Banach Fixed Point Theorem and Differential Equations

Question

Let $F:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be $L$-Lipschitz. Using the Banach fixed point theorem, I know we can prove that the differential equation $$ g'(t) = F(g(t)), \qquad g(0) = z$$ has a unique solution on the interval $[0,\varepsilon]$, where $\varepsilon$ is small enough so that $L\varepsilon < 1$.

I am told and have read that from here we go can from the interval $[0,\varepsilon]$ to obtain a solution on all of $\mathbb{R}$.

I understand that in common proof of this, we didn't have to start the integral at $0$, we could have obtained the solution on any interval of length $\varepsilon$. But I can't see how to put these solutions together.

Answer 1

Apply to your special case the theorem (adapted from Loomis, Advanced Calculus p. 267):

Let $A$ be an open subset of a Banach space $W$, let $I$ be an open interval in $\mathbb R$, and let $F$ be a continuous mapping from $I \times A$ to $W$ which is locally uniformly Lipschitz in its second variable. Then for any point $(t_0, z_0)$ in $I \times A$ for some neighborhood $U$ of $z_0$ and for any sufficiently small interval $J$ containing $t_0$, there is a unique function $f: J\to U$ which is a solution of the differential equation passing through the point $(t_0,z_0).$

$1).\ $ First prove that if $g_1$ and $g_2$ are solutions through $(t_0, z_0)$ on intervals $I_1\ni t_0$ and $I_2\ni t_0$, respectively, then $g_1 (t) = g_2(t)$ on $I_1\cap I_2$.

$2).\ $ Let $\mathscr F$ be the family of all solutions that pass through $(t_0,z_0).$ Then, the above shows that $f:=\bigcup \mathscr F$ is a function, and by construction, it is the unique maximal solution of the differential equation with the given initial condition.

Now that we know there $\text{is}$ a maximal solution,

$3).\ $ Suppose $g$ is a maximal solution on a finite interval $(a,b)$. We can choose an open interval $I$ containing $b$ and a $t_1$ so that $b - t_1 < 1/L.$ Put $z_1=g(t_1)$ and $M=\|F(z_1)\|.$ The fundamental theorem on existence/uniqueness now applies to produce a local solution $f$ that passes through $(t_1,z_1)$ in the $\delta$-interval of $t_1$ for any $\delta <r/(M + rL)$, where $r$ is a positive constant that depends on the open set $A$. But in our case, $A$ is the whole of $\mathbb R^n$ so $r$ is in fact, unrestricted. Thus, $\delta$ is bounded by $1/L$ and so because of how we chose $t_1$, we can choose $\delta$ so that $t_1+\delta>b$. But the maximal solution $g$ that passes through $(t_1,z_1)$ includes $f$, so we have a contradiction.