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This continues from How to understand the largest Lyapunov exponent?

It is said that we can differentiate the equation, $$\tau\frac{dh_i}{dt} = F_i = -h_i + \sum_{j=1}^N J_{ij} \phi(h_j),$$ against $h_j$, and get $\frac{\partial F_i}{\partial h_j}|_{t=t_s} -\delta_{ij} + J_{ij} \phi'(h_j(t_s))$ (we denote this Jacobian as $D_{ij}(t_s)$).

This is to get the Jacobian (the first order derivative of $f$ in $x'=f(x)$). Is this step (differentiating the right side against arbitrarily chosen variable) mathematically robust?


It is said that

Briefly, calculating the Lyapunov spectrum involves two steps: First, we evolve an initially orthonormal system Q in the tangent space along the trajectory using the Jacobian D. To this end, the variational equation $\tau$Q = D(t)Q has to be integrated..

Why is the equation called 'variational'? (Possibly because variation methods deal with small perturbation of an integrand of an integral, or other functionals, and Q is such a perturbation.)

Is 'integrating' the equation here the same as solving the equation (and therefore get the integral curve)?

Here is the orginal paper:

fig1
fig2

Charlie Chang
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  • Note that there is a dot in $\tau\dot Q=D(t)Q$, it partially overlaps the lower bow of the "g" above it. – Lutz Lehmann May 15 '22 at 16:36
  • Thanks for pointing that out. But then it confuses me that Q is perturbation (i.e. $dh_i$), then the left side of equ1 should be $\tau$Q, why $\dot Q$? – Charlie Chang May 15 '22 at 17:36
  • I see it is because if we use Taylor exp. then we have $\delta F(x)= F(x+\delta x) - F(x) = F'(x)\delta x$, not (as I had thought to be) $F(x) = F'(x) \delta x$ (I thought $F(x+\delta x) = F(x), F(x)=0$). So the left side should be $\dot {(x+\delta x)} - \dot x=\dot {\delta x} (= \dot Q c)$, where $c$ is the representation of direction. – Charlie Chang May 17 '22 at 09:31
  • Yes. (You notice the problem with using $δ$ as a symbol for a difference vs. as one of the infinitesimal parameters. When in doubt, I'd use it as a number and write $x=x_0+δ·x_1+δ^2·x_2+...$ for the perturbation series.) – Lutz Lehmann May 17 '22 at 09:40
  • Right. Above I used $\delta$ for a difference. If so, more strictly, it should be $\delta F(x)= F(x+\delta x) - F(x) = F'(x)\delta x+O((\delta x)^2) \ \quad = \frac {D(t)Qc +O((\delta x)^2)} \tau \ \quad =\dot {(x+\delta x)} - \dot x=\dot {\delta x} (= \dot Q c)$ – Charlie Chang May 17 '22 at 09:45

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For any differential equation $\dot x=F(x)$, $x_0=x_0$, the variation of the initial point $x_0+\delta·v_0$ produces a close-by solution $x(t)+δ⋅v(t)$. Here $v(t)$ will also depend on $δ$ internally. If one wants to compute the leading term of $v$, then one can use the linearized equation $$ \dot v=\frac{F(x+δ⋅v)-F(x)}{δ}\xrightarrow{δ\to 0}F'(x)·v. $$ (If $F$ is sufficiently smooth.) As this equation describes how variations of the initial point develop, it may be called the variational equation.

Lutz Lehmann
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  • So 1) it is because the tangent Q is perturbation / variation. 2) from $\dot x=F(x)$ we can (by replacing $x$ with $x+\delta x$ and using the linearity of $\frac{d}{dt}$) deduce an equation of $\dot {\delta x}$, i.e. $\delta F(x)$, which equals $F'(x)\delta x$ (Taylor expansion). $\n$ Let $\delta x =\epsilon v$.//As far as I know variation equation often involves a constant like $\epsilon$ times variation (right?), which does not explicitly exist in the original equ of Q. – Charlie Chang May 17 '22 at 08:50
  • Yes. But $Q$ is more the generating matrix for all perturbations, $δx=\varepsilon,Qc$ where the parameter vector $c$ selects the initial direction of the perturbation and $\varepsilon$ the size of the deviation from $x$. – Lutz Lehmann May 17 '22 at 08:56
  • Yes. The columns of $Q$ can be regarded as tangents/perturbations in various independent directions (or basis for tangents). $\hspace{1em}$ Why do we need $\epsilon$, which seems redundant to me, since varying the norm of $v$ would be exactly $\epsilon v$ – Charlie Chang May 17 '22 at 09:01
  • Because it is easier to think with only one infinitesimal quantity. The idea is that $c$ or $v(0)$ is fixed and the perturbation is approximated by partial sums of the perturbation series for $ε\to 0$, $$δx=εQ_1[c]+ε^2Q_2[c,c]+....$$ Again, limits in one variable are easier to think about than simultaneous limits in multiple directions. – Lutz Lehmann May 17 '22 at 09:09
  • ..Correction: 'by replacing with + and using the linearity of $\frac d{dt}$'.. and substracting $x$ from + – Charlie Chang May 17 '22 at 09:35