Lyapunov Indirect Method












1












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I have searched a reference (book/paper) where I could find a theorem related to the Lyapunov Indirect Method for any equilibrium point, but I have not found yet. I only found for the zero equilibrium point. Somebody can help me?



I am searching the following Theorem:



Let us consider the non-linear system $dot{x}=f(x)$ with an arbitrary equilibrium point $bar{x}$. This equilibrium point $bar{x}$ is:




  1. asymptotically stable if all eigenvalues of $Df(bar{x})$ have negative real part;

  2. unstable if at least one eigenvalue of $Df(bar{x})$ has positive real part;


where $Df(bar{x})$ is the Jacobin matrix of $f$ applied in the equilibrium point $bar{x}$.



Thank you very much!



Ana










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$endgroup$








  • 2




    $begingroup$
    You can always make a coordinate transformation $y = x - overline{x}$ and get a system with an equilibrium at the origin. This transformation preserves stability properties (as any orientation-preserving diffeomorphism, really), thus any conclusions that you draw for an equilibrium at the origin are valid for an equilibrium point of original system.
    $endgroup$
    – Evgeny
    Jan 23 at 12:30


















1












$begingroup$


I have searched a reference (book/paper) where I could find a theorem related to the Lyapunov Indirect Method for any equilibrium point, but I have not found yet. I only found for the zero equilibrium point. Somebody can help me?



I am searching the following Theorem:



Let us consider the non-linear system $dot{x}=f(x)$ with an arbitrary equilibrium point $bar{x}$. This equilibrium point $bar{x}$ is:




  1. asymptotically stable if all eigenvalues of $Df(bar{x})$ have negative real part;

  2. unstable if at least one eigenvalue of $Df(bar{x})$ has positive real part;


where $Df(bar{x})$ is the Jacobin matrix of $f$ applied in the equilibrium point $bar{x}$.



Thank you very much!



Ana










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You can always make a coordinate transformation $y = x - overline{x}$ and get a system with an equilibrium at the origin. This transformation preserves stability properties (as any orientation-preserving diffeomorphism, really), thus any conclusions that you draw for an equilibrium at the origin are valid for an equilibrium point of original system.
    $endgroup$
    – Evgeny
    Jan 23 at 12:30
















1












1








1





$begingroup$


I have searched a reference (book/paper) where I could find a theorem related to the Lyapunov Indirect Method for any equilibrium point, but I have not found yet. I only found for the zero equilibrium point. Somebody can help me?



I am searching the following Theorem:



Let us consider the non-linear system $dot{x}=f(x)$ with an arbitrary equilibrium point $bar{x}$. This equilibrium point $bar{x}$ is:




  1. asymptotically stable if all eigenvalues of $Df(bar{x})$ have negative real part;

  2. unstable if at least one eigenvalue of $Df(bar{x})$ has positive real part;


where $Df(bar{x})$ is the Jacobin matrix of $f$ applied in the equilibrium point $bar{x}$.



Thank you very much!



Ana










share|cite|improve this question











$endgroup$




I have searched a reference (book/paper) where I could find a theorem related to the Lyapunov Indirect Method for any equilibrium point, but I have not found yet. I only found for the zero equilibrium point. Somebody can help me?



I am searching the following Theorem:



Let us consider the non-linear system $dot{x}=f(x)$ with an arbitrary equilibrium point $bar{x}$. This equilibrium point $bar{x}$ is:




  1. asymptotically stable if all eigenvalues of $Df(bar{x})$ have negative real part;

  2. unstable if at least one eigenvalue of $Df(bar{x})$ has positive real part;


where $Df(bar{x})$ is the Jacobin matrix of $f$ applied in the equilibrium point $bar{x}$.



Thank you very much!



Ana







stability-in-odes non-linear-dynamics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 12:29







Ana

















asked Jan 23 at 12:23









AnaAna

284




284








  • 2




    $begingroup$
    You can always make a coordinate transformation $y = x - overline{x}$ and get a system with an equilibrium at the origin. This transformation preserves stability properties (as any orientation-preserving diffeomorphism, really), thus any conclusions that you draw for an equilibrium at the origin are valid for an equilibrium point of original system.
    $endgroup$
    – Evgeny
    Jan 23 at 12:30
















  • 2




    $begingroup$
    You can always make a coordinate transformation $y = x - overline{x}$ and get a system with an equilibrium at the origin. This transformation preserves stability properties (as any orientation-preserving diffeomorphism, really), thus any conclusions that you draw for an equilibrium at the origin are valid for an equilibrium point of original system.
    $endgroup$
    – Evgeny
    Jan 23 at 12:30










2




2




$begingroup$
You can always make a coordinate transformation $y = x - overline{x}$ and get a system with an equilibrium at the origin. This transformation preserves stability properties (as any orientation-preserving diffeomorphism, really), thus any conclusions that you draw for an equilibrium at the origin are valid for an equilibrium point of original system.
$endgroup$
– Evgeny
Jan 23 at 12:30






$begingroup$
You can always make a coordinate transformation $y = x - overline{x}$ and get a system with an equilibrium at the origin. This transformation preserves stability properties (as any orientation-preserving diffeomorphism, really), thus any conclusions that you draw for an equilibrium at the origin are valid for an equilibrium point of original system.
$endgroup$
– Evgeny
Jan 23 at 12:30












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