Can MATLAB solve systems of nonlinear equations?

During investigation of one problem, I find out a system of seven nonlinear algebraic equations, which analytic solution is need. Can MATLAB give me the analytic solution of that system?

Answer 1

[Walter Roberson]

If MATLAB can do it at all, it would have to be by way of the Symbolic Toolbox.

A lot of the time, simultaneous nonlinear equations have no known analytic solution method.

If you could give us examples, we might be able to estimate the possibility of success.

[Walter Roberson]

It looks to me like something the Symbolic Toolbox might reasonably be able to solve. My experience suggests, though, that with that many simultaneous equations, you will need a fair bit of memory and probably a number of hours of computing to solve it.

Are there any restrictions on the xi values? Do you only want the real-valued solutions? Only the non-negative solutions?

[Asatur Khurshudyan]

Unfortunately, I haven`t a computer of fair bit of memory, that`s why I couldn`t solve it yet.

No there are not any restrictions, but unknowns must be real-valued and they necessarily must be unique.

Answer 2

[Walter Roberson]

If you subs() in the definitions for A, B, C, D_ in to the other 7 equations, and simplify() that, and solve, you end up with 18 solutions of various complexities, with up to 4 free variables. The least complex single solution is:

[x1 = 0, x2 = x2, x3 = 0, x4 = x4, x5 = 0, x6 = (1-x2*c2-x4*c4)/c6, x7 = x7]

where x2 = x2 means that x2 is allowed to assume any value.

Here are the 18 solutions, in Maple notation. Be sure to read the comments following them:

[x1 = RootOf(_Z^2+x3^2+x5^2), x2 = (-RootOf(_Z^2+x3^2+x5^2)*c1-x3*c3-x4*c4-x5*c5-x6*c6+1)/c2, x3 = x3, x4 = x4, x5 = x5, x6 = x6, x7 = x7]

[x1 = RootOf(_Z^2+x3^2+x5^2), x2 = x2, x3 = x3, x4 = (-x5*(x5*c5+x2*c2+x3*c3-1)*RootOf(_Z^2+x3^2+x5^2)+(x3^2+x5^2)*(x5*c1-c6*x2))/RootOf(_Z^2+x3^2+x5
^2)/(c4*x5-x3*c6), x5 = x5, x6 = (x3*(x5*c5+x2*c2+x3*c3-1)*RootOf(_Z^2+x3^2+x5^2)-(x3^2+x5^2)*(x3*c1-c4*x2))/RootOf(_Z^2+x3^2+x5^2)/(c4*x5-x3*c6), x7
= x7]

[x1 = RootOf(_Z^2+c6^2+c4^2)*x3/c4, x2 = ((-x3*c6*c5+c4-x3*c4*c3)*RootOf(_Z^2+c6^2+c4^2)+c1*x3*(c6^2+c4^2))/c4/(RootOf(_Z^2+c6^2+c4^2)*c2+c4^2+c6^2),
x3 = x3, x4 = ((-c4^2*c1*x3-c1*x3*c6^2-x6*c6*c4*c2)*RootOf(_Z^2+c6^2+c4^2)-((x3*c3-1+x6*c6)*c4+x3*c6*c5)*(c6^2+c4^2))/c4^2/(RootOf(_Z^2+c6^2+c4^2)*c2
+c4^2+c6^2), x5 = x3*c6/c4, x6 = x6, x7 = x7]

[x1 = x1, x2 = (-x5*(x5*c5+x1*c1+x4*c4-1)*RootOf(_Z^2+x1^2+x5^2)+(x1^2+x5^2)*(c3*x5-c6*x4))/RootOf(_Z^2+x1^2+x5^2)/(-x1*c6+c2*x5), x3 = RootOf(_Z^2+
x1^2+x5^2), x4 = x4, x5 = x5, x6 = (x1*(x5*c5+x1*c1+x4*c4-1)*RootOf(_Z^2+x1^2+x5^2)+(x1^2+x5^2)*(-c3*x1+x4*c2))/RootOf(_Z^2+x1^2+x5^2)/(-x1*c6+c2*x5)
, x7 = x7]

[x1 = x1, x2 = ((-c2^2*c3*x1-x6*c6*c4*c2-c3*x1*c6^2)*RootOf(_Z^2+c6^2+c2^2)-(c6^2+c2^2)*((x1*c1-1+x6*c6)*c2+x1*c6*c5))/c2^2/(c6^2+RootOf(_Z^2+c6^2+c2
^2)*c4+c2^2), x3 = RootOf(_Z^2+c6^2+c2^2)*x1/c2, x4 = ((-x1*c6*c5+c2-c2*x1*c1)*RootOf(_Z^2+c6^2+c2^2)+c3*x1*(c6^2+c2^2))/c2/(c6^2+RootOf(_Z^2+c6^2+c2
^2)*c4+c2^2), x5 = x1*c6/c2, x6 = x6, x7 = x7]

[x1 = x1, x2 = (-x3*(x3*c3+x1*c1+x6*c6-1)*RootOf(_Z^2+x1^2+x3^2)-(x1^2+x3^2)*(c4*x6-x3*c5))/RootOf(_Z^2+x1^2+x3^2)/(-c4*x1+c2*x3), x3 = x3, x4 = (x1*
(x3*c3+x1*c1+x6*c6-1)*RootOf(_Z^2+x1^2+x3^2)+(x1^2+x3^2)*(-c5*x1+x6*c2))/RootOf(_Z^2+x1^2+x3^2)/(-c4*x1+c2*x3), x5 = RootOf(_Z^2+x1^2+x3^2), x6 = x6,
x7 = x7]

[x1 = x1, x2 = ((-c2^2*c5*x1-c5*x1*c4^2-c2*x4*c4*c6)*RootOf(_Z^2+c4^2+c2^2)-(c4^2+c2^2)*((x1*c1+x4*c4-1)*c2+x1*c3*c4))/c2^2/(c2^2+RootOf(_Z^2+c4^2+c2
^2)*c6+c4^2), x3 = c4*x1/c2, x4 = x4, x5 = RootOf(_Z^2+c4^2+c2^2)*x1/c2, x6 = ((-x1*c3*c4+c2-c2*x1*c1)*RootOf(_Z^2+c4^2+c2^2)+c5*x1*(c4^2+c2^2))/c2/(
c2^2+RootOf(_Z^2+c4^2+c2^2)*c6+c4^2), x7 = x7]

[x1 = x1, x2 = (-x5*(x5*c5+x1*c1+x4*c4-1)*RootOf(_Z^2+x1^2+x5^2)+(x1^2+x5^2)*(c3*x5-c6*x4))/RootOf(_Z^2+x1^2+x5^2)/(-x1*c6+c2*x5), x3 = RootOf(_Z^2+
x1^2+x5^2), x4 = x4, x5 = x5, x6 = (x1*(x5*c5+x1*c1+x4*c4-1)*RootOf(_Z^2+x1^2+x5^2)+(x1^2+x5^2)*(-c3*x1+x4*c2))/RootOf(_Z^2+x1^2+x5^2)/(-x1*c6+c2*x5)
, x7 = x7]

[x1 = 0, x2 = x2, x3 = 0, x4 = x4, x5 = 0, x6 = (1-x2*c2-x4*c4)/c6, x7 = x7]
[x1 = c2*x5/c6, x2 = x2, x3 = RootOf(_Z^2+c6^2+c2^2)*x5/c6, x4 = (-RootOf(_Z^2+c6^2+c2^2)*x5*c3-x5*c1*c2-x5*c5*c6+c6)/c6/(c4-RootOf(_Z^2+c6^2+c2^2)),
x5 = x5, x6 = (((x5*c5-1+x2*c2)*c6+x5*c1*c2)*RootOf(_Z^2+c6^2+c2^2)-c6*x2*c2*c4-c3*x5*c6^2-x5*c3*c2^2)/c6^2/(c4-RootOf(_Z^2+c6^2+c2^2)), x7 = x7]

[x1 = x1, x2 = (-x3*(x3*c3+x1*c1+x6*c6-1)*RootOf(_Z^2+x1^2+x3^2)-(x1^2+x3^2)*(c4*x6-x3*c5))/RootOf(_Z^2+x1^2+x3^2)/(-c4*x1+c2*x3), x3 = x3, x4 = (x1*
(x3*c3+x1*c1+x6*c6-1)*RootOf(_Z^2+x1^2+x3^2)+(x1^2+x3^2)*(-c5*x1+x6*c2))/RootOf(_Z^2+x1^2+x3^2)/(-c4*x1+c2*x3), x5 = RootOf(_Z^2+x1^2+x3^2), x6 = x6,
x7 = x7]

[x1 = c2*x3/c4, x2 = x2, x3 = x3, x4 = (((x2*c2+x3*c3-1)*c4+x3*c1*c2)*RootOf(_Z^2+c4^2+c2^2)-x3*c5*c4^2-c6*x2*c2*c4-c2^2*x3*c5)/c4^2/(-RootOf(_Z^2+c4
^2+c2^2)+c6), x5 = RootOf(_Z^2+c4^2+c2^2)*x3/c4, x6 = (-c5*RootOf(_Z^2+c4^2+c2^2)*x3-x3*c1*c2-x3*c4*c3+c4)/c4/(-RootOf(_Z^2+c4^2+c2^2)+c6), x7 = x7]

[x1 = x1, x2 = (-x3*(x3*c3+x1*c1+x6*c6-1)*RootOf(_Z^2+x1^2+x3^2)-(x1^2+x3^2)*(c4*x6-x3*c5))/RootOf(_Z^2+x1^2+x3^2)/(-c4*x1+c2*x3), x3 = x3, x4 = (x1*
(x3*c3+x1*c1+x6*c6-1)*RootOf(_Z^2+x1^2+x3^2)+(x1^2+x3^2)*(-c5*x1+x6*c2))/RootOf(_Z^2+x1^2+x3^2)/(-c4*x1+c2*x3), x5 = RootOf(_Z^2+x1^2+x3^2), x6 = x6,
x7 = x7]

[x1 = x1, x2 = (((x1*c1+x4*c4-1)*c2+x1*c3*c4)*RootOf(_Z^2+c4^2+c2^2)-c2^2*c5*x1-c5*x1*c4^2-c2*x4*c4*c6)/c2^2/(-RootOf(_Z^2+c4^2+c2^2)+c6), x3 = c4*x1
/c2, x4 = x4, x5 = RootOf(_Z^2+c4^2+c2^2)*x1/c2, x6 = (-c5*RootOf(_Z^2+c4^2+c2^2)*x1-x1*c3*c4+c2-c2*x1*c1)/c2/(-RootOf(_Z^2+c4^2+c2^2)+c6), x7 = x7]

[x1 = RootOf(_Z^2+x3^2+x5^2), x2 = (x3*RootOf(_Z^2+x3^2+x5^2)*c1+c3*x3^2+(x6*c6+x5*c5-1)*x3-c4*x5*x6)/(c4*RootOf(_Z^2+x3^2+x5^2)-c2*x3), x3 = x3, x4
= ((-x3*c3-x5*c5-x6*c6+1)*RootOf(_Z^2+x3^2+x5^2)+x3^2*c1+c1*x5^2+c2*x5*x6)/(c4*RootOf(_Z^2+x3^2+x5^2)-c2*x3), x5 = x5, x6 = x6, x7 = x7]

[x1 = RootOf(_Z^2+x3^2+x5^2), x2 = 1/2*((-2*c3*x3^2+(2-2*x5*c5+a*c6^2)*x3-c4*a*c6*x5)*RootOf(_Z^2+x3^2+x5^2)+2*x3^3*c1+(-x5*c6*a*c2+2*c1*x5^2)*x3+c4*
x5^2*a*c2)/x3/(c2*RootOf(_Z^2+x3^2+x5^2)+x3*c4+x5*c6), x3 = x3, x4 = 1/2*(-2*RootOf(_Z^2+x3^2+x5^2)*x3^2*c1+2*RootOf(_Z^2+x3^2+x5^2)*c2*a*c6*x5+a*c6^
2*x3^2+2*x3^2-2*x3^3*c3-2*c5*x3^2*x5+x5^2*c6^2*a-a*c2^2*x5^2)/x3/(c2*RootOf(_Z^2+x3^2+x5^2)+x3*c4+x5*c6), x5 = x5, x6 = 1/2*(((-c6*a*c2-2*x5*c1)*x3-
c2*x5*a*c4)*RootOf(_Z^2+x3^2+x5^2)+(-c4*c6*a-2*c3*x5)*x3^2+x5*(-2*x5*c5+2+a*c2^2)*x3-x5^2*a*c6*c4)/x3/(c2*RootOf(_Z^2+x3^2+x5^2)+x3*c4+x5*c6), x7 = 
x7]

[x1 = x1, x2 = (((x1*c1+x4*c4-1)*c2+x1*c3*c4)*RootOf(_Z^2+c4^2+c2^2)-c2^2*c5*x1-c5*x1*c4^2-c2*x4*c4*c6)/c2^2/(-RootOf(_Z^2+c4^2+c2^2)+c6), x3 = c4*x1
/c2, x4 = x4, x5 = RootOf(_Z^2+c4^2+c2^2)*x1/c2, x6 = (-c5*RootOf(_Z^2+c4^2+c2^2)*x1-x1*c3*c4+c2-c2*x1*c1)/c2/(-RootOf(_Z^2+c4^2+c2^2)+c6), x7 = x7]

[x1 = RootOf((-8*c6*c4*c3*c5+4*c4^2*c5^2+4*c6^2*c3^2+4*c2^2*c5^2+4*c6^2*c1^2+4*c1^2*c4^2+4*c2^2*c3^2-8*c1*c4*c2*c3-8*c2*c1*c6*c5)*_Z^2+c6^2+c4^2+(4*
c4*c2*c3-4*c1*c4^2-4*c6^2*c1+4*c6*c2*c5)*_Z)*(2+(c2^2+c4^2+c6^2)*a), x2 = ((((-2*c1*c2-2*c5*c6)*a-4*x6*c5)*c4+2*c3*((c6^2+c2^2)*a+2*x6*c6))*RootOf((-
8*c6*c4*c3*c5+4*c4^2*c5^2+4*c6^2*c3^2+4*c2^2*c5^2+4*c6^2*c1^2+4*c1^2*c4^2+4*c2^2*c3^2-8*c1*c4*c2*c3-8*c2*c1*c6*c5)*_Z^2+c6^2+c4^2+(4*c4*c2*c3-4*c1*c4
^2-4*c6^2*c1+4*c6*c2*c5)*_Z)+a*c4*c2)/((-4*c3*c2+4*c1*c4)*RootOf((-8*c6*c4*c3*c5+4*c4^2*c5^2+4*c6^2*c3^2+4*c2^2*c5^2+4*c6^2*c1^2+4*c1^2*c4^2+4*c2^2*
c3^2-8*c1*c4*c2*c3-8*c2*c1*c6*c5)*_Z^2+c6^2+c4^2+(4*c4*c2*c3-4*c1*c4^2-4*c6^2*c1+4*c6*c2*c5)*_Z)-2*c4), x3 = -2*((-c2*c5+c6*c1)*RootOf((-8*c6*c4*c3*
c5+4*c4^2*c5^2+4*c6^2*c3^2+4*c2^2*c5^2+4*c6^2*c1^2+4*c1^2*c4^2+4*c2^2*c3^2-8*c1*c4*c2*c3-8*c2*c1*c6*c5)*_Z^2+c6^2+c4^2+(4*c4*c2*c3-4*c1*c4^2-4*c6^2*
c1+4*c6*c2*c5)*_Z)-1/2*c6)*(a*c6^2+2+(c4^2+c2^2)*a)/(-2*c5*c4+2*c6*c3), x4 = (((2*c4*c2*c3+2*c6*c2*c5-2*c1*c4^2-2*c6^2*c1)*a-4*(-c2*c5+c6*c1)*x6)*
RootOf((-8*c6*c4*c3*c5+4*c4^2*c5^2+4*c6^2*c3^2+4*c2^2*c5^2+4*c6^2*c1^2+4*c1^2*c4^2+4*c2^2*c3^2-8*c1*c4*c2*c3-8*c2*c1*c6*c5)*_Z^2+c6^2+c4^2+(4*c4*c2*
c3-4*c1*c4^2-4*c6^2*c1+4*c6*c2*c5)*_Z)+(c6^2+c4^2)*a+2*x6*c6)/((-4*c3*c2+4*c1*c4)*RootOf((-8*c6*c4*c3*c5+4*c4^2*c5^2+4*c6^2*c3^2+4*c2^2*c5^2+4*c6^2*
c1^2+4*c1^2*c4^2+4*c2^2*c3^2-8*c1*c4*c2*c3-8*c2*c1*c6*c5)*_Z^2+c6^2+c4^2+(4*c4*c2*c3-4*c1*c4^2-4*c6^2*c1+4*c6*c2*c5)*_Z)-2*c4), x5 = 2*((-c3*c2+c1*c4
)*RootOf((-8*c6*c4*c3*c5+4*c4^2*c5^2+4*c6^2*c3^2+4*c2^2*c5^2+4*c6^2*c1^2+4*c1^2*c4^2+4*c2^2*c3^2-8*c1*c4*c2*c3-8*c2*c1*c6*c5)*_Z^2+c6^2+c4^2+(4*c4*c2
*c3-4*c1*c4^2-4*c6^2*c1+4*c6*c2*c5)*_Z)-1/2*c4)*(a*c4^2+2+(c6^2+c2^2)*a)/(-2*c5*c4+2*c6*c3), x6 = x6, x7 = x7]

If you examine these more carefully, the RootOf() that occur in solutions #1 through 8 and 10 through 17 can only have imaginary roots when the constants are real-valued and non-zero, and in each case at least one of the xi values has that imaginary root multiplied by a real value; other variables might have the imaginary number in a ratio combination where potentially the imaginary portions cancel out, but we do not need to work those out because at least one of the terms is provably imaginary in those cases.

Solution #9, the "least complex solution" noted above, is always real-valued provided that c6 is non-zero, and is thus a primary solution.

Solution #18 involves RootOf((-8*c6*c4*c3*c5+4*c4^2*c5^2+4*c6^2*c3^2+4*c2^2*c5^2+4*c6^2*c1^2+4*c1^2*c4^2+4*c2^2*c3^2-8*c1*c4*c2*c3-8*c2*c1*c6*c5)*_Z^2+c6^2+c4^2+(4*c4*c2*c3-4*c1*c4^2-4*c6^2*c1+4*c6*c2*c5)*_Z)

It is not immediately obvious whether this is real-valued or not. As it is a quadratic one can easily express the two roots directly. If one then factors those roots fully, it turns out that both of the roots have the term (-(-c5*c4+c6*c3)^2*(c2^2+c4^2+c6^2))^(1/2) . A small bit of examination shows that unless at least 2 of the constants involved are 0, that the expression will result in an imaginary number.

This leaves us with the primary solution being Solution #9, the one with several explicit 0s.

Is this the only solution? No.

If you examine the RootOf() expressions in solutions 1, 2, 4, 6, 8, 11, 13, 15, and 16, you will see RootOf() involving two xi values (per solution); that RootOf can escape having an imaginary root if the xi are both set to 0. In each of these cases, setting those xi to 0 reduces the solution to one similar to the easy solution. One could, if one so desired, term these to be primary solutions as well, as the fact that the RootOf() is generated instead of the xi values being deemed to be 0 is really just a weakness of the solver in finding real-only solutions.

If you examine the RootOf() expressions in solutions 3, 5, 7, 10, 12, 14, and 17, you will see RootOf() involving 2 constants (per solution); that RootOf() can escape having an imaginary root if the constants are both 0. However, in each case, the solutions are such that if those constants are 0, the solutions involve 0/0, suggesting that no real roots exist for those solutions. However, if one sets those constants to be 0 back in the original equations and re-solve, in each case 4 potential solutions will be generated. Three of those potential solutions will involve sqrt(-1), but in each of those situations, there is an associated xi value that can be set to 0 to remove the effect of the imaginary constant, leaving a solution of mostly 0's. The 4th of the potential solutions will involve a RootOf() that, similar to the above, can be made non-imaginary if two of the xi values are set to 0. None of these possibilities can rightfully be deemed primary solutions because they all involve the happenstance that two of the constants are 0.

The 18th solution, with the longer RootOf(), is a bit less tractable. I think I have shown that each possible zeroing of constants that directly removes the imaginary root results in a 0/0 in the solution, but if 0s are substituted for those constants back in the original expressions then the situation becomes akin to the above, where real roots are possible if particular xi values are 0; again these would not be primary roots in that they rely on the happenstance of some constants being 0.

So... there are 10 primary solutions, some of which might perhaps be duplicates (I haven't worked it out); all of them are relatively simple in form... once you have found them.

[Asatur Khurshudyan]

At first I want to thank you for your attention to my problem and for time that you vested for its solution.

At second I notice, that x7=a in my equations, but I can`t imagine, what is _Z?

Let assume, that none of constants are 0 (all the same, under the problem`s terms, they must be chosen by me).

At third I want to emphasize, that non of unknowns must be arbitrary. They all must be certain.

You can also do the following simplification. Define the unknown a from the second equation and put it in the rest equations. It must be give something.

[Asatur Khurshudyan]

It should be noted, that you must get at a pinch x7=x7, because it is a Lagrange`s multiplier and can be arbitrary. But for others you must get a certain value.

[Walter Roberson]

Odd, this time around I get a different number of solutions. I cannot cross-check the other system until Monday. I used the transcribed versions of the equations that I posted earlier, and I know that I didn't make typos because I copy-and-pasted from the other system in to what I posted.

With regard to _Z: in Maple notation RootOf(f(_Z)) where f(_Z) is an expression involving _Z, denotes that one must find the value of _Z which evaluates f(_Z) to be 0 (the roots of the equation); the value of the RootOf() expression is that value _Z. For example, RootOf(_Z^2-4) would be -2 and +2 as those are the values of _Z that evaluate _Z^2-4 to be 0.

MuPad's RootOf() notation is fairly similar, except that it generates a variable name, and puts that variable name as the second parameter of the RootOf(). For example, MuPad might instead generate RootOf(y59^2-4,y59) which would mean that the value of the RootOf would be the values of y59 for which y59^2-4 was 0.

In either variation, any one RootOf() may represent more than one possible value, and the bigger expression involving the RootOf() is valid for each possible value. For example, (5+RootOf(_Z^2-4)) would be valid for both 5+2 and 5+(-2).

Anyhow... Under the condition that *all* variables x1 through x7 must be completely defined, then if my analysis is correct, there are no solutions acceptable to you -- not unless you are willing to define some of the constants as being 0. And even if you are willing to define some of the constants as being 0, it is tricky (but not impossible) to find conditions which pin down *all* the variables: nearly all of the solutions involve at least one variable that is free to assume any real value.

An example of a completely pinned solution is if c1=0 and c2=0, then one of the possibilities is real valued only if all of the variables are 0 except that x4 = 1/c4 . This is, however, not the only solution when c1=0 and c2=0, merely one that does not have any free variables.

[Walter Roberson]

Even allowing x7 to vary, I do not find any fully-bound solutions unless some of the constants are defined as 0.

[Asatur Khurshudyan]

The fact is that unknowns x1,...,x6 defined forces, they can`t get any value, that`s why they must be unique.

In that case, you may take c1=c3=0 or c2=c4=0.

[Walter Roberson]

If c2=c4=0, then one of the solutions comes out as x6 = 1/c6, x7 free, and all the other values 0. This is not the only possible solution when those constants are 0, but it is one that fully defines the variables.

With c1=c3=0 ... sorry, it's late enough here that I am making mistakes and just wasted a fair bit of time. It looks like if there is a solution it is non-trivial, but I need to go back and try again after sleep.

[Asatur Khurshudyan]

Sure, I`m sorry for troubling you and thank you very much. Try that case later, when it`ll be convenient.

[Asatur Khurshudyan]

I did some simplifications last night, and obtain this system, where x7 is already reduced. Here it is.

http://depositfiles.com/files/8z6o6grf1

[Walter Roberson]

In future please use one of the systems that does not impose a large time delay to download. I am not going to pay $40 or more just for faster downloads for my volunteer work. You can find notations about download delays in the list of upload sites, http://www.mathworks.com/matlabcentral/answers/7924-where-can-i-upload-images-and-files-for-use-on-matlab-answers

[Asatur Khurshudyan]

Sorry, I didn`t know, that deposit files aren`t free for you. Don`t forget to give me your personal info, I want to say a word of thanks to you.

http://www.sendspace.com/file/8d4qh0

[Asatur Khurshudyan]

And so, did you find time for that system?

[Walter Roberson]

That site, depositfiles, is free for people to download from, but non-members have a delay of more than 60 seconds imposed on them before they can download the files. It is the site's way of pushing people to take out memberships. Their membership rates work out to about $75 per year... and, of course, are not interchangable with the numerous other fileshare sites such as rapidshare.

I did retrieve the file via sendspace. I did not, however, have a chance to work on the system of equations (I am trying to reduce the typing I do these days as my RSI (repetitive strain injury) is flaring up.)

[Asatur Khurshudyan]

OK, it`s all right.

Answer 3

[Asatur Khurshudyan]

In this pdf file is mentioned system of nonlinear algebraic equations. And what about the phrase analytic solution, I meant no numerical one.

http://depositfiles.com/files/py0hb9eq6

Thank you, for your attention!

[Walter Roberson]

Note: transcribing the equations gives

a*c1*D_^3+2*D_*((x3*x4+x5*x6)*A+(x2*x3-2*x1*x4)*B+(x2*x5-2*x1*x6)*C)-4*x1*(A^2+B^2+C^2) = 0

a*c2*D_^2-2*(x3^2+x5^2)*A+2*x1*x3*B+2*x1*x5*C = 0

a*c3*D_^3+2*D_*((x1*x4-2*x2*x3)*A+(x1*x2+x5*x6)*B+(x4*x5-2*x3*x6)*C)-4*x3*(A^2+B^2+C^2) = 0

a*c4*D_^2+2*x1*x3*A-2*(x1^2+x5^2)*B+2*x3*x5*C = 0

a*c5*D_^3+2*D_*((x1*x6-2*x2*x5)*A+(x3*x6-2*x4*x5)*B+(x1*x2+x3*x4)*C)-4*x5*(A^2+B^2+C^2) = 0

a*c6*D_^2+2*x1*x5*A+2*x3*x5*B-2*(x1^2+x3^2)*C = 0

x1*c1+x2*c2+x3*c3+x4*c4+x5*c5+x6*c6 = 1

D_ = x1^2+x3^2+x5^2

A = x1*(x3*x4+x5*x6)-x2*(x3^2+x5^2)

B = x3*(x1*x2+x5*x6)-x4*(x1^2+x5^2)

C = x5*(x1*x2+x3*x4)-x6*(x1^2+x3^2)

I used D_ instead of D because in Maple D is the differential operator.

[Asatur Khurshudyan]

Yes, they`re right!

Answer 4

[Eirini Gk]

Is there something that we can use -without MathToolbox-? for exaple is needed a function with a numerical method or something?

[Walter Roberson]

If you do not have a Symbolic Toolbox or equivalent, then one might be restricted to finding one solution, such as by using fsolve() (which requires the Optimization Toolbox).