1

Additional Exercises for Chapter 4

1. For each of the following systems, use a quadratic Lyapunov function candidate to show that the origin
is asymptotically stable. Then, investigate whether the origin is globally asymptotically stable.
(1) ẋ1 = −x1 + x22 , ẋ2 = −x2
(2) ẋ1 = (x1 − x2 )(x21 + x22 − 1), ẋ2 = (x1 + x2 )(x21 + x22 − 1)
(3) ẋ1 = −x1 + x21 x2 , ẋ2 = −x2 + x1
2. Using V (x) = x21 + x22 , study stability of the origin of the system
ẋ1 = x1 (k 2 − x21 − x22 ) + x2 (x21 + x22 + k 2 ), ẋ2 = −x1 (k 2 + x21 + x22 ) + x2 (k 2 − x21 − x22 )
when (a) k = 0 and (b) k 6= 0.
3. Using the variable gradient method, find a Lyapunov function V (x) that shows asymptotic stability of
the origin of the system
ẋ1 = x2 , ẋ2 = −(x1 + x2 ) − sin(x1 + x2 )
4. Consider the system
ẋ1 = x2 , ẋ2 = x1 − sat(2x1 + x2 )
Show that the origin is asymptotically stable, but not globally asymptotically stable.
5. Show that the origin of the following system is unstable.
ẋ1 = −x1 + x62 , ẋ2 = x32 + x61

6. Consider the system

m
X
ż = − ai yi , ẏi = −h(z, y)yi + bi g(z), i = 1, 2, . . . , m
i=1

where z is a scalar, y T = (y1 , . . . , ym ). The functions h(·, ·) and g(·) are continuously
Rz differentiable for
all (z, y) and satisfy zg(z) > 0, ∀ z 6= 0, h(z, y) > 0, ∀ (z, y) 6= 0, and 0 g(σ) dσ → ∞ as |z| → ∞.
The constants ai and bi satisfy bi 6= 0 and ai /bi > 0, ∀ i = 1, 2, . . . , m. Show that the origin is an
equilibrium point, and investigate its stability using a Lyapunov function candidate of the form
Z z Xm
V (z, y) = α g(σ) dσ + βi yi2
0 i=1

7. Consider the system

ẋ1 = x2 , ẋ2 = −x1 − x2 sat(x22 − x23 ), ẋ3 = x3 sat(x22 − x23 )
where sat(·) is the saturation function. Show that the origin is the unique equilibrium point, and use
V (x) = xT x to show that it is globally asymptotically stable.
8. The origin x = 0 is an equilibrium point of the system
ẋ1 = −kh(x)x1 + x2 , ẋ2 = −h(x)x2 − x31
Let D = {x ∈ R2 | kxk2 < 1}. Using V (x) = 14 x41 + 12 x22 , investigate stability of the origin in each of
the following cases.
(1) k > 0, h(x) > 0, ∀ x ∈ D; (2) k > 0, h(x) > 0, ∀ x ∈ R2 ;
(3) k > 0, h(x) < 0, ∀ x ∈ D; (4) k > 0, h(x) = 0, ∀ x ∈ D;
(5) k = 0, h(x) > 0, ∀ x ∈ D; (6) k = 0, h(x) > 0, ∀ x ∈ R2 .
2

9. Consider the system

ẋ1 = −x1 + g(x3 ), ẋ2 = −g(x3 ), ẋ3 = −ax1 + bx2 − cg(x3 )
where a, b, and c are positive constants and g(·) is a locally Lipschitz function that satisfies
g(0) = 0 and yg(y) > 0, ∀ 0 < |y| < k, k > 0
(a) Show that the origin is an isolated equilibrium point.
Rx
(b) With V (x) = 12 ax21 + 21 bx22 + 0 3 g(y) dy as a Lyapunov function candidate, show that the origin
is asymptotically stable.
(c) Suppose yg(y) > 0 ∀ y 6= 0. Is the origin globally asymptotically stable?
10. Consider the system
ẋ1 = x2 ,ẋ2 = −a sin x1 − kx1 − dx2 − cx3 , ẋ3 = −x3 + x2
Rx
where all coefficients are positive and k > a. Using V (x) = 2a 0 1 sin y dy + kx21 + x22 + px23 with some
p > 0, show that the origin is globally asymptotically stable.
11. Show that the system
1
ẋ1 = − x1 , ẋ2 = x1 − 2x2 , ẋ3 = x2 − 3x3
1 + x3
has a unique equilibrium point in the region xi ≥ 0, i = 1, 2, 3, and investigate stability of this point
using linearization.
12. For each of the following systems, use linearization to show that the origin is asymptotically stable.
Then, show that the origin is globally asymptotically stable.
ẋ1 = −x1 + x2 ẋ1 = −x31 + x2
(1) (2)
ẋ2 = (x1 + x2 ) sin x1 − 3x2 ẋ2 = −ax1 − bx2 , a, b > 0

13. Consider the system

ẋ1 = −x31 + α(t)x2 , ẋ2 = −α(t)x1 − x32
where α(t) is a continuous, bounded function. Show that the origin is globally uniformly asymptotically
stable. Is it exponentially stable?
14. Consider the system
ẋ1 = x2 , ẋ2 = −x1 − (1 + b cos t)x2
Find b > 0 such that the origin is exponentially stable for all |b| < b∗ .
∗

15. Consider the system

ẋ1 = x2 − g(t)x1 (x21 + x22 ), ẋ2 = −x1 − g(t)x2 (x21 + x22 )
where g(t) is continuously differentiable, bounded, and g(t) ≥ k > 0 for all t ≥ 0. Is the origin
uniformly asymptotically stable? Is it exponentially stable?
16. Consider two systems represented by
ẋ = f (x) (1)
ẋ = h(x)f (x) (2)
n n n
where f : R → R and h : R → R are continuously differentiable, f (0) = 0, and h(0) > 0. Show
that the origin of (1) is exponentially stable if and only if the origin of (2) is exponentially stable.
17. Investigate input-to-state stability of the system
ẋ1 = (x1 − x2 + u)(x21 + x22 − 1), ẋ2 = (x1 + x2 + u)(x21 + x22 − 1)
 1

Solution of Additional Exercises for Chapter 4

1. (1) Try V (x) = 12 (x21 + x22 ).

V̇ (x) = x1 (−x1 + x22 ) − x22 = −x21 − x22 + x1 x22

In the neighborhood of the origin, the term −(x21 + x22 ) dominates. Hence, the origin is asymptotically
stable. Moreover
Z t
£ ¤
x2 (t) = e−t x20 ⇒ x1 (t) = e−t x10 + e−(t−s) e−2s ds x220 = e−t x10 + e−t − e−2t x220
0

For all x0 , x(t) → 0 as t → ∞, which implies that the origin is globally asymptotically stable.
(2) Try V (x) = ax21 + bx22 , a > 0, b > 0.

V̇ (x) = 2ax1 (x1 −x2 )(x21 +x22 −1)+2bx2 (x1 +x2 )(x21 +x22 −1) = 2[ax1 (x1 −x2 )+bx2 (x1 +x2 )](x21 +x22 −1)

Let a = b.
V̇ (x) = −2a(x21 + x22 )[1 − (x21 + x22 )]
For x21 + x22 < 1, V̇ (x) is negative definite. Hence, the origin is asymptotically stable. It is not globally
asymptotically stable since there other equilibrium points on the unit circle.
(3) Try V (x) = 12 (ax21 + bx22 ), a > 0, b > 0.

V̇ (x) = ax1 (−x1 + x21 x2 ) + bx2 (−x2 + x1 ) = −ax21 + bx1 x2 − bx22 + ax31 x2 = −xT Qx + ax31 x2
· ¸
a −0.5b
where Q = . The matrix Q is positive definite when ab − b2 /4 > 0. Choose
−0.5b b
b = a = 1. Near the origin, the quadratic term −xT Qx dominates the fourth-order term x31 x2 . Thus,
V̇ (x) is negative definite and the origin is asymptotically stable. It is not globally asymptotically stable
since there are other equilibrium points at (1, 1) and (−1, −1).
2. Consider V (x) = x21 + x22 as a Lyapunov function candidate.

V̇ (x) = 2x21 (k 2 − x21 − x22 ) + 2x1 x2 (x21 + x22 + k 2 ) − 2x1 x2 (x21 + x22 + k 2 ) + 2x22 (k 2 − x21 − x22 )
= 2(x21 + x22 )(k 2 − x21 − x22 )

(a)
k = 0 ⇒ V̇ (x) = −2(x21 + x22 )2
The origin is globally asymptotically stable.
(b)
k 6= 0 ⇒ V̇ (x) > 0, for 0 < x21 + x22 < k 2
By Chetaev’s theorem, the origin is unstable.
· ¸
αx1 + βx2
3. Try g(x) = . To meet the symmetry requirement, take γ = β.
γx1 + δx2

V̇ (x) = (αx1 + βx2 )x2 − (βx1 + δx2 )[(x1 + x2 ) + sin(x1 + x2 )]

Take δ = β.
V̇ (x) = −βx21 + (α − 2β)x1 x2 − β(x1 + x2 ) sin(x1 + x2 )
Taking α = 2β and β > 0 yields

V̇ (x) = −βx21 − β(x1 + x2 ) sin(x1 + x2 )

2

which is negative definite in the set {|x1 + x2 | < π}. Now

· ¸ Z x
2 1 def 1
g(x) = β x = P x ⇒ V (x) = g T (y) dy = xT P x
1 1 0 2

where P is positive definite.

4. For |2x1 + x2 | ≤ 1, we have · ¸
0 1
ẋ = x
−1 −1
This matrix is Hurwitz. Hence, the origin is asymptotically stable. It is not globally asymptotically
stable because there is another equilibrium point at (1, 0).
5. Take V (x) = − 16 x61 + 41 x42 .

V̇ (x) = −x51 ẋ1 + x32 ẋ2 = x61 + x62 − x51 x62 + x32 x61

Near the origin ¯ 5 6 ¯ ¡ 6 ¢

¯−x1 x2 + x32 x61 ¯ ≤ 1
x1 + x62
2

Hence ¡ ¢¡ ¢
V̇ (x) ≥ 1 − 12 x61 + x62
which shows that V̇ (x) is positive definite. Application of Chetaev’s theorem shows that the origin is
unstable.
6. Since zg(z) > 0 for z 6= 0, we have g(z) P
> 0 for z > 0 and g(z) < 0 for z < 0. Since g(z) is continuous,
g(0) = 0. At the origin (z = 0, y = 0), ai yi = 0, h(·, ·)yi = 0, and g(0) = 0. Hence, the origin is an
equilibrium point. Consider the Lyapunov function candidate
Z z m
X
V (z, y) = α g(s) ds + βi yi2 , α > 0, βi > 0
0 i=1

V (z, y) is positive definite and radially unbounded.

m
X m
X m
X
V̇ = αg(z)ż + 2 βi yi ẏi = (−αai + 2βi bi )yi g(z) − 2 βi h(z, y)yi2
i=1 i=1 i=1

Choose βi = αai /2bi , to obtain

m
X
V̇ = −2 βi h(z, y)yi2 ≤ 0, ∀ (z, y)
i=1

Thus V̇ is negative semidefinite.

V̇ = 0 ⇒ yi (t) ≡ 0, ∀ i ⇒ ẏi (t) ≡ 0, ∀ i ⇒ g(z(t)) ≡ 0 ⇒ z(t) ≡ 0

By LaSalle’s theorem (Corollary 4.2), the origin is globally asymptotically stable.

7.

0 = x2
0 = −x1 − x2 sat(x22 − x23 ) ⇒ x1 = 0
0 = x3 sat(x22 − x23 ) ⇒ x3 sat(−x23 ) = 0 ⇒ x3 = 0
 3

Hence, the origin is the unique equilibrium point. consider

V (x) = xT x = x21 + x22 + x23

V̇ (x) = 2[x1 x2 − x1 x2 − x22 sat(x22 − x23 ) + x23 sat(x22 − x23 )] = −2(x22 − x23 )sat(x22 − x23 ) ≤ 0

V̇ is negative semidefinite.
V̇ = 0 ⇒ x22 (t) ≡ x23 (t) ⇒ ẋ3 (t) ≡ 0
Hence, both x2 (t) and x3 (t) must be constants. This implies that ẋ2 (t) ≡ 0. From the second
state equation we conclude that x1 (t) ≡ 0. Then, the first state equation implies that x2 (t) ≡ 0.
Consequently, x3 (t) ≡ 0. By LaSalle’s theorem (Corollary 4.2), the origin is globally asymptotically
stable.
8. V (x) = 14 x41 + 12 x22 is positive definite and radially unbounded.

V̇ (x) = x31 [−kh(x)x1 + x2 ] + x2 [−h(x)x2 − x31 ] = −kx41 h(x) − x22 h(x)

(1) k > 0, h(x) > 0 ∀ x ∈ D. In this case V̇ (x) is negative definite. Hence, the origin is asymptotically
stable.
(2) k > 0, h(x) > 0 ∀ x ∈ R2 . In this case V̇ (x) < 0, ∀ x 6= 0. Hence, the origin is globally
asymptotically stable.
(3) k > 0, h(x) < 0 ∀ x ∈ D. In this case V̇ (x) is positive definite. Hence, by Chetaev’s theorem, the
origin is unstable.
(4) k > 0, h(x) = 0 ∀ x ∈ D. In this case V̇ (x) = 0. Hence, the origin is stable. It is not asymptotically
stable because trajectories starting on the level surface V (x) = c remain on the surface for all future
time.
(5) k = 0, h(x) > 0 ∀ x ∈ D. In this case V̇ (x) = −x22 h(x) ≤ 0, ∀ x ∈ D. Moreover

V̇ (x) = 0 ⇒ x2 (t) ≡ 0 ⇒ ẋ2 (t) ≡ 0 ⇒ x1 (t) ≡ 0

Hence, by LaSalle’s theorem (Corollary 4.1), the origin is asymptotically stable.

(6) k = 0, h(x) > 0 ∀ x ∈ R2 . The same as part (5), except that all the conditions hold globally.
Hence, the origin is globally asymptotically stable.
9. (a)
0 = −x1 + g(x3 ), 0 = −g(x3 ), 0 = −ax1 + bx2 − cx3
From the properties of g(·) we know that g(x3 ) = 0 has an isolated solution x3 = 0. Substituting
x3 = 0 in the foregoing equations we obtain x1 = x2 = 0. Hence, the origin is an isolated equilibrium
point.
(b)
Z x3
a 2 b 2
V (x) = x1 + x2 + g(y) dy
2 2 0
V̇ (x) = ax1 [−x1 + g(x3 )] − bx2 g(x3 ) + g(x3 )[−ax1 + bx2 − cg(x3 )] = −ax21 − cg 2 (x3 ) ≤ 0

V̇ (x) = 0 ⇒ x1 (t) ≡ 0 and x3 (t) ≡ 0 ⇒ ẋ3 (t) ≡ 0

From the third state equation we see that x2 (t) ≡ 0. Hence, by LaSalle’s theorem (Corollary 4.1), the
origin is asymptotically stable.
(c) To conclude that the origin is globally asymptotically stable, we need to know that V (x) is radially
unbounded. But this is not guaranteed since
Z x
yg(y) > 0, ∀ |y| 6= 0 6⇒ g(y) dy → ∞ as |x| → ∞.
0
4

Consider, for example, g(y) = (1 − e−|y| )e−|y| sgn(y). For x > 0, we have
Z x
(1 − e−y )e−y dy = 1 − e−x − 12 (1 − e−2x ) → 12 as x → ∞
0

Thus we cannot conclude that the origin is globally asymptotically stable.

10.

V (x) = 2a(1 − cos x1 ) + kx21 + x22 + px23 ≥ kx21 + x22 + px23

Hence, V is positive definite and radially unbounded.

V̇ (x) = 2(−dx22 − cx2 x3 − px23 + px2 x3 )

Taking p = c, we obtain
V̇ = −2dx22 − 2px23 ≤ 0, ∀ x ∈ R3
V̇ ≡ 0 ⇒ x2 (t) ≡ 0 & x3 (t) ≡ 0 ⇒ a sin x1 (t) + kx1 (t) ≡ 0
Since k > a,
a sin x1 (t) + kx1 (t) ≡ 0 ⇒ x1 (t) ≡ 0
Using LaSalle’s theorem (Corollary 4.2), we conclude that the origin is globally asymptotically stable.

11.
1
0 = − x1
1 + x3
0 = x1 − 2x2 ⇒ x1 = 2x2
0 = x2 − 3x3 ⇒ x3 = 13 x2

Substitution of x1 and x3 in the first equation yields

√
−3 ± 15
2x22 + 6x2 − 3 = 0 ⇒ x2 =
2
Thus there is only one equilibrium point in the region xi ≥ 0; namely,
√ √
√ 15 − 3 15 − 3
x1 = 15 − 3, x2 = , x3 =
2 6
 ¯  
−1 0 −1 ¯ −1 0 −0.7621
∂f (1+x3 )2 ¯
=  1 −2 0 ¯ =  1 −2 0 
∂x ¯
0 1 −3 ¯ √ 0 1 −3
x = 15−3
3 6

The eigenvalues are −1.3671 ± j 0.449 and −3.2657. Hence, the equilibrium point is asymptotically
stable.

12. (1)

0 = −x1 + x2 ⇒ x1 = x2
0 = (x1 + x2 ) sin x1 − 3x2

Thus
x1 (2 sin x1 − 3) = 0 ⇒ x1 = 0 ⇒ x2 = 0
 5

Hence, the origin is the unique equilibrium point.

· ¸ ¯ · ¸
∂f −1 1 ∂f ¯¯ −1 1
= ; A= =
∂x sin x1 + (x1 + x2 ) cos x1 sin x1 − 3 ∂x ¯x=0 0 −3

A is Hurwitz; hence, the origin is asymptotically stable. To show global asymptotic stability, let
V (x) = 12 (x21 + x22 ).

V̇ (x) = −x21 + x1 x2 (1 + sin x1 ) − (3 − sin x1 )x22 ≤ −x21 + 2|x1 | |x2 | − 2x22

· ¸T · ¸ · ¸
|x1 | 1 −1 |x1 |
= − < 0, ∀ x 6= 0
|x2 | −1 2 |x2 |

Hence, the origin is globally asymptotically stable.

(2)

0 = −x31 + x2
b
0 = −ax1 − bx2 ⇒ x2 = − x1 ⇒ −x1 (x21 + a/b) = 0 ⇒ x1 = 0
a
Hence, the origin is the unique equilibrium point.
· ¸ ¯ · ¸
∂f −3x21 1 ∂f ¯¯ 0 1
= ; A= =
∂x −a −b ∂x ¯x=0 −a −b

A is Hurwitz; hence, the origin is asymptotically stable. To show global asymptotic stability, let
V (x) = 12 (x21 + αx22 ), α > 0.
V̇ (x) = −x41 + x1 x2 (1 − aα) − bαx22
Taking α = 1/a, we obtain
b
V̇ (x) = −x41 − x22 < 0, ∀ x 6= 0
a
Hence, the origin is globally asymptotically stable.

13. Take V (x) = 12 (x21 + x22 ).

V̇ = x1 [−x31 + α(t)x2 ] + x2 [−α(t)x1 − x32 ] = −x41 − x42

Hence, the origin is globally uniformly asymptotically stable. Linearization at the origin yields
· ¸
0 α(t)
A(t) =
−α(t) 0

The origin of the linear system is not exponentially stable since, with V (x) = 21 (x21 + x22 ), we have
V̇ = 0 which implies that V (x(t)) is constant along the solution. Thus, x(t) does not converge to the
origin as t tends to infinity. From Theorem 4.15, we conclude that the origin of the nonlinear system
is not exponentially stable.

14. View this system as a perturbation of the autonomous system

ẋ1 = x2 , ẋ2 = −x1 − x2

The matrix A of the autonomous system is Hurwitz. Find a Lyapunov function V (x) = xT P·x for the
¸
T 1 3 1
autonomous system by solving the Lyapunov equation P A + A P = −I, to obtain P = 2 .
1 2
6

Now use V (x) = xT P x = 32 x21 + x1 x2 + x22 as a Lyapunov function candidate for the perturbed system
with b 6= 0.

V̇ = −x21 − x22 − b cos t x1 x2 − 2b cos t x22 ≤ −x21 − x22 + |b| |x1 | |x2 | + 2|b| |x2 |2
· ¸T · ¸ · ¸
|x1 | 1 −|b|/2 |x1 |
= −
|x2 | −|b|/2 1 − 2|b| |x2 |

The right hand side is negative definite if (1 − 2|b| − b2 /4) > 0, which is the case if |b| < 0.472. Taking
b∗ = 0.472, we conclude that the origin is exponentially stable for all |b| < b∗ .
15. Try V (x) = 12 (x21 + x22 ).

V̇ = x1 x2 − g(t)x21 (x21 + x22 ) − x1 x2 − g(t)x22 (x21 + x22 ) = −g(t)(x21 + x22 )2 ≤ −4kV 2 (x)

Hence, V̇ (t, x) is negative definite and the origin is uniformly asymptotically stable. From the inequality
V̇ ≤ −4kV 2 , we cannot conclude exponential stability. Let us try linearization.
¯ · ¸¯ · ¸
∂f ¯ −3g(t)x21 − g(t)x22 1 − 2g(t)x1 x2 ¯ 0 1
A(t) = (t, x)¯¯ = ¯ =
∂x x=0
−1 − 2g(t)x1 x2 −3g(t)x22 − g(t)x21 ¯x=0 −1 0

A is a constant matrix that is not Hurwitz. Hence, by Theorem 4.15, we conclude that the origin is
not exponentially stable.
∂f
16. Let A1 = ∂x (0) be the linearization of (1). To find the linearization of (2), set g(x) = h(x)f (x). Then

∂gi ∂fi ∂h
= h(x) + fi (x)
∂xj ∂xj ∂xj

Hence
∂gi ∂fi ∂h ∂fi
(0) = h(0) (0) + (0)fi (0) = h(0) (0)
∂xj ∂xj ∂xj ∂xj
Hence
∂g
A2 =
(0) = h(0)A1
∂x
Since h(0) > 0, A1 is Hurwitz if and only if A2 is Hurwitz. By Theorem 4.15

(1) is exp. stable ⇔ A1 is Hurwitz ⇔ A2 is Hurwitz ⇔ (2) is exp. stable

17. It is not input-to-state stable because with u = 0 the origin is not globally asymptotically stable (notice
that the unforced system has equilibrium points on the unit circle).