Python-Scipy : ode module : issue enabling the step option of the solver

Question

我想在调用求解器时存储求解器本身采取的不同积分步骤:

solver1.integrate(t_end)

所以我做了一个 while 循环并启用了将其值设置为 True 的 step 选项:

while solver1.successful() and solver1.t < t0+dt:
    solver1.integrate(t_end,step=True)
    time.append(solver1.t)

然后我绘制 y，积分的结果，我的问题来了。我在定位区域出现不稳定:

我认为这是因为循环或类似的原因，所以我检查了删除 step 的结果:

while solver1.successful() and solver1.t < t0+dt:
    solver1.integrate(t_end)

令人惊讶的是......我得到了正确的结果:

这是一个非常奇怪的情况......如果你们中的任何人可以帮助我解决这个问题，我将不胜感激。

编辑:

要设置求解器，我会这样做:

solver1 = ode(y_dot,jac).set_integrator('vode',with_jacobian=True)
solver1.set_initial_value(x0,t0)

然后我使用 .append() 存储结果

Answer 1

当您设置 step=True 时你间接地给了 vode._integrator.runner (一个 Fortran 子程序)使用 itask=2 的指令, 默认为 itask=1 .您可以获得有关此的更多详细信息 runner做:

r._integrator.runner?

在 SciPy 0.12.0 文档中，您不会找到关于不同 itask=1 的解释。或 itask=2 , but you can find it here :

ITASK  = An index specifying the task to be performed.
!          Input only. ITASK has the following values and meanings.
!          1  means normal computation of output values of y(t) at
!             t = TOUT(by overshooting and interpolating).
!          2  means take one step only and return.
!          3  means stop at the first internal mesh point at or
!             beyond t = TOUT and return.
!          4  means normal computation of output values of y(t) at
!             t = TOUT but without overshooting t = TCRIT.
!             TCRIT must be input as RUSER(1). TCRIT may be equal to
!             or beyond TOUT, but not behind it in the direction of
!             integration. This option is useful if the problem
!             has a singularity at or beyond t = TCRIT.
!          5  means take one step, without passing TCRIT, and return.
!             TCRIT must be input as RUSER(1).