2
$\begingroup$

Determine the stability property of the critical point at the origin ($x=y=0$) for the following system:

I have an example: $$\begin{cases} & \mathrm{ } \dot{x}= \tan(y-x)\\ & \mathrm{ } \dot{y}= 2^y-2\cos(\dfrac{\pi}{3}-x) \end{cases} \tag{1}$$

  • I use the matrix $A=\dfrac{\partial F}{\partial x}\mid_{x=0}$ then we have $$A=\begin{pmatrix} & -1&1\\ & -\sqrt{3}&\ln 2 \end{pmatrix}$$

where $\tan(y-x)\approx y-x$

  • Since $\det(A-\lambda I)=0$ implies the equation: $$\lambda^2+(1-\ln 2)\lambda+\sqrt{3}-\ln 2=0$$

We have $a_1=1-\ln 2>0$ and $a_0=\sqrt{3}-\ln 2>0$.

By The Hurwitz criterion, $x \equiv 0$ is Asymptotically Stable.

================================================

I have a problem: Determine the stability property of $x=y=0$?

a/ $$\begin{cases} & \mathrm{ } \dot{x}= e^{x+2y}-\cos 3x\\ & \mathrm{ } \dot{y}= 2\sqrt{1+2x}-2e^y \end{cases} \tag{a}$$

b/ $$\begin{cases} & \mathrm{ } \dot{x}= -3x+4y+\sin^3x -y^2\\ & \mathrm{ } \dot{y}= -2x+\sin y+e^yx^2 \end{cases} \tag{b}$$

Any help will be appreciated. Thanks!

$\endgroup$
  • $\begingroup$ Where did these come from? See answer below. $\endgroup$ – Amzoti Sep 21 '13 at 17:30
6
$\begingroup$

For these systems, you may have to pull out several tools to figure them out.

For the first one, you guessed at the energy function, but there is no known way to derive those (for some physical systems, you can, but no general way).

What other tools might we use?

  • Find the critical points - for some pathological problems, this in itself can be very difficult - like some of your examples. For all of your examples, one of the critical points is $(0,0)$
  • Find the Jacobian matrix and evaluate the eigenvalues at the critical points - sometimes this may not work.
  • Plot the phase portrait and look at the qualitative behavior.

For the first problem, if we find the Jacobian matrix, evaluate it at the critical point $(0,0)$, we find the the eigenvalues are $-2$ and $0$, so what can you tell about the stability from that? You can try this approach for the others and see if it bears fruit. Note, for all of these, there is more than a single critical point and there are some strange behaviors going on, but that is to be expected given these systems.

Here are the four phase portraits for these systems (note, if you expand out the ranges, you can see wild behaviors).

enter image description here

enter image description here

enter image description here

enter image description here

Update

We are given:

$$ f(x, y) = x'= e^{x+2y}-\cos 3x\\ g(x,y) = y'= 2\sqrt{1+2x}-2e^y$$

It is clear that $(0,0)$ is a critical point.

Lets find the Jacobian matrix of this system, evaluate it at the critical point, determine the type of critical point and validate this behavior using the phase portrait. The Jacobian matrix is given by:

$$\displaystyle J(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\\\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{bmatrix} = \begin{bmatrix} e^{x+2y} + 3 \sin 3x & 2e^{x+2y} \\ \dfrac{2}{\sqrt{x^2+1}} & -2e^{y} \end{bmatrix}$$

Now, evaluated at the critical point $(0,0)$, we have:

$$J(0,0) = \begin{bmatrix} 1 & 2 \\ 2 & -2 \end{bmatrix}$$

The eigenvalues of this system are $2$ and $-3$, thus we have two real, distinct and opposite sign $\rightarrow$ a saddle point.

If you look at the coordinates $(0,0)$ of the third phase portrait, do you see the saddle point? What can you tell about the stability of a saddle point?

The phase portrait lets us see this qualitative behavior without having to do all that work.

$\endgroup$
  • 1
    $\begingroup$ You rock with your phase portraits! +1 $\endgroup$ – Namaste Sep 22 '13 at 0:09
  • $\begingroup$ Thanks Amzoti! I'm sorry when I say that Your picture is too difficult to understand for me. I mean, it would be great if you can post your guide/hints/solution to me with the steps for the form of this problem. Example: $$\begin{cases} & \mathrm{ } \dot{x}= e^{x+2y}-\cos 3x\\ & \mathrm{ } \dot{y}= 2\sqrt{1+2x}-2e^y \end{cases};(a)$$. I have trouble that whether I need a function approximation and how we do it? How we can find matrix $A$..so on.. Thanks! $\endgroup$ – kimtahe6 Sep 22 '13 at 3:35
  • $\begingroup$ @kimtahe6: Please see my updates for the third system. Look at the coordinates $(0,0)$ and see if can make out what the analysis tells us. Each green arrows is the direction of the derivative and each blue line is a solution curve with different initial conditions. What do you notice about all of the solution curves around the critical point? You should repeat this for the other systems and then compare those results to the phase portrait. Regards $\endgroup$ – Amzoti Sep 22 '13 at 4:24
  • $\begingroup$ Wow! It's great! Amzoti. Now, I see. Thank you very much! You're very good (y)! $\endgroup$ – kimtahe6 Sep 22 '13 at 4:30
  • $\begingroup$ Thanks! Amzoti! I haven't seen you for a long time, Hihi! I'm happy when you have comments :) . I think that the function is at ($x=y=0$) then eigenvalues of the Jacobian matrix doesn't change (We use the function approximation or not use), is it correct? $\endgroup$ – kimtahe6 Oct 4 '13 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.