# Continuous dependence on initial conditions: why $\|y(t)-z(t)\|\leq\delta+L\int_{t_0}^{t}\|y(x)-z(x)\|dx$?

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.

• 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. Commented Jul 12, 2019 at 10:58

$$y(t)-z(t)=y(t_0)-z(t_0)+\int_{t_0}^{t} (y'(s)-z'(s))\, ds$$ 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.