A.3 Local Existence

In the case of a nonlinear initial value problem

dxdt=f(x,t),x(a)=b, (35)

the hypothesis in Theorem 1 that f satisfies a Lipschitz condition on a slab (x, t) (t in I, all x) is unrealistic and rarely satisfied. This is illustrated by the following simple example.

Example 1

Consider the initial value problem

dydx=x2,x(0)=b>0. (36)

As we saw in Example 6, the equation x=x2 does not satisfy a “strip Lipschitz condition.” When we solve (36) by separation of variables, we get

x(t)=b1bt. (37)

Because the denominator vanishes for t=1/b, Eq. (37) provides a solution of the initial value problem in (36) only for t<1/b, despite the fact that the differential equation x=x2 “looks nice” on the entire real line—certainly the function ...

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