Question

i wrote a simple code to solve a steady state fermentation problem. It looks a lot because of so much variables.

function y = f (x)

mumax = 1.10; %h-1
Ksx = 1.32; %gl^-1
Pix=1.39; %gl^-1
Pmx=49.9; %gl^-1
Kix=304; %gl^-1
Pis=47.1; %gl^-1
Pms=95.5; %gl^-1
Kis=140; %gl^-1
qsmax=3.42; %gg^-1h^-1
Kss=2.05; %gl^-1
alp=0.39;
qpmax=3.02; %gg^-1h^-1
Ksp=2.05;  %gl^-1
Kip=140;  %gl^-1
Pip=47.1;  %gl^-1
Pmp=95.5;  %gl^-1
F=240;
S=40;
V=120;
D=0.5;
mu= (mumax*x(2)) / (Ksx+x(2));


  y=[x(1)*(-D+mu*(1-(x(3)-Pix)/(Pmx-Pix))*(Kix/(Kix+x(2))));
  D*(S-x(2))-(qsmax*(x(2)/(Kss+x(2)))*(1-((x(3)-Pis)/(Pms-Pis)))*(Kis/(Kis+x(2))))*x(1);
  -x(3)*D+x(1)*(-D+mu*(1-(x(3)-Pix)/(Pmx-Pix))*(Kix/(Kix+x(2))))*alp+qpmax*(x(2)/(Ksp+x(2)))*(1-(x(3)-Pip)/(Pmp-Pip))*x(1)*(Kip/(Kip+x(2)));]; 

endfunction


[x, fval,  info] = fsolve (@f, [2; 10; 30])

I defined D as 0.5 but actually i need the solutions for D in the interval between 0 and 1 and then plotting all x(1),x(2),x(3) vs D.
I tried something like

for i=0:0.1:1
D=num2str(i)

but its not working maybe i put it in a wrong way? Best would be to save all into one matrix to plot it easily.

Answer 1

There’s this workaround from the MATLAB help on fsolve for dealing with more than one input argument. Define D as a parameter:

function y = fffff (x,D)

then you can create an anonymous function to pry off the parameter for fsolve. try something like

ivals=linspace(0,1);
for j=ivals;D=j;
  [x, fval,  info] = fsolve (@(x) fffff(x,D), [2; 10; 30]);
  Q=[Q x];
end

If you want to test a lot of values for D (or if you are doing time evolution) then you would not build Q in this manner, but playing with fsolve like this should get the job done.