如何求解 R 中的二阶微分方程?
How do I solve a second order differential equation in R?
我正在学习 R 来求解二阶微分方程(可能使用 deSolve 包)。我在 python 中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,我想检查一下 R 如何解决这个问题。所以,如果有人请给我提供一个示例(基本上是代码翻译!),说明如何在 R 中解决此问题,我将不胜感激我可以尝试 R 中的另一个并学习一些 R(我知道这可能不是学习语言的理想方式)。
我明白这个问题可能没有什么建设性价值,但我只是R的新手,还请多多包涵!
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()
在R中直接计算M
有一个更快的方法,代码如下:
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()
我正在学习 R 来求解二阶微分方程(可能使用 deSolve 包)。我在 python 中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,我想检查一下 R 如何解决这个问题。所以,如果有人请给我提供一个示例(基本上是代码翻译!),说明如何在 R 中解决此问题,我将不胜感激我可以尝试 R 中的另一个并学习一些 R(我知道这可能不是学习语言的理想方式)。
我明白这个问题可能没有什么建设性价值,但我只是R的新手,还请多多包涵!
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()
在R中直接计算M
有一个更快的方法,代码如下:
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()