1. Introduction
Let f be a continuous function from Rn+1 to Rn , let x0 ∈ Rn and consider the
initial value problem
(1)
(2)
dx
(t) = f (t, x(t)),
dt
x(t0 ) = x0 .
We here wish to prove (given suitable assumptions concerning f ), that there is a
unique solution. In order to achieve this goal, we shall construct a sequence of approximations which we shall show converges to a solution. This is a situation which
arises naturally in many different contexts in analysis; in order to solve an equation
(or, e.g., define the integral of a function), we construct a sequence of approximations, and we wish to show that the sequence of approximations converges. In most
of these contexts, the underlying fact which makes it possible to prove covergence
is the following property of the real numbers.
Theorem 1. Let xi , i = 1, 2, . . . , be a sequence of real numbers such that for every
> 0, there is a positive integer N with the property that |xi − xj | ≤ for all
i, j ≥ N . Then there is a real number x such that
lim xi = x.
i→∞
A sequence with the properties assumed in Theorem 1 is called a Cauchy sequence,
and the theorem can be rephrased as saying that every Cauchy sequence of real
numbers converges. This property distinguishes the real numbers from the rational
numbers, and we shall take it for granted here. Those interested in a definition of
the real numbers (as well as a description of their properties) are referred to Walter
P. Rudin’s Principles of Mathematical Analysis.
2. Properties of continuous functions
In order to construct solutions to (1) and (2), we need to record some properties of
continuous functions. Let us begin with a definition.
Definition 1. Let K = [a, b], where a, b ∈ R, let n be a positive integer and let
x : K → Rn be a continuous function. Then the norm of x is defined by
kxk = max |x(t)|
t∈K
Remarks. The notation kxk does not give a reference to the interval K. Which
interval to use is supposed to be clear from the context. Recall that x : K → Rn
is continuous at t ∈ K if and only if the following statement holds: for every
> 0 there is a δ > 0 such that if s ∈ K and |t − s| < δ, then |x(t) − x(s)| ≤ .
Furthermore, x : K → Rn is said to be continuous if and only if x is continuous
at every t ∈ K. Finally, note that if x : K → Rn is continuous, then |x| is also
continuous. Thus |x| attains its maximum on the (compact) interval K, so that
kxk is well defined.
Let us now establish that continuous functions on compact intervals have a property
similar to that of the real numbers.
1
2
Lemma 1. Let K = [a, b], where a, b ∈ R, and let n be a positive integer. Let,
furthermore, xi : K → Rn , i = 1, 2, . . . , be a sequence of continuous functions such
that for every > 0 there is a positive integer N with the property that kxi −xj k ≤ for all i, j ≥ N . Then there is a continuous function x : K → Rn such that
lim kx − xi k = 0.
(3)
i→∞
Proof. To begin with, we need to define x(t) for t ∈ K. Fix, to this end, t ∈ K
and consider the sequence xi (t) of real numbers. Let > 0. Then, by assumption,
there is a positive integer N such that kxi − xj k ≤ for all i, j ≥ N . Thus, for all
i, j ≥ N , we have
|xi (t) − xj (t)| ≤ kxi − xj k ≤ (where the first inequality is an immediate consequence of the definition of kxi −xj k
and the second inequality is an immediate consequence of the choice of N ). Thus
xi (t), i = 1, 2, . . . , is a Cauchy sequence of real numbers. Due to Theorem 1, we
conclude that there is a real number, which we shall denote by x(t), such that
lim xi (t) = x(t).
i→∞
We thus have a function x : K → Rn . However, we do not know that it is continuous, nor do we know that (3) holds.
In order to prove that x is continuous at a given t ∈ K, let > 0 and N be large
enough that i, j ≥ N implies that kxi − xj k ≤ /3 (such an N exists due to the
assumptions). Let, furthermore, δ > 0 be small enough that if |s−t| < δ and s ∈ K,
then |xN (s) − xN (t)| ≤ /3 (that there is such a δ is an immediate consequence of
the fact that xN is a continuous function). Then, for |s − t| < δ, s ∈ K, we have
|x(t) − x(s)| ≤|x(t) − xN (t)| + |xN (t) − xN (s)| + |xN (s) − x(s)|
= lim |xi (t) − xN (t)| + |xN (t) − xN (s)| + lim |xN (s) − xi (s)|.
i→∞
i→∞
However,
|xi (t) − xN (t)| ≤ kxi − xN k ≤ /3
for i ≥ N , and the same estimate holds concerning |xi (s) − xN (s)|. Finally, δ is by
definition such that |xN (t) − xN (s)| ≤ /3. Adding up, we thus conclude that
|x(s) − x(t)| ≤ .
What we have proven is thus that if t ∈ K and > 0, then there is a δ > 0 such
that if |s − t| < δ, s ∈ K, then |x(t) − x(s)| ≤ . This is the definition of the concept
of continuity, and we conclude that x : K → Rn is continuous.
In order to prove that (3) holds, let > 0 and choose a positive integer N such that
kxi − xj k ≤ for all i, j ≥ N . If t ∈ K and i ≥ N , we then have
|xi (t) − x(t)| = lim |xi (t) − xj (t)| ≤ .
j→∞
Since the right hand side is independent of t, we conclude that
kxi − xk ≤ for all i ≥ N . In other words, for every > 0, there is an N such that kxi − xk ≤ for all i ≥ N . This statement is equivalent to (3).
3
3. Basic analysis inequalities
Before turning to the proof of existence and uniqueness of solutions, let us recall
some basic inequalities. To begin with, we shall need the Schwarz inequality:
|z · w| ≤ |z||w|,
for z, w ∈ Rn . If z = (z1 , . . . , zn ) and w = (w1 , . . . , wn ), this inequality can also be
written
!1/2 n
!1/2
n
n
X
X
X
2
2
zi wi ≤
zi
wi
.
i=1
i=1
i=1
Moreover, we shall need the following observation.
Lemma 2. Let K = [a, b], where a, b ∈ R, let n be a positive integer and let
φ : K → Rn be a continuous function. Then
Z
Z
b
b
φ(t)dt ≤
|φ(t)|dt.
a
a
Proof. Let the components of φ be denoted by φi , i = 1, . . . , n. In other words,
φ(t) = (φ1 (t), . . . , φn (t)).
Let y ∈ Rn be given by
b
Z
y=
φ(t)dt,
a
so that y = (y1 , . . . , yn ), where
Z
b
yi =
φi (t)dt
a
for i = 1, . . . , n. Then
|y|2 = y · y = y ·
Z
b
φ(t)dt =
a
n
X
i=1
Z
yi
b
Z
φi (t)dt =
a
n
bX
yi φi (t)dt.
a i=1
However, due to the Schwartz inequality, we have
n
X
yi φi (t) ≤ |y||φ(t)|,
i=1
so that
(4)
|y|2 ≤
Z
b
Z
|y||φ(t)|dt = |y|
a
b
|φ(t)|dt.
a
If y = 0, the desired inequality holds trivially. If y 6= 0, we can divide by |y| in (4)
in order to obtain the conclusion of the lemma.
4
4. Existence
Before turning to the question of existence, it is of interest to prove that if f :
Rn+1 → Rn is continuous and continuously differentiable with respect to its last n
variables, then (for s ∈ R and x, y ∈ Rn )
Z 1X
n
(5)
f (s, x) − f (s, y) =
(∂yi f )[s, τ x + (1 − τ )y](xi − yi )dτ.
0
i=1
Let, to this end,
γ(τ ) = τ x + (1 − τ )y.
Note that γ(1) = x and γ(0) = y. Furthermore, if γ = (γ1 , . . . , γn ), then
dγi
= xi − yi .
dτ
Consider s ∈ R and x, y ∈ Rn as fixed, and define
g(τ ) = f (s, γ(τ )).
Then
Z
f (s, x) − f (s, y) = g(1) − g(0) =
1
g 0 (τ )dτ.
0
On the other hand, via the chain rule,
g 0 (τ ) =
n
X
i=1
n
(∂yi f )(s, γ(τ ))
X
dγi
(τ ) =
(∂yi f )[s, τ x + (1 − τ )y](xi − yi ).
dτ
i=1
Combining these observations, we obtain (5).
We are now in a position to give a condition which ensures local existence.
Theorem 2 (Existence). Let f : Rn+1 → Rn be continuous and such that it is
continuously differentiable with respect to its last n variables. Consider the problem
(6)
(7)
dx
(t) = f (t, x(t)),
dt
x(t0 ) = x0 ,
where t0 ∈ R and x0 ∈ Rn . Then there is an > 0 and a continuously differentiable
x : (t0 − , t0 + ) → Rn satisfying (6) and (7).
Remark. By a slight modification of the proof, the more general statement found
in the book can also be proven.
Proof. As described earlier, the idea is to construct a sequence of approximate
solutions. We define the approximations recursively by
Z t
f (s, x0 )ds
(8)
x1 (t) = x0 +
t0
and
Z
(9)
t
xl (t) = x0 +
f (s, xl−1 (s))ds
t0
5
for l ≥ 2. Here x0 is the constant vector given in the statement of the theorem. We
wish to start by obtaining some rough control of the iterates xl . However, before
doing so, it is convenient to introduce the following terminology:
!1/2
n
X
2
2
max
|f (t, ξ)| +
|∂yi f (t, ξ)|
.
(10)
C0 =
|t−t0 |≤1, |ξ−x0 |≤1
i=1
Note that {(t, ξ) ∈ Rn+1 : |t − t0 | ≤ 1, |ξ − x0 | ≤ 1} is a compact set and that
!1/2
n
X
2
2
|f | +
|∂yi f |
i=1
is a continuous function on R
and finite.
n+1
. Thus the right hand side of (10) is well defined
Rough control. Let us prove that if > 0 is small enough, then |xl (t) − x0 | ≤ 1 for
|t − t0 | ≤ . Let us start by considering x1 . Due to (8), we have
Z t
Z t
(11)
|x1 (t) − x0 | = f (s, x0 )ds ≤
|f (s, x0 )|ds,
t0
t0
where we have used Lemma 2.
Remark: Note that we are here assuming that t ≥ t0 ; if t ≤ t0 , the right hand
side should be replaced by
Z t0
|f (s, x0 )|ds.
t
In what follows, we shall assume t ≥ t0 and leave the modifications necessary to
deal with the case t ≤ t0 to the reader.
For |s − t0 | ≤ 1, we have |f (s, x0 )| ≤ C0 due to the definition of C0 . Inserting this
information into (11), we conclude that
(12)
|x1 (t) − x0 | ≤ C0 |t − t0 |
for |t − t0 | ≤ 1. If we assume |t − t0 | ≤ 1/(C0 + 1), then |t − t0 | ≤ 1 (so that (12) is
applicable) and we have
1
≤ 1.
|x1 (t) − x0 | ≤ C0
C0 + 1
In order to prove that the same inequality holds with x1 replaced by xl , let us make
the inductive assumption that
|xl (t) − x0 | ≤ 1
for |t − t0 | ≤ 1/(C0 + 1). We already know that this statement holds for l = 1. We
thus only need to prove that if it holds for an l ≥ 1, then it holds for l + 1. Consider
Z t
Z t
(13)
|xl+1 (t) − x0 | = f (s, xl (s))ds ≤
|f (s, xl (s))|ds,
t0
t0
where we have assumed that t ≥ t0 and used Lemma 2. Due to the inductive
assumption and the definition of C0 , we know that the integrand on the far right
hand side of (13) is bounded by C0 in the relevant interval. This leads to the desired
conclusion: If |t − t0 | ≤ 1/(C0 + 1), then |t − t0 | ≤ 1 and |xl (t) − x0 | ≤ 1 for all
l. Note that due to the definition of C0 , we thus control f and its derivatives in
[t, xl (t)] for all l, assuming |t − t0 | ≤ 1/(C0 + 1).
6
Convergence. In order to prove convergence, consider the difference of two successive iterates:
Z t
(14)
xl+1 (t) − xl (t) =
[f (s, xl (s)) − f (s, xl−1 (s))]ds.
t0
Note that for |ξ − x0 | ≤ 1, |y − x0 | ≤ 1 and s such that |s − t0 | ≤ 1, we have
Z
n
1X
|f (s, ξ) − f (s, y)| = (∂yi f )[s, τ ξ + (1 − τ )y](ξi − yi )dτ 0
i=1
Z 1X
n
≤
|(∂yi f )[s, τ ξ + (1 − τ )y]||ξi − yi |dτ
0 i=1
(15)
!1/2
Z 1 X
n
2
≤
(∂yi f ) [s, τ ξ + (1 − τ )y]
|ξ − y|ds
0
i=1
≤C0 |ξ − y|,
where we have used (5), Lemma 2, the Schwartz inequality and the definition of
C0 . Inserting this information into (14), we obtain
Z t
(16)
|xl+1 (t) − xl (t)| ≤
C0 |xl (s) − xl−1 (s)|ds.
t0
Let us define by = 1/[2(C0 + 1)], K = [t0 − , t0 + ], and
kxk = max |x(t)|
t∈K
n
for continuous x : K → R . Note that all the iterates xl are continuous. Due to
(16), we obtain the conclusion that
kxl+1 − xl k ≤ C0 kxl − xl−1 k ≤
1
kxl − xl−1 k.
2
Iterating this estimate, we obtain
kxl+1 − xl k ≤
1
kx1 − x0 k.
2l
Assume that m > l. Then
kxm − xl k ≤
m−1
X
kxk+1 − xk k ≤
k=l
m−1
X
2−k kx1 − x0 k
k=l
≤2−l kx1 − x0 k
∞
X
2−j ≤ 2−l+1 kx1 − x0 k.
j=0
As a consequence, Lemma 1 is applicable to the sequence xl , so that there is a
continuous function x : K → Rn with the property that
lim kxl − xk = 0.
l→∞
Since xl converges to x, we conclude that |x(t) − x0 | ≤ 1 for all t ∈ K. By (15), we
conclude that
(17)
|f [t, x(t)] − f [t, xl (t)]| ≤ C0 kx − xl k.
7
This estimate can be used to conclude that
Z t
x(t) = x0 +
f [s, x(s)]ds
t0
(we leave this statement as an exercise). As a consequence, x is continuously
differentiable and it satisfies (6) and (7).
5. Uniqueness
The following lemma will be useful in the proof of uniqueness.
Lemma 3. Let t0 ∈ R and T be a positive real number. Let φ : [t0 , t0 + T ] → R
be a continuous and non-negative function. Assume there is a constant C ≥ 0 such
that
Z t
(18)
φ(t) ≤
Cφ(s)ds
t0
for all t ∈ [t0 , t0 + T ]. Then φ(t) = 0 for all t ∈ [t0 , t0 + T ].
Proof. Define
Z
t
h(t) =
φ(s)ds.
t0
Since φ is continuous, h is continuously differentiable, and we have
Z t
h0 (t) = φ(t) ≤
Cφ(s)ds = Ch(t),
t0
where we have used (18) and the definition of h. Letting
ĥ(t) = e−Ct h(t),
we thus have
ĥ0 (t) = e−Ct (h0 (t) − Ch(t)) ≤ 0.
Integrating this inequality, we obtain
e−Ct h(t) = e−Ct h(t) − e−Ct0 h(t0 ) = ĥ(t) − ĥ(t0 ) =
Z
t
ĥ0 (s)ds ≤ 0.
t0
Thus h(t) ≤ 0 for t ∈ [t0 , t0 + T ]. On the other hand, we know that h(t) ≥ 0. Thus
h(t) = 0. Due to (18), the definition of h and the fact that φ is non-negative, we
obtain the conclusion of the lemma.
Finally, we are in a position to prove uniqueness.
Theorem 3 (Uniqueness). Let f : Rn+1 → Rn be continuous and such that it is
continuously differentiable with respect to the last n variables. Consider the problem
(19)
(20)
dx
(t) = f (t, x(t)),
dt
x(t0 ) = x0 ,
where t0 ∈ R and x0 ∈ Rn . Assume x, y : (t− , t+ ) → Rn are two continuously
differentiable solutions to (19) and (20), where we assume t0 ∈ (t− , t+ ). Then
x = y on (t− , t+ ).
8
Proof. We have
Z
(21)
t
x(t) = x0 +
Z
f (s, x(s))ds,
t
y(t) = x0 +
t0
f (s, y(s))ds.
t0
Let t1 ∈ [t0 , t+ ). Since x and y are continuous, there is a constant M < ∞ such
that |x(t) − x0 | ≤ M and |y(t) − x0 | ≤ M for t ∈ [t0 , t1 ]. Let
!1/2
n
X
2
C0 =
sup
|∂yi f (t, ξ)|
.
t0 ≤t≤t1 , |ξ−x0 |≤M
i=1
Subtracting the two equations (21) from each other and using an estimate of the
form (15), we obtain
Z t
Z t
|x(t) − y(t)| ≤
|f [s, x(s)] − f [s, y(s)]|ds ≤ C0
|x(s) − y(s)|ds
t0
t0
for t ∈ [t0 , t1 ]. Applying Lemma 3 with φ(t) = |x(t) − y(t)| and C = C0 , we obtain
the conclusion that φ(t) = 0 for t ∈ [t0 , t1 ]. Since t1 ∈ [t0 , t+ ) was arbitrary, we
obtain x = y in [t0 , t+ ). The argument to prove that the functions coincide in
(t− , t0 ] is similar and is left to the reader.