POSITIVE SOLUTIONS OF SINGULAR STURM-LIOUVILLE...

Dynamics of Continuous, Discrete and Impulsive SystemsSeries A: Mathematical Analysis 8 (2001) 373-382Copyright c©2001 Watam Press

POSITIVE SOLUTIONS OF SINGULARSTURM-LIOUVILLE BOUNDARY VALUE

PROBLEMS

B.G. Zhang and Lingju KongDepartment of MathematicsOcean University of Qingdao

Qingdao 266003 P.R. China

Abstract. This paper studies the Sturm-Liouville boundary value problem(p(t)u′(t))′ + λa(t)f(t, u(t)) = 0, 0 < t < 1,αu(0)− βp(0)u′(0) = 0,γu(1) + δp(1)u′(1) = 0,

where λ > 0 and a is allowed to be singular at both end points t = 0 and t = 1. We shall

show the existence of this problem for λ on a suitable interval.

Keywords. Positive Solutions, boundary value problems, singular, fixed point theorem,

cone.

AMS (MOS) subject classification: 34B15

1 Introduction

Consider the Sturm-Liouville boundary value problem (p(t)u′(t))′ + λa(t)f(t, u(t)) = 0, 0 < t < 1,αu(0)− βp(0)u′(0) = 0,γu(1) + δp(1)u′(1) = 0,

(1)λ

where(H1) p(t) ∈ C([0, 1], [0,+∞)) and 0 <

∫ 1

0dtp(t) < +∞;

(H2) λ > 0, α, β, γ and δ are nonnegative, and βγ + αγ + αδ > 0;(H3) f(t, u) ∈ C([0, 1]× [0,+∞), R+) and a ∈ C((0, 1), [0,+∞));(H4) 0 <

∫ 1

0G(s, s)a(s)ds < +∞,

where

G(s, s) =1ρ(β + α

∫ s

0

dr

p(r))(δ + γ

∫ 1

s

dr

p(r)), 0 ≤ s ≤ 1,

and

ρ = αδ + αγ

∫ 1

0

dr

p(r)+ βγ.

For any t ∈ [0, 1], let

f0(t) = limu→0+

f(t, u)u

and f∞(t) = limu→+∞

f(t, u)u

.

374 B.G. Zhang and L. Kong

Consider the following cases:(L1) f0(t) = +∞, t ∈ [0, 1]; (L2) f∞(t) = +∞, t ∈ [0, 1];(L3) f0(t) = 0, t ∈ [0, 1]; (L4) f∞(t) = 0, t ∈ [0, 1];(L5) f0(t) = l1 > 0, t ∈ [0, 1]; (L6) f∞(t) = l2 > 0, t ∈ [0, 1].

For the special case a(t) = 1, the boundary value problem (1)λ has beeninvestigated in [1] and [3]. In this paper, we want to obtain some existenceresults of positive solutions to the singular boundary value problem (1)λ forλ on a suitable interval. Our results improve and generalize some results in[2, 4]. The other results, we refer to [6, 7]. Some of our ideas are borrowedfrom [5].

2 Preliminaries

It is well known that

G(t, s) =

1ρ (β + α

∫ s0

drp(r) )(δ + γ

∫ 1

tdrp(r) ), 0 ≤ s ≤ t ≤ 1,

1ρ (β + α

∫ t0

drp(r) )(δ + γ

∫ 1

sdrp(r) ), 0 ≤ t ≤ s ≤ 1,

(2)

is the Green’s function of the following boundary value problem (p(t)u′(t))′ = 0, 0 < t < 1,αu(0)− βp(0)u′(0) = 0,γu(1) + δp(1)u′(1) = 0,

where

ρ = αδ + αγ

∫ 1

0

dr

p(r)+ βγ.

ClearlyG(t, s) ≤ G(s, s), 0 ≤ t, s ≤ 1. (3)

By (H4), there exits t0 ∈ (0, 1) such that a(t0) > 0. We may choose θ ∈ (0, 12 )

such that t0 ∈ (θ, 1− θ).Define a cone Kθ as follows:

Kθ = u ∈ C([0, 1])| u(t) ≥ 0, minθ≤t≤1−θ

u(t) ≥Mθ||u||, (4)

where

Mθ = minδ + γ

∫ 1

1−θdrp(r)

δ + γ∫ 1

0drp(r)

,β + α

∫ θ0

drp(r)

β + α∫ 1

0drp(r)

and||u|| = sup

t∈[0,1]

|u(t)|.

We donote

φ(t) = δ + γ

∫ 1

t

dr

p(r), 0 ≤ t ≤ 1,

Positive Solutions of Boundary Value Problems 375

ψ(t) = β + α

∫ t

0

dr

p(r), 0 ≤ t ≤ 1.

Then

G(t, s) = 1

ρψ(s)φ(t), 0 ≤ s ≤ t ≤ 1,1ρψ(t)φ(s), 0 ≤ t ≤ s ≤ 1.

For θ ≤ t ≤ 1− θ, we have

G(t, s)G(s, s)

=

φ(t)φ(s) ≥

δ+γ∫ 1

1−θ

drp(r)

δ+γ∫ 1

0dr

p(r)

, s ≤ t,

ψ(t)ψ(s) ≥

β+α∫ θ

0dr

p(r)

β+α∫ 1

0dr

p(r)

, t ≤ s,

ThereforeG(t, s)G(s, s)

≥Mθ, θ ≤ t ≤ 1− θ.

i.e.,G(t, s) ≥MθG(s, s), θ ≤ t ≤ 1− θ. (5)

Define an operator Tλ on Kθ as follows:

Tλu(t) = λ

∫ 1

0

G(t, s)a(s)f(s, u(s))ds, λ > 0, u ∈ Kθ. (6)

Lemma 2.1. Tλ(Kθ) ⊂ Kθ.Proof. By (3), we have, for any t ∈ [0, 1] and u ∈ Kθ,

Tλu(t) = λ

∫ 1

0

G(t, s)a(s)f(s, u(s))ds

≤ λ

∫ 1

0

G(s, s)a(s)f(s, u(s))ds.

Hence

||Tλ|| ≤∫ 1

0

G(s, s)a(s)f(s, u(s))ds. (7)

By (5), we have

minθ≤t≤1−θ

Tλu(t) = minθ≤t≤1−θ

λ

∫ 1

0


≥Mθλ

∫ 1

0

G(s, s)a(s)f(s, u(s))ds. (8)

In view of (7) and (8), we have

minθ≤t≤1−θ

Tλu(t) ≥Mθ||Tu||, u ∈ Kθ.


Therefore Tλ(Kθ) ⊂ Kθ, The proof is complete.Lemma 2.2. Tλ : Kθ → Kθ is completely continuous.The proof is similar to that of Lemma 2 in [4], we omit it here.Lemma 2.3.([7],[8]) Let K be a cone in a Banach space E and Ω1, Ω2

be two bounded open sets in E such that 0 ∈ Ω1 and Ω1 ⊂ Ω2. Let T :K

⋂(Ω2\Ω1) → K be a completely continuous operator. If

||Tu|| ≤ ||u||, u ∈ K⋂∂Ω1 and ||Tu|| ≥ ||u||, u ∈ K

⋂∂Ω2,

or||Tu|| ≥ ||u||, u ∈ K

⋂∂Ω1 and ||Tu|| ≤ ||u||, u ∈ K

⋂∂Ω2,

Then T has at least one fixed point in K⋂

(Ω2\Ω1).Let

A = max0≤t≤1

∫ 1

0

G(t, s)a(s)ds

and

Bθ = minθ≤t≤1−θ

∫ 1−θ

θ

G(t, s)a(s)ds.

Lemma 2.4. Assume that (H1) − (H4) hold and there exist two differentpositive numbers a and b such that

max0≤t≤1,0≤u≤a

f(t, u) ≤ a

λA(9)

andmin

θ≤t≤1−θ,Mθb≤u≤bf(t, u) ≥ b

λBθ. (10)

Then the BV P (1)λ has at least one positive solution u ∈ Kθ and mina, b ≤||u|| ≤ maxa, b.Proof. Without loss of generality, we may assume that a < b.Let

Ωa = u ∈ C([0, 1])| ||u|| < a, Ωb = u ∈ C([0, 1])| ||u|| < b.

For any t ∈ [0, 1] and u ∈ Kθ

⋂∂Ωa, by (9), we have

f(t, u(t)) ≤ a

λA.

Hence

Tλu(t) = λ

∫ 1

0


≤ (λ∫ 1

0

G(s, s)a(s)ds)a

λA≤ a.

i.e.,||Tλu|| ≤ ||u||, u ∈ Kθ

⋂∂Ωa.


By (10), for any t ∈ [θ, 1− θ] and u ∈ Kθ

⋂∂Ωb, we have

f(t, u) ≥ b

λBθ.

Hence

Tλu(12) = λ

∫ 1

0

G(12, s)a(s)f(s, u(s))ds

≥ λ

∫ 1−θ

θ

G(12, s)a(s)f(s, u(s))ds

≥ (λ∫ 1−θ

θ

G(12, s)a(s)ds)

b

λBθ≥ b.

i.e.,

||Tλu|| ≥ Tλu(12) ≥ ||u||, u ∈ Kθ

⋂∂Ωb.

It follows from Lemma 2.3 that there exists a u ∈ Kθ

⋂(Ωb\Ωa ) such that

Tλu(t) = u(t) and ||u|| is between a and b. This means that u(t) is a solutionof the BV P (1)λ and mina, b ≤ ||u|| ≤ maxa, b. The proof is complete.

3 Main Results

In this section, we denote

λ1 =1A

supr>0

r

max0≤t≤1,0≤u≤r

f(t, u)

andλ2 =

1Bθ

infr>0

r

min0≤t≤1,Mθr≤u≤r

f(t, u).

Under conditions of (H1) − (H4), it is obvious that 0 < λ1 ≤ +∞ and0 ≤ λ2 < +∞.Theorem 3.1. Assume that (H1) − (H2), (L1) and (L2) hold. Then thereexists λ1 > 0 such that the BV P (1)λ has at least two positive solutions for0 < λ < λ1.Proof. Define

s(r) =r

A max0≤t≤1,0≤u≤r

f(t, u).

By (H3), we know that s : (0.+∞) → (0,+∞).Clearly, S is continuous in (0,+∞). In view of (L1) and (L2), we havelimr→0+

s(r) = limr→+∞

s(r) = 0. Hence there exists r0 ∈ (0,+∞) such that


s(r0) = maxr>0

s(r) = λ1. For 0 < λ < λ1, there exist two positive numbers a1

and a2 such that 0 < a1 < r0 < a2 < +∞ and

s(a1) = s(a2) = λ.

i.e.,a1

A max0≤t≤1,0≤u≤a1

f(t, u)=

a2

A max0≤t≤1,0≤u≤a2

f(t, u)= λ.

Hencemax

0≤t≤1,0≤u≤a1f(t, u) =

a1

λA

andmax

0≤t≤1,0≤u≤a2f(t, u) =

a2

λA.

On the other hand, by (L1) and (L2), there exist b1 and b2 such that 0 <b1 < a1 < r0 < a2 < b2 < +∞ and

f(t, u)u

≥ 1λMθBθ

, t ∈ [0, 1], u ∈ (0, b1)⋃

[Mθb2,+∞).

Hencef(t, u) ≥ u

λMθBθ≥ b1λBθ

, t× u ∈ [0, 1]× [Mθb1, b1],

andf(t, u) ≥ u

λMθBθ≥ b2λBθ

, t× u ∈ [0, 1]× [Mθb2, b2].

Therefore, we have

minθ≤t≤1−θ,Mθb1≤u≤b1

f(t, u) ≥ min0≤t≤1,Mθb1≤u≤b1

f(t, u) ≥ b1λBθ

andmin

θ≤t≤1−θ,Mθb2≤u≤b2f(t, u) ≥ min

0≤t≤1,Mθb2≤u≤b2f(t, u) ≥ b2

λBθ.

By Lemma 2.4, the BV P (1)λ has at least two positive solutions. The proofis complete.From the proof of Theorem 3.1, the following conclusion is obvious.Theorem 3.2. Assume that (H1) − (H4) and one of (L1) and (L2) hold.Then the BV P (1)λ has at least one positive solution for 0 < λ < λ1.Theorem 3.3. Assume that (H1) − (H4), (L3) and (L4) hold. Then thereexists λ2 ≥ 0 such that the BV P (1)λ has at least two positive solutions forλ2 < λ < +∞.Proof. Define

w(r) =r

Bθ min0≤t≤1,Mθr≤u≤r

f(t, u).


Clearly, w(r) is continuous in (0,+∞). From (L3) and (L4), we have

limr→0+

w(r) = limr→+∞

w(r) = +∞.

Hence there exists r0 ∈ (0,+∞) such that

w(r0) = minr>0

w(r) = λ2 ≥ 0.

Since λ2 < λ < +∞, we can find two positive numbers b1 and b2 such that0 < b1 < r0 < b2 < +∞ and

w(b1) = w(b2) = λ.

i.e.,

b1Bθ min

0≤t≤1,Mθb1≤u≤b1f(t, u)

=b2

Bθ min0≤t≤1,Mθb2≤u≤b2

f(t, u)= λ.

Hencemin

0≤t≤1,Mθb1≤u≤b1f(t, u) =

b1λBθ

andmin

0≤t≤1,Mθb2≤u≤b2f(t, u) =

b2λBθ

.

On the other hand, (L3) implies that there exists a1 ∈ (0, b1) such that

f(t, u)u

≤ 1λA

, t× u ∈ [0, 1]× [0, a1].

Hencef(t, u) ≤ a1

λA, t× u ∈ [0, 1]× [0, a1].

From (L4), there exists a ∈ (b2,+∞) such that

f(t, u)u

≤ 1λA

, t× u ∈ [0, 1]× [a,+∞).

Let M = max0≤t≤1,0≤u≤a

f(t, u). We can choose a2 > a such that a2 ≥ λMA.

Hencef(t, u) ≤ a2

λA, t× u ∈ [0, 1]× [0, a2].

By Lemma 2.4, the BV P (1)λ has at least two positive solutions for λ2 < λ <+∞. The proof is complete.From the proof of Theorem 3.3, we have the following results.Theorem 3.4. Assume that (H1) − (H4) and one of (L3) and (L4) hold.Then the BV P (1)λ has at least one positive solution for λ2 < λ < +∞.Corollary 3.1 Assume that (H1)−(H4) hold. Moreover, one of the following


conditions is true:(i) (L1) and (L4) hold;(ii) (L2) and (L3) hold.Then the BV P (1)λ has at least one positive solution for λ > 0.Proof. First, we prove the case (i).By Theorem 3.2, to prove the conclusion, we only need to prove λ1 = +∞.If

sup0≤t≤1,0≤u≤+∞

f(t, u) = M < +∞.

Thenλ1 ≥

1A

supr>0

r

M= +∞.

If f is unbounded, there exist tn ∈ [0, 1] and rn → +∞ such that

f(tn, rn) = max0≤t≤1,0≤u≤rn

f(t, u).

By (L4)

limrn→+∞

f(tn, rn)rn

= 0,

λ1 = maxr>0

s(r) ≥ maxrn>0

s(rn)

= maxrn>0

rnA max

0≤t≤1,0≤u≤rn

f(t, u)= maxrn>0

rnAf(tn, rn)

= +∞.

Hence λ1 = +∞.Now, we assume that (ii) holds. By Theorem 3.4, it is sufficient to proveλ2 = 0.By (L2), for t ∈ [0, 1], f(t, u) → +∞ as u→ +∞. There exist tn ∈ [0, 1] andrn → +∞ such that

f(tn,Mθrn) = min0≤t≤1,Mθrn≤u≤rn,

f(t, u).

Again by (L2),

limrn→0

f(tn,Mθrn)rn

= +∞,

λ2 = minr>0

w(r) ≤ minrn>0

w(rn)

= minrn>0

rnBθ min

0≤t≤1,Mθrn≤u≤rn

f(t, u)= minrn>0

rnf(tn,Mθrn)

= 0.

Hence λ2 = 0. The proof is complete.Remark 3.1. Corollary 3.1 improves and generalizes the results in [2] and[4].Similarly, we have


Corollary 3.2. Assume that (H1)− (H4) hold. Moreover, one of the follow-ing conditions is true:(i) (L1) and (L6) hold;(ii) (L2) and (L5) hold.Then the BV P (1)λ has at least one positive solution for 0 < λ < 1

Al1.

Corollary 3.3. Assume that (H1)− (H4) hold. Moreover, one of the follow-ing conditions is true:(i) (L3) and (L6) hold;(ii) (L4) and (L5) hold.Then the BV P (1)λ has at least one positive solution for 1

MθBθl2< λ < +∞.

Similarly to the proof of Corollary 3.1, we need only to prove λ1 ≥ 1Al1

forCorollary 3.2 and λ2 ≤ 1

MθBθl2for Corollary 3.3.

Example 3.1. Consider the singular boundary value problem: ((t12 + 1)u′(t))′ + λt−

12 (1− t)−

12 1+tu

u = 0, 0 < t < 1,u(0)− βu′(0) = 0,u(1) + 2δu′(1) = 0,

(11)λ

where p(t) = t12 + 1, a(t) = t−

12 (1 − t)−

12 , f(t, u) = (1+tu)

u , α = γ =1, β, δ ≥ 0. Then

∫ 1

0dtp(t) = 2 ln 2, ρ = β + δ + 2 ln 2, and G(s, s) =

(β+2 ln(1+s12 ))(δ+2 ln 2−2 ln(1+s

12 ))

β+δ+2 ln 2 . It is easy to check that (H1) − (H4) aresatisfied. Moreover, f0(t) = +∞ and f∞(t) = 0. Thus (L1) and (L4) hold.By Corollary 3.1, the boundary value problem (11)λ has at least one solutionfor λ > 0.Remark 3.2. Even if λ ≡ 1, the existence of positive solution of (11)λ cannot be obtained by the results in [2] and [4].

4 Acknowledgement

The research of this paper was supported by NNSF of China.

5 References

[1] R.P. Agarwal, Huei-Lin Hong and Chen-Chin Yeh, The existence of positive solu-tions for the Sturm-Liouville boundary value problems, Computers Math. Applic.,35, (1998), 89-96.

[2] L.H. Erbe and H. Wang, On the existence of positive solutions of ordinary differen-tial equations, Proc. Amer. Math. Soc., 120, (1994), 743-748.

[3] V. Anuradha, D.D. Hai and R. Shivaji, Existence results for superlinear semipositiveBVP’s, Proc. Amer. Math. Soc., 124, (1996), 757-763.

[4] Ruyun Ma, Positive solutions of singular second order boundary value problems,(in Chinese) Acta Mathematica Sinica, 41, (1998), 1225-1230.


[5] Qingliu Yao and Zhanbing Bai, Existence of positive solutions for boundary valueproblem of u(4)(t)−λh(t)f(u(t)) = 0, (in Chinese) Chinese Annals of Mathematics,20A:5, (1999), 575-578.

[6] W.C. Lian, F.H. Wong and C.C. Yeh, On the existence of positive solutions ofnonlinear second order differential equations, Proc. Amer. Math. Soc., 124, (1996),1117-1126.

[7] R.P. Agarwal and D. O’Regan, Singular boundary value problems for second orderordinary and delay differential equations, J. Differential Equations, 130, (1996),333-355.

[8] D. Guo and V. Lakshmikantham, Nonlinear Problems in Abstract Cones, AcademicPress, New York, 1988.

[9] M.A. Krasnosel’skii, Positive Solutions of Operator Equations, Noordhoff, Grohin-gen, The Netherlands, 1964.

Email: [email protected] June 2000; revised September 2000.http://monotone.uwaterloo.ca/∼journal

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