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,[email protected]
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,
Δutt
1 Nuut1, Δv
tt1 Nvvt1,
u0 a≥0, v0 b≥0, u1 c≥0, v1 d≥0,
P
whereλ, μare positive real parameters,Δu|tt1 ut1−ut1, andΔu|tt1 ut1−ut−1.
Throughout this paper, we assumef, g∈ CR2
,Rwithf0,0 0 g0,0andfu, v >
0, gu, v > 0 for all u, v/ 0,0, Iu, Iv ∈ CR,R satisfyingIu0 0 Iv0, Nu, Nv ∈
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 0,1×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.
D1
1
0s1−shisds <∞,fori1,2.
D2t1Nuu≤Iuu≤ −1−t1Nuuandt1Nvv≤Ivv≤ −1−t1Nvv.
D3uIuuandvIvvare nondecreasing.
D4Nu,∞limu→ ∞|Nuu|/u <1 andNv,∞limv→ ∞|Nvv|/v <1.
D5f∞limuv→ ∞fu, v/uv∞andg∞limuv→ ∞gu, v/uv∞.
D6f and g are nondecreasing on R2, that is,fu1, v1 ≤ fu2, v2 andgu1, v1 ≤
gu2, 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,0limu→0|Nuu|/u <1/2 andNv,0 limv→0|Nvv|/v <1/2,
D5f∞∞, g∞∞andf0limuv→0fu, 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.
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, αv∈C0,1×C 0,1is aG-lower solution ofHifαu, αv∈
C20,1×C20,1except at finite pointsτ
1, . . . , τnwith 0< τ1< . . . < τn<1 such that
L1at each τi, there exist αu τi−, αv τi−,αuτi, αvτi such that αu τi− ≤
αuτi, αvτi−≤αvτi,and
L2
αut Ft, αut, αvt≥0,
αvt Gt, αut, αvt≥0, t∈{ 0,1
τ1, . . . , τn}
,
αu0≤A, αu1≤C,
αv0≤B, αv1≤D.
2.1
We also say that βu, βv ∈ C 0,1 ×C 0,1 is a G-upper solution of the problem H if
βu, βv ∈ C20,1×C20,1 except at finite pointsσ1, . . . , σm with 0 < σ1 < · · · < σm < 1
such that
U1at each σi, there exist βuσj−, βvσj−,βuσj, βvσj such that βuσj− ≥
βu σj, βvσj−≥βvσj,and
U2
βut Ft, βut, βvt
≤0,
βvt Gt, βut, βvt
≤0, t∈ 0,1
{σ1, . . . , σn}
,
βu0≥A, βu1≥C,
βv0≥B, βv1≥D.
2.2
Lemma 2.2. LetF, G :D → Rbe continuous functions andD⊂0,1×R×R.Assume that there existhF, hG∈C0,1,Rsuch that
|Ft, u, v| ≤hFt, |Gt, u, v| ≤hGt, 2.3
for allt, u, v∈0,1×R×R,and
1
0
s1−shFsds
1
0
s1−shGsds <∞. 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, hG∈C0,1,Rsuch that
a2|Ft, u, v| ≤hFtand|Gt, u, v| ≤hGtfor allt, u, v∈Dαβ;
a3
1
0s1−shFsds
1
0s1−shGsds <∞;
a4Ft, u, v1 ≤Ft, u, v2,wheneverv1 ≤ v2andGt, u1, v ≤ Gt, u2, v,wheneveru1 ≤
u2.
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:
F∗t, u, v
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
Ft, βut, v
−u−βut
1 u
2 ifu > β
ut,
Ft, u, v ifαut≤u≤βut,
Ft, αut, v−
u−αut
1 u
2 ifu < α
F∗t, u, v ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
F∗t, u, βvt
ifv > βvt,
F∗t, u, v ifαvt≤v≤βvt,
F∗t, u, αvt ifv < αvt,
G∗t, u, v
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
Gt, βut, v
if u > βut,
Gt, u, v if αut≤u≤βut,
Gt, αut, v if u < αut,
2.6
G∗t, u, v
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
G∗t, u, βvt,
−v−βvt
1 v
2 ifv > β
vt,
G∗t, u, v ifαvt≤v≤βvt,
G∗t, u, αvt−
v−αvt
1 v
2 ifv < α
vt.
2.7
ThenF∗, G∗:0,1×R×R → Rare continuous and
|F∗t, u, v| ≤mαu, βu
hFt,
|G∗t, u, v| ≤mαv, βv
hGt,
2.8
for allt, u, v∈0,1×R×R,wheremα, β αβ1.For the problem
ut F∗t, ut, vt 0,
vt G∗t, 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.Letαu−ut0
cases. First, ifαvt0≤vt0,then byαut0> ut0and conditiona4,
0≥αu−ut0 αut0 F∗t0, ut0, vt0
αut0 Ft0, αut0, vt0−
ut0−αut0
1u2t 0
≥αut0 Ft0, αut0, αvt0− ut0−αut0
1u2t 0
≥ αut0−ut0
1u2t0 >0,
2.10
which is a contradiction. Next, ifαvt0> vt0,then by the definition ofF∗,
0≥αu−ut0 αut0 F∗t0, ut0, vt0
αut0 Ft0, αut0, αvt0−
ut0−αut0
1u2t 0
>0,
2.11
which is also a contradiction. Ift0τifor somei1, . . . , n,then sinceαu−uattains its positive
maximum atτi,
αu−u
τi−
≥0, αu−uτi≤0. 2.12
Ifαu−uτi−>0,then
0<αu−u
τi−
−αu−uτi αu
τi−
−αuτi. 2.13
This leads a contradiction to the definition ofG-lower solution. Ifαu−uτi− 0,then there
existsδ >0 such that for allt∈τi−δ, τi,
αu−ut>0, αu−ut≥0, αu−ut≤0. 2.14
Fort∈τi−δ, τi,ifαvt≤vt,then byαut> utand conditiona4,
0≥αu−ut αut F∗t, ut, vt
αut Ft, αut, vt−ut−αut
1u2t
≥αut Ft, αut, αvt−
ut−αut
1u2t
≥ αut−ut
1u2t >0,
which is a contradiction. Ifαvt> vt,then by definition ofF∗,
0≥αu−ut αut F∗t, ut, vt
αut Ft, αut, αvt−ut−αut
1u2t >0,
2.16
which is a contradiction. Ift00 or 1,then
0<αu−u0 αu0−A≤0,
0<αu−u1 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, Δutt
1Nuut1, Δv
tt1Nvvt1,
u0 a, v0 b, u1 c, v1 d,
S
whereF, G∈C0,1×R×R,R,Iu, Iv ∈CR,RsatisfyingIu0 0Iv0andNu, Nv ∈
CR,−∞,0.
Definition 2.4. αu, αv∈X∩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, Δαutt
1≥Nuαut1, Δα
vtt1 ≥Nvαvt1,
αu0≤a, αv0≤b, αu1≤c, αv1≤d.
2.18
We also define an upper solutionβu, βv∈X∩C2J×C2Jifβu, βvsatisfies the reverses
of the above inequalities.
The following existence theorem for upper and lower solutions method is proved in
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 inXwith0 ∈Ω1 andΩ1 ⊂ Ω2. LetT :K ∩Ω2\Ω1 → 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 ∩Ω2\Ω1.
Theorem 2.7. LetX be a Banach space,Ka cone inXand Ωbounded open inX.Let0 ∈ 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
Δutt
1Nuut1, Δv
tt1Nvvt1,
u0 a≥0, v0 b≥0, u1 c≥0, v1 d≥0,
P
where λ, μare positive real parameters, f, g ∈ CR2
, 0,∞ with f0,0 0, g0,0 0,
and fu, v > 0, gu, v > 0 for all u, v/ 0,0, Iu, Iv ∈ CR,R satisfying Iu0 0
Iv0, Nu, Nv ∈CR,−∞,0,andh1, h2 ∈C0,1,0,∞may be singular att0 and/or
We first set up an equivalent operator equatio for problemP. Let us defineAλ:X →
P C 0,1andBμ:X → P C 0,1by taking
Aλu, vta c−atλ
1
0
Kt, sh1sfus, vsdsWut, u,
Bμu, vtb d−btμ
1
0
Kt, sh2sgus, vsdsWvt, v,
3.1
where
Kt, s
⎧ ⎨ ⎩
t−Ivvt1−1−t1Nvvt1, 0≤t≤t1,
1−tIvvt1−t1Nvvt1, t1< t≤1.
Wut, ut
⎧ ⎨ ⎩
t−Iuut1−1−t1Nuut1, 0≤t≤t1,
1−tIuut1−t1Nuut1, t1< t≤1,
Wvt, ut
⎧ ⎨ ⎩
t−Ivvt1−1−t1Nvvt1, 0≤t≤t1,
1−tIvvt1−t1Nvvt1, t1< t≤1.
3.2
Also define
Tλ,μu, v
Aλu, v, Bμu, v
. 3.3
ThenTλ,μ : X → X 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∈t1,1|ut|, S0 t1/4,3t1/4, S1 3t11/4, t13/4,P{u, v∈
X | u, v≥0},andK{u, v∈ P | mint∈S0ut vt≥t1/4u0v0,mint∈S1ut
vt≥1−t1/4u1v1}.Thenumax{u0,u1}andP,Kare cones inX.By using
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, Δv
tt1Nvvt1,
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 |≤c1uforu ≤ m1and
|Nvv| ≤ c1vforv ≤ m1.Also chooseηλ andημ satisfying 0 < ηλ < 1−2c1/2λ
1
0s1−
sh1sdsand 0< ημ < 1−2c1/2μ
1
0s1−sh2sds.Sincef0 0 andg0 0,there exist
m2, m3>0 such thatfu, v≤ηλuvforuv≤m2andgu, v≤ημuvforuv≤m3.
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
s1−sh1sus vsds|Nuut1|
≤
ληλ
1
0
s1−sh1sdsc1
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
On the other hand, let us chooseη1andη2such that
1
η2
< μmin
⎧ ⎪ ⎨ ⎪ ⎩ t1
8mint∈S0
S0
Kt, sh2sds,
1−t1
8 mint∈S1
S1
Kt, sh2sds
⎫ ⎪ ⎬ ⎪ ⎭. 1 η2
< μmin
⎧ ⎪ ⎨ ⎪ ⎩ t1
8mint∈S0
S0
Kt, sh2sds,
1−t1
8 mint∈S1
S1
Kt, sh2sds
⎫ ⎪ ⎬ ⎪ ⎭. 3.8
Also byD5,we may chooseRf and Rg such thatfu, v ≥ η1uvforuv ≥ Rf and
gu, v ≥ η2uvfor uv ≥ Rg.LetΩ2 {u,v ∈ X | u, v < M2},where M2
max{8Rf/t1,8Rf/1−t1,8Rg/t1,8Rg/1−t1, 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
t∈S0,we have
ut vt≥ut≥ t1
82u0≥
t1
8uv
t1
8u, v ≥Rf. 3.9
Thusfut, vt≥η1ut vtfort∈S0.SinceWut, u≥0,we get fort∈S0,
Aλu, vt λ
1
0
Kt, sh1sfus, vsdsWut, u
≥λ
S0
Kt, sh1sfus, vsds
≥λη1
S0
Kt, sh1sus vsds
≥λη1
t1
8
S0
Kt, sh1sds2u02v0
≥λη1t1
8mint∈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 ∩Ω2\Ω1.
Second, consider caseλ > 0 andμ 0.Takingc1, ηλ, m1,andm2 as above and using
the same computation, we may show
Aλu, v ≤
1
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/1−t1, 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 ∩Ω2\Ω1.
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/1−t1, 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 ∩Ω2\Ω1and 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, Δv Δv
tt1 Nvvt1,
u0 a≥0, v0 b≥0, u1 c > a, v1 d > b.
Lemma 3.2. AssumeD5.LetRbe a compact subset ofR2\ {0,0}.Then there exists a constant
bR >0such 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
fort∈S0.Let us chooseη1>2π2/t21αh1,whereh1 mint∈S0h1t. Then byD5, there exists
Rf >0 such that
fu, v≥η1uv ∀uv≥Rf. 3.20
Sinceun0vn0 > 4/t1Rf for sufficiently largen,3.19impliesunt vnt > Rf for
t∈S0. Thus fort∈S0,
funt, vnt> η1unt vnt≥η1unt. 3.21
Hence we have fort∈S0,
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/αh1≥η1which is a contradiction to the choice ofη1.Supposeun1vn1 →
∞, then we also get a contradiction by a similar calculation withη2 > 2π2/1−t12 αh1,
whereh1 mint∈S1h1t.Finally, the caseβ >0 can also be proved by similar way using the
conditiong∞∞.
Lemma 3.3. AssumeD1,D3,and
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−t1
t−t1, t∈t1,1,
αvt
⎧ ⎪ ⎨ ⎪ ⎩
0, t∈ 0, t1,
d
1−t1
t−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,
Δutt
1Nuut1, 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∈0,1gβut, βvt.ThenMf, Mg > 0 and forλ∗, μ∗ 1/Mf,1/Mg,we
get
βuλ∗h1tf
βut, βvt
h1t
λ∗fβut, βvt
−1≤0,
βvμ∗h2tg
βut, βvt
h2t
μ∗gβut, 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∗, μ∗and the proof is done byLemma 3.3.
Lemma 3.5see9. 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 tn−t1Nuunt1 0,
Iuunt1 t−t1Nuunt1>0, on t1, tn,
Iuunt1 t−t1Nuunt1<0, ontn,1;
3.27
ifIuunt1<0,thentn∈ 0, t1and
Iuunt1 tn−t1Nuunt1 0,
Iuunt1 t−t1Nuunt1>0, on 0, tn,
Iuunt1 t−t1Nuunt1<0, ontn, t1;
3.28
ifIvvnt1>0,thensn∈t1,1and
Ivvnt1 sn−t1Nvvnt1 0,
Ivvnt1 t−t1Nvvnt1>0, on t1, sn,
Ivvnt1 t−t1Nvvnt1<0, onsn,1;
3.29
ifIvvnt1<0,thensn∈ 0, t1and
Ivvnt1 sn−t1Nvvnt1 0,
Ivvnt1 t−t1Nvvnt1>0, on 0, sn,
Ivvnt1 t−t1Nvvnt1<0, onsn, t1.
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,unt−1,vnt−1 unt1,vnt1,untn−,vnt−n≥untn,vntnand
uns−n,vns−n≥unsn,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−
byTheorem 2.3, problem 3.27 has a positive solution for all λn, μn. This contradicts to
Lemma 3.5and the proof is done.
Lemma 3.7. AssumeD1∼D6.Then every nonempty chain inAhas a unique supremum inA.
Proof. LetCbe a chain inA. Without loss of generality, we may choose a distinct sequence {λn, μn} ⊂ C such that λn, μn ≤ λn1, μn1. By Lemma 3.6, there exists λC, μC such
thatλn, μn → λC, μC.If we showλC, μC ∈ A, then the proof is done. Since{λn, μn}
is bounded, Lemma 3.2implies that there is a constantB such that un, vn < B, where
un, vnis a solution corresponding toλn, μn. By the compactness ofTλ,μ,{un, vn}has a
convergent subsequence converging to say,uC, vC. By Lebesgue Convergence theorem, we
see thatuC, vCis a solution of3.20atλC, μC. ThusλC, μC∈ A.
Theorem 3.8. AssumeD1∼D6.Then there exists a continuous curveΓsplittingR2\ {0,0}
into two disjoint subsetsO1 and O2 such that problem PT has at least one positive solution for
λ, μ∈ O1∪Γand no solution forλ, μ∈ O2.
Proof. λ∗, μ∗ is given inLemma 3.4. We know from Lemma 3.4that 3.20 has a positive solution at0, sfor all 0 < s ≤ μ∗. Thus{0, s|s > 0} ∩ Ais a nonempty chain inAand byLemma 3.7, it has unique supremum of the form0, s∗inA. This implies that3.20has a positive solution at0, sfor all 0 < s≤s∗and no solution at0, sfor alls > s∗. Similarly, there isr∗ ≥ λ∗ such that 3.20has a positive solution atr,0 for all 0 < r ≤ r∗ and no solution atr,0for allr > r∗.DefineL :R → R2by takingLt {r, s|srt}.Then
fort ∈ −r∗, s∗,Lt∩ Ais a nonempty chain inA.DefineΓtas the unique supremum ofLt∩ A. ThenΓis well defined on −r∗, s∗and as a consequence ofLemma 3.3, we see thatΓ is continuous on −r∗, s∗,Γ−r∗ r∗,0,and Γs∗ 0, s∗. Therefore, the curve
Γ Γ −r∗, s∗separatesR2\{0,0}into two disjoint subsetsO1andO2, whereO1is bounded
andO2is unbounded and we get the conclusion of this theorem forΓ,O1,andO2.
4. Multiplicity
In this section, we study existence of the second positive solution for two point boundary value problem3.20with λ, μin certain region of O1 appeared inTheorem 3.8. For the
computation of fixed point index, we need to consider problems 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,
Δutt
1 Nuut1, Δv
tt1 Nvvt1,
u0 aε, u1 cε,
v0 bε, v1 dε,
PTε
whereε >0, c > a≥0 andd > b≥0.Theorem 3.8implies that there exists a continuous curve
ΓεsplittingR2\ {0,0}into two disjoint subsetsOε,1andOε,2such that the problem4.1has
and lower solutions argument, we can easily show that if 0< ε < ε,thenOε,1∪Γε⊂ Oε,1∪Γε.
LetO ∪ε>0Oε,1∪Γε,thenO ⊂ O0,1.We state the main theorem for two point boundary
value problem3.20as follows.
Theorem 4.1. AssumeD1∼D6.Then there exists a continuous curveΓ0splittingR2\ {0,0} into two disjoint subsetsO0,1and,O0,2and there exists a subsetO ⊂ O0,1such that problem3.20has at least two positive solutions forλ, μ∈ O,at least one positive solution forλ, μ∈O0,1\ O∪Γ0, and no solution forλ, μ∈ O0,2.
Proof. LetO∪ε>0Oε,1∪Γεand letλ, μ∈ O.It is enough to prove that problem3.20has
the second solution atλ, μ.By the definition ofO,there existsε >0 such thatλ, μ∈ Oε,1∪Γε.
That is4.1has a positive solution atλ, μ.So letuε, vεbe a positive solution of problem
4.1atλ, μand letΩ {u, v∈X| −ε < ut< uεt,−ε < vt< vεtfort∈ 0,1, ut1<
uεt1, vt1 < vεt1}.ThenΩis bounded open inX,0 ∈Ω.Furthermore,Tλ,μ : K ∩Ω →
Kis condensing, since it is completely continuous. We show that Tλ,μu, v/νu, vfor all
u, v∈ K ∩∂Ωand allν≥1.If it is not true, then there existu, v∈ K ∩∂Ωandν0≥1 such
thatTλ,μu, v ν0u, v.Thusu, vis a positive solution of the following equation
ν0ut λh1tfut, vt 0, t∈0,1, t /t1,
ν0vt μh2tgut, vt 0, t∈0,1, t /t1,
ν0Δu|tt1Iuut1, ν0Δv|tt1 Ivvt1,
ν0Δutt1Nuut1, ν0Δvtt1Nvvt1,
ν0u0 a, ν0v0 b, ν0u1 c, ν0v1 d,
4.1
and we can consider the following two cases. The first case isut0 uεt0orvt0 vεt0
for somet0 ∈0,1. The second case isut1 uεt1orvt1 vεt1.First, let us consider
caseut0 uεt0for some t0 ∈ 0,1. One may prove similarly for casevt0 vεt0.
If t0 ∈ J,that is, t0/t1,let mt u−uεt,then mt ≥ 0 onJ, m0 < 0, m1 < 0,
and mt1 ≤ 0.Thus on one of intervals 0, t1 ort1,1containing t0,maximum principle
impliesm ≡ 0 and this contradicts to the facts of m0 < 0 and m1 < 0.Ift0 t1,then
ut1 uεt1, ut ≤ utand vt ≤ vεton 0,1.Thus by D6and D2, we get the
following contradiction:
ut1
1
ν0
Aλu, vt1
a b−at1
ν0
λ ν0
1
0
Kt1, sh1sfus, vsds
−t1Iuut1 1−t1Nuut1
ν0
< aε b−at1λ
1
0
Kt1, sh1sfuεs, vεsds
−t1Iuuεt1 1−t1Nuuεt1
uεt1 ut1.
Second, let us considerut1 uεt1. Sinceut≤utandvt≤vεt, we get
ut1 1 ν0
Aλu, v
t1 a b−at1 ν0
λ ν0
1
0
Kt1, sh1sfus, vsds
1−t1Iuut1−t1Nuut1
ν0
< aε b−at1λ
1
0
Kt1, sh1sfuεs, vεsds
1−t1Iuuεt1−t1Nuuεt1
uε
t1ut1.
4.3
One may show the contradiction similarly for casevt1 vεt1.This contradiction shows
Tλ,μu, v/νu, vfor allu, v∈ K ∩∂Ωandν≥1.Therefore, byTheorem 2.7, we obtain
iTλ,μ,K ∩Ω,K
1. 4.4
On the other hand, by Lemma 3.6, we know that there is λ1, μ1 such that 3.20has no
positive solution atλ1, μ1.Thus for any open setUinX, we get
iTλ1,μ1,K ∩ U,K
0. 4.5
LetRbe a compact rectangle containingλ, μandλ1, μ1.ByLemma 3.2, for allλ, μ∈ R,
there existsbR>0 such that all possible solutionsu, vofPTatλ, μsatisfyu, v< bR
andΩ⊂BbR.Defineh: 0,1×BbR∩ K → Kby
hτ,u, v Tτλ11−τλ,τμ11−τμu, v. 4.6
Thenh0,u, v Tλ,μu, v, h1,u, v Tλ1,μ1u, v, his completely continuous on 0,1×
K,andhτ,u, v/ u, vfor allτ,u, v∈ 0,1×∂BbR∩ K.By the property of homotopy invariance and4.5, we have
iTλ,μ, BbR∩ K,K
iTλ1,μ1, BbR∩ K,K
0. 4.7
By the additive property and4.4,4.7, we have
i
Tλ,μ,
B
bR
Ω
∩ K,K
−1. 4.8
5. Applications
In this section, we apply the results in previous sections to study the existence and multiplicity theorems of positive radial solutions for impulsive semilinear elliptic problems.
5.1. On an Annular Domain
Let us consider
Δuλk1|x|fu, v 0,
Δvμk2|x|gu, v 0, inΩl1, l2,|x|/r1,
Δu||x|r1Iu
u||x|r1, Δv||x|r1Iv
u||x|r1,
Δ∂u ∂r
|x|r1 −r
1−n
1 Nu
u||x|r1
m ,
Δ∂v ∂r
|x|r1 −r
1−n
1 Nv
v||x|r1
m ,
ux a≥0, vx b≥0 if|x|l1,
ux c > a, vx d > b if |x|l2,
PA
wheref0,0 0, g0,0 0,Δis the Laplacian ofu,0 < l1 < r1 < l2,andΩl1, l2 {x ∈
Rn| l
1 < |x| < l2} with n > 2. ∂u/∂r denotes the differentiation in the radial direction,
Δu||x|r1 ur1−ur1,Δ∂u/∂r||x|r1 ∂u/∂rr
1−∂u/∂rr1− andm −
l2
l1t
1−ndt.
Applying consecutive changes of variables,r |x|, s −l2
r t1−ndtandtm−s/m,we may
transform problem5.1into problem3.20, wheret1 r12−n−l12−n/l22−n−l21−nandhican
be written as
hit m2 rm1−t2n−1kirm1−t. 5.1
Ifki : l1, l2 → 0,∞are continuous, then hi : 0,1 → 0,∞are also continuous and
satisfiesD1. We may applyTheorem 4.1to obtain the following result.
Corollary 5.1. Assume D2 ∼ D6. Let ki ∈ C l1, l2,0,∞, i 1,2. Then there exists a continuous curveΓ0splittingR2\ {0,0}into two disjoint subsetsO0,1andO0,2 and there exists a subsetO ⊂ O0,1such that problem5.1has at least two positive solutions forλ, μ∈ O,at least one positive solution forλ, μ∈O0,1\ O∪Γ0,and no solution forλ, μ∈ O0,2.
Ifki : l1, l2 → 0,∞are continuous and singular atr l1 and/or l2,thenhi are also
singular att0and/or1.In this case, we assume
l2
l1
r2−n−l12−nl22−n−r2−nkirdr <∞, 5.2
Corollary 5.2. AssumeD2∼D6.If bothki∈Cl1, l2,0,∞satisfy
l2
l1
r2−n−l21−nl22−n−r2−nkirdr <∞. 5.3
Then the conclusion ofCorollary 5.1is valid.
5.2. On an Exterior Domain
Let us consider
Δuλk1|x|fu, v 0,
Δvμk2|x|gu, v 0, |x|> r0, |x|/r1,
Δu||x|r1Iu
u||x|r1, Δv||x|r1 Iv
v||x|r1,
Δ∂u ∂r
|
x|r1
n−2
r0
r1
r0
1−n
Nu
u||x|r1,
Δ∂v ∂r
|
x|r1
n−2
r0
r1
r0
1−n
Nv
v||x|r1,
ux a≥0, vx b≥0, if|x|r0,
ux → c > a, vx → d > b, as|x| → ∞,
PE
wheref0,0 0, g0,0 0,0 < r0 < r1,andn >2.Assume that bothki : r0,∞ → 0,∞
are continuous. Applying changes of variables,r|x|andt1−r/r02−n,we may transform
problem5.4into problem3.20, wheret11−r0/r1n−2andhiare written as
hit
r02n−1
n−221−t
−2n−1/n−2k
i
r01−t−1/n−2
. 5.4
We know that hi are singular at t 1 and can easily check that hi satisfyD1if ki satisfy
∞
r0 rkirdr <∞fori1,2.Thus byTheorem 4.1, we obtain the following result.
Corollary 5.3. AssumeD2∼D6.If bothki∈C r0,∞,0,∞satisfy
∞
r0
rkirdr <∞, 5.5
Acknowledgment
This work was supported for two years by Pusan National University Research Grant.
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