Solve Three non-linear equations for 3 variables system

Question

Deema Khunda 2022-7-27

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1768815-solve-three-non-linear-equations-for-3-variables-system

评论： Torsten 2022-7-27

采纳的回答： Walter Roberson

I have the followig three equations

eq1=72000==((B*4000)/(log(A*4000)-log(log(1+(1/Yse)))))

eq2=65000==((B*3500)/(log(A*3500)-log(log(1+(1/Yse)))))

eq3=55000==((B*3000)/(log(A*3000)-log(log(1+(1/Yse)))))

I would like to solve them for A,B, and Yse.

Every time I used Syms it comes back with empty cell and no symolic solutions found, What other functions do you recommened for solving them numerically ?

Thank you

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Answer 1

Walter Roberson 2022-7-27

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https://ww2.mathworks.cn/matlabcentral/answers/1768815-solve-three-non-linear-equations-for-3-variables-system#answer_1015930

编辑：Walter Roberson 2022-7-27

在 MATLAB Online 中打开

There are no real solutions. In order for there to be real solutions, some of the log() terms would have to cross zero, but when you study them (variables L1 and L2) they cannot do that. Instead the log terms are either always complex or else always positive in a way that does not permit balancing the equations.

syms A B Yse

eq1=72000==((B*4000)/(log(A*4000)-log(log(1+(1/Yse)))))

eq1 =

eq2=65000==((B*3500)/(log(A*3500)-log(log(1+(1/Yse)))))

eq2 =

eq3=55000==((B*3000)/(log(A*3000)-log(log(1+(1/Yse)))))

eq3 =

part_B = solve(eq1, B)

part_B =

eqn4 = subs([eq2, eq3], B, part_B)

eqn4 =

part_A = solve(eqn4(1), A)

Warning: Possibly spurious solutions.

part_A =

eqn5 = subs(eqn4(2:end), A, part_A)

eqn5 =

Ysea = vpasolve(eqn5(1))

Ysea = Empty sym: 0-by-1

Yseb = vpasolve(eqn5(2))

Yseb = Empty sym: 0-by-1

Z1 = simplify(lhs(eqn5(1)) - rhs(eqn5(1)))

Z1 =

L1 = simplify(cellfun(@(X) X, children(findSymType(Z1, 'log'),1)))

L1 =

Z2 = simplify(lhs(eqn5(2)) - rhs(eqn5(2)))

Z2 =

L2 = simplify(cellfun(@(X) X, children(findSymType(Z2, 'log'),1)))

L2 =

sol1 = arrayfun(@solve, L1, 'uniform', 0)

sol1 = 1×3 cell array

{0×1 sym} {0×1 sym} {0×1 sym}

sol2 = arrayfun(@solve, L2, 'uniform', 0)

sol2 = 1×3 cell array

{0×1 sym} {0×1 sym} {0×1 sym}

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Answer 2

Harshit Gupta 2022-7-27

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Hi, you need to specify the veriables to solve for and then this works just fine

syms A B Yse;

eq1=72000==((B*4000)/(log(A*4000)-log(log(1+(1/Yse)))))

eq1 =

eq2=65000==((B*3500)/(log(A*3500)-log(log(1+(1/Yse)))))

eq2 =

eq3=55000==((B*3000)/(log(A*3000)-log(log(1+(1/Yse)))))

eq3 =

Now you can store these equations in a structure array

eqns = [eq1, eq2, eq3]

eqns =

Now the input is complicated for solve() function but this can be further solved.

I recommend reading this part of the documentation: Troubleshoot Equation Solutions from solve Function

Hope that helps!

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Answer 3

Torsten 2022-7-27

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You might want to try a numerical solver, but the two I tested also didn't succeed.

fun = @(A,B,C)[72000-(B*4000)/(log(A*4000)-C);65000-(B*3500)/(log(A*3500)-C);55000-(B*3000)/(log(A*3000)-C)]
fun = function_handle with value:
    @(A,B,C)[72000-(B*4000)/(log(A*4000)-C);65000-(B*3500)/(log(A*3500)-C);55000-(B*3000)/(log(A*3000)-C)]
options = optimset('MaxFunEvals',1000000,'MaxIter',1000000);
sol = fsolve(@(x)fun(x(1),x(2),x(3)),[18 0 -1],options)
Solver stopped prematurely.

fsolve stopped because it exceeded the function evaluation limit,
options.MaxFunctionEvaluations = 1.000000e+06.
sol = 1×3
    0.0007   27.3825   -0.5611
%sol = lsqnonlin(@(x)fun(x(1),x(2),x(3)),[18 0 -1],[],[],options)
fun(sol(1),sol(2),sol(3))
ans = 3×1
1.0e+03 *

    5.1207
    1.2857
   -5.8480