Nonlinear Analysis 67 (2007) 3350–3356www.elsevier.com/locate/na

Existence of nodal solutions for Lidstone eigenvalue problems

Jia Xua,b,∗, XiaoLing Hana

a Mathematics and Information Science College, Northwest Normal University, Lanzhou 730070, Gansu, People’s Republic of Chinab College of Physical Education, Northwest Normal University, Lanzhou 730070, Gansu, People’s Republic of China

Received 31 August 2006; accepted 23 October 2006

Abstract

We are concerned with determining values of r for which there exist nodal solutions of the Lidstone eigenvalue problem

(−1)mu(2m)(x) = ra(x) f (u(x)), 0 < x < 1,

u(2i)(0) = u(2i)(1) = 0, i = 0, 1, 2, . . . , m − 1.

The proof of our main result is based upon eigenvalue theory and bifurcation techniques.c© 2007 Published by Elsevier Ltd

MSC: 34B16; 34B18

Keywords: Lidstone eigenvalue problem; Nodal solution; Existence; Bifurcation methods

1. Introduction

The Lidstone boundary value problem:

(−1)mu(2m)(x) = f (x, u(x)), 0 < x < 1, m ∈ N ,

u(2i)(0) = u(2i)(1) = 0, i = 0, 1, 2, . . . , m − 1

has been given considerable attention by many authors, See [1–4] and references therein. The discussions in thesepapers are mainly concerned with the existence and multiplicity of positive solutions addressed by using differentmethods, such as those of topology degree theory, critical point theory and the fixed point theorem in cones. However,the existence of sign-changing solutions for this problem have not been discussed.

Recently, in [5], Ma and Thompson were concerned with determining values of r for which there exist nodalsolutions of the boundary value problem

u′′+ ra(x) f (u(x)) = 0, 0 < x < 1, (1.1)

u(0) = u(1) = 0 (1.2)

∗ Corresponding address: Northwest Normal University, Department of Mathematics, 730070 Lanzhou, Gansu, People’s Republic of China.Tel.: +86 0931 3435212.

E-mail address: xujia@nwnu.edu.cn (J. Xu).

0362-546X/$ - see front matter c© 2007 Published by Elsevier Ltddoi:10.1016/j.na.2006.10.017

J. Xu, X. Han / Nonlinear Analysis 67 (2007) 3350–3356 3351

under the assumptions

(H1) f ∈ C(R, R) with s f (s) > 0 for s 6= 0;(H2) there exist f0, f∞ ∈ (0, ∞) such that

f0 = lim|s|→0

f (s)s

, f∞ = lim|s|→∞

f (s)s

;

(H3) a : [0, 1] −→ [0, ∞) is continuous and does not vanish identically on any subinterval of [0, 1].

Let µk be the kth eigenvalue of

u′′+ ra(x)u(x) = 0, 0 < x < 1,

u(0) = u(1) = 0.

By using bifurcation methods, they established the following result:

Theorem A. Let (H1), (H2) and (H3) hold. Assume that for some k ∈ N, eitherµk

f∞< r <

µk

f0,

orµk

f0< r <

µk

f∞.

Then (1.1) and (1.2) has two solutions u+

k and u−

k such that u+

k has exactly k − 1 simple zeros in (0, 1) and is positivenear 0, and u−

k has exactly k − 1 simple zeros in (0, 1) and is negative near 0.Inspired by [5], in this paper, we study the existence and multiplicity of nodal solutions of the Lidstone eigenvalue

problem

(−1)mu(2m)(x) = ra(x) f (u(x)), 0 < x < 1, (1.3)

u(2i)(0) = u(2i)(1) = 0, i = 0, 1, 2, . . . , m − 1 (1.4)

under the same conditions (H1), (H2) and (H3), where m ≥ 1 is an integer and r ∈ R is a parameter.Let λk be the kth eigenvalue of

(−1)mu(2m)(x) = ra(x)u(x), 0 < x < 1,

u(2i)(0) = u(2i)(1) = 0, i = 0, 1, 2, . . . , m − 1

and let ϕk be an eigenfunction corresponding to λk . By [6], we know that

0 < λ1 < λ2 < · · · < λk < λk+1 < · · · , limk→∞

λk = ∞

and that ϕk has exactly k − 1 simple zeros in (0, 1).Applying the bifurcation theorem of Rabinowitz [7,8], we will establish the following result:

Theorem 1.1. Let (H1), (H2) and (H3) hold. Assume that for some k ∈ N, either

λk

f∞< r <

λk

f0, (1.5)

or

λk

f0< r <

λk

f∞. (1.6)

Then (1.3) and (1.4) has two solutions u+

k and u−

k such that u+

k has exactly k − 1 simple zeros in (0, 1) and is positivenear 0, and u−

k has exactly k − 1 simple zeros in (0, 1) and is negative near 0.

Remark 1.1. For m = 1, Theorem 1.1 is the main result of [5] (Theorem A).

Remark 1.2. When the nodal number is zero (i.e. k = 1), the nodal solution is a positive solution.

3352 J. Xu, X. Han / Nonlinear Analysis 67 (2007) 3350–3356

2. Preliminary results

To prove our main result, we need following preliminary results.Assume that Gm(x, s) is the Green’s function of the homogeneous boundary value problem

(−1)mu(2m)(x) = 0, 0 ≤ b < x < c ≤ 1, (2.1)

u(2i)(b) = u(2i)(c) = 0, i = 0, 1, 2, . . . , m − 1, (2.2)

where b, c are constants in [0, 1].The Green’s function Gm(x, s) can be expressed as [4,9]

Gm(x, s) =

∫ c

bG(x, δ)Gm−1(δ, s)dδ, m ≥ 2, (2.3)

where

G1(x, s) = G(x, s) =1

c − b

{(x − b)(c − s), b ≤ x ≤ s ≤ c,(s − b)(c − x), b ≤ s ≤ x ≤ c. (2.4)

Lemma 2.1. For any (x, s) ∈ [b, c] × [b, c], we have

Gm(x, s) ≤

[∫ c

bG(δ, δ)dδ

]m−1

G(s, s), ∀m ∈ N . (2.5)

Proof. For any (x, s) ∈ [b, c] × [b, c], it is clear from Eq. (2.4) that G1(x, s) ≤ G(s, s), i.e. inequality (2.5) is truefor m = 1.

Assume that inequality (2.5) holds for m = k. Then, for m = k + 1, we have

Gk+1(x, s) =

∫ c

bG(x, δ)Gk(δ, s)dδ

≤

∫ c

bG(δ, δ)

[∫ c

bG(δ, δ)dδ

]k−1

G(s, s)dδ

=

[∫ c

bG(δ, δ)dδ

]k

G(s, s),

so inequality (2.5) is true for m = k + 1. �

Lemma 2.2. For any (x, s) ∈ [b +c−b

4 , c −c−b

4 ] × [b, c], we have

Gm(x, s) ≥

(c − b

4

)m[∫ c− c−b

4

b+c−b

4

G(δ, δ)dδ

]m−1

G(s, s), ∀m ∈ N . (2.6)

Proof. For any (x, s) ∈ [b +c−b

4 , c −c−b

4 ] × [b, c], it is clear from Eq. (2.4) that G1(x, s) ≥(c−b)

4 · G(s, s),i.e. inequality (2.6) is true for m = 1.

Assume that inequality (2.6) holds for m = k. Then, for m = k + 1, we have

Gk+1(x, s) ≥

∫ c− c−b4

b+c−b

4

G(x, δ)Gk(δ, s)dδ

≥

∫ c− c−b4

b+c−b

4

(c − b)

4G(δ, δ) ·

(c − b

4

)k[∫ c− c−b

4

b+c−b

4

G(δ, δ)dδ

]k−1

G(s, s)dδ

=

(c − b

4

)k+1[∫ c− c−b

4

b+c−b

4

G(δ, δ)dδ

]k

G(s, s),

J. Xu, X. Han / Nonlinear Analysis 67 (2007) 3350–3356 3353

so inequality (2.6) is true for m = k + 1. �

Lemma 2.3. Let a satisfy (H3), and let gn ∈ C([0, 1], [0, ∞)) be such that

gn(x) ≥ ρ, x ∈ [0, 1], (2.7)

for some ρ > 0. Suppose that the sequence {(µn, yn)} satisfies

(−1)m y(2m)n (x) = µna(x)gn(x)yn(x), y(2i)

n (0) = y(2i)n (1) = 0, i = 0, 1, 2, . . . , m − 1, (2.8)

with either

(yn|I )(x) > 0 for all n sufficiently large (2.9)

or

(yn|I )(x) < 0 for all n sufficiently large, (2.10)

where I := [α, β] with α < β is a given closed subinterval of (0, 1). Then

|µn| ≤ M0, (2.11)

for some positive constant M0.

Proof. We only deal with the case where (yn|I )(x) > 0 for all n sufficiently large. The other case can be treated in asimilar way. We may assume that (yn|I )(x) > 0 for all n ∈ N .

Let (αn, βn) be a subinterval of [0, 1] satisfying

(i) I ⊂ (αn, βn);(ii) y(2i)

n (αn) = y(2i)n (βn) = 0, i = 0, 1, 2, . . . , m − 1;

(iii) yn(x) > 0, ∀x ∈ (αn, βn).

Let Gm(x, s) be the Green’s function of

(−1)mu(2m)(x) = 0, x ∈ (αn, βn), (2.12)

u(2i)(αn) = u(2i)(βn) = 0, i = 0, 1, 2, . . . , m − 1; (2.13)

then

Gm(x, s) =

∫ βn

αn

G(x, δ)Gm−1(δ, s)dδ, (2.14)

where

G1(x, s) = G(x, s) =1

βn − αn

{(x − αn)(βn − s), αn ≤ x ≤ s ≤ βn,

(s − αn)(βn − x), αn ≤ s ≤ x ≤ βn(2.15)

and for any (x, s) ∈ [α +β−α

8 , β −β−α

8 ] × [α +β−α

4 , β −β−α

4 ], we have

G1(x, s) = G(x, s) ≥(β − α)2

32.

Further, we have

Gm(x, s) ≥(β − α)(3m−1)

32m · 2(m−1)=

(β − α)(3m−1)

2(6m−1). (2.16)

By (2.8), we have

yn(x) = µn

∫ βn

αn

Gm(x, s)a(s)gn(s)yn(s)ds. (2.17)

3354 J. Xu, X. Han / Nonlinear Analysis 67 (2007) 3350–3356

Through (2.16) and (2.17) and Lemmas 2.1 and 2.2, we have

minx∈[αn+

βn−αn4 ,βn−

βn−αn4 ]

[yn(x)] ≥ γ ‖(yn|[αn ,βn ])‖0, (2.18)

where γ =(β−α)(2m−1)

32(m−1)·4m =(β−α)(2m−1)

2(7m−5) .Since yn|I is concave down and positive, we have that

minx∈[α+

β−α4 ,β−

β−α4 ]

[yn(x)] ≥ γ ‖(yn|I )‖0 ≥ γ ‖(yn|[α+

β−α8 ,β−

β−α8 ]

)‖0. (2.19)

By (2.7), (2.16), (2.17) and (2.19), for any x ∈ [α +β−α

8 , β −β−α

8 ], we have that

yn(x) ≥ µn

∫ β−β−α

4

α+β−α

4

Gm(x, s)a(s)ρyn(s)ds

≥ µn ·(β − α)(3m−1)

32m · 2(m−1)· ρ · γ

∫ β−β−α

4

α+β−α

4

a(s)ds‖(yn|[α+

β−α8 ,β−

β−α8 ]

)‖0

= µn ·(β − α)(5m−2)

2(13m−6)· ρ

∫ β−β−α

4

α+β−α

4

a(s)ds‖(yn|[α+

β−α8 ,β−

β−α8 ]

)‖0.

Therefore

|µn| ≤ 2(13m−6)

(ρ(β − α)(5m−2)

∫ β−β−α

4

α+β−α

4

a(s)ds

)−1

. �

3. Proof of the main result

Let Cr[0, 1] = {u | u : [0, 1] −→ R is an r -times continuously differentiable function} with the norm

‖u‖r = max{‖u‖0, ‖u′‖0, . . . , ‖u(r)

‖0},

where ‖u‖0 = max{|u(x)| | 0 ≤ x ≤ 1}.Let Y = C0

[0, 1] with the norm ‖u‖0, E = {u ∈ C2m−1[0, 1] | u satisfy (1.4)} with the norm ‖u‖2m−1.

Define L : D(L) → Y by setting

Lu := (−1)mu(2m), u ∈ D(L),

where

D(L) = {u ∈ C2m[0, 1] | u satisfy (1.4)}.

Then L−1: Y −→ E is compact.

Let ζ, ξ ∈ C(R) be such that

f (u) = f0u + ζ(u); f (u) = f∞u + ξ(u). (3.1)

Clearly

lim|u|→0

ζ(u)

u= 0, lim

|u|→∞

ξ(u)

u= 0. (3.2)

Let

ξ̃ (u) = max0≤|s|≤u

|ξ(s)|;

J. Xu, X. Han / Nonlinear Analysis 67 (2007) 3350–3356 3355

then ξ̃ is nondecreasing and

limu→∞

ξ̃ (u)

u= 0. (3.3)

Let us consider

Lu − λa(x)r f0u = λa(x)rζ(u) (3.4)

as a bifurcation problem from the trivial solution u ≡ 0.Eq. (3.4) can be converted to the equivalent equation

u(x) =

∫ 1

0Gm(x, s)[λa(s)r f0u(s) + λa(s)rζ(u(s))]ds

:= λL−1[a(·)r f0u(·)](x) + λL−1

[a(·)rζ(u(·))](x).

Further we note that ‖L−1[a(·)ζ(u(·))]‖2m−1 = o(‖u‖2m−1) for u near 0 in E .

In what follows, we use the terminology of Rabinowitz [7,8]. Let E = R × E under the product topology. Let S+

kdenote the set of functions in E which have exactly k − 1 interior nodal (i.e. nondegenerate) zeros in (0, 1) and arepositive near t = 0, set S−

k = −S+

k , and Sk = S+

k⋃

S−

k . They are disjoint and open in E . Finally, let Φ±

k = R × S±

kand Φk = R × Sk .

The results of Rabinowitz [7,8] for (3.4) can be stated as follows: For each integer k ≥ 1, ν = {+, −}, there existsa continuum Cν

k ⊆ Φνk of solutions of (3.4), joining (

λkr f0

, 0) to infinity in Φνk . Moreover, Cν

k \(λkr f0

, 0) ⊂ Φνk .

Proof of Theorem 1.1. The proof is similar to the proof of Theorem 1.1 of [5]; the main difference is the norm‖ · ‖C2m−1 which we used in this paper instead of ‖ · ‖C1 in [5]. So we omit the proof here. �

4. Examples

Example 4.1. Consider the boundary value problem

y(4)(t) = r(y + tanh 2y), t ∈ (0, 1), (4.1)

y(0) = y(1) = y′′(0) = y′′(1) = 0. (4.2)

In this example, m = 2. By choosing a(t) ≡ 1, t ∈ (0, 1), f (y) = y + tanh 2y, we find f0 = 2, f∞ = 1. Clearly,conditions (H1), (H2) and (H3) can be satisfied.

For k = 1, we know λ1 = π4. By Theorem 1.1, when r ∈ (π4

2 , π4), the boundary value problem (4.1) and (4.2)has a positive solution u+

1 and a negative solution u−

1 , and u+

1 , u−

1 have no zeros in (0, 1).For k = 2, we know λ2 = 16π4. By Theorem 1.1, when r ∈ (8π4, 16π4), the boundary value problem (4.1) and

(4.2) has two solutions u+

2 and u−

2 , u+

2 has exactly one simple zero in (0, 1) and is positive near 0, u−

2 has exactly onesimple zero in (0, 1) and is negative near 0.

Example 4.2. Consider the boundary value problem

y(8)(t) = r(y3+ y), t ∈ (0, 1), (4.3)

y(0) = y(1) = y′′(0) = y′′(1) = y(4)(0) = y(4)(1) = y(6)(0) = y(6)(1) = 0. (4.4)

In this example, m = 4. By choosing a(t) ≡ 1, t ∈ (0, 1), f (y) = y3+ y, we find f0 = 1, f∞ = ∞. Clearly,

conditions (H1), (H2) and (H3) can be satisfied.For k = 1, we know λ1 = π8. By Theorem 1.1, when r ∈ (0, π8), the boundary value problem (4.3) and (4.4) has

a positive solution u+

1 and a negative solution u−

1 , and u+

1 , u−

1 have no zeros in (0, 1).For k = 2, we know λ2 = 256π8. By Theorem 1.1, when r ∈ (0, 256π8), the boundary value problem (4.3) and

(4.4) has two solutions u+

2 and u−

2 , u+

2 has exactly one simple zero in (0, 1) and is positive near 0, u−

2 has exactly onesimple zero in (0, 1) and is negative near 0.

3356 J. Xu, X. Han / Nonlinear Analysis 67 (2007) 3350–3356

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