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







share|cite|improve this question









$endgroup$

















    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











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 25 '18 at 15:41









      DmitryDmitry

      701618




      701618






















          1 Answer
          1






          active

          oldest

          votes


















          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052210%2fwhen-solving-2-odes-by-eliminating-time-is-valid%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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

          324k23214468




          324k23214468












          • $begingroup$
            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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052210%2fwhen-solving-2-odes-by-eliminating-time-is-valid%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
          Read more

          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

          Image
          0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
          Read more