Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Update

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.

share|improve this question

1 Answer 1

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 $f$.

share|improve this answer
    
I found sth. in a book. I will edit it. Pls have a look on that in some minutes. –  math12 May 4 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 at 11:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.