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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 $t = a .$ $t = a .$

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) = {[x_{1} (t) - x_{2} (t)]}^{2}

$ϕ (t) = {[x_{1} (t) - x_{2} (t)]}^{2}$ (40)

for which $ϕ (a) = 0,$ $ϕ (a) = 0,$ because $x_{1} (a) = x_{2} (a) = b .$ $x_{1} (a) = x_{2} (a) = b .$ We want to show that $ϕ (t) \equiv 0,$ $ϕ (t) \equiv 0,$ so that $x_{1} (t) \equiv x_{2} (t) .$ $x_{1} (t) \equiv x_{2} (t) .$ We will consider only the case $t ≧ a;$ $t ≧ a;$ the details are similar for the case $t ≦ a$ $t ≦ a$ .

If we differentiate each side in Eq. (40), we find that

\begin{array}{rcl} | ϕ ′ (t) | & = & | 2 [x_{1} (t) - x_{2} (t)] \cdot [x_{1}^{′} (t) - x_{2}^{′} (t)] | \\ = & | 2 [x_{1} (t) - x_{2} (t)] \cdot [f (x_{1} (t), t) - f (x_{2} (t), t)] | \\ ≦ & 2 k | x_{1} (t) - x_{2} (t) |^{2} = 2 k ϕ (t), \end{array}

$\begin{array}{rcl} | ϕ ′ (t) | & = & | 2 [x_{1} (t) - x_{2} (t)] \cdot [x_{1}^{′} (t) - x_{2}^{′} (t)] | \\ = & | 2 [x_{1} (t) - x_{2} (t)] \cdot [f (x_{1} (t), t) - f (x_{2} (t), t)] | \\ ≦ & 2 k | x_{1} (t) - x_{2} (t) |^{2} = 2 k ϕ (t), \end{array}$

by using the Lipschitz condition on f. Hence

ϕ ′ (t) ≦ 2 k ϕ (t) .

$ϕ ′ (t) ≦ 2 k ϕ (t) .$ (41)

Now let us temporarily ignore the fact that $ϕ (a) = 0$ $ϕ (a) = 0$ and compare $ϕ (t)$ $ϕ (t)$ with the solution of the differential equation

Φ ′ (t) = 2 k Φ (t)

$Φ ′ (t) = 2 k Φ (t)$ (42)

such that $Φ (a) = ϕ (a);$ $Φ (a) = ϕ (a);$ clearly

Φ (t) = Φ (a) e^{2 k (t - a)} .

$Φ (t) = Φ (a) e^{2 k (t - a)} .$ (43)

In comparing (41) with (42), it seems inevitable that

ϕ (t) ≦ Φ (t) for t ≧ a,

$ϕ (t) ≦ Φ (t) for t ≧ a,$ (44)

and this is easily proved (Problem 18). Hence

0 ≦ {[x_{1} (t) - x_{2} (t)]}^{2} ≦ {[x_{1} (a) - x_{2} (a)]}^{2} e^{2 k (t - a)} .

$0 ≦ {[x_{1} (t) - x_{2} (t)]}^{2} ≦ {[x_{1} (a) - x_{2} (a)]}^{2} e^{2 k (t - a)} .$

On taking square roots, we get

0 ≦ | x_{1} (t) - x_{2} (t) | ≦ | x_{1} (a) - x_{2} (a) | e^{k (t - a)} .

$0 ≦ | x_{1} (t) - x_{2} (t) | ≦ | x_{1} (a) - x_{2} (a) | e^{k (t - a)} .$ (45)

But $x_{1} (a) - x_{2} (a) = 0,$ $x_{1} (a) - x_{2} (a) = 0,$ so (45) implies that $x_{1} (t) \equiv x_{2} (t)$ $x_{1} (t) \equiv x_{2} (t)$ .

Example 1

The initial value problem

\frac{d x}{d t} = 3 x^{2/3}, x (0) = 0

$\frac{d x}{d t} = 3 x^{2/3}, x (0) = 0$ (46)

has both the obvious solution $x_{1} (t) \equiv 0$ $x_{1} (t) \equiv 0$ and the solution $x_{2} (t) = t^{3}$ $x_{2} (t) = t^{3}$ (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

| f (x, 0) - f (0, 0) | = | f_{x} (\bar{x}, 0) | \cdot | x - 0 |

$| f (x, 0) - f (0, 0) | = | f_{x} (\bar{x}, 0) | \cdot | x - 0 |$

for some $\bar{x}$ $\bar{x}$ between 0 and x. But $f_{x} (x, 0) = 2 x^{- 1/3}$ $f_{x} (x, 0) = 2 x^{- 1/3}$ is unbounded as $x \to 0,$ $x \to 0,$ so no Lipschitz condition can be satisfied.

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Table of Contents for A.4 Uniqueness of Solutions

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Table of Contents for
A.4 Uniqueness of Solutions