05 - Phase Paths

Trajectory or Phase Path

Let $x,y$ be solutions of the Plane Autonomous System of ODE
Given initial values $x(0)=a$ and $y(0)=b$ then there exists a unique solution called the

$$\text{trajectory or phase path}$$

in the phase plane

Phase Plane

The $(x,y)$ plane that the trajectories / phase path lie in

Used to visualise the paths

Trajectories are identical with time offsets

If $(x(t),y(t))$ is a solution of Plane Autonomous ODE
For any fixed number $t_0\in \mathbb{R}$ then

$$\tilde{x}(t):=x(t+t_0),\tilde{y}(t):=y(t+t_0)$$

solves the same ODE and they trace the same trajectory

Proof

$$\dot{\tilde{x}}=\dot{x}(t+t_0)=X(x(t+t_0),y(t+t_0))=X(\tilde{x}(t),\tilde{y}(t))$$

Similarly we have

$$\dot{\tilde{y}}(t)=Y(\tilde{x}(t),\tilde{y}(t))$$

Only works as the system is autonomous

Unique Trajectory

Through every point $(x_0,y_0)$ then there exists a

$$\text{UNIQUE trajectory}$$

with different trajectories never intersecting

However they may asymptote to same point $(x^{\ast },y^{\ast })$ as $t\to \inf$ or $t\to -\infty$

Proof - Tip

Relies on the fact that Picard guarantees the existence of a solution

Critical Point

Let $(x_0,y_0)$ be a point in the phase plane then

${(}_0,y_0)$ is a critical point if

$$X(x_0,y_0)=Y(x_0,y_0)=0$$

This results in a special trajectory where the solutions are constant in time

Closed Solutions

Trajectories in the phase plane that are closed in the phase plane (return to the same point)

Closed Solutions are Periodic

Assuming that they aren’t constant solutions then

Closed Solutions in the Phase plane are also Periodic Solutions

Proof

Suppose the trajectory is closed so for some finite value $t_0$ of $t$ then

$$(x(t_0),y(t_0))=(x(0),y(0))$$

while

$$(x(t),y(t))\ne (x(0),y(0))\text{ for }0<t<t_0$$

Defining

$$\overline{x}(t)=x(t+t_0),\underline{y}(t)=y(t+t_0)$$

So then as shown before by uniqueness then $(\underline{x}(t),\underline{y}(t))$ is another solution

As we have

$$\overline{x}(0)=x(t_0)=x(0)\text{ and }\overline{y}(0)=y(t_0)=y(0)$$

So by uniqueness of the solution (given Lipschitz) then

$$\begin{array}{rl}x(t+t_0)& =\overline{x}(t)=x(t)\\ y(t+t_0)& =\overline{y}(t)=y(t)\end{array}$$

for all $t$ hence closed trajectory corresponds to a periodic solution

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Nullclines

Curves where Curves where $X(x,y)=0$ or $Y(x,y)=0$

Note that critical points occur where the $X,Y-$nullclines intersect

Drawing Phase Diagrams

It is useful to draw the nullclines in order to find regions of the signs of $\frac{dx}{dt}$ and $\frac{dy}{dt}$
As this gives us 4 cases for the arrow direction

$$\frac{dx}{dt}>0,\frac{dy}{dt}>0⟹\text{right and up}$$

$$\frac{dx}{dt}>0,\frac{dy}{dt}<0⟹\text{right and down}$$

$$\frac{dx}{dt}<0,\frac{dy}{dt}>0⟹\text{left and up}$$

$$\frac{dx}{dt}<0,\frac{dy}{dt}<0⟹\text{left and down}$$

and then we have the trivial cases where either $X(x,y)=0$ or $Y(x,y)=0$

Bendixson's Criterion corollary

If

$$\frac{\partial X}{\partial x}+\frac{\partial Y}{\partial y}$$

has fixed sign in simply connected region $R$ then

There are no non-trivial closed trajectories lying entirely in $R$

Proof

Use Bendixson-Dulac Theorem with $\phi =1$ or $\phi =-1$