Lagrangian corresponding to these equations of motion












1












$begingroup$



I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$




I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.



$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$



for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?






homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
    $endgroup$
    – Aaron Stevens
    4 hours ago










  • $begingroup$
    It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
    $endgroup$
    – TheAverageHijano
    4 hours ago










  • $begingroup$
    You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
    $endgroup$
    – Ballanzor
    4 hours ago
















1












$begingroup$



I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$




I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.



$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$



for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?






homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
    $endgroup$
    – Aaron Stevens
    4 hours ago










  • $begingroup$
    It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
    $endgroup$
    – TheAverageHijano
    4 hours ago










  • $begingroup$
    You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
    $endgroup$
    – Ballanzor
    4 hours ago














1












1








1


1



$begingroup$



I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$




I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.



$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$



for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?






homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators







share|cite|improve this question











$endgroup$





I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$




I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.



$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$



for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?






homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators




homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Qmechanic♦

105k121911206




105k121911206










asked 4 hours ago









TheAverageHijanoTheAverageHijano

4759




4759












  • $begingroup$
    Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
    $endgroup$
    – Aaron Stevens
    4 hours ago










  • $begingroup$
    It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
    $endgroup$
    – TheAverageHijano
    4 hours ago










  • $begingroup$
    You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
    $endgroup$
    – Ballanzor
    4 hours ago


















  • $begingroup$
    Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
    $endgroup$
    – Aaron Stevens
    4 hours ago










  • $begingroup$
    It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
    $endgroup$
    – TheAverageHijano
    4 hours ago










  • $begingroup$
    You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
    $endgroup$
    – Ballanzor
    4 hours ago
















$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
4 hours ago




$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
4 hours ago












$begingroup$
It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
4 hours ago




$begingroup$
It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
4 hours ago












$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
4 hours ago




$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
4 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

I suppose the Lagrangian you found is actually
$$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
$$ddot{q}_2 - 2 q_1 q_2 = 0$$
There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
$$-ddot{q}_2 - 2 q_1 q_2 = 0$$
This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
$$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$






share|cite|improve this answer









$endgroup$













    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: "151"
    };
    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: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    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%2fphysics.stackexchange.com%2fquestions%2f463599%2flagrangian-corresponding-to-these-equations-of-motion%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









    4












    $begingroup$

    I suppose the Lagrangian you found is actually
    $$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
    Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
    $$ddot{q}_2 - 2 q_1 q_2 = 0$$
    There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
    $$-ddot{q}_2 - 2 q_1 q_2 = 0$$
    This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
    $$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      I suppose the Lagrangian you found is actually
      $$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
      Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
      $$ddot{q}_2 - 2 q_1 q_2 = 0$$
      There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
      $$-ddot{q}_2 - 2 q_1 q_2 = 0$$
      This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
      $$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        I suppose the Lagrangian you found is actually
        $$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
        Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
        $$ddot{q}_2 - 2 q_1 q_2 = 0$$
        There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
        $$-ddot{q}_2 - 2 q_1 q_2 = 0$$
        This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
        $$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$






        share|cite|improve this answer









        $endgroup$



        I suppose the Lagrangian you found is actually
        $$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
        Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
        $$ddot{q}_2 - 2 q_1 q_2 = 0$$
        There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
        $$-ddot{q}_2 - 2 q_1 q_2 = 0$$
        This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
        $$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        GRBGRB

        9551722




        9551722






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Physics 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%2fphysics.stackexchange.com%2fquestions%2f463599%2flagrangian-corresponding-to-these-equations-of-motion%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

            Olav Thon

            Image
            Olav Thon Fødd 29. juni 1923 Ål Yrke forretningsdrivande Olav Thon på Commons Olav Thon (fødd 29. juni 1923) er ein norsk, stort sett sjølvlært forretningsmann, fødd i Ål i Hallingdal, som har gjort store pengar på særleg kjøp, bygging og utvikling av fast eigedom, mellom anna kjøpesentre, hotell og restaurantar, i både Noreg og somme andre land. Han vart Ridder av 1. klasse av St. Olavs Orden i 2003, og er som ivrig friluftsmann æresmedlem i Den Norske Turistforening etter å ha ytt store økonomiske bidrag til foreininga sine hytter og ruter i fjellet. Han er kjend for alltid å gå med ei karakteristisk strikkelue. Olav Thon i samtale med Trond Helleland på sentralbanksjefens årstale i februar 2016 I ung alder reiste Olav Thon til Newcastle for å lære seg engelsk, og han var seinare ei tid i Sveits, der han gjorde forretningar i import og eksport. Deretter tok han til å kjøpe faste eigedomar i Noreg med stor økonomisk suksess, verksemda voks snøgt og v
            Read more

            Waikiki

            Image
            Denne artikkelen kan ha godt av ein språkvask , som reinskar opp målføringa og/eller innfører same språkstilen overalt. Denne artikkelen kan ha godt av ei opprydding for å nå ein høgare standard og/eller for å verta i tråd med standardoppsettet. Sjå korleis du redigerer ei side og stilmanualen for hjelp. Waikiki Land   USA Stat Hawaii Stad Honolulu Wikimedia Commons: Waikiki Utsyn over Waikiki Waikīkī er ein bydel i Honolulu, sør på øya Oahu, som er ei av Hawaii-øyane. Waikīkīstranda er ein av dei meste berømte strendene i verda. Waikīkī strekkjer seg langs stillehavskysten og har ei rekkje velrenommerte hotell. Bydelen er Hawaii-turismens høyborg. På Waikīkī Beach står ei statue av den lokale sportsstjerna Duka Kahanamoku. Kjelder | Denne artikkelen bygger på «Waikiki» frå Wikipedia på bokmål, den 4. november 2018. Denne artikkelen manglar kjelder eller referansar . Hjelp Wikipedia med å finna truverdige kjelder
            Read more

            Hudsonelva

            Image
            Hudson elv Land   USA Område New York Sideelvar Mohawk Lengd 507 km Middelvassføring 606 m³/s Munning Upper New York Bay i Atlanterhavet Elvene Hudson og Mohawk med nedbørsområdet markert med gult. Wikimedia Commons: Hudson River Hudsonelva ( Hudson River ) mohikansk Muh-he-kun-ne-tuk ) er ei elv i USA. Ho renn hovudsakleg gjennom delstaten New York, men markerer også ein del av grensa mellom New York og New Jersey. Hudsonelva har namn etter Henry Hudson, ein engelskmann som utforska elva i 1609. Hudsonelva sett frå Bear Mountain Bridge. Kjelda til elva er Tear-of-the-Clouds-sjøen i Adirondackfjella. Ved Albany renn Mohawk-elva saman med henne, og deretter renn ho nesten rett sørover til ho munnar ut i Atlanterhavet mellom Manhattan i New York og New Jersey. Eriekanalen, som blei opna den 26 oktober 1825, bind saman Hudsonelva og Eriesjøen, slik at byane ved Dei store sjøane fekk vasstilgang til A
            Read more