Matlab ode solver - putting condition in function

I use ode23s solver

and the results show that variables oscillate around zero plus and minus.

I want to fix the variables to zero when they first go under zero.

ode23s solver can't use odeset 'NonNegative' option,

so we should have to put some condition in the function.

Please give me help.

Answer 1

[Ameer Hamza]

The easiest option might be to do this after getting the solution from ode23s. For example

[t, y] = ode23s(..)

idx = find(y<0, 1);
y(idx:end) = 0

This assumes that your ODE is first-order and y is a column vector.

[kkh1897]

Thank you for your response.

But I should have to reflect every variables in each steps.

for example,

funtion dxdt = ode(t,x)
    dxdt = zeros(4,1)
    dxdt(1) = -x(1)*exp(a/x(4));
    dxdt(2) = -x(2)*exp(b/x(4));
    dxdt(3) = -x(3)*exp(c/x(4));
    dxdt(4) =  m1*dxdt(1) + m2*dxdt(2) + m3*dxdt(3);
end

(here, a,b,c,m1,m2,m3 are constant)

I want to put condition which makes 'x' go to zero at the moment it becomes under zero.

So I put condition inbetween, but it doesn't work.

funtion dxdt = ode(t,x)
    dxdt = zeros(4,1)
    dxdt(1) = -x(1)*exp(a/x(4));
    if x(1) < 0
        x(1) = 0;
    end
    dxdt(2) = -x(2)*exp(b/x(4));
    dxdt(3) = -x(3)*exp(c/x(4));
    dxdt(4) =  m1*dxdt(1) + m2*dxdt(2) + m3*dxdt(3);
end

[Ameer Hamza]

Can you show the value of variables and initial conditions used to run ode45.

[kkh1897]

I sent you a email. Thank you

[Ameer Hamza]

This is the code with ode45(). However, due to some reason, the result is different as compared to dsolve(). I am not sure why.

ode = @(t, x) [x(2);
               -sign(x(1)+x(2))];
IC = [1; 0];
tspan = [0 5];
ode(ode, tspan, IC)