When solving 2 ODEs by eliminating time is valid?












0












$begingroup$


When solving a 2nd order ODE, say
begin{equation}tag{*}begin{cases}frac{dx}{dt}=f(x,y)\frac{dy}{dt}=g(x,y),end{cases}end{equation}
it is common to eliminate time and solve the resulting 1st order ODE
begin{equation}tag{**}frac{dy}{dx}=frac{g(x,y)}{f(x,y)}end{equation}
that gives a dependence $y=y(x)$.



I wonder what are the conditions for this approach to be valid at an equilibrium point $(x^*,y^*)$? At this point, the fraction $frac{g(x^*,y^*)}{f(x^*,y^*)}=frac{0}{0}$. Potentially, this indeterminacy can be resolved using a sort of multivalued L'Hopital rule, but that is quite tricky.



Intuitively, I understand that the answer depends on the structure of the eigenvalues of the linearization of (*) at $(x^*,y^*)$, but I cannot formulate this quite well.



Let, say, the linearized system have a saddle at $(x^*,y^*)$. The equation ($**$) has two solutions corresponding to the stable and the unstable manifolds (they seem to be both unstable as they go away from $(0,0)$). Does it imply that the DE ($**$) isn't well posed?






ordinary-differential-equations dynamical-systems







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    0












    $begingroup$


    When solving a 2nd order ODE, say
    begin{equation}tag{*}begin{cases}frac{dx}{dt}=f(x,y)\frac{dy}{dt}=g(x,y),end{cases}end{equation}
    it is common to eliminate time and solve the resulting 1st order ODE
    begin{equation}tag{**}frac{dy}{dx}=frac{g(x,y)}{f(x,y)}end{equation}
    that gives a dependence $y=y(x)$.



    I wonder what are the conditions for this approach to be valid at an equilibrium point $(x^*,y^*)$? At this point, the fraction $frac{g(x^*,y^*)}{f(x^*,y^*)}=frac{0}{0}$. Potentially, this indeterminacy can be resolved using a sort of multivalued L'Hopital rule, but that is quite tricky.



    Intuitively, I understand that the answer depends on the structure of the eigenvalues of the linearization of (*) at $(x^*,y^*)$, but I cannot formulate this quite well.



    Let, say, the linearized system have a saddle at $(x^*,y^*)$. The equation ($**$) has two solutions corresponding to the stable and the unstable manifolds (they seem to be both unstable as they go away from $(0,0)$). Does it imply that the DE ($**$) isn't well posed?






    ordinary-differential-equations dynamical-systems







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      When solving a 2nd order ODE, say
      begin{equation}tag{*}begin{cases}frac{dx}{dt}=f(x,y)\frac{dy}{dt}=g(x,y),end{cases}end{equation}
      it is common to eliminate time and solve the resulting 1st order ODE
      begin{equation}tag{**}frac{dy}{dx}=frac{g(x,y)}{f(x,y)}end{equation}
      that gives a dependence $y=y(x)$.



      I wonder what are the conditions for this approach to be valid at an equilibrium point $(x^*,y^*)$? At this point, the fraction $frac{g(x^*,y^*)}{f(x^*,y^*)}=frac{0}{0}$. Potentially, this indeterminacy can be resolved using a sort of multivalued L'Hopital rule, but that is quite tricky.



      Intuitively, I understand that the answer depends on the structure of the eigenvalues of the linearization of (*) at $(x^*,y^*)$, but I cannot formulate this quite well.



      Let, say, the linearized system have a saddle at $(x^*,y^*)$. The equation ($**$) has two solutions corresponding to the stable and the unstable manifolds (they seem to be both unstable as they go away from $(0,0)$). Does it imply that the DE ($**$) isn't well posed?






      ordinary-differential-equations dynamical-systems







      share|cite|improve this question









      $endgroup$




      When solving a 2nd order ODE, say
      begin{equation}tag{*}begin{cases}frac{dx}{dt}=f(x,y)\frac{dy}{dt}=g(x,y),end{cases}end{equation}
      it is common to eliminate time and solve the resulting 1st order ODE
      begin{equation}tag{**}frac{dy}{dx}=frac{g(x,y)}{f(x,y)}end{equation}
      that gives a dependence $y=y(x)$.



      I wonder what are the conditions for this approach to be valid at an equilibrium point $(x^*,y^*)$? At this point, the fraction $frac{g(x^*,y^*)}{f(x^*,y^*)}=frac{0}{0}$. Potentially, this indeterminacy can be resolved using a sort of multivalued L'Hopital rule, but that is quite tricky.



      Intuitively, I understand that the answer depends on the structure of the eigenvalues of the linearization of (*) at $(x^*,y^*)$, but I cannot formulate this quite well.



      Let, say, the linearized system have a saddle at $(x^*,y^*)$. The equation ($**$) has two solutions corresponding to the stable and the unstable manifolds (they seem to be both unstable as they go away from $(0,0)$). Does it imply that the DE ($**$) isn't well posed?






      ordinary-differential-equations dynamical-systems




      ordinary-differential-equations dynamical-systems






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 25 '18 at 15:41









      DmitryDmitry

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          $begingroup$

          Indeed, it doesn't work at an equilibrium point, but you don't really need it there: $(x,y) = (x^*, y^*)$ is the solution with initial conditions $x(0)=x^*, y(0)=y^*$.

          It's also not defined on the curve $f(x,y) = 0$ (although there, when $g(x,y) ne 0$, you could look at $x$ as a function of $y$). However, it is OK everywhere else, and it can be useful to study limits of these solutions as $x to x^*$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Dear @Robert, I ask because I often saw this approach used for computing stable/unstable manifolds of a system of nonlinear DEs. In this case the solution goes exactly from the equilibrium. Sometimes this seems to be a trick rather than a formal approach.
            $endgroup$
            – Dmitry
            Dec 25 '18 at 16:24










          • $begingroup$
            I don't think so: the stable and unstable manifolds would have the same initial condition $y(x^*) = y^*$. The equilibrium should correspond to a limit. For example, the linear saddle-point system $dot{x} = y$, $dot{y} = x$ leads to $dfrac{dy}{dx} = dfrac{x}{y}$, not defined at $(0,0)$, but the solutions $y=x$ and $y = -x$ have limits $(0,0)$ as $x to 0$.
            $endgroup$
            – Robert Israel
            Dec 26 '18 at 0:52












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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          2












          $begingroup$

          Indeed, it doesn't work at an equilibrium point, but you don't really need it there: $(x,y) = (x^*, y^*)$ is the solution with initial conditions $x(0)=x^*, y(0)=y^*$.

          It's also not defined on the curve $f(x,y) = 0$ (although there, when $g(x,y) ne 0$, you could look at $x$ as a function of $y$). However, it is OK everywhere else, and it can be useful to study limits of these solutions as $x to x^*$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Dear @Robert, I ask because I often saw this approach used for computing stable/unstable manifolds of a system of nonlinear DEs. In this case the solution goes exactly from the equilibrium. Sometimes this seems to be a trick rather than a formal approach.
            $endgroup$
            – Dmitry
            Dec 25 '18 at 16:24










          • $begingroup$
            I don't think so: the stable and unstable manifolds would have the same initial condition $y(x^*) = y^*$. The equilibrium should correspond to a limit. For example, the linear saddle-point system $dot{x} = y$, $dot{y} = x$ leads to $dfrac{dy}{dx} = dfrac{x}{y}$, not defined at $(0,0)$, but the solutions $y=x$ and $y = -x$ have limits $(0,0)$ as $x to 0$.
            $endgroup$
            – Robert Israel
            Dec 26 '18 at 0:52
















          2












          $begingroup$

          Indeed, it doesn't work at an equilibrium point, but you don't really need it there: $(x,y) = (x^*, y^*)$ is the solution with initial conditions $x(0)=x^*, y(0)=y^*$.

          It's also not defined on the curve $f(x,y) = 0$ (although there, when $g(x,y) ne 0$, you could look at $x$ as a function of $y$). However, it is OK everywhere else, and it can be useful to study limits of these solutions as $x to x^*$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Dear @Robert, I ask because I often saw this approach used for computing stable/unstable manifolds of a system of nonlinear DEs. In this case the solution goes exactly from the equilibrium. Sometimes this seems to be a trick rather than a formal approach.
            $endgroup$
            – Dmitry
            Dec 25 '18 at 16:24










          • $begingroup$
            I don't think so: the stable and unstable manifolds would have the same initial condition $y(x^*) = y^*$. The equilibrium should correspond to a limit. For example, the linear saddle-point system $dot{x} = y$, $dot{y} = x$ leads to $dfrac{dy}{dx} = dfrac{x}{y}$, not defined at $(0,0)$, but the solutions $y=x$ and $y = -x$ have limits $(0,0)$ as $x to 0$.
            $endgroup$
            – Robert Israel
            Dec 26 '18 at 0:52














          2












          2








          2





          $begingroup$

          Indeed, it doesn't work at an equilibrium point, but you don't really need it there: $(x,y) = (x^*, y^*)$ is the solution with initial conditions $x(0)=x^*, y(0)=y^*$.

          It's also not defined on the curve $f(x,y) = 0$ (although there, when $g(x,y) ne 0$, you could look at $x$ as a function of $y$). However, it is OK everywhere else, and it can be useful to study limits of these solutions as $x to x^*$.






          share|cite|improve this answer









          $endgroup$



          Indeed, it doesn't work at an equilibrium point, but you don't really need it there: $(x,y) = (x^*, y^*)$ is the solution with initial conditions $x(0)=x^*, y(0)=y^*$.

          It's also not defined on the curve $f(x,y) = 0$ (although there, when $g(x,y) ne 0$, you could look at $x$ as a function of $y$). However, it is OK everywhere else, and it can be useful to study limits of these solutions as $x to x^*$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 16:00









          Robert IsraelRobert Israel

          331k23220475




          331k23220475












          • $begingroup$
            Dear @Robert, I ask because I often saw this approach used for computing stable/unstable manifolds of a system of nonlinear DEs. In this case the solution goes exactly from the equilibrium. Sometimes this seems to be a trick rather than a formal approach.
            $endgroup$
            – Dmitry
            Dec 25 '18 at 16:24










          • $begingroup$
            I don't think so: the stable and unstable manifolds would have the same initial condition $y(x^*) = y^*$. The equilibrium should correspond to a limit. For example, the linear saddle-point system $dot{x} = y$, $dot{y} = x$ leads to $dfrac{dy}{dx} = dfrac{x}{y}$, not defined at $(0,0)$, but the solutions $y=x$ and $y = -x$ have limits $(0,0)$ as $x to 0$.
            $endgroup$
            – Robert Israel
            Dec 26 '18 at 0:52


















          • $begingroup$
            Dear @Robert, I ask because I often saw this approach used for computing stable/unstable manifolds of a system of nonlinear DEs. In this case the solution goes exactly from the equilibrium. Sometimes this seems to be a trick rather than a formal approach.
            $endgroup$
            – Dmitry
            Dec 25 '18 at 16:24










          • $begingroup$
            I don't think so: the stable and unstable manifolds would have the same initial condition $y(x^*) = y^*$. The equilibrium should correspond to a limit. For example, the linear saddle-point system $dot{x} = y$, $dot{y} = x$ leads to $dfrac{dy}{dx} = dfrac{x}{y}$, not defined at $(0,0)$, but the solutions $y=x$ and $y = -x$ have limits $(0,0)$ as $x to 0$.
            $endgroup$
            – Robert Israel
            Dec 26 '18 at 0:52
















          $begingroup$
          Dear @Robert, I ask because I often saw this approach used for computing stable/unstable manifolds of a system of nonlinear DEs. In this case the solution goes exactly from the equilibrium. Sometimes this seems to be a trick rather than a formal approach.
          $endgroup$
          – Dmitry
          Dec 25 '18 at 16:24




          $begingroup$
          Dear @Robert, I ask because I often saw this approach used for computing stable/unstable manifolds of a system of nonlinear DEs. In this case the solution goes exactly from the equilibrium. Sometimes this seems to be a trick rather than a formal approach.
          $endgroup$
          – Dmitry
          Dec 25 '18 at 16:24












          $begingroup$
          I don't think so: the stable and unstable manifolds would have the same initial condition $y(x^*) = y^*$. The equilibrium should correspond to a limit. For example, the linear saddle-point system $dot{x} = y$, $dot{y} = x$ leads to $dfrac{dy}{dx} = dfrac{x}{y}$, not defined at $(0,0)$, but the solutions $y=x$ and $y = -x$ have limits $(0,0)$ as $x to 0$.
          $endgroup$
          – Robert Israel
          Dec 26 '18 at 0:52




          $begingroup$
          I don't think so: the stable and unstable manifolds would have the same initial condition $y(x^*) = y^*$. The equilibrium should correspond to a limit. For example, the linear saddle-point system $dot{x} = y$, $dot{y} = x$ leads to $dfrac{dy}{dx} = dfrac{x}{y}$, not defined at $(0,0)$, but the solutions $y=x$ and $y = -x$ have limits $(0,0)$ as $x to 0$.
          $endgroup$
          – Robert Israel
          Dec 26 '18 at 0:52


















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