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Class: Phase Field Crystal

In this class, we simulate a crystal using the phase-field crystal methodology. The phase-field crystal (PFC) model is a mesoscopic model that describes the dynamics of a crystal. The model is based on a free energy functional, which is minimized to find the equilibrium state of the system. The free energy functional is a function of the phase field \(\psi\), which is a real-valued field that describes the local density of the crystal. The PFC model is a powerful tool for studying the dynamics of crystals, and it has been used to study a wide range of phenomena, including the formation of defects, the dynamics of grain boundaries, and the growth of crystals.

file: comfit/models/phase_field_crystal.py 
class: PhaseFieldCrystal

Variables and parameters

The phase field \(\psi\).

pfc.psi

Basic model

The PFC methodology is based on postulating a free energy

The PFC free energy

\[ \mathcal F = \int d\boldsymbol{r} \tilde f(\psi, \nabla \psi, ...) \]

with the goal of the state that minimizes \(\mathcal F\) has a certain symmetry. In this documentation, we will present six such models to represent different crystalline structures in 1, 2 and 3 dimensions, all of which are on the form

The PFC free energy density

\[ \tilde f( \psi, \nabla \psi, ...) = \frac{1}{2} (\mathcal L(\nabla) \psi)^2 + \frac{1}{2} \texttt{r} \psi^2 + \frac{1}{3} \texttt t \psi^3 + \frac{1}{4} \texttt v \psi^4, \]

where \(\mathcal L (\nabla)\) is a gradient operator dependent on the dimension and target symmetry, listed below

Model Derivative operator \(\mathcal L\)
1D periodic \(\mathcal L_1 = (1+\nabla^2)\)
2D triangular \(\mathcal L_1 = (1+\nabla^2)\)
2D square \(\mathcal L_1 \mathcal L_2 = (1+\nabla^2) (2+\nabla^2)\)
3D bcc \(\mathcal L_1 = (1+\nabla^2)\)
3D fcc \(\mathcal L_1 \mathcal L_{4/3} = (1+\nabla^2) (4/3+\nabla^2)\)
3D simple cubic \(\mathcal L_1 \mathcal L_2 \mathcal L_3 = (1+\nabla^2) (2+\nabla^2) (3+\nabla^2)\)

Table: Phase-field crystal models. \(\mathcal L_X = (X+\nabla^2)\).

Historical context: model presented in Ref.1.

The primary field is \(\psi\), a real valued field

pfc.psi 

Crystal symmetry

The way in which this field can model a crystalline lattice is by the ground state having that particular symmetry. For instance, the figure below shows the (real) field \(\psi\) for a (a) 2D triangular, and a (b) 2D square lattice.

PFC symmetries PFC symmetrie

Figure: Ground state of the PFC model for (a) 2D triangular and (b) 2D square lattice.

The crystalline symmetry is described by a Bravais lattice, which consists of vectors that point to peaks on the lattice. The Fourier transform of a field that a Bravais lattice symmetry will itself have a lattice symmetry, which is called the reciprocal lattice, and is useful to express the field in terms of a Fourier series.

PFC Bravais and reciprocal lattice vectors PFC Bravais and reciprocal lattice vectors

Figure: Bravais lattices \(\mathcal B\) and their reciprocal lattices \(\mathcal R\) for square and triangular symmetry. In each case, \(\{\vec a^{(n)}\}_{n=1}^2\) are primitive lattice vectors and \(\{ \vec q^{(n)}\}_{n=1}^2\) primitive reciprocal lattice vectors (RLVs) that satisfy \(\vec a^{(n)} \cdot \vec q^{(m)} = 2\pi \delta_{nm}\). Amended and reprinted from Ref. [^Skogvoll2023SymmetryTopology] with permission.

Using the reciprocal lattice, one can express a field which has a Bravais lattice symmetry as a Fourier series in terms of the reciprocal lattice vectors

\[ \psi (\mathbf r) = \sum_{\vec q^{(n)} \in \mathcal R} \eta_n e^{\mathfrak i \vec q^{(n)} \cdot \vec r}, \]

the reciprocal lattice is \(d\) dimensions generated by \(d\) primitive RLVs \(\lbrace\boldsymbol{q}^{(n)}\rbrace_{n=1}^d\), satisfying

The primitive BLV-RLV orthogonality relation.

\[ \boldsymbol{q}^{(n)} \cdot \boldsymbol{a}^{(m)} = 2\pi \delta_{nm}, \quad n,m \leq d. \]

There are infinitely many BLVs and RLVs, but of particular interest are those with the smallest magnitude; which are sometimes named primary RLVs/BLVs. These values are encoded in the instance properties a,q, e.g.,

import comfit as cf

pfc = cf.PhaseFieldCrystal2DTriangular(1,1)
print(pfc.a)
print(pfc.q)
>> 
[[3.627, 6.283], [7.255, 0.0], [3.627, -6.283]]
[[0, 1], [0.866, -0.5], [-0.866, -0.5]]

Example: The 1D PFC - Crystal symmetry

The simplest example of a PFC model is the 1D periodic model, which simply looks like a sine wave.

1D PFC 1D PFC

The crystal symmetry in this case refers to \(\psi\) having a periodicity of \(a_0=2\pi\), and so the Bravais lattice is simply \(\{ a^{(n)} \} = \{...,-2\pi,-\pi,0,\pi,2\pi,... \}\). The primary BLVs are thus given by

\[ \mathcal B_{\textrm{per}}^{(1)} = \{ a^{(1)} = 2\pi,~ a^{(-1)} = -2\pi \} \]

The reciprocal lattice is in this case simply \(\{ q^{(n)} \} = \{...,q^{(-2)} =-2,~ q^{(-1)} = -1,~ q^{(0)} = 0,~ q^{(1)} = 1,~ q^{(2)} =2,... \}\), the primary set being

\[ \mathcal R_{\textrm{per}}^{(1)} = \{ q^{(1)} = 1,~ q^{(-1)} = -1 \} \]

and any field with that periodicity can be expressed as a Fourier series in terms of these RLVs by

\[ \psi(x) = \psi_0 + A (e^{\mathfrak iq^{(1)} x} + e^{\mathfrak i q^{(-1)} x}) + B (e^{\mathfrak iq^{(2)} x} + e^{\mathfrak i q^{(-2)} x}) + ..., \]

Cutting the Fourier series off at the primary RLVs mode is called the one-mode approximation

\[ \psi^{\textrm{eq}} \approx \psi_0 + A \sum_{q^{(n)} \in \mathcal R^{(1)}} e^{\mathfrak i q^{(n)} x}, \]

where \(\mathcal R_{\textrm{per}}^{(1)}\) as the primary RLVs.

Primitive vs primary LV

Note the subtle difference between a primitive LV and a primary LV. The primitive BLVs generate the lattice and satisfy the orthogonal relationship with the primitive RLVs which generate the reciprocal lattice. There are only \(d\) primitive LVs in \(d\) dimensions whereas there may be any number of primary lattice vectors. For instance, there are six primary BLVs for the triangular lattice and four primary lattice vectors for the 2D square lattice. In the following, the primary BLVs and RLVs have been chosen so that the first \(d\) primary LVs are primitive LVs that satisfy the orthogonality relation.

Below are all the lattice constants and primary RLVs and BLVs for the models we consider.

Lattice constant

\[ a_0 = 2\pi \]

Primary RLVs

\[ \mathcal R_{\textrm{per}}^{(1)} = \left \lbrace \begin{array}{l} q^{(1)} = 1 \\ q^{(-1)}= - q^{(1)} \end{array} \right \rbrace \]

Primary BLVs

\[ \mathcal B_{\textrm{per}}^{(1)} = \left \lbrace \begin{array}{l} a^{(1)} = a_0 \\ a^{(-1)} = -\mathbf a^{(1)} \end{array} \right \rbrace \]

Lattice constant

\[ a_0 = \frac{4\pi}{\sqrt 3} \]

Primary RLVs

\[ \mathcal R_{\textrm{tri}}^{(1)} = { \left \lbrace \begin{array}{l} \mathbf q^{(1)} = (\sqrt 3/2,-1/2) \\ \mathbf q^{(2)} = (0,1) \\ \mathbf q^{(3)} = (-\sqrt 3/2,-1/2) \\ \mathbf q^{(-n)} = - \mathbf q^{(n)}|_{n=1,2,3} \\ \end{array} \right \rbrace } \]

Primary BLVs

\[ \mathcal B_{\textrm{tri}}^{(1)} = { \left \lbrace \begin{array}{l} \mathbf a^{(1)} = a_0(1,0) \\ \mathbf a^{(2)} = a_0(1/2,\sqrt 3/2) \\ \mathbf a^{(3)} = a_0(1/2,-\sqrt 3/2) \\ \mathbf a^{(-n)}=-\mathbf a^{(n)}|_{n=1,2,3} \\ \end{array} \right \rbrace } \]

Lattice constant $$ a_0 = 2\pi $$

Primary RLVs

\[ \mathcal B_{\textrm{sq}}^{(1)} = \left \lbrace \begin{array}{l} \mathbf a^{(1)} = a_0(1,0) \\ \mathbf a^{(2)} = a_0(0,1) \\ \mathbf a^{(-1)},\mathbf a^{(-2)} \end{array} \right \rbrace \]

Primary Bravais Lattice vectors

\[ \mathcal R_{\textrm{sq}}^{(1)} = { \left \lbrace \begin{array}{l} \mathbf q^{(1)} = (1,0) \\ \mathbf q^{(2)} = (0,1) \\ \mathbf q^{(-n)} = - \mathbf q^{(n)} |_{n=1,2} \\ \end{array} \right \rbrace} \]
\[ \mathcal R_{\textrm{sq}}^{(2)} = \left \lbrace \begin{array}{l} \mathbf q^{(3)} = (1,-1) \\ \mathbf q^{(4)} = (1,1) \\ \mathbf q^{(-n)} = - \mathbf q^{(n)} |_{n=3,4} \\ \end{array} \right \rbrace \]

Lattice constant

\[ a_0 = 2\pi \sqrt {2} \]

Primary BLVs

\[ \mathcal B_{\textrm{bcc}}^{(1)} = { \left \lbrace \begin{array}{l} \mathbf a^{(1)} = a_0(-1,1,1)/2 \\ \mathbf a^{(2)} = a_0(1,-1,1)/2 \\ \mathbf a^{(3)} = a_0(1,1,-1)/2\\ \mathbf a^{(4)} = a_0(1,1,1)/2\\ \mathbf a^{(-n)}=-\mathbf a^{(n)}|_{n=1,...,4} \\ \end{array} \right \rbrace } \]

Primary RLVs

\[ \mathcal R_{\textrm{bcc}}^{(1)} = { \left \lbrace \begin{array}{l} \mathbf q^{(1)} = (0,1,1)/\sqrt 2 \\ \mathbf q^{(2)} = (1,0,1)/\sqrt 2 \\ \mathbf q^{(3)} = (1,1,0)/\sqrt 2\\ \mathbf q^{(4)} = (0,-1,1)/\sqrt 2\\ \mathbf q^{(5)} = (-1,0,1)/\sqrt 2\\ \mathbf q^{(6)} = (-1,1,0)/\sqrt 2\\ \mathbf q^{(-n)} = - \mathbf q^{(n)}|_{n=1,...6} \\ \end{array} \right \rbrace } \]

Lattice constant

\[ a_0 = 2\pi \sqrt 3 \]

Primary BLVs

\[ \mathcal B_{\textrm{fcc}}^{(1)} = { \left \lbrace \begin{array}{l} \mathbf a^{(1)} = a_0(0,1,1)/2 \\ \mathbf a^{(2)} = a_0(1,0,1)/2 \\ \mathbf a^{(3)} = a_0(1,1,0)/2\\ \mathbf a^{(4)} = a_0(0,-1,1)/2\\ \mathbf a^{(5)} = a_0(-1,0,1)/2\\ \mathbf a^{(6)} = a_0(-1,1,0)/2\\ \mathbf a^{(-n)}=-\mathbf a^{(n)}|_{n=1,...,6} \\ \end{array} \right \rbrace } \]

Primary RLVs

\[ \mathcal R_{\textrm{fcc}}^{(1)} = { \left \lbrace \begin{array}{l} \mathbf q^{(1)} = (-1,1,1)/\sqrt 3 \\ \mathbf q^{(2)} = (1,-1,1)/\sqrt 3 \\ \mathbf q^{(3)} = (1,1,-1)/\sqrt 3\\ \mathbf q^{(4)} = (1,1,1)/\sqrt 3\\ \mathbf q^{(-n)} = - \mathbf q^{(n)}|_{n=1,...,4}\\ \end{array} \right \rbrace } \]
\[ \mathcal R_{\textrm{fcc}}^{(4/3)} = { \left \lbrace \begin{array}{l} \mathbf q^{(5)} = (2,0,0)/\sqrt 3 \\ \mathbf q^{(6)} = (0,2,0)/\sqrt 3 \\ \mathbf q^{(7)} = (0,0,2)/\sqrt 3\\ \mathbf q^{(-n)} = - \mathbf q^{(n)} |_{n=5,6,7} \\ \end{array} \right \rbrace } \]

Lattice constant

\[ a_0 = 2\pi \]

Primary BLVs

\[ \mathcal B_{\textrm{sc}}^{(1)} = { \left \lbrace \begin{array}{l} \mathbf a^{(1)} = a_0(1,0,0) \\ \mathbf a^{(2)} = a_0(0,1,0) \\ \mathbf a^{(3)} = a_0(0,0,1)\\ \mathbf a^{(-n)} = - \mathbf a^{(n)}|_{n=1,2,3} \\ \end{array} \right \rbrace } \]

Primary RLVs

\[ \mathcal R_{\textrm{sc}}^{(1)} = { \left \lbrace \begin{array}{l} \mathbf q^{(1)} = (1,0,0) \\ \mathbf q^{(2)} = (0,1,0) \\ \mathbf q^{(3)} = (0,0,1)\\ \mathbf q^{(-n)} = - \mathbf q^{(n)} |_{n=1,2,3}\\ \end{array} \right \rbrace } \]
\[ \mathcal R_{\textrm{sc}}^{(2)} = { \left \lbrace \begin{array}{l} \mathbf q^{(4)} = (0,1,1) \\ \mathbf q^{(5)} = (1,0,1) \\ \mathbf q^{(6)} = (1,1,0)\\ \mathbf q^{(7)} = (0,-1,1) \\ \mathbf q^{(8)} = (-1,0,1) \\ \mathbf q^{(9)} = (-1,1,0)\\ \mathbf q^{(-n)} = - \mathbf q^{(n)}|_{n=4,...,9} \\\\ \end{array} \right \rbrace } \]
\[\mathcal R_{\textrm{sc}}^{(3)} = { \left \lbrace \begin{array}{l} \mathbf q^{(10)} = (-1,1,1) \\ \mathbf q^{(11)} = (1,-1,1) \\ \mathbf q^{(12)} = (1,1,-1)\\ \mathbf q^{(13)} = (1,1,1)\\ \mathbf q^{(-n)} = - \mathbf q^{(n)}|_{n=10,...,13} \\ \end{array} \right \rbrace } \]

The ground state

To find the ground state of a PFC, one inserts a particular mode approximation into the free energy density, average over a unit cell (coarse-grain) and minimizes it with respect to the amplitudes.

Example: The 1D PFC - Ground state

To find the ground state of the 1D PFC, we assume that we can express the field in the one-mode approximation, i.e.,

\[ \psi^{\textrm{eq}}(x) = \psi_0 + A (e^{\mathfrak i q x} + e^{\mathfrak i q x}), \]

where \(q\) is, for now, an arbitrary wave vector which we will show to be the primary RLV of the lattice. We want to insert this into the free energy density \(\tilde f\) and integrate over a unit cell (UC), which in this case is simply \([0,2\pi]\)

\[ \frac{1}{2\pi} \mathcal F_{UC} = \frac{1}{2\pi} \int_0^{2\pi} \tilde f(\psi^{\textrm{eq}}) dx \equiv \langle \tilde f(\psi^{\textrm{eq}}) \rangle_{UC} = \frac{1}{2} \langle (\mathcal L {\psi^{\textrm{eq}}})^2 \rangle_{UC} + \frac{1}{2} \texttt{r} \langle {\psi^{\textrm{eq}}}^2 \rangle_{UC} + \frac{1}{4} \langle {\psi^{\textrm{eq}}}^4 \rangle_{UC} \]

To insert this into the free energy density, we need to calculate some auxiliary quantities.

\[ \mathcal L \psi^{\textrm{eq}} = (1+\nabla^2) \psi^{\textrm{eq}} = \psi_0 + (1-q^2) A (e^{\mathfrak i q x} + e^{\mathfrak i q x}) \]

so

\[ \langle (\mathcal L \psi^{\textrm{eq}})^2 \rangle_{UC} = \langle (\psi_0 + (1-q^2) A (e^{\mathfrak i q x} + e^{\mathfrak i q x})(\psi_0 + (1-q^2) A (e^{\mathfrak i q x} + e^{\mathfrak i q x})) \rangle_{UC} \]

Now, to calculate this, we need to make use of a condition of resonance, which is that under the coarse-graining operation, terms with periodicity average to zero2. So we get

\[ \langle (\mathcal L \psi^{\textrm{eq}})^2 \rangle_{UC} = \psi_0^2 + 2 A^2 (1-q^2)^2 \]

Similarly, the other terms give

\[ \langle {\psi^{\textrm{eq}}}^2 \rangle_{UC} = \psi_0^2 + 2 A^2 \]
\[ \langle {\psi^{\textrm{eq}}}^3 \rangle_{UC} = \psi_0^3 + 6 \psi_0 A^2 \]
\[ \langle {\psi^{\textrm{eq}}}^4 \rangle_{UC} = \psi_0^4 + 12 \psi_0^2 A^2 + 6 A^4 \]

which gives

\[ \mathcal F_{UC} = 2\pi \left( \frac{1}{2} (\psi_0^2 + 2 A^2 (1-q^2)^2) + \frac{1}{2} \texttt{r} (\psi_0^2 + 2 A^2) + \frac{1}{3} \texttt t (\psi_0^3 + 6 \psi_0 A^2) + \frac{1}{4} \texttt v (\psi_0^4 + 12 \psi_0^2 A^2 + 6 A^4) \right ) \]

First of all, we see that the equilibrium value must have \(q=1\), since the free energy is minimized by this choice. Now, if the PFC is evolving according to conservative dynamics \(\partial_t \psi = \nabla^2 (\delta \mathcal F/\delta \psi)\) (which is the standard PFC evolution equation as explained further down in the document), the average value \(\psi_0\) will be conserved and therefore considered a simulation constant. In this case, we find the equilibrium value of \(A\), by differentiating wrt. \(A\) and setting to zero.

\[ \partial_A \mathcal F_{UC} = 2 \texttt{r} A + 4 \texttt t \psi_0 A + 6 \texttt v \psi_0^2 A + 6\texttt v A^3 = 0, \]

which gives \(A=0\) or

\[ \texttt{r} + 2 \texttt t \psi_0 + 3 \texttt v \psi_0^2 + 3 \texttt v A^2 = 0, \]
\[ A^2 = \frac{1}{3\texttt v} (-\texttt{r} - 2 \texttt t \psi_0 - 3\texttt v \psi_0^2) \]
\[ A = \pm \frac{1}{\sqrt{3\texttt v}} \sqrt{- \texttt{r} - 2 \texttt t \psi_0 - 3 \psi_0^2} \]

The sign does not matter in the case of the 1D PFC, since it will only affect the global sign of the amplitude, flipping the cosine wave. In higher dimensions, however, it requires more subtle treatment, and typically, one inserts both solutions into the free energy \(\mathcal F_{UC}\) to see which is lower

If instead, we are evolving the PFC under not conserved dynamics \(\partial_t \psi = - \delta \mathcal F/\delta \psi\), also the zero mode \(\psi_0\) will reach an equilibrium value. In this case, then, we need to solve the system of equations given by

\[ \left \lbrace \begin{array}{lrl} \partial_{\psi_0} \mathcal F_{UC} = 0: & \psi_0 + \texttt r \psi_0 + \texttt t \psi_0^2 + 2 \texttt t A^2 + \texttt v \psi_0^3 + 6\texttt v \psi_0 A^2 &=0 \\ \partial_{A} \mathcal F_{UC} = 0: & \texttt{r} + 2 \texttt t \psi_0 + 3 \texttt v \psi_0^2 + 3 \texttt v A^2 &= 0\\ \end{array} \right \rbrace \]

The pfc instance will be initialized with a list called eta0, which consists of the equilibrium values of the amplitudes to begin with.

Default resolution:

Default model parameters \((r,\psi_0)\): \((-0.3,-0.3)\)

Free energy per unit cell (calculation document):

\[ \mathcal F_{UC} = \frac{1}{2} \psi_0^2 + \frac{1}{2} \texttt{r} (\psi_0^2 + 2 A^2) + \frac{1}{3} \texttt t (\psi_0^3 + 6 \psi_0 A^2) + \frac{1}{4} \texttt v (\psi_0^4 + 12 \psi_0^2 A^2 + 6 A^4) \]

Equilibrium amplitude (conserved)

\[ \partial_{A} \mathcal F_{UC} = 0: \quad A =\pm \frac{1}{\sqrt {3\texttt v}} \sqrt{ -\texttt r - 2 \texttt t \psi_0 - 3 \texttt v \psi_0^2} \]

Equilibrium amplitude equations (unconserved)

\[ \left \lbrace \begin{array}{lrl} \partial_{\psi_0} \mathcal F_{UC} = 0: & 2 A^2 (\texttt t + 3 \texttt v \psi_0) + \psi_0 (1 + \texttt r + \texttt t \psi_0 + \texttt v \psi_0^2) &=0 \\ \partial_{A} \mathcal F_{UC} = 0: & \texttt r + 3 A^2 \texttt v + \psi_0 (2 \texttt t + 3 \texttt v \psi_0) &= 0\\ \end{array} \right \rbrace \]

Default resolution: \([7,12]^{-1}a_0\)

Default model parameters \((r,\psi_0)\): \((-0.3,-0.3)\)

Free energy per unit cell (calculation document.):

\[ \mathcal F_{UC} = \frac{\pi^2}{3\sqrt 3} \left (270 A^4 \texttt v + 48 A^3 (\texttt t + 3 \texttt v \psi_0) + \psi_0^2 (6 + 6 \texttt r + 4 \texttt t \psi_0 + 3 \texttt v \psi_0^2) + 36 A^2 (\texttt r + \psi_0 (2 \texttt t + 3 \texttt v \psi_0))\right) \]

Equilibrium amplitude equations (conserved)

\[ \left \lbrace \partial_{A} \mathcal F_{UC} = 0: A=0 ~ \textrm{or} ~ \quad A= \frac{1}{15\texttt v} \left ( -\texttt t - 3 \texttt v \psi_0 \pm \sqrt{ \texttt t^2 - 15 \texttt r \texttt v - 24 \texttt t \texttt v \psi_0 - 36 \texttt v^2 \psi_0^2} \right ) \right \rbrace \]

Equilibrium amplitude equations (unconserved)

\[ \left \lbrace \begin{array}{lrl} \partial_{\psi_0} \mathcal F_{UC} = 0: & 12 A^3 \texttt v + 6 A^2 (\texttt t + 3 \texttt v \psi_0) + \psi_0 (1 + \texttt r + \texttt t \psi_0 + \texttt v \psi_0^2) &=0 \\ \partial_{A} \mathcal F_{UC} = 0: & A(\texttt r + 15 A^2 \texttt v + 2 A (\texttt t + 3 \texttt v \psi_0) + \psi_0 (2 \texttt t + 3 \texttt v \psi_0)) &= 0\\ \end{array} \right \rbrace \]

Default resolution: \([7,7]^{-1}a_0\)

Default model parameters \((\texttt r,\psi_0)\): \((-0.3,-0.3)\)

Free energy per unit cell (calculation document):

\[ \mathcal F_{UC} = \frac{\pi^2}{3} (108 A^4 \texttt v + 108 B^4 \texttt v + \psi_0^2 (24 + 6 \texttt r + 4 \texttt t \psi_0 + 3 \texttt v \psi_0^2) + 24 B^2 (\texttt r + \psi_0 (2 \texttt t + 3 \texttt v \psi_0)) + 24 A^2 (\texttt r + 18 B^2 \texttt v + 4 B (t + 3 \texttt v \psi_0) + \psi_0 (2 \texttt t + 3 \texttt v \psi_0))) \]

Equilibrium amplitude equations

\[ \left \lbrace \begin{array}{lrl} \partial_{\psi_0} \mathcal F_{UC} = 0: & 4 B^2 (\texttt t + 3 \texttt v \psi_0) + \psi_0 (4 + \texttt r + \texttt t \psi_0 + \texttt v \psi_0^2) + 4 A^2 (\texttt t + 3 \texttt v (2 B + \psi_0)) &= 0 \\ \partial_{A} \mathcal F_{UC} = 0: & A (\texttt r + 9 A^2 \texttt v + 18 B^2 \texttt v + 2 \texttt t \psi_0 + 3 \texttt v \psi_0^2 + 4 B (\texttt t + 3 \texttt v \psi_0)) &= 0 \\ \partial_{B} \mathcal F_{UC} = 0: & 9 B^3 \texttt v + 2 A^2 (t + 3 \texttt v \psi_0) + B (\texttt r + 18 A^2 \texttt v + 2 \texttt \texttt t \psi_0 + 3 \texttt v \psi_0^2) &= 0\\ \end{array} \right \rbrace \]

Default resolution: \([7,7,7]^{-1}a_0\)

Default model parameters \((\texttt r,\psi_0)\): \((-0.3,-0.325)\)

Free energy per unit cell (calculation document):

\[ \mathcal F_{UC} = \frac{4 \sqrt 2 \pi^3}{3} \left (1620 A^4 \texttt v + 192 A^3 (t + 3 \texttt v \psi_0) + \psi_0^2 (6 + 6 \texttt r + 4 t \psi_0 + 3 \texttt v \psi_0^2) + 72 A^2 (\texttt r + \psi_0 (2 t + 3 \texttt v \psi_0)) \right ) \]

Equilibrium amplitude (conserved)

\[ \left \lbrace \partial_{A} \mathcal F_{UC} = 0: A=0 ~ \textrm{or} ~ \quad A = \frac{1}{45\texttt v} \left ( -2 \texttt t - 6 \texttt v \psi_0 \pm \sqrt{ 4 \texttt t^2 - 45 \texttt r \texttt v - 66 \texttt t \texttt v \psi_0 - 99 \texttt v^2 \psi_0^2} \right) \right \rbrace \]

Equilibrium amplitude equations (unconserved)

\[ \left \lbrace \begin{array}{lrl} \partial_{\psi_0} \mathcal F_{UC} = 0: & 48 A^3 \texttt v + 12 A^2 (\texttt t + 3 \texttt v \psi_0) + \psi_0 (1 + \texttt r + \texttt t \psi_0 + \texttt v \psi_0^2) &=0 \\ \partial_{A} \mathcal F_{UC} = 0: & A (\texttt r + 45 A^2 \texttt v + 4 A (\texttt t + 3 \texttt v \psi_0) + \psi_0 (2 \texttt t + 3 \texttt v \psi_0)) &= 0\\ \end{array} \right \rbrace \]

Default resolution: \([11,11,11]^{-1}a_0\)

Default model parameters \((\texttt r,\psi_0)\): \((-0.3,-0.325)\)

Free energy per unit cell (calculation document):

\[ \mathcal F_{UC} = \frac{2 \pi^3}{\sqrt 3} (1944 A^4 \texttt v + 810 B^4 \texttt v + \psi_0^2 (32 + 18 \texttt r + 12 \texttt t \psi_0 + 9 \texttt v \psi_0^2) + 108 B^2 (r + \psi_0 (2 \texttt t + 3 \texttt v \psi_0)) + 144 A^2 (\texttt r + 36 B^2 \texttt v + 6 B (t + 3 \texttt v \psi_0) + \psi_0 (2 \texttt t + 3 \texttt v \psi_0))) \]

Equilibrium amplitude equations

\[ \left \lbrace \begin{array}{lrl} \partial_{\psi_0} \mathcal F_{UC} = 0: & 54 B^2 (t + 3 \texttt v \psi_0) + \psi_0 (16 + 9 \texttt r + 9 t \psi_0 + 9 \texttt v \psi_0^2) + 72 A^2 (t + 3 \texttt v (3 B + \psi_0)) &= 0 \\ \partial_{A} \mathcal F_{UC} = 0: & A (\texttt r + 27 A^2 \texttt v + 36 B^2 \texttt v + 2 t \psi_0 + 3 \texttt v \psi_0^2 + 6 B (t + 3 \texttt v \psi_0)) &= 0 \\ \partial_{B} \mathcal F_{UC} = 0: & 15 B^3 \texttt v + 4 A^2 (t + 3 \texttt v \psi_0) + B (\texttt r + 48 A^2 \texttt v + 2 t \psi_0 + 3 \texttt v \psi_0^2) &= 0\\ \end{array} \right \rbrace \]

Default resolution: \([5,5,5]^{-1}a_0\)

Default model parameters \((\texttt r,\psi_0)\): \((-0.3,-0.325)\)

Free energy per unit cell (calculation document):

\[ \mathcal F_{UC} = \frac{2\pi^3}{3} \left (48 C^2 \texttt r + 270 A^4 \texttt v + 1620 B^4 \texttt v + 576 A^3 C \texttt v + 648 C^4 \texttt v + 96 C^2 \texttt t \psi_0 + 6 (36 + \texttt r + 24 C^2 \texttt v) \psi_0^2 + 4 \texttt t \psi_0^3 + 3 \texttt v \psi_0^4 + 192 B^3 (\texttt t + 3 \texttt v \psi_0) + 576 A B C (t + 3 \texttt v (3 B + \psi_0)) + 36 A^2 (\texttt r + 96 B^2 \texttt v + 36 C^2 \texttt v + 2 \texttt t \psi_0 + 3 \texttt v \psi_0^2 + 8 B (\texttt t + 3 \texttt v \psi_0)) + 72 B^2 (r + 54 C^2 \texttt v + \psi_0 (2 \texttt t + 3 \texttt v \psi_0)) \right ) \]

Equilibrium amplitude equations:

\[ \left \lbrace \begin{array}{lrl} \partial_{\psi_0} \mathcal F_{UC} = 0: & 8 C^2 \texttt t + 48 B^3 \texttt v + 144 A B C \texttt v + 36 \psi_0 + \texttt r \psi_0 + 24 C^2 \texttt v \psi_0 + \texttt t \psi_0^2 + \texttt v \psi_0^3 + 12 B^2 (t + 3 \texttt v \psi_0) + 6 A^2 (t + 3 \texttt v (4 B + \psi_0)) &= 0 \\ \partial_{A} \mathcal F_{UC} = 0: & 15 A^3 \texttt v + 24 A^2 C \texttt v + 8 B C (t + 3 \texttt v (3 B + \psi_0)) + A (\texttt r + 96 B^2 \texttt v + 36 C^2 \texttt v + 2 \texttt t \psi_0 + 3 \texttt v \psi_0^2 + 8 B (t + 3 \texttt v \psi_0)) &= 0 \\ \partial_{B} \mathcal F_{UC} = 0: & 45 B^3 \texttt v + 4 B^2 (t + 3 \texttt v \psi_0) + 2 A (A + 2 C) (t + 3 \texttt v \psi_0) + B (\texttt r + 48 A^2 \texttt v + 72 A C \texttt v + 54 C^2 \texttt v + 2 \texttt t \psi_0 + 3 \texttt v \psi_0^2) &= 0\\ \partial_{C} \mathcal F_{UC} = 0: & 27 C^3 \texttt v + C (\texttt r + 27 A^2 \texttt v + 81 B^2 \texttt v + 2 \texttt t \psi_0 + 3 \texttt v \psi_0^2) + 6 A (A^2 \texttt v + 9 B^2 \texttt v + B (t + 3 \texttt v \psi_0)) &= 0\\ \end{array} \right \rbrace \]

We refer to these amplitudes \(A,B,C\) as proto amplitudes and they are calculated by the functions calc_proto_amplitudes_conserved (by default) and calc_proto_amplitudes_unconserved in the pfc class. The proto amplitudes are saved in the pfc instance as eta0. The function calc_free_energy_from_proto_amplitudes calculates the free energy \(\mathcal F_{UC}\) from the proto amplitudes, including \(\psi_0\).

The amplitude approximation - deviations from the ground state

So far, we have only looked at the equilibrium state of the PFC, which has a specific symmetry, and the few-mode approximations that can be used to describe the PFC in this state. For small deviations from the equilibrium state, like in the presence of few dislocations and small strains, we can use the amplitude approximation, in which we assume that the field can be written as

The amplitude approximation

\[ \psi \approx \bar \psi + \sum_{\vec q^{(n)} \in \mathcal R^{(1)}} \eta_n(\vec r) e^{\mathfrak i \vec q^{(n)} \cdot \vec r}, \]

where \(\eta_n\) are slowly varying complex fields. These fields can be found by demodulating the field \(\psi\) with the primary RLVs, i.e.,

Demodulation

\[ \eta_n = \langle \psi e^{- \mathfrak i \boldsymbol{q}^{(n)} \cdot \boldsymbol{r}} \rangle \]

The figure below shows an example of the evolution of the PFC for a triangular lattice and the corresponding amplitude fields.

Figure: Snapshots of an example evolution of the 2D triangular PFC model at (top row) \(t=0\) and (bottom row) \(t=600\). Parameters used were \((r,\psi_0) = (-0.3,-0.3)\). The columns show the demodulated fields \(\bar \psi\) and \(\{ \eta_n \}_{n=1}^{3}\), where the complex fields are shown by their phase \(\theta_n\) and brightness corresponding to the magnitude \(|\eta_n|\). Taken from Ref. 2 with permission.

Elasticity

This quantity is the Poisson ratio, under some conditions. I think. To be fixed. \(\(\nu = \frac{\lambda}{(d-1)\lambda + 2\mu + \gamma}\)\)

Elastic constants are saved in these quantities

pfc.el_mu
pfc.el_lambda
pfc.el_gamma
pfc.el_nu

Stress tensor

The equations for calculating the stress tensor are found in Ref. 6, but we list them below, together wtih the elastic constants for each of the models. The stress tensor \(h_{ij}\) and its associated elastic constants interms of amplitudes (\(A,B,C\)) of the mode expansion for different PFCmodels. Here, \(\mathcal L_X = X+\nabla^2\). The elastic constants canbe expressed in Voigt notation by \(C_{11} = \lambda+2\mu+\gamma\),\(C_{12} = \lambda\), \(C_{44}=\mu\).

\[ \nabla \cdot h = \left \langle \tilde \mu_c \nabla \psi - \nabla \tilde f \right \rangle \]

Stress tensor

Elastic constants

\[ \lambda = \]
\[ \mu = \]
\[ \gamma = \]

Stress tensor

\[ h_{ij} = -2\left \langle (\mathcal L_1 \psi) \partial_{ij} \psi \right \rangle \]

Elastic constants

\[\lambda = 3A^2\]
\[\mu = 3A^2\]
\[\gamma = 0\]

Stress tensor $$ h_{ij} = -2\left \langle (\mathcal L_1 \mathcal L_2 \psi)(\mathcal L_1 + \mathcal L_2) \partial_{ij} \psi \right \rangle$$

Elastic constants

\[\lambda = 16B^2\]
\[\mu = 16B^2\]
\[\gamma = 8A^2-32B^2\]

Stress tensor

\[h_{ij} = -2\left \langle (\mathcal L_1 \psi) \partial_{ij} \psi \right \rangle\]

Elastic constants

\[\lambda = 4 A^2\]
\[\mu = 4 A^2\]
\[\gamma = -4A^2\]

Stress tensor

\[h_{ij} = -2\left \langle (\mathcal L_1 \mathcal L_{\frac 4 3} \psi)(\mathcal L_1 + \mathcal L_{\frac 4 3}) \partial_{ij} \psi \right \rangle\]

Elastic constants

\[\lambda = \frac{32}{81} A^2\]
\[\mu = \frac{32}{81} A^2\]
\[\gamma = \frac{32}{81} (2B^2 - A^2)\]

Stress tensor

\[h_{ij} = -2\left \langle (\mathcal L_1 \mathcal L_2 \mathcal L_3 \psi)(\mathcal L_2 \mathcal L_3 + \mathcal L_1 \mathcal L_3 + \mathcal L_1 \mathcal L_2) \partial_{ij} \psi \right \rangle\]

Elastic constants

\[\lambda = 16 B^2 + 128 C^2\]
\[\mu = 16 B^2 + 128 C^2\]
\[\gamma = 32 A^2 - 16 B^2 - 256 C^2\]

Strains

In order to calculate the strain of a PFC configuration, we follow a generalized procedure as that outlined in Ref. 2. The ground state of the PFC can be written as

\[ \psi = \psi_0 + \eta_1 \sum_{\vec q^{(n)} \in \mathcal R^{(1)}} e^{i \vec q^{(n)} \cdot \vec r} + \eta_2 \sum_{\vec q^{(n)} \in \mathcal R^{(2)}} e^{i \vec q^{(n)} \cdot \vec r} +..., \]

where \(\mathcal R^{(n)}\) is the nth closest modes, i.e., the full reciprocal lattice can be written as

\[ \mathcal R = \{\mathbf 0 \} \cup \left (\cup_{n=1}^\infty \mathcal R^{(n)} \right) \]

We define the following quantity

\[ \Phi = N_1q_1^2\eta_1^2 + N_2q_2^2\eta_2^2 + ... = \sum_{\vec q^{(n)} \in \mathcal R} \eta_{n,0}^2|\vec q^{(n)}|^2 \]

Assuming that the mode-approximations hold, we can calculate \(\Phi\) from the equilibrium amplitudes which are stored in the PhaseFieldCrystal instance, which is done routinely in the initialization of the class and stored in pfc.Phi. Consider now the structure function of the equilibrium state

\[ \mathcal S_{ij} = \langle \partial_i \psi \partial_j \psi\rangle \]
\[ = \eta_1^2 \sum_{\vec q^{(n)} \in \mathcal R^{(1)}} q^{(n)}_iq^{(n)}_j + \eta_2^2 \sum_{\vec q^{(n)} \in \mathcal R^{(2)}} q_i^{(n)}q_j^{(n)}+... \]
\[ = \eta_1^2 \frac{N_1 q_1^2}{d} \delta_{ij} + \eta_2^2 \frac{N_2 q_2^2}{d} \delta_{ij}+... \]
\[ = \frac{\Phi}{d}\delta_{ij} \]

The strain of the phase-field crystal can be calculated using the following formula The strain of the phase-field crystal can be calculated using the following formula2

\[ \varepsilon_{ij} = \frac{1}{2}\delta_{ij} - \frac{d}{2\Phi} \mathcal S_{ij}, \]

where \(\mathcal S = \langle \nabla \psi \nabla \psi \rangle\) is called the structure function 7 and is implemented in the calc_strain_tensor method of the PhaseFieldCrystal class. It should be noted that there is some residual strain due to the equilibrium periodicity of the lattice not matching exactly the periodicity of the simulation domain 10. It is therefore often useful to subtract the mean value for visualization purposes.

strain = pfc.calc_strain_tensor()
strain = strain - np.mean(strain, axis=0)

Dislocations

The Burgers vector is defined by

Burgers vector definition

\[ \oint_{\partial \mathcal M} d \boldsymbol{u} = -\boldsymbol{b}. \]

The minus sign in this convention reflects the fact that we consider the Burgers vector to be the disconnection error from the ending point to the starting point, when going an oriented path around the dislocation,

Burgers vector definition Burgers vector definition

Burgers vector definition: (a) The one-body density of a crystalline solid containing an edge dislocation in a 2D square lattice (superimposed), (b) a 3D simple cubic lattice with an edge dislocation (\(\vec b \perp \vec t\)), and (c) a 3D simple cubic lattice with a screw dislocation (\(\vec b \parallel \vec t)\). In all cases, a circulation (green) that is right-handed with respect to the tangent vector \(\vec t\), i.e., following a path around the dislocation, gives rise to a connection error: the Burgers vector \(\vec b\).
Reprinted from Ref. [^Skogvoll2023SymmetryTopology] with permission.

In three dimensions, the path defining the dislocation is given by the direction of the dislocation line tangent \(\boldsymbol{t}\). Multiplying this equation by the reciprocal lattice vector \(-\boldsymbol{q}^{(n)}\), we get

\[\oint_{\partial \mathcal M} d (-\boldsymbol{q}^{(n)} \cdot \boldsymbol{u}) = (\boldsymbol{b} \cdot \boldsymbol{q}^{(n)}) \equiv 2\pi s_n,\]

where \(s_n\) is the charge associated with the Burgers vector \(\boldsymbol{b}\)

Dislocation charge

\[ s_n = \frac{1}{2\pi} \boldsymbol{b} \cdot \boldsymbol{q}^{(n)}. \]

Using the primary BLVs we get the charges summarized below

Burgers vector \(\mathbf b\) \(s_1\) \(s_2\) \(s_3\)
\(\mathbf a^{(1)} = a_0 (1,0)\) \(\color{red} 1\) \(0\) \(\color{blue}-1\)
\(\mathbf a^{(2)} = a_0 (1/2,\sqrt 3/2)\) \(0\) \(\color{red} 1\) \(\color{blue}-1\)
\(\mathbf a^{(3)} = a_0 (1/2,-\sqrt 3/2)\) \(\color{red} 1\) \(\color{blue}-1\) \(0\)
Burgers vector \(\mathbf b\) \(s_1\) \(s_2\) \(s_3\) \(s_4\)
\(\mathbf a^{(1)} = a_0 (1,0)\) \(\color{red} 1\) \(0\) \(\color{red} 1\) \(\color{red} 1\)
\(\mathbf a^{(2)} = a_0(0,1)\) \(0\) \(\color{red} 1\) \(\color{blue}-1\) \(\color{red} 1\)
Burgers vector \(\mathbf b\) \(s_1\) \(s_2\) \(s_3\) \(s_4\) \(s_5\) \(s_6\)
\(\mathbf a^{(1)} = a_0/2 (-1,1,1)\) \(\color{red} 1\) \(0\) \(0\) \(0\) \(\color{red} 1\) \(\color{red} 1\)
\(\mathbf a^{(2)} = a_0/2 (1,-1,1)\) \(0\) \(\color{red} 1\) \(0\) \(\color{red} 1\) \(0\) \(\color{blue}-1\)
\(\mathbf a^{(3)} = a_0/2 (1,1,-1)\) \(0\) \(0\) \(\color{red} 1\) \(\color{blue}-1\) \(\color{blue}-1\) \(0\)
\(\mathbf a^{(4)} = a_0/2 (1,1,1)\) \(\color{red} 1\) \(\color{red} 1\) \(\color{red} 1\) \(0\) \(0\) \(0\)
Burgers vector \(\mathbf b\) \(s_1\) \(s_2\) \(s_3\) \(s_4\) \(s_5\) \(s_6\) \(s_7\)
\(\mathbf a^{(1)} = a_0/2(0,1,1)\) \(\color{red} 1\) \(0\) \(0\) \(\color{red} 1\) \(0\) \(\color{red} 1\) \(\color{red} 1\)
\(\mathbf a^{(2)} = a_0/2(1,0,1)\) \(0\) \(\color{red} 1\) \(0\) \(\color{red} 1\) \(\color{red} 1\) \(0\) \(\color{red} 1\)
\(\mathbf a^{(3)} = a_0/2 (1,1,0)\) \(0\) \(0\) \(\color{red} 1\) \(\color{red} 1\) \(\color{red} 1\) \(\color{red} 1\) \(0\)
\(\mathbf a^{(4)} = a_0/2(0,-1,1)\) \(0\) \(\color{red} 1\) \(\color{blue}-1\) \(0\) \(0\) \(\color{blue}-1\) \(\color{red} 1\)
\(\mathbf a^{(5)} = a_0/2 (-1,0,1)\) \(\color{red} 1\) \(0\) \(\color{blue}-1\) \(0\) \(\color{blue}-1\) \(0\) \(\color{red} 1\)
\(\mathbf a^{(6)} = a_0/2(-1,1,0)\) \(\color{red} 1\) \(\color{blue}-1\) \(0\) \(0\) \(\color{blue}-1\) \(\color{red} 1\) \(0\)
Burgers vector \(\mathbf b\) \(s_1\) \(s_2\) \(s_3\) \(s_4\) \(s_5\) \(s_6\) \(s_7\) \(s_8\) \(s_9\) \(s_{10}\) \(s_{11}\) \(s_{12}\) \(s_{13}\)
\(\mathbf a^{(1)} = a_0(1,0,0)\) \(\color{red} 1\) \(0\) \(0\) \(0\) \(\color{red} 1\) \(\color{red} 1\) \(0\) \(\color{blue}-1\) \(\color{blue}-1\) \(\color{blue}-1\) \(\color{red} 1\) \(\color{red} 1\) \(\color{red} 1\)
\(\mathbf a^{(2)} = a_0 (0,1,0)\) \(0\) \(\color{red} 1\) \(0\) \(\color{red} 1\) \(0\) \(\color{red} 1\) \(\color{blue}-1\) \(0\) \(\color{red} 1\) \(\color{red} 1\) \(\color{blue}-1\) \(\color{red} 1\) \(\color{red} 1\)
\(\mathbf a^{(3)} = a_0(0,0,1)\) \(0\) \(0\) \(\color{red} 1\) \(\color{red} 1\) \(\color{red} 1\) \(0\) \(\color{red} 1\) \(\color{red} 1\) \(0\) \(\color{red} 1\) \(\color{red} 1\) \(\color{blue}-1\) \(\color{red} 1\)

The dislocation density tensor

Given a PFC configuration, the dislocation density tensor may be calculated as 8

The dislocation density tensor

\[ \alpha_{ij} = \frac{2d}{N\eta_0^2} \sum_{n=1}^N D_i^{(n)} q_j^{(n)} = \frac{2 \pi d}{N} \sum_{n=1}^N \rho_i^{(n)} q_j^{(n)} \]

where \(\boldsymbol{D}^{(n)}\) is calculated from \(\boldsymbol{\psi }= (\Re(\eta_n), \Im(\eta_n))\) from the \(N\) primary reciprocal lattice vectors \(\boldsymbol{q}^{(n)}\) and

\[ \rho^{(n)}_i = \frac{1}{\pi \eta_0^2} D^{(n)}_i \]

In the language of this

Equations of motion

Conserved evolution

The basic equation motion presented in Ref. 1 can be derived by postulating a simple mechanism for free energy minimzation under the constraint of mass conservation.

The PFC evolution equation (evolve_PFC)

\[ \partial_t \psi = \nabla^2 \tilde \mu_c, \]

where

\[ \tilde \mu_c = \frac{\delta \mathcal F}{\delta \psi} = \left ( \mathcal L^2 \psi + \texttt r \psi + \psi^3 \right ) \]

The PFC chemical potential

which gives $$ {{\omega }{\mathfrak f}}= -\boldsymbol{k}^2 (\texttt r + {{\mathcal L}^2) $$}

\[ N = \nabla^2 (\psi^3) \]

Unconserved dynamics

The unconserved PFC evolution equation (evolve_PFC_unconserved)

\[ \partial_t \psi = - \tilde \mu_c \]

Evolution at mechanical equilibrium

The PFC evolution at mechanical equilibrium (evolve_PFC_mechanical_equilibrium)

Step 1:

\[\psi(t+\Delta t) = \textrm{Integrate($\Delta t$):} (\partial_t \psi)\]

Step 2:

\[\psi(t + \Delta t, \boldsymbol{r}) \leftarrow \psi(t + \Delta t, \boldsymbol{r} - \boldsymbol{u}^\delta),\]

where \(\boldsymbol{u}^\delta\) is the solution to

\[ \partial_{j} \mathcal C_{ijkl} \partial_l u_k = - g^\psi_i, \]

where

\[ g^\psi_i = -\partial_j \mathfrak h_{ij}^\psi, \]

and the elastic constants tensor \(\mathcal C\) is given by

\[ \mathcal C_{ijkl} = \lambda \delta_{ij}\delta_{kl} + 2\mu \delta_{k(i} \delta_{j)l} + \gamma \delta_{ijkl}. \]

This equation is solved in Fourier space by

\[ {u}_{\mathfrak f ~ i}^\delta = G_{\mathfrak f ~ ij} g_{\mathfrak f ~ j}^\psi, \]

where the Greens function \(G_{\mathfrak f ~ ij}\) is given in Ref. 9 as

\[ {G_{\mathfrak f ~ij}} (\mathbf k) =\frac{1}{\mathbf k^2}\left ( \frac{\delta_{ij}}{\mu + \gamma \kappa_{(i)}^2} - \frac{\kappa_i \kappa_j}{(\mu + \gamma \kappa_{(i)}^2 )(\mu + \gamma \kappa_{(j)}^2 )} \frac{\mu+\lambda}{1+ \sum_{l=1}^3 \frac{\mu + \lambda}{\mu+\gamma \kappa_l^2} \kappa_l^2 }\right ), \]

with \(\boldsymbol \kappa = \mathbf k/|\mathbf k|\) there is no implicit summation over indices \((i),(j)\).
By defining \(k_3=0\), this equation is also valid for the triangular and square symmetry in two dimensions.

Note that due to the asymmetry of the elastic constants, this method can only be used for small deviations of the lattice orientation.

Hydrodynamic PFC evolution

In Ref. 7, a hydrodynamic approach was derived. A simplified two-parameter model was proposed

The hydrodynamic PFC model (evolve_PFC_hydrodynamic)

\[ \partial_t \psi = \nabla^2 \tilde \mu_c - \boldsymbol{v} \cdot \nabla \psi \]
\[ \partial_t \boldsymbol{v} = \frac{1}{\rho_0} (\nabla \cdot h + \Gamma_S \nabla^2 \boldsymbol{v} + \boldsymbol{f}^{\textrm(ext)}) \]

We insert for \(\tilde \mu_c\) and write it in matrix form to emphasize the linear and non-linear parts

\[ \partial_t \begin{pmatrix} \psi \\ v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} \nabla^2 (r + \mathcal L^2) \psi \\ \frac{1}{\rho_0}\Gamma_S \nabla^2 v_1 \\ \frac{1}{\rho_0}\Gamma_S \nabla^2 v_2 \\ \frac{1}{\rho_0}\Gamma_S \nabla^2 v_3 \end{pmatrix} + \begin{pmatrix} \nabla^2 ( \psi^3) - \boldsymbol{v} \cdot \nabla \psi \\ \frac{1}{\rho_0} \left ( \left \langle \tilde \mu_c \partial_x \psi - \partial_x \tilde f \right \rangle + f_x^{(ext)}\right )\\ \frac{1}{\rho_0} \left ( \left \langle \tilde \mu_c \partial_y \psi - \partial_y \tilde f \right \rangle + f_y^{(ext)}\right ) \\ \frac{1}{\rho_0} \left ( \left \langle \tilde \mu_c \partial_z \psi - \partial_z \tilde f \right \rangle + f_z^{(ext)} \right ) \end{pmatrix} \]

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