We use $\dot{x}$ to represent the time derivative of $x,$ that is, $\dot{x} = dx/dt.$
We can interpret a @differential-equation of this form as a as a vector field. $f(x)$ tells us in which direction and at what magnitude we move at any point on the real line (where $f(x)$ is defined.) We can do this by plotting $\dot{x}$ vs $x.$
You can see that when $x$ is to the right of $0,$ the flow is to the right, and when $x$ is to the left of 0 the flow is to the left.
At some points, there is no flow.
A point at which the change in position of a system is $0$ is called a fixed point. We denote a fixed point at $x_p$ as $x^* = x_p.$
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A fixed point where nearby points flow towards it is said to be stable.
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A fixed point where nearby points flow away from it is said to be unstable.
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The convention, when drawing a dynamical system as a vector field is to draw filled in circles for stable fixed points and empty circles for unstable fixed points.
The starting point, $x_0,$ where we place a particle is called a phase point.
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The function describing the path taken by a particle starting at a phase point is called a trajectory and represents a solution to a @differential-equation with @initial-conditions $x = x_0.$
A drawing that shows the different trajectories taken from different phase points in a system is called a phase portrait.
We can also visualize trajectories by plotting $x(t)$ vs $t.$
Here's both views together.
A fixed point $x^*$ that is approached from any starting position on the real line (other than that at $x^*$ itself) is said to be globally stable.
Application: Population Growth
A very simple model of population growth is just exponential growth. You can model this as
$$ \dot{N} = r{N}, $$
where $r > 0.$ You can see by modeling this on the demo's above that population just goes to infinity with this. This is not realistic. A better model assumes there is a certain carrying capacity $K$, where if the population $N$ exceeds $K$, growth actually becomes negative. This is modeled using the logistic equation
$$ \dot{N} = rN(1 - \frac{N}{K}). $$
Linear Stability Analysis
While it's nice to have a visual intuition for whether a fixed point is stable or not, sometimes it's also nice to know analytically.
Let $x^*$ be a fixed point of $\dot{x} = f(x).$ Then, if $f'(x) \neq 0,$ if $f'(x^*)$ is negative, then $x^*$ is a stable fixed point . If $f'(x^*)$ is positive, then $x^*$ is an unstable fixed point.
This comes from letting $u(t) = x(t) - x^*$ be a small perturbation away from $x^*,$ differentiating it, writing its taylor series, then noticing that $f(x^*) = 0$ and terms greater than the linear term matter less than the linear term and writing
$$ \dot{u} = f'(x^*)u. $$
This is called the linearization about $x^*.$
It also only works if $f'(x) \neq 0.$ If that's not the case, the best bet is to fall back to graphical analysis or to solve explicitly if possible.
A fixed point $x^*$ where $f'(x) = 0$ and the stability depends on which side of $x^*$ $x_0$ lies on is called half-stable. It is denoted with a half-filled circle.
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Existence and Uniqueness
Consider the @initial-value-problem
$$ \dot{x} = f(x), \quad x(0) = x_0. $$
Suppose that $f(x)$ and $f'(x)$ are continuous on an open interval $R$ of the $x$-axis, and suppose that $x_0$ is a point in $R.$ Then, the initial value problem has a solution $x(t)$ on some time interval $(- \tau, \tau)$ about $t = 0,$ and the solution is unique.