Discretization of the free energy

Let, as before, $\Omega$ denote the domain occupied by the superconductor, and let $\Omega_H$ denote the hole (if any). The exact expression for the Gibbs' free energy is

$${\cal G} \int_{\Omega \,\cup\, \Omega_H} \left \{ \vert\psi\vert^2 \left (... ...rac{1}{1-T} \left \vert (-\imath \nabla - {\bf A}) \psi \right \vert^2 \right .$$ =

$$\left . + \frac{\kappa^2}{(1-T)^2} \left [ \vert\nabla \times {\b... ...ert^2 - 2 {\bf H}_e \cdot (\nabla \times {\bf A}) \right ] \right \} ~d\,\Omega$$ (31)

where the terms involving $\psi$ must be taken as identically zero in $\Omega_H$ (the hole).

From the previous definitions ${\cal G}$ in cell (i,j) is approximated by

$${\cal G}_{i,j} \frac{a_x a_y}{4} \left ( \frac{\vert\psi_{i,j}\vert^4}{2}-\vert\... ...j}\vert^2 +\frac{\vert\psi_{i+1,j}\vert^4}{2}-\vert\psi_{i+1,j}\vert^2 \right .$$ =

$$\left . +\frac{\vert\psi_{i+1,j+1}\vert^4}{2}-\vert\psi_{i+1,j+1}\vert^2 +\frac{\vert\psi_{i,j+1}\vert^4}{2}-\vert\psi_{i,j+1}\vert^2 \right )$$

$$+ \frac{a_x a_y}{2(1-T)} \left ( \frac{\vert U^x_{i,j} \psi_{i+1,... ...+ \frac{\vert U^x_{i,j+1} \psi_{i+1,j+1} - \psi_{i,j+1}\vert^2}{a_x^2} \right .$$

$$\left . +\frac{\vert U^y_{i,j} \psi_{i,j+1} - \psi_{i,j}\vert^2}{... ...+ \frac{\vert U^y_{i+1,j} \psi_{i+1,j+1} - \psi_{i+1,j}\vert^2}{a_y^2} \right )$$

$$+ \frac{\kappa^2a_x a_y}{(1-T)^2} \frac{\imath \ln L_{i,j}}{a_x a_y} \left ( \frac{\imath \ln L_{i,j}}{a_x a_y} - 2 H_e \right )$$ (32)

Notice that, since L_i,j approximates $\exp \left ( - \imath a_x a_y B_z \right )$ at cell (i,j), and since a_x, a_y and B_z are small (remember that distances are in units of $\xi (0)$ and fields in units of H_c2(0)), the imaginary part of $\ln L_{i,j}$ must lie in the branch that is closest to zero, i.e., between $-\pi$ and $\pi$. Again, the terms involving $\psi$ must be taken as identically zero if Bulk (i,j) is false.