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I have got the following system:

$\dot{z}_2 = - \gamma_2 \left ( \begin{bmatrix} \sin^2(x_1(t)) & \sin(x_1(t))x_2(t)\\ x_2(t)\sin(x_1(t)) & x_2(t)^2 \end{bmatrix} \right ) z_2$

which can be considered as a linear time-varying system. How can I proof the stability of the origin?

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  • $\begingroup$ Have you tried linearizing it around the origin and looking at the eigenvalues of the resulting Jacobian matrix? $\endgroup$ – Calculon Jul 30 '14 at 16:36
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Lyapunov methods work great (assuming $\gamma_2 > 0$). Let $V(z_2) = \dfrac{1}{2}\|z_2\|^2 = \dfrac{1}{2}(z_{21}^2+z_{22}^2)$. Then:

$\dot{V} = \dot{z}_{21}z_{21}+\dot{z}_{22}z_{22}$

$= (-\gamma_2\sin^2(x_1) z_{21} - \gamma_2\sin(x_1) x_2z_{22})z_{21} + (-\gamma_2\sin(x_1)x_2z_{21} - \gamma_2x_2^2z_{22})z_{22}$

$= -\gamma_2\left(z_{21}^2\sin^2(x_1) + 2z_{21}z_{22}\sin(x_1)x_2 + z_{22}^2x_{2}^2 \right)$

$= -\gamma_2\left(z_{21}\sin(x_1)+z_{22}x_2\right)^2$

$\le 0$.

This shows that $V(z_2) = \dfrac{1}{2}\|z_2\|^2$ is non-increasing. What does this tell you about the stability of the system near the origin?

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  • $\begingroup$ Because $V$ is descrescent and $\dot{V}$ is negative semi-definite, then the equilibrium $\boldsymbol{z} = 0$ is uniformly stable. Right? $\endgroup$ – Pietair Aug 4 '14 at 11:32

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