quality:correction
Lyapunov exponent for time-varying linearization

To introduce Lyapunov exponent consider a fundamental matrix X (t) {\displaystyle X(t)} (e.g., for linearization along a stationary solution x 0 {\displaystyle x_{0}} in a continuous system), the fundamental matrix is exp (d f t (x) d x | x 0 t) {\displaystyle \exp \left(\left.{\frac {df^{t}(x)}{dx}}\right|_{x_{0}}t\right)} consisting of the linearly-independent solutions of the first-order approximation of the system. The singular values { α j (X (t)) } 1 n {\displaystyle \{\alpha _{j}{\big (}X(t){\big)}\}_{1}^{n}} of the matrix X (t) {\displaystyle X(t)} are the square roots of the eigenvalues of the matrix X (t) ∗ X (t) {\displaystyle X(t)^{*}X(t)}.The largest Lyapunov exponent λ m a x {\displaystyle \lambda _{\mathrm {max} }} is as follows λ m a x = max j lim sup t → ∞ 1 t ln α j (X (t)). {\displaystyle \lambda _{\mathrm {max} }=\max \limits _{j}\limsup _{t\rightarrow \infty }{\frac {1}{t}}\ln \alpha _{j}{\big (}X(t){\big)}.} Lyapunov proved that if the system of the first approximation is regular (e.g., all systems with constant and periodic coefficients are regular) and its largest Lyapunov exponent is negative, then the solution of the original system is asymptotically Lyapunov stable.Later, it was stated by O. Perron that the requirement of regularity of the first approximation is substantial.