Phase lines · Differential Equations

Intuition

The direction field of an autonomous equation repeats sideways, so all the information in it lives on a single vertical line. Mark the equilibria on that line and put an arrow on each interval between them, pointing up where the rate is positive and down where it is negative. That one picture predicts the fate of every solution of the equation, and drawing it takes no integration at all.

A map of a river reduced to a single strip: where the current runs upstream, where it runs downstream, and where it is still. You do not need the whole map to know where a leaf dropped in will end up.

plane diagram: the-line-that-says-everything

The phase line of $y'=y(3-y)$ . Below zero the rate is negative and the arrow points down; between the equilibria it is positive; above three it is negative again. Every solution follows its arrow.

Drawing one, and what it tells you

The phase line of $y'=f(y)$ is the line of values of the unknown, with the equilibria marked and each interval between them carrying an arrow: upward where $f>0$ there, downward where $f<0$ . Since is continuous and does not vanish inside such an interval, the sign is the same throughout it, so one arrow describes the whole interval.

f>0\ \text{on an interval}\ \Longrightarrow\ \text{arrow up there}

Reading it

The arrows are found by testing one convenient value in each interval, because the sign cannot change without a root.
A solution starting in an interval moves the way the arrow points, monotonically, and approaches the equilibrium the arrow points at.
If the arrow points away from every neighbouring equilibrium the solution leaves the region — possibly to infinity, and possibly in finite input.
The phase line is a picture of the equation, not of a solution: it holds every solution at once, with the starting height left free.
Only an autonomous equation has one. For $y'=f(x,y)$ the arrow at a height would depend on where along the input it is drawn.

Applications

Practice

Test One Value in Each Interval

Between two neighbouring equilibria the sign of f cannot change, so one test value settles the arrow for the whole interval.

\begin{aligned}&y'=y(3-y):\ f(1)=2>0\\&\Rightarrow\ \text{arrow up on } (0,3)\end{aligned}

Try it

For $y'=y(3-y)$ , which way does the arrow point on the interval above $3$ ?