calculusdynamical systemsordinary differential equationsreal-analysis
Consider the system: $$\ddot x = x^3 -x$$
What is the method to follow to find a conserved quantity for this system?
So far what I have is:
$\dot x = y$ and $\dot y = x^3 – x$ and I can find the Jacobian of the system and find and classify fixed points.
How would I find a conserved quantity?
This equation is equivalent to a $2\times 2$ system $$ x_1'=x_2, \\ x_2'=-x_1-ax_1^2. $$ Critical point, is where the flux vanishes, i.e., $$ (x_1,x_2)=(0,0) \quad\text{and}\quad (x_1,x_2)=(0,-1/a). $$
Follow up answer to @lhf's post
The equation
$$(y-y^3) \, \partial_x H - (x+y^2) \, \partial_y H = 0, $$ as proposed by @lhf, is a linear first-order PDE for $H$. The method of characteristics then states that
$$ \frac{\mathrm{d}x}{y-y^3} = \frac{\mathrm{d}y}{-x-y^2} = \frac{\mathrm{d}H}{0}, $$
where the last fraction indicates that $H = c_1$ is a constant along the characteristic curve defined by the first equals sign:
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{x+y^2}{y(1-y^2)}$$
which defines $y$ as a function of $x$. The solution to this (homogeneous?) equation is given by Mathematica as:
$$ \small{ \left(\sqrt{3}+3\right) \log \left[\left(\sqrt{3}-1\right) y^2-2 x-\sqrt{3}-1\right] -\left(\sqrt{3}-3\right) \log \left[\left(\sqrt{3}+1\right) y^2+2 x-\sqrt{3}+1\right]=c_2 } $$ where I have absorbed a factor into the constant of integration. Put now $c_1$ as a function of $c_2$ to have $H = f(c_2)$, where $ f $ is an arbitrary function of its argument. Take for instance $f(\square) = \square$ as the identity function and you will thus find:
$$ \small{ H = \left(\sqrt{3}+3\right) \log \left[\left(\sqrt{3}-1\right) y^2-2 x-\sqrt{3}-1\right] -\left(\sqrt{3}-3\right) \log \left[\left(\sqrt{3}+1\right) y^2+2 x-\sqrt{3}+1\right] }$$
which, if I didn't make any mistake on the transcription, satisfies the original PDE.
Hope you find this useful!
Check with Mathematica:
Best Answer
For a second order ODE the following should do the trick: $$ \ddot x = x^3 - x \\ \ddot x\dot x = (x^3 - x)\dot x = \frac d{dt} \left( \frac{x^4}4 - \frac{x^2}2 \right) \\ \frac{\dot x ^2}2 - \frac{x^4}4 + \frac{x^2}2= const. $$