Symmetry non integrable Hamiltonian system We are in in open subset of $\mathbb{R}^6$ where the local coordinates
are $(q,p)$ where $q=(x,\theta_1,\theta_2)$ is the state and the costate is 
$p=(p_1,p_2,p_3)$.
We consider an Hamiltonian system and I compute 
numerically $2\pi$-periodic in time trajectories with respect to $(\theta_1,\theta_2)$, the others components are not necessarily periodic.
Projections on the plane $(\theta_1,\theta_2)$ are represented in figure below.
I would like to prove the symmetry on the $(\theta_1,\theta_2)$ plane.
 I should prove that (cf the manuscript figure) :
$A(t)=(\theta_1(t),\theta_2(t)),\; t\in[0,2\pi]$ 
is related with $C(t)=(2(x(t)-\theta_1(t))+\theta_1(t),2(\pi-x(t) -\theta_2(t))+\theta_2(t))$ by
$$
A(2\pi-t) = C(t)
$$
But problem : the Hamitnoian system is not integrable and I can't have 
explicit solutions...
Is there exists tehcniques to prove this (like Noether theorem but in the projective case)


 A: The projected orbits indeed seem to be reflexive symmetric in the line $\theta_2 = \pi - \theta_1$. The reflection in the line $\theta_2 = \pi-\theta_1$ is given by
\begin{equation}
 R: (\theta_1,\theta_2) \mapsto (\pi-\theta_1,\pi-\theta_2).
\end{equation}
However, if the orbit is symmetric, it doesn't mean that the solution is symmetric. In particular (assuming your system is nondegenerate), the direction of the flow reverses when you apply the reflection. So, a good way to test the symmetry is to see if the Hamiltonian system is invariant under simultaneous application of $R$ and time reversal. In other words, applying $R$ should have the same effect as reversing time, at least on the $\theta$-variables.
To be more concrete, let's look at the Hamiltonian system. We have
\begin{align}
 \frac{\text{d} \theta_{1,2}}{\text{d} t} &= \frac{\partial H}{\partial p_{2,3}},\tag{1}\\
 \frac{\text{d} p_{2,3}}{\text{d} t} &= -\frac{\partial H}{\partial \theta_{1,2}}.\tag{2}
\end{align}
Applying $R$ and time reversal to $\frac{\text{d} \theta_{1,2}}{\text{d} t}$ (in whatever order) leaves the left hand side of $(1)$ invariant. So, we obtain the requirement that $ \frac{\partial H}{\partial p_{2,3}}$ must be invariant with respect to those operations as well. Assuming your Hamiltonian $H$ is time-independent, this means that
\begin{equation}
  \frac{\partial H}{\partial p_{2,3}}(\theta_1,\theta_2) = \frac{\partial H}{\partial p_{2,3}}(\pi-\theta_1,\pi-\theta_2).
\end{equation}
Applying $t \mapsto -t$ to the left hand side of $(2)$ yields $-\frac{\text{d} p_{2,3}}{\text{d} t}$; therefore, we obtain a second condition
\begin{equation}
\frac{\partial H}{\partial \theta_{1,2}}(\theta_1,\theta_2) = -\frac{\partial H}{\partial \theta_{1,2}}(\pi-\theta_1,\pi-\theta_2).
\end{equation}
An example of an Hamiltonian obeying these conditions is
\begin{equation}
H = \frac{1}{2}(p_2^2+p_3^2) + \cos(\theta_1) + \cos(\theta_2).
\end{equation}
Note that I didn't do anything with $x$ or $p_1$. Keep in mind that the reflection $R$ might not leave $x$-values invariant; however, since you're projecting onto the $(\theta_1,\theta_2)$-plane anyway, this isn't relevant (but important to realise!).
