三维扩散/热方程的有限差分法

Question

我想用有限差分法来解三维的扩散方程。我想我的主回路有问题。具体而言，离散方程为：

使用Neumann边界条件（仅以一个面为例）：

现在代码是：

import numpy as np
from matplotlib import pyplot, cm
from mpl_toolkits.mplot3d import Axes3D ##library for 3d projection plots
%matplotlib inline

kx = 15     #Number of points
ky = 15
kz = 15
largx = 90  #Domain length.
largy = 90
largz = 90   

dt4 = 1/2 #Time delta (arbitrary for the time).
dx4 = largx/(kx-1)    #Position deltas.
dy4 = largy/(ky-1)
dz4 = largz/(kz-1) 

Tin = 25    #Initial temperature
kapp = 0.23

Tamb3d = 150 #Ambient temperature

#Heat per unit of area. One for each face.
qq1=0 / (largx*largz)
qq2=0 / (largx*largz)
qq3=0 / (largz*largy)
qq4=0 / (largz*largy)
qq5=0 / (largx*largy)
qq6=1000 / (largx*largy)


x4 = np.linspace(0,largx,kx)   
y4 = np.linspace(0,largy,ky)
z4 = np.linspace(0,largz,kz)


#Defining the function.
def diff3d(tt):
    
    w2 = np.ones((kx,ky,kz))*Tin         #Temperature array
    wn2 = np.ones((kx,ky,kz))*Tin
    
    for k in range(tt+2):
        wn2 = w2.copy()
        w2[1:-1,1:-1,1:-1] = (wn2[1:-1,1:-1,1:-1] + 
                        kapp*dt4 / dy4**2 *                                       
                        (wn2[1:-1, 2:,1:-1] - 2 * wn2[1:-1, 1:-1,1:-1] + wn2[1:-1, 0:-2,1:-1]) +  
                        kapp*dt4 / dz4**2 *                                       
                        (wn2[1:-1,1:-1,2:] - 2 * wn2[1:-1, 1:-1,1:-1] + wn2[1:-1, 1:-1,0:-2]) +
                        kapp*dt4 / dx4**2 *
                        (wn2[2:,1:-1,1:-1] - 2 * wn2[1:-1, 1:-1,1:-1] + wn2[0:-2, 1:-1,1:-1]))

        #Neumann boundary (dx=dy=dz for the time)
        w2[0,:,:] =   w2[0,:,:] + 2*kapp* (dt4/(dx4**2)) * (w2[1,:,:] - w2[0,:,:] - qq1 * dx4/kapp)
        w2[-1,:,:] =  w2[-1,:,:] + 2* kapp*(dt4/(dx4**2)) * (w2[-2,:,:] - w2[-1,:,:] + qq2 * dx4/kapp)
        w2[:,0,:] =   w2[:,0,:] + 2*kapp* (dt4/(dx4**2)) * (w2[:,1,:] - w2[:,0,:] - qq3 * dx4/kapp)
        w2[:,-1,:] =  w2[:,-1,:] + 2*kapp* (dt4/(dx4**2)) * (w2[:,-2,:] - w2[:,-1,:] + qq4 * dx4/kapp)
        w2[:,:,0] =   w2[:,:,0] + 2 *kapp* (dt4/(dx4**2)) * (w2[:,:,-1] - w2[:,:,0] - qq5 * dx4/kapp)
        w2[:,:,-1] =   w2[:,:,-1] + 2 *kapp* (dt4/(dx4**2)) * (w2[:,:,-2] - w2[:,:,-1] + qq6 * dx4/kapp)
       
    w2[1:,:-1,:-1] = np.nan    #We'll only plot the "outside" points.
    w2_uno = np.reshape(w2,-1)    

    #Plotting
    fig = pyplot.figure()
    X4, Y4, Z4 = np.meshgrid(x4, y4,z4)
    ax = fig.add_subplot(111, projection='3d')
    img = ax.scatter(X4, Y4, Z4, c=w2_uno, cmap=pyplot.jet())
    fig.colorbar(img)
    pyplot.show()

对于5000次迭代（qq6=1000/面积），我们得到：

我只在上面加热。不知何故，我最终的底部表面加热

除此之外，我还试图限制我加热的区域（只是一张脸的一小部分）。当我试着去做的时候，似乎热量只向一个方向传递，而忽略了所有其他方向。将（qq1=1000/面积）应用于一半正面（另一部分是绝热的，q=0），我们最终得出：

这很奇怪。我怀疑我在主循环（或者可能在边界条件）中遇到了一些我找不到的问题

Answer 1

也许您可以尝试一种更受测试驱动的方法。一个帮助是Pythonslice-命令。它在numpy数组访问中速度很快，并且在有限差分中使用时不太容易出错。试一试：

import numpy as np

def get_slices(mx, my, mz):
    jxw, jxp, jxe = slice(0,mx-2), slice(1,mx-1), slice(2,mx)  # west,  center, east
    jys, jyp, jyn = slice(0,my-2), slice(1,my-1), slice(2,my)  # south, center, north
    jzb, jzp, jzt = slice(0,mz-2), slice(1,mz-1), slice(2,mz)  # botoom,center, top
    return jxw, jxp, jxe,   jys, jyp, jyn,    jzb, jzp, jzt

nx, ny, nz = 15, 10, 15
jxw, jxp, jxe, \
jys, jyp, jyn, \
jzb, jzpc, jzt = get_slices(nx, ny, nz)

ax, ay, az = np.arange(nx), np.arange(ny), np.arange(nz)

print('axW', ax[jxw])
print('axP', ax[jxp])
print('axE', ax[jxe])

axW [ 0  1  2  3  4  5  6  7  8  9 10 11 12]
axP [ 1  2  3  4  5  6  7  8  9 10 11 12 13]
axE [ 2  3  4  5  6  7  8  9 10 11 12 13 14]

我不太清楚舒尔您代码中的图形显示了什么。因此，通过使用不同的T（或w2）初始化和不完全相同的kx, ky, kz值仔细检查它。注意np.meshgrid()命令中的indexing="ij"选项。使用np.nan在某种程度上是危险的，因为它改变了变量globaly

import numpy as np
from matplotlib import pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D

def get_grid(mx, my, mz, Lx,Ly,Lz):
    ix, iy, iz = Lx*np.linspace(0,1,mx), Ly*np.linspace(0,1,my), Lz*np.linspace(0,1,mz)
    x, y, z = np.meshgrid(ix,iy,iz, indexing='ij')
    print('ix', ix), print('iy', iy), print('iz', iz)
    return x,y,z

def plot_grid(x,y,z,T):
    def plot_boundary_only(x,y,z,T):
        mx, my, mz = x.shape
        x[1:-1, 1:-1, 1:-1],y[1:-1, 1:-1, 1:-1],z[1:-1, 1:-1, 1:-1],T[1:-1, 1:-1, 1:-1] = np.nan, np.nan, np.nan, np.nan     # remove interior
        x[1:-1, 1:,0], y[1:-1, 1:,0], z[1:-1, 1:,0],  T[1:-1, 1:,0] = np.nan, np.nan, np.nan, np.nan                         # remove bottom
        x[1:-1, my-1, 1:-1], y[1:-1, my-1, 1:-1], z[1:-1, my-1, 1:-1], T[1:-1, my-1, 1:-1] = np.nan, np.nan, np.nan, np.nan  # remove north face
        x[0, 1:, :-1], y[0, 1:, :-1], z[0, 1:, :-1], T[0, 1:, :-1] = np.nan, np.nan, np.nan, np.nan                          # remove west face
        return x,y,z,T
    
    x,y,z,T = plot_boundary_only(x,y,z,T)   
    fig = plt.figure(figsize=(15,15))
    ax = fig.add_subplot(111, projection='3d')
    img = ax.scatter(x,y,z, c=T.reshape(-1), s=150, cmap=plt.jet())
    fig.colorbar(img)
    
    #ax.zaxis.set_rotate_label(False) # To disable automatic label rotation
    ax.set_ylabel('Y')
    ax.set_xlabel('X', rotation=0)
    ax.set_zlabel('Z', rotation=0)
    plt.savefig("cube.png")
    plt.show()
    
def init_T(x,y,z):
    T = np.zeros_like(x)
    case = 'xz'
    if case == 'x': T = x
    if case == 'y': T = y
    if case == 'z': T = z
    if case == 'xz': T = x+z
    return T

nx, ny, nz = 35, 20, 30
Lx, Ly, Lz = nx-1, ny-1, nz-1
x,y,z = get_grid(nx, ny, nz, Lx,Ly,Lz)  # generate a grid with mesh size Δx = Δy = Δz = 1
T = init_T(x,y,z)
plot_grid(x,y,z,T)

这是运行热方程求解器的主程序。请注意，我们不能使用在曲面上显示点的图形来查看任何结果！Neumann条件位于我们看不到的底部，面上的值是不变的定值边界条件

def set_GradBC(u):
    # - Neuman BC at bottom boundary  -
    u[:, :, 0] = u[:, :, 1]
    return u


def dT(u,DU):
    DU[jxp,jyp,jzp] = (u[jxe,jyp,jzp] - 2.0*u[jxp,jyp,jzp] + u[jxw,jyp,jzp])*Dx + \
                      (u[jxp,jyn,jzp] - 2.0*u[jxp,jyp,jzp] + u[jxp,jys,jzp])*Dy + \
                      (u[jxp,jyp,jzt] - 2.0*u[jxp,jyp,jzp] + u[jxp,jyp,jzb])*Dz
    return DU

#mmmmmmmmmmmmm main mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
#   set the parameters             
dt = 0.01                                    # time step size
nIterations = 5000                           # number of iterations
nx, ny, nz = 40,20,30                       # number of grid points
Lx, Ly, Lz = nx-1, ny-1, nz-1                 # physical grid size
Kx, Ky, Kz = 1, 1, 1                          # diffusion coeficients
dx, dy, dz = Lx/(nx-1), Ly/(ny-1), Lz/(nz-1)  # mesh size
Dx, Dy, Dz = Kx/dx**2, Ky/dy**2, Kz/dz**2     # equation coefficients

#   initialize the field variables         -
x,y,z = get_grid(nx, ny, nz, Lx,Ly,Lz)       # generate the grid
T = init_T(x,y,z)                            # initialize the temperature
DT = np.zeros_like(T)                        # initialize the change of temperature
jxw, jxp, jxe, \
jys, jyp, jyn, \
jzb, jzp, jzt = get_slices(nx, ny, nz)       # slices for finite differencing

#   run the solving algorithm            
for jter in range(nIterations):
    T = set_GradBC(T)              # set Neumann boundary condition
    T = T + dt*dT(T,DT)            # update

下面是一个切片，沿JY1=常数显示结果：

def plot_y_slice(JY1, x,y,z,T):
    T2 = T[:,JY1,:]
    X =  x[:,JY1,:]
    Z =  z[:,JY1,:]

    fig = plt.figure(figsize=(15,15))
    ax = fig.add_subplot(111)
    ax.set_aspect('equal')
    plt.contourf(X,Z,T2, 40,alpha=0.99)
    plt.savefig("y_slice.png")
    plt.show()
    
JY1 = 10
plot_y_slice(JY1, x,y,z,T)

Answer 2

此代码有问题： w2[：，：，0]=w2[：，：，0]+2kapp（dt4/（dx4**2））*（w2[：，：，，-1]-w2[，：，：，0]-qq5*dx4/kapp） 请再检查一遍。 我最近在做一个类似的项目。你有时间和我分享一些经验吗