Using lsqcurvefit with fsolve

Dursman Mchabe

2021 9 4

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3 次查看（30 天）

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Hi everyone,

I am trying to solve a nonlinear equation using fsolve. The equation is:

for i = 1:length(t)
                        
            sol(i)    = fsolve(@(x)  (Param + x)-(K5/x +  2.*(ydata3(i).*K3.*K4./(x^2 + x.*K3 + K3.*K4)) +...
     (ydata3(i).*x.*K3./(x^2 + x.*K3 + K3.*K4))+ ...
     2.*(Param.*K1.*K2./(x^2 + x.*K1 + K1.*K2)) +...
     (ydata3(i).*x.*K1./(x^2 + x.*K1 + K1.*K2))+ ...
     (ydata2(i).*K6)./(K6 + x)), x0);
            
        end

However, the parameter, Param, is not known, I have to regress it from lsqcurve fitting using:

        Param0 = 0.035;
Param = lsqcurvefit(@chargeBal,Param0,t,ydata1);

Where the YDATA will have the size of 4 x 1, whereas the Param, has a size of 1 x 1.

How can I make this to work?

The complete code is:

function pH_Example
global t pH  K1  K2  K3 K4 K5 K6 ydata1 ydata2 ydata3 Param sol
pH = [11.48
                6.86
                6.72
                6.25];
            
           
 t       = [0
                30
                60
                100];
 ydata3 = [2.58E-01
                6.28E-02
                5.76E-02
                4.83E-02];
 
 ydata2 = [ 1.25E-02
                9.75E-03
                7.93E-03
                5.83E-03];
            
ydata1 = [3.31131E-12
        1.38038E-07
        1.38038E-07
        1.38038E-07];
            
            K1    = 1.09e-6;
            K2     = 3.69e-10;
            K3    = 6.2e-4;
            K4     = 3.11e-8;
            K5      = 3.02e-14;
            K6       = (1.91e-2 + 5.38e-4)/2;
            
            
    function H = chargeBal(Param,t)
        
        x0 = 10^(-11.48);
        
        for i = 1:length(t)
                        
            sol(i)    = fsolve(@(x)  (Param + x)-(K5/x +  2.*(ydata3(i).*K3.*K4./(x^2 + x.*K3 + K3.*K4)) +...
     (ydata3(i).*x.*K3./(x^2 + x.*K3 + K3.*K4))+ ...
     2.*(Param.*K1.*K2./(x^2 + x.*K1 + K1.*K2)) +...
     (ydata3(i).*x.*K1./(x^2 + x.*K1 + K1.*K2))+ ...
     (ydata2(i).*K6)./(K6 + x)), x0);
            
        end
        
        function sol = pHfun(t,x)
            
            K1    = 1.09e-6;
            K2     = 3.69e-10;
            K3    = 6.2e-4;
            K4     = 3.11e-8;
            K5        = 3.02e-14;
            K6       = (1.91e-2 + 5.38e-4)/2;
            
            
            x0 = 10^(-11.48);
            
            t       = [0
                30
                60
                100];
                       
            
            pH = [11.48
                6.86
                6.72
                6.25];
            
                                   
            
            sol = (Param + x)-(K5./x +  2.*(ydata3.*K3.*K4./(x^2 + x.*K3 + K3.*K4)) +...
                (ydata3.*x.*K3./(x^2 + x.*K3 + K3.*K4))+ ...
                2.*(Param.*K1.*K2./(x^2 + x.*K1 + K1.*K2)) +...
                (ydata3.*x.*K1./(x^2 + x.*K1 + K1.*K2))+ ...
                (ydata2.*K6)./(K6 + x));
            
        end
        H = sol(i);
    end
Param0 = 0.035;
Param = lsqcurvefit(@chargeBal,Param0,t,ydata1);
fprintf(1,'\tParameters:\n')
 for k1 = 1:length(Param)
     fprintf(1, '\t\tParam(%d) = %8.5f\n', k1, Param(k1))
 end
 
 
ph = 3 - log10(sol(i));
plot(t,ph,'r--')
hold on
plot (t,pH,'kd')
xlabel('Time (sec)')
ylabel('pH')
hold off
legend('pH-Mod','pH-Exp')
 
end