I would like to ask if I understand correctly the process of linearization for analyzing critical points.

I was given differential equation:

$\dot x = xy+1$

$\dot y = x+xy$

And my task was to linearize the system around stationary points.

What I've done is: calculated critical point (I think the result is point $T(1,-1)$) and then I calculated Jacobian matrix and it's eigenvalues and corresponding eigenvectors and after that I've got that the point $T$ is saddle and then I drew the picture.

I am not quite sure where can I check if steps that I've done are correct. Is there some program available for drawing phase portraits or online tool. Or if someone could solve the task and tell whether are my steps correct and if I'm missing something.

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    $\begingroup$ Yes, looks correct, the critical point is $(x, y) = (1, -1)$ and it is a saddle point. Try: bluffton.edu/homepages/facstaff/nesterd/java/slopefields.html for the phase portrait (click on system tab first). $\endgroup$ – Moo Jun 12 '18 at 22:24
  • $\begingroup$ If you know the nature of the fixed points - you can always try to plot the phase plane diagram for the ode. $\endgroup$ – Chinny84 Jun 12 '18 at 22:24

OK, here's a quick rundown on how I would do this; hopefully our OP Mirjan Pecenko can check his own work against what I do here.


$\dot x = xy + 1, \tag 1$

$\dot y = x + xy, \tag 2$

we have equilibria where

$\dot x = xy + 1 = \dot y = xy + x = 0; \tag 3$


$xy = -x \Longrightarrow -x + 1 = 0 \Longrightarrow x = 1, \tag 4$

and then

$xy + 1 = 0 \Longrightarrow xy = -1 \Longrightarrow y = (1)y = -1; \tag 4$

so it appears the only critical point is at $(1, -1)$. The Jacobean of the vector field

$\vec V(x, y) = \begin{pmatrix} xy + 1 \\ x + xy \end{pmatrix} \tag 5$

at $(1, -1)$ is

$J_V(1, -1) = \begin{bmatrix} (\partial(xy + 1)/\partial x & (\partial(xy + 1)/\partial y \\ (\partial(xy + x)/\partial x & (\partial(xy + x)/\partial y \end{bmatrix}_{(1, -1)}$ $= \begin{bmatrix} y & x \\ y + 1 & x \end{bmatrix}_{(1, -1)} = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix}; \tag 6$

it is now obvious that the eigenvalues of $J_V(1, -1)$ are $\pm 1$; therefore this point is a saddle; the eigenvectors at $(1, -1)$ are $(1, 0)^T$ for $-1$ and $(1/2, 1)^T$ for $1$; it is now easy to sketch a phase portrait for this system, a task I leave to my readers.

It should be noted that when sketching a phase portrait, it is often helpful to find those curves in $\Bbb R^2$ where $\dot x = 0$ and/or $\dot y = 0$. These curves are useful guides to finding the geometry of the solutions, since they show us where the tangent lines to the integral curves or vertical or horizontal, respectively. When combined with the shapes of the solutions near the critical point provided by the above analysis, we can get a pretty good idea of how the flow will appear. As with any hand-done graphical analysis, care must be taken to ensure that we draw accurately enough to capture only correct features of the trajectories.

Note Added in Edit, Thursday 14 June 2018 12:35 PM PST: It appears that the notion of isoclines, which proves to be most convenient as a guide to sketching phase portraits, may be generalized in a way which allows the extraction of more information about the integral curves and/or overall shape of the flow of a given vector field. In this problem, isoclines are only exploited in a rather peripheral way, since they are merely mentioned as a sort of after-thought in the comments. Nevertheless, they may be used much more extensively. Indeed, rather than restricting the use of the isocline method to determining the curves on which $\dot x = 0$ and/or $\dot y = 0$, we may if we so choose deploy it in an attempt to find just where $\dot x, \dot y$ take on any of their possible values. One technique which can help effect this aim is to use the gradient of the component functions, in this case $xy + 1$ and $xy + x$, to guide us towards regions of greater or lesser component magnitude; for example, since

$\dot x = xy + 1, \tag 7$

we have

$\nabla (\dot x) = (y, x)^T, \tag 8$

which indicates the direction in which $\dot x$ increases, so that we may, for example, find the directions in which points on the $\dot x = 0$ isocline must be moved to make $\dot x$ larger. By systematic application of such methodology, quite detailed phase portraits may be had. Unfortunately, at present I lack both the graphics tools and the time to illustrate what I am saying via pictorial means. End of Note.

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    $\begingroup$ Thanks for help! $\endgroup$ – Mirjan Pecenko Jun 12 '18 at 22:56
  • 1
    $\begingroup$ You're welcome; I hope it did help! $\endgroup$ – Robert Lewis Jun 12 '18 at 22:57
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    $\begingroup$ The only thing I don't quite understand. We always draw where the tangent lines are vertical/horizontal (what if x'=0 what is the value of y' or y'=0 what is value for x'). Is this step really necessary or it only provides more exact picture of solutions? $\endgroup$ – Mirjan Pecenko Jun 12 '18 at 22:59
  • $\begingroup$ @MirjanPecenko: I'll add a few words to my answer. Cheers! $\endgroup$ – Robert Lewis Jun 12 '18 at 23:19
  • $\begingroup$ @MirjanPecenko: There, does that help? $\endgroup$ – Robert Lewis Jun 12 '18 at 23:27

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