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A very simple integration method

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Yara Saavedra Ortega
回答： Devineni Aslesha 2020-5-18
So, I am trying to program a function which creates an array that contains a table with the value of four variables and its derivatives through some time instants. Here's what I've tried. However, it says one of my functions needs additional arguments.
function [integ] = sinteg(P,IC,t,dt)
%This function tries to employ a very simple integration method in order to
%get biomass, Carbon,nitrogen and product concentration.
% P represents a vector including the set of parameters necessary to
% calculate concentrations through the selected interval of time. CI
% represents another vector which contains a set of initial conditions
% for the process. t represents the total lenght of the interval of time
% that wishes to be evaluated. delta t corresponds to the size of the
% finite but next to infinitesimal time step.
inv=t/dt;
integ=zeros(inv,9);
integ(1,1)=0;
integ(1,2)= IC(1);
integ(1,3)= IC(2);
integ(1,4)= IC(3);
integ(1,5)= IC(4);
C=[IC(1),IC(2),IC(3),IC(4)];
aux=de_sys(C,P);
integ(1,6)=aux(1);
integ(1,7)=aux(2);
integ(1,8)=aux(3);
integ(1,9)=aux(4);
i=2;
while i<= t/dt
integ(i,1)=integ(i-1,1)+dt;
for j=1:1:5
integ(i,j)=integ(i-1,j)+integ(i-1,j+4)*dt;
end
aux=de_sys([integ(i,2),integ(i,3),integ(i,4)],P);
for j=6:1:9
integ(i,j)=aux(j-5);
end
i=i+1;
end
Here it is my other function
%This function calculates the value of each derivative for a specific point in time. It needs present conditions and a vector of parameters for the said point in time.
function dDV_dIV = de_sys(C,P)
mumax=P(1);
alpha=P(2);
Ksc=P(3);
Snm=P(4);
a1=P(5);
yxpsc=P(6);
mc=P(7);
Ksn=P(8);
Scm=P(9);
a2=P(10);
yxsn=P(11);
mn=P(12);
X=C(1);
Sc=C(2);
Sn=C(3);
dDV_dIV=zeros(1,4);
dDV_dIV(1)=(mumax*((Sc/(Ksc+Sc))*(Sn/(Ksn+Sn))*(1-(Sc/Scm)^a1)*(1-(Sn/Snm)^a2)))*X;
mu=(mumax*((Sc/(Ksc+Sc))*(Sn/(Ksn+Sn))*(1-(Sc/Scm)^a1)*(1-(Sn/Snm)^a2)));
dDV_dIV(2)=-(((1/yxsn)*mu)+mn)*X;
dDV_dIV(3)=-(((1/yxpsc)*mu)+mc)*X;
dDV_dIV(4)=(alpha*mu)*X;
end
  2 个评论
Yara Saavedra Ortega
Yara Saavedra Ortega 2020-5-16
Here's what the error says:
>> sinteg
Not enough input arguments.
Error in sinteg (line 11)
in=(t/dt);
Walter Roberson
Walter Roberson 2020-5-16
When you run sinteg that way, what value should it use for t ? Are you expecting that it will look in the current workspace (the base workspace) to find a variable named t and to use that ?

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采纳的回答

Devineni Aslesha
Devineni Aslesha 2020-5-18
The error is due to calling the function sinteg without passing any input arguments. Call the function sinteg as shown below.
P = 3*ones(1,12);
IC = [2 1 2 1];
t = 5;
dt = 1;
integ = sinteg(P,IC],t,dt);
For more information, refer the following link.
https://www.mathworks.com/help/matlab/learn_matlab/calling-functions.html

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