How can I solve this nonlinear differential equation.












0












$begingroup$


$$m cdot y''+k cdot y + u cdot sign(y') = 0 $$



Does this have an analytic solution? If not, how do you know?



I've tried to use the Laplace transform, but I can't figure out how to deal with the sign function of the derivative.



Also, since it's nonlinear, I can't use any of those techniques.






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What is $u$ ? ...
    $endgroup$
    – JJacquelin
    Dec 19 '18 at 7:42










  • $begingroup$
    k, u, and m are constants
    $endgroup$
    – Ion Sme
    Dec 19 '18 at 17:47
















0












$begingroup$


$$m cdot y''+k cdot y + u cdot sign(y') = 0 $$



Does this have an analytic solution? If not, how do you know?



I've tried to use the Laplace transform, but I can't figure out how to deal with the sign function of the derivative.



Also, since it's nonlinear, I can't use any of those techniques.






ordinary-differential-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What is $u$ ? ...
    $endgroup$
    – JJacquelin
    Dec 19 '18 at 7:42










  • $begingroup$
    k, u, and m are constants
    $endgroup$
    – Ion Sme
    Dec 19 '18 at 17:47














0












0








0


1



$begingroup$


$$m cdot y''+k cdot y + u cdot sign(y') = 0 $$



Does this have an analytic solution? If not, how do you know?



I've tried to use the Laplace transform, but I can't figure out how to deal with the sign function of the derivative.



Also, since it's nonlinear, I can't use any of those techniques.






ordinary-differential-equations







share|cite|improve this question









$endgroup$




$$m cdot y''+k cdot y + u cdot sign(y') = 0 $$



Does this have an analytic solution? If not, how do you know?



I've tried to use the Laplace transform, but I can't figure out how to deal with the sign function of the derivative.



Also, since it's nonlinear, I can't use any of those techniques.






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 6:08









Ion SmeIon Sme

11510




11510








  • 1




    $begingroup$
    What is $u$ ? ...
    $endgroup$
    – JJacquelin
    Dec 19 '18 at 7:42










  • $begingroup$
    k, u, and m are constants
    $endgroup$
    – Ion Sme
    Dec 19 '18 at 17:47














  • 1




    $begingroup$
    What is $u$ ? ...
    $endgroup$
    – JJacquelin
    Dec 19 '18 at 7:42










  • $begingroup$
    k, u, and m are constants
    $endgroup$
    – Ion Sme
    Dec 19 '18 at 17:47








1




1




$begingroup$
What is $u$ ? ...
$endgroup$
– JJacquelin
Dec 19 '18 at 7:42




$begingroup$
What is $u$ ? ...
$endgroup$
– JJacquelin
Dec 19 '18 at 7:42












$begingroup$
k, u, and m are constants
$endgroup$
– Ion Sme
Dec 19 '18 at 17:47




$begingroup$
k, u, and m are constants
$endgroup$
– Ion Sme
Dec 19 '18 at 17:47










2 Answers
2






active

oldest

votes


















0












$begingroup$

Let $z_1 = y$ and $z_2 = y'$, so $z_1' = z_2$ and $z_2' = -kz_1-utext{sign}(z_2)$. Then the equation can be written as $mathbf z' = F(mathbf z)$. This is a first order nonlinear equation. Since $F$ is continuously differentiable for $mathbf z$ in the upper or lower half planes, the usual existence and uniqueness theorems guarantee that initial conditions in either of those spaces will have unique solutions for some region surrounding it, not passing the $z_1$-axis, which will coincide with the solutions for the linear equations with $text{sign}(z_2) = pm z_2$ respectively. Orbits in the upper half plane will be upper half ellipses with center $(-frac{u}{k},0)$, and orbits in the lower half plane will be lower half ellipses with center $(frac{u}{k},0)$. These orbits are analytic in their respective half planes, but the connections on the $z_1$-axis have discontinuous second derivatives.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $$m cdot y''+k cdot y + u cdot text{sign}(y') = 0 $$



    In the domain where $y'>0qquad begin{cases}
    m cdot y''+k cdot y + u = 0quadtext{which general solution is :}\
    y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k}\
    y'(t)=-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t)
    end{cases}$



    In the domain where $y'<0qquad begin{cases}
    m cdot y''+k cdot y - u = 0quadtext{which general solution is :}\
    y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)+frac{u}{k}\
    y'(t)= -c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) osed form
    end{cases}$



    The general solution can be written on close form :
    $$y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k} text{sign}left(-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) right)$$



    The constants $c_1$ and $c_2$ have to be determined according to some boundary conditions. Since no boundary condition is specified in the wording of the question, no further calculus is possible.






    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: "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%2f3046057%2fhow-can-i-solve-this-nonlinear-differential-equation%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Let $z_1 = y$ and $z_2 = y'$, so $z_1' = z_2$ and $z_2' = -kz_1-utext{sign}(z_2)$. Then the equation can be written as $mathbf z' = F(mathbf z)$. This is a first order nonlinear equation. Since $F$ is continuously differentiable for $mathbf z$ in the upper or lower half planes, the usual existence and uniqueness theorems guarantee that initial conditions in either of those spaces will have unique solutions for some region surrounding it, not passing the $z_1$-axis, which will coincide with the solutions for the linear equations with $text{sign}(z_2) = pm z_2$ respectively. Orbits in the upper half plane will be upper half ellipses with center $(-frac{u}{k},0)$, and orbits in the lower half plane will be lower half ellipses with center $(frac{u}{k},0)$. These orbits are analytic in their respective half planes, but the connections on the $z_1$-axis have discontinuous second derivatives.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Let $z_1 = y$ and $z_2 = y'$, so $z_1' = z_2$ and $z_2' = -kz_1-utext{sign}(z_2)$. Then the equation can be written as $mathbf z' = F(mathbf z)$. This is a first order nonlinear equation. Since $F$ is continuously differentiable for $mathbf z$ in the upper or lower half planes, the usual existence and uniqueness theorems guarantee that initial conditions in either of those spaces will have unique solutions for some region surrounding it, not passing the $z_1$-axis, which will coincide with the solutions for the linear equations with $text{sign}(z_2) = pm z_2$ respectively. Orbits in the upper half plane will be upper half ellipses with center $(-frac{u}{k},0)$, and orbits in the lower half plane will be lower half ellipses with center $(frac{u}{k},0)$. These orbits are analytic in their respective half planes, but the connections on the $z_1$-axis have discontinuous second derivatives.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Let $z_1 = y$ and $z_2 = y'$, so $z_1' = z_2$ and $z_2' = -kz_1-utext{sign}(z_2)$. Then the equation can be written as $mathbf z' = F(mathbf z)$. This is a first order nonlinear equation. Since $F$ is continuously differentiable for $mathbf z$ in the upper or lower half planes, the usual existence and uniqueness theorems guarantee that initial conditions in either of those spaces will have unique solutions for some region surrounding it, not passing the $z_1$-axis, which will coincide with the solutions for the linear equations with $text{sign}(z_2) = pm z_2$ respectively. Orbits in the upper half plane will be upper half ellipses with center $(-frac{u}{k},0)$, and orbits in the lower half plane will be lower half ellipses with center $(frac{u}{k},0)$. These orbits are analytic in their respective half planes, but the connections on the $z_1$-axis have discontinuous second derivatives.






          share|cite|improve this answer









          $endgroup$



          Let $z_1 = y$ and $z_2 = y'$, so $z_1' = z_2$ and $z_2' = -kz_1-utext{sign}(z_2)$. Then the equation can be written as $mathbf z' = F(mathbf z)$. This is a first order nonlinear equation. Since $F$ is continuously differentiable for $mathbf z$ in the upper or lower half planes, the usual existence and uniqueness theorems guarantee that initial conditions in either of those spaces will have unique solutions for some region surrounding it, not passing the $z_1$-axis, which will coincide with the solutions for the linear equations with $text{sign}(z_2) = pm z_2$ respectively. Orbits in the upper half plane will be upper half ellipses with center $(-frac{u}{k},0)$, and orbits in the lower half plane will be lower half ellipses with center $(frac{u}{k},0)$. These orbits are analytic in their respective half planes, but the connections on the $z_1$-axis have discontinuous second derivatives.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 7:06









          AlexanderJ93AlexanderJ93

          6,193823




          6,193823























              0












              $begingroup$

              $$m cdot y''+k cdot y + u cdot text{sign}(y') = 0 $$



              In the domain where $y'>0qquad begin{cases}
              m cdot y''+k cdot y + u = 0quadtext{which general solution is :}\
              y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k}\
              y'(t)=-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t)
              end{cases}$



              In the domain where $y'<0qquad begin{cases}
              m cdot y''+k cdot y - u = 0quadtext{which general solution is :}\
              y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)+frac{u}{k}\
              y'(t)= -c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) osed form
              end{cases}$



              The general solution can be written on close form :
              $$y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k} text{sign}left(-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) right)$$



              The constants $c_1$ and $c_2$ have to be determined according to some boundary conditions. Since no boundary condition is specified in the wording of the question, no further calculus is possible.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$m cdot y''+k cdot y + u cdot text{sign}(y') = 0 $$



                In the domain where $y'>0qquad begin{cases}
                m cdot y''+k cdot y + u = 0quadtext{which general solution is :}\
                y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k}\
                y'(t)=-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t)
                end{cases}$



                In the domain where $y'<0qquad begin{cases}
                m cdot y''+k cdot y - u = 0quadtext{which general solution is :}\
                y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)+frac{u}{k}\
                y'(t)= -c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) osed form
                end{cases}$



                The general solution can be written on close form :
                $$y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k} text{sign}left(-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) right)$$



                The constants $c_1$ and $c_2$ have to be determined according to some boundary conditions. Since no boundary condition is specified in the wording of the question, no further calculus is possible.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$m cdot y''+k cdot y + u cdot text{sign}(y') = 0 $$



                  In the domain where $y'>0qquad begin{cases}
                  m cdot y''+k cdot y + u = 0quadtext{which general solution is :}\
                  y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k}\
                  y'(t)=-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t)
                  end{cases}$



                  In the domain where $y'<0qquad begin{cases}
                  m cdot y''+k cdot y - u = 0quadtext{which general solution is :}\
                  y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)+frac{u}{k}\
                  y'(t)= -c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) osed form
                  end{cases}$



                  The general solution can be written on close form :
                  $$y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k} text{sign}left(-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) right)$$



                  The constants $c_1$ and $c_2$ have to be determined according to some boundary conditions. Since no boundary condition is specified in the wording of the question, no further calculus is possible.






                  share|cite|improve this answer









                  $endgroup$



                  $$m cdot y''+k cdot y + u cdot text{sign}(y') = 0 $$



                  In the domain where $y'>0qquad begin{cases}
                  m cdot y''+k cdot y + u = 0quadtext{which general solution is :}\
                  y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k}\
                  y'(t)=-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t)
                  end{cases}$



                  In the domain where $y'<0qquad begin{cases}
                  m cdot y''+k cdot y - u = 0quadtext{which general solution is :}\
                  y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)+frac{u}{k}\
                  y'(t)= -c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) osed form
                  end{cases}$



                  The general solution can be written on close form :
                  $$y(t)=c_1cos(sqrt{frac{k}{m}}:t)+c_2sin(sqrt{frac{k}{m}}:t)-frac{u}{k} text{sign}left(-c_1sqrt{frac{k}{m}}sin(sqrt{frac{k}{m}}:t)+c_2sqrt{frac{k}{m}}cos(sqrt{frac{k}{m}}:t) right)$$



                  The constants $c_1$ and $c_2$ have to be determined according to some boundary conditions. Since no boundary condition is specified in the wording of the question, no further calculus is possible.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 19:03









                  JJacquelinJJacquelin

                  45.2k21856




                  45.2k21856






























                      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%2f3046057%2fhow-can-i-solve-this-nonlinear-differential-equation%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

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                      Read more

                      In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
                      Read more

                      How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

                      Image
                      1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
                      Read more