The Heat KernelHeat Kernel on an Unbounded Domain

# 3 Heat Kernel on an Unbounded Domain

This section is completely unrelated to the remainder of the article. Here we are going to further specialize to the case of the Laplacian $\Delta$ acting on functions. Our goal is to extend our results to all open manifolds.

The strategy we shall adopt is to consider an exhaustion $\Omega _1 \subseteq \Omega _2 \subseteq \cdots \subseteq M$ by relatively compact open submanifolds 1 . We take the heat kernels on $\bar{\Omega }_i$ and consider the limit as $i \to \infty$. We will show that this gives us a fundamental solution to the heat equation. Note that everything we have done in the previous sections apply to manifolds with boundary as well, as long as we require the initial conditions and solutions to vanish at the boundary.

We begin with a note on terminology. On an open manifold, the heat kernel is not necessarily unique. To avoid confusion, we will say a function $K_t(x, y)$ is a fundamental solution if for any continuous bounded $f_0(x)$, the function

$f(x, t) = \int _M K_t(x, y) f_0(y) \; \mathrm{d}y$

is a solution to the heat equation, and $f(x, t) \to f_0(x)$ as $t \to 0$ for all $x \in M$. The label “heat kernel” will be reserved for the one we explicitly construct, which we will show to be the smallest positive heat kernel.

The main property of the Laplacian that we will use is the maximum principle:

Theorem 3.1 (Maximum principle)

Let $M$ be a Riemannian manifold and $U \subseteq M$ a precompact open subset. Let $f$ be a continuous solution to the heat equation on $U_T = \bar{U} \times [0, T]$. Let $\partial ^* U_T$ be the subset of the boundary consisting of $\bar{U} \times \{ 0\} \cup (\partial U) \times [0, 1]$. Then

$\sup _{U_T} f = \sup _{\partial ^* U_T} f,\quad \inf _{U_T} f = \inf _{\partial ^* U_T} f.$

Proof
It suffices to show the statement for the supremum, as the infimum case follows by considering $-f$.

The idea is that at a maximum, the first derivatives all vanish, and the operator $\Delta + \frac{\partial }{\partial t}$ picks out information about the second derivative. The second derivative test then prevents the existence of maxima or minima.

After picking local coordinates, ellipticity means we can write

$\Delta + \frac{\partial }{\partial t} = \sum a^{ij}(x) \frac{\partial ^2}{\partial x^i \partial x^j} + b^i (x) \frac{\partial }{\partial x^i} + \frac{\partial }{\partial t},$

where $a^{ij}(x)$ is a negative definite matrix for all $x$ (one convinces oneself that there is no constant term since $\Delta + \frac{\partial }{\partial t}$ kills all constant functions).

The second derivative test is not very useful if the second derivatives vanish. Thus, we perform a small perturbation. For $\delta > 0$, we set

$g^\delta = f - t \delta .$

$\left(\Delta + \frac{\partial }{\partial t}\right)g^\delta = -\delta .$

It suffices to show that

$\sup _{U_T} g^\delta = \sup _{\partial ^* U_T} g^\delta .$

The result then follows from taking the limit $\delta \to 0$.

First suppose the supremum is achieved at some interior point $(x, t) \in U_T \setminus \partial U_T$. Then we know

$\sum a^{ij}(x) \frac{\partial ^2 f}{\partial x^i \partial x^j} = -\delta .$

But we know that $a^{ij}(x)$ is negative definite, so by elementary linear algebra, there must be some $v$ such that $v^i v^j \partial _i \partial _j f > 0$ (e.g. apply Sylvester's law of inertia to $a^{ij}$). So moving $x$ in the direction $v$ increases $f$, hence $g^\delta$. This contradicts the fact that $(x, t)$ is a maximum.

It remains to exclude the possibility that the supremum is attained when $t = T$. But if it is attained at $(x, T)$, then $\frac{\partial f}{\partial t}(x, T) \geq 0$, or else going back slightly in time will increase $f$ and hence $g^\delta$. So the same argument as above applies.

Proof
Note that here we only get a weak form of the maximum principle. We do not preclude the possibility that the supremum is attained at both the boundary and the interior. This is the price we have to pay for cheating by perturbing $u$ a bit to apply the second derivative test.

From this, we deduce the following bounds on the heat kernel:

Theorem 3.2

The heat kernel $H_t$ for a compact Riemannian manifold $M$ satisfies

1. $H_t(x, y) \geq 0$.

2. For every fixed $x \in M$ and $t > 0$, we have

$\int _M H_t(x, y)\; \mathrm{d}y \leq 1.$

Moreover, the integral $\to 1$ as $t \to 0$.

3. If $U \subseteq M$ is open, and $\bar{U}$ has heat kernel $K_t(x, y)$, then

$K_t(x, y) \leq H_t(x, y)$

for all $x, y \in U$ and $t > 0$. In particular, taking $\bar{U} = M$, the heat kernel is unique.

Proof

1. By the maximum principle, convolving any non-negative function with $H_t$ gives a non-negative function. So $H_t$ must itself be non-negative. (We would like to apply the maximum principle directly to $H_t$ but unfortunately $H_t$ is not continuous at $t = 0$)

2. $f(x, t) = \int _M H_t(x, y)\; \mathrm{d}y$ is the solution to the heat equation with initial condition $1$ everywhere.

3. Extend $K_t(x, y)$ by zero outside of $U$. For any non-negative function $f_0$, the convolution

$f(x, t) = \int _M (H_t(x, y) - K_t(x, y)) f_0(y) \; \mathrm{d}y$

is a solution to the heat equation on $U$ with initial conditions $0$. Moreover, when $x$ is on $\partial U$, the function $K_t(x, y)$ vanishes, so $f(x, t) \geq 0$. So by the maximum principle, $f \geq 0$ everywhere. It follows that $H_t(x, y) \geq K_t(x, y)$ everywhere.

Proof

These bounds allow us to carry out our initial strategy. Pick a sequence of exhausting relatively compact open submanifolds $\Omega _1 \subseteq \Omega _2 \subseteq \cdots \subseteq M$. Let $H_t^{\Omega _i}(x, y)$ be the heat kernel of $\bar{\Omega }_i$, extended to all of $M \times M$ by zero. We then seek to define a fundamental solution

$H_t(x, y) = \lim _{i \to \infty } H_t^{\Omega _i}(x, y).$

By (iii) above, we see that the limit exists pointwise, since it is an increasing sequence. To say anything more substantial than that, we need a result that controls the limit of solutions to the heat equation:

Lemma 3.3

Let $M$ be any Riemannian manifold and $a, b \in \mathbb {R}$. Suppose $\{ f_i\}$ is a non-decreasing non-negative sequence of solutions to the heat equation on $M \times (a, b)$ such that

$\int _M f_i(x, t)\; \mathrm{d}x \leq C$

for some constant $C$ independent of $i$ and $t$. Then

$f = \lim _{i \to \infty } f_i$

is a smooth solution to the heat equation and $f_i \to f$ uniformly on compact subsets together with all derivatives of all orders.

Proof
[Proof sketch] We only prove the case where $M$ is compact. Let $H_t$ be the heat kernel. Fix $[t_1, t_2] \subseteq (a, b)$. Then for any $x \in M$ and $t \in (t_1, t_2)$, we can write

$f_i(x, t) = \int _M H_{t - t_1}(x, y) f_i(y, t_1)\; \mathrm{d}y.\tag {*}$

By monotone convergence, we also have

$f(x, t) = \int _M H_{t - t_1} (x, y) f(y, t_1) \; \mathrm{d}y.$

So $f$ is a solution to the heat equation and is smooth. To show the convergence is uniform, simply observe that

$0 \leq (f - f_i)(x, t) \leq D \int _M \left[f(y, t_1) - f_i(y, t_1)\right]\; \mathrm{d}y$

for some constant $D$, and the right-hand side $\to 0$ by dominated convergence.

In the non-compact case, we multiply $f_i$ by a compactly supported bump function to reduce to the compact (with boundary) case, and replace $(*)$ by Duhamel's principle.

Proof

Theorem 3.4

If we define

$H_t(x, y) = \lim _{i \to \infty } H_t^{\Omega _i}(x, y),$

then $H_t$ is a smooth fundamental solution to the heat equation. Moreover, $H_t(x, y)$ is independent of the choices of $\Omega _i$. In fact,

$H_t(x, y) = \sup _{\Omega \subseteq M} H_t^\Omega (x, y).$

Moreover, $H_t(x, y)$ is the smallest positive heat kernel, i.e. $H_t(x, y) \leq H_t'(x, y)$ for any other positive heat kernel $H_t'$.

Proof
Since $H_t^{\Omega _i}$ are increasing, we know the pointwise limit $H_t(x, y)$ exists, but can possibly be infinite.

To see that $H_t(x, y)$ is in fact smooth, we apply Lemma 3.3. To do this, we observe that on any open subset of $\Omega _i$, the function $H_t^{\Omega _i}(x, y)$ is a solution to

$\left(\Delta _x + \Delta _y + 2\frac{\partial }{\partial t}\right) H_t^\Omega (x, y) = 0,$

which, after rescaling $t$, is the heat equation.

If we fix any relatively compact open $U \subseteq M$, then Theorem 3.2(ii) gives us a uniform bound

$\int _{U \times U} H_t^{\Omega _i}(x, y)\; \mathrm{d}x\; \mathrm{d}y \leq \operatorname{vol}U.$

So Lemma 3.3 tells us $H_t(x, y)$ is smooth on $U$. Since $U$ was arbitrary, $H_t(x, y)$ is a smooth function.

To see this is a fundamental solution, we again apply Lemma 3.3. By adding a constant, we may assume $f_0$ is positive. So by monotone convergence,

$f(x, t) = \lim _{i \to \infty } \int H_t^{\Omega _i}(x, y) f_0(y)\; \mathrm{d}y,\quad t > 0.$

Moreover, we see that

$\int H_t^{\Omega _i}(x, y) f_0(y)\; \mathrm{d}y \leq \left(\sup f_0\right) \int H_t^{\Omega _i}(x, y) \; \mathrm{d}y \leq \sup f_0.$

So $f(x, t)$ is bounded in the supremum, hence locally bounded in $L^1$. So $f$ is a solution to the heat equation.

The it remains to show that $f$ is in fact continuous as $t \to 0$. This will follow if we can show that for any $x \in M$ and any (relatively compact) open subset $U$ containing $x$, we have

$\lim _{t \to 0} \int _U H_t(x, y) \; \mathrm{d}y = 1.$

We know this limit is bounded above by $1$ by monotone convergence, and that it is equal to $1$ if we replace $H_t(x, y)$ by $H_t^U(x, y)$ by Theorem 3.2(ii). But this replacement only makes the integral smaller by the maximum principle applied to the difference. So we are done.

The rest is clear from the maximum principle.

Proof