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Suppose we have a two-dimensional system of ODEs, $$ \begin{array}{ccc} \dot{x} & = & f(x,y)\\ \dot{y} & = & g(x,y) \end{array} $$ What can one say about the solutions of this system, knowing the solution of the system, where on replaces either $x$ or $y$ above with a $0$, i.e. $$ \begin{array}{ccc} \dot{x} & = & f(0,y)\\ \dot{y} & = & g(0,y). \end{array} $$Does this mean something like "cutting out" a portion of the phase plane of the original system and analysing that ?

What properties of solutions from this last system are preserved in the ''full'' system above ? What can one infer about the full system from this one,

I realize that this is an open question. I'm not after a complete list of properties, just some basic answers that show what is retained. But please don't use any arguments that depend explicitly on the dimension of the phase plane, like Poincare-Bendixson, as I referred to two-dimensional systems only for brevity.

(A pathological example that shows what's not retained is $$ \begin{array}{ccc} \dot{x} & = & xy\\ \dot{y} & = & x^{2}, \end{array} $$ where for $x=0$ the only solution is the zero solution and obviously the full system has more rich behavior I believe -- but maybe this is just a pathological case, that should be excluded, I don't know.)

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  • $\begingroup$ IMHO I wouldn't expect anything very interesting to be inferred from such "restriction" of original system. Geometrically your construction means that you take sample of vector field at line $\lbrace x = 0 \rbrace$ and just "copy" it to all other lines $\lbrace x = {\rm const} \rbrace$. By the way such "restriction" never has isolated equilibriums, only full horizontal lines of equilibrium states. All dynamics lives in horizontal stripes between invariant lines that could be found from solving $g(0, y^\ast) = 0$. I agree with @RobertIsrael that only too simple facts could be deduced from here. $\endgroup$
    – Evgeny
    May 13, 2015 at 6:39

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The only connection I can see between the two systems is if $f(0,y) = 0$, in which case the solutions of your second system with $x=0$ are also solutions of the first system.

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  • $\begingroup$ ok, having a (negative) answer from authoritative user is itself a valuable information. thank you. $\endgroup$
    – temo
    May 13, 2015 at 9:57

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