Theorem: Existence and Uniqueness Theorem for 1st Order IVPs

Let $x' = \map f {t, x}$, $\map x {t_0} = x_0$ be an explicit ODE of dimension $n$.
Let there exist an open ball $V = \sqbrk {t_0 - \ell_0, t_0 + \ell_0} \times \map {\overline B} {x_0, \epsilon}$ of $\tuple {t_0, x_0}$ in phase space $\R \times \R^n$ such that $f$ is Lipschitz continuous on $V$.
{{explain|Notation needs to be explained: $\sqbrk {t_0 - \ell_0, t_0 + \ell_0}$ looks as though it should be an interval (and so needs to be written in Wirth interval notation $\closedint {t_0 - \ell_0} {t_0 + \ell_0}$ so as to abide by house style rules), and $\tuple {t_0, x_0}$ is probably an ordered pair. It's not clear enough. The immediate confusion arises because as $\closedint {t_0 - \ell_0} {t_0 + \ell_0}$ is a closed interval it is counter-intuitive for it to be one of the factors of an open ball expressed as a Cartesian product.}}
Then there exists $\ell < \ell_0$ such that there exists a unique solution $\map x t$ defined for $t \in \closedint {t_0 - \ell_0} {t_0 + \ell_0}$.
{{explain|what is an "IVP"?}}

Proof:
For $0 < \ell < \ell_0$, let $\XX = \map \CC {\closedint {t_0 - \ell_0} {t_0 + \ell_0}; \R^n}$ endowed with the sup norm be the Banach Space of Continuous Functions on Compact Space $\closedint {t_0 - \ell_0} {t_0 + \ell_0} \to \R^n$.
By Fixed Point Formulation of Explicit ODE it is sufficient to find a fixed point of the map $T: \XX \to \XX$ defined by:
:$\ds \map {\paren {T x} } t = x_0 + \int_{t_0}^t \map f {s, \map x s} \rd s$
{{explain|Notation not clear: what actually does $T x$ mean? And is $x$ that fixed point posited?}}
We also have Closed Subset of Complete Metric Space is Complete.
Therefore the Banach Fixed-Point Theorem it is sufficient to find a non-empty subset $\YY \subseteq \XX$ such that:
:$\YY$ is closed in $\XX$
:$T \YY \subseteq \YY$
:$T$ is a contraction on $\YY$
{{explain|Notation not clear: what does $T \YY$ mean?}}
First note that $V$ is closed and bounded, hence compact by the Heine-Borel Theorem.
{{questionable|For Heine-Borel Theorem to apply, $V$ needs to be demonstrated to be not only bounded, but also Definition:Totally Bounded Metric Space.}}
Therefore since $f$ is continuous, by the extreme value theorem, the maximum $\ds m = \sup_{\tuple {t, x} \mathop \in V} \size {\map f {t, x} }$ exists and is finite.
Let $\kappa$ be the Lipschitz constant of $f$.
Let:
:$\YY = \set {y \in \XX: \norm {\map y t - x_0} \le m \size {t - t_0}, t \in \closedint {t_0 - \ell_0} {t_0 + \ell_0} }$
be the cone in $\XX$ centred at $\tuple {t_0, x_0}$.
{{explain|Is this a specialised use of the term "cone" different from the 3-d geometrical construct? If so, it needs to be defined and linked to -- if not, it needs to be explained why it is a cone.}}
Clearly $\YY$ is closed in $\XX$.
Also for $y \in \YY$ we have:
{{begin-eqn}}
{{eqn | l = \size {\map {\paren {T y} } t - x_0}
      | o = \le
      | r = \size {\int_{t_0}^t \map f {s, \map y s} \rd s}
      | c = 
}}
{{eqn | o = \le
      | r = m \int_{t_0}^t \rd s
      | c = Absolute Value of Definite Integral
}}
{{eqn | r = m \set {t - t_0}
      | c = 
}}
{{end-eqn}}
Therefore $T \YY \subseteq \YY$.
Finally we must show that $T$ is a contraction on $\YY$ (we will find that this restricts our choice of $\ell$).
Let $y_1, y_2 \in \YY$.
We have:
{{begin-eqn}}
{{eqn | l = \size {\map {\paren {T y_1} } t - \map {\paren {T y_2} } t}
      | o = \le
      | r = \size {\int_{t_0}^t \map f {s, \map {y_1} s} - \map f {s, \map {y_2} s} \rd s}
      | c = 
}}
{{eqn | o = \le
      | r = \int_{t_0}^t \size {\map f {s, \map {y_1} s} - \map f {s, \map {y_2} s} } \rd s
      | c = Absolute Value of Definite Integral
}}
{{eqn | o = \le
      | r = \kappa \int_{t_0}^t \size {\map {y_1} t - \map {y_2} t} \rd s
      | c = Lipschitz Condition
}}
{{eqn | o = \le
      | r = \kappa \size {t - t_0} \norm {y_1 - y_2}_\sup
      | c = Estimation Lemma for Contour Integrals
}}
{{end-eqn}}
{{explain|Specific meaning of $\norm {y_1 - y_2}_\sup$ needs to be established.}}
Taking the supremum over $t$ we have:
:$\norm {T y_1 - T y_2}_\sup \le \kappa \ell \norm {y_1 - y_2}_\sup$
for all $y_1, y_2 \in \YY$.
Therefore choosing $\ell < \kappa^{-1}$, $T$ is a contraction on $\YY$ as required.
This completes the proof.
{{qed}}
Category:Differential Equations
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