Nonlinear oscillator and simple harmonic motion

[ra_forever8]

The nonlinear oscillator $y'' + f(y)=0$ is equivalent to the
Simple harmonic motion:
$y'= -z $,
$z'= f(y)$
the modified Symplectic Euler equation are
$$y'=-z+\frac {1}{2} hf(y)$$
$$y'=f(y)+\frac {1}{2} hf_y z$$
and deduce that the coresponding approximate solution lie on the family of curves
$$2F(y)-hf(y)y+z^2=constant$$
where $F_y= f(y)$.

ans =>
for the solution of the system lie on the family of curves, i was thinking

$$\frac{d}{dt}[2F(y)-hf(y)y+z^2]= y \frac{dy}{dt} + z \frac{dz}{dt}$$
$=y(-z+\frac{1}{2} hf(y)) +z(f(y)- \frac{1}{2} h f_y z)$
but I can not do anything after that to get my answer constant.


can any genius people please help me

 

[jvicens]

to get my answer constant.

To get the answer constant, we can use the fact that the system is equivalent to simple harmonic motion. This means that the energy of the system is conserved, so the total energy at any time $t$ is equal to the total energy at the initial time $t_0$. We can express the total energy as
$$E = \frac{1}{2}z^2 + F(y)$$
Since the energy is conserved, we can set $E = E_0$ where $E_0$ is the initial energy. This gives us
$$\frac{1}{2}z^2 + F(y) = E_0$$
We can rearrange this equation to get
$$2F(y) - hf(y)y + z^2 = 2E_0$$
So the solution of the system lies on the family of curves described by $2F(y) - hf(y)y + z^2 = 2E_0$. This is the same as the expression given in the forum post, with the constant $2E_0$ instead of just $constant$.