description/proof of sufficient conditions for existence of unique global solution on interval for Euclidean-normed Euclidean vectors space ODE
Topics
About: normed vectors space
About: differential equation
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description 1
- 2: Natural Language Description 1
- 3: Proof 1
- 4: Note 1
Starting Context
Target Context
- The reader will have a description and a proof of some sufficient conditions with which there is the unique solution on the whole interval with any initial condition for any Euclidean-normed Euclidean vectors space ordinary differential equation.
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: Structured Description 1
Here is the rules of Structured Description.
Entities:
\(\mathbb{R}^d\): \(= \text{ the Euclidean-normed Euclidean vectors space }\)
\(\mathbb{R}\): \(= \text{ the Euclidean-normed Euclidean vectors space }\)
\(J\): \(\subseteq \mathbb{R}\), \(= [t_1, t_e]\)
\(f\): \(: \mathbb{R}^d \times J \to \mathbb{R}^d\), \(\in \{\text{ the } C^0 \text{ maps }\}\)
\(\frac{dx}{dt} = f (x, t)\): = the ordinary differential equation with any initial condition, \(x (t_1) = x_{t_1}\)
//
Statements:
\(\forall t_0 \in J\)
(
\(\exists [t_0 - \epsilon_{t_0, 1}, t_0 + \epsilon_{t_0, 2}]\) that satisfies the conditions for the local solution existence of the equation for the initial condition with any initial value, \(x_{t_0}\), for \(t_0\), where \(\epsilon_{t_0, j}\) is not zero except \(\epsilon_{t_1, 1}\) and \(\epsilon_{t_e, 2}\) and does not depend on \(x_{t_0}\) although can depend on \(t_0\), which means that there is a map, \(x_{t_0} \mapsto x (t)\) for each \(t \in [t_0 - \epsilon_{t_0, 1}, t_0 + \epsilon_{t_0, 2}]\), and the map is bijective
)
\(\implies\)
The ordinary differential equation has the unique solution on the entire \(J\).
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2: Natural Language Description 1
For any Euclidean-normed Euclidean vectors space, \(\mathbb{R}^d\) and \(\mathbb{R}\), any closed interval, \(J \subseteq \mathbb{R} = [t_1, t_e]\), and any \(C^0\) map, \(f: \mathbb{R}^d \times J \to \mathbb{R}^d\), the ordinary differential equation, \(\frac{dx}{dt} = f (x, t)\) with any initial condition \(x (t_1) = x_{t_1}\), has the unique solution on the entire \(J\) if at each point, \(t_0 \in J\), there is a closed interval, \([t_0 - \epsilon_{t_0, 1}, t_0 + \epsilon_{t_0, 2}]\), that satisfies the conditions for the local solution existence of the equation for the initial condition with any initial value, \(x_{t_0}\), for \(t_0\), where \(\epsilon_{t_0, j}\) is not zero except \(\epsilon_{t_1, 1}\) and \(\epsilon_{t_e, 2}\), and does not depend on \(x_{t_0}\) although can depend on \(t_0\), which means that there is a map, \(x_{t_0} \mapsto x (t)\) for each \(t \in [t_0 - \epsilon_{t_0, 1}, t_0 + \epsilon_{t_0, 2}]\), and the map is bijective.
3: Proof 1
The open intervals, \((t_0 - \epsilon_{t_0, 1}, t_0 + \epsilon_{t_0, 2})\) for \(t_0 \neq t_1, t_e\), \([t_1, t_1 + \epsilon_{t_1, 2})\), and \((t_e - \epsilon_{t_e, 1}, t_e]\), cover compact \(J\), so, there is a finite subcover, \([t_1, t_1 + \epsilon_{t_1, 2}), (t_2 - \epsilon_{t_2, 1}, t_2 + \epsilon_{t_2, 2}), ..., (t_e - \epsilon_{t_e, 1}, t_e]\).
There is the local unique solution for \([t_1, t_1 + \epsilon_{t_1, 2})\) with the initial condition, \(\overline{x} (t_1) = x_{t_1}\), \(\overline{x}: [t_1, t_1 + \epsilon_{t_1, 2}) \to \mathbb{R}^d\). Let us define the restriction of \(x\), \(x \vert_{[t_1, t_1 + \epsilon_{t_1, 2})} = \overline{x}\).
There is an interval that intersects \([t_1, t_1 + \epsilon_{t_1, 2})\), \((t_{j_2} - \epsilon_{t_{j_2}, 1}, t_{j_2} + \epsilon_{t_{j_2}, 2})\), with an intersection point denoted as \(t_{1, j_2}\), and there is the local unique solution for it with the initial condition, \(\overline{x} (t_{j_2}) = x_{t_{j_2}}\), where \(x_{t_{j_2}}\) can be chosen such that \(\overline{x} (t_{1, j_2}) = x (t_{1, j_2})\), which is because the bijective map mentioned in Description is supposed to exist. In fact, \(\overline{x}\) coincides with \(x\) on the whole intersection area, because there is an interval around \(t_{1, j_2}\) and there is the unique solution on it with the initial condition, which has to agree with \(x\) and \(\overline{x}\) there, and if \(x\) and \(\overline{x}\) did not agree on the whole intersection area, \((t_{j_2} - \epsilon_{t_{j_2}, 1}, t_{j_2} + \epsilon_{t_{j_2}, 2})\) would have another solution by switching to \(x\) after passing \(t_{1, j_2}\) or \([t_1, t_1 + \epsilon_{t_1, 2})\) would have another solution by switching to \(\overline{x}\) after passing \(t_{1, j_2}\). Let us extend \(x\) to \([t_1, t_{j_2} + \epsilon_{t_{j_2}, 2})\) with \(\overline{x}\).
Thus, \(x\) can be extended to the whole \(J\). The extension is unique, because at each point, \(t_{j_k}\), \(x_{t_{j_k}}\) is determined uniquely.
4: Note 1
It is crucial that such a bijective map exists for this proposition. Just that there is a local solution around each point does not guaranteed the existence of any global solution, as is described in another article.
