# Looking for a Simple Argument for “Integral Curve Starting at A Singular Point is Constant”

Let $U$ be an open subset of $\mathbf R^n$ and $V:U\to\mathbf R^n$ be a differentiable vector field on $U$. Let $\mathbf p\in U$ be a singular point of $V$, that is, $V(\mathbf p)=\mathbf 0$.

Then the only integral curve which starts at $\mathbf p$ is the constant curve.

I know that one could simply use the theorem of “uniqueness of integral curves” and in fact adapt the same proof for this particular case.

But is anybody aware of a simple argument for this?

#### Solutions Collecting From Web of "Looking for a Simple Argument for “Integral Curve Starting at A Singular Point is Constant”"

Let $x$ be an integral curve through $p$:
$$x'(t) = V\bigl(x(t)\bigr),\quad x(0) = p.$$
Since $V$ is differentiable at $p$ and $V(p) = 0$, there exists an $M > 0$ such that $\|V(x)\| \leq M \|x – p\|$ for all $x$ in some neighborhood of $p$. By the flow equation and the triangle inequality,
\begin{align*}
\|x(t) – p\|
&= \left\lVert \int_{0}^{t} x'(s)\, ds\right\rVert \\
&= \left\lVert \int_{0}^{t} V\bigl(x(s)\bigr)\, ds\right\rVert \\
&\leq \int_{0}^{t} \left\lVert V\bigl(x(s)\bigr)\right\rVert\, ds \\
&\leq M\int_{0}^{t} \lVert x(s) – p\rVert\, ds\quad\text{for all $t \geq 0$.}
\end{align*}
That is, the continuous function $g(t) = \|x(t) – p\|$ satisfies
$$g(t) \leq M \int_{0}^{t} g(s)\, ds\quad\text{for all t \geq 0.} \tag{1}$$
It suffices to show $g = 0$ in some interval about $0$.

Let $t = \frac{1}{2M}$. Since $g$ is continuous on $[0, t]$, the extreme value theorem guarantees there exists a $t_{0}$ in $[0, t]$ such that
$$g(t_{0}) = \sup_{0 \leq s \leq t} g(s).$$
For later use, note that
$$Mt_{0} \leq Mt = \tfrac{1}{2},\qquad g(t_{0}) = \sup_{0 \leq s \leq t_{0}} g(s).$$

If $t_{0} = 0$, then $g(t_{0}) = 0$, and $g$ vanishes on $[0, t]$; otherwise, $0 < t_{0}$, and (1) implies
$$g(t_{0}) \leq M \int_{0}^{t_{0}} g(s)\, ds \leq Mt_{0} \sup_{0 \leq s \leq t_{0}} g(s) \leq \tfrac{1}{2} g(t_{0}).$$
That is, $g(t_{0}) = 0$, and again $g$ vanishes identically on $[0, t]$.

The same argument applied to $-V$ shows $x(t) = p$ on some open interval to the left of $0$.

Finally, let $x$ be an arbitrary integral curve through $p$, defined for all real $t$. The set of $t$ such that $x(t) = p$ is closed by continuity of $x$ and non-empty by hypothesis; the preceding argument shows this set is open, hence is the set of all reals.