Using Newton's method for 2 equations for finding concentration at different points

Question

Jared Dube 2022-11-7

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1845663-using-newton-s-method-for-2-equations-for-finding-concentration-at-different-points

编辑： Torsten 2022-11-7

Hi all, I'm currently trying to use Newton's method on two equations (I've attatched them) for finding the concentration of the lung and liver at three different R values 0.6, 0.8, and 0.9 from their intial guesses R = 0.6: Clung = 360.6 and Cliver = 284.6, R = 0.8: Clung = 312.25 and Cliver = 236.25, and R = 0.9: Clung = 215.5 and Cliver = 139.5. What I have so far is shown below, I'm hitting a wall and it wont allow me to create my J without giving an error if I use .*, etc. If I don't use .* in the J it gives me an error on dimensions instead. Any help would be greatly appreciated. Thank you

clearvars

clc

close all

%% Newton's Method to Systems

%% I have 2 equations, 3 sets of initials

maxiter = 100;

tol = 1e-6;

k = 0;

err(1) = 1;

%xold = [0.5; 0.25];

R_Old = [0.6; 0.8; 0.9];

C_Lung_Old = [360.6; 312.25; 215.5];

C_Liver_Old = [284.6; 236.25; 139.5];

Mat_Old = [R_Old,C_Lung_Old,C_Liver_Old];

while k<maxiter && err(k+1)>tol

k = k+1;

f1_1 = -780.*(1-R_Old(1))-R_Old(1).*(0.5.*C_Liver_Old(1)+1.5.*C_Lung_Old(1)) + (8.75.*C_Lung_Old(1)./(2.1 + C_Lung_Old(1))) + 2.*C_Lung_Old(1) ;

f1_2 = -780.*(1-R_Old(2))-R_Old(2).*(0.5.*C_Liver_Old(2)+1.5.*C_Lung_Old(2)) + (8.75.*C_Lung_Old(2)./(2.1 + C_Lung_Old(2))) + 2.*C_Lung_Old(2) ;

f1_3 = -780.*(1-R_Old(3))-R_Old(3).*(0.5.*C_Liver_Old(3)+1.5.*C_Lung_Old(3)) + (8.75.*C_Lung_Old(3)./(2.1 + C_Lung_Old(3))) + 2.*C_Lung_Old(3) ;

f2_1 = -0.5.*C_Lung_Old(1) + 0.322*(118*C_Liver_Old(1)./(7.0 +C_Liver_Old(1))) + 0.5*C_Liver_Old(1) ;

f2_2 = -0.5.*C_Lung_Old(2) + 0.322*(118*C_Liver_Old(2)./(7.0 +C_Liver_Old(2))) + 0.5*C_Liver_Old(2) ;

f2_3 = -0.5.*C_Lung_Old(3) + 0.322*(118*C_Liver_Old(3)./(7.0 +C_Liver_Old(3))) + 0.5*C_Liver_Old(3) ;

nF = [f1_1,f1_2,f1_3;f2_1,f2_2,f2_3];

J = [(C_Liver_Old+3*C_Lung_Old-1560)/2 (20*(3*R_Old-4)*C_Lung_Old*(5*C_Lung_Old+21)+1323*R_Old-5439)/(2*(10*C_Lung_Old+21).^2) 2*R_Old R_Old/2; ...

0 1/2 -((125*C_Liver_Old^2+1750*C_Liver_Old+72618)/(250*(C_Liver_Old+7)^2))];

xx = J\nF;

xnew = xx+Mat_Old;

f1 = -3*xnew(1).^2-xnew(2).^2 +10;

f2 = -xnew(1).^2-xnew(2)+1;

errtemp = [abs(f1) abs(f2)];

err(k+1) = max(errtemp);

Mat_Old = xnew;

end

semilogy(err)

return

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Answer 1

Torsten 2022-11-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1845663-using-newton-s-method-for-2-equations-for-finding-concentration-at-different-points#answer_1093608

编辑：Torsten 2022-11-7

在 MATLAB Online 中打开

Why don't you use "fsolve" ? Don't you have a license for the optimization toolbox ?

I don't understand J, f1 and f2 in your code.

maxiter = 100;
tol = 1e-6;
R_Old = [0.6; 0.8; 0.9];
C_Lung_Old = [360.6; 312.25; 215.5];
C_Liver_Old = [284.6; 236.25; 139.5];
sol = [];
for i = 1:3
    f = @(x) [-780.*(1-R_Old(i))-R_Old(i)*(0.5*x(2)+1.5*x(1)) + (8.75*x(1)/(2.1 + x(1))) + 2*x(1) ; ...
              -0.5*x(1) + 0.322*(118*x(2)/(7.0 +x(2))) + 0.5*x(2)] ;
    Mat_Old = [C_Lung_Old(i);C_Liver_Old(i)];
    err(1) = 1;
    k = 0;
    while k<maxiter && err(k+1)>tol   
        k = k+1;
        fv = f(Mat_Old);
        df = [(f(Mat_Old+1e-6*[1;0])-fv)/1e-6,(f(Mat_Old+1e-6*[0;1])-fv)/1e-6];
        xx = df\fv;  
        xnew = Mat_Old-xx;   
        err(k+1) = norm(abs(fv));
        Mat_Old = xnew;
    end
    sol = [sol,Mat_Old];
end
k = 3
k = 3
k = 4
sol
sol = 2×3
  351.3324  294.6213  185.6450
  277.2120  220.9628  114.0475