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Boundary Value Problems

Volume 2011, Article ID 212980,22pages doi:10.1155/2011/212980

Research Article

Multiple Positive Solutions of a Singular

Emden-Fowler Type Problem for Second-Order Impulsive Differential Systems

Eun Kyoung Lee and Yong-Hoon Lee

Department of Mathematics, Pusan National University, Busan 609-735, Republic of Korea

Correspondence should be addressed to Yong-Hoon Lee,yhlee@pusan.ac.kr Received 14 May 2010; Accepted 26 July 2010

Academic Editor: Feliz Manuel Minh ´os

Copyrightq2011 E. K. Lee and Y.-H. Lee. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

This paper studies the existence, and multiplicity of positive solutions of a singular boundary value problem for second-order differential systems with impulse effects. By using the upper and lower solutions method and fixed point index arguments, criteria of the multiplicity, existence and nonexistence of positive solutions with respect to parameters given in the system are established.

1. Introduction

In this paper, we consider systems of impulsive differential equations of the form ut λh1tfut, vt 0, t∈0,1, t /t1,

vt μh2tgut, vt 0, t∈0,1, t /t1, Δu|tt1Iuut1, Δv|tt1Ivvt1, Δu

tt1 Nuut1, Δv

tt1 Nvvt1,

u0 a≥0, v0 b≥0, u1 c≥0, v1 d≥0,

P

whereλ, μare positive real parameters,Δu|tt1 ut1ut1, andΔu|tt1 ut1ut1. Throughout this paper, we assumef, g∈ CR2,Rwithf0,0 0 g0,0andfu, v >

0, gu, v > 0 for all u, v/ 0,0, Iu, IvCR,R satisfyingIu0 0 Iv0, Nu, NvCR,−∞,0,andhiC0,1,0,∞, i 1,2.Here we denoteR 0,∞.We note that

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hi may be singular at t 0 and/or 1.Let J 0,1, J 0,1\ {0,1, t1}, P C 0,1 {u | u : 0,1 → Rbe continuous att /t1,left continuous att t1,and its right-hand limit at tt1exists}andXP C 0,P C 0,1.ThenP C 0,1andXare Banach spaces with norm usupt∈ 0,1|ut|andu, vuv,respectively. The solution of problemPmeans u, v∈X∩C2J×C2Jwhich satisfiesP.

Recently, several works have been devoted to the study of second-order impulsive differential systems. See, for example 1–6, and references therein. In Particular, E.K. Lee and Y.H. Lee 3studied problemPwhenfandgsatisfyf0,0>0 andg0,0>0.More precisely, let us consider the following assumptions.

D11

0s1shisds <∞,fori1,2.

D2t1Nuu≤Iuu≤ −1−t1Nuuandt1Nvv≤Ivv≤ −1−t1Nvv.

D3uIuuandvIvvare nondecreasing.

D4Nu,∞limu→ ∞|Nuu|/u <1 andNv,∞limv→ ∞|Nvv|/v <1.

D5flimuv→ ∞fu, v/uv∞andglimuv→ ∞gu, v/uv∞.

D6f and g are nondecreasing on R2, that is,fu1, v1fu2, v2 andgu1, v1gu2, v2wheneveru1, v1 ≤ u2, v2,where inequality onR2 can be understood componentwise.

Under the above assumptions, they proved that there exists a continuous curveΓ splitting R2\ {0,0} into two disjoint subsetsO1 and O2 such that problem3.20 has at least two positive solutions forλ, μ∈ O1,at least one positive solution forλ, μ∈Γ,and no solution forλ, μ∈ O2.

The aim of this paper is to study generalized Emden-Fowler-type problem for P, that is,f andg satisfyf0,0 0 andg0,0 0,respectively. In this case, we obtain two interesting results. First, for Dirichlet boundary condition, that is,abcd0,assuming D1,D2and

D4Nu,0limu0|Nuu|/u <1/2 andNv,0 limv→0|Nvv|/v <1/2,

D5f∞, g∞andf0limuv0fu, v/uv0, g0limuv→0gu, v/uv0, we prove that problemPhas at least one positive solution for allλ, μ∈R2\{0,0}.On the other hand, for two-point boundary condition, that is,c > aandd > b,assumingD1∼D6, we prove that there exists a continuous curveΓ0splittingR2\{0,0}into two disjoint subsets O0,1andO0,2and there exists a subsetO ⊂ O0,1such that problemPhas at least two positive solutions forλ, μ ∈ O,at least one positive solution for λ, μ ∈ O0,1 \ O∪Γ0,and no solution forλ, μ∈ O0,2.

Our technique of proofs is mainly employed by the upper and lower solutions method and several fixed point index theorems.

The paper is organized as follows: inSection 2, we introduce and prove two types of upper and lower solutions and related theorems, one for singular systems with no impulse effect and the other for singular impulsive systems and then introduce several fixed point index theorems for later use. InSection 3, we prove an existence result for Dirichlet boundary value problems and existence and nonexistence part of the result for two-point boundary value problems. In Section 4, we prove the existence of the second positive solution for two point boundary value problems. Finally, inSection 5, we apply main results to prove some theorems of existence, nonexistence, and multiplicity of positive radial solutions for impulsive semilinear elliptic problems.

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2. Preliminary

In this section, we introduce two types of fundamental theorems of upper and lower solutions method for a singular system with no impulse effect and an impulsive system and then introduce several well-known fixed point index theorems. We first give definition s of somewhat general type of upper and lower solutions for the following singular system:

ut Ft, ut, vt 0, t∈0,1, vt Gt, ut, vt 0, t∈0,1,

u0 A, u1 C, v0 B, v1 D,

H

whereF, G :0,1×R×R → Rare continuous.

Definition 2.1. We say thatαu, αvC 0,C 0,1is aG-lower solution ofHifαu, αvC20,1×C20,1except at finite pointsτ1, . . . , τnwith 0< τ1< . . . < τn<1 such that

L1at each τi, there exist αu τi, αv τiuτi, αvτi such that αu τiαuτi, αvτiαvτi,and

L2

αut Ft, αut, αvt≥0, αvt Gt, αut, αvt≥0, t∈ 0,1

1, . . . , τn}, αu0≤A, αu1≤C,

αv0≤B, αv1≤D.

2.1

We also say that βu, βvC 0,1 ×C 0,1 is a G-upper solution of the problem H if βu, βvC20,1×C20,1 except at finite pointsσ1, . . . , σm with 0 < σ1 < · · · < σm < 1 such that

U1at each σi, there exist βuσj, βvσjuσj, βvσj such that βuσjβu σj, βvσjβvσj,and

U2

βut F

t, βut, βvt

≤0, βvt G

t, βut, βvt

≤0, t∈ 0,1 {σ1, . . . , σn}, βu0≥A, βu1≥C,

βv0≥B, βv1≥D.

2.2

For the proof of the fundamental theorem on G-upper and G-lower solutions for problemH, we need the following lemma. One may refer to 7for the proof.

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Lemma 2.2. LetF, G :D → Rbe continuous functions andD⊂0,1×R×R.Assume that there existhF, hGC0,1,Rsuch that

|Ft, u, v| ≤hFt, |Gt, u, v| ≤hGt, 2.3

for allt, u, v∈0,1×R×R,and 1

0

s1shFsds 1

0

s1shGsds <∞. 2.4

Then problemHhas a solution.

Let Dβα {t, u, v | αut, αvt ≤ u, v ≤ βut, βvt, t ∈ 0,1}. Then the fundamental theorem of G-upper and G-lower solutions for singular problemHis given as follows.

Theorem 2.3. Letαu, αvand βu, βvbe a G-lower solution and a G-upper solution of problem H, respectively, such that

a1 αut, αvt≤βut, βvtfor allt∈ 0,1.

Assume also that there existhF, hGC0,1,Rsuch that

a2|Ft, u, v| ≤hFtand|Gt, u, v| ≤hGtfor allt, u, v∈Dαβ; a31

0s1shFsds1

0s1shGsds <∞;

a4Ft, u, v1Ft, u, v2,wheneverv1v2andGt, u1, vGt, u2, v,wheneveru1u2.

Then problemHhas at least one solutionu, vsuch that

αut, αvt≤ut, vt≤

βut, βvt

,t∈ 0,1. 2.5

Proof. Define a modified function ofFas follows:

Ft, u, v

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩ F

t, βut, v

uβut

1 u2 ifu > βut,

Ft, u, v ifαut≤uβut,

Ft, αut, v− uαut

1 u2 ifu < αut,

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Ft, u, v

⎧⎪

⎪⎪

⎪⎪

⎪⎩ F

t, u, βvt

ifv > βvt, Ft, u, v ifαvt≤vβvt, Ft, u, αvt ifv < αvt,

Gt, u, v

⎧⎪

⎪⎪

⎪⎪

⎪⎩ G

t, βut, v

if u > βut, Gt, u, v if αut≤uβut, Gt, αut, v if u < αut,

2.6

Gt, u, v

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎩ G

t, u, βvt,

vβvt

1 v2 ifv > βvt, Gt, u, v ifαvt≤vβvt, Gt, u, αvt−vαvt

1 v2 ifv < αvt.

2.7

ThenF, G:0,1×R×R → Rare continuous and

|Ft, u, v| ≤m αu, βu

hFt,

|Gt, u, v| ≤m αv, βv

hGt, 2.8

for allt, u, v∈0,1×R×R,wheremα, β αβ1.For the problem

ut Ft, ut, vt 0, vt Gt, ut, vt 0, t∈0,1

u0 A, u1 C, v0 B, v1 D,

M

Lemma 2.2guarantees the existence of solutions of problemM and thus it is enough to prove that any solutionu, vof problemMsatisfies

αut, αvt≤ut, vt≤

βut, βvt

, ∀t∈ 0,1. 2.9

Suppose, on the contrary,αu, αv/≤u, v, so we consider the caseαu /u.Letαuut0

maxt∈ 0,1αuut > 0.Ift0 ∈ 0,1\ {τ1, . . . , τn},thenαuut0 ≤ 0.We consider two

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cases. First, ifαvt0vt0,then byαut0> ut0and conditiona4, 0≥αuut0 αut0 Ft0, ut0, vt0

αut0 Ft0, αut0, vt0ut0αut0 1u2t0

αut0 Ft0, αut0, αvt0ut0αut0 1u2t0

αut0ut0 1u2t0 >0,

2.10

which is a contradiction. Next, ifαvt0> vt0,then by the definition ofF, 0≥αuut0 αut0 Ft0, ut0, vt0

αut0 Ft0, αut0, αvt0ut0αut0

1u2t0 >0, 2.11 which is also a contradiction. Ift0τifor somei1, . . . , n,then sinceαu−uattains its positive maximum atτi,

αuu τi

≥0, αuuτi≤0. 2.12

Ifαuuτi>0,then

0<αuu τi

−αuuτi αu τi

αuτi. 2.13

This leads a contradiction to the definition ofG-lower solution. Ifαuuτi 0,then there existsδ >0 such that for allt∈τiδ, τi,

αuut>0, αuut≥0, αuut≤0. 2.14

Fort∈τiδ, τi,ifαvt≤vt,then byαut> utand conditiona4, 0≥αuut αut Ft, ut, vt

αut Ft, αut, vt−utαut 1u2t

αut Ft, αut, αvt− utαut 1u2t

αut−ut 1u2t >0,

2.15

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which is a contradiction. Ifαvt> vt,then by definition ofF, 0≥αuut αut Ft, ut, vt

αut Ft, αut, αvt−utαut

1u2t >0, 2.16

which is a contradiction. Ift00 or 1,then

0<αuu0 αu0−A≤0,

0<αuu1 αu1−C≤0, 2.17

which is a contradiction. Similarly, we get contradictions for the caseαv/v.The proof for u, v≤βu, βvcan be done by similar fashion.

Now we introduce definition and fundamental theorem of upper and lower solutions for impulsive differential systems of the form

ut Ft, ut, vt 0, t /t1, t∈0,1, ut Gt, ut, vt 0, t /t1, t∈0,1, Δu|tt1Iuut1, Δv|tt1Ivvt1, Δu

tt1Nuut1, Δv

tt1Nvvt1,

u0 a, v0 b, u1 c, v1 d,

S

whereF, GC0,1×R×R,R,Iu, IvCR,RsatisfyingIu0 0Iv0andNu, NvCR,−∞,0.

Definition 2.4. αu, αvX∩C2J×C2Jis called a lower solution of problemSif αut Ft, αut, αvt≥0, t /t1,

αvt Gt, αut, αvt≥0, t /t1, Δαu|tt1Iuαut1, Δαv|tt1 Ivαvt1, Δαu

tt1Nuαut1, Δαv

tt1Nvαvt1, αu0≤a, αv0≤b, αu1≤c, αv1≤d.

2.18

We also define an upper solutionβu, βvX∩C2J×C2Jifβu, βvsatisfies the reverses of the above inequalities.

The following existence theorem for upper and lower solutions method is proved in 3.

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Theorem 2.5. Letαu, αvandβu, βvbe lower and upper solutions of problemS, respectively, satisfyinga1. Moreover, we assumea2∼a4andD3. Then problemShas at least one solution u, vsuch that

αut, αvt≤ut, vt≤

βut, βvt

,t∈ 0,1. 2.19

The following theorems are well known cone theoretic fixed point theorems. See Lakshmikantham 8for proofs and details.

Theorem 2.6. LetXbe a Banach space andKa cone inX. Assume thatΩ1andΩ2are bounded open subsets inXwith 0 ∈Ω1 andΩ1 ⊂ Ω2. LetT :K ∩Ω21 → Kbe a completely continuous such that either

iTu ≤ uforu∈ K ∩∂Ω1andTu ≥ uforu∈ K ∩∂Ω2or iiTu ≥ uforu∈ K ∩∂Ω1andTu ≤ uforu∈ K ∩∂Ω2. ThenT has a fixed point inK ∩Ω21.

Theorem 2.7. LetX be a Banach space,Ka cone inXand Ωbounded open inX.Let 0 ∈ Ωand T :K ∩Ω → Kbe condensing. Suppose thatTx /νx, for allx∈ K ∩∂Ωand allν≥1.Then

iT,K ∩Ω,K 1. 2.20

3. Existence

In this section, we prove an existence theorem of positive solutions for problemPwith Dirichlet boundary condition and the existence and nonexistence part of the result for problemPwith two-point boundary condition. Let us consider the following second-order impulsive differential systems.

ut λh1tfut, vt 0, t∈0,1, t /t1, vt μh2tgut, vt 0, t∈0,1, t /t1,

Δu|tt1Iuut1, Δv|tt1 Ivvt1 Δu

tt1Nuut1, Δv

tt1Nvvt1,

u0 a≥0, v0 b≥0, u1 c≥0, v1 d≥0,

P

where λ, μare positive real parameters, f, gCR2, 0,∞ with f0,0 0, g0,0 0, and fu, v > 0, gu, v > 0 for all u, v/ 0,0, Iu, IvCR,R satisfying Iu0 0 Iv0, Nu, NvCR,−∞,0,andh1, h2C0,1,0,∞may be singular att0 and/or 1.

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We first set up an equivalent operator equatio for problemP. Let us defineAλ:XP C 0,1andBμ:XP C 0,1by taking

Aλu, vta catλ 1

0

Kt, sh1sfus, vsdsWut, u,

Bμu, vtb dbtμ 1

0

Kt, sh2sgus, vsdsWvt, v,

3.1

where

Kt, s

⎧⎨

t−Ivvt1−1−t1Nvvt1, 0≤tt1, 1−tIvvt1t1Nvvt1, t1< t≤1.

Wut, ut

⎧⎨

t−Iuut1−1−t1Nuut1, 0≤tt1, 1−tIuut1t1Nuut1, t1< t≤1,

Wvt, ut

⎧⎨

t−Ivvt1−1−t1Nvvt1, 0≤tt1, 1−tIvvt1t1Nvvt1, t1< t≤1.

3.2

Also define

Tλ,μu, v

Aλu, v, Bμu, v

. 3.3

ThenTλ,μ : XX is well defined on X and problemPis equivalent to the fixed-point equation

Tλ,μu, v u, v inX. 3.4

Mainly due to D1, Tλ,μ is completely continuous see 3 for the proof. Let u0 supt∈ 0,t

1|ut|,u1supt∈ t

1,1|ut|, S0 t1/4,3t1/4, S1 3t11/4, t13/4,P{u, v∈ X | u, v≥0},andK{u, v∈ P | mint∈S0ut vtt1/4u0v0,mint∈S1ut vt≥1−t1/4u1v1}.Thenumax{u0,u1}andP,Kare cones inX.By using concavity ofTλ,μuwithu∈ P, we can easily show thatTλ,μP⊂ K.

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We now prove the existence theorem of positive solutions for Dirichlet boundary value problem

ut λh1tfut, vt 0, t∈0,1, t /t1, vt μh2tgut, vt 0, t∈0,1t /t1,

Δu|tt1Iuut1, Δv|tt1Ivvt1, Δutt

1Nuut1, Δvtt

1Nvvt1,

u0 0, v0 0, u1 0, v1 0.

PD

Theorem 3.1. Assume D1,D2,D4, and D5. Then problem PD has at least one positive solution for allλ, μ∈R2\ {0,0}.

Proof. First, we consider caseλ > 0 andμ > 0.By the factNu,0 < 1/2 andNv,0 < 1/2,we may choosec1, m1 > 0 such that max{Nu,0, Nv,0} < c1 < 1/2,|Nuu |≤c1uforum1and

|Nvv| ≤ c1vforvm1.Also chooseηλ andημ satisfying 0 < ηλ < 1−2c1/2λ1

0s1sh1sdsand 0< ημ < 1−2c1/2μ1

0s1sh2sds.Sincef0 0 andg0 0,there exist m2, m3>0 such thatfu, v≤ηλuvforuvm2andgu, vημuvforuvm3. LetΩ1 BM1 {u, v∈X | u, v< M1}withM1 min{m1, m2, m3}.Then foru, v∈ K ∩∂Ω1,we obtain by usingD2

Aλu, vt λ 1

0

Kt, sh1sfus, vsdsWut, u

ληλ 1

0

s1sh1sus vsds|Nuut1|

ληλ

1

0

s1sh1sdsc1

u, v

≤1 2u, v,

3.5

for allt∈ 0,1.Similarly, we obtain

Bμu, vt≤ 1

2u, v 3.6

for allt∈ 0,1.Thus

Tλ, μu, vAλu, vBμu, v≤ u, v. 3.7

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On the other hand, let us chooseη1andη2such that 1

η2 < μmin

⎧⎪

⎪⎩ t1

8min

t∈S0

S0

Kt, sh2sds,1−t1 8 min

t∈S1

S1

Kt, sh2sds

⎫⎪

⎪⎭.

1

η2 < μmin

⎧⎪

⎪⎩ t1

8min

t∈S0

S0

Kt, sh2sds,1−t1 8 min

t∈S1

S1

Kt, sh2sds

⎫⎪

⎪⎭.

3.8

Also byD5,we may chooseRf and Rg such thatfu, vη1uvforuvRf and gu, vη2uvfor uvRg.LetΩ2 {u,v ∈ X | u, v < M2},where M2 max{8Rf/t1,8Rf/1t1,8Rg/t1,8Rg/1t1, M11}.ThenΩ1⊂Ω2.Letu, v∈ K ∩∂Ω2, then we have the following four cases: u ≥ v and u u0,u ≥ v and u u1,u ≤ vandv v0,andu ≤ vandv v1.We consider the first case, the rest of them can be considered in a similar way. So letu ≥ vanduu0; then for tS0,we have

ut vtutt1

82u0t1

8uv t1

8u, v ≥Rf. 3.9 Thusfut, vtη1ut vtfortS0.SinceWut, u≥0,we get fortS0,

Aλu, vt λ 1

0

Kt, sh1sfus, vsdsWut, u

λ

S0

Kt, sh1sfus, vsds

λη1

S0

Kt, sh1sus vsds

λη1t1 8

S0

Kt, sh1sds2u02v0

λη1t1

8min

t∈S0

S0

Kt, sh1sdsu, v>u, v.

3.10

Therefore,

Tλ,μu, v≥ Aλu, v>u, v, 3.11 and byTheorem 2.6,Tλ,μhas a fixed point inK ∩Ω21.

Second, consider caseλ > 0 andμ 0.Takingc1, ηλ, m1,andm2 as above and using the same computation, we may show

Aλu, v ≤ 1

2u, v, 3.12

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for allu, v∈ K ∩∂Ω1,whereΩ1BM1withM1min{m1, m2}.Sinceμ0, Bμu, vt Wvt, v≤ |Nvvt1| ≤c1u, v ≤ 1

2u, v, 3.13 for allt∈ 0,1.Thus

Tλ,μu, v≤ Aλu, vBμu, v≤ u, v, 3.14 for u, v ∈ K ∩∂Ω1.Now, let us choose η1 andRf as above and letΩ2 {u, v ∈ X | u, v < M2},whereM2 max{8Rf/t1,8Rf/1t1, M11}.ThenΩ1 ⊂ Ω2 and we can show by the same computation as above,

Tλ,μu, v≥ Aλu, v>u, v, 3.15 foru, v∈ K ∩∂Ω2and thusTλ,μhas a fixed point inK ∩Ω21.

Finally, consider caseλ 0 andμ > 0.Takingc1,ημ, m1,andm3 as the first case, we may show by similar argument,

Bμu, v≤ 1

2u, v, Aλu, v ≤ 1

2u, v, 3.16

for allu, v∈ K ∩∂Ω1,whereΩ1BM1withM1min{m1, m3} .Thus

Tλ,μu, v≤ u, v, 3.17

foru, v∈ K ∩∂Ω1.Now, let us chooseη2andRgas the first case and letΩ2 {u, v∈X | u, v < M2},whereM2 max{8Rg/t1,8Rg/1t1, M11}.ThenΩ1 ⊂ Ω2 and we also show similarly, as before,

Tλ,μu, v≥Bμu, v>u, v, 3.18 foru, v∈ K ∩∂Ω2.Therefore,Tλ,μhas a fixed point inK ∩Ω21and this completes the proof.

Now let us consider two point boundary value problems given as follows:

ut λh1tfut, vt 0, t∈0,1, t/t1, vt μh2tgut, vt 0, t∈0,1, t /t1,

Δu|tt1Iuut1, Δv|tt1Ivvt1, Δutt

1Nuut1, ΔvΔvtt

1 Nvvt1,

u0 a≥0, v0 b≥0, u1 c > a, v1 d > b.

PT

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Lemma 3.2. AssumeD5.LetRbe a compact subset ofR2\ {0,0}.Then there exists a constant bR >0 such that for allλ, μ ∈ Rfor possible positive solutionsu, vof problem3.20atλ, μ, one hasu, v< bR.

Proof. Suppose on the contrary that there is a sequenceun, vnof positive solutions of3.20 at λn, μnsuch that λn, μn ∈ R for allnand un, vn → ∞.Since0,0/∈ R, there is a subsequence, say again{λn, μn},such thatαmin{λn}>0 orβ min{μn} >0. First, we assumeα >0. Fromun, vn → ∞, we knowun0vn0 → ∞orun1vn1 → ∞.

Supposeun0vn0 → ∞. Then by the concavity ofunandvn, we have unt vnt≥ t1

4un0vn0, 3.19

fortS0.Let us chooseη1>2/t21αh1,whereh1 mint∈S0h1t. Then byD5, there exists Rf >0 such that

fu, vη1uv ∀uvRf. 3.20

Sinceun0vn0 > 4/t1Rf for sufficiently largen,3.19impliesunt vnt > Rf for tS0. Thus fortS0,

funt, vnt> η1unt vnt≥η1unt. 3.21

Hence we have fortS0,

0unt λnh1tfunt, vnt> unt αh1η1unt. 3.22 If we multiply byφt sin2π/t1t−t1/4both sides in the above inequality and integrate onS0, then by the factsφt1/4>0, φ3t1/4<0 and integration by part, we obtain

0>

3t1/4

t1/4

untφtdtαh1η1

3t1/4

t1/4

untφtdt

≥ − 2π

t1

23t1/4

t1/4

untφtdtαh1η1 3t1/4

t1/4

untφtdt.

3.23

Thus2π/t12/αh1η1which is a contradiction to the choice ofη1.Supposeun1vn1

∞, then we also get a contradiction by a similar calculation withη2 >2/1t12 αh1, whereh1 mint∈S1h1t.Finally, the caseβ >0 can also be proved by similar way using the conditiong∞.

Lemma 3.3. AssumeD1,D3,and

Qfu, v1fu, v2,wheneverv1v2,gu1, vgu2, v,wheneveru1u2.

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If problem 3.20has a positive solution atλ, μ.Then the problem also has a positive solution at λ, μfor allλ, μ≤λ, μ.

Proof. Letu, vbe a positive solution of problem3.20atλ, μand letλ, μ∈R2\ {0,0}

withλ, μ≤λ, μ.Thenu, vis an upper solution of3.20atλ, μ.Defineαu, αvby

αut

⎧⎪

⎪⎩

0, t∈ 0, t1, c

1−t1t−t1, t∈t1,1,

αvt

⎧⎪

⎪⎩

0, t∈ 0, t1, d

1−t1t−t1, t∈t1,1.

3.24

Thenαu, αvis a lower solution of problem3.20atλ, μ.By the concavity ofu, v,u, v≥ αu, αv.Therefore,Theorem 2.5implies that problem3.20has a positive solution atλ, μ.

Lemma 3.4. AssumeD1 ∼ D4and Q.Then there existsλ, μ > 0,0such that problem 3.20has a positive solution for allλ, μ≤λ, μ.

Proof. It is not hard to see that the following problem:

ut h1t 0, t∈0,1, t /t1, vt h2t 0, t∈0,1, t /t1, Δu|tt1Iuut1, Δv|tt1Ivvt1, Δu

tt1Nuut1, Δv

tt1Nvvt1,

u0 a≥0, v0 b≥0, u1 c > a, v1 d > b

3.25

has a positive solution so letβu, βvbe a positive solution. LetMf supt∈ 0,1fβut, βvt andMg supt∈ 0,1ut, βvt.ThenMf, Mg > 0 and forλ, μ 1/Mf,1/Mg,we get

βuλh1tf

βut, βvt

h1t λf

βut, βvt

−1

≤0, βvμh2tg

βut, βvt

h2t μg

βut, βvt

−1

≤0.

3.26

This shows that βu, βv is an upper solution of 3.20 at λ, μ. On the other hand, αu, αv given inLemma 3.3is obviously a lower solution andαu, αv ≤ βu, βv.Thus by Theorem 2.5,3.20has a positive solution atλ, μand the proof is done byLemma 3.3.

We introduce a known existence result for a singular boundary value problem with no impulse effect.

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Lemma 3.5see 9. Consider,D1,D5andQ. For problem ut λh1tfut, vt 0, vt μh2tgut, vt 0, t∈0,1,

u0 a≥0, u1 c > a, v0 b≥0, v1 d > b,

UT

letAT {λ, μ∈R2\ {0,0} |3.27has a positive solution atλ, μ}.ThenAT,is bounded above.

DefineA {λ, μ∈R2\ {0,0} |3.20has a positive solution atλ, μ}.ThenA/by Lemma 3.4andA,≤is a partially ordered set.

Lemma 3.6. AssumeD1∼D6.ThenA,≤is bounded above.

Proof. Suppose on the contrary that there exists a sequenceλn, μn∈ Asuch that|λn, μn| →

∞.Letun, vnbe a positive solution of problem3.20atλn, μn.By conditionD2,we may choose sequencessn,tnin 0, t1∪t1,1such that ifIuunt1>0,thentn∈t1,1and

Iuunt1 tnt1Nuunt1 0, Iuunt1 t−t1Nuunt1>0, on t1, tn, Iuunt1 t−t1Nuunt1<0, ontn,1;

3.27

ifIuunt1<0,thentn∈ 0, t1and

Iuunt1 tnt1Nuunt1 0, Iuunt1 t−t1Nuunt1>0, on 0, tn, Iuunt1 t−t1Nuunt1<0, ontn, t1;

3.28

ifIvvnt1>0,thensn∈t1,1and

Ivvnt1 snt1Nvvnt1 0, Ivvnt1 t−t1Nvvnt1>0, on t1, sn, Ivvnt1 t−t1Nvvnt1<0, onsn,1;

3.29

ifIvvnt1<0,thensn∈ 0, t1and

Ivvnt1 snt1Nvvnt1 0, Ivvnt1 t−t1Nvvnt1>0, on 0, sn, Ivvnt1 t−t1Nvvnt1<0, onsn, t1.

3.30

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IfIuunt1>0,define

unt

⎧⎪

⎪⎪

⎪⎪

⎪⎩

unt, on 0, t1, unt−Iuunt1 t−t1Nuunt1, ont1, tn,

unt, on tn,1,

3.31

and ifIuunt1<0, define

unt

⎧⎪

⎪⎪

⎪⎪

⎪⎩

unt, on 0, tn, unt Iuunt1 t−t1Nuunt1, ontn, t1,

unt, ont1,1.

3.32

Moreover, ifIvvnt1>0, define

vnt

⎧⎪

⎪⎪

⎪⎪

⎪⎩

vnt, on 0, t1, vnt−Ivvnt1 t−t1Nvvnt1, ont1, sn,

vnt, on sn,1,

3.33

and ifIvvnt1<0, define

vnt

⎧⎪

⎪⎪

⎪⎪

⎪⎩

vnt, on 0, sn, vnt Ivvnt1 t−t1Nvvnt1, onsn, t1,

vnt, ont1,1.

3.34

Then we can easily see that un,vn ∈ C 0,1 × C 0,1 ∩ C20,1 × C20,1 except t1, tn, sn.Furthermore,unt1,vnt1 unt1,vnt1,untn,vntnuntn,vntnand unsn,vnsnunsn,vnsn.We also seeunt, vnt≥unt,vnton 0,1.Thus by D6,we get

unt λnh1tfunt,vnt unt λnh1tfunt,vnt λnh1t

funt,vnt−funt, vnt

≤0,

vnt μnh2tgunt,vnt vnt μnh2tgunt,vnt μnh2t

gunt,vnt−gunt, vnt

≤0.

3.35

We also getun0 un0 a,un1 un1 c,vn0 vn0 b,andvn1 vn1 d.

Thusun,vnis aG-upper solution of problemUTatλn, μn.IfIuunt1 0 orIvvnt1

0, then we consider un un or vn vn as a G-upper solution. Let αut,αvt c− ata,d−btb, then αuv is the G-lower solution of 3.27 at λn, μn.Therefore,

参照

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