how to find stable and unstable spiral point?
Ältere Kommentare anzeigen
Hi I am working on a differential equation given "0 == diff(x,2) + (z.*((1/3)*(diff(x))^3-diff(x)) + x)". To show wether (0,0) is a stable or unstable spiral point when z>0 and <0. also to show (0,0)is the center when z=0.
Antworten (2)
Saarthak Gupta
am 7 Sep. 2023
0 Stimmen
Hi Yixuan,
I understand you are trying to find the nature of an equilibrium point (0,0) under various conditions of the parameter z, for the second-order differential equation:
The following steps can help you achieve the desired result:
- Convert the second-order differential equation into a system of two first-order differential equations
- Calculate the Jacobian matrix for the system

- Find roots of the characteristic polynomial of the Jacobian at (0,0):

- The roots hence obtained will be in terms of z i.e,

An equilibrium point is called a stable spiral point if the Jacobian matrix has two complex eigenvalues with negative real parts.
We can evaluate both cases, i.e., z > 0 and z < 0 separately. For a case, if the eigenvalues obtained in step 4 are complex with negative real parts, we can conclude that (0,0) is a stable spiral point, otherwise it is unstable.
The Symbolic Math Toolbox can be used to calculate the Jacobian, and to compute the eigenvalues.
- The ‘jacobian’ function can be used to compute the Jacobian matrix.
- The ‘eig’ function can be used to compute the eigenvalues of the Jacobian matrix obtained above.
Please refer to the following MATLAB documentation for more details:
@Yixuan Qi, This is actually the Rayleigh Differential Equation. I assume you would like to utilize a graphical approach, as the terms "stable spiral" and "unstable spiral" suggest this. You can check out this link:
% Rayleigh Differential Equation
title('\mu < 0, Stable spiral')
title('\mu = 0, Center')
title('\mu > 0, Unstable spiral')
Kategorien
Mehr zu Matrix Computations finden Sie in Hilfe-Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!


