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Tsogerdene on 3 Aug 2011
I just wondering that why this code working wrong. Please help me! I can't manipulate the ODE function.
options=odeset('RelTol',10^-4);
[T,Y]=ode23('turshilt23',[0 300],[0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5],options);
plot(T,Y);
function dy=turshilt23(t,y,flag)
%here is some code
if(k==0)
y(1)=y(1)/2;
end
dy(1)=mu*y(1)*(1-(y(1)/m_max));
%here is some code
end

Arnaud Miege on 3 Aug 2011
Also, it seems that in your function turshilt23, you haven't defined what k, mu or m_max are. Finally you compute dy(1), but you call the ode solver with an initial condition vector of length 9. You need to compute dy(2), dy(3), ... dy(9) in your function as well.
Have a look at the documentation for the ode solvers, there are various examples you can inspire yourself from.
HTH,
Arnaud

Tsogerdene on 3 Aug 2011
function dy=turshilt23(t,y,flag)
global Global_CycB1;
global Global_CycB2;
global k;
global y_1;
dy=zeros(9,1);
% size of protein
mu=0.005;
m_max=10;
dy(1)=mu*y(1)*(1-(y(1)/m_max));
if Global_CycB2 >=0.099 & Global_CycB2<0.1
if Global_CycB2>Global_CycB1
y(1)=y(1)*(1/2);
end
end
%CycBT
k1=0.04;
k2_1=0.04;
k2_2=1;
k2_3=1;
dy(2)=k1-(k2_1+k2_2*y(3)+k2_3*y(5))*y(2);
Keq=1000;
beta=y(2)+y(7)+Keq^(-1);
Trimer=(2*y(2)*y(7))/(beta+(beta^2-4*y(2)*y(7))^(1/2));
CycB=y(2)-Trimer;
Global_CycB1=Global_CycB2;
Global_CycB2=CycB;
%Cdh1
k3_1=1;
k3_2=10;
k4_1=2;
k4=35;
J3=0.04;
J4=0.04;
dy(3)= (((k3_1+k3_2*y(5))*(1-y(3)))/(J3+1-y(3)))-(((k4*y(1)*CycB+k4_1*y(8))*y(3))/(J4+y(3)));
%Cdc20T
k5_1=0.005;
k5_2=0.2;
k6=0.1;
J5=0.3;
n=4;
dy(4)= k5_1+k5_2*((y(1)*CycB)^n/(J5^n+(y(1)*CycB)^n))-k6*y(4);
%Cdc20A
k6=0.1;
k7=1;
k8=0.5;
J7=1*(10^(-3));
J8=1*(10^(-3));
%IEP
k9=0.1;
k10=0.02;
dy(6)= k9*y(1)*CycB*(1-y(6))-k10*y(6);
%CKI
k11=1;
k12_1=0.2;
k12_2=50;
k12_3=100;
dy(7) = k11-(k12_1+k12_2*y(8) +k12_3*y(1)*CycB)*y(7);
%SK
k13_1=0;
k13_2=1;
k14=1;
dy(8)= k13_1+k13_2*y(9)-k14*y(8);
%TF
k15_1=1.5;
k15_2=0.05;
k16_1=1;
k16_2=3;
J15=0.01;
J16=0.01;
dy(9)= (((k15_1*y(1)+k15_2*y(8))*(1-y(9)))/(J15+1-y(9)))-(((k16_1+k16_2*y(1)*CycB)*y(9))/(J16+y(9)));
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 3 Aug 2011
Note: Using ~ in that way is not available in 2008 and earlier.

Tsogerdene on 8 Aug 2011
Thank you for helping. I just asking that y(1)=y(1)*(1/2)? I would like to divide actual value of y(1) when some conditions are satisfied. I used y(1)=y(1)*(1/2), but nothing is changed.
Walter Roberson on 8 Aug 2011
Typo correction: I meant for "future iterations"