R. Marz with coecient functions A(x t) L(R n R m ) d(x t) R n and b(x t) R m x D R m t I R, that are continuous in their arguments, and which have the

Nonlinear dierential-algebraic equations with properly formulated leading term R. Marz 1 Introduction In [BaMa], a uniform theory for investigating linear dierential-algebraic equations (DAEs) and their adjoint equations was proposed. By means of an additional coef- cient matrix it is exactly xed which derivatives of the solutions searched for are actually involved in the equation. Such a DAE is of the form A(t)(D(t)x(t)) 0 + B(t)x(t) q(t) ti (1.1) where the coecients A(t) and D(t) match well. As a nonlinear version, A(x(t) t)(d(t)x(t)) 0 + b(x(t) t)0 (1.) can be taken into account rst (e.g. [HiMa]). However, with somewhat more consistency, we obtain equations of the form A(x(t) t)(d(x(t) t)) 0 + b(x(t) t)0 (1.3) which we want to investigate in this paper. Note that it is just this sort of equations that result in the simulation of electric circuits, which is the origin of DAEs (e.g. [EsTi]). In this case, a transition to A(x(t) t)d x (x(t) t)x 0 (t)+b(x(t) t)+a(x(t) t)d t (x(t) t)0 is problematical. In the next section (x) we will determine what we mean by a properly formulated leading term. In x3 DAEs with index, f1 g will be characterized by algebraic criteria. x4 is devoted to linearization and perturbation theorems. A special structure, which is important for electric circuits for instance, will be analyzed in x5. Finally, constraint sets will be investigated in x6. Properly formulated leading terms and an equivalence theorem We investigate the equation A(x(t) t)(d(x(t) t)) 0 + b(x(t) t)0 (.1) 1

R. Marz with coecient functions A(x t) L(R n R m ) d(x t) R n and b(x t) R m x D R m t I R, that are continuous in their arguments, and which have the continuous partial derivatives A x d x b x. Denote D(x t) d x (x t). Denition.1 The leading term in (:1) is said to be properly formulated if ker A(x t) im D(x t) R n xd ti (.) and if there exists a projector function R C 1 (I L(R n )) such that R(t) R(t), ker A(x t) ker R(t) im D(x t) im R(t), and d(x t) R(t)d(x t) holds for x D ti. Consequently, the matrices A(x t) and D(x t) have constant rank for properly formulated leading terms. The subspaces A(x t) and im D(x t) are independent of x and have bases from the class C 1. It holds that A AR D RD. If A(x t) and D(x t) full the condition (.), but only ker A(x t) is independent of x and smooth, then we have, with a projector P A C 1 (I L(R n )), the relation A(x t) A(x t)p A (t) and, hence A(x(t) t)(d(x(t) t)) 0 A(x(t) t)(p A (t)d(x(t) t)) 0 ; A(x(t) t)p 0 A(t)d(x(t) t): Then, with A(x ~ t) : A(x t) d(x ~ t) : PA (t)d(x t) ~ b(x t) : b(x t); ;A(x t)p 0 A (t)d(x t), the equation ~A(x(t) t)( ~ d(x(t) t)) 0 + ~ b(x(t) t)0 (.3) has a proper leading term because of im ~ D(x t) im PA (t). We can proceed analogously if only im D(x t) is independent of x, or if the last condition of Denition.1 in not fullled. Hence, a proper formulation can be obtained if (.) holds and if one of the characteristic subspaces is independent of x and comes from C 1. I, is said to be a solution of Denition. A function x C(I x R m ) I x equation (.1) if x(t) D t I x and d(x(:) :) C 1 (I x R n ), and if equation (:1) is fullled pointwisely. Unfortunately, regularity conditions do not dene a linear function space here in general. In case d(x t) D(t)x is linear itself (cf. (1.)), a linear solution space is available by C 1 D : fx C : Dx C 1 g. Fortunately, it is relatively simple to transform equation (.1) into a (1.) form. This allows the application of standard-notions and -methods (dierentiability, linearization etc.) that are based on linear funtion spaces. We form the natural extension for equation (.1) with proper leading term A(x(t) t)(r(t)y(t)) 0 + b(x(t) t) 0 (.4) y(t) ; d(x(t) t) 0: (.5)

Nonlinear dierential-algebraic equations with properly formulated leading term 3 With x ; ; x y A A ; 0 d(x t)r(t)y D(x t)d(t)(0 R(t)) b(x t) b(x t) y;d(x t) we can write (.4), (.5) as A(x(t) t)(d(t)x(t)) 0 + b(x(t) t)0: (.6) Due to ker A(x t) ker A(x t), im D(t) im R(t) im D(x t) also (.6) has a properly formulated leading part with R(t) R(t) as the corresponding projector. Now the linear function space x C x 1 D y C : Dx Ry C 1 (.7) oers itself as solution space for (.6). For (.6) we seek functions x C 1 D whose function values lie in the domain of denition D of the coecients and full equation (.6) pointwisely. If x (:) is a solution of the original equation (.1), then the pair x (:) y (:) with y (t) : d(x (t) t) is obviously a solution of the class C 1 for (.6). If, reversely, a D pair x (:) y (:) from C 1 D forms a solution of (.6), then d(x (:) :) Rd(x (:) :) Ry C 1 also holds because of Ry C 1, and x (:) is a solution of equation (.1). Theorem.3 Let the leading term of (:1) be properly formulated. (i) The the leading term of the extension (:6) is properly formulated, too. (ii) The equations (:1) and (:6) are equivalent via the relation y(:) d(x(:) :): The sets and x M 0 (t) : fx y M 0 (t) :fx D: b(x t) im A(x t)g DR n : b(x t) im A(x t) y d(x t)g fx DR n : x M 0 (t) y d(x t)g: are the geometrical location of the solutions of (.1) and (.6), respectively, It always holds that x(t) M 0 (t) x(t) M 0 (t): The problem in how far these sets are lled with solutions leads to notions of indices and corresponding solvability statements. 3 Subspaces, matrix chain and index In this section we dene characteristic subspaces and matrix chains for (.1) and (.6). Further, let B(y x t) : (A(x t)y) x + b x (x t) for y R n x D t I and

4 R. Marz B(y x t) :(A(x t)y) 0 x + b x (x t) for y R n x DR n ti. More precisely, we have A(x t)y B(y x t) 0 x + bx (x t) 0 ;D(x t) I B(y x t) 0 ;D(x t) I For the original equation (.1) we form for x D ti y R n : G 0 (x t) A(x t)d(x t) N 0 (x t) ker G 0 (x t) S 0 (y x t) fz R m : B(y x t)z im G 0 (x t)g G 1 (y x t) G 0 (x t)+b(y x t)q 0 (x t) with a projector Q 0 (x t) L(R m ) onto N 0 (x t), P 0 (x t) I ; Q 0 (x t) N 1 (y x t) ker G 1 (y x t) S 1 (y x t) fz R m : B(y x t)p 0 (x t)z im G 1 (y x t)g: For (.6) this yields G 0 (x t) A(x t)d(x t) etc. for x DR n t I y R n. N 0 (t) R m ker R(t) depending on t only and being smooth is a special feature of (.6). Now, the relations among the subspaces of (.1) and (.6) are important, because the index will be dened via these subspaces later on. Lemma 3.1 Let equation (:1) have a proper leading term, then and N 0 (t) \ S 0 (y x t) (N 0 (x t) \ S 0 (y x t)) 0 (3.1) N 1 (y x t) \ S 1 (y x t) n ; Rm R n : Q 0 (x t) R(t) o : D(x t) ; + N 1 (y x t) \ S 1 (y x t) (3.) P roo f. We determine G 0 (x t) A(x t)d(x t) N 0 (t) 0 A(x t) 0 0 R m R n : A(x t) 0 R m ker R(t) S 0 (y x t) : B(y x t) im A(x t) ;D(x t) + 0 : S 0 (y x t) d(x t)

Nonlinear dierential-algebraic equations with properly formulated leading term 5 N 0 (t) \ S 0 (y x t) : R(t) 0 D(x t) S 0 (y x t) : N 0 (x t) \ S 0 (y x t) 0 : I 0 0 0 Furthermore, by Q 0 (t) P 0 I ; R(t) 0 (t) we also have 0 R(t) 0 0 B(y x t) A(x t) B(y x t)p 0 (t) G 0 R(t) 1 (y x t) : ;D(x t) I ; R(t) which implies S 1 (y x t) N 1 (y x t) Finally, we obtain 8 >< >:; : ; 0 R(t) im 6 4 B(y x t) A(x t) ;D(x t) I ; R(t) n ; : R(t) ;D(x t) +(I ; R(t)) 39 > 7 5 > with R m R n B(y x t) + A(x t) 0 n ; n ; n ; o : R(t) ;D(x t) with S0 (y x t) o : B(y x t) + A(x t) 0 D(x t) 0 R(t) : N0 (x t) R(t) 0 (A(x t)d(x t)+ o +B(y x t)q 0 (x t))(d(x t) ; + ) n ; o : Q0 (x t) R(t) D(x t) ; + N 1 (y x t) : Q 0 (x t) R(t) ;D(x t) S 0 (y x t) D(x t) ; + N 1 (y x t) for ; N 1 (y x t) \ S 1 (y x t). Thus, B(y x t)p 0 (x t)(d(x t) ; + ) B(y x t)p 0 (x t)d(x t) ; because of B(y x t) G 0 (x t)w, i.e., ;B(y x t)p 0 (x t) im G 1 (y x t) B(y x t)p 0 (x t) (G 0 (x t)+b(y x t)q 0 (x t))(p 0 (x t)w ; Q 0 (x t)): Conclusion 3. For (:1) and (:6) it holds that and dim(n 0 (t) \ S 0 (y x t)) dim(n 0 (x t) \ S 0 (y x t)) dim(n 1 (y x t) \ S 1 (y x t)) dim(n 1 (y x t) \ S 1 (y x t)): o

6 R. Marz Denition 3.3 An equation (:1) with properly formulated leading term is called a DAE of index 1 if or a DAE of index if N 0 (x t) \ S 0 (y x t) 0 for x D ti y R n dim(n 0 (x t) \ S 0 (y x t)) const N 1 (y x t) \ S 1 (y x t) 0 for x D ti y R n : Theorem 3.4 The original equation (:1) with proper leading term and its natural extension (:6) have the index f1 g simultaneously. Now we continue the above matrix chain with and G (y x t) :G 1 (y x t)+b(y x t)p 0 (x t)q 1 (y x t) G (y x t) :G 1 (y x t)+b(y x t)p 0 (t)q 1 (y x t) where Q 1 (y x t) L(R m ), Q 1 (y x t) L(R m+n ) are projectors onto N 1 (y x t) and N 1 (y x t), respectively. f1 g here, hence, We will investigate only problems (.1) with index N 1 (y x t) S 1 (y x t) R m (3.3) holds on principle, in fact for 1with N 1 (y x t) 0,S 1 (y x t) R m trivially, and for due to [GrMa], Theorem A.13. Hence, we may assume that Q 1 (y x t) projects onto N 1 (y x t) along S 1 (y x t). Analogously, let Q 1 (y x t) be the projector onto N 1 (y x t) along S 1 (y x t). Simple computation now yield Lemma 3.5 For f1 g it holds that 0 Q0 Q Q 1 1 D ; 0 DQ 1 D ; D P 1 D ; DP 1 D ; D Q 1 D ; DQ 1 D ; and D Q 1 G ;1 (DQ 1 G ;1 DQ 1 D ; ). 4 Linearization and perturbation theorems First, we consider the equation A(x(t) t)(d(t)x(t)) 0 + b(x(t) t)0 (4.1) i.e., the case that D(x t) D(t)x x D t I. Later on, we apply the obtained results via (.6) to the general form (.1). We x x C 1 D(I R m ) with x (t) D t I. Let I I be compact. For all

Nonlinear dierential-algebraic equations with properly formulated leading term 7 x from a suciently small neighbourhood U(x ) of x in C 1 D(I R m ) we can dene the map F : U(x ) CD(I 1 R m )! C(I R m ) F(x) : A(x(:) :)(Dx) 0 (:)+b(x(:) :) xu(x ) (4.) and we can write equation (4.1) as F(x) 0 where F is Frechet-dierentiable. The derivative is given by F x (x) L b (C 1 D(I R m ) C(I R m )) F x (x)4x A(x(:) :)(D4x) 0 (:)+B((Dx) 0 (:) x(:) :)4x(:) for 4x C 1 D(I R m ) (4.3) where we assume natural norms on C 1 D(I R m ) and C(I R m ). For applying the implicit function theorem (e.g.[kaak]), the property of the image im F x (x ) is of crucial importance. With A (t) A(x (t) t) B (t) B((D(t)x (t)) 0 x (t) t) ti, the linear DAE is nothing else but the equation A (t)(d(t)4x(t)) 0 + B (t)4x(t) q(t) ti (4.4) F x (x )4x q: Let S 0 (t) N 0 (t) G 1 (t) etc. denote the chain of subspaces and matrices generated for (4.4). Lemma 4.1 Let the DAE (4:1) have a properly formulatd leading term. DP 1 D ; DQ 1 D ; C 1 (I L(R n )). Let (i) The index--property, f1 g, transforms itself from (4:1) onto the linearization (4:4). (ii) For 1, F x (x ) is surjective. (iii) For it holds that im F x (x )C 1 DQ 1 (I G ;1 R m ). P r o o f. The assumption (i) immediately results from the construction of the subspaces and the matrices in x3. The assumptions (ii) and (iii) are concluded from the existence theorems for linear DAEs in [BaMa]. Conclusion 4. In the index- case, im F x (x ) is a non-closed proper subset in C(I R m ).

8 R. Marz If the DAE (4.1) has a dynamic degree of freedom, it has to be completed by initial or boundary conditions. Since A and D are singular, we cannot expect a degree of freedom m as in the case of regular DAEs, but a lower one. An IVP with the initial condition x(t 0 )x 0 R m is not solvable in general. Hence, x 0 has to be consistent to an extend. Basing on our experience with linear DAEs ([BaMa]), we impose an initial condition for (4.1) with t 0 I in the form D(t 0 )(x(t 0 ) ; x 0 )0 with x 0 R m (4.5) where L(R m ) will still have to be xed. By means of the mapping F IV P : U(x ) C 1 D(I R m )! C(I R m ) L IC L IC : im D(t 0 ) F IV P (x) : (Fx D(t 0 )x(t 0 )) x U(x ) (4.6) we can describe the IVP (4.1), (4.5) in a compact way by Now, the equation corresponds to the perturbed IVP F IV P (x) (0 D(t 0 )x 0 ): F IV P (x) (q D(t 0 )x 0 ) A(x(t) t)(d(t)x(t)) 0 + b(x(t) t)q(t) t I (4.7) D(t 0 )(x(t 0 ) ; x 0 )0: (4.8) Theorem 4.3 Let (4:1) be a DAE with proper leading term and index f1 g, and let x C 1 D(I R m ) be a solution of (4:1) and Further, for x U(x ), let DP 1 D ; DQ 1 D ; C 1 (I L(R n )) : P 1 (t 0 ): D(t)Q 1 (t)g (t) ;1 b(x(t) t) be continuously dierentiable w.r.t. t: (4.9) (i) If 1, then the IVP (4:7) (4:8) is uniquely solvable on I for arbitrary x 0 R m with jd(t 0 )(x 0 ; x (t 0 ))j, and q C(I R m ) with k q k 1, > 0 suciently small. For the solutions x C 1 D(I R m ) it holds that k x ; x k C 1 D const (jd(t 0)(x 0 ; x (t 0 ))j+ k q k 1 ): (ii) If, then the IVP (4:7) (4:8) is uniquely solvable on I for arbitrary x 0 R m with jd(t 0 )(x 0 ; x (t 0 ))j and q C ;1 DQ 1 G (I R m ) k q k 1 + k (DQ 1 G ;1 q)0 k 1 > 0 suciently small. For the solutions x CD(I 1 R m ) it holds that k x ; x k C 1 D const (jd(t 0)(x 0 ; x (t 0 ))j+ k q k 1 + k (DQ 1 G ;1 q)0 k 1 ):

Nonlinear dierential-algebraic equations with properly formulated leading term 9 (iii) The solution x of the IVP (4:1) (4:8) is continuously dierentiable w.r.t. x 0. The sensitivity matrix x x 0(t) :X(t) L(R m ) satises the IVP A(x(t) t)(d(t)x(t)) 0 + B((D(t)x(t)) 0 x(t) t)x(t) 0 t I D(t 0 )(X(t 0 ) ; I) 0: Remark: If we take into account that P 1 (t) I, Q 1 (t) 0 for 1, then the smoothness assumptions of Theorem 4.3. are always trivially given in this case. For the regularity condition (4.9) implies restrictions of the admissible structure of (4.1). The following condition is sucient for (4.9): D ; C 1 (I L(R n R m )) (x t) :D(t)Q 1 (t)g (t) ;1 b(x t) is continuously dierentiable and b 0 x(p 0 (t)x + sq 0 (t)x t)q 0 (t)x im G 1 (t) s [0 1] xd ti : (4.10) Namely, then (x t) (P 0 (t)x t) is true and for x(:) U(x ) it holds that d dt (x(t) t) d dt (P 0(t)x t) d dt (D(t); D(t)x(t) t) x (D(t) ; D(t)x(t) t) fd(t) ;0 D(t)x(t) +D(t) ; (D(t)x(t)) 0 g + t (D(t) ; D(t)x(t) t). The condition (4.10) is equivalent to W 0 (t)b 0 x(p 0 (t)x + sq 0 (t)x t)q 0 (t)x im W 0 (t)b 0 (t)q 0 (t): (4.11) If the derivative free part of (4.1) is linear in x or if (4.1) is a DAE in Hessenberg form, then (4.11) is given. P r oof of Theorem 4.3: F IV P x (x ) is a bijection from C 1 D(I R m ) onto C(I R m ) L IC for 1. For, F IV P x (x ) is injective but not surjective. According to Lemma 4.1, im F IV P x (x ) L IC C 1 DQ 1 G ;1 (I R m ) L IC : X. Equipped with a natural norm, X is a Banach space. We summarize the two cases that 1 by using Q 1 0 P 1 I for 1. For the mapping H(x d q) :F IV P (x) ; (q d) x U(x ) (q d) X it holds, due to the condition (4.9), that With d : D(t 0 )x (t 0 ) we obtain H(x d q) X: H(x d 0) 0 H x (x d 0) F IV P x (x ): H x (x d 0) is a homeomorphism for 1 as well as for. Due to the implicit function theorem there exists a uniquely determined continuously dierentiable mapping f : B(d ) B(0 ) X! C 1 D(I R m )

10 R. Marz with f(d 0) x, H(f(d q) d q)0,k f(d q);f(d 0) k C D 1 K(jd;d j+ k q k X ) for jd ; d j k q k X :k q k 1 + k (DQ 1 G ;1 q)0 k 1. Then the two assumption (i) and (ii) follow because of d D(t 0 )x 0. In particular, f(d 0) is continuously dierentiable w.r.t. d, i.e., the solution x(:) of the IVP (4.1) with D(t 0 )x(t 0 ) d has a continuous derivative w.r.t. d. With d D(t 0 )x 0 this implies x x 0(t) x d (t)d(t 0 ). As usually, the variational equation (4.10) now results by dierentiation w.r.t. x 0 : If we introduce a perturbation index for (4.1.) analogously to the standard case of DAEs, then the inequalities in (i) and (ii) mean that a DAE (4.1) with index has the perturbation index f1 g, too. For the extended system (.4), (.5) and for (.6), respectively, it holds with x ; x y, 4x ; 4x 4y that A(x(:) :)(R4y) F x (x)4x 0 (:)+B((Ry) 0 (:) x(:) :)4x(:) : 4y(:) ; D(x(:) :)4x(:) In particular, the equation F x (x )4x (q r) with x ; x y is nothing else but the linear DAE A(x (t) t)(r(t)4y(t)) 0 + B((R(t)y (t)) 0 x (t) t)4x(t) q(t) 4y(t) ; D(x (t) t)4x(t) r(t): (4.1) Now, let x (:) be a solution of the nonlinear equation (.1) and y d(x (:) :). Then x (:) solves the extended form (.4), (.5) with y Ry d(x (:) :). With the coecients (4.1) can be reduced to For r(t) 0 this yields A (t) A(x (t) t) D (t) D(x (t) t) B (t) B((d(x (t) t)) 0 x (t) t) (4.13) A (t)(r(t)4y(t)) 0 + B (t)4x(t) q(t) 4y(t) ; D (t)4x(t) r(t): (4.14) A (t)(d (t)4x(t)) 0 + B (t)4x(t) q(t) (4.15) which can be regarded as a linearization of the initial equation (.1). By Theorem 3.4 and Lemma 4.1 the index f1 g is transformed from (.1) to (4.14) and (4.15). Now, let N 0, S 0, G 1 etc. be the subspaces and matrices of the chain formed for A, D and B from (4.13). We investigate the perturbed IVP A(x(t) t)(d(x(t) t)) 0 + b(x(t) t)q(t) (4.16)

Nonlinear dierential-algebraic equations with properly formulated leading term 11 D (t 0 )D (t 0 ) ; (d(x(t 0 ) t 0 ) ; y 0 )0 (4.17) y 0 R n q C(I R m ): For I the initial condition (4.17) simplies to R(t 0 )(d(x(t 0 ) t 0 ) ; y 0 )0,i.e., d(x(t 0 ) t 0 )R(t 0 )y 0 : (4.18) Theorem 4.4 Let the DAE (:1) with proper leading part be of index 1. Let x (:) be the solution of (.1) on the interval I. Further, let I. Then the IVP (4:16) (4:18) is uniquely solvable with a solution x(:) dened on I for q C(I R m ), k q k 1, y 0 R n, jr(t 0 )(y 0 ; d(x (t 0 ) t 0 ))j, > 0 suciently small. x(:) depends continuously dierentiably on y 0. It holds that k x ; x k 1 + k d(x(:) :) ; d(x (:) :) k C 1 const(jr(t 0 )(d(x(t 0 ) t 0 ) ; d(x (t 0 ) t 0 ))j+ k q k 1 ): P r o o f of the theorem: We form the extended system for the DAE (4.16) and apply Theorem 4.3 for 1. In the nature of things DAEs of index require a higher regularity of some components. Here it is essential in how far we can assume (4.9) for the extended equation (.6). We have (cf. x) with n(x t) : D(t)Q 1 (t)g (t) ;1 b(x t) n(x t)+d (t)q 1 (t)d (t) ; (y ; d(x t)) (4.19) n(x t) :D (t)q 1 (t)g (t) ;1 b(x t): (4.0) In the case of a linear original equation (.1) with A(x t) A(t), b(x t) B(t)x + q(t), d(x t) D(t)x the expressions (4.19), (4.0) simplify to n(x t) D(t)Q 1 (t)g (t) ;1 q(t)+d(t)q 1 (t)d(t) ; y n(x t) D(t)Q 1 (t)x + D(t)Q 1 (t)g (t) ;1 q(t): It is natural to demand that DQ 1 G ;1 q C 1 for index- DAEs, and likewise that DQ 1 D ; C 1. Thus, condition (4.9), which requires that n(x(:) :) has to belong to the class C 1 for continuous x(:) y(:) and continuously dierentiable (Ry)(:), is fullled for (.1) in the linear case. Theorem 4.5 Let the DAE (:1) with proper leading term be of index. Let be continuously x C(I R m ) be the solution of (:1), and D P 1 D ;, D Q 1 D ; dierentiable. Moreover, let : P 1 (t 0 ). Let n(x(:) :) C 1 (I R n ) for all x C(I R m ) from a neighbourhood ofx and for y CR(I 1 R n ) from a neighbourhood of y d(x (:) :).

1 R. Marz (i) For q C 1 D Q 1 (I G ;1 R m ) kqk 1 + k(d Q 1 G ;1 q)0 k 1 y 0 R n jd P 1 D ; (y 0 ; d(x (t 0 ) t 0 ))j > 0 suciently small, the IVP (4:16) (4:17) is uniquely solvable and for the solution x C(I R m ) it holds that kx ; x k 1 + kd(x(:) :) ; d(x (:) :)k C 1 const fkqk 1 + k(d Q 1 G ;1 q)0 k 1 + jd (t 0 )D (t 0 ) ; (d(x(t 0 ) t 0 ) ; d(x (t 0 ) t 0 ))jg: (ii) The solution x(:) of the IVP depends contiuously dierentiably on y 0. Proof.We use the extended form (.6) for (.1) and write the IVP (4.16), (4.17) in the following way (cf. xx1,) A(x (t) t)(d(t)x(t)) 0 + b(x(t) t)q(t) (4.1) D(t 0 )P 1 (t 0 )(x(t 0 ) ; x 0 )0 (4.) with D(t 0 )P 1 (t 0 )(0D P 1 D ; )andq(t) ; q(t) 0. For (.6), the condition (4.9) is given by the assumption. q C 1 (I DQ 1 G ;1 R m+n ) holds if and only if q C 1 D Q 1 (I G ;1 R m ). Consequently, Theorem 4.3 yields the assertion. We formulate a further perturbation theorem, whose assumptions are possibly easier to be checked. Theorem 4.6 Let the DAE (:1) with proper leading term be of index. Let as well as D x x C(I R m ) be the solution of (:1), and let D P 1 D ;, D Q 1 D ; be continuously dierentiable. Let : P 1 (t 0 ). For all x CD 1 (I R m ) from a neighbourhood of x let n(x(:) :) and d(x(:) :) be continuously dierentiable. Then the assumptions from Theorem 4:5 remain true, where the solutions of the IVP (4:16) (4:17) are even from CD 1 (I R m ). Proof. We check immediately whether the linear index- DAE (4.1) has solutions (I R m ) r 4x C 1 D (I R m ), 4y C 1 R(I R n ) for right-hand sides q C 1 D Q 1 G ;1 C 1 R(I R n ). With X : C 1 D (I R m ) C 1 R(I R m ) and Y : C 1 D Q 1 G ;1 (I R m ) C 1 R(I R n )we obtain F x (x ) L b (X Y ) imf x (x )Y. Then the related mappping F IV P x (x ) acts bijectively between the spaces X and Y L IC. By assumption it holds that x ; x y X. The regularity conditions for n(x t) and d(x t) ensure that F(x) Y is always true for x X from a neighbourhood of x. We can further argue analogously to the proof of Theorem 4.3, where the mapping H(x d q) :F IV P (x) ; (q d) operates in the spaces X L IC Y and Y L IC now. The additional regularity conditions in case of DAEs of index, which shall

Nonlinear dierential-algebraic equations with properly formulated leading term 13 guarantee certain properties of the mappings (the images F(x) have to lie in the "right" space), imply restrictions on the admissible structure. An interesting special class of DAEs, which is important for applications, consists of DAEs for which N 0 (x t) does not depend on x, i.e., ker D(x t) N 0 (t) x D ti (4.3) and P 0 (t) is continuously dierentiable w.r.t. t. Quite often N 0 (x t) is even independent of x and t. Then it holds that d(x t) d(p 0 (t)x t) x D ti (4.4) and further d x (x t) d x (P 0 (t)x t), in particular, D (t) d x (P 0 (t)x (t) t): Lemma 4.7 Let (:1) be a DAE with proper leading term. Let (4:3) be valid and d C 1 (DI R n ), P 0 DAE (:1). C 1 (I L(R m )). Let x C(I R m ) be a solution of the (i) Then P 0 x D D ; and D x are continuously dierentiable on I. (ii) d(x(:) :) is continuously dierentiable on I for all x C 1 D (I R m ). P r o o f. Provided that P 0 x is continuously dierentiable, then D (t) d x (P 0 (t)x (t) t) is so, too. As a reexive generalized inverse with C 1 -projectors P 0 (t) D (t) ; D (t), R(t) D (t)d (t) ; also D (t) ; is continuously dierentiable. Furthermore, D x D P 0 x belongs to the class C 1. Then x CD 1 (I R m ) implies d(x(:) :)d(d (:) ; D (:)x(:) :) C 1 (I R n ): It remains to show that P 0 x is actually C 1. Therefore, we investigate the function K(x y t) :y ; d(x t) x D y R n ti: We have K(x (t) y (t) t) 0, and K x (x (t) y (t) t) ;D (t) acts bijectively between im P 0 (t) and im R(t). For each t I the equation K(x y t) 0 provides a solution function (: t) with K((y t) y t) 0, (y t) P 0 (t)(y t), (y t) (R(t)y t), P 0 (t)x (t) (y (t) t). Then the regularity of P 0 x results from that of, since y (t) is continuously dierentiable. For a better understanding let us remark that, in case of d(x t) D(t)x + (t), the equation y ; D(t)x ; (t) 0 leads to y R(t)y and P 0 (t)x D(t) ; y ; D(t) ;, i.e., (y t) D(t) ; y ; D(t) ; (t). For nonlinear DAEs (.1) satisfying the assumptions of Lemma 4.7, i.e., DAEs with a smooth function d and an only time-dependent smooth subspace N 0,itisconvenient to work with the function space C 1 P 0 (I R m )C 1 D (I R m ). (.1) can be left in the original form or we may take (d(x(t) t)) 0 d x (P 0 (t)x(t) t)(p 0 (t)x(t)) 0 + d t (P 0 (t)x(t) t):

14 R. Marz 5 Special systems of circuit simulation The modied nodal analysis (MNA) used in industrial simulation packages generates, for large classes of circuits, systems of the form (cf. [EsTi]) A c (q(a > 9 c e(t) t)0 +b c (e(t) j L (t) j V (t)) 0 ((j L (t) t)) 0 +b L (e(t) j L (t) j V (t)) 0 (5.1) b V (e(t) j L (t) j V (t)) 0 where e(t), j L (t), j V (t) denote the nodal potentials and the currents of inductances and the voltage sources, respectively. The functions q(z t) and (w t) arecontinuously dierentiable and the Jacobians q z (z t), w (w t) are positive denit. With 0 x : @ e j L j V 1 0 A A ~ : @ A c 0 0 I 0 0 1 A ~ d(x t) : q(a > c e t (j L t) 0 qz (A M(x t) : > c e t) 0 ~ b(x t) : @ 0 w (j L t) it holds that ~D : d x M ~ A > G 0 : ~ A ~ D ~ AM ~ A > im G 0 im ~ A ker G0 ker ~ D ker ~ A > : b c (e j L j V t) b L (e j L j V t) b V (e j L j V t The matrix G 0 (x t) has a nullspace N 0 that is independent of x and t. Obviously, (5.1) is nothing else but ~A( ~ d(x(t) t)) 0 + ~ b(x(t) t)0: (5.) If im D(x ~ t) is independent of x, then (5.) has a proper leading term. Because of the constant nullspace N 0, Lemma 4.7 is relevant. If im D(x ~ t) changes with x, we can put A ~ AP ~ A ~ in (5.) and shift the constant projector P with ker A ~ P A ~ ker A ~ below the derivative. The DAE ~A(P ~ A ~ d(x(t) t)) 0 + ~ b(x(t) t)0 (5.3) has a proper leading term with constant subspace N 0 and R ~ P A ~. Here, too, Lemma 4.7 may be applied. Moreover, P ~ A ~ D(x t) isalso constant now. On the other hand, we can dierentiate and investigate the equation ~A( ~ d(x(t) t)) 0 ~ AfM(x(t) t)( ~ A > x(t)) 0 + d t (x(t) t)g ~AM(x(t) t)( ~ A > x(t)) 0 + ~ b(x(t) t)+ ~ Adt (x(t) t)0: (5.4) If ker( ~ AM(x t)) is independent of x, (5.4) represents a DAE with proper leading term. If ker( ~ AM(x t)) depends on x, we proceed to ~AM(x(t) t)p >~ A ( ~ A > x(t)) 0 + ~ b(x(t) t)+ ~ Adt (x(t) t)0: (5.5) 1 A

Nonlinear dierential-algebraic equations with properly formulated leading term 15 Combined version are also possible, the choice depends, among other things, on the subspaces im D(x t), D(x t)s 1 (x t), D(x t)n 1 (x t) being constant in case of A(x(t) t)(d(x(t) t)) 0 + b(x(t) t)0: Both versions, (5.3) as well as (5.5), have a constant im D(x t). For them, it holds uniformly that S 1 (x t) fz : ~ b x (x t)p 0 z im ( ~ A ~ D(x t)+ ~ bx (x t)q 0 )g: Then wehave D(x t)s 1 (x t) P ~ A ~ D(x t)s 1 (x t) for (5.3), whereas D(x t)s 1 (x t) ~A > S 1 (x t) is true for (5.5). 6 Constraints Obviously, all solutions x C 1 D(I R m )ofthe DAE A(x(t) t)(d(t)x(t)) 0 + b(x(t) t)0 (6.1) have to satisfy the condition x(t) M 0 (t) ti, with the constraint set M 0 (t) :fx D: b(x t) im A(x t)g: (6.) For DAEs of index 1, M 0 (t) is completely lled with solutions and M 0 (t 0 ) is the set of consistent initial values at time t 0 I(cf. [HiMa]). Theorem 6.1 Let (6:1) be a DAE of index 1. Let t 0 I and x 0 M 0 (t 0 ). Then there exists a unique (maximal) solution x CD(I 1 R m ) with I 3 t 0, x (t 0 )x 0. P r oof.because of x 0 M 0 (t 0 ) there exists a y 0 R(t 0 ) y 0 R n, with A(x 0 t 0 )D(t 0 )D(t 0 ) ; y 0 + b(x 0 t 0 )0: With x D(t) ; D(t)x + Q 0 (t)x D(t) ; u + Q 0 (t)w, w : Q 0 (t)x + D(t) ; y, u : D(t)x we nd A(x t)d(t)d(t) ; y + b(x t) A(D(t) ; u + Q 0 (t)w t)d(t)w + b(d(t) ; u + Q 0 (t)w t) : F(u w t): It holds that F(u 0 w 0 t 0 ) 0 for u 0 : D(t 0 )x 0, w 0 : Q 0 (t 0 )x 0 + D(t 0 ) ; y 0. Moreover, F w (u 0 w 0 t 0 )G 1 (D(t 0 )w 0 D(t 0 ) ; u 0 + Q 0 (t 0 )w 0 t 0 )G 1 (y 0 x 0 t 0 ) is regular in case of index 1. Consequently, the implicit function theorem implies the local equivalence of the relations F(u w t) 0and w!(u t) with a

16 R. Marz continuous function!(u t), which has a continuous partial derivative! u (u t) and for which w 0!(u 0 t 0 ) is true. The regular IVP u 0 (t) ; R 0 (t)u(t) D(t)!(u(t) t) u(t 0 )u 0 (6.3) has a solution u C 1 (I R m ) for which u(t) R(t)u(t) t I, holds. Then the function x(t) :D(t) ; u(t)+q 0 (t)!(u(t) t) ti,isthe desired solution. Behind Theorem 6.1 is the idea that the dynamics of a DAE is dominated by an inherent regular dierential equation (here, in case of 1, the dierential equation (6.3)), and representations of the solution result from the constraints. In the case of DAEs (6.1) of index,hidden constraints have to be taken into account. Under the assumptions of the following lemma these hidden constraints can be described relatively easy. Lemma 6. For the DAE (6:1) of index let im G 1 (y x t) not depend on y and Q 0 (t)x. For a continuously dierentiable projector W 1 (x t) L(R m ), ker W 1 (x t) im G 1 (y x t), let the relation be valid. Then is also true. W 1 (x t) W 1 (P 0 (t)x t) xd ti (6.4) (W 1 b)(x t) (W 1 b)(p 0 (t)x t) xd ti (6.5) Proof. By construction, W 1 BQ 0 0 and W 1 (Ay) x Q 0 z 0 for all z R m, hence, Finally, we have (W 1 b)(x t) ; (W 1 b)(p 0 (t)x t) Z 1 0 Z 1 0 W 1 b x Q 0 W 1 (B ; (Ay) x )Q 0 0: (W 1 b) x (sx +(1; s)p 0 (t)x)q 0 (t)xds fw 1x (:::)Q 0 (t)xb(:::)+w 1 b x (:::)Q 0 (t)xgds 0: The structure introduced in Lemma 6. allows us to dierentiate the equation (W 1 b)(x t) 0 along a solution and thus to nd the hidden constraints. More precisely, let D(t) and D(t) ; be from C 1. Then the function (W 1 b)(x (t) t) is continuously dierentiable for each solution x C 1 D(I R m ). x (t) M 0 (t) implies (W 1 b)(x (t) t) 0 t I and (W 1 b)(d(t) ; D(t)x (t) t) 0 t I, respectively, and dierentiation yields (W 1 b) x (D(t) ; D(t)x (t) t)(d(t) ; D(t)x (t)) 0 +(W 1 b) 0 t(d(t) ; D(t)x (t) t)0: From (6.1) we obtain immediately D(t) ; (D(t)x (t)) ; ;D(t) ; A(x (t) t) ; b(x (t) t):

Nonlinear dierential-algebraic equations with properly formulated leading term 17 This makes clear that x (t) H 1 (t) ti with H 1 (t) : fx D :(W 1 b) x (x t)(d(t) ;0 D(t)x ; D(t) ; A(x t) ; b(x t)) +(W 1 b)(p 0 (t)x t) 0g must be true. Thus, H 1 (t) describes a hidden constraints. A(x t) ; is a reexive generalized inverse with A(x t) ; A(x t) R(t). It can be supposed that there are no further hidden constraints for and that M 1 (t) M 0 (t) \H 1 (t) ti represents the sharp geometrical location of the solution of the DAE (6.1) that is of index. Furthermore, it can be supposed that, under regularity conditions admitting the existence of the tangent space T x M 1 (t), the relations P 0 (t)t x M 1 (t) P 0 (t)s 1 (y x t) D(t)T x M 1 (t) D(t)S 1 (y x t) hold for x M 1 (t), y ;A(x t) ; b(x t). References [BaMa] K. Balla und R. Marz: A unied approach to linear dierential algebraic equations and their adjoint equations. Humboldt-Universitat Berlin, Institut fur Mathematik, Preprint 000-18. [EsTi] D. Estevez Schwarz und Caren Tischendorf: Structural analysis of electric circuits and Consequences for MNA. Int. Circ. Theor. Appl. 8 (000) 131-16. [GrMa] E. Griepentrog, R. Marz: Dierential-algebraic equations and their numerical tratment. Teubner, Leipzig, 1986. [HiMa] I. Higueras und R. Marz: Formulating dierential algebraic equations properly. Humboldt-Universitat Berlin, Institut fur Mathematik, Preprint 000-0. [KaAk] L.V. Kantorovich und G.P. Akilov: Funktsional'nyj analiz. Nauka, Moskva 1977.