E1 242: Nonlinear Systems and Control (2024j)

HW #2 Solutions

1. Consider the system ẋ = A(t)x, with

" #
3 cos(4t)
2 − 12 2 − 3 sin(4t)
2
A(t) := −3 sin(4t) .
2 −2 − 3 cos(4t)
2 − 12

Show that the eigenvalues of A(t) for any given t are

√
1 7
− ± i.
2 2
So, for any fixed t, the matrix A(t) is a Hurwitz or “stable” matrix. Verify that for this
LTV system, the state transition matrix is

e cos 2t e−2t sin 2t

 t 
Φ(t, 0) = .
−et sin 2t e−2t cos 2t

You may use the property that Φ(t, 0) is the unique solution of

dΦ(t, 0)
= A(t)Φ(t, 0), Φ(0, 0) = I,
dt
where I is the identity matrix. Show that there exists an initial condition x(0) that is
arbitrarily close to the origin but the resulting solution is unbounded and hence that the
origin is an unstable equilibrium point for the system ẋ = A(t)x.
Illustrate the instability of the origin through simulations.
You may use symbolic math tools for this problem.
Solution: Verification of the eigenvalues and state
 transition matrix is straight forward.
1
The solution with the initial condition x(0) = is the first column of the state transition
0
matrix, which grows unbounded as t → ∞. Hence, the origin is an unstable equilibrium
point for the system ẋ = A(t)x.

2. Consider the equation

E − e sin(E) = M,
where e and M are positive parameters. Suggest an iterative method to solve this equation
for E. Indicate any constraints on the parameters if required.
Solution:
E = e sin(E) + M =: f (E)
is a fixed point problem. For E1 and E2 ∈ R,

|f (E1 ) − f (E2 )| = e| sin(E1 ) − sin(E2 )| ≤ e|E1 − E2 |,

as the function sin(.) is continuously differentiable on R, with 1 as a uniform upper bound

on the magnitude of its derivative. Thus, if e ∈ [0, 1), f (E) is contraction mapping on R.
hence, there is a unique fixed point E ∗ , which solves the equation and can be be found by
the fixed point iteration.

3. Consider  
−x1 + |x2 |
f (x) = .
−2x1 + x22 − x1 x2

1
 Comment on continuous differentiability, local Lipschitzness, continuity and global Lips-
chitzness of the function f . Is f Lipschitz on compact sets? If so, please provide a Lipschitz
constant for f on the set where ||x||∞ ≤ 1.
Solution: f2 (x) is continuously differentiable everywhere. f1 (x) is continuously differ-
entiable everywhere except where x2 = 0. |x2 | is locally Lipschitz everywhere, including
at x2 = 0. So, f (x) is locally Lipschitz and hence also continuous everywhere. f is not
globally Lipschitz because of the quadratic terms in f2 (x), which means the || ∂f
∂x || is not
uniformly bounded in Rn .
Since f is locally Lipschitz everywhere it is also Lipschitz on compact sets. Then, for all
x, y ∈ W := {z | ||z||∞ ≤ 1},

|x1 x2 − y1 y2 | ≤ |x1 x2 − x1 y2 + x1 y2 − y1 y2 | ≤ |x1 | |x2 − y2 | + |y2 | |x1 − y1 | ≤ 1 ||x − y||1 .

So,

||f (x) − f (y)||1 ≤ |y1 − x1 | + | |x2 | − |y2 | | + 2|x1 − y1 | + |x22 − y22 | + |x1 x2 − y1 y2 |
≤ |y1 − x1 | + |x2 − y2 | + 2|x1 − y1 | + |x2 + y2 ||x2 − y2 | + |x1 x2 − y1 y2 |
≤ 4||x − y||1 , ∀x, y ∈ W.

4. HK 3.3
Solution: f1 : R → R and f2 : R → R are locally Lipschitz at x0 ∈ R. Then ∃r > 0, L1 >
0, L2 > 0 such that,

|f1 (x) − f1 (y)| ≤ L1 |x − y| and |f2 (x) − f2 (y)| ≤ L2 |x − y| ; ∀x, y ∈ Br (x0 ).

Also, ∃M ≥ 0 such that |f1 (x)| ≤ M , |f2 (x)| ≤ M ; ∀x, y ∈ Br (x0 ). Now,

|f1 (x) + f2 (x) − f1 (y) − f2 (y)| ≤ |f1 (x) − f1 (y)| + |f2 (x) − f2 (y)|
≤ (L1 + L2 )|x − y| , ∀x, y ∈ Br (x0 ).

So, f1 + f2 is locally Lipschitz at x0 .

Next,

|f1 (x)f2 (x) − f1 (y)f2 (y)| = |f1 (x)f2 (x) − f1 (x)f2 (y) + f1 (x)f2 (y) − f1 (y)f2 (y)|
≤ |f1 (x)| |f2 (x) − f2 (y)| + |f2 (y)| |f1 (x) − f1 (y)|
≤ M (L1 + L2 )|x − y| ; ∀x, y ∈ Br (x0 ).

So, f1 f2 is locally Lipschitz at x0 . Next, suppose f2 is Lipschitz at f1 (x0 ) with a Lipschitz

constant L2 on the ball Bρ (f1 (x0 )). As f1 is continuous at x0 , ∃δ > 0 such that,

|x − x0 | < δ =⇒ |f1 (x) − f1 (x0 )| < ρ.

Thus,

|f2 (f1 (x)) − f2 (f1 (y))| ≤ L2 |f1 (x) − f1 (y)| ; ∀f1 (x), f1 (y) ∈ Bρ (f1 (x0 ))
≤ L1 L2 |x − y| ; ∀x, y ∈ Bq (x0 ) where q = min{r, δ}

So, f2 (f1 (·)) is locally Lipschitz at x0 .

2
5. HK 3.4
Solution: Given f : Rn → Rn defined as,

f1 (x),
 if g(x)∥Kx∥ ≥ µ > 0
f (x) :=

f2 (x), if g(x)∥Kx∥ < µ


g(x)
where f1 (x) := g(x)∥Kx∥ Kx and f2 (x) := g(x)
µ Kx. Due to product property, the function
n
f2 (x) is locally Lipschitz at all x ∈ R . Now,

|∥Kx∥ − ∥Ky∥| ≤ ∥Kx − Ky∥ ≤ ∥K∥∥x − y∥ =⇒ ∥Kx∥ is globally Lipschitz.

This again implies that g(x)∥Kx∥ is locally Lipschitz on Rn .
The function h(x0 ) := x10 is Lipschitz on the set {x ∈ R | x ≥ µ > 0}.

1 1 |y − z| 1
|h(z) − h(y)| = − = ≤ 2 |y − z| ; ∀y, z ∈ {x ∈ R | x ≥ µ > 0}.
z y |zy| µ

So, using the “composition” and the “product” properties, we can say that f1 is locally
Lipschitz on the set {x ∈ Rn | g(x)∥Kx∥ ≥ µ > 0}. So, we can say that f is locally
Lipschitz at all x0 ∈ Rn except for x0 such that g(x0 )∥Kx0 ∥ = µ. We deal with this case
now. Let
1

 g(x)∥Kx∥ ,
 if g(x)∥Kx∥ ≥ µ > 0
q(x) :=

1
µ, if g(x)∥Kx∥ < µ
Then, f (x) = g(x)q(x)Kx. Now, suppose g(x)∥Kx∥ ≥ µ and g(y)∥Ky∥ < µ.

1 1 |g(x)∥Kx∥ − µ|
|q(x) − q(y)| = − ≤
g(x)∥Kx∥ µ µ2
1
≤ 2 |g(x)∥Kx∥ − g(y)∥Ky∥|
µ
≤ L1 |x − y| for some L1 and for arbitrary x, y satisfying the constraints.

So, we can conclude that f is locally Lipschitz on Rn . This implies that f is Lipschitz on
any compact subset of Rn .

6. Using the comparison lemma, show that the solutions of the system
2x2 2x1
ẋ1 = −x1 + , ẋ2 = −x2 +
1 + x22 1 + x21

satisfy the inequality √

||x(t)|| ≤ e−t ||x(0)|| + 2(1 − e−t ).
Further, show that for each finite x(0) ∈ R2 , the system has a unique solution for all t ≥ 0.

Solution: Given
2x2 2x1
ẋ1 = −x1 + , ẋ2 = −x2 + .
1 + x22 1 + x21

3
Let us denote the above dynamics as ẋ = f (x) where x := [x1 x2 ]T ∈ R2 .
Consider V (x) := ∥x∥2 = xT x.
 
2x1 x2 2x1 x2
V̇ = 2x1 x˙1 + 2x2 x˙2 = 2 −x21 − x22 + +
1 + x21 1 + x22
4|x1 ||x2 | 4|x1 ||x2 |
≤ −2V + +
1 + x21 1 + x22
 
|a| 1
≤ −2V + 2|x1 | + 2|x2 | ; as ≤
1 + a2 2
√ √  √ 
≤ −2V + 2 2 V ; as ∥x∥1 ≤ 2∥x∥2 ; ∀x ∈ R2
√
Now, let W := V = ∥x∥2 . For W ̸= 0,

V̇ √
Ẇ = √ ≤ −W + 2.
2 V
For W (t) = 0,
Z t+h
|W (t + h) − W (t)| |W (t + h)| ∥x(t + h)∥2 1
= = = f (x(τ ))dτ
h h h h t 2
1 t+h
Z
= f (x(t)) + [f (x(τ )) − f (x(t))] dτ
h t 2
1 t+h
Z
≤ ∥f (0)∥2 + ∥f (x(τ )) − f (x(t))∥2 dτ.
h t

Now, f (x(t)) is a continuous function of time. So, for each ϵ > 0, ∃δ > 0 such that
∀τ ∈ (t, t + δ), ∥f (x(τ )) − f (x(t))∥2 < ϵ. So, for all h < δ,
Z t+h Z t+h
1 1
∥f (x(τ )) − f (x(t))∥2 dτ < ϵ =⇒ lim ∥f (x(τ )) − f (x(t))∥2 dτ = 0.
h t h→0+ h t
√
So, if W (t) = 0, D+ W (t) ≤ ∥f (0)∥2 = 0 ≤ −W (t) + 2. Thus ∀t, we have
√
D+ W (t) ≤ −W (t) + 2.

Applying comparison lemma,

√
∥x(t)∥2 = W (t) ≤ exp−t ∥x(0)∥2 + 2(1 − exp−t )
√
≤ ∥x(0)∥2 + 2 ; ∀t ≥ 0.

Observation 1. All solutions starting from x(0)=(a,b) are confined to a compact set
∀t ≥ 0 and for as long as they exist. •

Note that f (x) is continuously differentiable at all x ∈ R2 . So, f (x) is locally Lipschitz
everywhere. This along with observation 1 implies that there exists a unique solution and
that it exists ∀t ≥ 0.