
The NLO BK equation in the case of zero active quark flavors
(\(n_f = 0\) ) reads [1]}
\(\partial _Y \left\langle S^{({{2}})}_{\ib {{{x,y}}}} \right\rangle = \frac{\alpha _\mathrm {s}N_\mathrm {c}}{2\pi ^2} K_1^\text{BC} \otimes \langle D_1 \rangle \\+ \frac{\alpha _\mathrm {s}^2 N_\mathrm {c}^2}{16\pi ^4} K_{2,1} \otimes \langle D_{2,1} \rangle + \frac{\alpha _\mathrm {s}^2 N_\mathrm {c}^2}{16\pi ^4} K_{2,2} \otimes \langle D_{2,2} \rangle \\+ \mathcal {O}(n_f),\) 

Уравнение NLO BK в случае нулевого количества активных кварковых флаворов (\(n_f = 0\)) записывается [1]:

\(\partial_Y \left\langle S^{({{2}})}_{\ib {{{x,y}}}} \right\rangle = \frac{\alpha_s N_c}{2\pi^2} K_1^\text{BC} \otimes \langle D_1 \rangle \\+ \frac{\alpha_s^2 N_c^2}{16\pi^4} K_{2,1} \otimes \langle D_{2,1} \rangle + \frac{\alpha_s^2 N_c^2}{16\pi^4} K_{2,2} \otimes \langle D_{2,2} \rangle \\+ \mathcal{O}(n_f),\