Journal of Applied Mathematics & Bioinformatics, vol.4, no.2, 2014, 37-45
ISSN: 1792-6602 (print), 1792-6939 (online)
Scienpress Ltd, 2014
Positive solutions of singular boundary
value problems for second order
impulsive differential equations
Ying He1
Abstract
This paper is devoted to study the positive solutions of nonlinear
singular two-point boundary value problems for second-order impulsive
differential equations.The existence of positive solutions is established
by using the fixed point theorem in cones.
Mathematics Subject Classification: 34B15
Keywords: Positive solution; Singular two-point boundary value problem;
Second-order impulsive differential equations; Fixed point theorem.
1
Introduction
This is the text of the introduction. Impulsive and singular differential
equations play a very important role in modern applied mathematics due to
1
School of Mathematics and Statistics, Northeast Petroleum University, Daqing163318,
P.R.China.
Project supported by Heilongjiang Province education department natural science
research item, China. E-mail: heying65338406@163.com
Article Info: Received : April 18, 2014. Revised : May 29, 2014.
Published online : June 15, 2014.
38
Positive solutions of singular boundary value problems...
their deep physical background and broad application. In this paper,we consider the existence of positive solutions of

−u00 = g(x, u),
x ∈ I 0,



 −∆u0 |
k = 1, 2, · · · , m,
x=xk = Ik (u(xk )),
(1.1)
0

R1 (u) = α1 u(0) + β1 u (0) = 0,



R2 (u) = α2 u(1) + β2 u0 (1) = 0,
here α1 , α2 , β2 ≥ 0, β1 ≤ 0, α12 + β12 > 0, α22 + β22 > 0, I = [0, 1], I 0 = I \
{x1 , x2 , · · · , xm }, 0 < x1 < x2 < · · · < xm < 1, R+ = [0, +∞), g ∈ C(I ×
0 −
0 +
R+ , R+ ), Ik ∈ C(R+ , R+ ), ∆u0 |x=xk = u0 (x+
k ) − u (xk ), u (xk ) (respectively
0
u0 (x−
k )) denotes the right limit (respectively left limit) of u (x) at x = xk ,g(x, u)
may be singular at u = 0.
In recent years, boundary problems of second-order differential equations
with impulses have been studied extensively in the literature(see for instance [19] and their references).In[1],Lin and Jiang studied the second-order impulsive
differential equation with no singularity and obtained two positive solutions
by using the fixed point index theorems in cone. However they did not consider the case when the function is singular.Motivated by the work mentioned
above,we study the positive solutions of nonlinear singular two-point boundary
value problems for second order impulsive differential equations (1.1) in this
paper.Our argument is based on the fixed point theorem in cones.
Moreover,for the simplicity in the following discussion,we introduce the
following hypotheses.
(H1 ) : There exists an ε0 > 0 such that g(x, u) and Ik (u) are nonincreasing in u ≤ ε0 ,for
each fixed x ∈ [0, 1]
(H2 ) :
For each fixed 0 < θ ≤ ε0
Z
1
α1 y − β1 α2 + β2 − α2 y
)θ)dy < ∞
α1 − β1
α 2 + β2
0
(H3 ) : φ1 (x) is the eigenfunction related to the smallest eigenvalue λ1 of
the eigenvalue
0<
g(y, (
problem −φ00 = λφ, R1 (φ) = R2 (φ) = 0.
m
P
(H4 ) :
g
∞
+
I ∞ (k)φ1 (xk )
k=1
R 1 α x−β α +β −α x
1
1 2
2
2 )φ (x)dx
1
0 ( α1 −β1
α2 +β2
< λ1 ,
39
Ying He
where g ∞ = lim sup max
u→+∞ x∈[0,1]
g(x,u)
,
u
I ∞ (k) = lim sup Iku(u) ,
u→+∞
Theorem 1.1. Assume that (H1 ) − (H4 ) are satisfied. Then problem (1.1)
has at least one positive solution u .Moreover,there exists a θ∗ > 0 such that
u(x) ≥ θ∗ (
2
α1 x − β1 α2 + β2 − α2 x
),
α1 − β1
α2 + β2
x ∈ [0, 1].
(1)
Preliminary Notes
In order to define the solution of (1.1) we shall consider the following space.
0
0 +
P C 0 (I, R) = {u ∈ C(I, R); u0 |(xk ,xk+1 ) ∈ C(xk , xk+1 ), u0 (x−
k ) = u (xk ), ∃ u (xk ), k =
1, 2, · · · , m} with the norm kukP C 0 = max{kuk, ku0 k}, here kuk = sup |u(x)|,
x∈[0,1]
ku0 k = sup |u0 (x)|. Then P C 0 (I, R) is a Banach space.
x∈[0,1]
Definition 2.1. A function u ∈ P C 0 (I, R) ∩ C 2 (I 0 , R) is a solution of (1.1)
if it satisfies the differential equation
u00 + g(x, u) = 0,
x ∈ I0
and the function u satisfies conditions ∆u0 |x=xk = −Ik (u(xk )) and R1 (u) =
R2 (u) = 0.
Let Q = I × I and Q1 = {(x, y) ∈ Q|0 ≤ x ≤ y ≤ 1}, Q2 = {(x, y) ∈
Q|0 ≤ y ≤ x ≤ 1}. Let G(x, y) is the Green’s function of the boundary value
problem
−u00 = 0, R1 (u) = R2 (u) = 0.
Following from [6], G(x, y) can be written by
(
G(x, y) :=
(α1 x−β1 )(α2 +β2 −α2 y)
,
ω
(α1 y−β1 )(α2 +β2 −α2 x)
,
ω
(x, y) ∈ Q1 ,
(x, y) ∈ Q2 .
where ω = α1 (α2 + β2 ) − β1 α2 > 0
It is easy to verify that G(x, y) has the following properties:
(i): G(x, y) ≥ 0, (x, y) ∈ [0, 1] × [0, 1]
(ii): G(x, y) ≤ G(y, y), (x, y) ∈ [0, 1] × [0, 1]
(2.1)
40
Positive solutions of singular boundary value problems...
1 α2 +β2 −α2 x
(iii): G(x, y) ≥ min{ αα11x−β
, α2 +β2 } G(y, y)
−β1
α1 x−β1 α2 +β2 −α2 x
≥ ( α1 −β1 α2 +β2 ) G(y, y)
(x, y) ∈ [0, 1] × [0, 1]
Consider the linear problem
−u00 (x) = λu(x),
R1 (u) = R2 (u) = 0.
By the theory of ordinary differential equations, we know that there exists
an eigenfunction φ1 (x) with respect to the first eigenvalue λ1 > 0 such that
φ1 (x) > 0 for x ∈ (0, 1).
Lemma 2.2. If u is a solution of the equation
Z 1
m
X
u(x) =
G(x, y)g(y, u(y))dy +
G(x, xk )Ik (u(xk )), x ∈ I.
0
(2.2)
k=1
then u is a solution of (1.1).
In fact by using inequalities (i), (ii) we have that
Z
1
kuk ≤
G(y, y)g(y, u(y))dy +
0
m
X
G(xk , xk )Ik (u(xk ))
k=1
and
α1 x − β1 α2 + β2 − α2 x
u(x) ≥ (
)
α1 − β1
α2 + β2
3
Z
1
G(y, y)g(y, u(y))dy
0
m
X
+ (
α1 x − β1 α2 + β2 − α2 x
)
G(xk , xk )Ik (u(xk ))
α1 − β1
α2 + β2
k=1
≥ (
α1 x − β1 α2 + β2 − α2 x
)kuk, x ∈ [0, 1].
α1 − β1
α2 + β2
Main Results
Lemma 3.1. Let E = (E, k · k) be a Banach space and let K ⊂ E be a cone
in E, and k · k be increasing with respect to K. Also, r, R are constants with
T
0 < r < R. Suppose that Φ : (Ω̄R \Ωr ) K → K ( ΩR = {u ∈ E, kuk < R} )
41
Ying He
is a continuous, compact map and assume that the conditions are satisfied:
T
(i) ||Φu|| > x, for u ∈ ∂Ωr K
T
(ii) u 6= µΦ(u), for µ ∈ [0, 1) and u ∈ ∂ΩR K
T
Then Φ has a fixed point in K {u ∈ E : r ≤ kuk ≤ R}.
Proof. In applications below, we take E = C(I, R) and define
K = {x ∈ C(I, R) : x(t) ≥ σkxk, t ∈ [0, 1]}.
One may readily verify that K is a cone in E. Now,let r > 0 be such that
Z 1
m
X
1
1
r < min{ε0 ,
G( , y)g(y, ε0 )dy +
G( , xk )Ik (ε0 )}
(3.1)
2
2
0
k=1
and let R > r be chosen large enough later.
T
Let us define an operator Φ : (Ω̄R \Ωr ) K → K by
Z 1
m
X
G(x, xk )Ik (u(xk )), x ∈ I.
(Φu)(x) =
G(x, y)g(y, u(y))dy +
0
k=1
T
First we show that Φ is well defined.To see this, notice that if u ∈ (Ω̄R \Ωr ) K
1 α2 +β2 −α2 x
1 α2 +β2 −α2 x
then r ≤ kuk ≤ R and u(x) ≥ ( αα11x−β
)kuk ≥ ( αα11x−β
)r, 0 ≤
−β1
α2 +β2
−β1
α2 +β2
x ≤ 1. Also notice by (H1 ) that
g(x, u(x)) ≤ g(x, (
α1 x − β1 α2 + β2 − α2 x
)r),
α1 − β1
α2 + β2
when 0 ≤ u(x) ≤ r,
and
g(x, u(x)) ≤ max max g(x, u)
when r ≤ u(x) ≤ R.
T
These inequalities with (H2 ) guarantee that Φ : (Ω̄R \Ωr ) K → K is well
defined.
T
T
Next we show that Φ : (Ω̄R \Ωr ) K → K .If u ∈ (Ω̄R \Ωr ) K, then we
have
Z 1
m
X
kΦuk ≤
G(y, y)g(y, u(y))dy +
G(xk , xk )Ik (u(xk ))
r≤u≤R 0≤x≤1
0
k=1
α1 x − β1 α2 + β2 − α2 x
(Φu)(x) ≥ (
)
α1 − β1
α2 + β2
Z
1
G(y, y)g(y, u(y))dy
0
m
X
+ (
α1 x − β1 α2 + β2 − α2 x
)
G(xk , xk )Ik (u(xk ))
α1 − β1
α2 + β2
k=1
≥ (
α1 x − β1 α2 + β2 − α2 x
)kΦuk, x ∈ [0, 1].
α1 − β1
α2 + β2
42
Positive solutions of singular boundary value problems...
T
i.e.Φu ∈ K so Φ : (Ω̄R \Ωr ) K → K.
It is clear that Φ is continuous and completely continuous.
We now show that
\
||Φu|| > kuk,
f or u ∈ ∂Ωr
K
(3.2)
T
1 α2 +β2 −α2 x
To see that,let u ∈ ∂Ωr K ,then kuk = r and u(x) ≥ ( αα11x−β
)r
−β1
α2 +β2
for x ∈ [0, 1]. So by (H1 ) and (3.1) we have
Z 1
m
X
1
1
1
(Φu)( ) =
G( , y)g(y, u(y))dy +
G( , xk )Ik (u(xk ))
2
2
2
0
k=1
Z 1
m
X
1
1
≥
G( , y)g(y, r)dy +
G( , xk )Ik (r)
2
2
0
k=1
Z 1
m
X
1
1
G( , y)g(y, ε0 )dy +
G( , xk )Ik (ε0 )
≥
2
2
0
k=1
> r = kuk.
so (3.2) is satisfied.
On the other hand, from (H4 ), there exist 0 < ε < λ1 − f ∞ and H > p
such that
Z 1
m
X
α1 x − β1 α2 + β2 − α2 x
∞
(I ∞ (k) + ε)φ1 (xk );
(λ1 − ε − g )
(
)φ1 (x)dx >
α
−
β
α
+
β
1
1
2
2
0
k=1
g(x, u) ≤ (g ∞ + ε)u, Ik (u) ≤ (I ∞ (k) + ε)u ∀ x ∈ [0, 1], u ≥ H.
m
P
Let C = max max g(x, u) +
max Ik (u), it is clear that
r≤u≤H 0≤x≤1
(3.3)
k=1 r≤u≤H
α1 x − β1 α2 + β2 − α2 x
)r) + C + (g ∞ + ε)u,
α1 − β1
α 2 + β2
α1 x − β1 α2 + β2 − α2 x
Ik (u) ≤ Ik ((
)r) + C + (I ∞ (k) + ε)u, ∀ x ∈ [0, 1], u ≥ 0.
α1 − β1
α2 + β2
T
Next we show that if R is large enough ,then µΦu 6= u for any u ∈ K ∂ΩR
T
and 0 ≤ µ < 1. If this is not true ,then there exist u0 ∈ K ∂ΩR and
0 ≤ µ0 < 1 such that µ0 Φu0 = u0 . Thus ||u0 || = R > r and u0 (x) ≥
1 α2 +β2 −α2 x
( αα11x−β
)R.Note that u0 (x) satisfies
−β1
α2 +β2
 00
u0 (x) + µ0 g(x, u0 (x)) = 0, x ∈ I 0 ,



 −∆u0 |
k = 1, 2, · · · , m,
0 x=xk = µ0 Ik (u0 (xk )),
(3.4)
0
 α1 u0 (0) + β1 u0 (0) = 0



α2 u0 (1) + β2 u00 (1) = 0.
g(x, u) ≤ g(x, (
43
Ying He
Multiply equation (3.4) by φ1 (x) and integrate from 0 to 1, using integration
by parts in the left side, notice that
Z
1
0
Z
φ1 (x)u000 (x)dx
x1
=
0
φ1 (x)du00 (x)
m−1
X
+
[φ1 (xk+1 )u00 (xk+1
− 0) −
0
Z
∆u00 (xk )φ1 (xk ) −
1
0
k=1
xk
Z
φ1 (x)du00 (x)
1
+
xm
u00 (x)φ01 (x)dx
Z
Z
+ φ1 (1)u00 (1) − φ1 (xm )u00 (xm + 0) −
= −
x1
φ1 (xk )u00 (xk
k=1
m
X
k=1
Z
= φ1 (x1 )u00 (x1 − 0) − φ1 (0)u00 (0) −
+
m−1
X Z xk+1
1
xm
xk+1
+ 0) −
xk
u00 (x)φ01 (x)dx]
u00 (x)φ01 (x)dx
φ01 (x)u00 (x)dx + φ1 (1)u00 (1) − φ1 (0)u00 (0).
Also notice that
Z
1
0
Z
φ01 (x)u00 (x)dx
1
=
0
φ01 (x)du0 (x)
Z
−
φ01 (0)u0 (0)
1
u0 (x)φ001 (x)dx
0
Z 1
0
0
= φ1 (1)u0 (1) − φ1 (0)u0 (0) + λ1
u0 (x)φ1 (x)dx.
=
φ01 (1)u0 (1)
−
0
thus,by the boundary conditions, we have
Z
1
0
φ1 (x)u000 (x)dx
= −
m
X
k=1
Z 1
− λ1
0
= −
m
X
k=1
=
m
X
k=1
∆u00 (xk )φ1 (xk ) − φ01 (1)u0 (1) + φ01 (0)u0 (0)
u0 (x)φ1 (x)dx + φ1 (1)u00 (1) − φ1 (0)u00 (0)
Z
∆u00 (xk )φ1 (xk )
1
− λ1
u0 (x)φ1 (x)dx
0
Z
1
µ0 Ik (u0 (xk ))φ1 (xk ) − λ1
u0 (x)φ1 (x)dx.
0
φ1 (x)du00 (x)
44
Positive solutions of singular boundary value problems...
So we obtain
Z 1
Z
m
X
λ1
u0 (x)φ1 (x)dx = µ0
Ik (u0 (xk ))φ1 (xk ) + µ0
0
1
g(x, u0 (x))φ1 (x)dx
0
k=1
m
m
X
X
∞
≤
(I (k) + ε)φ1 (xk )u0 (xk ) + C
φ1 (xk )
k=1
m
X
k=1
Z 1
α1 x − β1 α2 + β2 − α2 x
∞
+
Ik (
)r)φ1 (xk ) + (g + ε)
φ1 (x)u0 (x)dx
α1 − β1
α2 + β2
0
k=1
Z 1
Z 1
α1 x − β1 α2 + β2 − α2 x
)r)dx
+ C
φ1 (x)dx +
φ1 (x)g(x, (
α 1 − β1
α 2 + β2
0
0
Consequently,we obtain that
Z 1
m
X
∞
(I ∞ (k) + ε)φ1 (xk )u0 (xk )
u0 (x)φ1 (x)dx ≤
(λ1 − g − ε)
0
Z
k=1
Z 1
m
X
α1 x − β1 α2 + β2 − α2 x
φ1 (x)g(x, (
)r)dx + C(
φ1 (xk ) +
φ1 (x)dx)
α1 − β1
α2 + β2
0
k=1
1
+
0
m
X
α1 x − β1 α2 + β2 − α2 x
)r)φ1 (xk )
α1 − β1
α2 + β2
k=1
Z 1
m
X
α1 x − β1 α2 + β2 − α2 x
∞
φ1 (x)g(x, (
(I (k) + ε)φ1 (xk ) +
≤ ku0 k
)r)dx
α
α
1 − β1
2 + β2
0
k=1
Z 1
m
m
X
X
α1 x − β1 α2 + β2 − α2 x
Ik (
φ1 (xk ) +
φ1 (x)dx) +
)r)φ1 (xk )
+ C(
α
α
1 − β1
2 + β2
0
k=1
k=1
+
Ik (
We also have
Z 1
Z 1
α1 x − β1 α2 + β2 − α2 x
u0 (x)φ1 (x)dx ≥ ku0 k
(
)φ1 (x)dx
α1 − β1
α 2 + β2
0
0
Thus
R1
ku0 k ≤
0
φ1 (x)g(x,(
m
R
P
α1 x−β1 α2 +β2 −α2 x
)r)dx+C(
φ1 (xk )+ 01
α1 −β1
α2 +β2
k=1
(λ1 −g ∞ −ε)
R1
0
(
φ1 (x)dx)+
m
P
k=1
Ik (
α1 x−β1 α2 +β2 −α2 x
)r)φ1 (xk )
α1 −β1
α2 +β2
m
P
α1 x−β1 α2 +β2 −α2 x
)φ1 (x)dx−
(I ∞ (k)+ε)φ1 (xk )
α1 −β1
α2 +β2
k=1
=: R̄.
T
Let R > max{R, H}, then for any u ∈ K ∂ΩR and 0 ≤ µ < 1, we have
µΦu 6= u. Hence all the assumptions of Lemma 3.1 are satisfied,then Φ has
T
1 α2 +β2 −α2 x
)r
a fixed point u in K {u ∈ E : r ≤ ||u|| ≤ R}, u(x) ≥ ( αα11x−β
−β1
α2 +β2
∗
∀ x ∈ [0, 1]. Let θ := r, this we complete the proof of Theorem 1.
Ying He
45
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