How to solve system of 4th power 4 variables of Nonlinear equations?

Hello, everyone. I'm trying to solve this system:

Is it posible? I've been trying with 2 functions: solve() and vpasolve (). But the software throws me like 54 values per variable and I need just 1 per variable. My code below:

var= input ('Enter the domain of the function: \n');
syms x [1 var];
func= input ('Enter the function, i.e [x1+x2+...+xn]: \n');
y=[];

for s=x(1:end)
    dp= diff (func, [s]);
    y=[y, dp];
end

pc=solve (y, [x]);

Indeed the system comes from the variable "y".

Thanks in advance for the help. :)

Answer 1

[Walter Roberson]

I need just 1 per variable.

By inspection, we can see that the system is solved when x1 = x2 = x3 = x4 = 0.

Since you only need one value per variable, there is no need to do any calculation.

[Walter Roberson]

The complete set of real-valued solutions is approximately

[                                    0,                                   0,                                   0,                                   0]
[   0.25674925523236048688212710601821,                                   0, -0.32660452232118238176765750753045, 0.098032791181095773637864684531547]
[   0.36933305529055136638352799364547,                                   0, -0.47052794617768611744449480175281,  0.13921652175396515996694657183697]
[   -0.2782457129297294396534997431436,                                   0,  0.61790405255589594305967377482632,  0.81569653759507180313180090386021]
[    1.4044823522923983690020623417795,                                   0,   1.3868812332817306006796405989283,   1.4394640112584527657643248527928]
[ -0.001289775239042112164530435884057, -0.10971017290159475439376929659422, 0.012036322038097815970549471665703,  0.10974470867918817296310417252285]
[  -0.36368369901770668711483343456354,  0.67624127128071014678293567331449,  0.45730225698335101396497614817376, -0.15234752552832827044604349688265]
[   0.66951240188153039705999638020132,   1.0124836536331311292186545928505,   1.0251231488742942464794919026147, -0.23059133389871809227036382794927]
[  -0.25032592383300416183301799033127,   0.4526944940219641960544009466379,  0.20493230491780217723841009193105, -0.33288157059738164608584886038894]
[   0.21989670824056788554675290063111,  0.70418106204969133315019602379969,  0.49587096814943123550227091168513, -0.67840439112065772469837072933044]
[  -0.40270253520251506096238438462189,  0.91868426366202513747342931742426,  0.84398077630023732045277947176763,  0.96340314036558478581342141313912]
[    1.1915840432563093521494114082981,  -1.0664142221956572269620064613593,   1.1372392933011685829770128386268,   1.2977950341059828108496283612239]
[    1.5530761533036836195892644736812,   1.2502473712162513783378507297551,   1.5631184892331470755308131202637,   1.5329265467596766045843044727479]

[Abner Ojeda]

Thanks a lot, Walter. I'm going to assume the value of 0 for all the variables. I think that it's the best option instead of complicated my life. Just one more question, is that possible? I appreciate a lot your help, thanks.

[Walter Roberson]

eqn = [
    2*x(1)^2 - 2*x(3)^2 - 2*x(1)*x(4) + 2*x(1)^2;
    -4*x(2)*x(3) + 4*x(2)^3;
    -4*x(1)*x(3) + 4*x(3)^3 - 2*x(4) - 2*x(2);
    -4*x(4)*x(3) + 4*x(4)^3 - 2*x(1)^2
    ];

sol = solve(eqn);

mask_sym = imag(sol.x1) == 0 & imag(sol.x2) == 0 & imag(sol.x3) == 0 & imag(sol.x4) == 0;
mask = isAlways(mask_sym, 'unknown', 'false');
real_sol = [sol.x1(mask), sol.x2(mask), sol.x3(mask), sol.x4(mask)];

disp(double(real_sol))

[Abner Ojeda]

Awesome, Walter. Thanks for the help, it was very useful. :)