Maximum interval of the existence

[wu_weidong]

Homework Statement

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).

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
<br /> \[ y(t) = \left\{\begin{array}{ll}<br /> x(t) \,\, if \,\,t \in [t_0, \beta)\\<br /> y_0 \,\, if \,\,t = \beta \end{array} \right \]

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<br /> = 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

 

[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?

 

[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.