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function Test()

a = 2;
b = 1;
c = 0.5;
q = 0.001;
r = 10;

function F = Useful(x) %calculates existing values for x with size 11

eq1 = (1*(0.903*x(2))^(-1))-(0.903*x(1));  
eq2 = (1*(0.665*x(3))*(0.903*x(2))^(-1))-0.903*x(4); 
eq3 = (1*(0.399*x(5))*(0.903*x(2)))-0.665*x(6);
eq4 = (1*(0.399*x(5))*(0.903*x(2))^2)-0.903*x(7);
eq5 = (1*(0.399*x(5))*(0.903*x(2))^3)-1*x(8);
eq6 = (1*(0.665*x(3))*(0.399*x(5))*(0.903*x(2)))-1*x(9);
eq7 = (1*(0.665*x(3))*(0.399*x(5))*(0.903*x(2))^2)-0.903*x(10);
eq8 = (1*(0.665*x(3))*(0.399*x(5)))-0.903*x(11);
eq9 = x(3)+x(4)+x(9)+x(10)+x(11)-a;
eq10 = x(5)+x(6)+x(7)+x(8)+x(9)+x(10)+x(11)-b; 
eq11 = x(2)+x(6)+2*x(7)+3*x(8)+x(9)+2*x(10)-x(1)-x(4)-c;

F = [eq1;eq2;eq3;eq4;eq5;eq6;eq7;eq8; eq9; eq10; eq11];

end

Value(1,1) = 0;

 for d = 2:100 

    x = fsolve(@Useful,x0,options);   %Produces the x(1) to x(11) values

    Value(1,d) = (x(3)+x(5))*d+Value(1,d-1);  %Gives a new value after each iteration

    a = a-x(3);
    b = b-x(5);
    c = c-x(2);

 end

 function Zdot = rhs(t,z) %z = (e1,e2,e3,e4,e5)
 Zdot=zeros(5,1);

 Zdot(1) = -1*z(1);
 Zdot(2) = 1*z(1); 
 Zdot(3) = 1*z(1) - 1*z(2)*z(3);
 Zdot(4) = 1*1*z(1) - Value(1,100)*H(z(3))*z(4)*z(4); 
 Zdot(5) = Value(1,100)*H(z(3))*(z(4));
 end

 function hill = H(x)  
 hill = q/(q+x^r);
 end

 [T,Y] = ode15s(@rhs, [0, 120], [1, 0, 1, 0, 0]); %Solve second function with values giving z(1) to z(5)
 plot(T,Y(:,5))
end

I'm wondering, is it possible to pass on each Value obtained (Value (1), Value (2)... so on), into "function Zdot" or is only the final value possible to pass on? Essentially is this possible to implement:

 function Zdot = rhs(t,z) %z = (e1,e2,e3,e4,e5)
     Zdot=zeros(5,1);

     Zdot(1) = -1*z(1);
     Zdot(2) = 1*z(1); 
     Zdot(3) = 1*z(1) - 1*z(2)*z(3);
     Zdot(4) = 1*1*z(1) - Value(1,d)*H(z(3))*z(4)*z(4); 
     Zdot(5) = Value(1,d)*H(z(3))*(z(4));
     end

Any insights would be much appreciated and I would be extremely grateful. Thank you in advance!

Borys Ostapienko
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  • What exactly is `d` (supposed to be) at the time the solver calls for the evaluation of `rhs`? – Lutz Lehmann Jun 30 '18 at 12:24
  • d is simply the array position. The idea is that each time the solver runs, the position moves up by one. So... Run ODE with Value (1,1) then ODE with Value (1,2) and so on... – Borys Ostapienko Jul 02 '18 at 14:16
  • So you set either d or Value (1,d) as parameter of the ODE, which can be passed via anonymous function or inside an option structure via odeset. – Lutz Lehmann Jul 02 '18 at 14:47

0 Answers0