How can I use ODE45 continuously?
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Heejun Lee
el 28 de Jul. de 2021
This is the code I initially made.
[T,Y] = ode45(@hw,[0 100],[0.01 10]);
And this is function hw
//
function dy = hw(t,y)
D=0.01;
um=0.2;
Ks=0.05;
X0=0.01;
Sf=10;
dy = zeros(2,1);
dy(1) = -D*y(1) + (um*y(2))./(Ks+y(2))*y(1);
dy(2) = Sf * D - D*y(2) - (um*y(1)*y(2))./(0.4*(Ks+y(2)));
//
For this time I want to use D = 0.01 for t=0~50, and D =0.015 for t=50~100.
And value of X0, Sf for t=50~100 should be the value of X(which is y(1)), S(which is y(2)) at t=50.
How can I make this work?
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Respuesta aceptada
Chunru
el 28 de Jul. de 2021
Editada: Chunru
el 28 de Jul. de 2021
Just change D depending on t. Run the solver over the whole time period.
[Update 2: Change the ode options, ie. smaller time step. Three methods gives similar results. It seems that default time step is not good enough for this problem. The branching seems cause no problem as well.]
opts = odeset('MaxStep', 0.001);
[T,Y] = ode45(@hw1,[0 100],[0.01 10], opts);
h(1)=subplot(131); plot(T, Y);
title('D=0.01');
[T,Y] = ode45(@hw,[0 100],[0.01 10], opts);
h(2)=subplot(132); plot(T, Y);
title('D: Branching');
[T1,Y1] = ode45(@hw1,[0 50],[0.01 10], opts);
[T2 Y2] = ode45(@hw2,[50+eps 100],Y1(end,:), opts);
T=[T1;T2]; Y=[Y1; Y2];
h(3)=subplot(133); plot(T, Y)
title('Two ODEs')
%linkaxes(h, 'xy')
function dy = hw(t,y)
%D=0.01;
if t<=50
D = 0.01;
else
D = 0.015;
end
um=0.2;
Ks=0.05;
X0=0.01;
Sf=10;
dy = zeros(2,1);
dy(1) = -D*y(1) + (um*y(2))./(Ks+y(2))*y(1);
dy(2) = Sf * D - D*y(2) - (um*y(1)*y(2))./(0.4*(Ks+y(2)));
end
function dy = hw1(t,y)
D = 0.01;
um=0.2;
Ks=0.05;
X0=0.01;
Sf=10;
dy = zeros(2,1);
dy(1) = -D*y(1) + (um*y(2))./(Ks+y(2))*y(1);
dy(2) = Sf * D - D*y(2) - (um*y(1)*y(2))./(0.4*(Ks+y(2)));
end
function dy = hw2(t,y)
D = 0.015;
um=0.2;
Ks=0.05;
X0=0.01;
Sf=10;
dy = zeros(2,1);
dy(1) = -D*y(1) + (um*y(2))./(Ks+y(2))*y(1);
dy(2) = Sf * D - D*y(2) - (um*y(1)*y(2))./(0.4*(Ks+y(2)));
end
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