Double monod maximum rate estimation

I want to model a rate as a double monod, and I have the experimental data of two subtrates in time. I only need to estimate one parameter (maximum rate) and have been checking my notes from 10 years ago, browsing answers here, but none are similar to my case and I don't know how to proceed (I also forgot how these types of fittings are done). I need to use the data of both substrates in time for my rate to make sense, but I cannot understand how to implement that into ODE. Perhaps it's because I shouldn't be using ODE, but I'm just lost. Below are my equations.

[Torsten]

K_s,CH4, K_s,O2, F and MG are known constants ?

And r_max is to be fitted ?

[Maria Teresa Aguirrezabala Campano]

Yes, exactly

Answer 1

[Torsten]

% Initial guess for rmax
rmax0 = ...; 
% Time is a (nx1) vector, 
% Sdata is a (nx2) matrix with measurements of c_CH4 (1st column) and c_O2 (2nd column)
Time = ...;
Sdata = ...;
% Call fitting function
[B,Rsdnrm,Rsd,ExFlg,OptmInfo,Lmda,Jmat] = lsqcurvefit(@(x,xdata)MonodKinetics(x,xdata,Sdata),rmax0,Time,Sdata);
% Define kinetic equations
function S = MonodKinetics(rmax, Time, Sdata)
  x0 = [Sdata(1,1);Sdata(1,2)];  % (2x1) vector of initial conditions for c_CH4 and c_O2
  [T,S] = ode45(@DifEq, Time, x0);
  function xdot = DifEq(t, x)
    F = ...;
    MG = ...;
    K_CH4 = ...;
    K_O2 = ...;
    MTT = rmax*x(1)/(x(1)+K_CH4)*x(2)/(x(2)+K_O2);
    R = 0.2*MTT;        
    xdot = zeros(2,1);
    xdot(1) = MG - MTT - F;
    xdot(2) = R + MTT;
  end
end

[Maria Teresa Aguirrezabala Campano]

Thank you so much! It works, but the fit is not good. I guess my experimental data is not good enough to be trying this.

[Torsten]

You should plot the solution curves for several values of the parameter rmax in order to get a feeling in how far this parameter can influence the overall behaviour of C_CH4 and C_O2 if all other parameters remain constant.