\classheader{2013-02-15}

Today we'll talk about a method for reducing $n$th order differential equations to first order differential equations. Thus, for example, we can express acceleration as a function of position and velocity.

Let $\Omega\subset\R^n$ be open, and let $a:\Omega\times\R^n\to\R^n$ be a given function (representing acceleration). Then given $(x_0,y_0)\in\Omega\times\R^n$, there is a unique $\gamma:(-\epsilon,\epsilon)\to\Omega$ such that $\gamma(0)=x_0$, $\gamma'(0)=y_0$, and $\gamma''(t)=a(\gamma(t),\gamma'(t))$ for all $t\in(-\epsilon,\epsilon)$.

The trick was to introduce the space $\Omega\times\R^n$. Put $M=\Omega\times\R^n$, and let $\delta(t)=(\gamma(t),\gamma'(t))$. Any vector field on $M$ can be thought of as a function $M\to\R^n\times\R^n$; let $\widetilde{v}:M\to\R^n\times\R^n$ be the vector field defined by
\[\widetilde{v}(x,y)=(y,a(x,y)).\]
If $\gamma$ is as above, then we have $\delta(0)=(x_0,y_0)\in M$, and $\delta'(t)=(\gamma'(t),\gamma''(t))=\widetilde{v}(\delta(t))$. Thus, curves $\gamma$ satisfying our requirements are precisely integral curves of the vector field $\widetilde{v}$ on $M$.

\begin{theorem}
Let $M$ be a $\Cinfty$ manifold and let $v$ be $\Cinfty$ vector field on $M$. Let $p\in M$. Then there is a neighborhood $U(p)$ of $p$ in $M$, some $a>0$, and a $\Cinfty$ function $\gamma:U(p)\times(-a,a)\to M$ such that for all $x\in U(p)$, $\gamma_x(t)$ is the (unique) integral curve of $v$ such that $\gamma_x(0)=x$, where $\gamma_x(t):=\gamma(x,t)$.
\end{theorem}

A reference for this theorem is Hurewicz's \textit{Lectures on Ordinary Differential Equations}.

\begin{remark-N}
Let $\gamma$ be an integral curve of a $\Cinfty$ vector field. Then it is clear that $\gamma$ is $\Cinfty$, because we have that
\[\gamma'(t)=v(\gamma(t)),\]
so that if $\gamma$ is $C^k$, then $v\circ \gamma$ is $C^k$, hence $\gamma'$ is $C^k$, so that $\gamma$ is $C^{k+1}$.
\end{remark-N}

\begin{remark-N}
Assume that $v$ is a $C^1$ vector field. The contraction principle shows that $(x,t)\mapsto \gamma_x(t)$ is continuous in $(x,t)$ and defined on some neighborhood $U(p)\times(-a,a)$, as follows:

WLOG, let $p=0$, and let $M=\Omega$, an open subset of $\R^n$. Then the space of continuous functions
\[\{x\in\R^n\mid \|x\|\leq\alpha\}\times[-c,c]\;\;\longrightarrow\;\;\{x\in\R^n\mid \|x\|\leq R\}\]
with the $\|\cdot\|_\infty$ norm is a complete metric space.
\end{remark-N}
\begin{remark-N}
Assume that $v$ is a $C^2$ vector field on $\Omega$, an open subset of $\R^n$. We want to show that the partial derivatives
\[\frac{\partial}{\partial x_i}(\gamma_x(t))\]
exist. We can simply check that the differential equation defining them is satisfied. Let $v:\Omega\to\R^n$ be a vector field on $\Omega$, and so that for any $x\in\Omega$ we have $v'(x):\R^n\to\R^n$.

We introduce the (standard) notation $\phi_t(x)=\gamma_x(t)$. Thus, $\phi_t$ is defined at all $t\in(-a,a)$, and $\phi_t$ is a function $\phi_t:U(p)\to\Omega$, so that we have $\phi_t'(x):\R^n\to\R^n$ for any $x\in U(p)$, where $\phi_t'(x)$ is the derivative of $x\mapsto v(\phi_t(x))$.

Let $\widetilde{v}:M\to \R^n\times\M_n(\R)$ be the vector field on $M=\Omega\times\R^n$ defined by
\[\widetilde{v}(x,S)=(v(x),v'(x)S).\]
We claim that, for any fixed $t$, the map $x\mapsto(\phi_t(x),\phi_t'(x))$ is an integral curve for $\widetilde{v}$.

We have that
\[\frac{d}{dt}\phi_t(x)=v(\phi_t(x)),\]
so that
\[\widetilde{v}(\phi_t(x),\phi_t'(x))=\frac{d}{dt}(\phi_t(x),\phi_t'(x))=(v(\phi_t(x)),v'(\phi_t(x))\phi_t'(x)).\]
This ``reduces'' the question to integral curves of $\widetilde{v}$ on $\Omega\times\M_n(\R)$ (?).
\end{remark-N}


\begin{corollary}
Let $M$ be a $\Cinfty$ manifold, and let $v$ be a $\Cinfty$ vector field on $M$. For all $x\in M$, we've shown that there is a maximal integral curve $\gamma_x:I_x\to M$ such that $\gamma_x(0)=x$. Let
\[D=\{(x,t)\in M\times\R\mid t\in I_x\},\]
and write $\gamma_x(t)=\gamma(x,t)$ for all $(x,t)\in D$.

\begin{enumerate}
\item $D$ is an open subset of $M\times\R$.
\item If $\phi_t(x)$ is defined, then $\phi_s(x)$ is defined for all $0\leq s\leq t$ (or all $t\leq s\leq 0$).
\item If $\phi_t(x)$ is defined and if $\phi_s(\phi_t(x))$ is defined, then $\phi_{t+s}(x)$ is defined and
\[\phi_{t+s}(x)=\phi_s(\phi_t(x)).\]
\item Assume $M$ is compact. Then the theorem implies that there is an open interval $(-a,a)$ such that $(-a,a)\subset I_x$ for all $x\in M$, so that $D\supseteq M\times(-a,a)$; statement 3 then implies $D=M\times\R$. Furthermore, for a fixed $t$, the map $x\mapsto \phi_t(x)$ is $\Cinfty$ and induces an isomorphism on tangent spaces.
\item Observe that given a vector field $v$ on $M$, if a constant path $\gamma(t)=p$ is an integral curve of $v$, then $v(p)=0$. Conversely, if $v(p)=0$, then the constant curve to $p$ is an integral curve.
\end{enumerate}
\end{corollary}
\begin{proof}[Proof of 3]
We know $\gamma_x$ is defined on $[0,t+\epsilon)$, and that $\gamma_{\phi_t(x)}$ is defined on $[0,s+\epsilon')$. Now define
\[\delta(z)=\begin{cases}\gamma_x(z)&\text{ for all }z\in[0,t],\\
\gamma_{\phi_x(t)}(z-t)&\text{ for all }z\in[t,s+\epsilon).\end{cases}\]
Note that $\delta$ is an integral curve of $v$.

\end{proof}
\begin{exercise}
Let $\R\xrightarrow{\;\;h\;\;}\mathrm{Diffeo}(M)$ be a group homomorphism such that the map $M\times\R\to M$ defined by $(x,t)\mapsto h(t)(x)$ is $\Cinfty$. Prove that there is a $\Cinfty$ vector field $v$ on $M$ such that $h(t)=\phi_t$ for the vector field $v$. This does not require $M$ to be compact, just that the support of $v$ is compact.
\end{exercise}

\begin{remark}
If $v$ is a vector field on a manifold $M$ and $v(p)\neq 0$, then there is a neighborhood $U(p)$ of $p$ in $M$ and a diffeomorphism $f:U(p)\to \Omega$, where $\Omega$ is an open subset of $\R^n$, such that the vector field $v$ (restricted to $U(p)$) is carried by $f$ to the constant vector field $\frac{\partial}{\partial x_1}$ on $\Omega$.
\end{remark}
\begin{proof}[Proof of Remark]
Let's say that $Z$ is a slice of $M$ when $T_pZ\oplus \R v(p)=T_pM$. The hypotheses of the inverse function theorem hold at $(p,0)$, so that we can find coordinates in which the map from $Z\times(-\epsilon,\epsilon)\to M$ defined by $\phi_t(z)=\gamma_z(t)$ sends the vector field $v$ to a constant vector field, such as for example $\frac{\partial}{\partial x_1}$.
\end{proof}

Next time, we'll start with Lie brackets. It's motivated from two points of view; maybe 100 in fact.