In [5]:
from IPython.display import HTML

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 if (code_show){
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$( document ).ready(code_toggle);
<form action="javascript:code_toggle()"><input type="submit" value="Code Toggle"></form>''')
In [6]:
from IPython.display import HTML

<a href=""
<button type="submit">Download Notebook</button>

Benchmark Problem 1: Spinodal Decomposition

In [20]:
from IPython.display import HTML

HTML('''{% include jupyter_benchmark_table.html num="[1]" revision=1 %}''')
Revision History

See the full benchmark table and table key

See the journal publication entitled "Benchmark problems for numerical implementations of phase field models" for more details about the benchmark problems. Furthermore, read the extended essay for a discussion about the need for benchmark problems.


Spinodal decomposition is one of the oldest problems in the phase field canon, and its formulation in terms of continuum fields goes back to the seminal works by Cahn and Hilliard [1]. The Cahn-Hilliard equation thus predates the name "phase field" in this context, but the term has subsequently been adopted by the community. While spinodal decomposition may be one of the simplest problems to model, it is highly relevant, as a large number of phase field models include the diffusion of a solute within a matrix. Furthermore, precipitation and growth may also be modeled with the same formulation if the appropriate initial conditions are chosen. For the benchmark problem, we select a simple formulation that is numerically tractable so that results may be obtained quickly and interpreted easily, testing the essential physics while minimizing model complexity and the chance to introduce coding errors.

Free energy and dynamics

The spinodal decomposition benchmark problem has a single order parameter, $c$, which describes the atomic fraction of solute. The free energy of the system, $F$, is expressed as $$ F=\int_{V}\left(f_{chem}\left(c\right)+\frac{\kappa}{2}|\nabla c|^{2}\right)dV, $$ where $f_{chem}$ is the chemical free energy density and $\kappa$ is the gradient energy coefficient. For this problem, we choose $f_{chem}$ to have a simple polynomial form, $$ f_{chem}\left(c\right)=\varrho_{s}\left(c-c_{\alpha}\right)^{2}\left(c_{\beta}-c\right)^{2}, $$ such that $f_{chem}$ is a symmetric double-well with minima at $c_{\alpha}$ and $c_{\beta}$, and $\varrho_{s}$ controls the height of the double-well barrier. Because $f_{chem}$ is symmetric (Fig. 1), $c_{\alpha}$ and $c_{\beta}$ correspond exactly with the equilibrium atomic fractions of the $\alpha$ and $\beta$ phases.

Figure 1: free energy density

In [3]:
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np

c_alpha = 0.3
c_beta = 0.7
rho_s = 5.

c = np.linspace(0, 1, 1000)
plt.figure(figsize=(6, 5))
plt.plot(c, rho_s * (c - c_alpha)**2 * (c - c_beta)**2, lw=4, color='k')
plt.xlabel("Atomic fraction, $c$", fontsize=20)
plt.ylabel("Free energy density, $f$", fontsize=20)
plt.xticks([0, 0.5, 1.0], fontsize=15)
plt.yticks([0, 0.1, 0.2], fontsize=15)

Because $c$ must obey a continuity equation -- the flux of $c$ is conserved -- the evolution of $c$ is given by the Cahn-Hilliard equation [1], which is derived from an Onsager force-flux relationship [[2][balluffi2005kinetics]]: $$ \frac{\partial c}{\partial t}=\nabla\cdot\Bigg\{M\nabla\left(\frac{\partial f_{chem}}{\partial c}-\kappa\nabla^{2}c\right)\Bigg\} $$ where $M$ is the mobility of the solute. For simplicity, both the mobility and the interfacial energy are isotropic. We choose $c_{\alpha}=0.3$, $c_{\beta}=0.7$, $\varrho_{s}=5$, $M=5$, and $\kappa=2$. Because the interfacial energy, diffuse interface width, and free energy parameterization are coupled, we obtain the diffuse interface width of $l=7.071 \sqrt{\kappa/\varrho_s}=4.47$ units over which $c$ varies as $0.348<c<0.652$, and an interfacial energy $\sigma=0.01508\sqrt{\kappa \varrho_s}$ [3].

Parameter values

$c_{\alpha}$ 0.3
$c_{\beta}$ 0.7
$\varrho_{s}$ 5
$\kappa$ 2
$M$ 5
$\epsilon$ 0.01
$\epsilon_{\text{sphere}}$ 0.05
$c_0$ 0.5

Domain geometries and boundary conditions

Several boundary conditions, initial conditions and computational domain geometries are used to challenge different aspects of the numerical solver implementation. We test four combinations that are increasingly difficult to solve: two with square computational domains, see (a) and (b), with side lengths of 200 units, one with a computational domain in the shape of a "T," see (c), with a total height of 120 units, a total width of 100 units, and horizontal and vertical section widths of 20 units, and one in which the computational domain is the surface of a sphere with a radius of r = 100 units, see (d). While most codes readily handle rectilinear domains, a spherical domain may pose problems, such as having the solution restricted to a two-dimensional curved surface. The coordinate systems and origins are given in Fig. 2. Periodic boundary conditions are applied to one square domain, see (a), while no-flux boundaries are applied to the other square domain, see (b), and the "T"-shaped domain, see (c). Periodic boundary conditions are commonly used with rectangular or rectangular prism domains to simulate an infinite material, while no-flux boundary conditions may be used to simulate an isolated piece of material or a mirror plane. As the computational domain is compact for the spherical surface, no boundary conditions are specified for it. Note that the same initial conditions are used for the square computational domains with no-flux, see (b), and periodic boundary conditions, see (a), such that when periodic boundary conditions are applied, there is a discontinuity in the initial condition at the domain boundaries.

(a) Square periodic

A 2D square domain with $L_x = L_y = 200$ and periodic boundary conditions.

In [14]:

from IPython.display import SVG
    out = SVG(filename='../images/block1.svg')
    out = None

(b) Square no-flux

A 2D square domain with $L_x = L_y = 200$ and no flux boundary conditions.

(c) T-shape

A T-shaped region with zero flux boundary conditions and with dimensions, $a=b=100$ and $c=d=20$.

In [13]:

from IPython.display import SVG
    out = SVG(filename='../images/t-shape.svg')
    out = None

(d) Sphere

The domain is the surface of a sphere with radius 100.

Initial conditions

The initial conditions for the first benchmark problem are chosen such that the average value of $c$ over the computational domain is approximately $0.5$.

Initial conditions for (a), (b) and (c)

The initial value of $c$ for the square and "T" computational domains is specified by $$ c\left(x,y\right) = c_{0}+\epsilon\left[\cos\left(0.105x\right)\cos\left(0.11y\right)+\left[\cos\left(0.13x\right)\cos\left(0.087y\right)\right]^{2}\right.\nonumber \\ \left.+\cos\left(0.025x-0.15y\right)\cos\left(0.07x-0.02y\right)\right], $$ where $c_{0}=0.5$ and $\epsilon=0.01$.

Figure 2: initial $c$ for (a), (b) and (c)

In [2]:

import numpy as np
from bokeh.plotting import figure, show, output_file, output_notebook, gridplot
from bokeh.models import FixedTicker
from bokeh.palettes import brewer, RdBu11, Inferno256
from bokeh.models.mappers import LinearColorMapper
import matplotlib as plt
import as cm
import numpy as np
from bokeh.models import HoverTool, BoxSelectTool

def generate_colorbar(mapper, width, height, n_ticks):
    high, low = mapper.high, mapper.low
    pcb = figure(width=width,
                 x_range=[0, 1],
                 y_range=[low, high],
    pcb.image(image=[np.linspace(low, high, 400).reshape(400,1)],
              dh=[high - low],
    pcb.xaxis.major_label_text_color = None
    pcb.xaxis.major_tick_line_color = None
    pcb.xaxis.minor_tick_line_color = None
    pcb.yaxis[0].ticker=FixedTicker(ticks=np.linspace(low, high, n_ticks))
    return pcb

def generate_contour_plot(data, xrange, yrange, width, height, mapper):               
    p = figure(x_range=xrange, y_range=yrange, width=width, height=height, min_border_right=10)
    aa = p.image(image=[data],
                 dw=xrange[1] - xrange[0],
                 dh=yrange[1] - yrange[0],
    return p

def get_data(xrange, yrange, data_func):
    N = 300
    xx, yy = np.meshgrid(np.linspace(xrange[0], xrange[1], N),
                         np.linspace(yrange[0], yrange[1], N))
    data = data_func(xx, yy)
    return xx, yy, data

def get_color_mapper(mpl_cm, high, low):
    colormap =cm.get_cmap(mpl_cm)
    bokehpalette = [plt.colors.rgb2hex(m) for m in colormap(np.arange(colormap.N))]
    mapper = LinearColorMapper(high=high,
    return mapper

def get_data_square(data_func):
    return get_data((0, 200), (0, 200), data_func)

def get_data_tshape(data_func):
    xx, yy, data = get_data((-40, 60), (0, 120), data_func)
    mask = ((xx < 0) | (xx > 20)) & (yy < 100)
    data[mask] = 0.5
    return xx, yy, data

def get_plot_grid(data_xy_func, high, low, n_ticks):
    mapper = get_color_mapper(cm.coolwarm, high, low)
    width, height = 300, 300

    all_plots = []
    for data_func in get_data_square, get_data_tshape:
        xx, yy, data = data_func(data_xy_func)
        contour_plot = generate_contour_plot(data,
                                             (xx.min(), xx.max()),
                                             (yy.min(), yy.max()),
    all_plots.append(generate_colorbar(mapper, width // 4, height, n_ticks))
    return gridplot([all_plots]) 

def initial_concentration(x, y, epsilon=0.01, c_0=0.5):
    return c_0 + epsilon * (np.cos(0.105 * x) * np.cos(0.11 * y) + (np.cos(0.13 * x) * np.cos(0.087 * y))**2 \
                            + np.cos(0.025 * x - 0.15 * y) * np.cos(0.07 * x - 0.02 * y))

show(get_plot_grid(initial_concentration, 0.53, 0.47, 7),
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