Heat Ball and Heat Sphere Mean Value Property

Like solutions to the Laplace equation, (classical) solutions to the heat equation {\partial_{t}u-\Delta u=0} satisfy a mean value property. But instead of integrals over balls or spheres, the heat mean value property involves integrals over a heat ball. The heat ball of radius {r>0} centered at a point of {(x,t)} in {\mathbb{R}^{n}\times\mathbb{R}}, denoted {E(x,t;r)} is the set

\displaystyle E(x,t;r):=\left\{(y,s)\in\mathbb{R}^{n}\times\mathbb{R}:s\leq t \text{ and }\Phi(x-y,t-s)\geq r^{-n}\right\},

where {\Phi} denotes the heat kernel

\displaystyle \Phi(x-y,t-s)=\dfrac{1}{(4\pi(t-s))^{n/2}}e^{-\frac{\left|x-y\right|^{2}}{4(t-s)}}

When {(x,t)=(0,0)}, we sometimes write {E(r)} instead of {E(0,0;r)}.

After playing around with the inequality {\Phi(x-y,t-s)\geq r^{-n}}, one can verify that

\displaystyle E(x,y;r)=\left\{(y,s)\in\mathbb{R}^{n+1}:t-\frac{r^{2}}{4\pi}\leq s\leq t, \left|x-y\right|^{2}\leq 2n(t-s)\log\left(\frac{r^{2}}{4\pi (t-s)}\right)\right\}

and

\displaystyle \partial E(x,t;r)=\left\{(y,s)\in\mathbb{R}^{n+1}:t-\frac{r^{2}}{4\pi}\leq s\leq t,\left|x-y\right|^{2}=2n(t-s)\log\left(\dfrac{r^{2}}{4\pi (t-s)}\right)\right\}

From the above expressions, it is easy to see that {E(x,y;r)} is bounded and closed. Furthermore, the “center” {(x,t)} actually lies on the boundary, so it is somewhat of a misnomer. I leave it to the reader to verify that {E(x,t;r)=(x,t)+E(r)} and {E(r)} is obtained from {E(1)} by the parabolic scaling {(y,s)\mapsto(ry,r^{2}s)}.

The following theorem of N.A. Watson establishes the heat ball mean value property.

Theorem 1 Let {U\subset\mathbb{R}^{n}} be a bounded open set, and let {T>0}. Set {U_{T}:=U\times (0,T)}. If {u} is a {C_{1}^{2}(U_{T})} subsolution of the heat equation, then

\displaystyle u(x,t)=\dfrac{1}{4r^{n}}\iint_{E(x,t;r)}u(y,s)\dfrac{\left|x-y\right|^{2}}{(t-s)^{2}}dyds

for all {E(x,t;r)\subset U_{T}}, where equality holds if {u} is a solution of the heat equation.

After seeing the proof of this surprisingly recent theorem, I wondered if there was a corresponding result for the boundary of {\partial E(x,t;r)}, a heat sphere mean value property. There was no proof of reference for a proof of such a result in the book I was reading, and Watson’s paper makes no mention of such a result either. I asked a professor at Georgia Tech if he knew of mean value property for the heat sphere. He did not, but he gave me a very helpful suggestion which led to the following result; many thanks to Professor F. Bonetto.

Theorem 2 Let {U\subset\mathbb{R}^{n}} be a bounded open set, and let {T>0}. If {u} is a {C_{1}^{2}(U_{T})} solution of the heat equation on {U_{T}}, then

\displaystyle u(x,t)=-\dfrac{1}{2r^{n}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u(x+R\omega,t+s)\dfrac{R^{n}}{s}d\omega ds

for all {E(x,t;r)\subset U_{T}}.

It appears Theorem 2 is at least folklore, as I found the result contained in some very nice lecture notes of Professor G. Menon.

Interestingly, proving the mean value property for the heat equation is done in the reverse order of with the Laplace equation. With harmonic functions, one shows the mean value property holds over spheres and then uses integration in polar coordinates to show the property holds over balls. In contrast, we will show that the mean value property holds over heat balls and then use this to show that it holds over heat spheres.

To prove Theorem 1, I will follows [Evans].

Proof of Theorem 1: As the translate of {u} is again a solution of the heat equation (on a translated domain), we may assume without loss of generality that {(x,t)=(0,0)}. For {r>0}, define

\displaystyle \phi(r)=\dfrac{1}{r^{n}}\iint_{E(r)}u(y,s)\dfrac{\left|y\right|^{2}}{s^{2}}dyds=\iint_{E(1)}u(ry,r^{2}s)\dfrac{\left|y\right|^{2}}{s^{2}}dyds,

where we make the change of variable {(y,s)\mapsto (ry,r^{2}s)}. Since {u} is {C_{1}^{2}(U)} and {U} is a bounded domain, we can differentiate inside the integral to obtain

\displaystyle \begin{array}{lcl} \displaystyle\phi'(r)&=&\displaystyle\iint_{E(1)}\sum_{i=1}^{n}u_{y_{i}}y_{i}\dfrac{\left|y\right|^{2}}{s^{2}}+2ru_{s}\dfrac{\left|y\right|^{2}}{s}dyds\\ [2 em]&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}\sum_{i=1}^{n}u_{y_{i}}y_{i}\dfrac{\left|y\right|^{2}}{s^{2}}+2u_{s}\dfrac{\left|y\right|^{2}}{s}dyds\\ [2 em]&=:&\displaystyle A+B \end{array}

We introduce the function

\displaystyle \psi(y,s):=-\dfrac{n}{2}\log(-4\pi s)+\dfrac{\left|y\right|^{2}}{4s}+n\log r,

which can be obtained by taking the logarithm of {r^{n}\Phi(-y,-s)}. Observe that {\Phi(y,-s)=r^{-n}} on {\partial E(r)} and therefore {\psi=0} on {\partial E(r)}. We can write the expression {B} as

\displaystyle \begin{array}{lcl} \displaystyle B&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}4u_{s}\sum_{i=1}^{n}y_{i}\psi_{y_{i}}dyds\\ [1.5 em]&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}4nu_{s}\psi+4\sum_{i=1}^{n}u_{sy_{i}}y_{i}\psi dyds; \end{array}

where we integrate by parts with respect to {y} and use the fact that {\psi} vanishes on {\partial E(r)} for the boundary term. Integrating by parts now with respect to {s}, we obtain

\displaystyle \begin{array}{lcl} \displaystyle B&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4nu_{s}\psi+4\sum_{i=1}^{n}u_{y_{i}}y_{i}\psi_{s}dyds\\ [2 em]&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4nu_{s}\psi+4\sum_{i=1}^{n}u_{y_{i}}y_{i}\left(-\dfrac{n}{2s}-\dfrac{\left|y\right|^{2}}{4s^{2}}\right)dyds\\ [2 em]&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4nu_{s}\psi-\dfrac{2n}{s}\sum_{i=1}^{n}u_{y_{i}}y_{i}dyds-A \end{array}

Since {u} is a subsolution of the heat equation, we have the inequality

\displaystyle \begin{array}{lcl} \displaystyle\phi'(r)&=&\displaystyle A+B\\ [2 em] &=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4nu_{s}\psi-\dfrac{2n}{s}\sum_{i=1}^{n}u_{y_{i}}y_{i}dyds\\ [2 em]&\geq&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4n\Delta u\psi-\dfrac{2n}{s}\sum_{i=1}^{n}u_{y_{i}}y_{i}dyds\\ [2 em]&=&\displaystyle\sum_{i=1}^{n}\dfrac{1}{r^{n+1}}\iint_{E(r)}4nu_{y_{i}}\psi_{y_{i}}-\dfrac{2n}{s}u_{y_{i}}y_{i}dyds\\ [2 em]&=&\displaystyle\sum_{i=1}^{n}\dfrac{1}{r^{n+1}}\iint_{E(r)}\dfrac{2n}{s}u_{y_{i}}y_{i}-\dfrac{2n}{s}u_{y_{i}}y_{i}dyds=0 \end{array}

where we integrate by parts the first term (w.r.t. {y}) to obtain the penultimate expression. Therefore {\phi} is nondecreasing. Using the dominated convergence theorem and the continuity of {u}, we obtain

\displaystyle \begin{array}{lcl} \displaystyle\phi(r)\geq\lim_{t\rightarrow 0^{+}}\phi(t)&=&\displaystyle\iint_{E(1)}\left(\lim_{t\rightarrow 0^{+}}u(ty,t^{2}s)\right)\dfrac{\left|y\right|^{2}}{s^{2}}dyds\\ [2 em]&=&\displaystyle u(0,0)\iint_{E(1)}\dfrac{\left|y\right|^{2}}{s^{2}}dyds\\ [2 em]&=&\displaystyle 4u(0,0) \end{array}

The result {4=\iint_{E(1)}\left|y\right|^{2}s^{-2}dyds}, as it is an exercise in integration, but I have uploaded a PDF of the computation. Dividing both sides by {r} yields the desired inequality. In particular, if {u} is a solution then equality holds. \Box

Using the preceding result, we now prove Theorem 2. For convenience, we introduce the notation

\displaystyle R=R(r,s)=\left(-2ns\log\left(\dfrac{r^{2}}{-4\pi s}\right)\right)^{1/2}

Proof of Theorem 2: Since the translate of {u} is again a solution of the heat equation, we may assume without loss of generality that {(x,t)=(0,0)}. Using polar coordinates, we can write

\displaystyle \dfrac{1}{4r^{n}}\iint_{E(r)}u(y,s)\dfrac{\left|y\right|^{2}}{\left|s\right|^{2}}dyds=\dfrac{1}{4r^{n}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{0}^{R(r,s)}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}d\omega d\rho ds

Differentiating {\phi} with respect to {r}, we obtain

\displaystyle \begin{array}{lcl} \displaystyle\phi'(r)&=&\displaystyle-\dfrac{n}{4r^{n+1}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{0}^{R(r,s)}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}d\omega d\rho ds\\ [2 em]&+&\displaystyle\dfrac{1}{4r^{n}}\int_{0}^{R(r,-\frac{r^{2}}{4\pi})}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}d\omega d\rho\vert_{s=-\frac{r^{2}}{4\pi}}\\ [2 em]&+&\displaystyle\dfrac{1}{4r^{n}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}\cdot\dfrac{dR}{dr}d\omega\vert_{\rho=R(s,r)}ds \end{array}

Note that the second term in the expression above vanishes since {R(r,-\frac{r^{2}}{4\pi})=0}. We compute

\displaystyle \dfrac{dR}{dr}=\dfrac{(-2ns)^{1/2}}{r\log\left(\frac{r^{2}}{-4\pi s}\right)^{1/2}}

and note that

\displaystyle R\cdot\dfrac{dR}{dr}=\left(-2ns\log\left(\frac{r^{2}}{-4\pi s}\right)\right)^{1/2}\dfrac{(-2ns)^{1/2}}{r\left(\log\frac{r^{2}}{-4\pi s}\right)^{1/2}}=\dfrac{-2ns}{r}

If {u} is a (classical) solution to the heat equation, then our proof of the mean value property over the heat ball shows that {\phi'(r)=0}. Rearranging terms, we obtain

\displaystyle \begin{array}{lcl} \displaystyle\dfrac{n}{r}\iint_{E(r)}u(y,s)\dfrac{\left|y\right|^{2}}{\left|s\right|^{2}}dyds &=&\displaystyle\dfrac{n}{4r^{n+1}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{0}^{R(r,s)}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}d\omega d\rho ds\\ [2 em]&=&\displaystyle\dfrac{1}{4r^{n}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u\left(R\omega,s\right)\dfrac{R^{n+1}}{\left|s\right|^{2}}\cdot\dfrac{dR}{dr}d\omega ds\\ [2 em]&=&\displaystyle\dfrac{-2n}{4r^{n+1}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u(R\omega,s)\dfrac{R^{n}}{s}d\omega ds\\ [2 em] &=&\displaystyle\dfrac{-n}{2r^{n+1}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u(R\omega,s)\dfrac{R^{n}}{s}d\omega ds \end{array}

Multiplying both sides of the equality by {r/n} completes the proof. \Box

  1. L.C. Evans, Partial Differential Equations (Second Edition), AMS, 2010.
  2. G. Menov, “Lectures on Partial Differential Equations”, Accessed: 10/11/14.
  3. N.A. Watson, “A Theory of Subtemperatures in Several Variables”, Proc. Lon. Math. Soc. s3-26 (1973), 385-417.
  4. X. Yu, “Heat Equation – Maximum Principles”, Accessed: 10/1/14.
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