Mixing
Extrapolating value from function to a for loop and passing it back to the function
Extrapolating value from function to a for loop and passing it back to the function
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!
d
rhs
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
22 hours ago
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.
– LutzL
21 hours ago
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What exactly is
d(supposed to be) at the time the solver calls for the evaluation ofrhs?– LutzL
2 days ago