Is this a bug of ode function
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I find it's interesting if I change the initial value h0, it will affect the accuracy of "ode45"
close all
tspan=0:0.001:60;
h0=0;%!!!!!!!!!!!!!
[t,h]=ode45(@TankLvl, tspan, h0);
plot(t,h)
xlabel('t')
ylabel('y')
hold on
h2 = h0 + 0.005.*tspan - 0.05.*cos(tspan) + 0.05;
plot(tspan, h2,'r-')
function dh=TankLvl(t,h0)
A=2;
q_in=0.51+0.1*sin(t);
q_out=0.5;
dh=1/A*(q_in-q_out);
end
For example, if I change h0 as:
h0=10;
then the result will be:
It seems that the "ode" function can't deal with the initial value quite well. In order to deal with this issue, I can only set a pseudo initial value h0_pse=0, and use it as the input for "ode", then add the real initial value h0 to the computed h. Shown as following:
h0_pse=0;
h0=10;
[t,h]=ode45(@TankLvl, tspan, h0_pse);
h=h+h0;
1 个评论
madhan ravi
2019-3-6
Did you notice something when defining the function input argument?
回答（1 个）
SandeepKumar R
2019-3-6
编辑：SandeepKumar R
2019-3-6
It was a tolerance issue. Please define your funtions appropriately. Not a bug in MATLAB
close all
tspan=0:0.1:90;
h0=0;%!!!!!!!!!!!!!
opts = odeset('RelTol',1e-6,'AbsTol',1e-6);
[t,h]=ode45(@(t,z)TankLvl(t,z), [0,90], h0,opts);
plot(t,h,'*')
xlabel('t')
ylabel('y')
hold on
h2 = h0 + 0.005.*t - 0.05.*cos(t) + 0.05;
plot(t, h2,'r')
function dh=TankLvl(t,z)
A=2;
q_in=0.51+0.1*sin(t);
q_out=0.5;
dh=1/A*(q_in-q_out);
end
1 个评论
madhan ravi
2019-3-7
also did you notice the OP’s function argument while defining the function?
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2019-3-7
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