Solution Of A System Of Two Partial Differential Equations Of The Second Order Using Two Series
Hayder Jabbar Abood (Department of Mathematics, College of Education, Babylon University, Babylon,Iraq)
407 Introduction
Levenshtam (2009), considered systems of ordinary differential equations of the first order whose terms oscillate with a high frequency ω.For the problem on periodic solutions, the author justifies the averaging method and establishes a posteriori bounds for the error of partial sums of the complete asymptotic expansion for the solution. Asymptotic expansions of periodic solutions of second –and third- order equations and formal asymptotics of such solutions in the case of equations of arbitrary order were constructed in Abood1,2 (2004). The first-order asymptotic form was obtained and proved for the solution of a system of two partial differential equations with small parameters in the derivatives for the regular part, two boundary-layer parts and corner boundary part in Vasil'eva and Butuzob (1990). In Levenshtam and Abood (2005), an algorithm of asymptotic integration of the initial-boundary-value problem for the heat-conduction equation with minor terms (nonlinear sources of heat) in a thin rod
of thickness
1/2oscillating in time with frequency
1was proposed. In the present paper, we consider a system of partial differential equations of the second order and solve this system with the aid of two series and some conditions. Our paper continues the line of research initiated in Levenshtam and Abood (2005). Examples of applications of systems of second order partial differential equations in modelling can be found in elasticity in Nerantzaki and Kandilas (2008) and packed-bed electrode in Xiao-Ying Qin and Yan-Ping Sun (2009).
Statement Of The Problem
We study a system of second order partial differential equations with initial-boundary-value conditions
2
1 1 1
2
0 2
2 2 2
2
0
,
, , ,
,
,
, , ,
,
i i i
i i i
u
u
u
b x
a x t u
f
u x y t
t
y
x
v
v
v
b
x
a
x t v
f
u x y t
t
y
x
(2.1)
in the domain
( , , )
x y t
(0
x
1) (0
y
1) (0
t T
).
The initial boundary conditions are
Solution Of A System Of Two Partial Differential Equations Of The Second Order Using Two Series
Abstract
Author (s)
Hayder Jabbar Abood
Department of Mathematics, College of Education, Babylon University, Babylon-Iraq. E-mail: [email protected]
Key Words: Partial differential equations 1; System 2; Asymptotic 3.
For a system of two partial differential equations of second order, we obtain and justify two asymptotics solutions in the form of two series with respect to the small parameter
. We have proven the solution is unique and uniform in the domain
, and, further, each the asymptotics approximations are within
n1408
0 0,1 0 0,1
0,1 0,1
|
0,
|
0,
|
0,
|
0,
0,
0.
t x t x
y y
u
u
v
v
u
v
y
y
(2.2)
where
is a small parameter, and the functionsf u x y t
1( , , , , ),
andf u x y t
2( , , , , )
are continuous and infinitely differentiable with respect to each of their arguments. The construction is made under the following conditions.I. The functions
b x a x t
i
,
i,
andf i
i,
1, 2
have continuous derivatives of order
n
2
.II. We can assume that
b
2
1
. Ifb
2
b
2
x
, then making the change of the variable 21
0
x
z
b
d
, and weshall assume that
b
1
b x
0
andb
2
1
. Algorithm For Construction Of The AsymptoticsWe seek an asymptotics expansion of the solution of problem (2.1) and (2.2) as the following two series in the powers of
in the form0
0
( , , , )
[ ( , , )
( , , )
( , , )],
( , , , )
[ ( , , )
( , , )
( , , )],
i
i i i
i
i
i i i
i
u x y t
u x y t
p u x y
q u
y t
v x y t
v x y t
p v x y
q v
y t
(3.1)
where
u
i andv
i are coefficients of the regular part of the asymptotic,p u p v q u
i,
i,
i andp v
i are boundary-layer functions describing boundary layers near the initial instant of timet
0
and the ends of the rodx
0
andx
1
. The boundary-layer variables aret
, and
x
.Regular Parts Of The Asymptotics
u
Andv
409
2
1 1 1
2
0 2
2 2 2
2
0
, 0 0,1 , 0 0,1
0,1 0,1
,
, , ,
,
,
, , ,
,
|
0,
|
0, |
0,
|
0,
0,
0.
i i i
i i i
t x t x
y y
u
u
u
b x
a x t u
f
u x y t
t
y
x
v
v
v
b
x
a
x t v
f
v x y t
t
y
x
u
u
v
v
u
v
y
y
(4.1)
The following problems for the regular coefficients
u
0andv
0are obtained:
0
1 0 10 0
0
2 0 20 0
0 0
0 0
0,1 0,1
,
, , ,
,
,
, , ,
,
0, ,
0,
, , 0
0,
0,
0,
y y
u
b x
a x t u
f
u x y t
x
v
a
x t v
f
v x y t
x
u
y t
v
x y
u
v
y
y
(4.2)
By direct integration, it can be seen [Viik (2010), Viika and
Room
(2008), Tokovyy and Chien-Ching Ma,(2009), Toshiki, Son Shin, Murakami and Ngoc(2007)] system (4.2) is equivalent to system of integral equations
1
1
0 1 0 10
0
, exp , , , , ,
x x
u x t b p a p t dp b v t f y t d
(4.3)0
2
0
20
0
, exp , , , , .
t t
s
v x t a x p dp u x s f x y t ds
(4.4)Substituting (4.4) in (4.3), we arrive at an integral equation with respect tou0
x t, :0
0
0
0
0 0, , , , , , ,
x t
u x t
G x t s u s d ds g x twhereG x t0
, , , s
andg x t0( , ) are known functions. The solution of this integral equation can be expressed in terms of the resolvent 0
x t, , , s
of the Kernel G x t0
, , , s
(as in [Viik (2010), Viika andRoom
(2008), Tokovyy and Chien-Ching Ma,(2009), Toshiki, Son Shin, Murakami and Ngoc(2007)])
0 0 0 0
0 0
, , , , , , .
x t
u x t g x t
x t s g s d dsAfter that, the functionv0
x t, can be determined by (4.4).410
( )
(
) (
)
(
)
(
)
(
) (
)
(
)
(
)
(
)
(
)
(
)
(
)
1
1 1 10 0 0 11 0
2
0 0
2 1
2 1 20 0 0 21 0
2
0 0
2
1 1 1 1
1 1
0,1 0,1
,
,
, ,
, ,
,
,
,
, ,
, ,
,
0, ,
0, ,
,
, , 0
, , 0 ,
0,
0.
y y
u
b x
a x t u x t
f
u x y t u
f
u x y t
x
u
u
u
t
y
v
a x t v
x t
f
v x y t u
f
v x y t
t
v
u
v
y
x
u
y t
q u
y t
v
x y
p v x y
u
v
y
=y
=
¶
¶
=
+
+
-¶
¶
¶
¶
-¶
¶
¶
¶
=
+
+
-¶
¶
¶
¶
-¶
¶
= -
=
-¶
¶
=
=
¶
¶
(4.5)
From system (4.5) and direct integration, we have the system of integral equations
(
)
( ) (
)
( )
(
) (
)
(
)
(
)
1 1
1 1 2 1
0
2
0 0
10 0 0 11 0 2
,
exp
,
[
,
,
, ,
, ,
]
,
x x
u x t
b
p a p t dp b
a
t v
t
u
u
f
u x y t u
f
u x y t
d
u
t
y
s
s
s
s
s
-
-æ
ö÷
ç
÷
ç
=
ç
÷
÷
+
ç
÷
è
ø
¶
¶
¶
+
-
-¶
¶
¶
ò
ò
(4.6)
(
)
(
)
(
) (
)
(
)
(
)
1 2 1 1
0
2
0 0
20 0 0 21 0 2
,
exp
,
[
,
,
, ,
, ,
] .
t t
s
v
x t
a x p dp
a x s u x s
u
v
f
v x y t v
f
v x y t
ds
v
y
x
æ
ö÷
ç
÷
ç
=
ç
÷
÷
+
ç
÷
è
ø
¶
¶
¶
+
+
-
-¶
¶
¶
ò
ò
(4.7)
Substituting (4.7) in (4.6), we arrive at the integral equation with respect to u x t1
, :
1 1 1 1
0 0
, , , , , , , , ,
x t
u x t
G x y t s u s d ds g x y twhereG x y t1
, , , , s
andg x y t1( , , ) are known functions. The solution of this integral equation can be expressed in terms of the resolvent 1
x y t, , , , s
of the Kernel G x y t1
, , , , s
[Viik (2010), Viika andRoom
(2008), Tokovyy and Chien-Ching Ma,(2009), Toshiki, Son Shin, Murakami and Ngoc(2007)] and we have
1 1 1 1
0 0
, , , , , , , ,
x t
u x y t g x y t
x y s g s d ds . After that, the functionv x y t1
, ,
can be determined by(4.7).
411
( )
(
) (
)
(
)
(
)
(
)
1 0 0,1,
,
, ,
,
0, ,
0, ,
,
0.
n n n n n n y
u
b x
a x t u
x t
u x y t
x
u
u
y t
q u
y t
y
=¶
=
+ ¡
¶
¶
= -
=
¶
(4.8)
(
) (
)
(
)
(
)
(
)
2 0 0,1,
,
, ,
,
, , 0
, , 0 ,
0.
n n n n n n y
v
a x t v
x t
v x y t
t
v
v
x y
p v x y
y
=¶
=
+ ¡
¶
¶
= -
=
¶
(4.9)
where
(
)
(
) (
)
2
1 1
0 1 0 2 2
0
1
, ,
, ,
,
!
k n n nn k k k
k
u
u
u x y t
f
u x y t u
x t
k
u
t
y
- -=
¶
¶
¶
¡
=
-
-¶
¶
¶
å
and(
)
(
) (
)
2 0 00 2 0 2 2
0
1
, ,
, ,
,
!
k nn k k k
k
u
v
v x y t
f
v x y t v
x t
k
u
y
x
=
¶
¶
¶
¡
=
-
-¶
¶
¶
å
. This problem is quite similar toproblem (4.5) and can be solved by reduction to a system of integral equations.
Boundary Parts Of The Asymptotics
pu
Andpv
We construct the following group of coefficients of two series (3.1) that is the boundary-layer parts
, , ,
pu x y andpv x y
, , ,
. We consider problem (2.1) in a neighbourhood of the upper boundary
t0
of the domain and perform the change of variablest
; we obtain the system
2 1 1 2 2 2 2 2,
, , ,
,
,
, , ,
,
,
pu
pu
pu
b x
a x t pu
f
pu x y
y
x
pv
pv
pv
a
x t pv
f
pv x y
y
x
(5.1)
0,1 0,1, , 0
, , 0 ,
, , 0
, , 0 ,
0,
0.
y y
dpu
dpv
pu x y
u x y
pv x y
v x y
dy
dy
The boundary-layer functions arep u0 andp v0 . In this casep v0 0.
For
p u
0 we obtain the problem
0 0
1 0
0 0 0
0 0
0,1 0,1
, 0
, 0
,
0,
0, ,
0,
, , 0
, , 0 ,
0,
0.
y y
p u
p u
b x
a x
p u
x
X
x
p u
y
p u x y
u
x y
dp u
dp v
dy
dy
(5.2)412
1 1
0 1
0 0
(
), 0 exp
(
), 0
, 0
,
0,
,
u
x
a
x
s
ds
p u x
where
1
0
x
x
b
d
and
1
z
is the function inverse toz
x
.The system of equations for
p u
1 andp v
1 is
1 1
1 1 1
1 1 1
1 0,1
,
,
, , 0
, , 0 ,
0, ,
0,
0,
y
p u
p u
b x
a x t p u
x
x
p u x y
u x y
p u
y
dp u
dy
1
2 1 1
1 1
1 0,1
,
,
,
, , 0
, , 0
0.
y
p v
a
x t p v
x
x
p v x y
v x y
dp v
dy
Here
2 0 1
1
,
, 0
0,
10 0, , , 0, 0
2p u
a
x
x
p u x
f
p u x y
t
y
and
2
2 11
,
, 0
0,
20 0, , , 0, 0
2a
p v
x
x
p v x
f
p v x y
t
y
is smooth. In thesame way we obtain
2 0
1 2 0
, 0
,
,
0
, 0
,
0,
.
x
a
x
p v x s ds
x
p v
a
x
p v x s ds
x
1 1
1 1
0 1
1 1
(
), 0 exp
(
), 0
(
), 0
, 0
0,
.
u
x
a
x
s
p u
x
s
ds
x
x
413 The function
p u
n can be defined as the solution of the problemsn
n 1
,
, ,
,
n n
p u
p u
b x
a x t p u
x y
x
(5.3)
0,1
, , 0
, , 0 ,
0, ,
0,
0,
n n n
n
y
p u x y
u
x y
p u
y
dp u
dy
(5.4)where
1
1 1 0
0
2 1
1 0 2
0
1
, ,
[
, 0
,
, , , 0, 0
!
, , , 0, 0 ]
,
n
k k
n k k
k k n k k
a
x y
x
p u x
f
p u x y
k
t
p u
f
p u x y
p u
y
since (5.3) is a partial differential equation of the first order and first degree. Here the smooth function
n
x y
, ,
is known. Using the initial and boundary conditions (5.4), we obtain
1 1 1 0 1(
), 0 exp
(
), 0
(
), 0
, 0
0,
.
n
n n
u
x
a
x
s
p u
x
s
ds
x
x
For the function
p v
n , it is the solution of the problem
0,1
, ,
,
, , 0
, , 0
0,
n n n yp v
x y
p v x y
v x y
dpv
dy
where
22 1 1 0
0
2
1 1
1 0 2
0
1
, ,
,
[
, 0
,
, , , 0, 0
!
, , , 0, 0 ]
,
n
k k
n n k k
k k n n k k
a
x y
a
x t p v
x
p v x
f
p v x y
k
t
p v
p v
f
p v x y
p v
x
y
414 that satisfies the condition
p
n
x y
, ,
0,
where
n
x y
, ,
is a known continuous function vanishing for
x
and with partial derivative making jump on the characteristic line
x
. Integrating, we obtain
, ,
, 0
,
, ,
0,
,
n x n
x y r dr
x
p v x y
x
where the function
p v x y
n
, ,
is smooth everywhere.Boundary Parts Of The Asymptotics
qu
Andqv
Now, we find the coefficients
q u
i
, ,
y t
andq v
i
, ,
y t
. Consider problem (2.1) and the conditions (2.2) in a neighbourhood of the left boundary of the domain
and perform the change of variable
x
1. We obtain the system
2
1 1
2
0 2
2 2
2
1
1
0
,
,
, , ,
!
,
, , ,
,
, , ,
.
k k
k
qu
qu
qu
b
a
t qu
f qu
y t
t
y
k
qv
qv
qv
a
t qv
y t
f
qu
y t
t
y
For
q u
0
, ,
y t
we obtain the equationb
0
q u
00.
From this equation, taking into account the standard condition for boundary parts at infinity, that
means
q u
0
,
t
0
, and so we haveq u
0
, ,
y t
0
. Now, forq v
0
, ,
y t
we have the problem
0 0
2 0
0 0 0
0 0,1
0,
, ,
,
0, 0
,
0, ,
0, ,
,
, , 0
0,
0
y
q v
q v
a
t q v
y t
t
T
t
q v
y t
v
y t
q v
y
q v
y
Since this is a partial differential of first order and first degree, it’s solution have the form
0
2
0 0
0, ,
exp
0, ,
, 0
, ,
0,
.
v
y t
a
y s t
ds
t
T
q v
y t
t
415 It’s possible to find
q u
1 andq v
1 in the same way. Finally, we will continue to findq u
n andq v
n .For
q u
n we have the equation
b
0
q u
n n
, ,
y t
,
(6.1)where
1
1 1 0
0
2
1 1
1 0 2
0
1
, ,
[
0, ,
, 0, , , 0
!
, , , 0, 0 ]
,
n
k k
n k k
k
k
n n
k k
a
y t
y t q u
f
q u
y t
k
t
q u
q u
f
q u x y
q u
t
y
is known and continuous everywhere.
The function
n
, ,
y t
0
for
t
(is evident from our findings in the above), so we can seek this functionin the form
n
, ,
y t
z y t
n
,
t
, for
t
.
(6.2) The function
n
,
t
is smooth everywhere. By the conditionq u
n
, ,
y t
0
and integrating (6.1), we have
1
0
,
,
0
, ,
0,
.
n
n t
b
z y
t
s
s t ds
t
T
q u
y t
t
We can define the part
q v
n
, ,
y t
as the solution of the problemn n
, ,
,
n
q v
q v
y t
t
(6.3)
0,1
0, ,
0, ,
,
, , 0
0,
0
n n n
n
y
q v
y t
v
y t
q v
y
q v
y
(6.4)
416
2
2 1 1 0
0
2
1 1
1 0 2
0
1
, ,
0, ,
[
, 0
0, ,
, 0, , , 0
!
, 0, , , 0 ]
.
n
k k
n n k k
k k
n n
k k
a
y t
a
y t p v
x
p v
y t
f
p v
y t
k
t
p v
p v
f
p v
y t
p v
x
y
Also
, ,
n
y t
is a known continuous function that satisfies the conditionp v
n
, ,
y t
0,
vanishing for
t
and with partial derivative making jump on the characteristic line
t
. Solving (6.3) and using (6.4), we obtain
, ,
0, ,
exp
0
0, ,
, 0
0,
.
n n
n
v
y t
y s t
ds
t
T
q v
y t
t
where the function
q v
n
, ,
y t
is smooth everywhere.7. JUSTIFICATION OF THE ASYMOTOTICS
By
U
n
x y t
, , ,
andV
n
x y t
, , ,
we denote then
-th partial sums of the series (3.1).
Theorem. The solution u x y t
, , ,
,v x y t, , ,
of problem (2.1) and (2.2) admits the asymptotic solution ( , , , ) ( , , , )
n1 , ( , , , ) ( , , , )
n1n n
u x y t U x y t O v x y t V x y t O , (7.1)
uniformly in the domain
0 x 1
0 y 1
0 t T
.Proof. We set uUn1w1and vVn1w2. Substituting this in (1.1) and (2.1) we obtain the following problem for
the remainder terms w1and w2
2
1 1 1
1 1 1 1
2
2
2 2 2
2 2 2 2
2
1 0 1 0,1 2 0 2 0,1
1 2
0,1 0,1
, , , , ,
, , , , ,
| 0, | 0, | 0, | 0,
0, 0.
t x t x
y y
w w w
b x a x t w x y t
t y x
w w w
b x a x t w x y t
t y x
w w w w
w w
y y
Obviously, the inhomogeneous terms 1 and 2 can be estimated as
1, , , n
i x y t O
uniformly in. We
shall prove that the same estimate is valid for w1andw2. In view of the equalities
1
1 1 1
1 2
,
,
n
n n n n
n n
u U u U U U w O v V w O
417 We make the change of the variables k x y t
1, 2
i i
w re i , where kconst >0. We obtain the following system of equations for r1andr2:
2
1 1 1
1 1 1 1
2
2
2 2 2
2 2 2 2
2
ˆ
, , , , ,
ˆ
, , , , ,
r r r
b x c x t r x y t
t y x
r r r
b x c x t r x y t
t y x
(7.2)
with the homogeneous boundary conditions, as in the case ofr ii, 1, 2. Here
( ) 11 1 , , 2 2 , 1 , ˆ ( ).
k x t n
i i
c a x t k b x c a x t k e O Assume that r1 has a maximum at a point 1
x t1,1
of and r2 has a maximum at a point2
x t2, 2
, and assume thatr1
1 r2
2 . We shallconsider the first equation of (7.2) at1, (Ifr1
1 r2
2 , then we shall consider the first equation of (7.2) at2). We rewrite this equation as
2
1 1 1
1 1 1 1
2
ˆ
, , , , ,
r r r
b x c x t r x y t
t y x
(7.3)
Assume that the functionr1 is negative and has a minimum at1. (The case of a positive maximum can be
considered in the same way.) Then r1 0
t
and
1 0
r x
at this point (the inequality sign is possible only if 1lies on
the boundary of). Hence the left-hand side of (7.3) is negative at 1and is not larger thanr1
1 , while the right-hand side is of order
n1.
Consequently,
11 1 max 1 ,
n
r r x t O
.Since r1
1 r2
2 , it followsthat
12 2 max 2 ,
n
r r x t O
.Therefore
1, n
i
r x t O uniformly in. Hence we also have
1k x t n
i i
w r e O uniformly inG. The proof of Theorem is complete.
References
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