Gz3113501354

Samuel A. Iyase / International Journal of Engineering Research and Applications
(IJERA) ISSN: 2248-9622 www.ijera.com
Vol. 3, Issue 1, January -February 2013, pp.1350-1354
On The Existence Of Periodic Solutions Of Certain Fourth Order
Differential Equations With Dealy
Samuel A. Iyase
Department of Mathematics, Computer Science and Information Technology,
Igbinedion University, Okada, P.M.B. 0006, Benin City, Edo State, Nigeria.

Abstract
We derive existence results for the and Lk([0, 2 ]) of continuous, k times
periodic boundary value problem continuously differentiable or measurable real
x (iv )  a  f ( x)   cx  g (t , x( x   )  p(t )
x  x  functions whose kth power of the absolute value
is Labesgue integrable.
x(0) = x(2π), x (0) = x (2  ),  0) =  ( 2  ),
  x x
 (0) =  (2  )
x x We shall also make use of the sobolev space
using degree theoretic methods. The uniqueness of defined by
periodic solutions is also examined.
H 2k = { {x : [0,2 ]  R x, x are

Keywords and Phrases: “Periodic Solutions, absolutely continuous on [ 0,2 ] and,
Caratheodony Conditions, Fourth Order
 L2[ 0,2 ]
2
Differential Equations with delay. 2000 x with norm x H2 =
2
Mathematics Subject Classification: 34B15; 2
1 2 1 2
 
2
34C15, 34C25. ( x 2 (t )dt ) 2  x i (t ) dt .
2 o 2 i 1
o
1. Introduction i
d x
In this paper we study the periodic boundary value xi 
problem dt i
+ c x + g(t, x(t - )] = p(t)
( iv)
x + ax  f ( x )
  
x 
2. The Linear cases
(1.1) In this section we shall first consider the
x(0) = x(2π), x (0) = x (2 
  ),  0) =  ( 2  ),
x x equation:
 (0) =  (2  )
x x x(iv) (t) + a  (t) + b  (t) + c x (t) + dx(t- ) = 0
x x 
with fixed delay   [0,2 ) Where c ≠ 0 is a (2.1)
constant, p: [0, 2 ]→R and g: [0, 2 ]x  → x(0) = x(2  ), x (0) = x (2  ),  (0) =  (2  ),
  x x
 are 2π periodic in t and g satisfies certain  (0) =  (2  )
x x
Caratheodory conditions.
The unknown function x: [0, 2 ]→R is defined Where a, b, c, d, are constants
Lemma 2.1 Let c ≠ 0 and Let a/c < 0
for 0 < t <  by x(t -  ) = x( 2 -(t- ) Suppose that:
The differential equation x  a .+ b  +
( iv )
x x 0 < d/c < n, n ≥ 1 (2.2)
h(x) x + g(t, x(t -  )) = p(t)
 Then (2.1) has no non-trivial 2  periodic solution
(1.2) for any fixed   [0,2 ).
In which b < 0 is a constant was the object of a
recent study [6]. Proof
Results on the existence and uniqueness of t
By substituting x(t) = e with  = in, i2 = -1.
2 periodic solutions were established subject to We can see that the conclusion of the Lemma is
certain resonant conditions on g. Fourth order true if Ф(n,  ) = an 3 – cn + d Sin n  ≠ 0 for all n
differential equations with delay occur in a variety
of physical problems in fields such as Biology, ≥ 1 and   [0, 2 ) (2.3)
Physics, Engineering and Medicine. In recent year, By (2.2) we have
there have been many publications involving a d
differential equation with delay; see for example
c-1 ф(n, ) = n3-n + Sin n  ≤
c c
[1,2,4,5,6,8,9]. However, as far as we know, there
a 3 d a
are few results on the existence and uniqueness of n  n   n3  0
periodic solution to [1.1]. c c c
In what follows we shall use the spaces Therefore (n, ) ≠0 and the result follows
C([0, 2 ]), Ck([0, 2 ]) ‘

1350 | P a g e

Vol. 3, Issue 1, January -February 2013, pp.
L1 [0, 2 ] we shall write
If x  =
2 2 2
1 2 1 1
x
2 0
x(t )dt, ~(t )  x(t )  x  ~ (t )dt   (t )~x(t   )dt  2  (t ) x dt

x2  2
x x
2 o 0 0
So that 2
1
 (t ) x~(t   )dt
2

0
~ (t ) dt  0
x 2 o
x
We shall consider next the delay equation Using the identity
 a  b  cx  d (t ) x(t   ) = 0
( iv )
x x x  ( a  b)
2
a2 b2
ab   
(2.4) 2 2 2
We get
x(0) = x( 2 ), x (0) =,
 x ( 2 ),  (0) =  ( 2 ),
 x x =
 (0) =  ( 2 )
x x 1
2
1
2
1
2

 ~ (t )dt  2

x2  (t ) x dt  2  (t ) x~(t   )dt
2
Where a , b, c are constants and d 
1 x
L 2 2 0 0 0
Here the coefficient d in (2.4) is not necessarily
constant. We have he following results which
apart from being of independent interest are also
useful in the non-linear case involving (1.1)
1
+
Lemma 2.2 Let c ≠0 and let a/c < 0 Set Γ(t) = 2
2  
c-1d(t) 
2
[ x(t   )  ~(t )]2 ~ 2 ~ 2 (t   )
x x x
L  Suppose that
2
 
(t )  
0 < Γ(t)<1 (2.5) 0
2 2 2
Then for arbitrary constant b the equation (2.4)
x2 
 x~(t   )  dt
1
admits in H 2 only the trivial solution for every x
  [0, 2 ). 2
We note that a and c are not arbitrary. 1
=
2
Proof
 
2 2
If x  H 2 is a possible solution of (2.4) then on ~ 2 (t )dt  1 (t ) ~ 2  ~ 2 (t   ) dt +
 
 2 x x 
1
x
 2 0
multiplying (2.4) by x + ~ (t) and integrating over
x 0

 x(t   )  ~(t )
2
[0, 2 ] noting that (t )
 x 2 dt
1
  2
x
1
2 2 2
 (x  x(t))
 ~ 0

2 0
Using (2.5) we get


c 1 x (iv )  a  b   21
x x  a

2
2 (t )dt
x 0

 
c o 2 2
(t ) ~ 2 ~ 2
 x (t )dt  21
 

 1 ~2 x  x (t   ) dt
2
We have that 0 0
2
0=
2

From the periodicity of ~ we have that
x
(x  ~(t) c 1[ x (iv )  a  b   x  (t ) x(t   ) dt
1
2  
x x x 
2 2

 x (t )dt   x (t   )dt
0

~2 
~2
= - 0 0
2 2 Hence
 ( x  x (t ))x  (t ) x(t   )dt 
1 a ~2 1
  (t )dt  2 
x ~  2
2 c 0  [ ~
 (t   ) (t ) ~ 2 (t   )]dt
1 1 2
0 0 2 2 x x +
0
2 2

 (x
 
~ 2 (t )  (t ) ~ 2 (t   )]dt
  ( x  ~(t )){x(t )  (t ) x(t   )}dt
 1
1
2 x  2 [ 21 x
0 o
(2.6)
Using (2.5) we can see that the last expression is
non-negative hence

1351 | P a g e

2
x (iv )  a  f ( x)   cx  g (t , x(t   ))  p(t )
x  x 
0  1 [ 21
2   ( ~ (t   )  (t ) ~ (t   )]dt
0

x2 x2 (3.1)

  ~ H1 x(0)  x(2 ), x(0)  x(2 ), (0)  (2 ),(0)  (2 )
2
x   x x x x
2

By Lemma 1 of [8] where  > 0 is a constant.
where f :    is a continuous function and
This implies ~ = 0 a .e and that x = x .
x
But a constant map cannot be a solution of (2.4) g : [0,2 ]x   is such
since (t )  0 that g(. x) is a measurable on 0,2  for each
Thus x = 0 x   and g (t ,.)
is continuous on  for almost each t [0,2  ]
Theorem 2.1
Let all the conditions of Lemma 2.2 hold and let We assume moreover that for each r > 0
 be related to  by Lemma 2.2.Suppose there exists Yr
1
L
2
such that
V  L 2
2
further that satisfies g (t, x)  r (t ) (3.2)
0  V (t )  (t )   a.e t  0,2 where    > for a.e t  [0,2 ] and all x [r , r ] such a g is
0 then said to satisfy the
2

 ( x  ~(t ))c
Caratheodory’s condition.
[ x (iv )  a  f ( x) ] x  V (t ) x(t   ) dt
1  1
x x  x 
2 0 Theorem 3.1
2 Let c ≠ 0 and let a/c < 0
 (   ) ~
x 1
(2.7) Suppose that g is a caratheodory function
H 2
satisfying the inequalities
Proof
c 1 xg (t , x)  0 ( x  r) (3.3)
We have from the proof of Lemma 2.2 that
g (t , x )
2  (t )
( x  ~(t ))c 1 x (iv )  a  f ( x)   x  V (t ) x(t   ) dt
Lim sup (3.4)
1 cx 
x
2 

x x  x 
0

2 2

 ~ 2 (t   )  V (t ) ~ 2 (t   ) dt  1 ( 1
 (~Uniformly(ta.e. (t ))[0,2 ]
1 1   
 ( x x x (t )  V ) ~ t dt
2
x 2 where r > 0 is
2 2 0
2 2 0 constant and   L
2
is such that
2
0<
2 2
1
 ( 21  ( ~ 2 (t   )  (t ) ~ 2 (t   ))dt   ( 21

x x 2  ~ (t   ))dt
x 2
Suppose p  L2  is
2 such that
2 2
0 0 1
2
p
2 
0
p(t )dt = 0 then for arbitrary
1 1
 (x
 
+ ( ~ 2 (t )  (t ) ~ 2 (t ))dt   (
x 1 continuous function f the boundary value
2
2 2 2 problem (3.1) has at least one solution for every
  [0,2 )
0
2

 x (t )dt

~2
Proof
0
From condition (2.5) , Lemman 2.2 and Let be related to as in Lemma
Wirtinger’s inequality we have 2.2 so that by (3.3) and (3.4) there exists a
constant R1 such that for 1

x 
  ~ H1   ~
x   ~ H 1   ~ H 1  (   ) ~ H 1 0
x x x (3.5)
2 L2 
2 2 2 2
Define by

3. The Non Linear Case
We shall consider the non-linear delay
equation

1352 | P a g e

(cx) 1 g (t , x)  
~ 2  (   p )( x  ~ )
x H1 x
 1
2 2 2 2 2

~ (t , x)  (cR1 ) g (t , R1 )  ~ 2  ( x  ~
x H1 x H1 )
y  1 2 2 2
 (cR1 ) g (t , R1 ) Thus
(t )
 ~ 2  2 ( x  ~ 2 )
x H1 x H1 (3.11)
x  R1
2
 2

With  0 independent of x and 
0  x  R1 .Integrating (3.9) over [0,2π}We obtain
 R1  x  0 2 2
(1   )  (t ) x(t   )dt  c   g (t , x(t   ))
1
x0
0 0
(3.6)
(3.12)
Then
Since Γ(t) > 0 we derive that
0 < ~(t , x) ≤ Γ(t) +
y 2

(3.7)
1
2  (t )dt    0
0
(3.13)

for a .e t  [0,2 ] and all x   . Moreover the Hence if x(t) > r for all t  [0,2π], (3.3) and (3.12)
function ~(t , x) satisfy Caratheodory’s conditions
y implies that (1   )   0 contradicting   0.
and : [0,2 π] x  →  defined by Similarly if x(t) < - r for all t  [0,2π] we reach
a contradiction.
g (t , x(t   ))  g (t , x(t   ))  cx(t   ) ~(t , x(t   ))
~ y Thus there exists a, t1,  [0,2π] such that
(3.8)
x(t1 )  r. Le t2 be such that
is such that for a.e t  [0,2π] and all x   .
~(t, x(t -  ))   (t ) for some  (t ) 
g
t2

x = x(t2 )  x(t1 )   x( s)ds.
 This
Let   [0,1] be such that t1

+ a + λf( x )  ] + x +(1 - λ) Γ(t)x(t-τ) + λ ~ x  r  2 ~ H 1
(iv )
-1
c [x x  x  y implies that x
2
(t, x(t-τ)) x(t   ) Substituting this in (3.11) we get
x ~
 c 1 (1   )b  c 1 g (t , x(t   ))  c 1p(t )  0 ~2 c ~
x H1 1 x H1
2 2
(3.9)
For  = 0 we obtain (2.1) which by Lemma 2.2 or ~ H 1  c1 , c1  0
x (3.14)
2
admits only the trivial solution Now
For λ = 1 we get the original equation (1.1). To
prove that equation (3.1) has at least one solution,
we show according to the Leray-Shauder Method x H 1  x  ~ H 1  r  (2  1)c1  c 2
x
2 2
that the possible solution of the family of (3.15)
equations (3.9) are apriori bounded in C3[0,2π] Thus
independently of   [0,1].
x 2  c3 , c3  0
 (3.16)
Notice that by (3.5) one has
From (3.16)we have
0 < (1- λ) Γ(t) + λ ~ (t,x(t-τ)) < Γ(t) +
y (3.10)
x   c4 , c4  0 (3.17)
Then using Theorem 2.1 with V(t) = (1- λ)Γ(t) +
λ ~ (t, x(t-τ)) and Cauchy Schwarz inequality we
y 
Multiplying (3.9) by - x (t ) and integrating over
get [0,2π] we have
1
 2  a ( x 2  1   x 2 x    2 x 2  p 2 x 2 )
2 2
0 = x    
2
2
1
 ( x  ~(t )){c
 [ x (iv )  a  f ( x) ]  x 
1
x x  x 
2 0
Hence
~ ~
(1-λ)Γ(t) x(t-τ) +  (t,x(t-τ)+ g (t , x(t   )
 2  c5 , c5  0
x (3.18)

+(1-λ) b  - λp(t)}dt. And thus
x
x   c6 , c6  0
 (3.19)

1353 | P a g e

Multiplying (3.9) by  (t ) and integrating over
x Where  (t )  L2  is defined by
2
[0,2π]
 (t ) =
We get
g (t , x1 (t   ))  g (t , x 2 (t   ))
1
x 2  f ( )   2  1   2  2 x   c  2  2  p 2  2  b  2 c( x1  x 2 )
2 2
 x x x x x x
If u = x1 - x2 ≠ 0 and since 0 < β(t)  Γ(t) for
Thus a.e t  [0,2π] then using the arguments of
theorem 2.1 we have that u = 0 and thus x1
 2  c7 , c7  0
x (3.20) = x2 a. e.
And hence
   c8 , c8  0
x (3.21) REFERENCES
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
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x x  oscillation in neuromuscular systems.
Journal of Mathematical Biology 2 1975
x(0)  x(2 ), x(0)  x(2 ), (0)  (2 ),(0)  (2 )
  x x x x 87-105.
(4.1) [8] E.De. Pascal and R. Iannaci: Periodic
has art most one solution. solutions of generalized Lienard equations
with delay, Proceedings equadiff 82,
Proof Wurzburg (1982) 148 - 156
Let u = x1 –x2 for any two solutions x1, x2 of (4.1). [9] H.O. Tejumola, Existence of periodic
Then u satisfies the boundary value problem solutions of certain third order non-linear
c 1[u (iv )  a  bu]  u   (t )u(t   )  0
u   differential equations with delay. Journal
of Nigerian Mathematical Society Vol. 7
u(0)  u(2 ), u(0)  u(2 ), u(0)  u(2 ),(0)  (2 )
    u u (1988) 59-66

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