Four Lectures on Differential-Algebraic Equations Steffen Schulz Humboldt Universität zu Berlin June 3, 2003 Abstract Differential-algebraic equations DAEs arise in a variety of applications. Therefore their analysis and numerical treatment plays an important role in modern mathematics. This paper gives an introduction to the topic of DAEs. Examples of DAEs are considered showing their importance for practical problems. Several well known index concepts are introduced. In the context of the tractability index existence and uniqueness of solutions for low index linear DAEs is proved. Numerical methods applied to these equations are studied. Mathematics Subject Classification: 34A09, 65L80 Keywords: differential-algebraic equations, numerical integration methods Introduction In this report we consider implicit differential equations f x t, xt, t = 0 f x on an interval I R. If is nonsingular, then it is possible to formally solve for x f in order to obtain an ordinary differential equation. However, if x is singular, this is no longer possible and the solution x has to satisfy certain algebraic constraints. Thus equations where f x is singular are referred to as differentialalgebraic equations or DAEs. These notes aim at giving an introduction to differential-algebraic equations and are based on four lectures given by the author during his stay at the University of Auckland in 2003. The first section deals with examples of DAEs. Here problems from different kinds of applications are considered in order to stress the importance of DAEs when modelling practical problems. In the second section each DAE is assigned a number, the index, to measure it s complexity concerning both theoretical and numerical treatment. Several index notions are introduced, each of them stressing different aspects of the DAE considered. Special emphasis is given to the tractability index for linear DAEs. The definition of the tractability index in the second section gives rise to a detailed analysis concerning existence and uniqueness of solutions. The main tool is a procedure to decouple the DAE into it s dynamical and algebraic part. In section three this analysis is carried out for linear DAEs with low index as it was established by März [25]. The results obtained, especially the decoupling procedure, are used in the fourth section to study the behaviour of numerical methods when applied to linear DAEs. The material presented in this section is mainly taken from [8].
Examples of differential-algebraic equations Modelling with differential-algebraic equations plays a vital role, among others, for constrained mechanical systems, electrical circuits and chemical reaction kinetics. In this section we will give examples of how DAEs are obtained in these fields. We will point out important characteristics of differential-algebraic equations that distinguish them from ordinary differential equations. More information about differential-algebraic equations can be found in [2, 5] but also in [32].. Constrained mechanical systems Consider the mathematical pendulum in figure.. Let m be the pendulum s mass which is attached to a rod of length l [5]. In order to describe the pendulum in Cartesian coordinates we write down the potential energy Ux, y = mgh = mgl mgy. where xt, yt is the position of the moving mass at time t. The earth s acceleration of gravity is given by g, the pendulum s height is h. If we denote derivatives of x and y by ẋ and ẏ respectively, the kinetic energy is given by x ϕ y l h Figure.: The mathematical pendulum T ẋ, ẏ = 2 mẋ2 + ẏ 2..2 The term ẋ 2 + ẏ 2 describes the pendulum s velocity. The constraint is found to be 0 = gx, y = x 2 + y 2 l 2..3.-.3 are used to form the Lagrange function Lq, q = T ẋ, ẏ Ux, y λ gx, y. Here q denotes the vector q = x, y, λ. Note that λ serves as a Lagrange multiplier. The equations of motion are now given by Euler s equations d L L = 0, k =, 2, 3. dt q k q k We arrive at the system mẍ + 2λx = 0, mÿ mg + 2λy = 0, gx, y = 0..4 By introducing additional variables u=ẋ and v =ẏ we see that.4 is indeed of the form. 2
When solving.4 as an initial value problem, we observe that each initial value xt0, yt 0 = x 0, y 0 has to satisfy the constraint.3 consistent initialization. No initial condition can be posed for λ, as λ is determined implicitly by.4. Of course the pendulum can be modeled by the second order ordinary differential equation ϕ = g l sin ϕ when the angle ϕ is used as the dependent variable. However for practical problems a formulation in terms of a system of ordinary differential equations is often not that obvious, if not impossible..2 Electrical circuits Modern simulation of electrical networks is based on modelling techniques that allow an automatic generation of the model equations. One of the techniques most widely used is the modified nodal analysis MNA [7, 8]..2. A simple example To see how the modified nodal analysis works, consider the simple circuit in figure.2 taken from [39]. It consists of a voltage source v V = vt, a resistor with conductance G and a capacitor with capacitance C > 0. The layout of the circuit can be described by 0 A a = 0, 0 e U V G C e 2 Figure.2: A simple circuit where the columns of A a correspond to the voltage, resistive and capacitive branches respectively. The rows represent the network s nodes, so that and indicate the nodes that are connected by each branch under consideration. Thus A a assigns a polarity to each branch. By construction the rows of A a are linearly dependent. However, after deleting one row the remaining rows describe a set of linearly independent equations, The node corresponding to the deleted row will be denoted as the ground node. The matrix 0 A = 0 is called the incidence matrix. It is now possible to formulate basic physical laws in terms of the incidence matrix A [20]. Denote with i and v the vector of branch currents and voltage drops respectively and introduce the vector e of node potentials. For each node the node potential is it s voltage with respect to the ground node. Kirchhoff s Current Law KCL: Kirchhoff s Voltage Law KVL: For each node the sum of all currents is zero. For each loop the sum of all voltages is zero. } } Ai = 0 v = A T e 3
For the circuit in figure.2 KCL and KVL read and i V + i G = 0, i G + i C = 0.5a v V = e, v G = e e 2, v C = e 2.5b respectively. If we assume ideal linear devices the equations modelling the resistor and the capacitor are i G = Gv G, Finally we have v V = vt i C = C dv c dt..5c.5d for the independent source which is thought of as the input signal driving the system. The system.5 is called the sparse tableau. The equations of the modified nodal analysis are obtained from the sparse tableau by expressing voltages in terms of node potential via.5b and currents, where possible, by device equations.5c: i V + Ge e 2 = 0 Ge e 2 + C de 2 dt = 0 e = v 0 C 0 0 0 e e 2 i V G G + G G 0 0 0 The MNA equations reveal typical properties of DAEs: e e 2 i V 0 = 0.6 v i Only certain parts of x = e, e 2, i V T need to be differentiable. It is sufficient if e and i V are continuous. ii Any initial condition xt 0 = x 0 needs to be consistent, i.e. there is a solution passing through x 0. Here this means that we can pose an initial condition for e 2 or i V only. For.6 it is sufficient to solve the ordinary differential equation e 2t = C G vt + e 2 t. e 2 t can be thought of as the output signal. solution are uniquely determined as The remaining components of the e t = vt, i V t = G e t e 2 t. Another important feature that distinguishes DAEs from ordinary differential equations is that the solution process often involves differentiation rather than integration. This is illustrated in the next example. 4
.2.2 Another simple example If we replace the independent voltage in figure.2 source by a current source i I = it and the capacitor by an inductor with inductance L, we arrive at the circuit in figure.3. The sparse tableau now reads i I + i G = 0, i G + i L = 0,.7a v I = e, v G = e e 2, v L = e 2,.7b i G = Gv G, i I = it. v L = L di L dt,.7c.7d Thus the modified nodal analysis leads to e e 2 Ge e 2 = it Ge e 2 + i L = 0 L di L dt e 2 = 0 The solution is given by i L = it, e 2 = L di L dt = Ldit dt, e = e 2 + G it = L dit dt + G it,.8 I I G Figure.3: Another simple circuit under the assumption that the current it is differentiable. Notice that all component values are fixed. To solve for e 2 we need to differentiate the current i..3 A transistor amplifier We will now present a more substantial example adapted from [6]. Consider the transistor amplifier circuit in figure.4. P. Rentrop has received this example from K. Glashoff and H.J. Oberle and documented it in [34]. The circuit consists of eight nodes, U e t = 0. sin200πt is an arbitrary 00 Hz input signal and e 8, the node potential of the 8 th node, is the amplified output. The circuit contains two transistors. We model the behaviour of these semiconductor devices by voltage controlled current sources I gate = α ge gate e source, I drain = α ge gate e source, I source = ge gate e source with a constant α = 0.99, g is the nonlinear function g : R R, v gv = β exp v, β = 0 6, U F = 0.026. U F L 5
G 2 G 4 G 6 G 8 2 G C 0 G D S 4 3 C 3 5 G D S 7 8 6 C 5 output input Ue G G 3 C 2 G 5 G 7 C 4 G 9 Ub Figure.4: Circuit diagram for the transistor amplifier It is also possible to use PDE models partial differential equations to model semiconductor devices. This approach leads to abstract differential-algebraic systems studied in [23, 35, 40]. The modified nodal analysis can now be carried out as in the previous examples. Consider for instance the second node. KCL implies that 0 = i C i R i R2 i gate,2 = C v C v G G v G2 G 2 α g e 2 e 3 = C e2 e e2 G e 2 U b G2 + α g e 2 e 3 = C e e 2 e2 G + G 2 + Ub G 2 + α g e 2 e 3. U b = 6 is the working voltage of the circuit and the remaining constant parameters of the model are chosen to be G 0 = 0 3, G k = 9 0 3, k =,..., 9, C k = 0 6, k =,..., 5. A similar derivation for the other nodes leads to the quasi-linear system A Dxt = b xt.9 with A = D = C 0 0 0 0 C 0 0 0 0 0 C 2 0 0 0 0 0 C 3 0 0 0 0 C 3 0 0 0 0 0 C 4 0 0 0 0 0 C 5 0 0 0 0 C 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0,, bx = U e G 0 +e G 0 U b G 2 +e 2 G +G 2 α ge 2 e 3 ge 2 e 3 +e 3 G 3 U b G 4 +e 4 G 4 +αge 2 e 3 U b G 6 +e 5 G 5 +G 6 α ge 5 e 6 ge 5 e 6 +e 6 G 7 U b G 8 +e 7 G 8 +αge 5 e 6 e 8 G 9. A numerical solution of.9 can be calculated using Dassl or Radau5, see [6, 4]. 6
A mathematically more general version of.9 is A xt, t Dtxt = b xt, t.0 with a solution dependent matrix A. We identified x i with the node potential e i. Let us assume that N 0 t = ker A xt, t Dt does not dependent on x. We will follow [6] and investigate.0 in more detail. With fy, x, t = A xt, t y b xt, t,.0 can be written as f Dtxt, xt, t = 0.. Denote By, x, t = f xy, x, t and let Qt be a continuous projector function onto N 0 t. Calculate G y, x, t = Ax, tdt + By, x, tqt. For the transistor amplifier. in figure.4 this matrix is always nonsingular. We want to use this matrix in conjunction with the Implicit Function Theorem to derive an ordinary differential equation that determines the dynamical flow of.0. Let Dt be defined by DD D = D, DD = I 5, D DD = D, D D = P := I 8 Q. I k denotes the identity in R k and Dt is a generalized reflexive inverse of Dt. For more information on generalized matrix inverses see section 2.3. on page 8. For a solution x of. define ut = Dtxt, wt = Dt u t + Qtxt. Observe that ADx = ADw and x = P x + Qx = D Dx + Qx = D u + Qw. Thus it holds that Note that. ADw + b x, t F w, u, t := f Dw, D u + Qw, t = 0. u = R u + Dw, since Dw = DD u + DQx = Ru = u R u. The mapping F can be studied without requiring x to be a solution. Let y 0, x 0, t 0 R 5+8+, such that fy 0, x 0, t 0 = 0. For w 0 = Dt 0 y 0 + Qt 0 x 0, u 0 = Dt 0 x 0 it follows that F w 0, u 0, t 0 = f y 0, x 0, t 0 = 0, F ww 0, u 0, t 0 = G y0, x 0, t 0 is nonsingular. 7
Due to the Implicit Function Theorem there is a ϱ > 0 and a smooth mapping satisfying ω : B ϱ u0, t 0 I R m ωu 0, t 0 = w 0, F ωu, t, u, t = 0 u, t B ϱ u0, t 0. We use ω to define xt = Dt ut + Qtωut, t, t I. where u is the solution of the ordinary differential equation u t = R tut + Dtω ut, t, ut 0 = Dt 0 x 0..2 x is indeed a solution of.0, since f Dtxt, xt, t = f u, D u + Qωu, t, t = F ω, u, t = 0. This example shows that there is a formulation of the problem in terms of an ordinary differential equation.2 as was the case for the mathematical pendulum in the first example. However,.2 is available only theoretically as it was obtained using the Implicit Function Theorem. Thus we have to deal directly with the DAE formulation.0 when solving the problem. Nevertheless,.2 will play a vital part in analyzing.0 and in analyzing numerical methods applied to.0. In section 3 it will be shown how.2 can be obtained explicitly for linear DAEs. Section 4 is devoted to showing that there are numerical methods that, when applied directly to.0, behave as if they were integrating.2, given that.0 satisfies some additional properties. In this case results concerning convergence and order of numerical methods can be transferred directly from ODE theory to DAEs..4 The Akzo Nobel Problem The last example originates from the Akzo Nobel Central Research in Arnhem, the Netherlands, and is again taken from [6]. It describes a chemical process in which two species, FLB and ZLU, are mixed while carbon dioxide is continously added. The resulting species of importance is ZLA. The reaction equations are given in [5]. 2 FLB + 2 CO 2 k FLBT + H2 O ZLA + FLB k 2 /K k3 k FLB + 2 ZHU + 3 CO 2 FLB.ZHU + 2 CO 2 k FLBT + ZHU LB + nitrate ZLA + H2 O ZLB + ZHU FLB.ZHU The last equation describes an equilibrium where the constant K s = [FLB.ZHU] [FLB] [ZHU] plays a role in parameter estimation. Square brackets denote concentrations. 8
The chemical process is appropriately described by the reaction velocities r = k [FLB] 4 [CO 2 ] 2, r 2 = k 2 [FLBT] [ZHU], r 3 = k 2 [FLB] [ZLA], K r 4 = k 3 [FLB] [ZHU] 2, r 5 = k 4 [FLB.ZHU] 2 [CO 2 ] 2, see [6] for details. The inflow of carbon dioxide per volume unit is denoted by F in and satisfies pco2 F in = kla H [CO 2]. kla is the mass transfer coefficient, H the Henry constant and pco 2 is the partial carbon dioxide pressure [6]. It is assumed that pco 2 is independent of [CO 2 ]. The various constants are given by k = 8.7, k 4 = 0.42, K s = 5.83, k 2 = 0.58, K = 34.4, pco 2 = 0.9, k 3 = 0.09, kla = 3.3, H = 737. If we identify the concentrations [FLB], [CO 2 ], [FLBT], [ZHU], [ZLA], [FLB.ZHU] with x,..., x 6 respectively, we obtain the differential-algebraic equation 2r +r 2 r 3 r 4 2 r r 4 2 r 5 +F in x t = r r 2 +r 3 r 2 +r 3 2r 4..3 r 2 r 3 +r 5 0 K s x x 4 x 6 This DAE can be analyzed in a similar way as the previous example. The matrix 0 0 0 0 0 0 0 0 0 0.42x 6 x2 G = AD + BQ = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.84x 6 x2 0 0 0 0 0 is always nonsingular. Here, A = D = diag,,,,, 0 was chosen. 9
2 Index concepts for DAEs In the last section we saw that DAEs differ in many ways from ordinary differential equations. For instance the circuit in figure.3 lead to a DAE where a differentiation process is involved when solving the equations. This differentiation needs to be carried out numerically, which is an unstable operation. Thus there are some problems to be expected when solving these systems. In this section we try to measure the difficulties arising in the theoretical and numerical treatment of a given DAE. 2. The Kronecker index Let s take linear differential-algebraic equations with constant coefficients as a starting point. These equations are given as Ex t + F xt = qt, t I, 2. with E, F LR m. apriori clear. Even for 2. existence and uniqueness of solutions is not Example 2. For the DAE 0 x t + 0 0 0 xt = 0 0 0 a solution x = x 2 is given by x 2 t = gt and x t = t t 0 gs ds, where the function g CI, R can be chosen arbitrarily. In order to exclude examples like 2. we consider the matrix pencil λe +F. The pair E, F is said to form a regular matrix pencil, if there is a λ such that detλe+f 0. A simultaneous transformation of E and F into Kronecker normal form makes a solution of 2. possible. Theorem 2.2 Kronecker [9] Let E, F form a regular matrix pencil. Then there exist nonsingular matrices U and V such that UEV = I 0, UF V = 0 N C 0, 0 I where N = diagn,..., N k is a block-diagonal matrix of Jordan-blocks N i to the eigenvalue 0. The proof can be found in [9] or [5]. Notice that due to the special structure of N there is µ N such that N µ 0 but N µ = 0. µ is known as N s index of nilpotency. It does not depend on the special choice of U and V. We solve 2. by introducing the transformation x = V u, v at = Uqt. bt 0
Thus 2. is equivalent to UEV V xt + UF V V xt = Uqt I 0 ut + 0 N vt C 0 0 I The first equation is an ordinary differential equation u t + Cut = at for the u component. The second equation reads vt = bt Nv t ut vt = bt N b t Nv t = bt Nb t + N 2 v t µ = at. 2.2 bt = = N i b i t 2.3 i=0 determining the v component completely by repeated differentiation of the right hand side b. Since numerical differentiation is an unstable process, the index µ is a measure of numerical difficulty when solving 2.. Definition 2.3 Let E, F form a regular matrix pencil. The Kronecker index of 2. is 0 if E is nonsingular and µ, i.e. N s index of nilpotency, otherwise. 2.2 The differentiation index How can definition 2.3 be generalized to the case of time dependent coefficients or even to nonlinear DAEs? If we consider 2.3 again, it turns out that µ v t = N i b i+ t, i=0 meaning that exactly µ differentiations transform 2.2 into a system of explicit ordinary differential equations. This idea was generalized by Gear, Petzold and Campbell [4, 0, ]. The following definition is taken from [5]. Definition 2.4 The nonlinear DAE f x t, xt, t = 0 2.4 has differentiation index µ if µ is the minimal number of differentiations f x t, xt, t df x t, xt, t = 0, = 0,..., dµ f x t, xt, t dt dt µ = 0 2.5 such that the equations 2.5 allow to extract an explicit ordinary differential system x t = ϕ xt, t using only algebraic manipulations.
We now want to look at four examples to get a feeling of how to calculate the differentiation index. We always assume that the functions involved are smooth enough to apply definition 2.4. Example 2.5 For linear DAEs with constant coefficients forming a regular matrix pencil we have differentiation index µ if and only if the Kronecker index is µ. Example 2.6 Consider the system x = fx, y 2.6a 0 = gx, y. 2.6b The second equation yields 0 = dgx, y dt = g x x, yx + g y x, yy. If g y x, y is nonsingular in a neighbourhood of the solution, 2.6 is transformed to x = fx, y y = g y x, y g x x, yx = g y x, y g x x, yfx, y 2.6a 2.6b and the differentiation index is µ =. The DAE.6 modelling the circuit in figure.2 is of the form 2.6 with e x = e 2, y =, fx, y = G i V C e Ge e e 2 and gx, y = 2 i V. e + v V G Note that g y x, y = is nonsingular so that.6 is an index equation. 0 Example 2.7 The system x = fx, y 2.8a 0 = gx 2.8b can be studied in a similar way. 2.8b gives 0 = dgx dt = g x xx = g x xfx, y = hx, y. 2.8b Comparing with example 2.6 we know that 2.8a, 2.8b is an index system if h y x, y remains nonsingular in a neighbourhood of the solution. If this condition holds, 2.8 is of index 2, as two differentiations produce x = fx, y 2.8a y = h y x, y h x x, yfx, y = g x xf y x, y g xx x fx,y,fx,y +g x xf x x,yfx,y. 2.8b 2.8b defines the hidden constraint of the index 2 equation 2.8. 2
The DAE.8 modelling the circuit in figure.3 can be written as i L = L e 2 = fi L, e 2 2.0a 0 = i L i I = gi L. 2.0b The remaining variable e is determined by e = e 2 + G i I, where i I is the input current. 2.0 is of the form 2.8 with x = i L and y = e 2. h y x, y = g x f y = L is nonsingular and the index is 2. Example 2.8 Finally take a look at the system x = fx, y y = gx, y, z 2.a 2.b 0 = hx. 2.c Differentiation of 2.c yields and 0 = dhx dt x = x y = 0 = = h x xx = h x xfx, y = ĥx, y 2.c fx,y = fx, y 2.a and 2.b 2.2a gx,y,z ĥx, y = gx 2.c 2.2b is of the form 2.8 with x = x y and y = z. Define hx, y = gxxfx, y and compare with 2.8b to find that 2.2 is of the index 2 if hyx, y = gxxfyx, y = f z ĥ x ĥ y g z = 0 h xx f, + h x f x h xy f, + h x f y = h g x f y g z z remains nonsingular. This shows that 2. is an index 3 system if the matrix h x xf y x, yg z x, y, z is invertible in a neighbourhood of the solution x, y, z. Hidden constraints are given by 2.c but also by hx, y = gxxfx, y = h xx f, f + h x f x f + h x f y g = 0, which is condition 2.8b in terms of the index 2 system 2.2. Consider again the mathematical pendulum from section. in the formulation x = u = f x, y, u, v y = v = f 2 x, y, u, v u = 2 m λx = g x, y, u, v, λ v = +g 2 m λy = g 2x, y, u, v, λ 0 = x 2 + y 2 l 2 = hx, y. 2.3 For l > 0 the value h x,y f u,v g λ = 4 m x2 + y 2 is always nonsingular so that 2.3 is an index 3 problem. 3
2.3 The tractability index In definition 2.4 the function f is assumed to be smooth enough to calculate the derivatives 2.5. In applications this smoothness is often not given. For instance in circuit simulation input signals are continuous but often not differentiable. In this section we want to study the tractability index introduced by Griepentrog, März [3]. In fact we consider the generalization of the tractability index proposed by März [25]. The idea is to replace the smoothness requirements for the coefficients by the requirement on certain subspaces to be smooth. To define the tractability index we introduce linear DAEs with properly stated leading terms. A second matrix Dt is used when formulating the DAE as At Dtxt + Btxt = qt. 2.4 In contrast to the standard formulation Etxt + F txt = qt 2.5 the leading term in 2.4 precisely figures out which derivatives are actually involved. The formulation 2.4 was first used in [] to study linear DAEs and their adjoint equations. For 2.5 the adjoint equation E y F y = p is of a different type. For the more general formulation 2.4 the adjoint equation fits nicely into this general form: D A y B y = p. In this section we consider linear DAEs 2.4 with matrix coefficients A C I, LR n, R m, D C I, LR m, R n, B C I, LR m. Neither A nor D needs to be a projector function. Note that At and Dt are rectangular matrices in general. However, A and D are assumed to be well matched in the following sense. Definition 2.9 The leading term of 2.4 is properly stated if ker At im Dt = R n, t I, and there is a continuously differentiable projector function im Rt=im Dt, ker Rt=ker At t I. R C I, LR n with By definition At and Dt have a common constant rank if the leading term is properly stated [25]. Definition 2.0 A function x : I R m is said to be a solution of 2.4 if x C DI, R m = {x CI, R m Dx C I, R n } satisfies 2.4 pointwise. Let us point out that a solution x is a continuous function, but the part Dx : I R n is differentiable. 4
We now define a sequence of matrix functions and possibly time-varying subspaces. All relations are meant pointwise for t I. Let G 0 = AD, B 0 = B and for i 0 N i = ker G i, S i = { z R m B i z im G i } = { z R m Bz im G i }, Q i = Q 2 i, im Q i = N i, P i = I Q i, G i+ = G i + B i Q i, B i+ = B i P i G i+ D C i+dp 0 P i, C i+ = DP 0 P i+ D. 2.6 Here, D : I LR n, R m denotes the reflexive generalized inverse of D such that DD D = D, D DD = D, DD = R, D D = P 0. 2.7 Note that D is uniquely determined by 2.7 and depends only on the choice of Q 0. Section 2.3. contains more details about generalized matrix inverses. Definition 2. The DAE 2.4 with properly stated leading term is said to be a regular DAE with tractability index µ on the interval I if there is a sequence 2.6 such that G i has constant rank r i on I, Q i C I, LR m, DP 0 P i D C I, LR n, i 0, Q i+ Q j = 0, j = 0,..., i, i 0, 0 r 0 r µ < m and r µ = m. 2.4 is said to be a regular DAE if it is regular with some index µ. This index criterion does not depend on the special choice of the projector functions Q i [28]. As proposed in [24] the sequence 2.6 can be calculated automatically. Thus the index can be calculated without the use of derivative arrays [27]. Example 2.2 Consider the DAE t x t t x 2 t + t 0 0 x t x 2 t = 0 taken from [25]. With ker At = {0}, im Dt = R the leading term is properly stated. Calculate t t 2 G 0 t = AtDt = and N t 0 t = { z R 2 α R, z = α t } to find that N 0 t ker Bt. Independently of the choice of Q 0 in 2.6 we have G t = G 0 t + BtQt = G 0 t. 5
Similarly it follows that G i t = G 0 t for every i 0. This is not a regular DAE in the sense of definition 2.. Note that for every γ CI, R a solution is given by xt = γt t. Solutions are therefore not uniquely determined. This is the case in spite of the fact that for every t the local matrix pencil λad + B + AD of the reformulated DAE 0 = AtDtx t + Bt + AtD t t t 2 xt = x t + xt t is regular. The following lemma shows that definition 2. is indeed a generalization of the Kronecker index, i.e. in the case of constant coefficients, the Kronecker index and the tractability index for regular DAEs coincide. To show this, define the subspaces S EF = {z R m F z im E}, N E = ker E. for given matrices E, F LR m. Obviously for fixed t I we have N i t = N Gi t and S i t = S Gi tb i t in sequence 2.6. Lemma 2.3 For matrices E, F LR m the following statements are equivalent: N E S EF = {0} 2 For every projector Q E onto N E the matrix E + F Q E is nonsingular. 3 N E S EF = R m 4 E, F form a regular matrix pencil with Kronecker index. Proof: 2 E + F Q E z = 0 implies Q E z S EF. Since Q E z N E too, we have Q E z N E S EF = {0} and Q E z = 0. Thus 0 = Ez + F Q E z = Ez and z N E = im Q E. Therefore z = Q E z = 0. 2 3 G EF = E + F Q E is nonsingular. Show that Q = Q E G EF F is the projector onto N E along S EF. 3 4 There is exactly one projector Q onto N E along S EF. Since 3 2, we find Q = Q G EF F with G EF = E + F Q. Let P = I Q. Show that λ E+F is nonsingular for λ specp G EF F so that E, F form a regular matrix pencil. Due to theorem 2.2 there are nonsingular matrices U, V GLRm such that V E U = I N = Ē, V F U = C I = F. It follows that N Ē = ker Ē = U N E and S Ē F = {z Rm F z im Ē} = U S EF so that N Ē SĒ F = U N E S EF = {0}. 2.8 On the other hand N = { Ē z 2 R m z = 0, z 2 ker N} and S Ē F = { z 2 R m C z z 2 im Ē} = { z 2 R m z 2 im N}, 6
meaning that im N ker N = {0} and N = 0. Thus the Kronecker index is. 4 Kronecker index gives N = 0 and S Ē F = {0}, NĒ SĒ F = {0}. Use 2.8 to see N E S EF = UN Ē SĒ F = {0}. As in the previous section we now want to calculate the index of the DAEs modelling the electrical circuits in figure.2 and.3. 0 0 0 Example 2.4 For.6 we calculate G 0 = 0 C 0 and N 0 = R {0} R. Choose 0 0 0 0 0 G 0 Q 0 = 0 0 0 to find that G = G C 0 is nonsingular. For the circuit in figure 0 0 0 0.2 we therefore have index. Example 2.5 Equation.8 can be written as 0 0 0 0 e G G 0 e 2 + G G L i L 0 0 e it e 2 = 0.8 i L 0 0 0 0 0 0 leading to G 0 = 0 0 0, N 0 = R R {0}. With Q 0 = 0 0 it turns out that 0 0 L 0 0 0 G 0 0 G = G G 0 is singular and N = { z R 3 α R, z = z 2 = αl, z 3 = α }. 0 L Q = is a projector onto N satisfying Q Q 0 = 0. Finally G 2 = 0 0 L 0 0 L 0 0 G G 0 G G 0 L is nonsingular. Thus the index is 2. Note that the terms C i+ dissappear in 2.6 as Q 0 does not depend on t. Nevertheless, in general the derivatives of C i+ appearing in the definition of B i+ in sequence 2.6 are necessary in order to determine the index correctly. We will illustrate this in the next example which can be found in [25] as well. Example 2.6 The DAE x 2 = q x = fx 2.9a x 3 = q 2 + ηx 2 ηtq x = gx, x 2, x 3 2.9b 0 = q 3 ηtx 2 x 3 = hx 2, x 3 2.9c is easily checked to have differentiation index 3 as repeated differentiation of 2.9c yields 0 = q 3 q 2 + x 2 0 = q 3 q 2 + q x x = q 3 q 2 + q. The index does not depend on the value of η. We now write 2.9 as 0 ηt 0 0 0 0 x q x x 0 0 2 + 0 + η 0 x 2 = q 2 2.9 0 0 x 3 0 ηt x 3 q 3 7
with a properly stated leading term and calculate the sequence 2.6 G 0 = G 2 = 0 0 0 ηt 0 0 0 0 0 ηt+ 0 0 0, Q 0 =, Q 2 = 0 0 0 0 0 0 0 0, G =, G 3 = 0 ηt 0 ηt 0 ηtηt+ ηt+ 0 0 ηt 0 0 0 0 0 ηt+ 0 ηt, Q =. 0 0 0 0 0 ηt 0 Since det G 3 =, 2.9 is a regular DAE with index 3 independently of η. However, if we dropped the terms C i+ in 2.6 and defined G i+ = G i + B i Q i, B i+ = B i P i with G 0 = AD and B 0 = B we would obtain G 2 = 0 0 ηt++η 0 0 0, Q 2 = +η 0 ηt 0 ηt 0 ηt++ηηt ηt++η, G 3 = 0 0 ηt++η 0 η t det G 3 = + η shows that G 3 is singular for η =. Thus the use of the simpler version of B i would lead to an index criterion not recognizing the index properly. The previous example gives rise to investigating the relationship between G i and G i further. Due to G i P i = G i the matrix G i+ may be written as G i+ = G i + B i P i Q i I Pi D C idp 0 P i Q i. For low indices we thus find G 0 = G 0, G = G, G 2 = G 2 I P D C DP 0 Q with the nonsingular factor I P D C DP 0Q. The matrices G 2 and G 2 have therefore common rank and we had to choose an index 3 example in 2.6 to show the necessity of the second term in the definition of B i+. We don t have to restrict ourselves to linear DAEs 2.4. Nonlinear DAEs A xt, t dx, t, t + b xt, t = 0 2.20 can also be considered. For 2.20 the index µ is defined in such a way that all linearizations along solutions have the same index µ in the sense of definition 2.. The index case is studied extensively in [6]. We have already made use of this approach when investigating the transistor amplifier example in section.3. More information on nonlinear DAEs can be found in [27, 29]. 2.3. Some technical details In order to define the sequence 2.6 we introduced the generalized reflexive inverse D of D. Here we want to provide a short summary of the properties of generalized matrix inverses [4]. For a rectangular matrix M LR m, R n, a matrix M LR n, R m is called a generalized inverse of M if MM M = M.., 8
If the condition M MM = M holds as well, then M is called a reflexive generalized inverse of M. Observe that for any reflexive generalized inverse M of M the matrices M M 2 = M MM M = M M, MM 2 = MM MM = MM are projectors. Reflexive generalized inverses are not uniquely determined. Uniqueness is obtained if we require M M and MM to be special projectors. We could, for instance, require them to be ortho-projectors M M T = M M, MM T = MM. In this case M is called the Moore-Penrose inverse of M, often denoted by M +. In the case of DAEs with properly stated leading terms we appropriated the projectors P 0 t LR m and Rt LR n to determine D t R n, R m uniquely. D t is the reflexive generalized inverse of Dt defined by DD D = D, D DD = D, DD = R, D D = P 0. 2.2 If there was another generalized inverse D satisfying 2.2, then D = D D D = D R = D DD = P 0 DD = D DD = D. In definition 2. the condition Q i+ Q j = 0, j = 0,..., i, i 0 2.22 is required. We will show briefly that the projectors Q i in sequence 2.6 can always be chosen to satisfy 2.22. If for a given DAE 2.4 there was an index i such that N i + N i {0}, then 2.4 would not be a regular DAE as all G i would be singular. Thus N 0 N = {0} is a necessary condition for a regular DAE and the projector Q onto N can be chosen such that N 0 ker Q. For an index i let the projectors Q j for j =,..., i satisfy Q j Q k = 0, k = 0,..., j. Then N i+ N i = {0} implies N i+ N j = {0} for j =,..., i and Q i+ can be chosen such that N 0 N N i ker Q i+. 2.4 Other index concepts As seen in the previous sections a DAE can be assigned an index in several ways. In the case of linear equations with constant coefficients all index notions coincide with the Kronecker index. Apart from that, each index definition stresses different aspects of the DAE under consideration. While the differentiation index aims at finding possible reformulations in terms of ordinary differential equations, the tractability index is used to study DAEs without the use of derivative arrays. There are several other index concepts available. Here we want to introduce some of them briefly. 9
2.4. The perturbation index The perturbation index was introduced for nonlinear DAEs fx t, xt = 0 2.23 by Hairer, Lubich and Roche in [4]. 2.23 has perturbation index µ along a solution x on I = [0, T ] if µ is the smallest integer such that, for all functions ˆx having a defect fˆx t, ˆxt = δt, there exists on I an estimate ˆxt xt C ˆx0 x0 + max δξ + + max 0 ξ t 0 ξ t δµ ξ whenever the expression on the right-hand side is sufficiently small. Here C denotes a constant which depends only on f and the length of I. The perturbation index measures the sensitivity of solutions with respect to perturbations of the given problem [5]. 2.4.2 The geometric index Here we present the geometric index as it is introduced in [38]. Consider the autonomous DAE fx, x = 0 2.24 and assume that M 0 = f 0 is a smooth submanifold of R m R m. Then the DAE 2.24 can be written as x, x M 0. Each solution has to satisfy x W 0 = πm 0, where π : R m R m R m is the canonical projection onto the second component. If W 0 is a submanifold of R m, then x, x belongs to the tangent bundle T W 0 of W 0. In other words x, x M = M 0 T W 0. M is called the first reduction of M 0. Iterate this process to obtain a sequence M 0, M, M 2,... of manifolds where M i+ is the first reduction of M i and x, x i 0 M i. The geometric index is defined as the smallest integer µ such that M µ = M µ+. This index notion was introduced in [33] and studied extensively in [3] by Rabier and Rheinboldt. 20
2.4.3 The strangeness index This index notion is a generalization of the Kronecker index to DAEs Etx t + F txt = qt, t I R, 2.25 with time-dependent coefficients. It is due to Kunkel and Mehrmann [22]. The matrices U and V in theorem 2.2 now depend on t, i.e. 2.25 is transformed to UEV y + UF V UEV y = Uq Êy + ˆF y = ˆq. The pairs of matrix functions E, F and Ê, ˆF are said to be globally equivalent. For fixed t I define matrices T t, ˆT t, Zt and V t such that the column vectors of T t, ˆT t, Zt and V t span the subspaces ker Et, im E T, ker E T and imzt T NtT t, respectively. Use these matrices to define rt = rank Et, at = rank Zt T NtT t, st = rank V t T Zt T Nt ˆT t. dt = rt st, ut = m rt at st, We assume that the functions r, s and a are constant on I. Then E, F is globally equivalent to the pair 0 0 0 0 0 0 I d 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0, 0 F 2 0 F 4 F 5 0 0 0 F 24 F 25 0 0 I a 0 0 I s 0 0 0 0 0 0 0 0 0 = E, F. The proof can be found in [2]. The value s is called the strangeness of the pair E, F. Denote E, F by E 0, F 0 and s 0 = s. Similarly we define the strangeness s of the pair E, F. If we repeat the procedure described above, we arrive at a sequence of globally equivalent pairs E i, F i, i 0, each having strangeness s i. The strangeness index or s-index is then defined by µ = min{ i = 0,, 2,... s i = 0 }. Relations between the tractability index and the strangeness index are given in [36]. 2
3 Solvability of linear DAEs with properly stated leading term In this section we consider linear differential-algebraic equations At Dtxt + Btxt = qt, t I 3. with properly stated leading terms as in definition 2.9. A, B and D are continuous matrix functions with Dt LR m, R n, At LR n, R m Bt LR m, R m, qt R m A function x : I R m is said to be a solution of 3. if x C DI, R m = {x CI, R m Dx C I, R n } satisfies 3. pointwise. As in the previous section we define for t I pointwise G 0 = AD, B 0 = B and for i 0 N i = ker G i, S i = { z R m B i z im G i } = { z R m Bz im G i }, Q i = Q 2 i, im Q i = N i, P i = I Q i, G i+ = G i + B i Q i, B i+ = B i P i G i+ D C i+dp 0 P i, C i+ = DP 0 P i+ D. D is again the reflexive generalized inverse of D from section 2.3. For completeness we repeat the definition of index µ from the previous section. 3.2 Definition 3. The DAE 3. with properly stated leading term is said to be a regular DAE with tractability index µ on the interval I if there is a sequence 3.2 such that G i has constant rank r i on I, Q i C I, LR m, DP 0 P i D C I, LR n, i 0, Q i+ Q j = 0, j = 0,..., i, i 0, 0 r 0 r µ < m and r µ = m. 3. is said to be a regular DAE if it is regular with some index µ. The material presented here is mainly taken from [25], [] and [26]. 22
3. Decoupling of linear index- DAEs Let 3. be a regular index DAE with properly stated leading term. definition 3. the Matrix G is nonsingular. Due to Lemma 3.2 The matrices of sequence 3.2 satisfy a P 0 =D D, P 0 D =D, DP 0 =D, DP 0 D =DD =R, b RD = DD D = D, c A = AR = ADD, d Q 0 = G BQ 0, e P 0 = G AD, f P 0 x = P 0 y DP 0 x = DP 0 y a Dx = Dy. Proof: a and b are just the properties of the generalized reflexive inverse D. Remember that R C I, LR n is the smooth projector function realizing the decomposition ker At im Dt = R n provided by the properly stated leading term. ker A = ker R implies c. G Q 0 = ADQ 0 +BQ 2 0 = BQ 0 proves d. Similarly G P 0 = ADP 0 + BQ 0 P 0 = AD shows e. For f we only have to show. If DP 0 z = 0 then P 0 z ker D = ker AD = ker P 0 and thus P 0 z = 0. ker D =ker AD holds due to the properly stated leading term. Let s assume that x is a solution of the DAE 3.. Scaling with G yields Note that A Dx + Bx = q G A Dx + G Bx = G q. 3.3 G A Dx = c G AR Dx = G ADD Dx = e P 0 D Dx, G Bx = G BP 0x + G BQ 0x = G BP 0x + Q 0 x. d Thus multiplication of 3.3 by P 0 and Q 0 from the left shows that A Dx + Bx = q G A Dx + G Bx = G { q P0 D Dx + P 0 G BP 0x = P 0 G q Q 0 G BP 0x + Q 0 x = Q 0 G { q DP0 D Dx + DP0 G BP 0x = DP 0 G f Q 0 G BP 0x + Q 0 x = Q 0 G q a b a { R Dx + DG BP 0x = DG Q 0 G BP 0x + Q 0 x = Q 0 G q q q { Dx R Dx + DG BP 0x = DG Q 0 G BP 0x + Q 0 x = Q 0 G { Dx = R Dx DG BD Dx + DG Q 0 x = Q 0 G BD Dx + Q 0 G q q q q } } } } } 23
Every solution x can therefore be written as x = P 0 x + Q 0 x = D Dx + Q 0 x = D Dx Q 0 G BD Dx + Q 0 G q = I Q 0 G BD u + Q 0 G q 3.4 where u = Dx is a solution of the ODE u = R u DG BD u + DG q. 3.5 Definition 3.3 The explicit ordinary differential equation 3.5 is called the inherent regular ODE of the index- equation 3.. Lemma 3.4 i im D is a time varying invariant subspace of 3.5. ii 3.5 is independent of the choice of Q 0. Proof: i Because of im D = im R = keri R multiplication of 3.5 by I R gives I Ru = I RR u = I R Ru and v = I Ru satisfies the ODE v = I R v. If there is t I such that ut = Rt ut im Dt, then vt = 0. This means vt = 0 and thus ut = Rtut for every t. ii Let ˆQ 0 be another projector with im ˆQ 0 = N 0 and let ˆP 0, ˆD be defined as in 2.6. Then Ĝ = G I + Q 0 ˆQ0 P 0 implies Ĝ = I Q 0 ˆQ0 P 0 G and DĜ = DG. Finally note that DĜ B ˆD = DĜ B ˆP 0 D = DĜ BD DĜ B ˆQ 0 D a = d DG BD D ˆQ 0 D = DG BD. The decoupling procedure above and lemma 3.4 enable us to prove existence and uniqueness of solutions for the index DAE 3.. Theorem 3.5 Let 3. be a regular index DAE. For each d im Dt 0, t 0 I, the initial value problem At Dtxt + Btxt = qt, Dt0 xt 0 = d 3.6 is uniquely solvable in C D I, Rm. Proof: There is exactly one solution u C I, R m of the inherent ODE u = R u DG BD u + DG q satisfying the initial condition ut 0 = d. Lemma 3.4 shows that ut = Rtut for every t. Therefore x = I Q 0 G BD u + Q 0 G q C DI, R m is a solution of 3.6 satisfying Dx = u. The decoupling process shows the uniqueness. Note that the initial condition Dt 0 xt 0 = d for d im Dt 0 can be replaced by Dt 0 xt 0 = Dt 0 x 0, x 0 R m. 24
3.2 Decoupling of linear index-2 DAEs We now want to repeat the same argument for linear index 2 differential-algebraic equations. We assume that 3. is an index 2 DAE with properly stated leading term. Due to definition 3. and lemma 2.3 we have N t S t = R m. In this section we choose Q to be the canonical projector onto N along S. Lemma 2.3 also implies Q Q 0 = Q G 2 B Q 0 = 0 as required in definition 3.. For the sequence 3.2 to make sense we have to assume DP D C I, LR n. Then DQ D = DP D + DD = DP D + R is also smooth. Note that DQ D and DP D are projector functions. In addition to a, b, c and f from lemma 3.2 we now have Lemma 3.6 g Q = Q G 2 B, h G 2 AD = P P 0, i G 2 B = G 2 BP 0P + P D DP D DQ + Q + Q 0, j Q x = Q y DQ x = DQ y, k ΩΩ Ω = 0 for every projector function Ω C I, LR n. Proof: g follows from lemma 2.3, h can be proved similar to e in lemma 3.2, but i is a consequence of B = BP 0 + BQ 0 = BP 0 P + BP 0 Q + BQ 0 and G 2 Q 0 = BQ 0, G 2 Q = B Q, BP 0 Q = B Q + G 2 P D DP D DQ. To show j assume that DQ z = 0. Then Q z ker D = ker P 0 and Q z = Q 2 z = Q P 0 Q z = 0. Finally 0 = I ΩΩ implies 0 = I Ω Ω + I ΩΩ = Ω Ω + I ΩΩ. In order to decouple 3. in the index 2 case we again assume that x is a solution of the DAE. Since G 2 is nonsingular, we find A Dx + Bx = q G 2 A Dx + G 2 Bx = G 2 q 3.7 P D Dx +G 2 BP 0P x+p D DP D DQ x+q x+q 0 x=g 2 q using a, c, h and i. Due to I = P + Q = P 0 P + Q 0 P + Q we can decouple 3.7 by multiplying with P 0 P, Q 0 P and Q respectively. 3. is therefore equivalent to the system P 0 P D Dx +P 0 P G 2 BP 0P x+p 0 P D DP D DQ x=p 0 P G 2 q, 3.8a Q 0 P D Dx + Q 0 P G 2 BP 0P x +Q 0 P D DP D DQ x + Q 0 x=q 0 P G 2 q, 3.8b Q G 2 BP 0P x + Q x=q G 2 q. 3.8c 25
With a and f equation 3.8a takes the form DP D Dx + DP G 2 BP 0P x + DP D DP D DQ x = DP G 2 q. Use the product rule of differentiation to find DP D Dx = DP x DP D Dx. On the other hand DP D DP D DQ = DP D DQ as DP D DP P 0 Q = 0, so that 3.8a is equivalent to DP x DP D DP x + DP G 2 BD DP x = DP G 2 q. 3.8a A similar analysis involving g, j and k from lemma 3.6 yields Q 0 Q D DQ x + Q0 Q D DQ D DP x Each solution x of 3. can thus be written as where x = P 0 x + Q 0 x = D Dx + Q 0 x +Q 0 P G 2 BD DP x + Q 0 x = Q 0 P G 2 q, 3.8b DQ x = DQ G 2 q. 3.8c = D DP x + DQ x + Q 0 x 3.9 = KD u Q 0 Q D DQ D u+q 0 P +P 0 Q G 2 q+q 0Q D DQ G 2 q K = I Q 0 P G 2 B and u = DP x satisfies the ordinary differential equation u DP D u + DP G 2 BD u = DP G 2 q. As in the index case this ODE will be referred to as the inherent regular ODE. Definition 3.7 The explicit ordinary differential equation u = DP D u DP G 2 BD u + DP G 2 q 3.0 is called the inherent regular ODE of the index 2 equation 3.. For the index 2 case we now prove the lemma corresponding to lemma 3.4 from the previous section. Lemma 3.8 i im DP is a time varying invariant subspace of 3.0. ii 3.0 is independent of the choice of Q 0 and thus uniquely determined by the problem data. 26
Proof: To prove i, carry out a similar analysis as in the proof of lemma 3.4 but with R replaced by DP D. To see ii consider another projector ˆQ 0 with im ˆQ 0 = N 0 and the relation Ĝ = G I + Q 0 ˆQ0 P 0. The subspaces ˆN = I Q 0 ˆQ0 P 0 N and Ŝ = S are given in terms of N and S so that ˆQ = I + Q 0 ˆQ0 P 0 Q is the canonical projector onto ˆN along Ŝ. This implies D ˆP ˆD = DP D. Use the representation Ĝ 2 = I + Q 0 ˆP0 P P 0 G 2 to see that DP G 2 and DP G 2 BD are independent of the choice of Q 0. As in the previous section we are now able to prove existence and uniqueness of solutions for regular index 2 DAEs with properly stated leading terms. We make use of the function space C DQ G I, R m = { z CI, R m DQ G 2 z C I, R n }. 2 Theorem 3.9 Let 3. be a regular index 2 DAE with q C DQ G I, R m. For 2 each d im Dt 0 P t 0, t 0 I, the initial value problem At Dtxt + Btxt = qt, Dt0 P t 0 xt 0 = d 3. is uniquely solvable in C D I, Rm. Proof: Solve the inherent regular ODE 3.0 with initial value ut 0 = d. Lemma 3.8 yields ut im DtP t for every t and x = KD u Q 0 Q D DQ D u+q 0 P +P 0 Q G 2 q+q 0Q D DQ G 2 q is the desired solution of 3.. The initial condition Dt 0 P t 0 xt 0 = d can be replaced by Dt 0 P t 0 xt 0 = Dt 0 P t 0 x 0 for x 0 R m. 3.3 Remarks In sections.3 and.4 we presented examples of nonlinear differential-algebraic equations f Dx, x, t = 0, where the solution could be expressed as xt = Dt ut + Qtωut, t, t I. u was the solution of u t = R tut + Dtω ut, t, ut 0 = Dt 0 x 0 3.2 and ω was implicitly defined by F ω, u, t = f Dω, D u + Qω, t = 0. The ordinary differential equation 3.2 is thus only available theoretically. 27
In this section we made use of the sequence 3.2 established in the context of the tractability index in order to perform a refined analysis of linear DAEs with properly stated leading terms. We were able to find explicit expressions of 3.2 for these equations with index and 2. This detailed analysis lead us to results about existence and uniqueness of solutions for DAEs with low index. We were able to figure out precisely what initial conditions are to be posed, namely Dt 0 xt 0 = Dt 0 x 0 and Dt 0 P t 0 xt 0 = Dt 0 P t 0 x 0 in the index and index 2 case respectively. These initial conditions guarantee that solutions u of the inherent regular ODE 3.5 and 3.0 lie in the corresponding invariant subspace. Let us stress that only those solutions of the regular inherent ODE that lie in the invariant subspace are relevant for the DAE. Even if this subspace varies with t we know the dynamical degree of freedom to be rank G 0 and rank G 0 +rank G m for index and 2 respectively [25]. The results presented can be generalized for arbitrary index µ. The inherent regular ODE for an index µ DAE with properly stated leading term is given in [25]. There it is also proved that the index µ is invariant under linear transformations and refactorizations of the original DAE and the inherent regular ODE remains unchanged. Finally let us point out that we assumed A, D and B to be continuous only. The required smoothness of the coefficients in the standard formulation Ex + F x = q 3.3 was replaced by the requirement on certain subspaces to be spanned by smooth functions. Namely, the projectors R, DP D and DQ D are differentiable if DN and DS are spanned by continuously differentiable functions []. However, if the DAE ADx + Bx = q 3.4 is given with smooth coefficients and we orient on C -solutions, then comparisons with concepts for 3.3 can be made via ADx + B AD x = q. On the other hand, if E has constant rank on I and P E C I, LR m is a projector function onto ker E, we can reformulate 3.3 as EP E x + F EP Ex = q with a properly stated leading term. 28
4 Numerical methods for linear DAEs with properly stated leading term The last part is devoted to studying the application of numerical methods to linear DAEs of index µ = and µ = 2. From the previous section we know that 3.4 and 3.9 are representations of the exact the solution, respectively. In fact, it turns out that 3.4 is just a special cases of 3.9. To see this, observe that for µ = the matrix G is nonsingular so that Q = 0, P = I and G 2 = G. We therefore treat index and index 2 equations simultaneously in this section. We will show how to apply Runge-Kutta methods to DAEs At Dtxt + Btxt = qt 4. with properly stated leading terms. Results presented here follow the lines of [7, 8, 26]. Runge-Kutta methods for DAEs are also studied in [4]. When using the s-stage Runge-Kutta method c A β T, A = α ij LR s, c = Ae, β R s, e =,..., T R s, to solve an ordinary differential equation x t = F xt, t 4.2 numerically with stepsize h, an approximation x l to the exact solution xt l is used to calculate the approximation x l to xt l = xt l + h via s x l = x l + h β i X li i= where X li is defined by 4.3a X li = F X li, t li, i =,..., s, 4.3b and t li = t l + c i h are intermediate timesteps. The internal stages X i are given by X li = x l + h s α ij X lj. j= 4.3c Observe that 4.3a and 4.3c depend on the method and only 4.3b depends on the equation 4.2. If the ODE 4.2 is replaced by the DAE f x t, xt, t = 0 we also replace 4.3b by f X li, X li, t li = 0, i =,..., s 4.3b in the Runge-Kutta scheme. The matrix f x is singular. Therefore some components of the increments X li need to be calculated from 4.3c as seen in the following trivial example. 29
Example 4. If fx, x, t = x qt, then xt = qt. The numerical method 4.3a, 4.3b, 4.3c now reads s s x l = x l + h β i X li, qt li = X li = x l + h α ij X lj. i= This system can be solved if and only if A is nonsingular. In the following, we always assume A to be nonsingular. This leads to an expression of X li in terms of X lj. Lemma 4.2 Let A = α ij be nonsingular and A = α ij. Then Proof: X li =x l +h s α ij X lj, i=,..., s X li = h j= j= s α ij X lj x l, i=,..., s. If denotes the Kronecker product and e m =,..., T R m then = [ Xl ] h A I m e m x l Xl =e m x l +ha I m X ls X l X ls X l X ls j= X ls Now consider the linear DAE 4. with continuous matrix functions At LR n, R m, Dt LR m, R n, Bt LR m, R m. and a properly stated leading term. When applying the numerical scheme 4.3a,4.3b,4.3c we don t want to lose the additional information provided by the properly stated leading term. According to lemma 4.2 we therefore replace 4.3c by [DX] li = s α ij D lj X lj D l x l 4.3c h j= and solve the system A li [DX] li + B lix li = q li, i =,..., s 4.3b for X li. Here we write D l = Dt l, D li = Dt li, A li = At li and so on. Using this ansatz the output value s s x l =x l +h β i α ij X lj x l = β T A e s s x l + β i α ij X lj h i= j= i= j= is computed. For RadauIIA methods this expression simplifies considerably. Definition 4.3 The s-stage RadauIIA method is uniquely determined by requiring Cs, Ds, c s = and choosing c,..., c s to be the zeros of the Gauss-Legendre polynomial p s. For the conditions Cs, Ds see [3]. The Gauss-Legendre polynomial p s is orthogonal to every polynomial of degree less than s. RadauIIA methods are A- and L-stable and have order p = 2s. The last row of A coincides with β T [3, 5]. 30
Lemma 4.4 For the s-stage RadauIIA method β T A e = 0 holds and the output value computed by 4.3a, 4.3b, 4.3c, 4.3c is given by the last stage X ls. Proof: β T A e = Z s AA e = 0,..., 0, e = 0 and x l = β T A e x l + s s i= j= β i α ij X lj = X l 0,..., 0, I m = X ls. X ls To summarize these results we present the following algorithm for solving the DAE 4. using RadauIIA methods. Algorithm 4.5 Given an approximation x l to the exact solution xt l and a stepsize h, solve A li [DX] li + B lix li = q li, i =,..., s 4.3b for X li where [DX] li is given by [DX] li = h s α ij D lj X lj D l x l. j= 4.3c Return the output value x l = X ls as an approximation to xt l = xt l + h. The exact solution x of 4. satisfies xt M 0 t = {z R m Btz qt im AtDt } t. Since X li M 0 t li for every i and c s = we have x l = X ls M 0 t ls = M 0 t l for every RadauIIA method. Thus the RadauIIA approximation satisfies the algebraic constraint and RadauIIA methods are especially suited for solving DAEs [4, 5]. 4. Decoupling of the discretized equation Algorithm 4.5 replaces the DAE ADx + Bx = q 4. by the discretized problem A li [DX] li + B lix li = q li, i =,..., s. 4.4 As seen in section 3.2, the analytic solution x of index and index 2 equations 4. can be represented as x = KD u Q 0 Q D DQ D u+q 0 P +P 0 Q G 2 q+q 0Q D DQ G 2 q 4.5 3
where K = I Q 0 P G 2 B and the component u = DP x satisfies the inherent regular ordinary differential equation u DP D u + DP G 2 BD u = DP G 2 q. 4.6 If we applied the Runge-Kutta method directly to the inherent regular ODE, due to lemma 4.2 we would obtain s α ij Ulj u l DP D h li U li + DP G 2 BD U li = DP G 2 q 4.7 li j= for i =,..., s. Our aim is to show that the Runge-Kutta method, when applied to 4., behaves as if it was integrating the inherent regular ODE 4.6. We start by repeating the decoupling procedure from section 3.2 for the discretized equation 4.4. Doing so 4.4 is found to be equivalent to the system DP D li [DX] li + DP G 2 BP 0P li X li +DP D li D liq,li X li = DP G 2 q li Q 0 Q D li [DX] li + Q 0 P G 2 BP 0P li X li +Q 0 P D li DP D li D liq,li X li + Q 0,li X li = Q 0 P G 2 q li D li Q,li X li = DQ G 2 q li 4.8 for i =,..., s. The decoupled system 4.8 immediately implies the convergence of RadauIIA methods applied to 4. on compact intervals I if the stepsize h tends to zero [8]. A careful analysis of 4.8 also leads to the main result in this section. Theorem 4.6 Let 4. be an index µ equation, µ {, 2}. Let the subspaces D S and D N be constant. Then the difference between the exact solution and the solution obtained by using a RadauIIA method can be written as xt l x l = K l D l utl u l + Q 0 Q D { DQ G l 2 q l k DQ } α sj G 2 h q lj DQ G 2 q. l j=0 Here u l is exactly the RadauIIA approximation to the solution ut l of the inherent regular ODE 4.6. Note that k DQ h j=0 α sj G 2 q lj DQ G 2 q is exactly the Runge-Kutta l approximation to DQ G 2 q l. The proof of theorem 4.6 will use the following lemma. Lemma 4.7 DP D and DQ D are projector functions satisfying i DS = im DP = im DP D, DN = im DQ = im DQ D. If the subspaces DS and DN are constant, so that there are constant projectors V, W onto DS and DN respectively, then the following relations hold: ii DP D V = V, DP D W = 0, DQ D W = W, DQ D V = 0, iii DP D V =0, DP D W =0, DQ D W =0, DQ D V =0. 32
Proof: DP D and DQ D are projector functions due to lemma 3.2 and 3.6. The same lemmas imply i, so that DP D V = V and DQ D W = W hold as well. These relations together with i show ii. Finally use ii to prove iii by noting that V and W are constant projectors and therefore do not depend on t. Proof of theorem 4.6: The proof will be divided into four parts. In ➀ we analyze DP D li [DX] li and Q 0 Q D li [DX] li, so that we can find a representation of the numerical solution in part ➁. This representation will depend on U ls = D ls P,ls X ls. In ➂ we show that u l = U ls is exactly the RadauIIA solution of the inherent regular ODE. The poof will be completed by comparing the analytic and the numeric solution in part ➃. ➀ Analyze DP D li [Dx] li and Q 0 Q D li [Dx] li Write U li = D li P,li X li and u l = D l P,l x l. Then DP D li [Dx] li = DP D h li Use lemma 4.7 to see that = h DP D li s α ij Dlj X lj D l x l j= s α ij U lj + D lj Q,li X lj u l DQ x. l j= and DP D li U lj u l = DP D li V U lj u l = V U lj u l = U lj u l DP D li D ljq,li X lj DQ x l =DP D li W D ljq,li X lj DQ x l =0. We arrive at DP D li [Dx] li = h Similarly, lemma 4.7 implies Because of DQ D li [Dx] li = h s j= α ij U lj u l. s α ij D lj Q,li X lj DQ x. l j= Q0 Q D li = Q 0 Q P 0 Q D li = Q 0 Q D DQ D li, it follows that Q0 Q D li [DX] li = Q0 Q D h li s α ij D lj Q,li X lj DQ x. l j= 33
➁ Obtain a representation of the numerical solution x l The discretized system 4.8 now reads s h j= α ij Ulj u l + DP G 2 BD li U li +DP D li D liq,li X li = DP G 2 q li h Q0 Q D s li j= α ij Dlj Q,li X lj DQ x l + Q 0 P G 2 BD li U li +Q 0 P D li DP D li D liq,li X li + Q 0,li X li = Q 0 P G 2 q li D li Q,li X li = DQ G 2 q li but due to lemma 4.7 this reduces to s h j= α ij Ulj u l + DP G 2 BD li U li = DP G 2 q li h Q0 Q D s li j= α ij Dlj Q,li X lj DQ x l + Q 0 P G 2 BD li U li + Q 0,li X li = Q 0 P G 2 q li D li Q,li X li = DQ G 2 q li The numerical solution can thus be written as x l = X ls = P 0,ls X ls + Q 0,ls X ls = Dls Dls P,ls X ls + D ls Q,ls X ls + Q0,ls X ls = I Q 0 P G 2 B ls D l U ls + P 0 Q + Q 0 P l G 2 q 4.9 l + h Q0 Q D s DQ α l sj G 2 q lj DQ G 2 q. l j= The stage approximations U lj satisfy the recursion s h j= α ij Ulj u l + DP G 2 BD li U li = DP G 2 q li. 4.0 ➂ 4.0 is the RadauIIA method applied to the inherent regular ODE Again, lemma 4.7 implies DP D li U li = DP D li V U li = 0 in 4.7. This shows that 4.0 and 4.7 coincide. Therefore, and due to c s =, u l = U ls is exactly the Runge-Kutta solution of the inherent regular ODE 4.7. ➃ Compare the analytic and the numeric solution Use Lemma 4.7 to see, that in 4.5 DQ D l ut l = DQ D l V ut l = 0. Now the assertion follows by comparing 4.5 and 4.9. Theorem 4.6 is the central tool in analyzing the behaviour of RadauIIA methods when applied to DAEs 4.. In the case of index µ = theorem 4.6 shows that discretization and the decoupling procedure commute. 34
Corollary 4.8 Let the DAE 4. be of index. Assume that im Dt is constant. Then we have for any RadauIIA method xt l x l = K l D l utl u l, K = I Q0 G B. Proof: If the index is, we have Q = 0 and P = I. Thus N = {0} and S = R n. Since im Dt is constant, the subspaces DS and DN are constant as well. We can therefore apply theorem 4.6. Due to corollary 4.8 the following diagram commutes for index equations with constant im D. ADx + Bx = q 4. decoupling x = KD u + Q 0 G q u + DG BD u = DG RadauIIA discretization q discretization RadauIIA h A li [DX] li + B lix li = q li 4.4 decoupling x l = K l D l u l + Q 0l G l q l k j=0 α iju lj u l +D l G l B ld l u l =D l G l q l If the index is 2, we cannot expect the corresponding diagram to commute. However, the term h s j= DQ α sj G 2 q lj DQ G 2 q = [DQ l G 2 q] t l appearing in theorem 4.6 is exactly the RadauIIA approximation to DQ G lemma 4.2 so that 2 q l xt l x l = K l D l utl u l + Q0l Q l D l { DQ G 2 q l [DQ G 2 q] t l }. In this sense we have the following statement: When applying a RadauIIA method to problems of index µ {, 2} with constant subspaces DS and DN, then discretization and decoupling commute. Definition 4.9 The DAE 4. of index µ {, 2} is said to be numerically qualified, if µ = and im D is constant, µ = 2 and DS, DN are constant. 35
The commutativity of discretization and the decoupling process is the desired property for DAEs since it guarantees a good behaviour of the numerical method. Even though the numerical method is applied to the DAE directly, it behaves as if it was integrating the regular inherent ODE 4.6. In this case results concerning convergence on compact intervals I hold automatically. The RadauIIA method applied to a numerically qualified DAEs is convergent with the same order as for ODEs. Results obtained for ODEs concerning the reflexion of qualitative behaviour by the numerical solution can be transferred directly to DAEs using theorem 4.6. More information about stability preserving integration of index and 2 DAEs can be found in [7, 8]. However, the representation 4.9 shows that the Runge-Kutta scheme is weakly unstable when applied to index 2 DAEs. This is due to the inherent differentiation and becomes apparent for small stepsizes h. We focused on the application of RadauIIA methods. This restriction is not necessary. All results presented here can be proved in a similar way for BDF methods. The application of general linear methods to DAEs is currently being studied. 4.2 A numerical example Consider the index 2 example due to Gear and Petzold [2]. 0 0 x η t e t t + xt =, η R constant 4. η t 0 + η 0 { } x t + η t x 2 t = e t x t + η t x 2t + + η x 2 t = 0 In [2] it is shown that the BDF method fails completely for η = and is exponentially unstable for all other parameter values < η < 0.5. In [4] 4. is said to pose difficulties to every numerical method. Numerical results are given in figure 4.. 4. was solved on the interval [0, 3] using the implicit Euler method, the BDF 2 -formula and the RadauIIA method with two stages. The step-size used was h = 0.5. The exact solution is given by x t = η te t and x 2 t = e t, so that x 0 =, T is a consistent initial value. All numerical methods used fail even for moderate values of η due to the exponential instability. Consider the following reformulation 0 η t η t xt + 0 xt = e t 0. 4.2 4.2 now has a properly stated leading term and DN = R, DS = {0} show that the reformulated problem is numerically qualified. We therefore know discretization and the decoupling process commute. This means that solving the reformulated problem yields the correct numerical results as figure 4.2 shows. 36
0.8 exact solution Implicit Euler Method BDF2 RadauIIA Method 0.8 exact solution Implicit Euler Method BDF2 RadauIIA Method 0.6 0.6 0.4 0.4 0.2 0.2 0 0 0.5.5 2 2.5 3 0.8 0.6 η = 0.20 exact solution Implicit Euler Method BDF2 RadauIIA Method 0 0 0.5.5 2 2.5 3 0.8 0.6 η = 0.26 exact solution Implicit Euler Method BDF2 RadauIIA Method 0.4 0.4 0.2 0.2 0 0 0.5.5 2 2.5 3 η = 0.28 0 0 0.5.5 2 2.5 3 η = 0.52 Figure 4.: Numerical solutions 2. component of 4.. 0.8 exact solution Implicit Euler Method BDF2 RadauIIA Method 0.8 exact solution Implicit Euler Method BDF2 RadauIIA Method 0.6 0.6 0.4 0.4 0.2 0.2 0 0 0.5.5 2 2.5 3 0.8 0.6 η = 0.20 exact solution Implicit Euler Method BDF2 RadauIIA Method 0 0 0.5.5 2 2.5 3 0.8 0.6 η = 0.26 exact solution Implicit Euler Method BDF2 RadauIIA Method 0.4 0.4 0.2 0.2 0 0 0.5.5 2 2.5 3 η = 0.28 0 0 0.5.5 2 2.5 3 η = 0.52 Figure 4.2: Numerical solutions 2. component of 4.2. 37
Acknowledgements First of all I thank Prof John Butcher for inviting me to the University of Auckland. It was an honour for me to be part of his group. I learned a lot about mathematics but not only about mathematics. This report would not have been possible without the support of Prof Roswitha März. She and her group at the Humboldt Universität zu Berlin taught me all I know about differential-algebraic equations. I look forward to continuing my work with them. I greatly appreciate the help of Nicolette Moir and John Rugis who read this report and made very useful comments. The whole Numerical Analysis group at the University of Auckland was a very kind audience and I thank them for listening to me when I was presenting this material. In terms of financial assistance I d like to express my gratitude to the Studienstiftung des deutschen Volkes for supporting my stay here in Auckland. Finally I thank all my friends at the Carlton House and in Grafton Hall for turning my stay here into a wonderful time... but finally I thank you, Sabine, for letting me go for nine months but being always close to me at the same time. 38
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