Structures of matrix triples via constant equivalence transforma-

From (4.11) and (4.12), one sees thatx₁^(α)(0)=φ^(α)(0⁻) and hence,x₁^(α)(0)=x₀^(α)(τ) for allα=0, . . . ,`(ν−1). Lemma4.8therefore implies the continuity of the solutionxat the pointsiτ, 06<sup>i</sup> 6`, and hence the IVP (4.1) has a unique solution given by (4.5).

Remark 4.10. The conditions i)-ii) in Theorem4.9are only sufficient but not necessary to ensure the existence and uniqueness of a solution to the IVP (4.1). The reason is that in some cases, the actual smoothness requirement forφis more relaxed than i). This is illustrated in the following example.

Example 4.11. Consider the IVP consisting of the DDAE

•0 1 0 0

‚ •x(t)˙

˙ y(t)

‚

=

•1 0 0 1

‚ •x(t) y(t)

‚

−

•1 0 0 1

‚ •x(t−τ) y(t−τ)

‚ +

•f1(t) f₂(t)

‚

, (4.13)

for t∈I=[0,`τ), with an initial functionφ(t) :=

•φ1(t) φ2(t)

‚

, for t ∈[−τ, 0].

Applying Theorem4.9to(4.13), one sees thatν=2and the smoothness requirement for φis thatφ∈C^`+1([−τ, 0],C²). Nevertheless, one can derive from(4.13)that

(x(t) =x(t−τ)+y(t˙ −τ)−f₁(t)−f˙₂(t), y(t) =y(t−τ)−f₂(t),

and therefore, the necessary and sufficient smoothness requirement for an initial func-tion is only φ1∈C([−τ, 0],C)and φ2∈C¹([−τ, 0],C).

Remark 4.12. It is worth to note that the results presented in this section are based on the regularity of the matrix pair (E,A). Under this condition, even if the time intervalI is unbounded, all the results in this section are still valid.

4.3. Systems on unbounded time intervals 48 Definition 4.13. Two triples of matrices (E₁,A₁,B₁) and (E₂,A₂,B₂) in (C^m,n)³are called equivalent if there exist nonsingular matricesP∈C^m,m andQ∈C^n,nsuch that

(E₂,A₂,B₂)=(PE₁Q,P A₁Q,P B₁Q).

If this is the case, we write (E₁,A₁,B₁)∼(E₂,A₂,B₂).

This approach, however, has achieved only few results in special cases of the matrix triple (E,A,B), see e.g. [30, 31]. Within this subsection, we first recall these results and then we consider another special case, where the matrix coefficients of (4.1) are pairwise commutative. In this subsection, we again assume that the DDAE (4.1a) is square.

Lemma 4.14. Assume that the matrix triple(E,A,B)∈(C^n,n)³is regular. Then,

(E,A,B)∼ 0

@ 2 4

N₁ E₁₂ E₁₃ 0 E₂₂ E₂₃

0 0 N₂

3 5,

2 4

N₃ A₁₂ A₁₃ 0 A₂₂ A₂₃

0 0 N₄

3 5,

2 4

I 0 0

0 B₂₂ 0

0 0 I

3 5 1

A, (4.14)

where Ni, i =1, . . . , 4, are strictly upper triangular matrices, and E₂₂ and A₂₂ are such that ¡

range(E₂₂)+range(A₂₂)¢_⊥

and kernel(E₂₂)∩kernel(A₂₂) are both{0}. Further-more, it is allowed that any of the block rows (and hence, the corresponding block col-umn) may not be present in(4.14).

Proof. See [30], Theorem 2.

Let us assume thatE,A,B are already in the form (4.14). Partitioning the variablex and an inhomogeneityf correspondingly, we can rewrite the DDAE (4.1a) as

N₁x˙₁(t)=N₃x₁(t)+x₁(t−τ)−E₁₂x˙₂(t)−E₁₃x˙₃(t)+A₁₂x₂(t)+A₁₃x₃(t)+f₁(t), (4.15a) E₂₂x˙₂(t)=A₂₂x₂(t)+B₂₂x₂(t−τ)+¡

−E₂₃x˙₃(t)+A₂₃x₃(t)+f₂(t)¢

, (4.15b)

N2x˙3(t)=N4x3(t)+x3(t−τ)+f3(t). (4.15c) First one sees that the functionx₃is uniquely determined by equation (4.15c). The next lemma gives us the explicit solution formula.

Lemma 4.15. Consider the equation(4.15c). Furthermore, suppose that f₃is sufficiently smooth. Then(4.15c)has the unique solution

x₃(t)= −

n−1X

i=0

d

dtN₂−N₄

¶i

f₃(t+(i+1)τ), (4.16)

for all t∈[−τ,∞).

Proof. SetL :=_dt^dN₂−N₄to be the linear operator, which maps a (continuously differ-entiable) functionx₃toN₂x˙₃−N4x₃. SinceN₂,N₄are strictly upper triangular matrices of dimension at mostn, we can directly verify thatLⁿ=0.

Consider the shift forward operator∆−τ:x₃(t)7→x₃(t+τ), which commutes withN₂,

N₄, _dt^d, and hence, it commutes withL. Thus, equation (4.15c) in the operator form becomes

d

dtN₂∆_−τx₃(t−τ)−N₄∆_−τx₃(t−τ)=x₃(t−τ)+f₃(t), or equivalently, one has (I−L∆_−τ)x₃(t−τ)= −f₃(t) for allt≥0.

This leads us to the solution formula

x₃(t)= −(I−L∆−τ)⁻¹f₃(t+τ)= −

n−1X

i=0

Lⁱ∆ⁱ_−τf₃(t+τ),

for allt≥ −τ, that is exactly (4.16).

In the same way, one can uniquely determinex₁from (4.15a), provided thatx₂is already known. Thus, we are lead to the question whetherx₂is uniquely determined from (4.15b). The answer will be positive if the regularity of the matrix pair (E₂₂,A₂₂) is guaranteed and then we obtain the following proposition.

Proposition 4.16. Let the matrix triple (E,A,B) associated with the DDAE (4.1a) be regular and be given in the form (4.14). Furthermore, suppose that the matrix pair (E₂₂,A₂₂)as in(4.14)is also regular. Then, the decomposition(4.14)completely describes the dynamics of the DDAE(4.1a).

Proof. See [30], Theorem 2.

Remark 4.17. Unfortunately, the regularity of the matrix pair (E₂₂,A₂₂) is only a suffi-cient condition for the existence and uniqueness of a solutionxto the DDAE (4.1a), as shown in Example4.18below.

Example 4.18. [30] Consider the DDAE(4.1a)with the matrix coefficients

E= 2 4

1 0 0 0 0 1 0 0 0 3 5, A=

2 4

0 1 0 0 0 0 0 0 1 3 5,B =

2 4

0 0 1 1 0 0 0 1 0 3 5.

The inhomogeneity f is chosen so that one obtains any desired solution. One can see that the form(4.14)for such matrices E , A, B has only one non-empty block row and one non-empty block column E₂₂=E , A₂₂=A, B₂₂=B . Clearly, the matrix pair(E₂₂,A₂₂)is singular. However, the DDAE(4.1a)still has a unique solution.

Another attempt to explore the structure of matrix triple (E,A,B) can be made in the case that these three matrices are pairwise commutative.

Definition 4.19. LetE, A,B∈C^n,n. The matrix triple (E,A,B) is called acommutative tripleifE,A, andBpairwise commute.

Definition 4.20. i) LetR=[ri j]∈C^m,nandS∈C^p,q. Thempbynqmatrix R⊗S=[ri jS]

is called theKronecker productofRandS.

ii) Thevecoperator creates a column vector from a matrixR=£

r₁ r₂ . . . rn⁄

∈C^m,n

4.3. Systems on unbounded time intervals 50 by stacking the column vectors ofRbelow one another:

vec(R)= 2 6 6 6 4

r₁ r₂ ... rn

3 7 7 7 5 .

Theorem 4.21. Let R∈C^m,n, W ∈C^n,pand S∈C^p,q. Then, vec(RW S)=(S^T ⊗R) vec(W).

Proof. See [70], p. 254.

The next lemma provides a necessary and sufficient condition to transform a DDAE into another one with commutative coefficients.

Lemma 4.22. Consider the matrix triple(E,A,B)∈(C^n,n)³associated with the square DDAE(4.1a). The following assertions are equivalent.

i) There exist two nonsingular matrices P , Q∈C^n,nsuch that the triple(PEQ,P AQ,P BQ) is commutative.

ii) There exists a nonsingular matrix X ∈C^n,nwhich solves the linear system 2

4

E^T ⊗A−A^T ⊗E E^T ⊗B−B^T ⊗E A^T ⊗B−B^T ⊗A

3

5vec(X)= 2 4 0 0 0 3

5. (4.17)

Proof. The proof is based on a trivial observation that ifP, Q are two matrices that make the triple (PEQ,P AQ,P BQ) commutative thenI,QP also do, and the converse holds as well. Therefore,X =QP will solve the linear system

E X A−AX E=0 E X B−B X E=0 AX B−B X A=0.

(4.18)

Using the vec operator and the Kronecker product, see Theorem4.21, we obtain system (4.17).

Lemma 4.23. Consider a commutative matrix triple(E,A,B)∈(C^n,n)³. Then there exists a nonsingular matrix U such that

(U EU⁻¹,U AU⁻¹,U BU⁻¹)=

•J 0

0 N

‚ ,

•A₁₁ 0 0 A22

‚ ,

•B₁₁ 0 0 B22

‚¶

,

where J is nonsingular, N is nilpotent. Furthermore, the matrix triples(J,A₁₁,B₁₁)and (N,A₂₂,B₂₂)are commutative triples.

Proof. First, making use of the Jordan canonical form, we see that there exists a non-singular matrixU such that

(U EU⁻¹,U AU⁻¹,U BU⁻¹)=

•J 0

0 N

‚ ,

•A₁₁ A₁₂ A₂₁ A₂₂

‚ ,

•B₁₁ B₁₂ B₂₁ B₂₂

‚¶

,

whereN is nilpotent. Due to the commutativity of the matrix pair (E,A), we see that the pair (U EU⁻¹,U AU⁻¹) also commutes, and hence

N A₂₁=A₂₁J. (4.19)

LetνN be the nilpotency index ofN. Scaling (4.19) withN^νN−1, one has 0=N^νN−1A₂₁J, and due to the invertibility ofJ, it follows that 0=N^νN−1A21. Continuing in the same way, eventually one has 0=N A₂₁and henceA₂₁=0. Analogously, one obtainsA₁₂=0, B₂₁=0,B₁₂=0 and then the proof is finished.

In the following theorem, we transform commutative matrix triples into block di-agonal form.

Theorem 4.24. Suppose that (E,A,B)∈(C^n,n)³ is a commutative triple. Then, there exists a nonsingular matrix U such that,

(U EU⁻¹,U AU⁻¹,U BU⁻¹)

= 0 B B B

@ 2 6 6 6 4

J^E 0 0 0

0 N₂₂^E 0 0 0 0 N₃₃^E 0 0 0 0 N₄₄^E

3 7 7 7 5 ,

2 6 6 6 4

A₁₁ 0 0 0

0 J^A 0 0

0 0 N₃₃^A 0 0 0 0 N₄₄^A

3 7 7 7 5 ,

2 6 6 6 4

B₁₁ 0 0 0

0 B₂₂ 0 0

0 0 J^B 0

0 0 0 N₄₄^B 3 7 7 7 5 1 C C C A

, (4.20)

where J^E, J^A, J^B are nonsingular, N₂₂^E, N₃₃^E, N₄₄^E, N₃₃^A, N₄₄^A, N₄₄^B are nilpotent. Moreover, if the matrix triple(E,A,B)is regular then the last block row and the last block column are not present.

Proof. Applying Lemma 4.23to the triple (E,A,B), we find a nonsingular matrixU₁ such that

(U₁EU₁⁻¹,U₁AU₁⁻¹,U₁BU₁⁻¹)=

•J^E 0 0 E¹₂₂

‚ ,

•A¹₁₁ 0 0 A¹₂₂

‚ ,

•B₁₁¹ 0 0 B₂₂¹

‚¶

, whereJ^Eis nonsingular,E¹₂₂is nilpotent and the matrix triple¡

E₂₂¹ ,A¹₂₂,B₂₂¹ ¢

is commu-tative.

Now applying Lemma4.23to the triple¡

A¹₂₂,E₂₂¹ ,B₂₂¹ ¢

, we obtainU₂such that (U₂A¹₂₂U₂⁻¹,U₂E₂₂¹U₂⁻¹,U₂B₂₂¹U₂⁻¹)=

•J^A 0 0 A²₂₂

‚ ,

•E₁₁² 0 0 E₂₂²

‚ ,

•B₁₁² 0 0 B₂₂²

‚¶

, whereJ^Ais nonsingular,A²₂₂is nilpotent and the matrix triple¡

E₂₂² ,A²₂₂,B₂₂²¢

is commu-tative. Furthermore, sinceE₂₂¹ is nilpotent, so isU₂E¹₂₂U₂⁻¹. This follows thatE²₁₁and E₂₂² are also nilpotent.

Finally, applying Lemma4.23to the triple¡

B₂₂² ,E₂₂² ,A²₂₂¢

, we obtainU₃such that (U₃B₂₂²U₃⁻¹,U₃E₂₂²U₃⁻¹,U₃A²₂₂U₃⁻¹)=

•J^B 0 0 B₂₂³

‚ ,

•E₁₁³ 0 0 E₂₂³

‚ ,

•A³₁₁ 0 0 A³₂₂

‚¶

, where J^B is nonsingular,B₂₂³ is nilpotent. Since A²₂₂, E₂₂² are nilpotent, it follows that E₁₁³ ,E₂₂³ , A³₁₁,A³₂₂are also nilpotent.

4.3. Systems on unbounded time intervals 52

Set ˜U₂:=

•I 0 0 U₂

‚

∈C^n,n, ˜U₃:=

•I 0 0 U₃

‚

∈C^n,nandU:=U˜₃U˜₂U₁, we have (U EU⁻¹,U AU⁻¹,U BU⁻¹)

= 0 B B B

@ 2 6 6 6 4

J^E 0 0 0

0 E₁₁² 0 0 0 0 E₁₁³ 0 0 0 0 E₂₂³

3 7 7 7 5 ,

2 6 6 6 4

A¹₂₂ 0 0 0

0 J^A 0 0

0 0 A³₁₁ 0 0 0 0 A³₂₂

3 7 7 7 5 ,

2 6 6 6 4

B₂₂¹ 0 0 0 0 B₁₁² 0 0

0 0 J^B 0

0 0 0 B₂₂³ 3 7 7 7 5 1 C C C A ,

whereJ^E, J^A, J^B are nonsingular,E₁₁² ,E₁₁³ ,E₂₂³ , A³₁₁, A³₂₂,B₂₂³ are nilpotent. This is ex-actly the desired form (4.20).

To prove the second claim, we assume that the triple (E,A,B) is regular. Then the triple (U EU⁻¹,U AU⁻¹,U BU⁻¹) is also regular. By direct calculation of the characteristic ex-ponential polynomial, we have

det(λE−A−e^−λτB) = det(λJ^E−A11−e^−λτB11)·det(λN₂₂^E −J^A−e^−λτB22)·

·det(λN₃₃^E −N₃₃^A−e^−λτJ^B)·det(λN₄₄^E −N₄₄^A −e^−λτN₄₄^B).

Thus, to obtain the second claim, it suffices to prove that det(λN₄₄^E −N₄₄^A−e^−λτN₄₄^B)=0.

Note that this is not a trivial problem, since the matrix λN₄₄^E −N₄₄^A −e^−λτN₄₄^B is not necessarily triangular.

Let ˜νbe the size of the matrixN₄₄^E, then¡ N₄₄^E¢ν˜

=0,¡ N₄₄^A¢ν˜

=0,¡ N₄₄^B¢ν˜

=0. Due to the commutativity of the triple¡

N₄₄^E ,N₄₄^A,N₄₄^B¢

we see that

‡λN₄₄^E −N₄₄^A −e^−λτN₄₄^B·3 ˜ν

=

3 ˜ν

X

i,j,k=0

i+j+k=3˜ν

ˆi+j i

!ˆ 3 ˜ν i+j

!

¡λN₄₄^E¢i¡

−N₄₄^A¢j‡

−e^−λτN₄₄^B·k

and since there is at least one of three numbersi,j,kbigger than ˜ν−1, we have

‡λN₄₄^E −N₄₄^A −e^−λτN₄₄^B·3 ˜ν

=0.

Consequently, we have det(λN₄₄^E −N₄₄^A −e^−λτN₄₄^B)=0.

In analogy to Proposition4.16, we will use the decomposition (4.20) to derive the explicit solution formula of the DDAE (4.1a).

Assuming that the matrix triple (E,A,B) is already in the form (4.20) and furthermore, that this triple is regular, Theorem4.24implies that the last block column and the last block row do not occur in (4.20). Partitioning the variablexand the inhomogeneity f correspondingly, we obtain the following system

J^Ex˙₁(t)=A₁₁x₁(t)+B₁₁x₁(t−τ)+f₁(t), (4.21a) N₂₂^E x˙₂(t)=J^Ax₂(t)+B₂₂x₂(t−τ)+f₂(t), (4.21b) N₃₃^E x˙3(t)=N₃₃^Ax3(t)+J^Bx3(t−τ)+f3(t). (4.21c)

Since J^E is nonsingular, in fact (4.21a) is an implicit formulation of a retarded DDE in variablex₁, so one can use the method of steps (Section2.5) to compute x₁. On the other hand, the explicit representation forx₂andx₃are given by the next theorem.

Theorem 4.25. Consider the IVP(4.1)for the square DDAE(4.1a). We assume that the following conditions are satisfied.

1. The inhomogeneity function f is sufficiently smooth.

2. The matrix triple(E,A,B)as in(4.1a)is commutative and regular.

3. The DDAE(4.1a)is already in the form(4.21).

Then the following assertions hold.

i) The solution x₃of (4.21c)is given by x₃(t)= −

n−1X

i=0 i

X

j=0

ˆi j

!

¡N₃₃^E ¢j¡

−N₃₃^A¢i−j

(J^B)⁻⁽ⁱ⁺¹⁾f₃^(j)(t+(i+1)τ), (4.22a) for all t≥ −τ.

ii) The solution x₂of (4.21b)is given by x₂(t)= −

n−1X

i=0

(J^A)⁻⁽ⁱ⁺¹⁾¡ N₂₂^E¢i‡

B₂₂x⁽ⁱ⁾₂ (t−τ)+f₂⁽ⁱ⁾(t)

·

, (4.22b)

for all t≥0.

iii) Furthermore, if B₂₂ is also a nilpotent matrix then for t ≥nτone can directly eval-uate the function x₂(t)in terms of only the inhomogeneity f₂as

x₂(t)= −

n−1X

i=0

(J^A)⁻⁽ⁱ⁺¹⁾ Xi j=0

ˆi j

!

(N₂₂^E )^j(−B22)^i−jf₂^(j)(t−τ(i−j)). (4.22c)

Proof. First we observe that the two operators _dt^d and∆_τcommute with any constant matrix. Furthermore, the matricesN₂₂^E,N₃₃^E ,N₃₃^A are nilpotent matrices with nilpotency indices smaller thann.

i) Scaling equation (4.21c) with (J^B)⁻¹we obtain

(J^B)⁻¹N₃₃^E x˙3(t)=(J^B)⁻¹N₃₃^Ax3(t)+x3(t−τ)+(J^B)⁻¹f3(t).

Since the matrix triple¡

N₃₃^E,N₃₃^A,J^B¢

is commutative, this implies that the two matrix triples¡

(J^B)⁻¹N₃₃^E , (J^B)⁻¹N₃₃^A,I¢ and¡

(J^B)⁻¹,N₃₃^E,N₃₃^A¢

are also commutative triples.

Furthermore, both (J^B)⁻¹N₃₃^E and (J^B)⁻¹N₃₃^A are nilpotent matrices of nilpotency in-dices smaller thann. In analogy to Lemma4.15, we have

x₃(t) = −

n−1X

i=0

(J^B)⁻¹N₃₃^E d

dt−(J^B)⁻¹N₃₃^A

¶i

(J^B)⁻¹f₃(t+(i+1)τ),

= −

n−1

X

i=0 i

X

j=0

ˆi j

!

¡(J^B)⁻¹N₃₃^E¢j¡

−(J^B)⁻¹N₃₃^A¢i−j

(J^B)⁻¹f₃^(j)(t+(i+1)τ),

and since the triple¡

(J^B)⁻¹,N₃₃^E ,N₃₃^A¢

is commutative, we obtain (4.22a).

4.3. Systems on unbounded time intervals 54 ii) Since the matrix triple (N₂₂^E,J^A,B₂₂) is commutative, so is the triple (N₂₂^E, (J^A)⁻¹,B₂₂).

Furthermore, the matrix (J^A)⁻¹N₂₂^E is nilpotent of the same nilpotency index asN₂₂^E. By rewriting equation (4.21b) as

(J^A)⁻¹N₂₂^E x˙₂(t)=x₂(t)+(J^A)⁻¹B₂₂x₂(t−τ)+(J^A)⁻¹f₂(t), the formula (4.22b) is straightforward due to the identity

I−(J^A)⁻¹N₂₂^E d dt

¶₋1

=

n−1

X

i=0

¡(J^A)⁻¹N₂₂^E¢i d dt

¶i

, and the commutativity of the matrix triple (N₂₂^E , (J^A)⁻¹,B₂₂).

iii) In order to obtain (4.22c), we first rewrite (4.21b) as I−(J^A)⁻¹N₂₂^E d

dt +(J^A)⁻¹B₂₂∆_τ

¶

x₂(t)= −(J^A)⁻¹f₂(t).

SinceB₂₂ is nilpotent, due to the commutativity of the matrix triple (N₂₂^E, (J^A)⁻¹,B₂₂) one can directly verify that the operator ˜L :=(J^A)⁻¹N₂₂^E _dt^d−(J^A)⁻¹B22∆_τsatisfies ˜Lⁿ=0.

Therefore, the inverse of the operatorI−L˜ exists and is given by

n−1

X

i=0

L˜ⁱ =

n−1

X

i=0

(J^A)⁻ⁱ ˆ i

X

j=0

ˆi j

!

(N₂₂^E )^j(−B₂₂)^i−j d dt

¶j

(∆τ)^i−j

! . Thus, we have

x₂(t)= −

n−1X

i=0

(J^A)⁻⁽ⁱ⁺¹⁾

i

X

j=0

ˆi j

!

(N₂₂^E )^j(−B₂₂)^i−j d dt

¶j

(∆τ)^i−jf₂(t),

which is exactly (4.22c). Note that 06^i−j6n, so this representation forx₂only makes sense fort≥nτ.

Except for the two special cases presented above, using constant equivalent trans-formations to study the structure of matrix triple coefficients in DDAEs is still an open problem. The author believes that, to study the structure of more than two matrices, with more than one operator acting on them, the class of constant equivalent transfor-mations is too restricted. We will consider a richer class of transfortransfor-mations in the next subsection.

Im Dokument Analysis and numerical solutions of delay differential algebraic equations (Seite 69-76)