如何在微分方程 (SciPy) 中使用 if 语句？

Question

我正在尝试用 Python 求解微分方程。在这两个系统微分方程中，如果第一个变量 ( v) 的值大于阈值 (30)，则应将其重置为另一个值 (-65)。下面我把我的代码。问题是第一个变量的值在达到 30 后保持不变并且不会重置为 -65。这些方程描述了单个神经元的动力学。方程式取自本网站和此PDF 文件。

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.ticker import FormatStrFormatter
from scipy.integrate import odeint
plt.close('all')

a = 0.02
b = 0.2
c = -65
d = 8
i = 0

p = [a,b,c,d,i]

def fun(u,tspan,*p):
    du = [0,0]
    if u[0] < 30: #Checking if the threshold has been reached
        du[0] = (0.04*u[0] + 5)*u[0] + 150 - u[1] - p[4]
        du[1] = p[0]*(p[1]*u[0]-u[1])
    else:
        u[0] = p[2] #reset to -65    
        u[1] = u[1] + p[3] 

    return du


p = tuple(p)

y0 = [0,0]

tspan = np.linspace(0,100,1000)
sol = odeint(fun, y0, tspan, args=p)

 fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)         
plt.plot(tspan,sol[:,0],'k',linewidth = 5)
plt.plot(tspan,sol[:,1],'r',linewidth = 5)
myleg = plt.legend(['v','u'],\
    loc='upper right',prop = {'size':28,'weight':'bold'}, bbox_to_anchor=(1,0.9))

解决方案如下：

这是正确的解决方案Julia，这里u1代表v：

这是Julia代码：

using DifferentialEquations
using Plots

a = 0.02
b = 0.2
c = -65
d = 8
i = 0

p = [a,b,c,d,i]

function fun(du,u,p,t)
    if u[1] <30
        du[1] = (0.04*u[1] + 5)*u[1] + 150 - u[2] - p[5]
        du[2] = p[1]*(p[2]*u[1]-u[2])
    else
        u[1] = p[3]
        u[2] = u[2] + p[4]
    end
end

u0 = [0.0;0.0]
tspan = (0.0,100)
prob = ODEProblem(fun,u0,tspan,p)
tic()
sol = solve(prob,reltol = 1e-8)
toc()

plot(sol)

Answer 1

推荐方案

这使用事件并在每个不连续性之后单独积分。

import matplotlib.pyplot as plt
import numpy as np
from scipy.integrate import solve_ivp

a = 0.02
b = 0.2
c = -65
d = 8
i = 0

p = [a,b,c,d,i]

# Define event function and make it a terminal event
def event(t, u):
    return u[0] - 30
event.terminal = True

# Define differential equation
def fun(t, u):
    du = [(0.04*u[0] + 5)*u[0] + 150 - u[1] - p[4],
          p[0]*(p[1]*u[0]-u[1])]
    return du

u = [0,0]

ts = []
ys = []
t = 0
tend = 100
while True:
    sol = solve_ivp(fun, (t, tend), u, events=event)
    ts.append(sol.t)
    ys.append(sol.y)
    if sol.status == 1: # Event was hit
        # New start time for integration
        t = sol.t[-1]
        # Reset initial state
        u = sol.y[:, -1].copy()
        u[0] = p[2] #reset to -65    
        u[1] = u[1] + p[3]
    else:
        break

fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
# We have to stitch together the separate simulation results for plotting
ax.plot(np.concatenate(ts), np.concatenate(ys, axis=1).T)
myleg = plt.legend(['v','u'])

最小变化“解决方案”

看来您的方法对solve_ivp.

警告我认为在 Julia 和 中solve_ivp，处理这种事情的正确方法是使用事件。我相信下面的方法依赖于一个实现细节，即传递给函数的状态向量与内部状态向量是相同的对象，这允许我们就地修改它。如果它是副本，这种方法将不起作用。此外，在这种方法中不能保证求解器采取足够小的步长，以致于会踩到达到极限的正确点。使用事件将使这更正确，并且可以推广到其他微分方程，这些方程可能在不连续之前具有较低的梯度。

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.ticker import FormatStrFormatter
from scipy.integrate import solve_ivp
plt.close('all')

a = 0.02
b = 0.2
c = -65
d = 8
i = 0

p = [a,b,c,d,i]

def fun(t, u):
    du = [0,0]
    if u[0] < 30: #Checking if the threshold has been reached
        du[0] = (0.04*u[0] + 5)*u[0] + 150 - u[1] - p[4]
        du[1] = p[0]*(p[1]*u[0]-u[1])
    else:
        u[0] = p[2] #reset to -65    
        u[1] = u[1] + p[3] 

    return du

y0 = [0,0]

tspan = (0,100)
sol = solve_ivp(fun, tspan, y0)

fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)         
plt.plot(sol.t,sol.y[0, :],'k',linewidth = 5)
plt.plot(sol.t,sol.y[1, :],'r',linewidth = 5)
myleg = plt.legend(['v','u'],loc='upper right',prop = {'size':28,'weight':'bold'}, bbox_to_anchor=(1,0.9))

结果