Solution to Exercise 4.7: Consider the system
ẋ1 = x2 − x71 (x41 + 2x22 − 10),
ẋ2 = −x31 − 3x52 (x41 + 2x22 − 10).
Part (a): For which values of c is the set Ω = {x ∈ R2 |x41 + 2x22 ≤ c} invariant with respect to the
system dynamics? First, let us presents methods for proving that a set is invariant.
Geometric Interpretation: Let us define the boundary of set Ω as ∂Ω = {x ∈
R2 |x41 + 2x22 = c}. Because the trajectories of the underlying system are continuous
(and, therefore, cannot jump out of Ω directly), they must reach its boundary to
able to leave. Let us define g(x) = x41 + 2x22 . Checking the figure below, it is rather
trivial that if the angel between vector ∇g(x) and f (x) is greater than π/2 for all
x ∈ ∂Ω, then the trajectories of the system are pushed towards the interior of the set Ω
whenever they reach the boundary. Therefore, Ω is an invariant set if ∇g(x)> f (x) ≤ 0
for all x ∈ ∂Ω.
x2
∇g(x)
x ∈ ∂Ω
Ω:
∂Ω:
f (x)
x1
Algebraic Interpretation: Let us define g(x) = x41 + 2x22 and redefine set Ω as
Ω = {x ∈ R2 |g(x) ≤ c}. Evidently, we have
Z t
d
g(x(t)) = g(x(t1 )) +
g(x(τ )) dτ
t1 dτ
Z t
= g(x(t1 )) +
∇g(x(τ ))> ẋ(τ )dτ
t1
t
Z
∇g(x(τ ))> f (x(τ ))dτ.
= g(x(t1 )) +
t1
Again recall that because the trajectories of the underlying system are continuous
(and, therefore, cannot jump out of Ω directly), they must reach its boundary to able
to leave (i.e., a point at which g(x) = c). However, if ∇g(x)> f (x) ≤ 0 for all x ∈ ∂Ω,
whenever g(x(t)) = c, dg(x(τ ))/dτ becomes less than or equal to zero which makes
g(x(t)) smaller than c and, therefore, the trajectory is pushed back into the set Ω.
Now, notice that
∇g(x) =
4x31
4x2
and f (x) =
1
x2 − x71 (x41 + 2x22 − 10)
−x31 − 3x52 (x41 + 2x22 − 10)
,
and as a result,
∇g(x)> f (x) =
4x31
4x2
> x2 − x71 (x41 + 2x22 − 10)
−x31 − 3x52 (x41 + 2x22 − 10)
4
2
3
6 4
2
= 4x31 x2 − 4x10
1 (x1 + 2x2 − 10) − 4x2 x1 − 12x2 (x1 + 2x2 − 10)
6
4
2
= −4(x10
1 + 3x2 )(x1 + 2x2 − 10).
Therefore, for all x ∈ ∂Ω, that is, whenever x41 + 2x22 = c, we get
≥0
}|
{
z
6
∇g(x)> f (x) = −4 (x10
1 + 3x2 )(c − 10)
≤ 0,
if c ≥ 10. Therefore, we proved that Ω is an invariant set if c ≥ 10.
I skipped Part (b) in the lecture since we sadly didn’t have enough time to discuss it. Hopefully,
you can read this note and understand the solution easily. Otherwise, please do not hesitate to
come to my office for questions after you read it!
Part (b): First, let me present the statement of the theorem that we are going to use in this part
of the problem.
LaSalle’s Invariant Set Theorem: Let Ω ⊂ Rn be a compact set which is
invariant with respect to ẋ = f (x). Let V : Rn → R be a function in C 1 such that
V̇ (x) ≤ 0 for all x ∈ Ω. Let E be the set of points in Ω such that V̇ (x) = 0. If M
is the largest invariant set in E, then every solution with x(0) ∈ Ω approaches M as
t → ∞.
Well, we know Ω = {x ∈ R2 |g(x) ≤ c} for c ≥ 10 is an invariant set. We can easily see
that Ω is a bounded and
set p
(i.e., apcompact set) for any finite c. The boundedness is
√ closed
√
easy to see since Ω ⊆ [− 4 c, 4 c] × [− c/2, c/2] which is a bounded area. For closedness, let us
consider a sequence of points {xn }n∈N such that xn ∈ Ω, n ∈ N, and limn→∞ xn = x. Clearly,
g(xn ) ≤ c, ∀n ∈ N, implies that limn→∞ g(xn ) ≤ c. In addition, the continuity of g(·) implies that
limn→∞ g(xn ) = g(limn→∞ xn ) = g(x). Hence, g(x) ≤ c which shows that x ∈ Ω and, subsequently,
Ω is a closed set.
Let us use the mapping V : Rn → R such that V (x) = (x41 + 2x22 − 10)2 for all x ∈ Rn . Evidently,
V ∈ C 1 (since all polynomials are continuously differentiable). We have
V̇ (x) = 2(x41 + 2x22 − 10)(4x31 )ẋ1 + 2(x41 + 2x22 − 10)(4x2 )ẋ2
= 2(x41 + 2x22 − 10)(4x31 )[x2 − x71 (x41 + 2x22 − 10)]
+ 2(x41 + 2x22 − 10)(4x2 )[−x31 − 3x52 (x41 + 2x22 − 10)]
= −8 (x10
+ 3x6 ) (x4 + 2x22 − 10)2
| 1 {z 2} | 1
{z
}
≥0
≥0
≤ 0,
∀x ∈ Ω.
Now, we can define the set
n
o
E = x ∈ Ω|V̇ (x) = 0
= x ∈ R2 |x41 + 2x22 = 10 ∪ {(0, 0)}.
2
Notice that if x ∈ x ∈ R2 |x41 + 2x22 = 10 , then
d 4
x1 + 2x22 = 0,
dt
and, therefore, x(0) ∈ x ∈ R2 |x41 + 2x22 = 10 implies that x(t) ∈ x ∈ R2 |x41 + 2x22 = 10 for all
t ≥ 0. In addition, since f (0) = 0, x(t) = 0 for all t ≥ 0 if x(0) = 0. This proves that E is an invariant
set. Hence, M (i.e., the largest invariant set in E) is equal to E itself. Therefore, if x(0) ∈ Ω, then
x(t) → E as t → ∞. Unfortunately, this is not good for proving that we have a limit cycle since
we
know, at theend, that we converge to the origin {(0, 0)} or the trajectory described by
don’t
x ∈ R2 |x41 + 2x22 = 10 .
Part (c): Let us define Ω = {x ∈ R2 |c1 ≤ x41 + 2x22 ≤ c2 } for constants 0 < c1 ≤ 10 ≤ c2 (which is
the doughnut shape set that I defined in the lecture). Just remember that 0 ∈
/ Ω because we selected
c1 > 0. Let us show that this set is an invariant set.
Geometric Interpretation: Let us define the boundary of set Ω as ∂Ω = ∂Ω1 ∪
∂Ω2 where ∂Ωi = {x ∈ R2 |x41 + 2x22 = ci } for i = 1, 2. Notice that ∂Ω1 denotes the
inner boundary while ∂Ω2 is the outer layer. Recall that because the trajectories of
the underlying system are continuous (and, therefore, cannot jump out of Ω directly),
they must reach the boundary ∂Ω1 or ∂Ω2 to able to leave. Let us, as before, define
g(x) = x41 + 2x22 . Checking the figure below, it is rather trivial that (i ) if the angel
between vector ∇g(x) and f (x) is less than or equal π/2 for all x ∈ ∂Ω1 and (ii ) if the
angel between vector ∇g(x) and f (x) is greater than or equal π/2 for all x ∈ ∂Ω2 , then
the trajectories of the system are pushed towards the interior of the set Ω whenever
they reach the boundary. Therefore, Ω is an invariant set if ∇g(x)> f (x) ≥ 0 for all
x ∈ ∂Ω1 and ∇g(x)> f (x) ≤ 0 for all x ∈ ∂Ω2 .
x2
∇g(x)
x ∈ ∂Ω2
f (x)
∇g(x)
f (x)
x ∈ ∂Ω1
x1
Using this knowledge, now, we are ready to prove that Ω is indeed an invariant set. Let us
consider x ∈ ∂Ω1 , that is, x41 + 2x22 = c1 ≤ 10. We have
6
4
2
∇g(x)> f (x) = −4(x10
1 + 3x2 )(x1 + 2x2 − 10)
= −4 (x10
+ 3x6 ) (c1 − 10)
| 1 {z 2} | {z }
≥0
≥ 0.
3
≤0
Let us consider x ∈ ∂Ω2 , that is, x41 + 2x22 = c2 ≥ 10. We have
6
4
2
∇g(x)> f (x) = −4(x10
1 + 3x2 )(x1 + 2x2 − 10)
= −4 (x10
+ 3x6 ) (c2 − 10)
| 1 {z 2} | {z }
≥0
≥0
≤ 0.
Therefore, Ω is an invariant set. Let us show that Ω is a bounded and closedp
set ( hence,
p a compact
√ √
set). The boundedness follows from the fact that Ω ⊆ [− 4 c2 , 4 c2 ] × [− c2 /2, c2 /2] for any
finite c2 (because any subset of a bounded set is a bounded set). For closedness, let us consider a
sequence of points {xn }n∈N such that xn ∈ Ω, n ∈ N, and limn→∞ xn = x. Clearly, c1 ≤ g(xn ) ≤ c2 ,
∀n ∈ N, implies that c1 ≤ limn→∞ g(xn ) ≤ c2 . In addition, the continuity of g(·) implies that
limn→∞ g(xn ) = g(limn→∞ xn ) = g(x). Combining these two facts results in c1 ≤ g(x) ≤ c which
shows that x ∈ Ω and, subsequently, Ω is a closed set. With these results in hand, we have to select
a function V (·) to use LaSalle’s Invariant Set Theorem. In what follows, I present two versions. The
first choice for this function follows the hint in part (b) which is slightly different from the one that
I used during the lecture. After that, I also present the material presented during the class.
First Choice: Let V : Rn → R be such that V (x) = (x41 + 2x22 − 10)2 for all x ∈ Rn . Evidently, as
described in the last part, V ∈ C 1 . We have
V̇ (x) = 2(x41 + 2x22 − 10)(4x31 )ẋ1 + 2(x41 + 2x22 − 10)(4x2 )ẋ2
= 2(x41 + 2x22 − 10)(4x31 )[x2 − x71 (x41 + 2x22 − 10)]
+ 2(x41 + 2x22 − 10)(4x2 )[−x31 − 3x52 (x41 + 2x22 − 10)]
= −8 (x10
+ 3x6 ) (x4 + 2x22 − 10)2
{z
}
| 1 {z 2} | 1
≥0
≥0
≤ 0,
∀x ∈ Ω.
Now, we can define the set
n
o
E = x ∈ Ω|V̇ (x) = 0
= x ∈ R2 |x41 + 2x22 = 10 .
Notice that if x ∈ E, then
d 4
x + 2x22 = 0,
dt 1
which shows that x(t) ∈ E, for all t ≥ 0, whenever x(0) ∈ E. Hence, M (i.e., the largest invariant
set in E) is equal to E itself. Therefore, if x(0) ∈ Ω, then x(t) → E as t → ∞. Interestingly, in this
analysis, E only contains the trajectory that we believe is a limit cycle. However, to mathematically
prove that set E is actually a limit cycle, we need to use Poincaré–Bendixson theorem.
Poincaré–Bendixson Theorem: Any orbit of a continuous 2nd order system
that stays in a compact region of the phase plane approaches its ω-limit set, which
is either a fixed point (i.e., an equilibrium), a periodic orbit, or several fixed points
connected through homoclinic or heteroclinic orbits.
Evidently, Ω is a compact invariant set. It also does not contain any equilibrium (because you
can check yourself that the origin is the unique equilibrium of this system). Therefore, by Poincaré–
Bendixson Theorem, x(t) converges to a periodic orbit (a.k.a, a limit cycle). However, by our analysis,
we know that x(t) converges to E. As a result, there exists a stable periodic which is a subset of E.
4
Second Choice: Let V : Rn → R be such that V (x) = x41 + 2x22 − 10 for all x ∈ Rn . Notice that I
dropped the power of two in this definition, which makes some differences in our derivation but the
outcome is the same. Again, since dealing with polynomials, we get V ∈ C 1 . For this case, we have
V̇ (x) = (4x31 )ẋ1 + (4x2 )ẋ2
= (4x31 )[x2 − x71 (x41 + 2x22 − 10)] + (4x2 )[−x31 − 3x52 (x41 + 2x22 − 10)]
6
4
2
= −4(x10
1 + 3x2 )(x1 + 2x2 − 10).
Let us select c1 = 10. In this case, for all x ∈ Ω, we get x41 + 2x22 ≥ 10 which proves that
V̇ (x) = −4 (x10
+ 3x6 ) (x4 + 2x22 − 10)
{z
}
| 1 {z 2} | 1
≥0
≥0 if c1 =10
≤ 0.
Now, we can define the set
n
o
E = x ∈ Ω|V̇ (x) = 0
= x ∈ R2 |x41 + 2x22 = 10 .
Notice that if x ∈ E, then
d 4
x + 2x22 = 0,
dt 1
which shows that x(t) ∈ E, for all t ≥ 0, whenever x(0) ∈ E. Hence, M (i.e., the largest invariant
set in E) is equal to E itself. Therefore, if x(0) ∈ Ω, then x(t) → E as t → ∞.
5
Hint1 for (Extra) Exercise 4.6: Consider state-space model
ẋ1 = x2
ẋ2 = −2x1 /(1 + x21 )2
and candidate Lyapunov function
Z
x1
V (x) =
0
2y
1
dy + x22 .
(1 + y 2 )2
2
Rx
(i ) We have V ∈ C . To see this, note that 0 1 2y/(1+y 2 )2 dy is continuously differentiable because
2y/(1 + y 2 )2 is a continuous function! This is a direct result of the fundamental theorem of
calculus. In addition, 0.5x22 is continuously differentiable (because we know all polynomials are
indeed so). Finally, summation of any finite number of countinuosuly differential functions is
a continuously differentiable function.
1
(ii ) We have
Z
V (0) =
0
0
1
2y
dy + (0)2
2
2
(1 + y )
2
= 0.
(iii ) We have V (x) > 0 if x 6= 0. Notice that if x 6= 0, then either x1 6= 0 or x2 6= 0. Whenever
x2 6= 0, we can easily see that 0.5x22 > 0. What about the other term? Well, one can easily
note that
Z x1
2y
x21
dy
=
,
(1 + y 2 )2
1 + x21
0
which is positive if x1 6= 0. However, we can also note that if x1 > 0, we have
Z x1
2y
2y
dy > 0,
because
> 0, ∀y > 0.
2 )2
(1
+
y
(1
+
y 2 )2
0
In addition, if x1 < 0, we have
Z x1
Z 0
2y
2y
dy = −
dy
2
2
2 2
(1 + y )
0
x1 (1 + y )
>0
because
2y
< 0, ∀y < 0.
(1 + y 2 )2
(iv ) We have
2x1
ẋ1 + x2 ẋ2
(1 + x21 )2
2x1
2x1
=
x2 − x2
(1 + x21 )2
(1 + x21 )2
V̇ (x(t)) =
= 0.
Well, these properties are enough to say the origin is locally stable (i.e., ∀ > 0, ∃δ > 0 : kx(0)k ≤
δ ⇒ kx(t)k ≤ , ∀t ≥ 0). To do so, one just need to use Lyapunov theorem for local stability. But,
can we say something about the global stability (in a very weak sense that the trajectories of the
system stay bounded irrespective of their starting point)? Notice that
Z t
V (x(t)) = V (x(0)) +
V̇ (x(τ ))dτ
0
≤ V (x(0))
1 Not
because V̇ (x(t)) ≤ 0, ∀t ≥ 0
a complete solution!
6
Now, if the level sets of Lyapunov function V (·), that is, Ωc = {x ∈ R2 |V (x) ≤ c}, were compact
sets, we were done because the above inequality shows that the trajectories cannot leave the level set
that they started from and, hence, the trajectories of the system (no matter what initial condition
we select) are going to stay bounded for ever! However, the chosen Lyapunov function does
not have compact level sets. This is indeed the case because V (·) does not satisfy the radially
unbounded condition. To see this, let us set x = (x1 , 0) and let x1 to tend to infinity. This implies
that kxk = |x1 | goes to infinity. However, we have
Z x1
1
2y
dy + x22
V (x) =
lim
lim
x2 =0,x1 →∞
x2 =0,x1 →∞ 0
(1 + y 2 )2
2
1
x21
+ x22
=
lim
x2 =0,x1 →∞ 1 + x2
2
1
= 1,
and as a result,
[−∞, +∞] × {0} ⊆ {x ∈ R2 |V (x) ≤ c} = Ωc , ∀c ≥ 1.
Therefore, Ωc cannot be a compact set if c ≥ 1. Let us draw the level sets of V (·) for some c > 1.
x2
x1
The shaded area shows Ωc = {x ∈ R2 |V (x) ≤ c} and the thick lines demonstrate the boundary of
this set at which ∂Ωc = {x ∈ R2 |V (x) = c}. As you can see, we can indeed stay inside these sets
while having an unstable trajectory (which is very much the case in this example; see the proof for
the instability towards the end of your exercise books).
7