Boundary Value Problems
Volume 2009, Article ID 103867,34pages 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,
1Ravi P. Agarwal,
2, 3and Donal O’Regan
41Department of Applied Mathematics, Hohai University, Nanjing 210098, China
2Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901-6975, USA 3KFUPM Chair Professor, Mathematics and Statistics Department, King Fahd University of Petroleum and
Minerals, Dhahran 31261, Saudi Arabia
4Department of Mathematics, National University of Ireland, Galway, Ireland
Correspondence should be addressed to Haishen L ¨u,[email protected] 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−ugt, u λht, u, t∈0,1, u0 0u1, whereghmay be singular atu0,t0,1, and may change sign and be superlinear atu ∞. The approach is based on an approximation method together
with the theory of upper and lower solutions.
Copyrightq2009 Haishen L ¨u et al. 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.
1. Introduction
The singular boundary value problems of the form
−uft, u, t∈0,1,
u0 0u1
1.1
occurs in several problems in applied mathematics, see1–6and their references. In many papers, a critical condition is that
or there exists a constantL >0 such that for any compact setK ⊂0,1,there isε εK >0 such that
ft, r≥L ∀t∈K, r∈0, ε,
lim r→ ∞
ft, r
r 0 ∀t∈0,1.
1.3
We refer the reader to1–4. In the case, whenft, rmay change sign in a neighborhood of
r 0 and lim supr→∞ft, r/r ∞fort∈0,1, very few existence results are available in literature1.
In this paper we study positive solutions of the second boundary value problem
−ugt, u λht, u, t∈0,1,
u0 0u1;
1.4
hereg :0,1×0,∞ → Randh:0,1×0,∞ → 0,∞are continuous, so as a result, our nonlinearity may be singular att0,1 andu0.Also our nonlinearity may change sign and be superlinear atu ∞. Our main existence resultsTheorems1.1,1.2and1.4are newsee Remark 1.5, Examples3.1and3.2.
A functionuis a solution of the boundary value problem1.4if u : 0,1 → R, u
satisfies the differential equation1.4on0,1and the stated boundary data.
Let C0,1 denote the class of maps u continuous on 0,1, with norm |u|∞ maxt∈0,1|ut|. We put min{a, b} a∧b; max{a, b} a∨b.Given α, β ∈ C0,1,α ≤ β,
let
Dβα
v|v∈C0,1, α≤v≤β. 1.5
Let
M
h∈C0,1:
1
0
|hs|ds <∞with lim
t→0t|ht|<∞, tlim→1−1−t|ht|<∞
. 1.6
In this paper, we suppose the following conditions hold:
G1suppose there existgi:0,1×0,∞ → 0,∞ i1,2continuous functions such that
git,·is strictly decreasing fort∈0,1,
g1
·, rφ1·
, g2·, r∈M ∀r >0,
−g1t, r≤gt, r≤g2t, r fort, r∈0,1×0,∞,
1.7
H1there existhi:0,1×0,∞ → 0,∞ i1,2continuous functions such that
hit,·is increasing fort∈0,1,
h1·, r, h2·, r∈M forr >0,
h1t, r≤ht, r≤h2t, r fort, r∈0,1×0,∞;
1.8
H2there existsr >0 such thath1t, r>0 fort∈0,1.
The main results of the paper are the following.
Theorem 1.1. SupposeG1,H1,H2and the following conditions hold:
G2for allr2 > r1>0,there existsγ·∈Msuch thatg2·, r γ·r is increasing inr1, r2:
H3
lim r→ ∞
h1t, r
r 0 ∀t∈0,1; 1.9
H4there exists a sequence{Rj}j∞1such that limj→ ∞Rj ∞and
lim j→ ∞
h2
s, Rja1
Rj 0
, 1.10
wherea11 10g2s,1ds.
Then there exists λ∗1 > 0 such that for every λ ≥ λ∗1,1.4has at least one positive solutionu∈C0,1∩C10,1andu >0 fort∈0,1.
Theorem 1.2. SupposeG1,H1,H2and the following conditions hold:
G3for allr2 > r1>0 there existsγ·∈Msuch thatgt, r γtr is increasing inr1, r2;
G4there existsc1>0such that
0≤gt, r, t∈0,1, 0< r < c1; 1.11
G5there existsc2∈0, c1,0< β <1such that for allr ∈0, c2
1
0
t1−tg1t, rltdt≥rπ, 1.12
where
gmt, r min
gt, r,m rβ
form≥1, 1.13
Then there existsλ∗2 >0 such that
iif 0 < λ < λ∗2,1.4has at least one solutionu ∈ C0,1∩C10,1and u > 0 for
t∈0,1;
iiifλ > λ∗2,1.4has no solutions.
Remark 1.3. Notice thatgmt, rsatisfiesG1, G3, G4and for fixedm≥1,
1
0
t1−tgmt, rltdt≥rπ forr ∈0, c2,
gt, r≥gmt, r≥g1t, r fort∈0,1, r ∈0,∞.
1.14
Theorem 1.4. SupposeG1,H1,H2and the following conditions hold:
G6there existsτ ≥τ1such that
lim r→0
τrg−t, r
ht, r 0, 1.15
where τ1 is defined in Lemma 2.1 and gt, r max{0, gt, r}, g−t, r max{0,
−gt, r};
H5for allr2 > r1>0,there existsγ·∈M such thatht, r γtr is increasing inr1, r2.
Then there existsλ∗3 >0 such that
iif 0 < λ < λ∗3,1.4has at least one solutionu ∈ C0,1∩C10,1and u > 0 for
t∈0,1;
iiifλ > λ∗3,1.4has no solutions.
Remark 1.5. In 5, 6, the authors consider the boundary value problem 1.4 under the conditions
lim r→ ∞
h2t, r
r 0. 1.16
In Section 3, we give two examples see Examples 3.1 and 3.2 which satisfy the conditions inTheorem 1.1orTheorem 1.2but they do not satisfy the conditions in1–5.
2. Proof of Main Results
2.1. Some Lemmas
Lemma 2.1. Consider the following eigenvalue problem
−uτut, t∈0,1,
Then the eigenvalues are
τm mπ2 form1,2, . . ., 2.2
and the corresponding eigenfunctions are
φmt sinmπt form1,2, . . . . 2.3
LetGt, sbe the Green’s function for the BVP:
−u 0 fort∈0,1,
u0 u1 0. 2.4
Then
Gt, s ⎧ ⎨ ⎩
s1−t, 0≤s < t≤1,
t1−s, 0≤t < s≤1.
2.5
Also for allt, s∈0,1×0,1, define
Nt, s ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
Gt, s φ1t
if t /0,1,
1−s
π if t0,
s
π if t1.
2.6
It follows easily that
0< Gt, s≤t1−t fort, s∈0,1×0,1,
s1−s
2π ≤Nt, s≤
1
2 fort, s∈0,1×0,1.
2.7
Define the operatorA, B:M → C0,1by
Axt 1
0
Gt, sxsds,
Bxt 1
0
Nt, sxsds.
2.8
The following four results can be found in5 notice limr→ ∞h2t, r/r 0 is not
Lemma 2.2. SupposeG1andH1hold. Letn0 ∈ N.Assume that for everyn > n0,there exist
an,δn,δ∈Msuch that
0≤ant, |δnt| ≤δt, lim
n→ ∞δnt 0, fort∈0,1 2.9
and there existu,un,un,u∈C0,1such that
0< ut≤unt≤unt≤ut fort∈0,1, 2.10
andu0 u1 0. If
−unt antunt
≤g
t,1
nv
λht, v δnt antvt fort∈0,1,
−unt antunt
≥g
t,1
nv
λht, v δnt antvt fort∈0,1,
2.11
whereλ≥0 andv∈Dun
un, then1.4has a solutionu∈C0,1∩C
10,1such thatut≤ut≤ut
fort∈0,1.
Lemma 2.3. Letψ :0,1×0,∞ → 0,∞be a continuous function with
ψt,·is strictly decreasing,
ψ·, r∈M ∀r >0.
2.12
Then the problem
−ωt ψ
t, ωt 1
n
fort∈0,1,
ω0 ω1 0
2.13
has a solutionωn∈C0,1such that
ωnt≤ωn1t≤1ω1t≤1
1
0
ψs,1ds fort∈0,1, n∈N. 2.14
If we letωt limn→ ∞ωntfort∈0,1,then
ω∈C0,1, ωt>0 fort∈0,1,
−ωt ψt, ωt fort∈0,1,
ω0 ω1 0.
Next we consider the boundary value problem
−uatut ft, t∈0,1,
u0 0u1, 2.16
wherea, f ∈M,at≥0 fort∈0,1.
Lemma 2.4. The following statements hold:
ifor anyf∈M,2.16is uniquely solvable and
uAau Af; 2.17
iiifft≥0fort∈0,1,then the solution of 2.16is nonnegative.
Corollary 2.5. LetΦ : M → C0,1∩C10,1be the operator such thatΦfis the solution of
2.16. Then we have
iiff1t≤f2tfort∈0,1,thenΦf1t≤Φf2tfort∈0,1;
iiletE ⊂Mandβ ∈ M.If|ft| ≤βt,t∈ 0,1for allf ∈ E,thenΦEis relatively compact with respect to the topology ofC0,1.
Lemma 2.6see2. Letf∈M,f≥0,f /≡0,u∈C0,1∩C10,1satisfy
−uf in 0,1,
u0 u1 0. 2.18
Then there existmmf>0,MMf>0such that
mlt≤ut≤Mlt fort∈0,1. 2.19
2.2. The Proof of
Theorem 1.1
Claim 1see5. There existsλ∗1 > 0,c >0,independent ofλ,such that for allλ ≥λ∗1there existRλ> c,u∈C0,1,withcφ1t≤ut≤Rλφ1tand
−ut −g1t, ut λh1t, ut, fort∈0,1,
with
g1·, u·, h1·, u·∈M. 2.21
Letλ∗1>0, c >0 andu∈C0,1be defined inClaim 1. Define
ψt, r g2t, r fort∈0,1. 2.22
FromG1notice thatψsatisfies the assumptions ofLemma 2.3, so there existω, ωn∈
C0,1,ωnt>0, ωt>0 fort∈0,1such that
−ωnt g2
t,1
nωn
fort∈0,1,
ωn0 ωn1 0,
ωnt≤ωn1t≤1ω1t≤a1 fort∈0,1, n∈N,
ωt lim
n→ ∞ωnt fort∈0,1,
−ωt g2t, ωt fort∈0,1,
ω0 ω1 0,
2.23
wherea11 10g2s,1ds.
Letλ≥λ∗1,n∈Nbe fixed. We consider the following boundary value problem:
−vt λh2t, vωn λh1t, u fort∈0,1,
v0 v1 0. 2.24
ByH4,there exist{Rj}∞j1such that limj→ ∞Rj∞and
lim j→ ∞
h2
t, Rja1
Rj
0 for t∈0,1, 2.25
so
lim j→ ∞
λh2
t, Rja1
λh1t, ut
Rj
There existsj0∈Nsuch that
λh2
t, Rj0a1
λh1t, ut≤Rj0. 2.27
Ifv∈C0,1and 0≤vt≤Rj0φ1tfort∈0,1,then
1
0
Nt, sλh2s, vs ωns λh1s, uds
≤
1
0
Nt, sλh2s, vs a1 λh1s, uds
≤
1
0
Nt, sλh2
s, Rj0φ1s a1
λh1s, u
ds
≤
1
0
Nt, sλh2
s, Rj0a1
λh1s, u
ds
≤ Rj0
2 , fort∈0,1,
2.28
and so
0≤
1
0
Gt, sλh2s, vs ωns λh1s, usds≤Rj0φ1t fort∈0,1. 2.29
LetΦ:C0,1 → C0,1be the operator defined by
Φvt:
1
0
Gt, sλh2s, vs ωns λh1s, usds forv∈C0,1, t∈0,1.
2.30
It is easy to see thatΦis a continuous and completely continuous operator. Also if 0≤vt≤ Rj0φ1tfort ∈ 0,1,then 0 ≤ Φvt ≤ Rj0φ1tfor t ∈ 0,1, so Schauder’s fixed point
theorem guarantees that there existsv∈0, Rj0φ1such thatΦv v, that is,
−vt λh2t,vt ωns λh1t, ut,
v1 v1 0. 2.31
Let
Thenun∈C0,1,un1 un1 0,and
−unt −ωnt−vnt
g2
t,1
nωn
λh2t, ωnvn λh1t, u
≥g2
t,1
nun
λh1t, u λh2t,un fort∈0,1.
2.33
Let
ut ωt Rj0φ1t fort∈0,1, 2.34
so
0≤unt≤ut fort∈0,1. 2.35
FromClaim 1, we obtain
−ut −g1t, u λh1t, u
≤λh1t, u
≤λh1t, u g2
t,1
nun
λh2t,un
≤ −unt fort∈0,1,
2.36
that is,
−u−unt≤0 fort∈0,1. 2.37
A standard argument yields
ut≤unt fort∈0,1. 2.38
FromG2,there existsγ ∈ Msuch thatr → g2t,1/nr γtr is increasing on
0,|u|∞. Letunu.From2.35and2.38,we have
Also forv∈Dun
un we have
−unt γtunt
−g1t, un λh1t, un γtunt
≤ −g1t, v λh1t, v γtvt
≤ −g1
t,1
nv
λh1t, v γtvt
≤g
t,1
nv
λht, v γtvt fort∈0,1,
−unt γtunt
≥g2
t,1
nun
λh1t, u λh2t,un γtunt
≥g2
t,1
nun
γtunt λh2t,un
≥g2
t,1
nv
γtvt λh2t, vt
≥g
t,1
nv
λht, v γtvt fort∈0,1.
2.40
NowLemma 2.2withδn≡0,n∈Nguarantees that there exists a solutionu∈C0,1to1.4 with
ut≤ut≤ut fort∈0,1. 2.41
2.3. The Proof of
Theorem 1.2
Let
Λ λ∈R|1.4has at least one positive solution. 2.42
Claim 2. Let
λ∗ 1
maxt∈0,1 10Nt, sh2
s, a2φ1
here
a21
1 4
1
0
⎡ ⎢ ⎣g2t,1
1
utβ
et ⎤ ⎥ ⎦dt,
ut c2lt fort∈0,1,
et ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
sup r∈c1,1c2/2
g−t, r, if c1<1
c2
2,
0, if c1≥1c2
2.
2.44
Then0, λ∗∈Λ.
Proof ofClaim 2. Letn≥1 be fixed. Lemma 2.86implies that there existsαn,1 ∈C0,1such
that
ut≤αn,1t≤ut, 2.45
−αn,1t g1
t,1
nαn,1t
δnt fort∈0,1,
αn,10 αn,11 0,
2.46
whereg1is defined inG5,and
δnt g1
t, ut−g1
t,1
nut
, 2.47
ut clt fort∈0,1,
cmax
⎧ ⎪ ⎨ ⎪
⎩c1, πtsup∈0,1 ⎡ ⎢ ⎣2B
⎛ ⎜
⎝ 1
u·β ⎞ ⎟
⎠t Bet ⎤ ⎥ ⎦
⎫ ⎪ ⎬ ⎪ ⎭.
2.48
which does not depend onn.
On the other hand, let
ψt, r g2t, r
1
From G1 notice ψ satisfies the assumptions of Lemma 2.3, so there exist ω, ωn ∈
C0,1such that
−ωnt g2
t,1
nωn
1
utβet fort∈0,1,
ωn0 ωn1 0,
ωnt≤ωn1t≤1ω1t≤a2 fort∈0,1, n∈N,
ωt lim
n→ ∞ωnt fort∈0,1,
−ωt g2t, ωt
1
utβ et fort∈0,1,
ω0 ω1 0.
2.50
Next we consider the boundary value problem
−vnt λh2t, ωnvn fort∈0,1,
vn0 vn1 0,
2.51
whereλ∈0, λ∗.
LetΦ:C0,1 → C0,1be the operator defined by
Φvt:λ 1
0
Gt, sh2s, ωnvds forv∈C0,1, t∈0,1. 2.52
It is easy to see thatΦis a continuous and completely continuous operator. Also, if 0≤vt≤ φ1tfort∈0,1,then
0≤Φvt λ 1
0
Gt, sh2s, ωnvds
≤λ∗ 1
0
Gt, sh2
s, a2φ1
ds
φ1t
1
0Nt, sh2
s, a2φ1
ds
maxt∈0,1 10Nt, sh2
s, a2φ1
ds
≤φ1t fort∈0,1.
Thus Schauder fixed point theorem guarantees that there existsvn∈0, φ1such thatΦvn
vn,that is,
−vnt λh2t, ωnvn,
vn0 vn1 0.
2.54
Let
unt ωnt vnt, ut ωt φ1t fort∈0,1. 2.55
Thenun,u∈C0,1,un0 un1 0,u0 u1 0,
0≤unt≤ut fort∈0,1, 2.56
−unt −ωnt−vnt
g2
t,1
nωn
1
utβ et λh2t, ωnvn
≥g2
t,1
nun
1
utβet λh2t,un fort∈0,1, λ∈0, λ
∗.
2.57
Now let us consider the problem
−ut g
t,1
nu
λht, u δnt fort∈0,1, λ∈0, λ∗
u0 u1 0,
2.58
whereδnis defined in2.47.
We will proveαn,1is a lower solution of2.58andunis an upper solution of2.58. Now2.46and the positivity ofht, simplies that
−αn,1t g1
t,1
nαn,1t
δnt
≤g
t,1
nαn,1t
λht, αn,1t δnt,
soαn,1is a lower solution of2.58. On the other hand, from the definition ofg1 andu, we
have
g1t, umin
⎧ ⎨ ⎩g
t, u, 1
utβ ⎫ ⎬
⎭≤
1
utβ fort∈0,1,
−g1
t,1
nu
−min
⎧ ⎪ ⎨ ⎪ ⎩g
t,1
nu
, 1
1/nuβ ⎫ ⎪ ⎬ ⎪
⎭g
−t,1
nu
≤g−
t,1
nu
≤et fort∈0,1,
2.60
so
δnt≤ 1
utβ et fort∈0,1. 2.61
Consequently, we have
−unt≥g2
t,1
nun
1
utβet λh2t,un
≥g
t,1 nun
1
utβet λht,un
≥g
t,1 nun
λht,un δnt,
2.62
sounis an upper solution of2.58. We next prove that
αn,1t≤unt fort∈0,1. 2.63
Suppose2.63is not true. Letyt αn,1t−untand letσ∈0,1be the point where
ytattains its maximum over0,1.We have
On the other hand, sinceαn,1σ>unσ,we have
−αn,1σ g1
σ,1
nαn,1σ
δnσ
≤g
σ,1
nαn,1σ
δnσ
≤g
σ,1
nαn,1σ
1
uσβeσ
≤g2
σ,1
nαn,1σ
1
uσβeσ
< g2
σ,1
nunσ
1
uσβ eσ λh2σ,unσ
≤ −unσ,
2.65
so
yσ αn,1σ−unσ>0, 2.66
and this is a contradiction.
From G3,there exists γ ∈ Msuch that r → gt,1/nr γtr is increasing in 0,|u|∞.Letut≡ut,unt αn,1t.From2.45,2.56, and2.63, we have
0< ut≤unt≤unt≤ut fort∈0,1. 2.67
Also forv∈Dun
un,we have
−unt γtunt
≤g
t,1
nun
γtunδnt
≤g
t,1
nv
γtvδnt λht, v
≤g
t,1 nun
γtunδnt λh2t,un
≤ −unt γtunt.
On the other hand, by2.61
|δnt| ≤ 1
utβet≡δt,
lim
n→ ∞δnt 0 fort∈0,1.
2.69
NowLemma 2.2guarantees that there exists a solutionu∈C0,1∩C10,1to1.4with
ut≤ut≤ut fort∈0,1. 2.70
Thus1.4has a solution forλ ∈ 0, λ∗soClaim 2 holds. In particular,Λ/∅ and supΛ >
0.
Claim 3. Ifλ∈Λ,then0, λ∈Λ.
Proof ofClaim 3.
Step 1. We may assume thatλ >0.Letχbe a positive solution of1.4,that is,
−χ gt, χλht, χ, t∈0,1,
χ0 0χ1. 2.71
We prove that there existsρ >0 such that
χt≥ρlt fort∈0,1. 2.72
ByG4,gt, r≥0 fort∈0,1,r ∈0, c1.From the continuity ofχandχ0 0 χ1,it
follows that there is 0< δ <1/2 such that
0≤χt< c1 fort∈0, δ∪1−δ,1. 2.73
Then
−χ ≥λht, χ fort∈0, δ∪1−δ,1. 2.74
Letv∈C10, δ∩C0, δso that
−vt ht, χ fort∈0, δ,
v0 vδ 0. 2.75
It follows thatλvt≤χtfort∈0, δ.Lemma 2.6implies that there existsm >0 so that
The same reason implies that
minf{tδ−1,1−t} ≤vt fort∈1−δ,1. 2.77
It follows that
mλlt≤χt fort∈ '
0,δ
2
(
∪
'
1−δ
2 ,1
(
. 2.78
Moreover,
inf
χt
lt :t∈
0,δ
2
∪
1−δ
2 ,1
>0. 2.79
On the other hand, we easily have
inf
χt
lt :t∈ '
δ
2, 1−δ
2
(
>0, 2.80
so
inf
χt
lt :t∈0,1
ρ >0, 2.81
and thus
χt≥ρlt fort∈0,1. 2.82
Step 2. Letrρ∧c2andut rlt.Then
ut≤A
gm
·,1
nu∧χ
δn
t fort∈0,1, m, n≥1, 2.83
where
δnt g1
t, u∧χ−g1
t,1
nu∧χ
. 2.84
Notice
FromG5, we have
Ag1·, u∧χt 1
0
Gt, sg1s, uds
φ1t
1
0
Nt, sg1s, uds
≥ φ1t
2π 1
0
s1−sg1s, rlsds
≥ rφ1t
2 ≥ut fort∈0,1,
2.86
so
A
gm
·,1
nu∧χ
δn
t
1
0
Gt, s '
gm
s,1
nu∧χ
−g1
s,1
nu∧χ
g1s, u∧χ(ds
≥
1
0
Gt, sg1s, u∧χds
≥ut fort∈0,1.
2.87
Step 3. Let 0< μ < λ. For eachm≥1,there existsrm> r,independent ofn.Let
umt rmlt fort∈0,1. 2.88
Then
A
gm
·,1
nv∧χ
δnμh2
·, v∧χt≤umt fort∈0,1, v∈Duum, n≥1.
2.89
Letv∈C0,1∩C10,1such that
−vλh2
t, χ fort∈0,1,
v0 v1 0. 2.90
ByLemma 2.6, there existsM >0 such that
Let
rm>max
⎧ ⎪ ⎨ ⎪
⎩Mπtsup∈0,1B ⎛ ⎜
⎝ m
u∧χβ
1
uβe
⎞ ⎟ ⎠t, r
⎫ ⎪ ⎬ ⎪
⎭. 2.92
Noteu≤usincerm> r.Letv≥uand noticenoteg−·, r 0 if 0< r < c1fromG4
A
gm
·,1
nv∧χ
δn t 1 0
Gt, s '
gm
s,1
nv∧χ
−g1
s,1
nu∧χ
g1s, u∧χ(ds
≤
1
0
Gt, s )
m
1/nv∧χβ g
−s,1
nu
g1s, u*ds
≤
1
0
Gt, s ⎡
⎣ m
v∧χβ
1
uβ e
⎤ ⎦ds
≤
1
0
Gt, s ⎡ ⎢
⎣ m
u∧χβ
1
uβ e
⎤ ⎥ ⎦ds
φ1t
⎡ ⎢ ⎣B
⎛ ⎜
⎝ m
u∧χβ
1
uβ e
⎞ ⎟ ⎠ ⎤ ⎥ ⎦t
≤π ⎡ ⎢ ⎣B ⎛ ⎜
⎝ m
u∧χβ
1
uβ e
⎞ ⎟ ⎠ ⎤ ⎥
⎦t·lt fort∈0,1.
2.93
On the other hand,
Aμh·, v∧χt μ
1
0
Gt, shs, v∧χds
≤λ 1
0
Gt, sh2
s, χds
vt≤Mlt fort∈0,1,
so
A
gm
·,1
nv∧χ
δnμh
·, v∧χt
≤A
gm
·,1
nv∧χ
δn
t Aμh·, v∧χt
≤π ⎡ ⎢ ⎣B ⎛ ⎜
⎝ m
u∧χβ
1
uβ e
⎞ ⎟ ⎠ ⎤ ⎥
⎦t·lt Mlt
≤umt fort∈0,1, v∈
+ u, um
, , n≥1.
2.95
Step 4. Let 0< μ < λ.Letn, m≥1 be fixed. There existsβn,m∈C0,1such that
ut≤βn,mt≤umt,
−βn,mt gm
t,1
nβn,m∧χ
μht, βn,m∧χ
δnt fort∈0,1,
βn,m0 βn,m1 0.
2.96
Letn, m >1 be fixed. FromRemark 1.3, there existγn ∈M,γn ≥0 such thatgmt, r
γntris increasing in1/n,1/nrm/2.We easily prove that
gmt, r∧χγntr is increasing in
-1 n, 1 n rm 2 .
. 2.97
Letγt γn. We havegmt,1/nr∧χ γtr is increasing in0, rm/2.From 2.83and
2.89, we have for fixedv∈C0,1,ut≤vt≤umtthat
ut Aγut≤A
gm
·,1
nu∧χ
δn
t Aγut
≤A
gm
·,1
nu∧χ
γuδnμh
·, v∧χt
≤A
gm
·,1
nv∧χ
γvδnμh
·, v∧χt
≤umt A
γ um
t.
Fixv ∈ C0,1with ut ≤ vt ≤ umt.FromLemma 2.4, there existsΨv ∈C0,1such that
−Ψvt γtΨvt
gm
t,1
nv∧χ
γtvt δnt μh
t, v∧χ fort∈0,1
Ψv0 Ψv1 0.
2.99
Then
Ψvt AγΨvt A
gm
·,1
nv∧χ
γvδnμh
·, v∧χ t fort∈0,1,
2.100
so2.98implies that
ut Aγut≤Ψvt AγΨvt
≤umt A
γ um
t fort∈0,1.
2.101
FromCorollary 2.5, we have
ut≤Ψvt≤umt fort∈0,1. 2.102
Also,
// //gm
t,1
nv∧χ
γvδnμh
t, v∧χ////
≤g1
t,φ1t
n
g2
t,1
n
γ///um///
∞|δnt|λh2
t,//χ//∞
≡βt∈M fort∈0,1.
2.103
NowΨ:Dum
u → Duum is compact, so Schauder’s fixed point theorem implies that there exists
βn,m∈C0,1such thatut≤βn,mt≤umtandΨβn,mt βn,mtfort∈0,1:
−βn,m t gm
t,1
nβn,m∧χ
μht, βn,m∧χ
δnt fort∈0,1,
βn,m0 βn,m1 0,
// //gm
t,1
nβn,m∧χ
μht, βn,m∧χ
δnt
// //≤3g2
t, u∧χλh2
t, χ.
Let m ≥ 1 be fixed. We consider the sequence {βn,m}∞n1.Fix n0 ∈ {2,3, . . .}. Let us
look at the interval1/2n01,1−1/2n01.The mean value theorem implies that there exists τ ∈1/2m01,1−1/2m01with|β
n,mτ| ≤8/3supt∈0,1umt.As a result
βn,mt
∞
nn01 is bounded,equicontinuous family on '
1 2n01,1−
1 2n01
(
. 2.105
The Arzela-Ascoli theorem guarantees the existence of subsequenceNn0 of integers and a
functionzn0,m∈1/2
n01,1−1/2n01withβ
n,mconverging uniformly tozn0,mon1/2
n01,1−
1/2n01asn → ∞throughN
n0.Similarly,
βn,m
∞
nn01 is bounded,equicontinuous family on '
1 2n02,1−
1 2n02
(
, 2.106
so there is a subsequenceNn01 ofNn0and a functionzn01,m ∈C1/2
n02,1−1/2n02with βn,mconverging uniformly tozn01,mon1/2
n02,1−1/2n02asn → ∞throughN
n01.Note zn01,m zn0,m on 1/2n01,1 −1/2n01since Nn01 ⊆ Nn0. Proceed inductively to obtain
subsequences of integersNn0 ⊇ Nn01 ⊇ · · · ⊇ Nk ⊇ · · · and functionszk,m ∈ C1/2
k1,1−
1/2k1withβ
n,mconverging uniformly tozk,mon1/2k1,1−1/2k1asn → ∞throughNk, andzk,mzk−1,mon1/2k,1−1/2k.
Define a functionum :0,1 → 0,∞byumt zk,mton1/2k1,1−1/2k1and
um0 um1 0.Noticeumis well defined andut≤umt≤umtfort∈0,1.Next, fix
t∈0,1 without loss of generality assumet /1/2and letn∗∈ {n0, n01, . . .}be such that
1/2n∗1< t <1−1/2n∗1.LetN∗
n∗ {i∈Nn :i≥n∗}.Nowβn,m, n∈N∗n∗satisfies the integral
equation
βn,mt βn,m
1 2 βn,m 1 2 t−1
2
t
1/2
s−t
gm
s,1
nβn,m∧χ
μhs, βn,m∧χ
δns
ds,
2.107
fort∈1/2n1,1−1/2n1.Noticetaket2/3 saythat{β
n,m1/2}, n∈Nn∗∗,is a bounded sequence sinceut ≤ βn,mt ≤ umtfort ∈0,1.Thus{βn,m1/2}n∈N∗n∗ has a convergent
subsequence; for convenience we will let{βn,m1/2}n∈N∗
n∗ denote this subsequence also, and
letτ ∈Rbe its limit. Now for the above fixedt,and letn → ∞throughNn∗∗to obtain
gm
t,1
nβn,m∧χ
−→gm
t, zk,m∧χ
,
ht, βn,m∧χ
−→ht, zk,m∧χ
,
δn−→0.
As a result,
zk,mt zk,m
1 2
τ
t−1
2
t
1/2
s−tgms, zk,m∧χ
μhs, zk,m∧χ
ds, 2.109
that is,
umt um
1 2
τ
t−1
2
t
1/2
s−tgms, um∧χ
μhs, um∧χ
ds. 2.110
We can do this argument for eacht∈0,1and so
−umt gmt, um∧χ
μht, um∧χ
fort∈0,1. 2.111
It remains to show thatumis continuous at 0 and 1.
Letε >0 be given. Sinceum∈C0,1there existsδ >0 withumt< ε/2 fort∈0, δ. As a resultut≤βn,mt≤umt< ε/2 fort∈0, δ.Consequently,ut≤umt≤ε/2< εfor
t∈0, δand soumis continuous at 0.Similarly,umis continuous at 1.As a resultum∈C0,1 and
−umt gmt, um∧χ
μht, um∧χ
fort∈0,1,
um0 um1 0.
2.112
Next we prove
umt≤χt fort∈0,1. 2.113
Suppose 2.113is not true. Letyt umt−χtand σ ∈ 0,1be the point whereyt attains its maximum over0,1.We have
yσ>0, yσ≤0. 2.114
On the other hand, sinceumσ> χσ,we have
yσ umσ−χσ
−gmσ, um∧χ
−μhσ, um∧χ
gσ, χλhσ, χ
−gmσ, χσ−μhσ, χσgσ, χσλhσ, χσ
≥λ−σhσ, χσ>0.
2.115
Thus we have
−umgmt, um μht, um,
um0 um1 0,
ut≤umt≤χt fort∈0,1.
2.116
By the same reason as above, we obtain subsequences of integersNm0 ⊇Nm01⊇ · · · ⊇ Nk ⊇ · · · and functionszm ∈ C1/2k1,1−1/2k1with um converging uniformly tozk on
1/2k1,1−1/2k1asm → ∞throughN
k,andzkzk−1on1/2k,1−1/2k.
Define a functionu:0,1 → 0,∞byut zkton1/2k1,1−1/2k1andu0
u1 0.Noticeuis well defined andut ≤ ut ≤ χtfor t ∈ 0,1.Next fix t ∈ 0,1 without loss of generality assumet /1/2and letm∗∈ {m0, m01, . . .}be such that 1/2m∗1<
t <1−1/2m∗1.LetN∗
m∗ {k∈Nm∗:k≥m∗}.Nowum, m∈Nm∗∗satisfies the integral equation
umt um
1 2
um
1 2
t−1
2
t
1/2
s−tgms, um μhs, um
ds 2.117
fort∈1/2m∗1,1−1/2m∗1.Noticetaket2/3 saythat{u
m1/2},m∈Nm∗∗is a bounded
sequence since ut ≤ umt ≤ χtfor t ∈ 0,1.Thus {um1/2}m∈N∗m∗ has a convergent
subsequence; for convenience we will let{um1/2}m∈Nm∗∗ denote this subsequence also, and
letτ ∈Rbe its limit. Now for the above fixedt,and lettingm → ∞throughNk∗to obtain
ut u
1 2
τ
t−1
2
t
1/2
s−tgs, u μhs, uds. 2.118
we can do this argument for eacht∈0,1and so
−ut gt, u μht, u fort∈0,1. 2.119
Also reasoning as before we have thatuis continuous at 0 and 1.
Thus we have
−ugt, u μht, u,
u0 u1 0. 2.120
Now letλ∗2supΛ>0. Then
iif 0 < λ < λ∗2,1.4has at least one solutionu ∈ C0,1∩C10,1and u > 0 for
t∈0,1;
2.4. The Proof of
Theorem 1.4
Claim 4. Let
λ∗ 1
maxt∈0,1 10Nt, sh2
s, a3φ1
ds >0; 2.121
here
a311
4
1
0
-g2s,1 h2
s,1
2φ1s
φ1s
2//φ1//∞
.
ds. 2.122
Then0, λ∗∈Λ.
Proof ofClaim 4. Letλ∈0, λ∗be fixed. From assumptionG6, it follows that there isτ ≥τ1
andc3∈0,1,such that ifn >2/c3, 0< k < c3/2<1,we have
0< k//φ1//∞<
c3
2, 0< 1
nkφ1t< c3,
τ1/nkφ1t
g−t,1/nkφ1t
ht,1/nkφ1t
≤λ.
2.123
Thus,
τkφ1t g−
t,1/nkφ1t
ht,1/nkφ1t
≤λ. 2.124
Then, forn >2/c3,
τkφ1t g−
t,1
nkφ1t
≤λh
t,1
nkφ1t
, 2.125
and we have
τkφ1t≤λh
t,1
nkφ1t
−g−
t,1
nkφ1t
≤g
t,1
nkφ1t
−g−
t,1
nkφ1t
λh
t,1
nkφ1t
g
t,1
nkφ1t
λh
t,1
nkφ1t
−λht, kφ1t
λht, kφ1t
g
t,1
nkφ1t
λht, kφ1t
δnt,
where
δnt λh
t,1
nkφ1t
−λht, kφ1t
. 2.127
Letut kφ1t.We have
−ut τ1kφ1t≤τkφ1t≤g
t,1
nut
λht, ut δnt fort∈0,1. 2.128
Let
ψt, s g2t, s λh2
t,1
2 φ1t
φ1t
2//φ1//∞
. 2.129
FromG1notice thatψsatisfies the assumptions ofLemma 2.3, so there existω, ωn∈
C0,1such that
−ωnt g2
t,1
nωn
λh2
t,1
2 φ1t
φ1t
2//φ1//∞
fort∈0,1,
ωn0 ωn1 0,
ωt lim
n→ ∞ωnt fort∈0,1,
ωnt≤1 1 4
1
0
-g2s,1 h2
s,1
2φ1s
φ1s
2//φ1//∞
.
dsa3 fort∈0,1, n∈N.
2.130
Consider the boundary value problem
−vt λh2t, ωnv fort∈0,1,
v0 v1 0.
2.131
LetΦ:C0,1 → C0,1be the operator defined by
Φvt:λ 1
0
It is easy to see thatΦis a continuous and completely continuous operator. Also if 0≤vt≤ φ1tfort∈0,1,then
0≤Φvt λ 1
0
Gt, sh2s, ωnvds
≤λ∗ 1
0
Gt, sh2
s, a3φ1
ds
φ1t
1
0Nt, sh2
s, a3φ1
ds
maxt∈0,1 10Nt, sh2
s, a3φ1
ds
≤φ1t fort∈0,1.
2.133
Thus Schauder’s fixed point theorem guarantees that there exists vn ∈ 0, φ1 such that
Φvn vn,that is,
−vnt λh2t, ωnvn,
vn0 vn1 0.
2.134
Let
unt ωnt vnt, ut ωt φ1t fort∈0,1. 2.135
Thenun,u∈C0,1,un0 un1 0, u0 u1 0, 0≤unt≤utfort∈0,1, and
−unt −ωnt−vnt
g2
t,1
nωn
λh2
t,1
2φ1t
φ1t
2//φ1//
λh2t, ωnvn
≥g2
t,1
nun
λh2
t,1
2 φ1t
φ1t
2//φ1//
λh2t,un fort∈0,1.
2.136
We next prove that
ut≤unt fort∈0,1. 2.137
Suppose2.137is not true. Letyt ut−untandσ ∈ 0,1be the point where
ytattains its maximum over0,1.We have
On the other hand, sinceuσ>unσ,we have
−uσ≤g
σ,1
nuσ
λhσ, uσ δnσ
g
σ,1
nuσ
λh
σ,1
nuσ
≤g2
σ,1
nuσ
λh2
σ,1
2 φ1σ
< g2
σ,1
nunσ
λh2
σ,1
2 φ1σ
φ1σ 2//φ1//
λh2σ,unσ
≤ −unσ.
2.139
Thusyσ uσ−unσ>0, and this is a contradiction. As a result,2.137is true. On the other hand, we have
|δnt| ≤λ
// //h
t,1
nkφ1t
−ht, kφ1t////
≤2λh2
t,1
2 //φ1//
2.140
forn >2/c3.Consequently, fort∈0,1, δn → 0 andn → ∞.
From assumptionsG2andH5,there exists aγ, τ∈M,n >2/c3,so thatgt,1/n
r ht, r atris increasing in0,|u|∞,whereat γt τt.Letunut. Forv∈Duunn, we have
−unt atunt
≤g
t,1
nunt
λht, unt δnt atunt
≤g
t,1
nunt
λht, unt atunt
λh
t,1
nunt
−λht, unt
≤g
t,1
nunt
λht,unt atunt λh
t,1
nunt
≤g2
t,1
nunt
λh2t,unt
φ1t
2//φ1//∞
λh2
t,1
2φ1t
atunt
≤ −unt antunt.
Reasoning as in the proof of Theorem 1.1, Lemma 2.2 guarantees that 1.4 has a solutionu∈C0,1∩C10,1.
Thus 1.4 has a solution for λ ∈ 0, λ∗ so Claim 4 holds. In particular, Λ/∅ and supΛ>0.
Claim 5. Ifλ∈Λ,then0, λ∈Λ.
Proof ofClaim 5. We may assume thatλ >0.Letχbe a positive solution of1.4,that is,
−χ gt, χλht, χ, t∈0,1,
χ0 0χ1. 2.142
We first prove that there existsρ >0 such that
χt≥ρlt fort∈0,1. 2.143
ByG6,there existsσ >0 such that for allr∈0, σ,we have
τrg−t, r
ht, r ≤λ, 2.144
that is,
τr≤λht, r−g−t, r fort∈0,1, r ∈0, σ. 2.145
From the continuity ofχandχ0 0χ1,it follows that there is 0< δ <1/2 such that
χt< σ for t∈0, δ∪1−δ,1. 2.146
Then
−χgt, χλht, χ
gt, χλht, χ−g−t, χ
≥λht, χ−g−t, χ
≥τχt fort∈0, δ∪1−δ,1.
2.147
The next part is similar to the proof of2.72, that is, there existsρ >0 such that
We consider the boundary value problem
−ugt, u∧χμht, u∧χ,
u0 u1 0, 2.149
whereμ ∈0, λ.Letg1t, 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 of6, Theorem 1.2are satisfied so
2.149has a positive solutionu∈C10,1∩C0,1.We next prove that
ut≤χt fort∈0,1. 2.150
Suppose2.150is not true. Letyt ut−χtandσ∈0,1be the point whereyt
attains its maximum over0,1.We have
yσ>0, yσ≤0. 2.151
On the other hand, sinceuσ> χσ,we have
yσ uσ−χσ
−gσ, u∧χ−μhσ, u∧χgσ, χλhσ, χ
−gσ, χσ−μhσ, χσgσ, χσλhσ, χσ
λ−μhσ, χσ
>0.
2.152
This is a contradiction, so
ut≤χt fort∈0,1. 2.153
Thus we have
−ugt, u μht, u,
u0 u1 0. 2.154
Letλ∗3supΛ>0. Then
iif 0 < λ < λ∗3,1.4has at least one solutionu ∈ C0,1∩C10,1and u > 0 for
t∈0,1;
3. Example
Example 3.1. Consider the boundary value problem
−u−√1
uλqu ∀0< t <1,
u0 u1 0,
3.1
whereλ >1.
Define{xn}∞n1asx12, x2nx42n−1, x2n1x2n1,and
qr ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
r2, ifr∈0,2,
x22n−1, ifr∈x2n−1, x2n,
x2
2n1− √x2n
x2n1−x2n
r−x2n √x2n, ifr∈x2n, x2n1.
3.2
Then,Theorem 1.1implies that there existsλ∗1>0 such that for everyλ≥λ∗1,3.1has at least one positive solutionu∈C0,1∩C10,1andu >0 fort∈0,1.
To see this, let
g1t, r g2t, r √1
r fort, r∈0,1×0,∞,
h2t, r qr fort, r∈0,1×0,∞,
h1t, r
⎧ ⎨ ⎩
√
r fort, r∈0,1×16,∞,
qr fort, r∈0,1×0,16.
3.3
It is easy to see thatG1,H1,H2,andH3are satisfied.
For allr2 > r1 > 0,let γt 1/2r1√r1.Theng2t, r 1/2r1√r1r is increasing in
r1, r2.
On the other hand,a11 101/√sds3 and letRjx2j−3,so we have
lim j→ ∞
h2
s, Rja1
Rj
lim j→ ∞
h2
s, x2j
x2j ·
x2j
x2j−3
lim j→ ∞
√x
2j
x2j ·
x2j
x2j−3
0.
3.4
ThusG2andH4are satisfied. ThenTheorem 1.1implies that there existsλ∗1>0 such that for everyλ ≥ λ∗1,3.1has at least one positive solutionu∈ C0,1∩C10,1andu > 0 for
Example 3.2. Consider the boundary value problem
−ugt, u λht, u, t∈0,1,
u0 0u1, 3.5
where
gt, r ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
1
rα
// //sin1
r //
//, 0< r ≤ 1 π,
−1
rαsin 1
r,
1
π < r,
ht, r r2,
3.6
withα >0.ThenTheorem 1.2guarantees that there existsλ∗2>0 such that
iif 0 < λ < λ∗2,3.5has at least one solutionu ∈ C0,1∩C10,1and u > 0 for
t∈0,1;
iiifλ > λ∗2,3.5has no solutions.
To see this, letβ min{1/2, α/2}, g1t, r 1/rβπα,andg2t, r 1/rα,fort, r∈
0,1×0,∞,andh1t, r h2t, r r2,fort, r∈0,1×0,∞. Notice thatG1,H1,and
H2are satisfied.
For allr2> r1>0,let
γt≡sup r∈Λ
//
//∂g∂r////1<∞, 3.7
whereΛ r1, r2\ {nπ |n∈N}, so we havegt, r γtr is increasing inr1, r2.
Letc11/πand we have
0≤gt, r, t∈0,1, 0< r < c1. 3.8
Letn0be fixed such that
21/α−β< n0π
π
6. 3.9
Letc2∈0, c1be such that
c32< 1
22−βπ1−β ∞
0
nn0
) 6
6n1
2−β
−
6 6n5
2−β*
and we have forn≥n0, r∈0, c2,
1 rtα
// //sin 1
rt // //≥ 1
rtβ fort∈ '
1
rnπ5π/6, 1
rnππ/6
( .
3.11
Also we have
1
0
t1−tgmt, rltdt≥ 1/2
0
t1−tgmt, rltdt
≥ 1 2
1/2
0
tgmt, rltdt
≥ 1 2
∞
0
nn0
1/rnππ/6
1/rnπ5π/6t
1 rtβ dt
≥ 1 2rβ
∞
0
nn0
1/rnππ/6
1/rnπ5π/6t
1−βdt
1
2r22−βπ2−β ∞
0
nn0
) 6
6n1
2−β
−
6 6n5
2−β*
≥rπ.
3.12
ThusG5is satisfied.
Acknowledgment
This research is supported by NNSF of China10871059.
References
1 R. P. Agarwal and D. O’Regan,Singular Differential and Integral Equations with Applications, Kluwer Academic Publishers, Dordrecht, The Netherlands, 2003.
2 R. P. Agarwal and D. O’Regan, “Twin solutions to singular Dirichlet problems,”Journal of Mathematical Analysis and Applications, vol. 240, no. 2, pp. 433–445, 1999.
3 D. O’Regan,Theory of Singular Boundary Value Problems, World Scientific, River Edge, NJ, USA, 1994.
4 P. Habets and F. Zanolin, “Upper and lower solutions for a generalized Emden-Fowler equation,” Journal of Mathematical Analysis and Applications, vol. 181, no. 3, pp. 684–700, 1994.
5 H. L ¨u, D. O’Regan, and R. P. Agarwal, “An approximation approach to eigenvalue intervals for singular boundary value problems with sign changing nonlinearities,”Mathematical Inequalities and Applications, vol. 11, no. 1, pp. 81–98, 2007.
