We define an equilibrium point as a point $\mathbf{a}\in\mathbb R^n$ such that for the dynamical system $ \frac{d\bf x}{dt}=\bf{F(x)}$ we have $\bf{F(a)}=\mathbf{0}$.
Claim: A solution $\mathbf{x}(t)$ such that $\mathbf{x}(0)=\mathbf a$ is just the constant function $\mathbf{x}(t)=\mathbf{a}$.
I've been trying to show myself why this is true algebraically. Intuitively, it makes sense that if you start at a point where there is no tendency to change, then your solution willl just be a constant at that equilibrium point. However I'm not sure how to "show" this?
Can someone please explain why this is true? (I tried to go for an explanation like "since the derivative is zero, then the solution must be a constant" - but this doesn't make sense to be since the derivative is only zero at one point, how can we infer a constant solution (without intuition) only given the initial condition?
Thanks in advance
In the context of dynamical systems you normally assume $F$ to be at least continuously differentiable and then Picard-Lindelöf gives you uniqueness of the solution. As the constant function is clearly a solution to the problem, it has to be the only one.
Of course, you are right: If $F$ is does not satisfy the conditions needed for Picard-Lindelöf (Lipschitz-continuity) then this is not true:
$$\dot{x} = \sqrt{\left|x\right|}$$ $$x(0) = 0$$
has two solutions: The constant function $x(t) = 0$ but also $x(t) =\frac{t²}{4}$