28: Why Local Solution Existence Does Not Guarantee Global Existence for Euclidean-Normed Euclidean Vectors Space ODE

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A description of why the local solution existence does not guarantee the global solution existence for Euclidean-normed Euclidean vectors space ODE

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Topics

About: normed vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Note

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Starting Context

• The reader knows a definition of Euclidean-normed Euclidean vectors space.
• The reader admits the local unique solution existence for a closed interval domain for a Euclidean-normed Euclidean vectors space ordinary differential equation.

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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.

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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.

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Main Body

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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, $[t_0-{ϵ}_{t_0,x_0,1},t_0+{ϵ}_{t_0,x_0,2}]$, where ${ϵ}_{t_0,x_0,i}$ 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, $x(t):[t_1-{ϵ}_{t_1,x_1,1},t_1+{ϵ}_{t_1,x_1,2}]\to {\mathbb{R}}^d$, by the initial condition at $t_1$, then get the connected next local solution by the initial condition at $t_2=t_1+{ϵ}_{t_1,x_1,2}$ with the value of $x(t_2)$, getting the extended solution, $x(t):[t_1,t_2+{ϵ}_{t_2,x(t_2),2}]$, and so on. But the issue is that there is no guarantee that $t_i+{ϵ}_{t_i,x(t_i),2}$ eventually reaches $t_e$, because it may converge to a value, $t_l<t_e$, 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 $[t_l-{ϵ}_{t_l,x_{t_l},1},t_l+{ϵ}_{t_l,x_{t_l},2}]$? ... 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).

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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 $[t_l-{ϵ}_{t_l,x_l,1},t_l+{ϵ}_{t_l,x_l,2}]$ local solution to be able to be connected to the not-fully-extended solution.

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References

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