2

A

tI,

3

A

ttI

, and

2

f tf

t I

1

Bt

0nn

are n ×

n matrices. Note that,

and I are the zero and the

identity matrices, respectively.

3)

T

1

,0 ,,

n

tgtgt

rxx x

1

0

,

n

t

gx

which is a 2m1

vector.

For ,1,2,,ij n

, let is an arbitrary fixed

and let

00t

C

opyright © 2012 SciRes. IJMNTA

M. A. RAMADAN, S. M. EL-KHOLY

126

11 012010

,,

21 022020

0

,,

10 0

,, ,

,, ,

,

,,

m

ijin j

m

nijnin j

mm

m

zttzttz tt

ZZ

zttzttztt

Ztt ZZ

zttz tt

(3)

be a fundamental matrix solution to the linear system:

A

t

zz (4)

which equal to the identity matrix for .

0

Consider with

tt

00z0

z small enough, and

let us denote by

00t

00

,,ttzz

0

z the unique solution of Equa-

tion (2) which equal to

at . By hypotheses (1)

0

tt

and (2),

0

z

0

,,ttz is defined on a maximal right inter-

val,

0,tl, and satisfies the following integral equation:

0

1

000 0000000

,,,,,,,, ,,d

t

t

ttZ ttZ ttZstBsstssts

zzzzzrzz

This gives us the following integral inequality:

0

1

000 0000000

,,,,,e,,,,,d

t

t

ttZ ttZ ttZstBsstssts

zzzzzrzz (5)

where .

T

010e

Equations (3) and (4) give us the following differential

equations:

,,ijij nij

zftzz

,

,

,

,

(6)

,,nijij nij

ztzftz

(7)

,,injinj ninj

zftzz

(8)

,,nin jin jninj

ztzftz

(9)

for .

,1,2,,ij n

Since is a decreasing function, so Equations (6)

and (7) lead to:

t

22 22

,, ,,

1

2ij nijij nij

tz zfttzz

0

2d

22

,, 0

e

t

t

f

uu

ij nij

tz zt

(10)

By the same way we can obtain from Equations (8)

and (9) the following:

0

2d

22

,,

e

t

t

f

uu

in jnin j

tz z

(11)

For

T

12 12

,,,,,,, m

nn

x

xxyyy Rz, consider

the norm 2

1

m

i

i

z

z, where is the norm defined

in .

m

R

For

T

001 02001020

,,,,,,, m

nn

x

xxyyy Rz we

have:

,, ,0,0

0

00

,, ,0,0

0

22

,0,0,0, 0

11 1

,ijin jijin j

nijnin jnijnin j

nn n

ijjinjjnijjnin jj

ij j

ZZZZ

Ztt ZZZ Z

zx zyzx zy

xy

x

zxy

y

satisfies (4), then we get: Using Shwartz inequality, Minkowski inequality [8] and

suitable assumptions lead to:

0

2d

00 00

,31e

t

tfu u

Zttn t

z

z (12)

We have also:

00tst

1

00

T

1020 0

,,e

,, ,,,,

m

Ztt Zst

tst tsttst

1

00

0

1

22

00

1

,,e

,, ,,

i

i

n

ii

n

i

Ztt Zst

ttsttst

Since

2,,

mtst

t

is a decreasing function of , we get:

t

Copyright © 2012 SciRes. IJMNTA

M. A. RAMADAN, S. M. EL-KHOLY 127

22 2 2

0000

,,,,,,,,e S

2d

t

f

uu

in

i ini

t tsttsts sstsst

So,

d

1

00 0

,, e

t

S

fu u

Ztt Zstent

(13)

As mentioned the system satisfies the integral inequality (5), then inequalities (12) and (13) give

0

1

0000000000

,,,,,e,,,,,d

t

t

ttZ ttZ ttZstBsstssts

zzzzzrzz

0

dd

2

0000 000

,,31ee

tt

ttS

fu utfuu

ttntn tnfs

0

t

,,,df sstss

zz z

For all t we can replace g by another function say g

defined as follows. By hypothesis (4) it follows that there

exists a

zzgx (14)

0

such that if

x, then

,tfto

gx x

We defined the function by:

:nn

RRR

g

,if

,,if

gt

tgt

ave

xx

gx xx

and we h

x, t0.

ry ,x

It is clear that for eveR

tR

n

,tgx fto

x

is of class and is locally Lipschitzian

g

in

n

CR R

n12

,,,

x

xx

ginal function

. So, we w

g satisfies

ill admit from now that the

all the properties of the ori

g.

Since 1

f

CR

we get from inequality (14) that:

0

00 0 0

,,31e

t

tfu

t

ttn t

D

0

00

,, d

t

s

t s

du

z

zz

u

g

zz

0,ttl , with positive constant D [9,10]. This givess

the followin

0000 0

,,31 ,,

Dh

ttntett l

zzz (15)

Thus

00

,,tt

z

zas well as

00

,,tt

z

z are bounded

on and so

0,tl

00

,,tt

z

z can be extended to the

right of l. This contradicts the maximality of l. This

means that the solution of Equation (1) is bonded. The

proof of theorem 1 is complete.

Theorem 2

If the hypotheses of theorem 1 are hold and

lowing assumption is satisfied:

1) There exist two constants such that:

u

the fol-

,0hk

2,ftf tkftth

then the zero solution of Equation (1) is uniformly stable

solution.

If in addition

2)

0

dft t

holds, thzero solution en the of

Equation (1) is asymptotically stable.

Proof

n

Our stability question is reduced to the stability of the

zero solutio

0t

z

will be divided into two intervals. In the

to the system (2). The interval of

tfirst interval

we have

0,tth and in the second interval

,th

.

lh

and si

on a ma

Firstly, with nce the hypotheses of theorem

1 are hold, and then the solution

,tz

of Equation

00

,tz

l right interval (1xima) is defined

0,tl. It is

proved in theorem 1 that the solution is bounded and

00

,,tt

z

z cantended to the right of l. Therefore, be ex

00

,,tt

z

z exists on

0,tl with lh.

Assume hl

. We are going to find an estimate

for

00

,,tt

z

z on the interval

,hl . From hypothesis

(1) of theorem 2, we have for thl :

,

,,, d

t

unk fsstgsszz x

dd

000 0

,,31e,e

t

SS

t

fuufu

ttnhhtnh

zzz z

0

,

h

00

dd

0

, ,

t

hS

fu u

nhnkfs fs

z z

00 00

h

00 00

e,,,, d

t

fu ustst s

zzxz ,3 1e,

t

tthtn h

z z

Copyright © 2012 SciRes. IJMNTA

M. A. RAMADAN, S. M. EL-KHOLY

128

with

1

0, nk

nh

(16)

00

d

000

d

00

,,

31e ,,

e

t

h

t

S

fuu

tfuu

h

tt

nh ht

nhnkfs sts

zz

zz

zz

,

,d

(17)

d

000

d

00

31e ,,

e,

t

h

t

S

fuu

tfuu

h

vt hht

nhnkfssts

zz

zz

So

,d

1,vtnh nkftvtthl

.

By integration we get:

1d

00

,,e ,

,

t

h

nhnk fuu

ttt h

thl

vv

zz (18

with

)

00

31,,vh nhht

zz

From inequality (18) we can see that

00

,,tt

z

z is

bounded since

00

,,tt

z

z is also bounded, it folls

that For

ow

l .0

we can get:

2

e

13013

Dh

nh

From (15) it follows that

00

,,

13

tt nh

zz

for all

0,tth provided that 0

zore,

from (16) and (1t

. Theref

, we ge8)

00

,,ttvh

zz for

all th o, if h, then the solution . S00

t

starting

00

,,ttzz

from any point 0

z

, with 0

z, exists on

0

t

, and satisfies

00

,,tt

zz for al

usly we btain that

l 0

tt. If

and:

0

th, then analogool

0

1d

0

e ,

t

nhnkfuu

(19

,th

Therefore, with the same

00 0

,,31ttnt

zz z)

t

as before, 0

z im-

plies again

00

,,tt

zHence the

If in

z for all 0

tt.

zero solution is uniformly stable.

addition

0

dft t

is satisfied, then by (16), (18) and (19) it follows tha

zero solution to (1) is asymptotically stable. The proof of

theorem 2 is complete.

3. Examples

We confirm the results of the introduced theorems by

considering two numerical examples for which the func-

tions

t the

f

t,

t

and

tg satisfy theorems assump-

tions.

For the system

2,ft t

t0

xxxgx

where

,

12

,

x

xx

.

Example 1

21

t

ft t

,

11tt

,

12

112 2

,, 1

x

xt

gtxxt

and

22

12

212 2

,, 1

x

xt

gtxx t

.

(a)

(b)

Figure 1. Numerical solutions if Example 1, x1 component (a)

and x2 component (b).

Copyright © 2012 SciRes. IJMNTA

M. A. RAMADAN, S. M. EL-KHOLY 129

(a)

(b)

Figure 2. Numerical solutions if example 2, x1 component (a)

and x2 component (b).

Example 2

2

1

ft t

, ,

2

1e

t

t

12

112 2

,,

x

x

gtxx t

and

22

12

2

212

,,

x

x

t

.

gt

xx

We solved these two examples numerically using

th order Runge-Kutta method. The results of example 1

are drawn as shown in Figure 1 and the results of exam-

own in Figure 2. The curves are

rawn for different initial values of

d 2 demonstrate time increases

all the compents of the solutions tends to zero.

means, that te aptotically stable. W

verify the rightness of our proved theorems.

4. Conclusion

ifo ly stable as well

as

REFERENCES

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four-

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d0

x.

that, as the

Figures 1 an

on This

he solutions arsymhich

We introduced two theorems which provide the sufficient

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ear damped vectorial oscillator and the conditions for the

stability of the zero solution to be unrm

asymptotically stable. We verified our theoretical re-

sults by solving two examples satisfying the assumptions

of the two proved theorems.

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Copyright © 2012 SciRes. IJMNTA