N = 3 + P | Name of the solution | Restriction conditions |
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P = 0 | {BSM} | ${k}_{1}={k}_{2}^{* }={k}_{1r}+{\rm{i}}{k}_{1i},\quad {p}_{1}={p}_{2}^{* }\,=\,{p}_{1r}+{\rm{i}}{p}_{1i}$, | | | $\tfrac{{k}_{1r}}{{k}_{3}}=\tfrac{{k}_{1r}\ {p}_{1r}-{k}_{1i}\ {p}_{1i}}{{k}_{3}\ {p}_{3}}=\tfrac{{k}_{1r}^{3}-3{k}_{1r}\ {k}_{1i}^{2}}{{k}_{3}^{3}}$ | P = 1 | {BSM, S} | the same as above | P = 2 | {BSM, B} | ${k}_{1}={k}_{2}^{* }={k}_{1r}+{\rm{i}}{k}_{1i},\quad {p}_{1}={p}_{2}^{* }\,=\,{p}_{1r}+{\rm{i}}{p}_{1i}$, | | | ${k}_{4}={k}_{5}^{* }={k}_{4r}+{\rm{i}}{k}_{4i},\quad {p}_{4}={p}_{5}^{* }={p}_{4r}+{\rm{i}}{p}_{4i}$ | | | | $\tfrac{{k}_{1r}}{{k}_{3}}=\tfrac{{k}_{1r}\ {p}_{1r}-{k}_{1i}\ {p}_{1i}}{{k}_{3}\ {p}_{3}}=\tfrac{{k}_{1r}^{3}-3{k}_{1r}\ {k}_{1i}^{2}}{{k}_{3}^{3}}$, | P = 2 | {BSM, L} | ${k}_{1}={k}_{2}^{* }={k}_{1r}+{\rm{i}}{k}_{1i},\quad {p}_{1}={p}_{2}^{* }\,=\,{p}_{1r}+{\rm{i}}{p}_{1i}$, | | | ${k}_{4}={k}_{5}^{* }={k}_{4r}+{\rm{i}}{k}_{4i},\quad {p}_{4}={p}_{5}^{* }={p}_{4r}+{\rm{i}}{p}_{4i}$, | | | $\tfrac{{k}_{1r}}{{k}_{3}}=\tfrac{{k}_{1r}\ {p}_{1r}-{k}_{1i}\ {p}_{1i}}{{k}_{3}\ {p}_{3}}=\tfrac{{k}_{1r}^{3}-3{k}_{1r}\ {k}_{1i}^{2}}{{k}_{3}^{3}}$, | | | ${k}_{4r}^{2}+{k}_{4i}^{2}\to 0$ | P = 3 | {BSM, B, S} | the same as what's used in {BSM, B} | P = 3 | {BSM, L, S} | the same as what's used in {BSM, L} | P = 3 | {BSM, BSM} | ${k}_{1}={k}_{2}^{* }={k}_{1r}+{\rm{i}}{k}_{1i},\quad {p}_{1}={p}_{2}^{* }\,=\,{p}_{1r}+{\rm{i}}{p}_{1i}$, | | | ${k}_{4}={k}_{5}^{* }={k}_{4r}+{\rm{i}}{k}_{4i},\quad {p}_{4}={p}_{5}^{* }\,=\,{p}_{4r}+{\rm{i}}{p}_{4i}$, | | | $\tfrac{{k}_{1r}}{{k}_{3}}=\tfrac{{k}_{1r}\ {p}_{1r}-{k}_{1i}\ {p}_{1i}}{{k}_{3}\ {p}_{3}}=\tfrac{{k}_{1r}^{3}-3{k}_{1r}\ {k}_{1i}^{2}}\ {{k}_{3}^{3}}$, | | | $\tfrac{{k}_{4r}}{{k}_{6}}=\tfrac{{k}_{4r}\ {p}_{4r}-{k}_{4i}\ {p}_{4i}}{{k}_{6}\ {p}_{6}}=\tfrac{{k}_{4r}^{3}-3{k}_{4r}\ {k}_{4i}^{2}}{{k}_{6}^{3}}$ |
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