This question stems from the confusion that I feel after reading this popular blog post by Sabine Hossenfelder. It is based on this paper which is paywalled, unfortunately.

The claim is the following:

Rather, as presented in his 1969 Tellus paper, Lorenz intended the phrase to describe the existence of an absolute finite-time predicability barrier in certain multi-scale fluid systems, implying a breakdown of continuous dependence on initial conditions for large enough forecast lead times.

The breakdown of continuity came as a complete surprise to me, and in fact, here's my draft of the proof of the converse:

  1. Consider the linear vector space of functions with compact support over the phase space, augmented with the usual $L_2$ norm.
  2. Define the action of the (classical) Hamiltonian on the functions by $ \hat{H} A = - i \left\{H, A \right\}_{PB}$. Observe that with this definition, the operator is self-adjoint (can be proven by an argument involving integration by parts).
  3. By Stone's theorem, there must be a strongly continuous 1-parametric group of time translations, hence the breakdown of continuous dependence on initial conditions is impossible.

Since my conclusion apparently contradicts the conclusion in the abstract of Palmer et al., I would like to know what exactly goes wrong that can lead to a finite-time predictability barrier in systems exhibiting the "real butterfly effect".

Update: a much simpler argument by a friend of mine: take $U(T/2)$ (evolution operator associated to time interval $T/2$ where $T > 0$ is the supposed predictability barrier). By construction it is continuous, as $T / 2 < T$. Hence, $U(T/2)^4 = U(2T)$ is also continuous. This allows us to see into the future that is more distant than the barrier with a continuous operator, which is in contradiction with the original assumption. We conclude that $T$ must be either infinite or zero.

  • $\begingroup$ What guarantees that there is only one parameter that is being varied? $\endgroup$ – probably_someone Sep 6 '20 at 20:46
  • $\begingroup$ @probably_someone I’m not sure I understand your question, but typically physical time is the only evolution parameter in classical mechanics. $\endgroup$ – Prof. Legolasov Sep 6 '20 at 20:49
  • $\begingroup$ But when you're talking about sensitivity to initial conditions, then those initial conditions are the parameters that are being varied, correct? And the notion of continuity that you need to demonstrate is not solely with respect to time, but also with respect to movement in the parameter space of initial conditions. $\endgroup$ – probably_someone Sep 6 '20 at 20:53
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    $\begingroup$ @probably_someone the continuity is in the phase space topology, so any parameter would do actually. $\endgroup$ – Prof. Legolasov Sep 6 '20 at 20:54
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    $\begingroup$ I just skimmed the referenced paper - what Lorenz "intended" to say is not what actually happens. They show in this paper that the system Lorenz described does, in fact, show continuous dependence on initial conditions. In fact, the paper gave no definitive counterexample; the only possible counterexample suggested was Navier-Stokes in three dimensions, and whether that exhibits continuous dependence on initial conditions is unproven so far. $\endgroup$ – probably_someone Sep 7 '20 at 0:38

There are some issues with the post on the website. I cannot understand how to pass form Navier-Stokes equations, which are PDEs, to some ODE system (Lorenz' equations?) which seem to be the subject of the discussion (also looking at the formulation of the question by Prof.Legolasov). Unfortunately I do not have access to the paper.

From a pure mathematical viewpoint everything is clear, so that the problem must concern the physical modeling: some breakdown of the mathematical description should take place at some level for physical reasons, but without reading the paper it is difficult to discuss it.

Here is a brief description of the mathematical scenario regarding the dependence on initial data.

Every (autonomous) ODE system as Lorenz' one can be written as $$\frac{dx}{dt} = F(x(t))\tag{1}$$ where $x\in M$, $M$ being some $C^k$ manifold and $F$ a $C^k$ vector field on $M$ with $k\geq 1$.

Now it is a standard result of ODE theory that, if $x=x(t|x_0)$ is the maximal solution of (1) with initial condition $x(0)= x_0\in M$, thus defined in an open interval $I_{x_0}\ni 0$, then

(1) the set $$D:= \left\{(t,x_0) \in \mathbb{R}\:\left|\: t \in I_{x_0}, x_0 \in M \right.\right\}$$ is open in $\mathbb{R}\times M$

(2) the map $$\Phi : D\ni (t,x_0) \mapsto \Phi_t(x_0):= x(t,x_0) \in M$$ is jointly $C^k$ (and $C^{k+1}$ in the variable $t$).

Hence, in the full domain of the solution there is a $C^k$ (hence continuous) dependence form initial data. In particular, when fixing some $T$, the map $$M_T \ni x \mapsto \Phi_T(x)\:,$$ where $M_T = \{x \in M \:|\: (x,T) \in D\}$, is necessarily continuous.

If instead we are dealing with proper PDEs, Navier-Stokes equations precisely, things are much more delicate and, as is very well known. Just the proof of an existence and uniqueness theorem for given initial data (which are now functions) is problematic. Continuous dependence form initial date is even more problematic.

Regarding Prof.Legolasov's suggestion I have some problems with it.

(a) Are we referring to Lorenz' ODE system? That is non Hamiltonian as the manifold has odd dimension, so what is $H$?

(b) Are we instead using some non-symplectic Poisson structure?

(c) Next, even referring to $L^2(M)$ (what measure?) over a Poisson manifold $M$, $$-i\{H, \cdot\} : C^\infty_c(M) \to L^2(M)$$ is symmetric for instance in $\mathbb{R}^{2n}$ referring to the natural Euclidean structure and the standard symplectic structure in orthonormal Cartesian coordinates, but it is not necessarily essentially selfadjoint (essentially selfadjointness of PDE on Riemannian manifolds is a delicate issue and general results are known for elliptic operators and $-i\{H, \cdot\}$ is not elliptic in general [what Riemannian metric in general when we are endowed with a Poisson structure only?]).

(d) Finally, even if we produce a unitary strongly continuous group generated by some selfadjoint extension of $-i\{H, \cdot\}$, I cannos see the relation with the continuous dependence from the initial data of the ODE associated to $H$.

  • $\begingroup$ (a) good point, (c) Louisville measure, I believe such $H$ is essentially self-adjoint though I'll need to check that again when I have the chance, (d) the idea is that if $U$ is strongly continuous, then for an approximation of initial data by a converging sequence of probability distributions the corresponding time-translated sequence will converge, too; hence continuity w.r.t. initial data. $\endgroup$ – Prof. Legolasov Sep 7 '20 at 16:04
  • $\begingroup$ Il the manifold is symplectic and you use the Liouville measure, then you have that your operator is in fact symmetric, but I do not know if it admits selfadjoint extensions. I find a bit involved your procedure, especially because one has standard results on odes at disposal , however it could work. $\endgroup$ – Valter Moretti Sep 7 '20 at 16:22

Here is what is probably missing.

That predictability limit required the existence of a feedback mechanism which operates during each time step in the simulation. The feedback cumulatively amplifies the sensitivity to initial conditions and magnifies the effects of rounding and discretization errors in the algorithms and thereby causes the system to go divergent after a certain number of iterations. This effect is not seen if the feedback path is broken.

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    $\begingroup$ I don't see how this addresses the argument above. It seems like you're using a different (but fuzzy) definition of "go divergent" or "finite-time limit of predictability". Could you try to explain more rigorously what you mean, and contrast it with the definitions used in the question? $\endgroup$ – Daniel Sep 7 '20 at 1:09
  • $\begingroup$ @daniel, this is the limit of my understanding of this topic. why not post your own answer? then I'll delete mine. -NN $\endgroup$ – niels nielsen Sep 7 '20 at 4:37
  • $\begingroup$ I would if I thought I knew the answer. It's possible that we're really talking about differential equations with solutions which diverge in finite time. In that case the problem with the argument given in the post is that $U(T/2)$ is not actually everywhere continuous, I think. But I'm not sure that's what's going on with the original "butterfly effect." $\endgroup$ – Daniel Sep 7 '20 at 22:24

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