Volume 2010, Article ID 376471,14pages doi:10.1155/2010/376471
Research Article
Triple Positive Solutions for a Type of
Second-Order Singular Boundary Problems
Jiemei Li
1, 2and Jinxiang Wang
11Department of Mathematics, Northwest Normal University, Lanzhou 730070, China 2The School of Mathematics, Physics Software Engineering, Lanzhou Jiaotong University,
Lanzhou 730070, China
Correspondence should be addressed to Jiemei Li,[email protected]
Received 7 April 2010; Accepted 26 August 2010
Academic Editor: Wenming Zou
Copyrightq2010 J. Li and J. Wang. 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 deals with the existence of triple positive solutions for a type of second-order singular boundary problems with general differential operators. By using the Leggett-Williams fixed point theorem, we establish an existence criterion for at least three positive solutions with suitable growth conditions imposed on the nonlinear term.
1. Introduction
In this paper, we study the existence of triple positive solutions for the following second-order singular boundary value problems with general differential operators:
ut atut btut htft, ut 0, t∈0,1,
u0 u1 0, 1.1
wherea∈C0,1∩L10,1,b∈C0,1,−∞,0, andh∈C0,1,0,∞with
1
0
t1−t|bt|dt <∞, 1
0
t1−thtdt <∞. 1.2
Whenab≡0 ora≡0 andb /≡0, the two kinds of singular boundary value problems have been discussed extensively in the literature; see1–10and the references therein. Hence, the problem that we consider is more general and is different from those in previous work.
Furthermore, we will see in the later that the presence ofatut btutbrings us three main difficulties:
1the Green’s function cannot be explicitly expressed;
2the equivalence between BVP1.1and its associated integral equation has to be proved;
3the compactness of associated integral operator has to be verified.
We will overcome the above mentioned difficulties in Section 2. Also, although the Leggett-William fixed point theorem is used extensively in the study of triple positive differential equations, the method has not been used to study this type of second-order singular boundary value problem with general differential operators. We are concerned with solving these problems in this paper.
To state our main tool used in this paper, we give some definitions and notations. LetEbe a real Banach space with a coneP. A mapα : P → 0,∞is said to be a nonnegative continuous concave functional onPifαis a continuous and
αtx 1−ty≥tαx 1−tαy, 1.3
for allx, y∈Pandt∈0,1. Leta, bbe two numbers such that 0< a < bandαa nonnegative continuous concave functional onP. We define the following convex sets:
Pa{x∈P: x < a},
Pα, a, b {x∈P :a≤αx, x ≤b}. 1.4
Theorem 1.1 Leggett-Williams fixed point theorem. Let T : Pc → Pc be completely continuous, and letαbe a nonnegative continuous concave functional onPsuch thatα≤ x for all x∈Pc. Suppose that there exist0< d < a < b≤csuch that
i{x∈Pα, a, b:αx> a}/∅andαTx> a, forx∈Pα, a, b;
ii Tx < d, for x < d;
iiiαTx> a, forx∈Pα, a, cwith Tx > b.
ThenT has at least three fixed pointsx1, x2, x3inPcsatisfying x1 < d, a < αx2, x3 > dand
αx3< a.
Remark 1.2. We note the existence of triple positive solutions of other kind of boundary value problems; see He and Ge11, Zhao et al.12, Zhang and Liu13, Graef et al.14, and the references therein.
2. Preliminaries and Lemmas
Throughout this paper, we assume the following:
H1a∈C0,1∩L10,1;
H2b∈C0,1,−∞,0, and01s1−s|bs|ds <∞;
H3h:0,1 → 0,∞is continuous and does not vanish identically on any subinterval of0,1, and01s1−shsds <∞;
H4f :0,1×0,∞ → 0,∞is continuous.
Lemma 2.1. Suppose that (H1) and (H2) hold. Then
ithe initial value problem
ut atut btut 0, t∈0,1,
u0 0, u0 1, 2.1
has a unique solutionϕ∈AC0,1∩C10,1andϕ∈ACloc0,1; iithe initial value problem
ut atut btut 0, t∈0,1,
u1 0, u1 −1, 2.2
has a unique solutionψ∈AC0,1∩C10,1andψ∈ACloc0,1.
Proof. We only provei.iican be treated in the same way.
Suppose thatϕ∈AC0,1∩C10,1andϕ∈ACloc0,1is a solution of2.1, that is,
ϕt atϕt btϕt 0, t∈0,1,
ϕ0 0, ϕ0 1.
2.3
Let
pt exp
t
0
asds
. 2.4
Multiplying both sides of2.3bypt, then
ptϕt−ptbtϕt, t∈0,1. 2.5
Sinceϕ∈C10,1andϕ∈ACloc0,1, integrating2.5on0, t,0≤t <1, we have
ptϕt 1−
t
0
Moreover, integrating2.6on0, t,0≤t≤1, we have
ptϕt t t
0
psasϕsds− t
0
s
0
pτbτϕτdτ ds. 2.7
Let
vt ⎧ ⎪ ⎨ ⎪ ⎩
ptϕt
t , t∈0,1, p0ϕ0, t0.
2.8
Clearly,v∈C0,1, and2.7reduces to
vt 11
t t
0
sasvsds−1 t
t
0
s
0
τbτvτdτ ds. 2.9
By using Fubini’s theorem, we have
t
0
s
0
τbτvτdτds t
0
t−ssbsvsds. 2.10
Therefore,
vt 11
t t
0
sasvsds− 1 t
t
0
t−ssbsvsds, 2.11
which implies thatvis a solution of integral equation2.11.
Conversely, ifv∈C0,1is a solution of2.11withv0 1, by reversing the above argument we could deduce that the functionϕt tvt/ptis a solution of2.1and satisfy
ϕ ∈ AC0,1 ∩ C10,1and ϕ ∈ AC
loc0,1. Therefore, to prove that2.1 has a unique
solution,ϕ∈AC0,1∩C10,1, andϕ∈ACloc0,1is equivalent to prove that2.11has a unique solutionv∈C0,1.
To do this, we endow the following norm inC0,1:
v ∗ sup
t∈0,1
vtexp
− t
0
|as|s1−s|bs|ds. 2.12
LetK:C0,1 → C0,1be operator defined by
Kvt 11
t t
0
sasvsds−1 t
t
0
Since
1
t t
0
sasvsds≤ t
0
|vsas|ds−→0, t−→0,
1
t t
0
st−sbsvsds≤ t
0
s1−s|bsvs|ds−→0, t−→0,
2.14
then,Kis well defined. Set
Dt exp
−
t
0
|as|s1−s|bs|ds
. 2.15
Then, for anyz, w∈C0,1,
|Kzt−Kwt|
1
t t
0
sas−st−sbszs−wsds
≤ t
0
|as|s1−s|bs||zs−ws|Ds 1 Dsds
≤ t
0
|as|s1−s|bs| 1
Dsds· z−w ∗
1
Dt−1
z−w ∗,
2.16
and subsequently,
Dt|Kzt−Kwt| ≤1−Dt· z−w ∗. 2.17
Thus,
Kz−Kw ∗≤1−Dt· z−w ∗. 2.18
Since 0≤1−Dt<1,Khas a unique fixed pointv∈C0,1by Banach contraction principle. That is,2.11has a unique solutionv∈C0,1.
Remark 2.2. Lemma 2.1generalizes Theorem 2.1of1, wherea≡0.
Lemma 2.3. Suppose that (H1) and (H2) hold. Then
iϕis nondecreasing in0,1;
iiψis nonincreasing in0,1.
Suppose on the contrary thatϕ is not nondecreasing in0,1. Then there exists τ0 ∈ 0,1such that
ϕτ0 maxϕt|t∈0,1>0, ϕτ0 0, ϕτ0≤0. 2.19
This together with the equationϕt atϕt btϕt 0 implies that
ϕτ0 −bτ0ϕτ0>0, 2.20
which is a contradiction!
Remark 2.4. From Lemmas2.1and2.3, there exist positive constantsC1,C2,D1, andD2such
that
D1t≤ϕt≤C1t, D21−t≤ψt≤C21−t, t∈0,1. 2.21
In fact, since
lim
t→0
ϕt t ϕ
0 1>0, 2.22
we have thatϕt/t∈C0,1andϕt/t≥0, t∈0,1. Then, there exist constantsC1>0 and
D1≥0, such that
D1≤ ϕt
t ≤C1, t∈0,1, 2.23
that is
D1t≤ϕt≤C1t, t∈0,1. 2.24
In the following, we will show thatD1>0. Suppose on the contrary, if there existt∗∈0,1,
such that
ϕt∗ t∗ tmin∈0,1
ϕt
t D10, 2.25
then,ϕt∗ 0, which is a contradiction!
The other inequality can be treated in the same manner.
Lemma 2.5. Suppose that (H1), (H2), and (H3) hold. Then
lim
t→0ϕt
1
t
ψshsds0, lim
t→1ψt
t
0
Proof. We only prove the first equality; the other can be treated in the same way. From Remark 2.4andH3, we have
0≤ϕt 1
t
ψshsds≤C2ϕt
1
t
1−shsds. 2.27
Lemma 2.1of2together with the facts thatϕ∈C10,1andH3implies that
lim
t→0ϕt
1
t
1−shsds0. 2.28
Combining2.27and2.28, we have
lim
t→0ϕt
1
t
ψshsds0. 2.29
Lemma 2.6. Suppose that (H1), (H2), and (H3) hold. Then the problem
ut atut btut ht 0, t∈0,1,
u0 u1 0 2.30
has a unique solution
ut 1
0
Gt, sqshsds, 2.31
where
qs exp
s
1/2
aτdτ
,
Gt, s 1 ρ
⎧ ⎨ ⎩
ϕsψt, 0≤s≤t≤1,
ϕtψs, 0≤t≤s≤1,
ρϕ
1 2
ψ
1 2
−ϕ
1 2
ψ
1 2
.
2.32
Moreover,ut>0on0,1.
Proof. ByLemma 2.3and2.32, we have
Gt, s≤Gs, s, t, s∈0,1. 2.33
Now we check that the function
ut t
0
1
ρϕsψtqshsds 1
t
1
ρψsϕtqshsds 2.34
satisfies2.30. In fact,
ut ψt t
0
1
ρϕsqshsdsϕ
t1
t
1
ρψsqshsds,
ut ψt t
0
1
ρϕsqshsdsψ
t1
ρϕtqtht
ϕt 1
t
1
ρψsqshsds−ϕ
t1
ρψtqtht.
2.35
Therefore,
ut atut btut 1
ρqtht
ϕt ψt ϕt ψt
1
ρqtht ϕ 1 2 ψ 1 2 ϕ 1 2 ψ 1 2 exp − t
1/2
asds
−ht.
2.36
Equation2.34andLemma 2.5imply that
u0 lim
t→0ut 0, u1 limt→1ut 0. 2.37
SinceGt, s>0 fort, s∈0,1, then
ut 1
0
Gt, sqshsds >0, t∈0,1. 2.38
LetY C0,1with the norm
u ∞ max
t∈0,1|ut|, 2.39
and letPbe a cone inC0,1defined by
Lemma 2.7. Suppose that (H1)–(H3) hold anduis a positive solution of 2.30. Then
ut≥ u γt, t∈0,1, 2.41
where
γt min
t∈0,1
ϕt ϕ,ψt
ψ
. 2.42
Furthermore, for anyδ∈0,1/2, there exists correspondingγδ>0such that
ut≥γδ u , t∈δ,1−δ. 2.43
Proof. In fact, if 0< t≤s <1, then
Gt, s Gs, s
ϕt ϕs ≥
ϕt
ϕ, 2.44
and if 0< s≤t <1, then
Gt, s Gs, s
ψt ψs ≥
ψt
ψ. 2.45
Combining this andGt, s≤Gs, s, t, s∈0,1, we have
ut 1
0
Gt, sqshsds≥γt 1
0
Gs, sqshsds
≥γt u , t∈0,1.
2.46
Take
γδmin
γt:t∈δ,1−δ. 2.47
ThenLemma 2.3guarantees thatγδ>0, andLemma 2.7guarantees that2.43holds.
Remark 2.8. FromLemma 2.7andRemark 2.4, we have
Gt, s≥γtGs, s≥ D1D2
ρ γts1−s, t, s∈0,1. 2.48
Now, for anyu∈P, we can define the operatorT :C0,1 → C0,1by
Tut 1
0
Lemma 2.9. Let (H1)–(H4) hold. ThenT :P → Pis a completely continuous operator.
Proof. From H3 and H4 and Lemma 2.6, it is easy to see that T : P → P, and T is continuous by the Lebesgue’s dominated convergence theorem.
LetB⊆Pbe any bounded set. ThenH3andH4imply thatTBis a bounded set inP.
Since
Tut ψt t
0
1
ρϕsqshsfs, usds
ϕt 1
t
1
ρψsqshsfs, usds,
2.50
then this together with the similar proof of Lemma 2.1of2yields
Tu∈L10,1. 2.51
From this fact, it is easy to verify thatTBis equicontinuous. Therefore, by the Arzela-Ascoli theorem,T :P → Pis a completely continuous operator.
3. Main Result
Letα:P → 0,∞be nonnegative continuous concave functional defined by
αu min
t∈δ,1−δut, u∈P. 3.1
We notice that, for eachu∈P,αu≤ u , and also that byLemma 2.6,u∈P is a solution of
1.1if and only ifuis a fixed point of the operatorT.
For convenience we introduce the following notations. Let
mρ·
C1C2
1
0
s1−sqshsds −1
,
η 1
ρD1D2t∈minδ,1−δγt 1−δ
δ
s1−sqshsds.
3.2
Theorem 3.1. Assume that (H1)–(H4) hold. Let0< a < b < b/γδ≤c, and suppose thatfsatisfies the following conditions:
S1ft, u< ma, fort∈0,1, u∈0, a;
S2ft, u> b/η, fort∈δ,1−δ, u∈b, b/γδ;
S3ft, u< mc, fort∈0,1, u∈0, c.
Then the boundary value problem1.1has at least three positive solutionsu1, u2, u3inPcsatisfying
Proof. From Lemma 2.9, T : P → P is a completely continuous operator. Ifu ∈ Pc, then
u ≤c, and assumptionS3implies thatft, u< mc, t∈0,1. Therefore
Tu max
t∈0,1|Tut|tmax∈0,1 1
0
Gt, sqshsfs, usds
≤ C1C2
ρ mc 1
0
s1−sqshsds
≤c.
3.3
Hence,T :Pc → Pc. In the same way, ifu∈Pa, thenT :Pa → Pa. Therefore, conditioniiof
Leggett-williams fixed-point theorem holds.
To check conditioniof Leggett-Williams fixed-point theorem, chooseut b/γδ, t∈
0,1. It is easy to see thatu∈Pα, b, b/γδandαu αb/γδ> b. so,
u∈P
α, b, b
γδ
:αu> b
/
∅. 3.4
Hence, ifu ∈Pα, b, b/γδ, thenb ≤ ut ≤ b/γδ, t ∈ δ,1−δ. From assumptionS2and
Remark 2.8, we have
min
t∈δ,1−δTut t∈minδ,1−δ
1
0
Gt, sqshsfs, usds
≥ min
t∈δ,1−δ 1−δ
δ
Gt, sqshsfs, usds
> b
η·t∈minδ,1−δ 1−δ
δ
Gt, sqshsds
> bD1D2
ηρ t∈minδ,1−δγt
1−δ
δ
s1−sqshsds
b.
3.5
Finally, we assert that ifu∈Pα, b, cand Tu > b/γδ, thenαTu> b. To see this, suppose
thatu∈Pα, b, cand Tu > b/γδ, then
min
t∈δ,1−δTut t∈minδ,1−δ
1
0
Gt, sqshsfs, usds
≥γδ
1
0
Gs, sqshsfs, usds
≥γδmax t∈0,1
1
0
Gs, sqshsfs, usds
γδ Tu > b.
To sum up, all the conditions of Leggett-williams fixed-point theorem are satisfied. Therefore,
T has at least three fixed points, that is, problem 1.1 has at least three positive solutions
u1, u2, u3inPcsatisfying u1 < a, b < αu2, u3 > aandαu3< b.
Theorem 3.2. Assume that (H1)–(H4) hold. Let0 < a1 < b1 < b1/γδ < a2 < b2 < b2/γδ < · · ·<
an, n∈N, and suppose thatfsatisfies the following conditions:
A1ft, u< mai, fort∈0,1, u∈0, ai, 1≤i≤n;
A2ft, u> bi/η, fort∈δ,1−δ, u∈bi, bi/γδ, 1≤i≤n−1.
Then the boundary value problem1.1has at least2n−1positive solutions.
Proof. Whenn1, it follows from conditionA1thatT :Pa1 → Pa1, which means thatThas
at least one fixed pointu1∈Pa1by the Schauder fixed-point Theorem. Whenn2, it is clear
thatTheorem 3.1holdswithc1 a2. Then we can obtain at least three positive solutions
u1, u2, andu3satisfying u1 ≤a1, ψu2> b1and u3 > a1withψu3< b1. Following this
way, we finish the proof by the induction method.
4. Example
Consider the following boundary value problem:
ut−2ut−3ut htfut 0, t∈0,1,
u0 u1 0, 4.1
where
ht 1 t1−t,
fu ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ u
4, u∈0,1, 41u
4 −10, u∈1,2, 21
2 , u∈2,30,
u−912u
231u , u≥30.
4.2
Then, by computation, we have
ϕt 1
4
et−e−3t, ψt 1
4
e1−t−e−31−t,
ρ 1
8e3
e2−1e23, qt e−2t1.
Furthermore, fort∈0,1,
√ 3
3 t≤ϕt≤t, √
3
3 1−t≤ψt≤1−t. 4.4
In fact, letαt et−e−3t−4√3/3t, thenα0 0, andαt et3e−3t−4√3/3. It is easy to
compute that
min
t∈0,1
et3e−3t 4
√ 3
3 . 4.5
Then,αt≥0, t∈0,1, that is
ϕt≥
√ 3
3 t, t∈0,1. 4.6
The other inequalities in4.4can be proved by the same method. Thus, we can choose thatC1C21,D1D2
√
3/3 andδln√2. By computation, we have
γt 4e
3
e4−1min
ϕt, ψt,
γδ min t∈δ,1−δγt
3√2e3
4e4−1 ≈0.39747. . . .
4.7
Leta1/4, b2, andc20. Then, we can compute
m e
33
4e2 ≈0.78107· · ·,
η
√
2e5e2−4
e2−1e4−1e23 ≈0.19994· · ·.
4.8
Consequently,
fu≤ 1
4 · 1 4 <
1
4mma, u∈
0,1
4
,
fu 21
2 > 2
η b η, u∈
2, 2 γδ
⊂2,10,
fu≤ 21
2 <20m, u∈0,20.
Therefore, all the conditions ofTheorem 3.1are satisfied, then problem4.1has at least three positive solutionsu1, u2, andu3satisfying
u1 ≤
1
4, ψu2>2, u3 > 1
4 with ψu3<2. 4.10
Acknowledgment
The first author was partially supported by NNSF of China10901075.
References
1 D. O’Regan, “Singular Dirichlet boundary value problems. I. Superlinear and nonresonant case,” Nonlinear Analysis. Theory, Methods & Applications, vol. 29, no. 2, pp. 221–245, 1997.
2 H. Asakawa, “Nonresonant singular two-point boundary value problems,”Nonlinear Analysis. Theory, Methods & Applications, vol. 44, no. 6, pp. 791–809, 2001.
3 R. Dalmasso, “Positive solutions of singular boundary value problems,”Nonlinear Analysis. Theory, Methods & Applications, vol. 27, no. 6, pp. 645–652, 1996.
4 D. R. Dunninger and H. Y. Wang, “Multiplicity of positive radial solutions for an elliptic system on an annulus,”Nonlinear Analysis. Theory, Methods & Applications, vol. 42, no. 5, pp. 803–811, 2000.
5 K. S. Ha and Y.-H. Lee, “Existence of multiple positive solutions of singular boundary value problems,”Nonlinear Analysis. Theory, Methods & Applications, vol. 28, no. 8, pp. 1429–1438, 1997.
6 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.
7 C.-G. Kim and Y.-H. Lee, “Existence and multiplicity results for nonlinear boundary value problems,” Computers & Mathematics with Applications, vol. 55, no. 12, pp. 2870–2886, 2008.
8 F. H. Wong, “Existence of positive solutions of singular boundary value problems,”Nonlinear Analysis. Theory, Methods & Applications, vol. 21, no. 5, pp. 397–406, 1993.
9 X. Xu and J. P. Ma, “A note on singular nonlinear boundary value problems,”Journal of Mathematical Analysis and Applications, vol. 293, no. 1, pp. 108–124, 2004.
10 X. J. Yang, “Positive solutions for nonlinear singular boundary value problems,”Applied Mathematics and Computation, vol. 130, no. 2-3, pp. 225–234, 2002.
11 X. M. He and W. G. Ge, “Triple solutions for second-order three-point boundary value problems,” Journal of Mathematical Analysis and Applications, vol. 268, no. 1, pp. 256–265, 2002.
12 D. X. Zhao, H. Z. Wang, and W. G. Ge, “Existence of triple positive solutions to a class ofp-Laplacian boundary value problems,”Journal of Mathematical Analysis and Applications, vol. 328, no. 2, pp. 972– 983, 2007.
13 B. G. Zhang and X. Y. Liu, “Existence of multiple symmetric positive solutions of higher order Lidstone problems,”Journal of Mathematical Analysis and Applications, vol. 284, no. 2, pp. 672–689, 2003.
