Linear Systems of ODE's: Eigenvalues and Stability

[Niles]

Homework Statement

Hi all.

I am given by following linear system:

$$\begin{array}{l} \dot x = dx/dt = ax \\ \dot y = dy/dt = - y \\ \end{array}$$

The eigenvalue of the matrix of this system determines the stability of the fixpoint (0,0):

$$A=\left( {\begin{array}{*{20}c} a & 0 \\ 0 & { - 1} \\ \end{array}} \right) \quad \Rightarrow \quad \lambda_{1,2}= 0, a.$$

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.Niles.

 

[Dick]

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

 

[Niles]

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?

 

[Dick]

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.

 

[dirk_mec1]

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.-a<0 or a>0.

Then it's called a degenerate critical point. Only phase plane analysis can give you info about the stability. The Blow-up method always works in this case but it may happen that you have to blow up several times. You can also look at Morse Lemma or in rare cases transforming to another coordinate system (like polar) can give you the desired result.

 

[Niles]

Lets say I have the following non-linear system:

$$\begin{array}{l} dx/dt = - xy \\ dy/dt = - y + x^2 \\ \end{array}.$$

We will look at one of the fixpoints, namely (x,y) = (0,0). The Jacobian is given by:

$$\[ \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),$$

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?

 

[Niles]

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.

 

[1Drew1]

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 non-zero. Therefore, we can conclude that y must go to zero. Using a well-known 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.

 

[ufk1988]

Lets say I have the following non-linear system:

$$\begin{array}{l} dx/dt = - xy \\ dy/dt = - y + x^2 \\ \end{array}.$$

We will look at one of the fixpoints, namely (x,y) = (0,0). The Jacobian is given by:

$$\[ \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),$$

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?

Can anyone please tell me, what are the corresponding eigenvectors