How Is a Nonlinear Oscillator Equivalent to a System of First-Order Equations?

[wel]

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 $$z'' + z =0$$
I am 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.

 

[HallsofIvy]

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
$$2F(y)+ z^2 = \textrm{constant}$$
where $F_y= f(y)$. verify that if $f(y)=y$ the curves are circle.

=>
nonlinear oscillator $y" + f(y) =0$

where
$$y'= -z,$$
$$z'= f(y)$$

This shows that the single equation $y''+ f(y)= 0$ can be written as the two equations $y'= -z$ and $z'= f(y)$. To show they are equivalent you must also show the other way: that if $y'= -z$ and $z'= f(y)$ then $y''+ f(y)= 0$.

$$y'' + f(y) =0$$
$$-y+f(y) =0$$ since ($y''=-y$)

where did you get that $y''= -y$? That would be true ony iin the very simple case that $f(y)= -y$. If this is a "nonlinear oscillator then $f$ must be nonlinear.

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

$$\frac{d}{dt}[F(y)^2+z^2]= y \frac{dy}{dt} + z \frac{dz}{dt}$$

This isn't true. You are missing a factor of $2$. Further you have "lost" F on the right. $\frac{dF}{dy}$ is not necessarily equal to $y$.

$2F(y)+z^2 = -yz+ zy$

Where did the $-yz+ zy$ come from? What you had before was a single expression that was not equal to anything. Why do

$$2F(y)+ z^2 = \textrm{constant}$$

if $f(y)=y$, then the differential equation is $y'' + y =0$, meaning that $y=A\cos x +B\sin x$ and $z=-y'= - A \sin x +B \cos x$

No, if $y= A \cos(x)+ B \sin(x)$ then $z= -y'= A sin(x)- B cos(x)$.

can someone please check my first,second and last part of answer.

 

[micromass]

Dear wel,

I have corrected your LaTeX code. The delimiter $ is not valid here. Thus things like $F$ will not yield correct LaTeX code. Please read our FAQ: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[wel]

i have edited my answers for the first and second part. Would you someone don't mind checking it please.

 

[wel]

correction for last part of answer
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?