A description of why the local solution existence does not guarantee the global solution existence for Euclidean-normed Euclidean vectors space ODE
Topics
About: normed vectors space
The table of contents of this article
Starting Context
Target Context
- The reader will understand why the local solution existence for Euclidean-normed Euclidean vectors space ordinary differential equation does not guarantee the global solution existence for the entire domain interval.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
Suppose a Euclidean-normed Euclidean vectors space ordinary differential equation, \(\frac{dx}{dt} = f (x, t)\), satisfies the conditions for having local unique solution around every point, \(t_0\), in the interval, \([t_1, t_e]\), with the initial condition, \(x (t_0) = x_0\) where \(x_0\) is any value in \(\mathbb{R}^d\), for a closed interval, \([t_0 - \epsilon_{t_0, x_0, 1}, t_0 + \epsilon_{t_0, x_0, 2}]\), where \(\epsilon_{t_0, x_0, i}\) means that it depends on \(t_0\) and \(x_0\). That does not guarantee the existence of a global solution for the entire \([t_1, t_e]\). Why? Get the local solution, \(x (t): [t_1 - \epsilon_{t_1, x_1, 1}, t_1 + \epsilon_{t_1, x_1, 2}] \to \mathbb{R}^d\), by the initial condition at \(t_1\), then get the connected next local solution by the initial condition at \(t_2 = t_1 + \epsilon_{t_1, x_1, 2}\) with the value of \(x (t_2)\), getting the extended solution, \(x (t): [t_1, t_2 + \epsilon_{t_2, x (t_2), 2}]\), and so on. But the issue is that there is no guarantee that \(t_i + \epsilon_{t_i, x (t_i), 2}\) eventually reaches \(t_e\), because it may converge to a value, \(t_l \lt t_e\), failing to extend the solution to the entire \([t_1, t_e]\). . . . But, why do we not get a local solution by an initial condition at the limit point \(t_l\) having an interval \([t_l - \epsilon_{t_l, x_{t_l}, 1}, t_l + \epsilon_{t_l, x_{t_l}, 2}]\)? ... But what is really \(x_{t_l}\)? As \(x\) has not been extended to \(t_l\), it cannot be chosen to be \(x (t_l)\), and there is no guarantee that there is a \(x_{t_l}\) that makes the local solution coincide with the so-far-extended solution on the so-far-extended domain (note that the local existence guarantees that any initial value at \(t_l\) can be realized, not that any value at another point can be realized).
2: Note
A sufficient condition to guarantee the global solution existence is the bijective-ness of the map, \(x (t_i) \mapsto x (t_j)\), between the set of possible values at \(t_i\) and the set of possible values at \(t_j\) where \(t_i\) and \(t_j\) are any points in \([t_1, t_2]\), which guarantees the \([t_l - \epsilon_{t_l, x_l, 1}, t_l + \epsilon_{t_l, x_l, 2}]\) local solution to be able to be connected to the not-fully-extended solution.
