WWJMRD 2018; 4(3): 105-112 www.wwjmrd.com
International Journal Peer Reviewed Journal Refereed Journal Indexed Journal UGC Approved Journal Impact Factor MJIF: 4.25 E-ISSN: 2454-6615 Habeeb Kareem Abdulah Department of Mathematics, Faculty of Education for girls, University of Kufa, Najaf, Iraq Amal Khalaf Haydar
Department of Mathematics, Faculty of Education for girls, University of Kufa, Najaf, Iraq Kawther Reyadh Obead Department of Mathematics, Faculty of Education for girls, University of Kufa, Najaf, Iraq
Correspondence: Habeeb Kareem Abdulah Department of Mathematics, Faculty of Education for girls, University of Kufa, Najaf, Iraq
Solving Of Third Order Retarded Dynamical System
via Lambert W Function and Stability Analysis
Habeeb Kareem Abdulah, Amal Khalaf Haydar, Kawther Reyadh
Obead
Abstract
In this paper, we use the Lambert W function to derive a solution third order dynamical systems of retarded type. As well as we discuss analytical stability through illustrative examples.
Keywords:
Delay differential equations, Lambert W function, stability
1. IntroductionThe Lambert W function grows in the modeling of phenomena in numerous fields of science and engineering. In physics, it is used for the purpose of quantum theory, plasma physics and solar physics and many other applications. (See, [6], [3] [7] and [4].)
Many researchers’ works on the Lambert W function to obtain analytical solutions of delay differential equations (see, [8], [3], and [2]). Also, many researchers are concerned with studying the stability of delay differential equations besides studying their solutions their (see, [1], [8], [9], [5]) the infinite branches the Lambert W function are defined in the following series [3]
0 1
1 k
)) ( (ln
))) ( (ln(ln ))
( ln(ln )
( ln ) ( W
i m
m k
m k im
k k
z z C
z z
z (1)
Where
ln
k(
z
)
are thek
th logarithm branch and the coefficientsC
im can be expressed in terms of nonnegative sterling numbers of first kind.The principal branch (i.e., k = 0) of the Lambert W function can be represented by the following power series
1
1 0
!
)
(
)
(
W
n
n n
z
n
n
z
Where
z
C
[3].Theorem (1.1) [3]. For any given
z
R
, the principal branchW
0(z)
of the Lambert W function defined by
e
z
) where
,
(
y
iy,
gy
-y
e
-where z
,
-x
x,
z
W
1
0
cot
1
1
)
(
0
Have the following properties:
1.
W
0(
z
)
Is real and increasing if
(
1
,
)
e
z
.~ 106 ~
2.
W
0(
z
)
Is complex valued and decreasing negative realpart if , 1)
2 (
e z
.3.
W
0(
z
)
Is complex and purely imaginary parts if2
z .
4.
W
0(
z
)
Is complex valued with decreasing positivereal part if )
2 , (
z .
2. Solution of Retarded Dynamical Systems of Third Order and their stability
Consider the following third order retarded dynamical system with constant coefficients
) ( ) ( ) ( )
( )
(
0 1
2 2
2 3
3
T t by t y a t y dt
d a t y dt
d a t y dt
d (2)
With the initial delay condition
.
0
)
(
)
(
t
φ
t
, -T
t
y
If we assume that
a
2
3
,
a
1
3
2 anda
0
3, thenequation (2) become (3)
We are concerned in finding the solution beside stability analysis for the equation (3) that’s a time delay
T
.Let
λ
be a complex number and the solution of equation(3) be of the formula
y
(
t
)
e
λt
.
Substitutingλt
e
t
y
(
)
in equation (3) gives (4) Which represents the transcendental characteristic equation of delay differential equation (3)? Multiplyingboth sides ofequation (4) by
e
λT
yieldsb
λT
e
α
λ
)
3
(
. (5)Taking the cubic root on both sides of equation (5) gives
3
3
)
(
b
λT
e
α
λ
. (6)Since, every function
W
(
λ
)
which satisfies
()
)
(
e
WW
can be expressed in terms of the LambertW
function [2]. Then the roots of the characteristic equation (4) can be found by writing equation (6) as follows:.
3
3
3
)
3
3
(
)
3
3
(
b
Tα
e
T
Tα
Tλ
e
Tα
Tλ
(7)
Then
3
3
)
3
3
3
(
b
Tλ
Tα
Tα
e
T
W
(8)Thus, the characteristic roots of equation (4) is
α b
α T e T W T
λ 3 3 )
3 (
3 (9)
It is clear that the characteristic root
λ
depends on of the parametersα, T
,
b
(in terms ofa
2, a
1 anda
0). Hence, the stability property of DDE (3) depends on those parameters. We have the general solution of the delay differential equation (3) as follows
i
t
i
λ
e
i
K
y(t)
0 ]) 3 3 3 ( 3 [
-T, , t i
t α b α T e T i W T e i
K
(10)
Since, the solution
y
(
t
)
is equal to the initial delay condition
(
t
)
fort
-T,
0
, and then we have
0 ]) 3 3 3 ( 3 [ )
( , t -T,
i
t α b α T e T i W T e i K
t
(11)
Now, we find the coefficients
K
i by using the function)
(
t
as follows:The interval
-T,
0
can be partitioned into 2M parts such that M is a sufficiently large positive integer number.] 0 2 [
] 2
3 2
2 [ ] 2
2 2
[ ] 2 [ ] 0 , [
, M -T
M T ,-T M T -T M T ,-T M T -T M T -T,-T
T
Then, we find the values of the initial delay function
φ
(
t
)
at the endpoints of the above subintervals using (11) as follows)
0
(
)
0
(
1
1
)
0
(
0
0
)
0
(
1
1
)
0
(
)
0
(
K
M
y
M
K
y
K
y
K
y
K
M
y
M
φ
)
2
(
)
2
(
1
1
)
2
(
0
0
)
2
(
1
1
)
2
(
)
2
(
M
T
M
y
M
K
M
T
y
K
M
T
y
K
M
T
y
K
M
T
M
y
M
K
M
T
φ
)
(
)
(
1
1
)
(
0
0
)
(
1
1
)
(
)
(
T
M
y
M
K
T
y
K
T
y
K
T
y
K
T
M
y
M
K
T
φ
We can write the above equations in matrix form as:
M K
M K
M K
T M y T M y
M T M y M
T M y
M y M
y
T φ
M T φ φ
1 )
( )
(
) 2 ( )
2 (
) 0 ( )
0 (
) (
) 2 (
) 0 (
). ( ) ( )
(T,M Y T,M K M
φ
Thus
)).
(
)
(
1
)
(
M
Y
T,
M
T,
M
K
The coefficients
K
i are given by . )) ( ) ( 1 (lim Y T,M T,M i i
K
M
(12)
To find the initial delay function
φ(t)
, we assume that-T
, t
t
y
(
)
0
, theny(t)
0
,
t
-T,
0
. Then the equation (3) gives the following third order ODE0 0 ) ( 3 ) ( 2 3 ) ( 2 2 3 ) ( 3 3
yt yt , t
dt d t
y dt d t y dt
d (13)
We shall use the general solution of equation (13) to be the required initial delay function
φ(t)
in
-T
,
0
.3. Illustrative Examples
Example (3.1): Consider the following third order DDE
0 ), 1 ( ) ( 8 ) ( 12 ) ( 2 2 6 ) ( 3 3
y t y t yt t
dt d t y dt d t y dt
d (14)
]
0
,
1
[
),
(
)
(
t
t
t
y
By comparison with equation (3), we have
b
1
, α
2
andT
1
. To consider the stability margin, we consider1
0
, t
y(t)
theny
(
t-
1
)
0
,
t
[
-
1
,
0
]
.Hence the equation (14) becomes the third order ODE.
0 ) ( 8 ) ( 12 ) ( 2 2 6 ) ( 3 3
t y t y dt
d t
y dt
d t y dt
d (15)
The general solution of equation (15) is given as
t
e
C
Bt
At
y(t)
(
2
)
2
(16)Assuming
y
(
0
)
0
, y
(
1
)
1
andy
(
2
)
1
yields521
12
91
19
.
, B
-
.
A
andC
0
. Hence the initial delay function of the equation (14) is]
0
,
1
[
,
2
)
521
12
2
91
19
(
.
t
.
t
e
t
t
(t)
φ
(17)The general solution of third order DDE (14) is as follows
10 ]2 ) 3 1 3 2
3 1 ( 3 [ )
( , t - ,
i
t e
i W e i K t
y
(18)
Where
i
K unknown constants and the characteristic roots are are given by equation (9) as
.
2
)
3 1
3
2
3
1
(
3
W
i
e
(19)
Case (1)
10
] 2 ) 3 2 3 1 ( 3 [ )
( , t - ,
i
t e
i W e i K t
y
,
2 ) 3 2 3 1 (
3
Wi e
i
To find the coefficients
i
K we partition the interval
-
1
,
0
into 2 subintervals as follows.
-
1,0
[
1,
-
0.5]
[-0.5,
0]
Then
)
0
(
)
0
(
)
0
(
)
0
(
K
1y
1K
0y
0K
1y
1φ
)
5
0
(
)
5
0
(
)
5
0
(
)
5
0
(
.
K
1y
1.
K
0y
0.
K
1y
1.
φ
)
1
(
)
1
(
)
1
(
)
1
(
K
1y
1
K
0y
0
K
1y
1
φ
The above system can be written matrix form as follows
1 0 1
1 0
1
1 0
1
1 0
1
)
1
(
)
1
(
)
1
(
)
5
0
(
)
5
0
(
)
5
0
(
)
0
(
)
0
(
)
0
(
)
1
(
)
5
0
(
)
0
(
K
K
K
y
y
y
.
y
.
y
.
y
y
y
y
φ
.
φ
φ
-
)
1
(
)
5
0
(
)
0
(
)
1
(
)
1
(
)
1
(
)
5
0
(
)
5
0
(
)
5
0
(
)
0
(
)
0
(
)
0
(
11 0
1
1 0
1
1 0
1 1
0 1
φ
.
φ
φ
y
y
y
.
y
.
y
.
y
y
y
y
K
K
K
-
6345
239
5481
30
0
02
5204
7
03
7294
2
0671
2
02
5204
7
03
7294
2
1313
7
7282
52
4377
1
1313
7
7282
52
1
1
1
1
.
.
.
i
e+
.
-e+
.
.
i
e+
.
+
e+
.
i
.
-.
.
i
.
+
.
~ 108 ~
i . - .
i . - .
-i . + .
8570 1 5560 0
0000 0 1120 1
8570 1 5560 0
Therefore the solution of the equation (14) is given as
t
i
.
+
.
-e
t
.
e
t
i
.
-.
e
t
y
i
.
.
i
.
-
.
-i
.
+
.
)
8352
12
9484
7
(
(
)
7261
0
(
)
(
)
8352
12
9484
7
(
)
(
)
(
)
8570
1
5560
0
0000
0
1120
1
8570
1
5560
0
The characteristic roots
for the branches,
,
,
,
i
0
1
2
3
are as follows:0.7261
0
;
i
i
.
+
.
-; λ
i
1
7
9484
12
8352
i
.
-.
-; λ
i
1
7
9484
12
8352
i
.
.
; λ
i
2
10
5183
32
2111
i
.
.
; λ
i
2
10
5183
32
2111
i
.
+
.
-; λ
i
3
11
8656
51
2659
i
.
-.
-; λ
i
3
11
8656
51
2659
From the above characteristic roots we note that
)
0
,
Re(
, then equation (14) is stable.Case (2)
10 ]2 ) ) 2
3 2
1 ( 3 2
3 1 ( 3 [ )
( , t - ,
i
t i e
i K t
y
e
i
W
,
. 2 )) 2
3
2 1 ( 3 2
3 1 (
3
ii
W
i
e
)
1
(
)
5
0
(
)
0
(
)
1
(
)
1
(
)
1
(
)
5
0
(
)
5
0
(
)
5
0
(
)
0
(
)
0
(
)
0
(
11 0
1
1 0
1
1 0
1 1
0 1
φ
.
φ
φ
y
y
y
.
y
.
y
.
y
y
y
y
K
K
K
-
6345
239
5481
30
0
1
003
3546
4
003
6127
7
0531
6
6514
1
001
8549
5
002
8494
3
0569
24
5067
90
9907
1
5203
1
4878
1
6762
19
1
1
1
.
.
.
i
e+
.
-e+
.
i
.
-.
-i
e+
.
-e+
.
i
.
+
.
-i
.
-.
i
.
+
.
-
i . + .
i . + .
i . - .
-K K -K
0064 0 0092 0
0485 0 2245 0
4647 0 8423 1
1 0 1
Therefore the solution of the equation (14) is given as
t
i
.
+
.
-e
i
.
+
.
t
i
.
+
.
-e
i
.
+
.
t
i
.
-.
-e
i
.
-
.
-t
y
)
3691
19
0791
9
(
)
0064
0
0092
0
(
)
8371
1
8365
1
(
)
0485
0
2245
0
(
)
1322
6
9645
5
(
)
4647
0
8423
1
(
)
(
The characteristic roots
λ
for the branches,
,
,
,
i
0
1
2
3
are as follows:i
.
+
.
-; λ
i
0
1
8365
1
8371
i
.
+
.
-; λ
i
1
9
0791
19
3691
i
.
-
.
-; λ
i
1
5
9645
6
1322
i
.
.
; λ
i
2
11
0383
38
5795
i
.
.
; λ
i
2
9
8866
25
8140
i
.
+
.
-; λ
i
3
12
2066
57
5933
i
.
-.
-; λ
i
3
11
4806
44
9292
Since
Re(
)
0
,
, then equation (14) is stable.Case (3)
10 ]2 ) ) 2
3 2
1 ( 3 1 ( 3 [ )
(
3 2
, -, t i
t i e
e i K t
y
i
W
,
2 ) ) 2
3 2
1 ( 3 1 (
3 3
2
i e
i
W
i
) 1 (
) 5 0 (
) 0 (
) 1 ( )
1 ( )
1 (
) 5 0 ( ) 5 0 ( ) 5 0 (
) 0 ( )
0 ( )
0
( 1
1 0
1
1 0
1
1 0
1 1
0 1
φ . φ
φ
y y
y
. y . y . y
y y
y
-
6345
239
5481
30
0
1
001
8549
5
002
8494
3
0531
6
6514
1
003
3546
4
003
6127
7
4878
1
6762
19
9907
1
5203
1
0569
24
5067
90
1
1
1
.
.
.
i
e+
.
+
e+
.
i
.
+
.
-i
e+
.
+
e+
.
i
.
-.
-i
.
+
.
i
.
-.
-
i
.
+
.
-i
.
-
.
i
.
-
.
4647
0
8423
1
4012
0
7503
1
0064
0
0092
0
Therefore the solution of the equation (14) is given as
t
i
.
+
.
-e
i
.
+
.
t
i
.
-
.
-e
i
.
-
.
t
i
.
-.
-e
i
.
-
.
t
y
)
1322
6
9645
5
(
)
4647
0
8423
1
(
)
8371
1
8365
1
(
)
4012
0
7503
1
(
)
3691
19
0791
9
(
)
0064
0
0092
0
(
)
(
The characteristic roots
λ
i for the branches,
,
,
,
i
0
1
2
3
are as follows:i
.
-
.
-; λ
i
0
1
8365
1
8371
i
.
+
.
-; λ
i
1
5
9645
6
1322
i
.
-.
-; λ
i
1
9
0791
19
3691
i
.
+
.
-; λ
i
2
9
8866
25
8140
i
.
-.
-; λ
i
2
11
0383
38
5795
i
.
+
.
-; λ
i
3
11
4806
44
9292
i
.
-.
-; λ
i
3
12
2066
57
5933
From the above characteristic roots we note that
)
0
,
Re(
, then equation (14) is stable.Example (4.2): Consider the following third order DDE
0
),
2
(
2
)
(
27
)
(
27
)
(
9
)
(
22 3
3
y
t
y
t
y
t-
t
dt
d
t
y
dt
d
t
y
dt
d
(18)
]
0
,
2
[
),
(
)
(
t
t
t
y
We have
b
2
,
-3
andT
2
. To consider the stability margin, we considery
(
t
)
0
, t
-
2
then]
0
2
[
0
)
2
(
t-
,
t
-
,
y
.Therefore the equation (18) becomes the third order ODE.
0 ) ( 27 ) ( 27 ) ( 9 ) (
2 2 3
3
t y t
y dt
d t
y dt
d t y dt
d
(19)
The general solution of equation (19) is given as t
e
C
Bt
At
t
y
(
)
(
2
)
3 (20)
Assuming that
y
(
0
)
0
, y
(
0
.
5
)
1
, andy
(
1
)
2
we getA
0
.
6934
, B
0
.
7929
andC
0
. Hence the initial delay function of the equation (18) ist
e
t
.
t
.
t
)
(
0
6934
20
7929
)
3(
(21)
The general solution of the third order DDE (18) by as follows by equation (10), we get
2
0
]
3
)
3 2
2
3
2
(
2
3
[
)
(
, t
-
,
i
t
e
i
W
e
i
K
t
y
Where
i
K unknown constants and the characteristic roots are are given by equation (9) as
3
)
3 2
2
3
2
(
e
i
W
i
(19)
Case (1)
20 ]3 )) 2
3 2 1 ( 3 2 2 3 2 ( 2 3 [ )
( , t - ,
i
t i e
i W e i K t
y
,
))
3
2
3
2
1
(
3 2
2
3
2
(
2
3
e
i
i
W
i
To find the coefficients
i
K we partition the interval
-
2
,
0
in to subintervals as follows.
-
2
,
0
[
2
, -
1
.
5
]
[
-
1
.
5
, -
1
]
[
-
1
,-
0
,
5
]
[
-
0
,
5
,
0
]
Then
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
)
0
(
K
2y
2K
1y
1K
0y
0K
1y
1K
2y
2φ
)
5
0
(
)
5
0
(
)
5
0
(
)
5
0
(
)
5
0
(
)
5
0
(
.
K
2y
2.
K
1y
1.
K
0y
0.
K
1y
1.
K
2y
2.
φ
) 1 ( )
1 ( )
1 ( )
1 ( )
1 ( )
1
( K2y2 K1y1 K0y0 K1y1 K2y2
φ
) 5 1 ( )
5 1 ( )
5 1 ( )
5 1 ( )
5 1 ( )
5 1
( . K 2y 2 . K1y 1 . K0y0 . K1y1 . K2y2 .
φ
) 2 ( )
2 ( )
2 ( )
2 ( )
2 ( 2
( )K2y2 K1y1 K0y0 K1y1 K2y2
~ 110 ~
The above system can be written matrix form as follows:
2 1 0 1 2 2 1 0 1 2 2 1 0 1 2 2 1 0 1 2 2 1 2 1 0 0 1 1 2 2)
2
(
)
2
(
)
2
(
)
2
(
)
2
(
)
5
1
(
)
5
1
(
)
5
1
(
)
5
1
(
)
5
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
5
0
(
)
5
0
(
)
0
(
)
0
(
)
5
0
(
)
0
(
)
5
0
(
)
0
(
)
5
0
(
)
0
(
)
2
(
)
5
1
(
)
1
(
)
5
0
(
)
0
(
K
K
K
K
K
y
y
y
y
y
.
y
.
y
.
y
.
y
.
y
y
y
y
y
y
.
y
.
y
y
y
.
y
y
.
y
y
.
y
y
φ
.
φ
φ
.
φ
φ
-
)
2
(
)
5
1
(
)
1
(
)
5
0
(
)
0
(
)
2
(
)
2
(
)
2
(
)
2
(
)
2
(
)
5
1
(
)
5
1
(
)
5
1
(
)
5
1
(
)
5
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
1
(
)
5
0
(
)
5
0
(
)
0
(
)
0
(
)
5
0
(
)
0
(
)
5
0
(
)
0
(
)
5
0
(
)
0
(
1 2 1 0 1 2 2 1 0 1 2 2 1 0 1 2 2 1 2 1 0 0 1 1 2 2 2 1 0 1 2φ
.
φ
φ
.
φ
φ
y
y
y
y
y
.
y
.
y
.
y
.
y
.
y
y
y
y
y
y
.
y
.
y
y
y
.
y
y
.
y
y
.
y
y
K
K
K
K
K
-
0108
.
0
0305
0
0740
0
1271
.
0
0
1
03
3673
1
03
0627
3
02
9285
3
02
9979
1
02
9947
1
02
1844
4
8789
89
6137
43
6422
56
0700
12
7877
4
9154
5
1
9999
20
7492
4
8966
2
6249
3
1
0005
0
0020
0
0019
0
0095
0
02
5488
1
01
4129
5
4213
6
3904
45
02
6190
4
03
9139
1
02
6879
1
02
4264
2
0059
0
0450
0
4450
10
4141
7
0610
44
2416
5
0140
0
2126
0
1
1799
3
6424
1
1
9806
4
4232
4
1
.
.
-i
e+
.
+
e+
.
-i
e+
.
-e+
.
i
e+
.
-e+
.
-i
.
+
.
i
.
+
.
i
.
-
.
-i
.
-.
i
.
+
.
i
.
-.
i
.
-.
i
e+
.
+
e+
.
-i
.
-.
i
e+
.
-e+
.
-i
e+
.
+
e+
.
-i
.
-.
i
.
-.
-i
.
+
.
-i
.
-.
i
.
+
.
-i
.
+
.
i
.
+
.
-i
.
+
.
-i
.
-
.
-i
.
+
.
i
.
-
.
0018
0
0004
0
0178
0
0331
0
0522
0
0379
0
0325
0
0676
0
0000
0
0038
0
Therefore the solution of the equation (18) is given as
t
i
.
+
.
-e
i
.
+
.
t
i
.
+
.
-e
i
.
+
.
-t
i
.
+
.
e
i
.
-.
t
i
.
-
.
-e
i
.
+
.
t
i
.
-.
-e
i
.
-.
t
y
)
4887
17
0590
4
(
)
0018
0
0004
0
(
)
6316
7
0695
3
(
)
0178
0
0331
0
(
)
1313
0
0920
3
(
)
0522
0
0379
0
(
)
0951
4
5501
2
(
)
0325
0
0676
0
(
)
2556
14
7926
3
(
)
0000
0
0038
0
(
)
(
The characteristic roots
λ
for the branches,
,
,
,
i
0
1
2
3
are as follows:i
.
+
.
; λ
i
0
3
0920
0
1313
i
.
+
.
-; λ
i
1
3
0695
7
6316
i
.
-
.
-; λ
i
1
2
5501
4
0951
i
.
+
.
-; λ
i
2
4
0590
17
4887
i
.
-.
-; λ
i
2
3
7926
14
2556
i
.
+
.
-; λ
i
3
4
6591
27
0754
i
.
-.
-; λ
i
3
4
4839
23
8920
From the above characteristic roots we note that
0
)
Re(
for some
then the equation (18) is unstable [3].Case (2)
