As we all learned in a first course in PDEs, for "reasonable" initial data $u_0$, the solution to the linear heat equation \begin{equation} \partial_t u = \triangle u \end{equation} exists classically and converge to zero uniformly.

The proof can be given using the explicit Gaussian kernel \begin{equation} K_t(x) = \frac{1}{(4\pi t)^{d/2}} \exp( -|x|^2 / 4t ) \end{equation} for which \[ u(t,x) = K_t \star u_0 \] solves the heat equation. The fact that $K_t$ is a Schwartz function for every $t > 0$, and that it depends smoothly on $t$, means that as long as $u_0$ is locally integrable and has no more than polynomial growth (for example), then $u(t,x)$ is well-defined everywhere and smooth.

Furthermore, by H\"older's inequality we see \begin{equation} |u(t,x)| \leq |K_t|_{L^p} |u_0|_{L^q} \end{equation} where $p$ and $q$ are conjugate exponents. Observe that by explicit computation, $K_t$ always have $L^1$ norm $1$, but has decreasing $L^p$ norm for every $p > 1$. This shows that, in particular, if $u_0 \in L^q$ for any $q\in [1,\infty)$, then the classical solution converges uniformly to zero.

Notice that the theory leaves out $L^\infty$. In particular, we see easily that if $u_0 \equiv C$, then the convolution defines the classical solution $u(t,x) \equiv C$ as the solution to the heat equation, and this does not converge to $0$.

An obvious question is: is it the case that for bounded, locally integrable functions $u_0$, that convergence to a constant is always the case for solutions to the linear heat equation?

## On compact manifolds

Our case is bolstered by the situation on compact manifolds. When solving the heat equation on compact manifolds, if the initial data is bounded, it is also in $L^2$. So we can decompose the initial data using the eigen-expansion relative to the Laplace-Beltrami operator $\triangle$. The system then diagonalizes into decoupled ordinary differential equations which, except for the kernel of $\triangle$,have solutions that are exponentially decaying.

In fact, this is sufficient to tell us that on compact manifolds, solutions to the heat equation converge *exponentially* to its mean.

We immediately however see a difference with the case on $\mathbb{R}^d$.
On Euclidean space, the solution frequently only decay *polynomially* to zero, when the data is in $L^p$. One can check this by letting $u_0$ be a Gaussian function, for which the heat equation solution decay at rate $t^{-d/2}$.

## Solutions are asymptotically locally constant

On the positive side, we do have that, for every $u_0$ that is bounded, the solutions are asymptotically locally constant *at every time*.
This is because if $u_0$ is bounded (say by the number $b$), we have that
\[
u(t,x) - u(t,y) = \int (K_t(x-z) - K_t(y-z))u_0(z) ~\mathrm{d}z
\]
so
\begin{equation}
|u(t,x) - u(t,y)| \leq b \int |K_t(x-z) - K_t(y-z) | ~\mathrm{d}z
\end{equation}
Performing an explicit change of variables we get
\begin{equation}
|u(t,x) - u(t,y)| \leq \frac{b}{\pi^{d/2}} \int | \exp(- |z|^2) - \exp(- |z - (y-x)/ (2\sqrt{t})|^2) | ~\mathrm{d}z.
\end{equation}
For $(y-x)/\sqrt{t}$ sufficiently small, the integral can be very roughly bounded by $O(|y-x| / \sqrt{t})$. This shows that on every compact set $K$,
\begin{equation}
\sup_K u(t,\cdot) - \inf_K u(t,\cdot) \lesssim t^{-1/2}
\end{equation}
giving that the solutions are asymptotically locally constant.

This argument, however, says nothing about whether the "constant" value is the same between different times.

## Oscillatory solutions

In fact, it is fairly easy to construct a bounded initial data for which the solution, evaluated at $x = 0$, fails to converge as $t \to \infty$.
Performing the explicit change of variable we get
\begin{equation}
u(t,0) = \int \frac{1}{(4\pi)^{d/2}} \exp( - |z|^2/4) u_0( - \sqrt{t} z) \mathrm{d}z.
\end{equation}
Denote by $\mu_G$ the Gaussian measure $(4\pi)^{-d/2} \exp( - |z|^2/4) \mathrm{d}z$ (which we observe has total mass 1)
we see that we can write $u(t,0)$ as the integral of a *rescaling* of $u_0$ against $\mu_G$.

First choose $\epsilon \ll 1$. We can choose $0 < \lambda < \Lambda < \infty$ such that \[ \int_{ |z| \leq \lambda} \mu_G < \epsilon, \quad \int_{|z| \geq \Lambda} \mu_G < \epsilon. \] Let $\kappa = \Lambda / \lambda$. Let $u_0$ be defined by \begin{equation} u_0(x) = \begin{cases} 0 & |x| = \frac12 \newline 1 & |x| \in [\lambda \kappa^n, \lambda \kappa^{n+1}), n \text{ is even}\newline 0 & |x| \in [\lambda \kappa^m, \lambda \kappa^{m+1}), m \text{ is odd} \end{cases} \end{equation} Observe that this function satisfies \[ u_0(\kappa^{2n} x) = u_0(x) \] and \[ u_0(x) + u_0(\kappa x) \equiv 1.\] By construction, there exists a value $\gamma \in [1 - 2\epsilon, 1]$ such that \[ \int u_0(x) \mu_G = \gamma, \quad \int u_0(\kappa x) \mu_G = 1 - \gamma.\] Returning to the formula for $u(t,0)$, we see that \begin{equation} u(\kappa^{2n},0) = \begin{cases} \gamma & n \text{ is even} \newline 1 - \gamma & n \text{ is odd} \end{cases} \end{equation} This shows an oscillatory behavior at $x = 0$, showing that the limit as $t\to \infty$ of $u(t,0)$ to not exist.

On the other hand, this particular solution has the nice property that, on any compact set $K$, and for any fixed $t_0 > 0$, the sequence of functions $v_n(x) = u(t_0 \kappa^{4n}, x)$ converges uniformly to $u(t_0,0)$ as $n \to \infty$.