I want to find the five unknown of a system of five nonlinear equations. Thank you !

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Thierry Rebetez 2022 年 1 月 12 日
コメント済み: Thierry Rebetez 2022 年 1 月 16 日
Here the code:
clc;
clear;
close all;
clearvars;
syms ET i1 i2 i3 i4 i5 x1 x2 x3 x4 x5 % constants defined below
syms k1 k2 k3 k4 k7 % unknown variables
ET=1
i1=0.2
i2=0.32
i3=0.43
i4=0.3
i5=0.15
x1=100
x2=200
x3=400
x4=600
x5=800
eq1 = 0==-i1+k7*k3/(k4+k7)*(-(k2+k1*x1-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x1-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x1+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x1))*(ET-(-(k2+k1*x1-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x1-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x1+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x1)))/(1+2*k3/(k4+k7)*(-(k2+k1*x1-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x1-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x1+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x1)));
eq2 = 0==-i2+k7*k3/(k4+k7)*(-(k2+k1*x2-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x2-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x2+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x2))*(ET-(-(k2+k1*x2-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x2-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x2+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x2)))/(1+2*k3/(k4+k7)*(-(k2+k1*x2-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x2-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x2+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x2)));
eq3 = 0==-i3+k7*k3/(k4+k7)*(-(k2+k1*x3-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x3-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x3+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x3))*(ET-(-(k2+k1*x3-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x3-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x3+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x3)))/(1+2*k3/(k4+k7)*(-(k2+k1*x3-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x3-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x3+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x3)));
eq4 = 0==-i4+k7*k3/(k4+k7)*(-(k2+k1*x4-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x4-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x4+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x4))*(ET-(-(k2+k1*x4-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x4-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x4+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x4)))/(1+2*k3/(k4+k7)*(-(k2+k1*x4-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x4-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x4+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x4)));
eq5 = 0==-i5+k7*k3/(k4+k7)*(-(k2+k1*x5-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x5-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x5+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x5))*(ET-(-(k2+k1*x5-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x5-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x5+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x5)))/(1+2*k3/(k4+k7)*(-(k2+k1*x5-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x5-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x5+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x5)));
eq1=subs(eq1); eq2=subs(eq2); eq3=subs(eq3); eq4=subs(eq4); eq5=subs(eq5);
[k1,k2,k3,k4,k7]=fsolve( [eq1, eq2, eq3, eq4, eq5], [k1,k2,k3,k4,k7])

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回答 (1 件)

Torsten 2022 年 1 月 13 日

function main
x0 = rand(5,1);
x = fsolve(@fun,x0)
end
function res = fun(x)
k1 = x(1);
k2 = x(2);
k3 = x(3);
k4 = x(4);
k7 = x(5);
ET=1;
i1=0.2;
i2=0.32;
i3=0.43;
i4=0.3;
i5=0.15;
x1=100;
x2=200;
x3=400;
x4=600;
x5=800;
res(1) = -i1+k7*k3/(k4+k7)*(-(k2+k1*x1-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x1-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3 /(k4+k7)*x1+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x1))*(ET-(-(k2+k1*x1-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x1-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x1+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x1)))/(1+2*k3/(k4+k7)*(-(k2+k1*x1-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x1-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x1+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x1)));
res(2) = -i2+k7*k3/(k4+k7)*(-(k2+k1*x2-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x2-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x2+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x2))*(ET-(-(k2+k1*x2-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x2-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x2+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x2)))/(1+2*k3/(k4+k7)*(-(k2+k1*x2-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x2-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x2+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x2)));
res(3) = -i3+k7*k3/(k4+k7)*(-(k2+k1*x3-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x3-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x3+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x3))*(ET-(-(k2+k1*x3-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x3-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x3+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x3)))/(1+2*k3/(k4+k7)*(-(k2+k1*x3-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x3-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x3+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x3)));
res(4) = -i4+k7*k3/(k4+k7)*(-(k2+k1*x4-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x4-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x4+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x4))*(ET-(-(k2+k1*x4-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x4-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x4+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x4)))/(1+2*k3/(k4+k7)*(-(k2+k1*x4-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x4-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x4+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x4)));
res(5) = -i5+k7*k3/(k4+k7)*(-(k2+k1*x5-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x5-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x5+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x5))*(ET-(-(k2+k1*x5-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x5-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x5+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x5)))/(1+2*k3/(k4+k7)*(-(k2+k1*x5-k7*k3/(k4+k7)*ET)+sqrt((k2+k1*x5-k7*k3/(k4+k7)*ET)^2+4*k2*ET*(2*k1*k3/(k4+k7)*x5+k7*k3/(k4+k7))))/(2*k3/(k4+k7)*(k7+2*k1*x5)));
end
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Thierry Rebetez 2022 年 1 月 16 日
Hi Torsten
I think it is better to continue to chat on this platform if you want :) I think indeed that these equations are correct. The complete equation we have tried to solve is a combination of these three equations here:
h1 = (-(k2+k1.*x-k7.*K1.*ET)+sqrt((k2+k1.*x-k7.*K1.*ET).^2+4.*k2.*ET.*(2.*k1.*K1.*x+k7.*K1)))./(2.*K1.*(k7+2.*k1.*x));
i1=k7.*K1.*h1.*(ET-h1)./(1+2.*K1.*h1);
K1 = k3./(k4+k7);
with all k's positive or null.
I've tried with other values of ET, i1 to i5 and x1 to x5. It gives the following result:
ET =
0.0100
Equation solved, inaccuracy possible.
The vector of function values is near zero, as measured by the default value
of the function tolerance. However, the last step was ineffective.
<stopping criteria details>
x =
4.8110
1.9126
136.9516
-0.0000
43.0017
ET =
0.0100
res =
-0.0058 0.0065 -0.0034 0.0014 -0.0003
It is better but not perfect. What do you think about it?
Another question: the x which are given above (x =4.8110...) must be squared to get the k's, right?
Thank you very much for your help!

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R2018a

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