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







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$endgroup$








  • 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






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oldest

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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









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      2 Answers
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      2 Answers
      2






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      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






























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