Differential equation in matlab Announcing the arrival of Valued Associate #679: Cesar...

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Differential equation in matlab



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Stochastic predator-preymatlab differential equationSolving 2nd order ODE with 2 independent parameters(over finite intervals), with bounds on solutionAnalytic solution to the one-compartment modelMatlab differential equation verificationIncluding a time delay term for a differential equationFinding parameters using two data sets for the same ODE systemmass balance from a system of nonlinear differential equationsDeciding the parameter value in a ODE systemHow can I solve/integerate the One-To-One kinetic binding model system of differential equations analytically?












1












$begingroup$


A pathogen population ($P(t)$) under the effect of a drug($A(t)$) is modelled as



${dPover dt}=rP-{E_{max}A^nover C_{50}^n+A^n}P$ and $A(t)$ is modelled as



$A(t)={Dover V}(e^{-k(t-t_0)}-e^{-k_a(t-t_0)})$.



From literature I know the values of $D,V,t_0$(lag time), $k,k_a$. So, $A(t)$ can be obtained at the respective time points the ODE is run.
$r,E_{max},n,C_{50}$ are also known.



I want to solve the ODE ${dPover dt}$ using ode45 of Matlab. (Without obtaining the analytical solution of $P(t)$ as this is only a simplified system and the original system cannot be solved analytically).



Can someone please let me know if the below code is right in solving $P(t)$. I am not sure about the way $A(t)$ is included into the ODE equation.



y0=10;
[t,y]=ode45(@Model,1:200,y0);

function s= Model(t,y)
r=0.4;
Emax=50;
n=1;
c_50=0.1;
k=0.25;
D=40;
V=6.6;
t0=1;
k_a=0.5;
s=zeros(1,1);
A=(D/V)*(exp(-k*(t-t0))-exp(-k_a*(t-t0)));
s(1)=r*y(1)-((Emax*(A^n))/((c_50^n)+(A^n)))*y(1);
end









share|cite|improve this question









$endgroup$












  • $begingroup$
    Matlab has a very comprehensive help with several examples. I urge you to check it.
    $endgroup$
    – Jon
    Mar 25 at 11:06
















1












$begingroup$


A pathogen population ($P(t)$) under the effect of a drug($A(t)$) is modelled as



${dPover dt}=rP-{E_{max}A^nover C_{50}^n+A^n}P$ and $A(t)$ is modelled as



$A(t)={Dover V}(e^{-k(t-t_0)}-e^{-k_a(t-t_0)})$.



From literature I know the values of $D,V,t_0$(lag time), $k,k_a$. So, $A(t)$ can be obtained at the respective time points the ODE is run.
$r,E_{max},n,C_{50}$ are also known.



I want to solve the ODE ${dPover dt}$ using ode45 of Matlab. (Without obtaining the analytical solution of $P(t)$ as this is only a simplified system and the original system cannot be solved analytically).



Can someone please let me know if the below code is right in solving $P(t)$. I am not sure about the way $A(t)$ is included into the ODE equation.



y0=10;
[t,y]=ode45(@Model,1:200,y0);

function s= Model(t,y)
r=0.4;
Emax=50;
n=1;
c_50=0.1;
k=0.25;
D=40;
V=6.6;
t0=1;
k_a=0.5;
s=zeros(1,1);
A=(D/V)*(exp(-k*(t-t0))-exp(-k_a*(t-t0)));
s(1)=r*y(1)-((Emax*(A^n))/((c_50^n)+(A^n)))*y(1);
end









share|cite|improve this question









$endgroup$












  • $begingroup$
    Matlab has a very comprehensive help with several examples. I urge you to check it.
    $endgroup$
    – Jon
    Mar 25 at 11:06














1












1








1





$begingroup$


A pathogen population ($P(t)$) under the effect of a drug($A(t)$) is modelled as



${dPover dt}=rP-{E_{max}A^nover C_{50}^n+A^n}P$ and $A(t)$ is modelled as



$A(t)={Dover V}(e^{-k(t-t_0)}-e^{-k_a(t-t_0)})$.



From literature I know the values of $D,V,t_0$(lag time), $k,k_a$. So, $A(t)$ can be obtained at the respective time points the ODE is run.
$r,E_{max},n,C_{50}$ are also known.



I want to solve the ODE ${dPover dt}$ using ode45 of Matlab. (Without obtaining the analytical solution of $P(t)$ as this is only a simplified system and the original system cannot be solved analytically).



Can someone please let me know if the below code is right in solving $P(t)$. I am not sure about the way $A(t)$ is included into the ODE equation.



y0=10;
[t,y]=ode45(@Model,1:200,y0);

function s= Model(t,y)
r=0.4;
Emax=50;
n=1;
c_50=0.1;
k=0.25;
D=40;
V=6.6;
t0=1;
k_a=0.5;
s=zeros(1,1);
A=(D/V)*(exp(-k*(t-t0))-exp(-k_a*(t-t0)));
s(1)=r*y(1)-((Emax*(A^n))/((c_50^n)+(A^n)))*y(1);
end









share|cite|improve this question









$endgroup$




A pathogen population ($P(t)$) under the effect of a drug($A(t)$) is modelled as



${dPover dt}=rP-{E_{max}A^nover C_{50}^n+A^n}P$ and $A(t)$ is modelled as



$A(t)={Dover V}(e^{-k(t-t_0)}-e^{-k_a(t-t_0)})$.



From literature I know the values of $D,V,t_0$(lag time), $k,k_a$. So, $A(t)$ can be obtained at the respective time points the ODE is run.
$r,E_{max},n,C_{50}$ are also known.



I want to solve the ODE ${dPover dt}$ using ode45 of Matlab. (Without obtaining the analytical solution of $P(t)$ as this is only a simplified system and the original system cannot be solved analytically).



Can someone please let me know if the below code is right in solving $P(t)$. I am not sure about the way $A(t)$ is included into the ODE equation.



y0=10;
[t,y]=ode45(@Model,1:200,y0);

function s= Model(t,y)
r=0.4;
Emax=50;
n=1;
c_50=0.1;
k=0.25;
D=40;
V=6.6;
t0=1;
k_a=0.5;
s=zeros(1,1);
A=(D/V)*(exp(-k*(t-t0))-exp(-k_a*(t-t0)));
s(1)=r*y(1)-((Emax*(A^n))/((c_50^n)+(A^n)))*y(1);
end






ordinary-differential-equations matlab mathematical-modeling






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asked Mar 25 at 10:17









sam_roxsam_rox

52321021




52321021












  • $begingroup$
    Matlab has a very comprehensive help with several examples. I urge you to check it.
    $endgroup$
    – Jon
    Mar 25 at 11:06


















  • $begingroup$
    Matlab has a very comprehensive help with several examples. I urge you to check it.
    $endgroup$
    – Jon
    Mar 25 at 11:06
















$begingroup$
Matlab has a very comprehensive help with several examples. I urge you to check it.
$endgroup$
– Jon
Mar 25 at 11:06




$begingroup$
Matlab has a very comprehensive help with several examples. I urge you to check it.
$endgroup$
– Jon
Mar 25 at 11:06










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your time vector likely has too large steps for solver. Try replace



1:200


with



linspace(0,0.125,256)


enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for the answer.
    $endgroup$
    – sam_rox
    Mar 25 at 13:07



















0












$begingroup$

I copied your code to Octave (it runs most matlab code) and the results seem wrong, as the solution "explodes". This could be a precision in ode45. I attach the Wolfram Mathematica solution to the same problem. The solution decays to zero almost immediately.



enter image description here






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    It works just fine in Matlab if you just don't have too large delta t steps. 1:200 will give delta t of 1 everywhere which may be too large for many ODE solvers.
    $endgroup$
    – mathreadler
    Mar 25 at 12:00














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Your time vector likely has too large steps for solver. Try replace



1:200


with



linspace(0,0.125,256)


enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for the answer.
    $endgroup$
    – sam_rox
    Mar 25 at 13:07
















1












$begingroup$

Your time vector likely has too large steps for solver. Try replace



1:200


with



linspace(0,0.125,256)


enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for the answer.
    $endgroup$
    – sam_rox
    Mar 25 at 13:07














1












1








1





$begingroup$

Your time vector likely has too large steps for solver. Try replace



1:200


with



linspace(0,0.125,256)


enter image description here






share|cite|improve this answer









$endgroup$



Your time vector likely has too large steps for solver. Try replace



1:200


with



linspace(0,0.125,256)


enter image description here







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 25 at 11:57









mathreadlermathreadler

15.6k72263




15.6k72263












  • $begingroup$
    thank you for the answer.
    $endgroup$
    – sam_rox
    Mar 25 at 13:07


















  • $begingroup$
    thank you for the answer.
    $endgroup$
    – sam_rox
    Mar 25 at 13:07
















$begingroup$
thank you for the answer.
$endgroup$
– sam_rox
Mar 25 at 13:07




$begingroup$
thank you for the answer.
$endgroup$
– sam_rox
Mar 25 at 13:07











0












$begingroup$

I copied your code to Octave (it runs most matlab code) and the results seem wrong, as the solution "explodes". This could be a precision in ode45. I attach the Wolfram Mathematica solution to the same problem. The solution decays to zero almost immediately.



enter image description here






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    It works just fine in Matlab if you just don't have too large delta t steps. 1:200 will give delta t of 1 everywhere which may be too large for many ODE solvers.
    $endgroup$
    – mathreadler
    Mar 25 at 12:00


















0












$begingroup$

I copied your code to Octave (it runs most matlab code) and the results seem wrong, as the solution "explodes". This could be a precision in ode45. I attach the Wolfram Mathematica solution to the same problem. The solution decays to zero almost immediately.



enter image description here






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    It works just fine in Matlab if you just don't have too large delta t steps. 1:200 will give delta t of 1 everywhere which may be too large for many ODE solvers.
    $endgroup$
    – mathreadler
    Mar 25 at 12:00
















0












0








0





$begingroup$

I copied your code to Octave (it runs most matlab code) and the results seem wrong, as the solution "explodes". This could be a precision in ode45. I attach the Wolfram Mathematica solution to the same problem. The solution decays to zero almost immediately.



enter image description here






share|cite|improve this answer









$endgroup$



I copied your code to Octave (it runs most matlab code) and the results seem wrong, as the solution "explodes". This could be a precision in ode45. I attach the Wolfram Mathematica solution to the same problem. The solution decays to zero almost immediately.



enter image description here







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 25 at 11:12









PierreCarrePierreCarre

2,253215




2,253215








  • 1




    $begingroup$
    It works just fine in Matlab if you just don't have too large delta t steps. 1:200 will give delta t of 1 everywhere which may be too large for many ODE solvers.
    $endgroup$
    – mathreadler
    Mar 25 at 12:00
















  • 1




    $begingroup$
    It works just fine in Matlab if you just don't have too large delta t steps. 1:200 will give delta t of 1 everywhere which may be too large for many ODE solvers.
    $endgroup$
    – mathreadler
    Mar 25 at 12:00










1




1




$begingroup$
It works just fine in Matlab if you just don't have too large delta t steps. 1:200 will give delta t of 1 everywhere which may be too large for many ODE solvers.
$endgroup$
– mathreadler
Mar 25 at 12:00






$begingroup$
It works just fine in Matlab if you just don't have too large delta t steps. 1:200 will give delta t of 1 everywhere which may be too large for many ODE solvers.
$endgroup$
– mathreadler
Mar 25 at 12:00




















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