EL8223 Homework 4 Solution
1. With the given vector fields f1 and τ , we find the Lie bracket [f1 , τ ] to be
[f1 , τ ] = [0, −2x2 , −2x3 ]T
(1)
Since [f1 , τ ] is not in span{f1 }, we define f2 = [f1 , τ ] and define ∆1 = span{f1 , f2 }. Now,
computing [f2 , τ ], we find
[f2 , τ ] = [0, 2, 2]T
(2)
Note that [f2 , τ ] is not in span{f1 , f2 }. To see this, note that the matrix [f1 (x), f2 (x)] has
zeros in the first row and the determinant of the matrix formed by the second and third rows
of [f1 (x), f2 (x)] is −2(x2 − x3 )(x2 x3 − 1). Hence, considering, for instance, a point x with
x2 = 0.5 and x3 = 2, we get f1 (x) = [0, 1.25, 5]T and f2 (x) = [0, −1, −4]T . If we were able
to write the vector field [f2 , τ ] we found above as c1 (x)f1 (x) + c2 (x)f2 (x) with some scalar
real-valued functions c1 (x) and c2 (x), then at a point x such that x2 = 0.5 and x3 = 2, we
would need to have 1.25c1 (x) − c2 (x) = 2 and 5c1 (x) − 4c2 (x) = 2, which is easily seen to be
not possible with any numbers c1 (x) and c2 (x). Hence, since the vector field [f2 , τ ] is not in
span{f1 , f2 }, we define f3 = [f2 , τ ] and define ∆2 = span{f1 , f2 , f3 }. Now, computing [f3 , τ ],
we find [f3 , τ ] = [0, 0, 0]T . Hence, the distribution ∆2 = span{f1 , f2 , f3 } is invariant under
the vector field τ . From the construction above, ∆2 is the smallest distribution that contains
the given distribution ∆ and is invariant under the given vector field τ . Looking at the vector
fields f1 , f2 , and f3 , we see that f2 is in span{f1 , f3 }. Hence, we can simply write ∆2 as
span{f1 , f3 }.
To check if ∆2 is involutive, we compute the Lie bracket [f1 , f3 ] as
[f1 , f3 ] = [0, −4x2 , −4x3 ]
(3)
We see that [f1 , f3 ] belongs to ∆2 . Hence, ∆2 is involutive and is the smallest involutive
distribution that contains the given distribution ∆ and is invariant under the given vector
field τ .
2. We are given a distribution ∆ = span{f1 , f2 }. To find the involutive closure of this distribution, we compute the Lie bracket [f1 , f2 ] as [f1 , f2 ] = [0, 0, x1 + 1] . This vector field
does not belong to the distribution ∆. To see this, try writing the vector field [f1 , f2 ] as
c1 (x)f1 (x) + c2 (x)f2 (x) with some scalar real-valued functions c1 (x) and c2 (x). It is easily
seen that it is definitely not possible to write the vector field [f1 , f2 ] in this form. Hence, we
define f3 = [f1 , f2 ] and define ∆1 = span{f1 , f2 , f3 }. Now, computing the Lie brackets [f1 , f3 ]
and [f2 , f3 ], we see that [f1 , f3 ] = [0, 0, 1]T and [f2 , f3 ] = [0, 0, −(x1 + 1)]T . The vector field
[f2 , f3 ] is seen to be the same as −[f1 , f2 ]. The vector field [f1 , f3 ] does not belong to ∆1 since
for an x at which x1 = −1, we would have f3 (x) = [0, 0, 0]T and [f1 , f3 ] = [0, 0, 1]T cannot
be written as a linear combination of f1 (x) and f2 (x). Hence, define f4 = [f1 , f3 ] and define
∆2 = span{f1 , f2 , f3 , f4 }. Now, computing the Lie brackets [f1 , f4 ], [f2 , f4 ], and [f3 , f4 ], we
see that
[f1 , f4 ] = [0, 0, 0]T
(4)
[f2 , f4 ] = [0, 0, −1]T
(5)
T
[f3 , f4 ] = [0, 0, 0] .
1
(6)
Hence, the vector fields [f1 , f4 ], [f2 , f4 ], and [f3 , f4 ] belong to the distribution ∆2 . Therefore,
∆2 is involutive. From the construction above, we see that ∆2 is the smallest involutive
distribution that contains ∆ and is therefore the involutive closure of ∆. Looking at the
vector fields f3 and f4 , we see that f3 (x) is simply (x1 + 1)f4 (x). Hence, we can simply write
∆2 as span{f1 , f2 , f4 }. Looking at the vector fields f1 , f2 , and f4 , we see that this is simply
R3 .
3. To check if the given distribution ∆ = span{f1 , f2 } is involutive, we compute the Lie bracket
[f1 , f2 ] as
[f1 , f2 ] = [0, 0, 0]T .
(7)
Hence, the vector field [f1 , f2 ] belongs to the distribution ∆. Therefore, ∆ is an involutive
distribution of dimension 2. To use ∆ to perform the change of coordinates, we first need to
find a function λ1 such that λ1 spans ∆⊥ . For the given vector fields f1 and f2 , we can easily
see that the covector field [−x1 , −1, 1] is orthogonal to f1 and f2 . This covector field is the
gradient of the scalar real-valued function λ1 = − 21 x21 − x2 + x3 . Alternatively, this function
λ1 could also have been found by using the construction in the proof of Frobenius theorem.
We next need to pick two scalar real-valued functions µ1 and µ2 such that the mapping
from x to Φ(x) = [µ1 (x), µ2 (x), λ1 (x)]T is a local diffeomorphism (around the origin). Here,
we can choose, for example, µ1 (x) = x1 and µ2 (x) = x2 . Then, the mapping from x to
z = Φ(x) = [µ1 (x), µ2 (x), λ1 (x)]T is a global diffeomorphism. To write the dynamics of the
given system in the new coordinates z, we need to transform the vector fields f and g given
below:


 
x2 + x21
0
f (x) =  −x1 x2 + x3  ; g(x) =  1 
(8)
x31 + x3
1
With these vector fields f and g, the dynamics of the given system are of the form ẋ =
f (x) + g(x)u. In the new coordinates, the vector fields f and g can be written as f (z) and
g(z) defined as



x2 + x21
1
0 0


∂Φ
1 0   −x1 x2 + x3 
f (z) =
f (x)
=  0


∂x
x=Φ−1 (x)
−x1 −1 1
x31 + x3
x=Φ−1 (x)


z2 + z12
=  −z1 z2 + z3 + 21 z12 + z2 
(9)
0

  
 
1
0 0
0 
0

∂Φ
1 0  1 
g(z) =
g(x)
=  0
=  1  . (10)


∂x
x=Φ−1 (x)
−x1 −1 1
1
0
x=Φ−1 (x)
Therefore, denoting z = [z1 , z2 , z3 ]T , the dynamics of the given system when written in the z
coordinates are:
ż1 = z2 + z12
1
ż2 = −z1 z2 + z3 + z12 + z2 + u
2
ż3 = 0.
2
(11)