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 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.

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 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.

Proof

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
[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.

Proof

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.

Proof