optimization of delayed differential equations (dde)

2 views (last 30 days)
Hello,
I try to optimise (finding the parameter's value) in the system of dde, unfortunately I can't find any example how to do it. I used to optimization of ode using ode45 solver with lsqlin for instance. I will appreciate the help. The dde23 seems not to be working with lsqlin. I will appreciate any help.
  1 Comment
Torsten
Torsten on 9 May 2022
Edited: Torsten on 9 May 2022
Do you have a code for your model equations set up with dde23 ?
If yes, you should include it and tell which parameter(s) you are trying to optimize on the basis of which input data.

Sign in to comment.

Accepted Answer

Torsten
Torsten on 9 May 2022
Edited: Torsten on 9 May 2022
Make the best of it.
pTrue = [1 1 1 1 1 1 1 1 1];
time = 0:7;
Y = ...;
Am1=[1,1.1,1.4,0.7,0.8,1.6,2,1.5];
Bm1=[1.5,1.0,0.4,1.7,0.9,1.3,1.5,1.4];
Am=@(t)interp1(time,Am1,t)
Bm=@(t)interp1(time,Bm1,t)
p = lsqnonlin(@(p) Errors(p,time,Y,Am,Bm),0.8*pTrue)
function res = Errors(p,time,Y,Am,Bm)
lags=[1,2];
[T,SOL]=dde23(@(t,y,Z)ddefunEx(t,y,Z,p,Am,Bm), lags, [1,1,1,1], time);
res = Y-SOL;
res = res(:);
end
function dydt = ddefunEx(t,y,Z,p,Am,Bm)
ylag1 = Z(:,1);
ylag2 = Z(:,2);
E1p=y(1)
E2p=y(2)
A1=y(3)
B1=y(4)
n_A=p(1);%0.6;
n_B=p(2);%0.1;
KE1=p(3);%0.2;
KE2=p(4);%0.2;
Vp_A=p(5);%1.2;
Vp_B=p(6);%1.5;
alpha_p=p(7);%0.4;
kA=p(8);%0.4;
kB=p(9);%1.2;
%Auxiliary equations
ALPHA1=n_A*E1p*A1/(KE1*(1+A1))-A1;
ALPHA2=n_B*E2p*B1/(KE2*(1+B1))-B1;
dydt= [Vp_A*Am(t)-alpha_p*ylag1(1);
Vp_B*Bm(t)-alpha_p*ylag1(2);
kA*Am(t)-ylag2(3)-ALPHA1+ALPHA2;
kB*Bm(t)-ylag2(4)+ALPHA1-ALPHA2];
end
  4 Comments
Priya Verma
Priya Verma on 14 Mar 2024
how to plot graphs between lags and variables for dde ?...please reply ..
Torsten
Torsten on 14 Mar 2024
What do you mean by your question ? The line
[T,SOL]=dde23(@(t,y,Z)ddefunEx(t,y,Z,p,Am,Bm), lags, [1,1,1,1], time);
gives you a solution SOL at times T that you can plot. How do you think that the lags come into play ?

Sign in to comment.

More Answers (1)

Malgorzata Wieteska
Malgorzata Wieteska on 9 May 2022
% I'm embedding simplified version of the system. I run it the function ddedunEx to get simulated data (Y) and then try to optimise for the values of the parameters against obtained earlier data.
function dydt = ddefunEx(t,y,Z,p)
%global p
ylag1 = Z(:,1);
ylag2 = Z(:,2);
time=0:7;
E1p=y(1)
E2p=y(2)
A1=y(3)
B1=y(4)
Am1=[1,1.1,1.4,0.7,0.8,1.6,2,1.5];
Bm1=[1.5,1.0,0.4,1.7,0.9,1.3,1.5,1.4];
Am=interp1(time,Am1,t)
Bm=interp1(time,Bm1,t)
n_A=p(1);%0.6;
n_B=p(2);%0.1;
KE1=p(3);%0.2;
KE2=p(4);%0.2;
Vp_A=p(5);%1.2;
Vp_B=p(6);%1.5;
alpha_p=p(7);%0.4;
kA=p(8);%0.4;
kB=p(9);%1.2;
%Auxiliary equations
ALPHA1=n_A*E1p*A1/(KE1*(1+A1))-A1;
ALPHA2=n_B*E2p*B1/(KE2*(1+B1))-B1;
dydt= [Vp_A*Am-alpha_p*ylag1(1);
Vp_B*Bm-alpha_p*ylag1(2);
kA*Am-ylag2(3)-ALPHA1+ALPHA2;
kB*Bm-ylag2(4)+ALPHA1-ALPHA2;];
end
%lags=[1,2];
%tspan=0:7;
%sol = dde23(@ddefunEx, lags, [1,1,1,1], tspan);
%figure(2)
%plot(sol.x,sol.y)
%%%%%%%%%%%%% Obtaining data
%Y=sol.y+0.05*randn(size(sol.x));
function res=Errors(p,Y)
tspan=0:7;;
x0=[4;1];
lags=[1,2];
[T,X]=dde23(@ddefunEx, lags, [1,1,1,1], tspan);
res1=(X(:,1)-Y(:,1));
res2=(X(:,2)-Y(:,2));
res3=(X(:,3)-Y(:,3));
res4=(X(:,4)-Y(:,4));
res=abs(res1)+abs(res2) +abs(res3)+abs(res4);
%pTrue=[0.6;0.1;0.2;0.2];
%[pOpt,resnorm,res,exitflag,~,lambda,J]=...
% lsqnonlin(@(p) Errors(p,Y),0.8*pTrue);

Categories

Find more on Startup and Shutdown in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!