Boundary Value Problems
Volume 2010, Article ID 458015,16pages doi:10.1155/2010/458015
Research Article
Existence of Positive Solutions of
a Singular Nonlinear Boundary Value Problem
Ruyun Ma
1and Jiemei Li
1, 21College of Mathematics and Information Science, Northwest Normal University, Lanzhou 730070, China 2The School of Mathematics, Physics & Software Engineering, Lanzhou Jiaotong University,
Lanzhou 730070, China
Correspondence should be addressed to Ruyun Ma,ruyun [email protected]
Received 21 May 2010; Accepted 11 August 2010
Academic Editor: Vicentiu Radulescu
Copyrightq2010 R. Ma and J. Li. 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.
We are concerned with the existence of positive solutions of singular second-order boundary value problemut ft, ut 0,t∈0,1,u0 u1 0, which is not necessarily linearizable. Here, nonlinearityf is allowed to have singularities att 0,1. The proof of our main result is based upon topological degree theory and global bifurcation techniques.
1. Introduction
Existence and multiplicity of solutions of singular problem
uft, u 0, t∈0,1,
u0 u1 0, 1.1
theory in cones. Recently, Ma 7studied the existence of nodal solutions of the singular boundary value problem
uratfu 0, t∈0,1,
u0 u1 0, 1.2
by applying Rabinowitz’s global bifurcation theorem, whereais allowed to have singularities att0,1 andfis linearizable at 0 as well as at∞. It is the purpose of this paper to study the existence of positive solutions of1.1, which is not necessarily linearizable.
LetXbe Banach space defined by
X
φ∈L1loc0,1| 1
0
t1−tφtdt <∞
, 1.3
with the norm
φX
1
0
t1−tφtdt. 1.4
Let
Xφ∈X|φt≥0, a.e. t∈0,1,
Xp
φ∈X|
1
0
t1−tφtdt >0
.
1.5
Definition 1.1. A functiong :0,1×R → Ris said to be anL1loc-Carath´eodory function if it satisfies the following:
ifor eachu∈R,g·, uis measurable; iifor a.e.t∈0,1,gt,·is continuous; iiifor anyR >0, there existshR ∈Xp,such that
gt, u≤hRt, a.e. t∈0,1,|u| ≤R. 1.6
In this paper, we will prove the existence of positive solutions of 1.1by using the global bifurcation techniques under the following assumptions.
H1Letf : 0,1×0,∞ → 0,∞be an L1
loc-Carath´eodory function and there
exist functionsa0·,a0·,c∞·, andc∞·∈Xp,such that
for someL1loc-Carath´eodory functionsξ1, ξ2defined on0,1×0,∞with
ξ1t, u ◦a0tu, ξ2t, u ◦ a0tu
, asu−→0, 1.8
uniformly for a.e.t∈0,1, and
c∞tu−ζ1t, u≤ft, u≤c∞tuζ2t, u, 1.9
for someL1loc-Carath´eodory functionsζ1, ζ2defined on0,1×0,∞with
ζ1t, u ◦c∞tu, ζ2t, u ◦c∞tu, asu → ∞, 1.10
uniformly for a.e.t∈0,1.
H2ft, u>0 for a.e.t∈0,1andu∈0,∞. H3There exists functionc1·∈Xp,such that
ft, u≥c1tu, a.e. t∈0,1, u∈0,∞. 1.11
Remark 1.2. Ifa0·,a0·,c∞·, andc∞·∈C0,1,0,∞, then1.8implies that
ξ1t, u ◦u, ξ2t, u ◦u, asu → 0, 1.12
and1.10implies that
ζ1t, u ◦u, ζ2t, u ◦u, asu → ∞. 1.13
The main tool we will use is the following global bifurcation theorem for problem which is not necessarily linearizable.
Theorem ARabinowitz,8. Let V be a real reflexive Banach space. LetF : R×V → V be
completely continuous, such thatFλ,0 0, for allλ∈R. Leta, b∈R a < b, such thatu0is an isolated solution of the following equation:
u−Fλ, u 0, u∈V, 1.14
forλ aandλ b, wherea,0,b,0are not bifurcation points of 1.14. Furthermore, assume that
dI−Fa,·, Br0,0/dI−Fb,·, Br0,0, 1.15
whereBr0is an isolating neighborhood of the trivial solution. Let
then there exists a continuum (i.e., a closed connected set)CofScontaininga, b× {0}, and either iCis unbounded inV×R, or
iiC ∩R\a, b× {0}/∅.
To state our main results, we need the following.
Lemma 1.3see1, Proposition 4.7. Leta∈Xp, then the eigenvalue problem
uλatu0, t∈0,1,
u0 u1 0
1.17
has a sequence of eigenvalues as follows:
0< λ1a< λ2a<· · ·< λka< λk1a<· · ·, lim
k→ ∞λka ∞. 1.18
Moreover, for eachk∈N,λkais simple and its eigenfunctionψk ∈C10,1has exactlyk−1zeros in0,1.
Remark 1.4. Note that ψk ∈ C10,1andψk0 ψk1 0 for eachk ∈ N. Therefore, there exist constantsMk>0, such that
ψkt≤Mkt1−t, t∈0,1. 1.19
Our main result is the following.
Theorem 1.5. Let (H1)–(H3) hold. Assume that either
λ1c∞<1< λ1 a0
1.20
or
λ1a0<1< λ1c∞, 1.21
then1.1has at least one positive solution.
Remark 1.6. For other references related to this topic, see9–14and the references therein.
2. Preliminary Results
Lemma 2.1see15, Proposition 4.1. For anyh∈X, the linear problem
ut ht 0, t∈0,1,
has a unique solutionu∈W1,10,1andu∈AC
loc0,1, such that
ut
1
0
Gt, shsds, 2.2
where
Gt, s
⎧ ⎨ ⎩
s1−t, 0≤s≤t≤1,
t1−s, 0≤t≤s≤1.
2.3
Furthermore, ifh∈X, then
ut≥0, t∈0,1. 2.4
LetY C0,1be the Banach space with the normumaxt∈0,1|ut|, and
E{u∈C0,1|u0 u1 0}. 2.5
LetL:DL⊂Y → Xbe an operator defined by
Lu−u, u∈DL, 2.6
where
DL u∈W1,10,1|u∈X, u0 u1 0. 2.7
Then, fromLemma 2.1,L−1:X → C0,1is well defined.
Lemma 2.2. Leta∈Xpandψ1be the first eigenfunction of1.17. Then for allu∈DL, one has
1
0
utψ1tdt 1
0
utψ1tdt. 2.8
Proof. For anyδ∈0,1/2, integrating by parts, we have 1−δ
δ
utψ1tdtuψ11δ−δ− uψ1 1−δ
δ
1−δ
δ
utψ1tdt. 2.9
Sinceu∈DLandψ1∈C10,1, then
lim δ→0uδψ
Therefore, we only need to prove that
lim δ→0u
δψ
1δ 0, lim
δ→0u
1−δψ
11−δ 0. 2.11
Let us deal with the first equality, the second one can be treated by the same way. Note that
u∈DL, then
tututu∈L10, δ, 2.12
which implies thattut∈AC0, δ. Thentutis bounded on0, δ. Now, we claim that
lim t→0t
ut0. 2.13
Suppose on the contrary that limt→0t|ut|a >0, then forδsmall enough, we have
tut≥ a
2, t∈0, δ. 2.14
Therefore,
∞>
δ
0
utdt≥
δ
0
a
2tdt∞, 2.15
which is a contradiction. Combining1.19with2.13, we have
uδψ
1δ≤M11−δδuδ−→0, δ → 0. 2.16
This completes the proof.
Remark 2.3. Under the conditions ofLemma 2.2, for the later convenience,2.8is equivalent to
Lu, ψ1
u, Lψ1
. 2.17
Lemma 2.4see1, Lemma 2.3. For everyρ∈X, the subsetKdefined by
KL−1φ∈X|φt≤ρt, a.e.t∈0,1 2.18
is precompact inC0,1.
LetΣ⊂R×Ebe the closure of the set of positive solutions of the problem
We extend the functionfto anL1
loc-Carath´eodory functionfdefined on0,1×Rby
ft, u
⎧ ⎨ ⎩
ft, u, t, u∈0,1×0,∞,
ft,0, t, u∈0,1×−∞,0.
2.20
Thenft, u≥ 0 foru∈ Rand a.e.t ∈ 0,1. Forλ ≥ 0, letube an arbitrary solution of the problem
Luλft, u. 2.21
Sinceλft, ut ≥ 0 for a.e. t ∈ 0,1,Lemma 2.2yieldsut ≥ 0 for t ∈ 0,1. Thus,uis a nonnegative solution of2.19, and the closure of the set of nontrivial solutionsλ, uof2.21 inR×Eis exactlyΣ.
Let g : 0,1×R → Rbe an L1
loc-Carath´eodory function. Let N : E → X be the
Nemytskii operator associated with the functiongas follows:
Nut gt, ut, u∈E. 2.22
Lemma 2.5. Letgt, u≥0on0,1×R. Letu∈DLbe such thatLu≥λNuin0,1,λ≥0. Then,
ut≥0, t∈0,1. 2.23
Moreover,ut>0, t∈0,1, wheneveru /≡0.
LetN:E → Xbe the Nemytskii operator associated with the functionfas follows:
Nut ft, u, u∈E. 2.24
Then2.21, withλ≥0, is equivalent to the operator equation
uλL−1Nu, u∈E, 2.25
that is,
ut λ
1
0
Gt, sNusds, u∈E. 2.26
Lemma 2.6. Let (H1) and (H2) hold. Then the operatorL−1N : C0,1 → C0,1is completely
Proof. From1.10inH1, there existsR >0, such that, for a.e.t∈0,1and|u|> R,
|ζ1t, u| ≤ 1
2c∞tu, |ζ2t, u| ≤ 1 2c
∞tu. 2.27
Sincefis anL1
loc-Carath´eodory function, then there existshR∈Xp, such that, for a.e.t∈0,1
and|u| ≤R,|ft, u| ≤hRt. Therefore, for a.e.t∈0,1andu∈R, we have
ft, u≤ 3 2c
∞tuhRt. 2.28
For convenience, let T L−1N. We first show that T : C0,1 → C0,1 is continuous.
Suppose thatum → uinC0,1asm → ∞. Clearly,ft, um → ft, uasm → ∞for a.e.
t∈0,1and there existsM >0 such thatum ≤Mfor everym∈N. It is easy to see that
|Tumt−Tut| ≤
1
0
s1−sfs, ums−fs, usds,
fs, ums−fs, us≤3c∞sM2hRs, a.e. s∈0,1.
2.29
By the Lebesgue dominated convergence theorem, we have thatTum → Tu inC0,1 as
m → ∞. Thus,L−1Nis continuous.
LetDbe a bounded set inC0,1.Lemma 2.4together with2.28shows thatTDis precompact inC0,1. Therefore,Tis completely continuous.
In the following, we will apply the Leray-Schauder degree theory mainly to the mappingΦλ:E → E,
Φλu u−λL−1Nu. 2.30
ForR >0, letBR {u∈E:u< R}, let degΦλ, BR,0denote the degree ofΦλonBRwith respect to 0.
Lemma 2.7. LetΛ ⊂ R be a compact interval withλ1a0, λ1a0∩Λ ∅, then there exists a numberδ1>0with the property
Φλu/0, ∀u∈Y : 0<u ≤δ1,∀λ∈Λ. 2.31
Setvn un/un. ThenLvn μnun−1Nun μnun−1ft, unandvn 1. Now, from conditionH1, we have the following:
a0tun−ξ1t, un≤ft, un≤a0tunξ2t, un, 2.32
and accordingly
μn
a0tvn−ξ1t, un
un
≤μn
ft, un
un ≤μn
a0tvn ξ2t, un
un
. 2.33
Let ϕ0 and ϕ
0 denote the nonnegative eigenfunctions corresponding to λ1a0 and
λ1a0, respectively, then we have from the first inequality in2.33that
μn
a0tvn−
ξ1t, un
un
, ϕ0
≤
μn
ft, un
un
, ϕ0
Lvn, ϕ0
. 2.34
FromLemma 2.2, we have that
Lvn, ϕ0
vn, Lϕ0
λ1a0
vn, a0tϕ0
. 2.35
Sinceun → 0 inE, from1.12, we have that
ξ1t, un
un −→
0, asun −→0. 2.36
By the fact thatvn1, we conclude thatvn vinE. Thus,
vn, a0tϕ0
−→v, a0tϕ0
. 2.37
Combining this and2.35and lettingn → ∞in2.34, it follows that
μ∗a0tv, ϕ0
≤λ1a0
a0tϕ0, v
, 2.38
and consequently
μ∗≤λ1a0. 2.39
Similarly, we deduce from second inequality in2.33that
λ1 a0
≤μ∗. 2.40
Corollary 2.8. Forλ∈0, λ1a0andδ∈0, δ1,degΦλ, Bδ,0 1.
Proof. Lemma 2.7, applied to the intervalΛ 0, λ, guarantees the existence ofδ1 >0,such
that forδ∈0, δ1,
u−τλL−1Nu/0, u∈E: 0<u ≤δ, τ ∈0,1. 2.41
This together withLemma 2.6implies that for anyδ∈0, δ1,
degΦλ, Bδ,0 degI, Bδ,0 1, 2.42
which ends the proof.
Lemma 2.9. Supposeλ > λ1a0, then there existsδ2>0such that for allu∈Ewith0<u ≤δ2, for allτ ≥0,
Φλu/τϕ0, 2.43
whereϕ0is the nonnegative eigenfunction corresponding toλ1a0.
Proof. Suppose on the contrary that there existτn≥0 and a sequence{un}withun>0 and
un → 0 inEsuch thatΦλun τnϕ0for alln∈N. As
LunλNun τnλ1a0a0tϕ0 2.44
andτnλ1a0a0tϕ0≥0 in0,1, it concludes fromLemma 2.2that
unt≥0, t∈0,1. 2.45
Notice thatun∈DLhas a unique decomposition
unwnsnϕ0, 2.46
wheresn ∈Randwn, a0tϕ00. Sinceun≥0 on0,1andun>0, we have from2.46 thatsn>0.
Chooseσ >0,such that
σ < λ−λ1a0
λ . 2.47
ByH1, there existsr1 >0, such that
Therefore, for a.e.t∈0,1, u∈0, r1,
ft, u≥a0tu−ξ1t, u≥1−σa0tu. 2.49
Sinceun → 0, there existsN∗>0, such that
0≤un≤r1, ∀n≥N∗, 2.50
and consequently
ft, un≥1−σa0tun, ∀n≥N∗. 2.51
Applying2.51, it follows that
snλ1a0
ϕ0, a0tϕ0
un, Lϕ0
Lun, ϕ0
λNun, ϕ0
τnλ1a0
a0tϕ0, ϕ0
≥λNun, ϕ0
≥λ1−σa0tun, ϕ0
λ1−σa0tϕ0, un
λ1−σsn
a0tϕ0, ϕ0
.
2.52
Thus,
λ1a0≥λ1−σ. 2.53
This contradicts2.47.
Corollary 2.10. Forλ > λ1a0andδ∈0, δ2,degΦλ, Bδ,0 0.
Proof. Let 0< δ≤δ2, whereδ2is the number asserted inLemma 2.9. AsΦλis bounded inBδ, there existsc >0 such thatΦλu/cϕ0, for allu∈Bδ. ByLemma 2.9, one has
Φλu/τcϕ0, u∈∂Bδ, τ ∈0,1. 2.54
This together withLemma 2.6implies that
degΦλ, Bδ,0 deg
Φλ−cϕ0, Bδ,0
0. 2.55
Proposition 2.11. λ1a0, λ1a0is a bifurcation interval from the trivial solution for2.30. There exists an unbounded componentCof positive solutions of2.30which meetsλ1a0, λ1a0× {0}. Moreover,
C ∩ R\λ1 a0
, λ1a0
× {0}∅. 2.56
Proof. For fixedn ∈ Nwithλ1a0−1/n > 0, let us take thatan λ1a0−1/n,bn
λ1a01/nandδmin{δ1, δ2}. It is easy to check that, for 0< δ <δ, all of the conditions
of Theorem A are satisfied. So there exists a connected componentCn of solutions of2.30 containingan, bn× {0}, and either
iCnis unbounded, or
iiCn∩R\an, bn× {0}/∅.
ByLemma 2.7, the caseiican not occur. Thus,Cnis unbounded bifurcated froman, bn×{0} inR×E. Furthermore, we have fromLemma 2.7that for any closed intervalI ⊂ an, bn\ λ1a0, λ1a0, ifu ∈ {y ∈E |λ, y ∈ Σ, λ ∈I}, thenu → 0 inEis impossible. SoCn must be bifurcated fromλ1a0, λ1a0× {0}inR×E.
3. Proof of the Main Results
Proof ofTheorem 1.5. It is clear that any solution of2.30of the form1, uyields solutionsu
of1.1. We will show thatCcrosses the hyperplane{1} ×EinR×E. To do this, it is enough to show thatCjoinsλ1a0, λ1a0× {0}toλ1c∞, λ1c∞× {∞}. Letηn, yn∈ Csatisfy
ηnyn−→ ∞. 3.1
We note that ηn > 0 for alln ∈ N since0,0 is the only solution of2.30 forλ 0 and C ∩{0} ×E ∅.
Case 1. consider the following:
λ1c∞<1< λ1 a0
. 3.2
In this case, we show that the interval
λ1c∞, λ1 a0
⊆ {λ∈R|λ, u∈ C}. 3.3
We divide the proof into two steps.
Step 1. We show that{ηn}is bounded.
Sinceηn, yn∈ C,Lynηnft, yn. FromH3, we have
Letϕdenote the nonnegative eigenfunction corresponding toλ1c1.
From3.4, we have
Lyn, ϕ
≥ηn
c1tyn, ϕ
. 3.5
ByLemma 2.2, we have
λ1c1
yn, c1tϕ
yn, Lϕ
≥ηn
c1tϕ, yn
. 3.6
Thus,
ηn≤λ1c1. 3.7
Step 2. We show thatCjoinsλ1a0, λ1a0× {0}toλ1c∞, λ1c∞× {∞}.
From3.1and3.7, we have thatyn → ∞. Notice that2.30is equivalent to the integral equation
ynt ηn 1
0
Gt, sfs, ynsds, 3.8
which implies that
ηn 1
0
Gt, sc∞syns ζ2
s, ynsds≥ynt
≥ηn 1
0
Gt, sc∞syns−ζ1
s, ynsds.
3.9
We divide the both sides of3.9byynand setvn yn/yn. Sincevn is bounded inE, there exist a subsequence of{vn}andv∗∈Ewithv∗≥0 andv∗/≡0 on0,1, such that
ηn−→η∗, vn vω ∗ inE, 3.10
relabeling if necessary. Thus,3.9yields that
η∗
1
0
Gt, sc∞sv∗sds≥v∗t≥η∗
1
0
Letϕ∞ andϕ∞ denote the nonnegative eigenfunctions corresponding toλ1c∞and
λ1c∞, respectively, then it follows from the second inequality in3.11that
λ1c∞c∞ϕ∞, v∗Lϕ∞, v∗−ϕ∞, v∗− 1
0
ϕ∞tv∗tdt
≥ − 1
0
ϕ∞tη∗
1
0
Gt, sc∞sv∗sdsdt
−η∗
1
0
c∞sv∗s
1
0
Gt, sϕ∞tdtds
η∗
1
0
c∞sv∗sϕ∞sds
η∗c∞ϕ∞, v∗,
3.12
and consequently
η∗≤λ1c∞. 3.13
Similarly, we deduce from the first inequality in3.11that
λ1c∞≤η∗. 3.14
Thus,
λ1c∞≤η∗≤λ1c∞. 3.15
SoCjoinsλ1a0, λ1a0× {0}toλ1c∞, λ1c∞× {∞}.
Case 2. λ1a0<1< λ1c∞.
In this case, ifηn, yn∈ Cis such that
lim n→ ∞
ηnyn∞,
lim
n→ ∞ηn∞,
3.16
then
λ1a0, λ1c∞⊆ {λ∈0,∞|λ, u∈ C}, 3.17
and moreover,
Assume that{ηn}is bounded, applying a similar argument to that used inStep 2of Case1, after taking a subsequence and relabeling if necessary, it follows that
ηn−→η∗∈λ1c∞, λ1c∞, yn−→ ∞, as n−→ ∞. 3.19
AgainCjoinsλ1a0, λ1a0× {0}toλ1c∞, λ1c∞× {∞}and the result follows.
Remark 3.1. Lomtatidze13, Theorem 1.1proved the existence of solutions of singular two-point boundary value problems as follows:
ut gt, u,
ua 0, ub 0, 3.20
under the following assumptions: A1
gt, x≤h1tx, 0< x < δ,
gt, x≥h2tx, x >
1
δ,
3.21
wherehi:a, b → Ri1,2satisfies the following condition:
b
a
t−ab−t|hit|dt <∞ i1,2, 3.22
A2Fori1,2, letvibe the solution of singular IVPs
vt hitv, va 0, va 1, 3.23
satisfyingv1has at least one zero ina, bandv2has no zeros ina, b.
It is worth remarking thatA1-A2imply Condition1.21inTheorem 1.5. However, Condition 1.21is easier to be verified than A1-A2since λ1c∞ and λ1a0 are easily
estimated by Rayleigh’s Quotient.
The language of eigenvalue of singular linear eigenvalue problem did not occur until Asakawa1in 2001. The first part ofTheorem 1.5is new.
Acknowledgments
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