Given a smooth map f:Rm→Rm and initial value v0∈Rm, let’s say Jn:=Dfn(v0)∈Rm×m. For k=1,⋯,m, consider the length of the k-th longest axis of the ellipsoid JnN, which is the unit sphere N:={x∈Rm:∥x∥2=1} in m dimensions, as rk(n). Now, the k-th Lyapunov numberLk of v0 is defined as follows:
Lk:=n→∞lim(rk(n))1/n
The k-th Lyapunov exponent of v0 is defined as in hk:=lnLk.
In rk(n), the superscript (n) does not mean power or number of derivatives, but implies that the map f has been applied n times.
Obviously, in definitions, L1≥L2≥⋯≥Lm, and h1≥h2≥⋯≥hm, usually only the largest h1 that is meaningfully comparable with 0 is considered. This is similar to comparing L1 with 1, but with less clarity than the Lyapunov exponent which only requires consideration of sign, thus it’s not commonly used. However, when studying the concept, the Lyapunov number is more intuitive and thus helpful.
Description
The Lyapunov number for a multidimensional map is literally an extension of the 1-dimensional map’s Lyapunov number, defined using the axes of the ellipsoid. Here, the axis formed by the Jacobian of map fn means that fn is moving the point in the direction of that axis. If the size of this axis is larger than 1, it implies expansion, and if smaller, contraction in that direction. Considering that the Lyapunov number of a 1-dimensional map is defined based on whether the magnitude of the derivative is greater or smaller than 1, i.e., based on increment or decrement, this generalization to include nonlinear maps is quite reasonable.
Moreover, reconsidering the Lyapunov number from the perspective of numerical calculation, we inevitably think of the singular value decomposition of matrices as the value of n increases, leading to extremely large singular values σ1 and very small singular values σm. Unlike humans, computers have limitations in storing such numbers, and even without that, calculating Jn=Dfn(v0) is a challenge itself. Therefore, it’s generally better to avoid directly calculating JnN and to use a numerically smarter method instead.
For some U(i), if the size of the axis of the ellipsoid JnN and JnU(i) is the same, it does not matter if we calculate JnU(i). According to the chain rule,
JnU(0)=Df(vn−1)⋯Df(v0)N
we will calculate from the right-hand side term Df(v0)N in order towards the left. Given that N is a unit sphere, if we consider N as the orthogonal basis N=[w1(0)⋯wm(0)],
z1=Df(v0)w1(0)⋮zm=Df(v0)wm(0)
then,
JnU(0)=Df(vn−1)⋯Df(v1)[z1⋯zm]
By applying Gram-Schmidt orthogonalization to the obtained [z1⋯zm], we get the orthogonal basis [y1(1)⋯ym(1)]. Considering not to acquire too large values, if we set it as [w1(1)⋯wm(1)]:=[y1(1)2y1(1)⋯ym(1)2ym(1)], for some U(1),
JnU(1)=Df(vn−1)⋯Df(v1)[w1(1)⋯wm(1)]
Repeating this calculation yields,
JnU(n)=[w1(n)⋯wm(n)]
Therefore, the length of the k-th axis of ellipsoid JnN is similar to wk(n)21/n. Meanwhile, yk(i)2 represents the increment or decrement in the k-th direction with each repetition, so the length of the axis becomes rk(n)≈yk(1)2⋯yk(n)2. Thus, for sufficiently large n,
hk=≈≈=lnn→∞lim(rk(n))1/nln(rk(n))1/nln(yk(1)2⋯yk(n)2)1/nn1i=1∑nlnyk(i)2