# Maximum interval of the existence

1. Sep 8, 2007

### wu_weidong

1. The problem statement, all variables and given/known data

Suppose f(t,x) is a continuous vector valued function on $$\mathbb{R} \times \mathbb{R}^n$$. If f is locally Lipschitz with respect to x with the property that $$\|f(t,x)\| \le C \|x\|$$ for some positive constant C > 0, then prove that the maximum interval of the existence of the initial value problem x' = f(t,x) with $$x(t_0) = x_0$$ is equal to $$(-\infty, \infty)$$.

3. The attempt at a solution

f is locally Lipschitz with respect to x with the property that $$\|f(t,x)\| \le C \|x\|$$ for some positive constant C > 0. From the local existence and uniqueness theorem, the system has a unique solution defined on $$[t_0 - a, t_0 + a]$$.
Let
$$\alpha$$ = inf{$$l < t_0$$: there is a unique solution on $$[l, t_0]$$}
$$\beta$$ = sup{$$r > t_0$$: there is a unique solution on $$[t_0, r]$$}

By the definition of $$\alpha$$ and $$\beta$$, and the existence-uniqueness theorem, there is a unique solution x(t) defined on $$(\alpha, \beta)$$.

Assume there is a compact set K $$\in \mathbb{R}^n$$ s.t. the solution $$x(t) \in K$$ for all $$t \in J$$, J = $$[t_0, \infty)$$, that is, the solution is defined for all $$t > t_0, t \in \mathbb{R}$$.

Assume J = $$[t_0, \infty)$$ does not hold. Thus, x is defined on $$[t_0,\beta), \beta < \infty$$, but there is no solution on $$[t_0, \beta]$$, and $$\beta \in \mathbb{R}$$.

Consider the function
$$$y(t) = \left\{\begin{array}{ll} x(t) \,\, if \,\,t \in [t_0, \beta)\\ y_0 \,\, if \,\,t = \beta \end{array} \right$$$

where $$y_0$$ is any element in K. Thus, $$\| f(t,y) \| \le C \| y \|$$.

Let g(t) = max{$$C \|x\|$$} and $$\| f(t,y) \| \le g(t)$$. F(t,y) is bounded by g(t), which is an integrable function on $$[t_0, \beta]$$ and is therefore integrable itself.

Define $$x(\beta) = x_0 + \int^\beta_{t_0} f(s, y(s)) \, ds$$

Since this is the limit of the integral $$\int^t_{t_0}$$ as $$t \to \beta$$, and K is closed, $$x(\beta) \in K$$ and in particular $$x(\beta) \in \mathbb{R}^n$$. Also, since f(s, y(s)) = f(s, x(s)) almost everywhere, x satisfies
$$x(t) = x_0 + \int^t_{t_0} f(s, x(s)) \, ds = x_0 + \int^t_{t_0} f(s, y(s)) \, ds$$
on the inteval $$[t_0, \beta]$$. Thus, it is absolutely continuous (from the first equality) and it satisfies the IVP (second equality). We defined an extension to $$[t_0, \beta]$$, contradicting maximality.

Did I make sense?

Thank you.

Regards,
Rayne

2. Sep 8, 2007

### EnumaElish

I am unfamiliar with some of the concepts you used. How is "maximum interval" defined? How does existence over a compact interval contradict maximality?

3. Sep 9, 2007

### EnumaElish

You assume x(t) is in compact K for all t in J. Next you posit that if J was not right, then the solution must be defined over a right-open interval but not over its closure. You then prove that this cannot be the case.

Then you argue that this contradicts maximality.

Have I gotten it right so far?

It makes sense to me up until "contradicting maximality." And that's because I am unfamiliar with the notion of a maximum interval.

Last edited: Sep 9, 2007