06 - Stability

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 $l^{\infty }$

Unstable Critical Point

Critical point that is not stable

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

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

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 $0$

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)

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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