Time derivative of parameters within ODE solvers

조회 수: 4 (최근 30일)
Seyed Ali Baradaran Birjandi
Seyed Ali Baradaran Birjandi 2018년 11월 22일
편집: Torsten 2018년 11월 23일
I have an ODE which has a parameter whose 1st and 2nd order time derivatives are also included in the ODE:
function dy = ODE(t,y)
f1 = myfun(y(t));
dy = y + f1 + df1/dt + ddf1/dt^2;
end
Unfortunately, the function of analytical derivatives of myfun is not available. Therefore, df1 and ddf1 can be computed numerically, only. Given that the time step in Matlab ode solvers is not fixed, I wonder if there is a way to numerically compute df1 and ddf1.

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Torsten
Torsten 2018년 11월 23일
편집: Torsten 2018년 11월 23일
function dy = ODE(t,y)
dt = 1e-8;
fm = myfun(t-dt);
f = myfun(t);
fp = myfun(t+dt);
df = (fp - fm) / (2 * dt);
ddf = (fp - 2 * f + fm) / dt^2;
dy = y + f + df + ddf;
end
  댓글 수: 6
Seyed Ali Baradaran Birjandi
Seyed Ali Baradaran Birjandi 2018년 11월 23일
It depends on y only, i.e: y^2.
Torsten
Torsten 2018년 11월 23일
편집: Torsten 2018년 11월 23일
Let
z = f(y)
the value that "myfun" returns for argument y.
Then
dz/dt = df/dy * dy/dt
d^2z/dt^2 = d^2f/dy^2 * (dy/dt)^2 + df/dy * d^2y/dt^2
Inserting into your differential equation gives
dy/dt = y + f + df/dy * dy/dt + d^2f/dy^2 * (dy/dt)^2 + df/dy * d^2y/dt^2
or
df/dy * d^2y/dt^2 + (df/dy - 1) * dy/dt + d^2f/dy^2 * (dy/dt)^2 + y + f = 0
Now you can approximate df/dy and d^2f/dy^2 as I suggested above and solve the system (z1 = y, z2 = dy/dt)
z1' = z2
z2' = -((df/dy - 1) * z2 + d^2f/dy^2 * (z2)^2 + z1 + f)/(df/dy)
using ODE45, e.g.

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