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## Main Question or Discussion Point

The relevant equations:

(1) [tex]\dot{x}[/tex] = v(x), x [tex]\in[/tex] U

(2) [tex]\varphi[/tex] = x

Let x

|v(x)| < k|x| for sufficiently small |x| =/= 0, where k > 0 is a positive constant.

Huh? i was good up until the inequality.

If somebody has the book, it is section 2.8 of the first chapter.

(1) [tex]\dot{x}[/tex] = v(x), x [tex]\in[/tex] U

(2) [tex]\varphi[/tex] = x

_{0}, t_{0}[tex]\in[/tex] R x_{0}[tex]\in[/tex] ULet x

_{0}be a stationary point of a vector field v, so that v(x_{0}) = 0. Then, as we now show, the solution of equation (1) satistying the initial condition (2) is unique, ie., if [tex]\varphi[/tex] is any solution of (1) such that [tex]\varphi[/tex](t_{0}) = x_{0}, then [tex]\varphi[/tex](t)[tex]\equiv[/tex] x_{0}. There is no loss of generality in assuming that x_{0}= 0. Since the field v is differentiable and v(0) = 0, we have:|v(x)| < k|x| for sufficiently small |x| =/= 0, where k > 0 is a positive constant.

Huh? i was good up until the inequality.

If somebody has the book, it is section 2.8 of the first chapter.