Can't the heat equation be inverted?

Question

I have heard people say that the heat equation isn't invertible because it smooths out irregularities that can not be recovered by backwards time evolution. But is this so? I will now argue that it can be inverted assuming that its Fourier modes decay quickly enough.
For a function $f(t, x)$, the heat equation is
$$ frac{partial}{partial t} f =  frac{partial^2}{partial x^2} f.$$
We may now write
$$
f(0, x) = int_{- infty}^infty d k ; tilde{f}(k) e^{i k x }
$$
and using the linearity of the heat equation, solve
$$
f(t, x) = int_{- infty}^infty d k ; tilde{f}(k) e^{i k x - k^2 t}.
$$
One might be concerned that the above integral might not converge for negative $t$. However, if we demand that $tilde{f}(k)$ decays fast enough, namely faster than $e^{- k^2 t}$ for all $t$, then this isn't a problem.
Assuming the above argument is correct, let's ask if it's possible to construct two functions $f(0, x)$ and $g(0, x)$ which aren't equal, but are exactly equal by some finite time $t$? (Such functions would violate the aforementioned fall off condition.) Wouldn't this violate the linearity of the heat equation? The function $f(0, x) - g(0, x)$ would evolve from being non-zero to exactly $0$.
So I guess my general question is, can the heat equation be inverted or not?

knzhou · Accepted Answer

Yes, the heat equation can be inverted in theory, if you assume strong enough conditions on the initial values. When people say it can't be, they just mean that it can't be done practically because doing so is numerically ill-conditioned, and also that if you use "typical" initial conditions you can evolve to a non-smooth function.

Correct answer by knzhou on January 6, 2021

Kasper · Answer

You are correct that on the restricted function space you propose, a solution will exist and the solution will be unique. However, notice that the inverted heat equation is not continuous with respect to the initial conditions. Given an initial data function $f_0(x)$, consider $g_0(x) = f_0(x) + epsilon e^{ikx}$. Although $f_0$ and $g_0$ can be taken as close together as you want, $||f_0 - g_0||_infty = epsilon$, their respective solutions $f(x,t)$ and $g(x,t)$ will grow apart without bound $||f-g||_infty = epsilon e^{k^2 t}nearrow+infty$.
This isn't necessary for the equation to be solvable, but for this reason we call the inverted heat equation an ill-posed problem.

Can't the heat equation be inverted?

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