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Let $G\subset\mathbb{R}^n$ be open, $f\colon\mathbb{R}\times G\to\mathbb{R}^n$ continious and locally Lipschitz-continious in $x$, consider the IVP $$ \begin{cases}\dot{x}=f(t,x)\\x(t_0)=x_0\end{cases} $$ with $t_0\in\mathbb{R}, x_0\in G$. Without loss of generality we can assume that $f(t,0)=0$.

Hello! I do not see why we can assume wlog that $f(t,0)=0$. Does anybody see that and can explain it to me?

I guess it is so easy, but I am too blind...

With greetings


In a book I found this, maybe you can explain this to me.

Let $\tilde{x}(t), t\geqslant t_0$, be a marked solution of the ODE with initial value $\tilde{x}(t_0)=\tilde{x}_0$. And let $x(t)$ be the solution of the IVP above. Set $y(t):=x(t)-\tilde{x}(t)$. Then it is $$ \dot{y}(t)=f(t,x(t))-f(t,\tilde{x}(t))=f(t,y(t)+\tilde{x}(t))-f(t,\tilde{x}(t)). $$ Set $$ g(t,y(t)):=f(t,y(t)+\tilde{x}(t))-f(t,\tilde{x}(t)). $$ So one gets the ODE $\dot{y}=g(t,y)$. It is $g(t,0)=0$, so $\dot{y}=g(t,y(t))$ has the trivial solution $y_*\equiv 0$ to the starting value $y_*(t_0)=0$.

Therefore its enough to consider $f(t,0)=0$.

Maybe you can explain that to me because I do not understand this argumentation.

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If $f(t,0)=\mathbf{v} \neq \mathbf{0}$, then we consider $\tilde{f}(t,\mathbf{x}) = f(t,\mathbf{x})-\mathbf{v}$. It is easily checked that $\tilde{f}$ has the same properties as $f$.

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I found sth. in a book. I will edit it. Pls have a look on that in some minutes. – math12 May 4 '14 at 10:28
It is totally trivial in your case. You start with a function with some properties, and you say that it is not restrictive to assume that its value at a given point is zero. If not, just consider a translation of the function. – Siminore May 4 '14 at 11:00

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