绘制扩散方程时的量纲问题

Question

我重新post回答了这个问题，因为我在第一个 post 中犯了一个错误。我正在研究扩散方程。但是，当我尝试绘制它时，我收到一条与维度相等性相关的错误消息

x and y must have same first dimension, but have shapes (401,) and (2001, 401)

x 轴应给出粒子的位置，而 y 轴应给出给定位置的粒子数。我不确定在这种情况下如何使尺寸相等。

我的代码：

import numpy as np
import matplotlib.pyplot as plt

D = 0.5
M = 2000
L = 200
tmax = 2000
dx = 1
dt = 1
s = D*dt/dx**2 #diffusion number
nx = int(2*L/dx)
nt = int(tmax/dt)
nx = nx + 1
nt = nt + 1
x = np.linspace(-L,L,nx)
t = np.linspace(0,tmax,not)

def InitCond(x, M):
    if x == -50:
        c0 = M
    elif x == 50:
        c0 = M
    else:
        c0 = 0
    return c0

def LeftEnd(t):
    return 0

def RightEnd(t):
    return 0

def DiffusionSolver(x,t,s,L,c0,left,right):
    nt = np.size(t)
    nx = np.size(x)
    u = np.zeros((nt,nx))
    # Use the initial condition to define values of 'u' at the 
#zeroeth timestep:
    for j in range(nx):
        u[0,j] = c0(x[j], M)
    # Use the boundary condition to fix the string on its ends:
    u[:,-L] = left(t)
    u[:,L] = right(t)
# Entering the double loop to calculate the solution at every space 
#point at every timestep:
    for k in range(0,nt-1):
        for j in range(1,nx-1):
            u[k+1,j] = s*(u[k,j+1] + u[k,j-1]) + (1 - 2*s)*u[k,j]
    return u
u_num = DiffusionSolver(x,t,s,L,InitCond,LeftEnd,RightEnd)
plt.plot(x, u_num)

Answer 1

您只需转换 u_num 数组，使其成为正确的形状。像这样：

import numpy as np
import matplotlib.pyplot as plt

D = 0.5
M = 2000
L = 200
tmax = 2000
dx = 1
dt = 1
s = D*dt/dx**2 #diffusion number
nx = int(2*L/dx)
nt = int(tmax/dt)
nx = nx + 1
nt = nt + 1
x = np.linspace(-L,L,nx)
t = np.linspace(0,tmax,nt)

def InitCond(x, M):
    if x == -50:
        c0 = M
    elif x == 50:
        c0 = M
    else:
        c0 = 0
    return c0

def LeftEnd(t):
    return 0

def RightEnd(t):
    return 0

def DiffusionSolver(x,t,s,L,c0,left,right):
    nt = np.size(t)
    nx = np.size(x)
    u = np.zeros((nt,nx))
    # Use the initial condition to define values of 'u' at the 
#zeroeth timestep:
    for j in range(nx):
        u[0,j] = c0(x[j], M)
    # Use the boundary condition to fix the string on its ends:
    u[:,-L] = left(t)
    u[:,L] = right(t)
# Entering the double loop to calculate the solution at every space 
#point at every timestep:
    for k in range(0,nt-1):
        for j in range(1,nx-1):
            u[k+1,j] = s*(u[k,j+1] + u[k,j-1]) + (1 - 2*s)*u[k,j]
    return u
u_num = DiffusionSolver(x,t,s,L,InitCond,LeftEnd,RightEnd)
u_num = u_num.T #HERE!
plt.plot(x, u_num)
plt.show()

这对我有用。

绘制扩散方程时的量纲问题

Dimension problem when plotting diffusion equation

python

plot