APM2614/202/1/2018
Tutorial letter 202/1/2018
APPLIED DYNAMICAL SYSTEMS
APM2614
Semester 1
Department of Mathematical Sciences
Ths tutorial letter contains solutions for assignment 02.
BARCODE
Define tomorrow.
university
of south africa
The solutions of Assignment 02 follow below. Please study the solutions carefully together with
your marked assignment.
1.
u2
x1
x1
_x
2
_
x2
x2
+
2
1
2 x1
+
_
x2
u1
_
1
y
u1
2. (a) The matrix form of the system equations for S1 is
[ẋ1 ] = [−1] x1 + [1] u
[y] = [α] x1 .
Thus
A = [−1] ,
B = [1] ,
C = [α] .
The controllability matrix for S1 is
M = [B] = [1]
which has rank 1 so S1 is completely controllable. The observability matrix for S1 is
S = [C] = [α]
which has rank 1 provided that α 6= 0. So S1 is completely observable for all α ∈ R\ {0} .
The matrix form of the system equations for S2 is
ẋ2
2
1
x1
1
=
+
w
ẋ3
1 −2
x2
1
x2
z = [1 α]
.
x3
The controllability matrix for S2 is
M = [B, AB] =
2
1
3
1 −1
APM2614/202/1/2018
for which rank(M) = 2 and therefore the system S2 is completely controllable.
The observability matrix for S2 is
C
1
α
S=
=
1+α 3−α
CA
for which
det (M) = − α2 + 2α − 3 = − (α + 3) (α − 1) .
Hence det (M) 6= 0 provided that α ∈
/ {1; −3} . The system is therefore completely observable for all α ∈ R\ {−3; 1} .
(b) The system equations for S3 are
ẋ1
ẋ2
ẋ3
z
=
=
=
=
−x1 + u
2x2 + x3 + αx1
x2 − 2x3 + αx1
x2 + αx3
In matrix form these are


  


−1 0
0
1
ẋ1
x1


 ẋ2  =  α 2


1
x2 + 0  u
ẋ3
x3
α 1 −2
0


x1
z = [0 1 α]  x2  .
x3
The controllability matrix for S3 is


1
−1
1
2α 
M = B, AB, A2 B =  0 α
0 α −2α
for which det (M) = −4α2 6= 0 for all α 6= 0. The system S3 is therefore completely
controllable for all α ∈ R\ {0} .
The observability matrix for S3 is

 

C
0
1
α
S =  CA  =  α2 + α α + 2 1 − 2α 
2α − 2α2
5
5α
CA2
√
for which det (S) = 2α (α − 1) (α2 + 4α − 1) 6= 0 for all α ∈
/ 0, 1, −2 ± 5 . The system
√
S3 is therefore completely observable for all α ∈ R\ 0, 1, −2 ± 5 .
(c) The system equations for S4 are
ẋ1
ẋ2
ẋ3
y
=
=
=
=
−x1 − (x2 + αx3 )
2x2 + x3 − (w + αx1 )
x2 − 2x3 − (w + αx1 )
αx1 .
3
In matrix form these are




 

ẋ1
−1 −1 −α
x1
0
 ẋ2  =  −α
2
1   x2  +  −1  w
ẋ3
−α
1 −2
x3
−1


x1

y = [α 0 0] x2  .
x3
The controllability matrix for S4 is


0 α+1
2 − 2α
M = B, AB, A2 B =  −1 −3 −α2 − α − 5 
−1
1
−α2 − α − 5
for which det (M) = 8 (α − 1) 6= 0 for α 6= 1.
The system is therefore completely controllable for all α ∈ R\ {1} .
The observability matrix is

 

C
α
0
0

−α
−α
−α2
S =  CA  = 
2
2
α (α + α + 1) −α (α + 1) α (3α − 1)
CA
√
for which det (S) = −α3 (α2 + 4α − 1) 6= 0 for α ∈
/ 0, −2 ± 5 .
√
The system S4 is therefore completely observable for all α ∈ R\ 0, −2 ± 5 .
3. We solve
3
ẋ = 0 ⇒ y = x2
4
ẏ = 0 ⇒ y = x
to find the singular points. This gives the points
(0, 0)
and
4 4
,
3 3
.
The Jacobian matrix is given by
J=
6x −4
1 −1
.
At (0, 0):
J (0, 0) =
4
0 −4
1 −1
APM2614/202/1/2018
for which
σ = −1
∆ = 4
σ 2 − 4∆ = 1 − 16 = 15 < 0
so that (0, 0) is locally a stable focus.
At 43 , 34 :
J
4 4
,
3 3
=
8 −4
1 −1
for which
σ = 7
∆ = −4 < 0
so that
4 4
,
3 3
is locally a saddle point.
The phase plane diagram is given by
4
3
4
3
2. (a) ẋ = 0 ⇒ x3 = y
ẏ = 0 ⇒ x = y 3 .
(1)
(2)
5
3
Substitute y = x3 (from (1)) into (2): x = (x3 ) = x9 .
This gives x (1 − x8 ) = 0. Therefore x = 0 or x = ±1.
The critical points therefore are (0, 0) , (1, 1) and (−1, −1) .
The Jacobian matrix is given by
2
3x
−1
J=
.
1 −3y 2
J (0, 0) =
0 −1
1
0
for which
σ = 1
∆ = 1
2
σ − 4∆ = −4 < 0
and so (0, 0) is a centre.
3 −1
J (1, 1) = J (−1, −1) =
for which
1 −3
σ = 0
∆ = −9 + 1 = −8 < 0
and so both (1, 1) and (−1, −1) are saddle points.
(b) ẋ = 0 ⇒ y = x2 − 2.
(1)
1
ẏ = 0 ⇒ y (2x − 1) = 0 ⇒ y = 0 or x = 2 .
√
Now let y = 0 in (1). This gives x2 = 2 so that x = ± 2.
Next let x = 21 in (1). This gives y = 14 − 2 = − 74 . The critical points therefore are
√ √ 1 7
,−
2, 0 , − 2, 0
and
.
2 4
The Jacobian matrix is
J=
J
√
2, 0 =
−2x
1
2y 2x − 1
.
√
−2 2 √ 1
for which
0
2 2−1
√
∆=2 2−8<0
√ so that
2, 0 is a saddle point.
Similarly
√ 2√2
1
√
J − 2, 0 =
0 −2 2 − 1
so that
6
√
∆ = −2 2 − 8 < 0
APM2614/202/1/2018
√ and therefore − 2, 0 is also a saddle point.
−1 1
1
7
J 2, −4 =
− 72 0
for which
σ = −1
7
∆ =
2
σ 2 − 4∆ = 1 − 14 = −13 < 0
and so 12 , − 74 is a stable spiral point.
(c) ẋ = 0 ⇒ x = 0 or 10 − x − 21 y = 0.
(1)
ẏ = 0 ⇒ y = 0 or 16 − y − x = 0.
(2)
Clearly (0, 0) is a critical point. Now let x = 0 into 16 − y − x = 0. This gives y = 16.
Therefore (0, 16) is also a critical point. Now let y = 0 into 10 − x − 21 y = 0. This gives
x = 10 and so (10, 0) is another critical point. Finally let x = 10 − 21 y (from (1)) into (2).
This gives (4, 12) . The critical points therefore are
(0, 0) , (10, 0) , (0, 16)
The Jacobian is
J=
J (0, 0) =
10 0
0 16
and
(4, 12) .
− 12 x
10 − 2x − 21 y
−y
16 − 2y − x
.
for which
σ = 26
∆ = 160
2
σ − 4∆ = 36
and so (0, 0) is an unstable node.
−10 −5
J (10, 0) =
for which
0
6
∆ = −60 < 0
and so (10, 0) is a saddle point.
2
0
J (0, 16) =
for which
−16 −16
∆ = −32 < 0
and so (0, 16) is also a saddle point.
−4 −2
J (4, 12) =
for which
−12 −12
σ = −16
∆ = 48 − 24 = 24
2
σ − 4∆ = 256 − 96 = 160
and therefore (4, 12) is a stable node.
7
4. (a) Let ẋ = y. Then ẏ = −2y −
therefore
4x
.
1+x2
The corresponding plane autonomous system is
ẋ = y
ẏ = −
4x
− 2y
1 + x2
and the only critical point is (0, 0) . The Jacobian is
#
"
0
1
J=
1−x2
−2
−4 (1+x
2 )2
so that
J (0, 0) =
0
1
−4 −2
for which
σ = −2
∆ = 4
2
σ − 4∆ = 4 − 16 = −12 < 0
and so (0, 0) is a stable spiral point.
(b) Let ẋ = y. Then ẏ = −x − ε 31 y 2 − y . The corresponding plane autonomous system is
therefore
ẋ = y
ẏ = −x − εy
1 2
y −1
3
and the only critical point is (0, 0) . The Jacobian is
0
1
J=
.
−1 −εy 2 + ε
0 1
J (0, 0) =
so that
−1 ε
σ = ε
∆ = 1>0
2
σ − 4∆ = ε2 − 4 = (ε − 2) (ε + 2) .
Here we have σ 2 − 4∆ < 0 for −2 < ε < 2 and σ 2 − 4∆ > 0 for ε > 2 and ε < −2. Hence
ε>2
0<ε<2
ε=0
−2 < ε < 0
ε < −2
:
:
:
:
:
(0, 0)
(0, 0)
(0, 0)
(0, 0)
(0, 0)
is an unstable node
is an unstable focus
is a centre
is a stable focus
is a stable node.
If ε = ±2, σ 2 − 4∆ = 0 and all we know is that (0, 0) is unstable if ε = 2 and stable if
ε = −2.
8
APM2614/202/1/2018
5. (a) Let g (x) = 3.5x − 5x2 .
Having a 2–cycle (a1 , a2 ) means that
(1)
a2 = g (a1 ) , a1 = g (a2 ) ⇒ a1 = g (g (a1 )) .
We therefore solve
0 = g (g (x)) − x
(2)
with g (x) as defined in (1).
From (2) we have that
2
0 = 3.5 3.5x − 5x2 − 5 3.5x − 5x2 − x
5x (100x3 − 140x2 + 63x − 9)
=
.
3
(3)
Since x = g (x (x)) is also solved by the solutions of x = g (x) we already know that
x=0
and
x=
1
2
are solutions of (3) because
x = g (x) ⇒ 3.5x − 5x2 − x = 2.5x − 5x2
= 5x (0.5 − x) .
We can therefore divide (3) by
x (2x − 1) .
This gives
50x2 − 45x + 9 = (10x − 3) (5x − 3) = 0
so that
x=
3
10
or
3
x= .
5
The 2–cycle is therefore given by
(a1 , a2 ) =
3 3
,
10 5
.
(b) The 2–cycle will be an attractor only if
|f 0 (a1 ) f 0 (a2 )| < 1
where
f (x) = 3.5x − 5x2 .
Hence from
f 0 (x) = 3.5 − 10x
9
we have
3
f
= 3.5 − 3 = 0.5
10
3
0
= 3.5 − 6 = −2.5
f
5
0
so that
f
0
3
10
3
1
5
5
f
−
=
= − > 1.
5
2
2
4
0
The 2–cycle is therefore not an attractor.
6. Given
V (x1 , x2 ) = ax41 + bx21 + cx1 x2 + dx22
(1)
and the system
ẋ1 = x2
ẋ2 = x2 − x31
we calculate
V̇ (x1 , x2 ) = 4ax31 ẋ1 + 2bx1 ẋ1 + cẋ1 x2 + cx1 ẋ2 + 2dx2 ẋ2
= 4ax31 x2 + 2bx1 x2 + cx22 + cx1 x2 − x31 + 2dx2 x2 − x31
= (4a − 2d) x31 x2 + (2b + c) x1 x2 + (c + 2d) x22 − cx41 .
For V̇ to be positive definite, we choose
and
Then a = 12 , b =
1
2
4a − 2d
2b + c
c + 2d
c
=
=
=
=
0
0
1
−1.
and d = 1 so that
V̇ (x1 , x2 ) = x41 + x22
is clearly positive definite. Furthermore
V (x1 , x2 ) =
=
1 4 1 2
x + x − x1 x2 + x22
2 1 2 1
1 4 1
1
x1 + (x1 − x2 )2 + x22
2
2
2
is also positive definite.
[V (x1 , x2 ) is positive definite if V (x1 , x2 ) > 0 for all (x1 , x2 ) 6= (0, 0) .]
The origin is therefore unstable since
(0, 0) .
10
d
dt
(V (x1 , x2 )) ≥ 0 with equality if and only if (x1 , x2 ) =
APM2614/202/1/2018
7. The fixed points are found by solving
x∗ x2∗ − 4λ = 0.
This gives
x∗ = 0 or
√
x∗ = ±2 λ
(if λ ≥ 0).
Let
g (x) = x x2 − 4λ .
Then
g 0 (x) = 3x2 − 4λ.
At
x∗ = 0, g; (0) = −4λ =
> 0 if λ < 0 i.e. unstable
< 0 if λ > 0 i.e. stable.
√
At x∗ = −2 λ, g 0 (x∗ ) = 8λ > 0 for λ > 0 (unstable).
√
At x∗ = 2 λ, g 0 (x∗ ) = 8λ > 0 for λ > 0 (unstable).
The bifurcation diagram is therefore
x =2 λ
x
able
unst
unstable
stable
unst
able
λ
x=_2 λ
11