[test] 板書メモ: 初期値問題の解の存在

Assume there exist a solution $x$ like this:
$$ x(t) = x_0 + \int_0^t f(t,x)\,dt $$
and let $x_k$ be:
$$x_k(t) = \begin{cases} x_0 & (k=0) \\ x_0+\int_{t_0}^t f(t,x_{k-1})\,dt & (k>0) \end{cases} $$
which converges to $x$ as $n\to\infty$.
Let $\Omega := \left\{(t, x) \mid \|x-x_0\|\leq M|t-t_0|,\,|t-t_0|\leq\delta\right\}$. if $\delta$ is small enough, $\Omega\subset G=[t_0-a,t_0+a]\times D$.

The graph of $x_k(t)$ is contained in $\Omega$ for $I_\delta$.
We show the graph of $x^k(t)$ $(k=1,2,\ldots)$ is contained in $G$.
[$k=0$]: Apparent.
[$k\geq 1$]: Suppose the graph of $x_{k-1}(t)$ is contained in $\Omega$. i.e. $(t,x_{k-1}(t))\in\Omega\;(\forall t\in I_\delta)$.
$$\|x_k(t)-x_0\| = \left|\int_{t_0}^t f(t,x_{k-1}(t))\,dt\right| \leq M\int_{t_0}^t |dt| = M|t-t_0|.$$
$\therefore (t,x_k(t))\in\Omega\subset G$.

Next, we show $\{x_k(t)\}$ converges to a certain function over $\Omega$.
\begin{equation} x_k(t) = x_0(t) + \sum_{l=1}^k \left(x_l(t)-x_{l-1}(t)\right). \end{equation}
\begin{eqnarray} x_k(t) - x_{k-1}(t) &=& \int_{t_0}^t f(s,x_{k-1}(s))-f(s,x_{k-2}(s))\,ds \\ \left\|x_k(t) - x_{k-1}(t)\right\| &=& \left\|\int_{t_0}^t f(s,x_{k-1}(s))-f(s,x_{k-2}(s))\,ds\right\| \\ &\leq& L \int_{t_0}^t \left\| x_{k-1}(s) - x_{k-2}(s)\right\| dt \\ &\leq& L \sup_{t_0\leq s\leq t} \left\| x_{k-1}(s) - x_{k-2}(s)\right\| \cdot |t-t_0| \\ &\leq& L\delta \sup_{t_0\leq s\leq t} \left\| x_{k-1}(s) - x_{k-2}(s)\right\|. \\ \end{eqnarray}
By taking $\delta\leq\frac{1}{2L}$,

\begin{eqnarray} \left\|x_k(t) - x_{k-1}(t)\right\| &\leq& \frac{1}{2} \sup_{s\in I_\delta} \left\| x_{k-1}(s) - x_{k-2}(s)\right\| \\ \sup_{t\in I_\delta}\left\|x_k(t) - x_{k-1}(t)\right\| &\leq& \frac{1}{2} \sup_{t\in I_\delta} \left\| x_{k-1}(t) - x_{k-2}(t)\right\| \\ &\leq& \cdots \\ &\leq& \frac{1}{2^{k-1}} \sup_{t\in I_\delta} \left\| x_1(t) - x_0(t)\right\| = \text{Const.} \\ \end{eqnarray}

$$\therefore\quad \lim_{k\to\infty} x_k(t)-x_{k-1}(t)=0.$$
By Weierstrass dominated convergence theorem:
$$\|x_k(t)\|\leq C_k,\;\sum_{k=0}^\infty C_k<\infty \implies \sum_{k=0}^\infty x_k(t)\; \text{converges,}$$
thus $\sum_{k=1}^\infty \left(x_k(t)-x_{k-1}(t)\right)$ converges uniformly in t.
So $x$, namely the limit of $x_k(t)$
$$x(t) = \lim_{k\to\infty}x_k(t) = x_0 + \lim_{k\to\infty}\sum_{l=1}^k\left(x_l(t)-x_{l-1}(t)\right)$$
exists.
\begin{eqnarray} \| f(t,x_k(t))-f(t,x(t)) \| &\leq& L \|x_k(t)-x(t)\| \\ &\leq& L \sup_{t\in I_\delta}\|x_k(t)-x(t)\| \\ \sup_{t\in I_\delta} \| f(t,x_k(t))-f(t,x(t)) \| &\leq& L \sup_{t\in I_\delta}\|x_k(t)-x(t)\| \to 0\quad(k\to\infty)\\ \end{eqnarray}
So $f(t,x_k(t))\to f(t,x(t))\quad(k\to\infty)$ uniformly in $t$.

\begin{eqnarray} x(t) &=& x_0 + \lim_{k\to\infty} \int_{t_0}^t f(s,x_{k-1}(s))\,ds \\ &=& x_0 + \int_{t_0}^t \lim_{k\to\infty} f(s,x_{k-1}(s))\,ds \quad(\because \text{uniform convergence})\\ &=& x_0 + \int_{t_0}^t f(s,x(s))\,ds \\ \end{eqnarray}

By assumption, $f(s,x(s))$ is continuous in $s$. So $\int_{t_0}^t f(s,x(s))\,ds$ is a $C^1$ function.
We can then differentiate $x$ and we have
$$\dot{x} = \frac{d}{dt}x_0 + \frac{d}{dt}\int_{t_0}^t f(s,x(s))\,ds = f(s,x(s)).$$