Interval bounds on the solutions of semi-explicit index-one DAEs. Part 1: analysis

Abstract

This article presents two methods for computing interval bounds on the solutions of nonlinear, semi-explicit, index-one differential-algebraic equations (DAEs). Part 1 presents theoretical developments, while Part 2 discusses implementation and numerical examples. The primary theoretical contributions are (1) an interval inclusion test for existence and uniqueness of a solution, and (2) sufficient conditions, in terms of differential inequalities, for two functions to describe componentwise upper and lower bounds on this solution, point-wise in the independent variable. The first proposed method applies these results sequentially in a two-phase algorithm analogous to validated integration methods for ordinary differential equations. The second method unifies these steps to characterize bounds as the solutions of an auxiliary system of DAEs. Efficient implementations of both are described using interval computations and demonstrated on numerical examples.

Appendix A: Uniqueness proofs

Lemma 7.1

Let $E\subset \mathbb R ^n$ be connected and let $\psi :E\rightarrow \mathbb R$ be continuous. If the set $\{\varvec{\xi }\in E: \psi (\varvec{\xi }) =0\}$ is nonempty and open with respect to $E$, then $\psi (\varvec{\xi })=0, \forall \varvec{\xi }\in E$.

Proof

Let $E_1=\{\varvec{\xi }\in E: \psi (\varvec{\xi }) =0\}$ and $E_2=\{\varvec{\xi }\in E: \psi (\varvec{\xi }) \ne 0\}$, and note that $E_1\cap E_2=\emptyset$ and $E_1\cup E_2=E$. Since $E$ is connected, it cannot be written as the disjoint union of two nonempty open (w.r.t. $E$) sets. But $E_1$ is nonempty and open w.r.t. $E$ by hypothesis, and $E_2$ is open w.r.t. $E$ because it is the inverse image of an open set under a continuous mapping on $E$. Hence, $E_2=\emptyset$ and $E_1=E$. $\square$

Lemma 7.2

Let $(\mathbf{x},\mathbf{y})\in C^1(I\times P,D_x)\times C^1(I\times P,D_y)$ and $(\mathbf{x}^*,\mathbf{y}^*)\in C^1(\widetilde{I}\times \widetilde{P},D_x)\times C^1(\widetilde{I}\times \widetilde{P},D_y)$ be solutions of (1a) on $I\times P$ and $\widetilde{I}\times \widetilde{P}$, respectively, and suppose that $(\mathbf{x},\mathbf{y})$ is regular. Then

• 1. For any $(t^{\prime },\mathbf{p}^{\prime })\in I\times P$, there exists an open ball around $(t^{\prime },\mathbf{p}^{\prime }), U^{\prime }\subset D_t\times D_p$, an open ball around $(t^{\prime },\mathbf{p}^{\prime },\mathbf{x}(t^{\prime },\mathbf{p}^{\prime })), V^{\prime }\subset D_t\times D_p\times D_x$, an open ball around $\mathbf{y}(t^{\prime },\mathbf{p}^{\prime }), Q^{\prime }\subset D_{y}$, and a function $\mathbf{h}\in C^1(V^{\prime },Q^{\prime })$ satisfying $(t,\mathbf{p},\mathbf{x}(t,\mathbf{p}))\in V^{\prime }$ and $\mathbf{y}(t,\mathbf{p}) = \mathbf{h}(t,\mathbf{p},\mathbf{x}(t,\mathbf{p}))\in Q^{\prime }, \forall (t,\mathbf{p})\in U^{\prime }\cap (I\times P)$.

• 2. If $\hat{P}\subset P\cap \widetilde{P}$ is connected and $\exists (t^{\prime },\hat{\mathbf{p}}) \in (I\cap \widetilde{I})\times \hat{P}$ such that $\mathbf{x}(t^{\prime },\mathbf{p})$ $=\mathbf{x}^*(t^{\prime },\mathbf{p}), \forall \mathbf{p}\in \hat{P}$, and $\mathbf{y}(t^{\prime },\hat{\mathbf{p}})=\mathbf{y}^*(t^{\prime },\hat{\mathbf{p}})$, then $\mathbf{y}(t^{\prime },\mathbf{p})=\mathbf{y}^*(t^{\prime },\mathbf{p}), \forall \mathbf{p}\in \hat{P}$.

Proof

Choose any $(t^{\prime },\mathbf{p}^{\prime })\in I\times P$. Since $(\mathbf{x},\mathbf{y})$ is a regular solution of (1a) on $I\times P, (t^{\prime },\mathbf{p}^{\prime },\mathbf{x}(t^{\prime },\mathbf{p}^{\prime }),\mathbf{y}(t^{\prime },\mathbf{p}^{\prime }))\in \mathcal G \cap \mathcal G _R$. Then, by Theorem 2.2, there exists an open ball around $(t^{\prime },\mathbf{p}^{\prime },\mathbf{x}(t^{\prime },\mathbf{p}^{\prime })), V^{\prime }\subset D_t\times D_p\times D_x$, an open ball around $\mathbf{y}(t^{\prime },\mathbf{p}^{\prime }), Q^{\prime }\subset D_{y}$, and a function $\mathbf{h}\in C^1(V^{\prime },Q^{\prime })$ such that $\mathbf{h}(t^{\prime },\mathbf{p}^{\prime },\mathbf{x}(t^{\prime },\mathbf{p}^{\prime }))$ $=\mathbf{y}(t^{\prime },\mathbf{p}^{\prime })$ and, for each $(t,\mathbf{p},\mathbf{z}_x)\in V^{\prime }, \mathbf{h}(t,\mathbf{p},\mathbf{z}_x)$ is the unique element of $Q^{\prime }$ satisfying $\mathbf{g}(t,\mathbf{p},\mathbf{z}_x,\mathbf{h}(t,\mathbf{p},\mathbf{z}_x))=\mathbf{0}$. Now, by continuity, there exists an open ball $U^{\prime }$ around the point $(t^{\prime },\mathbf{p}^{\prime })$ small enough that $(t,\mathbf{p},\mathbf{x}(t,\mathbf{p}))\in V^{\prime }$ for every $(t,\mathbf{p})\in U^{\prime }\cap (I\times P)$, and it follows that

$$\begin{aligned} \mathbf{g}(t,\mathbf{p},\mathbf{x}(t,\mathbf{p}),\mathbf{h}(t,\mathbf{p},\mathbf{x}(t,\mathbf{p})))=\mathbf{0}, \quad \forall (t,\mathbf{p})\in U^{\prime }\cap (I\times P). \end{aligned}$$

(48)

Again by continuity, it is possible to choose $U^{\prime }$ small enough that $\mathbf{y}(t,\mathbf{p})\in Q^{\prime }$ for all $(t,\mathbf{p})\in U^{\prime }\cap (I\times P)$, which implies, by the uniqueness property of $\mathbf{h}$ in $Q^{\prime }$, that

$$\begin{aligned} \mathbf{y}(t,\mathbf{p})=\mathbf{h}(t,\mathbf{p},\mathbf{x}(t,\mathbf{p})), \quad \forall (t,\mathbf{p})\in U^{\prime }\cap (I\times P). \end{aligned}$$

(49)

This establishes the first conclusion of the lemma.

To prove the second conclusion, choose any $\hat{P}, \hat{\mathbf{p}}$ and $t^{\prime }$ as in the hypothesis of the lemma and define

$$\begin{aligned} R\equiv \{\mathbf{p}\in \hat{P}: \Vert \mathbf{y}(t^{\prime },\mathbf{p})-\mathbf{y}^*(t^{\prime },\mathbf{p})\Vert =0\}. \end{aligned}$$

(50)

By hypothesis, $\hat{\mathbf{p}}\in R$ so that $R$ is nonempty. It will be shown than $R$ is open with respect to $\hat{P}$. Choose any $\mathbf{p}^{\prime }\in R$ and, corresponding to the point $(t^{\prime },\mathbf{p}^{\prime })$, let $U^{\prime }, V^{\prime }, Q^{\prime }$ and $\mathbf{h}$ be as in the first conclusion of the lemma. By hypothesis, $(t^{\prime },\mathbf{p}^{\prime },\mathbf{x}^*(t^{\prime },\mathbf{p}^{\prime }))=(t^{\prime },\mathbf{p}^{\prime },\mathbf{x}(t^{\prime },\mathbf{p}^{\prime }))\in V^{\prime }$, and by the definition of $R, \mathbf{y}^*(t^{\prime },\mathbf{p}^{\prime })=\mathbf{y}(t^{\prime },\mathbf{p}^{\prime })\in Q^{\prime }$, so continuity implies that we may choose an open all around $\mathbf{p}^{\prime }, J_{\mathbf{p}^{\prime }}$, small enough that $J_{\mathbf{p}^{\prime }}\times \{t^{\prime }\}\subset U^{\prime }$, and $(t^{\prime },\mathbf{p},\mathbf{x}^*(t^{\prime },\mathbf{p}))\in V^{\prime }$ and $\mathbf{y}^*(t^{\prime },\mathbf{p})\in Q^{\prime }$, for all $\mathbf{p}\in J_{\mathbf{p}^{\prime }}\cap \widetilde{P}$. Then the first conclusion of the theorem gives

$$\begin{aligned} \mathbf{y}(t^{\prime },\mathbf{p})=\mathbf{h}(t^{\prime },\mathbf{p},\mathbf{x}(t^{\prime },\mathbf{p})), \quad \forall \mathbf{p}\in J_{\mathbf{p}^{\prime }}\cap \hat{P}, \end{aligned}$$

(51)

and an identical argument shows that

$$\begin{aligned} \mathbf{y}^*(t^{\prime },\mathbf{p}) =\mathbf{h}(t^{\prime },\mathbf{p},\mathbf{x}^*(t^{\prime },\mathbf{p})), \quad \forall \mathbf{p}\in J_{\mathbf{p}^{\prime }}\cap \hat{P}. \end{aligned}$$

(52)

But $\mathbf{x}^*(t^{\prime },\mathbf{p})=\mathbf{x}(t^{\prime },\mathbf{p}), \forall \mathbf{p}\in \hat{P}$ by hypothesis, so this implies that $\mathbf{y}^*(t^{\prime },\mathbf{p})$ $=\mathbf{y}(t^{\prime },\mathbf{p}), \forall \mathbf{p}\in J_{\mathbf{p}^{\prime }}\cap \hat{P}$. Thus $R$ is open with respect to $\hat{P}$. Now, since $\hat{P}$ is connected by hypothesis and $R$ is nonempty and open with respect to $\hat{P}$, Lemma 7.1 shows that $R=\hat{P}$; i.e. $\mathbf{y}^*(t^{\prime },\mathbf{p})=\mathbf{y}(t^{\prime },\mathbf{p}), \forall \mathbf{p}\in \hat{P}$. $\square$

Lemma 7.3

Let $(\mathbf{x},\mathbf{y})\in C^1(I\times P,D_x)\times C^1(I\times P,D_y)$ and $(\mathbf{x}^*,\mathbf{y}^*)\in C^1(\widetilde{I}\times \widetilde{P},D_x)\times C^1(\widetilde{I}\times \widetilde{P},D_y)$ be solutions of (1a) on $I\times P$ and $\widetilde{I}\times \widetilde{P}$, respectively, and suppose that $(\mathbf{x},\mathbf{y})$ is regular. If $\hat{P}\subset P\cap \widetilde{P}$ is connected and compact and $\exists (\hat{t},\hat{\mathbf{p}}) \in (I\cap \widetilde{I})\times \hat{P}$ such that $\mathbf{x}(\hat{t},\mathbf{p})=\mathbf{x}^*(\hat{t},\mathbf{p}), \forall \mathbf{p}\in \hat{P}$, and $\mathbf{y}(\hat{t},\hat{\mathbf{p}})=\mathbf{y}^*(\hat{t},\hat{\mathbf{p}})$, then $\mathbf{x}(t,\mathbf{p}) = \mathbf{x}^*(t,\mathbf{p})$ and $\mathbf{y}(t,\mathbf{p}) = \mathbf{y}^*(t,\mathbf{p}), \forall (t,\mathbf{p})\in (I\cap \widetilde{I})\times \hat{P}$.

Proof

Choose any $\hat{P}, \hat{\mathbf{p}}$ and $\hat{t}$ as in the hypothesis of the lemma and define

$$\begin{aligned} R\equiv \{t\in I\cap \widetilde{I}: \max _{\mathbf{p}\in \hat{P}}\left(\Vert \mathbf{x}(t,\mathbf{p})-\mathbf{x}^*(t,\mathbf{p})\Vert \right)+\Vert \mathbf{y}(t,\hat{\mathbf{p}})-\mathbf{y}^*(t,\hat{\mathbf{p}})\Vert =0\}. \end{aligned}$$

(53)

$R$ is nonempty since it contains $\hat{t}$. It will be shown that $R$ is open with respect to $I\cap \widetilde{I}$. Choose any $t^{\prime }\in R$. Applying the second conclusion of Lemma 7.2, we have $\mathbf{y}^*(t^{\prime },\mathbf{p})=\mathbf{y}(t^{\prime },\mathbf{p}), \forall \mathbf{p}\in \hat{P}$. Choose any $\mathbf{p}^{\prime }\in \hat{P}$ and, corresponding to the point $(t^{\prime },\mathbf{p}^{\prime })$, let $U^{\prime }, V^{\prime }, Q^{\prime }$ and $\mathbf{h}$ be as in the first conclusion of Lemma 7.2. By the definition of $R, (t^{\prime },\mathbf{p}^{\prime },\mathbf{x}^*(t^{\prime },\mathbf{p}^{\prime }))=(t^{\prime },\mathbf{p}^{\prime },\mathbf{x}(t^{\prime },\mathbf{p}^{\prime }))\in V^{\prime }$ and, by the argument above, $\mathbf{y}^*(t^{\prime },\mathbf{p}^{\prime })=\mathbf{y}(t^{\prime },\mathbf{p}^{\prime })\in Q^{\prime }$. Then continuity implies that there exists an open ball around $t^{\prime }, J_{t^{\prime }}$, and an open ball around $\mathbf{p}^{\prime }, J_{\mathbf{p}^{\prime }}$, such that $J_{t^{\prime }}\times J_{\mathbf{p}^{\prime }}\subset U^{\prime }$, and $(t,\mathbf{p},\mathbf{x}^*(t,\mathbf{p}))\in V^{\prime }$ and $\mathbf{y}^*(t,\mathbf{p})\in Q^{\prime }$, for all $(t,\mathbf{p})\in (J_{t^{\prime }}\times J_{\mathbf{p}^{\prime }})\cap (\widetilde{I}\times \widetilde{P})$. From Lemma 7.2, we have

$$\begin{aligned} \mathbf{y}(t,\mathbf{p})&=\mathbf{h}(t,\mathbf{p},\mathbf{x}(t,\mathbf{p})), \quad \forall (t,\mathbf{p})\in (J_{t^{\prime }}\times J_{\mathbf{p}^{\prime }})\cap (I\times \hat{P}), \end{aligned}$$

(54)

and an identical argument using the uniqueness property of $\mathbf{h}$ in $Q^{\prime }$ shows that

$$\begin{aligned} \mathbf{y}^*(t,\mathbf{p})=\mathbf{h}(t,\mathbf{p},\mathbf{x}^*(t,\mathbf{p})), \quad \forall (t,\mathbf{p})\in (J_{t^{\prime }}\times J_{\mathbf{p}^{\prime }})\cap (\widetilde{I}\times \hat{P}). \end{aligned}$$

(55)

Then, by definition,

$$\begin{aligned} \dot{\mathbf{x}}(t,\mathbf{p})&= \mathbf{f}(t,\mathbf{p},\mathbf{x}(t,\mathbf{p}),\mathbf{h}(t,\mathbf{p},\mathbf{x}(t,\mathbf{p}))), \quad \forall (t,\mathbf{p})\in (J_{t^{\prime }}\times J_{\mathbf{p}^{\prime }})\cap (I\times \hat{P}), \end{aligned}$$

(56)

$$\begin{aligned} \dot{\mathbf{x}}^*(t,\mathbf{p})&\!= \mathbf{f}(t,\mathbf{p},\mathbf{x}^*(t,\mathbf{p}),\mathbf{h}(t,\mathbf{p},\mathbf{x}^*(t,\mathbf{p}))), \quad \forall (t,\mathbf{p})\in (J_{t^{\prime }}\times J_{\mathbf{p}^{\prime }})\cap (\widetilde{I}\!\times \! \hat{P}).\nonumber \\ \end{aligned}$$

(57)

But $\mathbf{f}$ and $\mathbf{h}$ are continuously differentiable and hence the mapping $(t,\mathbf{p},\mathbf{z}_x)\mapsto \mathbf{f}(t,\mathbf{p},\mathbf{h}(t,\mathbf{p},\mathbf{z}_x))$ is Lipschitz on $V^{\prime }$ by Lemma 2.1. The definition of $R$ gives $\mathbf{x}(t^{\prime },\mathbf{p})=\mathbf{x}^*(t^{\prime },\mathbf{p}), \forall \mathbf{p}\in \hat{P}$, so a standard application of Gronwall’s inequality shows that $\mathbf{x}(t,\mathbf{p})=\mathbf{x}^*(t,\mathbf{p}), \forall (t,\mathbf{p})\in (J_{t^{\prime }}\times J_{\mathbf{p}^{\prime }})\cap ((I\cap \widetilde{I})\times \hat{P})$. Furthermore, this implies that $\mathbf{y}(t,\mathbf{p})=\mathbf{h}(t,\mathbf{p},\mathbf{x}(t,\mathbf{p}))=\mathbf{h}(t,\mathbf{p},\mathbf{x}^*(t,\mathbf{p}))=\mathbf{y}^*(t,\mathbf{p}), \forall (t,\mathbf{p})\in (J_{t^{\prime }}\times J_{\mathbf{p}^{\prime }})\cap ((I\cap \widetilde{I})\times \hat{P})$.

Now, since $\mathbf{p}^{\prime }\in \hat{P}$ was chosen arbitrarily, the preceding construction applies to every $\mathbf{p}\in \hat{P}$. Thus, to every $\mathbf{q}\in \hat{P}$, there corresponds an open ball around $t^{\prime }, J_{t^{\prime }}(\mathbf{q})$, and an open ball around $\mathbf{q}, J_{\mathbf{q}}$, such that $(\mathbf{x},\mathbf{y})(t,\mathbf{p})=(\mathbf{x}^*,\mathbf{y}^*)(t,\mathbf{p}), \forall (t,\mathbf{p})\in (J_{t^{\prime }}(\mathbf{q})\times J_{\mathbf{q}})\cap ((I\cap \widetilde{I})\times \hat{P})$. Noting that the $J_{\mathbf{q}}$ constructed in this way form an open cover of $\hat{P}$, compactness of $\hat{P}$ implies that there exist finitely many elements of $\hat{P}, \mathbf{q}_1,\ldots ,\mathbf{q}_n$, such that $\hat{P}$ is covered by $J_{\mathbf{q}_1}\cup \ldots \cup J_{\mathbf{q}_n}$. Let $J_{t^{\prime }}^*\equiv J_{t^{\prime }}(\mathbf{q}_1)\cap \ldots \cap J_{t^{\prime }}(\mathbf{q}_n)$. Then, for every $\mathbf{p}\in \hat{P}$, there exists $i\in \{1,\ldots ,n\}$ such that $\mathbf{p}\in J_{\mathbf{q}_i}$, which implies that $(\mathbf{x},\mathbf{y})(t,\mathbf{p})=(\mathbf{x}^*,\mathbf{y}^*)(t,\mathbf{p}), \forall t\in J_{t^{\prime }}^*\cap (I\cap \widetilde{I})$. Therefore, $J^*_{t^{\prime }}\cap (I\cap \widetilde{I})$ is contained in $R$, so that $t^{\prime }$ is an interior point of $R$ when viewed as a subset of $I\cap \widetilde{I}$, and since $t^{\prime }\in R$ was chosen arbitrarily, $R$ is open with respect to $I\cap \widetilde{I}$. Since $I\cap \widetilde{I}$ is connected and $R$ is nonempty and open with respect to $I\cap \widetilde{I}$, Lemma 7.1 shows that $R=I\cap \widetilde{I}$. But by definition, this implies that $\mathbf{x}(t,\mathbf{p})=\mathbf{x}^*(t,\mathbf{p})$ and $\mathbf{y}(t,\hat{\mathbf{p}})=\mathbf{y}^*(t,\hat{\mathbf{p}}), \forall (t,\mathbf{p})\in (I\cap \widetilde{I})\times \hat{P}$. Finally, the second conclusion of Lemma 7.2 implies that $\mathbf{y}(t,\mathbf{p})=\mathbf{y}^*(t,\mathbf{p}), \forall (t,\mathbf{p})\in (I\cap \widetilde{I})\times \hat{P}$. $\square$

Theorem 7.1

Let $(\mathbf{x},\mathbf{y})\in C^1(I\times P,D_x)\times C^1(I\times P,D_y)$ and $(\mathbf{x}^*,\mathbf{y}^*)\in C^1(\widetilde{I}\times \widetilde{P},D_x)\times C^1(\widetilde{I}\times \widetilde{P},D_y)$ be solutions of (1a) on $I\times P$ and $\widetilde{I}\times \widetilde{P}$, respectively, and suppose that $(\mathbf{x},\mathbf{y})$ is regular. If $\hat{P}\subset P\cap \widetilde{P}$ is connected and $\exists (\hat{t},\hat{\mathbf{p}}) \in (I\cap \widetilde{I})\times \hat{P}$ such that $\mathbf{x}(\hat{t},\mathbf{p})=\mathbf{x}^*(\hat{t},\mathbf{p}), \forall \mathbf{p}\in \hat{P}$, and $\mathbf{y}(\hat{t},\hat{\mathbf{p}})=\mathbf{y}^*(\hat{t},\hat{\mathbf{p}})$, then $\mathbf{x}(t,\mathbf{p}) = \mathbf{x}^*(t,\mathbf{p})$ and $\mathbf{y}(t,\mathbf{p}) = \mathbf{y}^*(t,\mathbf{p}), \forall (t,\mathbf{p})\in (I\cap \widetilde{I})\times \hat{P}$.

Proof

Choose any $\mathbf{p}\in \hat{P}$. Clearly, $\{\mathbf{p}\}\subset P\cap \widetilde{P}$ is compact and connected, and Lemma 7.2 guarantees that $\mathbf{y}(\hat{t},\mathbf{p})=\mathbf{y}^*(\hat{t},\mathbf{p})$. Then Lemma 7.3 shows that $\mathbf{x}(t,\mathbf{p})=\mathbf{x}^*(t,\mathbf{p})$ and $\mathbf{y}(t,\mathbf{p})=\mathbf{y}^*(t,\mathbf{p}), \forall t\in I\cap \widetilde{I}$. $\square$

Corollary 3.1 is a simple consequence of these developments. By the definition of a solution of (1), we have $\mathbf{x}(t_0,\mathbf{p})=\mathbf{x}^*(t_0,\mathbf{p}), \forall \mathbf{p}\in \hat{P}$, and $\mathbf{y}(t_0,\hat{\mathbf{p}})=\mathbf{y}^*(t_0,\hat{\mathbf{p}})$ by hypothesis. Since $\hat{P}$ is connected, the result follows from Theorem 7.1.