STATE TRANSFORMATION FOR EULER-LAGRANGE SYSTEMS

M. Mabrouk

Inria-Lorraine, projet CONGE, ISGMP Bt. A, Ile du Saulcy, 57 045 Metz Cedex 01, France.

J. C. Vivalda

Inria-Lorraine, projet CONGE, ISGMP Bt. A, Ile du Saulcy, 57 045 Metz Cedex 01, France.

Keywords:

Euler-Lagrange systems, state transformation, afﬁne forms.

Abstract:

The transformation of an Euler-Lagrange system into a state afﬁne system in order to solve some interesting

problem as the design of observer, the output tracking control, is considered in this paper. A necessary and a

sufﬁcient condition is given as well as a method to compute this transformation.

1 INTRODUCTION

Euler- Lagrange systems with n generalized conﬁgu-

ration coordinates q = (q

, ..., q

)

⊤

are described by

equations of the form

˙q = v,

M(q) ˙v + C(q, v)v + V (q) = τ,

(1)

where M(q) denotes the inertia matrix, while

C(q, v)v, with v = ˙q = ( ˙q

, ..., ˙q

)

⊤

the gener-

alized velocities, denotes the centrifugal and Corio-

lis forces, V (q) consists of the gravity terms and τ

is the vector of input torques. This celebrated fam-

ily of systems has been the subject of an important

literature over half a century, because the equations

of many physical devices belong to this family: (M.

Spong and Vidyasagar, 1989). When these systems

are fully-actuated, they are globally feedback lineariz-

able. But feedback linearization can be performed

only when all the variables are measured. Unfortu-

nately in practice, very often the variables of veloc-

ity cannot be measured. Therefore, the global output

feedback stabilization of these systems with y = q as

output is challenging from a practical point of view.

But, from a theoretical point of view, it is one of the

most difﬁcult problems in the ﬁeld of nonlinear con-

trol: Indeed, the matrix C(q, v)v is a nonafﬁne func-

tion of the unmeasured part of the state v: this fact

precludes from applying most of the classical tech-

niques. For instance, the methods of (L. Praly and

Z.P. Jiang, 1993), (J.P. Gauthier and I. Kupka, 1994)

and (J-B. Pomet, R. M. Hirschorn, W. A. Cebuhar,

1993). For more explanations on the obstacles which

are due to the presence of terms nonafﬁne with respect

to the unmeasured variables, see the introduction of

(F. Mazenc and J.C. Vivalda, 2002).

Most recently, in (Besancon, G. 2000) an elegant

alternative for one-degree-of -freedom systems was

reported. The author presented a reduced order ob-

server converging exponentially . This observer is

based upon a global nonlinear change of coordinates

which makes the system afﬁne in the unmeasured part

of the state. This is crucial to deﬁne a very simple

controller to solve the problem of tracking trajectory.

So a very natural question arises: which conditions

ensure that an Euler -Lagrange systems (1) can be

transformed, with the help of a change of coordinates,

into some structure afﬁne in the unmeasured part of

the state.

This question has been addressed in (Besancon, G.

2000) and (Loria, A and Pantely, E, 1999). However

the questions of existence and computation of the re-

quired solution were not answered. In the present

paper, we address these question: we show that this

problem can be brought back to the resolution of a set

of partial differential equation for which an explicit

solution is given.

The paper is organized as follows. The main result

is stated and proved in Section 3. Section 4 is devoted

to example. Section 5 contains concluding remarks.

Preliminary

Euler-lagrange systems are such that:

• The matrix M(q) is symmetric positive deﬁnite for

all q.

Mabrouk M. and C. Vivalda J. (2005).

STATE TRANSFORMATION FOR EULER-LAGRANGE SYSTEMS.

In Proceedings of the Second International Conference on Informatics in Control, Automation and Robotics - Robotics and Automation, pages 43-48

DOI: 10.5220/0001169200430048

 SciTePress

• The equality

M(q) = C

⊤

(q, v) + C(q, v). holds.

Also, throughout the paper,

• M

(R) denotes the set of n-square real matrices.

• GL

(R) denotes the set of n-square real invertible

matrices.

• For S ∈ M

(R) symmetric positive deﬁnite S

denotes the square root of S.

2 PROBLEM STATEMENT

We consider the family of Euler-Lagrange systems

described by the equations (1) where the output is

q = (q

... q

)

⊤

∈ R

, and the input is τ =

(τ

... τ

)

⊤

∈ R

. The unmeasured part of the state

is v = ( ˙q

... ˙q

)

⊤

As it is pointed in the introduction the difﬁculty

to stabilize or to construct observers for system (1)

mainly stems from the fact that Coriolis and centrifu-

gal forces vector in (1), has a quadratic growth in the

generalized velocities v, which are not measured. The

global change of coordinates introduced in (Besan-

con, G. 2000) for one-degree-of freedom (i.e. n = 1)

systems overcomes this problem by rewriting the dy-

namics with functions which are linear in the unmea-

sured velocities. As it is discussed in (Besancon, G.

2000), the design procedure might be extended to the

case of systems with more degrees of freedom, as

soon as the same kind of change of coordinates can

be found, that is to say if we can select an invertible

matrix T(q) such that

d T(q)

= T(q)M

−1

(q)C(q, v) (2)

Condition (2) is necessary but not sufﬁcient. A nec-

essary and sufﬁcient condition is the existence of a

nonsingular matrix T such that

d T(q)

v = T(q)M

−1

(q)C(q, v)v (3)

Remak 2.1. Since matrix M (q) is positive deﬁnite,

there exists an uniformly bounded matrix ∆(q) such

that M(q) = ∆

⊤

(q)∆(q). It can be easily checked

that in the case one degree of freedom, ∆(q) is a so-

lution of (3) and (2). But in the general case, this

decomposition of M(q) does not provide a solution of

(3) or (2).

Looking for some more general factorization of

M(q), as investigated in (Loria, A and Pantely, E,

1999) for robot manipulators, may lead to solutions.

Remak 2.2. One can remark that if (2) admits a solu-

tion, then (3) admits also (the same) solution. But the

converse is false as shown in the following example.

Consider the following inertia matrix M(q)

M(q) =



−q

0 1



Using the Christoffel symbols of the ﬁrst kind (M.

Spong and Vidyasagar, 1989), matrix C is given by

C(q, v) =

−q



−v



and an easy calculation shows that the matrix

T(q) =

−q

(4)

satisﬁes equation (3), but not (2). In fact, equation (2)

does not admit any solution (as we will see later).

3 MAIN RESULT

In this section we present the main contribution of the

paper, namely the solution of problem (2) or (3). We

begin with the study of the equation (2). We provide

necessary and sufﬁcient conditions for the existence

of a change of coordinates as well as method to com-

pute this solution (Corollary 3.3).

Theorem 3.1. Consider the nonlinear system (1).

Equation (2) admits a solution if and only if

∂C

∂q

−

∂C

∂q

= C

⊤

−1

− C

⊤

−1

, (5)

for all 1 ≤ i, j ≤ n. Where the matrices C

are such

C(q, v) =

i=n

i=1

(q)v

(6)

To establish Theorem 3.1, we need to prove the fol-

lowing preliminary lemma, and its corollary.

Lemma 3.2. (Isidori, 1989) Let x

, ..., x

denote co-

ordinates of a point x in R

and y

, ..., y

coordi-

nates of a point y in R

. Let M

, ..., M

be smooth

functions

: R

−→ R

n×n

(7)

such that

∂M

∂x

−

∂M

∂x

+ M

− M

= 0. (8)

Consider the set of partial differential equations

∂y(x)

∂x

= M

(x)y(x), 1 ≤ i ≤ m. (9)

Given a point (x

, y

) ∈ R

× R

, there exist a

neighborhood U of x

and a unique smooth function

y(x) which satisﬁes (9) and is such that y(x

) = y

ICINCO 2005 - ROBOTICS AND AUTOMATION

Corollary 3.3. Let M

(q), . . . , M

(q) be smooth

functions

Consider the set of partial differential equations

∂T

∂q

(q) = T(q)M

(q), ∀i = 1, . . . n. (10)

Given any matrix T

∈ GL

(R), q

∈ R

, there

exists a unique smooth matrix T(q) which satisﬁes

(10) and is such that T(q

) = T

if and only if the

functions M

(q), . . . , M

(q) satisfy the conditions

∀i < j ≤ n; M

− M

∂M

∂q

−

∂M

∂q

. (11)

Proof. Necessity:

The necessity follows from the assumption that a

solution T(q) exists. Then from the property

∂

∂q

∂

∂q

(12)

one has

∂(T M

)

∂q

∂(T M

)

∂q

(13)

Expanding the derivatives on both sides we obtain

T (M

∂M

∂q

) = T (M

∂M

∂q

)

(14)

which, due to the fact that T(q) is invertible, yields

the condition (11).

Sufﬁciency:

• Using Lemma 3.2.

Let T

∈ GL

(R) and note T

−1

= (Γ

, ...., Γ

being the columns of T

−1

Conditions (11) ensure the existence of a family of

functions Γ

such that for all k we have

∂Γ

∂q

= −M

, Γ

) = Γ

(15)

The matrix Γ = (Γ

, ..., Γ

) satisﬁes

∂Γ

∂x

= −M

Γ, Γ(q

) = T

−1

. (16)

Since Γ(q

) = T

−1

which is non singular, we con-

clude that there exists neighborhood U of q

such

that Γ is non singular and as a solution of (10), we

take the matrix T(q ) = Γ

−1

(q).

The previous proof gives a condition of existence,

but not a method allowing construction of the so-

lution. However, the control implementation needs

the knowledge of a matrix T(q).

In the sequel we will give another proof of the

corollary 3.3, based on a reasoning by induction

which provides an explicit solution of (2). More-

over this solution is deﬁned on the whole domain

of deﬁnition of the matrices M

• By induction.

By induction on n we show that if (11) holds, then we

have the following property P(n).

P(n): n ≥ 1, there exists an invertible matrix T(q)

such that equations (10) holds.

1. For n = 1: Equation (10) becomes

∂T(q

)

∂q

= T

′

) = T(q

)M(q

) (17)

which admits solutions for all matrix M

∈

(R).

2. For n = 2 : Let M

, q

), M

, q

) ∈ M

.(R)

Denote by Φ

) the solution of the non au-

tonomous differential equation

dΦ

) = Φ

(18)

Now, let us construct a particular solution T(q ) of

the system (10) in the form T(q) = Ψ(q

)Φ

Since T(q) is a solution of (10), one can prove that

∂T

∂q

dΨ(q

)

+ Ψ(q

)

∂Φ

∂q

(19)

It follows that

dΨ(q

)

= Ψ(q

)(Φ

−1

−

∂Φ

∂q

−1

)

Let

K(q) = Φ

−1

−

∂Φ

∂q

−1

(20)

Then Ψ(q

) exists if we have

∂K(q)

∂q

= 0.

Now

∂K(q)

∂q

= Φ

−1

+ Φ

∂M

∂q

−1

−Φ

−1

−

∂

∂q

−1

∂Φ

∂q

= Φ

∂M

− M

−

∂M

)Φ

−1

= 0

(21)

So P(2) is true.

STATE TRANSFORMATION FOR EULER-LAGRANGE SYSTEMS

3. Assume that P(n) is true, where n1.

Now using the induction hypothesis, we must prove

that P(n + 1) is true ?

Let M

, . . . , M

n+1

∈ M

(R) such that

− M

∂M

∂q

−

∂M

∂q

(22)

The induction hypothesis implies that there exists

an invertible matrix T

n+1

, q

, . . . , q

) such that

∂T

n+1

∂q

= T

n+1

, ∀i = 1, . . . , n.

Let us show that there are solutions of the form T =

n+1

Observe that

∂T

∂q

= Ψ

n+1

)

∂T

n+1

∂q

= Ψ

n+1

= TM

and

∂T

∂q

n+1

dΨ

n+1

+ Ψ

∂T

n+1

∂q

n+1

= TM

n+1

= Ψ

n+1

So we get

dΨ

n+1

= Ψ

n+1

−

∂T

n+1

∂q

n+1

−1

n+1

(23)

) exists if the term (T

n+1

−

∂T

n+1

∂q

n+1

−1

n+1

is independent of q

, q

, . . . , q

Easy calculations show that for i = 1, 2, . . . , n; we

have

∂[T

n+1

−1

n+1

−

∂T

n+1

∂q

n+1

−1

n+1

]

∂q

= 0

(24)

This concludes the proof.

Proof of Theorem 3.1

Returning to the problem of the existence of the

solution of (2), note that we have

T =

i=n

i=1

∂T

∂q

Moreover the matrix C(q, v) is linear with respect

to v

(M. Spong and Vidyasagar, 1989). This yields

for all q

C(q, v) =

i=n

i=1

(q)v

(25)

where matrices C

are such that, C

+ C

⊤

∂M

∂q

Therefore one can deduce easily that T satisﬁes

∂T

∂q

= TM

−1

. (26)

According to Corollary 3.3, we deduce that a solu-

tion of (2) exists if and only if

∂T

∂q

= TM

, ∀i = 1

, . . . n where M

= M

−1

Now,

∂M

∂q

−

∂M

∂q

= −M

−1

+ C

⊤

−1

∂C

∂q

− M

−1

∂C

∂q

−1

+ C

⊤

−1

= M

− M

−1

(

∂C

∂q

−

∂C

∂q

−C

⊤

−1

+ C

⊤

−1

It follows that a necessary and sufﬁcient condition

for the existence of T(q) such that (2) is given by:

∂C

∂q

−

∂C

∂q

= C

⊤

−1

− C

⊤

−1

Which allows as to conclude. 

The preceding theorem gives an algebraic char-

acterization of a family of Euler-Lagrange systems

which can be transformed, with the help of a change

of coordinates into some structure, afﬁne in the un-

measured part of the state v = ˙q.

The following gives another characterization of

such a solution of the differential equation (2),

Theorem 3.4. Consider an Euler-Lagrange system

(1). The following conditions are equivalent .

1. There exists a matrix T(q) such that ( 2) holds.

2. There exists a matrix N (q) such that M(q) =

⊤

(q)N(q) and N

⊤

(q)

N(q) = C(q, v).

3. There exists a function Θ(q) : R

→ R

and N(q)

nonsingular such that M (q) = N

⊤

(q)N(q) and

Jacobian(Θ) = N (q).

Proof. Due to the space limitation we only give the

proof of the two ﬁrst properties.

Suppose that (2) admits a solution.

Calculating

, where

M = T

⊤−1

−1

ICINCO 2005 - ROBOTICS AND AUTOMATION

We have

= −T

⊤−1

′

−1

⊤

⊤−1

−T

⊤−1

−1

(TM

−1

C)T

−1

⊤−1

−1

= −T

⊤−1

⊤

+ C −

−1

= 0

(27)

since C

⊤

+ C =

So,

M = T

⊤−1

−1

= S, where S is a symmet-

ric positive deﬁnite matrix .

Let N = S

T, then we obtain

M(q) = N

⊤

(q)N(q) (28)

and

N(q)

N(q) = C(q, v). (29)

For sufﬁciency assume that there exists N (q) such

that M(q) = N

⊤

(q)N(q) and N

⊤

(q)

N(q) =

C(q, v).

So we get, N

⊤

(q)

∂N

∂q

(q) = C

(q).

One can check easily that conditions (31) are satis-

ﬁed, which concludes the proof.

Now, we present our ﬁnal result on reduction of

Euler-Lagrange systems.

Let us consider the problem of ﬁnding T such that

(3) is satisﬁed.

Observe ﬁrst that the matrix M

−1

C(q, v)v is

quadratic in v with coefﬁcients depending only on q.

i.e there exists R

such that

−1

(q)C(q, v)v =

i=1

v (30)

then one can easily show the following

Theorem 3.5. Consider an Euler-Lagrange system

(1). A necessary and sufﬁcient condition for (3) to

admit a solution is that there exist R

such that

∀i < j ≤ n; R

− R

∂R

∂q

−

∂R

∂q

. (31)

Proof. The proof follows from similar reasoning as

in Theorem 3.1 and Corollary 3.3 gives an explicit

solution of the problem.

Remak 3.6. One can remark that when n = 1, a

solution always exists (See (Besancon, G. 2000)).

In the case of higher order system, the problem (2)

can admit no solution. Indeed, taking again the ex-

ample of the Remark 2.2, a solution exists if and only

if e

−q

= 0 which is impossible.

But if we take



0 0

−q



, R



−1

−q

0 0



(32)

one can check easily that conditions (30) and (31) are

satisﬁed. So (3) admits a solution.

Observe also that there is no function Θ(q) such

that Jacobian(Θ) = T(q).

4 EXAMPLE

As an example, consider the inverted pendulum. The

dynamics obtained by Euler-Lagrange formulation

are:

(M + m)¨x + ml

θ cos θ − ml

sin θ = τ

ml¨x cos θ + ml

θ − mlg sin θ = 0,

where (M, x) are mass and position of the cart which

is moving horizontally, (m, l; θ) are mass, length

and angular derivation from the upward vertical po-

sition for the pendulum which is pivoting around a

point ﬁxed on the cart. We denote the state vec-

tor (θ x

θ ˙x)

⊤

as (q

)

⊤

. The output is

y = (q

, q

)

⊤

The inertia matrix is:

M(q) =



cos q



with a

= M + m, a

= ml and a

= ml

Using Christoffel symbols, we obtain

=(0)



0 −a

sin(q

)

0 0



One can check easily that condition (11) are veriﬁed

so according to Theorem 3.1, equation (2) admits a

solution.

Let us consider Φ

) =



1 0

0 1



as a solution

of the differential equation

dΦ

) = Φ

= 0. (33)

Let

Ψ(q

) =



) Ψ

)

) Ψ

)



(34)

STATE TRANSFORMATION FOR EULER-LAGRANGE SYSTEMS

the solution of the differential equation

dΨ(q

)

= Ψ(q

)(Φ

−1

−

∂Φ

∂q

−1

)

= Ψ(q

−1

(35)

Using Mathematica 4.1, one can check easily that

a solution of Equation (35) with initial condition

Ψ(0) = I



1 0

0 1



Ψ(q

) =

β(0) cos(q

)−a

β(q

)

β(0)

β(q

)

β(0)

(36)

where β(q

) =

− a

cos(q

)

Thanks to the Corollary 3.3, a solution T(q

) of

Equation (2) with T(0) = T

= I

, is

T(q

) = Ψ(q

)Φ

)

β(0) cos(q

)−a

β(q

)

β(0)

β(q

)

β(0)

(37)

Therefore, the following change of coordinates

= q

β(0) cos(s) − a

β(s)

β(0)

ds,

β(s)

β(0)

ds, p = T (q) ˙q,

transforms the dynamics of the Cart-Pendulum into a

double integrator

Θ = p, (38)

˙p = T(q )M

−1

(q)τ = u. (39)

Clearly This system is linear in the unmeasured

part of the state and a high gain observer can be con-

structed (M. Mabrouk, F. Mazenc and J.C. Vivalda,

2004).

5 CONCLUSION

A necessary and a sufﬁcient condition for determin-

ing a state change of coordinate which transform an

Euler-Lagrange system into an afﬁne system in the

unmeasured part of state was given. Obviously in the

case of one degree of freedom , a solution always ex-

ists. A case of higher order system, is for instance,

that of the cart-pendulum system (F. Mazenc and J.C.

Vivalda, 2002), the Tora system (Z. P. Jiang and I.

Kanellakopoulos, 2000) and the overhead crane (B. d

Andra-Novel and J. Lvine, 1990). We conjecture the

result several others problems in nonlinear control.

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