Ich habe es geschafft, weil es nur wenige Programme im Netz gab, um die Translokationsgleichung zu lösen, und es gab kein methodisiertes, als ich es auf Japanisch und Englisch (oder einem fehlerhaften) suchte.
#https://github.com/christian512/SemiLagPy/blob/master/advection1D.ipynb
#Ich habe darauf hingewiesen.
import numpy as np
import matplotlib.pyplot as plt
class SemiLagrangian1D:
"""
SemiLagrangian scheme for periodic 1D problems.
"""
def __init__(self,x,y,u,dt):
"""
Initizalizes the SemiLagrangian object given initial configuration
x : x positions (must be equidistant)
y : variable at all x positions at time t
u : speed of advection
dt: time step
"""
# Check dimensions
assert x.shape[0] == y.shape[0], "x and y have different length"
# Get the x stepsize
dx = x[1] - x[0]
# check if x is equidistant
assert x[-1] == x[0] + dx*(x.shape[0]-1), "x array not equidistant"
# Set class variables
self.x = x
self.y = y
self.u = u
self.dt = dt
self.dx = dx
def cubic_evolve(self,nt=1):
"""
Propagates the variable using a cubic interpolation scheme.
nt: number of time stepsize
"""
# loop through time steps
for l in range(nt):
# temporary array
y_temp = np.zeros(self.y.shape[0])
# loop through array
for i in range(self.y.shape[0]):
# idx left to departure point
x_dep = self.x[i]-self.u*self.dt
j = int(np.floor(x_dep/self.dx))
# alpha
a = (self.x[i]-self.u*self.dt - j*self.dx)/self.dx
# calculate next time step
f = lambda x: x % self.y.shape[0] if x >= self.y.shape[0] else x
y_temp[i] = - a * (1-a)*(2-a)/6 * self.y[f(j-1)]
y_temp[i] += (1-a**2)*(2-a)/2 * self.y[f(j)]
y_temp[i] += a*(1+a)*(2-a)/2 * self.y[f(j+1)]
y_temp[i] -= a*(1-a**2)/6 * self.y[f(j+2)]
self.y = np.copy(y_temp)
return self.y
def linear_interporation(self,nt=1):
for l in range(nt):
# temporary array
y_tmp = np.zeros(self.y.shape[0])
for i in range(1,NX-1):
x_tmp = self.x[0]+i*self.dx
x_tmp -= self.u*self.dt
ib = int((x_tmp-self.x[0])/self.dx)
if ib < 0: ib = 0
if ib >= len(self.x): ib = NX-2
distance = (x_tmp - ib*self.dx)/DX
y_tmp[i] = (1-distance)*self.y[ib]+distance*self.y[ib+1]
self.y = np.copy(y_tmp)
return self.y
L = 5 # Domain length
NX = 1000 # Number of grid points
X = np.linspace(0,L,num=NX) # Array of grid points
DX = X[1] - X[0] # spatial grid point distance
DT = 0.01 # time step size
NT = 1000 # number of time steps
CA = (1 + 0.02*np.sqrt(130)) # advection speed
print('Courant= ' + str(CA*DT/DX))
# Sine function
F_ini2 = np.zeros(X.shape[0])
F_ini2[0:int(X.shape[0]/5)] = np.sin(np.pi*X[0:int(X.shape[0]/5)])
# Plotting
# plt.plot(X,F_ini2,label='Sine function')
# plt.legend()
# plt.show()
# Create array for storing evolution of F(x,t)
F_arr = np.zeros([NT,X.shape[0]])
F_arr[0,:] = np.copy(F_ini2)
# Create SemiLagrangian object
sl = SemiLagrangian1D(X,F_arr[0,:],CA,DT)
# Calculate the variable for all time steps using semi-lagrangian
for i in range(1,NT):
F_arr[i,:] = sl.cubic_evolve(nt=1)
#F_arr[i,:] = sl.linear_interporation(nt=i)
#ans = sl.linear_interporation(nt=300)
# Plot initial vs. evolution
# print(F_arr[300,:])
print("f(0.5,t_0) = {}".format(F_arr[0,int(X.shape[0]/10)]))
print("f(x1,t1) = {}".format(F_arr[300,int(X.shape[0]/10) + int(-CA*X.shape[0]/5)]))
plt.plot(X,F_arr[0,:],label='Initial')
plt.plot(X,F_arr[300,:],label='Evolved')
plt.legend()
plt.show()
Recommended Posts