124 CHAPTER 6. EXISTENCE, UNIQUENESS, AND DYNAMICS The smallestKsatisfying this inequality is denoted by Lip(f) and is called theLipschitz constant of f.
More generally,f islocally Lipschitz continuous if every point inU has a neighborhood such that f restricted to that neighborhood is Lipschitz continuous.
The proof of the following proposition is left to the reader.
Proposition 6.1.3. Let U ⇢Rn open and let f 2C1(U,Rn). Then f is locally Lipschitz continuous.
We will use the contraction mapping theorem to prove the existence and uniqueness of solutions. Thus the first step is to identify an appropriate complete metric.
Proposition 6.1.4. Let J = [a, b]⇢R . Consider C0(J,R) with the norm kx yk1:= sup
t2Jkx(t) y(t)k. Then, C0(J,R) is a Banach space.
Proof. We begin by remarking thatC0(J,R)is a vector space overRsince iff, g2C0(J,R) and a2R, thenf+g2C0(J,R) and af 2C0(J,R).
To show that C0(J,R) is a complete metric space we need to show that if {fn} ⇢ C0(J,R) is a Cauchy sequence, then there exists ˆf 2 C0(J,R) to which the sequence converges. We provide a proof in three steps. First, we define a function ˆf: [a, b] ! R. Then, we show that ˆf is continuous. Finally we show that limn!1kfn fˆk1= 0.
Observe that for everyx2[a, b]
|fn(x) fm(x)| sup
z2[a,b]|fn(z) fm(z)|=kfn fmk1. (6.1) As a consequence, the assumption that{fn}is Cauchy implies that{fn(x)}⇢Ris Cauchy for each x2[a, b]. SinceRis complete we can set ˆf(x) := limn!1fn(x).
To show that ˆf is continuous it suffices to show that ˆf is continuous at each ˆx2[a, b].
To this end let{xk}⇢[a, b] be a sequence such that limk!1xn= ˆx. We need to show that limk!1fˆ(xk) = ˆf(ˆx) or equivalently given✏ >0, there exitsK 2N such that if k K, then|fˆ(xk) fˆ(ˆx)|<✏. Combining (6.1) with the assumption that{fn} is Cauchy allows us to conclude that for m and nsufficiently large
|fn(x) fm(x)|< ✏ 3
independent of x2[a, b]. In addition, since this is true for all nsufficiently large
n!1lim |fn(x) fm(x)|=|fˆ(x) fm(x)|< ✏ 3.
6.1. EXISTENCE AND UNIQUENESS OF SOLUTIONS TO ODES 125 Finally, since fm is continuous, given✏< 3✏, there existsK such that if k K, then
|fm(xk) fm(ˆx)|< ✏ 3.
Using these three inequalities we obtain that given ✏>0, for sufficiently large kand m
|fˆ(xk) fˆ(ˆx)|<|fˆ(xk) fm(xk)|+|fm(xk) fm(ˆx)|+|fm(ˆx) fˆ(ˆx)|<✏, which is the desired result.
We leave it to the reader to check that limn!1kfn fˆk1= 0.
Theorem 6.1.5 (Existence and Uniqueness). If f: U ! Rn is a locally Lipschitz continuous function defined on an open set U ⇢ Rn and x0 2 U, then there exists a solution ': ( a, a)!U for some a >0 to the initial value problem
˙
x=f(x), x(0) =x0. (6.2)
Furthermore, ' is the unique solution in the sense that if : ( b, b) ! U is a solution, then 'and agree on their common domain.
Proof. Since f is locally Lipschitz continuous there exists ✏ > 0 such that f is Lipschitz continuous onB✏(x0)⇢U. SinceB✏(x0) is compact, it follows thatf is bounded onB✏(x0).
Therefore, there are constants K and M with
kf(x) f(y)k Kkx yk and kf(x)k M for all x, y2B✏(x0). Choosea >0 such that
a <min
⇢ ✏ 2M, 1
K , (6.3)
and set J := [ a, a].
By Lemma 6.1.1 it is sufficient to prove the existence of'satisfying '(t) =x0+
Z t
0
f('(s))ds, (6.4)
for all t2J.
To this end setX := B✏(x0) ⇢C0(J,Rn) where in an abuse of notation we use x0 to denote the constant function taking the value x0. DefineT:X!C0(J,Rn) by
T(↵)(t) :=x0+ Z t
0
f(↵(s))ds.
Therefore, by the Contraction Mapping Theorem to obtain a fixed point ofT it is sufficient to prove (i) thatT(X)⇢X and (ii) that T is a contraction.
