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.

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?

Re: Numerical solution of differential equations

Quote:

Originally Posted by

**grandy** 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.

Quote:

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

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

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