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Introduction to Differential Algebraic Equations

Dr. Abebe Geletu

Ilmenau University of TechnologyDepartment of Simulation and Optimal Processes (SOP)

Winter Semester 2011/12


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4.1 Definition and Properties of DAEs

A system of equations that is of the form

F (t, x, x) = 0

is called a differential algebraic equation (DAE) if the Jacobianmatrix ∂F

∂x is singular (non-invertible); where, for each t, x(t) ∈ Rn

and

F (t, x(t), x(t)) =

F1(t, x(t), x(t))F2(t, x(t), x(t))

...Fn(t, x(t), x(t))

.


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4.1 Definition and Properties of DAEs ...Example: The system

x1 − x1 + 1 = 0 (1)

x1x2 + 2 = 0 (2)

is a DAE. To see this, determine the Jacobian ∂F∂x of

F (t, x, x) =

(x1 − x1 + 1x1x2 + 2

)with x =

(x1x2

), so that

∂F

∂x=

(∂F1∂x1

∂F1∂x2

∂F2∂x1

∂F2∂x2

)=

(−1 0x2 0

), ( see that, det

(∂F

∂x

)= 0).

⇒ the Jacobian is a singular matrix irrespective of the values of x2.Observe that: in this example the derivative x2 does not appear.


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4.1 Definition and Properties of DAEs ...

Solving for x1 from the first equation x1 − x1 + 1 = 0 we getx1 = x1 + 1. Replace this for x1 in the second equation x1x2 + 2 = 0to wire the DAE in equations (23) & (23) equivalently as:

x1 = x1 + 1 (3)

(x1 + 1)x2 + 2 = 0 (4)

In this DAE:• equation (3) is a differential equation; while• equation (4) is an algebraic equation.⇒ There are several engineering applications that have such modelequations.


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4.1 Definition and Properties of DAEs ...Suppose F (t, x, x) = A(t)x+B(t)x+ d(t).Hence, for the system F (t, x, x) = 0 the Jacobian will be ∂F

∂x = A(t).I If A(t) is a non-singular (an invertible) matrix, then

[A(t)]−1 (A(t)x(t) +B(t)x(t) + d(t)) = [A(t)]−1 0

⇒ x(t) + [A(t)]−1B(t)x(t) + [A(t)]−1 d(t)) = 0

⇒ x(t) = − [A(t)]−1B(t)x(t)− [A(t)]−1 d(t)).

This is an ordinary differential equation.

Remark

In general, if the Jacobian matrix ∂F∂x is non-singular (invertible),

then the system F (t, x, x) = 0 can be transformed into an ordinarydifferential equation (ODE) of the form x = f(t, x). Some numericalsolution methods for ODE models have been already discussed.

I Therefore, the most interesting case is when ∂F∂x is singular.


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4.2. Some DAE models from engineeringapplications

• There are several engineering applications that lead DAE modelequations.Examples: process engineering, mechanical engineering andmechatronics ( multibody Systems eg. robot dynamics, car dynamics,etc), electrical engineering (eg. electrical network systems, etc), waterdistribution network systems, thermodynamic systems, etc.

Frequently, DAEs arise from practical applications as:I differential equations describing the dynamics of the process, plusI algebraic equations describing:• laws of conservation of energy, mass, charge, current, etc.• mass, molar, entropy balance equations, etc.• desired constraints on the dynamics of the process.


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4.2. Some DAE models from engineeringapplications ... CSTRAn isothermal CSTR

A B → C.

Model equation:

V = Fa − F (5)

CA =Fa

V(CA0 − CA)−R1 (6)

CB = −Fa

VCB +R1 −R2 (7)

CC = −Fa

VCC +R2 (8)

0 = CA −CB

Keq(9)

0 = R2 − k2CB (10)


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4.2. Some DAE models from practical applications...a CSTR...• Fa-feed flow rate of A• CA0-feed concentration of A• R1, R2 - rates of reactions• F - product withdrawal rate• CA, CB, CC - concentration of species A, B and C, resp., in themixture.Definining

x = (V,CA, CB, CC)>

z = (R1, R2)>

the CSTR model equation can be written in the form

x = f(x, z)

0 = g(x, z).


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Some DAE models from practical applications...asimple pendulum

Newton’s Law:mx = −F

l xmy = mgF

l yConservation of mechanicalenergy: x2 + y2 = l2

(DAE) x1 = x3

x2 = x4

x3 = − F

m lx1

x4 = gF

lx2

0 = x2 + y2 − l2.Introduction to Differential Algebraic Equations

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Some DAE models from practical applications...anRLC circuit

Kirchhoff’s voltage and current lawsyield:

conservation of current:

iE = iR, iR = iC , iC = iL

conservation of energy:

VR + VL + VC + VE = 0

Ohm’s Laws:

CVC = iC , LVL = iL, VR = RiR


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Some DAE models from practical applications...anRLC circuit...

After replacing iR with iE and iC with iL we get a reducedDAE:

VC =1

CiL (11)

VL =1

LiL (12)

0 = VR +RiE (13)

0 = VE + VR + VC + VL (14)

0 = iL − iE (15)

(16)

Define x(t) = (VC , VL, VR, iL, iE)


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Some DAE models from practical applications...anRLC circuit...The RLC system can be written as:

x =

1C 0 0 0 00 1

L 0 0 00 0 0 0 00 0 0 0 00 0 0 0 0

x (17)

0 =

0 0 1 0 R1 1 1 0 00 0 0 1 −1

x+

010

VE (18)

which is of the form

x = Ax (19)

0 = Bx+Dz. (20)


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4.3. Classification of DAEs• Frequently, DAEs posses mathematical structure that are specific toa given application area.• As a result we have non-linear DAEs, linear DAEs, etc.• In fact, a knowledge on the mathematical structure of a DAEfacilitates the selection of model-specific algorithms and appropriatesoftware.Nonlinear DAEs:In the DAE F (t, x, x) = 0 if the function F is nonlinear w.r.t. any oneof t, x or x, then it is said to be a nonlinear DAE.

Linear DAEs: A DAE of the form

A(t)x(t) +B(t)x(t) = c(t),

where A(t) and B(t) are n× n matrices, is linear. If A(t) ≡ A andB(t) ≡ B, then we have time-invariant linear DAE.


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4.3. Classification of DAEs...Semi-explicit DAEs:A DAE given in the form

x = f(t, x, z) (21)

0 = g(t, x, z) (22)

• Note that the derivative of the variable z doesn’t appear in the DAE.• Such a variable z is called an algebraic variable; while x is called adifferential variable.• The equation 0 = g(t, x, z) called algebraic equation or aconstraint.

Examples:• The DAE model given for the RLC circuit, the CSTR and the simplependulum are all semi-explicit form.


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4.3. Classification of DAEs...A fully-implicit DAEs:The DAE

F (t, x, x) = 0

is in fully-implicit form.

Examples:(i) F (t, x, x) = Ax+Bx+ b(t) is a fully-implicit DAE.(ii) The equation (see equations (23) & (23) )

x1 − x1 + 1 = 0

x1x2 + 2 = 0

is a fully-implicit DAE.

I Any fully-implicit DAE can be always transformed into asemi-explicit DAEs .


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4.3. Classification of DAEs...Example (transformation of a fully-implicit DAE into a semi-explicitDAE):• Consider the linear time-invariant DAE Ax+Bx+ b(t) = 0 , whereλA+B is nonsingular, for some scalar λ. Then there are non-singularn× n matrices G and H such that:

GAH =

[Im OO N

]and GBH =

[J OO In−m

](23)

where Im the m×m identity matrix (here m ≤ n), N is an(n−m)× (n−m) nilpotent matrix; i.e., there is a positive integerp such that Np = 0, J ∈ Rm×m and In−m is the (n−m)× (n−m)identity matrix.I Now, we can write Ax+Bx+ b(t) = 0 equivalently as

(GAH) (H−1)x+ (GBH) (H−1)x+Gb(t) = 0. (24)


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4.3. Classification of DAEs...• Use the block decomposing given in equation (23) to write[

Im OO N

](H−1)x+

[J OO In−m

](H−1)x+Gb(t) = 0. (25)

• Use the variable transformation w(t) = H−1x(t) to write[Im OO N

]w +

[J OO In−m

]w +Gb(t) = 0. (26)

• Decompose the vector w(t) as w(t) =

[w1(t)w2(t)

], with w1(t) ∈ Rm

and w2(t) ∈ Rn−m and correspondingly the vector Gb(t) =

[b1(t)b2(t)

]so

that:

w1(t) + Jw1(t) + b1(t) = 0 (27)

Nw1(t) + w2(t) + b2(t) = 0. (28)


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4.3. Classification of DAEs...

• Use now the nilpotent property of the matrix N ; i.e., multiply thesecond set of equation by Np−1 to get

w1(t) + Jw1(t) + b1(t) = 0 (29)

Npw1(t) +Np−1w2(t) +Np−1b2(t) = 0. (30)

From this it follows that (since Npw1(t) = 0)

w1(t) = −Jw1(t)− b1(t) (31)

0 = −Np−1w2(t)−Np−1b2(t). (32)

Therefore, we have transformed the fully-implicit DAEAx+Bx+ b(t) = 0 into a semi-explicit form.• Similarly, using mathematician manipulations, any nonlinearfully-implicit DAE can be transformed into a semi-explicit DAE.


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4.4. Index of a DAE• DAEs are usually very complex and hard to be solved analytically.⇒ DAEs are commonly solved by using numerical methods.

Question:Is it possible to use numerical methods of ODEs for the solution ofDAEs?

Idea:

Attempt to transform the DAE into an ODE.

• This can be chieved through repeated derivations of the algebraicequations g(t, x, z) = 0 with respect to time t.

Definition

The minimum number of differentiation steps required to transform aDAE into an ODE is known as the (differential) index of the DAE.


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4.4. Index of a DAE...Example:

(DAE) x1 = x1 + 1

(x1 + 1)x2 + 2 = 0.

Here x2 the algebraic variable (i.e. z = x2). Differentiateg(x1, x2) = 0 to find a description for the time-derivative x2 of thealgebraic variable. So

d

dt[g(x1, x2)] =

d

dt[0]⇒ d

dt[(x1 + 1)x2 + 2] = 0

⇒ x1x2 + (x1 + 1)x2 = 0.

⇒ x2 = − x1x2(x1 + 1)

= −(x1 + 1)x2(x1 + 1)

= −x2

I Only one differentiation step is required to describe x2. So, theDAE is of index 1.


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4.4. Index of a DAE...• The CSTR model is of index 2.• The DAE model for the simple penudulum is of index 3.

I DAEs with index greater than 1 are commonly known as higherindex DAEs.I The higher the index, the more difficult will be the DAE to solve.I Transformation of a higher index DAE into a lower index DAE (orto an ODE) is commonly known as index reduction. In general, indexreduction for higher index DAE simplifies computational complexities.

Two serious issues to consider when solving DAEs

• The solutions of the lower index DAE may not be a solution of theoriginal DAE. This is known as a drift off effect.• Finding initial conditions that satisfy both the differential andalgebraic parts of a DAE may not be trivial, known as consistency ofinitial conditions.


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4.4. Index of a DAE...Therefore, computational algorithms for a DAE should:

be cable of identifying consistent initial conditions to the DAE;as well as,provide automatic index reduction mechanisms to simplify theDAE.

I Modern software like Sundials, Modelica, use strategies forindex-reduction coupled with methods of consistent initialization.In the following we consider only index 1 semi-explicit DAEs:

(DAE) x = f(t, x, z) (33)

0 = g(t, x, z). (34)

Since ∂g∂t + ∂g

∂xdxdt + ∂g

∂zdzdt = 0. Hence, a semi-explicit DAE is of

index 1 iff[∂g∂z

]−1exists. That is, one differentiation step yields the

differential equation: dzdt = −

[∂g∂z

]−1 [∂g∂x

](f(t, x, z))−

[∂g∂z

]−1∂g∂t .


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4.5. An overview of numerical Methods for DAEs• Using mathematical principles and transformation of variables,fully-implicit DAEs can be transformed to semi-explicit DAEs.• Note that, in the semi-explicit DAE

x = f(t, x, z) (35)

0 = g(t, x, z). (36)

if Jacobian matrix[∂g∂z

]singular (non-invertible), then the DAE is of

higher index.I Since many applications have model equations as semi-explicitDAEs, the discussion next is restricted to this form.

• In the DAE above, if both f and g do not explicitly depend on timet; i.e. f(t, x, z) = f(x, z) and g(t, x, z) = g(x, z), then the model isan autonomous DAE.


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4.5. An overview of numerical Methods for DAEs...I BDF and collocation methods are two most widely used methodsfor numerical solution of DAEs.

(I) BDF for DAEs:Consider the initial value DAE

x = f(t, x, z), x(t0) = x0 (37)

0 = g(t, x, z). (38)

Ideas of BDF:I Select a time step h so that ti+1 = ti + h, i = 0, 1, 2, . . .I Given xi = x(ti) and zi = z(ti), determine the valuexi+1 = x(ti+1) by using (extrapolating) values xi, xi−1, . . . , xi−m+1

of the current and earlier time instants of x(t).I Simultaneously compute zi+1 = z(ti+1).


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4.5. An overview of numerical Methods forDAEs...BDF• There is a unique m-th degree polynomial P that interpolates them+ 1 points

(ti+1, xi+1), (ti, xi), (ti−1, xi−1), . . . , (ti+1−m, xi+1−m).

• This interpolating polynomials P can be written as

P (t) =

m∑j=0

xi+1−jLj(t)

with the Lagrange polynomial

Lj(t) =

m∏l = 0l 6= j

[t− ti+1−l

ti+1−j − ti+1−l

], j = 0, 1, . . . ,m.


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4.5. ....BDF for DAEs...• Observe that

P (ti+1−j) = xi+1−j , j = 0, 1, . . . ,m.

In particular P (ti+1) = xi+1.• Thus, replace xi+1 by P (ti+1) to obtain

P (ti+1) = f(ti+1, xi+1). (∗)

But

P (ti+1) =

m∑j=0

xi+1−jLj(ti+1) = xi+1L0(ti+1) +

m∑j=1

xi+1−jLj(t)

Putting this into (*) we get:

xi+1 = −m∑j=1

xi+1−jLj(ti+1)

L0(ti+1)+

1

L0(ti+1)f(ti+1, xi+1)


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4.5. ....BDF for DAEs...Define

aj =Lj(ti+1)

L0(ti+1), bm =

1

hL0(ti+1).

• The values aj , j = 1, . . . ,m and bm can be read from lookup tables.

An m-step BDF Algorithm (BDF m) for a DAE:

Given a1, . . . , am,bm

xi+1 = −m∑j=1

ajxi+1−j + bmhf(ti+1, xi+1, zi+1) (39)

0 = g(ti+1, xi+1, zi+1). (40)

• Each iteration of the BDF requires a Newton algorithm for thesolution a system of nonlinear equations. Hence, the Jacobian ∂g

∂wneeds to be well-conditioned, where w = (t, x, z).


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4.5. ....BDF for DAEs...• Given x0 = x(t0), BDF requires consistent initial conditions to beobtained by solving

g(t0, x0, z0) = 0.

to determine z0 = z(t0).

The m-step BDF algorithm converges if m ≤ 6; i.e.

xi − x(ti) ≤ O(hm), zi − z(ti) ≤ O(hm)

for consistent initial conditions.

• BDF is commonly used to solve DAEs of index 1 or 2

Software based on BDF for DAEs:� Matlab: ode15i

� Open source: DASSL ; CVODE, CVODES, and IDA (underSundials); ODEPACK solvers, etc.


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Orthogonal Collocation• To collocate a function x(t) through another function p(t) = tocaptur the properties of x(t) by using p(t).• In general, we use a simpler function p(t) to collocate x(t); eg., p(t)can be a polynomial, a trigonometric function, etc.⇒ polynomial and trigonometric functions are usually simpler to workwith. (The discussion here is restricted to polynomilas)

Weirstraß’ Theorem

If x(t) is a continuous function on [a, b], then for any given ε > 0,there is a polynomial p(t) such that

maxa≤t≤b

|x(t)− p(t)| < ε

• Hence, we use the polynomial p(t) instead of x(t).• However, Weirstraß’ theorem doese not specifye how to constructthe approximating polynomial p(t) .


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• Suppose p(t) = a0 + a1t+ . . .+ amtm is the approximating

polynomial to x(t) on [a, b].Note: If the coefficients a0, a1, . . . , am are given, then theapproximating polynomial is exactly known.Question: How to determine a0, a1, . . . , am?Question: How to construct the apprximating polynomial?• There are several ways to construct p(t) to approximates x(t)according to Weirstraß’ theorem.Here we require p(t) to satisfy the following property:• p(ti) = x(ti) =: xi, i = 1, . . . , N.for some selected time instants t1, t2, . . . , tN from the interval [a, b].• This property relates p(t) and x(t) and is known as interpolatoryproperty.

Uniqueness of an interpolating polynamial

There is a unique interpolatory polynomial p(t) for the N data points(t1, x1),(t2, x2), . . .,(tN , xN ) with degree deg(p) = m = N − 1.


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• In the following we use N = m+ 1.Figure


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According to the interpolatory property, we have

x1 = p(t1) = a0 + a1t1 + . . .+ amtm1 (41)

x2 = p(t2) = a0 + a1t2 + . . .+ amtm2 (42)

... (43)

xm+1 = p(tm+1) = a0 + a1tm+1 + . . .+ amtmm+1. (44)

⇒

x1x2...

xm+1

=

1 t1 t21 . . . tm11 t2 t22 . . . tm2...

...... . . .

...1 tm+1 t2m+1 . . . tmm+1

a0a1...am

Thus, if we known x1, x2, . . . , xm+1 then we can computea0, a1, . . . , am and vice-versa.• But, since x(t) is not yet known, x1, x2, . . . , xm+1 are unknown.


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Hence, both

x1x2...

xm+1

and

a0a1...am

are unkowns.

• To avoid working with two unknown vectors, we can define thepolynamial p(t) in a better way as:

p(t) =

m+1∑i=1

xiLi(t) (45)

where Li(t) are Lagrange polynomials given by

Li(t) =

m+1∏j = 1j 6= i

t− tjti − tj

=(t− t1) (t− t2) . . . (t− ti−1) (t− ti+1) . . . (t− tm+1)

(ti − t1) (ti − t2) . . . (ti − ti−1) (ti − ti+1) . . . (ti − tm+1).


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Properties:

Li(tj) =

1, if i = j0, if i 6= j

? (satisfaction of the interpolatory property)

p(ti) = xi, i = 1, 2, . . . ,m+ 1.

? (approximation of x(t) by p(t))The polynomial p(t) in equation (45) can be made to satisfy theWeirstraß’ theorem by taking sufficiently large number of timeinstants t1, t2, . . . , tm+1 from [a, b].

Question: What is the best choice for t1, t2, . . . , tm+1?


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Determination of collocation Points

An idea to determine collocation points:

Select values τ1, τ2, . . . , τN from the interval [0, 1] and define the ti’sas follows:

ti = a+ τi(b− a), i = 1, . . . , N.

• This is a good idea, since the same values τ1, τ2, . . . , τN from [0, 1]can be used to generate collocation points on various intervals [a, b].Example:(i) If [a, b] = [2, 10], then ti = 2 + τi (10− 2) , i = 1, . . . , N ;(ii) If [a, b] = [0, 50], then ti = 0 + τi (50− 0) , i = 1, . . . , N ; etc.

Question: What is the best way to select the valuesτ1, τ2, . . . , τN from [0, 1]?


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Determination of collocation Points...

I Recall that: solution of a differential equation (or DAEs) is anintegration process.⇒ Numerical methods for one-dimensional integrals can give usinformation on how to select τ1, τ2, . . . , τN .⇒ Gauss quadrature rules are one of the best numerical integrationmethods.To approximate the integral

I[f ] =

∫ 1

0f(τ)dτ,

by the (iterpolatory) quadrature rule QN [f ] =∑N

k=1wkf(τk), where:• the integration nodes τ1, τ2, . . . , τN ∈ [0, 1]• weights w1, w2, . . . , wN

are constructed based on the interval [0, 1] .


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Determination of collocation points...Thus we would like to obtain the approximation

I[f ] =

∫ 1

0f(τ)dτ ≈ QN [f ] =

N∑i=1

wif(τi).

• Once a quadrature rule QN [·] is constructed it can be used toapproximate integrals of various functions.

Question: How to determine the 2N unknowns τ1, τ2, . . . , τN ,w1, w2, . . . , wN?

We require the quadrature rule to satisfy the equality:∫ b

aW (τ)p(τ)dτ =

N∑i=1

wip(τi)

for all polynomials p with degree deg(p) ≤ 2N .


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That is, QN [·] should integrate each of the polynomials

p(t) = 1, τ, τ2, . . . , τ2N

exactly. This implies ∫ 1

01dτ =

∑i=1

wi (46)∫ 1

0τdτ =

∑i=1

wiτi (47)∫ 1

0τ2dτ =

∑i=1

wiτ2i (48)

... (49)∫ 1

0τ2Ndτ =

∑i=1

wiτ2Ni . (50)


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We need to solve the system of 2N nonlinear equations:

1 = w1 + w2 + . . .+ wN (51)1

2= w1τ1 + w2τ2 + . . .+ wNτN (52)

(53)1

3= w1τ

21 + w2τ

22 + . . .+ wNτ

2N (54)

... (55)1

2N= w1τ

2N1 + w2τ

2N2 + . . .+ wNτ

2NN . (56)

From the equation 1 = w1 + w2 + . . .+ wN we can solve for w1 andreplace for it in the remaining equations.⇒ It remains to determine the 2N − 1 unknowns τ2, . . . , τN andw1, w2, . . . , wN from the reduced set of 2N − 1 equations.• Unfortunately, this system of equations is difficult to solve.


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Orthogonal polynomials

•There is a simpler way if we use orthogonal polynomials.I The concept of orthogonality requires a definition of scalarproduct.• The scalar product of functions h and g with respect on theinterval [0, 1] is

〈h, g〉 =

∫ 1

0h(τ)g(τ)dτ.

• Two functions h and g are orthogonal on the interval [0, 1] if

〈h, g〉 =

∫ 1

0h(τ)g(τ)dτ = 0.

• In this lecture, we are only interested in set of polynomials thatare orthogonal to each other.


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Orthogonal polynomials...• Orthogonal polynomials on [0, 1] (as defined above) are known asshifted Legendre polynomials.

Theorem (Three-term recurrence relation)

Suppose {p0, p1, . . .} the set of shifted Legendre orthogonalpolynomials on [0, 1] with deg(pn) = n and leading coefficient equalto 1. The shifted Legendre polynomials are generated by the relation

pn+1(τ) = (τ − an)pn(τ)− bnpn−1(τ)

with p0(τ) = 1 and p−1(τ) = 0, where the recurrence coefficients aregiven as

an = 12 , n = 0, 1, 2, . . .

bn = n2

4(4n2−1) , n = 0, 1, 2, . . .


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Orthogonal polynomials...The first four shifted Legendre orthogonal polynomials arep0(τ) = 1, p1(τ) = 2τ − 1, p2(τ) = 6τ2 − 6τ + 1, p3(τ) =20τ3 − 30τ2 + 12τ − 1.

Further important properties of orthogonal polynomials:Given any set of orthogonal polynomials {p0, p1, p2, . . .} the followinghold true:• Any finite set of orthogonal polynomials {p0, p1, . . . , pN−1} islinearly independent.• The polynomial PN is orthogonal to each of p0, p1, . . . , pN−1 .• Any non-zero polynomial q with degree deg(q) ≤ N − 1 can bewritten as a linear combination:

q(τ) = c0p0(τ) + c1p1(τ) + . . . , cN−1pN−1(τ)

where at least one of the scalars c0, c1, . . . , cN−1 is non-zero.Introduction to Differential Algebraic Equations

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Orthogonal polynomials...• We demand the quadrature rule is exact for polynomials up todegree 2N − 1.• Let P (τ) be any polynomial of degree 2N − 1.Hence,

I[P ] = QN [P ]⇒∫ 1

0P (τ)dτ =

N∑i=1

wiP (τi).

• Given the N -th degree orthogonal polynomial pN , the polynomial Pcan be written as

P (τ) = pN (τ)q(τ) + r(τ)

where q(τ) and r(τ) are polynomials such that 0 < deg(r) ≤ N − 1and• deg(q) = N − 1, since

2N − 1 = deg(P ) = deg(pN ) + deg(q) = N + deg(q).


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Orthogonal polynomials...• Now, from the equation

∫ 10 P (τ)dτ =

∑Nk=1wkP (τk) it follows that∫ b

a(pN (τ)q(τ) + r(τ)) dτ =

N∑i=1

wi (pN (τi)q(τk) + r(τi)) .

⇒∫ b

apN (τ)q(τ)dτ︸︷︷︸

=0

+��

��∫ b

ar(τ)dτ =

N∑i=1

wipN (τi)q(τi) +

��

��

��N∑i=1

wir(τi).

Observe the following:• Using polynomial exactness:

∫ ba r(τ)dτ =

∑Ni=1wi r(τi).

• Orthogonality implies < pN , pk >= 0, k = 1, . . . , N − 1. Thus,∫ ba pN (τ)q(τ) =∫ b

apN (τ)

N−1∑i=1

ckpk(τ)dτ =N−1∑k=1

ck

∫ b

apN (τ)pk(τ)dτ︸︷︷︸

=0

= 0.


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Orthogonal polynomials...It follows that

∑Nk=1wkpN (τk)q(τk) = 0. (?)

Theorem

If the quadrature nodes τ1, τ2, . . . , τN are zeros of the N−th degreeshifted Legendre polynomial pN (τ), then• all the roots τ1, τ2, . . . , τN lie inside (0, 1);• the quadrature weights are determined from

wi =

∫ 1

0Li(τ)dτ,

where Li(τ), i = 1, . . . , N is the Lagrange function defined usingτ1, τ2, . . . , τN and wi > 0, i = 1, . . . , N .• The quadrature rule QN [·] integrates polynomials degree up to2N − 1 exactly.

Hence, equation (?) holds true if p1(τ1) = pN (τ2) = . . . , p(τN ) = 0.Introduction to Differential Algebraic Equations

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Orthogonal polynomials...

• Therefore, the quadrature nodes τ1, τ2, . . . , τN are chosen as thezeros of the N -th degree shifted Legendre orthogonal polynomial.

Question:

Is there a simple way to compute the zeros τ1, τ2, . . . , τN of anorthogonal polynomial pN and the quadrature weightsw1, w2, . . . , wN?

The answer is give by a Theorem of Welsch & Glub (see next slide).


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Orthogonal polynomials...

Theorem (Welsch & Glub 1969 )

The quadrature nodes τ1, τ2, . . . , τN and weights w1, w2, . . . , wN canbe computed from the spectral factorization of

JN = V >ΛV ; Λ = diag(λ1, λ2, . . . , λN ), V V > = IN ;

of the symmetric tri-diagonal Jacobi matrix

JN =

a0√b1√

b1 a1√b2

√b2

. . .. . .

. . . aN−2√

bN−1√bN−1 aN−1

where

a0, an, bn, n = 1, . . . , N − 1 are the known coefficients from therecurrence relation. In particular ...


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Theorem ...

τk = λk, k = 1, . . . , N ; (57)

wk =(e>V ek

)2, k = 1, . . . , N, (58)

where e1, ek are the 1st and the k−the unit vectors of length N .

A Matlab program:

n = 5; format short

beta = .5./sqrt(1-(2*(1:n-1)).^(-2)); % recurr. coeffs

J = diag(beta,1) + diag(beta,-1) % Jacobi matrix

[V,D] = eig(J); % Spectral decomp.

tau = diag(D); [tau,i] = sort(tau); % Quad. nodes

w = 2*V(1,i).^2; % Quad. weights


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• There are standardised lookup tables for the integration nodesτ1, τ2, . . . , τN and corresponding weights w1, w2, . . . , wN .

The collocation points t1, t2, . . . , tN ∈ [a, b] can be determined byusing the quadrature points τ1, τ2, . . . , τN ∈ [0, 1] through therelation:

ti = a+ τi(b− a).

Example: Suppose we would like to collocate x(t) on [a, b] = [0, 10]using a polynomial x(t) with deg(x) = m = 3.• Required number of collocation points = N = m+ 1 = 3 + 1 = 4.⇒ First, determine the zeros of the 4-th degree shifted Legendrepolynomial p4(τ); i.e., find τ1, τ2, τ3, τ4 .• These value can be determined to be:

τ1 = 0.0694318442, τ2 = 0.3300094783,τ3 = 0.6699905218, τ4 = 0.9305681558.


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• Next, determine the collocation points :t1 = 0 + (10− 0) ∗ τ1 = 0.694318442, t2 = 0 + (10− 0) ∗ τ2 =

3.300094783,t3 = 0 + (10− 0) ∗ τ3 = 6.699905218,t4 = 0 + (10− 0) ∗ τ4 = 9.305681558.


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The collocation polynomial is p(t) =∑4

i xiLi(t). where

L1(t) =(t− t2)(t− t3)(t− t4)

(t1 − t2)(t1 − t3)(t1 − t4)(59)

L2(t) =(t− t1)(t− t3)(t− t4)

(t2 − t1)(t2 − t3)(t2 − t4)(60)

L3(t) =(t− t1)(t− t2)(t− t4)

(t3 − t1)(t3 − t2)(t3 − t4)(61)

L4(t) =(t− t1)(t− t2)(t− t3)

(t4 − t1)(t4 − t2)(t4 − t3). (62)

Hence, we approximate x(t) using the third degree polynomial

p(t) = x(t) = x1L1(t) + x2L2(t) + x3L3(t) + x4L4(t)

It remains to determine the coefficients x1, x2, x3, x4.


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Discretization of DAEs using orthogonal collocationsGiven the DAE:

x = f(t, x, z), t0 ≤ t ≤ tf (63)

0 = g(t, x, z) (64)

where x(t)> = (x1(t), x2(t), . . . , xn(t)) andz(t)> = (z1(t), z2(t), . . . , zm(t)).• Determine the a set of collocation points t1, t2, . . ., tN .• corresponding to each differential and algebraic variable definecollocation polynomials:

xk(t) −→ p(k)(t) =

N∑i=1

x(k)i Li(t), k = 1, . . . , n; (65)

zj(t) −→ p(j)(t) =

N∑i=1

z(j)i Li(t), j = 1, . . . ,m. (66)


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... orthogonal collocation of DAEs• In the DAE, replace xk(t) and zj(t) by q(k)(t) and p(j)(t) so that

p(1)(t) = f1(t,(p(1)(t), p(2)(t), . . . , p(n)(t)

),(q(1)(t), q(2)(t), . . . , q(m)(t)

))(67)

p(2)(t) = f2(t,(p(1)(t), p(2)(t), . . . , p(n)(t)

),(q(1)(t), q(2)(t), . . . , q(m)(t)

))(68)

... (69)

p(n)(t) = f1(t,(p(1)(t), p(2)(t), . . . , p(n)(t)

),(q(1)(t), q(2)(t), . . . , q(m)(t)

))(70)

0 = g1(t,(p(1)(t), p(2)(t), . . . , p(n)(t)

),(q(1)(t), q(2)(t), . . . , q(m)(t)

))(71)

0 = g2(t,(p(1)(t), p(2)(t), . . . , p(n)(t)

),(q(1)(t), q(2)(t), . . . , q(m)(t)

))(72)

... (73)

0 = gm(t,(p(1)(t), p(2)(t), . . . , p(n)(t)

),(q(1)(t), q(2)(t), . . . , q(m)(t)

))(74)


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... orthogonal collocation of DAEs• Next discretize the system above using t1, t2, . . . , tN to obtain:

p(1)(ti) = f1

(ti,(x(1)i , x

(2)i , . . . , x

(n)i

),(z(1)i , z

(2)i , . . . , z

(m)i

))p(2)(ti) = f2

(ti,(x(1)i , x

(2)i , . . . , x

(n)i

),(z(1)i , z

(2)i , . . . , z

(m)i

))...

p(n)(ti) = fn

(ti,(x(1)i , x

(2)i , . . . , x

(n)i

),(z(1)i , z

(2)i , . . . , z

(m)i

))0 = g1

(ti,(x(1)i , x

(2)i , . . . , x

(n)i

),(z(1)i , z

(2)i , . . . , z

(m)i

))0 = g2

(ti,(x(1)i , x

(2)i , . . . , x

(n)i

),(z(1)i , z

(2)i , . . . , z

(m)i

))...

0 = gm

(ti,(x(1)i , x

(2)i , . . . , x

(n)i

),(z(1)i , z

(2)i , . . . , z

(m)i

)),

i = 1, 2, . . . , N.Introduction to Differential Algebraic Equations

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... orthogonal collocation of DAEs• Solve the system of (n+m)×N equations to determine the

(n+m)×N unknwons x(k)i , k = 1, . . . , n, i = 1, . . . , N and

z(j)i , j = 1, . . . ,m, i = 1, . . . , N .

Example: Use a four point orthogonal collocation to solve the DAE

x1 = x1 + 1, 0 ≤ t ≤ 1.

0 = (x1 + 1)x2 + 2.

Solution: we use the collocation polynomials

p(1)(t) =

4∑i=1

x(1)i (t)Li(t) and p(2)(t) =

4∑i=1

x(2)i (t)Li(t)

to collocate x1(t) and x2(t), respectively.Introduction to Differential Algebraic Equations

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Advantages and disadvantages collocation methods

• can be used for higher index DAEs.• efficient for both initial value as well as boundary values DAEs.• more accurate

Disadvantages:• Can be computationally expensive• The approximating polynomials may display oscillatory properties,impacting accuracy• computational expenses become high if t1, t2, . . . , tN−1 areconsidered variables.


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Resources - LiteratureReferences:• U. M. Ascher, L. R. Petzold : Computer Methods for OrdinaryDifferential Equations and Differential-Algebraic Equations. SIAM1998.• K. E. Brenan, S. L. Campbell, and L.R. Petzold: (1996). NumericalSolution of Initial-Value Problems in Differential-Algebraic Equations.,SIAM, 1996.• Campbell, S. and Griepentrog, E., Solvability of General DifferentialAlgebraic Equations, SIAM J. Sci. Comput. 16(2) pp. 257-270, 1995.• M. S. Erik and G. Sderlind: Index Reduction inDifferential-Algebraic Equations Using Dummy Derivatives. SIAMJournal on Scientific Computing, vol. 14, pp. 677-692, 1993.• C. W. Gear : Differential-algebraic index transformation. SIAM J.Sci. Stat. Comp. Vol. 9, pp. 39 – 47, 1988.• Walter Gautschi: Orthogonal polynomials (in Matlab). Journal ofComputational and Applied Mathematics 178 (2005) 215– 234• Lloyd N. Trefethen: Is Gauss Quadrature Better thanClenshaw.Curtis? SIAM REVIEW, Vol. 50, No. 1, pp. 67 – 87, 2008.


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Resources - Literature .. Software• Pantelides Costas. (1988) ”The consistent initialization ofdifferential algebraic systems.” SIAM Journal on Scientific andStatistical Computing, vol. 9: 2, pp. 213–231, 1988.• L. Petzold: Differential/Algebraic Equations Are Not ODEs, SIAMJ. Sci. Stat. Comput. 3(3) pp. 367-384, 1982.Open source software:• DASSL - - FORTRAN 77:http://www.cs.ucsb.edu/ cse/software.html• DASSL Matlab interface:http://www.mathworks.com/matlabcentral/fileexchange/16917-dassl-mex-file-compilation-to-matlab-5-3-and-6-5• Sundials - https://computation.llnl.gov/casc/sundials/main.html• Modelica - https://modelica.org/• Odepack - http://www.cs.berkeley.edu/ kdatta/classes/lsode.html• Scicos- http://www.scicos.org/


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