Difference between revisions of "Non-Linear Systems of ODEs" - Rhea

Consider the system of ODEs in 4.0,

$ \frac{dx_1}{dt}=f_1(t,x_1,x_2,...x_n) $

$ \frac{dx_2}{dt}=f_2(t,x_1,x_2,...x_n) $

...

$ \frac{dx_n}{dt}=f_n(t,x_1,x_2,...x_n) $

When the $ n $ ODEs are not all linear, this is a nonlinear system of ODE. Consider an example,

$ \frac{dx}{dt}=x(1-2x-3y) $,

$ \frac{dy}{dt}=2y(3-x-2y) $.

In this tutorial, we will analyse this system in different aspects to build up a basic completed concept.

6.1 Equilibrium Point

An equilibrium point is a constant solution to a differential equation. Hence, for an ODE system, an equilibrium point is going to be a solution of a pair of constants. Set all of the differential terms equal to $ 0 $ to find the equilibrium point.

In the example in 6.0, we set $ \frac{dx}{dt}=\frac{dy}{dt}=0 $, hence $ x(1-2x-3y)=2y(3-x-2y)=0 $. Solve this system of normal equations.

· When $ x=2y=0 $, then $ x=y=0 $.

· When $ x=3-2x-2y=0 $, then $ x=0 $, $ y=\frac{3}{2} $.

· When $ 1-2x-3y=2y=0 $, then $ x=\frac{1}{2} $, $ y=0 $.

· When $ 1-2x-3y=3-x-2y=0 $, then $ x=-7 $, $ y=5 $.

Hence, the equilibrium points of this nonlinear system are $ (x=0,y=0) $, $ (x=0,y=\frac{3}{2}) $, $ (x=\frac{1}{2},y=0) $, and $ (x=-7,y=5) $. This means in a xy-coordinate, macroscopically, the graph of the solution of ODE (a function) will keep a dynamic equilibrium, near which the sum of velocity (measured in both direction and speed) of each point on the graph is $ 0 $.

6.2 Linearisation

Macroscopically, the whole system is nonlinear, but we still need a linear system for further analysis. So here comes a method of linearisation near the equilibrium points. We linearise the graph of the solution for details to sketch a global phase portrait. A similar concept we can refer to is the expansion of Taylor series. It is not a linearisation, but using a method to approach the exact function, which is kinda like using local phase portraits to approach the global one. Linearisation here is a method to identify the local phase portraits.

Suppose $ (x_0,y_0) $ is an equilibrium point and define two deviation variables $ \epsilon (t)=x(t)-x_0 $, $ \mu (t)=y(t)-y_0 $, where $ x(t)→x_0 $ and $ y(t)→y_0 $. The "deviation variables" mean that we start slightly differently from $ (x_0,y_0) $ and measure the difference to be more accurate. Hence $ \epsilon(t) $ and $ \mu (t) $ are approaching to $ 0 $.

Then we differentiate the deviation variables to have

$ \frac{d\epsilon}{dt}=\frac{dx}{dt}=f(x_0+\epsilon,y_0+\mu) $,

$ \frac{d\mu}{dt}=\frac{dy}{dt}=g(x_0+\epsilon,y_0+\mu) $.

By the expansion of Taylor series for two-variable functions, we have

$ \frac{d\epsilon}{dt}=f(x_0,y_0)+\epsilon \frac{\partial f}{\partial x}|_{(x_0,y_0)}+\mu \frac{\partial f}{\partial y}|_{(x_0,y_0)}+... $,

$ \frac{d\mu}{dt}=g(x_0,y_0)+\epsilon \frac{\partial g}{\partial x}|_{(x_0,y_0)}+\mu \frac{\partial g}{\partial y}|_{(x_0,y_0)}+... $

Here we converted the nonlinear system to a linear one. As $ f(x_0,y_0)≈g(x_0,y_0)→0 $, so we can write the system into the matrix form, which is the linearisation near $ (x_0,y_0) $: $ \begin{bmatrix} \frac{d\epsilon}{dt} \\ \frac{d\mu}{dt} \end{bmatrix}=\begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{bmatrix} \begin{bmatrix} \epsilon \\ \mu \end{bmatrix} $.

This is called Jacobian matrix, usually denoted as $ J $.

Still considering the example in 6.0, we will have the general linearisation

$ J=Df(x,y)=\begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{bmatrix}=\begin{bmatrix} 1-4x-3y & -3x \\ -2y & 6-2x-8y \end{bmatrix} $ for this nonlinear system.

Hence, plug all the equilibrium points from 6.1 into this general linearisation, we will have:

· $ J_{(0,0)}=\begin{bmatrix} 1 & 0 \\ 0 & 6 \end{matrix} $. The eigenvalues are $ \lambda_1=1 $, $ \lambda_2=6 $, and the corresponding eigenvectors are $ \bold{v_1}=\begin{bmatrix} 1 \\ 0 \end{bmatrix} $, $ \bold{v_2}=\begin{bmatrix} 0 \\ 1 \end{bmatrix} $.

6.3 Exercises

6.4 References

Institute of Natural and Mathematical Science, Massey University. (2017). 160.204 Differential Equations I: Course materials. Auckland, New Zealand.