我正在学习 R 来解决二阶微分方程(可能使用deSolve包)。我将它写成python,将其写成两个一阶微分方程,并在下面给出
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
def fun(X, t):
y , dy , z = X
M = np.sqrt (1./3. * (1/2. * dy **2 + 1./2. * y **2))
dz = (M*z) # dz/dt
ddy = -3.* M * dy - y # ddy/dt
return [dy ,ddy ,dz]
y0 = 1
dy0 = -0.1
z0 = 1.
X0 = [y0, dy0, z0]
M0 = np.sqrt (1./3. * (1./2. * dy0 **2. + 1./2.* y0 **2))
t = np.linspace(0., 100., 10001.) # time spacing
sol = odeint(fun, X0, t)
y = sol[:, 0]
dy = sol[:, 1]
z = sol[:, 2]
M = np.sqrt (1./3. * (1./2. * dy**2. + 1./2.* y **2))
#Graph plotting
plt.figure()
plt.plot(t, y)
plt.plot(t, z)
plt.plot(t, M)
plt.grid()
plt.show()
Python 很容易解决这个问题,但是对于另一个类似但复杂的问题,python显示错误。我也在python中尝试过ode(vode / bdf),但问题仍然存在。现在,我想查看 R 如何解决问题。所以,如果有人请给我一个如何在 R 中解决此问题的示例(基本上是代码翻译!),我将非常感激,这样我就可以尝试其他一个在 R 中,并且还学习了一些 R (我知道这可能不是学习a的理想方式式语言)。
我知道这个问题可能没什么建设性价值,但我只是 R 的新手,所以请耐心等待我!
答案 0 :(得分:2)
这应该是Python代码到R
的翻译library(deSolve)
deriv <- function(t, state, parameters){
with(as.list(c(state, parameters)),{
M <- sqrt(1/3 * (1/2 * dy^2 + 1/2 * y^2))
dz <- M*z # dz/dt
ddy <- -3* M * dy - y # ddy/dt
list(c(dy, ddy, dz))
})
}
state <- c(y = 1,
dy = -0.1,
z = 1)
times <- seq(0, 100, length.out = 10001)
sol <- ode(func = deriv, y = state, times = times, parms = NULL)
y <- sol[, "y"]
dy <- sol[, "dy"]
z <- sol[, "z"]
M <- sqrt(1/3 * (1/2 * dy^2 + 1/2* y^2))
plot(times, z, col = "red", ylim = c(-1, 18), type = "l")
lines(times, y, col = "blue")
lines(times, M, col = "green")
grid()
使用此代码可以更快捷地直接计算 R 中的M
:
library(deSolve)
deriv <- function(t, state, parameters){
with(as.list(c(state, parameters)),{
M <- sqrt(1/3 * (1/2 * dy^2 + 1/2 * y^2))
dz <- M*z # dz/dt
ddy <- -3* M * dy - y # ddy/dt
list(c(dy, ddy, dz), M = M)
})
}
state <- c(y = 1,
dy = -0.1,
z = 1)
times <- seq(0, 100, length.out = 10001)
sol <- ode(func = deriv, y = state, times = times, parms = NULL)
## save to file
write.csv2(sol,file = "path_to_folder/R_ODE.csv")
## plot
matplot(sol[,"time"], sol[,c("y", "z", "M")], type = "l")
grid()