# Existence of Solutions for Sturm-Liouville Boundary Value Problems with Impulses DOI : http://dx.doi.org/10.17577/IJERTV8IS100305 Text Only Version

#### Existence of Solutions for Sturm-Liouville 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

Abstract-In the paper, we investigate the Sturm-Liouville 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.

1. PRELIMINARIES

In order to prove our Theorems, we need the following definition and Lemmas.

KeywordsSturm-Liouville; 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

Sturm-Liouville 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  studied a second-order

nonlinear differential equation subject to Sturm-Liouville type boundary conditions and impulsive conditions. The

Further by  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  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.() 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

1. 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,

2. there exists u U and v (0,1) with u vSu .

j j j j j j

LEMMA 2. () Let M be a closed convex and nonempty subset of a Banach space X . Let A, B be two operators such that

1. Ax By M whenever x, y M.

u(0) p(0)u(0) 0,

u(1) p(1)u(1) 0. (3.2)

where

2. A is a compact and continuous.

3. 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.() Let X be a Banach space and W PC(J , X ) . If the following conditions are satisfied:

1. 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,

1. 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.() 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

2. 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 Leray-Schauder 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 Arzela-Ascoli 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.

3. 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)

1. 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.

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