Centre for Mathematical Sciences
Mathematics, Faculty of Science
Solutions Exercise set 1
1. (2.6 in the book)
(i) f (x) = 1/(1−x2 ) is Lipschitz continuous near 0 with Lipschitz constant L = 2a/(1−a2 )2
on the interval [−a, a], 0 < a < 1.
This can be seen using the mean value theorem: f (x) − f (y) = f 0 (ξ)(x − y) where ξ is
some number between x and y and |f 0 (ξ)| = |2ξ/(1 − ξ 2 )2 | ≤ 2a/(1 − a2 )2 for |ξ| < a.
Alternatively, one can estimate |f (x) − f (y)|/|x − y| directly.
(ii) f (x) = |x|1/2 is not Lipschitz continuous near 0. We have |x|1/2 /|x| = |x|−1/2 → ∞ as
x → 0.
(iii) f (x) = x2 sin x1 is Lipschitz continuous near 0 (although f 0 is not continuous at 0).
L = 3 is a Lipschitz constant for R (not the best). This can be seen by using the mean value
theorem. We have
(
2x sin x1 − cos x1 , x 6= 0,
0
f (x) =
0,
x = 0,
where the definition of the derivative is used to calculate f 0 (0). Note that we can apply the
mean value theorem even if f 0 is not continuous at 0. We have
|f 0 (x)| ≤ 2|x|| sin
1
x
| + | cos
1
x
| ≤ 2 + 1 = 3,
since | sin y| ≤ |y| for all y ∈ R.
2.
a) f (t, x) = t2 +x2 satisfies a Lipschitz condition in [0, 1]×[0, 1], because |f (t, x)−f (t, y)| =
|x2 − y 2 | = |x + y||x − y| ≤ 2|x − y|. One can also use the fact that f is C 1 .
b) |f (t, x)−f (t, y)| = | sin t·cos x−sin t·cos y| = | sin t|| cos x−cos y| ≤ | cos x−cos y| =
| sin ξ||x − y| ≤ |x − y| by the mean value theorem. Again one can also use that f is C 1 .
c) |f (t, x) − f (t, y)| = ||t − x| − |t − y|| ≤ |(t − x) − (t − y)| = |x − y| by the (reverse)
triangle inequality. Note that f is not C 1 near the line x = t.
d) f (t, x) = |x|α is Lipschitz for α ≥ 1. For α > 1 it is C 1 . For α = 1 one can do as in (c).
For α ∈ (0, 1) it is not Lipschitz, since |f (t, x)|/|x| = |x|α−1 → ∞ as |x| → 0.
3. (2.7 in the book)
n
X
tk
.
xn (t) =
k!
k=0
4. (2.8 in the book)
Please, turn over!
(
0,
xn (t) = 2
t ,
n even,
n odd,
n = 0, 1, 2, . . . .
The Picard iteration does not converge.
5. We take t0 = 0, T = 1 and δ = 1 in the proof of Picard-Lindelöf. Note that
f (t, x) = t2 + e−x
2
is continuously differentiable on R2 and hence locally Lipschitz continuous in x (uniformly in t).
Setting V = [0, 1] × [−1, 1], we find that M = supV |f (t, x)| = 2 and T0 = min(T, δ/M ) =
min(1, 1/2) = 1/2. Thus, the solution of the IVP is defined for 0 ≤ t ≤ 1/2. By construction,
the solution satisfies |x(t)| = |x(t) − 0| ≤ δ = 1.
Alternatively, note that |f (t, x) − f (t, y)| = |fx0 (t, ξ)||x − y| for some ξ between x and y by
2
the mean value theorem.
We have that fx0 (t, x) = −2xe−x . Since |fx0 (t, x)| is bounded for all
p
(t, x) ∈ R2 (by 2/e), we have that f (t, x) is globally Lipschitz continuous in x on R2 . Hence,
the solution is defined on the whole of R by Corollary 2.6.
6. Let
f (t, x) = sin t + ln(1 + x2 ).
We have that
2x
1 + x2
with |fx0 (t, x)| ≤ 1 (the arithmetic-geometric mean inequality). By the mean value theorem
fx0 (t, x) =
|f (t, x) − f (t, y)| = |fx0 (t, ξ)||x − y| ≤ |x − y|,
so f is Lipschitz with constant L = 1. The result follows from Corollary 2.6.
7. Let f (t, x) = sin(tx) and g(t, x) = tx. Then
1
|f (t, x) − f (t, y)| ≤ |t||x − y| ≤ |x − y|,
2
1
0≤t≤ .
2
Hence, f is Lipschitz with constant L = 1/2 in this set. Let x(t) be the solution of x0 = f (t, x),
x(0) = 0.2 and y(t) the solution of y 0 = g(t, y), y(0) = 0.2. We can solve the second equation,
2
yielding y(t) = 0.2et /2 . By the theorem on continuous dependence we know that
|x(t) − y(t)| ≤
M L|t−t0 |
(e
− 1),
L
where we can take M = sup0≤t≤1/2 |f (t, y(t)) − g(t, y(t))| (see the remark on p. 5 of the
lecture notes for lecture 2). Using the hint we have that
|f (t, y(t)) − g(t, y(t))| = | sin(t y(t)) − t y(t)| ≤
Thus we can take
M=
Hence,
|x(0.5) − y(0.5)| ≤
e3/8
e3/8
|t|3 |y(t)|3
≤ 3 3
=
.
3!
2 ·5 ·6
6000
e3/8
.
6000
e3/8 1/4
(e − 1) ≈ 10−4 .
3000
8. To prove Theorem 2.4, we repeat the proof of the contraction principle. Set xn = K n (x), where
x ∈ C is arbitrary. Then
kxn − xm k ≤
n
X
n
X
kxj − xj−1 k =
j=m+1
kK j−1 (x1 ) − K j−1 (x0 )k
j=m+1

n
X
≤

θj−1  kx1 − x0 k
j=m+1

≤
∞
X

θj  kx1 − x0 k
j=m
→0
as m → ∞, where n > m. Hence, {xn } is a Cauchy sequence and has a limit x̄. To prove the
error estimate, we simply let n → ∞ in the above inequality, giving

kx̄ − K m (x)k ≤ 
∞
X

θj  kK(x) − xk
j=m
Note that K n is a contraction for sufficiently large n (since θn → 0). Since any fixed point of K
is a fixed point of K n , this shows that x̄ is unique.
For the second part, suppose that K n (x) = x. Then K n (K(x)) = K n+1 (x) = K(K n (x)) =
K(x). Hence K(x) is a fixed point of K n and therefore K(x) = x by uniqueness.
δ
}. The Banach
9. We take C as before (defined using the sup norm), but with T0 = min{T, M
n
space is X = C([0, T0 ], R ), equipped with the norm k · kα . Note that if kxn − xkα → 0 then
kxn − xk → 0 as well, since kxk = sup0≤t≤T0 {eαt e−αt |x(t)|} ≤ eαT0 kxkα . Hence, C is a
closed subset of X. To show that K is a contraction, note that
Z
t
Z
t
|f (s, x(s)) − f (s, y(s))| ds ≤ L
0
|x(s) − y(s)| ds
0
Z
t
eαs e−αs |x(s) − y(s)| ds
0
Z t
≤ Lkx − ykα
eαs ds
=L
0
L
= (eαt − 1)kx − ykα .
α
It follows that
e−αt |K(x)(t) − K(y)(t)| ≤
L
(1 − e−αt )kx − ykα
α
and hence
kK(x) − K(y)kα ≤ θkx − ykα ,
with
θ=
L
< 1.
α
Please, turn over!
10. We find that
R0 (t) = 2(x(t) − y(t)) · (f (t, x(t)) − f (t, y(t)))
≤ 2C|x(t) − y(t)|ρ(|x(t) − y(t)|)
p
p
≤ 2C R(t)ρ( R(t)).
If ρ(r) vanishes in some interval [0, r0 ), r0 > 0, we find from this that R0 ≤ 0 when R is
sufficiently small, showing that R(t) ≡ 0. Assume therefore that ρ(r) > 0 for r > 0 in what
follows. Then
Z t3
R0 (t)
p
p
dt ≤ 2C(t3 − t1) < ∞,
t1 2 R(t)ρ( R(t))
where t1 < t3 < t2 . But on the other hand,
p
"
# Z
Z t3
R(t3 )
s = R(t)
R0 (t)
1
p
p
dt =
=
ds = ∞.
R0 (t)
√
ds
=
dt
ρ(s)
0
t1 2 R(t)ρ( R(t))
2 R(t)
Hence, R(t) ≡ 0.