A.4 Uniqueness of Solutions
It is possible to establish the existence of solutions of the initial value problem in (35) under the much weaker hypothesis that f(x, t) is merely continuous; techniques other than those used in this section are required. By contrast, the Lipschitz condition that we used in proving Theorem 1 is the key to uniqueness of solutions. In particular, the solution provided by Theorem 3 is unique near the point t=a.
We will outline the proof of Theorem 4 for the 1-dimensional case in which x is a real variable. A generalization of this proof to the multivariable case can be found in Chapter 6 of Birkhoff and Rota.
Let us consider the function
ϕ(t)=[x1(t)−x2(t)]2
(40)
for which ϕ(a)=0, because x1(a)=x2(a)=b. We want to show that ϕ(t)≡0, so that x1(t)≡x2(t). We will consider only the case t≧a; the details are similar for the case t≦a.
If we differentiate each side in Eq. (40), we find that
by using the Lipschitz condition on f. Hence
ϕ'(t)≦2kϕ(t).
(41)
Now let us temporarily ignore the fact that ϕ(a)=0 and compare ϕ(t) with the solution of the differential equation
Φ'(t)=2kΦ(t)
(42)
such that Φ(a)=ϕ(a); clearly
Φ(t)=Φ(a)e2k(t−a).
(43)
In comparing (41) with (42), it seems inevitable that
ϕ(t)≦Φ(t)for t≧a,
(44)
and this is easily proved (Problem 18). Hence
On taking square roots, we get
0≦|x1(t)−x2(t)|≦|x1(a)−x2(a)|ek(t−a).
(45)
But x1(a)−x2(a)=0, so (45) implies that x1(t)≡x2(t).
The initial value problem
dxdt=3x2/3,x(0)=0
(46)
has both the obvious solution x1(t)≡0 and the solution x2(t)=t3 (which is readily found by separation of variables). Hence the function f(x, t) must fail to satisfy a Lipschitz condition near (0, 0). Indeed, the mean value theorem yields
for some x¯ between 0 and x. But fx(x,0)=2x−1/3 is unbounded as x→0, so no Lipschitz condition can be satisfied.