I'm very new to ODE in $\Bbb{R}^n$, so it might take some time to get used to proofs in this area. For now, I'm faced with this problem:
Let $\Vert\;\Vert$ be the Euclidean norm on $\Bbb{R}^n$. Let $x:[0,a]\to \Bbb{R}^n$ be differentiable. We consider the following O.D.E
\begin{align}\begin{cases}x'(t)=f(t,x(t)) & t\geq 0,\\x(0)=x_0\in \Bbb{R}^n&\end{cases}\end{align} where \begin{align}f:\Bbb{R}^{+}\times \Bbb{R}^{n}\to \Bbb{R}^{n}\end{align}is continuous such that $\langle x,f(t,x)\rangle\leq 0,\;\forall\;t\geq 0,\;x\in\Bbb{R}^n.$ So, how do I prove that the solution of $f$ exists on $\Bbb{R}^+?$ Thanks for your time and help!
Existence for a short time is given by Peano's Theorem, see https://en.wikipedia.org/wiki/Peano_existence_theorem
Existence for any $t\geq0$ usually needs $f$ to be locally Lipschitz, see e.g. https://en.wikipedia.org/wiki/Ordinary_differential_equation#Global_uniqueness_and_maximum_domain_of_solution Let us assume you have that. The only way the solution ceases to exist at a time $T_0$, is by 'blowing up', i.e. $\|x(t)\|\rightarrow\infty$ for $t\rightarrow T_0$. Let us take a look at $$\frac{d}{dt}(\|x(t)\|^2)=2\langle x',x\rangle=2\langle f(t,x(t)),x(t)\rangle\leq 0.$$ Hence $\|x(t)\|$ is monotonically decreasing and therefore $\|x(t)\|\leq \|x(0)\|$. Hence it cannot blow up.