Page 29 - IJOCTA-15-3

Using infinitesimal symmetries for determining the first Maxwell time of geometric control problem on SH(2)
Moreover, the Lie derivative of the metric g has
1
˙
h 3 = sin(2α), ˙ α = h 3 . the form:
2 ∂f 1 ∂f 3
L v (g) =2(cosh z − sinh z )dx 2
Next, we introduce another change of variables: ∂x ∂x
∂f 3 ∂f 1 2
+ 2(cosh z − sinh z )dy
γ = 2α ∈ R/4πZ, c = 2h 3 ∈ R. ∂y ∂y
∂f 2 2
+ 2( )dz
Finally, we obtain that the equation describing ∂z
our vertical system corresponds to a mathemati- ∂f 1 ∂f 3
+ (2 cosh z − 2 sinh z
cal pendulum given by: ∂y ∂y
∂f 1 ∂f 3
− 2 sinh z + 2 cosh z )dxdy
˙ γ = c, ˙ c = − sin(γ). ∂x ∂x
∂f 1
+ (2f 1 sinh z + 2 cosh z − 2f 3 cosh z
The total energy integral of the pendulum ob- ∂z
2
2
2
tained is given as: E = c 2 −cos(γ) = 2h −h +h , ∂f 3 ∂f 2
2 3 1 2 − 2 sinh z + 2 )dxdz
according to this energy, the cylinder C can be ∂z ∂x
S 5
decomposed in the following way. C = i=1 C i , + (−2 sinh z ∂f 1 − 2f 1 cosh z
where ∂z
∂f 3 ∂f 2
C 1 = {λ ∈ C | E ∈ (−1, 1)} + 2 cosh z + 2f 3 sinh z + 2 )dzdy
∂z ∂y
C 2 = {λ ∈ C | E ∈ (1, ∞)}
Now the relation L v (g) = 0 implies:
C 3 = {λ ∈ C | E = 1, c ̸= 0}
C 4 = {λ ∈ C | E = −1} ∂f 1 ∂f 3
cosh z − sinh z = 0, (17)
C 5 = {λ ∈ C | E = 1} ∂x ∂x
∂f 3 ∂f 1
For a detailed derivation of the solutions to our cosh z ∂y − sinh z ∂y = 0, (18)
5
system, refer to Butt( ).
∂f 2
= 0, (19)
∂z
and
4.3. Infinitesimal symmetries ∂f 1 ∂f 3
2 cosh z − 2 sinh z
∂y ∂y
In this subsection, we compute the symmetry al- ∂f 1 ∂f 3
gebra of the control system Equation (11). −2 sinh z + 2 cosh z = 0,
∂x ∂x
and
Proposition 3. Infinitesimal symmetries of the
control system Equation (11) form a Lie algebra 2f 1 sinh z + 2 cosh z ∂f 1 − 2f 3 cosh z
generated (over R) by the vector fields: ∂z
∂f 3 ∂f 2
−2 sinh z + 2 = 0,
v 1 = −x∂ y − y∂ x − ∂ z , ∂z ∂x
v 2 = ∂ x , and
v 3 = ∂ y . ∂f 1 ∂f 3
−2 sinh z − 2f 1 cosh z + 2 cosh z
∂z ∂z
Proof. It is clear that ∆ constitutes a contact +2f 3 sinh z + 2 ∂f 2 = 0.
distribution. Indeed, ∆ = ker(ω), where ∂y
1
ω = cosh(z)dy − sinh(z) ∈ Λ (M) and we have ∂f 3
Using equation Equation (16), we have f 1 = .
dω ∧ ω ̸= 0. Let ∂z
By substituting f 1 into Equations (17) and (18),
v = f 1 (x, y, z)X 1 +f 2 (x, y, z)X 2 +f 3 (x, y, z)X 3 be
we obtain the following:
a vector field. The conditions ω(L v (X 1 )) = 0 and
ω(L v (X 2 )) = 0, lead to the following system: ∂ f 3 ∂f 3
2
cosh z − sinh z = 0,
∂f 3 ∂f 3 ∂z∂x ∂x
f 2 + cosh z + sinh z = 0, (15) (20)
2
∂x ∂y ∂f 3 ∂ f 3
cosh z − sinh z = 0.
∂f 3 ∂y ∂z∂y
f 1 − = 0. (16)
∂z
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