02. Plane Autonomous Systems of ODEs

Plane Autonomous System of ODEs

Pair of ODEs in the form

$$\begin{array}{r}\frac{dx}{dt}=X(x,y)\\ \frac{dy}{dt}=Y(x,y)\end{array}$$

Autonomous meaning there is no $t-$dependence in $X$ or $Y$
Plane meaning just two equations i.e. $(x,y)$

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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

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Phase Plane

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

Used to visualise the paths

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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

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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

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2.1 Critical Points and Closed Trajectories

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

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Closed Solutions

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

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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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Refer to Lecture Notes for more Examples! (pg ~ $30$)

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2.2 Stability and Linearisation

Stable Critical Point

Critical point $(a,b)$ is stable if for $ϵ>0$ there exists $\delta >0$ and $t_0$ such that for

Any solution $(x(t),y(t))$ for which $\sqrt{(x(t_0)-a{)}^2+(y(t_0)-b{)}^2}<\delta$ then

$$\sqrt{(x(t)-a{)}^2+(y(t)-b{)}^2}<ϵ,\forall t>t_0$$

Note that you can use other norms such as $l^1$ or $l^{\infty }$

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Unstable Critical Point

Critical point that is not stable

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General Solution of the Linearised System at a Critical Point

Suppose $P=(a,b)$ is a critical point for Plane Autonomous ODE so linearising around $P$

Let $\underline{Z}(t)=\left(\begin{array}{c}\zeta \\ \eta \end{array}\right)$ with equation $\dot{\underline{Z}}=M\underline{Z}$ where

$$M=\left(\begin{array}{cc}{X}_x{|}_P& X_y{|}_P\\ Y_x{|}_P& Y_y{|}_P\end{array}\right)$$

Let ${\lambda }_1,{\lambda }_2$ be the eigenvalues of $M$ then

For ${\lambda }_1\ne {\lambda }_2$ we have general solution

$$\underline{Z}(t)=c_1{\underline{Z}}_1{e}^{{\lambda }_{1}t}+c_2{\underline{Z}}_2{e}^{{\lambda }_{2}t}$$

for constants $c_i$

For ${\lambda }_1={\lambda }_2=\lambda \in \mathbb{R}$ we have general solutions

1. If $M-\lambda I=0$ then $M=\lambda I$ so we have solution

$$\underline{Z}(t)=\underline{C}{e}^{\lambda t}$$

for any constant vector $\underline{C}$
2) If $M-\lambda I=0$ then there exists constant vector ${\underline{Z}}_1$ with

$$\underline{Z}(t)=(c_1{\underline{Z}}_1+(c_0+c_{1}t){\underline{Z}}_0)e^{\lambda t}$$

where ${\underline{Z}}_0$ is the one linearly independent eigenvector of $M$ so we get general solution
for constants $c_i$

Proof

We do this by linearising around the point

Suppose $P=(a,b)$ is a critical point for Plane Autonomous ODE so

$$X(a,b)=0=Y(a,b)$$

Hence $x=a$ and $y=b$ is a constant solution for Plane Autonomous ODE
So linearise by setting

$$x=a+\zeta (t);y=b+\eta (t)$$

where $\zeta$ and $\eta$ are small

So using Taylor’s Theorem

$$\begin{array}{rl}\dot{x}=\dot{\zeta }=X(a+\zeta ,b+\eta )& =X(a,b)+\zeta X_x{|}_P+\eta X_y{|}_P+\text{ h.o. terms}\\ \dot{y}=\dot{\eta }=Y(a+\zeta ,b+\eta )& =Y(a,b)+\zeta Y_x{|}_P+\eta Y_y{|}_P+\text{ h.o. terms}\end{array}$$

By neglecting higher order terms we can write it in matrix form

$$\left(\begin{array}{c}\dot{\zeta }\\ \dot{\eta }\end{array}\right)=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)\left(\begin{array}{c}\zeta \\ \eta \end{array}\right)$$

where

$$M:=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)=\left(\begin{array}{cc}{X}_x{|}_P& X_y{|}_P\\ Y_x{|}_P& Y_y{|}_P\end{array}\right)$$

By setting $\underline{Z}(t)=\left(\begin{array}{c}\zeta \\ \eta \end{array}\right)$ then we have equation

$$\dot{\underline{Z}}=M\underline{Z}$$

Solving with eigen-vectors and eiven-values then
${\underline{Z}}_0{e}^{\lambda t}$ is a solution with constant vector ${\underline{Z}}_0$ and constant scalar $\lambda$ if

$$\lambda {\underline{Z}}_0=M{\underline{Z}}_0$$

That is, ${\underline{Z}}_0$ is an eigen-vector of $M$ with eigen-value $\lambda$

Suppose the eigen-values of the $2\times 2$ matrix $M$ is ${\lambda }_1,{\lambda }_2$

For ${\lambda }_1\ne {\lambda }_2$ we have general solution

$$\underline{Z}(t)=c_1{\underline{Z}}_1{e}^{{\lambda }_{1}t}+c_2{\underline{Z}}_2{e}^{{\lambda }_{2}t}$$

for constants $c_i$

For ${\lambda }_1={\lambda }_2=\lambda \in \mathbb{R}$ then by Cayley Hamilton Theorem since ${\chi }_M(x)=(x-\lambda {)}^2$ we have

$$(M-\lambda I)=0\text{ or }(M-\lambda I)\ne 0$$

1. If $M-\lambda I=0$ then $M=\lambda I$ so we have solution

$$\underline{Z}(t)=\underline{C}{e}^{\lambda t}$$

for any constant vector $\underline{C}$
2) If $M-\lambda I=0$ then there exists constant vector ${\underline{Z}}_1$ with

$${\underline{Z}}_0:=(M-\lambda I){\underline{Z}}_1\ne 0$$

but

$$(M-\lambda I){\underline{Z}}_0=(M-\lambda I{)}^2{\underline{Z}}_1=0$$

where ${\underline{Z}}_0$ is the one linearly independent eigenvector of $M$ so we get general solution

$$\underline{Z}(t)=(c_1{\underline{Z}}_1+(c_0+c_{1}t){\underline{Z}}_0)e^{\lambda t}$$

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2.3 Classification of Critical Points

Non-Degenerate Critical Point

Critical Point $(a,b)$ is non-degenerate if eigenvalue of $M=\left(\begin{array}{cc}{X}_x{|}_P& X_y{|}_P\\ Y_x{|}_P& Y_y{|}_P\end{array}\right)$ are all not $0$

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Classification of Critical Points

Assuming critical point is non-degenerate
If that happens we would have to need to have more terms in the Taylor Expansion making it much harder

Let ${\lambda }_1,{\lambda }_2$ be the eigenvalues of $M=\left(\begin{array}{cc}{X}_x{|}_P& X_y{|}_P\\ Y_x{|}_P& Y_y{|}_P\end{array}\right)$ with ${\lambda }_1,{\lambda }_2\ne 0$

Note that a centre in linearisation does not imply a centre in the non-linear system (rest are fine)

Case 1: $0<{\lambda }_1<{\lambda }_2$ (both real)

Node Stability: UNSTABLE

Proof

So from the General Solution of the Linearised System at a Critical Point then

As $t\to -\infty$ then $\underline{Z}(t)\to 0$ and $\underline{Z}(t)\sim c_1{\underline{Z}}_1{e}^{{\lambda }_{1}t}$ unless $c_1=0$ then $\underline{Z}(t)\sim c_2{\underline{Z}}_2{e}^{{\lambda }_{2}t}$
As $t\to \infty$ then $\underline{Z}(t)\to 0$ and $\underline{Z}(t)\sim c_2{\underline{Z}}_2{e}^{{\lambda }_{2}t}$ unless $c_2=0$ then $\underline{Z}(t)\sim c_1{\underline{Z}}_1{e}^{{\lambda }_{1}t}$

So the trajectories converge onto critical point into past but go to infinity in the future
Hence unstable

Case 2: ${\lambda }_1<{\lambda }_2<0$ (both real)

Node Stability: STABLE

Proof

Identical to Case $0<{\lambda }_1<{\lambda }_2$ but with $t\to -t$ and roles of ${\underline{Z}}_1,{\underline{Z}}_2$ switched
So the trajectories converge onto critical point into future but come from infinity in the past
Hence stable

Case 3: ${\lambda }_1={\lambda }_2=\lambda$

Node Stability: $\{\begin{array}{ll}\lambda <0& \text{STABLE STAR}\\ \lambda >0& \text{UNSTABLE INFLECTED NODE}\end{array}$

Proof

If $M-\lambda I=0$ then we have a star as $Z=\underline{C}{e}^{\lambda t}$

If $M-\lambda I\ne 0$ then we have inflected node as $\underline{Z}(t)=(c_1{\underline{Z}}_1+(c_0+c_{1}t){\underline{Z}}_0)e^{\lambda t}$

Case 4: ${\lambda }_1<0<\lambda 2$ (both real)

Node Stability: SADDLE (UNSTABLE)

Proof

We have general form $\underline{Z}(t)=c_1{\underline{Z}}_1{e}^{{\lambda }_{1}t}+c_2{\underline{Z}}_2{e}^{{\lambda }_{2}t}$
If $c_1=0$ then

$$\begin{array}{rlrlrl}Z(t)& \to \infty & & \text{ along }{\underline{Z}}_2& & \text{ as }t\to \infty \\ Z(t)& \to 0& & \text{ along }{\underline{Z}}_2& & \text{ as }t\to -\infty \end{array}$$

If $c_2=0$ then

$$\begin{array}{rlrlrl}Z(t)& \to 0& & \text{ along }{\underline{Z}}_1& & \text{ as }t\to \infty \\ Z(t)& \to \infty & & \text{ along }{\underline{Z}}_1& & \text{ as }t\to -\infty \end{array}$$

If $c_1,c_2\ne 0$ then

$$\begin{array}{rlrlrl}Z(t)& \to \infty & & \text{ along }{\underline{Z}}_2& & \text{ as }t\to \infty \\ Z(t)& \to \infty & & \text{ along }{\underline{Z}}_1& & \text{ as }t\to -\infty \end{array}$$

So we have that most trajectories come in approximately parallel to $\pm {\underline{Z}}_1$
and then asymptotic to $\pm {\underline{Z}}_2$

Case 5: ${\lambda }_1=-i\nu ,{\lambda }_2=i\nu$ for some $\nu \in \mathbb{R}$

Node Stability: CENTRE (STABLE)

Proof

We have that $c_i{Z}_i$ are a conjugate pair so put $c_1=r{e}^{i\theta }$ and ${\underline{Z}}_1=(1,k{e}^{i\varphi }{)}^T$ then

$$c_1{\underline{Z}}_1=\left(\begin{array}{c}r{e}^{i\theta }\\ rk{e}^{i(\varphi +\theta )}\end{array}\right)$$

So that

$$\underline{Z}(t)=\left(\begin{array}{c}2r\cos (\nu t-\theta )\\ 2rk\cos (\nu t-(\varphi +\theta ))\end{array}\right)$$

So $\underline{Z}(t)$ is periodic and hence stable

Case 6: ${\lambda }_1=\mu -i\nu ,{\lambda }_2=\mu +i\nu$ for some $\mu \ne 0,\nu \in \mathbb{R}$

Node Stability: SPIRAL $\{\begin{array}{ll}\mu <0& \text{STABLE}\\ \mu >0& \text{UNSTABLE}\end{array}$

Proof

We have that $c_i{Z}_i$ are a conjugate pair so put $c_1=r{e}^{i\theta }$ and ${\underline{Z}}_1=(1,k{e}^{i\varphi }{)}^T$ then

$$c_1{\underline{Z}}_1=\left(\begin{array}{c}r{e}^{i\theta }\\ rk{e}^{i(\varphi +\theta )}\end{array}\right)$$

So that

$$\underline{Z}(t)=e^{\mu t}\left(\begin{array}{c}2r\cos (\nu t-\theta )\\ 2rk\cos (\nu t-(\varphi +\theta ))\end{array}\right)$$

If $\mu >0$ then $|\underline{Z}(t)|\to \infty$ as $t\to \infty$ so trajectory spirals out into future
Hence we have an unstable spiral

If $\mu <0$ we have a time reversal of previous case so it spirals in
Hence we have a stable spiral

Note that we can do case $5,6$ using matrices (refer to page $36-38$ in lecture notes)

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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

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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 $Y(x,y)=0$

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Refer to Lecture Notes for more Examples! (pg ~ $39-47$)

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2.4 The Bendixson-Dulac Theorem

06 - Bendixson-Dulac Theorem

Bendixson-Dulac Theorem

Consider system $\dot{x}=X(x,y)$ and $\dot{y}=Y(x,y)$ with $X,Y\in C^1$
If there exists a function $\phi (x,y)\in C^1$ with

$$\rho :=\frac{\partial }{\partial x}(\phi X)+\frac{\partial }{\partial y}(\phi Y)>0$$

in a simply connected region $R$ then

There can be no non-trivial closed trajectory lying entirely in $R$

Proof

Note that a non-trivial trajectory that is not just a fixed point

Suppose $C$ is a closed trajectory lying entirely in $R$
Let $D$ be the disc (lies entirely in $R$, as $R$ is simply connected) whose boundary is $C$

Applying Green’s Theorem in the Plane then consider integral

$$\begin{array}{rl}{\iint }_D\rho dxdy& ={\iint }_D\left[\frac{\partial }{\partial x}(\phi X)+\frac{\partial }{\partial y}(\phi Y)\right]dxdy\\ & ={\oint }_C-\phi Ydx+\phi Xdy\\ & ={\oint }_C-\phi (-\dot{y}dx+\dot{x}dy)\end{array}$$

However on $C$ then $dx=\dot{x}dt$ and $dy=\dot{y}dt$ which is zero
This contradicts positivity of $\rho$ so there can be no such $C$

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2.4.1 Corollary

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 $\phi =-1$

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