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I wrote the following in Mathematica code but the result was not the expected. As you can see it seems like just half of an elipse is there, what can I do to see the complete elipse?

StreamPlot[{2 A (1 - .0001 A) - .01 A L, -.5 L + .0001 A L}, {A, 0, 
1000}, {L, 0, 1000}]

enter image description here

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  • $\begingroup$ I also tried to modify the A and L range but still, no complete elipse. $\endgroup$ – user441848 Jan 24 '18 at 4:28
  • $\begingroup$ Never use upper-case letters to start the name of a variable in Mathematica as it is likely to conflict with internal names. $\endgroup$ – David G. Stork Jan 24 '18 at 4:38
  • $\begingroup$ Why do you think there should be a full ellipse? $\endgroup$ – David G. Stork Jan 24 '18 at 4:39
  • $\begingroup$ @DavidG.Stork I used upper-case letter because that's how was written in the DE model. But for the next time i'll use lower-case letters. $\endgroup$ – user441848 Jan 24 '18 at 4:51
  • $\begingroup$ @DavidG.Stork intuition, why should not it be an ellipse or a circle? Why just half? :) $\endgroup$ – user441848 Jan 24 '18 at 4:54
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Well, as David already pointed out you haven't explained why you think it should resemble an ellipse. You can take an ellipse equation and calculate the partial derivatives and rotate them by 90 degrees:

ellipse = x^2/8 + y^2/2 - 1;
RotationMatrix[Pi/2].(D[ellipse, #] & /@ {x, y})

StreamPlot[%, {x, -10, 10}, {y, -10, 10}]

Mathematica graphics

Since your working with DE anyway, you might know that this is equivalent to taking the Curl

Curl[ellipse, {x, y}]
(* {-y, x/4} *)

Now, do the same for the generalized ellipse I gave in this answer in expr and calculate the curl. If you expand everything, you see that there is no term that contains x^2 or x*y. Your example however contains such terms. Again, why do you think this should be an ellipse?

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  • $\begingroup$ Is not an ellipse then? $\endgroup$ – user441848 Jan 24 '18 at 5:22
  • $\begingroup$ @Anneliset. Look, you present something and you claim it should give an ellipse in the vector plot. Everyone says, why you think this should be the case. I show you, how you can get the general form of such an ellipse vector field and that your example does not have the same form. Everyone really tries to help, although the tone is sometimes very direct. We cannot guess what you try to do and my answer should have shown you, how you can create the vector field of an ellipse. I assumed that since you have A^2 a AL terms in your form, you somehow only took the ellipse equation.. $\endgroup$ – halirutan Jan 24 '18 at 13:50
  • $\begingroup$ .. and tried to put it into StreamPlot, but I can only make wild guesses. You have to clear the situation. $\endgroup$ – halirutan Jan 24 '18 at 13:51
  • $\begingroup$ I see, perhaps I should not have claim that it is an ellipse, as I said before my intuition made me think about an ellipse or something circular. $\endgroup$ – user441848 Jan 24 '18 at 18:36

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