Lyapunov and behavior of a linearized system

In summary: V}{\partial y}\frac{\partial V}{\partial y} = \frac{y}{y^2} = \frac{1}{y}Therefore, we have shown that \frac{\partial V}{\partial y} = \frac{1}{y}. Similarly, we can show that \frac{\partial V}{\partial x} = \frac{x^3}{x^4} = \frac{1}{x}. So our final expression for V(x,y) is:V(x,y) = \frac{1}{x}x^3 + \frac{1}{y}y^2Simplifying, we get:V(x,y) = x^2
  • #1
dprimedx
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Homework Statement



Examine the behavior in the neighborhood of the origin. Derive the given Lyapunov V(x,y). Show the region [itex]\frac{dV}{dt} = 0[/itex] and relate it to your analysis.

Homework Equations



[itex]\dot{x} = x^3 - xy [/itex]
[itex]\dot{y} = -y + y^2 + xy - x^3[/itex]

As given by the text:
[itex]V(x,y) = \frac{x^4}{4} + \frac{y^2}{2}[/itex]
[itex]\dot{V}(x,y) = x^6 + yx^3 (1+x)+y^2 (1-x-y)[/itex]

The Attempt at a Solution


I've been stumped on this problem for a few days now. I spoke with my professor and he said to just "do the best you can and use what we learned in class! If it doesn't work, it doesn't work!"

For the first part of the question, I attempted to linearize the system using the Jacobian evaluated at the origin which gives the following:
{0,0}
{0,-1}

Sorry. I'm not used to LaTeX and I couldn't figure out how to enter a matrix!

To determine what sort of equilibrium that is locally, I tried two methods. The first was to check the trace (-1) and determinant (0) and check it against a chart he had given us. From this, I would guess that it would be a saddle point equilibrium. However, when I try by checking the eigenvalues (0 and -1 again), I get a line field. Am I doing something wrong here?

Part two was the difficult part. My professor instructed us to simply do as much as we can. The method he had shown us won't work for this case. Here's what I tried:

To find V(x,y), use the following:
[itex] V(\textbf{x}) = \textbf{x}^{T}.B.\textbf{x}[/itex]
[itex] A^{T}.B+B.A=-Identity[2][/itex]

Typically we use Mathematica for assignments. Using this method in Mathematica, you can't solve it. Doing it by hand results in B = {{0,0},{0,1/2}} (again, sorry for the matrices!)

Now, this shouldn't be a problem. However, when you plug B into the equation for V, you get [itex]\frac{y^2}{2}[/itex] which isn't what the book gives. My professor said just to explain this and continue onward. I'd really like to know what it doesn't work though. Is the method correct, but the book wrong? Is there another method that would result in what the book gave?

My final issue is what is holding me back from completing this. For part c, we never covered what dV/dt = 0 represents and I'm unable to reason it myself. I know the Lyapunov, V, is the basin of attraction, but can't what the change in the basin of attraction really mean.

I'm also asked to graph everything, but it doesn't seem right to me. V doesn't fit with the global (non-linear) system and I can't find a linear system to graph. And dV/dt=0 looks rather odd and doesn't look like it relates to V at all.

Mathematica file: https://www.dropbox.com/sh/ccb10ai10kbk7ev/9Ntunu1801/p286.9.18.nb
It's apparently too large to attach!

And here's the textbook information:
Dynamics and Bifurcations, Hale and Kocak. Page 286, #9.18.

Thank you for your help and thanks for this wonderful community! I may hardly ever post, but I do lurk around almost daily to get my dose of nerdgasms.
 
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  • #2

Hi there,

Thank you for your detailed post and for your interest in this problem. I understand that you have put a lot of effort into solving this and that it can be frustrating when things don't seem to work out. Let me try to help you with your questions and hopefully provide some insights that will help you complete this problem.

Firstly, for the linearization at the origin, you are correct in obtaining a saddle point equilibrium. The eigenvalues of the Jacobian matrix are indeed 0 and -1, indicating that the equilibrium is a saddle point. However, the line field you obtained is also correct. This is because the linearization only gives you information about the behavior of the system near the equilibrium point, and not globally. The line field shows the behavior of the system for all values of x and y, not just near the origin. So both methods are correct, they just give different information about the system.

Moving on to the derivation of V(x,y), I'm not sure what method your professor is referring to, but here is how I would approach it. Since we are given the Lyapunov function V(x,y), we can use the definition of Lyapunov functions to show that it satisfies the Lyapunov equation:

\dot{V}(x,y) = \frac{\partial V}{\partial x}\dot{x} + \frac{\partial V}{\partial y}\dot{y}

Substituting in the given equations for \dot{x} and \dot{y}, we get:

\dot{V}(x,y) = x^3\frac{\partial V}{\partial x} - xy\frac{\partial V}{\partial x} - y\frac{\partial V}{\partial y} + y^2\frac{\partial V}{\partial y} + xy\frac{\partial V}{\partial x} - x^3\frac{\partial V}{\partial x}

Simplifying, we get:

\dot{V}(x,y) = -y\frac{\partial V}{\partial y} + y^2\frac{\partial V}{\partial y}

Since we want to show that \dot{V} is equal to 0, we can set this expression equal to 0 and solve for \frac{\partial V}{\partial y}:

0 = -y\frac{\partial V}{\partial y} + y^2
 

Related to Lyapunov and behavior of a linearized system

1. What is the Lyapunov stability criterion?

The Lyapunov stability criterion is a mathematical tool used to analyze the stability of a nonlinear system. It states that if there exists a function, called a Lyapunov function, that satisfies certain conditions, then the system is stable.

2. How does linearization affect the behavior of a system?

Linearization is the process of approximating a nonlinear system with a linear one. This can simplify the analysis of the system, but it can also change the behavior of the system. In some cases, the linearized system may behave similarly to the original nonlinear system, but in other cases, it may behave very differently.

3. What are the limitations of using linearization to analyze a system?

Linearization is only applicable to systems that are close to a stable equilibrium point. It also assumes that the system is small enough for the linear approximation to be accurate. If the system is highly nonlinear or far from equilibrium, linearization may not provide an accurate analysis.

4. How can Lyapunov functions be used to determine stability?

Lyapunov functions are used to prove the stability of a system by showing that they satisfy certain conditions. These conditions ensure that the function decreases along the trajectories of the system, indicating stability. If a Lyapunov function can be found for a system, it is a strong indication that the system is stable.

5. Can linearization be used to analyze all types of systems?

No, linearization is only applicable to systems that can be approximated as linear. This means that the system must have certain properties, such as being close to equilibrium and having small perturbations. Nonlinear systems cannot be linearized, and therefore, other methods must be used to analyze their behavior.

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