I need some manipulation with ODE45. Please help me!

2011 8 3
3 个回答
4 次查看（30 天）

0 个投票

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

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回答（3 个）

Arnaud Miege
Arnaud Miege 2011-8-3

2 个投票

Please format your code.
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
Tsogerdene 2011-8-3

0 个投票

Thank you for your answer. I wrote all of code.
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));
Mad=1;
dy(5)= ((k7*y(6)*(y(4)-y(5)))/(J7+y(4)-y(5)))-((k8*Mad*y(5))/(J8+y(5)))-k6*y(5);
%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)));
I defined all of them. But doesn't working. Please help me!

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Arnaud Miege
Arnaud Miege 2011-8-3
It works for me. What error message do you get?
Some comments:
* Your function is a function of t, y and flag but t and flag are not used. Consider replacing those with ~
* Your function defines k and y_1 as global variables, but those aren't used in the function. You can remove those two lines.
The only changes I made were therefore:
function dy=turshilt23(~,y,~)
global Global_CycB1;
global Global_CycB2;
% global k;
% global y_1;
I used 0.1 for the other two global variables Global_CycB1 and Global_CycB2. The solver solved OK.
Walter Roberson
Walter Roberson 2011-8-3
Note: Using ~ in that way is not available in 2008 and earlier.

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Tsogerdene
Tsogerdene 2011-8-8

0 个投票

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.

3 个评论
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Walter Roberson
Walter Roberson 2011-8-8
When you modify y(1) in your code, you are only modifying the _local_ y(1) . It is not possible to permanently change the y for figure iterations.
Tsogerdene
Tsogerdene 2011-8-8
Oh I see. Thank you very much!. I understood that.
Walter Roberson
Walter Roberson 2011-8-8
Typo correction: I meant for "future iterations"

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