A tutorial on the Frobenious Theorem, one of the most important results in differential geometry, with emphasis in its use in nonlinear control theory. All results are accompanied by proofs, but for a more thorough and detailed presentation refer to the book of A. Isidori.

1. THE FROBENIUS THEOREM
SOPASSAKIS PANTELIS
Abstract. This text aims to be a self-contained report on the Frobenius
theorem. First, some necessary definitions are given and basic facts about distributions are stated. The dual objects of distributions - the codistributions are introduced. We try to describe these notions and engage them to involutiveness and complete integrability, to state and prove the Frobenius Theorem
which is of great importance in nonlinear control theory.
Ferdinand Georg Frobenious (1849-1917) was a German mathematician well known for his contributions to differential equations,
linear algebra, group theory and differential geometry. He was
the first man to prove the Cayley-Hamilton theorem. He studied
mathematics at the University of Berlin where he completed his
doctorinary thesis on the solution of differential equations under
Weierstrass.After its completion in 1870, he taught in Berlin for a
few years before receiving an appointment at the Polytechnicum in
Zurich (now ETH Zurich). In 1893 he returned to Berlin, where
he was elected to the Prussian Academy of Sciences. Group theory
was one of Frobenius’ principal interests in the second half of his
career. One of his first notable contributions was the proof of the
Sylow theorems for abstract groups.
Despite being named for Ferdinand Georg Frobenius, the theorem we discuss within the following pages was first proven by
Alfred Clebsch and F. Deahna. Deahna was the first to establish
the sufficient conditions for the theorem, and Clebsch developed
the necessary conditions. Frobenius is responsible for applying the
theorem to Pfaffian systems, thus paving the way for its usage in
differential topology.
1. Introduction
A Distribution is a central notion in control theory. Hereinafter when using the
term smooth we refer to functions of class C ∞ . We give the definition straight
away:
Definition 1. A distribution is a mapping D : n → P ( n ) that assigns a subspace
of n to every x ∈ n . That is D : n → P ( n ) , x → D (x) ⊂ n (Here P
stands for the powerset) according to D(x) = span {f1 (x), f2 (x), . . . , fd (x)}, where
f1 , f2 , . . . , fd are not necessarily considered to be linearly independent.
1991 Mathematics Subject Classification. Interdisciplinary Program of Post Graduate Studies in Applied Mathematical Sciences - School of Applied Mathematical and Physical Sciences,
National Technical University, Athens, Greece.
Key words and phrases. Distributions, Involutive, Frobenious Theorem.
1

2. 2
SOPASSAKIS PANTELIS
♦
Considering of the distribution as a new mathematical object, we define the binary operations of addition and intersection of two distributions as:
(1.1)
(D1 + D2 )(x) = D1 (x) + D2 (x)
(1.2)
(D1 ∩ D2 )(x) = D1 (x) ∩ D2 (x)
Apparently D1 (x) + D2 (x) and D1 (x) ∩ D2 (x) are once again distributions. Let
D0 = span 0. Then D0 is the unitary element for addition and intersection.Let
D( n ) be the space of all distributions D with domain the eucleidian space n .
Then (D, +) and (D, ∩) are rendered semigroups with unity the distribution D0 .
Indeed for every D ∈ D, one has that D + D0 = span {f1 , . . . , fd } + span {0} = D.
We say that D1 ⊆ D2 , whenever for every x ∈ n we have D1 (x) ⊆ D2 (x). The
dimension of a distribution for a fixed x ∈ n is the dimension of the subspace
D(x), that is dim D(x) = dim span {f1 (x), f2 (x), . . . , fd (x)}. We now move on to
some examples:
Example 1. Let D ∈ D( n ) and D(x) = span {f1 (x), . . . , fd (x)} = span {F (x)},
where F (x) ∈ Mn×d ( ). Thus D = Im F (x). Let now:


x1
0
0

0
(1.3)
F : 3 → M3×3 ( ), x → F (x) =  0 x1 + x2
0
0
x1 + x2 + x3
It is easy to verify that rank(F (x)) = 0 if and only if x = 0, rank(F (x)) = 3 if and
only if x1 = 0,x2 = −x1 and x1 + x2 + x3 = 0. The dimension of D in this example
is a function of x.
•
Definition 2. A distribution D ∈ D( n ) is said to be of constant degree k if for
every x ∈ n , dim(D(x)) = k. Such a distribution is sometimes called nonsingular.
♦
Example 2. Consider of the distribution D(x) = Im F (x) ∈ D( 4 ), with:
 2

x4 + 1
x2 + 2
x2 + 1 x1
2
4
 0
x2 + x1 + 2
1
2
1

(1.4)
F (x) = 
x1 + x2
0
−x4 − 1 0 
3
1
1
1
1
Then dim(D(x)) = 3 for every x ∈
constant degree 3.
4
and thus D is said to be a distribution of
•
Definition 3. A distribution D is said to be smooth if there exist smooth vectors
f1 , . . . , fd : n → such that span {f1 , . . . , fd } = D
♦

3. NONLINEAR CONTROL THEORY
3
A smooth distribution D can be considered as vector space over the field of
smooth functions. That is, every vector field f ∈ D = span {f1 , . . . , fd } is expressed
as a (finite) linear combination:
d
f (x) =
(1.5)
aj (x)fj (x)
j=1
where f, fj , a(x) ∈ C ∞ (U).
Proposition 1. Let D ∈ D(U) be a nonsingular smooth distribution of constant
degree d and let Y : U → n be a smooth vector field with Y (x) ∈ D(U) for all
x ∈ U. Then there exist k real smooth functions m1 , m2 , . . . , mk : S → , such
that:
(1.6)
Y (x) = m1 (x)X1 (x) + m2 (x)X2 (x) + . . . + mk (x)Xk (x)
for x ∈ U and where X1 , X2 , . . . , Xk are vector fields of D(U).
Proof.
We suppose that there exist X1 , X2 , . . . , Xk and m1 , m2 , . . . , mk : S → such that
Y (x) = m1 (x)X1 (x) + m2 (x)X2 (x) + . . . + mk (x)Xk (x) holds for all x ∈ U and we
show that m1 , . . . , mk are smooth functions. We have:
Y (x) = m1 (x)X1 (x) + m2 (x)X2 (x) + . . . + mk (x)Xk (x)





Xk,1 (x)
X1,1 (x)
Y1 (x)
 Xk,2 (x) 
 X1,2 (x) 
 Y1 (x) 












.
.
.
.
.
.






.
.
.






 Yk (x)  = m1 (x) ·  X1,k (x)  + . . . + mk (x) ·  Xk,k (x) 
⇔





Xk,k+1 (x)
X1,k+1 (x)
Yk + 1(x)












.
.
.
.
.
.






.
.
.

Yn (x)
X1,n (x)
Xk,n (x)
Now, introducing the notation:






X1,1 (x)
Xk,1 (x)
Y1 (x)
X1,2 (x)
Xk,2 (x)



 . 

y =  .  , ξ1 (x) = 
 , . . . , ξk (x) = 

.
.
.
.
.




.
.
Yk (x)
X1,k (x)
Xk,k (x)
We suppose without loss of generality that for every x the corresponding ξi = ξi (x),
for i = 1, 2, . . . , k are linearly independent. Furthermore we have that y = m1 ξ1 +
. . . + mk ξk = [ξ1 ξ2 . . . ξk ] · [m1 m2 . . . mk ]T . The matrix Ξ = [ξ1 ξ2 . . . ξk ] is
k × k and invertible; therefore we have:
 
m1
 m2 
 
(1.7)
 .  = Ξ−1 · y(x)
 . 
.
smooth smooth
mk
Consequently, m1 , . . . , mk are (linearly independent) smooth functions.

4. 4
SOPASSAKIS PANTELIS
Proposition 2. Let D1 , D2 be two smooth distributions, then:
(1) D1 + D2 is a smooth distribution
(2) D1 ∩ D2 is not necessarily smooth.
We skip the proof of 1 as it is beyond the purpose of this essay. For 2, we give
a counter example. So let:
(1.8)
D1 = span
1
1
and
(1.9)
1 + x1
1
D2 = span
D1 and D2 are smooth distributions and their intersection is given by:
(1.10)
(1.11)
(D1 ∩ D2 )(x) = span
1
1
= D1 , for x1 = 0
(D1 ∩ D2 )(x) = {0} , for x1 = 0
Definition 4. Let D ∈ D(U) be a distribution. We call x0 ∈ U a regular point of
D if there exists an open neighborhood U0 of x0 such that D is nonsingular in U0
♦
The following propositions states that every distribution, either if it is nonsingular of not, has ”‘many”’ regular points.
Proposition 3. The set RD = {x0 ∈ U : x0 is a regular point of D} is an open
dense subset of U
C
In fact the (Lebesgue) measure of nonregular points is 0. µL (RD ) = 0
Proposition 4. Let D1 and D2 be two smooth distributions in U, D2 is nonsingular
and D1 ⊂ D2 in a dense subset of U. Then D1 ⊂ D2 in U.
Proposition 5. Let D1 and D2 be two smooth distributions in U, D1 is nonsingular
in U and D1 ⊂ D2 in U and D1 = D2 in a dense subset of U. Then D1 = D2 in U
The following result is very important as it postulates a sufficient condition for
the intersection of two distributions to be locally smooth.
Proposition 6. Let x0 be a regular point of D1 , D2 and D1 ∩ D2 of D(U). Then
there exists a neighborhood U0 of x0 such as the restriction of D1 ∩ D2 in U0 is a
smooth distribution.

5. NONLINEAR CONTROL THEORY
5
A common regular point for D1 , D2 and D1 ∩ D2 does exist because the set RD
of regular points is dense in U. The proof of this proposition is realized by showing
that D1 ∩ D2 is spanned by smooth vector fields, but this is easy to show since x0
is a regular point of all D1 , D2 and D1 ∩ D2 .
Before proceeding to the definition of the notion of involutive distributions, we
define the Lie bracket of two vector fields. In general a Lie bracket is a binary
operation which satisfies certain properties.
Definition 5. Let L be a vector field over the field F . A Lie algebra over F is an
F-vector space with a bilinear map, the Lie bracket:
[·, ·] : L × L → L, (x, y) → [x, y]
(1.12)
satisfying the following properties:
(1.13)
[x, x] = 0, for all x ∈ L
(1.14)
[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0, for all x, y, z ∈ L
The second property is known as the Jacobi identity. The Lie Bracket is sometimes
referred to as commutator.
♦
When studying smooth (in the sense of at least C 1 ) vector fields, we define the
Lie bracket to be the mapping:
(1.15)
C 1(
n
;
m
) × C 1(
n
;
m
) → C 1(
n
;
m
), (x(t), y(t)) → [x, y](t) =
∂x
∂y
x(t) −
y(t)
∂t
∂t
We define the derived algebra of a Lie algebra L.
Definition 6. Let (L, [·, ·]) be a Lie algebra. Then the set:
(1.16)
L1 = [L, L] = {z : ∃x, y ∈ L : [x, y] = z}
with the restriction of the Lie bracket of L, on L1 is called the derived algebra or L.
♦
The following important proposition guarantees that the restriction of the Lie
bracket of L, on L1 has a meaning since L1 is a subset of L.
Proposition 7. L1 is an ideal of L, hence L1 ⊆ L
To a fixed x0 ∈ U, we assign a vector space D(x0 ) ∈ D(U) which becomes a Lie
algebra with the aforementioned Lie bracket. By means of the the above proposition (7), one has that [D(x0 ), D(x0 )] ⊆ D(x0 ). However this does not imply that
for every x0 , y0 ∈ U there exists a z0 ∈ U such that [D(x0 ), D(y0 )] ⊆ D(z0 ). If
a distribution has this property we say that is involutive. Let us define [D, D] =
{[D(x), D(y)], x, y ∈ U}. Then we may give a formal definition for involutive distributions.
Definition 7. We say that a distribution D is involutive if [D, D] ⊆ D, that is for
every f, g ∈ D it follows that [f, g] ∈ D. An involutive distribution is a []-invariant
distribution.
♦

6. 6
SOPASSAKIS PANTELIS
Let D = span {f1 , . . . , fd } be a nonsingular distribution. Then the smooth functions f, g can be expressed as a finite linear combination of {f1 , . . . , fd }. So we
have:
d
f (x) =
(1.17)
aj (x)fj (x)
j=1
d
g(x) =
(1.18)
bj (x)fj (x)
j=1
Thus:
d
d
[f, g](x) = [
(1.19)
aj (x)fj (x),
j=1
d
d
(1.20)
bj (x)fj (x)]
j=1
ai (x)bj (x)[fi (x), fj (x)]
=
i=1 j=1
Hence for a distribution to be involutive, it is sufficient (and necessary) that
[fi , fj ] ∈ D, for 1 ≤ i, j ≤ d
(1.21)
Note that for i = j, from the definition of the Lie bracket, [fi , fj ] = 0, and from
the skew-commutativity property for i = j we have [fi , fj ] = −[fj , fi ]. If D =
span {f1 , f2 } then it follows that for every f, g ∈ D, we have:
d
(1.22)
det
[f, g] =
i=1,j=1,i=j
ai
bi
aj
[fi , fj ]
bj
For example, if d = 3, then:
[f, g] = det
a1
b1
a2
a
[f1 , f2 ] + det 2
b2
b2
a3
a
[f2 , f3 ] + det 1
b3
b1
a3
[f1 , f3 ]
b3
Employing a well known result from linear algebra, we state the following proposition:
Proposition 8. A nonsingular distribution D ∈ D(U) is involutive if and only if
(1.23)
rank {f1 (x), . . . , fd (x)} = rank {f1 (x), . . . , fd (x), [fi , fj ](x)}
for every x ∈ U and 1 ≤ i = j ≤ d
Example 3. Consider of the distribution D = span {f1 , f2 }, defined on a subset U
of 3 , namely U = ∗ × ∗ × and suppose that f1 , f2 are defined as:


x1
f1 (x) = x1 + x2 
(1.24)
1

(1.25)

x2
f1 (x) = x2 + 1
3
1

7. NONLINEAR CONTROL THEORY
7
then the rank of F (x) = [f1 (x) f2 (x)] is 3 for all x ∈ U and the distribution D
is of constant rank k = 2 (nonsingular). Employing the abovementioned criterion
(Proposition 8), we first compute the Lie bracket of f1 with f2 .
(1.26)
df1 (x)
df2 (x)
f1 (x) −
f2 (x)
dx  
 dx

1 0 0
x1
0
= 1 1 0 x1 + x2  − 0
0 0 0
1
0


2
x1 + x3 + 1
= 2x3 (x2 + 1)
3
0
[f1 , f2 ] =
(1.27)
(1.28)
1
0
0


0
x2
2x3  x2 + 1
3
0
1
We now have:

(1.29)
x1
[f1 , f2 , [f1 , f2 ]](x) = x1 + x2
1
x2
x2 + 1
3
1

x2 + 1
3
2x3 (x2 + 1)
3
0
This matrix has constant rank 3 for all x ∈ U, therefore this distribution is not
involutive.
•
It is now clear that is a distribution defined on U with dim(U) = d, is nonsingular with constant rank k = d, is also involutive. In the special case that D is one
dimensional, we have D = span {f1 }. Then [D, D] is the zero-dimensional subspace
of U and by the above criterion (Proposition 8), every one dimensional distribution is involutive. The following proposition, easily follows from the definition of
involutiveness.
Proposition 9. Suppose that D1 , D2 are two involutive distributive, the same holds
for D1 ∩ D2 , however D1 + D2 might not be involutive.
We state a counter-example to prove the second statement. As mentioned before, every one-dimensional distribution is involutive. Furthermore, every twodimensional distribution can be expressed as a sum of two one-dimensional distributions. That is D = span {f1 , f2 } = span {f + 1} + span {f2 } = D1 + D2 and not
every two-dimensional distribution is involutive. However every distribution can
be decomposed in a sum of involutive ones. We now give the definition of covector
fields and codistributions - a couple of crucial and important notions in control
theory.
Definition 8. A covector field in U ⊆ n is a map w : n → ( n )∗ , where ( n )∗
is the dual space of n , i.e. the space of linear continuous functionals φ : n → .
A codistribution is a map Ω : n → P(( n )∗ ), x → Vx such that Vx is a subspace
of ( n )∗ .
♦
Hereinafter we will denote the set of all codistributions on U by D∗ (U). The
notions of smoothness and involutiveness are defined analogously for codistributions. The binary operations of addition and intersection and the binary relation of

8. 8
SOPASSAKIS PANTELIS
inclusion are also defined the same way. Given a distribution D we may correlate
to that a codistribution D⊥ called the annihilator of D and defined to be:
(1.30)
D⊥ (x) = {w∗ ∈ (
n ∗
) | w∗ , u = 0, ∀u ∈ D(x)}
Here ·, · , stands for the eucleidian inner product in n . For every x ∈ U ⊆ n ,
D⊥ (x) is a subspace of ( n )∗ . The annihilator of a distribution is a codistribution and since the second dual of n is isomorphic to n , the annihilator of a
codistribution is again a distribution. In that sense we define the annihilator of a
codistribution to be:
Ω⊥ (x) = {u ∈
(1.31)
n
| w∗ , u = 0, ∀w∗ ∈ Ω(x)}
Remark 1. Let D ∈ D(U) be a smooth distribution and D⊥ ∈ D∗ (U) its annihilator. Then D⊥ is not necessarily smooth. Indeed, if we consider of the following
one-dimensional distribution D ∈ D( n )
D(x) = span {x}
(1.32)
then we have that:
(1.33)
D⊥ (x) =
{0}
( )∗
if x = 0
if x = 0
Observe that D is smooth while D⊥ is not.
The annihilator of a distribution D, is a complementary object since the following
dimension identity holds:
(1.34)
dim(D) + dim(D⊥ ) = dim(U)
Additionaly, some fine properties of annihilators are presented:
(1.35)
⊥
⊥
D1 ⊆ D2 ⇔ D1 ⊇ D2
(1.36)
⊥
⊥
[D1 ∩ D2 ]⊥ = D1 + D2
The above properties can be easily checked. If we consider of a smooth distribution
as a matrix F (with smooth entries), in terms of D is spanned by the columns of
⊥
F , then DF can be realized as a ”‘distribution”’ spanned by some rows w∗ . In that
sense, we can write the annihilator of DF as:
(1.37)
⊥
DF (x) = {w∗ ∈ M1×d ( )|w∗ · F (x) = 0, x ∈ U}
Indeed ( n )∗ is isomorphic to M1×d ( ) therefore every k-subspace of ( n )∗ is in
one-to-one correspondence with a matrix in Mk×d ( ). Vice versa, supposing that
a smooth codistribution ΩG is spanned by the rows of a matrix G(x) (with smooth
entries), the annihilator of ΩG is:
(1.38)
Ω⊥ (x) = {u ∈ Md×1 ( )|G(x) · u = 0x ∈ U}
G
That is
(1.39)
Ω⊥ (x) = ker(G(x))
G
At this point we have obtained a better realization of codistributions. We state the
following important proposition:
Proposition 10. Let x0 ∈ U be a regular point of a smooth distribution D ∈ D(U).
Then x0 is a regular point of D⊥ and there exists a neighborhood S x0 such that
D⊥ |S is a smooth codistribution.

9. NONLINEAR CONTROL THEORY
9
Let D be a smooth distribution defined on U ⊆ n and nonsingular with constant
degree d. In a neighborhood U0 x0 of every x0 ∈ U, by definition there exist d
smooth functions f1 , . . . , fd that span D.
(1.40)
D = span(f1 , . . . , fd )
According to the last proposition, x0 are regular points for D⊥ = Ω and Ω|U0 is a
smooth nonsingular codistribution. Therefore by the dimension theorem dim(Ω) =
n − d in U0 and consequently there are n − d covector fields w1 , . . . , wd that span
Ω.
(1.41)
Ω = span(w1 , . . . , wd )
By definition wj , fi = 0 for 1 ≤ j ≤ n − d and 1 ≤ j ≤ d. Set F (x) =
[f1 (x) f2 (x) . . . fd (x)]. Then wj (x)·F (x) = 0 for every x ∈ U, with rank(F (x)) = d
for all x ∈ U. We seek solutions to wj (x) · F (x) = 0 with 1 ≤ j ≤ n − d subject to
span wj = D. In particular we seek for solutions in the form:
(1.42)
wj =
∂λj
∂x
Then, it is easy to verify that:
∂λj
· F (x) = 0 ⇔ dλj (x), fi (x) = 0
∂x
We want to solve the above system of partial differential equations for λj and find
n − d linearly independent solutions. The Frobenius theorem provides some handy
(sufficient and necessary) conditions for these functions to exist. We have first to
give the following definition:
(1.43)
wj (x) · F (x) = 0 ⇔
Definition 9. A nonsingular d-dimensional distribution D, defined on U ⊆ ( n )
is said to be completely integrable if for every x0 ∈ U there exist a neighborhood
U0 x0 and n − d real smooth functions λ1 , λ2 , . . . , λn−d defined on U0 , such that
span(dλ1 , dλ2 , . . . , dλn−d ) = D⊥
♦
It is now time for us to state and prove the Frobenious Theorem:
Theorem 1. [Frobenius] A nonsingular distribution is completely integrable if and
only if it is involutive.
Before giving the proof of the Frobenius theorem we have to recall some properties of the Lie bracket and directional derivative. The derivative of g along f is
defined as:
∂g T T
∂f
Lf g(x) = f T (x)[
(1.44)
] + g(x)
∂x
∂x
where g is a covector field and f is a vector field. The result of this operation
is again a covector field. An alternative definition is adopted by some authors,
according to which:
(1.45)
(Lf g)(x) = dg(x), f (x) = g(f )|x
where g(f )|x is a convenient symbolic representation of the directional derivative
and should not be confused with the synthesis (g ◦ f )(x) which is something completely different. Employing this notation, we have the following properties:

10. 10
SOPASSAKIS PANTELIS
Lemma 1. Let A, B, C be vector fields of proper smoothness, φ a real valued smooth
function and l1 , l2 ∈ . Then:
(1) [A, B](φ) = A(B(φ)) − B(A(φ))
(2) [A, B] = −[B, A]
(3) [A, l1 B + l2 C] = l1 [A, B] + l2 [A, C]
(4) [A, φ · B] = B(φ)A + φ · [A, B]
(5) A( B, C ) = A(B), C + B, A(C)
(6) A(dφ) = dA(φ)
We now come to the proof of the Frobenius Theorem:
Proof.
We first prove the that complete integrability ⇒ involutiveness. So suppose that
a nonsingular distribution D of constant degree d is completely integrable. By
definition of complete integrability, there are n − d covector fields λ1 , . . . , λn − d
defined on neighborhoods U0 x0 such that span(dλ1 , dλ2 , . . . , dλn−d ) = D⊥ , or
equivalently:
(1.47)
∂λj
fi (x) = 0
∂x
⇔ dλj (x), fi (x) = 0
(1.48)
⇔ Lfi λj (x) = 0
(1.49)
⇔ λj (fi )|x = 0
(1.46)
For x ∈ U0 and i = 1, 2, . . . , d. Now differentiating λj along [fi , fk ] we have:
(1.50)
[fi , fk ](λj )|x = fi (fk (λj ))|x − fk (fi (λj ))|x = 0
Applying this operation for j = 1, 2, . . . , n − d we obtain:

 

dλ1
[fi , fk ](λ1 )|x
 [fi , fk ](λ2 )|x   dλ2 

 

(1.51)

 =  .  [fi , fk ]|x = 0
.
.

  . 
.
.
[fi , fk ](λn−d )|x
dλn−d
For x ∈ U0 . By assumption the covector fields {dλ1 , . . . , dλn−d } span D⊥ . Therefore [fi , fk ]|x is in D for every index i, k = 1, 2, . . . , d and we conclude that D is
invloutive. We now prove that Involutiveness ⇒ Complete Integrability. So we
suppose that D is a nonsingular involutive distribution of constant degree d defined
on some subset U of n . From a proposition stated before (Proposition 1), we know
that in a neighborhood U0 of every point x0 ∈ U there exist d smooth vector fields
defined on U0 such that they span D(x). That is:
(1.52)
D(x) = span(f1 (x), . . . , fd (x)), x ∈ U0
Now let fd+1 , . . . , fn x be a complementary set of vector fields defined on U0 such
that:
(1.53)
span(f1 (x), . . . , fd (x)) ⊕ span(fd+1 (x), . . . , fn (x)) =
n
, x ∈ U0
Let Φf (x) stand for the flow of the vector field f , that is the function Φf (x0 ) assigns
t
t
to every initial point x0 for the differential equation x = f (x), a function φ(t; x0 )
˙

11. NONLINEAR CONTROL THEORY
11
that is a solution to the IVP (initial value problem) x = f (x), with x(0) = x0 =
˙
φ(0; x0 ). So φ(t; x0 ) = Φf (x0 ). By definition we have:
t
∂ f 0
Φ (x ) = (f ◦ Φf )|x0
t
∂t t
f
0
Φ0 (x ) = x0
(1.54)
(1.55)
Formally, we say that Φf is a mapping:
t
(1.56)
Φf : S → C 1 (I), x0 → Φf (x0 ) = φ(t, x0 )
t
t
for x0 ∈ S and t ∈ I, and f is Lipschitz with respect to x in S and the solution of
the related IVP exists for every t ∈ I. Φf (x) is a local diffeomorphism. One can
t
easily verify that Φ has the semigroup property, that is:
Φf |x0 = Φf ◦ Φf |x0
t
t+T
T
(1.57)
To every function f1 , f2 , . . . , fd , fd+1 , . . . , fn we assign a flow Φfi |x for i = 1, 2, . . . , n
t
and for all x ∈ U0 . We define the mapping:
(1.58)
Claim 1. For
F :B (
n
)→
n
1
1
, z = (z1 , . . . , zn ) → Φf1 ◦ · · · Φf1 |x0
z
z
sufficiently small, F has the following properties:
(1) Is defined for all z ∈ B and is a diffeomorphism
(2) For all z ∈ B , the first d columns of the matrix ∂F /∂x are linearly independent vectors in D(F (z)).
We prove the first claim: From the Picard-Lindel¨f theorem, we have that for all
o
x ∈ n and sufficiently small |t|, the flow Φf (x) of a (Locally Lipschitz) vector field
t
f is well defined and therefore F is well defined for all z ∈ B ( n ) for sufficiently
small . In order to show that F is a local diffeomorphism, it is sufficient to show
that rank(F ) = n for all z ∈ B ( n ). To this end, we define for convenience:
(1.59)
M =
∂M
∂x
We have that:
(1.60)
(1.61)
(1.62)
∂F
∂
1
i−1
n
= (Φf1 ) · · · (Φfi−1 )
(Φfi ◦ · · · ◦ Φfn )|x0
z
z
z
∂zi
∂zi zi
1
i−1
i
n
= (Φf1 ) · · · (Φfi−1 ) · fi ◦ Φfi ◦ · · · ◦ Φfn |x0
z
z
z
z
f
i−1
1
i−1
= (Φf1 ) · · · (Φfi−1 ) · fi ◦ Φ−zi−1 ◦ · · · ◦ Φfi 1 |F (z)
z
z
−z
For z = 0 we have F (0) = x0 and
(1.63)
∂F
|0 = fi |x0
∂zi
The vector fields f1 (x0 ), · · · , fn (x0 )are linearly independent and tanget to F , hence
the n columns of (F ) are linearly independent at z = 0 and due to the fact that
F is continuous, the martix F is of full rank at 0 and in a neighborhood of 0.
◦
We now prove the second claim: We have already shown that for all z ∈ B
(
n)
,

12. 12
SOPASSAKIS PANTELIS
∂F
∂zi
Instead of proving the claim (1.2) we formulate and prove an other one. If the
following claim holds, so does the claim (1.2).
(1.64)
f
i−1
fi−1
1
(Φf1 ) · · · (Φzi−1 ) · fi ◦ Φ−zi−1 ◦ · · · ◦ Φfi 1 |x=F (z) =
z
−z
Claim 2. For all x ∈ U 0 , where U 0 is a neighborhood of x0 , for |t| sufficiently
small and for any two vector fields τ, ϑ in D the following holds:
(Φϑ ) τ ◦ (Φϑ )|x ∈ D(x)
t
−t
(1.65)
This claim imlies that (Φϑ ) τ ◦ (Φϑ )|x is a locally defined vector space in D and
t
−t
(1.2) follows. So we prove the claim (2). Let ϑ be a vector space in D and let us
define:
Vi (t) = (Φϑ ) fi ◦ (Φϑ )|x
t
−t
(1.66)
for i = 1, 2, . . . , d. Observe that:
I = (Φϑ ) (Φϑ )
−t
t
d ϑ
(Φ ) (Φϑ )
t
dt −t
d
∂ϑ
◦ Φϑ |x
⇒ 0 = (Φϑ ) ◦ (Φϑ )|x + (Φϑ )star
t
−t
t
dt −t
∂x
∂ϑ
d
◦ Φϑ |x
⇒ (Φϑ ) ◦ (Φϑ )|x = −(Φϑ )star
t
−t
t
dt −t
∂x
⇒0=
Furthermore:
d
∂fi
(fi ◦ Φϑ |x ) =
ϑ ◦ Φϑ |x
t
t
dt
∂x
The functions Vi (t) defined earlier satisfy:
dVi
(1.68)
= (Φϑ ) [ϑ, fi ] ◦ Φϑ
−t
t
dt
By assumtion D is involutive and both ϑ and fi belong to D, thus their Lie bracket
belongs to D so there exist functions λij defined locally in U 0 such that:
(1.67)
d
(1.69)
λij fj ∈ D
[ϑ, fi ] =
j=1
Hence we can express dVi /dt as:
d
(1.70)
dVi
= (Φϑ ) [ϑ, fi ] ◦ Φϑ = (Φϑ ) (
λij fj ) ◦ Φϑ
−t
t
−t
t
dt
j=1
d
(1.71)
λij (Φϑ |x )Vj (t)
t
=
j=1
So the functions Vi (t) are the solutions of a system of linear diffenential equations,
and for that reason we may set:
(1.72)
[V1 (t) · · · Vd (t)] = [V1 (0) · · · Vd (0)] · X(t)
Where X(t) ∈ Md×d ( ) in (1.72) is the fundamental matrix of solutions. We
multiply (1.72) on the left both sides by (Φϑ ) and we get
t
(1.73)
[f1 ◦ Φϑ |x · · · fd ◦ Φϑ |x ] = [((Φϑ ) )f1 |x · · · ((Φϑ ) )fd |x ]X(t)
t
t
t
t

13. NONLINEAR CONTROL THEORY
13
and we replace x by Φϑ |x to get
−t
(1.74)
[f1 (x) f2 (x) . . . fd (x)] = [(Φϑ ) f1 |x ◦ Φϑ |x · · · (Φϑ ) fd |x ◦ Φϑ |x ]
t
−t
t
−t
X(t) is nonsingular for all t with |t| sufficiently small, therefore for i = 1, 2, · · · , d
we have:
(1.75)
(Φϑ ) fi |x ◦ Φϑ |x ∈ span {f1 |x , · · · , fd |x }
t
t
Or equivalently,
(1.76)
(Φϑ ) fi |x ◦ Φϑ |x ∈ D(x)
t
−t
Employing the fact that the arbitary τ ∈ D is written as:
d
(1.77)
τ=
aj fj
j=1
(where aj are functions from the field of C ∞ functions), the proof of the claim (2)
is complete.
◦
We now use the results from the claims we have proven so far to complete the proof
of the Frobenius’ theorem. So let U be the image of the mapping F . Then U is
an open neighborhood of x0 for F (0) = x0 . Since F is a (local) diffeomorphism
(We know that from claim 1), it is one-to-one, and therefore it has an inverse
F −1 : U → B ( n ) and is smooth. Set:


φ1 (x)
 φ2 (x) 


(1.78)
 .  = F −1 (x)
 . 
.
φn (x)
where φ1 (x), . . . , φn (x) are real-valued smooth functions fi (x) : U → n . We claim
that the last n − d of these functions defined in (1.78) are independent solutions of
(1.43). By definition of F in (1.58), we have:
(1.79)
∂F −1
∂x
x=F (z)
∂F
∂z
=I
for all z ∈ B ( n ) that is for all x ∈ U. By the claim (1.2), the first d columns
of the second factor on the left-hand side of (1.79) form a basis of D at any point
x ∈ U. Hence the vectors {dφd+1 , . . . , dφn } are annihilated by the vectors of D at
each x ∈ U and therefore form a basis for D⊥ . The proof of the Frobenius’ Theorem
is now complete.
The proof of the sufficiency of involutiveness for a nonsingular distribution to be
completely integrable is constructive and offers us a frame for the construction of
functions such that their gradients form a basis for the annihilator of a given distribution. We give an example straihtforward to explain the constructive procedure.

14. 14
SOPASSAKIS PANTELIS
Example 4. Consider of a distribution D = span(f1 , f2 ) defined on U = x ∈
with:




2x3
−x1
f1 =  −1  , f1 = −2x2 
(1.80)
0
x3
3
: x2 + x2 = 0
1
3
We define:

2x3
G(x) = [f1 (x) f2 (x)] =  −1
0
(1.81)

−x1
−2x2  ,
x3
Since rank G(x) = 2, for all x ∈ U, D is nonsingular with constant degree d = 2.
The Lie Bracket of f1 with f2 is:
(1.82)
(1.83)
(1.84)
[f1 , f2 ] = (f2 ) f1 − (f1 ) f2

 

0
−1 0 0 2x3
=  0 −2 0  −1  − 0
0
0
0 1
0


−4x3
= 2 
0
0
0
0


−x1
2
0 −2x2 
0
x3
the matrix

2x3
[f1 f2 [f1 , f2 ]] =  −1
0
(1.85)
−x1
−2x2
x3

−4x3
2 
0
has rank 2 for all x ∈ U and the distribution is involutive (According to proposition
8). Set f3 (x) = e3 . We calculate the flows of f1 , f2 and f3 .






2z1 x3 + x1
exp(−z2 )x1
z3 + x1
1
2
3
(1.86) Φf1 (x) =  −z1 + x2  , Φf2 (x) = exp(−2z2 )x2  , Φf3 (x) =  x2  .
z
z
z
x3
exp(z2 )x3
x3
The mapping F is defined to be:
(1.87)
(1.88)
1
2
3
F (z)|x=x0 = Φf1 ◦ Φf2 ◦ Φf3 |x0
z
z
z


2z1 exp(z2 )x0 + exp(−z2 )(z3 + x0 )
3
1

−z1 + exp(−2z2 )x0
=
2
exp(z2 )x0
3
The partial differential equation:
(1.89)
∂λ
G(x) = [0 0]
∂x
has the solution:
(1.90)
λ(x) = z3 (x) = (x1 + 2x2 x3 )x3
•

15. NONLINEAR CONTROL THEORY
15
References
[1] Isidori A., Nonlinear Control Systems - An introduction, Springer Verlag Editions International, 2nd edition, 1989.
[2] Erdmann K., Wildon J., K., Introduction to Lie Algebras, Springer Verlag Editions International, 2006.
[3] Wikipedia, the free encyclopedia, available online: http://www.wikipedia.com
E-mail address, Sopassakis P.: chvng@mail.ntua.gr