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where $\mathbf{z}\left(t\right):\mathbb{R} \mapsto \mathbb{R}^m, ~\mathbf{f}\left(\mathbf{z}\right): \mathbb{R}^m \to \mathbb{R}^m$.
For this system the variational equation has the following form:
where $\mathbf{J}_\mathbf{f}\left(\mathbf{z}\right): \mathbb{R} \mapsto \mathbb{R}^{m\times m}$ is Jacobian of the function $\mathbf{f}\left(\mathbf{z}\right)$
and $\boldsymbol{\Phi}\left(t\right): \mathbb{R} \mapsto \mathbb{R}^{m\times m}$ is variational matrix.
To find out what happens to the variations, you need to solve the variational equation and the system equation simultaneously.
To do this, you work with a new augmented state vector $\mathbf{z}_\ast\left(t\right):\mathbb{R}\mapsto\mathbb{R}^{m + m^2}$ of length $m + m^2$:
The calculation of the Lyapunov exponent was based on the QR decomposition method, the application of which can be viewed via the script odeExplicitSolversLyapunovExponents.m.
odefun: function handle that defines the augmented system of ODEs to be integrated. This function must compute the derivatives for both the original system state and the flattened variational matrix;
tspan: interval of integration, specified as a two-element vector;
tau: fixed time discretization step;
incond: vector of initial conditions. This vector must consist of the initial state vector concatenated with the elements of the initial orthogonal matrix (usually the flattened identity matrix);
Method (Name-Value Pair): string specifying the explicit Runge-Kutta method to be used. Available options:
t: column vector of evaluation points used to perform the integration;
zsol: matrix in which each row corresponds to the full solution vector (state + variations) at the value returned in the corresponding row of t;
lyap_exp: matrix of Lyapunov exponents in which each row contains the spectrum $[\lambda_1, \dots, \lambda_m]$ calculated at the value returned in the corresponding row of t;
dzdt_eval: (optional) matrix of the evaluated derivative vectors at each time step.
The Kaplan—Yorke Dimension
Let the Lyapunov exponents be sorted in descending order $\lambda _{1}\geq \lambda _{2}\geq \dots \geq \lambda _{m}$, then