对于一个 n n n 维连续时间系统
d Z d t = F ( Z , t ) (1) \dfrac{\rm{d}\bold{Z}}{
{\rm d}{t}}=\bold{F}(\bold{Z},t)\tag{1} dtdZ=F(Z,t)(1)
其中 Z = ( Z 1 , Z 2 , ⋯ , Z n ) T \bold{Z}=(Z_1,Z_2,\cdots,Z_n)^T Z=(Z1,Z2,⋯,Zn)T ， F = ( F 1 , F 2 , ⋯ , F n ) T \bold{F}=(F_1,F_2,\cdots,F_n)^T F=(F1,F2,⋯,Fn)T
给定一个初始条件 Z ( 0 ) = ( Z 1 ( 0 ) , Z 2 ( 0 ) , ⋯ , Z n ( 0 ) ) T \bold{Z}(0)=\Big(Z_1(0),Z_2(0),\cdots,Z_n(0)\Big)^T Z(0)=(Z1(0),Z2(0),⋯,Zn(0))T
d δ Z d t = J δ Z (2) \dfrac{\rm{d}\delta\bold{Z}}{ {\rm d} t}={\bold J}\delta{\bold Z}\tag{2} dtdδZ=JδZ(2)
其中 J \bold J J 是 n × n n\times n n×n 的 Jacobian矩阵， J i j = ∂ F i ∂ Z j J_{ij}=\dfrac{\partial F_i}{\partial Z_j} Jij=∂Zj∂Fi
方程（2）的解可以写成
δ Z ( t ) = M ( Z ( t ) , t ) δ Z ( 0 ) \delta {\bold Z}(t)={\bold M}({\bold Z}(t),t)\delta{\bold Z}(0) δZ(t)=M(Z(t),t)δZ(0)
M ( Z ( t ) , t ) {\bold M}({\bold Z}(t),t) M(Z(t),t) 为切线映射，它的推导很简单：
d M d t = J M \dfrac{
{\rm d}{\bold M}}{
{\rm d}t}=\bold J\bold M dtdM=JM
令 λ 1 , λ 2 , ⋯ , λ n \lambda_1,\lambda_2,\cdots,\lambda_n λ1,λ2,⋯,λn 为系统的 n n n 个 Lyapunov指数， λ 1 ≥ λ 2 ≥ ⋯ ≥ λ n \lambda_1\geq\lambda_2\geq\cdots\geq\lambda_n λ1≥λ2≥⋯≥λn
选择 n n n 个正交向量作为（2）的初始状态，先令其为
e 1 ( 0 ) = ( 1 , 0 , 0 , ⋯ ) ; e 2 ( 0 ) = ( 0 , 1 , 0 , ⋯ ) , ⋯ ; e n ( 0 ) = ( 0 , 0 , ⋯ , 1 ) {\bold e}_1(0)=(1,0,0,\cdots);{\bold e}_2(0)=(0,1,0,\cdots),\cdots;{\bold e}_n(0)=(0,0,\cdots,1) e1(0)=(1,0,0,⋯);e2(0)=(0,1,0,⋯),⋯;en(0)=(0,0,⋯,1)
将其代入，得到时间 τ \tau τ 后一组向量 v 1 ( τ ) , v 2 ( τ ) , ⋯ , v n ( τ ) \pmb{v}_1(\tau),\pmb{v}_2(\tau),\cdots,\pmb{v}_n(\tau) v1(τ),v2(τ),⋯,vn(τ)，再对这些向量作施密特正交化，有
e 1 ( τ ) = v 1 ( τ ) ∣ ∣ v 1 ( τ ) ∣ ∣ , e 2 ( τ ) = v 2 ( τ ) − < v 2 ( τ ) , e 1 ( τ ) > e 1 ( τ ) ∣ ∣ v 2 ( τ ) − < v 2 ( τ ) , e 1 ( τ ) > e 1 ( τ ) ∣ ∣ , ⋮ e n ( τ ) = v n ( τ ) − < v n ( τ ) , e 1 ( τ ) > e 1 ( τ ) − ⋯ − < v n ( τ ) , e n − 1 ( τ ) > e n − 1 ( τ ) ∣ ∣ v n ( τ ) − < v n ( τ ) , e 1 ( τ ) > e 1 ( τ ) − ⋯ − < v n ( τ ) , e n − 1 ( τ ) > e n − 1 ( τ ) ∣ ∣ \bold e_1(\tau)=\dfrac{\pmb v_1(\tau)}{||\pmb v_1(\tau)||},\\[2ex]\bold e_2(\tau)=\dfrac{\pmb v_2(\tau)-<\pmb v_2(\tau),\pmb e_1(\tau)>\pmb e_1(\tau)}{||\pmb v_2(\tau)-<\pmb v_2(\tau),\pmb e_1(\tau)>\pmb e_1(\tau)||},\\[2ex]\vdots\\[2ex]\\\bold e_n(\tau)=\dfrac{\pmb v_n(\tau)-<\pmb v_n(\tau),\pmb e_1(\tau)>\pmb e_1(\tau)-\cdots-<\pmb v_n(\tau),\pmb e_{n-1}(\tau)>\pmb e_{n-1}(\tau)}{||\pmb v_n(\tau)-<\pmb v_n(\tau),\pmb e_1(\tau)>\pmb e_1(\tau)-\cdots-<\pmb v_n(\tau),\pmb e_{n-1}(\tau)>\pmb e_{n-1}(\tau)||} e1(τ)=∣∣v1(τ)∣∣v1(τ),e2(τ)=∣∣v2(τ)−<v2(τ),e1(τ)>e1(τ)∣∣v2(τ)−<v2(τ),e1(τ)>e1(τ),⋮en(τ)=∣∣vn(τ)−<vn(τ),e1(τ)>e1(τ)−⋯−<vn(τ),en−1(τ)>en−1(τ)∣∣vn(τ)−<vn(τ),e1(τ)>e1(τ)−⋯−<vn(τ),en−1(τ)>en−1(τ)
记分母部分分别为 P 1 ( 1 ) , P 2 ( 1 ) , ⋯ , P n ( 1 ) \pmb P_1(1),\pmb P_2(1),\cdots,\pmb P_n(1) P1(1),P2(1),⋯,Pn(1)。
r r r 次迭代计算后，Lyapunov 指数可以估计为：
λ i ≈ ∑ m = 1 r log 2 P i ( m ) r τ ( i = 1 , ⋯ , n ) \lambda_i\approx\dfrac{\sum_{m=1}^r\log_2\pmb P_i(m)}{r\tau}\quad\quad(i=1,\cdots,n) λi≈rτ∑m=1rlog2Pi(m)(i=1,⋯,n)