optimization of delayed differential equations (dde)

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Malgorzata Wieteska
评论： Torsten 2024-3-14
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 个评论
Torsten
Torsten 2022-5-9
编辑：Torsten 2022-5-9
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.

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采纳的回答

Torsten
Torsten 2022-5-9
编辑：Torsten 2022-5-9
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
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显示 2更早的评论
Priya Verma
Priya Verma 2024-3-14
how to plot graphs between lags and variables for dde ?...please reply ..
Torsten
Torsten 2024-3-14
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 ?

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更多回答（1 个）

Malgorzata Wieteska
% 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);

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