PENG9570 Lecture 3

PENG9570 — Lecture 3 (February 26th)

Phase line plots

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

$$u^{\prime }=u(1-u)-h$$

Bifurcations

Bifurcation diagram:

Problem

• Sketch the phase line plot of

$$p^{\prime }=p(1-p)-hp$$

for different values of the parameter $h$.

• Identify stable and unstable equilibria (sinks and sources).

Example

Phase line plot of

$$\begin{array}{rl}{p}^{\prime }& =\stackrel{f(p)}{\overbrace{p(1-p)-hp}},h>0\\ & =p(1-p-h)\\ & =p(1-h-p)\end{array}$$

Equilibria ($p^{\prime }=0$): $p^{\ast }=0$ and $p^{\ast }=1-h$.

I. $h>1$: $p^{\prime }=p(1-h-p)$

II. $h<1$: $p^{\prime }=p(1-h-p)$

(If $h=1$: $p^{\prime }=-p^2$, $p^{\ast }=0$.)

Bifurcation diagram:

Two-variable ODE systems

$$\begin{array}{rl}& x^{\prime }=P(x,y),x=x(t)\\ & y^{\prime }=Q(x,y),y=y(t)\end{array}$$

Solution curves:

Example

$$x^{\prime \prime }=-x\iff x^{\prime \prime }+x=0$$

Characteristic polynomial:

$$\begin{array}{r}{\lambda }^2+0\cdot \lambda +1=0\\ {\lambda }^2=-1\\ \lambda =\pm i\end{array}$$

General solution

$$x(t)=c_1\cos t+c_2\sin t$$

Alternatively: Introduce $y=x^{\prime }$. Then

$$\begin{array}{rl}& x^{\prime }=y\\ & \underset{=y^{\prime }}{\underbrace{x^{\prime \prime }}}=-x\end{array}$$

System of ODEs:

$$\begin{array}{rl}& x^{\prime }=y\\ & y^{\prime }=-x\end{array}$$

We know that

$$x(t)=C_1\cos t+C_2\sin t$$ $$\begin{array}{rl}\Rightarrow y& =x^{\prime }\\ & =-c_1\sin t+c_2\cos t\end{array}$$

Then,

$$\begin{array}{rl}{x}^2+y^2& ={\left(c_1\cos t+c_2\sin t\right)}^2\\ & +{\left(-c_1\sin t+c_2\cos t\right)}^2\\ =c_1^2{\cos }^{2}t& +2c_1{c}_2\cos t\sin t\\ +& c_2^2{\sin }^{2}t\\ +& c_1^2{\sin }^{2}t-2c_1{c}_2\sin t\cos t\\ +& +c_2^2{\cos }^{2}t\end{array}$$ $$\begin{array}{rl}& =c_1^2(\stackrel{=1}{\overbrace{c^{2}t+{\sin }^{2}t}})\\ & +c_2^2(\underset{=1}{\underbrace{{\cos }^{2}t+{\sin }^{2}t}})\\ & =c_1^2+c_2^2=c^2\\ & \Rightarrow x^2+y^2=c^2\end{array}$$

$$\text{ 2. }\begin{array}{rl}\frac{dy}{dx}& =\frac{dy/dt}{dx/dt}=\frac{y^{\prime }}{x^{\prime }}\\ & =\frac{x}{-y}\\ \Rightarrow \frac{dy}{dx}& =-\frac{x}{y}\end{array}$$

(Separable ODE)

$$\begin{array}{rl}\int ydy& =-\int xdx\\ \frac{1}{2}{y}^2& =-\frac{1}{2}{x}^2+C_1\end{array}$$ $$\begin{array}{c}\Rightarrow x^2+y^2=C\end{array}\text{\hspace{1em}(1)}$$

I.e. circles with radius $\sqrt{c}$. 3. $\frac{d}{dt}\left(x^2+y^2\right)$

$$=\frac{d}{dt}{x}^2+\frac{d}{dt}{y}^2$$ $$=2xy-2yx=0$$

Derivative of $x^2+y^2$ is zero means

$$\begin{array}{rl}& x^2+y^2=\text{ conot }\\ & x^2+y^2=c\end{array}$$

Solution curves are circles with radius $\sqrt{c}$.

Linear systems of ODEs

$$\begin{array}{rl}& x^{\prime }=ax+by\\ & y^{\prime }=cx+dy\end{array}$$

($a,b,c,d$ are constants.) Coefficient matrix:

$$A=\left[\begin{array}{ll}a& b\\ c& d\end{array}\right],\det A=ad-bc\ne 0$$

Assumption: $A$ is non-singular. System in matrix–vector form:

$${\vec{x}}^{\prime }=A\vec{x},\vec{x}=\left[\begin{array}{l}x\\ y\end{array}\right],{\vec{x}}^{\prime }=\left[\begin{array}{l}{x}^{\prime }\\ y^{\prime }\end{array}\right]$$

Solution

We look for solutions of the form

$$\vec{x}=\vec{v}{e}^{\lambda t}$$

Inserting this into ${\vec{x}}^{\prime }=A\vec{x}$ we get:

$$\begin{array}{rl}& \vec{v}{e}^{\lambda t}\lambda =A\vec{v}{e}^{\lambda t}\\ \iff & (A-\lambda I)\vec{v}=\vec{0}\end{array}$$

Here $\lambda$ is an eigenvalue of $A$ and $\vec{v}$ the corresponding eigenvector. Non-trivial solutions require

$$\det (A-\lambda I)=0\text{(characteristic equation)}$$

For $A=\left[\begin{array}{ll}a& b\\ c& d\end{array}\right]$:

$$\left|\begin{array}{cc}a-\lambda & b\\ c& d-\lambda \end{array}\right|=0$$ $$\begin{array}{rl}& (a-\lambda )(d-\lambda )-bc=0\\ \Rightarrow & {\lambda }^2-\underset{\mathrm{tr}A}{\underbrace{(a+d)}}\lambda +\underset{\det A}{\underbrace{ad-bc}}=0\\ \Rightarrow & {\lambda }^2-\mathrm{tr}A\cdot \lambda +\det A=0\end{array}$$

Three different types of solution pairs:

I. Two real and unequal $\lambda$‘s II. One real $\lambda$ (skipped) III. Two complex conjugate $\lambda$‘s: $\lambda =\alpha \pm i\beta$

Case I

${\lambda }_1,{\lambda }_2$ are both real and ${\lambda }_1\ne {\lambda }_2$. Assume ${\vec{v}}_1$ and ${\vec{v}}_2$ are the corresponding eigenvectors. Then,

$$\vec{x}(t)=\underset{\text{general solution }}{\underbrace{C_1{\vec{v}}_1{e}^{{\lambda }_{1}t}+C_2{\vec{v}}_2{e}^{{\lambda }_{2}t}}}$$

Case III

${\lambda }_1=\alpha +i\beta$, ${\lambda }_2=\alpha -i\beta$, with $\alpha ,\beta \in \mathbb{R}$. The vectors $\vec{w}+i\vec{v}$ and $\vec{w}-i\vec{v}$ are the corresponding eigenvectors. Then, using Euler’s formula $e^{i\theta }=\cos \theta +i\sin \theta$:

$$\begin{array}{rl}\vec{x}& =C_1(\vec{w}+i\vec{v})e^{(\alpha +i\beta )t}+C_2(\vec{w}-i\vec{v})e^{(\alpha -i\beta )t}\\ & =\underset{\text{general solution }}{\underbrace{C_1{e}^{\alpha t}(\vec{w}\cos \beta t-\vec{v}\sin \beta t)+C_2{e}^{\alpha t}(\vec{w}\sin \beta t+\vec{v}\cos \beta t)}}\end{array}$$

Types of equilibria

Center:

• Source: all orbits move away from $(0,0)$.
• Sink: all orbits approach $(0,0)$ as $t\to \infty$ (or as $t\to -\infty$ for a source).
• Saddle (node): orbits approach $(0,0)$ except for orbits on the unstable manifold.

Example

$$\begin{array}{rl}& x^{\prime }=3x-2y\\ & y^{\prime }=2x-2y\end{array}A=\left[\begin{array}{ll}3& -2\\ 2& -2\end{array}\right]$$

Characteristic equation:

$$\begin{array}{rl}\det (A-\lambda I)& =\left|\begin{array}{cc}3-\lambda & -2\\ 2& -2-\lambda \end{array}\right|\\ & =(3-\lambda )(-2-\lambda )-(-2)\cdot 2\\ & ={\lambda }^2-\lambda -2=0\end{array}$$ $$\begin{array}{rl}\Rightarrow \lambda & =\frac{1\pm \sqrt{1-4(-2)}}{2}\\ & =\frac{1\pm \sqrt{9}}{2}=\{\begin{array}{c}2\\ -1\end{array}\\ {\lambda }_1& =2,{\lambda }_2=-1\end{array}$$

Real eigenvalues: one positive, one negative — a saddle equilibrium.

Eigenvector corresponding to ${\lambda }_1=2$:

$$\begin{array}{rl}& (A-2I)\left[\begin{array}{l}{V}_1\\ V_2\end{array}\right]=\left[\begin{array}{l}0\\ 0\end{array}\right]\\ & \left[\begin{array}{cc}3-2& -2\\ 2& -2-2\end{array}\right]\left[\begin{array}{l}{V}_1\\ V_2\end{array}\right]=\left[\begin{array}{l}0\\ 0\end{array}\right]\end{array}$$ $$\begin{array}{c}{v}_1-2v_2=0\Rightarrow v_1=2v_2\\ 2v_1-4v_2=0\\ \left[\begin{array}{l}{v}_1\\ v_2\end{array}\right]=\left[\begin{array}{c}2v_2\\ v_2\end{array}\right]=v_2\left[\begin{array}{l}2\\ 1\end{array}\right]\end{array}$$

We choose ${\vec{v}}_1=\left[\begin{array}{l}2\\ 1\end{array}\right]$

Similarly we find

$${\vec{V}}_2=\left[\begin{array}{l}1\\ 2\end{array}\right]$$

The general solution:

$$\begin{array}{rl}\vec{x}(t)& =c_1{\vec{v}}_1{e}^{{\lambda }_{1}t}+c_2{\vec{v}}_2{e}^{{\lambda }_{2}t}\\ & =c_1\left[\begin{array}{l}2\\ 1\end{array}\right]e^{2t}+c_2\left[\begin{array}{l}1\\ 2\end{array}\right]e^{-t}\\ x(t)& =2c_1{e}^{2t}+c_2{e}^{-t}\\ y(t)& =c_1{e}^{2t}+2c_2{e}^{-t}\end{array}$$

Nonlinear systems of ODEs

$$\begin{array}{rl}& x^{\prime }=P(x,y)\\ & y^{\prime }=Q(x,y)\end{array}$$

Assume that $(a,b)$ is an equilibrium: $x^{\prime }=y^{\prime }=0$ at $(x,y)=(a,b)$, that is

$$\begin{array}{rl}& P(a,b)=0\\ & Q(a,b)=0\end{array}$$

Reminder: linearisation

The linearisation of a function $f$ at $x=a$ is

$$y=f(a)+f^{\prime }(a)(x-a)$$

The linearisation of a function $F=F(x,y)$ at $(x,y)=(a,b)$ is

$$z=F(a,b)+\frac{\partial F}{\partial x}(a,b)(x-a)+\frac{\partial F}{\partial y}(a,b)(y-b)$$

This means that, near $(a,b)$,

$$\begin{array}{rl}P(x,y)& \approx P(a,b)+P_x(a,b)(x-a)+P_y(a,b)(y-b)\\ Q(x,y)& \approx Q(a,b)+Q_x(a,b)(x-a)+Q_y(a,b)(y-b)\end{array}$$

To simplify we redefine variables:

$$\begin{array}{rl}\bar{x}& =x-a,\bar{y}=y-b\\ \Rightarrow P(x,y)& \approx P_x(a,b)\bar{x}+P_y(a,b)\bar{y}\\ Q(x,y)& \approx Q_x(a,b)\bar{x}+Q_y(a,b)\bar{y}\end{array}$$

Using these approximations we get the linearised system

$$\underset{\text{linearised ODE system }}{\underbrace{\left[\begin{array}{l}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\left[\begin{array}{ll}{P}_x(a,b)& P_y(a,b)\\ Q_x(a,b)& Q_y(a,b)\end{array}\right]\left[\begin{array}{l}x\\ y\end{array}\right]}}$$

where the Jacobian of the ODE system is

$$J=\left[\begin{array}{ll}{P}_x(a,b)& P_y(a,b)\\ Q_x(a,b)& Q_y(a,b)\end{array}\right]$$

Does the linearised system behave in the same way as the nonlinear system near the equilibrium? Yes, but with one limitation. The Hartman–Grobman theorem states that if ${\lambda }_1,{\lambda }_2$ are not pure imaginary numbers, then the orbits of the original system near the equilibrium behave the same way as orbits of the linearised system near the equilibrium.

Example

$$\begin{array}{rl}& x^{\prime }=-y-x^3=P(x,y)\\ & y^{\prime }=x=Q(x,y)\end{array}$$

Equilibrium: $(0,0)$. Linearisation:

$$\begin{array}{c}{P}_x(0,0)={\left[-3x^2\right]}_{x=0}=0,P_y(0,0)=-1\\ P(x,y)\approx 0\cdot x-1\cdot y=-y\\ x^{\prime }=-y,y^{\prime }=x\end{array}$$

Jacobian:

$$J=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$$

Eigenvalues of $J$:

$$\begin{array}{c}\det (J-\lambda I)=0\\ \left|\begin{array}{cc}0-\lambda & -1\\ 1& 0-\lambda \end{array}\right|=0\\ {\lambda }^2+1=0\\ \lambda =\pm i\end{array}$$

Purely imaginary eigenvalues of the Jacobian, so the Hartman–Grobman theorem does not apply.

$$\begin{array}{rl}& x^{\prime }=-y-x^3\\ & y^{\prime }=x\end{array}$$

• Linear system: center at $(0,0)$.
• Nonlinear system: stable spiral at $(0,0)$.

Nullclines

The nullclines of a system of ODEs are the curves obtained when $x^{\prime }$ and $y^{\prime }$ are set equal to zero:

$$\begin{array}{rl}& x^{\prime }=P(x,y)=0\text{(nullcline for }x\text{)}\\ & y^{\prime }=Q(x,y)=0\text{(nullcline for }y\text{)}\end{array}$$

Example

Nullclines of

$$\begin{array}{rl}& x^{\prime }=x-y\\ & y^{\prime }=-x+2y\end{array}\begin{array}{rl}& y=x\\ & y=\frac{1}{2}x\end{array}$$

$(0,0)$ is an unstable equilibrium.