Re: maximum interval of the existence

On 09.09.2007 07:38, lancer6238@xxxxxxxxx wrote:
So all I needed to do was to show there is always a bound for the
solution?

Yepp! Because it is a standard result in ODE theory that the solution
explodes if you approach the *finite* boundary of the maximal interval
where the solution exists.

Here's my revised proof:

f(t,x) is a continuous vector valued function on R x R^n. By the
existence-uniqueness theorem, there exists an w > 0 s.t. x' = f(t,x)
has a unique solution on the interval [t_0 - w, t_0 + w]. For any x in
R^n, there is a maximal open interval (a,b) on which the initial value
problem has a unique solution x(t).

Moreover, if a or b is finite, the solution x(t) is unbounded in (a,
b).

Therefore, we need to show there is a bound for the solution on every
bounded interval.

x(t) = x_0 + INTEGRATE(from t_0 to t) f(s, x(s)) ds for t_0 <= t <
b
||x(t)|| = ||x_0 + INTEGRATE(from t_0 to t) f(s, x(s)) ds||
<= ||x_0|| + ||INTEGRATE(from t_0 to t) f(s, x(s)) ds||
<= ||x_0|| + INTEGRATE(from t_0 to t) ||f(s, x(s))|| ds
<= ||x_0|| + C INTEGRATE(from t_0 to t) ||x|| ds
By Gronwall's equality,

rather: inequality ;)

||x(t)|| <= ||x_0|| exp(C(t-t_0))
<= ||x_0|| exp(Ct) for t_0 <= t < b

Therefore, x(t), t in [t_0, b) is bounded:
||x(t)|| < r for some r > 0 and t_0 <= t < b

Similar arguments follow for a < t <= t_0 to show
x(t), t in (a, t_0] is bounded:
||x(t)|| < d for some d > 0 and a < t <= t_0

Since there is a bound for the solution on every bounded interval, the
maximal interval of existence is (-infinity, infinity).

Am I correct now?

Seems fine. :-)

--
Best wishes,
J.
.

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