D.3 Local Solutions
Lemma D.3.1 Let D
≡ and suppose U is an open set containing
D such that f :
U → F^{n} is C^{1}. (Recall this means all partial derivatives of f exist and
are continuous.) Then for K =
Mn, where M denotes the maximum of for z ∈ D,
it follows that for all x,y ∈ D,
f (x)− f (y) ≤ K x− y.


Proof: Let x,y ∈ D
and consider the line segment joining these two points,
x+
t
for
t ∈. Letting
h =
f for
t ∈, then
∫
1 ′
f (y)− f (x) = h(1)− h(0) = 0 h (t)dt.


Also, by the chain rule,
n
′ ∑ ∂f
h (t) = ∂xi (x+t (y− x))(yi − xi).
i=1


Therefore,
 
∫ 1∑n ∂f 
 ∂xi (x+t (y− x))(yi − xi)dt
∫ 0 in=1 
1∑ ∂f 
≤ 0 ∂xi (x+t (y− x))yi − xidt
i=n1
≤ M ∑ yi − xi ≤ M nx − y . ■
i=1
Now consider the map, P which maps all of ℝ^{n} to D
given as follows. For
x ∈ D,
Px =
x. For
xD, Px will be the closest point in
D to
x. Such a closest point
exists because
D is a closed and bounded set. Taking
f ≡, it follows
f is a
continuous function defined on
D which must achieve its minimum value by the extreme
value theorem from calculus.
Lemma D.3.2 For any pair of points, x,y ∈ F^{n},
≤.
Proof: The above picture suggests the geometry of what is going on. Letting z ∈ D
, it
follows that for all
t ∈,
x − Px2 ≤ x− (Px + t(z− Px))2


= x − P x2 + 2tRe ((x − Px )⋅(P x − z))+ t2z− P x2


Hence
2tRe ((x − Px)⋅(P x− z))+ t2z− Px2 ≥ 0


and this can only happen if
Re ((x − Px )⋅(P x − z)) ≥ 0.


Therefore,
Re ((x − P x)⋅(Px − P y)) ≥ 0
Re((y − P y)⋅(Py − P x)) ≥ 0
and so
Re (x − Px − (y− P y)) ⋅(P x− P y) ≥ 0


which implies
Re(x − y)⋅(Px − Py) ≥ P x− P y2


Then using the Cauchy Schwarz inequality it follows
■
With this here is the local existence and uniqueness theorem.
Theorem D.3.3 Let
be a closed interval and let U be an open subset of F^{n}. Let
f :
× U → F^{n} be continuous and suppose that for each t ∈, the map x → is
continuous. Also let x_{0} ∈ U and c ∈. Then there exists an interval, I ⊆ such that c ∈ I
and there exists a unique solution to the initial value problem,
′
x = f (t,x),x(c) = x0 (4.6)
 (4.6) 
valid for t ∈ I.
Proof: Consider the following picture.
The large dotted circle represents U and the little solid circle represents D
as indicated.
Here
r is so small that
D is contained in
U as shown. Now let
P denote the projection map
defined above. Consider the initial value problem
x′ = f (t,Px),x (c) = x0. (4.7)
 (4.7) 
From Lemma D.3.1 and the continuity of x →
, there exists a constant,
K such that if
x,y ∈ D, then
≤ K for all
t ∈. Therefore, by Lemma
D.3.2
f (t,P x)− f (t,P y) ≤ K Px− Py ≤ K x − y .


It follows from Theorem D.1.2 that 4.7 has a unique solution valid for t ∈
. Since
x is
continuous, it follows that there exists an interval,
I containing
c such that for
t ∈ I,
x ∈ D. Therefore, for these values of
t, f =
f and so there is a unique
solution to
4.6 on
I. ■
Now suppose f has the property that for every R > 0 there exists a constant, K_{R} such that for
all x,x_{1} ∈B
,
f (t,x)− f (t,x1) ≤ KR x− x1. (4.8)
 (4.8) 
Corollary D.3.4 Let f satisfy 4.8 and suppose also that
→ f is continuous. Suppose
now that x_{0} is given and there exists an estimate of the form < R for all t ∈ [0
,T)
where
T ≤∞ on the local solution to
x′ = f (t,x),x(0) = x0. (4.9)
 (4.9) 
Then there exists a unique solution to the initial value problem, 4.9 valid on [0,T).
Proof: Replace f
with
f where
P is the projection onto
B. Then by
Theorem
D.1.2 there exists a unique solution to the system
′
x = f (t,Px ),x(0) = x0


valid on
for every
T_{1} < T. Therefore, the above system has a unique solution on [0
,T) and
from the estimate,
Px =
x.
■