how to draw graphs of ODE's In order to solve this question How to calculate $\omega$-limits I'm trying to learn how to draw graphs of ODE's. For example, let $p\in \mathbb R^2$ in the case of the field $Y=(Y_1, Y_2)$, given by:
$Y_1=-y_2+y_1(y_1^2+y_2^2)\sin\left(\dfrac{\pi}{\sqrt{y_1^2+y_2^2}}\right)$
$Y_2=y_1+y_2(y_1^2+y_2^2)\sin\left(\dfrac{\pi}{\sqrt{y_1^2+y_2^2}}\right)$
I need help.
Thanks so much
 A: This is a very strange system indeed, note I am assuming$\left(Y_1 = \dfrac{dy_1}{dt},~ Y_2 = \dfrac{dy_2}{dt}\right)$.
If we look at a phase portrait, lets see if it sheds any light on matters.

Well, it looks like there are some closed orbits near the origin. Lets peer in a little closer to the origin.

You can use a CAS like Maple, Mathematica, SAGE, Maxima or Autograph or others to do more analysis .
A: I did the plot of direction field of your system of ODEs by using Maple. I feel, doing this by hand is a bit hard at least for this system. I hope the codes help you to find the points you are looking for in a convenient way. 
  [> with(DEtools):
  [> dfieldplot([diff(x(t), t) = -y(t)+x(t)*(x(t)^2+y(t)^2)*sin(Pi/sqrt(x(t)^2+y(t)^2)), diff(y(t), t) = x(t)+y(t)*(x(t)^2+y(t)^2)*sin(Pi/sqrt(x(t)^2+y(t)^2))], [x(t), y(t)], t = -20 .. 20, x = -1 .. 1, y = -1 .. 1, color = blue, dirfield = [20, 20]);


A: This is a nice example of what a nonlinear term can do to a stable, but not asymptotically stable, equilibrium. It helps to introduce the polar radius $\rho=\sqrt{y_1^2+y_2^2}$, because this function satisfies the ODE 
$$\frac{d\rho }{dt} = \frac{y_1}{\rho}\frac{dy_1}{dt}+\frac{y_2}{\rho}\frac{dy_2}{dt} = \rho^3\sin \frac{\pi}{\rho} \tag1$$
The analysis of equilibria of this ODE tells you  about the orbits of the original system: 


*

*There are stable closed orbits of radius $\rho=\dfrac{1}{2k }$, $k=1,2,\dots$ 

*There are unstable closed orbits of radius $\rho=\dfrac{1}{2k-1 }$, $k=1,2,\dots$ 

*In between, the orbits are spirals converging to the nearest stable closed orbit. 

*Outside, in the region $\rho>1$, the orbits go off into infinity in a hurry (in such a hurry that they get there in finite time).


The plot given by Amzoti  illustrates all of the above points.
A: ( Explanation of yesterday's post implication reg constant of integration)
These are Limit Cycles in Control theory. There are three limiting (asymptotic) loci as sketched by Amzoti. However, caution is needed in choosing boundary conditions if the integrate is desired to be in a neat, closed or elegant form involving circular trig functions. Else, the graphs would look displaced from what might be considered their "natural" position. I mention this as OP asked how to draw the integrands or phase portraits.  
(I am using $ y_1 = x$, and $y_2 = y$).
if x(0) is given, then better not to choose y(0) arbitrarily because: 
$(y'-y) = (x'-x) $ and $(x'^2 + y'^2) = (x^2 + y^2) (1+R^2)$,
where $R = (x^2 + y^2) \sin (\pi/ \sqrt{x^2 + y^2})$
together define functional dependance between x and y. ( The f(x,y)= 0 result can be posted here later on by the OP to gain further clarity). Constant of integration is to be chosen for the most "natural"  or elegant form. In general of course, the constants can be chosen quite arbitrarily.
I give a simpler example of what I mean.
Take $y' = \sin (x)/m$ and $x'= \sin (y)$. Integrating, $\cos (x) = m \cos (y) +$ arbitrary constant,  which should be set to zero if the phase portraits do not get an asymmetrical look displaced away from argument values $0, \pi/2, \pi, 2 \pi$ etc. periods. This may be "desirable" but not essential for generality. Here $\cos (x)/\cos (y)= m$ looks more elegant. The advantage becomes apparent further on when trying to form higher order decoupled equations  $F( x, x'' ) = 0$,  $G(y, y'' ) = 0$, towards a fuller or more comprehensive situation.
