Does it look like the correct 3D phase portrait?

Question

What I am talking about: https://en.m.wikipedia.org/wiki/Phase_space

You can test it by yourself in my site(I have made it on three.js),you can move it and scale

https://imaimachi.sumy.ua/levels

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I have a system of kinetic equations:

$$\dfrac{dx}{dt}= 2x+32-z $$ $$\dfrac{dy}{dt}= 5y+x+17 $$ $$\dfrac{dz}{dt}= 6x-2z+19y $$

And I had an idea to build a phase portrait of it. I made it using isoclines method. I found isocline equations for each angle from 1 to 360 degrees. I drawn it using three js in browser.

What did I get:

I have never done a 3D phase portraits and actually never worked with 3D graphics.

So, does it look like the correct phase portrait for this system?

Update. What are the steps literally:

I make a system

$$\dfrac{dz}{dx}= \dfrac{6x-2z+19y}{2x+32-z} $$ $$\dfrac{dz}{dy}= \dfrac{6x-2z+19y}{5y+x+17} $$

Equals constant for each angle:

$$\dfrac{6x-2z+19y}{2x+32-z} = tg \alpha $$ $$\dfrac{6x-2z+19y}{5y+x+17} = tg \alpha $$

For each $\alpha$ from 1 to 360 I get solution for this system in form

$$y=a(b*x+c)$$ $$z=d(e*x+h)$$

And then draw the isocline parametric equation $(x,\space a(b*x+c),\space d(e*x+h))$

Update 2. The second test

I have drawn a phase portrait for this system

$$\dfrac{dx}{dt}= 5y+3 $$ $$\dfrac{dy}{dt}= x+16y-z-9 $$ $$\dfrac{dz}{dt}= 12x-18y+z $$

Phase portrait

At this time it looks like of the type of equilibrium point

But the problem is that the system has only one equilibrium point $x=\dfrac{3}{5},\space y=-\dfrac{3}{5},\space z=-18$. In the graph I made it seems like (0,0,0)

Answer 1

A general methodology for such systems :

Write your differential system under the form:

$$V'=AV+B \tag 1 $$

for an appropriate matrix $A$ and vector $B$.

The RHS is affine. Transform it into a linear system :

$$W'=AW \ \ \text{where} \ \ W:=V-V_0$$

by substracting to (1) equation :

$$0=AV_0+B$$

[Such a constant vector $V_0$ exists due to the invertibility of $A$] .

then express $W$ in the form of the exponential of a matrix applied to a vector :

$$W=\exp(tA) W_{ini}$$

where $W_{ini}$ is obtained through initial conditions, finally giving :

$$V=\exp(tA)(V_{ini}-V_0)+V_0$$

(of course : $V_{ini}$ are initial conditions on $V$).

Numerical application :

$$A=\begin{pmatrix}2 &0 &-1\\1& 5 &0\\6&19& -2\end{pmatrix}, \ \ B=\begin{pmatrix}32\\17\\0\end{pmatrix} , \ \ V_0=\begin{pmatrix}-643/9\\98/9\\-998/9\end{pmatrix} $$

then use a Computer Algebra System for obtaining $\exp(tA)$ using eigenvalues $ 3.8920, 2.1725, -1.0644$ of $A$.

Answer 2

Yes, it is correct.

The first one is exception - it’s wrong, because I rotated each part of integral curve due to z axis, instead of x, and some logics in my code was wrong.

The second one is perfectly correct, just look:

X - RED

Y - GREEN

Z - BLUE

Look at the x equation- it depends only on y, and the dependence is linear. Now look on the “2D” case of x/y to see better:

I think it’s obvious that all is correct.

So, now I can see how the system behaves, haven’t solved it directly - this is the phase portrait.

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This is how z depends on x