Drawing phase portraits · Differential Equations

Intuition

Put the pieces together and a nonlinear portrait can be drawn without solving anything. Find the equilibria, classify each through its Jacobian, draw the nullclines and the direction of motion between them, and connect: trajectories leave one equilibrium along its unstable directions and arrive at another. The curves that separate different long-run fates — the separatrices — run into the saddles along their stable directions.

Mapping a river basin from a few surveyed points: the springs, the lakes and the passes. Once they are placed, the rivers can be sketched between them, and the watersheds run over the passes.

plane diagram: two-basins-and-a-separatrix

$x'=x-x^{3}$ , $y'=-y$ : stable nodes at $(\pm1,0)$ and a saddle at the origin. Every start to the right of the vertical axis ends at $(1,0)$ , every start to its left at $(-1,0)$ ; the axis itself, the saddle's stable direction, is the separatrix between the two basins.

From equilibria to a whole picture

To sketch the portrait of $x'=f$ , $y'=g$ : find every equilibrium; classify each by the eigenvalues of its Jacobian, drawing the eigenvector directions of saddles and nodes; draw the nullclines and mark the direction of motion between them; then join the pieces, using that trajectories never cross. The stable directions of a saddle continue as separatrices, which divide the plane into basins of attraction: the sets of starting points that end at the same equilibrium.

\text{basin of}\ \mathbf{p}=\bigl\{\mathbf{x}_{0}:\ \mathbf{x}(t)\to\mathbf{p}\ \text{as}\ t\to\infty\bigr\}

Joining the pieces

$x'=x-x^{3}$ , $y'=-y$ : equilibria , and . The Jacobian gives stable nodes at and a saddle at .

Which basin a start belongs to

\begin{aligned}&x'=x-x^{3},\ y'=-y,\ x(0)>0\\&\Longrightarrow\ (x,y)\to(1,0)\end{aligned}

The two equations are decoupled. The second makes y decay to zero whatever happens. The first is an autonomous equation in x alone, whose phase line has equilibria at −1, 0 and 1, and to the right of zero it points towards 1 from both sides. So every start with positive x ends at (1, 0), every start with negative x at (−1, 0), and the vertical axis, which stays put, separates the two basins.

Proof steps

y=y_{0}e^{-t}\to0

The second equation is decoupled, and its solutions decay.

x'=x(1-x)(1+x)

The first is autonomous: a phase line, as in chapter two.

0<x<1\Rightarrow x'>0,\qquad x>1\Rightarrow x'<0

Between its equilibria the sign is fixed.

x(0)>0\ \Longrightarrow\ x(t)\to1

So every positive start moves monotonically to the stable equilibrium at 1.

Applications

Practice

The Recipe

Equilibria first, then their Jacobians, then the nullclines with directions, and finally join the pieces without crossing.

f=0,\ g=0\ \longrightarrow\ J\ \longrightarrow\ \text{nullclines}

Try it

What is the first step in sketching a nonlinear phase portrait?