我想解决两个化合物A和B的瞬态扩散方程，如图所示。我认为图像是显示我的问题的更好方法。Diffusion equations and boundary conditions.
您可以看到，反应仅在表面发生，并且A的通量等于B的通量。因此，这两个方程仅在表面处耦合。边界条件类似于ROBIN边界条件，在Fipy手册中进行了说明。但是，主要区别在于边界条件中第二个变量的存在。有人知道如何在Fipy中制定此边界条件吗？我想我需要在ROBIN边界条件上添加一些额外的术语，但我无法弄清楚。
非常感谢您的帮助。
这是用x = 0的ROBIN边界条件求解上述方程的代码。
-D（dC_A / dx）= -kC_A
-D（dC_B / dx）= -kC_B
在这种情况下，我可以轻松地使用ROBIN边界条件来求解方程。对于该边界条件，结果似乎是合理的。
"""
Question for StackOverflow
"""
#%%
from fipy import Variable, FaceVariable, CellVariable, Grid1D, TransientTerm, DiffusionTerm, Viewer, ImplicitSourceTerm
from fipy.tools import numerix
#%%
##### Model parameters
L= 8.4853e-4 # m boundary layer thickness
dx= 1e-8 # mesh size
nx = int(L/dx)+1 # number of meshes
D = 1e-9 # m^2/s diffusion coefficient
k = 1e-4 # m/s reaction coefficient R = k [c_A],
c_inf = 0. # ROBIN general condition, once can think R = k ([c_A]-[c_inf])
c_init = 1. # Initial concentration of compound A, mol/m^3
#%%
###### Meshing and variable definition
mesh = Grid1D(nx=nx, dx=dx)
c_A = CellVariable(name="c_A", hasOld = True,
mesh=mesh,
value=c_init)
c_B = CellVariable(name="c_B", hasOld = True,
mesh=mesh,
value=0.)
#%%
##### Right boundary condition
valueRight = c_init
c_A.constrain(valueRight, mesh.facesRight)
c_B.constrain(0., mesh.facesRight)
#%%
### ROBIN BC requirements, defining cellDistanceVectors
## This code is for fixing celldistance via this link:
## https://stackoverflow.com/questions/60073399/fipy-problem-with-grid2d-celltofacedistancevectors-gives-error-uniformgrid2d
MA = numerix.MA
tmp = MA.repeat(mesh._faceCenters[..., numerix.NewAxis,:], 2, 1)
cellToFaceDistanceVectors = tmp - numerix.take(mesh._cellCenters, mesh.faceCellIDs, axis=1)
tmp = numerix.take(mesh._cellCenters, mesh.faceCellIDs, axis=1)
tmp = tmp[..., 1,:] - tmp[..., 0,:]
cellDistanceVectors = MA.filled(MA.where(MA.getmaskarray(tmp), cellToFaceDistanceVectors[:, 0], tmp))
#%%
##### Defining mask and Robin BC at left boundary
mask = mesh.facesLeft
Gamma0 = D
Gamma = FaceVariable(mesh=mesh, value=Gamma0)
Gamma.setValue(0., where=mask)
dPf = FaceVariable(mesh=mesh,
value=mesh._faceToCellDistanceRatio * cellDistanceVectors)
n = mesh.faceNormals
a = FaceVariable(mesh=mesh, value=k, rank=1)
b = FaceVariable(mesh=mesh, value=D, rank=0)
g = FaceVariable(mesh=mesh, value= k * c_inf, rank=0)
RobinCoeff = (mask * Gamma0 * n / (-dPf.dot(a)+b))
#%%
#### Making a plot
viewer = Viewer(vars=(c_A, c_B),
datamin=-0.2, datamax=c_init * 1.4)
viewer.plot()
#%% Time step and simulation time definition
time = Variable()
t_simulation = 4 # seconds
timeStepDuration = .05
steps = int(t_simulation/timeStepDuration)
#%% PDE Equations
eqcA = (TransientTerm(var=c_A) == DiffusionTerm(var=c_A, coeff=Gamma) +
(RobinCoeff * g).divergence
- ImplicitSourceTerm(var=c_A, coeff=(RobinCoeff * a.dot(-n)).divergence))
eqcB = (TransientTerm(var=c_B) == DiffusionTerm(var=c_B, coeff=Gamma) -
(RobinCoeff * g).divergence
+ ImplicitSourceTerm(var=c_B, coeff=(RobinCoeff * a.dot(-n)).divergence))
#%% A loop for solving PDE equations
while time() <= (t_simulation):
time.setValue(time() + timeStepDuration)
c_B.updateOld()
c_A.updateOld()
res1=res2 = 1e10
viewer.plot()
while (res1 > 1e-6) & (res2 > 1e-6):
res1 = eqcA.sweep(var=c_A, dt=timeStepDuration)
res2 = eqcB.sweep(var=c_B, dt=timeStepDuration)
可以将其解决为完全隐式的系统。下面的代码将问题简化为具有统一的域大小和扩散系数。 k设置为0.2。它带有一些警告，很好地捕获了分析解决方案（请参阅下文）。
from fipy import (
CellVariable,
TransientTerm,
DiffusionTerm,
ImplicitSourceTerm,
Grid1D,
Viewer,
)
L = 1.0
nx = 1000
dx = L / nx
konstant = 0.2
coeff = 1.0
mesh = Grid1D(nx=nx, dx=dx)
var_a = CellVariable(mesh=mesh, value=1.0, hasOld=True)
var_b = CellVariable(mesh=mesh, value=0.0, hasOld=True)
var_a.constrain(1.0, mesh.facesRight)
var_b.constrain(0.0, mesh.facesRight)
coeff_mask = ~mesh.facesLeft * coeff
boundary_coeff = konstant * (mesh.facesLeft * mesh.faceNormals).divergence
eqn_a = TransientTerm(var=var_a) == DiffusionTerm(
coeff_mask, var=var_a
) - ImplicitSourceTerm(boundary_coeff, var=var_a) + ImplicitSourceTerm(
boundary_coeff, var=var_b
)
eqn_b = TransientTerm(var=var_b) == DiffusionTerm(
coeff_mask, var=var_b
) - ImplicitSourceTerm(boundary_coeff, var=var_b) + ImplicitSourceTerm(
boundary_coeff, var=var_a
)
eqn = eqn_a & eqn_b
for _ in range(5):
var_a.updateOld()
var_b.updateOld()
eqn.sweep(dt=1e10)
Viewer((var_a, var_b)).plot()
print("var_a[0] (expected):", (1 + konstant) / (1 + 2 * konstant))
print("var_b[0] (expected):", konstant / (1 + 2 * konstant))
print("var_a[0] (actual):", var_a[0])
print("var_b[0] (actual):", var_b[0])
input("wait")
注意以下内容：
正如所写，边界条件仅是一阶精确的，这对于这个问题并不重要，但是可能会在较高维度上对您造成伤害。可能有一些方法可以解决此问题，例如在边界附近有一个小单元格或为边界条件添加显式的二阶校正。
等式在此处耦合。如果解耦，则可能需要迭代负载才能达到平衡。
它确实需要进行几次迭代才能达到平衡，但不是必须的。这可能是由于求解器没有几次尝试就无法充分收敛。耦合方程可能存在一些不良条件。