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I'm going to plot the phase portrait of this system:

$\dfrac{dx}{dt}=-x^2 + 4 y^2$

$\dfrac{dy}{dt}=-8 - 4 y + 2 x y$

The singular point $(x,y)$ can be found from the system:

$-x^2 + 4 y^2=0$

$-8 - 4 y + 2 x y=0$

Seems this system doesn't have roots wolframalfa. What does it mean? I have integral curves(phase trajectories) that do not intersect? If yes, hence, obviously I can't determine type of the singular point ?

By the way, I heard from my lector, that it can be drawn in Wolframalfa. Can you provide an example respectively to my system?

Stdugnd4ikbd
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1 Answers1

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The system that you have given has indeed two singular points: $$x_1 = -2,\qquad y_1 = -1,$$ and $$X_2 = 4, \qquad y_2 = 2;$$ You can verify this by checking that $dx/dt$ and $dy/dt$ are both zero.

To check what type of singular points these are, we can study the system linearized around a stable point: $$ \dot x \approx A x $$ where $A = \nabla f (\bar x)$. In this case, we get that $$ A(x,f) = \nabla f(x,y) = \begin{bmatrix} -2x & 8y \\ 2y & -4+2x \end{bmatrix}.$$

This matrix has two eigenvalues, $$ \lambda_{1,2}(x,y) = -2 \pm \sqrt{4-8x + 4x^2 + 16x^2},$$ of which, in the case of the two given stable points, one is positive: $$ \lambda_1(x_1,y_1) \approx 5.21$$ $$ \lambda_1(x_2,y_2) = 8 $$ Because the linearized system has positive eigenvalues, the singular points are unstable.

To draw the phase portrait you can use a vector field plot.

Riccardo Sven Risuleo
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  • Thank You very much. I understood stuff around linear systems, but didn't understand how to get lambda from nonlinear. In addition I want to ask You about two more things. At first, I found two equations for lambda, it is $\lambda=\dfrac{a+d}{2}+-\sqrt{\dfrac{(a+d)^2-4(ad-bc)}{4}}$, and the second was the same but with $4detA$ under the root. Which is true? – Stdugnd4ikbd Feb 13 '19 at 17:19
  • And the second one is how to linearize the system to get a,b,c,d? It was the step, where I stuck. – Stdugnd4ikbd Feb 13 '19 at 17:24
  • You linearize the nonlinear system by taking the first approximation in the taylor expansion. The entry $i,j$ of the matrix $A$ is the partial derivative of the $i$th equation with respect to the $j$th variable. As for the eigenvalues, the formulas you show are the same, as $detA = ad-bc$ :) – Riccardo Sven Risuleo Feb 14 '19 at 08:31
  • Understood yet, thank You – Stdugnd4ikbd Feb 15 '19 at 11:36