Discretization of the TDGL equations

In the following, discrete approximations for each term of (3)-(4) are derived, maintaining second order accuracy in space.

Term

$(- \imath \nabla - {\bf A})^2 \psi$ : From the identity

$$(-\imath \nabla - {\bf A} )^2 \psi = - {\overline {\cal{U}}}... ...) - {\overline {\cal{U}}}^y \partial^2_{yy} ({\cal{U}}^y \psi)$$

a second order approximation at (x_i,y_j) reads

$$\left . (-\imath \nabla - {\bf A} )^2 \psi \right \vert _{(x... ... \psi_{i,j} + {\overline U}^{x}_{i-1,j} \psi_{i-1,j}}{a_x^2}$$

$$- \frac {U^{y}_{i,j} \psi_{i,j+1} - 2 \psi_{i,j} + {\overline U}^{y}_{i,j-1} \psi_{i,j-1}}{a_y^2} + {\cal{O}}(a_x^2+a_y^2)$$ (10)

Term

$( \vert\psi\vert^2-1) \psi$ : It is readily approximated by

$$({\overline \psi}_{i,j} \psi_{i,j} - 1) \psi_{i,j}$$ (11)

Term

${\rm Re} \left[ {\overline \psi} \left( -\imath\nabla -{\bf A}\right) \psi \right]$ : From the identity

$$\left ( - \imath \,\partial_x - A_x \right ) \psi = - \imath\, {\overline {\cal{U}}^x} \partial_x ({\cal{U}}^x \psi )$$

it follows that

$$\left . {\rm Re} \left[ {\overline \psi} \left( -\imath\, \p... ...},y_j}= ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

$$= {\rm Im} \left ( \, \frac{ {\overline {\cal{U}}^x_{i,j}} ... ...{\cal{U}}^x_{i,j} \psi_{i,j}}{a_x} \right ) + {\cal{O}}(a_x^2)$$

$$= \frac{1}{a_x} {\rm Im} \left ( {\overline \psi_{i,j}} {\ov... ...{i,j}} { U}^x_{i,j} \psi_{i+1,j} \right ) + {\cal{O}}(a_x^2)$$ (12)

and analogously for the y component.

Term

$\nabla \times \nabla \times {\bf A}$ ( $=\nabla \times {\bf B}$) : We introduce as auxiliary variable

$$L_{i,j} = U^x_{i,j} U^y_{i+1,j} {\overline U}^x_{i,j+1} {\overline U}^y_{i,j}$$ (13)

In the program, the corresponding array is bloop(i,j). From this and Stokes' identity it follows that

$$L_{i,j}= \exp \left ( - \imath a_x a_y B_z (x_i+\frac{a_x}{... ...y}{2}) \right ) \left ( 1 + {\cal{O}}(a_x^4 + a_y^4) \right )$$ (14)

so that, since ${\bf B} = (0,0,B_z)$ and thus $\nabla \times {\bf B} = (\partial_y B_z, - \partial_x B_z, 0)$, we can use the approximations

$$\partial_y B_z (x_i+\frac{a_x}{2},y_j) \frac{\imath}{a_x a_y^2} \left ( {\overline L_{i,j-1}} L_{i,j} - 1 \right ) + {\cal{O}}(a_x^2+a_y^2)$$ = (15)

$$-\partial_x B_z (x_i,y_j+\frac{a_y}{2}) \frac{\imath}{a_x^2 a_y} \left ( {\overline L_{i,j}} L_{i-1,j} - 1 \right ) + {\cal{O}}(a_x^2+a_y^2)$$ = (16)

Term

${\partial_t {\bf A}}$ : From

$${\partial_t} \left [ {\overline {\cal{U}}}^x(x,y,t) {\cal{U... ...delta,y,t) \int_x^{x+\delta} {\partial_t A_x} (\xi,y,t) ~d\xi$$

$$=- \imath \,\delta~{\overline {\cal{U}}}^x(x,y,t) {\cal{U}}^... ...\partial_t A_x} (x+\frac{\delta}{2},y,t) + {\cal{O}}(\delta^2)$$

it follows that

$${\partial_t A_x} (x_i+\frac{a_x}{2},y_j,t) = \frac{\imath}{a... ...\overline U}^{x}_{i,j} \partial_t U_{i,j}^x + {\cal{O}}(a_x^2)$$ (17)

and similarly for $\partial_t A_y$.

Collecting the previous results, the numerical method for interior nodes reads:

$$\partial_t \psi_{i,j} \frac {U^{x}_{i,j} \psi_{i+1,j} - 2 \psi_{i,j} + {\overline U}^{x... ...si_{i,j+1} - 2 \psi_{i,j} + {\overline U}^{y}_{i,j-1} \psi_{i,j-1}}{\eta a_y^2}$$ =

$$- \frac{1-T}{\eta} ({\overline \psi}_{i,j} \psi_{i,j} - 1) \psi_{i,j} +\tilde{f}_{i,j}$$ (18)

$$\partial_t U^x_{i,j} - \imath (1-T) U^x_{i,j} {\rm Im} \left ( {\overline \psi_{i,j}} ... ...frac{\kappa^2}{a^2_y} U^x_{i,j} \left ( {\overline L_{i,j-1}}L_{i,j}-1 \right )$$ = (19)

$$\partial_t U^y_{i,j} - \imath (1-T) U^y_{i,j} {\rm Im} \left ( {\overline \psi_{i,j}} ... ...frac{\kappa^2}{a^2_x} U^y_{i,j} \left ( {\overline L_{i,j}}L_{i-1,j}-1 \right )$$ = (20)

Finally, a simple forward-Euler scheme is adopted to discretize the time variable with step $\Delta t$, namely

$$\psi_{i,j}(t+\Delta t) = \psi_{i,j}(t) + \Delta t \, \partial_t \,\psi_{i,j}(t)$$ (21)

and analogously for U^x_i,j and U^y_i,j. Notice that the random force $\tilde{f}$ is also treated as a vertex variable. At each vertex it is selected from a Gaussian distribution with zero mean and standard deviation $\sigma$ given by

$$\sigma =\sqrt{(\pi E_0/6 \Delta t)(T/T_c)}$$

as done in Ref. [5].