# How is heat propagation a semi-deterministic process?

Arnol’d in his Ordinary Differential Equations, states that

The propagation of heat is a semideterministic process: the future is determined by the present but the past is not.

How is this so? If there is a differential equation to determine the future, can’t we use the same equation to backtrack the past?

• math.stackexchange.com/q/628720/87420 – Kyle Kanos Aug 12 at 11:50
• @KyleKanos I’m not asking the time reversibility, but the determinacy your of the equation on the reverse direction. – Atom Aug 12 at 12:05
• Isn't that link saying the same thing as your accepted answe here? The two equations are different, so why would you expect the same solution to apply to both? – Kyle Kanos Aug 12 at 14:37
• @KyleKanos I think, it is saying that the equation is not rendered the same upon time-reversal. What I’m asking is whether a given initial condition, why we can’t employ the time-reversed to get the origin of the initial condition. – Atom Aug 12 at 14:52
• And the answer to that question is as above: it's a different equation so why would you expect the same solution to different equations? – Kyle Kanos Aug 12 at 14:55

I did not check the reference, so I may be missing something.

However, here is what I think. Many different "pasts" can lead to the present solution, and there is only a single far future.

Take a metallic rod in a room at temperature $$T_\text{ambiant}$$. No matter where nor by how much you heat it, the final state (which theoretically is never reached but in practice is) will be the same: the rod will have an homogeneous temperature, that of the room $$T_\text{ambiant}$$. From that state, you cannot go back and guess how the initial temperature was.

In your question you mention to use the same equation to backtrack the past, but you cannot do that. The equation contains an initial condition, and this part differs if you start in the final state compared to starting in an earlier state. So while the heat equation that describes the dynamics of the temperature change is the same in both cases, the initial conditions aren't the same. And it turns out that infinitely many different prior states can lead to the same present (or future) states.

Clarifying my last paragraph a bit more, as suggested in a comment. Take the rod again, this time we take a snapshot of the temperature distribution. We notice that the temperature is uniform and the same everywhere. From this, we cannot conclude that this was always the case. In fact, it could be that it was always like that, that the rod is in thermal equilibrium. But there is also infinitely many possibilities that lead to this state, for example if the rod was heated anywhere by any (reasonable) amount. The two totally different initial conditions lead to the same final state.

If the snapshot shows a Gaussian-like temperature distribution, you cannot know whether it is due to a Dirac delta-like initial perturbation from an otherwise uniform temperature distribution that lead to this state, or whether the rod was heated all along its length in such a way as to make a Gaussian-like temperature distribution. Here again, we have a present state, but we cannot know what prior states lead to the present state. However we do know how the present state will evolve, thanks to the heat equation and our initial conditions (the snapshot temperature distribution).

Another perspective is that the heat equation is not invariant under time reversal. It means that if one changes $$t$$ to $$-t$$, then the heat equation is not valid anymore. In other words, it is not able to describe the dynamics of the evolution of temperature when time goes backward. From that perspective, one can realize that one cannot retrieve information about the past, i.e. one cannot describe how the temperature changed with time to arrive at a particular temperature distribution.

• Great answer! Can you just clarify the last paragraph a bit more? – Atom Aug 12 at 13:15
• So this means that three form that the equation takes for a given setup also depends on the initial temperature profile of the object? – Atom Aug 12 at 14:08
• @Atom No, the heat equation is what it is. It's the initial conditions that depend on the time one sets as initial. And one solve for the equation + initial (and boundary) conditions. – thermomagnetic condensed boson Aug 12 at 15:06
• But what is the "thing" (physical reason) which causes heat propagation to not have a unique history given an initial setup? – Atom Aug 12 at 15:12
• SE posts are version controlled, so please do not make your post look like a revision table, instead just seamlessly integrate the new material into the post. There is an edit history button at the bottom of the post for those interested in seeing what changed. – Kyle Kanos Aug 16 at 17:12