$, which will be useful in later analysis for the estimate of its growth.

\[l.MaxPrin\] For all $ t {\geqslant}0 $ and $ -N {\leqslant}k {\leqslant}N $, we have $${\mathfrak{v}}_k(t) {\geqslant}0, \ \ \ |{\mathfrak{v}}_k(t)| {\leqslant}\max_k |{\mathfrak{v}}_k(0)|,\ \ \ |{\mathfrak{u}}_k(t)| {\leqslant}{\mathfrak{v}}_k(t).$$

Note that for the coefficients in the summation part of equation we have $ \frac{1}{(x_l - x_k)^2} >0 $. Therefore for $ f(t) := \min_k {\mathfrak{v}}_k(t) $, we have $$f'(t) {\geqslant}\dfrac{1}{2} \left( 1-\dfrac{1}{\xi} \right) \dfrac{1}{x_1^2} f(t).$$ Combined with the fact $ {\mathfrak{v}}_k(0) {\geqslant}0 $, this gives us the first claim. For the second claim, since $ {\mathfrak{v}}_k $’s are nonnegative, we know that $ \frac{{\mathrm{d}}}{{\mathrm{d}}t}\max_{k} {\mathfrak{v}}_k  {\leqslant}0 $ and this yields the desired result. The third claim follows from linearity. We know that both $ {\mathfrak{v}}+ {\mathfrak{u}}$ and $ {\mathfrak{v}}- {\mathfrak{u}}$ satisfy the equation , and we also have $ ({\mathfrak{v}}+ {\mathfrak{u}})(0) {\geqslant}0 $ and $ ({\mathfrak{v}}- {\mathfrak{u}})(0) {\geqslant}0 $. Similarly to the first claim, this gives us $ {\mathfrak{v}}_k(t) + {\mathfrak{u}}_k(t) {\geqslant}0 $ and $ {\mathfrak{v}}_k(t) - {\mathfrak{u}}_k(t)  {\geqslant}0 $, which completes the proof.

Due to our choice for the initial value of $ {\mathfrak{v}}$, we will have that $ \{{\mathfrak{v}}_k\}_{-N {\leqslant}k {\leqslant}N} $ are symmetric with respect 