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Global Bifurcation Results for General Laplacian Problems Fixed Point Theory and Applications 2012, 2012:7 doi:10.1186/1687-1812-2012-7 Eun Kyoung Lee eunkyoung165@gmail.comYong-Hoon Lee

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Global Bifurcation Results for General Laplacian Problems

Fixed Point Theory and Applications 2012, 2012:7 doi:10.1186/1687-1812-2012-7

Eun Kyoung Lee (eunkyoung165@gmail.com)Yong-Hoon Lee (yhlee@pusan.ac.kr)Byungjae Son (mylife1882@hanmail.net)

ISSN 1687-1812

Article type Research

Submission date 23 December 2010

Acceptance date 18 January 2012

Publication date 18 January 2012

Article URL http://www.fixedpointtheoryandapplications.com/content/2012/1/7

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Global bifurcation results for general

Laplacian problems

Eun Kyoung Lee1, Yong-Hoon Lee2∗ and Byungjae Son2

1Department of Mathematics Education, Pusan National University,

for the following general Laplacian problem,

(P )

where f : [0, 1] × R × R → R is continuous and ϕ, ψ : R → R are odd increasing

homeomorphisms of R, when ϕ, ψ satisfy the asymptotic homogeneity conditions.

(P )

where f : [0, 1] × R × R → R is continuous with f(t, u, 0) = 0 and ϕ, ψ : R → R are

odd increasing homeomorphisms of R with ϕ(0) = ψ(0) = 0 We consider the following

conditions;

(Φ1) limt →0 ϕ(σt) ψ(t) = σ p −1 , for all σ ∈ R+, for some p > 1.

(Φ2) lim|t|→∞ ϕ(σt) ψ(t) = σ q −1 , for all σ ∈ R+, for some q > 1.

(F1) f (t, u, λ) = ◦(|ψ(u)|) near zero, uniformly for t and λ in bounded intervals.

(F2) f (t, u, λ) = ◦(|ψ(u)|) near infinity, uniformly for t and λ in bounded intervals.

(F3) uf (t, u, λ) ≥ 0.

We note that ϕ r (t) = |t| r −2 t, r > 1 are special cases of ϕ and ψ We first prove following

global bifurcation result

Theorem 1.1 Assume (Φ1), (Φ2), (F1), (F2) and (F3) Then for any j ∈ N, there exists

a connected component C j of the set of nontrivial solutions for (P ) connecting (0, λ j (p)) to

(∞, λ j (q)) such that (u, λ) ∈ C j implies that u has exactly j − 1 simple zeros in (0, 1), where

λ j (r) is the j-th eigenvalue of (ϕ r (u ′ (t))) ′ + λϕ r (u(t)) = 0 and u(0) = u(1) = 0.

By the aid of this theorem, we can prove the following existence result of solutions

Theorem 1.2 Consider problem

(A)

where g : [0, 1] × R × R → R is continuous and ϕ is odd increasing homeomorphism of R,

which satisfy (Φ1) and (Φ2) with ϕ = ψ Also ug(t, u) ≥ 0 and there exist positive integers

k, n with k ≤ n such that µ = lim s →0 g(t,s) ϕ(s) < λ k (p) ≤ λ n (q) < lim |s|→∞ g(t,s) ϕ(s) = ν uniformly

in t ∈ [0, 1] Then for each integer j with k ≤ j ≤ n, problem (A) has a solution with exactly

j − 1 simple zeros in (0, 1) Thus, (A) possesses at least n − k + 1 nontrivial solutions.

In [1], the authors studied the existence of solutions and global bifurcation results for

The main purpose of this article is to derive the same result for N = 1 case with Dirichlet

boundary condition which was not considered in [1]

For p-Laplacian problems, i.e., ϕ = ψ = ϕ p, many authors have studied for the existenceand multiplicity of nontrivial solutions [2–6] In [2, 5, 6], the authors used fixed point theory

or topological degree argument Also global bifurcation theory was mainly employed in [3, 4].Moreover, there are some studies related to general Laplacian problems [3, 7, 8], but most

of them are about ϕ = ψ case In [3], the authors proved some results under several kinds

of boundary conditions and in [7], the authors considered a system of general Laplacianproblems In [8], the author studied global continuation result for the singular problem

In this paper, we mainly study the global bifurcation phenomenon for general Laplacian

problem (P ) and prove the existence and multiplicity result for (A).

This article is organized as follows: In Section 2, we set up the equivalent integral operator

of (P ) and compute the degree of this operator In Section 3, we verify the existence of global

bifurcation having bifurcation points at zero and infinity simultaneously In Section 4, we

introduce an existence result as an application of the previous result and give some examples.

where a : L1(0, 1) → R is a continuous function which sends bounded sets of L1 into bounded

sets of R and satisfying

0[0, 1] is continuous and maps equi-integrable sets of

L1(0, 1) into relatively compact sets of C1

0[0, 1] One may refer Man´asevich-Mawhin [4, 3]

and Garcia-Huidobro-Man´asevich-Ward [7] for more details If we define the operator T λ

then (P ) is equivalently written as u = T λ

ϕψ (u) Now let us consider p-Laplacian problem

p are completely continuous

The main purpose of this section is to compute the Leray-Schauder degree of I − T λ

ϕψ

Following Lemma is for the property of ϕ and ψ with asymptotic homogeneity condition (Φ1)

and (Φ2), which is very useful for our analysis The proof can be modified from Proposition4.1 in [9].

Lemma 2.1 Assume that ϕ, ψ are odd increasing homeomorphisms of R which satisfy (Φ1)

and (Φ2) Then, we have

To compute the degree, we will make use of the following well-known fact [10]

Lemma 2.2 If λ is not an eigenvalue of (E p ), p > 1 and r > 0, then

Now, let us compute deg(I − T λ

ϕψ , B(0, r), 0) when λ is not an eigenvalue of (E p)

(Φ1) and (Φ2) Then,

(i) The Leray-Schauder degree of I − T λ

ϕψ is defined for B(0, ε), for all sufficiently small ε.

(ii) The Leray-Schauder degree of I − T λ

ϕψ is defined for B(0, M ), for all sufficiently large

0[0, 1] for all small ε Indeed,

suppose there exist sequences {u n }, {τ n } and {ε n } with ε n → 0 and ∥u n ∥0 = ε n such that

Now, we show that {v ′

n } is uniformly bounded Since ∥v n ∥0 = 1, ∫t

≥ A, for all n > N0 This

implies that 2λ ≥ ϕ(Aε n)

ψ(ε n) for all n > N0 However, ϕ(Aε n)

ψ(ε n) → ϕ p (A) as n → ∞ This is a

contradiction Thus by the above inequality, we get

for some C2 > 0 Therefore, {v ′

n } is uniformly bounded By the Arzela-Ascoli Theorem, {v n }

has a uniformly convergent subsequence in C[0, 1] relabeled as the original sequence so let

limn →∞ v n = v Now, we claim that q n (t) → q(t), where

Since |a(−λψ(u n))| ≤ λψ(ε n), a( −λψ(u n))

ϕ(ε n) has a convergent subsequence Without loss ofgenerality, we say that the sequence { a( −λψ(u n))

ϕ(ε n) } converges to d Also by the facts that

ds = 0 and by the definition of a p , d = a p(−λϕ p (v)).

Therefore, we can easily see that

Consequently, v is a solution of (E p ) Since λ / ∈ {λ n (p) }, v ≡ 0 and this fact yields a

contradiction By the properties of the Leray-Schauder degree, we get

We begin with this section recalling what we mean by bifurcation at zero and at infinity Let

X be a Banach space with norm ∥ · ∥, and let F : X × I → X be a completely continuous

operator, where I is some real interval Consider the equation

Definition 3.1 Suppose that F(0, λ) = 0 for all λ in I, and that ˆ λ ∈ I We say that (0, ˆλ)

is a bifurcation point of (9) at zero if in any neighborhood of (0, ˆ λ) in X × I, there is a nontrivial solution of (9) Or equivalently, if there exist sequences {x n ̸= 0} and {λ n } with

(∥x n ∥, λ n)→ (0, ˆλ) and such that (x n , λ n ) satisfies (9) for each n ∈ N.

Definition 3.2 We say that ( ∞, ˆλ) is a bifurcation point of (9) at infinity if in any borhood of ( ∞, ˆλ) in X × I, there is a nontrivial solution of (9) Equivalently, if there exist sequences {x n ̸= 0} and {λ n } with (∥x n ∥, λ n) → (∞, ˆλ) and such that (x n , λ n ) satisfies (9)

completely continuous operator

Lemma 3.3 (i) Assume (Φ1) and (F1) If (0, ˆ λ) is a bifurcation point of (P ), then ˆ λ = λ n (p)

for some p ∈ N.

(ii) Assume (Φ2) and (F2) If ( ∞, ˆλ) is a bifurcation point of (P ), then ˆλ = λ n (q) for some

q ∈ N.

Proof: We prove assertion (i) Suppose that (0, ˆ λ) is a bifurcation point of (P ) Then there

exists a sequence {(u n , λ n)} in C1

0[0, 1] × R with (u n , λ n) → (0, ˆλ) and such that (u n , λ n)

satisfies u n = F(u n , λ n ) for each n ∈ N Equivalently, (u n , λ n) satisfies

0 −λ n ψ(u n (ξ)) − f(ξ, u n , λ n )dξ Since f (t, u, λ) = ◦(|ψ(u)|) near zero,

uniformly for t and λ, for some constants K1 and K2

→ ∞ as n → ∞, then for arbitrary

A > 0, there exists N0 ∈ N such that

This implies that 2K2 ≥ ϕ(Aε n)

ψ(ε n) , for all n ≥ N0 This is impossible Thus

n } is uniformly bounded and by the Arzela-Ascoli Theorem, {v n } has a

uniformly convergent subsequence in C[0, 1] Let v n → v in C[0, 1] Now claim that

Since a( −λ n ψ(u n)−f(·,u n ,λ n))

ψ(ε n) is bounded, considering a subsequence if necessary, we may assume

that sequence { a( −λ n ψ(u n)−f(·,u n ,λ n))

ψ(ε n) } converges to d as n → ∞ This implies that

The converse of first part of Theorem 3.3 is true in our problem

Lemma 3.4 Assume (Φ1) and (F1) If µ is an eigenvalue of (E p ), then (0, µ) is a bifurcation

point.

Proof: Suppose that (0, µ) is not a bifurcation point of (P ) Then there is a neighborhood

of (0, µ) containing no nontrivial solutions of (P ) In particular, we may choose an ε-ball B ε such that there are no solutions of (P ) on ∂B ε × [µ − ε, µ + ε] and µ is the only eigenvalue of

(E p ) on [µ − ε, µ + ε] Let Φ(u, λ) = u − F(u, λ) Then deg(Φ(·, λ), B(0, ε), 0) is well-defined

for λ with |λ − µ| ≤ ε Moreover, from the homotopy invariance theorem,

deg(Φ(·, λ), B(0, ε), 0) ≡ constant, for all λ with |λ − µ| ≤ ε.

Now, we claim that

deg(Φ(·, µ − ε), B(0, ε), 0) = deg(Φ p(·, µ − ε), B(0, ε), 0),

We know that F(·, µ−ε) and T µ−ε

p are completely continuous To apply the homotopy

invari-ance theorem, we need to show that 0 / ∈ u − H µ −ε (u, τ )(∂B

Hence, we obtain that

and we see that {v ′

n } is uniformly bounded Therefore, by the Arzela-Ascoli Theorem, {v n }

has a uniformly convergent subsequence in C[0, 1] Without loss of generality, let v n → v.

Moreover, using the fact that

This implies v ≡ 0 and this is a contradiction Consequently, deg(I − H µ −ε(·, τ), B(0, ε), 0)

is well defined Therefore, by the homotopy invariance theorem,

deg(Φ(·, µ − ε), B(0, ε), 0) = deg(Φ p(·, µ − ε), B(0, ε), 0).

Similarly,

deg(Φ(·, µ + ε), B(0, ε), 0) = deg(Φ p(·, µ + ε), B(0, ε), 0).

Let µ is k-th eigenvalue of (E p) Then by Lemma 2.2, we get

deg(Φ(·, µ − ε), B(0, ε), 0) = (−1) k −1 and deg(Φ(·, µ + ε), B(0, ε), 0) = (−1) k

.

This is a contradiction to the fact deg(Φ(·, µ − ε), B(0, ε), 0) = deg(Φ(·, µ + ε), B(0, ε), 0).

Thus (0, µ) is a bifurcation point of (P ). Now, we shall adopt Rabinowitz’s standard arguement [11] Let S denote the closure of

the set of nontrivial solutions of (P ) and S+

k denote the set of u ∈ C1

0[0, 1] such that u has exactly k − 1 simple zeros in (0,1), u > 0 near 0, and all zeros of u in [0,1] are simple Let

Moreover, let C k denote the component of S which meets (0, µ k ), where µ k = λ k (p) By

the similar argument of Theorem 1.10 in [11], we can show the existence of two types ofcomponentsC emanating from (0, µ) contained in S, when µ is an eigenvalue of (E p); either

it is unbounded or it contains (0, ˆ µ), where ˆ µ( ̸= µ) is an eigenvalue of (E p) The existence

of a neighborhood O k of (0, µ k ) such that (u, λ) ∈ S ∩ O k and u ̸≡ 0 imply u ∈ S k is alsoproved in [11] Actually, only the first alternative is possible as shall be shown next

Lemma 3.5 Assume (Φ1), (Φ2), and (F1) Then, C k is unbounded in S k × R.

Proof: Suppose C k ⊂ (S k × R) ∪ {(0, µ k)} Then since S k ∩ S j =∅ for j ̸= k, it follows from

the above facts,C kmust be unbounded inS k ×R Hence, Lemma 3.5 will be established once

we showC k ̸⊂ (S k ×R)∪{(0, µ k)} is impossible It is clear that C k ∩O k ⊂ (S k ×R)∪{(0, µ k)}.

Hence ifC k ̸⊂ (S k ×R)∪{(0, µ k)}, then there exists (u, λ) ∈ C k ∩(∂S k ×R) with (u, λ) ̸= (0, µ k)

and (u, λ) = lim n →∞ (u n , λ n ), u n ∈ S k If u ∈ ∂S k , u ≡ 0 because u dose not have double

zero Henceforth λ = µ j , j ̸= k But then, (u n , λ n) ∈ (S k × R) ∩ O j for large n which is

impossible by the fact that (u n , λ n)∈ S ∩ O j implies u n ∈ S j The proof is complete. 

Lemma 3.6 Assume (Φ1), (Φ2), (F1), and (F3) Then for each k ∈ N, there exists a constant M k ∈ (0, ∞) such that λ ≤ M k for every λ with (u, λ) ∈ C k

Proof: Suppose it is not true, then there exists a sequence {(u n , λ n)} ⊂ C k such that

λ n → ∞ Let ρ j n be the jth zero of u n Then there exists j ∈ {1, , k − 1} such that

|ρ (j+1) n − ρ j n | ≥ 1

k Thus for each n, there exists σ j n ∈ (ρ j n , ρ (j+1) n ) such that u ′ n (σ j n) = 0

Let u n (t) > 0 for all t ∈ (ρ j n , ρ (j+1) n ) Suppose σ j n ∈ (ρ j n , ρ jn +3ρ (j+1)n

n → ∞ This is impossible Now, if σ j n ∈ ( ρ jn +3ρ (j+1)n

4 , ρ (j+1) n ), then by integrating the equation in (P ) from t ∈ [ρ j n , σ j n ] to σ j n , we see that u n satisfies

λ n

Also if σ j n ∈ ( ρ jn +3ρ (j+1)n

4 , ρ (j+1) n ), then we have

ϕ(2k |u n (t) |) ψ( |u n (t) |) ≥

λ n

Since both (13) and (14) are impossible, there is no sequence {(u n , λ n)} ⊂ C k satisfying

λ n → ∞ Consequently, there exists an M k ∈ (0, ∞) such that λ ≤ M k 

Proof of Theorem 1.1

By Lemmas 3.3, 3.4, and 3.5, for any j ∈ N, there exists an unbounded connected component

C j of the set of nontrivial solutions emanating from (0, λ j (p)) such that (u, λ) ∈ C j implies u has exactly j −1 simple zeros in (0,1) From Lemma 3.6, there is an M j such that (u, λ) ∈ C j

implies that λ ≤ M j , and there are no nontrivial solutions of (P ) for λ = 0, it follows that for any M > 0, there is (u, λ) ∈ C j such that ∥u∥1 > M Hence, we can choose subsequence {(u n , λ n)} ⊂ C j such that λ n → ˆλ and ∥u n ∥1 → ∞ Thus, (∞, ˆλ) is a bifurcation point and

Put f (t, u, λ) = −µϕ(u) + g(t, u) We can easily see that f(t, u, λ) = ◦(|ϕ(u)|) near zero

uniformly for t and λ in bounded intervals The equation in (A g) can be equivalently changedinto the following equation

(A f)

By the similar argument in the proof of Theorem 1.1, for each k ≤ j ≤ n, there is a

connected branch C j of solutions to (A f ) emanating from (0, λ j (p) − µ) which is unbounded

in C01[0, 1] ×R and such that (u, λ) ∈ C j implies that u has exactly j −1 simple zeros in (0,1).

From the fact ug(t, u) ≥ 0, it can be proved that there is an M j > 0 such that (u, λ) ∈ C j

implies that λ ≤ M j, by the same argument as in the proof of Lemma 3.6 Since there is

a constant K g > 0 such that g(t, s) ≤ K g ϕ(s) for all (t, s) ∈ [0, 1] × R, if (u, λ) ∈ C j , then

λ > −K g Hence C j will bifurcate from infinity also, which can only happen for λ = λ j (q) −ν.

Since λ j (q) − ν < 0 < λ j (p) − µ and C j is connected, there exists u ̸= 0 such that (u, 0) ∈ C j

This u is a solution of (A) Since this is true for every such j, (A) has at least n − k + 1