Table 7 The eigenvalues at each equilibrium point of Eq. (8).

From: Government regulatory policies for digital transformation in small and medium-sized manufacturing enterprises: an evolutionary game analysis

Equilibrium points

Eigenvalues

E1(0, 0, 1)

\({\it{\lambda }}_1^1 = {\it{a}} - {\it{b}}\), \({\it{\lambda }}_1^2 = {\it{d}} - {\it{C}}_{\it{T}} + \frac{{\it{\lambda }}}{2}{\it{\delta }}^{\it{2}}\), \({\it{\lambda }}_1^3 = - {\it{(f}} + {\it{g}} - \frac{{\it{\lambda }}}{2}{\it{\delta }}^{\it{2}}{\it{)}}\)

E2(0, 0, 0)

\({\it{\lambda }}_2^1 = {\it{a}}\), \({\it{\lambda }}_2^2 = - {\it{C}}_{\it{T}}\), \({\it{\lambda }}_2^3 = {\it{f}} + {\it{g}} - \frac{{\it{\lambda }}}{2}{\it{\delta }}^{\it{2}}\)

E3(0, 1, 1)

\({\it{\lambda }}_3^1 = {\it{a}} - {\it{b}}\), \({\it{\lambda }}_3^2 = {\it{C}}_{\it{T}} - {\it{d}} - \frac{{\it{\lambda }}}{{\it{2}}}{\it{\delta }}^{\it{2}}\), \({\it{\lambda }}_3^3 = - {\it{(f}} + {\it{g)}}\)

E4(0, 1, 1)

\({\it{\lambda }}_4^1 = {\it{a}}\), \({\it{\lambda }}_4^2 = {\it{C}}_{\it{T}}\), \({\it{\lambda }}_4^3 = {\it{f}} + {\it{g}}\)

E5(1, 0, 1)

\({\it{\lambda }}_5^1 = - {\it{(a}} - {\it{b)}}\), \({\it{\lambda }}_5^2 = {\it{c}} + {\it{d}} - {\it{C}}_{\it{T}} + \frac{{\it{\lambda }}}{2}{\it{\delta }}^2\), \({\it{\lambda }}_5^3 = - {\it{(e}} + {\it{f}} + {\it{g}} - \frac{{\it{\lambda }}}{2}{\it{\delta }}^2{\it{)}}\)

E6(1, 0, 0)

\({\it{\lambda }}_6^1 = - {\it{a}}\), \({\it{\lambda }}_6^2 = {\it{c}} - {\it{C}}_{\it{T}}\), \({\it{\lambda }}_6^3 = {\it{e}} + {\it{f}} + {\it{g}} - \frac{{\it{\lambda }}}{2}{\it{\delta }}^2\)

E7(1, 1, 0)

\({\it{\lambda }}_7^1 = - {\it{a}}\), \({\it{\lambda }}_{\it{7}}^{\it{2}} = {\it{C}}_{\it{T}} - {\it{c}}\), \({\it{\lambda }}_7^3 = {\it{e}} + {\it{f}} + {\it{g}}\)

E8(1, 1, 1)

\({\it{\lambda }}_8^1 = - {\it{(a}} - {\it{b)}}\), \({\it{\lambda }}_8^2 = {\it{C}}_{\it{T}} - {\it{c}} - {\it{d}} - \frac{{\it{\lambda }}}{2}{\it{\delta }}^2\), \({\it{\lambda }}_8^3 = - {\it{(e}} + {\it{f}} + {\it{g)}}\)

E9, E10, E11, E12

At least one eigenvalue is greater than zero.