2018 2nd International Conference on Modeling, Simulation and Optimization Technologies and Applications (MSOTA 2018) ISBN: 978-1-60595-594-0
On Initial Value and Boundary Value Problem of Linear Differential
Algebra System
Shu-yan YAN
University of Sheffield United Kingdom
Keywords: Linear differential algebraic system, Initial value problem, Boundary value problem, Analytical solution.
Abstract. The problem of solving the initial value problem of linear differential algebra system and the analytical solution of two-point boundary value problem is studied. In three cases, the expressions of the initial value problem and the analytical solution of the boundary value problem are given respectively. Example illustrates the good results of the proposed method.
Introduction
In the study of practical problems, it is often required to solve the initial value problem[1-5]
0 ( ) (0)
Ax Bx b t x x
(1) and two-point boundary value problem of linear differential algebra systems[6-7].
( )
(0) ( )
Ax Bx b t x x T
(2) Where A B, Rn n ,xR b tn, ( )R tn, R T, 0 。In this paper, the reversibility of matrix is analyzed. The analytical solution of initial value problem and two-point boundary value problem of linear differential algebra system are given in three cases.
Main Conclusions
Theorem 1 If the matrix Ais reversible, the analytical solution of the initial value problem of the linear differential algebra system (1) is
1 1 ( ) 1
0 0
( ) A Bt t A B t z ( )
x t e x
e A b z dz .If the matrixA and (eA BT1 I) are reversible, the analytical solution of the two-point boundary value problem of linear differential algebraic system (2) is
1 1 1 1 1 1 ( ) 1
0 0
( ) A Bt( A BT ) T A Bz ( ) t A B t z ( )
x t e e I
e A b z dz
e A b z dz .Proof Because of the matrix Ais reversible, the first formula in equation (1) can be transformed into the ordinary differential equations
) (
1 1
t b A Bx A
x . (3)
The general solution of equation (3) is
1 1 1
( ) A Bt A Bt ( )
x t e
e A b t dt C.1 1 ( ) 1
0 0
( ) A Bt t A B t z ( )
x t e x
e A b z dz .And when(eA BT1 I) is also reversible, the solution of the two-point boundary value problem (2) is:
1 1 1 1 1 1 ( ) 1
0 0
( ) A Bt( A BT ) T A Bz ( ) t A B t z ( )
x t e e I
e A b z dz
e A b z dzWhere I is the unit matrix.
Before giving the following conclusions, the concept of a nilpotent matrix is given.
Definition 1 For matrix N , If there is a positive integer k1 , make Nk 0 and Nj0 (j1, 2, ,k1), the matrix N is called the nilpotent matrix.
Definition 2 If the matrix A is odd, and the matrix Bis reversible, Then the analytical solution of the initial value problem of linear differential algebra system (1) is
1 1
0
0
( ) ( ) 1
0
1 ( )
0
( ) ( ) ,
( ) ( ).
t
D t t D t z
t k
j j
j
v t e v e D e z dz
w t N d t
If the matrixBand I eD1Tare reversible, when
1
0
) ( )
(
0 )) ( )
0 ( (
k
j
j j
j
T d d
N , then the analytical
solution of the two-point boundary value problem of linear differential algebraic system (2) is
1 1 1 1 1 1( ) 1
0 0
1 ( )
0
( ) ( ) ( ) ( ) ,
( ) ( ),
T T
D t D T D z D t z
k
j j
j
v t e e I e D e z dz e D e z dz
w t N d t
whereDRm m ,NR(n m ) (n m), ( )v t Rm, ( )w t Rn m .
Proof Because the matrix A is vertible matrix, and the matrix Bis reversible. Therefore, the first equation in equation (1) can be changed to
) (
1 1
t B B x x A
B (4)
Reversible matrix S exists by standard form Jordan theory, let
1 1 0
0
D SB AS
N
,
whereDism m invertible matrix (1 m n)andNis (n m ) ( n m)nilpotent matrix. Let
v Sx
w
,
1 ( )
( )
( )
e t SB b t
d t
,
herevis m-dimensional vector andwis (n m )-dimensional vector. ( )e t is m-dimensional vector and ( )d t is (n m )-dimensional vector.
Corresponding
0 0
0
v Sx
w
( 0 , 0 )
m n m
v R w R .
( ), ( ).
Dv v e t Nw w d t
(5) The initial value problem (1) becomes
0 0
0 0
( ), ( ) , ( ), ( ) .
Dv v e t v t v Nw w d t w t w
(6) The first one in equation (5) is the ordinary differential equations. According to Theorem 1, the general solution of the first equation in(6)is
] )
(
[ 1 1
1
C dt t e D e e
v D t D t .
Therefore, the solution to the first initial value problem in equation (6) is
1 1
0
0
( ) ( ) 1
0
( ) D t t t D t z ( )
t
v t e v
e D e z dzIn equation (6) , since the coefficient matrix Nof the second equation is a nilpotent matrix, the
following solution is available. j
j j
dt t w d t
w( )( ) ( )and
dt t d d t d
j
j ( )
) (
)
(
are the j derivative of
) (t
w and d(t), respectively. The second differential equation in equation (6) can be changed to
) (t d w N
w
Derived on both sides of the formula can obtain
) 1 , , 2 , 1 ( ), ( )
( )
( ( 1) ( )
)
(
k j
t d t Nw t
w j j j .
Then substituting one can have
) ( ) ( )
( )
( ) ( )
(t Nw(1) t d t N2w(2) t Nd(1) t d t
w
3 (3) 2 (2) (1)
( ) 1 ( 1) (1)
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
k k k k
N w t N d t Nd t d t
N w t N d t Nd t d t
BecauseNk 0, therefore
1
0
) (
) ( )
(
k
j
j j
t d N t
w .
This is the solution of the second differential equation in equation (6). From this equation, we can see that w(t) is uniquely determined by the derivatives of d(t)andd(t). It does not contain any constants in the general solution of ordinary differential equations. This point indicates that the differential algebraic equations are quite different from the ordinary differential equations.
The solution of the second equation in equation (6) must satisfy its initial condition, that is, it must be
1
0
0 ) ( 0
0 ( ) ( )
k
j
j j
t d N t
w
w (7)
Comment 1 When equation (7) be satisfied, there is a solution of equation (6). If equation (7) does not hold, the solution of equation (6) does not exist. Equation (7) must be established firstly. d(t)
must be a derivative up to k1 (kis the number of Ntimes of a nilpotent matrix), d(t) are some linear combinations of b(t).Therefore, it is required b(t) have a k1 derivative.
In fact, because there is obviously
1 1 1
( ) ( ) 1 ( )
2 0 2 2 2 2 2
0 0 0
( ) ( ) ( ) ( )
k k k
j j j j j
j j j
S w S N d t S N d t S NS S d t
whereS2Rm m andS2is vertible.
Thus acting 1 1 0
0
D
B A S S
N
on both sides of
1 0
2
0 0
S S
S
.(Note that the location N is the same, This is because the standard form Jordan of the similar matrix class is unique in the standard type Jordan except for the order of the blocks.)The formula (7) is established or not non-affected by the influence of the transformation.
The two-point boundary value problem of equation (2) also becomes
( ), (0) ( ) ( ), (0) ( )
Dv v e t v v T Nw w d t w w T
(8)
And for the first two-point boundary value problem of equation (8): whenI eD1T is invertible, have solution and the solution is
1 1 1 1 1 1( ) 1
0 0
( ) D t( D T ) T D z ( ) T D t z ( )
v t e e I
e D e z dz
e D e z dzFor the second two-point boundary value problem of equation (8), when
1
0
) ( )
(
0 )) ( )
0 ( (
k
j
j j
j
T d d
N , the only solution of second equation of system (8)is
1
0
) (
) ( )
(
k
j
j j
t d N t
w .
Therefore, both matrixs B and I eD1T are reversible, and under the assumption of
1
0
) ( )
(
0 )) ( )
0 ( (
k
j
j j
j
T d d
N , the solution of equation (8) is
1 1 1 1 1 1( ) 1
0 0
1 ( )
0
( ) ( ) ( ) ( ) ,
( ) ( ).
T T
D t D T D z D t z
k
j j
j
v t e e I e D e z dz e D e z dz
w t N d t
In fact, there is a third situation.: the matrixA andBare both singular. This situation can be categorized into the first two cases in two methods.
Mathod1. Assume rank(A)r (1r n), therefore, there are reversible matrix n n
SR and
n n
TR , make
0
0 0
r
I
SAT
(Irisrorder identity matrix.)
Thus the first differential equation in equation (1) becomes
( )
SATxSBTxSb t
According to the block matrix rule
1 2 1
3 4 2
( ) 0
( ) 0 0
r r
r
n r n r
x U U x u t
I
x U U x u t
Where 1
r r
U R , ) 2
r n r
U R （ ,.,., 1( )
r u t R , 2( )
n r
u t R .therefore
1 2 1
3 4 2
( ),
0 ( ).
r r n r
r n r
x U x U x u t U x U x u t
whenU4is reversible, we have
)] (
[ 3 2
1
4 U x u t
U
xnr r (9) therefore
) ( )
( )
(U1 U2U41U3 x u1 t U2U41u2 t
xr r (10)
r
x can be got by equation (10), re-substituting (9) can be solvedxnr, solution can be got.When is
4
U singular , the following method 2 can be used.
Method2. Assumrank(B)s,(1sn), therefore, the reversible matrix are exist Q,RRnn, make
0
0 0
s
I
QBR
(Isis s order identity matrix).
Thus the first differential equation in equation (1) can be changed to
) (t Qb QBRy y
QAR
According to the block matrix rule
1 2
3 4
( ) 0
( ) 0 0
s s
s
n s n s
V V I y u t
V V y u t
Where 1
s s
VR , ) ( ) ( ) ( )
2 3 , 4 , ( ) , ( )
s n s n s s n s n s s n s
s n s
V R （ ，V R V R u t R u t R .therefore
) (
) (
4 3
2 1
t u y V y V
t u y y V y V
s n s n s
s s s n s
WhenV4 is reversible, we can have
) (
1 4 3 1
4 V x V v t
V
xns s ns . (11)
Therefore
) ( )
( )
(V1V2V41V3 xs xs vs t V2V41vns t (12)
At this time, the equation (12) becomes the situation of theorem 2 and can be solved.
Application
Example Finding the solution of linear differential algebraic equations
1 1
2 2
3 3
2 1 3 0 2 3 sin
1 1 1 0 2 3 0
1 1 1 1 1 3 cos
x x t
x x
x x t
. (13)
2 1 3 0 2 3
1 1 1 2, 0 2 3 2
1 1 1 1 1 3
rankA rank rankB rank
,
let
1 1 0 1 0 2
1 2 0 , 0 1 1 ,
0 1 1 0 0 1
S T x Ty
Easy to verify that
1 0 0 0 0 0 sin
0 1 0 , 0 2 1 , ( ) sin
0 0 0 1 3 1 cos
t
SAT SBT Sb t t
t
.
At this time, the equation (13) becomes
1 1
2 2
3 3
1 0 0 0 0 0 sin
0 1 0 0 2 1 sin
0 0 0 1 3 1 cos
y y t
y y t
y y t
1 1
2 2
1
3 2
1 0 0 0 sin
0 1 0 2 sin
0 1 3 (1) cos
y y t
y y t
y
y t
y
So
1
2 2 3
1 2 3
sin
2 sin
0 3 cos
y t
y y y t y y y t
That is
1 1 2 1 2 3 2 1
1
cos , (sin cos ), 3
2
t
y tC y C C e t t y y C .
The general solution of system (13) is
1
2 2 1
3
2 1
cos 1
1
(sin cos )
2 3
(sin cos ) 3 2
2
t
t
t C
y
y t t C e C
y
t t C e C
1 1 1 2 1
2 2 2 2 1
3 3 3
2 1
1 0 2 3sin 4 cos 6 3
0 1 1 2(sin cos ) 4 3
0 0 1 3
(sin cos ) 3 2
2
t
t
t
x y y t t C e C
x T y y t t C e C
x y y
t t C e C
Solve with Method 2 Let
0 0 1 1 2 3
0 1 0 , 0 1 3
1 1 0 0 1 2
Q R
.
Easy to verify
1 0 0 1 0 2 cos
0 1 0 , 1 0 2 , ( ) 0
0 0 0 1 0 1 sin
t
QBR QAR Qb t
t
At this time, the equation (13) becomes
1 1
2 2
3 3
1 0 2 1 0 0 cos
1 0 2 0 1 0 0
1 0 2 0 0 0 sin
y y t
y y
y y t
That is
1 1
3
2 2
1
3 2
1 0 2 1 0 cos
1 0 2 0 1 0
1, 0 ( 1) sin
y y t
y
y y
y
y t
y
1 1
2 2
3 1
1 0 cos 2sin
1 0 2sin
sin
y y t t
y y t
y y t
The solution of system (13) is
1 1
2 1
3 1 2
1 3
( ) sin cos
2 2
1 1
( ) sin cos
2 2
1 1
( ) sin cos
2 2
t
t
t
y t t t C e
y t t t C e
y t t t C e C
1 1
2 2
3 3
x y
x R y
x y
1 2
2 2
1 2
3sin 4 cos 6 3
2(sin cos ) 4 3
3 3
sin cos 3 2
2 2
t
t
t
t t C e C
t t C e C
t t C e C
.
Reference
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