Linear Systems of ODE's: Eigenvalues and Stabilityby Niles Tags: eigenvalues, linear, stability, systems 

#1
Feb1809, 06:44 AM

P: 1,863

1. The problem statement, all variables and given/known data
Hi all. I am given by following linear system: [tex] \begin{array}{l} \dot x = dx/dt = ax \\ \dot y = dy/dt =  y \\ \end{array} [/tex] The eigenvalue of the matrix of this system determines the stability of the fixpoint (0,0): [tex] A=\left( {\begin{array}{*{20}c} a & 0 \\ 0 & {  1} \\ \end{array}} \right) \quad \Rightarrow \quad \lambda_{1,2}= 0, a. [/tex] So there are two eigenvalues given by 0 and a. When a<0, both eigensolutions decay, and the fixpoint (0,0) is stable. When a>0, we have a saddle point. But what happens when a=0? How can I determine the stability there? Thanks in advance. Best regards, Niles. 



#2
Feb1809, 08:26 AM

Sci Advisor
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P: 25,168

Zero isn't an eigenvalue of that matrix (unless a=0). 1 is the other eigenvalue. ??




#3
Feb1809, 08:54 AM

P: 1,863

Duh, you are right. The eigenvalues are 1 and a.
So when a=0, one of the eigenvalues are 0. How do I determine the stability in this case? 



#4
Feb1809, 09:20 AM

Sci Advisor
HW Helper
Thanks
P: 25,168

Linear Systems of ODE's: Eigenvalues and Stability
Well, x'=a*x for a>0 has exponentially growing solutions (unstable), exponentially shrinking solutions for a<0 and for a=0? Linear solutions, right? I'm actually not sure what (if anything) you call that. I tried looking it up and my references just talk about a<0 or a>0.




#5
Feb1809, 09:23 AM

P: 664





#6
Feb1909, 02:43 PM

P: 1,863

Lets say I have the following nonlinear system:
[tex] \begin{array}{l} dx/dt =  xy \\ dy/dt =  y + x^2 \\ \end{array}. [/tex] We will look at one of the fixpoints, namely (x,y) = (0,0). The Jacobian is given by: [tex] \[ \left( {\begin{array}{*{20}c} {  y} & {  x} \\ {2x} & {  1} \\ \end{array}} \right) \quad \Rightarrow \quad \left( {\begin{array}{*{20}c} 0 & 0 \\ 0 & {  1} \\ \end{array}} \right), [/tex] where I have evaluated it in the fixpoint (0,0). The eigenvalues are 0 and 1. In this case, how do I use phase plane analysis to determine whether (0,0) is stable of unstable? 



#7
Feb2009, 01:16 AM

P: 1,863

There's no other way than to draw the phase plane? Doing this, I find that it is unstable.
I haven't heard of Morse Lemma. 



#8
Mar1410, 02:30 PM

P: 1

One way to prove the point is stable is using Lyapunov stability theory, a common practice in control engineering. Consider the function
V = 1/2 * x^2 + 1/2 * y^2 which we differentiate along the trajectories of x_dot, y_dot to find V_dot =  y^2, which implies that x and y are always bounded, and V always decreasing while y is nonzero. Therefore, we can conclude that y must go to zero. Using a wellknown lemma known as Barbalats lemma, if the second derivative of V is bounded, then we can say that (d/dt y) also goes to zero. If (d/dt y) goes to zero, and we know that y must go to zero, then x must also go to zero. Therefore, the equilibrium soln (x,y) = (0,0) is globally, asymptotically stable. 



#9
May110, 08:22 PM

P: 1




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