I want to verify if I made this proof correctly (original version in spanish so please excuse my translation mistakes)...

Let $f: \mathbb{R} \to\mathbb{R}^n$ continuous and Lipschitz on second variable. Show that for each $(t_0, x_0) \in \mathbb{R} \times \mathbb{R}^n $ Cauchy Problem has unique solution in $\mathbb{R}$.


As $f: \mathbb{R} \to\mathbb{R}^n $ is continuous and Lipschitz, let's take a compact interval $\textrm{I}=[a, b]: a, b \in \mathbb{R}$ , where $f$ is continuous and lipschitz, and $\lvert t-t_0\rvert \leq (b-a)$, with $\lvert f\rvert \leq M$ in $\textrm{I}\times \mathbb{R}^n$ and let $\textrm{X} = \mathscr{C}(\textrm{I}, \mathbb{R}^n) $ the metric space of continuous functions $\varphi: \textrm{I} \to \mathbb{R}^n$ with uniform metric $d(\varphi_1,\varphi_2)=max\{\lvert \varphi_1(t)-\varphi_2(t)\rvert: t \in \mathbb{R}\}$.

For $\varphi \in \textrm{X}$ we define $\textrm{F}:\textrm{X} \to \textrm{X}$ so that $\textrm{F}(\varphi (t))=x_0+\int_{t_0}^{t} f(s, \varphi (s)) ds$

  1. $\textrm{F}(\varphi (t)) \subseteq \textrm{X} $ because it's defined as $\textrm{F}:\textrm{X} \to \textrm{X}$ and $\lvert \textrm{F}(\varphi(t))-x_0\rvert = \lvert \int_{t_0}^{t}f(s, \varphi(s)) ds\rvert \leq M \lvert t-t_0 \rvert \leq M(b-a)$

  2. It has to be shown that $\textrm{F}^n$ is a contraction. We have that for each $\varphi_1,\varphi_2 \in \textrm{X}$ and all $n\geq 0 $ $\lvert{\textrm{F}^{n}(\varphi_1(t))-\textrm{F}^{n}(\varphi_2(t))}\rvert \leq \frac{K^{n}\lvert t-t_0 \rvert^{n}}{(n)!}d(\varphi_1(t) , \varphi_2(t)), \ \ \ t\in \textrm{I}$

For $n=0$
$\lvert{\textrm{F}^{0}(\varphi_1(t))-\textrm{F}^{0}(\varphi_2(t))}\rvert = \lvert \varphi_1(t) , \varphi_2(t) \rvert \leq \frac{K^{0}\lvert t-t_0 \rvert^{0}}{(0)!}d(\varphi_1(t) , \varphi_2(t))=d(\varphi_1(t) , \varphi_2(t))$

Thus inequality is satisfied. Now assume that it is valid for $r$

Then consider $\lvert{\textrm{F}^{r+1}(\varphi_1(t))-\textrm{F}^{r+1}(\varphi_2(t))}\rvert $

So $\lvert{\textrm{F}^{r+1}(\varphi_1(t))-\textrm{F}^{r+1}(\varphi_2(t))}\rvert = \lvert{\textrm{F}(\textrm{F}^r(\varphi_1(t)))-\textrm{F}(\textrm{F}^{r}(\varphi_2(t)))}\rvert \leq$ $\leq \lvert{\int_{t_0}^{t}{\lvert{f(s,\textrm{F}^{r}(\varphi_1(s)))-f(s,\textrm{F}^{r}(\varphi_2(s)))}\rvert ds}}\rvert \leq$ $\leq \lvert{\int_{t_0}^{t}{K \lvert{\textrm{F}^{r}(\varphi_1(s))-\textrm{F}^{r}(\varphi_2(s))} \rvert ds}}\rvert \leq \frac{K^{r+1}\lvert t-t_0 \rvert^{r+1}}{(r+1)!}d(\varphi_1(t) , \varphi_2(t))$

Let's take $n=r+1$ then:

$d(\textrm{F}^{n}(\varphi_1(t))-\textrm{F}^{n}(\varphi_2(t))) \leq \frac{K^{n}(b-a)^{n}}{(n)!}d(\varphi_1(t) , \varphi_2(t)) $ For $n \to +\infty \ \ : \lim\limits_{n \to +\infty}\frac{K^{n}(b-a)^{n}}{(n)!}= 0$

So for $n$ great, $\textrm{F}^{n}$ is a contraction. Applying Banach's Theorem for $F^n$ contractions, there is a unique fixed point for $\textrm{F}$ so that there's a unique solution $\varphi (t) \in \mathbb{R} : \textrm{F}(\varphi(t))=\varphi(t)$

Now, for $\mathbb{R}$ let's take $\mathbb{R}= \displaystyle\bigcup_{n}\ {\textrm{I}_n }:\ t_0\in\textrm{I}_n \ \ ;\ \ \textrm{I}_n \subseteq \textrm{I}_{n+1}$, the union de infinite compact intervals containing $t_0$. If $\varphi_n$ is the unique solution through $(t_0, x_0)$ in $\textrm{I}_n$, then for uniqueness of solution, we have that $\varphi_{n+1}|\ _{\textrm{I}_n}=\varphi_{n}$, so $\varphi(t)=\varphi_n(t), \ t\in\mathbb{R}$, defined in $\mathbb{R}$ and it is the unique solution through $(t_0, x_0)$. \ \ $\square$

  • $\begingroup$ Note that $F^{r+1}(x)=F(F^r(x)))$, not $F^r(F^r(x)))$. You might have to say a sentence more on why the fixed point for $F^n$ is also a fixed point for $F$. $\endgroup$ – LutzL Apr 21 '15 at 16:15
  • $\begingroup$ Thanks! that was a typing error... I'll extend comments about Banach Theorem for $F^n$ contraction $\endgroup$ – Ralexar Apr 21 '15 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.