee a,∗ , Inbo Sim b a Department of Mathematics, Pusan National University, Busan 46241, Republic of Korea b Department of Mathematics, University of Ulsan, Ulsan 44610, Republic of Korea

Received 30 September 2018; revised 21 May 2019; accepted 23 October 2019

Abstract We investigate the existence and multiplicity of nodal radial solutions for a prescribed mean curvature problem on the exterior domain of a ball using the global bifurcation theory. Moreover, we establish the asymptotic behavior of the solutions on the subcontinua, which bifurcate from a trivial branch. All results are obtained by considering a radially equivalent second order problem with a singular weight. © 2019 Elsevier Inc. All rights reserved. MSC: 34B09; 34B16; 34C23 Keywords: Mean curvature; Exterior domain; Singular weight; Global bifurcation; Nodal radial solutions; Asymptotic behavior

1. Introduction In this study, we are concerned with the existence of nodal radial solutions of the following problem:   ⎧ ⎨−div φN (∇v(x)) = λk(|x|)f (v(x)), ⎩v|∂ = 0,

lim v(x) = 0,

in , (Eλ )

|x|→∞

* Corresponding author.

E-mail addresses: [email protected] (R. Yang), [email protected] (Y.-H. Lee), [email protected] (I. Sim). https://doi.org/10.1016/j.jde.2019.10.035 0022-0396/© 2019 Elsevier Inc. All rights reserved.

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2

 y , y ∈ RN , | 1−|y|2 ∈ RN : |x| > R}, R

where φN (y) =

· | denotes the Euclidean norm in RN ,  denotes the exte-

> 0, N ≥ 3, λ a nonnegative real parameter, f : R → R a rior domain {x continuous and odd function satisfying f (s)s > 0 for s = 0, and the function k satisfies (H K) k ∈ L1 ([R, ∞), [0, ∞)) is radially symmetric and k ≡ 0 on any subinterval in (R, ∞) ∞ with R rk(r)dr < ∞. Problem (Eλ ) is related to the mean curvature problems on a flat Minkowski space with a Lorentzian metric in differential geometry and the theory of classical relativity (see [1–3] and the references therein). To study the existence of radial solutions of problem (Eλ ) on the exterior of a ball, we transform problem (Eλ ) into the one dimensional prescribed mean curvature problem via consecutive transformations |x| = r, v(x) = u(r), and t = ( Rr )−(N−2) as follows ⎧  

⎨− β(t)φ 1 u (t) = λh(t)f u(t) , t ∈ (0, 1), β(t) ⎩u(0) = 0 = u(1), where φ(y) = as

 y 1−y 2

(Pλ )

with y ∈ (−1, 1), β and h can be obtained from the transformations

β(t) =

1

R − N−1 t N−2 , h(t) = β 2 (t)k Rt − N−2 . N −2

Let us introduce a new condition on function h : (0, 1] → [0, ∞). (H H ) h ∈ L1loc ((0, 1], [0, ∞)) and h ≡ 0 on any subinterval in (0, 1) with 1 τ h(τ )dτ < ∞. 0

We note that if h in problem (Pλ ) satisfies (H H ), then the corresponding k in problem (Eλ ) satisfies (H K). Studies on the existence of positive solutions for prescribed mean curvature problems on bounded or exterior domains using the fixed point theory and/or topological methods (mostly bifurcation technique) have received much attention [4–7,9–13,18]. However, so far, there have been only a few results on the existence of nodal solutions for prescribed mean curvature problems. In the study of radial solutions for a prescribed mean curvature problem, it is more difficult to analyze the solutions of problem (Pλ ) on the exterior of a ball than that on a bounded ball because of the singular coefficient β and singular weight h [18]. One of the difficulties is that we do not know the boundedness of the solutions of problem (Pλ ) in C 1 -norm, while those when β ≡ 1, all solutions are bounded in C 1 -norm. The other difficulty is that the concave-convex properties of solutions are uncertain owing to the influence of the mean curvature operator and coefficient

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Fig. 1. Solutions of problem (Pλ0 ) with respect to R, N , and λ.

function β. Some solutions for the following auxiliary problem ⎧   ⎨− β(t)φ 1 u (t) = λ, t ∈ (0, 1), β(t) ⎩u(0) = 0 = u(1),

(Pλ0 )

are shown in Fig. 1. This figure shows that the concave-convex properties of the solutions of problem (Pλ0 ) may depend on R, N , and λ (see (Sλ ) in Section 3). From now on, we assume that f satisfies the following assumptions with f0  lim f (s) s and s→0

f (s) . |s|→∞ s

f∞  lim

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(F ) 0 ≤ f0 < ∞, (F1 ) 0 < f0 < ∞, (F2 ) f∞ = 0. Let us list some previous studies which motivates this paper. Coelho-Corsato-Obersnel-Omari [7] showed positive solutions of the following one-dimensional Minkowski-curvature problem ⎧   ⎨− φ(u (t)) = λf (t, u(t)), t ∈ (0, T ), ⎩u(0) = 0 = u(T ), by using the global bifurcation. Coelho-Corsato-Rivetti [8] investigated the following problem on the domain of a ball B(R) = {x ∈ RN : |x| < R}   −div φN (∇v(x)) = f (|x|, v(x)),

in B(R),

v = 0,

on ∂B(R),



(1.1)

which is equivalent to non-singular one-dimensional problem with a mixed boundary condition (u (0) = 0 = u(R)). This extends [7], [9] and [10] in some direction. In the domain of a ball, the authors of [12,13] gave the global structure (existence, nonexistence and multiplicity) according to a positive parameter λ of the set of positive radial solutions for the problem   −div φN (∇v(x)) = λf (|x|, v(x)),

in B(R),

v = 0,

on ∂B(R),



(1.2)

under suitable conditions on nonlinear function f applying global bifurcation technique with/without topological argument. On the same domain, Dai-Wang [14] studied nodal radial solutions of problem (1.2). Assum= f0 a(r) for f0 > 0 and some a ∈ C[0, R] with a(r) > 0 for r ∈ (0, R], they ing lim f (r,s) s s→0

concluded that there exist two unbounded continua Ck+ and Ck− (k ∈ N) of S, the closure of the set of nontrivial solution pairs of (1.2) in R × X, where X := {u ∈ C 1 [0, R] : u (0) = u(R) = 0} with the norm u = u ∞ , bifurcating from (λk (a)/f0 , 0), where λk (a) should be the k-th eigenvalue of −(r N−1 u ) = λr N−1 a(r)u, u (0) = u(R) = 0.

r ∈ (0, R),

Moreover, for (λ, u) ∈ Ck± , lim ||u|| = 1. λ→∞

Very recently, on an exterior domain, Yang-Lee [18] showed C 1 -regularity of nonnegative solutions of problem (Pλ ) under condition (F ) and obtained the existence of positive radial solutions of problem (Eλ ) under (F1 ) and (F2 ) with f ∈ C([0, ∞), [0, ∞)). In this work, we aim to study the existence (Theorem 1.1 and Theorem 1.2) of nodal solutions of (Pλ ) and the asymptotic behavior of these solutions (Theorem 1.3).

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Theorem 1.1. Assume (H H ), (F1 ), and (F2 ). Then, for each k ∈ N, there exists λ∗ ∈ (0, λkf(h) ] 0 such that problem (Pλ ) has no (k − 1)-nodal solution for all λ ∈ (0, λ∗ ) and at least two (k − 1)-nodal solutions for all λ ∈ ( λkf(h) , ∞). 0 Here, λk (h) is an eigenvalue of problem (3.2) given in Section 3. Conditions (F1 ) and (F2 ) are not enough to verify whether λ∗ = λkf(h) or not. But if we add more conditions on f , we may 0 obtain the existence result in a global sense. Theorem 1.2. Assume (H H ), (F1 ), (F2 ), and f (s) s ≤ f0 for all s = 0. Then, for each k ∈ N, ] and at least two (k − 1)-nodal problem (Pλ ) has no (k − 1)-nodal solution for all λ ∈ (0, λkf(h) 0 , ∞). solutions for all λ ∈ ( λkf(h) 0

Remark 1.1. When k = 1, the above theorems tell us that one of solutions is positive and the other is negative.

It is appropriate to determine the asymptotic behavior of uβ ∞ as λ approaches infinity to compare it with the asymptotic behavior of u ∞ in [14]. Theorem 1.3. For each k ∈ N, if (λ, u) ∈ Ck± , which is satisfied in Theorem 3.2, then lim

λ→∞

u

∞ = 1. β

For the proofs, we apply the modified Gronwall-Bellman inequality and the modified Picone identity to obtain some critical lemmas (see Lemma 2.3 and 2.5), which are used to overcome the difficulties in dealing with nodal cases. This paper is organized as follows. In Section 2, we list some preliminary lemmas and give the modified Gronwall-Bellman inequality and Picone identity and their applications. In Section 3, we obtain the unbounded continuum by applying Rabinowitz type global bifurcation theory. In Section 4, we give the proof of Theorem 1.3 and the asymptotic behavior of the solutions of problem (Pλ ). Section 5 is devoted to proving Theorem 1.1 and Theorem 1.2 and giving an example of non-odd function f to illustrate the existence results. 2. Preliminaries In this section, we list the related properties of the solutions of problem (Pλ ) [18] and give the modified Gronwall-Bellman inequality and Picone identity and their applications to deal with the nodal solutions of problem (Pλ ).



We say u a solution of problem (Pλ ) if u ∈ C[0, 1] ∩ C 1 (0, 1], uβ ∞ < 1, βφ uβ is absolutely continuous in any compact subinterval of (0, 1), and u satisfies the equation and the boundary conditions of problem (Pλ ). If h ∈ L1 (0, 1), it is easy to check that all solutions of problem (Pλ ) belong to C 1 [0, 1]. On the other hand, when h holds (H H ), h may not be integrable near 0 and the C 1 -regularity of the solutions may not be general.

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Remark 2.1. (Example 2.1 in [18]) For example, consider the problem ⎧   ⎨− β(t)φ 1 u (t) = h(t), t ∈ (0, 1), β(t) ⎩u(0) = 0 = u(1), where   N N−2 (ln t + 1)2 ln t + 2N−3 (N − 1)Rt N−1 R h(t) = +

⎛ ⎞ .  N−1 2  N−1 2 3 N−2 (ln t+1) N−2 (ln t+1) 2(N − 2)t 1 − t ⎠ 8(N − 2)2 ⎝ 1 − t 2 2

Rt ln t Note that h satisfies condition (H H ) but is not integrable near 0 and a solution u(t) = − 2(N−2) ∈ / 1 C [0, 1].

The C 1 -regularity of the solutions is useful to apply the Rabinowitz type global bifurcation theory. The C 1 -regularity of non-negative solutions was established for problem (Pλ ) in [18]. Here, we show the C 1 -regularity of all the solutions of problem (Pλ ).

Proposition 2.1. Assume (H H ) and (F ). Also assume u ∈ C[0, 1] ∩ C 1 (0, 1], uβ ∞ < 1,   1,1 (0, 1), and u satisfies the equation and the boundary conditions of problem (Pλ ). βφ uβ ∈ Wloc   1 Then, u ∈ C [0, 1] and βφ uβ ∈ W 1,1 (0, 1). Proof. Let u ≡ 0 on [0, 1] and t ∗ = inf{t ∈ (0, 1) : u (t) = 0}. We first show t ∗ > 0. Suppose t ∗ = 0. Then, there exists a sequence {tn } with u (tn ) = 0 satisfying tn → 0 as n → ∞. By (F ), we obtain, for some c  c( u ∞ ) > 0,

|f u(t) | ≤ c|u(t)|, for t ∈ [0, 1].

(2.1)

By (H H ), we consider a sufficiently small constant γ0 > 0 such that γ0 τ h(τ )dτ < 1.

λc

(2.2)

0

We choose a sufficiently large N0 ∈ N such that 0 < tn < γ0 for all n ≥ N0 . By integrating the first equation in problem (Pλ ) over (t, tn ) for t ∈ (0, tn ), we obtain 1

β(t)φ u (t) = β(t)

tn t

which is followed by



λh(τ )f u(τ ) dτ,

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⎛ 1 u (t) = β(t)φ −1 ⎝ β(t)

tn

7

⎞

λh(τ )f u(τ ) dτ ⎠ .

t

By integrating the above equation over (0, t), we get t u(t) =

⎛ 1 β(s)φ −1 ⎝ β(s)

tn

⎞

λh(τ )f u(τ ) dτ ⎠ ds.

s

0

Using (2.1), for t ∈ (0, tn ), and by setting u tn ,∞  max |u(t)|, we have 0≤t≤tn

 t tn |u(t)| ≤ λc

h(τ )dτ ds u tn ,∞ 0

s

tn tn ≤ λc

h(τ )dτ ds u tn ,∞ 0

s

tn ≤ λc

τ h(τ )dτ u tn ,∞ . 0

Using (2.2), we obtain u tn ,∞ ≡ 0 for all n ≥ N0 , i.e., u is identically zero on [0, tn ] for all n ≥ N0 . Thus, we may assume that u(tn ) = u (tn ) = 0 for a sufficiently large n. For t ∈ [tn , 1], we obtain  t ⎞  ⎛   s  

−1 ⎝ 1  ⎠ |u(t)| =  β(s)φ λh(τ )f u(τ ) dτ ds  β(s)   tn

t ≤

⎛ 1 β(s)φ −1 ⎝ β(s)

tn

tn

s

⎞ 

 λh(τ ) f u(τ )  dτ ⎠ ds

tn

 t s ≤



 λh(τ ) f u(τ )  dτ ds

tn tn

t =λ



 (t − τ )h(τ ) f u(τ )  dτ

tn

t ≤λ tn



 (1 − τ )h(τ ) f u(τ )  dτ

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t ≤ cλ

(1 − τ )h(τ )|u(τ )|dτ. tn

By the Gronwall-Bellman inequality, we obtain u(t) = 0 for t ∈ [tn , 1]. Thus, u ≡ 0 on [0, 1]. This is a contradiction. Hence, t ∗ ≥ γ > 0 with a constant γ . Without loss of generality, we assume u > 0 near 0. The case u < 0 near 0 can be proved by a similar argument. By the continuity of u and definition of t ∗ , we have u > 0 in (0, t ∗ ). Even though the remaining part of the proof is the same as that mentioned in [18], for the readers’ convenience, we complete the C 1 -regularity. By integrating the first equation of problem (Pλ ) over (t, t ∗ ) for t ∈ (0, t ∗ ), we get ⎛ 1 u (t) = β(t)φ −1 ⎝ β(t)

t ∗

⎞

λh(τ )f u(τ ) dτ ⎠ .

t

Thus, u (t) > 0 for t ∈ (0, t ∗ ). The first equation in problem (Pλ ) can be rewritten as N N−2

u

+ (N−1)(N−2)t u 3 R2 −  = λh(t)f (u(t)). 3   1−

u (t) β(t)

2

From the above equation, we can see that u is concave on [0, t ∗ ]. By (H H ), for a fixed ε > 0, we can choose a sufficiently small δ ∈ (0, t ∗ ) such that δ τ h(τ )dτ < ε. 0

By integrating the first equation of problem (Pλ ) over (s, t) for 0 < s < t < δ, we get 1

1

β(s)φ u (s) = β(t)φ u (t) + β(s) β(t)

t



λh(τ )f u(τ ) dτ.

s

Using (2.1) and the fact that

u(τ ) τ

decreases for τ ∈ [0, t ∗ ], we obtain

1

u (s) < β(t)φ u (t) + cλ β(t)

t τ h(τ )

u(τ ) dτ τ

s

1

u(s) ≤ β(t)φ u (t) + cλ β(t) s

t τ h(τ )dτ. s

(2.3)

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On multiplying both sides by

s u(s) ,

9

we obtain

su (s) su (t)  + cλε. ≤ 1 u(s) u(s) 1 − ( β(t) u (t))2

We consider m0  uβ ∞ < 1. Then, we have su (s) su (t)  + cλε. ≤ u(s) u(s) 1 − m20

(2.4)

On the other hand, from (2.1) and (2.3), we get 1

u (s) < β(t)φ u (t) + cλu(t) β(t)

t h(τ )dτ. s

By integrating the above equation over (0, t) for t < t ∗ with respect to s, we obtain 1

u(t) < tβ(t)φ u (t) + cλu(t) β(t)

t t h(τ )dτ ds 0

=

s

t

tu (t) 1 1 − ( β(t) u (t))2

+ cλu(t)

τ h(τ )dτ. 0

By choosing a sufficiently small t such that t cλ

1 τ h(τ )dτ < , 2

0

we get u(t) < 

2tu (t) 1 1 − ( β(t) u (t))2

2tu (t) ≤ . 1 − m20 Hence,  tu (t) u(t)

>

1 − m20 2

.

(2.5)

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Suppose on the contrary u ∈ / C 1 [0, 1], then Because lim

s→0+

u(s) s

diverges to ∞ as s → 0+ . Let us recall (2.4).

su (t)  = 0, u(s) 1 − m20

it follows that lim sup s→0+

su (s) ≤ cλε. u(s)

By the arbitrariness of ε, the above inequality reduces to lim

t→0+

tu (t) = 0, u(t)

(2.6)

  which contradicts (2.5). Therefore, u ∈ C 1 [0, 1]. Next, we show that βφ uβ ∈ W 1,1 (0, 1). Us    t ing (2.1) and the fact that |u(t)| =  0 u (τ )dτ  ≤ t u ∞ for t ∈ [0, 1], we get

|h(t)f u(t) | ≤ cth(t) u ∞ .  



From (H H ), h(·)f u(·) belongs to L1 (0, 1) and βφ uβ ∈ L1 (0, 1) is satisfied. Thus,   βφ uβ ∈ W 1,1 (0, 1). 2 We define the space E := {u ∈ C 1 [0, 1] : u(0) = 0 = u(1)} endowed with the norm u = u ∞ + u ∞ . Next, we state the well-known Gronwall-Bellman inequality for later use. Lemma 2.1. Let ε > 0 and m ∈ L1 (a, b) with a < b satisfying m ≥ 0 a.e. in (a, b). Suppose u ∈ C[a, b] and t u(t) ≤ ε +

m(s)u(s)ds, a

for all t ∈ [a, b]. Then, u(t) ≤ εe

t a

m(s)ds

for all t ∈ [a, b].

We also require a modified Gronwall-Bellman inequality to show that all zeros of the solutions of problem (Pλ ) are simple. Lemma 2.2. (Lemma 2.2 in [15]) Let ε > 0 and m ∈ L1 (0, T ) satisfy m ≥ 0 a.e. in (0, T ). Suppose u ∈ C[0, T ] and

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T u(t) ≤ ε +

m(s)u(s)ds, t

for all t ∈ [0, T ]. Then, u(t) ≤ εe

T t

m(s)ds

for all t ∈ [0, T ].

The following lemma is valid for positive solutions (see Lemma 2.5 in [18]). Here, we prove that it is also valid for nodal solutions. Lemma 2.3. Assume (H H ) and (F ). If u ∈ E is a solution of problem (Pλ ) with u(t0 ) = u (t0 ) = 0 at some t0 ∈ [0, 1], then u is identically zero on [0, 1]. Proof. By integrating the first equation of problem (Pλ ) over (t0 , t) for t ∈ (0, 1), we obtain ⎛ u (t) = −β(t)φ −1 ⎝

t

1 β(t)

⎞

λh(τ )f u(τ ) dτ ⎠ .

t0

By integrating the above equation over (t0 , t) again, we get t u(t) = −

⎛

s

1 β(s)φ −1 ⎝ β(s)

t0

⎞

λh(τ )f u(τ ) dτ ⎠ ds.

t0

We divide the problem into the following two cases: Case 1. t0 = 0. First, we consider t ∈ (0, t0 ). This holds that  t ⎞  ⎛   s  

1 |u(t)| =  β(s)φ −1 ⎝ λh(τ )f u(τ ) dτ ⎠ ds  β(s)   t0

t0 ≤

⎛ 1 β(s)φ −1 ⎝ β(s)

t

t0

t0

⎞ 

 λh(τ ) f u(τ )  dτ ⎠ ds

s

t0 t0 ≤



 λh(τ ) f u(τ )  dτ ds

s

t

t0 =λ



 (τ − t)h(τ ) f u(τ )  dτ

t

t0 ≤λ t



 τ h(τ ) f u(τ )  dτ

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t0 ≤ cλ

τ h(τ )|u(τ )|dτ. t

In the last inequality, we have used (2.1). By (H H ) and Lemma 2.2, we obtain u ≡ 0 on [0, t0 ]. Next, we consider t ∈ (t0 , 1). By a similar computation, we obtain t



 (t − τ )h(τ ) f u(τ )  dτ

|u(t)| ≤ λ t0

t



 (1 − τ )h(τ ) f u(τ )  dτ

≤λ t0

t (1 − τ )h(τ )|u(τ )|dτ.

≤ cλ t0

Because h ∈ L1 (t0 , 1), by applying Lemma 2.1, we obtain u ≡ 0 on [t0 , 1]. Note that the former argument is sufficient to show that u ≡ 0 on [0, 1] when t0 = 1. Case 2. t0 = 0. Consider 0 <  < 1. Let · ,∞ denote the usual supremum norm defined on [0, ]. We notice that h(·)f (u(·)) ∈ L1 (0, 1) by Proposition 2.1. Using (2.1) and the fact t |u(t)| ≤ 0 |u (τ )|dτ ≤ t u t,∞ , we get, for all t ∈ [0, ], ⎛ 1 |u (t)| ≤ β(t)φ −1 ⎝ β(t)

t

⎞ 

 λ h(τ )f u(τ )  dτ ⎠

0

t ≤



 λh(τ ) f u(τ )  dτ

0

t ≤ λc

h(τ )|u(τ )|dτ 0

t ≤ λc

τ h(τ )dτ u ,∞ ,

0

which gives

t

u ,∞ ≤ λc 0

τ h(τ )dτ u ,∞ .

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t We may consider a sufficiently small  such that λc 0 τ h(τ )dτ < 1. Then, u ,∞ = 0, which shows u ≡ 0 on [0, ]. Thus,  is a double zero of u. By Case 1, we can get u ≡ 0 on [, 1]. Hence, the proof is complete. 2 We finally introduce the modified Picone identity, which will be used to determine the block of an unbounded subcontinuum. We define 



y

l1 [y] = β(t)φ β(t)



+ b1 (t)y,

L2 [z] = z

+ b2 (t)z. Lemma 2.4. (Lemma 2.6 in [18]) Let b1 (t) and b2 (t) be functions on an interval I. Let y and z  (t)  y

 < 1 for be functions such that y, β(t)φ β(t) , and z are differentiable on I , z(t) = 0 and  yβ(t) all t ∈ I . Then, the modified Picone identity can be written as   y 2 z

y − yβ(t)φ = (b1 − b2 )y 2 z β(t)     y y2 2yy z y 2 z 2 − y β(t)φ − + 2 − yl1 [y] + L2 [z]. β(t) z z z

d dt



(2.7)

Remark 2.2. 

 y

2yy z y 2 z 2 y β(t)φ − + 2 ≥ 0. β(t) z z

This equality holds if and only if y =

yz

z

= 0.

The next lemma plays an important role to compare the solutions of (Pλ ) with the eigenfunctions of (3.2). Lemma 2.5. Let b1 (t), b2 (t) > 0 for t ∈ (a, b) and y, z ∈ C 1 [a, b] satisfy the following equalities: 

 β(t)φ

y

β(t)



+ b1 (t)y = 0, in (a, b),

z

+ b2 (t)z = 0, in (a, b). Let β(·)φ

y (·)

β(·)

   (t)  and z be differentiable in (a, b) and  yβ(t)  < 1 for all t ∈ [a, b]. If b1 y, b2 z ∈

L1 (a, b), y(a) = y(b) = 0, and z = 0 in (a, b), we assume either (i) or (ii) as mentioned below. (i) z(t) = 0 at t = a and t = b, (ii) z(t) has at least one zero point at t = a and t = b.

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14

Then, b (b1 (t) − b2 (t)) y 2 dt ≥ 0,

(2.8)

a

and the equality holds if and only if y = Proof. Because

y 2 z

z

 and yβ(t)φ

y

β(t)

yz

z

= 0.

 are differentiable in (a, b), we obtain

     

b  y (t) y (t) y dt = lim y(t)β(t)φ − lim y(t)β(t)φ yβ(t)φ β(t) β(t) β(t) t→b− t→a + a

= lim  t→b−

y(t)y (t) y(t)y (t) − lim   2 t→a +  2 (t) (t) 1 − yβ(t) 1 − yβ(t)

= 0, and b 

y 2 z

z



dt = lim

t→b−

y 2 (t)z (t) y 2 (t)z (t)  =Eb − Ea . − lim z(t) z(t) t→a +

a

For (i), we have Ea = Eb = 0. By integrating (2.7) over (a, b), we get (2.8). For (ii), without loss of generality, we may assume z(a) = 0 (the argument would be similar if z(b) = 0). If z > 0 in 2

(t) (a, b), we have z (a + ) > 0 and y (t)z ≥ 0 for t near a from the right hand side. Thus, Ea ≥ 0. z(t) On the other hand, by L’Hospital rule, we get

Ea = lim

t→a +

y 2 (t)z (t) y 2 (t)z

(t) + 2y(t)y (t)z (t) = lim z(t) z (t) t→a + = lim

t→a +

−b2 (t)y 2 (t)z(t) + 2y(t)y (t)z (t) z (t)

≤ lim 2y(t)y (t) t→a +

= 0. Thus, Ea = 0. If z < 0 in (a, b), similarly, we can show Ea = 0. Finally, by integrating (2.7) over (a, b), we get (2.8). 2

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3. Bifurcations In this section, we prove the existence of unbounded continuum Ck of problem (Pλ ) by applying Rabinowitz type global bifurcation theory [16]. To describe the Rabinowitz global bifurcation theorem, let X be a real Banach space with the norm || · || and let us consider the operator equation u = λL(u) + H(λ, u),

(3.1)

where L : X → X is compact linear and H : R × X → X is completely continuous. Assume that H = o( u ) near u = 0 uniformly on bounded λ intervals. By modifying Theorem 1.3 in [16], we obtain the following theorem. Theorem 3.1. If μ is a characteristic value of L with odd algebraic multiplicity, then there exists a subcontinuum C in S, the closure of the set of nontrivial solutions of (3.1), bifurcating from {(λ, 0) : λ ∈ R} at (μ, 0) and either (i) C is unbounded in R × X, or (ii) C ∩ ((R \ {μ}) × X) = ∅. Let us rewrite problem (Pλ ) into problem (Sλ ) [18] to apply Theorem 3.1 as ⎧    2 3 ⎪ ⎨

u 1 − β(t) + −u = λh(t)f (u) ⎪ ⎩ u(0) = 0 = u(1).

N

(N−1)(N−2)t N−2 R2

u 3 , t ∈ (0, 1),

(Sλ )

To investigate the nonlinear problem (Sλ ), we need to know a related linear problem

−u

(t) = λh(t)u(t), t ∈ (0, 1), u(0) = 0 = u(1).

(3.2)

 : E → E be defined by Let L  Lu(t) =

1 G(t, s)h(s)u(s)ds, 0

with G(t, s) =

(1 − t)s, 0 ≤ s ≤ t ≤ 1, (1 − s)t, 0 ≤ t ≤ s ≤ 1.

(3.3)

 is compact linear, problem (3.2) is equivalent to u = λLu,  and the Then, it is easy to show that L  characteristic values of L are the corresponding eigenvalues of problem (3.2). The next lemma [17] gives information on the eigenvalues and eigenfunctions of problem (3.2).

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 is a countable set Lemma 3.1. Assume (H H ). Then, the set of all characteristic values of L  is {λn (h) : n ∈ N} satisfying 0 < λ1 (h) < · · · < λn (h) < · · · → ∞. For each n, ker(I − λn (h)L) a subspace of C 1 [0, 1] and its dimension is 1. Let un be a corresponding characteristic function to λn (h), then the number of interior zeros of un is n − 1. Let us rewrite (Sλ ) into (Wλ ) as ⎧    2 3 ⎪ ⎨

u 1 − β(t) + −u = λh(t) (f0 + ξ(u)) u ⎪ ⎩ u(0) = 0 = u(1),

N

(N−1)(N−2)t N−2 R2

u 3 , t ∈ (0, 1),

(Wλ )

where ξ : R → R is a continuous function and f (s) = (f0 + ξ(s)) s. Then, it follows from (F1 ) that lim ξ(s) = 0. Next, we define the operator equation which is equivalent to (Wλ ) as follows s→0

u = λLu + H(λ, u),

(Dλ )

where L : E → E is defined by 1 Lu(t) = f0

G(t, s)h(s)u(s)ds, 0

and H : R × E → E is defined by ⎧ ⎡ ⎛ ⎤ ⎞ ⎪  2 3 1 ⎨ u (s) ⎠ ⎢ ⎥ H(λ, u(t)) = G(t, s) λf0 h(s)u(s) ⎣⎝ 1 − − 1⎦ + ⎪ β(s) ⎩ 0

⎛ λh(s)ξ(u(s))u(s) ⎝ 1 −



u (s) β(s)

2

⎞3

⎠ + (N − 1)(N − 2)s R2

N N−2

⎫ ⎪ ⎬

u (s)3 ds, ⎪ ⎭

with G(t, s) given in (3.3). It is not difficult to check that L is compact linear in E, H is completely continuous in R × E, and H = o( u ) near u = 0 uniformly on bounded λ intervals [18]. We denote S = {(λ, u) ∈ R × E : u is a nontrivial solution of (Dλ ) with λ > 0}. By employing Theorem 3.1 and Lemma 3.1, we obtain the main theorem of this section. However, the proof is provided later. Theorem 3.2. Assume (H H ) and (F1 ). Then, for each k ∈ N, there exist two unbounded sub, 0). Furthermore, Ck± ∩ ([0, ∞) × {0}) = continua Ck± in S bifurcating from R × {0} at ( λkf(h) 0   λk (h) λk (h) + − , 0)}, and if (λ, u) ∈ C \ {( , 0)} C \ {( , 0)} , then u is a (k − 1)-nodal solu{( λkf(h) k k f0 f0 0 tion in (0, 1) satisfying u (0+ ) > 0 (u (0+ ) < 0).

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Let Nk+ (k ∈ N) denote the set of u ∈ E such that u has exactly k − 1 simple interior zeros in (0, 1), u(t) > 0 near 0, and all zeros of u on [0, 1] are simple. Let Nk− = −Nk+ and Nk = Nk− ∪ Nk+ . Then, we see that Nk ∩ Nj = ∅ if k = j . Lemma 3.2. Nk+ is an open set of E. Proof. Let u0 ∈ Nk+ . We choose a neighborhood Nδ (u0 ) = {u ∈ E : u − u0 < δ} of u0 . It suffices to prove that Nδ (u0 ) ⊂ Nk+ . We first consider k ≥ 2. From the definition of Nk+ , u0 has exactly k − 1 simple zeros, i.e., t1 , t2 , . . . , tk−1 in (0, 1) with u 0 (0+ ) = 0, u 0 (1− ) = 0, u 0 (ti ) = 0 for i ∈ {1, 2, . . . , k − 1}. Let m0 = min{|u 0 (0+ )|, |u 0 (1− )|, |u (t1 )|, |u (t2 )|, . . . , |u (tk−1 )|} > 0. We denote by $ ti the positive maximum or negative minimum point of u0 in the interval (ti−1 , ti ) for i = 1, . . . , k, t0 = 0, tk = 1, respectively. Given δ0 > 0, we denote I0 = [0, δ0 ), Ik = (1 − δ0 , 1], and Ii = (ti − δ0 , ti + δ0 ) for i ∈ {1, 2, . . . , k − 1}. We choose δ0 suitably small such that k % Ii and intervals Ii for i ∈ {0, 1, . . . , k} are disjoint and do not contain the |u 0 (t)| > m20 for t ∈ i=0

points $ ti . Then, [$ ti , $ ti+1 ] ∩ Ii = Ii for i ∈ {1, 2, . . . , k − 1}. Claim 1. Every function u ∈ Nδ (u0 ) has no more than one zero in Ii for i ∈ {0, 1, . . . , k} and k % δ < m20 . Thus, it follows that u has no more than k + 1 simple zeros in Ii . i=0

Suppose on the contrary that there exists ti∗ ∈ Ii with u (ti∗ ) = 0. Then, m0 m0 > δ > |u(ti∗ ) − u0 (ti∗ )| + |u (ti∗ ) − u 0 (ti∗ )| > |u 0 (ti∗ )| > . 2 2 This is a contradiction. Claim 2. Every function u ∈ Nδ (u0 ) has no zero on J := [0, 1]/

k %

Ii for δ <

i=0

min |u0 (t)| > 0.

l0 2,

where l0 =

t∈J

By the fact that |u(t)| ≥ |u0 (t)| − |u0 (t) − u(t)| > l0 − δ >

l0 , ∀t ∈ J, 2

we obtain u(t) = 0 for all t ∈ J . ti , $ ti+1 ] for i ∈ {1, 2, . . . , k − Claim 3. Every function u ∈ Nδ (u0 ) has at least one simple zero in [$ 1} and δ < l1 := min |u0 ($ ti )| > 0. Thus, it follows that u has no less than k − 1 simple zeros tk ]. in [$ t1 , $

i=1,2,...,k

ti ,$ ti+1 ] for i ∈ {1, 2, . . . , k − 1}. Suppose on the We show that u ∈ Nδ (u0 ) changes its sign in [$ contrary that u does not change its sign in this interval, then it satisfies either |u($ ti ) − u0 ($ ti )| ≥ l1

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or |u($ ti+1 ) − u0 ($ ti+1 )| ≥ l1 , which contradicts the fact |u(t) − u0 (t)| ≤ u − u0 < δ < l1 for t ∈ [$ ti ,$ ti+1 ]. Now, by choosing 0 < δ < min{ m20 , l20 , l1 } and using Claim 2 and 3, for u ∈ Nδ (u0 ), u has at k k−1 % % least k − 1 simple zeros in [$ t1 , $ tk ] ∩ Ii , i.e., u has at least k − 1 simple zeros in Ii . Thus, i=0

i=1

it follows from Claim 1 and 2 that u has exactly k − 1 simple zeros in (0, 1). It is obvious from Claim 1 that u has a unique zero point 0 in I0 and 1 in Ik and they are simple zeros because |u (0+ )| ≥ |u 0 (0+ )| − |u 0 (0+ ) − u (0+ )| ≥ m0 − δ > m20 and |u (1− )| ≥ |u 0 (1− )| − |u 0 (1− ) − u (1− )| > m20 . Hence, all zeros of u on [0, 1] are simple, and it is easy to check that u (0+ ) > 0. Consequently, u ∈ Nk+ . For the case k = 1, after slight modifications in Claim 1 and 2, we can conclude that u only has zero points 0 and 1 on [0, 1] and they are simple zeros with u (0+ ) > 0. This completes the proof. 2 Similarly, Nk− is open in E. Consequently, Nk is also open in E. By similar arguments as in Lemma 3.3 and 3.4 ([18]), we can get the following lemmas. Lemma 3.3. Let C be a continuum in S bifurcating from {(λ, 0) : λ ∈ R}. Then, C ∩ R × {0} ⊂ %∞ λj (h) j =1 {( f0 , 0)}. Lemma 3.4. For each k ∈ N, assume that there exists a neighborhood Ok of ( λkf(h) , 0) such that 0 , 0) or u ≡ 0 implies either u ∈ Nk+ or u ∈ Nk− . (λ, u) ∈ S ∩ Ok . Then, (λ, u) = ( λkf(h) 0

It is not difficult to check that ∂Nk+ ⊂ {u ∈ E : u ≡ 0 on [0, 1]} ∪ {u ∈ E : u has j (0 ≤ j ≤ k − 1) simple zeros in (0, 1), u(t) > 0 near 0, and has at least one double zero on [0, 1]}. We denote Ck as a continuum in S bifurcating from R × {0} at ( λkf(h) , 0). 0 Lemma 3.5. Assume (λ, u) ∈ Ck , u ∈ ∂Nk+ , and there exists a sequence {(λn , un )} ⊂ S ∩ ([0, ∞) × Nk+ ) converging to (λ, u) in R × E. Then, (λ, u) = ( λkf(h) , 0). 0 Proof. By applying Lemma 2.3, we have u ≡ 0 on [0, 1]. Thus, by Lemma 3.3, λ is a characλ (h) teristic value of L. We now prove λ = λkf(h) . Suppose on the contrary λ = jf0 for j = k. Then, 0 λ (h)

by Lemma 3.4, there exists a neighborhood Oj of ( jf0 , 0) such that if (λ, u) ∈ Oj is a solution of problem (Dλ ) and u ≡ 0, then u ∈ Nj . This contradicts the fact that there exists a sequence {(λn , un )} ⊂ S ∩ ([0, ∞) × Nk+ ) converging to (λ, 0) in R × E. 2 By a similar argument as in Lemma 3.5, we can easily show the following lemma. Lemma 3.6. Assume (λ, u) ∈ Ck and u ∈ ∂Nk− . Also assume that there exists a sequence {(λn , un )} ⊂ S ∩ ([0, ∞) × Nk− ) converging to (λ, u) in R × E. Then, (λ, u) = ( λkf(h) , 0). 0 Lemma 3.7. Ck ⊂ ([0, ∞) × Nk+ ) ∪ ([0, ∞) × Nk− ) ∪ ( λkf(h) , 0). 0 Proof. Suppose on the contrary Ck ⊂ ([0, ∞) × Nk+ ) ∪ ([0, ∞) × Nk− ) ∪ ( λkf(h) , 0). Then, without 0 + loss of generality, from Lemma 3.4, there exist (λ, u) ∈ Ck ∩ ([0, ∞) × ∂Nk ) and a sequence

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19

{(λn , un )} in Ck ∩ ([0, ∞) × Nk+ ), which converges to (λ, u) in R × E. By Lemma 3.5, we have (λ, u) = ( λkf(h) , 0). This is a contradiction. 2 0 Proof of Theorem 3.2. It follows from Theorem 3.1 and Lemma 3.1 that for each k ∈ N, there exists a continuum Ck in S bifurcating from R × {0} at ( λkf(h) , 0), which is either unbounded or 0 $

$

λ(h) $ contains a pair ( λ(h) f0 , 0) for a characteristic value f0 of L with λ(h) = λk (h). From Lemma 3.7, the latter cannot occur so that the former is the only possible alternative, i.e., Ck is unbounded. Now, because of the oddness of function f , we can decompose Ck into two maximal unbounded subcontinua Ck± such that Ck± only meet ( λkf(h) , 0) and Ck+ ∪ Ck− = Ck with Ck+ ⊂ ([0, ∞) × Nk+ ) ∪ 0

, 0)} and Ck− ⊂ ([0, ∞) × Nk− ) ∪ {( λkf(h) , 0)} and this completes the proof. 2 {( λkf(h) 0 0

4. Proof of the asymptotic behavior of solutions

In this section, we show the asymptotic behavior of uβ ∞ . For this, we require the following lemma. Lemma 4.1. Assume that (λ, u) ∈ Ck± \ {( λkf(h) , 0)}. Then for any sufficiently large, there exists 0

a positive b0 = b0 ( ) such that u ∞ ≥ b0 , for all λ > . Proof. Suppose on the contrary that there exists a sequence {(λn , un )} ⊂ Ck+ \ {( λkf(h) , 0)} satis0 fying u n ∞ → 0 as λn → ∞. Then, together with the fact that |un (t)| ≤ t u n ∞ for t ∈ [0, 1], it follows that lim (λn , un ) = (∞, 0).

n→∞

On the other hand, we know that {(λn , un )} satisfies problem (Dλ ) and thus by Lemma 3.6, we can obtain lim λn = λkf(h) , which contradicts lim λn = ∞. 2 0 n→∞

n→∞



Proof of Theorem 1.3. To show the asymptotic behavior of uβ ∞ as λ → ∞ for (λ, u) ∈

, 0)}, without loss of generality, we choose a sequence {(λn , un )} in Ck+ \ {( λkf(h) , 0)} Ck± \ {( λkf(h) 0 0 u

satisfying λn → ∞. It suffices to prove lim βn ∞ = 1. From Lemma 4.1, we may choose n→∞

δ > 0 and N0 ∈ N such that u n ∞ ≥ δ for all n ≥ N0 . From now on, we consider the sequence {(λn , un )} only for all n ≥ N0 . By the definition of Nk+ , un has k − 1 simple zeros on (0, the zeros on [0, 1] by  1). We denote  un ∗∗  u n 0 1 k−1 k ∗∗ 0 = tn < tn < · · · < tn < tn = 1. Let tn ∈ [0, 1] satisfy  β (tn ) = β ∞ . Then, there exists j +1

j ∈ {0, 1, · · · , k − 1} such that tn∗∗ ∈ [tn , tn j

u n

]. From the fact that u n ∞ ≥ δ and the continuity

of u n , it follows β ∞ > 0 and consequently tn∗∗ = 0. j

j +1

Assume that un is positive on (tn , tn ). We may prove for the negative case in a similar j j +1 fashion. Obviously, u n has at least one zero tn∗ on (tn , tn ). We first show that the zero of u n on j j +1 (tn , tn ) exists uniquely. For this, by integrating the first equation in problem (Pλ ) from tn∗ to t j j +1 for t ∈ [tn , tn ], we get

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⎛ u n β

⎜ 1 (t) = −φ −1 ⎝ β(t)

t

⎞

⎟ λn h(τ )f un (τ ) dτ ⎠ .

(4.1)

tn∗

u n u n ∗ j +1 ∗ j +1 β is decreasing on [tn , tn ] and negative on (tn , tn ]. Moreover, we see β u

j j is positive on [tn , tn∗ ), we do not know the monotonicity of βn on [tn , tn∗ ] which is dependent

u j j +1 on β, h, f . Thus, tn∗ is the unique zero of βn on (tn , tn ) and this implies u n has a unique j j +1 zero tn∗ on (tn , tn ). It is not hard to see that {tn∗∗ } has either a subsequence {tn∗∗k } satisfying j j +1 tn∗∗k ∈ [tnk , tn∗k ] for all k, or a subsequence {tn∗∗m } satisfying tn∗∗m ∈ [tn∗m , tnm ] for all m. Thus, writing j the subsequences again {tn∗∗ }, we divide the rest of the argument into two cases: (i) tn∗∗ ∈ [tn , tn∗ ] j +1 for all n and (ii) tn∗∗ ∈ [tn∗ , tn ] for all n. We prove for Case (i), then the proof for Case (ii) can u

be done after some suitable modifications due to the monotonicity of βn . In Case (i), we note u n ∗∗ u n ∗∗ u n u n j ∗ β (tn ) = | β (tn )| = β ∞ , since β is nonnegative on [tn , tn ].

We see that

Define

u n (t), s ∈ [tn∗∗ , tn∗ ]. t∈[s,tn ] β

Fn (s) = max∗

Then, Fn is nonincreasing on [tn∗∗ , tn∗ ]. Furthermore, there exists δ0 > 0 such that 



 u

 u

Fn (tn∗∗ ) =  n (tn∗∗ ) = n ∞ β

β

≥ δ0 .

The fact u n (tn∗ ) = 0 implies Fn (tn∗ ) = 0. Now, we show that Fn is continuous on [tn∗∗ , tn∗ ]. We prove right continuity of Fn on the interval [tn∗∗ , tn∗ ) first. Let t0 ∈ [tn∗∗ , tn∗ ). Then for t > 0 with t0 + t ∈ [tn∗∗ , tn∗ ), we have 0 ≥ Fn (t0 + t) − Fn (t0 ) =

u n u n (t) − max (t). t∈[t0 +t,tn∗ ] β t∈[t0 ,tn∗ ] β max

Take tnM ∈ [t0 , tn∗ ] satisfying u n M u

(tn ) = max∗ n (t). t∈[t0 ,tn ] β β When tnM ∈ [t0 + t, tn∗ ], then Fn (t0 + t) − Fn (t0 ) = 0. When tnM ∈ [t0 , t0 + t], then by the monotonicity of Fn ,

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|Fn (t0 + t) − Fn (t0 )| = Fn (t0 ) − Fn (t0 + t) ≤

u

u n M (tn ) − n (t0 + t) → 0, as t → 0, β β

u

because βn is continuous. Similarly, we can show that Fn is left continuous in (tn∗∗ , tn∗ ], and thus Fn is continuous on [tn∗∗ , tn∗ ]. Therefore by Intermediate Value Theorem, we may take ρn ∈ j (tn∗∗ , tn∗ ) ⊂ (tn , tn∗ ) such that Fn (ρn ) =

δ0 . 2

(4.2)

Now we set ρ∗ = lim inf ρn , n→∞

∗ = lim inf tn∗ , t∞ n→∞

∗∗ t∞ = lim inf tn∗∗ , n→∞

F (s) = lim sup Fn (s). n→∞

∗∗ ≤ ρ ≤ t ∗ . By considering a subsequence if necessary, we have Then, it is obvious that t∞ ∗ ∞

F (ρ∗ ) = lim Fn (ρn ) = n→∞

δ0 , 2

and ∗ F (t∞ ) = lim Fn (tn∗ ). n→∞

∗ , then again considering a subsequence, we have If ρ∗ = t∞

δ0 ∗ ) = lim Fn (tn∗ ) = 0, = F (t∞ n→∞ 2 ∗ . Similarly, if ρ = t ∗∗ , then and this contradiction shows ρ∗ < t∞ ∗ ∞

δ0 ∗∗ ) = lim Fn (tn∗∗ ) ≥ δ0 , = F (t∞ n→∞ 2 ∗∗ . Thus, t ∗∗ < ρ < t ∗ . and this contradiction shows ρ∗ > t∞ ∗ ∞ ∞ ∗ − ρ ), there exists a constant σ such that u (t ∗ − ρ Claim. For any given ρ  ∈ (0, t∞ ) ≥ σ0 for ∗ 0 n ∞ a sufficiently large n.

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∗ −ρ Suppose on the contrary un (t∞ ) → 0 as n → ∞, then for any sufficiently small ρ ∈ j ∗∗ ∗∗ + ρ, t ∗ − ρ (0, ρ∗ − t∞ ) and sufficiently large n, we have [t∞ ] ⊂ [tn∗∗ , tn∗ ] ⊂ [tn , tn∗ ]. ∞ j j +1 By integrating the first equation in problem (Pλ ) from tn∗ to t for t ∈ [tn , tn ], we get

⎛ ⎜ 1 u n (t) = −β(t)φ −1 ⎝ β(t)

t

⎞

⎟ λn h(τ )f un (τ ) dτ ⎠ .

tn∗ j +1

It is seen that un is increasing on (tn , tn∗ ) and decreasing on (tn∗ , tn ). By virtue of the monoj ∗∗ + ρ, t ∗ − ρ tonicity of un on (tn , tn∗ ), we get un (t) → 0 as n → ∞ for all t ∈ [t∞ ]. It follows ∞ that j

t

∗∗ u n (τ )dτ = un (t) − un (t∞ + ρ) → 0,

∗∗ +ρ t∞

∗∗ + ρ, t ∗ − ρ as n → ∞ for all t ∈ [t∞ ]. From the Fatou’s lemma, we obtain ∞

t 0≤ ∗∗ +ρ t∞

lim inf u n (τ )dτ n→∞

∗∗ + ρ, t ∗ − ρ for t ∈ [t∞ ]. In particular, ∞ the interval, we get

t ≤ lim inf n→∞

u n (τ )dτ = 0,

∗∗ +ρ t∞

∗ −  t∞ ρ

∗∗ +ρ lim inf un (τ )dτ t∞ n→∞

= 0. Since u n is nonnegative on

∗∗ ∗ lim inf u n ≡ 0, a.e. on [t∞ + ρ, t∞ −ρ ]. n→∞

∗∗ + ρ, t ∗ − ρ Combining the facts ρ∗ ∈ (t∞ ), lim ρn = ρ∗ and introducing a subsequence if ∞ n→∞ necessary, we may choose N1 large enough satisfying ∗∗ ∗ ρn ∈ (t∞ + ρ, t∞ −ρ ), for all n > N1 ,

and ∗∗ u n → 0 pointwise a.e. on [t∞ + ρ, ρn ]. ∗∗ + ρ, ρ ] satisfying lim u ($ Therefore, we may choose $ t ∈ [t∞ n n t) = 0, i.e., for given ε > 0, there n→∞

exists N2 (> N1 ) such that |u n ($ t)| < ε, for all n > N2 . Let us consider ε =

R 4(N−2) δ0

on [tn , tn∗ ], we obtain j

where δ0 is given in (4.2). Then using the fact that u n is decreasing

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u n N −2

N −2 N −2 δ0 t) < (t) ≤ max u n (t) = un ($ ε= , ∗ ∗ R t∈[$t,tn ] R R 4 t∈[$ t,tn ] β max

(4.3)

for all n > N2 . Since Fn is nonincreasing on [tn∗∗ , tn∗ ] and $ t ≤ ρn for n > N2 , (4.3) implies that δ0 u

δ0 t) = max n (t) < . = Fn (ρn ) ≤ Fn ($ 2 4 t∈[$ t,tn∗ ] β This contradiction completes the proof of Claim. j j +1  ∗∗ ), then [t ∗∗ + ρ , t ∗ − ρ ∗∗ ∗ It is not hard to see that if ρ ∈ (0, ρ∗ − t∞ ∞ 2 ∞ 2 ] ⊂ [tn , tn ](⊂ [tn , tn ]) for ρ  ∗∗ + ρ, t ∗ − ρ ∗∗ + , t ∗ − ρ sufficiently large n. It follows from (4.1) that for all t ∈ [t∞ ](⊂ [t∞ ∞ 2 ∞ 2 ]), ⎞  ⎛   t      un 

λ n ⎜ ⎟  −1  (t) = φ ⎝  h(τ )f un (τ ) dτ ⎠ . β   β(t)   tn∗  ∗ −ρ ∗ −ρ ∗ − ρ Particularly at the point t = t∞  in the above equality, using the fact [t∞ , t∞ 2] ⊂ ∗ ∗ , tn ], we obtain [t∞ − ρ

⎞ ⎛ tn∗     u ∗ 

⎟ λn u

⎜ 1 ≥ n ∞ ≥  n (t∞ −ρ ) = φ −1 ⎝ ∗ h(τ )f un (τ ) dτ ⎠ β β β(t∞ − ρ ) ⎛ ⎜ λn ≥ φ −1 ⎜ ⎝ β(t ∗ − ρ ) ∞

 ∗ −ρ t∞ 2



∗ − t∞ ρ

⎞



⎟ h(τ )f un (τ ) dτ ⎟ ⎠.

(4.4)

∗ − t∞ ρ

∗ ) and for The above claim shows that there exists σ0 > 0 such that un (s) ≥ σ0 for s ∈ (ρ∗ , t∞ ∗ ∗ ∗ sufficiently large n. Since ρ∗ < t∞ − ρ , we get un (s) ≥ σ0 for s ∈ (t∞ − ρ , t∞ − ρ 2 ). Thus ρ  ∗ −  t∞

2 t ∗ − ρ h(τ )f un (τ ) dτ is bounded away from zero for sufficiently large n. This implies ∞

λn ∗ β(t∞ − ρ )

 ∗ −ρ t∞ 2





h(τ )f un (τ ) dτ → ∞ as λn → ∞,

∗ − t∞ ρ

and by the property of φ −1 , we get ⎛ ⎜ φ −1 ⎜ ⎝

λn −ρ )

∗ β(t∞

⎞

 ∗ −ρ t∞ 2





⎟ h(τ )f un (τ ) dτ ⎟ ⎠ → 1, as n → ∞.

∗ − t∞ ρ

u

Consequently, from (4.4), we obtain lim βn ∞ = 1 and the proof is done. n→∞

2

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5. Proofs of other main results In this section, to determine the shape of an unbounded continuum Ck under condition f∞ = 0, we obtain a couple of existence results of the nodal solutions of problem (Pλ ). It is sufficient to solve problem (Pλ ) instead of (Eλ ) because the solutions of (Pλ ) can be transformed into radial solutions of (Eλ ). We emphasize that for problems defined on a bounded domain, all solutions are bounded in the || · || direction [7,9,10,12–14]. This implies that the subcontinuum is unbounded in the λ direction. On the other hand, as mentioned in Introduction, we do not know the boundedness of solutions of problem (Pλ ) for λ due to the presence of β. Therefore, we use the following lemma for an a priori estimate of solutions. The proof can be obtained after suitable modifications to Lemma 3.7 in [18]. Lemma 5.1. Assume (H H ), (F1 ) and (F2 ). Then, for any > 0, there exists M > 0 such that for all solutions u ∈ E of problem (Pλ ) with λ ∈ (0, ), we get u ≤ M . It follows from (F1 ) and (F2 ) that we may choose a constant Lf0 ≥ f0 such that f (s) s ≤ Lf0 for all s = 0. We now make a lower bound of the subcontinuum in the λ direction by using of Lemma 2.5. Lemma 5.2. Assume (H H ), (F1 ) and (F2 ). Let u ∈ E be a (k − 1)-nodal solution of problem (Pλ ) at λ. Then, λ > λLk (h) . f 0

Proof. We consider χ as an eigenfunction corresponding to the eigenvalue λk (h) of problem (3.2). Let λ ≥ 0 and u ∈ E be a (k − 1)-nodal solution of problem (Pλ ). Let {ti } and {si } be the simple zeros of u and χ with 0 = t0 < t1 < · · · < tk = 1 and 0 = s0 < s1 < · · · < sk = 1, respectively. Then, there exist i, j ∈ N such that (ti−1 , ti ) ⊂ (sj −1 , sj ). We rewrite (ti−1 , ti ) by (a, b). By considering y = u, b1 (t) = λh(t)f (u(t))/u(t), z = χ , and b2 (t) = λk (h)h(t) in Lemma 2.5, we have b (b1 (t) − b2 (t)) |u|2 dt > 0. a

Since b1 (t) = λh(t)f (u(t))/u(t) ≤ λLf0 h(t), we get b

b 2

λLf0 a

a

This implies that λ> and the proof is complete.

2

h(t)|u|2 dt.

h(t)|u| dt > λk (h)

λk (h) , Lf0

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Proof of Theorem 1.1. Note that Theorem 3.2 implies the existence of two unbounded continua Ck±in the closure of the set of (k − 1)-nodal solutions of problem (Pλ ) bifurcating from  λk (h) f0 , 0

. By Lemma 5.2, problem (Pλ ) has no (k − 1)-nodal solution for λ ≤

λk (h) Lf0 .

Define

λ∗ = sup{λ > 0 : S ∩ Ck± ⊂ (λ, ∞) × E}. Then we deduce that the problem has no (k − 1)-nodal solution for all λ ∈ (0, λ∗ ). From , Lemma 5.1, we also deduce that Ck± is unbounded in the λ direction. Therefore, for all λ > λkf(h) 0 + − problem (Pλ ) has at least one (k − 1)-nodal solution in Nk and Nk . This completes the proof. 2 Proof of Theorem 1.2. Suppose on the contrary that there exists a (k − 1)-nodal solution u at . Let {ti } and {si } be the zeros of u and χ with 0 = t0 < t1 < · · · < tk = 1 and 0 = λ ≤ λkf(h) 0 s0 < s1 < · · · < sk = 1, respectively. Then, there exist i, j ∈ N such that (ti−1 , ti ) ⊂ (sj −1 , sj ). We rewrite (ti−1 , ti ) by (a, b). By considering y = u, b1 (t) = λh(t)f (u(t))/u(t), z = χ , and b2 (t) = λk (h)h(t) in Lemma 2.5, we get b (b1 (t) − b2 (t)) |u|2 dt > 0. a

On the other hand, since b1 (t) = λh(t)f (u(t))/u(t) ≤ λf0 h(t), we get b (b1 (t) − b2 (t)) |u|2 dt a

b =

(λf (u(t))/u(t) − λk (h)) h(t)|u|2 dt a

b ≤

(λf0 − λk (h)) h(t)|u|2 dt ≤ 0. a

This is a contradiction.

2

Remark 5.1. The main results are valid even if we drop the oddness on f . From Dancer’s unilateral global bifurcation theorem [19], we can decompose Ck into two maximal subcontinua Ck+ and Ck− , which are both unbounded in the proof of Theorem 3.2. Specifically, we conclude it by eliminating the case Ck+ ∩ Ck− = {( λkf(h) , 0)} because Ck+ ⊂ ([0, ∞) × Nk+ ) ∪ {( λkf(h) , 0)} and 0 0 , 0)} by Lemma 3.7 and 3.2. Ck− ⊂ ([0, ∞) × Nk− ) ∪ {( λkf(h) 0

Finally, we provide an example of the case when f is not odd to illustrate the main results.

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Example 5.1. Let us consider the problem  ⎧ ⎨−div 

∇v 1−|∇v|2

⎩v|∂ = 0,



3

= λk(|x|) 2+vv+v 4 +sin v ,

in , (E1 )

lim v(x) = 0,

|x|→∞

where  = {x ∈ RN : |x| > R}, R > 0, N ≥ 3, λ a positive real parameter, and k satisfies (H K). f (s) 1+s 2 Since f (s) s = 2+s 4 +sin s , we get f0 = 0.5, f∞ = 0, and Lf0 = max s ≈ 0.97286. By applys∈R

ing the results of Theorem 1.1 and Lemma 5.2, problem (E1 ) has no (k − 1)-nodal radial solution for all λ ∈ (0, λLk (h) ], and at least two (k − 1)-nodal radial solutions for all λ ∈ (2λk (h), ∞) in f Nk+ and Nk− .

0

Acknowledgment The authors are very grateful to the anonymous referees and editors for their valuable comments and very professional suggestions that improves the original manuscript. The second author was supported by the National Research Foundation of Korea, Grant funded by the Korea Government (MEST) (NRF2016R1D1A1B04931741). The third author was supported by the National Research Foundation of Korea, Grant funded by the Korea Government (MEST) (NRF2018R1D1A3A03000678). References [1] R. Bartnik, L. Simon, Spacelike hypersurfaces with prescribed boundary values and mean curvature, Commun. Math. Phys. 87 (1982) 131–152. [2] A.E. Treibergs, Entire spacelike hypersurfaces of constant mean curvature in Minkowski space, Invent. Math. 66 (1982) 39–56. [3] C. Gerhardt, H-surfaces in Lorentzian manifolds, Commun. Math. Phys. 89 (1983) 523–553. [4] C. Bereanu, P. Jebelean, J. Mawhin, Radial solutions for Neumann problems involving mean curvature operators in Euclidean and Minkowski spaces, Math. Nachr. 283 (2010) 379–391. [5] C. Bereanu, P. Jebelean, J. Mawhin, Multiple solutions for Neumann and periodic problems with singular ϕ-Laplacian, J. Funct. Anal. 261 (2011) 3226–3246. [6] C. Bereanu, P. Jebelean, J. Mawhin, Radial solutions of Neumann problems involving mean extrinsic curvature and periodic nonlinearities, Calc. Var. 46 (2013) 113–122. [7] I. Coelho, C. Corsato, F. Obersnel, P. Omari, Positive solutions of the Dirichlet problem for one-dimensional Minkowski-curvature equation, Adv. Nonlinear Stud. 12 (2012) 621–638. [8] I. Coelho, C. Corsato, S. Rivetti, Positive radial solutions of the Dirichlet problem for the Minkowski-curvature equation in a ball, Topol. Methods Nonlinear Anal. 44 (2014) 23–39. [9] C. Bereanu, P. Jebelean, P.J. Torres, Positive radial solutions for Dirichlet problems with mean curvature operators in Minkowski space, J. Funct. Anal. 264 (2013) 270–287. [10] C. Bereanu, P. Jebelean, P.J. Torres, Multiple positive radial solutions for a Dirichlet problem involving the mean curvature operator in Minkowski space, J. Funct. Anal. 265 (2013) 644–659. [11] C. Bereanu, P. Jebelean, J. Mawhin, Radial solutions for some nonlinear problems involving mean curvature operators in Euclidean and Minkowski spaces, Proc. Am. Math. Soc. 137 (2009) 161–169. [12] R. Ma, H. Gao, Y. Lu, Global structure of radial positive solutions for a prescribed mean curvature problem in a ball, J. Funct. Anal. 270 (2016) 2430–2455. [13] G. Dai, Bifurcation and positive solutions for problem with mean curvature operator in Minkowski space, Calc. Var. 55 (2016) 1–17. [14] G. Dai, J. Wang, Nodal solutions to problem with mean curvature operator in Minkowski space, Differ. Integral Equ. 30 (2017) 463–480.

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