Exact solution used in MES runs

We would like to MES the operation (in a cylindrical geometry)

$$ J\nabla \cdot \left(S_n\frac{\nabla_\perp \phi}{B}\right) $$

As we have a homogenenous $B$-field, we have normalized it out, and remain with

$$ J\nabla \cdot \left(S_n\nabla_\perp \phi\right) $$

In [1]:
%matplotlib notebook

from sympy import init_printing
from sympy import S
from sympy import sin, cos, tanh, exp, pi, sqrt

from boutdata.mms import x, y, z, t
from boutdata.mms import Delp2, DDX, DDY, DDZ

import os, sys
# If we add to sys.path, then it must be an absolute path
common_dir = os.path.abspath('./../../../../common')
# Sys path is a list of system paths
sys.path.append(common_dir)
from CELMAPy.MES import get_metric, make_plot, BOUT_print

init_printing()

Initialize


In [2]:
folder = '../twoGaussians/'
metric = get_metric()

Define the variables


In [3]:
# Initialization
the_vars = {}

Define manifactured solutions

We have that

$$JS = J\nabla\cdot(S_n\nabla_\perp\phi) = JS_n\nabla_\perp^2\phi + J\nabla S_n\cdot \nabla_\perp \phi = JS_n\nabla_\perp^2\phi + J\nabla_\perp S_n\cdot \nabla_\perp \phi$$

We will use the Delp2 operator for the perpendicular Laplace operator (as the y-derivatives vanishes in cylinder geometry). We have

Delp2$(f)=g^{xx}\partial_x^2 f + g^{zz}\partial_z^2 f + 2g^{xz}\partial_x\partial_z f + G^1\partial_x f + G^3\partial_z f$

Using the cylinder geometry, we get that

Delp2$(f)=\partial_x^2 f + \frac{1}{x^2}\partial_z^2 f + \frac{1}{x}\partial_x f$

Further on, due to orthogonality we have that $$\nabla_\perp S_n\cdot \nabla_\perp \phi = \mathbf{e}^i\cdot \mathbf{e}^i(\partial_i S_n)(\partial_i \phi) = g^{xx}(\partial_x S_n)(\partial_x \phi) + g^{zz}(\partial_z S_n)(\partial_z \phi) = (\partial_x S_n)(\partial_x \phi) + \frac{1}{x^2}(\partial_z S_n)(\partial_z \phi)$$

This gives

$$JS = \nabla\cdot(S_n\nabla_\perp\phi) = xS_n\partial_x^2 \phi + S_n\frac{1}{x}\partial_z^2 \phi + S_n\partial_x \phi + (\partial_x S_n)(\partial_x \phi) + \frac{1}{x}(\partial_z S_n)(\partial_z \phi)$$

We will use this to calculate the analytical solution.

NOTE:

  1. z must be periodic
  2. The field $f(\rho, \theta)$ must be of class infinity in $z=0$ and $z=2\pi$
  3. The field $f(\rho, \theta)$ must be single valued when $\rho\to0$
  4. The field $f(\rho, \theta)$ must be continuous in the $\rho$ direction with $f(\rho, \theta + \pi)$
  5. Eventual BC in $\rho$ must be satisfied

In [4]:
# We need Lx
from boututils.options import BOUTOptions
myOpts = BOUTOptions(folder)
Lx = eval(myOpts.geom['Lx'])

In [5]:
# Two normal gaussians

# The gaussian
# In cartesian coordinates we would like
# f = exp(-(1/(2*w^2))*((x-x0)^2 + (y-y0)^2))
# In cylindrical coordinates, this translates to
# f = exp(-(1/(2*w^2))*(x^2 + y^2 + x0^2 + y0^2 - 2*(x*x0+y*y0) ))
#   = exp(-(1/(2*w^2))*(rho^2 + rho0^2 - 2*rho*rho0*(cos(theta)*cos(theta0)+sin(theta)*sin(theta0)) ))
#   = exp(-(1/(2*w^2))*(rho^2 + rho0^2 - 2*rho*rho0*(cos(theta - theta0)) ))

w = 0.8*Lx
rho0 = 0.3*Lx
theta0 = 5*pi/4
the_vars['phi'] = exp(-(1/(2*w**2))*(x**2 + rho0**2 - 2*x*rho0*(cos(z - theta0)) ))

w = 0.5*Lx
rho0 = 0.2*Lx
theta0 = 0
the_vars['S_n'] = exp(-(1/(2*w**2))*(x**2 + rho0**2 - 2*x*rho0*(cos(z - theta0)) ))

Calculate the solution


In [6]:
the_vars['S'] = metric.J*(  the_vars['S_n']*Delp2(the_vars['phi'], metric=metric)\
                          + metric.g11*DDX(the_vars['S_n'], metric=metric)*DDX(the_vars['phi'], metric=metric)\
                          + metric.g33*DDZ(the_vars['S_n'], metric=metric)*DDZ(the_vars['phi'], metric=metric)\
                          )

Plot


In [7]:
make_plot(folder=folder, the_vars=the_vars, plot2d=True, include_aux=False)

In [8]:
BOUT_print(the_vars, rational=False)