The Heat Kernel — Heat Kernel for the Laplacian

2 Heat Kernel for the Laplacian

In the previous section, our solutions f(x,t)f(x, t) had the property that f(,t)f0f(-, t) \to f_0 in L2L^2. One might hope that, for example, if f0f_0 is continuous, then this converges in C0C^0. Another shortcoming of our previous approach is that we have no control over what HtH_ t looks like as a function as the whole construction is manifestly global.

In this section, we will construct the heat kernel via an alternative method. This method involves starting with an initial “guess” of what the heat kernel should be. We then show that if our initial guess is “good enough”, then there is an iterative procedure that gives us the actual heat kernel. Moreover, this iterative procedure is sufficiently explicit that we can relate the asymptotic behaviour of HtH_ t as t0t \to 0 back to that of our original guess.

We will now restrict to the case of the Laplace–Beltrami operator, since we can use the explicit solution for Rd\mathbb {R}^ d as an inspiration for our initial guess. We fix a pp and consider the Laplace–Beltrami operator on Ωp\Omega ^ p.

Lemma 2.1

For any NN, there exists a smooth section Kt(x,y)=Kt(y,x)TΓ(Ωp(Ωp))K_ t(x, y) = K_ t(y, x)^ T \in \Gamma (\Omega ^ p \boxtimes (\Omega ^ p)^*) such that

  1. Kt(x,y)δ(x,y) as t0. K_ t(x, y) \to \delta (x, y)\text{ as }t \to 0.

    More precisely,

    limt0MKt(x,y)α(y)  dy=α(x) in C0 \lim _{t \to 0} \int _ M K_ t(x, y)\, \alpha (y)\; \mathrm{d}y = \alpha (x)\text{ in }C^0

    for all continuous αΩp(M)\alpha \in \Omega ^ p(M).

  2. There is a constant CC and an exponent NN such that

    (tΔx)Kt(x,y)CtN \left|\left(\frac{\partial }{\partial t} - \Delta _ x\right) K_ t(x, y)\right| \leq C t^ N

    for all x,yMx, y \in M.

  3. For any other continuous section Ft(x,y)C0(Ωp(Ωp)×[0,))F_ t(x, y) \in C^0(\Omega ^ p \boxtimes (\Omega ^ p)^* \times [0, \infty )), the convolution

    (KF)t(x,y)=0tdsMdz  Kts(x,z)Gs(z,y) (K * F)_ t(x, y) = \int _0^ t \mathrm{d}s\int _ M \mathrm{d}z\; K_{t - s} (x, z)\, G_ s(z, y)

    is a C2C^2 section of Ωp(Ωp)×(0,)\Omega ^ p \boxtimes (\Omega ^ p)^* \times (0, \infty ).

To motivate why one should care about the convolution, notice that Duhamel's principle essentially says that the heat kernel HH satisfies

(ddt+Δx)HF=F \left(\frac{\mathrm{d}}{\mathrm{d}t} + \Delta _ x\right) H * F = F

for all FF. The condition that the convolution is C2C^2 is a technical point that will be useful later on.

Proof

Write r(x,y)r(x, y) for the geodesic distance between x,yMx, y \in M. The very first naive guess for Kt(x,y)K_ t(x, y) might just be

1(4πt)d/2er2/4t. \frac{1}{(4\pi t)^{d/2}} e^{-r^2/4t}.

This definitely satisfies (i). To see how good an attempt we made, we compute t+Δx\frac{\partial }{\partial t} + \Delta _ x of this.

To do so, we fix a yMy \in M and express xx in geodesic polar coordinates about yy. Then there is a formula

Δx(F(r)α)=(d2Fdr2+d1rdFdr+12gdgdrdFdr)α+2rdFdrrrα+FΔxα \Delta _ x (F(r) \alpha ) = \left(\frac{\mathrm{d}^2 F}{\mathrm{d}r^2} + \frac{d - 1}{r} \frac{\mathrm{d}F}{\mathrm{d}r} + \frac{1}{2g} \frac{\mathrm{d}g}{\mathrm{d}r} \frac{\mathrm{d}F}{\mathrm{d}r}\right)\alpha + \frac{2}{r} \frac{\mathrm{d}F}{\mathrm{d}r} \nabla _{r \frac{\partial }{\partial r}} \alpha + F \Delta _ x \alpha

where gg is the determinant of the metric (note that rrr\frac{\partial }{\partial r} is defined at r=0r = 0, but r\frac{\partial }{\partial r} is not, hence the funny notation). So we find that

(t+Δx)er2/4t(4πt)d/2=r4gtdgdrer2/4t(4πt)d/2. \left(\frac{\partial }{\partial t} + \Delta _ x\right) \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}} = -\frac{r}{4g t} \frac{\mathrm{d}g}{\mathrm{d}r} \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}}.

If we were in flat space, then dgdr=0\frac{\mathrm{d}g}{\mathrm{d}r} = 0, and so this is in fact a legitimate fundamental solution, as one would expect. However, when dgdr̸=0\frac{\mathrm{d}g}{\mathrm{d}r} \not= 0, then we can't quite bound this as t0t \to 0.

To solve this, we introduce some correction terms, which are small enough to preserve the property (i). We shall find ui(x,y)Γ(Ωp(Ωp))u_ i(x, y) \in \Gamma (\Omega ^ p \boxtimes (\Omega ^ p)^*) with f0(x,x)=If_0(x, x) = I such that

KN,t(x,y)=er2/4t(4πt)d/2(i=0Ntiui(x,y)) K_{N, t} (x, y) = \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}} \left(\sum _{i = 0}^ N t^ i u_ i(x, y)\right)

satisfies the equality

(t+Δx)KtN(x,y)=er2/4t(4πt)d/2tNΔxuN(x,y). \left(\frac{\partial }{\partial t} + \Delta _ x\right) K^ N_ t(x, y) = \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}}\, t^ N \Delta _ x u_ N(x, y).

The precise shape of the right-hand side does not matter. It is just what happens to be convenient. The point is that the exponent of tt is Nd2N - \frac{d}{2}, which is eventually positive. Then we simply have to take K=KNK = K_ N for large enough NN (note that this NN is different from the exponent NN in (ii) by d2\frac{d}{2}). Actually, we will only succeed in finding uNu_ N in a neighbourhood of the diagonal, since we work in geodesic polars. But we can simply take K=KNφ(x,y)K = K_ N \varphi (x, y), where φ\varphi is a bump function supported near the diagonal. ((iii) is left as an exercise)

To find the uiu_ i, we assume KNK_ N is of the given form, and compute

(tΔx)KN,t(x,y)=er2/4t(4πt)d/2i=0N[(it+r4gtdgdr)tiui(x,y)+ti1rrui+tiΔxui(x,y)]. \left(\frac{\partial }{\partial t} - \Delta _ x\right) K_{N, t}(x, y) = \frac{e^{-r^2/4t}}{(4\pi t)^{d/2}} \sum _{i = 0}^ N \left[\left( \frac{i}{t} + \frac{r}{4gt} \frac{\mathrm{d}g}{\mathrm{d}r}\right) t^ i u_ i(x, y) + t^{i - 1} \nabla _{r \frac{\partial }{\partial r}} u_ i + t^ i \Delta _ x u_ i(x, y)\right].

To obtain the desired result, we need the coefficient of er2/4t(4πt)d/2ti1\frac{e^{-r^2/4t}}{(4\pi t)^{d/2}} t^{i - 1} to vanish for iNi \leq N. So we need to inductively solve

(i+r4gdgdr)ui(x,y)+rrui(x,y)=Δxui1(x,y). \left(i + \frac{r}{4g} \frac{\mathrm{d}g}{\mathrm{d}r}\right) u_ i(x, y) + \nabla _{r \frac{\partial }{\partial r}} u_ i(x, y) = \Delta _ x u_{i - 1}(x, y).

The diligent reader will observe that the left-hand side is exact, so we can write this as

rr(rig1/4ui(x,y))=rig1/4Δxui1(x,y). \nabla _{r\frac{\partial }{\partial r}} \left(r^ i g^{1/4} u_ i(x, y)\right) = r^ i g^{1/4} \Delta _ x u_{i - 1}(x, y).

The existence and uniqueness of a local solution then follows easily (one has to be slightly careful about r=0r = 0, where the original equation gives an initial condition iui=Δxui1i u_ i = \Delta _ x u_{i - 1}).

For future reference, we note that rr=0\nabla _{r \frac{\partial }{\partial r}} = 0 when r=0r = 0, so

Kt(x,x)=consti=0Ntid/2ui(x,x), K_ t(x, x) = \text{const} \cdot \sum _{i = 0}^ N t^{i - d/2} u_ i(x, x),

where ui(x,x)u_ i(x, x) is a rational function in gijg^{ij} and its derivatives.

We will now fix an NN large enough and take KK to be the function given by the previous lemma.

To find the actual fundamental solution, we have to understand the convolution a bit better. Recall that we expect a genuine fundamental solution HH to satisfy (ddt+Δx)(HF)=F\left(\frac{\mathrm{d}}{\mathrm{d}t} + \Delta _ x\right) (H * F) = F. Using the fact that Kt(x,y)δ(x,y)K_ t(x, y) \to \delta (x, y) as t0t \to 0, we can calculate

Lemma 2.2
(t+Δx)(KF)=F+[(t+Δx)K]F. \left(\frac{\partial }{\partial t} + \Delta _ x\right) (K * F) = F + \left[\left(\frac{\partial }{\partial t} + \Delta _ x\right)K\right] * F.\fakeqed

So at least formally, the heat kernel should be given by

H=m=0(1)mK[(t+Δx)K]m. H = \sum _{m = 0}^\infty (-1)^ m K * \left[\left(\frac{\partial }{\partial t} + \Delta _ x\right)K\right]^{*m}.

Barring convergence issues, the formula above also shows that (ddt+Δx)H=0\left(\frac{\mathrm{d}}{\mathrm{d}t} + \Delta _ x\right) H = 0.

To check convergence, the naive bound

FGLCtvol(M)FLGL \| F * G\| _{L^\infty } \leq C t \operatorname{vol}(M) \| F\| _{L^\infty } \| G\| _{L^\infty }

is not quite enough. We set

G=(t+Δx)K. G = \left(\frac{\partial }{\partial t} + \Delta _ x \right) K.

We then have GtLCtN\| G_ t\| _{L^\infty } \leq C t^ N. We claim that

Gt(j+1)LCj+1vol(M)j(N+j)(N+2)(N+1)tN(j+1)+j. \| G^{*(j + 1)}_ t\| _{L^\infty } \leq \frac{C^{j + 1} \operatorname{vol}(M)^ j}{(N + j) \cdots (N + 2)(N + 1)} t^{N(j + 1) + j}.

The terms in the denominator are the interesting bits that come from doing the bound more carefully. They ensure the sum above converges by the ratio test. This is a standard inductive calculation

Gt(j+1)(x,y)0tdsMdz  Gts(x,z)Gsj(z,y)0tdsMdz  CtNCjvol(M)j1(N+j1)(N+1)sNj+j1Cj+1vol(M)j(N+j1)(N+1)tNj0tsN+j1  ds=Cj+1vol(M)j(N+j)(N+1)tN(j+1)+j. \begin{aligned} |G^{*(j + 1)}_ t(x, y)| & \leq \int _0^ t \mathrm{d}s\int _ M \mathrm{d}z\; |G_{t - s}(x, z)| |G_ s^{*j}(z, y)|\\ & \leq \int _0^ t \mathrm{d}s\int _ M \mathrm{d}z\; C t^ N \cdot \frac{C^ j \operatorname{vol}(M)^{j - 1}}{(N + j - 1) \cdots (N + 1)} s^{Nj + j - 1}\\ & \leq \frac{C^{j + 1} \operatorname{vol}(M)^ j}{(N + j - 1) \cdots (N + 1)} t^{Nj} \int _0^ t s^{N + j - 1}\; \mathrm{d}s\\ & = \frac{C^{j + 1} \operatorname{vol}(M)^ j}{(N + j) \cdots (N + 1)} t^{N(j + 1) + j}. \end{aligned}

Adding these up and using the ratio test, we find that

H=K+Km=1(1)Gm H = K + K * \sum _{m = 1}^\infty (-1) G^{*m}

converges, and we have the following theorem:

Theorem 2.3

There is a fundamental solution Ht(x,y)H_ t(x, y), and it satisfies

Ht(x,y)=Kt(x,y)+O(tN+1). H_ t(x, y) = K_ t(x, y) + O(t^{N + 1}).

In particular, we have

Ht(x,y)δ(x,y) as t0. H_ t(x, y) \to \delta (x, y) \text{ as }t \to 0.\fakeqed

From this, we can read off the leading term of Ht(x,x)H_ t(x, x) as t0t \to 0, which is, unsurprisingly, er2/4t(4πt)d/2\frac{e^{-r^2/4t}}{(4\pi t)^{d/2}}. As a recap, to produce this estimate, we used this as an initial approximation to the heat kernel, and then perform some procedures to turn this into an actual heat kernel. We then observe that all the correction terms we add are of higher order, so we are happy.

Note that all we used about KtK_ t was the properties stated in Lemma 2.1, and the procedure is pretty general. One can interpret the success of this procedure as testifying to the locality of the heat kernel. The properties of KtK_ t we required were always local conditions, namely that KtK_ t looks locally like a heat kernel for small tt. We then get to produce an actual heat kernel HtH_ t, whose difference from KtK_ t is something we can control very well, and in particular vanishes in the limit t0t \to 0.

In general, if our manifold is written as a union of two open submanifolds, and we have produced heat kernels on each submanifold (suitably interpreted), then we can attempt to glue them together using a partition of unity. This will not give us an actual heat kernel, but the failure to be a heat kernel decreases exponentially as t0t \to 0 (since Ht(x,y)H_ t(x, y) falls exponentially as t0t \to 0 whenever x̸=yx \not= y). The above argument then lets us produce a heat kernel whose difference from the “fake” one vanishes to all orders as t0t \to 0.