solution of 3d nonlinear equation

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ali hassan 2020-9-29

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https://ww2.mathworks.cn/matlabcentral/answers/601690-solution-of-3d-nonlinear-equation

编辑： Walter Roberson 2020-9-30

x_p, y_p, z_p=(4, 5, 2)
x_1, y_1, z_1=(8, 9, 5)
x_2, y_2, z_2=(2, 5, 1)
x_3, y_3, z_3=(6, 1, 3)
t_1=5.692820*10^-9
t_2=-2.924173*10^-9
t_3=-12.010097*10^-9
c=3.0*10^8

and my three equations are

eqn1 = sqrt((x(s)-x_p)^2+(y(s)-y_p)^2+(z(s)-z_p)^2)-sqrt((x(s)-x_1)^2+(y(s)-y_1)^2+(z(s)-z_2)^2)-(c*t_1)
eqn2 = sqrt((x(s)-x_p)^2+(y(s)-y_p)^2+(z(s)-z_p)^2)-sqrt((x(s)-x_2)^2+(y(s)-y_2)^2+(z(s)-z_2)^2)-(c*t_1)
eqn3 = sqrt((x(s)-x_p)^2+(y(s)-y_p)^2+(z(s)-z_p)^2)-sqrt((x(s)-x_3)^2+(y(s)-y_3)^2+(z(s)-z_3)^2)-(c*t_1)

where only x(s), y(s), z(s) are unknown are rest all are known

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Walter Roberson 2020-9-29

编辑：Walter Roberson 2020-9-29

Perhaps the first equation should involve t1 instead of t_1, and second equation should involve t2 instead of t_1, and the third should involve t3 instead of t_1 ? You do not use t1, t2, or t3 after you define them.

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Answer 1

Walter Roberson 2020-9-29

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https://ww2.mathworks.cn/matlabcentral/answers/601690-solution-of-3d-nonlinear-equation#answer_502099

在 MATLAB Online 中打开

x_p = 4; y_p = 5; z_p = 2;
x_1 = 8; y_1 = 9; z_1 = 5;
x_2 = 2; y_2 = 5; z_2 = 1;
x_3 = 6; y_3 = 1; z_3 = 3;
t1 = 5.692820*10^-9;
t2 = -2.924173*10^-9;
t3 = -12.010097*10^-9;
syms xs ys zs     %our unknowns
syms c            %constant not given in question 
eqn1 = sqrt((xs-x_p)^2+(ys-y_p)^2+(zs-z_p)^2)-sqrt((xs-x_1)^2+(ys-y_1)^2+(zs-z_2)^2)-(c*t1);
eqn2 = sqrt((xs-x_p)^2+(ys-y_p)^2+(zs-z_p)^2)-sqrt((xs-x_2)^2+(ys-y_2)^2+(zs-z_2)^2)-(c*t2);
eqn3 = sqrt((xs-x_p)^2+(ys-y_p)^2+(zs-z_p)^2)-sqrt((xs-x_3)^2+(ys-y_3)^2+(zs-z_3)^2)-(c*t3);
sol = solve([eqn1, eqn2, eqn3], [xs, ys, zs]);
disp(sol.xs)
disp(sol.ys)
disp(sol.zs)

the values will be parameterized in c, which you indicate is a known value, but which you did not provide a value for.

There are two solutions for each variable.

You should probably re-substitute the solutions and verify that the values work, as MATLAB warns that some of the solutions produced might no be true solutions.

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Walter Roberson 2020-9-29

在 MATLAB Online 中打开

First of all, you used a vector of values in place of one of the constants. The vector was probably two elements long. For example you might have c as a vector of two values. That would have expanded the three equations into six equations.

When you use solve() with a vector of equations, MATLAB tries to find values of the variables that satisfy all of the equations at the same time. You cannot do something like

eqn = [x^2 + y == c, x - y^2 == 2*c];
solve( subs(eqn, c, [5 6]) )

and expect to get out a solution for one set of solutions for c == 5 and another set of solutions for c == 6. If you want to solve a system of equations while varying a parameter, then you should solve() with each of the values of the variable used in turn.

Secondly, the only way that the system I set up for you could have had 4 variables was if either you left c in place instead of substituting a constant, or else what you substituted for c itself had an unresolved variable in it.

I suspect the error about cell syntax and so on has to do with the fact that it did not find a solution.

ali hassan 2020-9-30

编辑：ali hassan 2020-9-30

https://www.mathworks.com/matlabcentral/answers/601690-solution-of-3d-nonlinear-equation#comment_1030756

THANKYOU SO MUCH SIR. i ran the code on matlab 2019 and i have done some tweaks and i am getting the solution but i want to filter my solution

this is a set of possible solutions i get from my code.but i only need three values but i get 6 possible solutions.i know that my solution can neither be negative nor it can be complex and it should show only accepted answer after ignoring other solution

there are 6 possible solutions but only three are right. now how to use loop maybe to ignore left entries as it is negative and it should only display right entries as solution

CODE:

x_p = 4; y_p = 5; z_p = 2;

x_1 = 8; y_1 = 9; z_1 = 5;

x_2 = 2; y_2 = 5; z_2 = 1;

x_3 = 6; y_3 = 1; z_3 = 3;

c=3.0*10^8;

t1 = 5.692820*10^-9;

t2 = -2.924173*10^-9;

t3 = -12.010097*10^-9;

syms xs ys zs %our unknowns

eqn1 = sqrt((xs-x_p)^2+(ys-y_p)^2+(zs-z_p)^2)-sqrt((xs-x_1)^2+(ys-y_1)^2+(zs-z_2)^2)-(c*t1);

eqn2 = sqrt((xs-x_p)^2+(ys-y_p)^2+(zs-z_p)^2)-sqrt((xs-x_2)^2+(ys-y_2)^2+(zs-z_2)^2)-(c*t2);

eqn3 = sqrt((xs-x_p)^2+(ys-y_p)^2+(zs-z_p)^2)-sqrt((xs-x_3)^2+(ys-y_3)^2+(zs-z_3)^2)-(c*t3);

sol = solve([eqn1, eqn2, eqn3], [xs ys zs]);

m = 1;

for n = 1:length(sol.xs)

possibleSol(1,m) = double(sol.xs(n));

possibleSol(2,m) = double(sol.ys(n));

possibleSol(3,m) = double(sol.zs(n));

m= m+1;

end

Walter Roberson 2020-9-30

编辑：Walter Roberson 2020-9-30

在 MATLAB Online 中打开

That is not 6 possible solutions, that is two solutions with three components each.

possibleSol(:, all(possibleSol>0 & imag(possibleSol)==0, 1))

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solution of 3d nonlinear equation

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