Differential Algebraic Equations

Abstract

Let H be a Hilbert space and $\nu \in \mathbb {R}$. We saw in the previous chapter how initial value problems can be formulated within the framework of evolutionary equations. More precisely, we have studied problems of the form

$$\displaystyle \begin{aligned} \begin {cases} \left (\partial _{t,\nu }M_{0}+M_{1}+A\right )U=0 & \text{ on }\left (0,\infty \right ),\\ M_{0}U(0{\scriptstyle {+}})=M_{0}U_{0} \end {cases} \end{aligned} $$

for U ₀ ∈ H, M ₀, M ₁ ∈ L(H) and $A\colon \operatorname {dom}(A)\subseteq H\to H$ skew-selfadjoint; that is, we have considered material laws of the form

$$\displaystyle M(z)\mathrel{\mathop:}= M_{0}+z^{-1}M_{1}\quad (z\in \mathbb {C}\setminus \{0\}). $$

Let H be a Hilbert space and $\nu \in \mathbb {R}$. We saw in the previous chapter how initial value problems can be formulated within the framework of evolutionary equations. More precisely, we have studied problems of the form

(10.1)

for U ₀ ∈ H, M ₀, M ₁ ∈ L(H) and $A\colon \operatorname {dom}(A)\subseteq H\to H$ skew-selfadjoint; that is, we have considered material laws of the form

Here, the initial value is attained in a weak sense as an equality in the extrapolation space H ⁻¹(A). The first line is also meant in a weak sense since the left-hand side turned out to be a functional in $H_{\nu }^{-1}(\mathbb {R};H)\cap L_{2,\nu }(\mathbb {R};H^{-1}(A))$. In Theorem 9.4.3 it was shown that the latter problem can be rewritten as

$$\displaystyle \begin{aligned} \left(\partial_{t,\nu}M_{0}+M_{1}+A\right)U=\delta_{0}M_{0}U_{0}. \end{aligned}$$

In this chapter we aim to inspect initial value problems a little closer but in the particularly simple case when A = 0. However, we want to impose the initial condition for U and not just M ₀ U. Thus, we want to deal with the problem

(10.2)

for two bounded operators M ₀, M ₁ and an initial value U ₀ ∈ H. This class of differential equations is known as differential algebraic equations since the operator M ₀ is allowed to have a non-trivial kernel. Thus, (10.2) is a coupled problem of a differential equation (on $(\operatorname {ker} M_{0})^{\bot }$) and an algebraic equation (on $\operatorname {ker} M_{0}$). We begin by treating these equations in the finite-dimensional case; that is, $H=\mathbb {C}^{n}$ and $M_{0},M_{1}\in \mathbb {C}^{n\times n}$ for some $n\in \mathbb {N}$.

10.1 The Finite-Dimensional Case

Throughout this section let $n\in \mathbb {N}$ and $M_{0},M_{1}\in \mathbb {C}^{n\times n}$.

Definition

We define the spectrum of the matrix pair (M ₀, M ₁) by

and the resolvent set of the matrix pair (M ₀, M ₁) by

Remark 10.1.1

• (a) It is immediate that σ(M ₀, M ₁) is closed since the mapping $z\mapsto \det (zM_{0}+M_{1})$ is continuous.

• (b) Note in particular that the spectrum (the set of eigenvalues) of a matrix A corresponds in this setting to the spectrum of the matrix pair (1, −A).

In contrast to the case of the spectrum of one matrix, it may happen that $\sigma (M_{0},M_{1})=\mathbb {C}$ (for example we can choose M ₀ = 0 and M ₁ singular). More precisely, we have the following result.

Lemma 10.1.2

The set σ(M ₀, M ₁) is either finite or equals the whole complex plane $\mathbb {C}$ . If σ(M ₀, M ₁) is finite then $\operatorname {card}(\sigma (M_{0},M_{1}))\leqslant n.$

Proof

The function $z\mapsto \det (zM_{0}+M_{1})$ is a polynomial of order less than or equal to n. If it is constantly zero, then $\sigma (M_{0},M_{1})=\mathbb {C}$ and otherwise $\operatorname {card}(\sigma (M_{0},M_{1}))\leqslant n$. □

Definition

The matrix pair (M ₀, M ₁) is called regular if $\sigma (M_{0},M_{1})\ne \mathbb {C}$.

The main problem in solving an initial value problem of the form (10.2) is that one cannot expect a solution for each initial value $U_{0}\in \mathbb {C}^n$ as the following simple example shows.

Example 10.1.3

Let $M_{0}=\begin {pmatrix} 1 & 1\\ 0 & 0 \end {pmatrix}, \,M_{1}=\begin {pmatrix} 1 & 0\\ 0 & 1 \end {pmatrix}$ and let $U_{0}\in \mathbb {C}^{2}$. We assume that there exists a solution $U\colon \mathbb {R}_{\geqslant 0}\to \mathbb {C}^{2}$ satisfying (10.2); that is,

The second and third equation yield that the second coordinate of U ₀ has to be zero. Then, for $U_{0}=(x,0)\in \mathbb {C}^{2}$ the unique solution of the above problem is given by

$$\displaystyle \begin{aligned} U(t)=\big(U_{1}(t),U_{2}(t)\big)=(x\mathrm{e}^{-t},0) \quad (t\geqslant0). \end{aligned}$$

Definition

We call an initial value $U_{0}\in \mathbb {C}^{n}$ consistent for (10.2) if there exists ν > 0 and $U\in C(\mathbb {R}_{\geqslant 0};\mathbb {C}^{n})\cap L_{2,\nu }(\mathbb {R}_{\geqslant 0};\mathbb {C}^{n})$ such that (10.2) holds. We denote the set of all consistent initial values for (10.2) by

Remark 10.1.4

It is obvious that $\operatorname {IV}(M_{0},M_{1})$ is a subspace of $\mathbb {C}^{n}$. In particular, $0\in \operatorname {IV}(M_{0},M_{1})$.

It is now our goal to determine the space $\operatorname {IV}(M_{0},M_{1})$. One possibility for doing so uses the so-called quasi-Weierstraß normal form.

Proposition 10.1.5 (Quasi-Weierstraß Normal Form )

Assume that (M ₀, M ₁) is regular. Then there exist invertible matrices $P,Q\in \mathbb {C}^{n\times n}$ such that

$$\displaystyle \begin{aligned} PM_{0}Q=\begin{pmatrix} 1 & 0\\ 0 & N \end{pmatrix},\quad PM_{1}Q=\begin{pmatrix} C & 0\\ 0 & 1 \end{pmatrix}, \end{aligned}$$

where $C\in \mathbb {C}^{k\times k}$ and $N\in \mathbb {C}^{(n-k)\times (n-k)}$ for some k ∈{0, …, n}. Moreover, the matrix N is nilpotent; that is, there exists $\ell \in \mathbb {N}$ such that N ^ℓ = 0.

Proof

Since (M ₀, M ₁) is regular we find $\lambda \in \mathbb {C}$ such that λM ₀ + M ₁ is invertible. We set and obtain

Let now $P_{2}\in \mathbb {C}^{n\times n}$ such that

for some invertible matrix $J\in \mathbb {C}^{k\times k}$ and a nilpotent matrix $\widetilde {N}\in \mathbb {C}^{(n-k)\times (n-k)}$ (use the Jordan normal form of M _0,1 here). Then

Now, by the nilpotency of $\widetilde {N}$, the matrix $(1-\lambda \widetilde {N})$ is invertible by the Neumann series. We set

and obtain

$$\displaystyle \begin{aligned} P_{3}M_{0,2}=\begin{pmatrix} 1 & 0\\ 0 & (1-\lambda\widetilde{N})^{-1}\widetilde{N} \end{pmatrix},\text{ and }\quad P_{3}M_{1,2}=\begin{pmatrix} J^{-1}-\lambda & 0\\ 0 & 1 \end{pmatrix}. \end{aligned}$$

Note that $(1-\lambda \widetilde {N})^{-1}\widetilde {N}$ is nilpotent, since the matrices commute and $\widetilde {N}$ is nilpotent. Thus, the assertion follows with , , P = P ₃ P ₂ P ₁, and $Q=P_{2}^{-1}$. □

It is clear that the matrices P, Q, C and N in the previous proposition are not uniquely determined by M ₀ and M ₁. However, the size of N and C as well as the degree of nilpotency of N are determined by M ₀ and M ₁ as the following proposition shows.

Proposition 10.1.6

Let $P,Q\in \mathbb {C}^{n\times n}$ be invertible such that

$$\displaystyle \begin{aligned} PM_{0}Q=\begin{pmatrix} 1 & 0\\ 0 & N \end{pmatrix},\quad PM_{1}Q=\begin{pmatrix} C & 0\\ 0 & 1 \end{pmatrix}, \end{aligned}$$

where $C\in \mathbb {C}^{k\times k}$, $N\in \mathbb {C}^{(n-k)\times (n-k)}$ for some k ∈{0, …, n}, and N is nilpotent. Then (M ₀, M ₁) is regular and

• (a) k is the degree of the polynomial $z\mapsto \det (zM_{0}+M_{1})$.

• (b) N ^ℓ = 0 if and only if

  $$\displaystyle \begin{aligned} \sup_{|z|\geqslant r}\left\Vert z^{-\ell+1}(zM_{0}+M_{1})^{-1} \right\Vert <\infty \end{aligned}$$

  for one (or equivalently all) r > 0 such that $B\left (0,r\right )\supseteq \sigma (M_{0},M_{1})$.

Proof

First, note that

$$\displaystyle \begin{aligned} \det(zM_{0}+M_{1})=\frac{1}{\det P\,\det Q}\det\begin{pmatrix} z+C & 0\\ 0 & zN+1 \end{pmatrix}=\frac{1}{\det P\,\det Q}\det(z+C) \end{aligned}$$

for all $(z\in \mathbb {C})$. Hence, (M ₀, M ₁) is regular and

$$\displaystyle \begin{aligned} k=\deg\det((\cdot)+C)=\deg\det((\cdot)M_{0}+M_{1}), \end{aligned}$$

which shows (a). Moreover, we have ρ(M ₀, M ₁) = ρ(−C) and

$$\displaystyle \begin{aligned} \left(zM_{0}+M_{1}\right)^{-1}=Q\begin{pmatrix} (z+C)^{-1} & 0\\ 0 & (zN+1)^{-1} \end{pmatrix}P\quad (z\in\rho(M_{0},M_{1})), \end{aligned}$$

and hence, for r > 0 with $B\left (0,r\right )\supseteq \sigma (M_{0},M_{1})$ we have

$$\displaystyle \begin{aligned} \left\Vert (zM_{0}+M_{1})^{-1} \right\Vert \leqslant K_1\left\Vert (zN+1)^{-1} \right\Vert \quad (|z|\geqslant r) \end{aligned}$$

for some K ₁⩾0, since $\sup _{|z|\geqslant r}\left \Vert (z+C)^{-1} \right \Vert <\infty$. Now let $\ell \in \mathbb {N}$ such that N ^ℓ = 0. Then

$$\displaystyle \begin{aligned} \left\Vert (zN+1)^{-1} \right\Vert =\left\Vert \sum_{k=0}^{\ell-1}(-1)^{k}z^{k}N^{k} \right\Vert \leqslant K_2\left\vert z \right\vert ^{\ell-1}\quad (\left\vert z \right\vert \geqslant r) \end{aligned}$$

for some constant K ₂⩾0 and thus,

$$\displaystyle \begin{aligned} \left\Vert (zM_{0}+M_{1})^{-1} \right\Vert \leqslant K_1K_2\left\vert z \right\vert ^{\ell-1}\quad (|z|\geqslant r). \end{aligned}$$

Assume on the other hand that

$$\displaystyle \begin{aligned} \sup_{|z|\geqslant r}\left\Vert z^{-\ell+1}(zM_{0}+M_{1})^{-1} \right\Vert <\infty \end{aligned}$$

for some $\ell \in \mathbb {N}$ and r > 0 with $\sigma (M_{0},M_{1})\subseteq B\left (0,r\right )$. Then there exist $\widetilde {K}_{1},\widetilde {K}_{2}\geqslant 0$ such that

$$\displaystyle \begin{aligned} \left\Vert (zN+1)^{-1} \right\Vert \leqslant\left\Vert \begin{pmatrix} (z+C)^{-1} & 0\\ 0 & (zN+1)^{-1} \end{pmatrix} \right\Vert \leqslant\widetilde{K}_{1}\left\Vert \left(zM_{0}+M_{1}\right)^{-1} \right\Vert \leqslant\widetilde{K_{2}}\left\vert z \right\vert {}^{\ell-1} \end{aligned}$$

for all $z\in \mathbb {C}$ with $\left \vert z \right \vert \geqslant r$. Now, let $p\in \mathbb {N}$ be minimal such that N ^p = 0. We show that $p\leqslant \ell$ by contradiction. Assume p > ℓ. Then we compute

$$\displaystyle \begin{aligned} 0& =\lim_{n\to\infty}\frac{1}{n^{\ell}}(nN+1)^{-1}N^{p-\ell-1}=\lim_{n\to\infty}\sum_{k=0}^{p-1}(-1)^{k}n^{k-\ell}N^{k+p-\ell-1}\\ & = \lim_{n\to\infty}\sum_{k=0}^{\ell-1}(-1)^{k}n^{k-\ell}N^{k+p-\ell-1} + (-1)^{\ell}N^{p-1} \\ & = (-1)^{\ell}N^{p-1}, \end{aligned} $$

which contradicts the minimality of p. □

Theorem 10.1.7

Let (M ₀, M ₁) be regular and $P,Q\in \mathbb {C}^{n\times n}$ be chosen according to Proposition 10.1.5 . Let $k=\deg \det ((\cdot )M_{0}+M_{1})$ . Then

$$\displaystyle \begin{aligned} \operatorname{IV}(M_{0},M_{1})=\left\{ U_{0}\in\mathbb{C}^{n} \,;\, Q^{-1}U_{0}\in\mathbb{C}^{k}\times\{0\} \right\}. \end{aligned}$$

Moreover, for each $U_{0}\in \operatorname {IV}(M_{0},M_{1})$ the solution U of (10.2) is unique and satisfies $U\in C(\mathbb {R}_{\geqslant 0};\mathbb {C}^{n})\cap C^{1}(\mathbb {R}_{>0};\mathbb {C}^{n})$ as well as

Proof

Let $C\in \mathbb {C}^{k\times k}$ and $N\in \mathbb {C}^{(n-k)\times (n-k)}$ be nilpotent as in Proposition 10.1.5. Obviously U is a solution of (10.2) if and only if both is continuous on $\mathbb {R}_{\geqslant 0}$ and solves

(10.3)

Clearly, if $Q^{-1}U_{0}=(x,0)\in \mathbb {C}^{k}\times \{0\}$ then V  given by for t⩾0 is a solution of (10.3) for ν > 0 large enough. On the other hand, if V  given by $V(t)=(V_{1}(t),V_{2}(t))\in \mathbb {C}^{k}\times \mathbb {C}^{n-k}$ ($t\geqslant 0$) is a solution of (10.3) then we have

$$\displaystyle \begin{aligned} \partial_{t,\nu}NV_{2}+V_{2}=0\quad \text{on }\left(0,\infty\right). \end{aligned}$$

Since N is nilpotent, there exists $\ell \in \mathbb {N}$ with N ^ℓ = 0. Hence,

$$\displaystyle \begin{aligned} N^{\ell-1}V_{2}(t) = -N^{\ell-1}\partial_{t,\nu}NV_{2}(t) = \partial_{t,\nu}N^\ell V_{2}(t) =0\quad (t>0), \end{aligned}$$

which in turn implies ∂ _t,ν N ^ℓ−1 V ₂ = 0 on $\left (0,\infty \right )$. Using again the differential equation, we infer N ^ℓ−2 V ₂(t) = 0 for t > 0. Inductively, we deduce V ₂(t) = 0 for t > 0 and by continuity , which yields $V_{0}=Q^{-1}U_{0}\in \mathbb {C}^{k}\times \{0\}$. The uniqueness follows from Proposition 10.2.7 below. □

10.2 The Infinite-Dimensional Case

Let now M ₀, M ₁ ∈ L(H). Again, it is our aim to determine the space of consistent initial values for the problem

(10.4)

Here, consistent initial values are defined as in the finite-dimensional setting:

Definition

We call an initial value U ₀ ∈ H consistent for (10.4) if there exist ν > 0 and $U\in C(\mathbb {R}_{\geqslant 0};H)\cap L_{2,\nu }(\mathbb {R}_{\geqslant 0};H)$ such that (10.4) holds. We denote the set of all consistent initial values for (10.4) by

Before we try to determine $\operatorname {IV}(M_{0},M_{1})$ we prove a regularity result for solutions of (10.4).

Proposition 10.2.1

Let ν > 0, U ₀ ∈ H and $U\in C(\mathbb {R}_{\geqslant 0};H)\cap L_{2,\nu }(\mathbb {R}_{\geqslant 0};H)$ be a solution of (10.4). Then and

Proof

We extend U to $\mathbb {R}$ by 0. First, observe that is continuous, since U is continuous and . By Lemma 9.4.2 (with A = 0), we obtain

Since ∂ _t,ν is closed and M ₀ is bounded, ∂ _t,ν M ₀ is closed as well. Since M ₁ is bounded, therefore also ∂ _t,ν M ₀ + M ₁ is closed. Thus, and therefore , and

□

We now come back to the space $\operatorname {IV}(M_{0},M_{1})$. Since we are now dealing with an infinite-dimensional setting, we cannot use normal forms to determine $\operatorname {IV}(M_{0},M_{1})$ without dramatically restricting the class of operators. Thus, we follow a different approach using so-called Wong sequences.

Definition

We set

and for $k\in \mathbb {N}_0$ we set

The sequence $(\operatorname {IV}_{k})_{k\in \mathbb {N}_0}$ is called the Wong sequence associated with (M ₀, M ₁).

Remark 10.2.2

By induction, we infer $\operatorname {IV}_{k+1}\subseteq \operatorname {IV}_{k}$ for each $k\in \mathbb {N}_0$.

As in the matrix case, we denote by

the resolvent set of (M ₀, M ₁) .

Lemma 10.2.3

Let $k\in \mathbb {N}_0$ . Then:

• (a) M ₁(zM ₀ + M ₁)⁻¹ M ₀ = M ₀(zM ₀ + M ₁)⁻¹ M ₁ for each z ∈ ρ(M ₀, M ₁).

• (b) $(zM_{0}+M_{1})^{-1}M_{0}[\operatorname {IV}_{k}]\subseteq \operatorname {IV}_{k+1}$ for each z ∈ ρ(M ₀, M ₁).

• (c) If $x\in \operatorname {IV}_{k}$ we find x ₁, …, x _k+1 ∈ H such that for each z ∈ ρ(M ₀, M ₁) ∖{0}

  $$\displaystyle \begin{aligned} (zM_{0}+M_{1})^{-1}M_{0}x=\frac{1}{z}x+\sum_{\ell=1}^{k}\frac{1}{z^{\ell+1}}x_{\ell}+\frac{1}{z^{k+1}}(zM_{0}+M_{1})^{-1}x_{k+1}. \end{aligned}$$

• (d) If $\rho (M_{0},M_{1})\ne \varnothing$ then $M_{1}^{-1}[M_{0}[\overline {\operatorname {IV}_{k}}]]\in \overline {\operatorname {IV}_{k+1}}$.

Proof

The proofs of the statements (a) to (c) are left as Exercise 10.6. We now prove (d). If k = 0 there is nothing to show. So assume that the statement holds for some $k\in \mathbb {N}_0$ and let $x\in M_{1}^{-1}\left [M_{0}\left [\overline {\operatorname {IV}_{k+1}}\right ]\right ]$. Since $\overline {\operatorname {IV}_{k+1}}\subseteq \overline {\operatorname {IV}_{k}}$, we infer $x\in M_{1}^{-1}\left [M_{0}\left [\overline {\operatorname {IV}_{k}}\right ]\right ]\subseteq \overline {\operatorname {IV}_{k+1}}$ by induction hypothesis. Hence, we find a sequence $(w_{n})_{n\in \mathbb {N}}$ in $\operatorname {IV}_{k+1}$ with w _n → x. Let now z ∈ ρ(M ₀, M ₁). Then, by (b), we have $(zM_{0}+M_{1})^{-1}M_{0}w_{n}\in \operatorname {IV}_{k+2}$ for each $n\in \mathbb {N}$ and hence, $(zM_{0}+M_{1})^{-1}M_{0}x\in \overline {\operatorname {IV}_{k+2}}$. Moreover, since $M_{1}x\in M_{0}\left [\overline {\operatorname {IV}_{k+1}}\right ]$, we find a sequence $(y_{n})_{n\in \mathbb {N}}$ in $\operatorname {IV}_{k+1}$ with M ₀ y _n → M ₁ x. Setting now

(where, again, we have used (b)) for $n\in \mathbb {N}$, we derive

$$\displaystyle \begin{aligned} x_{n} & =(zM_{0}+M_{1})^{-1}zM_{0}x+(zM_{0}+M_{1})^{-1}M_{0}y_{n}\\ & =x-\left(zM_{0}+M_{1}\right)^{-1}(M_{1}x-M_{0}y_{n}) \to x \end{aligned} $$

as n →∞ and thus, $x\in \overline {\operatorname {IV}_{k+2}}$. □

The importance of the Wong sequence becomes apparent if we consider solutions of (10.4).

Lemma 10.2.4

Assume that $\rho (M_{0},M_{1})\ne \varnothing$ . Let ν > 0 and $U\in L_{2,\nu }(\mathbb {R}_{\geqslant 0};H)\cap C(\mathbb {R}_{\geqslant 0};H)$ be a solution of (10.4). Then $U(t)\in \bigcap _{k\in \mathbb {N}_0}\overline {\operatorname {IV}_{k}}$ for each t⩾0.

Proof

We prove the claim, $U(t)\in \overline {\operatorname {IV}_k}$ for all $t\geqslant 0$ and $k\in \mathbb {N}_0$, by induction. For k = 0 there is nothing to show. Assume now that $U(t)\in \overline {\operatorname {IV}_{k}}$ for each t⩾0 and some $k\in \mathbb {N}_0$. By Proposition 10.2.1 we know that

and thus, in particular,

$$\displaystyle \begin{aligned} M_{0}U(t)-M_{0}U_{0}+\int_{0}^{t}M_{1}U(s)\,\mathrm{d} s=0\quad (t\geqslant0). \end{aligned}$$

Let now t⩾0 and h > 0. Then we infer

$$\displaystyle \begin{aligned} M_{0}U(t+h)-M_{0}U(t)+M_{1}\int_{t}^{t+h}U(s)\,\mathrm{d} s=0 \end{aligned}$$

and hence,

$$\displaystyle \begin{aligned} \int_{t}^{t+h}U(s)\,\mathrm{d} s\in M_{1}^{-1}\big[M_{0}[\overline{\operatorname{IV}_{k}}]\big]\subseteq\overline{\operatorname{IV}_{k+1}} \end{aligned}$$

by Lemma 10.2.3 (d). Since U is continuous, the fundamental theorem of calculus implies $U(t)\in \overline {\operatorname {IV}_{k+1}}$, which yields the assertion. □

In particular, the space of consistent initial values has to be a subspace of $\bigcap _{k\in \mathbb {N}_0}\overline {\operatorname {IV}_{k}}$. We now impose an additional constraint on the operator pair (M ₀, M ₁), which is equivalent to being regular in the finite-dimensional setting (cf. Proposition 10.1.6).

Definition

We call the operator pair (M ₀, M ₁) regular if there exists ν ₀⩾0 such that

• (a) $\mathbb {C}_{\operatorname {Re}>\nu _{0}}\subseteq \rho (M_{0},M_{1})$, and

• (b) there exist C⩾0 and $\ell \in \mathbb {N}$ such that for all $z\in \mathbb {C}_{\operatorname {Re}>\nu _{0}}$ we have $\left \Vert (zM_{0}+M_{1})^{-1} \right \Vert \leqslant C\left \vert z \right \vert ^{\ell -1}$.

Moreover, we call the smallest $\ell \in \mathbb {N}$ satisfying (b) the index of (M ₀, M ₁) , which is denoted by $\operatorname {ind}(M_{0},M_{1})$.

Remark 10.2.5

Note that for matrices M ₀ and M ₁ the index equals the degree of nilpotency of N in the quasi-Weierstraß normal form by Proposition 10.1.6.

From now on, we will require that (M ₀, M ₁) is regular. First, we prove an important result on the Wong sequence in this case.

Proposition 10.2.6

Let (M ₀, M ₁) be regular, $k\in \mathbb {N}_0$ , and $k\geqslant \operatorname {ind}(M_{0},M_{1})$ . Then

$$\displaystyle \begin{aligned} \overline{\operatorname{IV}_{k}}=\overline{\operatorname{IV}_{\operatorname{ind}(M_{0},M_{1})}}. \end{aligned}$$

Proof

We show that $\overline {\operatorname {IV}_{k}}=\overline {\operatorname {IV}_{k+1}}$ for each $k\geqslant \operatorname {ind}(M_{0},M_{1}).$ Since the inclusion “⊇” holds trivially, it suffices to show $\operatorname {IV}_{k}\subseteq \overline {\operatorname {IV}_{k+1}}$. For doing so, let $k\geqslant \operatorname {ind}(M_{0},M_{1})$ and $x\in \operatorname {IV}_{k}$. By Lemma 10.2.3 (c) we find x ₁, …, x _k+1 ∈ H such that

$$\displaystyle \begin{aligned} (zM_{0}+M_{1})^{-1}M_{0}x=\frac{1}{z}x+\sum_{\ell=1}^{k}\frac{1}{z^{\ell+1}}x_{\ell}+\frac{1}{z^{k+1}}(zM_{0}+M_{1})^{-1}x_{k+1} \end{aligned}$$

for each $z\in \mathbb {C}_{\operatorname {Re}>\nu _{0}}$. Since $k\geqslant \operatorname {ind}(M_{0},M_{1}),$ we derive

$$\displaystyle \begin{aligned} z(zM_{0}+M_{1})^{-1}M_{0}x\to x\quad (\operatorname{Re} z\to\infty), \end{aligned}$$

and since the elements on the left-hand side belong to $\operatorname {IV}_{k+1}$, by Lemma 10.2.3 (b), the assertion immediately follows. □

We now prove that in case of a regular operator pair (M ₀, M ₁) the solution of (10.4) for a consistent initial value U ₀ is uniquely determined.

Proposition 10.2.7

Let (M ₀, M ₁) be regular, $U_{0}\in \operatorname {IV}(M_{0},M_{1})$ , and ν > 0 such that a solution $U\in C(\mathbb {R}_{\geqslant 0};H)\cap L_{2,\nu }(\mathbb {R}_{\geqslant 0};H)$ of (10.4) exists. Then this solution is unique. In particular

$$\displaystyle \begin{aligned} (\mathcal{L}_{\rho}U)(t)=\frac{1}{\sqrt{2\pi}}\big((\mathrm{i} t+\rho)M_{0}+M_{1}\big)^{-1}M_{0}U_{0}\quad (\mathit{\text{a.e. }}t\in\mathbb{R}) \end{aligned}$$

for each $\rho >\max \{\nu ,\nu _{0}\}$.

Proof

By Proposition 10.2.1 we have and

Applying the Fourier–Laplace transformation, $\mathcal {L}_{\rho }$, for $\rho >\max \{\nu ,\nu _{0}\}$ we deduce

$$\displaystyle \begin{aligned} (\mathrm{i} t+\rho)M_{0}\big(\mathcal{L}_{\rho}U(t)-\frac{1}{\sqrt{2\pi}}\frac{1}{\mathrm{i} t+\rho}U_{0}\big)+M_{1}\mathcal{L}_{\rho}U(t)=0\quad (\text{a.e. }t\in\mathbb{R}) \end{aligned}$$

which in turn yields

$$\displaystyle \begin{aligned} \mathcal{L}_{\rho}U(t)=\frac{1}{\sqrt{2\pi}}\big((\mathrm{i} t+\rho)M_{0}+M_{1}\big)^{-1}M_{0}U_{0}\quad (\text{a.e. }t\in\mathbb{R}) \end{aligned}$$

and, in particular, proves the uniqueness of the solution. □

Remark 10.2.8

Let U be a solution of (10.4) for a consistent initial value U ₀. Then the formula in Proposition 10.2.7 shows that $U\in \bigcap _{\rho >\nu _0}L_{2,\rho }(\mathbb {R};H)$ and hence, we also have . If ν ₀ > 0 then we even obtain $U\in L_{2,\nu _0}(\mathbb {R};H)$ since $\sup _{\rho >\nu _0}\left \Vert U \right \Vert { }_{L_{2,\rho }(\mathbb {R};H)}=\sup _{\rho >\nu _0} \left \Vert \mathcal {L}_{\rho }U \right \Vert { }_{L_2(\mathbb {R};H)} < \infty$ (cp. Lemma 8.1.1), and therefore also .

One interesting consequence of the latter proposition is the following.

Corollary 10.2.9

Let (M ₀, M ₁) be regular. Then the operator $M_{0}\colon \operatorname {IV}(M_{0},M_{1})\to H$ is injective.

Proof

Let $U_{0}\in \operatorname {IV}(M_{0},M_{1})$ with M ₀ U ₀ = 0. By Proposition 10.2.7, the solution U of (10.4) with satisfies

$$\displaystyle \begin{aligned} \mathcal{L}_{\rho}U(t)=\frac{1}{\sqrt{2\pi}}\big((\mathrm{i} t+\rho)M_{0}+M_{1}\big)^{-1}M_{0}U_{0}=0 \end{aligned}$$

and hence, U = 0, which in turn implies . □

We now want to determine the space $\operatorname {IV}(M_{0},M_{1})$ in terms of the Wong sequence.

Proposition 10.2.10

Let (M ₀, M ₁) be regular. Then

$$\displaystyle \begin{aligned} \operatorname{IV}_{\operatorname{ind}(M_{0},M_{1})}\subseteq\operatorname{IV}(M_{0},M_{1})\subseteq\overline{\operatorname{IV}_{\operatorname{ind}(M_{0},M_{1})}}. \end{aligned}$$

Proof

The second inclusion follows from Lemma 10.2.4 and Proposition 10.2.6. Let now $U_{0}\in \operatorname {IV}_{\operatorname {ind}(M_{0},M_{1})}$ and set

Let . By Lemma 10.2.3 (c) we find x ₁, …, x _k+1 ∈ H such that

$$\displaystyle \begin{aligned} V(z)=\frac{1}{\sqrt{2\pi}}\left(\frac{1}{z}U_{0}+\sum_{\ell=1}^{k}\frac{1}{z^{\ell+1}}x_{\ell}+\frac{1}{z^{k+1}}\left(zM_{0}+M_{1}\right)^{-1}x_{k+1}\right)\quad (z\in\mathbb{C}_{\operatorname{Re}>\nu_{0}}). \end{aligned}$$

In particular, we read off that $V\in \mathcal {H}_2(\mathbb {C}_{\operatorname {Re}>\nu };H)$ for all ν > ν ₀. Now, let ν > ν ₀. By the Theorem of Paley–Wiener (more precisely by Corollary 8.1.3) there exists $U\in L_{2,\nu }(\mathbb {R}_{\geqslant 0};H)$ such that

$$\displaystyle \begin{aligned} \left(\mathcal{L}_{\rho}U\right)(t)=V(\mathrm{i} t+\rho)\quad (\text{a.e. }t\in\mathbb{R},\rho>\nu). \end{aligned}$$

Moreover,

$$\displaystyle \begin{aligned} zV(z)-\frac{1}{\sqrt{2\pi}}U_{0}=\frac{1}{\sqrt{2\pi}}\left(\sum_{\ell=1}^{k}\frac{1}{z^{\ell}}x_{\ell}+\frac{1}{z^{k}}\left(zM_{0}+M_{1}\right)^{-1}x_{k+1}\right)\quad (z\in\mathbb{C}_{\operatorname{Re}>\nu}) \end{aligned}$$

and hence $\left (z\mapsto zV(z)-\frac {1}{\sqrt {2\pi }}U_{0}\right )\in \mathcal {H}_2(\mathbb {C}_{\operatorname {Re}>\nu };H)$ as well. Since

we infer and, thus, is continuous by Theorem 4.1.2. Hence, $U\in C(\mathbb {R}_{\geqslant 0};H)$ and since $\operatorname {\mathrm {spt}} U\subseteq \mathbb {R}_{\geqslant 0}$ we derive . Finally, by the definition of V ,

$$\displaystyle \begin{aligned} M_{0}\left(zV(z)-\frac{1}{\sqrt{2\pi}}U_{0}\right)=-\frac{1}{\sqrt{2\pi}}M_{1}(zM_{0}+M_{1})^{-1}M_{0}U_{0}=-M_{1}V(z) \end{aligned}$$

for all $z\in \mathbb {C}_{\operatorname {Re}>\nu }.$ Hence,

from which we see that U solves (10.4). □

Finally, we treat the case when $\operatorname {IV}(M_{0},M_{1})$ is closed.

Theorem 10.2.11

Let (M ₀, M ₁) be regular and $\operatorname {IV}(M_{0},M_{1})$ closed. Then the operator $S\colon \operatorname {IV}(M_{0},M_{1})\to C(\mathbb {R}_{\geqslant 0};H)$ , which assigns to each initial state, $U_0\in \operatorname {IV}(M_{0},M_{1})$ , its corresponding solution, $U\in C(\mathbb {R}_{\geqslant 0};H)$ , of (10.4) is bounded in the sense that

$$\displaystyle \begin{aligned} S_{n}\colon \operatorname{IV}(M_{0},M_{1})\to C([0,n];H),\quad U_{0}\mapsto SU_{0}|{}_{[0,n]} \end{aligned}$$

is bounded for each $n\in \mathbb {N}$.

Proof

By Proposition 10.2.10 we infer that $\operatorname {IV}(M_{0},M_{1})=\overline {\operatorname {IV}_{k}}$ with . Let ν > ν ₀⩾0. By Proposition 10.2.7 and Corollary 8.1.3, there exists C⩾0 such that

$$\displaystyle \begin{aligned} \sqrt{2\pi}\left\Vert \partial_{t,\nu}^{-k}SU_{0} \right\Vert {}_{L_{2,\nu}(\mathbb{R}_{\ge{0}};H)}& = \left\Vert \left(z\mapsto z^{-k}(zM_{0}+M_{1})^{-1}M_{0}U_{0}\right) \right\Vert {}_{\mathcal{H}_2(\mathbb{C}_{\operatorname{Re}>\nu};H)} \\ & \leqslant C\sqrt{\frac{\pi}{\nu}}\left\Vert M_{0}U_{0} \right\Vert {}_{H} \end{aligned} $$

for each $U_{0}\in \operatorname {IV}(M_{0},M_{1})$, where we have used the regularity of (M ₀, M ₁) and

$$\displaystyle \begin{aligned} \left\Vert (z\mapsto z^{-1}M_{0}U_{0}) \right\Vert {}_{\mathcal{H}_2(\mathbb{C}_{\operatorname{Re}>\nu};H)}=\sqrt{\frac{\pi}{\nu}}\left\Vert M_{0}U_{0} \right\Vert {}_{H}. \end{aligned}$$

In particular, $S\colon \operatorname {IV}(M_{0},M_{1})\to H^{-1}(\partial _{t,\nu }^k)$ is bounded. Since $L_{2,\nu _{0}}(\mathbb {R}_{\geqslant 0};H)\hookrightarrow H^{-1}(\partial _{t,\nu }^k)$ continuously, we infer that $S\colon \operatorname {IV}(M_{0},M_{1})\to L_{2,\nu _{0}}(\mathbb {R}_{\geqslant 0};H)$ is bounded by the closed graph theorem. Hence, also

$$\displaystyle \begin{aligned} S_{n}\colon \operatorname{IV}(M_{0},M_{1})\to L_2([0,n];H),\quad U_{0}\mapsto SU_{0}|{}_{[0,n]} \end{aligned}$$

is bounded for each $n\in \mathbb {N}$ and since C([0, n];H)↪L ₂([0, n];H) continuously, we infer that S _n is bounded with values in C([0, n];H) again by the closed graph theorem. □

Remark 10.2.12

The variant of the closed graph theorem used in the proof above is the following: Let X, Y  be Banach spaces and Z a Hausdorff topological vector space (e.g. a Banach space) such that Y ↪Z continuously. Let T : X → Z be linear and continuous with T[X] ⊆ Y . Then T ∈ L(X, Y ). Indeed, by the closed graph theorem it suffices to show that T : X → Y  is closed. For doing so, let (x _n)_n be a sequence in X with x _n → x and Tx _n → y for some x ∈ X, y ∈ Y . Then Tx _n → Tx in Z by the continuity of T and Tx _n → y in Z by the continuous embedding. Hence, y = Tx and thus, T is closed.

10.3 Comments

The theory of differential algebraic equations in finite dimensions is a very active field. The main motivation for studying these equations comes from the modelling of electrical circuits and from control theory (see e.g. [28] and Exercise 10.5). The main reference for the statements presented in the first part of this chapter is the book by Kunkel and Mehrmann [57]. Of course, also in the finite-dimensional case Wong sequences can be used to determine the consistent initial values, see Exercise 10.1. For instance, in [13] the connection between Wong sequences and the quasi-Weierstraß normal form for matrix pairs is studied. Of course, the theory is not restricted to linear and homogeneous problems. Indeed, in the non-homogeneous case it turns out that the set of consistent initial values also depends on the given right-hand side.

The theory of differential algebraic equations in infinite dimensions is less well studied than the finite-dimensional case. We refer to [114], where the theory of C ₀-semigroups is used to deal with such equations. Moreover, we refer to [97, 98], where sequences of projectors are used to decouple the system. Moreover, there exist several references in the Russian literature, where the equations are called Sobolev type equations (see e.g. [111]). The results on infinite-dimensional problems presented here are based on [121, 124, 125]. In [124] the focus was on systems with index 0 with an emphasis on exponential stability and dichotomy.

We also add the following remark concerning the result in Theorem 10.2.11. By Corollary 10.2.9 we know that $M_{0}\colon \operatorname {IV}(M_{0},M_{1})\to H$ is injective. If $\operatorname {IV}(M_{0},M_{1})$ is closed, it follows that the operator $C\colon \operatorname {dom}(C)\subseteq \operatorname {IV}(M_{0},M_{1})\to \operatorname {IV}(M_{0},M_{1})$ given by

is well-defined and closed. Using this operator, C, Theorem 10.2.11 states that if $\operatorname {IV}(M_{0},M_{1})$ is closed then − C generates a C ₀-semigroup on $\operatorname {IV}(M_{0},M_{1})$. The precise statement can be found in [121, Theorem 5.7]. Moreover, C is bounded if $\operatorname {IV}_{\operatorname {ind}(M_{0},M_{1})}$ is closed (cf. Exercise 10.7).

Exercises

Exercise 10.1

Let $M_{0},M_{1}\in \mathbb {C}^{n\times n}$ such that (M ₀, M ₁) is regular and define the Wong sequence $(\operatorname {IV}_{j})_{j\in \mathbb {N}_0}$ associated with (M ₀, M ₁). Moreover, let $P,Q\in \mathbb {C}^{n\times n}$, $C\in \mathbb {C}^{k\times k},$ and $N\in \mathbb {C}^{(n-k)\times (n-k)}$ be as in the quasi-Weierstraß normal form for (M ₀, M ₁) with N nilpotent (cf. Proposition 10.1.5). We decompose a vector $x\in \mathbb {C}^{n}$ into and $\widehat {x}\in \mathbb {C}^{n-k}$ such that . Prove that

$$\displaystyle \begin{aligned} x\in\operatorname{IV}_{j}\Leftrightarrow \widehat{Q^{-1}x} \in\operatorname{ran} N^{j}\quad (j\in\mathbb{N}_0). \end{aligned}$$

Moreover, show that for each z ∈ ρ(M ₀, M ₁) we have

$$\displaystyle \begin{aligned} \operatorname{IV}_{j}=\operatorname{ran}\left((zM_{0}+M_{1})^{-1}M_{0}\right)^{j}\quad (j\in\mathbb{N}_0). \end{aligned}$$

Exercise 10.2

Let $E\in \mathbb {C}^{n\times n}$. We set , where 1 denotes the identity matrix in $\mathbb {C}^{n\times n}$. A matrix $X\in \mathbb {C}^{n\times n}$ is called a Drazin inverse of E if the following properties hold:

• EX = XE,

• XEX = X,

• XE ^k+1 = E ^k.

Prove that each matrix $E\in \mathbb {C}^{n\times n}$ has a unique Drazin inverse.

Hint: For the existence consider the quasi-Weierstraß form for (E, 1).

Exercise 10.3

Let $M_{0},M_{1}\in \mathbb {C}^{n\times n}$ with (M ₀, M ₁) regular and M ₀ M ₁ = M ₁ M ₀. Denote by $M_{0}^{\mathrm {D}}$ the Drazin inverse of M ₀ (see Exercise 10.2). Prove:

• (a) $M_{0}^{\mathrm {D}}M_{1}=M_{1}M_{0}^{\mathrm {D}}$,

• (b) $\operatorname {ran} M_{0}^{\mathrm {D}}M_{0}=\operatorname {IV}(M_{0},M_{1})$,

• (c) For all $U_{0}\in \operatorname {IV}(M_{0},M_{1})$ the solution U of (10.2) is given by

  $$\displaystyle \begin{aligned} U(t)=\mathrm{e}^{-tM_{0}^{\mathrm{D}}M_{1}}U_{0}\quad (t\geqslant0). \end{aligned}$$

Exercise 10.4

Let $M_{0},M_{1}\in \mathbb {C}^{n\times n}$ with (M ₀, M ₁) regular. Prove that there exist two matrices $E,A\in \mathbb {C}^{n\times n}$ with (E, A) regular and EA = AE such that

• $\operatorname {IV}(E,A)=\operatorname {IV}(M_{0},M_{1})$,

• U solves the initial value problem (10.2) for the matrices M ₀, M ₁ if and only if U solves the initial value problem (10.2) for the matrices E, A with the same initial value $U_{0}\in \operatorname {IV}(M_{0},M_{1})$.

Exercise 10.5

We consider the following electrical circuit (see Fig. 10.1) with a resistor with resistance R > 0, an inductor with inductance L > 0 and a capacitor with capacitance C > 0. We denote the respective voltage drops by v _R, v _L and v _C. Moreover, the current is denoted by i. The constitutive relations for resistor, inductor and capacitor are given by

$$\displaystyle \begin{aligned} Ri & =v_{R},\\ Li' & =v_{L},\\ Cv_{C}^{\prime} & =i, \end{aligned} $$

respectively. Moreover, by Kirchhoff’s second law we have

$$\displaystyle \begin{aligned} v_{R}+v_{C}+v_{L}=0. \end{aligned}$$

Write these equations as a differential algebraic equation and compute the index and the space of consistent initial values. Moreover, compute the solution for each consistent initial value for R = 2 and C = L = 1.

Fig. 10.1

Electrical circuit

Exercise 10.6

Prove the assertions (a) to (c) in Lemma 10.2.3.

Exercise 10.7

Let M ₀, M ₁ ∈ L(H).

• (a) Assume that $\rho (M_{0},M_{1})\ne \varnothing$. Prove that for each $k\in \mathbb {N}$ the space $\operatorname {IV}_{k}$ is closed if and only if $M_{0}\left [\operatorname {IV}_{k-1}\right ]$ is closed.

• (b) Assume that (M ₀, M ₁) is regular with $\operatorname {ind}(M_{0},M_{1})\geqslant 1$. Prove that if $\operatorname {IV}_{\operatorname {ind}(M_{0},M_{1})}$ is closed then the operator

  $$\displaystyle \begin{aligned} M_{0}|{}_{\operatorname{IV}_{\operatorname{ind}(M_{0},M_{1})}} \colon \operatorname{IV}_{\operatorname{ind}(M_{0},M_{1})}\to M_{0}\left[\operatorname{IV}_{\operatorname{ind}(M_{0},M_{1})-1}\right] \end{aligned}$$

  is an isomorphism.