# Conversion methods for improving structural analysis of differential-algebraic equation systems

## Abstract

Structural analysis (SA) of a system of differential-algebraic equations (DAEs) is used to determine its index and which equations to be differentiated and how many times. Both Pantelides’s algorithm and Pryce’s $$\varSigma$$-method are equivalent: if one of them finds correct structural information, the other does also. Nonsingularity of the Jacobian produced by SA indicates success, which occurs on many problems of interest. However, these methods can fail on simple, solvable DAEs and give incorrect structural information including the index. This article investigates $$\varSigma$$-method’s failures and presents two conversion methods for fixing them. Under certain conditions, both methods reformulate a DAE system on which the $$\varSigma$$-method fails into a locally equivalent problem on which SA is more likely to succeed. Aiming at achieving global equivalence between the original DAE system and the converted one, we provide a rationale for choosing a conversion from the applicable ones.

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1. 1.

The colon notation $$p\,{:}\,q$$ for integers pq denotes either the unordered set or the enumerated list of integers i with $$p\le i\le q$$, depending on context.

2. 2.

Throughout this article, “derivatives of $$x_j$$” include $$x_j$$ itself as its 0th derivative: $$x_j^{(l)}=x_j$$ if $$l=0$$.

3. 3.

When we present a DAE example, we also present its signature matrix $$\varSigma$$, the canonical offset pair $$(\mathbf {c};\mathbf {d})$$, and the associated System Jacobian $$\mathbf {J}$$.

4. 4.

We consider it with parameters $$\beta =\epsilon =1$$, $$\alpha _1=\alpha _2=\delta =1$$, and $$\gamma =-1$$, and we use subscripts for parameter indices. The equations $$g_1,g_2$$ are renamed $$f_3, f_4$$ and the variables $$y_1,y_2$$ are renamed $$x_3, x_4$$.

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## Acknowledgements

The authors acknowledge with thanks the financial support for this research: GT was supported in part by the McMaster Centre for Software Certification through the Ontario Research Fund, Canada, NSN was supported in part by the Natural Sciences and Engineering Research Council of Canada, and JDP was supported in part by the Leverhulme Trust, the UK. The authors thank the anonymous reviewers for providing valuable suggestions on improving this article.

## Author information

Authors

### Corresponding author

Correspondence to Guangning Tan.

Communicated by Anne Kværnø.

## Preliminary results and proof of Lemma 4.4

### Preliminary results and proof of Lemma 4.4

Let the notation be as at the start of §4.2. We prove a lemma first and then Lemma 4.4.

### Lemma 7.1

Let $$r\in J{\setminus }\bigl \{\,l\,\bigr \}$$, $$w_1=y_r+(v_r/v_l)\cdot x_l^{(d_l-\overline{c})}$$, and

\begin{aligned} w_2 = w_1^{(\overline{c}-c_i)} = \left( y_r+(v_r/v_l)\cdot x_l^{(d_l-\overline{c})}\right) ^{(\overline{c}-c_i)}. \end{aligned}
(7.1)

Then

\begin{aligned} \sigma \left( x_j,w_2\right) = \left\{ \begin{array}{ll} < d_j-c_i&{} \quad \text {if }j\in J{\setminus }\bigl \{\,l\,\bigr \}\\ \le d_j-c_i &{} \quad \text {otherwise}. \end{array}\right. \end{aligned}
(7.2)

### Proof

Obviously $$\sigma \left( x_l,w_1\right) =d_l-\overline{c}$$ when $$j=l\in J$$. Now consider the case $$j\ne l$$. Since $$x_j$$ can occur only in $$v_r$$ and $$v_l$$ in $$w_1$$, we have $$\sigma \left( x_j,w_1\right) \le \sigma \left( x_j,\mathbf {v}\right) \le d_j-\overline{c}$$.

Noting that $$\overline{c}=\max _{i\in M} c_i$$, we have $$\overline{c}-c_i\ge 0$$ for all $$i\in M$$. Then (7.2) results from connecting $$\sigma \left( x_j,w_2\right) =\sigma \left( x_j,w_1\right) +(\overline{c}-c_i)$$ with (4.12) and the results in the previous paragraph.

The proof of Lemma 4.4 uses the two assumptions preceding it.

### Proof

Write $$\overline{{\varSigma }}$$ in Fig. 1 into the following $$2\times 3$$ block form: We aim to verify below the relations between $$\overline{\sigma }_{ij}$$ and $${\widetilde{d}}_j-{\widetilde{c}}_i$$ in each block.

1. (1)

$$\overline{{\varSigma }}_{11}$$. Consider $$j,r\in J{\setminus }\bigl \{\,l\,\bigr \}$$. By (4.10), we substitute $$w_2$$ in (7.1) for every $$x_r^{(d_r-c_i)}$$ in $$f_i$$ for all $$i=1\,{:}\,n$$. By (7.2), $$\sigma \left( x_j, w_2\right) < d_j- c_i$$ for all $$i\in M$$. So these expression substitutions do not introduce $$x_r^{(d_r-c_i)}$$ in $$\overline{f}_i$$, where $$r\in J{\setminus }\bigl \{\,l\,\bigr \}$$. Given M in (4.8), we have $$d_j-c_i>\sigma _{ij}$$ for all $$i\notin M$$ and $$j\in J$$. Hence

\begin{aligned} \sigma \left( x_j,\overline{f}_i\right) < d_j-c_i \qquad \text {for }j\in J{\setminus }\bigl \{\,l\,\bigr \}, \ i=1\,{:}\,n. \end{aligned}
(7.3)

What remains to show is the case $$j=l$$. From (4.9), $$x_r^{(d_r-\overline{c})}=y_r+ (v_r/v_l) \cdot x_l^{(d_l-\overline{c})}$$. Taking the partial derivatives of both sides with respect to $$x_l^{(d_l-\overline{c})}$$ and applying Griewank’s Lemma (4.1) with $$w=x_{r}^{(d_{r}-\overline{c})}$$ and $$q=\overline{c}-c_i\ge 0$$ for all $$i\in M$$, we have

\begin{aligned} \frac{v_{r}}{v_l} = \frac{\partial x_{r}^{(d_{r}-\overline{c})}}{\partial x_l^{(d_l-\overline{c})}} = \frac{\partial x_{r}^{(d_{r}-\overline{c}+\overline{c}-c_i) }}{\partial x_l^{(d_l-\overline{c}+\overline{c}-c_i)}} = \frac{\partial x_{r}^{(d_{r}-c_i)}}{\partial x_l^{(d_l-c_i)}}. \end{aligned}
(7.4)

Then

\begin{aligned} \frac{\partial \overline{f}_i}{\partial x_l^{(d_l-c_i)}}= & {} \frac{\partial f_i}{\partial x_l^{(d_l-c_i)}} + \sum _{r\in J{\setminus }\{l\}} \frac{\partial f_i}{\partial x_r^{(d_r-c_i)}} \cdot \frac{\partial x_r^{(d_r-c_i)}}{\partial x_l^{(d_l-c_i)}} \text { by the chain rule} \\= & {} J_{il} + \sum _{r\in J{\setminus }\{l\}} J_{ir} \cdot \frac{v_r}{v_l} = \frac{1}{v_l} \sum _{r\in J} J_{ir} v_r= \frac{1}{v_l}(\mathbf {J} \mathbf {v})_i = 0\,\, \text {by }(6.4) \hbox { and } \mathbf {J} \mathbf {v}=\mathbf {0}. \end{aligned}

This gives $$\sigma \left( x_l,\overline{f}_i\right) < d_l-c_i$$ for all $$i=1\,{:}\,n$$. Together with (7.3) we have proved the “<” part in $$\overline{{\varSigma }}_{11}$$.

2. (2)

$$\overline{{\varSigma }}_{12}$$. The substitutions do not affect $$x_j$$, for all $$j\notin L$$. By (7.2), such an $$x_j$$ occurs in every $$w_2$$ of order $$\le d_j-c_i$$, where $$i\in M$$. Hence also $$\sigma \left( x_j,\overline{f}_i\right) \le d_j-c_i$$ for all $$i=1\,{:}\,n$$ and $$j\notin L$$.

3. (3)

$$\overline{{\varSigma }}_{13}$$. Consider $$r\in J{\setminus }\bigl \{\,l\,\bigr \}$$. For an $$i\in M$$, $$y_r$$ occurs of order $$\overline{c}-c_i$$ in $$w_2$$ in (7.1). For all $$i=1\,{:}\,n$$, if a substitution occurs for an $$x_r^{(d_r-c_i)}$$ in $$f_i$$, then $$\sigma \left( y_r,\overline{f}_i\right) =\overline{c}-c_i$$; otherwise $$\sigma \left( y_r,\overline{f}_i\right) =-\infty$$. In either case $$\sigma \left( y_r,\overline{f}_i\right) \le \overline{c}-c_i$$.

4. (4)

$$\overline{{\varSigma }}_{21}$$. Equalities hold on the diagonal and in the $$l{\text {th}}$$ column, as $$y_r^{(d_r-\overline{c})}$$ and $$y_l^{(d_l-\overline{c})}$$ occur in $$g_l$$, where $$r\in J$$. What remains to show is the “<” part. Assume that $$j\,,r\,,l\in J$$ are distinct. Then by (4.9) and (4.12),

\begin{aligned} \sigma \left( x_{j},g_{r}\right)&=\sigma \left( x_{j}, y_{r}-x_{r}^{(d_{r}-\overline{c})} + \frac{v_{r}}{v_l} \cdot x_l^{(d_l-\overline{c})}\right) \le \sigma \left( x_{j},\mathbf {v}\right) < d_j- \overline{c}. \end{aligned}
(7.5)
5. (5)

$$\overline{{\varSigma }}_{22}$$. Assume again that $$j\,,r\,,l$$ are distinct, where $$r\in J$$ and $$j=s+1\,{:}\,n$$. Then replacing the “<” in (7.5) by “$$\le$$” proves the “$$\le$$” part in $$\overline{{\varSigma }}_{22}$$.

6. (6)

$$\overline{{\varSigma }}_{23}$$. Consider $$r, j\in J$$. By $$0=g_l= -y_l+ x_l^{(d_l-\overline{c})}$$ and (4.9), $$y_j$$ occurs in $$g_r$$ only if $$j=r$$, and $$\sigma \left( y_j,g_j\right) =0$$. Hence, on the diagonal lie zeros, and everywhere else is filled with $$-\infty$$.

Also worth noting is that in the $$y_l$$ column is only one finite entry $$\sigma _{n+l,n+l}=0$$, and that in the $$g_l$$ row are only two finite entries $$\sigma _{n+l,n+l}=0$$ and $$\sigma _{n+l,l}=d_l-\overline{c}$$.

Recalling (4.13) for the formulas of $${\widetilde{c}}_i$$ and $${\widetilde{d}}_j$$ of $$\overline{{\varSigma }}$$, we can summarize that the above items (1)–(6) verify the relations between $$\overline{\sigma }_{ij}$$ and $${\widetilde{d}}_j-{\widetilde{c}}_i$$ in $$\overline{{\varSigma }}$$ for all $$i,j=1\,{:}\,n+s$$; see Fig. 1. $$\square$$

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