Footnotes

... symmetry¹

This proposition assumes that $N_1$ is even. A similar statement can be made when $N_1$ is odd.

... discretizing²

Symmetric difference equations were used to discretize the continuous derivatives. For example, let $\Delta_1 = \frac 1 N_1$ and $\Delta_2 = \frac 1 N_2$ and then $\frac {\partial^2 f}{\partial x_1^2}$ was replaced by $\frac {f(x_1+\Delta_1,x_2,x_3) - f(x_1-\Delta_1,x_2,x_3)}{\Delta_1^2}$ and $\frac {\partial^2 f}{\partial x_1 \partial x_2}$ was replaced by $\frac {f(x_1+\Delta_1,x_2+\Delta_2,x_3) - f(x_1-\Delta_1,x_2+\Delta_2,x_3) + f... ...lta_1,x_2-\Delta_2,x_3) - f(x_1+\Delta_1,x_2-\Delta_2,x_3)} {\Delta_1 \Delta_2}$.

... equation³

The notation $\frac {\partial C}{\partial a}$ where $C$ is a scalar-valued function and $a$ is a $3 \times 1$ vector is defined as the $3 \times 1$ vector $[\frac {\partial C}{\partial a_1}, \frac {\partial C}{\partial a_2}, \frac {\partial C}{\partial a_3}]^T$.

... derivatives⁴

The coefficients $\tilde{\mu}[k]$ and $\tilde{\eta}[k]$ are assumed to be constant when computing the partial derivatives.

... equations⁵

The gradient of $T_d$ evaluated at $n$ is computed as $[\frac {N_1}{2}(T_d[n_1+1,n_2,n_3] - T_d[n_1-1,n_2,n_3]), \frac {N_2}{2}(T_d[... ..._d[n_1,n_2-1,n_3]), \frac {N_3}{2}(T_d[n_1,n_2,n_3+1] - T_d[n_1,n_2,n_3-1])]^T$

... units⁶

The minimum was computed as $256 \times 0.002234$ and the maximum as $320 \times 0.002234$.

... units⁷

The minimum was computed as $192 \times 0.004538$ and the maximum as $256 \times 0.004538$.

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