### Reversing the Heat Equation

Today's class on Partial differential equations gave me a whole new perspective on the ubiquitous heat equation (whose forms in various coordinate systems, I "solved" numerically for my Masters thesis). I'm going to share that perspective here.Let's start with something simple and 1d. A rod (with temperature independent conductivity), perhaps. You're given a rod with a non-steady state temperature profile. One end of the rod has been maintained at a constant high temperature for 20 minutes - and you are handed over the rod after the 20 minutes. The entire rod is convecting (and has always been) with the ambient (with a constant heat transfer coefficient). Eventually, it would have reached a steady state - but that would have taken more than 20 mintues. So, essentially, you're given a rod whose temperatur distribution still being governed by the heat equation rather than the Laplace equation. In other words, the profile is still evolving.

Is it possible to predict the initial temperature profile, given an instantaneous temperature profile as detailed above?

The answer, well, is yes and no.

Yes, because, if a solution does exist, it is unique - fact that can be proven by some math [1]. So, essentially, for a given final temperature profile T_f(x), the ambient temperature (T_inf), the boundary conditions (T_o and T_(x=L)) . Starting with the T_f(x) and going back in time (by solving the reverse heat equation), one can, in principle, recover the initial temperature profile.

No, because, well, it diverges. A primitive variable seperable closed-form expression for the solution makes things quite clear.

where, the a's are the Fourier coefficients of T_f(x). In the case of the forward heat equation, the exponent would have been negative, rendering the higher harmonics more or less neglegible But, in the reverse case, this term would tend to blow up at higher values of n. If T_f(x) were purely sinusoidal (only one a_n), then the equation would be quite friendly. The equation poses no problem for a small number of harmonics.

But suppose there were a "little" noise. Noise usually has many frequencies, some quite high. The corresponding terms would blow up. Evaluating the above series is not going to be easy.

This gives us another elegant insight: the "sharper" a profile (a profile with a discontinuity, for instance would yield infinite terms in the fourier space), the more difficult it is to reconstruct the original temperature profile. This figures: the larger the local temperature gradient, the faster it tends to smooth itself out! Minor spatial fluctuations in temperature are removed faster by the diffusion equation.

Let us now consider a 5 cm x 5cm x 5cm copper block. (Copper because it's always at more or less the same temperature at ambient heat transfer coefficients). Suppose we heat it up to a temperature and cool it in air, and measure its temperature by a thermocouple with an accuracy of 0.1C. Suppose we measure time with a digital watch with an accuracy of 1 sec. Suppose, also, that we know the heat transfer coefficients accurately. Given the time since the start of convective cooling (measured by the stop-watch, actually) and the final temperature , can the initial temperature be determined?

Yes, you would say. Exponenital cooling curve. Easy to fit it.

Suppose we look at a situation where the instruments would say that a steady state has been reached. There's no way to get back the original temperature --- there's many possible solutions!!!!!!! The only way out is it use more accurate devices.

And now, add to this the complexity of temperature profiles. Suppose we were using steel piece (lower conductivity) , and somehow "magically" obtained its temperautre profile at a certain time throughout its volume. Suppose the profile was somewhat jagged (but steadying out gradually) ......... Well, with a lot of computers one can arrivie at an estimate ... with an error bar.

And this wisdom can be applied to the Navier stokes too. The only process in the entire N/S equations that "loses" information is, actually, viscous dissipation. Turbulence generates entropy faster by wrinkling the instantaneous temperature and velocity profiles to bring about high amounts of heat transfer and viscous dissipation respectively.

Maybe I'm getting way too carried away here.....