# Derivation of the inversion stencil using a non-symmetric forward-backward scheme

Derivation of a non-symmetric stencil of

$$b = \nabla\cdot(A\nabla_\perp f)+Bf$$

using a forward stencil on $\nabla\cdot(A\nabla_\perp f)$, and a backward stencil on $\nabla_\perp f$.

The stencil will not be symmetric as $f(x-h_x)$, $f(x)$ and $f(x+h_x)$ would be multiplied with $J(x,y)$. Symmetry requires that $f(x-h_x)$ is multiplied with $J(x-h_x)$, $f(x)$ with $J(x)$ and $f(x+h_x)$ with $J(x+h_x)$.

For symmetric version, see ForwardsBackwards.ipynb and BackwardsForwards.ipynb



In [1]:

from IPython.display import display
from sympy import init_printing
from sympy import symbols, expand, together, as_finite_diff, collect
from sympy import Function, Eq, Subs
from collections import deque

init_printing()




In [2]:

def finiteDifferenceOfOneTerm(factors, wrt, stencil):
"""
Finds the finite different approximation of a term consisting of several factors

Input:
factors - An iterable containing the factors of the term
wrt     - Take the derivative of the term with respect to this variable
stencil - An iterable containing the points to be used in the stencil

Output
term - The finite difference approximation of the term
"""
# Take the derivative
factorsDiff = []
for factor in factors:
factorsDiff.append(as_finite_diff(factor.diff(wrt), stencil))

# Putting together terms
term = 0
# Make object for cyclic permutation
cyclPerm = deque(range(len(factors)))
for perm in range(len(cyclPerm)):
# Initialize a dummy term to store temporary variables in
curTerm = factorsDiff[cyclPerm[0]]
for permNr in range(1,len(factors)):
curTerm *= factors[cyclPerm[permNr]]
# Make a cyclic premutation
cyclPerm.rotate(1)
term += curTerm
return term




In [3]:

def fromFunctionToGrid(expr, sym):
"""
Change from @(x,z) to @_xz, where @ represents a function

Input:
expr - The expression to change
sym  - symbols('@_xz, @_xp1z, @_xm1z, @_xzp1, @_xzm1')
xp1 = x+hx
zm1 = z-hz
etc.
"""
curFun = str(syms[0]).split('_')[0]
for sym in syms:
curSuffix = str(sym).split('_')[1]
if curSuffix == 'xz':
expr = expr.subs(Function(curFun)(x,z), sym)
elif curSuffix == 'xp1z':
expr = expr.subs(Subs(Function(curFun)(x,z), x, x+hx).doit(), sym)
elif curSuffix == 'xm1z':
expr = expr.subs(Subs(Function(curFun)(x,z), x, x-hx).doit(), sym)
elif curSuffix == 'xzp1':
expr = expr.subs(Subs(Function(curFun)(x,z), z, z+hz).doit(), sym)
elif curSuffix == 'xzm1':
expr = expr.subs(Subs(Function(curFun)(x,z), z, z-hz).doit(), sym)

return expr




In [4]:

x, z, hx, hz = symbols('x, z, h_x, h_z')
hx, hz = symbols('h_x, h_z', positive=True)

f = Function('f')(x, z)
A = Function('A')(x, z)
B = Function('B')(x, z)
gxx = Function('g^x^x')(x, z)
gzz = Function('g^z^z')(x, z)
J = Function('J')(x, z)

# Dummy function
g = Function('g')(x,z)

# Stencils
backwardX = [x-hx, x]
forwardX  = [x, x+hx]
backwardZ = [z-hz, z]
forwardZ  = [z, z+hz]



We are here discretizing the equation

$$b = \nabla\cdot(A\nabla_\perp f)+Bf \simeq \frac{1}{J}\partial_x \left(JAg^{xx}\partial_x f\right) + \frac{1}{J}\partial_z \left(JAg^{zz}\partial_z f\right) + Bf$$

where the derivatives in $y$ has been assumed small in non-orthogonal grids.

We will let $T$ denote "term", the superscript $^F$ denote a forward stencil, and the superscript $^B$ denote a backward stencil.

NOTE:

sympy has a built in function as_finite_diff, which could do the derivation easy for us. However it fails if

• Non derivative terms or factors are present in the expression
• If the expression is a Subs object (for example unevaluated derivatives calculated at a point)

We therefore do this in a sligthly tedious way.

## Calculating the first term

### Calculate the finite difference approximation of $\partial_x f$



In [5]:

fx = f.diff(x)
fxB = as_finite_diff(fx, backwardX)
display(Eq(symbols('f_x'), fx))
display(Eq(symbols('f_x^B'), together(fxB)))



### Calculate the finite difference approximation of $\frac{1}{J}\partial_x \left(JAg^{xx}\partial_x f\right)$

We start by making the substitution $\partial_x f \to g$ and calulate the first term of the equation under consideration



In [6]:

# Define the factors
factors = [J, A, gxx, g]
term1 = finiteDifferenceOfOneTerm(factors, x, forwardX)
term1 /= J
display(Eq(symbols('T_1^F'), term1))



We now back substitute $g\to \partial_x f$



In [7]:

term1 = term1.subs(Subs(g,x,x+hx).doit(), Subs(fxB,x,x+hx).doit())
term1 = term1.subs(g, fxB)
display(Eq(symbols('T_1^F'), term1))



## Calculating the second term

### Calculate the finite difference approximation of $\partial_z f$



In [8]:

fz = f.diff(z)
fzB = as_finite_diff(fz, backwardZ)
display(Eq(symbols('f_z'), fz))
display(Eq(symbols('f_z^B'), together(fzB)))



### Calculate the finite difference approximation of $\frac{1}{J}\partial_z \left(JAg^{zz}\partial_z f\right)$

We start by making the substitution $\partial_z f \to g$ and calulate the second term of the equation under consideration



In [9]:

# Define the factors
factors = [J, A, gzz, g]
term2 = finiteDifferenceOfOneTerm(factors, z, forwardZ)
term2 /= J
display(Eq(symbols('T_2^F'), term2))




In [10]:

term2 = term2.subs(Subs(g,z,z+hz).doit(), Subs(fzB,z,z+hz).doit())
term2 = term2.subs(g, fzB)
display(Eq(symbols('T_2'), term2))



## Calculating the third term



In [11]:

term3 = B*f
display(Eq(symbols('T_3^F'), term3))



### Collecting terms



In [12]:

b = term1 + term2 + term3
display(Eq(symbols('b'), b))




In [13]:

# Converting to grid syntax
functions = ['f', 'A', 'J', 'g^x^x', 'g^z^z', 'B']
for func in functions:
curStr = '{0}_xz, {0}_xp1z, {0}_xm1z, {0}_xzp1, {0}_xzm1'.format(func)
syms = symbols(curStr)
b = fromFunctionToGrid(b, syms)




In [14]:

# We must expand before we collect
b = collect(expand(b), symbols('f_xz, f_xp1z, f_xm1z, f_xzp1, f_xzm1'), exact=True)
display(Eq(symbols('b'),b))