Is this a bug of ode function

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;

[madhan ravi]

Did you notice something when defining the function input argument?

Answer 1

[SandeepKumar R]

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

[madhan ravi]

also did you notice the OP’s function argument while defining the function?