Positive definite Hessian implies Lyapunov stability

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The Lagrange-Dirichlet Theorem is partially reversed by the following result:



  1. if the costraints are holonomous, bilateral, ideal,

  2. if $q^*$ is a critical point for the potential energy $U$,

then the equilibrium $q^*$ is Lyapunov stable IFF the (quadratic form associated to the) Hessian matrix of $U$ is positive definite.



It seems intuitive, but I don't know how to prove it.
Should I use first and/or second Lyapunov method?







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    up vote
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    down vote

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    The Lagrange-Dirichlet Theorem is partially reversed by the following result:



    1. if the costraints are holonomous, bilateral, ideal,

    2. if $q^*$ is a critical point for the potential energy $U$,

    then the equilibrium $q^*$ is Lyapunov stable IFF the (quadratic form associated to the) Hessian matrix of $U$ is positive definite.



    It seems intuitive, but I don't know how to prove it.
    Should I use first and/or second Lyapunov method?







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      The Lagrange-Dirichlet Theorem is partially reversed by the following result:



      1. if the costraints are holonomous, bilateral, ideal,

      2. if $q^*$ is a critical point for the potential energy $U$,

      then the equilibrium $q^*$ is Lyapunov stable IFF the (quadratic form associated to the) Hessian matrix of $U$ is positive definite.



      It seems intuitive, but I don't know how to prove it.
      Should I use first and/or second Lyapunov method?







      share|cite|improve this question














      The Lagrange-Dirichlet Theorem is partially reversed by the following result:



      1. if the costraints are holonomous, bilateral, ideal,

      2. if $q^*$ is a critical point for the potential energy $U$,

      then the equilibrium $q^*$ is Lyapunov stable IFF the (quadratic form associated to the) Hessian matrix of $U$ is positive definite.



      It seems intuitive, but I don't know how to prove it.
      Should I use first and/or second Lyapunov method?









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 20 at 16:17









      enzotib

      5,73121430




      5,73121430










      asked Aug 20 at 5:40









      Gauss the Mauss

      261




      261




















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          First of all let's me say that a stable equilibrium position is every isolated local minimum of $U$, even if the minimum cannot be deducted from the Hessian of $U$, begin this null (consider the 2D example $U = q_1^4+q_2^4$.)



          In the case the Hessian of $U$ is not null and is not positive definite, this means that at least one of its eigenvalues is negative, then in the study of small oscillations around $q^*$, the fundamental motion related to this eigenvalues is not a sinusoidal oscillation, but is an exponential solution in which the system depart from $q^*$. This is in some way related to the first Lyapunov method.






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            First of all let's me say that a stable equilibrium position is every isolated local minimum of $U$, even if the minimum cannot be deducted from the Hessian of $U$, begin this null (consider the 2D example $U = q_1^4+q_2^4$.)



            In the case the Hessian of $U$ is not null and is not positive definite, this means that at least one of its eigenvalues is negative, then in the study of small oscillations around $q^*$, the fundamental motion related to this eigenvalues is not a sinusoidal oscillation, but is an exponential solution in which the system depart from $q^*$. This is in some way related to the first Lyapunov method.






            share|cite|improve this answer
























              up vote
              0
              down vote













              First of all let's me say that a stable equilibrium position is every isolated local minimum of $U$, even if the minimum cannot be deducted from the Hessian of $U$, begin this null (consider the 2D example $U = q_1^4+q_2^4$.)



              In the case the Hessian of $U$ is not null and is not positive definite, this means that at least one of its eigenvalues is negative, then in the study of small oscillations around $q^*$, the fundamental motion related to this eigenvalues is not a sinusoidal oscillation, but is an exponential solution in which the system depart from $q^*$. This is in some way related to the first Lyapunov method.






              share|cite|improve this answer






















                up vote
                0
                down vote










                up vote
                0
                down vote









                First of all let's me say that a stable equilibrium position is every isolated local minimum of $U$, even if the minimum cannot be deducted from the Hessian of $U$, begin this null (consider the 2D example $U = q_1^4+q_2^4$.)



                In the case the Hessian of $U$ is not null and is not positive definite, this means that at least one of its eigenvalues is negative, then in the study of small oscillations around $q^*$, the fundamental motion related to this eigenvalues is not a sinusoidal oscillation, but is an exponential solution in which the system depart from $q^*$. This is in some way related to the first Lyapunov method.






                share|cite|improve this answer












                First of all let's me say that a stable equilibrium position is every isolated local minimum of $U$, even if the minimum cannot be deducted from the Hessian of $U$, begin this null (consider the 2D example $U = q_1^4+q_2^4$.)



                In the case the Hessian of $U$ is not null and is not positive definite, this means that at least one of its eigenvalues is negative, then in the study of small oscillations around $q^*$, the fundamental motion related to this eigenvalues is not a sinusoidal oscillation, but is an exponential solution in which the system depart from $q^*$. This is in some way related to the first Lyapunov method.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 20 at 16:33









                enzotib

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