I am reading Milnor's Topology from the Differential Viewpoint, p.23. The paragraph below is extracted from the book:
Let $\varphi : \Bbb R^n \to \Bbb R$ be a smooth function. Given any fixed unit vector $c=(c_1,...,c_n)\in S^{n-1}$ and a point $x_0 \in \Bbb R^n$, consider the differential equations $$ \frac{dx_i}{dt}(t)=c_i \cdot \varphi (x_1(t),...,x_n(t))~(i=1,...,n),~~ ~x(0)=x_0.$$ These equations have a unique solution $x=x(t)=(x_1(t),...,x_n(t))$ defined for all real numbers.
I know that existence and uniqueness follow from a theorem in ODE, but how can we assure that the domain of $x$ is the whole $\Bbb R$?
No global existence.
For example, $n=2$, $\varphi(x,y)=x^2+y^2$, $(x_0,y_0)=(1,1)$ and $c=(3/5,4/5)$.
Then $$ x'=3(x^2+y^2)/5, \quad y'=4(x^2+y^2)/5, $$ and hence $$ x'\ge \frac{3x^2}{5}\quad\Longrightarrow\quad \frac{x'}{x^2}\ge \frac{3}{5} \quad\Longrightarrow\quad \frac{1}{x(0)}-\frac{1}{x(t)}\ge \frac{3t}{5} \quad\Longrightarrow\quad x(t)\ge \frac{1}{1-\frac{3t}{5}}=\frac{5}{5-3t}. $$ I.e., blow-up before $t=5/3$.
However, if $|\varphi(x)|\le a|x|+b$, then the system of ODEs possesses a global solution.