Linear Systems of ODE's: Eigenvalues and Stability

In summary, the conversation discusses a linear system and its eigenvalues, specifically focusing on the stability of a fixpoint (0,0). It is determined that when a<0, the fixpoint is stable, and when a>0, it is a saddle point. The conversation then delves into what happens when a=0 and the use of phase plane analysis and other methods to determine stability in this case. It is ultimately concluded that the point (0,0) is unstable.
  • #1
Niles
1,866
0

Homework Statement


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.Niles.
 
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  • #2
Zero isn't an eigenvalue of that matrix (unless a=0). -1 is the other eigenvalue. ??
 
  • #3
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
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
Dick said:
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.
 
  • #6
Lets say I have the following non-linear 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
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
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.
 
  • #9
Niles said:
Lets say I have the following non-linear 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?

Can anyone please tell me, what are the corresponding eigenvectors
 

Related to Linear Systems of ODE's: Eigenvalues and Stability

1. What is a linear system of ODEs?

A linear system of ODEs is a set of ordinary differential equations that can be written in the form of a matrix equation. It includes multiple equations, each representing the rate of change of a different dependent variable, and all variables are linearly related to each other.

2. What are eigenvalues and eigenvectors of a linear system of ODEs?

Eigenvalues and eigenvectors are special values and corresponding vectors that represent the behavior of a linear system of ODEs. Eigenvalues are the values of the system's parameters that determine the behavior of the system, while eigenvectors are the corresponding vectors that represent the direction of the system's solutions.

3. How do eigenvalues and eigenvectors affect the stability of a linear system of ODEs?

The stability of a linear system of ODEs is determined by the values of its eigenvalues. If all eigenvalues have negative real parts, the system is stable and will approach an equilibrium state. If any eigenvalues have positive real parts, the system is unstable and will not reach an equilibrium state. Complex eigenvalues with negative real parts represent oscillating behavior.

4. How can eigenvalues and eigenvectors be calculated for a linear system of ODEs?

To calculate eigenvalues and eigenvectors for a linear system of ODEs, the system's matrix must first be formed. Then, the characteristic polynomial of the matrix is found and solved. The resulting values are the eigenvalues, and the corresponding eigenvectors can be found by plugging in each eigenvalue into the system's matrix equation and solving for the corresponding vector.

5. What is the significance of eigenvalues and eigenvectors in real-world applications?

Eigenvalues and eigenvectors have various applications in fields such as physics, engineering, and economics. They are used to analyze the stability and behavior of systems, such as electrical circuits, mechanical systems, and economic models. They also play a role in image and signal processing algorithms, such as the famous eigendecomposition technique used in principal component analysis.

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