How should I show that all solutions are periodic?

[Math100]

Homework Statement

Consider the differential equation $\ddot{x}+x+\frac{3}{2}\beta x\lvert x \rvert=0, \beta\geq 0$.

a) Show that all solutions are periodic.
b) Deduce whether or not these solutions are Poincare stable.

Relevant Equations

None.

Proof: a)

Consider the differential equation $\ddot{x}+x+\frac{3}{2}\beta x\lvert x \rvert=0$
for $\beta\geq 0$.
By definition, the system of $\ddot{x}+x+\frac{3}{2}\beta x\lvert x \rvert=0$ for
$\beta\geq 0$ is conservative because $\ddot{x}=f(x)$ such that $f(x)$
is a function of $x$ only and can be integrated once to give
$\frac{1}{2}\dot{x}^2+V(x)=E$, where $E$ is a constant and $f(x)=-V'(x)$.
Then we have $\ddot{x}+x+\frac{3}{2}\beta x\lvert x \rvert=0\implies \\dot{x}=-x-\frac{3}{2}\beta x\lvert x \rvert$, so we get
$f(x)=-V'(x)=-x-\frac{3}{2}\beta x\lvert x \rvert$.
This gives $V(x)=-\int f(x)dx=\int (x+\frac{3}{2}\beta x\lvert x \rvert)dx =\frac{x^2}{2}+\frac{\beta}{2}x^3\operatorname{sgn}(x)+C$ where
$\operatorname{sgn}(x)$ is the signum function of the real variable $x$
defined as $\operatorname{sgn}(x)=1$ for $x>0$,
$\operatorname{sgn}(x)=-1$ for $x<0$.
Hence, $E=\frac{1}{2}\dot{x}^2+V(x)=\frac{\dot{x}^2+\beta x^3\operatorname{sgn}(x)+x^2}{2} =\frac{1}{2}\dot{x}^2+\frac{1}{6}(3\beta x^2\lvert x \rvert+3x^2)$.
Since $\frac{1}{2}\dot{x}^2+V(x)=E\implies \frac{1}{2}\dot{x}^2+V(x)=C$
where $\dot{x}=y, \dot{y}=f(x)$, it follows that $\frac{1}{2}\dot{x}^2=E-V(x)\implies \dot{x}^2=2(E-V(x))\implies y=\pm\sqrt{2(E-V(x))}$.
Thus, $y=\pm\sqrt{2(E-V(x))}=\pm\sqrt{2(\frac{1}{2}\dot{x}^2+\frac{1}{6}(3\beta x^2\lvert x \rvert+3x^2))} =\pm\sqrt{\dot{x}^2+\beta x^2(\lvert x \rvert-x\operatorname{sgn}))}$.

From the work/proof shown above, how should I show that all solutions are periodic? Also for part b), how should I determine whether or not these solutions are Poincare stable?

 

[Orodruin]

What are the classical turning points (ie, points where $\dot x = 0$)? What can you say about the time required to move between those points?

 

[Math100]

I've got $\dot{x}=y=0$. But what does this mean?

 

[Orodruin]

What are those points in $x$. Are they finite?

What happens at those points? Does the solution just pause and continue through or does it turn around?

 

[Math100]

From the system of $\dot{x}=y, \dot{y}=-x-\frac{3}{2}\beta x\lvert x \rvert$, after setting $\dot{x}=y=0$ and $\dot{y}=-x-\frac{3}{2}\beta x\lvert x \rvert=0$, we get that the fixed point occurs at $(x, y)=(0, 0)$ and is stable centre.

 

[Orodruin]

I was talking about the turning points, not the fixed point.

 

[Math100]

How should I find the turning points then?

 

[Orodruin]

Solve this:

I've got $\dot{x}=y=0$.

 

[renormalize]

where $\dot{x}=y,$ ...
Thus, $y=$ ... $\pm\sqrt{\dot{x}^2+\beta x^2(\lvert x \rvert-x\operatorname{sgn}))}$.

Are you sure? If you replace $y$ in the second equation using the first equation and square both sides, $\dot{x}^2$ cancels and you're left with $0=\beta x^2(\lvert x \rvert-x\operatorname{sgn}))$. Do you think that's true?

 

[Math100]

Are you sure? If you replace $y$ in the second equation using the first equation and square both sides, $\dot{x}^2$ cancels and you're left with $0=\beta x^2(\lvert x \rvert-x\operatorname{sgn}))$. Do you think that's true?

Sorry, I think I've made some huge mistakes above from my initial/first post. The equation of the phase paths should be $y=\pm\sqrt{2(C-V(x))}$ where $\dot{x}=y, \dot{y}=f(x)$. So I have $y=\pm\sqrt{2(C-(\frac{x^2}{2}+\frac{\beta}{2}x^3\operatorname{sgn}(x)))}$ and $y=0$ for $C=\frac{x^2}{2}+\frac{\beta}{2}x^3\operatorname{sgn}(x)$. But how does this shows that all solutions are periodic?

 

[Orodruin]

What are those points in $x$. Are they finite?

What happens at those points? Does the solution just pause and continue through or does it turn around?