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I'm given the system of differential equations \begin{align*} \begin{cases} x'(t) &= y(1 + x^2 + y^2) \\ y'(t) &= x (1 + x^2 + y^2) \end{cases} \end{align*} I need to plot (with Maple) the phase portrait of this, and also the direction field. From this I need to determine the nature and stability of the critical point $(0, 0)$. Now, for this critical point we have the linearized system (which results when you set all the non-linear terms zero) \begin{align*} \begin{pmatrix} \tilde{x}'(t) \\ \tilde{y}'(t) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \tilde{x}(t) \\ \tilde{y}(t) \end{pmatrix} \end{align*} I used $\tilde{x} $ notation to make clear this is an approximation and not the same system as the original one. The eigenvalues of this matrix are $\lambda = \pm 1$, which means this critical point is a saddle point, and this is always unstable.

Now, I was wondering if there some trick to provide 'nice' initial conditions, such that the phase portrait shows clearly the solution curves? I entered this in maple:

DEplot([diff(x(t), t) = y(t)*(1+x(t)^2+y(t)^2), diff(y(t), t) = x(t)*(1+x(t)^2+y(t)^2)], 
    [x(t), y(t)], t = 0 .. 5, [[x(0) = -2, y(0) = -2], [x(0) = 3, y(0) = 0]], 
    x = -3 .. 3, y = -5 .. 5, arrows = medium, stepsize = 0.05, linecolor = blue)

The result I got was:

enter image description here

But I'm not satisfied with this, because the solution curves are not nicely shown. How does one specify the initial conditions such that this happens? Is it just by trial and error?

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Try and choose initial conditions "earlier" along the direction curves.

Eg: Something like $x(0)=-3$, $y(0)=2$ should get you a good picture of a solution curve on the left of the origin.

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Are you after some visualisation similar to the diagram below?

enter image description here

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  • $\begingroup$ yes, exactly. How can I get this in Maple? $\endgroup$ – Kamil Oct 17 '15 at 13:41
  • $\begingroup$ Dunno, I specialize in free resources... :-) $\endgroup$ – Did Oct 17 '15 at 13:43

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