Dear Khalil Henchi, I attach a PDF with an analysis and solution to your problem.

Given the large number of parameters involved, the solution can only be expressed subject to several complex conditions.

I hope you find it useful.

Best regards from Spain.

you may use fsolve function in MATLAB, which solves system of nonlinear equations.

Dear Khalil Henchi,

thanks to the last 2 equations, x1, y1 can be considered as sine or cosine of the same parameter, analogously for x2, y2:

x1=sin(t), y1=cos(t)

x2=sin(T), y2=cos(T).

Thus your first 3 equations will have only 2 unknowns t, T.

Without further information about a,b,... I cannot help more.

Maybe some geometric intuition can help. 2 last equations are circles in 2-dim space, or cylinders in 4-dim space. The 1st equation is a line in (x1, y1) space. But let us move algebraically now. Note that you have 5 equations for 4 variables, a bit too many to have a solution. Indeed, numerical simulations might not help.

You can solve first (1) and (4) together in (x1,y1) space analytically. There will probably be just 2 solutions.

Take the 1st solution (just a number) and substitute in eq. (2,3). With (x1,y1) being fixed, (2) and (3) are just linear equations in (x2,y2) space. Generically such a system of 2 linear equations have just one solution. And this solution is unlikely to belong to a circle given by eq. (5). So most likely your system does not have any solution, except for very special cases of parameters.

The system is decomposed into two parts. From the eqn.1 and 4 depending on C1 we can obtain two, one or no solutions for x1,y1. Then we can multiply eqn.2 by x2, eqn.3 by y2, and add them. Taking into account eqn.5 we obtain two equations of the second order for x2,y2:

2bx1x2y2 = -d+C2x2+C3y2,

x2*x2+y2*y2 = 1

( x1 we have already found). This system can be solved analitically. And of course prof. Yegorov is right, this system may have no solutions.

Dear Khalil,

In addition to the nice answers, I suggest a method to get the solutions

for the system, according to the given constants.

Or you may control them to have solutions.

The details are in the attached file.

Best regards

Its my great pleasure to contribute here, the best methods to solve this type of problems with four or more variables is the Newton, Quasi-newton's and conjugate gradient methods. These methods are simple and easy to implement always most especially conjugate gradient methods.

For more information, you can visit some of research on researchgate.

Best wishes,

Jamilu Sabi'u

@ Jamilu,

There are many elementary and advanced numerical methods to solve any linear and nonlinear systems.

Usually, the researcher starts with the available elementary methods

to find the solutions as in our case, no need to any advanced numerical method, just solving quadratic equations directly that provide exact solutions under suitable conditions over the coefficients.

Your mentioned method(s) are much complicated in this type of problems.

You can use Wolfram Mathematica for the immediate solution, which is quite easier or Matlab.

@Khalil Henchi

Use matlab or specify the constants .

So that others will help you .

B.Rath

Dear Khalil Henchi, I attach a PDF with an analysis and solution to your problem.

Given the large number of parameters involved, the solution can only be expressed subject to several complex conditions.

I hope you find it useful.

Best regards from Spain.

@ Jesus,

Well done.

Do you think your answer is different than an earlier one (answer 6), 5 days ago?

Best regards From Lebanon

A very good and detailed answer by Jesus! I have predicted similar solution structure but was too lazy to get into details. (Still it is not my scientific work).

I got a negative surprise. About half of the participants believe that going to numerical program will always give an answer. Maybe there are good programs, but I think that most of them will get stuck working on the system with no solutions. This requires analytical thinking. But I feel that only the old generation of mathematicians (who started to study math before computers became available for all) have tried to solve such systems by hand. What will happen with math in 20-30 years?

I want to thank you all, for yours answers, for the constants a, b, d, c1, c2, and c3 they are parameters that are defined as follow :

a^2+b^2+c^2=1 and c1^2+c2^2+c3^2=1. But for the moment, I am not considering the constraints on these parameters.

@ Yuri,

I agree computers will never think to start with the best choice of which equations should consider first.

In the current system the right strategy is to start with the equations ( 1 and 4) followed by ( 2 and 5 ). 4 equations in 4 unknowns, with the obvious geometrical interpretation as intersections of lines and circles.

Finally, the obtained solutions should satisfy (3) to have solutions, and this will decide if the system is consistent or not.

All these ideas explained by Peter, Viera, Victor, you and me.

All answers are very close, with a little bit different approach.

( check the previous pdf files)

Jesus arranged them in a nice format.

But we should confess that fast computations using available machines made almost everything is possible.Let me remind you about your previous answer to Is there "Applied Mathematics" without "Computer"?

You urge researchers to use computers and me too.

Best

The noteworthy particular case may be interesting

a = 0,

d = ±sqrt (b^2-C1^2), abs(b/C1)>=1),

C2 = C3 = 0,

for which due to coincidense (3) and (4) the system may have maximum allowable number of isolated roots, namely, four.

About 30 years ago I was able to program similar things in FORTRAN. But first we tried to lower the dimension; ideally complex algebraic or even transcendental equation, but in one dimension. There were several numerical methods to solve 1-dim equation for a continuous function on a finite interval (for example, divide it by 2). If the area is multidimensional and infinite - there is a problem, where to start search and how to continue (you cannot cover infinity in finite time). If the function is discontinuous - this is the 2nd problem. If the system has no solutions - this is the 3rd problem. FORTRAN normally got stuck if a programmer did not work out an escape by theoretical considerations for a particular function, translated into an algorithm.

Yes, computers became a lot smarter in 30 years. But I think that no program can cover all the tricky cases described above.

P.S. Just try to test a program: Find all solutions to the equation (1dim, continuous!)

sin (x)=1+eps- exp(-x), for a sequence of eps=0.1, 0.01, 0.001... The trick is that for eps=0 there is an infinite number of solutions, while for any small eps it is finite, but can be very large.

Sorry,

...due to coincidense (2) and (3)...

@Khalil Henchi.

Your equations make absolutely no physical sense to me! I think there must be some sign errors.

Could it perhaps be that you meant to write the equations

a*x1 + b*y1 = C1

-a*y1*y2 + b*x1*y2 + c*x2 = C2

a*x2*y1 - b*x1*x2 + c*y2 = C3

x12 + y12 = x22 + y22 = 1,

with a2 + b2 + c2 = C12 + C22 + C32 = 1?

These at least correspond to a sensible physical problem, with a reasonably nice solution. In contrast, garbage equations have garbage solutions, from which nothing can be learned.

Agree with Kåre Olaussen. Amir

@Kåre Olaussen,

Your system has a very nice geometrical illustration (the green and yellow squares are equal). But I have failed to find any physical interpretation.

Victor

The physics/geometry of my problem is to find a rotation which transform one 3-dimensional unit vector into another, by first rotating around the x-axis, and next around the z-axis. Or maybe the other way around --- I don't have the details in my head just now. That has a solution under certain conditions, but not always.

Unfortunately, my set of equations are somewhat different from those of the original question, to which sufficient mathematical analysis has already been posted. My problem is that I cannot grasp the meaning of those equations, even with the (mostly non-existing) solution. The problem in many ways SHOUTS about rotations, but this does not quite match with the signs. Which is why I wondered about misprints...

The last two equations imply that (x1,y1) and (x2,y2) are complex numbers of unit modulus: (x1,y1)=(cosθ,sinθ) and (x2,y2)=(cosφ,sinφ).

The first equation then becomes acosθ+bsinθ =C1, which can be immediately solved by setting a = rcosα and b = rsinα and writing it as

rcos(θ-α)=C1, whence one can find θ, thus (x1,y1).

From any one of the two remaining equations one can then find φ and, thus, (x2,y2)-but, since there are more equations than unknowns, there is a consistency condition that must be satisfied. So it can transpire that the system, in fact, doesn't have a solution at all.

So what? This is a high school exercise in algebra/trigonometry, hardly an obstacle for anything beyond that...

Stam ≪So what? This is a high school exercise in algebra/trigonometry, hardly an obstacle for anything beyond that...≫

If you look at the original question, and subsequent posts, it certainly does not qualify as a "high school exercise".

My whole point was to interpret the problem in a way which indeed turns the algebra into a simple exercise of trigonometry. The original problem, as formulated by the OP, is just a garbage problem which in general has no solutions, only for very restrictive conditions on the coefficients. And even when such conditions are fulfilled, the solution just looks like an ugly mess. What else can one expect? Garbage in gives garbage out. The analysis provided by Jesus Lobo, quite sensibly does not try to complete it to the bitter end; the solution just looks too ugly to be interesting or interpretable.

When you interpret the problem as rotating one unit vector into another, you still have three equations in two unknown. Which looks like imposing too many conditions. However, when you use two equations to solve for the two unknown rotation angles, you almost (modulo a sign) automatically satisfy the last equation, because rotations preserve the length of the vector. The problem does not even require trigonometry to understand, just play around with two pencils to figure out how you can rotate one into the other. Certain conditions must still be satisfied, but they are much less restrictive.

I don't understand the poster of the original problem, who writes

Khalil ≪a^2+b^2+c^2=1 and c1^2+c2^2+c3^2=1. But for the moment, I am not considering the constraints on these parameters≫

That it just ridiculous! Without such kind of restrictions, the problem is an overdetermined arbitrary mess. The same happens if you don't have all signs right in the equations. In respect to all participants of this thread, conditions like those cited, and other context of relevance, should have been spelled out in the opening post.

KO> The mathematical problem is an exercise in algebra and trigonometry-the five equations lead to three conditions for two angles, that's all.

In its present form there's just one constraint. Nothing complicated.

So the question as posed is at high school level-and so's the answer.

@ KhaliI,

I think you are an undergraduate student.

I am happy to share your questions.

I assumed RG is a platform to share knowledge for all.

All are involved, from the undergraduate students to full professors.

One can follow any question as a volunteer.No rules and constraints.

If I don't like some question, no one obliged me to share the thread.

I assumed to respect all other members regardless of their qualifications. I will try my best to give helpful answers as possible as I can.

I can't guarantee that my answer is correct or the best but sharing others may help to learn from each other.

Best regards