On the existence of solutions of two differential equations with a nonlocal conditionAcademic research paper on "Mathematics"

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Abstract of research paper on Mathematics, author of scientific article — E.M.A. Hamd-Allah

Abstract In this paper we study the existence of solutions of two Cauchy problems of two nonlinear differential equations with nonlocal condition. The continuous dependence of the solutions on the coefficients of the nonlocal condition will be studied.

Academic research paper on topic "On the existence of solutions of two differential equations with a nonlocal condition"

﻿JID: JOEMS

Journal of the Egyptian Mathematical Society (2015) 000, 1-6

[m;December 17, 2015;7:32]

Egyptian Mathematical Society Journal of the Egyptian Mathematical Society

www.etms-eg.org www.elsevier.com/locate/joems

Review Paper

On the existence of solutions of two differential equations with a nonlocal condition

E.M.A. Hamd-Allah*

Faculty of Science, Alexandria University, Alexandria, Egypt

Received 2 August 2015; revised 17 September 2015; accepted 3 October 2015 Available online xxx

Keywords

Nonlocal conditions; Existence of solution; Fixed point theorem; continuous dependence

Abstract In this paper we study the existence of solutions of two Cauchy problems of two nonlinear differential equations with nonlocal condition. The continuous dependence of the solutions on the coefficients of the nonlocal condition will be studied.

2010 Mathematics Subject Classifications: Primary 32A55; 11D09; Secondary 45D05; 60G22; 33E30,

Copyright 2015, Egyptian Mathematical Society. Production and hosting by Elsevier B.V.

This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

1. Introduction

Problems with nonlocal conditions have been extensively studied by several authors in the last two decades. The reader is referred to [1-5] and [6-13] and references therein. Consider the two nonlinear differential equations

dx(t ) dt

= f[t, x(t)

dx(t ) dt

t e (0, T],

* Tel.:+201224420145.

Peer review under responsibility of Egyptian Mathematical Society.

dx(t) dt

= £ t, x(t)

dx(t) dt

a.e.t e (0, T],

with the nonlocal condition

Y2akx(rk) = x0, Tk e (0, T).

Our aim here is to study the existence of solutions for the two problems (1) with the nonlocal condition (3) and (2) with the nonlocal condition (3). Moreover, the continuous dependence of the solutions of the above two problems on x0 and the nonlocal coefficients ak will be studied.

2. Functional integral equations

Lemma 2.1. LetY^m= 1 ak = 0. The solution of the nonlocal problem (1) and (3) can be expressed by the integral equation

S1110-256X(15)00078-4 Copyright 2015, Egyptian Mathematical Society. Production and hosting by Elsevier B.V. This is an open access article

http://dx.doi.org/10.1016/j.joems.2015.10.002

/ ™ fTk \ f

x(t) = A I x0 ak I y(s)ds I + I y(s)ds,

\ k=1 j0 J j0

A=(t•) •

where y is the solution of the functional integral equation

y(t) = f[t, Ax0 — A 2 ak f y(s)ds +f y(s)ds, y(t)\, k=i j0 j0 t e [0, T]. (5)

Proof. Let dXr = y(t) in Eq. (1), then we obtain

y(t) = f(t, x(t), y(t)) where

x(t) = x(0) + i y(s)ds. (6)

Letting t = Tkin (6), we obtain

m m m „

^2 akx(rk) = ^ akx(Q) ak y(s)ds.

7,_ 1 7,_ 1 7,_ 1 ^ 0

c(0) = a^xq - y(s)ds^

where A = (£m=i ak)-1. And we obtain

/ ™ f Tk \ r t

x(t ) = A xQ — 2_\ ak I y(s)ds I + I y(s)ds,

k=1 0 0

where y is the solution of the functional integral equation

/ f Tk [ t \

y(t) = f\t, Ax0 — Ay^akl y(s)ds + I y(s)ds,y(t )l,

k=1 0 0

t e [0, T].

By similar way, the following lemma can be proved. □

Lemma 2.2. Let^2k= 1 ak = 0. The solution of the nonlocal problem (2) and (3) can be expressed by the integral equation

(t ) = A i xq — V ak f y(s)ds) + i y(s)ds, \ k=i Jq f Jq

where y is the solution of the functional integral equation y(t) = g^t, Axo — A 22 ak y(s)ds + y(s)ds, y(t^,

2.1. Existence of solutions

Consider the functional integral equations (5) and (10) with the following assumptions:

(i) f: [0, T] x R x R ^ R is continuous and satisfies Lipschitz condition

| f (t, Ui, U2 ) — f (t, Vi, V2 )| < Mi (|ui — Vl|+|U2 — V2|),

(ii) g: [0, T] x R x R ^ R is measurable in t e [0, T] for any (ui, u2) e R x R and satisfies Lipschitz condition

|g(t, ui, u2 ) — g(t, Vi, V2 )| < M2 (|ui — Vi|+|u2 — V21 ),

[ gt, 0, 0)|dt < N

(iii) M* = Mi(2T + i) < i

(iv) M** = M2(2T + i) < i.

Now we have the following theorem

Theorem 2.1. Let the assumptions (i) and (iii) be satisfied. Then the functional integral equation (5) has a unique solution y e C[0, T].

Proof. Define the operator H by

Hy(t ) = f[t, Axq — A V ak f y(s) ds +1 y(s) ds, y(t )),

t y(s) ds, y(t) ,

t e [0, T]. Let y e C[0, 7], then |Hy(t2 ) — Hy(ti )|

m r Tk r t2

f ( t2, AxQ — Ay^akl y(s) ds + I y(s) ds, y(t2) k=i Jq Jq m /• Tk r t2

— f[t2, Axq — A 22 ak I y(s) ds +1 y(s) ds, y(ti ) k=i Jq Jq ™ r Tk r t2

f ( t2, AxQ — A YJ ak I y(s) ds +1 y(s) ds, y(ti k=i Jq Jq

™ ¡-Tk rh

— f ( t2, AxQ — A2_,akl y(s) ds + I y(s) ds, y(ti ) k=i Jq Jq m /• Tk r ti

f ( t2, Axq — A /"J ak i y(s) ds +/ y(s) ds, y(ti k=i Jq Jq

™ pTk çh

— f [ ti, AxQ — A2_,akl y(s) ds + I y(s) ds, y(ti )

k=i Jq Jq !• t2

< Mi ^(s^ds + Miyt2) — y(ti)| Jti

m r Tk r ti

f ( t2, Axq — AJ2 ak I y(s) ds +1 y (s) ds, y(ti k=i Jq Jq

t e [0, T].

m ¡-Tk çh \

— f [ ti, AxQ — A2_,akl y(s) ds + I y(s) ds, y(ti ) k=i Jq Jq )

JID: JOEMS

On the existence of solutions of two differential equations with a nonlocal condition

which implies that the operator H maps C[0, T] into itself, i.e. H: C[0, T] ^ C[0, T].

Now, let u, v e C[0, T] then we have

Hu(t) — Hv(t) = fit, Ax0 — A

m ç Tk ç t

' I u(s) ds + I u(s) ds,u(t)

k=1 j0

—f[t, Ax0 — A^^ak J v(s) ds + / v(s) ds, v(t )

\Hu(t) — Hv(t)| < MijAj

k=1 j0

\u(s) — v(s)\ds

+ M1 [ \u(s) — v(s)\ds + M1\u(t) — v(t)\,

\\Hu(t) - Hv(t)|| < M1 (2 T + 1)||u - v|| = M*||u - v||.

And M* < 1, which proves that the operator H: C[0, T] ^ C[0, T] is contraction.

Applying Banach contraction fixed point [14], then the functional equation (5) has a unique fixed point y e C[0, T]. Now, consider the nonlocal problem (1) and (3). □

Theorem 2.2. Let the assumptions of Theorem 2.1 be satisfied. Then the nonlocal problem (1) and (3) has a unique solution x e C1[0, T].

Proof. From Lemma 2.1 and Theorem 2.1, the solution of the problem (1) and (3) is given by (4) where y is given by (5). To complete the proof, we prove that the integral equation (5) satisfies nonlocal problem (1) and (3). Differentiating (4), we get

- = ,(t) = fU x(t), -

Let t = Tk in (4), we get

m m / m . Tk \

ak x(Tk) = ^2 ak A I xo — ^ ak y(s)ds I

k=1 k=1 V k=1 '

y(s)ds

= x0 —

™ P Tk ™ P Tk

y]ak y(s) ds ^Y] ak y(s) ds

This completes the proof. □

Now we have the following theorem:

Theorem 2.3. Let the assumptions (ii) and (iv) be satisfied. Then the functional integral equation (10) has a unique solution y e L1[0, T].

Proof. Define the operator G by

m /• Tk r t

Gy(t) = g[t, Ax0 - ^aJ y(s) ds +/ y(s) ds, y(t)),

t G [0, T].

Let y g L1[0, T], then

\\Gy\\i1 = i \(Gy)(t )\dt

y(s) ds

Ax0 — A ak y(s) ds k=1 Jo

+ \y(t)\Jdt + \g(t, 0, 0)\ dt. (13) From condition (ii) we have

\ g(t, u, v)\ — \g(t, 0, 0) \ < \ g(t, u, v) — g(t, 0, 0) \ < M-( \u\ + \ v \ ), then

\ g(t, u, v)\ < M- ( \u\ + \ v \ ) + \g(t, 0, 0) \ .

Substitute into (13), we get

\ \ Gy\ \ Ll < M- T\ A\ x0 + NT + M- (2T + 1) \\ y\ \ ¿1.

Then \ \ Gy\ \ l1 g L1, which implies that the operator G maps ¿1[0, T] into itself, i.e. G: ¿1[0, T] ^ ¿1[0, T]. Now, let u, v g L1 [0, T] then we have

Gu(t ) — Gv(t ) = g(t, Ax0 — A

m /• Tk f t

2_,ak I u(s) ds + I u(s) ds,u(t )

k=1 0 0

t v(s) ds, v(t)

—g| t, Ax0 — A2_\akl v(s) ds + I v(s) ds, v(t

\ Gu(t) — Gv(t)\ < M2

m /. T

(u(s) — v(s)) ds

+ I (u(s) — v(s)) ds

+ M2 \ u(t ) — v(t ) \

< M2 \ A \

■A f Tk

k=1 j0

u(s) — v(s) ds

+ M2 f \ u(s) — v(s)\ds + M2\u(t) — v(t)\ ,

| | Gu(t) - Gv(t)11 l1 < (M2 T + M2 T + M2) 11 u - v | | l1 < M2 (2T + 1) 11 u - v | | l = M**||u - v||l,.

And M** < 1, which proves that the operator G: L1[0, T] ^ L1[0, T] is contraction.

Applying Banach contraction fixed point [14], then the functional equation (10) has a unique fixed point y e L1[0, T] such that y(t) = g(t, x(t), y(t)), a.e.t e (0, T].

Now, consider the nonlocal problem (2) and (3). □

Theorem 2.4. Let the assumptions of Theorem 2.3 be satisfied. Then the nonlocal problem (2) and (3) has a unique absolutely continuous solution x e AC[0, T].

Proof. From Lemma 2.2 and Theorem 2.3, we deduce that there exists a unique absolutely continuous solution x e AC[0, T] of

the integral equation (10). To complete the proof, we prove that the integral equation (10) satisfies the nonlocal problem (2) and

Differentiating (9), we get

dx ( dx

dt = ^) = i1'x(t)' dt

Let t = Tk in (9) ,we get

22 ak x(Tk) = xq.

This completes the proof.

This implies that there exists a unique absolutely continuous solution x e AC[0, T] of the non local problem (2) and (3). □

3. Continuous dependence

Here we study the continuous dependence of the solution of the problem (1) and (3)

Definition 1. The solution x e Ci[0, T] of the problem (i) and (3) is called continuously dependent on xQ if, for every e > 0 thereexists 8(e) > Osuchthat |xQ — xQ| <S implies ||x — x|| < e where x, x are the solutions of the problems (i) with the condition (3) and (1) with the condition

22 ak x(Tk) = Xo' Tk e (0, T).

Theorem 3.1. Let the assumptions of Theorem 2.1 be satisfied. Then the solution of the problem (1) and (3) depends continuously on xq.

Proof. Since the solution of the problem (i) and (3) is the solution of the integral equation

x(t ) = A I x0 —

m r Tk \ rt

■/_,akl y(s) ds I + I y(s) ds

k=i Jo J Jo

and the solution of the problem (1) and (14) is the solution of the integral equation

x(t) = A | x0 — '^/ak J y(s) ds \ + I y(s)ds.

Then, for the two corresponding solutions x and x we have

|x(t) — X(t)| < |A||xo — x0| + |A|

+ [ |y(s) — y^ds. Jo

J2ak /

|y(s) — yÇs^ds

||x(t) — X(t)|| < |A||xo — Xo| + T|A| + T ||y — y||

||y — y||

< |A||xo — xo|+ 2 T||y — y||.

||y(t ) — y(t )|| < Mi|A||xo — xo|+Mi (1 + 2 T )||y — y|| Mi|A|

1 — (1 + 2 T ) M1

|xo — xo |.

Substitute in (15), we get

2 TM1|A|

||x(t) — x(t)|| < |A||xo — xo| + --, ^ ,, |xo — xo|

1 — M1

1 — (1 + 2 T ) M1

1 — (1 + 2 T ) Mi |A||xo — xo|

||x(t) — x(t)|| < (1 — M1 )(1 — M*)-[\A\S = e.

which completes the proof of the theorem. □

Definition 2. The solution x e C1[0, T] of the problem (1) and (3) is called continuously dependent on the coefficients J]m=1 ak if, for every e > 0 there exists 8(e) > 0 such that £m= 1 a — X | < S implies ||x — x|| < e where x and x are the solutions of the problems (1) with the condition (3) and (1) with the condition

22 xkx(Tk) = xo, Tk e (o, T).

Theorem 3.2. Let the assumptions of Theorem 2.1 be satisfied. Then the solution of the problem (1) and (3) depends continuously on the coefficients i ak.

Proof. Since the solution of the problem (i) and (3) is the solution of the integral equation

x(t ) = A(xo — 22 ak j y(s) ds J + I y(s)ds

and the solution of the problem (1) and (16) is the solution of the integral equation

x ~ [Tk „ f t

x(t ) = x (xo — 22 xk X(s) ds) + I x(s) ds k=i Jo Jo

Then, for the two corresponding solution x and xx we have

x(t) — x(t) = (A — A) xo — A

ak , Jo

y(s)ds

m f Tk f t

22 xk y(s) ds + I (y(s) — x(s))ds.

,, Jo Jo

è ak f

y(s) ds

k=1 — Ay

■A f Tk .A f T'

22 xk y(s) ds = A22ak

Jo ,, Jo

y(s) ds

■A f Tk____A f Tk

22 ak I X(s) ds + A 22 ak I X(s) ds

Jo ,, Jo

JID: JOEMS

On the existence of solutions of two differential equations with a nonlocal condition

™ _ f Tk____™ _ r Tk _

y àk I y(s) ds + A àk \ ~(s) ds

ak ~(s) ds k=1 j0 m f Tk = A Y]ak (y(s) — y(s)) ds k=1 j0 ■A f Tk

+ A Y] (ak — àk) ~(s) ds k=1 •/0

à A [ Tk + (A — A) Y^ àk à(s) ds. k=1 •/0

Substitute from (18) in (17), we obtain

~ A [ t

x(t) — ~(t) = (A — A) x0 — A Y] ak /

!• Tk

(y(s) — ~(s) ) ds

— A V^ (ak — ~ ) / ~(s) ds k=1 •/0

— (A — A)

e ~k i

k=1 •/0

~(s) ds

+ f (y(s) — ~(s))ds.

\ \ x(t ) — ~(t ) \ \ < \A — ~\ \ x0 \ + T \ \y — ~\ \ + T \ A

+ T \ A — A \

+ T \ \y — ~\ \

ak — aàk ày

< \A — A \ \ x0 \ + T\\~\ \|\ A + 2 T \ \y — ~\ \

ak — ~k \ + \ A — A \

< \ A \. \ A

ak — ~k \ \ x0 \ + T \ A

ak — aàk ày

+ T \ A \.\A \

\ ak — a~k \ \ \~\ \ + 2 T \ \ y — ~\ \

< \ A Now

ak — àk \ ( \ A \ \ x0 \ +2 T \ \ ~\ \ ) + 2 T \ \y — ~\ \ . (19)

y(t) — ~(t) = f t, Ax0 — A

y(s) ds

+ y(s) ds, y(t)

— fit, A x0 — A

m~ fT'

~(s) ds

+ i) ~(s) ds, ~(t)

\ y(t) — ~(t) \ < M1 \ A — A \ \ x0 \ +M1

m f t,

y(s) ds

~(s) ds

+M1 T\y(s) — ~(s) \ +M1 \y(s) — ~(s) \ . (20)

Substitute from (18) into (20), we get

\y(t) — ~(t)\ < M1 \ A — A\ \ x0 \

+ M1 A

J2ak /

rrf J 0

\ y(s) — ~(s) \ ds

™ /• Tk

+ M1 \ A\ V\ak — à \ / \ ~(s) \ds

+ M1 A — Aà

XI àk /

\ ~(s) \ ds

+ M1 T \ y(s) — ~(s) \ +M1 \ y(s) — ~(s) \ .

\ \y(t) — à(t)\ \ < M1 \ A — A \ \ x0 \ + M1 T \ \ à \

\ A \ X \ ak — àk \ + \ A — A \ X àk k=1 k=1 + 2 M1 T y — ày + M1 y — ày .

\ \y(t) — à(t)\ \ <

1 — (1 + 2T) M1

xJ2 \ ak — àk\ ( \A\ \ x0 \ +2 T\ \ à\ ).

Substitute from (21) into (19),we get

\ \ x(t ) — à(t ) \ \ < \ Aak — àk \ ( \ A \ \ x0 \ +2 T \ \à \ )

2 TM1 \ A\ m à

^-„ . E\ ak — ak \( \A \ \ x0 \ +2T\\ à \\ )

1 — (1 + 2T) M1

(1 — M1 ) \ A \ ~ A

< —-^^ ( \ A \ \ x0 \ +2 T \ \à \ \ ^ \ ak — àk \

1 — (1 + 2T ) M1

I I x(t) - ) | | < (1 - Mi)(1 - M*)-11 A|

x ( |A|| X0 | + 2T||jT|| )S = e.

which completes the proof of the theorem. □

Now for the continuous dependence of the solution of the problem (2) and (3) with respect to x0 and the coefficients Y^k= 1 ak we have the following

Theorem 3.3. Let the assumptions of Theorem 2.3 be satisfied. Then the solution of the problem (2) and (3) depends continuously on x0.

Proof. Since the solution of the problem (2) and (3) is the solution of the integral equation

x(t) = A — '<m,ak J y(s) + I y(s) ds

and the solution of the problem (2) and (14) is the solution of the integral equation

X(t ) = AI X0 —

(Xo — X ak f X(s) d^ + i y(s)ds.

\ k=i Jo f Jo

I I x(t) — X(t)| | < | A| | Xo — XXo |

+ | A |£]| ak | | | y — X | L +| | У — X | Li

< | A | | xo — Xo | + 2 11 y — X | Li.

| | y(t ) — X(t ) 11 Li < M2 T | A | | xo — Xo | + M2 (1 + 2 T ) 11 y — X | Li M2 T | A |

1 — (1 + 2 T ) M2

|xo — xo|.

Substitute in (22), we get

2 TM2|A|

||x(t) — x(t)|| < |A||xo — xo| + --„ , ' . _. |xo — xo|

1 — M2

1 — (1 + 2 T ) M2

1 — (1 + 2 T ) M2 |A||xo — xo|.

||x(t) - x(t)|| < (1 - M2)(1 - M**)-1|A|5 = e.

which completes the proof of the theorem. □

Theorem 3.4. Let the assumptions of Theorem 2.3 be satisfied. Then the solution of the problem (2) and (3) depends continuously on the coefficientsY^,m= 1 ak.

Proof. The proof straights forward as the proof of Theorem 3.2. □

Acknowledgment

valuable remarks that helped improve the quality of this

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The authors would like to thank the referee for the time taken to review this paper and for his/her suggestions and