How should I show that all solutions of this equation are bounded?

[Math100]

Homework Statement

Consider the equation $\ddot{x}+x+\epsilon(\alpha x^2\operatorname{sgn}(x)+\beta x^{3})=0, 0<\epsilon<<1, \alpha>0, \beta>0$, where $\alpha$ and $\beta$ are constants, and the signum function, $\operatorname{sgn}(x)$, is defined by $\operatorname{sgn}(x)=1$ for $x>0$, $\operatorname{sgn}(x)=0$ for $x=0$ and $\operatorname{sgn}(x)=-1$ for $x<0$.

a) Determine the potential energy $V(x)$ of the system, and show that all solutions of the equation are bounded and periodic, and that all the phase curves are invariant under reflections in both the x-axis and the $\dot{x}$-axis of the phase diagram. (You don't need to provide the phase diagram.)

b) Hence, with initial condition $x(0)=0$, show that solutions with angular frequency $\omega$ can be expressed as a Fourier series of the form $x(t)=\sum_{k=1}^{\infty}B_{2k-1}sin((2k-1)\omega t)=B_{1}sin\omega t+B_{3}sin 3\omega t+...$.

c) Assuming that for the Fourier series in part (b), $\omega$ and $B_{n}, n=1, 3, 5, ...$, can be expanded as a power series in $\epsilon$, show that as $\epsilon\to 0, \omega=1+O(\epsilon), B_{1}=O(1)$ and $B_{n}=O(\epsilon)$ for $n=3, 5, 7, ...$.

Relevant Equations

The signum function is defined by $\operatorname{sgn}(x)=1$ for $x>0$, $\operatorname{sgn}(x)=0$ for $x=0$ and $\operatorname{sgn}(x)=-1$ for $x<0$.

a) Proof:

By definition, the potential energy $V(x)$ is given by $F(x)=-\frac{dV}{dx}$.
Note that $\ddot{x}=-\frac{dV}{dx}$ where $\ddot{x}=-x-\epsilon(\alpha x^2\operatorname{sgn}(x)+\beta x^{3})$.
This gives $\frac{dV}{dx}=x+\epsilon(\alpha x^2\operatorname{sgn}(x)+\beta x^{3})\implies V(x)=\int (x+\epsilon(\alpha x^{2}\operatorname{sgn}(x)+\beta x^{3}))dx$, so we have $V(x)=\frac{x^2}{2}+\frac{\epsilon\alpha}{3}x^{3}\operatorname{sgn}(x)+\frac{\beta x^{4}}{4}+C$.
Hence, the potential energy $V(x)$ of the system is
$V(x)=\frac{x^2}{2}+\frac{\epsilon\alpha}{3}x^{3}\operatorname{sgn}(x)+\frac{\beta x^4}{4}$.
Now we will find all solutions of the equation $\ddot{x}+x+\epsilon(\alpha x^{2}\operatorname{sgn}(x)+\beta x^{3})=0, 0<\epsilon<<1, \alpha>0, \beta>0$ where $\alpha$ and $\beta$ are constants and the signum function $\operatorname{sgn}(x)$ is defined by $\operatorname{sgn}(x)=1$ for $x>0$, $\operatorname{sgn}(x)=0$ for $x=0$ and $\operatorname{sgn}(x)=-1$ for $x<0$.
Let $\epsilon=0$.
Then the unperturbed equation is $\ddot{x}+x=0$ and this has root $x=x_{0}$ such that the general solution is
$x_{0}(t)=A\cos(t)+B\sin(t)$, where $A$ and $B$ are constants.
Applying the perturbation method produces $x(t)=x_{0}(t)+\epsilon x_{1}(t)+\epsilon^{2} x_{2}(t)+...$ for $0<\epsilon<<1$.

From here, what should I do in order to find all solutions of the given equation and show/prove that all solutions of this equation are bounded and periodic and that all the phase curves are invariant under reflections in both the x-axis and the $\dot{x}$-axis of the phase diagram? Also, is the work shown above correct up to here?

 

[pasmith]

It is rare for one to be able to solve a non-linear ODE explicitly in closed form for arbitrary initial conditions, such that one can show directly that all solutions take a particular form. Intead you must adopt a qualitative approach. The entire purpose of a course on nonlinear ODEs is to equip you with techniques to get information about the behaviour of solutions without needing to actually solve them.

Think about your system. It is of the form $\ddot x + f(x) = 0$. This has the immediate conserved quantity $\frac12 \dot x^2 + \int f(x)\,dx$, and in fact the system is Hamiltonian.

Where are its fixed points? Are there any other than the origin?

What sorts of trajectory are possible in a 2D Hamiltonian system?

Both $\dot x^2$ and $V(x)$ are non-negative. Is that consistent with an unbounded trajectory?

Once you've concluded that only periodic orbits enclosing the origin are possible, you know that $x(t)$ is periodic, so can be written as a fourier series. Do the symmetries of the trajectories and the initial condition $x(0) = 0$ require any terms of the fourier series to vanish?

 

[anuttarasammyak]

The former two terms of ODE are of harmonic oscillator. The third term is suppressing |x|. I do not find factors to enhance |x|.

 

[Math100]

It is rare for one to be able to solve a non-linear ODE explicitly in closed form for arbitrary initial conditions, such that one can show directly that all solutions take a particular form. Intead you must adopt a qualitative approach. The entire purpose of a course on nonlinear ODEs is to equip you with techniques to get information about the behaviour of solutions without needing to actually solve them.

Think about your system. It is of the form $\ddot x + f(x) = 0$. This has the immediate conserved quantity $\frac12 \dot x^2 + \int f(x)\,dx$, and in fact the system is Hamiltonian.

Where are its fixed points? Are there any other than the origin?

What sorts of trajectory are possible in a 2D Hamiltonian system?

Both $\dot x^2$ and $V(x)$ are non-negative. Is that consistent with an unbounded trajectory?

Once you've concluded that only periodic orbits enclosing the origin are possible, you know that $x(t)$ is periodic, so can be written as a fourier series. Do the symmetries of the trajectories and the initial condition $x(0) = 0$ require any terms of the fourier series to vanish?

So $f(x)=\epsilon(\alpha x^{2}\operatorname{sgn}(x)+\beta x^{3})+x$ where the fixed points occur when both $\ddot{x}=0$ and $\dot{x}=0$. This implies that $f(x)=0$ and $x=0$ forces that the fixed point is $(0, 0)$. Since both $\dot{x}^{2}$ and the potential energy $V(x)$ are non-negative, it follows that all trajectories are bounded. I think that the only fixed point is $(0, 0)$ because there doesn't exist any nonzero values of $x$ that satisfy $f(x)=0$ for $0<\epsilon<<1$. Only the bounded trajectories are consistent with periodic orbits around the origin. Thus, the solution $x(t)$ is periodic and can be written as a Fourier series. Are these correct though?

 

[anuttarasammyak]

I observe not periodic but dissipative.