Existence and Uniqueness Theorems
John Stalker∗
March 3, 2015
In these notes, Br (ξ) denotes the open ball of ra- By the Fundamental Theorem of the Calculus,
dius r about ξ and B r (ξ) is the closed ball of radius
Z 1
∂f
r about ξ. This notation will be used in Euclidean f (1, x, y , y ) − f (0, x, y , y ) =
(t, x, ya , yb ) dt.
a b
a b
spaces of any dimension, including dimension 1.
0 ∂t
Theorem 1 Suppose that ξ ∈ R, η ∈ Rn , r > 0 and By the Chain Rule,
s > 0. Suppose
∂f
∂F
∂u
u(t, x, ya , yb ) =
(x, u(t, ya , yb )) (t, ya , yb ).
n
∂t
∂y
∂t
F : B r (ξ) × B s (η) → R
From the definition of u we see that
satisfies
∂F
∂y (x, y) ≤ M
∂u
(t, ya , yb ) = (yb − ya ).
∂t
for each x ∈ B r (ξ) and y ∈ B s (η), where the ∂F/∂y Since the matrix norm is submultiplicative
is a matrix of partial derivatives and the norm is the
∂f
usual matrix norm. Then
u(t, x, ya , yb ) ≤ M kya − yb k.
∂t
kF (x, ya ) − F (x, yb )k ≤ M kya − yb k
Integrating this inequality gives
for all x ∈ B r (ξ) and ya , yb ∈ B s (η).
kF (x, ya ) − F (x, yb )k ≤ M kya − yb k
Before beginning the proof, note that the norm is
continuous and B r (ξ) × B s (η) is closed and bounded, as required.
so if ∂F/∂y is continuous then there must be such
From now on we assume that
an M .
F : B r (ξ) × B s (η) → Rn
Proof: Let
satisfies
u(t, ya , yb ) = tya + (1 − t)yb .
kF (x, ya ) − F (x, yb )k ≤ M kya − yb k
and
f (t, x, ya , yb ) = F (x, u(t, ya , yb )).
for all x ∈ B r (ξ) and ya , yb ∈ B s (η) and that
Then
kF (x, η)k ≤ N
F (x, ya ) − F (x, yb ) = f (1, x, ya , yb ) − f (0, x, ya , yb ).
∗ School
for all x ∈ B r (ξ). If F is continuous then there is
such an N .
of Mathematics, TCD, Copyright 2015
1
Theorem 2 Suppose 0 < σ < s. There is a ρ > 0 Now
such that if y0 ∈ Bσ (η) and y: Bρ (ξ) → Rn satisfies
kz(ξ)k = ky(ξ) − ηk = ky0 − ηk < σ
y ′ (x) = F (x, y(x))
and
for x ∈ Bρ (ξ) and
|x − ξ| < ρ′ ,
y(ξ) = y0
so
then
kz(x)k ≤
y(x) ∈ Bs (η)
for all x ∈ Bρ (ξ).
N
N
σ+
< s.
exp(M ρ′ ) −
M
M
The Bootstrap Lemma from the preceding chapter
then allows us to conclude that
Proof: It will be shown that this holds with
1
Ms + N
ρ = min r,
.
log
M
Mσ + N
kz(x)k < s
for all x ∈ Bρ (ξ).
By continuity there is a ρ′ > 0 such that
Theorem 3 If ya and yb satisfy the differential equation
y ′ (x) = F (x, y(x))
y(x) ∈ Bs (η)
for all x ∈ Bρ′ (ξ). Suppose that x ∈ Bρ′ (ξ). The
derivative of the the norm of a matrix valued function
is bounded by the norm of the derivative. We apply for all x ∈ Bρ (ξ) with initial conditions
this to
ya (ξ) = za , yb (ξ) = zb
z(x) = y(x) − η,
obtaining
d
kz(x)k ≤ kz ′ (x)k = ky ′ (x)k = kF (x, y(x))k.
dx
then
kya (x) − yb (x)k ≤ kza − zb k exp(M |x − ξ|).
Proof: If ya and yb are solutions then
d
kya (x) − yb (x)k ≤ ky ′ (x) − y ′ (x)k
a
b
dx
= kF (x, ya (x)) − F (x, yb (x))k
≤ M kya (x) − yb (x)k.
By the triangle inequality
kF (x, y(x))k ≤ kF (x, y(x)) − F (x, η)k + kF (x, η)k.
By hypothesis
kF (x, y(x)) − F (x, η)k ≤ M ky(x) − ηk = M kz(x)k
The last step is justified because, by the preceding
theorem,
ya (x), yb (x) ∈ Bs (η)
and
kF (x, η)k ≤ N.
Thus
d
kz(x)k ≤ M kz(x)k + N.
dx
for all x ∈ Bρ (ξ). Applying Gronwall again,
kya (x) − yb (x)k ≤ kya (ξ) − yb (ξ)k exp(M |x − ξ|).
By Gronwall’s inequality
N
N
exp(M |x − ξ|) −
.
kz(x)k ≤ kz(ξ)k +
M
M
There are two important corollaries.
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Now
Theorem 4 The initial value problem
y ′ (x) = F (x, y(x)),
kz − ηk < σ
y(ξ) = y0
and
has at most one solution in the interval Bρ (ξ).
kF (t, η)k ≤ N
Assume there are two solutions, ya and yb . Then the
preceding theorem, with za = zb = y0 gives Proof:
so
ky1 (x) − y0 (x)k ≤ σ + N |x − x0 |.
By generalised induction we show that for all n ≥ 1
and all x ∈ Bρ (ξ)
•
kya (x) − yb (x)k = 0
and hence ya = yb .
Theorem 5 Suppose that for each z ∈ Bσ (x) there
is a solution to the initial value problem
y ′ (x) = F (x, y(x)),
kyn (x) − yn−1 (x)k ≤ gn (x)
where
y(ξ) = y0
in Bρ (ξ). This solution depends continuously on z,
uniformly in x.
gn (x) =
σM n−1 |x − x0 |n−1 M n−1 N |x − x0 |n
+
,
(n − 1)!
n!
•
Proof: Suppose ǫ > 0. Let δ = ǫ exp(−M ρ). Then
if
kza − zb k < δ
kyn (x) − ηk ≤
n
X
gm (x).
m=1
and ya and yb are the solutions with those initial conditions then, by the theorem,
In case n = 1 both of these are just the statement
kya (x) − yb (x)k ≤ δ exp(M |x − ξ|) < ǫ
ky1 (x) − y0 (x)k ≤ σ + N |x − x0 |.
for all x ∈ Bρ (ξ). This is the definition of continuity.
It is uniform because the δ we found depends only on
ǫ, not on x.
To make the preceding theorem non-trivial we still
need to show that solutions exist.
proved above. For the induction step we assume the
two inequalities above and the corresponding inequalities with n replaced by any smaller integer. Using
these, we prove the corresponding inequalities with n
replaced by n + 1. First of all, gm (x) is non-negative,
so
Theorem 6 For any z ∈ Bσ (η) there is a contin
∞
nously differentiable y: Bρ (ξ) such that
X
X
g
(x)
≤
gm (x)
m
y ′ (x) = F (x, y(x)), y(ξ) = z.
m=1
m=1
N
N
exp(M |x − x0 |) −
=
σ
+
Proof: We define, inductively,
M
M
Z x
< s.
F (t, yn (t)) dt.
y0 (x) = η, yn+1 (x) = z +
x0
Thus
Z x
yn (x) ∈ Bs (η)
(y1′ (t) − y0′ (t)) dt
y1 (x) − y0 (x) = y1 (ξ) − y0 (ξ) +
for all x ∈ Bρ (ξ). Since we also have the correspondx0
Z x
ing inequality with n replaced by n − 1. From this it
F (t, y0 (t)) dt.
=z−η+
follows
that
Zxx0
F (t, η) dt.
=z−η+
kF (x, yn (x)) − F (x, yn−1 (x)k ≤ M kyn (x) − yn−1 (x)
x0
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for all x ∈ Bρ (ξ). By the other part of the induction
hypothesis the right hand side is bounded by M gn (x).
Since
Z x
(F (x, yn (x)) − F (x, yn−1 (x)) dx
yn+1 (x)−yn (x) =
x0
we see that
kyn+1 (x) − yn (x)k ≤
Z
x
M gn (t) dt = gn+1 (x).
x0
That’s half of what we needed to prove for the induction step. The other half follows because, by the
triangle inequality,
kyn+1 (x) − ηk = kyn+1 (x) − yn (x)k + kyn (x) − ηk .
For the first norm on the right we use the inequality
just obtained and for the second we use the induction
hypothesis. The result is
yn+1 (x)−yn (x) ≤ gn+1 (x)+
n
X
m=1
gm (x) =
n+1
X
gm (x).
m=1
This is the other half of what we needed to prove,
thus completing the induction.
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