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Hindawi Publishing Corporation
International Journal of Differential Equations
Volume 2012, Article ID 838947, 21 pages
doi:10.1155/2012/838947
Research Article
Multiple Solutions for Nonlinear Doubly
Singular Three-Point Boundary Value Problems
with Derivative Dependence
R. K. Pandey and A. K. Barnwal
Department of Mathematics, Indian Institute of Technology, Kharagpur 721302, India
Correspondence should be addressed to R. K. Pandey, rkp@maths.iitkgp.ernet.in
Received 25 May 2012; Accepted 12 July 2012
Academic Editor: Yuji Liu
Copyright q 2012 R. K. Pandey and A. K. Barnwal. 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 study the existence of multiple nonnegative solutions for the doubly singular threepoint boundary value problem with derivative dependent data function −pty t qtft, yt, pty t, 0 < t < 1, y0 0, y1 α1 yη. Here, p ∈ C0, 1 ∩ C1 0, 1 with pt > 0
on 0, 1 and qt is allowed to be discontinuous at t 0. The fixed point theory in a cone is applied
to achieve new and more general results for existence of multiple nonnegative solutions of the
problem. The results are illustrated through examples.
1. Introduction
In this paper, we consider the following three-point boundary value problem of SturmLiouville type:
− pty t qtf t, yt, pty t ,
0 < t < 1,
1.1
with boundary conditions
y0 0,
y1 α1 y η ,
1.2
t
where 0 < α1 < h1/hη and ht 0 1/pxdx.
Throughout this paper, we assume the following conditions on the functions pt, qt,
and ft, y, py :
E1 p ∈ C0, 1 ∩ C1 0, 1 with pt > 0 on 0, 1 and 1/p ∈ L1 0, 1;
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E2 qt > 0, q is not identically zero on 0, 1 and q ∈ L1 0, 1;
E3 f ∈ C0, 1 × 0, ∞ × R, 0, ∞ and f is not identically zero.
Note that condition E2 allows qt be discontinuous at t 0, and if p0 0, then the
differential equation 1.1 is called doubly singular 1.
Nonlocal boundary value problem have variety of applications in the area of applied
mathematics and physical sciences. The design of a large size bridge with multipoint supports
can be considered as an application of these types of boundary value problem 2. Some more
applications can be found in 3–5 and the references therein. Recently, motivated by the wide
application of boundary value problems in physical and applied mathematics, the study of
multipoint boundary value problems has received increasing interest see 2, 6–12 and the
references therein.
Nonsingular multipoint boundary value problems have been extensively studied in
literature, see 13–16 for derivative dependent data function ft, y, z and 8, 10, 12 for
derivative independent data function ft, y, z ft, y.
Some attention has been devoted to singular multipoint boundary value problems see
17, 18 and the references therein. When pt 1 and qtft, y, z may have singularity
at t 0, t 1, y 0 and z 0, differential equation 1.1 with boundary conditions
y 0 0, y1 α1 yη is considered by Chen et al. 17 and Agarwal et al. 18. Chen
et al. proved the existence of at least one positive solution while Agarwal et al. established
that this problem may have at least two positive solutions and also may have no positive
solutions under some conditions on qt and ft, y, z.
Bai and Ge 19 have generalized the Leggett-Williams fixed point theory and applied
to
y qtf t, y, y 0,
y0 0,
t ∈ 0, 1,
y1 0,
1.3
to achieve at least three positive solutions of the two-point boundary value problem.
In this work, we consider the problem 1.1-1.2 with unbounded coefficient of y
along with singularity in the data function ft, y, z.
Existence of nonnegative solutions of the problem 1.1-1.2 may be established
either directly or by reducing the problem to
y qtf t, y, y 0,
1.4
and applying the existing results. But direct consideration of the problem provides better
results, especially as the order of singularity increases. This may be demonstrated by the
following simple linear three-point boundary value problem:
r 5
r−1
2r 1t − r ,
− t y t t
4
y0 0,
0.5 ≤ r < 1, t ∈ 0, 1,
1
y1 y
.
4
1.5
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3
The problem 1.5 can be reduced to the following boundary value problem:
−v x 5
2r−1/1−r
1/1−r
r
,
x
−
2r
1x
4
1 − r2
1−r 1
v0 0,
v1 v
,
4
1
x ∈ 0, 1,
1.6
t
by change of variable x 1 − r 0 τ −r dτ t1−r .
Now we apply the result Theorem 4.2 of this work to the problem 1.5 and conclude
that the problem has at least one nonnegative solution yt with
3 31 − r 1−r/r 3
r
2
.
sup yt ≤
2 4 − 4r
4
t∈0,1
1.7
Further, for pt 1, Theorems 4.2 and 4.3 may be regarded as extension of Theorem
3.1 in 19 for three-point singular boundary value problem. Now applying Theorem 4.2
with pt 1 to the reduced problem 1.6, we get that the problem 1.5 has at least one
nonnegative solution yt with
sup yt sup |vt| ≤
t∈0,1
t∈0,1
31 − r 1−r/r 3
3
r
2
.
4 − 4r 4 − 4r
4
1.8
Now as r approaches to one, that is, the order of singularity increases, the upper bound
for supt∈0,1 |yt| in 1.8 approaches to ∞ while in 1.7 approaches to 4.125, which can be
seen from Figure 1. As smaller upper bound for supt∈0,1 |yt| will enable to find nonnegative
solutions faster and hence will be helpful in constructing efficient numerical algorithms
to find multiple nonnegative solutions, thus it is justified to consider the singular problem
directly. A detailed working is given in Example 5.1.
In this work, we are concerned with existence of multiple nonnegative solutions of
the three-point doubly singular boundary value problem 1.1-1.2. To achieve this, we use
generalized Leggett-Williams fixed point theorem established by Bai and Ge 19.
For this purpose, we first establish certain properties of Green’s function of the
corresponding homogeneous boundary value problem. Then fixed point theorem of
functional type generalized Leggett-Williams fixed point theorem is applied to yield
multiple nonnegative solutions for the boundary value problem 1.1-1.2.
We organize this work as follows. In Section 2, we present some definitions and basic
results required for this work. Section 3 deals with nonnegativity of Green’s function and
some basic properties. Section 4 is devoted to existence of at least one and three or odd
number of nonnegative solutions. In Section 5, we demonstrate the results through examples.
2. Background and Definitions
The proof of main results is based on fixed point theorem of functional type in a cone given
by Bai and Ge 19, which deals with three fixed points of completely continuous nonlinear
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24
21
Bounds for y
18
1
15
12
9
6
3
2
0
0.5
0.6
0.7
0.8
0.9
r
(1) When singular problem is reduced to regular problem
(2) When singular problem is solved directly
Figure 1: Variation of bounds for y in both cases.
operators defined in a cone of an ordered Banach space. In this section, we provide some
background material from the theory of cone in Banach spaces to make the paper selfcontained.
Definition 2.1. A subset D of Banach space E is said to be retract of E if ∃ a continuous map
r : E → D such that rx x for every x ∈ D.
Corollary 2.2. Every close convex set of a Banach space is a retract of Banach space.
Definition 2.3. Let E be a Banach space, P ⊂ E is nonempty convex, closed set, P is said to be
cone provided that
1 λu ∈ P for all λ ≥ 0, u ∈ P , and
2 u ∈ P, −u ∈ P implies u 0.
Note. From Corollary 2.2, a cone P of Banach Space E is retract of E.
Definition 2.4. A subset R of Banach space X is called relatively compact if R closure of R is
compact.
Definition 2.5. Consider two Banach spaces X and Y , a subset Ω of X, and a map T : Ω → Y .
Then T is said to be completely continuous operator if
1 T is continuous, and
2 T maps bounded subset of Ω into relatively compact sets.
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5
Definition 2.6. The map α is said to be a nonnegative continuous convex functional on P
provided that α : P → 0, ∞ is continuous and
α tx 1 − ty ≤ tαx 1 − tα y ,
2.1
for all x, y ∈ P and 0 ≤ t ≤ 1. Similarly, the map γ is said to be a nonnegative continuous
concave functional on P provided that γ : P → 0, ∞ is continuous and
γ tx 1 − ty ≥ tγx 1 − tγ y ,
2.2
for all x, y ∈ P , and 0 ≤ t ≤ 1.
Definition 2.7. Suppose α, β : P → 0, ∞ are two continuous convex functionals satisfying
x ≤ M max αx, βx ,
for x ∈ P,
2.3
where M is positive constant, and
Ω x ∈ P : αx < r, βx < L /
φ
for r > 0, L > 0.
2.4
From 2.3 and 2.4, Ω is a bounded nonempty open subset of P .
Definition 2.8. Let r > a > 0, L > 0 be given constants, α, β : P → 0, ∞ two nonnegative
continuous convex functionals satisfying 2.3 and 2.4, and ψ a nonnegative continuous
concave functional on the cone P . Then bounded convex sets are defined as
P αr , βL x ∈ P : αx < r, βx < L ,
P αr , βL x ∈ P : αx ≤ r, βx ≤ L ,
P αr , βL , ψa x ∈ P : αx < r, βx < L, ψx > a ,
2.5
P αr , βL , ψa x ∈ P : αx ≤ r, βx ≤ L, ψx ≥ a .
Theorem 2.9 see 20. Let X be retract of real Banach space E. Then for every bounded relatively
open subset U of X and every completely continuous operator T : U → X which has no fixed point
on ∂U (relative to X), there exists an integer iT, U, X such that if iT, U, X / 0, then T has at least
one fixed point in U. Moreover, iT, U, X is uniquely defined.
Theorem 2.10 see 20. Let E be Banach space, X retract of E, X1 a bounded convex retract of X,
and U ⊂ X nonempty open subset, such that U ⊂ X1 . If T : X1 → X is completely continuous,
T X1 ⊂ X1 , such that there is no fixed point of T in X1 \ U, then iT, U, X 1.
Theorem 2.11 see 19 fixed point theorem of functional type. Let E be Banach space,
P ⊂ E a cone, and c ≥ b > a > d > 0, L2 ≥ L1 > 0 given constants. Assume that α, β
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are nonnegative continuous convex functionals on P such that 2.3 and 2.4 are satisfied. ψ is a
nonnegative continuous concave functional on P such that ψx ≤ αx for all x ∈ P αc , βL2 and let
T : P αc , βL2 → P αc , βL2 be a completely continuous operator. Suppose that
1 {x ∈ P αb , βL2 , ψa : ψx > a} /
Φ and ψT x > a for x ∈ P αb , βL2 , ψa ,
2 αT x < d, βT x < L1 for all x ∈ P αd , βL1 ,
3 ψT x > a for all x ∈ P αc , βL2 , ψa with αT x > b.
Then T has at least three fixed points x1 , x2 , x3 ∈ P αc , βL2 such that
x1 ∈ P αd , βL1 ,
x2 ∈ x ∈ P αc , βL2 , ψa : ψx > a ,
\ P αc , βL2 , ψa ∪ P αd , βL1 .
x3 ∈ P α , β
c
2.6
L2
3. Some Preliminary Results
In this section, we construct the Green’s function and establish some properties, required to
establish the main results in Section 4.
Lemma 3.1. The Green’s function for the following boundary value problem:
− pty t 0
,
B.C. : y0 0, y1 α1 y η
t ∈ 0, 1
3.1
is given by
⎧
⎪
G1 t, s,
⎪
⎪
⎪
⎨G t, s,
2
Gt, s ⎪G3 t, s,
⎪
⎪
⎪
⎩
G4 t, s,
0 ≤ s ≤ min t, η < 1;
0 ≤ t ≤ s ≤ η;
η ≤ s ≤ t ≤ 1;
0 < max η, t ≤ s ≤ 1.
3.2
Here
G1 t, s hs
δ − ht α1 ht,
δ
G2 t, s ht
δ − hs α1 hs,
δ
G3 t, s 1
ht α1 h η − hs δhs ,
δ
G4 t, s δ h1 − α1 h η > 0,
ht
h1 − hs,
δ
t
1
dx,
ht px
0
h1
0 < α1 < .
h η
3.3
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7
Proof. Consider the following linear differential equation:
− pty t qtFt,
3.4
where F ∈ C0, 1, 0, ∞. Integrating the above differential equation twice first from t to 1
and then from 0 to t, changing the order of integration, and applying the boundary conditions,
we get
t
yt hsqsFsds ht
1
0
qsFsds
t
η
1
1
ht
α1
hsqsFsds − hsqsFsds α1 h η
qsFsds .
δ
0
0
η
3.5
For t ∈ 0, η, yt can be written as
yt t
hs
δ − ht α1 htqsFsds
0 δ
η
1
ht
ht
δ − hs α1 hsqsFsds h1 − hsqsFsds,
δ
t
η δ
3.6
or
yt t
G1 t, sqsFsds 0
η
G2 t, sqsFsds 1
t
G4 t, sqsFsds.
3.7
η
Similarly, for t ∈ η, 1, yt can be written as
yt η
hs
δ − ht α1 htqsFsds
δ
0
1
t
1
ht
ht α1 h η − hs δhs qsFsds h1 − hsqsFsds,
δ
η δ
t
3.8
or
yt η
G1 t, sqsFsds 0
t
G3 t, sqsFsds 1
η
G4 t, sqsFsds.
3.9
t
From 3.7 and 3.9, we may write
yt 1
Gt, sqsFsds,
0
3.10
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where Gt, s is given in the lemma. It is easy to see that Gt, s satisfies all the properties
of Green’s function. Hence Gt, s is the Green’s function for the boundary value problem
3.1.
Lemma 3.2. The Green’s function Gt, s satisfies the following properties:
i maxt∈0,1 pt∂Gt, s/∂t < ∞,
ii Gt, s ≥ 0 for all t, s ∈ {0, 1 × 0, 1},
iii there exist a constant λ in (0,1) such that mint∈η,1 Gt, s ≥ λ maxt∈0,1 Gt, s for s ∈
0, 1, where
⎧
α1 h η
α1 h1 − h η
⎪
⎪
⎪
,
⎪
⎨min h1 − α1 hη , h1
λ
⎪
h η
h1 − α1 h η
⎪
⎪
⎪min
,
,
⎩
α1 h1
α1 h1 − h η
0 < α1 ≤ 1,
h1
1 ≤ α1 < .
h η
3.11
Proof. i
⎧
hs
⎪
⎪
α1 − 1,
⎪
⎪
δ
⎪
⎪
⎪
⎪
⎪
1
⎪
⎪
⎪ δ − hs − α1 hs,
∂Gt, s ⎨ δ
pt
⎪1
∂t
⎪
⎪
⎪
α1 h η − hs ,
⎪
⎪
δ
⎪
⎪
⎪
⎪
⎪
⎪
⎩ 1 h1 − hs,
δ
0 ≤ s ≤ min t, η < 1;
0 ≤ t ≤ s ≤ η;
3.12
η ≤ s ≤ t ≤ 1;
0 < max η, t ≤ s ≤ 1.
Since pt∂Gt, s/∂t is independent of t, therefore maxt∈0,1 pt∂Gt, s/∂t < ∞.
ii For t < η, α1 ht − hη > h1/hηht − hη and
h1
1
G1 t, s ≥ hsht − 1 ≥ 0,
δ
h η
3.13
it is easy to show that G1 t, s ≥ 0, for t ≥ η.
Next we show G2 t, s ≥ 0, G3 t, s ≥ 0 and G4 t, s ≥ 0 as follows:
G2 t, s ≥
ht ≥ 0;
hs h1 − h η
δh η
G3 t, s 1
hs{h1 − ht} α1 h η {ht − hs} ≥ 0,
δ
G4 t, s ht
h1 − hs ≥ 0.
δ
Thus Gt, s ≥ 0 for all t, s ∈ {0, 1 × 0, 1}.
3.14
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9
iii We prove the inequality for the following cases:
a s ∈ 0, η and b s ∈ η, 1,
a for s ∈ 0, η, we further divide this case in two parts as follows.
1 When 0 < α1 ≤ 1, 2 when 1 ≤ α1 < h1/hη.
Case 1 For 0 < α1 ≤ 1. It is easy to see that
G1 t, s hs h1 − α1 h η htα1 − 1
δ
3.15
implies
max G1 t, s t∈0,1
hs h1 − α1 h η ,
δ
α1 hs h1 − h η .
δ
min G1 t, s t∈η,1
3.16
Next,
G2 t, s ht h1 − α1 h η hsα1 − 1 ,
δ
hs ≤
h1 − α1 h η ,
δ
3.17
as ht ≤ hs.
Thus for 0 < α1 ≤ 1,
max Gt, s t∈0,1
hs h1 − α1 h η ,
δ
min Gt, s t∈η,1
α1 hs h1 − h η .
δ
3.18
hs h1 − h η .
δ
3.19
Case 2 For 1 ≤ α1 < h1/hη. It is easy to see that
max G1 t, s t∈0,1
α1 hs h1 − h η ,
δ
min G1 t, s t∈η,1
Next,
G2 t, s ht h1 − α1 h η hsα1 − 1 ,
δ
α1 hs ≤
h1 − h η ,
δ
as ht ≤ hs ≤ h η .
3.20
Thus for 1 ≤ α1 < h1/hη,
max Gt, s t∈0,1
α1 hs h1 − h η ,
δ
min Gt, s t∈η,1
hs h1 − h η .
δ
3.21
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International Journal of Differential Equations
Combining 3.18 and 3.21, we may write for s ∈ 0, η,
⎧ hs ⎪
⎪
h1 − α1 h η , 0 < α1 ≤ 1,
⎪
⎨ δ
max Gt, s α1 hs h1
⎪
h1 − h η , 1 ≤ α1 < ,
t∈0,1
⎪
⎪ δ
h η
⎩
⎧
α1 hs ⎪
⎪
h1 − h η ,
0 < α1
⎪
⎨
δ
min Gt, s ⎪
hs t∈η,1
⎪
⎪
h1 − h η , 1 ≤ α1 <
⎩
δ
3.22
≤ 1,
h1
.
h η
b For s ∈ η, 1. For this case, G3 t, s and G4 t, s are considered. From 3.2, it can be easily
seen that for s ∈ η, 1,
h1
max Gt, s h1 − hs,
t∈0,1
δ
h η
min Gt, s h1 − hs.
t∈η,1
δ
3.23
Thus from 3.22 and 3.23, we get
⎧
⎧
hs ⎪
⎪
⎪
⎪
h1 − α1 h η , 0 < α1 ≤ 1,
⎪
⎪
⎪
⎪
⎪
δ
⎪
⎪
⎨
⎪
⎪
⎪ s ∈ 0, η : α1 hs ⎨
h1
⎪
h1 − h η , 1 ≤ α1 < ,
⎪
⎪
max Gt, s ⎪ δ
h η
⎪
⎪
t∈0,1
⎩
⎪
⎪
⎪
⎪
⎪
⎪
h1
h1
⎪
⎪
h1 − hs, 0 < α1 < .
⎩ s ∈ η, 1 :
δ
h η
⎧
⎧
α1 hs ⎪
⎪
⎪
⎪
h1 − h η , 0 < α1 ≤ 1,
⎪
⎪
⎪
⎨
δ
⎪
⎪
⎪
⎪
⎨ s ∈ 0, η : ⎪ hs h1
⎪
⎪
h1 − h η ,
1 ≤ α1 < ,
min Gt, s ⎩
δ
⎪
h
η
t∈η,1
⎪
⎪
⎪
⎪
h
η
h1
⎪
⎪
⎪
h1 − hs, 0 < α1 < .
⎩ s ∈ η, 1 :
δ
h η
3.24
From 3.24,
⎧
α1 h η
α1 h1 − h η
⎪
⎪
⎪
max Gt, s,
⎪
⎨min h1 − α1 hη , h1
t∈0,1
min Gt, s ≥
⎪
t∈η,1
h η
h1 − α1 h η
⎪
⎪
⎪min
max Gt, s,
,
⎩
α1 h1 t∈0,1
α1 h1 − h η
0 < α1 ≤ 1,
h1
1 ≤ α1 < .
h η
3.25
International Journal of Differential Equations
11
Consequently, setting
⎧
α1 h η
α1 h1 − h η
⎪
⎪
⎪
, 0 < α1 ≤ 1,
⎪
⎨min h1 − α1 hη , h1
λ
⎪
h η
h1 − α1 h η
h1
⎪
⎪
⎪
, 1 ≤ α1 < ,
,
⎩min
α
h1
α1 h1 − h η
h η
1
3.26
min Gt, s ≥ λmax Gt, s.
3.27
there holds
t∈η,1
t∈0,1
It can be easily seen that 0 < λ < 1. This completes the proof.
4. Existence of Multiple Nonnegative Solutions
Let X C0, 1 C2 0, 1 be endowed with ordering x ≤ y if xt ≤ yt for all t ∈ 0, 1 and
x max{x1 , x 1 }, where
x1 sup |xt|,
t∈0,1
4.1
x sup ptx t.
1
t∈0,1
Let E {x : x ∈ X, x < ∞} be bounded subset of X. E is Banach Space.
Now define a cone P ⊂ E as
P
x ∈ E : xt ≥ 0, min xt ≥ λmax xt, ptx t ≤ 0 .
t∈η,1
t∈0,1
4.2
The boundary value problem 1.1-1.2 has a solution yt if and only if yt solves the
following operator equation:
yt T yt,
4.3
where the operator T : P → P is given by
T y t 1
Gt, sqsf s, ys, psy s ds,
0 ≤ t ≤ 1.
4.4
0
Here Gt, s is the Green’s function of the problem 3.1 defined in Lemma 3.1.
Lemma 4.1. Let (E1)–(E3) hold, then the operator T : P → P is well defined and is completely
continuous.
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International Journal of Differential Equations
Proof. First we show that the operator T is well defined. For this, we take y ∈ P . From E2,
E3, and Gt, s ≥ 0, it follows that T yt ≥ 0.
Now applying Lemma 3.2, we get
min T yt min
t∈η,1
t∈η,1
min T yt ≥ λ
t∈η,1
1
1
Gt, sqsf s, ys, psy s ds,
0
max Gt, sqsf s, ys, psy s ds,
0 t∈0,1
λmax
1
t∈0,1
4.5
Gt, sqsf s, ys, psy s ds,
0
λmax T y t.
t∈0,1
It is easy to show that ptT y t ≤ 0. Thus T is well defined.
We now show that T is completely continuous. Let {yn } be a sequence in P and y0 ∈ P
with limn → ∞ yn y0 . Then, there exists a constant k1 > 0 such that yn < k1 for all n ∈
N ∪ {0}. Thus yn − y0 → 0 as n → ∞ implies supt∈0,1 |yn − y0 t| and supt∈0,1 |ptyn −
y0 t| → 0 as n → ∞. So yn t → y0 t and ptyn t → pty0 t as n → ∞.
Since f is continuous on {0, 1 × 0, k1 × −k1 , k1 }, so
1
T yn t − T y0 t Gt, sqs f s, yn s, psyn s − f s, y0 s, psy s ds
0
0
−→ 0 as n −→ ∞.
⇒ T yn − T y0 1 −→ 0
as n −→ ∞.
1
∂Gt, s
pt T yn t − T y t pt
qs f s, yn s, psyn s
0
∂t
0
−f s, y0 s, psy0 s ds
4.6
4.7
−→ 0 as n −→ ∞.
⇒ T yn − T y0 −→ 0 as n −→ ∞.
1
From 4.6 and 4.7,
T yn − T y0 −→ 0
Hence T : P → P is a continuous operator.
as n −→ ∞.
4.8
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13
Next we prove that T maps every bounded subset of P into relatively compact set. Let
B {x ∈ P : x ≤ k2 , k2 is a positive constant} be any bounded subset of P . For y ∈ B,
T yt 1
Gt, sqsf s, ys, psy s ds
0
≤ max
t∈0,1
≤
1
Gt, sqsf s, ys, psy s ds
0
1
max
s,u,v∈0,1×0,k2 ×−k2 ,k2 fs, u, vmax
t∈0,1
4.9
Gt, sqsds
0
⇒ T yt < ∞.
Therefore T B is uniformly bounded. Further, equicontinuity of T B follows from
1 T yt1 − T yt2 Gt, s|tt1 − Gt, s|tt2 qsf s, ys, psy s ds
0
−→ 0 as t1 −→ t2 ,
1
∂Gt, s ∂Gt, s pt T y t1 − T y t2 pt
qsf s, ys, psy s ds
−
∂t tt1
∂t tt2
0
t2
∂Gt, s · sup pt
≤ 2 max
qsf s, ys, psy s ds,
∂t
t1
t∈0,1
−→ 0 as t1 −→ t2 .
4.10
Thus from Arzela-Ascoli Theorem, T B is relatively compact subset of P and also T : P → P
is completely continuous.
Next, define functionals α, β, ψ : P → 0, ∞ such that
α y sup yt,
t∈0,1
β y sup pty t,
t∈0,1
4.11
ψ y min yt.
t∈η,1
Clearly, α, β are nonnegative continuous convex functionals such that y max{αy, βy}
satisfying 2.3 and 2.4, and ψ is nonnegative concave functional with ψy ≤ αy.
14
International Journal of Differential Equations
Let
C min
1
Gt, sqsds,
t∈η,1
L sup
η
1
t∈0,1
Gt, sqsds,
0
4.12
1
∂Gt, s
N sup pt
qsds.
∂t
t∈0,1 0
Now we state the main results of this work.
Theorem 4.2. Suppose that (E1)–(E3) are satisfied and ft, y, z satisfies the following condition.
H1 if there exist real constants d > 0 and L1 > 0 such that ft, y, z < min{d/L, L1 /N} for
t, y, z ∈ {0, 1 × 0, d × −L1 , L1 },
then boundary value problem 1.1-1.2 has at least one nonnegative solution y1 such that
supt∈0,1 |y1 t| < d with supt∈0,1 |pty1 t| < L1 .
Theorem 4.3. Suppose that (E1)–(E3) are satisfied. There exist real constants a, c, d, L1 , and L2 with
c ≥ a/λ b > a > d > 0, L2 ≥ L1 > 0 such that a/C ≤ min{c/L, L2 /N} and ft, y, z satisfies
following conditions.
H1 ft, y, z < min{d/L, L1 /N} for t, y, z ∈ {0, 1 × 0, d × −L1 , L1 },
H2 ft, y, z > a/C for t, y, z ∈ {η, 1 × a, a/λ × −L2 , L2 },
H3 ft, y, z ≤ min{c/L, L2 /N} for t, y, z ∈ {0, 1 × 0, c × −L2 , L2 }.
Then boundary value problem 1.1-1.2 has at least three nonnegative solutions y1 , y2 , and y3 in
P αc , βL2 such that
sup y1 t < d,
t∈0,1
sup pty1 t < L1 ,
t∈0,1
a < min y2 t ≤ sup y2 t ≤ c,
t∈η,1
sup pty2 t ≤ L2 ,
t∈0,1
t∈0,1
a
sup y3 t ≤ with sup pty3 t ≤ L2 .
λ
t∈0,1
t∈0,1
4.13
International Journal of Differential Equations
15
Proof of Theorem 4.2. Let U P αd , βL1 be open subset of P . We now show that T U ⊂ U.
For y ∈ U,
α T y sup T yt
t∈0,1
1
sup
Gt, sqsf s, ys, psy s ds,
0
t∈0,1
d
< sup
L t∈0,1
1
4.14
Gt, sqsds < d, from condition H1 0
implies that αT y < d.
Consider that
β T y sup ptT y t,
t∈0,1
1
∂Gt, s
sup pt
qsf s, ys, psy s ,
∂t
0
t∈0,1
1
L1
∂Gt, s
<
sup pt
qsds, from condition H1 ,
N t∈0,1 0
∂t
4.15
implies that βT y < L1 .
Thus T U ⊂ U. Next, we show that T has no fixed point on ∂U U \ U. On contrary,
suppose there exists a fixed point yt on ∂U such that T yt yt. Then from 4.14 and
4.15, αy < d and βy < L1 , which are not possible. So the operator T has no fixed point
on ∂U and from Theorem 2.10 iT, U, P 1. Thus the operator T has at least one fixed point
in U and also the boundary value problem 1.1-1.2 has at least one nonnegative solution
y1 such that supt∈0,1 |y1 t| < d with supt∈0,1 |pty1 t| < L1 .
Proof of Theorem 4.3. It is easy to see that ψy ≤ αy for each y ∈ P αc , βL2 . We now show
that T : P αc , βL2 → P αc , βL2 is well defined. For y ∈ P αc , βL2 ,
α T y sup T yt,
t∈0,1
sup
t∈0,1
≤
1
0
c
sup
L t∈0,1
≤ c.
Gt, sqsf s, ys, psy s ds,
1
0
4.16
Gt, sqsds, from assumption H3 ,
16
International Journal of Differential Equations
β T y sup ptT y t,
t∈0,1
1
∂Gt, s
sup pt
qsf s, ys, psy s ,
∂t
0
t∈0,1
1
L2
∂Gt, s
≤
sup pt
qsds, from assumption H3 ,
N t∈0,1 0
∂t
4.17
≤ L2 .
From 4.16 and 4.17,
T y ∈ P αc , βL2 .
4.18
Thus, T : P αc , βL2 → P αc , βL2 is well defined, and by Lemma 4.1, it is completely
continuous. Now Condition 2 of Theorem 2.11 can be proved by similar manner. Choose
yt a/λ ∈ P , 0 ≤ t ≤ 1, then αy supt∈0,1 |yt| a/λ, βy supt∈0,1 |pty t| 0 <
Φ. Further if
L2 , ψx mint∈η,1 |yt| a/λ > a. Thus, {y ∈ P αba/λ , βL2 , Ψa : Ψy > a} /
a/λ L2
y ∈ P α , β , Ψa , then a ≤ yt ≤ a/λ for η ≤ t ≤ 1. Then by definition of ψ and assumption
H2 , we have
ψ T y min T yt
t∈η,1
min
t∈η,1
≥ min
t∈η,1
>
1
Gt, sqsf s, ys, psy s ds
0
1
4.19
Gt, sqsf s, ys, psy s ds
η
a
·C a
C
Thus, Condition 1 of Theorem 2.11 is satisfied. We finally show that condition 3 of
Theorem 2.11 holds, too. Suppose y ∈ P αc , βL2 , Ψa with αT y > b. Then by definition of
ψ and T y ∈ P , we have
Ψ T y min T yt
t∈η,1
≥ λmax T yt
t∈0,1
≥ λα T y
a.
4.20
International Journal of Differential Equations
17
So, Condition 3 of Theorem 2.11 is also satisfied. Therefore, Theorem 2.11 yields that
boundary value problem 1.1-1.2 has at least three nonnegative solutions y1 , y2 , and y3
in P αc , βL2 such that
sup y1 t < d,
t∈0,1
sup pty1 t < L1 ;
t∈0,1
a < min y2 t ≤ sup y2 t ≤ c,
t∈η,1
t∈0,1
a
sup y3 t ≤ ,
λ
t∈0,1
sup pty2 t ≤ L2 ;
t∈0,1
4.21
sup pty3 t ≤ L2 .
t∈0,1
Corollary 4.4. Suppose that (E1)–(E3) are satisfied. If there exist constants 0 < d1 < a1 <
a1 /λ < d2 < a2 < a2 /λ < d3 < · · · < dn , 0 < L1 ≤ L2 ≤ L3 ≤ · · · ≤ Ln n ∈ N, with
ai /C < min{di1 /L, Li1 /N}, 1 < i ≤ n − 1 such that f satisfies the following conditions:
M1 ft, y, z < min{di /L, Li /N}, for t, y, z ∈ {0, 1 × 0, di × −Li , Li }, 1 ≤ i ≤ n,
M2 ft, y, z > ai /C, for t, y, z ∈ {η, 1 × ai , ai /λ × −Li1 , Li1 }, 1 ≤ i ≤ n − 1,
then boundary value problem 1.1-1.2 has at least 2n − 1 nonnegative solutions.
Proof. When n 1, the result follows from Theorem 4.2. When n 2, it is clear that all the
conditions of Theorem 4.3 hold with c d2 , d d1 , a a1 . Thus the boundary value
problem 1.1-1.2 has at least three positive solutions y1 , y2 , and y3 . Following this way, we
complete the proof by induction method.
Finally, we demonstrate these results through examples.
5. Example
In Example 5.1, we demonstrate the detailed working of the boundary value problem 1.5
mentioned in the introduction. Example 5.2 verifies our results.
Example 5.1. Consider the boundary value problem 1.5.
Here,
5
f t, y, py 2r 1t − r,
4
qt tr−1 ,
3
max f t, y, py r 2.
4
t∈0,1
5.1
31 − r
.
r4 − 4r 5.2
Following the notations of this work, it is easy to see that
3 31 − r 1−r/r
L
,
2 4 − 4r
N
Now for d ≥ 3/231 − r/4 − 4r 1−r/r 3/4r 2 and L1 ≥ 31−r3/4r 1/r4−
4 ,ft, y, py ≤ min{d/L, L1 /N}.
r
18
International Journal of Differential Equations
Then from Theorem 4.2, the problem has at least one nonnegative solution yt with
3 31 − r 1−r/r 3
r
2
,
sup yt ≤
2 4 − 4r
4
t∈0,1
31 − r 3
sup pty t ≤
r
2
.
r4 − 4r 4
t∈0,1
5.3
Next we reduce the problem and then apply Theorem 4.2 for pt 1. Using the
t
transformation x 1−r 0 1/xr dx t1−r , the boundary value problem 1.5 can be reduced
to regular boundary value problem as
−v x 5
2r−1/1−r
1/1−r
r
,
x
−
2r
1x
4
1 − r2
1−r 1
v0 0,
v1 v
.
4
1
x ∈ 0, 1,
5.4
Here,
5
2r−1/1−r
1/1−r
r
,
t
−
2r
1t
4
1 − r2
5
1
max F x, v, v 21 r − r .
4
x∈0,1
1 − r2
F x, v, v 1
5.5
Now following the notation of this work for pt 1,
L
31 − r2 31 − r 1−r/r
,
4 − 4r r4 − 4r N
31 − r2
.
r4 − 4r 5.6
Now for d ≥ 3/4 − 4r 31 − r/4 − 4r 1−r/r 3/4r 2 and L1 ≥ 3/r4 − 4r 3/4r 2,ft, v, v ≤ min{d/L, L1 /N}. So the problem has at least one nonnegative solution vt
with
sup |vt| ≤
t∈0,1
31 − r 1−r/r 3
3
r
2
,
4 − 4r 4 − 4r
4
sup v t ≤
t∈0,1
3
3
r
2
.
r4 − 4r 4
5.7
Hence the boundary value problem 1.5 has at least one nonnegative solution yt with
sup yt ≤
t∈0,1
3
31 − r 1−r/r 3
r
2
,
4 − 4r 4 − 4r
4
sup y t ≤
t∈0,1
3
3
r
2
.
r4 − 4r 4
5.8
Now in this case it is easy to show that if r approaches one, that is, the order of
singularity increases, upper bound for supt∈0,1 |yt| approaches ∞ while in case of direct
solving, upper bound for supt∈0,1 |yt| approaches 4.125. As smaller upper bound for
supt∈0,1 |yt| will enable to find nonnegative solutions faster and hence will be helpful
in constructing efficient numerical algorithms to find multiple nonnegative solutions.
International Journal of Differential Equations
19
Example 5.2. Consider the following boundary value problem:
− t1/2 y t t−1/2 f t, yt, t1/2 y t ,
y0 0,
y1 0 < t < 1,
5.9
3
1
y
.
2
3
i If
⎧
z 2
sin t 10y2 8
⎪
⎪
⎨ 3 400 300 ,
f t, y, z ⎪
z 2
sin t 21
⎪
⎩
,
3
50
300
y ≤ 4;
5.10
y ≥ 4;
then the boundary value problem 5.9 has at least one nonnegative solution.
ii Further, if
⎧
⎪
sin
⎪
⎪
⎪
⎪
⎪
⎪ 3
⎨ sin
f t, y, z ⎪
⎪ 3
⎪
⎪
⎪
sin
⎪
⎪
⎩
3
2
10y2 8
z
,
400
4000 2
z
t 9y2 24
,
400
40002
z
t 328353
,
400
4000
t
y≤4
4 ≤ y ≤ 191
5.11
y ≥ 191,
then the boundary value problem 5.9 has at least three nonnegative solutions.
Proof. Here, α1 3/2 and η 1/3. After simple calculation, we get L 5.464, N 3.732,
C 1.5396, and λ 0.1786.
i At least one nonnegative solution: we choose d 3 and L1 10. Here,
min{d/L, L1 /N} 0.549;
f t, y, z < 0.549,
for 0 ≤ t ≤ 1, 0 ≤ y ≤ 3, −10 ≤ z ≤ 10.
5.12
Thus, condition H1 is satisfied. Now from Theorem 4.2 the problem has at least one
nonnegative solution y1 such that supt∈0,1 |y1 t| < 3 with supt∈0,1 |pty1 t| < 10.
ii At least three nonnegative solutions: we choose constants d 4, a 34, c 4800,
10,
and L2 3500. Here, min{d/L, L1 /N} 0.732, min{c/L, L2 /N} 878.45, a/λ L1
190.34, and a/C 22.084;
f t, y, z < 0.732,
f t, y, z > 22.084
f t, y, z < 878.45
for
for 0 ≤ t ≤ 1, 0 ≤ y ≤ 4, −10 ≤ z ≤ 10;
1
≤ t ≤ 1, 34 ≤ y ≤ 190.34, −3500 ≤ z ≤ 3500;
3
for 0 ≤ t ≤ 1, 0 ≤ y ≤ 4800, −3500 ≤ z ≤ 3500.
5.13
20
International Journal of Differential Equations
Thus, conditions H1 , H2 , and H3 are satisfied. Now from Theorem 4.3 the problem has
at least three nonnegative solutions y1 , y2 , and y3 such that
sup y1 t < 4,
t∈0,1
sup pty1 t < 10;
t∈0,1
34 < min y2 t ≤ sup y2 t < 4800,
t∈η,1
t∈0,1
sup y3 t ≤ 190.34,
sup pty2 t ≤ 3500,
t∈0,1
5.14
sup pty3 t ≤ 3500.
t∈0,1
t∈0,1
Remark 5.3. For α1 1, the problem 1.1-1.2 can be regarded as two-point boundary value
problem with boundary conditions as
y0 0,
y 1 0,
5.15
in the limiting case η → 1− . Thus the results established in this work are also valid for the
two-point boundary value problem 21, 22.
Remark 5.4. Theorem 4.3 and Corollary 4.4 extend Theorem 3.1 and Corollary 3.1 of 19 to
doubly singular three-point boundary value problem.
Acknowledgments
This work is supported by DST, New Delhi, and UGC, New Delhi, India.
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