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I have found all equilibria, studied their nature, and have been able to make a parametric plot of the following non-linear system along a time axis:

$$r'(t)=i-l.r(t)-\text{ux}. r(t). x(t)-\text{uy}. r(t). y(t) \\ x'(t)=\text{ex}. \text{ux}. r(t). x(t)-\text{mx}. x(t)\\y'(t)=\text{ey}.\text{uy}. r(t). y(t)-\text{my}. y(t)$$

$i,l,ux,uy,mx,my,ex,ey$ are parameters.I am currently reading about dynamical systems and the authour mention uncoupling the systems, for example this one:

$$\dot x =x,\;\; \dot y=y,\;\;\dot z=-z$$

I do not understand what this uncoupling is. Is it about setting one of the variable as constant ?

Is it possible to uncouple all three dimensional autonomous systems (mine for example) ?

How to draw the xy phase plane portrait, from this xyz system, for example?

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    $\begingroup$ Notice for the uncoupled system, you can solve for $x'$ by itself with no dependence on $y'$ or $z'$. You can also solve for $y'$ or $z'$ by themselves. For the second system, the solutions are each some exp-$ke^t$. A 3D phase portrait is done w/Mathematica (NDSolve/ParametricPlot3D or users.dimi.uniud.it/~gianluca.gorni/Mma/Mma.html). In Maple, see mapleprimes.com/questions/35774-Phase-Portrait-In-3D. You can also plot the solutions for $x, y, z$ individually to understand the behavior of the system. I am not sure about your equation. Are those constants $e, u, m$? Regards $\endgroup$
    – Amzoti
    Aug 13, 2013 at 15:11
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    $\begingroup$ I edited. So uncoupling is possible only when there is no dependence between the equations ? $\endgroup$ Aug 13, 2013 at 15:23
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    $\begingroup$ Uncoupling is possible when you can get one equation to not have dependence on the others. So, if could solve for $x'$ by itself, and the other equations depended on it, you can substitute $x$ into the other equations and make them easier too. So, even if you can uncouple just one of the equations, it can be very helpful. Regards $\endgroup$
    – Amzoti
    Aug 13, 2013 at 15:32
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    $\begingroup$ Also, we can simply state this as: the terminology uncoupled means that each differential equation in the system depends on exactly one variable (like your second system). In a coupled system, one of the equations must involve two or more variables. $\endgroup$
    – Amzoti
    Aug 13, 2013 at 15:37

2 Answers 2

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Here is an example of a 3D Phase Portrait.

enter image description here

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  • $\begingroup$ Fascinating: 3D! +1 $\endgroup$
    – amWhy
    Aug 14, 2013 at 1:52
  • $\begingroup$ It's awesome! :D $\endgroup$
    – amWhy
    Aug 14, 2013 at 2:23
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If your system is written as $\dot{x}=Ax$ (i.e. it is a linear system), $x\in\mathbb R^n$, then you can under some conditions transform your system to $\dot{y}=By$, where $B$ is diagonal matrix, hence uncoupling the system. Here $B=U^{-1}AU$, $U$ is the eigenvector matrix of A, and $x=Uy$.

In case you have a nonlinear system $\dot{x}=f(x)$, you can do a similar transformation by replacing A by Jacobian of $f$, i.e. $A=Df$. Once you carry out the transformation as above, the linear part of the system will be diagonalized, but ofcourse you would have coupling in the higher order terms.

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  • $\begingroup$ Ok. Let's suppose I do linearize the first system I wrote, then how do I draw the xy phase plane portrait for example ? Do I just set r to be at the equilibrium value and then do the phase portrait as if in a 2D system? $\endgroup$ Aug 13, 2013 at 18:11
  • $\begingroup$ Ok I am going to learn to do projections then, sorry I am confused about it, thanks ! $\endgroup$ Aug 13, 2013 at 18:16

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