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Let $D\subset[a,b]\times\mathbb{R}^n$ open, $f:D\rightarrow\mathbb{R^n}$ continuous and locally lipschitz-continuous w.r.t $y$ and let $(t_0,y_0)\in D$. If the solution of

$y'(t)=f(t,y(t)),\hspace{1cm} y(t_0)=y_0\in\mathbb{R}^n$

f.a $t\in [a,b]$ exists, then f.a $\varepsilon>0$ there is >$\delta=\delta(\varepsilon)>0$, such that:

$(i)$ If $\|y_0-z_0\|<\delta$ then there exists also the solution of

$z'(t)=f(t,z(t)),\hspace{1cm}z(t_0)=z_0\in\mathbb{R}^n$ for $t\in[a,b]$

Proof

Since $D$ is open there exists $\bar{\delta}>0$ and a compact set $K=\{(t,z(t)):t\in[a,b],\|y(t)-z(t)\|\leq\bar{\delta}\}\subset D$. f is lipschitz continuous w.r.t $y$ on $K$ with lipschitz constant $L$. Let $\delta<\bar{\delta}$ and $\|y_0-z_0\|<\delta$. Then for all $t_0,t\in [a,b]$ it holds that

$\|y(t)-z(t)\|\leq\delta+L\int_{t_0}^{t}\|y(x)-z(x)\|dx$?

But why? I can't really see where the integral is comming from.

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  • $\begingroup$ You mean ${\cal D}\supset [a,b]\times \Bbb R^n$ as open superset, else finding solutions on $[a,b]$ with no regard to any restrictions to $\cal D$ makes no sense. $\endgroup$ Commented Jul 12, 2019 at 10:58

1 Answer 1

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$y(t)-z(t)=y(t_0)-z(t_0)+\int_{t_0}^{t} (y'(s)-z'(s))\, ds$ so $\|y(t)-z(t)\|\leq \|y(t_0)-z(t_0)\|+\int_{t_0}^{t} \|(f(s,y(s))-f(s,z(s)))\|\, ds$. Now apply Lipschitz condition.

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