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Given the following system of nonlinear ODEs,

$$x_1'=-x_1-x_2$$ $$x_2'=2x_1-x_2^3$$

I need to use the quadratic Lyapunov function

$$V(x) = x^TQx$$

where $Q$ is a positive definite matrix such that

$$A^TQ+QA=-I$$

and where $A=Df(0,0)$, to find a neighbourhood $U$ around the origin, as large as possible, such that $U$ is positively invariant and all solutions starting in $U$ tend to the origin as $t\to\infty$.

I've carefully examined the course notes and the book by Perko (Dynamical Systems and Differential Equations), but couldn't find a clue how to do this problem.

I'd appreciate it very much if someone could please explain:

  1. How is this neighbourhood found?

  2. How is $V$ related to all this?

Rodrigo de Azevedo
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sequence
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    Sure you cannot find $(a,b,c)$ such that $V(x)=ax_1^2+bx_2^2+cx_1x_2$ is positive definite and $\dot V(x)<0$ in a neighbourhood of $(0,0)$ minus the point $(0,0)$ and $\dot V(0,0)=0$? One usually starts by checking if some $V(x)=ax_1^2+bx_2^2$ would not do the job... – Did Mar 04 '16 at 06:55

1 Answers1

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Write $$\dot{x}=Ax+g(x)=\left[\matrix{-1 & -1\\ 2 & 0}\right]x+\left[\matrix{0\\-x_2^3}\right]$$ The symmetric positive definite solution to $$QA+A^TQ=-I$$ is (by direct calculation) $$Q=\frac{1}{2}\left[\matrix{3 & 1\\ 1 & 2}\right]$$ Now for the Lyapunov function candidate $V=x^TQx$ we have $$\dot{V}=\dot{x}^TQx+x^TQ\dot{x}=(Ax+g(x))^TQx+x^TQ(Ax+g(x))\\ =x^T(QA+A^TQ)x+2x^TQg(x)\\ =-\|x\|^2+2x^TQg(x)\\ \leq -\|x\|^2+2\lambda_{\max}(Q)\|x\|\|g(x)\|$$ Note that $\|g(x)\|=|x_2|^3\leq \|x\|^3$ and therefore $$\dot{V}\leq -\|x\|^2+2\lambda_{\max}(Q)\|x\|^4\\ \leq -\|x\|^2+2\frac{\lambda_{\max}(Q)}{\lambda_{\min}(Q)}V\|x\|^2\\ \leq -\|x\|^2\left(1-2\frac{\lambda_{\max}(Q)}{\lambda_{\min}(Q)}V\right)$$ Thus $\dot{V}< 0$ for all $$V< \frac{\lambda_{\min}(Q)}{2\lambda_{\max}(Q)}$$ with $V\neq 0$ and therefore all trajectories staring within the neighborhood $$U:=\left\{x: V(x)< \frac{\lambda_{\min}(Q)}{2\lambda_{\max}(Q)}\right\}$$ remain in $U$ for all time thereafter and tend to the origin as $t\rightarrow\infty$.

RTJ
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  • Where does the $g(x) = \begin{bmatrix} 0\ -x_2^3 \end{bmatrix}$ come from? Also, why is $\dot{V}=x^T(QA+A^TQ)x+2x^TQg(x)$? Thanks – sequence Mar 05 '16 at 01:21
  • I didn't yet completely understand your approach, but can you please take a peek at what I did, maybe this way is also acceptable?

    $V(x) = 3/2 x_1^2 + x_1x_2 + x_2^2 > (1/2 x_1+x_2)^2 \ge 0$. $\dot{V} = 3x_1 x_1' + x_1'x_2 + x_1x_2' +2x_2x_2' =-x_1^2-x_2^2-x_1x_2^3-2x_2^4 \le -\lvert x \lvert^2 +3\lvert x\lvert^4= \lvert x\lvert^2 (-1 + 3\lvert x\lvert^2) \implies \lvert x \lvert \le 1/\sqrt{3}.$ [Where $\lvert \cdot \lvert$ denotes the vector norm]. @CTNT

     – sequence Mar 05 '16 at 03:01
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    @sequence The above reasoning is not fully correct. Note that $V\leq c_1$ is the area of an ellipse while $|x| \leq c_2$ is a disk. What you have proved is that $V$ is decreasing for all points of some disk. Thus, the positively invariant region $U$ is the largest area defined by $V \leq c_1$ and included within the disk not the disk itself . – RTJ Mar 05 '16 at 07:35
  • Can you please clarify how you got your $g(x)$, and the derivative involving $-I = QA+A^TQ$? @CTNT – sequence Mar 05 '16 at 19:57
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    @sequence $g(x)$ is the vector that contains all nonlinearities in the ODEs. The first one $\dot{x}_1=-x_1-x_2$ has no nonlinear term and therefore the first term is zero. The second term is the nonlinearity in $\dot{x}_2=\ldots$ i.e. $-x_2^3$. For the derivative I will edit my answer to explain in more detail. – RTJ Mar 05 '16 at 22:34
  • This is awesome, thanks :) Can you also please clarify what is $\lambda_{max}(Q)$? I guess it is related to the norm of $Q$, but how exactly? And why is $V = \frac{\lVert x \lVert^2}{\lambda_{min}(Q)}$? @CTNT – sequence Mar 06 '16 at 00:00
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    @sequence $\lambda_{\max}(Q)$ is the largest eigenvalue of $Q$. If you want to compare $V$ with $|x|^2$ you can use the eigenvalues of $Q$. Specifically it holds true that $$\lambda_{\min}(Q)|x|^2\leq V\leq \lambda_{\max}(Q)|x|^2$$ – RTJ Mar 06 '16 at 00:04
  • Which matrix norm are you using for $Q$ in $\lVert Q\lVert\le \lambda_{max}(Q) $? @CTNT – sequence Mar 06 '16 at 02:19
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    @sequence For a symmetric, positive definite matrix $Q$ the induced 2-norm $|Q|2$ (which in the general case is $\lambda{max}^{1/2}(Q^T Q)$) is equal to $\lambda_{\max}(Q)$ – RTJ Mar 06 '16 at 08:44
  • But if it's equal, then where's the last inequality in $$\dot{V}=\dot{x}^TQx+x^TQ\dot{x}=(Ax+g(x))^TQx+x^TQ(Ax+g(x))\ =x^T(QA+A^TQ)x+2x^TQg(x)\ =-|x|^2+2x^TQg(x)\ \leq -|x|^2+2\lambda_{\max}(Q)|x||g(x)|$$ coming from? @CTNT – sequence Mar 09 '16 at 02:14
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    @sequence In general for a symmetric, positive matrix $Q$ we have $x^TQz\leq \lambda_{\max}(Q)|x||z|$ . The equality holds only if both $x$, $z$ point to the direction of the corresponding eigenvector. – RTJ Mar 09 '16 at 07:25