Solving nonlinear equation using fsolve.

I have nonlinear coupled equation. I have T varied from 300 to 1300, each T give coupled-equation. How to solve each equation using fsolve? I know to solve coupled equation if only there is one coupled-equation, but do I have to solve one by one? 1000 equation total. Thank you

Answer 1

[Torsten]

Use a for-loop in which you call fsolve (or fzero) 1000 times with T varying from 300 to 1300.

Best wishes

Torsten.

[kusniar deny permana]

Yes, I have tried using for-loop, but the function must be saved in another file so i can call it. but, do i have to save 1000 function/equation? as you can see in:

http://www.mathworks.com/help/optim/ug/fsolve.html?searchHighlight=fsolve#butbm05-1

and

http://www.dam.brown.edu/people/sahn/html/fsolveDemo.html

Can I add fsolve directly in my loop? pls help me with the coding. thank you.

[Torsten]

If I understand you correctly, the equation you are trying to solve vary with T.

Thus my suggestion is to call fsolve somehow like

x0 = 1;
for i = 1:1001
  T = 300+(i-1);
  sol(i) = fsolve(@(x)fun(x,T),x0);
  x0 = sol(i);
end

This way, you get a solution array sol(i) depending on the temperature T(i).

Best wishes

Torsten.

[kusniar deny permana]

Here is my equation and code:

 syms x 
 %R,Regnault constant (gas constant) = 8.3144598 J/(K.mol)
 R=8.3144598
 %temperature
 T=300
 Z=1
 while T<=1300  

     %Ga1, Gibbs standard for Mg alpha(hcp)
     if T>=298 & T<=923
        Ga1(Z)= (-8367.34)+(143.675547.*T)-(26.1849782.*T.*log(T))+(0.4858*(10.^-3).*(T.^2))-(1.393669*10.^-6.*T.^3)+(78950.*T.^-1)
    else T>923 & T<=1300
        Ga1(Z)=(-14130.185)+(204.716215.*T)-(34.3088.*T.*log(T))+(1038.192.*10.^25.*T.^-9)
    end

    %Gb1, Gibbs standard for Li alpha(hcp)
    if T>=200 & T<=453
        Gb1(Z)=(-10737.817)+(219.637482.*T)-(38.940488.*T.*log(T))+(35.466931.*10.^-3.*T.^2)-(19.869816.*10.^-6.*T.^3)+(159994.*T.^-1)
    elseif T>453 & T<=500
        Gb1(Z)=(-559733.123)+(10549.879893.*T)-(1702.8886493.*T.*log(T))+(2258.3294448.*10.^-3.*T.^2)-(571.066077.*10.^-6.*T.^3)+(33885874.*T.^-1)
    else T>500 & T<=1300
        Gb1(Z)=(-9216.994)+(181.278285.*T)-(31.2283718.*T.*log(T))+(2.633221.*10.^-3.*T.^2)-(0.438058.*10.^-6.*T.^3)-(102387.*T.^-1)
    end

    %Ga2, Gibbs standard for Mg beta(bcc)
    if T>=298 & T<=923
        Ga2(Z)=(-5267.34)+(141.575547.*T)-(26.1849782.*T.*log(T))+(0.4858.*10.^-3.*T.^2)-(1.393669.*10.^-6.*T.^3)+(78950.*T.^-1)
    else T>923 & T<=1300
        Ga2(Z)=(-11030.185)+(202.616215.*T)-(34.3088.*T.*log(T))+(1038.192*10.^25.*T.^-9)
    end

    %Gb2, Gibbs standard for Li beta(bcc)
    if T>=200 & T<=453
        Gb2(Z)=(-10583.817)+(217.637482.*T)-(38.940488.*T.*log(T))+(35.466931.*10.^-3.*T.^2)-(19.869816.*10.^-6.*T.^3)+(159994.*T.^-1)
    elseif T>453 & T<=500
        Gb2(Z)=(-559579.123)+(10547.879893.*T)-(1702.8886493.*T.*log(T))+(2258.329444.*10.^-3.*T.^2)-(571.066077.*10.^-6.*T.^3)+(33885874.*T.^-1)
    else T>500 & T<=1300
        Gb2(Z)=(-9062.994)+(179.278285.*T)-(31.2283718.*T.*log(T))+(2.633221.*10.^-3.*T.^2)-(0.438058.*10^-6.*T.^3)-(102387.*T.^-1)
    end

    %Xb=mol fraction Li, Xa=mol fraction Mg, x defined as Xb so Xa=(1-x)

    Gm1= (1-x).*Ga1+x.*Gb1+R.*T.*((1-x).*log(1-x)+x.*log(x))+(1-x).*x.*((-6856)+4000.*(1-2.*x)+4000.*(1-2.*x).^2)
    Gm2= (1-x).*Ga2+x.*Gb2+R.*T.*((1-x).*log(1-x)+x.*log(x))+(1-x).*x.*((-18335)+8.49.*T+3481.*(1-2.*x)+(2658-0.114.*T).*(1-2.*x).^2)

    %assumption Gm1=Gm2 to find roots x0, intersection two phase alpha-beta
    %G1 is difference of Gibbs Li and Mg
    Gm=Gm2-Gm1

    %roots of Gm, x0
        for i=1:length(Gm)
        Gma=Gm(i)
        Gma==0
        Gmb=solve(Gma,x)

        Gmd(i)=Gmb(2)
    end

    F1=diff(Gm1)
    F2=diff(Gm2)

    F = [F1-F2; %in x1 and x2
        F1.*(x(2)-x(1))-(Gm2(x(2))-Gm1(x(1)))];

    % loop for next T value
    T=T+100
    Z=Z+1
end

The equation I want to solve is coupled equation "F", which is roots x1 and x2. I found x0 as Gmd. So i have a coupled equation and x0 as T change.

I need to solve coupled-equation as T change (on while-loop). So it would be Ti,x1i,x2i.......Tn,x1n,x2n. Thank you Torsten.