solving Z + 2 simultaneous nonlinear equations numerically (Newton) with sum

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khaled elbatawy 2020-11-7

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编辑： khaled elbatawy 2020-12-6

Hello ,

I am trying to solve Z + 2 simultaneous nonlinear System but i have problems with that because of the sum . I do not know how to deal with that in Matlab

I have 3 Equations with 3 Unknowns the equations as follows

The Unknowns are

,

and

Thank you very much for any Help

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Answer 1

Amanpreetsingh Arora 2020-11-10

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You can solve the system of equations using "fsolve". Implement a function that takes the Z+2 dimensional unknowns as input and returns Z+2 dimensional vector with the values of LHS of the Z+2 equations. Then pass this function with an initial guess to "fsolve". The documentation for "fsolve" is as follows.

https://in.mathworks.com/help/optim/ug/fsolve.html

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khaled elbatawy 2020-11-10

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Thank you for your answer .

I know the function and i have created also Code to lose with Newton-Raphson

But still my Problem with Sum how to express it in the right way , because the sum here is for the indices

Here is it my code but always error with sum

I have created for loop inside the function to take every Value of Vartheta because it is array with the number of Z = 14

But still the error and the problem with the applying of the sum

fun=@root3d;
xo=[0.0001,0.0001,40*pi/180];
x = fsolve(fun,xo,foptions);
function F = root3d(x) 
F_A=500;F_R=1000;alpha_o=40*pi/180;n=3/2;Z=14;
for g=1:Z
    Vartheta(g)=(2*pi/Z)*(g-1);
end
for h=1:Z
Geo.Dw=11.86;
K=3.583822434736866;
E=1.019359284554166;
t=0.111984000000000;
dm=59.959999999999994;
Geo.fi=0.520236087689713;
Geo.fo=0.532883642495784;
gamma = (Geo.Dw*cos(x(3)))/dm;
A = (Geo.fi + Geo.fo-1)*Geo.Dw;
F_i = ((1/Geo.fi)+((2*gamma)/(1-gamma)))/(4-(1/Geo.fi)+ ((2*gamma)/(1-gamma)));
F_o = ((1/Geo.fo)-((2*gamma)/(1+gamma)))/(4-(1/Geo.fo)- ((2*gamma)/(1+gamma)));
a_st = ((K-E)/((pi/4)*(1-t^2)*(1-F_i)))^(1/3);
b_st = ((t*E - K*t^3)/((pi/4)*(1-t^2)*(1+F_i)))^(1/3);
Geo.roi = (1/Geo.Dw)*(4-(1/Geo.fi)+ ((2*gamma)/(1-gamma)));
Geo.roo = (1/Geo.Dw)*(4-(1/Geo.fo)- ((2*gamma)/(1+gamma)));
Cdi = 2.79 * 10^(-4)*(2*K/(pi*a_st))*(Geo.Dw*Geo.roi)^(1/3);
Cdo = 2.79 * 10^(-4)*(2*K/(pi*a_st))*(Geo.Dw*Geo.roo)^(1/3);
Cd = Geo.Dw^(1/2)/(Cdi + Cdo)^(3/2);
F(1)=x(1)-x(2)*tan(x(3))*cos(Vartheta(h))-A*sin(x(3)-alpha_o)/cos(x(3)); 
F(2)=F_A-symsum(Cd*sin(x(3)(h))*(A*((cos(alpha_o)/cos(x(3)(h)))-1)+(x(2)(h)*cos(Vartheta)/cos(x(3)))).^n,h,1,Z); 
F(3)=F_R-symsum(Cd*cos(x(3)(h))*cos(Vartheta)*(A*((cos(alpha_o)/cos(x(3)(h)))-1)+(x(2)(h)*cos(Vartheta)/cos(x(3)))).^n,h,1,Z);
end
end

Amanpreetsingh Arora 2020-11-11

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How many equations and variables do you have? Your original post mentions Z+2 but the function only has 3 equations and 3 unknowns.

As per your original post, your function input "x" needs to be Z+2 dimensional. For example, you can represent your unknowns as follows.

x(1) =

x(2) =

x(3:end) = vector

which can be mapped to

F(1:Z) = The first equation (which is collection of Z equations)

F(Z+1) = 2nd eqn, F(Z+2) = 3rd eqn.

For summation, just keep adding values to F(Z+1) and F(Z+2) in the loop, as follows.

% Inside the loop
F(Z+1)=F(Z+1) + <summation term>;
F(Z+2)=F(Z+2) + <summation term>;

You don't need symsum for this.

khaled elbatawy 2020-12-6

编辑：khaled elbatawy 2020-12-6

在 MATLAB Online 中打开

Thank you for your explaination , And sorry for the Late answer which was because of strong reasons .

My Z in my Case is 14 and every one of the 14 has

,

and

, and from the first equation shows that

and

are dependent on the angle

that is mean

x(1)=

x(2)=

and so on till Z=14

i have not understood well what do you mean with this F(1:Z) = The first equation (which is collection of Z equations)

F(Z+1) = 2nd eqn, F(Z+2) = 3rd eqn. ?

here is it the code that i have created , if you can see and suggest me if it is wrong or better

for g=1:Z
   Vartheta(g)=(2*pi/Z)*(g-1);
end
xo=[0.0001,0.0001,40*pi/180];
[x,fval] = fsolve(@(x)root3d(x,F_R_A(v),F_A_A(v),Varth,Vartheta,h,Result,K,E,t,ChRing),xo);
Result(h, :) = abs(x);
function [F] = root3d(x,F_R,F_A,Varth) 
alpha_o=40*pi/180;n=3/2;Z=14;
Cd = Geo.Dw^(1/2)/(Cdi + Cdo)^(3/2);
F(1)=x(1)-x(2)*tan(x(3))*cos(Varth(1))-A*sin(x(3)-alpha_o)/cos(x(3));
F(2)=x(4)-x(5)*tan(x(6))*cos(Varth(2))-A*sin(x(6)-alpha_o)/cos(x(6));
F(3)=x(7)-x(8)*tan(x(9))*cos(Varth(3))-A*sin(x(9)-alpha_o)/cos(x(9));
F(4)=x(10)-x(11)*tan(x(12))*cos(Varth(4))-A*sin(x(12)-alpha_o)/cos(x(12));
F(5)=x(13)-x(14)*tan(x(15))*cos(Varth(5))-A*sin(x(15)-alpha_o)/cos(x(15));
F(6)=x(16)-x(17)*tan(x(18))*cos(Varth(6))-A*sin(x(18)-alpha_o)/cos(x(18));
F(7)=x(19)-x(20)*tan(x(21))*cos(Varth(7))-A*sin(x(21)-alpha_o)/cos(x(21));
F(8)=x(22)-x(23)*tan(x(24))*cos(Varth(8))-A*sin(x(24)-alpha_o)/cos(x(24));
F(9)=x(25)-x(26)*tan(x(27))*cos(Varth(9))-A*sin(x(27)-alpha_o)/cos(x(27));
F(10)=x(28)-x(29)*tan(x(30))*cos(Varth(10))-A*sin(x(30)-alpha_o)/cos(x(30));
F(11)=x(31)-x(32)*tan(x(33))*cos(Varth(11))-A*sin(x(33)-alpha_o)/cos(x(33));
F(12)=x(34)-x(35)*tan(x(36))*cos(Varth(12))-A*sin(x(36)-alpha_o)/cos(x(36));
F(13)=x(37)-x(38)*tan(x(39))*cos(Varth(13))-A*sin(x(39)-alpha_o)/cos(x(39));
F(14)=x(40)-x(41)*tan(x(42))*cos(Varth(14))-A*sin(x(42)-alpha_o)/cos(x(42));
F_2sum=0;
F_3sum=0;
F_1sum=0;
for m=1:Z
    F_2= Cd*sin(x(m))*(A*((cos(alpha_o)/cos(x(m)))-1)+(x(m)*cos(Vartheta(m))/cos(x(m))))^n;
    F_3=Cd*cos(x(m))*cos(Vartheta(m))*(A*((cos(alpha_o)/cos(x(m)))-1)+(x(m)*cos(Vartheta(m))/cos(x(m))))^n;
    
    F_2sum=F_2sum+F_2;
    F_3sum=F_3sum+F_3;
    
end
F(15)=F_A-real(F_2sum);
F(16)=F_R-real(F_3sum);
end

Theoritically the start Value for all

,

and

are the same for all

thank you in advance

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solving Z + 2 simultaneous nonlinear equations numerically (Newton) with sum

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solving Z + 2 simultaneous nonlinear equations numerically (Newton) with sum

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