Non Linear Theory

General #

$$ \begin{align*} \frac{dx}{dt}=F(x,y)\\ \frac{dy}{dt}=G(x,y) \end{align*} $$

Solution to this is a directed curve.

The critical points can be found by solving $F(x,y)=0,G(x,y)=0$

Types of critical points

1. Node
   1. Stable
   2. Unstable
2. Saddle point
3. Center
4. Spiral

Critical Points and Stability for linear system #

Depends on auxilary roots of system of linear odes

$$ \begin{align*} \frac{dx}{dt}&= a_{1}x+b_{1}y\\ \frac{dy}{dt}&= a_{2}x+b_{2}y\\ \\ \text{Auxilary quadratic equation}\\ m^{2}-(a_{1}+b_{2})m+(a_{1}b_{2}-a_{2}b_{1})=0\\ x=Ae^{mt};y=Be^{mt} \end{align*} $$

Since it has (0,0) critical point which has to be isolated, $a_{1}b_{2}-a_{2}b_{1}\ne 0$

Implies that 0 cannot be root.

Real #

Distinct #

Same sign (Case A) #

Node

Negative #

Asymptotically stable

Positive #

Unstable

Opposite sign (Case B) #

Saddle Point

Unstable

Same (Case D) #

Node

m>0 : unstable m<0 : asymptotically stable

Complex #

Pure Im (Case E) #

Center

Stable

Mixed (Case C) #

Spiral

Asymptotically stable

Critical Points and Stability Proof #

Case A #

If the roots $m_{1},m_{2}$ are real and distinct and of same sign

Negative #

Let $m_{1}<m_{2}<0$

$$ \begin{align*} \text{Solution:}\\ x&= c_{1}A_{1}e^{m_{1}t}+c_{2}A_{2}e^{m_{2}t}\\ y&= c_{1}B_{1}e^{m_{1}t}+c_{2}B_{2}e^{m_{2}t} \end{align*} $$

If $c_{2}=0$, then the path has slope $\frac{B_{1}}{A_{1}}$, if $c_{1}=0$ it has slope $\frac{B_{2}}{A_{2}}$. Otherwise, since $m_{1}-m_{2}<0$

$$

\frac{y}{x}= \frac{\left(\frac{c_{1}B_{1}}{c_{2}}\right)e^{(m_{1}-m_{2})t}+B_{2}}{(\frac{c_{1}A_{1}}{c_{2}})e^{(m_{1}-m_{2})t}+A_{2}}

$$

Other paths approach slope $\frac{B_{2}}{A_{2}}$ as $t\rightarrow \infty$ and enter (0,0)

Critical point: Node Asymptotically stable

Positive #

Consider $m_{1}>m_{2}>0$

Exact same situation but happens as $t\rightarrow -\infty$

Critical point: Node unstable

Case B #

Roots are real, distinct and of opposite sign

Let $m_{1}<0,m_{2}>0$

$$ \begin{align*} \text{Solution:}\\ x&= c_{1}A_{1}e^{m_{1}t}+c_{2}A_{2}e^{m_{2}t}\\ y&= c_{1}B_{1}e^{m_{1}t}+c_{2}B_{2}e^{m_{2}t} \end{align*} $$

If $c_{2}=0$ the half-line paths enter (0,0) as $t\rightarrow \infty$ (second line) and when $c_{1}=0$ they enter (0,0) as $t\rightarrow - \infty$ (first line)

Otherwise, none of the paths approach (0,0) as $t\rightarrow \pm \infty$ and when $t\rightarrow \infty$ they approach the first line and when $t\rightarrow - \infty$ they approach the second line

Critical point: Saddle Unstable

Case C #

Roots are conjugate complex but not purely imaginary

Discriminant of the auxilary equation is negative $$ \begin{align*} (a_{1}+b_{2})^{2}-4(a_{1}b_{2}-a_{2}b_{1})\\ = (a_{1}-b_{2})^{2}+4a_{2}b_{1}<0 \end{align*} $$

Roots: $a\pm i b$ imaginary $$ \begin{align*} x&= e^{at}(c_{1}(A_{1}\cos bt-A_{2}\sin bt)+c_{2}(A_{1}\sin bt+A_{2}\cos bt))\\ y&= a^{at}(c_{1}(B_{1}\cos bt-B_{2}\sin bt)+c_{2}(B_{1}\sin bt+B_{2}\cos bt)) \end{align*} $$

If $a<0\Leftrightarrow (a_{1}+b_{2})<0$ then they approach the origin, but not enter it but wind around it in like a spiral-like manner.

$$ \begin{align*} \frac{d\theta}{dt}&= \frac{xy’-yx’}{x^{2}+y^{2}}\\ &= \frac{a_{2}x^{2}+(b_{2}-a_{1})xy-b_{2}y^{2}}{x^{2}+y^{2}} \end{align*} $$

The determinant of numerator is <0

Therefore it is always positive when $a_{2}>0$ and negative when $a_{2}<0$

Critical point: Spiral Asymptotically stable

The spiral might tend off to infinity if $a>0$

Case D #

Two subcases

$a_{1}=b_{2}=a,a_{2}=b_{1}=0$

$$ \begin{align*} x(t)=c_{1}e^{mt}\\ y(t)=c_{2}e^{mt} \end{align*} $$

Other cases

$$ \begin{align*} x(t)=c_{1}Ae^{mt}+c_{2}(A_{1}+A)e^{mt}\\ y(t)=c_{1}Be^{mt}+c_{2}(B_{1}+B)e^{mt} \end{align*} $$

Both these cases, it’s easy to see that it’s a node

• m<0: node, stable
• m>0: node, unstable

Case E #

Same as case C with real part = 0

Solution is an ellipse centered at origin

Critical point: Center stable

Stability Theorems #

Asymptotically stable iff both the roots have non-postitive real parts and it is unstable iff both have positive real roots.

The critical point of the linear system is asymptotically stable iff the coefficients of auxiliary equation are both +ve

$$ -(a_{1}+b_{2}),(a_{1}b_{2}-a_{2}b_{1})>0 $$

Lyapunovs Method #

For studying the stability problem in a broader context

$$ \begin{align*} \frac{dx}{dt}=F(x,y)\\ \frac{dy}{dt} =G(x,y) \end{align*} $$

Let E(x,y) be a function that is continuous and has continuous first partial derivatives. E can be regarded as a function of t along solution $(x(t),y(t))$

Rate of change of E wrt t

$$ \frac{dE}{dt}= \frac{\partial E}{\partial x} \frac{dx}{dt}+ \frac{\partial E}{\partial y} \frac{dy}{dt}= \frac{\partial E}{\partial x}F+ \frac{\partial E}{\partial y}G $$

A function E(x,y) of positive type and with $\frac{\partial E}{\partial x}F+ \frac{\partial E}{\partial y}G$ of negative semi-definite type is called a Lyapunov function for the system.

If there exists a Lyapunov function, then critical point is stable If additionally that the $\frac{\partial E}{\partial x}F+ \frac{\partial E}{\partial y}G$ is negative, then the critical point is asymptotically stable

types #

• Positive: E is 0 at origin and E>0 otherwise
  • $E(x,y)=ax^{2m}+by^{2n};a,b>0$
• Negative: E is 0 at origin and E<0 otherwise
  • $E(x,y)=ax^{2m}+by^{2n};a,b<0$
• Positive semi-definite: E is 0 at origin and $E\ge 0$ otherwise
  • $E(x,y)=x^{2m},y^{2n}, (x-y)^{2m}$
• Negative semi-definite: E is 0 at origin and $E\le 0$ otherwise

Simple Critical Points of Nonlinear systems #

The process of replacing a general system by simpler linear system is called linearization

$$ \begin{align*} \frac{dx}{dt}=a_{1}x+b_{1}y+f(x,y)\\ \frac{dy}{dt}=a_{2}x+b_{2}y+g(x,y) \end{align*} $$

Assume $a_{1}b_{2}-a_{2}b_{1}\ne0$

Linearization by considering f,g=0 can be possible when

$$ \begin{align*} \lim_{(x,y)\rightarrow (0,0)} \frac{f(x,y)}{\sqrt{x^{2}+y^{2}}}=0\\ \lim_{(x,y)\rightarrow (0,0)} \frac{g(x,y)}{\sqrt{x^{2}+y^{2}}}=0 \end{align*} $$

• Implies that (0,0) is a critical point of the original system.
• Would also be of same type as the same case it would with its corresponding linearized system
• If it’s asymptotically stable in the linearized system, it’s also asymptotically stable in the non-linear version.