The Heat Kernel — Heat 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 Δ\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 Ω1Ω2M\Omega _1 \subseteq \Omega _2 \subseteq \cdots \subseteq M by relatively compact open submanifolds 1 . We take the heat kernels on Ωˉi\bar{\Omega }_ i and consider the limit as ii \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 Kt(x,y)K_ t(x, y) is a fundamental solution if for any continuous bounded f0(x)f_0(x), the function

f(x,t)=MKt(x,y)f0(y)  dy 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)f0(x)f(x, t) \to f_0(x) as t0t \to 0 for all xMx \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 MM be a Riemannian manifold and UMU \subseteq M a precompact open subset. Let ff be a continuous solution to the heat equation on UT=Uˉ×[0,T]U_ T = \bar{U} \times [0, T]. Let UT\partial ^* U_ T be the subset of the boundary consisting of Uˉ×{0}(U)×[0,1]\bar{U} \times \{ 0\} \cup (\partial U) \times [0, 1]. Then

supUTf=supUTf,infUTf=infUTf. \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-f.

The idea is that at a maximum, the first derivatives all vanish, and the operator Δ+t\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

Δ+t=aij(x)2xixj+bi(x)xi+t, \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 aij(x)a^{ij}(x) is a negative definite matrix for all xx (one convinces oneself that there is no constant term since Δ+t\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 δ>0\delta > 0, we set

gδ=ftδ. g^\delta = f - t \delta .

Then we instead have

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

It suffices to show that

supUTgδ=supUTgδ. \sup _{U_ T} g^\delta = \sup _{\partial ^* U_ T} g^\delta .

The result then follows from taking the limit δ0\delta \to 0.

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

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

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

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

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 uu 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 HtH_ t for a compact Riemannian manifold MM satisfies

  1. Ht(x,y)0H_ t(x, y) \geq 0.

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

    MHt(x,y)  dy1. \int _ M H_ t(x, y)\; \mathrm{d}y \leq 1.

    Moreover, the integral 1\to 1 as t0t \to 0.

  3. If UMU \subseteq M is open, and Uˉ\bar{U} has heat kernel Kt(x,y)K_ t(x, y), then

    Kt(x,y)Ht(x,y) K_ t(x, y) \leq H_ t(x, y)

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

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

  2. f(x,t)=MHt(x,y)  dyf(x, t) = \int _ M H_ t(x, y)\; \mathrm{d}y is the solution to the heat equation with initial condition 11 everywhere.

  3. Extend Kt(x,y)K_ t(x, y) by zero outside of UU. For any non-negative function f0f_0, the convolution

    f(x,t)=M(Ht(x,y)Kt(x,y))f0(y)  dy 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 UU with initial conditions 00. Moreover, when xx is on U\partial U, the function Kt(x,y)K_ t(x, y) vanishes, so f(x,t)0f(x, t) \geq 0. So by the maximum principle, f0f \geq 0 everywhere. It follows that Ht(x,y)Kt(x,y)H_ t(x, y) \geq K_ t(x, y) everywhere.

These bounds allow us to carry out our initial strategy. Pick a sequence of exhausting relatively compact open submanifolds Ω1Ω2M\Omega _1 \subseteq \Omega _2 \subseteq \cdots \subseteq M. Let HtΩi(x,y)H_ t^{\Omega _ i}(x, y) be the heat kernel of Ωˉi\bar{\Omega }_ i, extended to all of M×MM \times M by zero. We then seek to define a fundamental solution

Ht(x,y)=limiHtΩi(x,y). 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 MM be any Riemannian manifold and a,bRa, b \in \mathbb {R}. Suppose {fi}\{ f_ i\} is a non-decreasing non-negative sequence of solutions to the heat equation on M×(a,b)M \times (a, b) such that

Mfi(x,t)  dxC \int _ M f_ i(x, t)\; \mathrm{d}x \leq C

for some constant CC independent of ii and tt. Then

f=limifi f = \lim _{i \to \infty } f_ i

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

Proof sketch

We only prove the case where MM is compact. Let HtH_ t be the heat kernel. Fix [t1,t2](a,b)[t_1, t_2] \subseteq (a, b). Then for any xMx \in M and t(t1,t2)t \in (t_1, t_2), we can write

()fi(x,t)=MHtt1(x,y)fi(y,t1)  dy. 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)=MHtt1(x,y)f(y,t1)  dy. f(x, t) = \int _ M H_{t - t_1} (x, y) f(y, t_1) \; \mathrm{d}y.

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

0(ffi)(x,t)DM[f(y,t1)fi(y,t1)]  dy 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 DD, and the right-hand side 0\to 0 by dominated convergence.

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

Theorem 3.4

If we define

Ht(x,y)=limiHtΩi(x,y), H_ t(x, y) = \lim _{i \to \infty } H_ t^{\Omega _ i}(x, y),

then HtH_ t is a smooth fundamental solution to the heat equation. Moreover, Ht(x,y)H_ t(x, y) is independent of the choices of Ωi\Omega _ i. In fact,

Ht(x,y)=supΩMHtΩ(x,y). H_ t(x, y) = \sup _{\Omega \subseteq M} H_ t^\Omega (x, y).

Moreover, Ht(x,y)H_ t(x, y) is the smallest positive heat kernel, i.e. Ht(x,y)Ht(x,y)H_ t(x, y) \leq H_ t'(x, y) for any other positive heat kernel HtH_ t'.

Proof

Since HtΩiH_ t^{\Omega _ i} are increasing, we know the pointwise limit Ht(x,y)H_ t(x, y) exists, but can possibly be infinite.

To see that Ht(x,y)H_ t(x, y) is in fact smooth, we apply Lemma 3.3. To do this, we observe that on any open subset of Ωi\Omega _ i, the function HtΩi(x,y)H_ t^{\Omega _ i}(x, y) is a solution to

(Δx+Δy+2t)HtΩ(x,y)=0, \left(\Delta _ x + \Delta _ y + 2\frac{\partial }{\partial t}\right) H_ t^\Omega (x, y) = 0,

which, after rescaling tt, is the heat equation.

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

U×UHtΩi(x,y)  dx  dyvolU. \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 Ht(x,y)H_ t(x, y) is smooth on UU. Since UU was arbitrary, Ht(x,y)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 f0f_0 is positive. So by monotone convergence,

f(x,t)=limiHtΩi(x,y)f0(y)  dy,t>0. 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

HtΩi(x,y)f0(y)  dy(supf0)HtΩi(x,y)  dysupf0. \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)f(x, t) is bounded in the supremum, hence locally bounded in L1L^1. So ff is a solution to the heat equation.

The it remains to show that ff is in fact continuous as t0t \to 0. This will follow if we can show that for any xMx \in M and any (relatively compact) open subset UU containing xx, we have

limt0UHt(x,y)  dy=1. \lim _{t \to 0} \int _ U H_ t(x, y) \; \mathrm{d}y = 1.

We know this limit is bounded above by 11 by monotone convergence, and that it is equal to 11 if we replace Ht(x,y)H_ t(x, y) by HtU(x,y)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.