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 <math>\epsilon(t) $ and $ \mu $ are approaching to $ 0 $.

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.