3
$\begingroup$

Could someone please help?

The question reads: For which real numbers $k$ is the zero state a stable equilibrium of the dynamic system $x_{t+1} = Ax_t$?

$A = \begin{bmatrix} 0.1 &k \\ 0.3 & 0.3 \end{bmatrix}$

So, my thought is I need to find the eigenvalues. In order to do this I calculated the characteristic polynomial as

$x^2 - 0.4x + 0.03 - 0.3k = 0$, with $x$ representing eigenvalues.

Using the quadratic formula I found that the (real) eigenvalues are $$x = \frac{2 \pm \sqrt{1+30k}}{10}$$ and for the zero state to be in stable equilibrium $\sqrt{1+30k}<8$. Hence, $k<21/10$ (for stable equilibrium).

My question is how do I figure out the values for $k$ if the eigenvalues are complex?

Do I solve the inequality $\sqrt{-1-30k} < 8$?

Thanks!

  • 0
    Just so you know, I reformatted your question using LaTeX for easier viewing.2011-01-06
  • 1
    For the real case, you also need $k\geq \cfrac{-1}{30}$ to keep them real.2011-01-06
  • 0
    You're quite welcome. Glad it was helpful.2011-01-06

1 Answers 1

4

In general, the zero state of a dynamical system of the form $x_{t+1} = Ax_t$ is a stable equilibrium when all of the eigenvalues of $A$ are inside the unit circle in the complex plane. So if you have an eigenvalue of the form $a+bi$, you get stability when $a^2 + b^2 < 1$. Since this is a homework question, I'll let you work out the rest of the problem from here. (This isn't going to be equivalent to solving $\sqrt{-1-30k} < 8$, though.)

  • 0
    Good answer. Just to be clear, though, one has *asymptotic* stability if $a^2 + b^2 <1$. The system is generally considered stable if $a^2 + b^2 = 1$.2011-01-06
  • 0
    So, a = 2 and b= sqrt(-1-30k)? With this I get: 4 + -1 - 30k < 10 which leads to 3 - 30k < 10. Therefore, if the eigenvalues are complex k < -7/30 ????2011-01-06
  • 0
    sorry.... k > -7/302011-01-06
  • 1
    @user5430: you lost the denominator of 10 in your expressions for a and b. As Mike said, you then need to compare a^2+b^2 with 1.2011-01-06
  • 0
    @user5430: You should also ask the question what happens if $k=-7/30$.2011-01-06
  • 0
    Oops - I didn't notice the lost factor of 10.2011-01-06
  • 0
    @Ross: Can I not multiply both sides by 10? Am I being stupid here? Very possible; I forget the last time I had more than 4hrs of sleep at once :)2011-01-06
  • 1
    @user5430: but the 10 gets squared when you square a and b2011-01-06
  • 0
    Thanks Ross! Then I get sqrt((2/10)^2+(1/10 sqrt(-1-30 k))^2) < 1 which leads to 1/10 sqrt(3-30 k) < 1 and in the end k > -97/30.2011-01-06
  • 0
    @Ross: Thanks for handling the OP's questions while I was away from my computer.2011-01-06
  • 0
    @user5430: Writing 1/10 sqrt(3-30 k) is a good way to get confused about what is divided by what. sqrt(3-30 k)/10 is clearer. k>-97/30 is correct, but you need k small enough that the roots are complex. See my comment above about the real case.2011-01-06
  • 0
    @Mike: you're welcome.2011-01-06