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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.

- Integral inequality (Cauchy-Schwarz)
- Show that $\lim\limits_{n \to \infty} \sup\limits_{k \geq n} \left(\frac{1+a_{k+1}}{a_k}\right)^k \ge e$ for any positive sequence $\{a_n\}$
- The preimage of continuous function on a closed set is closed.
- Fundamental Theorem of Calculus for distributions.
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- Pointwise vs. Uniform Convergence

But is anybody aware of a simple argument for this?

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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.

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