M2AA1 Diffferential Equations: Problem Sheet 1
Question 1
It is a straightforward calculation with the quotient rule for derivatives, to show
that the given solution actually satisfies x0 = x(1 − x). Since x(1 − x) is bounded
and Lipschitz on any bounded set U, one can use the uniqueness part of Picard’s
Theorem to say that solutions are unique: first, locally, but the argument can be
repeated at the end points of any interval to continue solutions uniquely to larger
intervals.
Showing that φt+s (x) = φs (φt (x)), where φt (x) = x/(x + (1 − x)e−t ), is a simple
calculation.
Question 2 Part (a) follows by substitution: note that both solutions are differentiable. Similarly, for each T
0
for t ≤ T
x(t) =
3
(t − T ) /27 for t ≥ T
is a solution (note that the right hand side is differentiable). So we have infinitely
many solutions. More generally, if T1 ≤ 0 ≤ T2 then

 (1/27)(t − T1 )3 if t < T1 ;
0
if t ∈ [T1 , T2 ];
x(t) =

3
(1/27)(t − T2 )
if t > T2 .
is a solution. How did we get this solution? Let us consider the case x > 0 first. Of
course, provided x 6= 0 one can apply separation of variables, i.e. write (1/x1/3 )dx =
dt which gives 3x(1/3) = t + c0 and so x(t) = (1/27)(t + c0 )3 with (t + c0 ) > 0. Taking
c0 = −T2 gives x(t) = (1/27)(t − T2 )3 when t > T2 .
Similarly, if we consider x < 0 then (1/(−x)1/3 )dx = dt which gives −3(−x)(1/3) =
t + c1 and so x(t) = (1/27)(t + c1 )3 with (t + c1 ) < 0. Taking c1 = −T1 , gives
(1/27)(t − T1 )3 when t < T1 .
When x = 0 then x(t) ≡ 0 is a solution for all t ∈ [A, B] for any choice of A < B.
Patching these solutions gives the solution.
Note that x(T2 ) = 0 and x(t) > 0 for t > T2 . So the solution ‘jumps’ from
somewhere where ẋ = 0 to somewhere where ẋ > 0 in finite time. In Question 4 we
show this is impossible when the right hand side is Lipschitz. In the current situation
the right hand side is non-Lipschitz: |y 3/2 − x2/3 |/|y − x| is not bounded (take for
example x = 0 and let y ↓ 0).
1
The non-uniqueness of the solution does not contradict Picard’s Theorem, since
the Lipschitz assumption is not satisfied at 0 : |x(2/3) − 0|/|x − 0| = |1/x(1/3) | → ∞
as x → 0.
By the way: for any α ∈ (0, 1) the following initial value problem ẋ = |x|α , x(0) =
0 has infinitely many solutions. This is of course related to the fact that 1/|x|α is
not integrable at 0, and therefore the solution for the initial value problem ẋ =
|x|α , x(0) = x0 with x0 6= 0 hits the line x = 0 in finite time; compare question 6
below.
Question 3 Consider ẋ = (1 + x2 ) with x(0) = x0 . The solution is x(t) =
tan(t+c0 ), as can be seen from applying the quotient rule to sin(t+c0 )/ cos(t+c0 ), and
in order to satisfy x(0) = x0 , one can choose c0 = arctan x0 . Note that tan(t) → ±∞
for t → ±π/2, the solutions go to ±∞ in finite time (as t → ±π/2 − arctan(x0 )).
Question 4
Consider ẋ = f (x) on R, with f Lipschitz continuous on [a, b], f (x) > 0 on (a, b),
f (a) = f (b) = 0. Let x(t) be the solution with initial value x(0) = x0 ∈ (a, b). Then
R x(t) 1
ds = t. t 7→ x(t) is strictly increasing because x0 (t) = f (x(t)) > 0 on (a, b).
f (s)
x0
R x(t) 1
One has x0 f (s)
ds = t. Since t 7→ x(t) is monotone, denote x 7→ x(t) its inverse.
Rx 1
Then one obtains x0 f (s)
ds = t(x). Note that f is Lipschitz so |f (x)| = |f (x) −
R b− 1
R b− 1
f (b)| < K|x − b|. So t(b − ) = x0 f (s)
ds ≥ K1 x0 |x−b|
ds = log(b − x)|b−
x0 → ∞
as ↓ 0. So the time it takes to travel from x0 to b − tends to infinity as ↓ 0.
Similarly, near a. So a < x(t) < b for each t ∈ R. It is impossible to reach a zero of f
in finite time! Here you see the difference with the situation considered in Question
4. There solutions can reach a zero of f in finite time.
Question 5
a) x0 ≡ 1, R
t
x1 (t) = 1 + 0 f (x0 (s)) ds = 1 + t
Rt
Rt
x2 (t) = 1 + 0 f (x1 (s)) ds = 1 + 0 (1 + s)2 ds = 1 + (1/3)(1 + t)3 − 1/3 =
1 + t + t2 + tR3 /3.
Rt
t
x3 = 1 + 0 f (x2 (s)) ds = 1 + 0 [1 + s + s2 + s3 /3]2 ds =
4
5
6
7
= 1 + t + t2 + t3+ 2/3t
+ 1/3t + 1/9t
+ 1/63t
.
x
0 1
0
0
b) Let p = x and z =
. Then ż =
z and z0 =
.
p
−1
0
1
0 1
Write A =
. Then A2 = −I, A3 = −A and A4 = I. So z1 =
−1 0
Rt
0
1 t
0
t
z0 + 0 Az0 (s) ds = (I + tA)
=
=
, z2 = z0 +
1
−t 1
1
1
2
Rt
0
t
ds
=
+
.
z
=
z
+
Az2 (s) =
Az
(s)
ds
=
z
+
3
0
1
0
2
0
0
0
1 −t /2
R t 1 − s2 /2
t − t3 /6
0
ds =
It follows that x0 = 0, x1 = t, x2 = t,
+ 0
−s
1 − t2 /2
1
x3 = t − t3 /6. (If one keeps going one gets more and more terms from the Taylor
expansion of sin(t).)
Rt
Rt
1
−s
Question 6
f (x) = x + 1/x. Since f 0 (x) = 1 − 1/x2 ∈ (0, 1) when x ∈ [1, ∞) one has
|f (y) − f (x)| < |y − x| by the Mean Value Theorem. Of course there are no solutions
to the equation f (p) = p.
So it is NOT enough to have |f (y)−f (x)| < |y −x|. In order to apply the Banach
fixed point theorem one needs that there exists λ ∈ (0, 1) so that |f (y) − f (x)| ≤
λ|y − x| for all x, y.
WRONG answer to why |f (y)−f (x)| < |y−x| is the following: |x+1/x−y−1/y| ≤
|x − y| + |1/x − 1/y| < |x − y|.
3