PENG9570 Lecture 2

PENG9570 — Lecture 2 (February 25th)

Slope fields

$$\frac{dy}{dx}=f(x,y)$$

Slope field (direction field)

• Differential equation: $y^{\prime }=f(x,y)$
• The derivative $y^{\prime }$ is the slope of the tangent of the solution curve
• A slope field is defined by the magnitude of the derivatives, indicated by short straight lines
• An ODE has (most often) infinitely many solutions

The initial value problem

$$\frac{dy}{dx}=f(x,y),y(0)=y_0$$

Normally one solution per initial value!

Solution with initial condition $y(0)=0.2$.

Slope fields, problem

Which of the following ODEs would produce the slope field shown below?

• A. $\frac{dy}{dx}=|y|\cdot |x|$
• B. $\frac{dy}{dx}=y\cdot x$
• C. $\frac{dy}{dx}=y-|x|$
• D. $\frac{dy}{dx}=|y|\cdot x$
• E. $\frac{dy}{dx}=|y|\cdot |x|$

(On the slide, options A, B, D and E are crossed out; the answer is C.)

ODEs: stability and bifurcations

$$u^{\prime }=f(u)$$

First-order, autonomous, one-variable ODE.

Equilibria are points where

$$u^{\prime }=0\iff f(u)=0$$

(aka steady states, critical points, stationary points). Equilibria are calculated using this equation.

Equilibria: sinks and sources

Three types:

• Sinks (a.k.a. attractors or asymptotically stable equilibria)
• Sources (a.k.a. repellors or unstable equilibria)
• Nodes (semistable equilibria)

Definition

If $u(t)$ is initially “close enough” to the sink $u=u^{\ast }$ and if ${\lim }_{t\to \infty }u(t)=u^{\ast }$, then $u^{\ast }$ is a locally asymptotically stable equilibrium (i.e. a sink).

Slope field of $u^{\prime }=\left(y-\frac{1}{2}\right)(y-1)(2-y)$:

• $u^{\ast }=0.5$: unstable equilibrium (source)
• $u^{\ast }=1$: locally asymptotically stable equilibrium (sink)
• $u^{\ast }=2$: unstable equilibrium (source)

Criterion for asymptotic stability

We introduce $U(t)$ by

$$u(t)=u^{\ast }+U(t)$$

(Note that $u^{\prime }=U^{\prime }$.) Then, by linearisation:

$$\begin{array}{rl}{U}^{\prime }=u^{\prime }=f(u)=f\left(u^{\ast }+U\right)& \approx f\left(u^{\ast }\right)+f^{\prime }\left(u^{\ast }\right)U\\ \iff U^{\prime }& \approx \underset{\text{constant}}{\underbrace{f^{\prime }\left(u^{\ast }\right)}}U\\ \iff U& \approx C{e}^{f^{\prime }\left(u^{\ast }\right)t}\end{array}$$

For $u(t)$ to converge to $u^{\ast }$, we need $U(t)\to 0$ as $t\to \infty$. This requires $e^{f^{\prime }\left(u^{\ast }\right)t}\to 0$ as $t\to \infty$.

$U(t)\to 0$ as $t\to \infty$ when $f^{\prime }\left(u^{\ast }\right)<0$.

Asymptotic stability when $f^{\prime }\left(u^{\ast }\right)<0$.

An equilibrium from which $u(t)$ escapes as $t\to \infty$ is unstable and is called a source (or a repellor). A node is a sink or a source depending on circumstances; it is sometimes called a semistable equilibrium.

Phase line plots

Phase line plots show the graph of $f(u)$ (where $u^{\prime }=f(u)$) together with arrows that point in the direction in which the solution grows.

Example: $u^{\prime }=\underset{f(u)}{\underbrace{u(1-u)}}$

Generic phase line plot:

$$u^{\prime }=f(u)$$

Example: logistic growth with harvesting

$$P^{\prime }=rP\left(1-\frac{P}{K}\right)-\underset{\begin{array}{c}\text{harvesting}\\ \text{rate}\end{array}}{\underbrace{H}}$$

Scaling (dimensionless):

$$u=\frac{P}{K},\tau =rt\Rightarrow \frac{du}{d\tau }=u(1-u)-h,h=\frac{H}{rK}$$

Scaled ODE:

$$\frac{du}{d\tau }=u(1-u)-h$$

Equilibria ($\frac{du}{d\tau }=0$):

$$\begin{array}{rl}& u(1-u)-h=0\\ \Rightarrow & u^2-u+h=0\\ \Rightarrow & u^{\ast }=\frac{1}{2}\pm \frac{\sqrt{1-4h}}{2}\end{array}$$

Three cases:

I. $1-4h<0$: no equilibrium II. $1-4h=0$: one equilibrium III. $1-4h>0$: two equilibria

Phase line plots:

I. $1-4h<0\iff h>\frac{1}{4}$

II. $1-4h=0\iff h=\frac{1}{4}$

III. $1-4h>0\iff 0<h<\frac{1}{4}$

Visualising the dependence on the parameter $h$ — the bifurcation diagram

Equilibria $u^{\ast }$ and $h$ are related by

$$u^{\ast }\left(1-u^{\ast }\right)-h=0\iff h=u^{\ast }\left(1-u^{\ast }\right)$$

Change axes:

Bifurcation diagram for the logistic model with harvesting.