Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By romsek

Math Help - Numerical solution of differential equations

  1. #1
    Junior Member
    Joined
    Mar 2014
    From
    uk
    Posts
    51

    Numerical solution of differential equations

    Show that the nonlinear oscillator $y" + f(y) =0$ is equivalent to the system
    $y'= -z $,
    $z'= f(y)$
    and that the solutions of the system lie on the family of curves
    $2F(y)+ z^2 = constant $
    where $F_y= f(y)$. verify that if $f(y)=y$ the curves are circle.
    =>
    nonlinear oscillator $y''+ f(y) =0$
    where
    $y'= -z $,
    $z'= f(y)$
    so that means
    $z''+z =0$

    for the solution of the system lie on the family of curves, i was thinking
    $\frac{d}{dt}[2F(y(t))+z^2(t)]= 2F \frac{dy}{dt} + 2z \frac{dz}{dt}$
    $=-2Fz +2zf(y)$
    $=-2f(y)z+2zf(y)$
    $\frac{d}{dt}[2F(y)+z^2]=0$
    $2F(y)+ z^2 = constant $

    for last part
    if $f(y)=y$ , then the differential equation is $y'' + y =0$, meaning that
    $y=A cosx +B Sinx$ and $z=-y'= - A sinx +B cosx$.
    And I also know $z'' + z =0$
    im trying to connect the above equations.
    So, I can get $cos^2x + sin^2x =1$ which show the curves on the circles.

    can someone please check my first,second and last part of answer.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2014
    From
    uk
    Posts
    51

    Re: Numerical solution of differential equations

    correction for last part:
    if $f(y)=y$ , then the differential equation is $y'' + y =0$, meaning that

    $y=A cosx +B Sinx$ and $z=-y'= - A sinx +B cosx$


    are the rotate axes.


    $pA^2+qAB+rB^2=1$


    $p,q,r$ depends on $x$


    choose $x$ such that $q=0$


    $pA^2+rB^2=1$


    what can i do after that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,237
    Thanks
    851

    Re: Numerical solution of differential equations

    Quote Originally Posted by grandy View Post
    Show that the nonlinear oscillator $y" + f(y) =0$ is equivalent to the system
    $y'= -z $,
    $z'= f(y)$
    and that the solutions of the system lie on the family of curves
    $2F(y)+ z^2 = constant $
    where $F_y= f(y)$. verify that if $f(y)=y$ the curves are circle.
    =>
    nonlinear oscillator $y''+ f(y) =0$
    where
    $y'= -z $,
    $z'= f(y)$
    so that means
    $z''+z =0$

    for the solution of the system lie on the family of curves, i was thinking
    $\frac{d}{dt}[2F(y(t))+z^2(t)]= 2F \frac{dy}{dt} + 2z \frac{dz}{dt}$
    $=-2Fz +2zf(y)$
    $=-2f(y)z+2zf(y)$
    $\frac{d}{dt}[2F(y)+z^2]=0$
    $2F(y)+ z^2 = constant $
    This all looks good.

    for last part
    if $f(y)=y$ , then the differential equation is $y'' + y =0$, meaning that
    $y=A cosx +B Sinx$ and $z=-y'= - A sinx +B cosx$.
    And I also know $z'' + z =0$
    im trying to connect the above equations.
    So, I can get $cos^2x + sin^2x =1$ which show the curves on the circles.

    can someone please check my first,second and last part of answer.
    There's an easier way.

    $f(y)=y$

    $F(y)=\frac 1 2 y^2 -\frac {d^2} 2$ here $\frac {d^2} 2$ is just an integration constant

    $2F(y)=y^2 - d^2$

    $2F(y)+z^2=y^2 - d^2 + z^2=C^2$

    $y^2 + z^2 = C^2 + d^2 = r^2$

    $y^2 + z^2 = r^2$ and this is clearly a circle in the symplectic plane
    Thanks from grandy
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2014
    From
    uk
    Posts
    51

    Re: Numerical solution of differential equations

    What will happen if $2F(y)-hf(y)y+z^2=constant$ what are these curves when $f(y)=y$?
    => Then I follow your above method:
    $2F(y)-hy^2+z^2$
    after that what I can do with $hy^2$ if I follow your method to get $y^2+z^2=r^2$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2014
    From
    uk
    Posts
    51

    Re: Numerical solution of differential equations

    There's an easier way.

    $f(y)=y$

    $F(y)=\frac 1 2 y^2 -\frac {d^2} 2$ here $\frac {d^2} 2$ is just an integration constant

    $2F(y)=y^2 - d^2$

    $2F(y)+z^2=y^2 - d^2 + z^2=C^2$

    $y^2 + z^2 = C^2 + d^2 = r^2$

    $y^2 + z^2 = r^2$ and this is clearly a circle in the symplectic plane[/QUOTE]

    What will happen if $2F(y)-hf(y)y+z^2=constant$ what are these curves when $f(y)=y$?
    => Then I follow your above method:
    $2F(y)-hy^2+z^2$
    after that what I can do with $hy^2$ if I follow your method to get $y^2+z^2=r^2$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Advance numerical solution of differential equations
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 11th 2014, 05:46 AM
  2. Advanced numerical solution of differential equation
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: April 9th 2014, 07:03 PM
  3. Advanced Numerical solution of differential equations
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: March 19th 2014, 01:06 PM
  4. Advanced Numerical solution of differential equations
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: March 15th 2014, 12:43 AM
  5. Numerical solution of differential equations (Euler method)
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: October 27th 2008, 03:12 PM

Search Tags


/mathhelpforum @mathhelpforum