Can the phase portrait of SHO rotate counter-clockwise? or is it the case that there can be no physical motion corresponding to that?

Question

Framing the question

In the case for Simple Harmonic Oscillation, we have the equation: $$\ddot{x}+x=0 \tag{1} \label{1}$$ (say, we put all the coefficients to be 1)

Now, if we try to solve it in phase space, we can do that easily by introducing $\dot x = y$ as a dimension, and thus, we have two 1st order ODE: $$ \begin{align} \dot x &= y \tag{2} \label{2}\\ \dot y &= -x \end{align} $$

And, if we plot y vs x, we get the phase diagram :

i.e. we get a clockwise rotating vector field which is reasonable.

But, when we solve the same 2nd order differential equation ($\ddot x + x = 0$) by introducing $\dot x = -y$ then we get $$ \begin{align} \dot x &= -y \tag{3} \label{3}\\ \dot y &= x \end{align} $$

and, solving this would give me counter-clockwise rotating vector field:

The question

If we try to visualize what is happening with the pendulum, then, for the 2nd case:
Thinking about the dynamics of the pendulum based on equation \eqref{3} i.e. figure 2:
Take for example the point $x = x_{max}$, now the solution suggests that the dynamics would evolve as - in the next time step $y = \dot x > 0$, which should mean that x should increase, right? But, we can see the solution suggests the otherwise. (I can't really understand this, as we are getting that vector field solution (figure 2) based on those equations \eqref{3}, which is getting contradicted by itself)

Also, thinking about the pendulum, when it's at the furthest right point (i.e. $x_{max}$) x would tend to decrease, rather the solution/equations suggest otherwise.

So, by physically thinking about the pendulum or oscillator situation, it seems that the \eqref{3} solution is un-physical. And not only that the dynamics contradicts the equations itself (not sure where I'm going wrong - but there is no analogy/physics I used to understand that issue, rather the issue is purely math - should mean that every system described by those equations should have this issue)

So, the question is, is there any dynamics that is represented by that counter-clockwise rotating phase portrait? Or is it always the case that every physical centers(in the context of linear dynamics of 2D systems) in the phase portraits should be rotating clockwise (since for any general equations, that issue will occur in the near vicinity of the endpoints for the counter-clockwise rotating phase portrait)?

Answer 1

In the second case you are tracking negative velocity; in the first case you plot $y=v$ but in the second case you plot $y=-v$ where $v$ is the actual velocity of the pendulum. Let's call the initial conditions $(x_0,v_0)$ with $v_0>0$. In your first plot that point is located at $(x,y)=(x_0,v_0)$ but in the second plot it's located at $(x,y)=(x_0,-v_0)$. If you carefully consider this you will find that $x$ will decrease at $x=x_\text{max}$.

Answer 2

With equations (3) you're merely plotting a mirror image (reflected over the $x$ axis) of the usual, position vs. momentum, phase space.

So it's really just a change of variables, without any physical meaning or consequence.