# Questions about uncoupling dynamical systems and phase plane portraits of the uncoupled systems.

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?

• 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 Aug 13, 2013 at 15:11
• I edited. So uncoupling is possible only when there is no dependence between the equations ? Aug 13, 2013 at 15:23
• 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 Aug 13, 2013 at 15:32
• 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. Aug 13, 2013 at 15:37

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.