There is also an old example. This system reflects a real physical process. As far as I remember, it is connected with electricity.
It would be nice to show the complete solution, or an approach to obtaining the complete solution.

1 - 2cos(x1) + 2cos(x2) - 2cos(x3) - x4 = 0;
1- 2cos(5x1) + 2cos(5x2) - 2cos(5x3) = 0;
1 - 2cos(7x1) + 2cos(7x2) - 2cos(7x3) = 0;

Yes, for a specific task x4 can be considered as something like a parameter. But the beauty of the system is precisely in this form. In other words, the real system became the basis for this one.
Also, please excuse me, I do not own or use SMath Studio. I think I can perceive the algorithm (idea of the solution) in words. If it's not difficult for you, of course. And, if you want, I will share my approach to solving it.

Hello Alvaro.
I think that in this case it is not very convenient to immediately apply Draghilev’s method. I tried two ways to get the solution. One way is to work directly with the system itself, that is, add some equation to the system to get the starting point, for example x1+x2+x3+x4=0. After this, we calculate the line in R^4. This line will be one of the subsets of the system solution. The coordinates of points x1, x2, x3 with a period of 2Pi will define an infinite set of subsets with coordinate x4. By changing the additional equation, we can get an idea of the set of such subsets. They are all solutions of the system.
Since the solution is numerical, it is probably impossible to say that we have found a complete solution. But practically it is possible.
As for visualization, we can display on a graph a combination of 3 coordinates out of 4: x1,x2,x3; x1,x2,x4: x1,x3,x4; x2,x3,x4. (This produces an infinite set of closed curves.)

The second way is to reduce the original system to a system of polynomial equations by means of the following replacement

cos(x1) = x1, cos(x2) = x2, cos(x3) = x3.

1-2*x1+2*x2-2*x3-x4 = 0
-32*x1^5+32*x2^5-32*x3^5+40*x1^3-40*x2^3+40*x3^3-10*x1+10*x2-10*x3+1=0
-128*x1^7+128*x2^7-128*x3^7+224*x1^5-224*x2^5+224*x3^5-112*x1^3+112*x2^3-112*x3^3+14*x1-14*x2+14*x3+1=0

In this case, we get more opportunities to identify all the original subsets of solutions, because for any one auxiliary equation, according to the theory, all solutions are found (Gröbner bases). Then we make a reverse replacement, take into account the period, look at the graphs and, on a practical basis, assert that we have obtained a solution to the original system.
It's definitely a lot easier for me because Maple does all the work for me. Both Gröbner bases and powerful procedures for solving ODEs are implemented there. It takes literally seconds to obtain a solution in the form of a set of curves.



Of course, because SMath does not have the corresponding functions.

Alvaro, if you're interested, I can share Maple's texts with you. The texts, of course, are not at a programmer level, but for the Method and the specific task they are quite understandable. For example, this can be done through my VK profile. Or suggest your own way.

Hi Alexey. I only use Maple v 11, which actually runs on Windows 98 and with classic worksheets. This is because it is very easy to install and to explain the software in class without losing almost any of the computing power. I think my email is public in this forum: it is razonar at outlook dot com. The method I use for Draghilev's method is based on the procedure that Viacheslav posted somewhere here on the forum. The main modification was to optimize it for two or three variables, and subsequently change the way of entering the time span and the initial conditions.

Best regards.
Alvaro.

Do I need also make replacement for left part of differential equation?
Something like that
cos(x1) = y1
dy1/dt=dcos(x1)/dt
dy1/dt=-sin(x1)*dx1/dt
dy1/dt=-sqrt(1-y^2)*dx1/dt
dx1/dt=-1/sqrt(1-y^2)*dy1/dt

Меня смущает, что в новой системе нет никаких ограничений на значение косинусов. Ведь если их модуль больше единицы, мы получим комплексные числа. Решение исходной системы находится в вещественной области.

[RUS]В новой системе, где произведена замена cos(xi) на xi, часть решений совпадает с исходной системой на некотором участке при cos меньших единицы. При значениях больше единицы получаем отклонения от исходной системы и комплексные числа, которые тоже являются решениями в исходной системе. Кружочком обозначил точку, где начинается отклонение.[/RUS]
[ENG]In the new system, where cos(xi) has been replaced by xi, part of the solutions coincide with the original system at some part at cos less than unity. For values greater than one, we obtain deviations from the original system and complex numbers, which are also solutions in the original system. The circle indicates the point where the deviation begins.[/ENG]

Some solutions to check out:
0 + 0.405718i 0 + 0.392443i 0.261484 -0.943054
0 + 0.405880i 0 + 0.392625i 0.261456 -0.943057
0 + 0.406042i 0 + 0.392806i 0.261427 -0.943061
0 + 0.406205i 0 + 0.392988i 0.261399 -0.943065
0 + 0.406367i 0 + 0.393169i 0.261370 -0.943069
0 + 0.406529i 0 + 0.393351i 0.261342 -0.943072
0 + 0.406691i 0 + 0.393532i 0.261313 -0.943076
0 + 0.406853i 0 + 0.393713i 0.261285 -0.943080
0 + 0.407015i 0 + 0.393895i 0.261256 -0.943083
0 + 0.407177i 0 + 0.394076i 0.261228 -0.943087
0 + 0.407339i 0 + 0.394257i 0.261200 -0.943091
0 + 0.407501i 0 + 0.394438i 0.261171 -0.943095
0 + 0.407662i 0 + 0.394618i 0.261143 -0.943098
0 + 0.407824i 0 + 0.394799i 0.261115 -0.943102
0 + 0.407986i 0 + 0.394980i 0.261087 -0.943106
0 + 0.408147i 0 + 0.395160i 0.261059 -0.943110
0 + 0.408309i 0 + 0.395340i 0.261030 -0.943113
0 + 0.408470i 0 + 0.395521i 0.261002 -0.943117
0 + 0.408631i 0 + 0.395701i 0.260974 -0.943121
0 + 0.408793i 0 + 0.395881i 0.260946 -0.943124
0 + 0.408954i 0 + 0.396061i 0.260918 -0.943128

Edited by user 09 March 2024 01:53:31(UTC)  | Reason: Not specified

In fact, in this case, this approach is the most reliable for identifying patterns in the location of solutions in space. I took some time and tried several types of auxiliary equations. I think, given symmetry, three main types of closed polynomial curves emerged. And therefore closed trigonometric curves with a period of 2Pi will represent the entire solution after the transition to the original variables.

An example of visualization of solving a system of 3 equations with 3 variables using the Draghilev method. The system has 6 solutions. All solutions are found for one initial point.

4(x1^2 + x2 - 11)*x1 + 2x1 + 2x2^2 + 2cos(x1) - 14 = 0;
2x1^2 + 2x2 + 4(x2^2 + x1 - 7)*x2 - sin(x2) - 22 = 0;
x1^2 + x3^2 - 1 = 0;

Edited by user 17 March 2024 08:52:04(UTC)  | Reason: Not specified