Nonlinear multiple equations = fsolve?

Hello,

I have a problem with finding a solution for multiple nonlinear equations. I tried using 2 different forms of the equations, but I get 2 different problems in each case.

1) first attempt - I only get the solution of [0;0;0] which is true to fulfill the equation, but there is/there are other solutions which i can't find.

2) second attempt - it doesn't find the solution with info:

No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,

but the vector of function values is not near zero as measured by the

default value of the function tolerance.

Here is the function.

function F = myfun(x,a,b,c,alfa,beta,gamma)
F= [ 
%first attempt
  alfa*(sqrt(x(1)^2+x(2)^2+x(3)^2)*sqrt(a(1)^2+a(2)^2+a(3)^2))-((x(1)*a(1)+x(2)*a(2)+x(3)*a(3)));
  beta*(sqrt(x(1)^2+x(2)^2+x(3)^2)*sqrt(b(1)^2+b(2)^2+b(3)^2))-((x(1)*b(1)+x(2)*b(2)+x(3)*b(3)));
  gamma*(sqrt(x(1)^2+x(2)^2+x(3)^2)*sqrt(c(1)^2+c(2)^2+c(3)^2))-((x(1)*c(1)+x(2)*c(2)+x(3)*c(3)));

%second attempt  
%   alfa-((x(1)*a(1)+x(2)*a(2)+x(3)*a(3))/(sqrt(x(1)^2+x(2)^2+x(3)^2)*sqrt(a(1)^2+a(2)^2+a(3)^2)));
%  beta-((x(1)*b(1)+x(2)*b(2)+x(3)*b(3))/(sqrt(x(1)^2+x(2)^2+x(3)^2)*sqrt(b(1)^2+b(2)^2+b(3)^2)));
%   gamma-((x(1)*c(1)+x(2)*c(2)+x(3)*c(3))/(sqrt(x(1)^2+x(2)^2+x(3)^2)*sqrt(c(1)^2+c(2)^2+c(3)^2)));
];

And running it:

x0 = [10;10;10];           

a=[0,1,1.73];
b=[0,1.73,1];
c=[-1,0,1.73];
alfa=0.8926;
beta=0.8183;
gamma=0.5603;

[out,fval] = fsolve(@(x) myfun(x,a,b,c,alfa,beta,gamma),x0)

In fact I try to solve such a problem:

I've got 3 vectors (in 3D) and 3 angles - in fact cosinus of them. This are angles between each of those vectors and other vector which i try to find.

Has anybody idea how to help me?

Regards,

Piotrek

Answer 1

[Matt J]

The problem is equivalent to the constrained least squares problem

where

A=[a/norm(a);b/norm(b);c/norm(c)];
d=[alfa;beta;gamma];

The solution can be found analytically using my trustregprob routine (Download)

x=trustregprob(A.'*A,A.'*d,1,1);

The solution I find has a slight 4% relative error, meaning there is no exact solution to the original problem, but that's likely because the input data alfa,beta,gamma,a,b,c was posted with only 4 decimal places precision.

>> x, relativeError=norm(A*x-d)/norm(d)



x =

    0.3403
    0.4507
    0.8253


relativeError =

    0.0392

Answer 2

[KALYAN ACHARJYA]

Is this the solution?

Equation solved.

fsolve completed because the vector of function values is near zero as measured by the default value of the function tolerance, and the problem appears regular as measured by the gradient.

<stopping criteria details>

out =
   1.0e-06 *
    0.4311
    0.1091
    0.7377
fval =
   1.0e-06 *
    0.1510
    0.4820
    0.1193

>> 

Answer 3

[Piotr Ogórek]

Yes, it is.

But the out ~=[0;0;0] doesn't satisfy me. I'm looking for another solution which exists.

[madhan ravi]

what's your expected solution ? please clarify

[Piotr Ogórek]

okay, so for this parameters:

I would expect that the vector should be [1;2;3]

x0 = [10;10;10];           
a = [2,3,1];
b = [0,2,1];
c = [1,0,1];

alfa=11/14;
beta=sqrt(7/10);
gamma=2/sqrt(7);
[out,fval] = fsolve(@(x) myfun(x,a,b,c,alfa,beta,gamma),x0)

[Piotr Ogórek]

Actually, all vectors with direction [1;2;3] or multiplied by a scalar fullfil the equations:

But how to find it?

myfun([1;2;3],a,b,c,alfa,beta,gamma)

ans =

     0
     0
     0

>> myfun([10;20;30],a,b,c,alfa,beta,gamma)

ans =

     0
     0
     0

>> myfun([20;40;60],a,b,c,alfa,beta,gamma)

ans =

     0
     0
     0