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Volume 2009, Article ID 103867, 34 pagesdoi:10.1155/2009/103867 Research Article An Approximation Approach to Eigenvalue Intervals for Singular Boundary Value Problems with Sign Changing

Volume 2009, Article ID 103867, 34 pages

doi:10.1155/2009/103867

Research Article

An Approximation Approach to Eigenvalue

Intervals for Singular Boundary Value Problems with Sign Changing and Superlinear Nonlinearities

Haishen L ¨u,1 Ravi P Agarwal,2, 3 and Donal O’Regan4

1 Department of Applied Mathematics, Hohai University, Nanjing 210098, China

2 Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901-6975, USA

3 KFUPM Chair Professor, Mathematics and Statistics Department, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia

4 Department of Mathematics, National University of Ireland, Galway, Ireland

Correspondence should be addressed to Haishen L ¨u,haishen2001@yahoo.com.cn

Received 25 June 2009; Accepted 5 October 2009

Recommended by Ivan T Kiguradze

This paper studies the eigenvalue interval for the singular boundary value problem−u gt, u 

λh t, u, t ∈ 0, 1, u0  0  u1, where g  h may be singular at u  0, t  0, 1, and may change sign and be superlinear at u ∞ The approach is based on an approximation method togetherwith the theory of upper and lower solutions

Copyrightq 2009 Haishen L ¨u et al This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited

1 Introduction

The singular boundary value problems of the form

−u ft, u, t ∈ 0, 1,

papers, a critical condition is that

or there exists a constant L > 0 such that for any compact set K ⊂ 0, 1, there is ε  ε K > 0

r  0 and lim supr→ ∞ft, r/r  ∞ for t ∈ 0, 1, very few existence results are available

in literature1

In this paper we study positive solutions of the second boundary value problem

−u gt, u  λht, u, t ∈ 0, 1,

Remark 1.5, Examples3.1and3.2

satisfies the differential equation 1.4 on 0, 1 and the stated boundary data.

maxt ∈0,1 |ut| We put min{a, b}  a ∧ b; max{a, b}  a ∨ b Given α, β ∈ C0, 1, α ≤ β,

In this paper, we suppose the following conditions hold:

H1 there exist h i:0, 1 × 0, ∞ → 0, ∞ i  1, 2 continuous functions such that

h i t, · is increasing for t ∈ 0, 1,

h1·, r, h2·, r ∈ M for r > 0,

h1t, r ≤ ht, r ≤ h2t, r for t, r ∈ 0, 1 × 0, ∞;

1.8

H2 there exists r > 0 such that h1t, r > 0 for t ∈ 0, 1.

The main results of the paper are the following

Theorem 1.1 Suppose G1, H1, H2 and the following conditions hold:

G2 for all r2 > r1> 0, there exists γ · ∈ M such that g2·, r  γ·r is increasing in r1, r2:

j1such that limj→ ∞R j  ∞ and

Then there exists λ∗1 > 0 such that for every λ ≥ λ∗

1,1.4 has at least one positive

solution u ∈ C0, 1 ∩ C10, 1 and u > 0 for t ∈ 0, 1.

Theorem 1.2 Suppose G1, H1, H2 and the following conditions hold:

G3 for all r2 > r1> 0 there exists γ · ∈ M such that gt, r  γtr is increasing in r1, r2;

G4 there exists c1> 0 such that

Then there exists λ∗2 > 0 such that

Theorem 1.4 Suppose G1, H1, H2 and the following conditions hold:

G6 there exists τ ≥ τ1such that

In Section 3, we give two examples see Examples 3.1 and 3.2 which satisfy the

2 Proof of Main Results

2.1 Some Lemmas

Lemma 2.1 Consider the following eigenvalue problem

−u τut, t ∈ 0, 1,

Then the eigenvalues are

The following four results can be found in5 notice limr→ ∞h2t, r/r  0 is not

needed in the proofs there

Lemma 2.2 Suppose G1 and H1 hold Let n0 ∈ N Assume that for every n > n0, there exist

a n , δ n , δ ∈ M such that

0≤ a n t, |δ n t| ≤ δt, lim

and there exist u, u n , u n , u ∈ C0, 1 such that

and u0  u1  0 If

Next we consider the boundary value problem

−u atut  ft, t ∈ 0, 1,

where a, f ∈ M, at ≥ 0 for t ∈ 0, 1.

Lemma 2.4 The following statements hold:

i for any f ∈ M, 2.16 is uniquely solvable and

u  Aau  Af

ii if ft ≥ 0 for t ∈ 0, 1, then the solution of 2.16 is nonnegative.

Corollary 2.5 Let Φ : M → C0, 1 ∩ C10, 1 be the operator such that Φf is the solution of

2.16 Then we have

i if f1t ≤ f2t for t ∈ 0, 1, then Φf1t ≤ Φf2t for t ∈ 0, 1;

ii let E ⊂ M and β ∈ M If |ft| ≤ βt, t ∈ 0, 1 for all f ∈ E, then ΦE is relatively

compact with respect to the topology of C 0, 1.

Lemma 2.6 see 2 Let f ∈ M, f ≥ 0, f /≡ 0, u ∈ C0, 1 ∩ C10, 1 satisfy

−u f in 0, 1,

Then there exist m  mf > 0, M  Mf > 0 such that

Claim 1see 5 There exists λ∗

1 > 0, c > 0, independent of λ, such that for all λ ≥ λ∗

1there

exist R λ > c, u ∈ C0, 1, with cφ1t ≤ ut ≤ R λ φ1t and

−ut  −g1t, ut  λh1t, ut, for t ∈ 0, 1,

Let λ∗1> 0, c > 0 and u ∈ C0, 1 be defined inClaim 1 Define

C 0, 1, ω n t > 0, ωt > 0 for t ∈ 0, 1 such that

There exists j0∈ N such that

R j0φ1t for t ∈ 0, 1, then 0 ≤ Φvt ≤ R j0φ1t for t ∈ 0, 1, so Schauder’s fixed point

theorem guarantees that there existsv ∈ 0, R j0φ1 such that Φv  v, that is,

−vt  λh2t, vt  ω n s  λh1t, ut,

Let

0, |u|∞ Let u n  u From 2.35 and 2.38, we have

which does not depend on n.

On the other hand, let

ψ t, r  g2t, r  1

From G1 notice ψ satisfies the assumptions of Lemma 2.3, so there exist ω, ω n ∈

Thus Schauder fixed point theorem guarantees that there existsv n ∈ 0, φ1 such that Φv n 

so α n,1is a lower solution of2.58 On the other hand, from the definition of g1 and u, we

Suppose2.63 is not true Let yt  α n,1 t− u n t and let σ ∈ 0, 1 be the point where

On the other hand, since α n,1 σ > u n σ, we have

and this is a contradiction

0, |u|∞ Let ut ≡ ut, u n t  α n,1 t From 2.45, 2.56, and 2.63, we have

On the other hand, by2.61

NowLemma 2.2guarantees that there exists a solution u ∈ C0, 1 ∩ C10, 1 to 1.4 with

Thus1.4 has a solution for λ ∈ 0, λ∗ soClaim 2 holds In particular,Λ / ∅ and sup Λ > 0.

The same reason implies that

Let n, m > 1 be fixed FromRemark 1.3, there exist γ n ∈ M, γ n ≥ 0 such that g m t, r 

γ n tr is increasing in 1/n, 1/n  r m /2  We easily prove that

g m

t, r ∧ χ γ n tr is increasing in

1

Let γ t  γ n We have g m t, 1/n  r ∧ χ  γtr is increasing in 0, r m /2  From 2.83 and

2.89, we have for fixed v ∈ C0, 1, ut ≤ vt ≤ u m t that

Fix v ∈ C0, 1 with ut ≤ vt ≤ u m t FromLemma 2.4, there existsΨv ∈ C0, 1 such

β n,m ∈ C0, 1 such that ut ≤ β n,m t ≤ u m t and Ψβ n,m t  β n,m t for t ∈ 0, 1 :

Let m ≥ 1 be fixed We consider the sequence {β n,m}∞n1 Fix n0 ∈ {2, 3, } Let us

look at the interval1/2 n0 1, 1 − 1/2 n0 1 The mean value theorem implies that there exists

2n0 1, 1− 1

2n0 1

(

function z n0,m ∈ 1/2 n0 1, 1 − 1/2 n0 1 with β n,m converging uniformly to z n0,mon1/2 n0 1, 1−

2n0 2, 1− 1

2n0 2

(

so there is a subsequence N n0 1of N n0and a function z n01,m ∈ C1/2 n0 2, 1 − 1/2 n0 2 with

β n,m converging uniformly to z n01,mon1/2 n0 2, 1 − 1/2 n0 2 as n → ∞ through N n0 1 Note

z n01,m  z n0,m on 1/2 n0 1, 1 − 1/2 n0 1 since N n0 1 ⊆ N n0 Proceed inductively to obtain

subsequences of integers N n0 ⊇ N n0 1 ⊇ · · · ⊇ N k ⊇ · · · and functions z k,m ∈ C1/2 k1, 1−

1/2 k1 with β n,m converging uniformly to z k,mon1/2 k1, 1 − 1/2 k1 as n → ∞ through N k ,

and z k,m  z k −1,mon1/2 k , 1 − 1/2 k .

Define a function u m :0, 1 → 0, ∞ by u m t  z k,m t on 1/2 k1, 1 − 1/2 k1 and

u m 0  u m 1  0 Notice u m is well defined and ut ≤ u m t ≤ u m t for t ∈ 0, 1 Next, fix

t ∈ 0, 1 without loss of generality assume t / 1/2  and let n∗∈ {n0, n0 1, } be such that 1/2 n∗1< t < 1 − 1/2 n∗1 Let N n∗∗  {i ∈ N n : i ≥ n∗} Now β n,m , n ∈ N∗

n∗satisfies the integralequation

β n,m t  β n,m

12



 β

n,m

12



t−12

subsequence; for convenience we will let{β n,m 1/2} n ∈N∗

As a result,

z k,m t  z k,m

12



 τ



t−12



 τ



t−12

Let ε > 0 be given Since u m ∈ C0, 1 there exists δ > 0 with u m t < ε/2 for t ∈ 0, δ.

As a result ut ≤ β n,m t ≤ u m t < ε/2 for t ∈ 0, δ Consequently, ut ≤ u m t ≤ ε/2 < ε for

t ∈ 0, δ and so u m is continuous at 0 Similarly, u m is continuous at 1 As a result u m ∈ C0, 1

N k ⊇ · · · and functions z m ∈ C1/2 k1, 1 − 1/2 k1 with u m converging uniformly to z k on

1/2 k1, 1 − 1/2 k1 as m → ∞ through N k , and z k  z k−1on1/2 k , 1 − 1/2 k .

Define a function u : 0, 1 → 0, ∞ by ut  z k t on 1/2 k1, 1 − 1/2 k1 and u0 

without loss of generality assume t / 1/2  and let m∗∈ {m0, m01, } be such that 1/2 m∗1<

t < 1 −1/2 m∗1 Let N m∗∗  {k ∈ N m∗: k ≥ m∗} Now u m , m ∈ N∗

m∗satisfies the integral equation

u m t  u m

12



 u

m

12



t−12

u t  u

12



 τ



t−12

2.4 The Proof of Theorem 1.4

Proof of Claim 4 Let λ ∈ 0, λ∗ be fixed From assumption G6, it follows that there is τ ≥ τ1

and c3∈ 0, 1, such that if n > 2/c3, 0 < k < c3/2 < 1, we have

It is easy to see thatΦ is a continuous and completely continuous operator Also if 0 ≤ vt ≤

Suppose2.137 is not true Let yt  ut − u n t and σ ∈ 0, 1 be the point where

On the other hand, since uσ > u n σ, we have

n σ > 0, and this is a contradiction As a result, 2.137 is true

On the other hand, we have

for n > 2/c3 Consequently, for t ∈ 0, 1, δ n → 0 and n → ∞.

From assumptionsG2 and H5, there exists a γ, τ ∈ M, n > 2/c3, so that g t, 1/n 

r   ht, r  atr is increasing in 0, |u|∞, where at  γt  τt Let u n  ut For v ∈ D u n

Reasoning as in the proof of Theorem 1.1, Lemma 2.2 guarantees that 1.4 has a

We consider the boundary value problem

−u gt, u ∧ χ μht, u ∧ χ,

where μ ∈ 0, λ Let g1t, u  g1t, u ∧ χ, g2t, u  g2t, u ∧ χ, h1t, u  h1t, u ∧ χ, and

h2t, u  h2t, u ∧ χ We easily prove that the conditions of 6, Theorem 1.2 are satisfied so

2.149 has a positive solution u ∈ C10, 1 ∩ C0, 1 We next prove that

Suppose2.150 is not true Let yt  ut − χt and σ ∈ 0, 1 be the point where yt

one positive solution u ∈ C0, 1 ∩ C10, 1 and u > 0 for t ∈ 0, 1.

To see this, let

For all r2 > r1 > 0, let γ t  1/2r1√

r1 Then g2t, r  1/2r1√

r1r is increasing in

r1, r2.

On the other hand, a1 1  1

01/√s ds  3 and let R j  x 2j − 3, so we have

Example 3.2 Consider the boundary value problem

To see this, let β  min{1/2, α/2}, g1t, r  1/r β  π α , and g2t, r  1/r α , for t, r ∈

0, 1 × 0, ∞, and h1t, r  h2t, r  r2, for t, r ∈ 0, 1 × 0, ∞ Notice that G1, H1, and

whereΛ  r1, r2 \ {nπ | n ∈ N}, so we have gt, r  γtr is increasing in r1, r2.

6n 5

2−β*

and we have for n ≥ n0, r ∈ 0, c2,

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