 Open Access
 Authors : Qiaoluan Li , Shuang Zhang , Lina Zhou , Weihua Jiang
 Paper ID : IJERTV8IS100305
 Volume & Issue : Volume 08, Issue 10 (October 2019)
 Published (First Online): 04112019
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Existence of Solutions for SturmLiouville Boundary Value Problems with Impulses
Qiaoluan Li, Shuang Zhang, Lina Zhou College of Mathematics & Information Science Hebei Normal University,
Shijiazhuang, 050024, China
Weihua Jiang (corresponding author) * School of Science,
Hebei University of Science and Technology, Shijiazhuang, 050018,China
AbstractIn the paper, we investigate the SturmLiouville boundary value problem. By using fixed point methods, we establish sufficient conditions to guarantee the existence of solutions. At the end of the paper, two examples are given to illustrate our main results.
results, two examples are given in Section 4.

PRELIMINARIES
In order to prove our Theorems, we need the following definition and Lemmas.
KeywordsSturmLiouville; Impulses; Existence
Let PC1 (J, R) {u : J R,u  ,u ' 
(t t ) (t t )
(t t ) (t t )
k , k1 k , k1
C(tk ,tk 1 ),
1. INTRODUCTION
Consider the impulsive differential equation with
u(t ) u(t ), u '(t ),u(t ),u '(t )} with the norm
k k k k k
k k k k k
 u  max{sup  u(t) , sup  u '(t) } . A function u is called
SturmLiouville boundary conditions
t[0,1]
t[0,1]
( p(t)u '(t)) ' f (t, u,Tu, Su) 0, t J , t t j
j j j j
j j j j
p(t )u(t ) I (u(t )),
a solution of Eq.(1.1) if u PC1 (J , R) satisfies Eq.(1.1).
It is easy to know that u is the solution of Eq.(1.1) if and only if u satisfies the integral equation
p(t )u '(t ) J (u(t
)),
j=1,2, n, 1 n
j j j j
u(t) 0 G(t, s) f (s,u,Tu, Su)ds G(t,tj )J j (u(t j ))
u(0) p(0)u(0) 0,
u(1) p(1)u(1) 0. (1.1)
n
n
j 1
j 1
H (t, t j )I j (u(t j )) ,
Where
p C[0,1], p(t) 0, J [0,1],
Where
t 1 ( 1 d )( s d ), 0 s t 1,
Tu 0 k (t, s)u(s)ds, Su 0 k1 (t, s)u(s)ds,
k 0 ,
1
G(t, s)
t p( )
0 p( )
k1 0 , , , , 0 ,
1 d 0 ,
0 p( )
(
t d )(
0 p( )
1 d ), 0 t s 1,
s p( )
f , I , J are continuous, z(t ) z(t ) z(t ) .
(2.1)
j j j j j
Recently, the research of impulsive initial and boundary value problems is extensive and there is increasing interest
and
( 1 d ), 0 s t 1,
on the existence of impulsive differential equations.
1 p(s)
H (t, s)
t p( )
Numerous papers have been published on this class of
( t d ), 0 t s 1.
equations and good results were obtained [1,3,6,7,10]. For
p(s)
0 p( )
instance, in 2008, Kaufmann [4] studied a secondorder
nonlinear differential equation subject to SturmLiouville type boundary conditions and impulsive conditions. The
Further by [11] we know that
k (t)k (s) G(t, s) 1 k (s) (or k (t)),
authors used Krasnoselskii's fixed point theorem to obtain 2 2
2 2
the existence of solutions. In 2012, Wang [8] studied
where k (t) (
t d
)(
1 d ),
impulsive fractional differential equation. Some sufficient 2
conditions for existence of the solutions are obtained by
using fixed point methods.
0 p( )
t p( )
.
1 1
1 1
( d )( d )
This paper is organized as follows. In Section 2, we shall offer some basic definitions, preliminary results and
0 p( )
0 p( )
Lemmas.
In section 3, we prove the main results. To illustrate our
The following Lemmas are needed.
LEMMA 1.([2]) Let X be a Banach space with X be
closed and convex. Assume U is a relatively open subset of
with 0U, and let S :U be a compact, continuous
( p(t)u '(t)) ' f (t, (u v)* ,T (u v)* , S(u v)* )
maps. Then either

S has a fixed point in U , or
g(t) 0,
t J ,
t t j
p(t )u '(t ) J (u(t ) v(t )) a ,
j=1,2, n,

there exists u U and v (0,1) with u vSu .
j j j j j j
LEMMA 2. ([5]) Let M be a closed convex and nonempty subset of a Banach space X . Let A, B be two operators such that

Ax By M whenever x, y M.
u(0) p(0)u(0) 0,
u(1) p(1)u(1) 0. (3.2)
where

A is a compact and continuous.

B is a contraction mapping.
x* (t) x(t),
0,
0,
x(t) 0,
x(t) 0.
Then there exits a z M such that z Az Bz . LEMMA 3.([9]) Let X be a Banach space and W PC(J , X ) . If the following conditions are satisfied:

W is uniformly bounded subset of PC(J , X ) ,


(ii) W is equicontinuous in (tk ,tk 1 ), k 0,1, 2 m where
In the following Theorem 1, we shall prove (3.2) has a solution u(t) v(t) and u(t) v(t) is a nonnegative solution
of Eq.(1.1).
THEOREM 1. Assume (H1) hold and I j 0,

J (x) M . There exist R , a 0, j 1, 2, , n such that
t0 0, tm1 1,
j 2 0 j
1
(iii) W(t)={u(t):
u W , t J \{t1 ,…,tm
}},
0 G(t, s) f (s, u,Tu, Su)ds 0, for
t [0,1], 0 u R0 , and
W (tk ) {u(tk ) : u W} and W (tk ) {u(tk ) : u W}
are relatively compact set of X . Then W is a relatively compact subset of PC(J , X ) .
1
0 2 0
0 2 0
R k (s) max ( f (s, z,Tz, Sz) g(s))ds
0s1,0 zR0
n n
n n
M 2 k2 (t j ) k2 (t j )aj .
(3.3)
LEMMA 4.([8]) Let X be a Banach spaces and F : X X
j 1
j 1
be a completely continuous operator. If the set
E(F) {y X : y Fy for some [0,1]}
Then Eq. (1.1) has at least one positive solution.
1 * * *
1 * * *
PROOF. Let
is bounded, then F has at least a fixed point.
Fu 0 G(t, s)( f (s,(u v) ,T (u v) , S (u v) ) g(s))ds


MAIN RESULTS
n n
n n
G(t, t j )J j (u(t j ) v(t j )) G(t, t j )aj .
2
2
We make the following assumptions:
j 1
j 1
(H1) There exists a positive function
g C[0,1] such
We define P {x PC1 (J , R) : x(t) k (t)  x }.
that
f (t,u,Tu, Su) g(t), t (0,1), u [0, ) .
It is easy to know Fu 0 for u P
and
(H2)  f (t,u, v, w) f (t,u, v , w)  m(t)( u u   v v 
Fu k (t)( 1 k (s)( f (s,(u v)* ,T (u v)* , S (u v)* )
 w w ), where m(t) C[0,1] .
(H3) There exist L , L , M , M such that
2 0 2
n n
n n
g(s))ds k2 (t j )J j (u(t j ) v(t j )) k2 (t j )aj ) .
1 2 1 2
 Ik (u) Ik (v)  L1  u v ,  Ik (u)  M1 ,
Furthermore
j 1
j 1
 Jk (u) Jk (v)  L2  u v ,  Jk (u)  M2 .
(H4) There exists a constant L such that
Fu 1 ( 1 k (s)( f (s,(u v)* ,T (u v)* , S(u v)* g(s))ds)
n n
n n
2
2
0
3
 f (t,u,Tu Su)  L3 (1  u   Tu   Su ) .
k2 (t j )J j (u(t j ) v(t j )) k2 (t j )aj ),
Assume v(t) is the solution of the following equation
j 1
j 1
( p(t)v '(t)) ' g(t) 0, t J , t t
then
Fu k2 (t)  Fu  . Hence F (P) P .
j
j
Since f and J j
are continuous, we get F is continuous.
p(tj )v '(t j ) a j ,
v(0) p(0)v '(0) 0,
j=1,2, n,
We will prove F is uniformly bounded.
Let D P be bounded, i.e. there exists
L 0
such that
v(1) p(1)v '(1) 0, (3.1)
 u  L for u D .
then v satisfies
Let A
max  f (t, y,Ty, Sy) g(t)  ,
t[0,1], y[0, L]
1 n M
max
G(t, s) ,
v(t) G(t, s)g(s)ds G(t,tj )aj ,
(t ,s)[0,1][0,1]
0 j 1 1 1
* * *
* * *
where a j
will be defined in Theorem 1, G t, s is defined as
 Fu 
0 k2 (s)( f (s,(u v) ,T (u v) , S (u v) )
(2.1).
Consider
g(s))ds M 2
n n
n n
G(t,t j ) G(t,t j )aj
j 1 j 1
A 1 n
 f (s, 0, 0, 0)  ds nM } .
1
1
0 k2 (s)ds nMM 2 M aj .
0 2
j 1
Therefore F(D) is bounded.
So, F maps PC1[0,1] into the following
It is easy to know F : P P is equicontinuous. Hence F
is completely continuous.
PC1 [0,1] {x PC1[0,1] : 0
k
k
2
2
2
x(t) k2 (t)}.
such that
k2 (t)
Let U {u P and  u  R0 }, notice that
Define  x k inf{ 0 : k2 (t) x(t) k2 (t)} .
F :U PC1 (J , R) is continuous and completely
We know PC1 [0,1] is a subspace of PC1[0,1] and PC1 [0,1]
continuous. Choose u U and (0,1) such that u Fu .
k2 k2
is an Banach space with the norm x . Let u,v PC1 [0,1] ,
We claim that  u  R . If not, then  u  R .
k2 k2
0
 u  FU 
0  (Fu)(t) (Fv)(t) 
1
1
 u v  ( G(t, s)m(s)(k (s)
k (s, )k ( )d
s
s
1 ( 1 k (s) max { f (s, z,Tz, Sz) g(s)}ds
k2 0
2 0 2
0 2
0 s1,0 z R0
1 n
)
)
k (s, )k ( )d ds L G(t,t )k (t ))
n n
n n
M 2 k2 (t j ) k2 (t j )aj )
0 1 2 4
j 1
j 2 j
j 1
j 1
Mk2 (t)  u v k .
2
2
which implies  u  R0 . By the nonlinear alternative theorem
So  Fu Fv 
k
k
2
M  u v  . From M 1 , the operator F
k
k
2
of LeraySchauder type, F has a fixed point u U .
is a contraction. By Banach's contraction principle, F has a
n
n
1 unique fixed point. The proof is complete.
u(t)
0 G(t, s)g(s)ds G(t,tj )aj
j 1
THEOREM 3. Assume that (H2), (H3) hold,
n
n
G(t, t j )J j (u(t j ) v(t j ))
L :
sup  H (t, s) ,
(t ,s )[0,1][0,1]
2 1
2 1
j 1 1 1
1 * * *
1 * * *
0 G(t, s) f (s,(u v) ,T (u v) , S (u v) )ds
sup k (s)m(s)(1 T1 S1)ds 1,
t[0,1] 0
(3.5)
v(t) .
L2 n
Let x(t) u(t) v(t) 0 . For t t j we get
( p(t)x '(t)) ' f (t, x(t),Tx(t), Sx(t)) ,
k2 (t j ) nLL1 1,
j 1
then the Eq.(1.1) has at least one solution.
(3.6)
and p(tj )x(t j ) J j (x(t j )).
PROOF. Choose Br
{u PC1 (J , R),  u  r} , where
Furthermore we obtain x(0) p(0)x(0) 0 ,
and x(1) p(1)x(1) 0. So Eq.(1.1) has a positive solution.
r 2 , 11
The proof is complete.
1 k (s)  f (s, 0, 0, 0)  ds 2
1 M n
1 M n
k (t ) nLM ,
2
j
1
2
j
1
THEOREM 2. Assume that (H2) holds, I j 0,
2 0 2
j 1
 J (u) J (v)  L  u v ,  J (u)  M ,
M<1 ,
and define on Br the operator , by
j j 4 j 2
where
(u)(t) 0 G(t, s) f (s, u,Tu, Su)ds
1
1
1
1
s
s
M inf{a 0 : G(t, s)m(s)(k (s) k (s, )k ( )d
and
0 2 0 2
1 n
1 n
n n
k (s, )k ( )d )ds L G(t,t )k (t )
(u)(t) G(t, t j )J j (u(t j )) H (t, t j )I j (u(t j )).
0 1 2 4
j 1
j 2 j
j 1
j 1
ak2 (t)} , (3.4)
then Eq.(1.1) has a unique continuous solution.
Let us observe that if u, v Br then u v Br
1
1
 (u) (v) 
. Indeed
PROOF. Define F : PC1 (J , R) PC1 (J , R) as follows:
0 G(t, s)  f (s, u,Tu, Su) f (s, 0, 0, 0)  ds
1 n 1 n
(Fu)(t) 0 G(t, s) f (s, u,Tu, Su)ds G(t, t j )J j (u(t j )).
G(t, s)  f (s, 0, 0, 0)  ds G(t,tj )  J j (u(t j )) 
For u PC1 (J , R) , we have
1
j 1
0
n
n
H (t, t j )  I j (u(t j )) 
j 1
 (Fu)(t) 
1
0 G(t, s)  f (s, u,Tu, Su) f (s, 0, 0, 0)  ds
n
j 1
1
k (s)m(s)(r Tr Sr)ds
1
1
G(t, s)  f (s, 0, 0, 0)  ds G(t,tj )  J j (u(t j )) 
0 2 2
0 j 1
r1 2 r .
k2 (t) {1 m(s)( u   Tu   Su )ds
It is easy to see that is a contraction mapping. Since f is
0 continuous, we get is continuous and  u  r.
It is easy to see that is equicontinuous on interval
L3 1
M 2 n
(tk , tk 1 ] . So is relatively compact on Br . Hence by PC
where A 0 k2 (s)ds
k2 (t j ) nLM1.
j 1
We obtain
type ArzelaAscoli Theorem, is compact on Br . By A
Lemma 2, Eq.(1.1) has at least one solution on J . The proof
 u 
1 B
. By Lemma 4, we deduce F has a fixed point.
is complete.
THEOREM 4. Assume (H3), (H4) hold and
The proof is complete.

EXAMPLES
1
1
s
s
1
1
B L3 k (s)(1 k (s, )d k (s, )d )ds 1,
(3.7)
In this section we give two examples to illustrate our
0 2 0
0 1
main results.
then the Eq.(1.1) has at least one solution. EXAMPLE 1. Consider the following equation
PROOF. Define 1
Fu(t)

n
G(t, s) f (s,u,Tu, Su)ds G(t,t )J (u(t ))
u ''(t) f (t, u,Tu, Su) 0, t [0,1], t 2
0
j 1
j j j
1 1 1
n
n
j 1
H (t,t j )I j (u(t j )) .
u '( 2) 10  sin u( 2) ,
u(0) u '(0) 0,
We first prove F is continuous. Let {um } be a sequence
u(1) 0, (4.1)
such that um
u in PC1 (J , R) . For t J ,
 (Fum )(t) (Fu)(t)  7 t
1 1 where f (t,u,Tu,Su) u (t) 2u(t) su(s)ds
2
k (s)ds  f (., u ,Tu , Su ) f (., u,Tu, Su)  8 0
0 2
m m m
1 1 3
1 n n

( su(s)ds)2 . Choose g(t) for u 0 ,

L k (t )  u u  LL   u
2
2
2
2
u , 2 0 16
j m
j 1
1 m
j 1 1
where L : sup  H (t, s)  . By the continuous of f ,
(t ,s )[0,1][0,1]
f (t,u,Tu,Su) g(t) (u 1)2
16
we get F is continuous.
Second, we prove F maps bounded sets into bounded set
su(s)ds 1 ( su(s)ds) 0 .
t 1 2
t 1 2
0 2 0
in PC1 (J , R) .
We choose R 2 , a 10 , it s easy to see that (3.3) holds.
For any 0,
 (Fu)(t) 
B {u PC1 (J , R) :  u  } ,
0 3 1 27
By Theorem 1, Eq.(4.1) has at least one positive solution.
EXAMPLE 2. Consider the following equation
L3
k (s)(1  u   Tu   Su )ds
1
1
0 2
t 1 1

M 2 n
k (t ) LnM
2
j
2
j
u ''(t) u 0 tsu(s)ds 0 sin s u(s)ds,
t [0,1], t
2
1
1
1
1
j 1
1
s
1
s
u( 1 ) 1 sin u( 1 ),
L3
(1
k(s, )d
k (s, )d )k(s)ds
2 2 2
0 0
M n
0 1
u '( 1 ) 1 cos u( 1 ),
1
1
2 k (t ) LnM
2
j
2
j
: l .
2 3 2
j 1
u(0) u(0) 0,
Which implies that  Fu  l .
It is easy to know that F is equicontinuous on the interval
(t j , t j 1 ] . By Lemma 3, F is continuous and complete
continuous.
u(1) 0. (4.2)
Let E(F ) {u PC1 (J , R), u F (u), (0,1)}.
 u(t)  Fu(t) 
where f (t,u, v, w) u v w , Tu 0 tsu(s)ds ,
t
t
1
L 1 L  u  1 s
Su 0 sins u(s)ds. It is easy to know that (H2), (H3) hold,
2 k2 (s)(1
k (s, )d
0 0 0
and L 1, L 1 , L 1 ,
1 M n
1 M n
k (s, )d )ds 2 k (t ) nLM
1 2 2 3
0 1 2 j 1
j 1
1 1 5 3cos1
k (s)m(s)(1 T1 S1)ds 1,
1
1
s
s
1
1
: A L3 k (s)(1 k (s, )d
k (s, )d )ds  u  ,
0 2 6
0 2 0
0 1
and (3.6) holds. By Theorem 3, the Eq. (4.2) has at least one solution.
REFERENCES

S. D. Akgol and A. Zefer, Asymptotic integration of secondorder impulsive differential equations, Appl. Math. Lett.,Vol.76, pp. 17, 2018.

C.Z. Bai, Positive solutions for nonlinear fractional differential equations with coefficient that changes sign,Nonlinear Anal., Vol.64, pp. 677685, 2006.

D. Josef, Positive solutions of nonlinear delayed differential equations with impulses, Appl.Math.Lett., Vol.72, pp. 1622, 2017.

E.R. Kaufmann, N. Kosmatov and Y.N. Raffoul, A secondorder boundary value problem with impulsive effects on an unbounded domain, Nonlinear Anal.,Vol.69, pp. 29242929, 2008.

M. A. Krasnoselskii, Topological methods in the theory

of nonlinear integral equations, New York: Pergamon Press, 1964.

Q. Y. Li, Y. M. Zhou, F. Z. Cong and H. Liu, Positive solutions to superlinear attractive singular impulsive differential equation, Appl. Math. Comput.,Vol.338, pp. 822827, 2018.

Y. Tian, S. Shang and Q. Huo, Antiperiodic solutions of fouth order impulsive differential equations, Math. Methods Appl. Sci.,Vol.41, pp. 769780, 2018.

J. R. Wang, X. Z. Li and W. Wei, On the natural solution of an impulsive fractional differential equation of order q (1, 2) ,
Commun Nonlinear Sci Numer Simulat.,Vol.17, pp. 43844394, 2012.

W. Wei, X. Xiang and Y. Peng, Nonlinear impulsive integro differential equation of mixed type and optimal controls, Optimization., Vol.55, pp. 141156, 2006.

Z. L. Wei, Periodic boundary value problems for second order impulsive integrodifferential equations of mixed type in Banach spaces, J. Math. Anal. Appl.,Vol. 195, pp. 214229, 1995.

X. Zhang, L. Liu and H. Zou, Eigenvalues of fourthorder singular SturmLiouville boundary value problems, Nonlinear Anal., Vol.68, pp. 384392, 2008.