Theoretical Mathematics & Applications, vol.4, no.2, 2014, 31-44
ISSN: 1792-9687 (print), 1792-9709 (online)
Scienpress Ltd, 2014
Multiple Positive solutions of Sturm-Liouville
problems for second order singular and
impulsive differential equations
Ying He1
Abstract
In this paper,we study the existence of multiple positive solutions of
nonlinear singular two-point boundary value problems for second-order
impulsive differential equations.The proof is based on the theory of fixed
point index in cones.
Mathematics Subject Classification: 34B15
Keywords: Multiple positive solutions; Singular two-point boundary value
problem; Second-order impulsive differential equations; Fixed point index in
cones.
1
School of Mathematics and Statistics, Northeast Petroleum University, Daqing 163318,
P.R.China.
Project supported by Heilongjiang province education department natural science
research item, China. (12541076).
E-mail: heying65338406@163.com
Article Info: Received : April 20, 2014. Revised : June 4, 2014.
Published online : June 30, 2014.
32
1
Multiple Positive solutions of Sturm-Liouville problems...
Introduction
Impulsive differential equations play a very important role in modern applied mathematics due to their deep physical background and broad application. In this paper,we consider the existence of multiple positive solutions of
two-point boundary value problems for nonlinear second-order singular and
impulsive differential equations:


−Lu = h(x)g(x, u),
x ∈ I 0,



0


 −∆(pu )|x=xk = Ik (u(xk )), k = 1, 2, · · · , m,
(1.1)
∆(pu)|x=xk = I k (u(xk )), k = 1, 2, · · · , m,


0

R1 (u) = αu(0) − βu (0) = 0,



 R (u) = γu(1) + δu0 (1) = 0,
2
here Lu = (p(x)u0 )0 + q(x)u is sturm-liouville operator, I = [0, 1], I 0 =
I \ {x1 , x2 , · · · , xm } and 0 < x1 < x2 < · · · < xm < 1 are given, R+ =
[0, +∞), g ∈ C(I ×R+ , R+ ), Ik , I k ∈ C(R+ , R+ ), ∆(pu0 )|x=xk = p(xk )u0 (x+
k )−
+
−
+
0 −
0 +
0 −
p(xk )u (xk ), ∆(pu)|x=xk = p(xk )u(xk )−p(xk )u(xk ) u (xk ), u(xk ) ( u (xk ), u(x−
k ))
0
denote the right limit ( left limit) of u (x) and u(x) at x = xk respectively,h(x) ∈
C(I, R+ ) and may be singular at x = 0 or x = 1.
Throughout this paper, we always suppose that
(S1 ) p(x) ∈ C 1 ([0, 1], R), p(x) > 0, q(x) ∈ C([0, 1], R), q(x) ≤ 0, α, β, γ, δ ≥
0, ρ = βγ + αγ + αδ > 0.
It is well known that there are abundant results about the existence of
positive solutions of boundary value problems for second order impulsive differential equations. Some works can be found in [1, 3, 6 − 9] and references
therein.They,mainly investigated the case p(x) = 1 and q(x) = 0. In this paper ,we will consider the case p(x) 6= 1,and q(x) 6= 0. Here we also mention
that second order dynamic inclusions on time scales with impulses has been
studied in [2] .We obtain the existence results of positive solutions,by means
of the fixed point index theorem in cones under some conditions on g(x, u)
concerning the first eigenvalue corresponding to the relevant linear operator.
To conclude the introduction,we introduce the following notation:
Ik (u)
I k (u)
g(x, u)
, I0 (k) = lim inf
, I 0 (k) = lim inf
+
+
x∈[a,b]
u→0
u→0
u
u
u
g0 = lim inf
min
+
u→0
Ik (u)
g(x, u)
I k (u)
, I∞ (k) = lim inf
, I ∞ (k) = lim inf
;
u→+∞
u→+∞
x∈[a,b]
u
u
u
g∞ = lim inf min
u→+∞
33
Ying He
Moreover,for the simplicity in the following discussion,we introduce the
following hypotheses.
(H1 ) :
m
P
σ (I0 (k)φ1 (xk ) + I 0 (k)φ01 (xk ))
g0 + k=1
> λ1 ,
Rb
φ (x)h(x)dx
a 1
m
P
σ (I∞ (k)φ1 (xk ) + I ∞ (k)φ01 (xk ))
g∞ + k=1
> λ1 .
Rb
φ
(x)h(x)dx
1
a
here σ = min{ m(a)
, n(b) } (see section 2), and φ1 (x) is the eigenfunction related
m(1) n(0)
to the smallest eigenvalue λ1 of the eigenvalue problem −Lφ = λφh, R1 (φ) =
R2 (φ) = 0.
(H2 ) :
There is a p > 0 such that 0 ≤ u ≤ p and 0 ≤ x ≤ 1 implies
g(x, u) ≤ ηp, Ik (u) ≤ ηk p, I¯k (u) ≤ η̄k p
here η, ηk , η̄k ≥ 0, η +
m
P
(ηk + η̄k ) > 0, η
k=1
R1
0
G(y, y)h(y)dy +
m
P
G(xk , xk )(ηk +
k=1
η̄k ) < 1 and G(x, y) is the Green’s function of boundary value problem −Lu =
0, R1 (u) = R2 (u) = 0 (see section 2).
R0
(H3 ) :
0 < 1 G(y, y)h(y)dy < +∞
2
Preliminary Notes
In this paper, we shall consider the following space
+
P C(I, R) = {u ∈ C(I, R); u|(xk ,xk+1 ) ∈ C(xk , xk+1 ), u(x−
k ) = u(xk ), ∃ u(xk ),
k = 1, 2, · · · , m} P C 0 (I, R) = {u ∈ C(I, R); u0 |(xk ,xk+1 ) ∈ C(xk , xk+1 ), u0 (x−
k) =
0
0 +
u (xk ), ∃ u (xk ), k = 1, 2, · · · , m} with the norm kukP C = sup |u(x)|,
x∈[0,1]
kukP C 0 = max{kukP C , ku0 kP C }, Then P C(I, R),P C 0 (I, R) are Banach spaces.
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
Lu + h(x)g(x, u) = 0,
x ∈ I0
and the function u satisfies conditions ∆(pu0 )|x=xk = −Ik (u(xk )), ∆(pu)|x=xk =
I k (u(xk )) and R1 (u) = R2 (u) = 0.
34
Multiple Positive solutions of Sturm-Liouville problems...
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
−Lu = 0, R1 (u) = R2 (u) = 0.
Following from [4], G(x, y) can be written by
(
m(x)n(y)
, (x, y) ∈ Q1 ,
ω
G(x, y) :=
m(y)n(x)
, (x, y) ∈ Q2 .
ω
(2.1)
Lemma 2.2. Suppose that (S1 ) holds, then the Green’s function G(x, y),
defined by (2.1), possesses the following properties:
(i): m(x) ∈ C 2 (I, R) is increasing and m(x) > 0, x ∈ (0, 1].
(ii): n(x) ∈ C 2 (I, R) is decreasing and n(x) > 0, x ∈ [0, 1).
(iii): (Lm)(x) ≡ 0, m(0) = β, m0 (0) = α.
(iv): (Ln)(x) ≡ 0, n(1) = δ, n0 (1) = −γ.
(v): ω is a positive constant. Moreover, p(x)(m0 (x)n(x) − m(x)n0 (x)) ≡ ω.
(vi): G(x, y) is continuous and symmetrical over Q.
(vii): G(x, y) has continuously partial derivative over Q1 , Q2 .
(viii): For each fixed y ∈ I, G(x, y) satisfies LG(x, y) = 0 for x 6= y, x ∈ I.
Moreover,
R1 (G) = R2 (G) = 0 for y ∈ (0, 1).
0
(viiii): Gx has discontinuous point of the first kind at x = y and
G0x (y + 0, y) − G0x (y − 0, y) = −
1
, y ∈ (0, 1).
p(y)
Following from Lemma2.2, it is easy to see that:
G(x, y) ≤ G(y, y) =
m(y)n(y)
, x, y ∈ [0, 1]
ω
G(x, y) ≥ σG(y, y), x ∈ [a, b], y ∈ [0, 1],
(2.2)
where a ∈ (0, t1 ], b ∈ [tm , 1), 0 < σ = min{ m(a)
, n(b) } < 1
m(1) n(0)
Consider the linear Sturm-Liouvile problem
−(Lu)(x) = λu(x)h(x),
R1 (u) = R2 (u) = 0.
By the Sturm-Liouvile 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).
35
Ying He
Let E be a Banach space and K ⊂ E be a closed convex cone in E. For
r > 0, let Kr = {u ∈ K : ||u|| < r} and ∂Kr = {u ∈ K : ||u|| = r}. The
following two Lemmas are needed in our argument.
Lemma 2.3. Let Φ : K → K be a continuous and completely continuous
mapping and Φu 6= u for u ∈ ∂Kr . Thus the following conclusions hold:
(i) If ||u|| ≤ kΦuk for u ∈ ∂Kr , then i(Φ, Kr , K) = 0;
(ii) If kuk ≥ kΦuk for u ∈ ∂Kr , then i(Φ, Kr , K) = 1.
Lemma 2.4. Let Φ : K → K be a continuous and completely continuous
mapping. Suppose that the following two conditions are satisfied:
(i) inf ||Φu|| > 0;
u∈∂Kr
(ii) µΦu 6= u for every u ∈ ∂Kr and µ ≥ 1.
Then, i(Φ, Kr , K) = 0.
In applications below, we take E = C(I, R) and define
K = {u ∈ C(I, R) : u(x) ≥ σkuk, x ∈ [a, b]}.
One may readily verify that K is a cone in E.
Define an operator Φ : K → K by
Z
1
(Φu)(x) =
G(x, y)h(y)g(y, u(y))dy+
0
X
G(x, xk )(Ik (u(xk ))+I k (u(xk ))), x ∈ I.
0<xk <x
It follows form (H3 ) that φ is well defined.
Lemma 2.5. If (H3 ) is satisfied,then Φ : K → K is continuous and completely continuous, Moreover, Φ(K) ⊂ K.
Proof By the property of continuous of g(x, u), Ik (x), I k (x), it is easy to see
that Φ : K → K is continuous and completely continuous. Thus we only need
to show Φ(K) ⊂ K. In fact, for u ∈ K, by using inequalities (2.2) and (H3 ),
we have that
Z
1
kΦuk ≤
G(y, y)h(y)g(y, u(y))dy+
0
X
0<xk <x
G(xk , xk )(Ik (u(xk ))+I¯k (u(xk ))) < +∞
36
Multiple Positive solutions of Sturm-Liouville problems...
On the other hand ,for any x ∈ [a, b],by (2.2) we obtain
Z 1
X
(Φu)(x) =
G(x, y)h(y)g(y, u(y))dy +
G(x, xk )(Ik (u(xk )) + I k (u(xk )))
Z
0
0<xk <x
X
a
≥
G(x, y)h(y)g(y, u(y))dy +
b
G(x, xk )(Ik (u(xk )) + I k (u(xk )))
0<xk <x
ÃZ
0
≥ σ
G(y, y)h(y)g(y, u(y))dy +
1
X
!
G(xk , xk )(Ik (u(xk )) + I k (u(xk )))
0<xk <x
≥ σkΦuk
Thus, Φ(K) ⊂ K.
Lemma 2.6. If u is a fixed point of the operator Φ, then u is a solution of
problem (1.1).
3
Main Results
Lemma 3.1. If (H2 ) is satisfied, then i(Φ, Kp , K) = 1.
Proof Let u ∈ K with kuk = p. It follows from (H2 ) that
Z
1
kΦuk ≤
G(y, y)h(y)g(y, u(y))dy +
0
m
X
G(xk , xk )(Ik (u(xk )) + I k (u(xk )))
k=1
Z
1
≤ p[η
G(y, y)h(y)dy +
0
m
X
G(xk , xk )(ηk + η k )] < p = kuk.
k=1
Thus
kΦuk < kuk,
∀ u ∈ ∂Kp .
It is obvious that Φu 6= u for u ∈ ∂Kp . Therefore, i(Φ, Kp , K) = 1, here we
use Lemma2.3.
Theorem 3.2. Assume that (H1 ) − (H3 ) are satisfied. Then problem (1.1)
has at least two positive solutions u1 and u2 with
0 < ||u1 || < p < ||u2 ||.
37
Ying He
Proof According to Lemma 3.1, we have that
i(Φ, Kp , K) = 1.
(3.1)
Since (H1 ) holds, then there exists 0 < ε < 1 such that
m
P
σ
(I0 (k)φ1 (xk ) + I 0 (k)φ01 (xk ))
] > λ1 ,
Rb
φ
(x)h(x)dx
1
a
k=1
(1 − ε)[g0 +
σ
m
P
(I∞ (k)φ1 (xk ) + I ∞ (k)φ01 (xk ))
] > λ1 .
Rb
φ (x)h(x)dx
a 1
k=1
(1 − ε)[g∞ +
(3.2)
By the definitions of g0 , I0 , one can find 0 < r0 < p such that
g(x, u) ≥ g0 (1−ε)u, Ik (u) ≥ I0 (k)(1−ε)u, I k (u) ≥ I 0 (k)(1−ε)u ∀ x ∈ [a, b], 0 < u < r0 .
Let r ∈ (0, r0 ), then for u ∈ ∂Kr , we have
u(x) ≥ σkuk = σr > 0. x ∈ [a, b]
Thus
1
(Φu)( ) =
2
Z
1
0
Z
0<xk < 2
b
≥
a
X
1
1
G( , y)h(y)g(y, u(y))dy +
G( , xk )(Ik (u(xk )) + I k (u(xk )))
2
2
1
X
1
1
G( , xk )(Ik (u(xk )) + I k (u(xk )))
G( , y)h(y)g(y, u(y))dy +
2
2
1
0<xk < 2
Z
b
≥ g0 (1 − ε)
a
+ (1 − ε)
X
X
1
1
G( , y)h(y)u(y)dy + (1 − ε)
G( , xk )I0 (k)u(xk )
2
2
1
0<xk < 2
1
G( , xk )I 0 (k)u(xk )
2
1
0<xk < 2


Z
b
≥ (1 − ε)σr g0
a
X
1
1
G( , y)h(x)dy +
G( , xk )(I0 (k) + I 0 (k)) > 0
2
2
1
0<xk < 2
from which we see that inf ||Φu|| > 0, namely, hypothesis (i) of Lemma 2.4
u∈∂Kr
holds. Next we show that µΦu 6= u for any u ∈ ∂Kr and µ ≥ 1.
38
Multiple Positive solutions of Sturm-Liouville problems...
If this is not true, then there exist u0 ∈ ∂Kr and µ0 ≥ 1 such that µ0 Φu0 =
u0 . Note that u0 (x) satisfies


Lu0 + µ0 h(x)g(x, u0 (x)) = 0,
x ∈ I 0,



0


 −∆(pu0 )|x=xk = µ0 Ik (u0 (xk )), k = 1, 2, · · · , m,
(3.3)
∆(pu0 )|x=xk = µ0 I k (u0 (xk )), k = 1, 2, · · · , m,



αu0 (0) − βu00 (0) = 0,



 γu (1) + δu0 (1) = 0.
0
0
Multiply equation (3.3) by φ1 (x) and integrate from a to b, note that
Z b
Z x1
0
0
φ1 (x)[(p(x)u0 (x)) + q(x)u0 (x)]dx =
φ1 (x)[(p(x)u00 (x))0 + q(x)u0 (x)]dx
+
a
m−1
X Z xk+1
xk
k=1
b
Z
+
=
a
φ1 (x)[(p(x)u00 (x))0 + q(x)u0 (x)]dx
φ1 (x)[(p(x)u00 (x))0 + q(x)u0 (x)]dx
xm
Z x1
Z
0
0
0
φ1 (x1 )p(x1 )u0 (x1 − 0) −
p(x)u0 (x)φ1 (x)dx +
a
+
[φ1 (xk+1 )p(xk+1 )u00 (xk+1 − 0) − φ1 (xk )p(xk )u00 (xk + 0)
−
xk
=
Z
p(x)u00 (x)φ01 (x)dx
φ1 (xm )p(xm )u00 (xm
−
m
X
xk+1
+
q(x)u0 (x)φ1 (x)dx]
xk
Z b
+ 0) −
xm
∆(p(xk )u00 (xk ))φ1 (xk )
Z
p(x)u00 (x)φ01 (x)dx
Z
.
Also note that
Z b
Z
0
0
p(x)φ1 (x)u0 (x)dx =
a
Z
+
=
b
a
x1
q(x)u0 (x)φ1 (x)dx
xm
Z
p(x)φ01 (x)u00 (x)dx
p(x)φ01 (x)du0 (x)
a
b
+
+
b
+
q(x)φ1 (x)u0 (x)dx
a
m−1
X Z xk+1
k=1
xk
p(x)φ01 (x)du0 (x)
p(x)φ01 (x)du0 (x)
xm
m
X
−
−
b
−
k=1
=
q(x)u0 (x)φ1 (x)dx
a
m−1
X
Zk=1xk+1
−
x1
k=1
m
X
k=1
Z
∆(p(xk )u0 (xk ))φ01 (xk )
b
−
a
Z
∆(p((xk )u0 (xk ))φ01 (xk )
u0 (x)(p(x)φ01 (x))0 dx
Z
b
+
b
u0 (x)q(x)φ1 (x)dx + λ1
a
h(x)φ1 (x)u0 (x)dx
a
39
Ying He
Z
b
a
+
m
X
φ1 (x)[(p(x)u00 (x))0 + q(x)u0 (x)]dx = −
=
b
h(x)φ1 (x)u0 (x)dx
a
k=1
m
X
∆(p(xk )u00 (xk ))φ1 (xk )
k=1
Z
∆(p(xk )u0 (xk ))φ01 (xk ) − λ1
m
X
Z
I k (u0 (xk ))φ01 (xk ))
µ0 (Ik (u0 (xk ))φ1 (xk ) +
=
a
m
X
u0 (x)h(x)φ1 (x)dx
a
k=1
So we obtain
Z b
λ1
u0 (x)h(x)φ1 (x)dx
b
− λ1
µ0 (Ik (u0 (xk ))φ1 (xk ) + I k (u0 (xk ))φ01 (xk ))
k=1
Z
b
+ µ0
φ1 (x)h(x)g(x, u0 (x))dx
a
≥
(1 − ε)
m
X
u0 (xk )(I0 (k)φ1 (xk ) + I 0 (k)φ01 (xk ))
k=1
Z
b
φ1 (x)u0 (x)h(x)dx
+ (1 − ε)g0
a
Since u0 (x) ≥ σ||u0 || = σr, we have
Rb
a
φ1 (x)u0 (x)h(x)dx > 0 and
m
P
u0 (xk )(I0 (k)φ1 (xk )+
k=1
I 0 (k)φ01 (xk )) > 0. So from the above inequality we see that λ1 > (1 − ε)g0 .
Thus
Z b
m
X
[λ1 − (1 − ε)g0 ]
u0 (x)h(x)φ1 (x)dx ≥ (1 − ε)
(I0 (k)φ1 (xk ) + I 0 (k)φ01 (xk ))u0 (xk )
a
k=1
≥ (1 − ε)σr
m
X
(I0 (k)φ1 (xk ) + I 0 (k)φ01 (xk )).
k=1
Rb
Rb
Since a u0 (x)h(x)φ1 (x)dx ≤ r a φ1 (x)h(x)dx, we have
Z
b
[λ1 − (1 − ε)g0 ]
h(x)φ1 (x)dx ≥ (1 − ε)σ
a
m
X
(I0 (k)φ1 (xk ) + I 0 (k)φ01 (xk )),
k=1
which contradicts (3.2) again. Hence Φ satisfies the hypotheses of Lemma 2.4
in Kr . Thus
i(Φ, Kr , K) = 0.
(3.4)
40
Multiple Positive solutions of Sturm-Liouville problems...
On the other hand, from (H1 ), there exists H > p such that for any x ∈
[a, b], u ≥ H.
g(x, u) ≥ g∞ (1 − ε)u, Ik (u) ≥ I∞ (k)(1 − ε)u, I k (u) ≥ I ∞ (k)(1 − ε)u, (3.5)
Let
m
P
max |Ik (u) − I∞ (k)(1 − ε)u| +
C = max max |g(x, u) − g∞ (1 − ε)u| +
k=1 0≤u≤H
0≤u≤H a≤x≤b
m
P
max |I k (u) − I ∞ (k)(1 − ε)u|. It is clear that for any x ∈ [a, b], u ≥ 0.
k=1 0≤u≤H
g(x, u) ≥ g∞ (1−ε)u−C, Ik (u) ≥ I∞ (k)(1−ε)u−C, I k (u) ≥ I ∞ (k)(1−ε)u−C,
(3.6)
H
Choose R > R0 := max{ σ , p} and let u ∈ ∂KR . Since u(x) ≥ σ||u|| = σR >
H for x ∈ [a, b], from (3.5) we see that
g(x, u(x)) ≥ g∞ (1 − ε)u(x) ≥ σg∞ (1 − ε)R, ∀ x ∈ [a, b].
Ik (u(xk ) ≥ σI∞ (k)(1 − ε)R, I k (u(xk ) ≥ σI ∞ (k)(1 − ε)R.
Essentially the same reasoning as above yields inf ||Φu|| > 0. Next we show
u∈∂KR
that if R is large enough, then µΦu 6= u for any u ∈ ∂KR and µ ≥ 1. In fact,
if there exist u0 ∈ ∂KR and µ0 ≥ 1 such that µ0 Φu0 = u0 , then u0 (x) satisfies
equation (3.3) .
Multiply equation (3.3) by φ1 (x) and integrate from a to b, using integration by parts in the left side to obtain
Z b
m
X
µ0 (Ik (u0 (xk ))φ1 (xk ) + I k (u0 (xk ))φ01 (xk ))
λ1
u0 (x)h(x)φ1 (x)dx =
a
Z
k=1
b
+ µ0
φ1 (x)h(x)g(x, u0 (x))dx
a
≥
Z b
m
X
0
(1 − ε)
(I∞ (k)φ1 (xk ) + I ∞ (k)φ1 (xk ))u0 (xk ) + (1 − ε)g∞
u0 (x)φ1 (x)h(x)dx
à m k=1
!
Z b
X
0
φ1 (x)h(x)dx .
− C
(φ1 (xk ) + φ1 (xk )) +
a
a
k=1
If g∞ ≤ λ1 , then we have
à m
!
Z b
Z b
X
[λ1 − (1 − ε)g∞ ]
u0 (x)h(x)φ1 (x)dx + C
φ1 (x)h(x)dx
(φ1 (xk ) + φ01 (xk )) +
a
≥ (1 − ε)
m
X
k=1
k=1
(I∞ (k)φ1 (xk ) + I ∞ (k)φ01 (xk ))u0 (xk ).
a
41
Ying He
thus
Z
à m
X
b
ku0 k[λ1 − (1 − ε)g∞ ]
h(x)φ1 (x)dx + C
a
≥ (1 − ε)σku0 k
Z
(φ1 (xk ) + φ01 (xk )) +
!
b
φ1 (x)h(x)dx
a
k=1
m
X
(I∞ (k)φ1 (xk ) + I ∞ (k)φ01 (xk )).
k=1
and
µ
C
ku0 k ≤
m
P
(1 − ε)σ
m
P
(φ1 (xk ) +
k=1
φ01 (xk ))
+
Rb
a
¶
φ1 (x)h(x)dx
(I∞ (k)φ1 (xk ) + I ∞ (k)φ01 (xk )) − [λ1 − (1 − ε)g∞ ]
k=1
Rb
a
=: R̄.
φ1 (x)h(x)dx
(3.7a )
If g∞ > λ1 , we can choose ε > 0 such that (1 − ε)g∞ > λ1 , then we have
!
à m
Z b
Z b
X
0
(φ1 (xk ) + φ1 (xk )) +
φ1 (x)h(x)dx
≥ [(1 − ε)g∞ − λ1 ]
φ1 (x)u0 (x)h(x)dx
C
a
k=1
a
Z
b
φ1 (x)h(x)dx.
≥ [(1 − ε)g∞ − λ1 ]σku0 k
a
Thus
µ
C
ku0 k ≤
m
P
(φ1 (xk ) +
k=1
φ01 (xk ))
[(1 − ε)g∞ − λ1 ]σ
+
Rb
a
Rb
a
¶
φ1 (x)h(x)dx
φ1 (x)h(x)dx
=: R̄.
(3.7b )
Let R > max{p, R̄}, then for any u ∈ ∂KR and µ ≥ 1, we have µΦu 6= u.
Hence hypothesis (ii) of Lemma 2.4 is satisfied and
i(Φ, KR , K) = 0.
(3.8)
In view of (3.1), (3.4) and (3.8), we obtain
i(Φ, KR \ K̄p , K) = −1, i(Φ, Kp \ K̄r , K) = 1.
Then Φ has fixed points u1 and u2 in Kp \ K̄r and KR \ K̄p , respectively, which
means u1 (x) and u2 (x) are positive solution of the problem (1.1) and 0 <
ku1 k < p < ku2 k.
42
Multiple Positive solutions of Sturm-Liouville problems...
Corollary 3.3. The conclusion of Theorem 3.2 is valid if (H1 ) is replaced
by
(H1∗ )
g0 = ∞ or
m
X
I0 (k)φ1 (xk ) = ∞ or
m
X
k=1
I 0 (k)φ01 (xk ) = ∞;
k=1
and
g∞ = ∞ or
m
X
I∞ (k)φ1 (xk ) = ∞ or
k=1
m
X
I ∞ (k)φ01 (xk ) = ∞.
k=1
Example 3.4.

1

Lu + 21 (u 3 + u3 ) = 0, x ∈ I 0 ,



0


 −∆(pu )|x=xk = ck u(xk ), ck ≥ 0,
∆(pu)|x=xk = dk u(xk ), dk ≥ 0,



R1 (u) = αu(0) − βu0 (0) = 0,



 R (u) = γu(1) + δu0 (1) = 0,
2
(3.9)
here Lu = (p(x)u0 )0 + q(x)u. Assume that (S1 ) is satisfied. Then problem (3.9)
has at least two positive solutions u1 and u2 with
0 < ||u1 || < 1 < ||u2 ||
provided
1
1<
d
Ã
1−
m
X
!
Z
1
G(xk , xk )(ck + dk ) , d =
G(y, y)dy.
0
k=1
Proof To see this we will apply Theorem 3.2 (or Corollary 3.3).
By (3.10), η > 0 is chosen such that
m
X
1
G(xk , xk )(ck + dk )).
1 < η < (1 −
d
k=1
Set
1 1
g(x, u) = (u 3 + u3 ).
2
Note
g0 = ∞,
so (H1 ) (or (H1∗ )) holds.
g∞ = ∞,
(3.10)
43
Ying He
Let ηk = ck , η̄k = dk , then η, ηk , η̄k satisfy
Z
1
η
G(y, y)dy +
0
m
X
G(xk , xk )(ηk + η k ) < 1.
k=1
Let p = 1, then for 0 ≤ u ≤ p, we have
1 1
1 1
g(x, u) = (u 3 + u3 ) ≤ + < ηp = η,
2
2 2
and
Ik (u) = ck u = ηk u ≤ ηk p,
I k (u) = dk u = η k u ≤ η k p
thus (H2 ) holds. The result now follows from Theorem 3.2 (or Corollary3.3).
References
[1] R.P.Agarwal , D. O’Regan, Multiple nonnegative solutions for second
order impulsive differential equations. Appl. Math. Comput., 114, (2000),
51-59.
[2] W. Ding , M. Han, Periodic boundary value problem for the second order
impulsive functional differential equations. Appl. Math. Comput, 155 A,
(2004), 709-726.
[3] D. Guo and V. Lakshmikantham, Nonlinear Problems in Abstract Cones,
Academic Press, Boston, MA, 1988.
[4] X. Liu, D. Q. Jiang, Z. Wei, Periodic boundary value problems for second order impulsive integrodifferential equations of mixed type in Banach
spaces.J. Math. Anal. Appl., 195, (1995), 214-229.
[5] S. G. Hristova , D. D. Bainov, Monotone-iterative techniques of V.
Lakshmikantham for a boundary value problem for systems of impulsive
differential-difference equations. J. Math. Anal. Appl., 1997, (1996), 1-13.
[6] V. Lakshmik ntham, D.D. Bainov and P.S. Simeonov, Theory of Impulsive
Differential Equations. World Scientific, Singapore, 1989.
44
Multiple Positive solutions of Sturm-Liouville problems...
[7] E. Lee, Y. Lee, Multiple positive solutions of singular two point boundary value problems for second order impulsive differential equation, Appl.
Math. Comput., 158, ( 2004), 745-759
[8] X. Liu, D. Q. Jiang, Multiple positive solutions of Dirichlet boundary
value problems for second order impulsive differential equations. J. Math.
Anal. Appl., 321, (2006), 501-514.
[9] X. Liu, D. Q. Jiang, Z. Wei, Periodic boundary value problems for second order impulsive integrodifferential equations of mixed type in Banach
spaces.J. Math. Anal. Appl., 195, (1995), 214-229.
[10] I. Rachunkova, J. Tomecek, Impulsive BVPs with nonlinear boundary
conditions for the second order differential equations without growth restrictions. J. Math. Anal. Appl., 292, (2004), 525-539.