A Shooting Method for Fully Implicit Index{2 Dierential{Algebraic Equations
Rene Lamour
Abstract
A shooting method for two{point{boundary value problems for fully implicit index{
1 and {2 dierential{algebraic equations is presented. A combination of the shooting equations with a method of the calculation of consistent initial values leads to a system of nonlinear algebraic equations with nonsingular Jacobian. Examples are given.
AMS(MOS) Classication: 65L10
Keywords: DAE, BVP, Index 2, consistent initial value
1 Introduction
In this paper we consider the fully implicit index{2 system
f
(x
0(t
)x
(t
)t
) = 0t
2a b
] (1.1)with the boundary condition
g
(x
(a
)x
(b
)) = 0:
(1.2)Such problems arise as models of electrical networks, chemical reactions or index{reduced sys- tems of mechanical motions. The possibility of the direct solution of given index{2 problems is very useful because
the index reduction changes the stability behaviour of the DAE
it is easier to reduce an index{3 system by only one step than by two steps.
The realization of the shooting method is strongly connected with an integration method that integrates index{2 problems well. This is given if
ker(
f
y0) =N
= const,e.g. for the BDF{method. The presented shooting method links a procedure for the cal- culation of consistent initial values (this procedure alone is very useful) with the shooting equations, and the Jacobian of the whole method becomes nonsingular.
In Chapter 2 we introduce some projectors which are useful for the description of index{2 DAE's. We dene a Green function for the explicit representation of of the solution of a linear index{2 DAE (Chapter 3). The numerical solution of (1.1),(1.2) by a shooting method is presented in Chapter 4 and some remarks to the numerical realization you can nd in Chapter 5. Numerical examples complete the paper (Chapter 6).
2 Index determination and projectors
We investigate the nonlinear DAE
f
(x
0(t
)x
(t
)t
) = 0 (2.1)as an IVP or BVP. For the numerical approximation of (2.1) it is necessary to know which index the DAE has.
Let
x
? be a solution of the considered problem (2.1) andA
(t
) :=f
y0(x
0?(t
)x
?(t
)t
)B
(t
) :=f
x0(x
0?(t
)x
?(t
)t
):
(2.2) We dene the chain of matrix functions Ma87]A
0 :=A B
0 :=B
;AP
0A
i+1 :=A
i+B
iQ
iB
i+1 := (B
i;A
i+1(P
0P
1P
i+1)0P
0P
i;1)P
i:
(2.3)Q
i is dened to be a projector ontoN
i := ker(A
i(t
))P
i :=I
;Q
i fori
0 andP
0 =:P
. Then the following denition is given.Definition. Ma91] The ordered pair f
A B
g of continuous matrix functions is said to be index{{tractable if all matricesA
j(t
)j
= 0:::
;1, within the chain (2.3) are singular with smooth nullspaces, andA
(t
) remains nonsingular.The nonlinear DAE (2.1) is said to be index{1{tractable locally around
x
? if the pair of the linearization (2.2) is so, too.The nonlinear DAE (2.1) is said to be index{
{tractable locally aroundx
? for>
1 if the pair of the linearization (2.2) is so in a neighbourhood of the solution.We are interested in the index{2{tractable case under the assumption that ker(
f
y0) =N
=const
, i.e.P
0= 0:
The following situation is given
A
1 =A
0+B
0Q B
1 = (B
0;A
1(PP
1)0)P
(2.4)A
2 =A
1+B
1Q
1 = (A
+BQ
+BPQ
1)(I
;P
1(PP
1)0PQ
1):
(2.5)Lemma 2.1
Denote by Q an arbitrary projector, and P :=I
;Q. Then the matrixM:=
I
;QZ
Pis nonsingular and its inverse is given by
M
;1=
I
+QZ
P:
Proof. We consider the equation
M
z
= 0:
It followsz
=QZ
Pz
)Pz
= 0 and we havez
= 0. 2Using Lemma 2.1 it is clear that
A
2 is nonsingular,A
~2 :=A
+BQ
+BPQ
1 (2.6)is nonsingular. However for arbitrary projectors
Q
andQ
1 we can test the nonsingularity ofA
~2 pointwise without any knowledge of the derivative of a projector as inA
2.Lemma 2.2
The following relations are valid ifP
0 = 0 :a:
)A
;12A
=P
1P
if
P
0 = 0 andQ
1Q
= 0 :b:
)A
;12A
=P
;Q
1c:
)A
;12BQ
=Q
d:
)A
;12BPQ
1 =Q
1;P
1(PQ
1)0PQ
1(2.7)
Proof. For
P
0 = 0 (PQ
1)0 =;(PP
1)0 is fullled. We obtain the relations a.) and b.) multiplying (2.5) from the right byP
1P
, and c.) byQ
. For d.) we multiply (2.5) byA
;12 and use the relation b.) and c.). 2Sometimes it is very useful to choose a special structure of the projectors. We focus our interest on the so{called canonical projector
Q
1 withQ
1 =Q
1A
;12BP
(2.8)(cf. Ma91b]). This projector fulls the condition
Q
1Q
= 0 and it is really calculable, becauseQ
1 =Q
1A
;12BP
=Q
1A
~;12BP:
(2.9)For our further investigations we assume that
P
0= 0 (2.10)and
Q
1Q
= 0 (2.11)where
Q
1 represents the canonical projector given in (2.9).3 Representation of the solution for a linear index{2 BVP
In this chapter we present a solution of the linear system
A
(t
)x
0(t
) +B
(t
)x
(t
) =q
(t
) (3.1)D
ax
(a
) +D
bx
(b
) =:
(3.2)First we consider the IVP (3.1) with the initial condition
P
(s
)P
1(s
)(x
(s
);) = 0:
(3.3)For the projector
P
1 :=I
;Q
1 we prefer now the canonical projectorQ
1 given in (2.9).(3.1),(3.3) is uniquely solvable for all
q
withfq
2C
:Q
1A
;12q
2C
1g(cf. Ma89]).For better understanding the index{2 case we split the solution
x
into three parts:u
:=PP
1x v
:=PQ
1x
andw
:=Qx:
(With
Q
1Q
= 0 alsoPP
1 andPQ
1 are projectors.)Multiplying (3.1) by
PP
1A
;12QP
1A
;12 andPQ
1A
;12 we obtainu
0;(PP
1)0u
+PP
1A
;12Bu
=PP
1A
;12q
;
v
0;PQ
1)0u
+QP
1A
;12Bu
+w
=QP
1A
;12q
v
=PQ
1A
;12q:
(3.4)The Fundamental Matrix of a DAE is given by the solution of the homogeneous IVP
AX
0+BX
= 0 (3.5)P
(s
)P
1(s
)(X
(s s
);I
) = 0:
(3.6)Using (3.4) we have
V
:=PQ
1X
0,U
:=PP
1X
=PP
1Y
whereY
solvesY
0= ((PP
1)0;PP
1A
;12B
)Y Y
(s s
) =I
and
W
:=QX
=PQ
1)0U
;QP
1A
;12BU
. With (3.6) andQW
=W
,PP
1U
=U
we createX
(t s
) =M
(t
)Y
(t s
)P
(s
)P
1(s
) (3.7) withM
(t
) := (I
+Q
(t
)PQ
1)0(t
) ;P
1(t
)A
;12B
(t
)]P
(t
)P
1(t
)). Using Lemma 2.1 it is easy to verify thatPP
1M
=PP
1M
;1 =PP
1:
Recall thatP
(t
)P
1(t
)Y
(t s
)P
(s
)P
1(s
) =Y
(t s
)P
(s
)P
1(s
) (3.8) (cf. Ma89]).Remark 1
Using the special projectorQ
=QP
1A
;12B
the splitted system (3.4) and also the matrixM
of the fundamental matrixX
look a little bit easier. However, this projector is very dicult to calculate, because of the derivatives inA
2.Now we are looking for a representation of the solution of the IVP (3.1),(3.3). Using the fundamental matrix
Y
(t s
) we have for the component uu
(t
) =Y
(t s
)(PP
1+ZstY
(s
)PP
1A
;12qd
):
(3.9)With (3.8) we transform (3.9) into
u
(t
) =Y
(t s
)P
(s
)P
1(s
)(+ZstM
(s
)Y
(s
)PP
1()h
()d
) withh
(t
) :=PP
1A
;12q
. For the other components we havev
=PQ
1A
;12q
andw
=QP
1A
;12q
+PQ
1)0u
;QP
1A
;12Bu
+v
0 and thereforex
=u
+v
+w
=
Mu
+q
(t
)with
q
(t
) := (PQ
1+QP
1)A
;12q
+PQ
1A
;12q
)0.x
(t
) =X
(t s
)(+ZstX
(s
)h
()d
)+q
(t
) (3.10) represents the solution of the IVP (3.1),(3.3). Now we consider the solutionx
(t
) int
=a
andt
=b
.x
(a
) =X
(a a
)+q
(a
) andx
(b
) =X
(b a
)(+RabX
(a
)h
()d
)+q
(b
) with unknown. The boundary condition (3.2) requiresS
:= (D
aX
(a a
)+D
bX
(b a
))=
;D
bX
(b a
)RabX
(a
)h
()d
=: ~ (3.11) with :=;(D
aq
(a
) +D
bq
(b
)).Theorem 3.1
Ma91] Let (3.1), (3.2) be a tractable index{2 equation and the projectors full (2.10) and (2.11). Then, for arbitrary right{hand sidesq
withq
2C a b
]PQ
1A
;12q
2C
1a b
] and 2im(D
aD
b) (3.1), (3.2) have a unique solution iker(
S
) = im(I
;PP
1) (3.12)im(
S
) = im(D
aD
b):
(3.13)Proof. The unique solution of (3.1), (3.2) is related to the solution of the IVP (3.1), (3.3).
Then it becomes clear that only
PP
1 inuences the solution. Hence, we require for =PP
1:
(3.14)We are looking for solutions of (3.11) in the set P := f
z
jz
2 im(PP
1)g:
The right{hand side of (3.11) fulls 2im(D
aD
b). This means that (3.13) is a necessary condition for the solvability of (3.11). The structure ofX
(t s
) providesS
=SP
(a
)P
1(a
) or ker(S
)im(I
;P
(a
)P
1(a
)):
(3.15)! Let
2P be a solution of (3.11), then also + 2P solves (3.11) with 2ker(S
). The uniqueness requires that P\ker(S
) = f0g) ker(S
) im(I
;PP
1). With (3.15) formulae (3.12) follows.Let (3.12) be valid, and
1 and 2 2P denote two solutions of (3.11). Then 1 ;2 2 ker(S
), but ker(S
)\P=f0gand 1 =2. 2S
; denotes the generalized reexive inverse ofS
withS
;SS
;=S
;SS
;S
=S
and
S
;S
=P
(a
)P
1(a
):
(3.16)This representation of
S
; is possible if (3.12) is valid. We multiply (3.11) byS
;:P
(a
)P
1(a
)=S
;:
~With (3.10) we have
x
(t
) =X
(t a
)S
;+Rat
X
(t
)h
()d
;RabX
(t a
)S
;D
bX
(b
)h
()d
+q
(t
):
Using (3.16)S
;D
bX
(b a
) =PP
1;S
;D
aX
(a a
) (3.17)is valid and, therefore,
X
(t
);X
(t a
)(PP
1;S
;D
aX
(a a
))X
(a
) =X
(t a
)S
;D
aX
(a
):
Now we introduce the Green's functionG
(t s
) :=( +
X
(t a
)S
;D
aX
(a s
)s > t
;
X
(t a
)S
;D
bX
(b s
)s < t
(3.18) and the following Theorem holds.Theorem 3.2
Let Theorem 3.1 be valid andS
;denotes a reexive inverse ofS
withS
;S
=PP
1(a
). The solution of the BVP (3.1),(3.2) has the representationx
(t
) =X
(t a
)S
;+ZabG
(t
)h
()d
+q
(t
):
4 Numerical solution by shooting method
The solution of BVP's by shooting methods requires that we are able to integrate the consid- ered equation. For DAE's this means that we have to make available consistent initial values.
We nd dierent ideas for the calculation of consistent initial values.
Numerical dierentiation
is used in the code DASSL (see also LPG91]),Formel manipulation
is proposed by Han90],Special structure
of the DAE is used by AP91].We use a general approach taking advantage of the given subspaces by using special projectors.
A further disadvantage of shooting methods for DAE's is the singularity of the Jacobian. This problem we overcome as in the index 1 case (cf. Lam91]) by the combination of the shooting equation with the equation for the calculation of consistent initial values.
4.1 Consistent initial values
We consider the nonlinear DAE
f
(x
0(t
)x
(t
)t
) = 0:
(4.1)For a better understanding of the index{2 case let us consider the transferable or index{1 case. The assumption ker(
f
y0) =N
(t
) allows us to transform (4.1) intof
((Px
)0(t
);P
0x
(t
)x
(t
)t
) = 0 (4.2) (see GM87]).Index 1:
We are looking for consistent initial values for the IVP (4.2) withP
(s
)(x
(s
);) = 0:
(4.3)We split
x
=Px
+Qx
=:u
+v
and denotey
:= (Px
)0;P
0x
(recall thatPy
=y
). Let us dene :=y
+v
then (4.2) considered int
=s
is written as followsf
(P u
+Q s
) = 0 (4.4)with known
u
=P
and searched . The Jacobian of (4.4) is given byf
y0P
+f
x0Q
(=A
1)which is nonsingular for the index{1 case.
Index 2:
Now we transfer this technique to the index{2 case. Here we assume the stronger condition thatker(
f
y0) =N
const. (i.e.P
= const,P
0= 0) (4.5) This assumption does not restrict the class of numerically solvable problems. We know (compare with the example in GP84]) that only assumption (4.5) saves numerical success for index{2 problems. With (4.5), (4.2) has the structuref
((Px
)0x t
) = 0:
(4.6)We represent (4.6) in a more detailed way
f
((PP
1x
)0+ (PQ
1x
)0x t
) = 0 orf
((PP
1x
)0;(PP
1)0PP
1x
+ (PQ
1x
)0;(PQ
1)0PP
1x x t
) = 0 (4.7) with the initial condition(
PP
1)(s
)(x
(s
);) = 0:
(4.8)We split x into the components
x
=PP
1x
+PQ
1x
+Qx
=:u
+v
+w
and we deney
:= (PP
1x
)0;(PP
1)0PP
1x
and:=
y
+v
+w:
(4.9)Here
PP
1y
=y
is valid. Now (4.7) readsf
(PP
1 +v
0;(PQ
1)0u u
+ (PQ
1+Q
)s
) = 0:
(4.10) The trouble in formula (4.10) is caused by the unknown termv
0. Therefore, we consider (4.10) in a neighbouring points
+h
and replacev
0by the nite dierencev
0v
h;v
We use the symbol (
h :
)h= ()(s
+h
).f
:=f
(PP
1 + (PQ
1 )h;PQ
1h
;(
PQ
1)0u u
+ (PQ
1+Q
)s
) = 0 (4.11)f
h :=f
((PP
1 )h+ (PQ
1 )h;PQ
1h
;f(
PQ
1)0u
ghu
h+f(PQ
1+Q
) ghs
+h
) = 0 (4.12) withu
h=u
+hu
0=u
+h
(PP
1 + (PP
1)0u
):
Theorem 4.1
Let the projectorQ
1 depend onu
=PP
1x
andt
only,P
0= 0 andQ
1Q
= 0.Then the system (4.11),(4.12) has a nonsingular Jacobian in the point(
y
?+v
?0;(PQ
1)0u
?x
?s
) ifh
is suciently small, where (:
)? is the part ofx
?.Proof. The Jacobian is given by
J
=0
B
B
B
B
@
@ f
@ @ f
@
h@ @ f
h@ f
h@
h1
C
C
C
C
A
:
with@ f
@
=A
(PP
1;1hPQ
1) +B
(PQ
1+Q
)@ f
@
h = 1hA
fPQ
1gh@ f
h@
= ;1hA
hPQ
1;h
fA
(PQ
1)0;B
ghPP
1@ f
h@
h = fA
(PP
1+h1PQ
1) +B
(PQ
1 +Q
)gh(4.13)
with
A
:=f
y0 andB
:=f
x0. Using the identities given in Lemma 2.2 and (P
;Q
1)PP
1 =PP
1(
P
;Q
1)PQ
1 = ;P
1P
= (P
1;Q
) (4.14)for
P
0= 0 we haveJ
=
A
2 0 0A
h2!
(4.15)with
=
I
+P
1((1 +h1)PQ
1)0)PQ
1h1f
P
;Q
1gh(PQ
1);h
fA
;12 (A
(PQ
1)0;B
)ghPP
1;
h1(
P
;Q
1)fPQ
1ghI
+fP
1((1 + 1h)PQ
1)0)PQ
1gh!
:
To show the nonsingularity of , rst we consider the equation1 2
!
= 0 (4.16)
with
=
I
+P
1((1 + 1h)PQ
1)0)PQ
1 ;1h(P
;Q
1)fPQ
1gh1hf
P
;Q
1gh(PQ
1)I
+fP
1((1 + h1)PQ
1)0)PQ
1gh!
:
The result of the rst equation of (4.16) after multiplying byQ
1 isQ
11 = 0 (4.17)and the second equation multiplied by
Q
h1 yieldsQ
h12 = 0:
(4.18)Using (4.17) and (4.18) in (4.16) gives
1 =2 = 0:
Now we consider the matrix
~ := ;0 0
h
fA
;12 (A
(PQ
1)0;B
)ghPP
1 0!
:
~ depends continuously onh
and ~ = forh
=h
, i.e. ifh
is suciently small, then is nonsingular. 2Remark 2
We consider (4.11), (4.12) for a linear DAEA
(t
)x
0(t
) +B
(t
)x
(t
) =q
(t
)and obtain
A
(s
)(PP
1 +fPQ
1 gh;PQ
1h
; (PQ
1)0u
)+
B
(s
)(u
+ (PQ
1+Q
) ) =q
(s
) (4.19)A
(s
+h
)fPP
1 gh+fPQ
1 gh;PQ
1h
; f(PQ
1)0u
gh)+
B
(s
+h
)(u
h+f(PQ
1+Q
) gh) =q
(s
+h
):
(4.20)We multiply (4.19) by
A
;12 (s
) and (4.20) byA
;12 (s
+h
) and obtain (P
;Q
1)(PP
1 + (PQ
1 )h;PQ
1h
; (PQ
1)0u
)+
A
;12 (s
)B
(s
)(u
+ (PQ
1+Q
) ) =A
;12 (s
)q
(s
) (4.21)f(
P
;Q
1)(PP
1 )gh+ (PQ
1 )h;PQ
1h
; f(PQ
1)0u
gh)+
A
;12 (s
+h
)B
(s
+h
)fu
+ (PQ
1+Q
) gh) = fA
;12q
gh:
(4.22) The multiplication of (4.21) and (4.22) byPQ
1 and fPQ
1gh respectively, yieldsPQ
1 =PQ
1A
;12q
(s
)f
PQ
1 gh = fPQ
1A
;12 ghq
(s
+h
) orhlim!0
f
PQ
1 gh;PQ
1h
=v
0= (PQ
1A
;12q
)0 (4.23)Using (4.23) we consider (4.19) for
h
!0 and realize exactly the linear DAE becausePP
1 + (PQ
1 )0;(PQ
1)0u
= (Px
)0and
u
+ (PQ
1+Q
) =x:
The accuracy of the numerical solution depends essentially on the condition of the matrix
. We investigate the condition of the matrix . Using Lemma 2.1 andPQ
1(P
;Q
1) = 0 the inverse of is given by ;1=
I
;P
1((1 + h1)PQ
1)0)PQ
1 ; 1h(P
;Q
1)fPQ
1gh1hf
P
;Q
1gh(PQ
1)I
;fP
1((1 + h1)PQ
1)0)PQ
1gh!
:
(4.24)We introduce the constants
K
1 :=kK
2 :=kP
1(PQ
1)0PQ
1kCab]:
Using the Taylor expansion (
PQ
1)h =PQ
1+h
(PQ
1)0+O
(h
2) we obtaink
k (1 +K
1+K
2) + 2hK
1+O
(h
)k
;1k (1 +K
1+K
2) + 2hK
1+O
(h
):
This proves theCorollary
The condition of is bounded by cond()((1 +K
1+K
2) + 2hK
1+O
(h
))2:
Remark 3
The essential part of the estimation shows that cond()O
(h
;2) in the worst case. This is not surprising because of the numerical dierentiation.4.2 The shooting method
We consider now a boundary value problem
f
(x
0(t
)x
(t
)t
) = 0t
2a b
] (4.25)g
(x
(a
)x
(b
)) = 0:
(4.26)The idea of shooting is well known. We subdivide the interval
a b
] intom
subintervalsa
=t
0< t
1<
< t
m =b
and we look for the initial values
z
i :=x
(t
i)i
= 0::: m
;1.x
(t s z
) denotes the solution of the IVP (4.25) withPP
1(s
)(x
(s
);z
) = 0:
The
z
i have to full the boundary conditiong
(z
0x
(t
mt
m;1z
m;1)) = 0 (4.27)and the matching condition
(
PP
1)i(z
i;x
(t
it
i;1z
i;1)) = 0:
(4.28) (The symbol ()i reads like ()(t
i)).The disadvantage of the system (4.27),(4.28) is the singularity of the Jacobian (as in the index{1 case). However we use the same idea that solves this problem in the index 1 case (cf. Lam91]), too. We combine the shooting equation with the equations (4.11),(4.12) for the determination of the initial values. For this aim we split the variable
z
i into the partsz
i= (PP
1)iz
i+ (PQ
1)iz
i+Q
iz
i =:u
i+v
i+w
iand with i :=
y
i+v
i+w
i (cf.(4.9)) we havez
i =u
i+(PQ
1+Q
)i i. The shooting equations are given byg
(u
0+ (PQ
1+Q
)0 0x
(t
mt
m;1u
m;1)) = 0 (4.29)u
i;(PP
1)ix
(t
it
i;1u
i;1) = 0 i=1:::m;1 (4.30) and the equations for the determination of the initial values int
i readf
i :=f
((PP
1)i i+fPQ
1 ghi;(PQ
1)i ih
;(PQ
1)0iu
iu
i+ (PQ
1+Q
)i it
i) = 0 (4.31)f
hi :=f
(fPP
1 ghi+fPQ
1 ghi;(PQ
1)i ih
;f(PQ
1)0iu
ighu
hi+f(PQ
1+Q
)i ight
i+h
) = 0i
= 0::: m
;1:
For the variableu
we have to ensure thatPP
1u
=u:
This is valid for
t
ii
= 1::: m
;1 by using (4.30). Foru
0 we extend (4.29) tog
(u
0+ (PQ
1+Q
)0 0x
(t
mt
m;1u
m;1)) +K
;1(I
;PP
1)0u
0= 0 (4.32)where
K
is a nonsingular matrix withim(
g
0xag
x0b)Mim(K
;1(I
;PP
1)0) =R
n:
For the calculation of the unknowns
u
0u
m;1 we have to solve the equations (4.30) and (4.32). But in (4.32) also 0 is engaged. This means that we extend our system to the equations for the determination of the initial values (4.31) in the pointt
0.Theorem 4.2
Let the assumptions of Theorem 3.1 and 4.1 be fullled andQ
1=Q
1(t
) only, in this case system (4.32),(4.30) and (4.31) (fori
= 0) has a nonsingular Jacobian.Proof. We order the variables in the following way:
= (u
0::: u
m;1 0 h0) then the Jacobian
J
is given byJ
=0
B
B
B
B
B
B
B
B
B
B
B
B
@
G
auG
bu ...G
av 0M
0I
...... ... ...
M
m;2I
...:::::::::::::::::::::::::::::::::::
F
u0 ...J
01
C
C
C
C
C
C
C
C
C
C
C
C
A
(4.33)
where we have used the following abbreviations
G
au :=g
x0a +K
;1(I
;PP
1)0G
bu:=g
x0bX
(t
mt
m;1)G
av :=g
x0a(PQ
1+Q
)0M
i := ;(PP
1)i+1X
(t
i+1t
i)F
u0 :=
;
A
0(PQ
1)00+ (BPP
1)0f;
A
0(PQ
1)00+ (BPP
1)0gh(I
+O
(h
))!
J
0 denotes the matrix (4.13) int
0:
We investigate the equationJ
= 0:
(4.34)J
denotes the matrix with the structure ofJ
, but the matricesJ
0 are replaced byJ
0 :=
A
2 0 0A
h2!
(cf. (4.15)). The second to m{th equation of (4.34) is given byu
i+1= (PP
1)i+1X
(t
i+1t
i)u
ii
= 0::: m
;2:
(4.35) This leads tou
m;1= (PP
1)m;1X
(t
m;1t
0)u
0:
The latter 2n{dimensional equations are given by
J
0
0h
0
!
=
;
A
0(PQ
1)00+ (BPP
1)0f;
A
0(PQ
1)00+ (BPP
1)0gh(I
+O
(h
))!
u
0:
(4.36)We multiply the rst equation of (4.36) by
A
;12 and the second one by (A
h2);1. Using (4.24) and (2.8) we have0 = ((
P
;Q
1)(PQ
1)0;A
;12BPP
1)u
0 (4.37)h
0 = f(
P
;Q
1)(PQ
1)0;A
;12BPP
1gh(I
+O
(h
))u
0:
(4.38) Setting (4.37) in the rst equation of (4.34) yields(
g
x0a+K
;1(I
;PP
1)0)u
0+g
x0bX
(t
mt
m;1)u
m;1+g
x0a(PQ
1+Q
)0 0= (
g
x0a+K
;1(I
;PP
1)0+g
x0bX
(t
mt
0) +g
x0a(;Q Q
1(PQ
1)0;A
;12B
]PP
1)u
0= (
g
x0a(I
;Q Q
1(PQ
1)0;A
;12B
]PP
1) +g
x0bX
(t
mt
0) +K
;1(I
;PP
1)0))u
0= (
g
x0aX
(t
0t
0) +g
x0bX
(t
mt
0) +K
;1(I
;PP
1)0)u
0:
Usingu
0=PP
1u
0 we obtain(
g
0xaX
(t
0t
0) +g
0xbX
(t
mt
0))u
0=Su
0= 0and with (3.12) it follows that
u
0 2ker(S
) oru
0 = (I
;PP
1)z
and this givesu
0 = 0. With (4.35) we haveu
i = 0 and, using (4.36), 0and 0h= 0. The matrixJ
is a regular perturbation ofJ
, i.e. forh
suciently small alsoJ
is nonsingular. 2Remark 4
To determine the unknownsu
0::: u
m;1 0 h0 and i hi i=1:::m;1
we have to solve the systemsf(4.32), (4.30),(4.31) with
i
= 0gand (4.31) fori
= 1m
;1.All these systems have a nonsingular Jacobian. Consequently, for the solution of the BVP (1.1), (1.2) the represented shooting method is realizable with a common Newton{like method (taking into consideration the structure of the Jacobian, of course).
However a part of the unknowns of the nonlinear algebraic system are partially projected vectors (
u
=PP
1u
). The question is:Does the Newton method save this condition ? Or, in other words :
Is the correction
u
also aPP
1 projection?To answer this question let us consider the nonlinear system
S
g :=g
(u
0+ (PQ
1+Q
)0 0x
(t
mt
m;1u
m;1)) +K
;1(I
;PP
1)0u
0 (4.39)S
iu :=u
i;(PP
1)ix
(t
it
i;1u
i;1) i=1:::m;1 (4.40)S
0 :=8
>
>
>
>
<
>
>
>
>
:
f
((PP
1)0 0+ fPQ1gh0;(h PQ1)00;(
PQ
1)00u
0u
0+ (PQ
1+Q
)0 0t
0)f
(fPP
1 gh0+ fPQ1gh0;(h PQ1)00;f(
PQ
1)00u
0ghu
h0+f(PQ
1+Q
)0 0ght
0+h
)(4.41) The Newton{correction
:= (