- As I see it, the problem is at t = 0, y(1,2) = 0
ODE solver - division by zero at time boundaries
7 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Bastian Andersen
am 12 Dez. 2019
Beantwortet: darova
am 16 Dez. 2019
Hi
I have a problem when solving my system of ODEs. I simplified it to what I find essential for the problem.
As I see it, the problem is at t = 0, y(1,2) = 0 , whereas dy(2:end,2) = NaN AND at t = p.t_empty, y(1,1) = 0, whereas dy(2:end,1) = NaN
This is however the time interval I am interested in the solution for. It is essentially a system of two tanks. A reaction occurs in tank 1 and the matter is transferred to tank 2 meanwhile.
Do you have any suggestions to how I can solve this?
p.Q = 40;
p.Q_R = 150;
p.V = 20;
p.n = 101;
p.t_empty = p.V/p.Q;
y0 = zeros(p.n*2+2,1);
y0(1:2) = [p.V, 1];
options = odeset('RelTol',1e-12,'AbsTol',1e-12);
[t,y] = ode45('ODE_tank',[0,p.t_empty],y0,options,p);
figure(1)
plot((0:p.n-1),y(end,2:p.n+1))
figure(2)
plot((0:p.n-1),y(end,p.n+3:end))
function dy = ODE_tank(t,y,options,p)
y = reshape(y,[],2);
dy = zeros(size(y));
dy(1,1) = -p.Q;
dy(2,1) = -p.Q_R/y(1,1)*y(2,1);
dy(3:p.n,1) = p.Q_R/y(1,1)*(y(2:p.n-1,1) - y(3:p.n,1));
dy(p.n+1,1) = p.Q_R/y(1,1)*y(p.n,1);
dy(1,2) = p.Q;
dy(2:end,2) = p.Q/y(1,2)*(y(2:end,1)-y(2:end,2));
dy = reshape(dy,[],1);
end
2 Kommentare
Akzeptierte Antwort
darova
am 16 Dez. 2019
- As I see it, the problem is at t = 0, y(1,2) = 0
Can you replace 0 with 1e-3?
0 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Ordinary Differential Equations finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!