Showing that $\ddot{x}=x-x^3$ is a conservative system

Question

I am aware that there are multiple ways of showing that the system $\,\ddot{x}=x-x^3$ is Conservative. One of which is applying Newton's law from physics $$\frac{dE(t)}{dt}=\frac{d}{dt}\left[\frac12m\dot{x}^2+V(x)\right]=0$$ where, in this case, $$-\frac{dV(x)}{dx}=x-x^3\quad\text{and}\quad m=1.$$

However, I was wondering why the following method would not work. For a consertive system the curl is $0$ (assuming the domain of definition is simply connected), in other words, $\nabla\times\boldsymbol{F}=\boldsymbol{0}.$

The given differential equation can be written as $$\begin{cases} \dot{x(t)}=y(t) \\ \dot{y(t)}=x(t)-x^3(t)\end{cases}$$ where $\boldsymbol{F}=y\hat{i}+(x-x^3)\hat{j}.$

But $\nabla\times\boldsymbol{F}=-3x^2\hat{k}\neq\boldsymbol{0}.$

What is wrong with this reasoning?

Answer 1

You sort of confused the notion of "conservative force field in 3D" and the notion of "conservative force field in 1D". The theorem

a smooth force field $F$ on $\Bbb R^3$ is conservative if and only if $\nabla\times F=\vec 0$

is only valid in $\Bbb R^3$.

But your problem is in 1D. You should not really embed your problem into 2D or 3D (well, you can, but with care).

If you intend on embedding the problem into 3D, I first point out your error. When you want to apply the above theorem, $y$ should mean the $y$-coordinate of a position. Your quantity $y(t)$ is not the $y$-coordinate of a position, but instead is the $x$-component of velocity. The correct way to embed the problem into 3D is to define two extra equations

$$\begin{cases} \ddot{x}(t)=x(t)-(x(t))^3 \\ \ddot{y}(t)=0 \\ \ddot{z}(t)=0 \end{cases}$$

So the force field is $F(x,y,z)=(x-x^3,0,0)$, which you can calculate that $\nabla\times F=\vec 0$.

As a note, there is a generalisation of the above theorem in quote to other dimensions, including for 1D, 2D, and any dimensions higher than 3D. It goes like this:

A smooth vector field $F:\Bbb R^n\to\Bbb R^n$ is given. Write $F(x_1,x_2,\dots,x_n)=(F_1(x_1,x_2,\dots,x_n),F_2(x_1,x_2,\dots,x_n),\dots,F_n(x_1,x_2,\dots,x_n))$. We calculate the partial derivatives $$\frac{\partial F_i}{\partial x_j}-\frac{\partial F_j}{\partial x_i}$$ at each point, for all $i,j=1,2,\dots,n$. Then $F$ is a conservative force field if and only if all the numbers $\frac{\partial F_i}{\partial x_j}-\frac{\partial F_j}{\partial x_i}$ are zero everywhere.

It should be an easy verification that when $n=3$, this theorem is the same as the theorem stating a force field $F$ is conservative if and only if curl of $F$ is zero.

Also, pay special attention that the domain of $F$ has to be the whole $\Bbb R^n$ but not anything else. If the domain is something else, the theorem usually does not work.

So if you ever want to do something like calculating curl to see if a force field is conservative, you should check if the dimension of your problem is 3. If it is not, use this generalised theorem instead.

(If you are ever interested in this generalised theorem, the keyword is Poincaré lemma)

Answer 2

If $\mathbf F$ means the vector of force, then you can't compute its $\mathrm{curl}$ because it is 1-dimensional. If it's the flow direction in the phase space, then $\nabla \times\mathbf F=0$ does not hold. Just compute this for a harmonic oscillator.

Your given differential equations are 2 first-order equations, but force involves second-order differentials.

Answer 3

Another way of looking into the same problem:

In a conservative system with a single force, the work of force $F$ has to be a function of starting and ending positions only (exact trajectory and motion durations are unimportant). In this particular case, the work of force $F$ is simply an integral of the form:

$$A=\int_{x_0}^{x_1}F(x)dx$$

...and the end result will be a function of ${x_0,x_1}$ only. A one-dimensional system is a conservative one for all forces of the form $F=F(x)$.

Answer 4

Multiplying by $\dot x$ both sides we get at

$$ \frac 12\frac{d}{dt}(\dot x)^2 = \frac{d}{dt}\left(\frac 12 x^2-\frac 14 x^4\right) $$

or

$$ \dot x^2 +\frac 12x^4-x^2 = C_0 $$

that with $C_0 > 0$ is a beautiful oscillator. Concluding, the movement is conservative.

Attached a phase plot for $C_0 = 1$