27: Local Unique Solution Existence for Euclidean-Normed Euclidean Vectors Space ODE

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of the local unique solution existence for Euclidean-normed Euclidean vectors space ordinary differential equation

---

Topics

About: normed vectors space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of Euclidean-normed Euclidean vectors space.
• The reader knows a definition of map.
• The reader knows a definition of derivative of normed vectors spaces map.
• The reader knows a definition of map between arbitrary subsets of Euclidean $C^{\mathrm{\infty }}$ manifolds $C^k$ at point, where $k$ excludes $0$ and includes $\mathrm{\infty }$.
• The reader knows a definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.
• The reader admits the contraction mapping principle.
• The reader admits the proposition that derivative of $C^1$, Euclidean-normed Euclidean vectors spaces map is the Jacobian.

---

Target Context

• The reader will have a description and a proof of the local unique solution existence for a closed interval domain for a Euclidean-normed Euclidean vectors space ordinary differential equation, and will have a clarification on the solution domain area.

---

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

For any Euclidean-normed Euclidean vectors spaces, ${\mathbb{R}}^d$ and $\mathbb{R}$, any $x_0$-centered $K$-radius open ball, $B\subseteq {\mathbb{R}}^d$, any interval, $I\subseteq \mathbb{R}:[t_0-{ϵ}_1,t_0+{ϵ}_2]$, and any $C^0$ map, $f:B\times I\to {\mathbb{R}}^d$, that satisfies the Lipschitz estimate, $\Vert f(x_1,t)-f(x_2,t)\Vert \le L\Vert x_1-x_2\Vert$ for any $x_1,x_2\in B$ and any $t\in I$ and satisfies also the 2 inequalities, ${ϵ}_i\le \frac{K}{M}$ (where $M$ is any number that satisfies $su{p}_{x\in B,t\in I}\Vert f(x,t)\Vert \le M$) and ${ϵ}_i<L^{-1}$, the initial-valued ordinary differential equation, $\frac{dx}{dt}=f(x,t)$ on $B\times I$ with the initial condition $x(t_0)=x_0$, has the unique $C^1$ solution, $x:I\to B$, on the entire $I$. It is called "local", because $B$ and $I$ are usually chosen for the solution for the local area in a wider area for which the equation really is.

Note that if $f$ satisfies the Lipschitz estimate on $B\times I$ and $\Vert f(x,t)\Vert$ is finite there, ${ϵ}_i$ can be always chosen to satisfy the inequalities, because if the inequalities do not hold for $I$ after $L$ and $M$ are chosen, a narrower interval $I^{\prime }$ can be chosen to satisfy the inequalities without those $L$ and $M$ moved, because $L$ and $M$ do not need to be bigger for the narrower domain.

---

2: Note 1

While the domain of $x$ is not open on $\mathbb{R}$, $C^1$-ness of $x$ is by the definition of map between arbitrary subsets of Euclidean $C^{\mathrm{\infty }}$ manifolds $C^k$ at point, where $k$ excludes $0$ and includes $\mathrm{\infty }$, and the derivative at each of the boundary points is $\frac{dx}{dt}=\frac{d{x}^{\prime }}{dt}$, where $x^{\prime }:(t_0-{ϵ}_j-ϵ,t_0-{ϵ}_j+ϵ)\to {\mathbb{R}}^d$ is a $C^1$ extension of $x$ such that $x^{\prime }{|}_{(t_0-{ϵ}_j-ϵ,t_0-{ϵ}_j+ϵ)\cap I}=x{|}_{(t_0-{ϵ}_j-ϵ,t_0-{ϵ}_j+ϵ)\cap I}$.

$C^1$-ness of $x$ at each of the boundary points equals the existence of the one-sided derivative with the continuousness; the derivative of $x$ at each of the boundary points equals the one-sided derivative, by the proposition that $C^k$-ness of any map from any (possibly half) closed interval into any subset of any Euclidean $C-\mathrm{\infty }$ manifold at any closed boundary point equals the existence of the one-sided derivatives with continuousness, and the derivatives are the one-sided the derivatives, where $k$ excludes $0$ and $\mathrm{\infty }$.

---

3: Proof

The ordinary differential equation equals $x(t)=x_0+{\int }_{t_0}^{t}f(x(s),s)ds$, because if $x(t)$ satisfies one, it will satisfy the other: suppose that $\frac{dx}{dt}=f(x,t)$, then, $x(t)-x(t_0)={\int }_{t_0}^t\frac{dx}{dt}(s)ds={\int }_{t_0}^{t}f(x,s)ds$; suppose that $x(t)=x_0+{\int }_{t_0}^{t}f(x(s),s)ds$, then, $\frac{dx}{dt}=f(x,t)$.

For the set, $Y:\{y|I\to B\text{ such that }y(t_0)=x_0\}$, of all the $C^0$ maps, define the map, $T:Y\to Y$, by $(Ty)(t)=x_0+{\int }_{t_0}^{t}f(y(s),s)ds$, which is certainly into $Y$, because $\Vert (Ty)(t)-x_0\Vert =\Vert {\int }_{t_0}^{t}f(y(s),s)ds\Vert \le |{\int }_{t_0}^t\Vert f(y(s),s)\Vert ds|\le |{\int }_{t_0}^{t}Mds|\le M{ϵ}_i\le K$, and $Ty$ is continuous.

Let us make $Y$ a complete metric space with $dist(y_1,y_2)=su{p}_{t\in I}\Vert y_1(t)-y_2(t)\Vert$ for any $y_1,y_2\in Y$. That is indeed a metric: for any $y_1,y_2,y_3\in Y$, 1) $dist(y_1,y_2)\ge 0$ with the equality holding if and only if $y_1=y_2$; 2) $dist(y_1,y_2)=dist(y_2,y_1)$; 3) $dist(y_1,y_3)=su{p}_{t\in I}\Vert y_1(t)-y_3(t)\Vert =su{p}_{s\in I}\Vert y_1(t)-y_2(t)+y_2(t)-y_3(t)\Vert \le su{p}_{t\in I}(\Vert y_1(t)-y_2(t)\Vert +\Vert y_2(t)-y_3(t)\Vert )\le su{p}_{t\in I}\Vert y_1(t)-y_2(t)\Vert +su{p}_{t\in I}\Vert y_2(t)-y_3(t)\Vert =dist(y_1,y_2)+dist(y_2,y_3)$. That is indeed complete: for any Cauchy sequence, $y_1,y_2,...$, for any positive $ϵ$, there is an $N$ such that $dist(y_j,y_k)<ϵ$ for each $N<j,k$, but for any $t\in I$, $\Vert y_j(t)-y_k(t)\Vert \le su{p}_{t\in I}\Vert y_k(t)-y_j(t)\Vert <ϵ$, which implies that the sequence converges point-wise with the limit, $y:I\to {\mathbb{R}}^d$, but $\Vert y(t)-x_0\Vert =\Vert y(t)-y_j(t)+y_j(t)-x_0\Vert \le \Vert y(t)-y_j(t)\Vert +\Vert y_j(t)-x_0\Vert =\Vert y(t)-y_j(t)\Vert +K-ϵ$, but $\Vert y(t)-y_j(t)\Vert <ϵ$ for an $N<j$, so, $\Vert y(t)-y_j(t)\Vert +K-ϵ<ϵ+K-ϵ=K$, so, $y$ is into $B$; $\Vert y_k(t)-y_j(t)\Vert \le su{p}_{t\in I}\Vert y_k(t)-y_j(t)\Vert <ϵ$, which means that the sequence converges uniformly, so, the limit is continuous.

Now, $\Vert (T{y}_1)(t)-(T{y}_2)(t)\Vert =\Vert {\int }_{t_0}^{t}f(y_1(s),s)-f(y_2(s),s)ds\Vert \le |{\int }_{t_0}^t\Vert f(y_1(s),s)-f(y_2(s),s)\Vert ds|\le |{\int }_{t_0}^{t}L\Vert y_1(s)-y_2(s)\Vert ds|\le L{ϵ}_{i}su{p}_{t\in I}\Vert y_1(t)-y_2(t)\Vert =L{ϵ}_{i}dist(y_1-y_2)$ where $L{ϵ}_i<1$. So, $dist(T{y}_1,T{y}_2)=su{p}_{t\in I}\Vert (T{y}_1)(t)-(T{y}_2)(t)\Vert \le L{ϵ}_{i}dist(y_1-y_2)$, so, $T$ is a contraction, and by the contraction mapping principle, there is the unique fixed element, $x\in Y$ such that $Tx=x$. As $x=x_0+{\int }_{t_0}^{t}f(y(s),s)ds$, $x$ is $C^1$ over $I$, and that $x(t)$ is the unique (as $x$ is unique among the continuous maps, it is even more unique among the $C^1$ maps) $C^1$ solution of the ordinary differential equation.

---

4: Note 2

It is important to know how ${ϵ}_i$ is determined: it especially depends on the initial condition, which is the reason why the local existences at every point in an interval does not guarantee the global solution existence for the entire interval (refer to another proposition).

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>