[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:
$$x_k(t)=\{\begin{array}{ll}{x}_0& (k=0)\\ x_0+{\int }_{t_0}^{t}f(t,x_{k-1})dt& (k>0)\end{array}$$
which converges to $x$ as $n\to \mathrm{\infty }$.
Let $\mathrm{\Omega }:=\{(t,x)\mid \parallel x-x_0\parallel \le M|t-t_0|,|t-t_0|\le \delta \}$. if $\delta$ is small enough, $\mathrm{\Omega }\subset G=[t_0-a,t_0+a]\times D$.

The graph of $x_k(t)$ is contained in $\mathrm{\Omega }$ for $I_{\delta }$.
We show the graph of $x^k(t)$ $(k=1,2,\dots )$ is contained in $G$.
[$k=0$]: Apparent.
[$k\ge 1$]: Suppose the graph of $x_{k-1}(t)$ is contained in $\mathrm{\Omega }$. i.e. $(t,x_{k-1}(t))\in \mathrm{\Omega }(\mathrm{\forall }t\in I_{\delta })$.
$$\parallel x_k(t)-x_0\parallel =|{\int }_{t_0}^{t}f(t,x_{k-1}(t))dt|\le M{\int }_{t_0}^t|dt|=M|t-t_0|.$$
$\therefore (t,x_k(t))\in \mathrm{\Omega }\subset G$.

Next, we show $\{x_k(t)\}$ converges to a certain function over $\mathrm{\Omega }$.
$$x_k(t)=x_0(t)+\sum _{l=1}^k(x_l(t)-x_{l-1}(t)).$$
$$\begin{array}{rcl}{x}_k(t)-x_{k-1}(t)& =& {\int }_{t_0}^{t}f(s,x_{k-1}(s))-f(s,x_{k-2}(s))ds\\ \Vert x_k(t)-x_{k-1}(t)\Vert & =& \Vert {\int }_{t_0}^{t}f(s,x_{k-1}(s))-f(s,x_{k-2}(s))ds\Vert \\ & \le & L{\int }_{t_0}^t\Vert x_{k-1}(s)-x_{k-2}(s)\Vert dt\\ & \le & L\underset{t_0\le s\le t}{sup}\Vert x_{k-1}(s)-x_{k-2}(s)\Vert \cdot |t-t_0|\\ & \le & L\delta \underset{t_0\le s\le t}{sup}\Vert x_{k-1}(s)-x_{k-2}(s)\Vert .\end{array}$$
By taking $\delta \le \frac{1}{2L}$,

$$\begin{array}{rcl}\Vert x_k(t)-x_{k-1}(t)\Vert & \le & \frac{1}{2}\underset{s\in I_{\delta }}{sup}\Vert x_{k-1}(s)-x_{k-2}(s)\Vert \\ \underset{t\in I_{\delta }}{sup}\Vert x_k(t)-x_{k-1}(t)\Vert & \le & \frac{1}{2}\underset{t\in I_{\delta }}{sup}\Vert x_{k-1}(t)-x_{k-2}(t)\Vert \\ & \le & \cdots \\ & \le & \frac{1}{2^{k-1}}\underset{t\in I_{\delta }}{sup}\Vert x_1(t)-x_0(t)\Vert =\text{Const.}\end{array}$$

$$\therefore \underset{k\to \mathrm{\infty }}{lim}{x}_k(t)-x_{k-1}(t)=0.$$
By Weierstrass dominated convergence theorem:
$$\parallel x_k(t)\parallel \le C_k,\sum _{k=0}^{\mathrm{\infty }}{C}_k<\mathrm{\infty }⟹\sum _{k=0}^{\mathrm{\infty }}{x}_k(t)\text{converges,}$$
thus $\sum _{k=1}^{\mathrm{\infty }}(x_k(t)-x_{k-1}(t))$ converges uniformly in t.
So $x$, namely the limit of $x_k(t)$
$$x(t)=\underset{k\to \mathrm{\infty }}{lim}{x}_k(t)=x_0+\underset{k\to \mathrm{\infty }}{lim}\sum _{l=1}^k(x_l(t)-x_{l-1}(t))$$
exists.
$$\begin{array}{rcl}\parallel f(t,x_k(t))-f(t,x(t))\parallel & \le & L\parallel x_k(t)-x(t)\parallel \\ & \le & L\underset{t\in I_{\delta }}{sup}\parallel x_k(t)-x(t)\parallel \\ \underset{t\in I_{\delta }}{sup}\parallel f(t,x_k(t))-f(t,x(t))\parallel & \le & L\underset{t\in I_{\delta }}{sup}\parallel x_k(t)-x(t)\parallel \to 0(k\to \mathrm{\infty })\end{array}$$
So $f(t,x_k(t))\to f(t,x(t))(k\to \mathrm{\infty })$ uniformly in $t$.

$$\begin{array}{rcl}x(t)& =& x_0+\underset{k\to \mathrm{\infty }}{lim}{\int }_{t_0}^{t}f(s,x_{k-1}(s))ds\\ & =& x_0+{\int }_{t_0}^t\underset{k\to \mathrm{\infty }}{lim}f(s,x_{k-1}(s))ds(\because \text{uniform convergence})\\ & =& x_0+{\int }_{t_0}^{t}f(s,x(s))ds\end{array}$$

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
$$\dot{x}=\frac{d}{dt}{x}_0+\frac{d}{dt}{\int }_{t_0}^{t}f(s,x(s))ds=f(s,x(s)).$$