EE6415: Nonlinear Systems Analysis Tutorial 5 solution

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1. Define T : C[0, 1] → C[0, 1] by T(x)(t) = 1 + R t
0
x(s)ds where the metric in C[0, 1] is defined as
d(f, g) = maxxı[0,1] |f(x) − g(x)|.
(a) Is T a contraction?
(b) If the space is changed to C[0,
1
2
] will T be a contraction?

2. Let f : [0, 1] → [0, 1] be given by f(x) = 1
1+x
. Answer the following questions:
(a) Is the map a contraction?
(b) Does the function f has a unique fixed point?

3. For each of the functions f(x) given, find whether f is (a) continuously differentiable (b) locally
Lipschitz (c) continuous (d) globally Lipschitz.
(a) f(x) = (
x
2
sin( 1
x
), x ̸= 0
0, x = 0
(b) f(x) = x
3
3 + |x|
(c) f(x) = 
−x1 + a|x2|
−(a + b)x1 + bx2
1 − x1x2


4. Let ∥.∥α and ∥.∥β be two different p-norms on R
n. Show that f : R
n → R
m is Lipschitz in ∥.∥α iff
it is Lipschitz in ∥.∥β

5. The following result is known as the Gronwall-Bellman inequality. Prove the result.
Let I denote an interval of the real line of the form [a, ∞) or [a, b] or [a, b), with a < b. Let β and
u be real-valued continuous functions defined on I. If u is differentiable in the interior Io of I and
satisfies the differential inequality
u˙(t) ≤ β(t)u(t), t ∈ Io
1
then, u(t) is bounded by the solution of the corresponding differential equation ˙v(t) = β(t)v(t).

u(t) ≤ u(a) exp(Z t
a
β(s)ds)
for all t ∈ I.

6. Let f(t, x) be piecewise continuous in t, locally Lipschitz in x, and
∥f(t, x)∥ ≤ k1 + k2∥x∥, ∀ (t, x) ∈ [t0, ∞) × R
n
(a) Show that the solution of
x˙ = f(t, x), x(t0) = x0
satisfies
∥x(t)∥ ≤ ∥x0∥ exp(k2(t − t0)) + k1
k2
(exp(k2(t − t0)) − 1), ∀t ≥ t0
for which the solution exists. [Hint: Use Gronwall-Bellman inequality]

(b) Can the solution have a finite escape time

7. If the system ˙x = f(t, x), x(t0) = x0 = [a, b]
T
is given by
x˙ 1 = −x1 +
2×2
1 + x
2
2
x˙ 2 = −x2 +
2×1
1 + x
2
1
show that the state equation has a unique solution defined for all t ≥ 0.

8. The following result is known as the comparison lemma. Prove the lemma.
Consider the scalar differential equation
u˙ = f(t, u), u(t0) = u0
where f(t, u) is locally Lipschitz in u, and continuous in t, for all t ≥ 0 and for all u ∈ J ⊆ R. Let
[t0, T) be the interval of existence for a solution u(t) ∈ J, for all t ∈ [t0, T). Let v(t) be a continuous
function whose upper right hand derivative is denoted by D+v(t), and satisfies,
D+v(t) ≤ f(t, v(t)), v(t0) ≤ u0
with v(t) ∈ J for all t ∈ [t0, T). Then, v(t) ≤ u(t) for all t ∈ [t0, T).

9. Using the comparison lemma, show that the solution of the state equation
x˙ 1 = −x1 +
2×2
1 + x
2
2
x˙ 2 = −x2 +
2×1
1 + x
2
1
satisfies the inequality
∥x(t)∥2 ≤ e
−t
∥x(0)∥2 +

2(1 − e
−t
)