Gradient Estimates for Elliptic Operators with Second-Order Discontinuous Coefficients

G. Metafune; L. Negro; C. Spina

doi:10.1007/s00009-019-1415-x

Mediterranean Journal of Mathematics

December 2019, 16:138 | Cite as

Gradient Estimates for Elliptic Operators with Second-Order Discontinuous Coefficients

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Authors and affiliations

G. Metafune
L. Negro
C. Spina

Article

First Online: 15 October 2019

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Abstract

We consider the second-order elliptic operator

$$\begin{aligned} L =\Delta +(a-1)\sum _{i,j=1}^N\frac{x_ix_j}{|x|^2}D_{ij}+c\frac{x}{|x|^2}\cdot \nabla -\frac{b}{|x|^{2}}, \end{aligned}$$

$a>0$, $b,\ c \in \mathbb {R}$ and we prove gradient estimates for the heat kernel.

Keywords

Elliptic operators discontinuous coefficients analytic semigroups kernel estimates

Mathematics Subject Classification

47D07 35B50 35J25 35J70

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Notes

Appendix

Radial and Angular Derivatives

In this section, we study further properties about the spherical harmonics and the Laplace–Beltrami operator $\Delta _0$ on the sphere. For $x\in \mathbb {R}^N$, we use spherical coordinates to write $x=r\omega $, where $r:=|x|$, $\omega :=\frac{x}{|x|}\in S^{N-1}$. For every $u\in C^2(\mathbb {R}^n)$, we denote by $D_r u$, $D_{rr} u$ the radial derivatives of u and by $\nabla _{\tau }u$ the tangential component of its gradient. They are defined through the formulas

$$\begin{aligned} D_r u&=\sum _{i=1}^{N} D_iu\,\frac{x_i}{r},\qquad D_{rr} u=\sum _{i,j=1}^N D_{ij}u\,\frac{x_i x_j}{r^2},\qquad \nabla u=D_ru \frac{x}{|x|}+\frac{\nabla _\tau u}{r}.\nonumber \\ \end{aligned}$$

(27)

We recall that in spherical coordinates, we have the following relation between the Laplace operator and $\Delta _0$:

$$\begin{aligned} \Delta =D_{rr}+\frac{N-1}{r}D_r+\frac{\Delta _0}{r^2}. \end{aligned}$$

The operator $A:=rD_r$ is the Euler operator $\sum _{i=1}^nx_iD_i$ and, for a given function u, the Euler’s theorem implies that u is $\alpha $-homogeneous if and only if $Au=\alpha u$.

For $i,j=1,\dots ,N$ we introduce, moreover, the angular operators $S_{ij}$ defined as

$$\begin{aligned} S_{ij}:=x_iD_j-x_jD_i. \end{aligned}$$

They are first-order differential operators and have a central role in the analysis on the sphere since they allow to decompose the Laplace–Beltrami operator into a sum of second-order angular derivatives. Obviously, $S_{ii}=0$ and $S_{ij}=-S_{ji}$ so it is sufficient to consider $S_{ij}$ for $i<j$.

We collect in the next proposition some basic properties about the angular derivatives $S_{ij}$.

Proposition 4.1

Denote by $\left[ A,B\right] =AB-BA$ the commutator of A, B and by $\delta _i^j$ the Kronecker delta. The following properties hold.

(i)
For every $i<j$ and $h<k$ we have
$$\begin{aligned} \left[ S_{ij},S_{hk}\right] =-\delta _i^hS_{jk}+\delta _i^kS_{jh}+\delta _j^hS_{ik}-\delta _j^kS_{ih}. \end{aligned}$$
In particular, $S_{ij}$ and $ S_{hk}$ commute if and only if $(i,j)=(h,k)$ or $\left\{ i,j\right\} \cap \left\{ h,k\right\} =\emptyset $.
(ii)
For every radial function $u=u(r)$, with $u\in C^2(\mathbb {R}^N)$ and for every $v\in C^2(\mathbb {R}^N)$ , we have $S_{ij}u=0$ and $S_{ij}(uv)=u S_{ij}v$.
(iii)
For every radial differential operator $E=\sum _{k=0}^n a_k(r)\frac{d^k}{dr^k}$, we have $S_{ij}E=ES_{ij}$.
(iv)
$\Delta \,S_{ij}=S_{ij}\,\Delta $ and $\Delta _0\,S_{ij}=S_{ij}\,\Delta _0$.

Proof

(i) easily follows by a straightforward computation.

Let $u,v\in C^2(\mathbb {R}^N)$ with $u=u(r)$; we have $S_{ij}u=x_iu'(r)\frac{x_j}{r}-x_ju'(r)\frac{x_i}{r}=0$. Moreover, $S_{ij}(uv)=uS_{ij}v+(S_{ij}u)v=uS_{ij}v$ that proves (ii).

To prove (iii), we observe that for $E=a(r)D_r$ we have, using (ii),

$$\begin{aligned} S_{ij}(a(r)D_r)= & {} S_{ij}\left( \frac{a(r)}{r}\sum _{k=1}^Nx_kD_k\right) =\frac{a(r)}{r}\left( x_iD_j-x_jD_i\right) \sum _{k=1}^Nx_kD_k\\= & {} \frac{a(r)}{r}\left[ x_iD_j+x_i\sum _{k=1}^Nx_kD_{jk}-x_jD_i-x_j\sum _{k=1}^Nx_kD_{ik}\right] \\= & {} \frac{a(r)}{r}\left[ S_{ij}+\sum _{k=1}^Nx_k(x_iD_{jk}-x_jD_{ik})\right] . \end{aligned}$$

On the other hand,

$$\begin{aligned} (a(r)D_r)S_{ij}= & {} \left( \frac{a(r)}{r}\sum _{k=1}^Nx_kD_k\right) S_{ij}=\frac{a(r)}{r}\sum _{k=1}^Nx_kD_k\left( x_iD_j-x_jD_i\right) \\= & {} \frac{a(r)}{r}\left[ x_iD_j+\sum _{k=1}^Nx_kx_iD_{jk}-x_jD_i-\sum _{k=1}^Nx_kx_jD_{ik}\right] \\= & {} \frac{a(r)}{r}\left[ S_{ij}+\sum _{k=1}^Nx_k(x_iD_{jk}-x_jD_{ik})\right] . \end{aligned}$$

Comparing the last expressions, we have $S_{ij}E=ES_{ij}$; the general case follows easily by induction.

Finally, to prove (iv), let us suppose, without losing generality, $i=1,\,j=2$. We have

$$\begin{aligned} \Delta S_{12}= & {} \sum _{i=1}^ND_{ii}\left( x_1D_2-x_2D_1\right) \\= & {} \sum _{i=1}^N\left( x_1D_2D_{ii}+2D_ix_1D_iD_2-x_2D_1D_{ii}-2D_ix_2D_iD_1\right) \\= & {} \sum _{i=1}^N\left( x_1D_2-x_2D_1\right) D_{ii}+2D_1D_2-2D_2D_1= S_{12}\Delta \end{aligned}$$

and so, $S_{12}$ commutes with $\Delta $.

Using property (ii), we have also

$$\begin{aligned} S_{12}(r^2\Delta )=r^2S_{12}\Delta =(r^2\Delta )S_{12}. \end{aligned}$$

Since $r^2\Delta =r^2D_{rr}+(N-1)rD_r+\Delta _0$, it follows immediately from property (iii) and the fact that $S_{12}$ commutes with $r^2\Delta $, that $\Delta _0\,S_{ij}=S_{ij}\,\Delta _0$.

$\square $

Before stating the main results, it is worth noting that every function $f\in C^2(S^{N-1})$ is extendible on a neighbourhood of $S^{N-1}$ preserving the same degree of regularity; therefore, in what follows, we can always suppose f defined on an open set of $\mathbb {R}^N$ containing $S^{N-1}$. In particular, $\nabla f$, $\Delta f$ and $S_{ij}f$ are well defined. For instance, if we define $\tilde{f}$ on $\mathbb {R}^N{\setminus }\{0\}$ by $\tilde{f}(y):=f(\frac{y}{|y|})$, being $\tilde{f}$ constant along every radial direction, it follows immediately by the decomposition of $\nabla $ and $\Delta $ that

$$\begin{aligned} \nabla _\tau f(x)=\nabla \tilde{f}(x),\quad \Delta _0 f(x)=\Delta \tilde{f}(x),\quad \text {for every}\; x\in S^{N-1}. \end{aligned}$$

Let now $\omega \in S^{N-1}$ and $v\in T_\omega (S^{N-1})$; the tangential derivative of f in $\omega $ in the direction v is given by

$$\begin{aligned} (\nabla f(\omega ),v)=(\nabla _\tau f(\omega ),v). \end{aligned}$$

Setting $\omega =(\omega _1,\ldots ,\omega _N)\in S^{N-1}$, let us define for $i<j$

$$\begin{aligned} \omega _{ij}:=(0,\ldots ,-\omega _j,0,\ldots ,\omega _i,0,\ldots )\in T_\omega (S^{N-1}), \end{aligned}$$

i.e. the vector which has $-\omega _j$ and $\omega _i$ as, respectively, the ith and the jth components and zeros in the other entries. $\left( \omega _{ij}\right) _{1\le i<j\le N}$ is, by construction, a system of generators of $T_\omega (S^{N-1})$.

The next propositions clarify the role and the interaction among the operators introduced so far. We begin by a geometric Lemma.

Lemma 4.2

Let $\omega \in S^{N-1}$ and $x,y\in \omega ^\perp $. Then

$$\begin{aligned} (x,y)=\sum _{i<j}(x,\omega _{ij})(y,\omega _{ij}),\qquad |x|^2=\sum _{i<j}|(x,\omega _{ij})|^2. \end{aligned}$$

(28)

Proof

Since $(\omega _{ij},x)=\omega _ix_j-\omega _jx_i,\; (\omega _{ij},y)=\omega _iy_j-\omega _jy_i$, we have

$$\begin{aligned} \sum _{i<j}(x,\omega _{ij})(y,\omega _{ij})= & {} \sum _{i<j}\left( \omega _ix_j-\omega _jx_i\right) \left( \omega _iy_j-\omega _jy_i\right) \\= & {} \sum _{i<j}\left( \omega _i^2x_jy_j+\omega _j^2x_iy_i-\omega _j\omega _ix_iy_j-\omega _i\omega _jx_jy_i\right) \\= & {} \sum _{i\ne j}\omega _i^2x_jy_j-\sum _{i\ne j}\omega _j\omega _ix_iy_j. \end{aligned}$$

Adding and subtracting, in the last formula, the term $\sum _{i=1}^N\omega _i^2x_iy_i$, we obtain

$$\begin{aligned} \sum _{i<j}(x,\omega _{ij})(y,\omega _{ij})= & {} \sum _{i, j=1}^N\omega _i^2x_jy_j-\sum _{i, j=1}^N\omega _j\omega _ix_iy_j =(x,y)|\omega |^2-(\omega ,x)(\omega ,y)\\= & {} (x,y). \end{aligned}$$

Choosing $x=y$, we get $\sum _{i<j}|(x,\omega _{ij})|^2=|x|^2$. $\square $

Proposition 4.3

Let f, g be $C^1$ functions defined on a neighbourhood of $S^{N-1}$. The following properties hold.

(i)
$S_{ij}f(\omega )$ is the tangential derivative of f in $\omega $ in the direction $\omega _{ij}$, i.e.
$$\begin{aligned} S_{ij}f(\omega )= (\nabla f(\omega ),\omega _{ij})=(\nabla _\tau f(\omega ),\omega _{ij}). \end{aligned}$$
(ii)
The jth component of $\nabla _{\tau } f$ satisfies
$$\begin{aligned} (\nabla _{\tau } f(\omega ))_j=\sum _{i\ne j}\omega _i \,S_{ij}f(\omega ),\quad \omega \in S^{N-1}. \end{aligned}$$
(iii)
For every $\omega \in S^{N-1}$
$$\begin{aligned} (\nabla _{\tau } f(\omega ),\nabla _{\tau } g(\omega ))=\sum _{i<j}S_{ij}f(\omega )S_{ij}g(\omega ),\quad |\nabla _{\tau } f(\omega )|^2=\sum _{i<j}|S_{ij}f(\omega )|^2. \end{aligned}$$

Proof

The first assertion is an immediate consequence of the definition of $S_{ij}$. For the second sentence, we observe that for $\omega \in S^{N-1}$ it follows from (27) that

$$\begin{aligned} (\nabla _{\tau } f(\omega ))_j= & {} D_jf(\omega )-\omega _jD_rf(\omega )=D_jf(\omega )\sum _{i=1}^N\omega _i^2-\omega _j\sum _i\omega _iD_if(\omega )\\= & {} \sum _{i=1}^N\omega _i \left[ \omega _i D_jf(\omega )-\omega _j D_if(\omega )\right] =\sum _{i\ne j}\omega _i \,S_{ij}f(\omega ). \end{aligned}$$

(iii) is an immediate consequence of (i) and (28). $\square $

We can now show the announced decomposition of the Laplace–Beltrami operator as a sum of second-order angular derivatives.

Proposition 4.4

$$\begin{aligned} \sum _{i<j}S_{ij}^2f=\Delta _0f,\quad \text {for every}\quad f\in C^2(\mathbb {R}^N), \end{aligned}$$

where using the spherical coordinates $x=r\omega $, $\Delta _0f(r\omega )$ acts on the $\omega $-variable.

In particular, on $S^{N-1}$ we have the decomposition ${\sum _{i<j}S_{ij}^2=\Delta _0}$.

Proof

We observe preliminary that

$$\begin{aligned} r^2D_{rr}=\sum _{i,j=1}^Nx_ix_jD_{ij}=2\sum _{i<j}x_ix_jD_{ij}+\sum _{i=1}^Nx_i^2D_{ii} \end{aligned}$$

(29)

and

$$\begin{aligned} S_{ij}^2= & {} (x_iD_j-x_jD_i)(x_iD_j-x_jD_i)=x_i^2D_{jj}+x_j^2D_{ii}-x_iD_i-x_jD_j\\&-2x_ix_jD_{ij}. \end{aligned}$$

Summing over $i<j$ the last expression and using (29) we obtain

$$\begin{aligned} \sum _{i<j}S_{ij}^2= & {} \sum _{i<j}\left( x_i^2D_{jj}+x_j^2D_{ii}\right) -\sum _{i<j}\left( x_iD_i+x_jD_j\right) -2\sum _{i<j}x_ix_jD_{ij}\\= & {} \sum _{i\ne j}x_j^2D_{ii}-\sum _{i\ne j}x_iD_i-r^2D_{rr}+\sum _{i=1}^Nx_i^2D_{ii}\\= & {} \sum _{i=1}^N\left( r^2-x_i^2\right) D_{ii}-(N-1)rD_r-r^2D_{rr}+\sum _{i=1}^Nx_i^2D_{ii}\\= & {} r^2\Delta -(N-1)rD_r-r^2D_{rr}. \end{aligned}$$

Recalling that $r^2\Delta =r^2D_{rr}+(N-1)rD_r+\Delta _0$ we get the conclusion. $\square $

Next we have the following integration by parts formula.

Proposition 4.5

For every $i<j$ and for every $u,v\in C^1(S^{N-1})$

$$\begin{aligned} \int _{S^{N-1}} (S_{ij}u)v\,\mathrm{{d}}\sigma =-\int _{S^{N-1}} u(S_{ij}v)\,\mathrm{{d}}\sigma . \end{aligned}$$

Proof

Let $i<j$ and $u,v\in C^1(S^{N-1})$. We have obviously $S_{ij}(uv)=S_{ij}u\,v+u\,S_{ij}v$ and so

$$\begin{aligned} \int _{S^{N-1}} (S_{ij}u)v\,\mathrm{{d}}\sigma&=-\displaystyle \int _{S^{N-1}}u\,S_{ij}v\,\mathrm{{d}}\sigma +\displaystyle \int _{S^{N-1}}S_{ij}(uv)\,\mathrm{{d}}\sigma . \end{aligned}$$

Using the Gauss–Green theorem, the claim immediately follows by observing that

$$\begin{aligned} \int _{S^{N-1}}S_{ij}(uv)\,\mathrm{{d}}\sigma= & {} \int _{S^{N-1}}\omega _iD_j(uv)-\omega _jD_i(uv)\,\mathrm{{d}}\sigma \\= & {} \int _{B(0,1)}D_{ij}(uv)-D_{ji}(uv)\,\mathrm{{d}}x=0. \end{aligned}$$

$\square $

Let $\mathcal {H}_n$ be the set of the spherical harmonics of order n. We recall that from [9, Proposition 2.1 (iii)], we have

$$\begin{aligned} \Vert \phi \Vert _\infty \le \sqrt{\mathrm{{dim}}\mathcal H_n}\,\Vert \phi \Vert _{L^2(S^{N-1)}},\quad \text {for every}\;\phi \in \mathcal {H}_n. \end{aligned}$$

The following proposition shows that $\mathcal {H}_n$ is preserved by $S_{ij}$.

Proposition 4.6

Let $\phi \in \mathcal {H}_n$. Then, for every $i<j$, $S_{ij}\phi \in \mathcal {H}_n$. Moreover, if $-\lambda _n=-n(n+N-2)$ is the eigenvalue associated with $\phi $, then

$$\begin{aligned} \Vert S_{ij}\phi \Vert _\infty\le & {} \sqrt{\lambda _n \mathrm{{dim}}\mathcal H_n}\,\Vert \phi \Vert _{L^2(S^{N-1)}}, \end{aligned}$$

(30)

$$\begin{aligned} \Vert \nabla _\tau \phi \Vert _\infty\le & {} \sqrt{\frac{N(N-1)}{2}\lambda _n \mathrm{{dim}}\mathcal H_n} \,\Vert \phi \Vert _{L^2(S^{N-1)}}. \end{aligned}$$

(31)

Proof

Let $p(x)=r^n\phi (\omega )$; p is a homogeneous harmonic polynomial by construction and $S_{ij}p$ is a polynomial of the same degree by the definition of $S_{ij}$. Moreover, using (iv) of Proposition 4.1, we have

$$\begin{aligned} \Delta S_{ij}p=S_{ij}\Delta p=0 \end{aligned}$$

and so $S_{ij}p$ is harmonic since from (ii) of Proposition 4.1 $S_{ij}P(r\omega )=r^nS_{ij}\phi (\omega )$, we have $S_{ij}\phi \in \mathcal {H}_n$. In particular,

$$\begin{aligned} \Delta _0 S_{ij}\phi =S_{ij}\Delta _0 \phi =-\lambda _n S_{ij}\phi \end{aligned}$$

and

$$\begin{aligned} \Vert S_{ij}\phi \Vert _\infty \le \sqrt{\mathrm{{dim}}\mathcal H_n}\,\Vert S_{ij}\phi \Vert _{L^2(S^{N-1)}}. \end{aligned}$$

(32)

Now, using propositions 4.5 and 4.4,

$$\begin{aligned} \sum _{i<j}\Vert S_{ij}\phi \Vert ^2_{L^2(S^{N-1)}}= & {} \sum _{i<j}\int _{S^{N-1}}S_{ij}\phi \, S_{ij}\phi \,\mathrm{{d}}\sigma =-\sum _{i<j}\int _{S^{N-1}}(S_{ij}^2\phi )\phi \,\mathrm{{d}}\sigma \\= & {} -\int _{S^{N-1}}\sum _{i<j}(S_{ij}^2\phi )\phi \,\mathrm{{d}}\sigma =-\int _{S^{N-1}}\Delta _0 \phi \, \phi \,\mathrm{{d}}\sigma \\= & {} \lambda _n\Vert \phi \Vert ^2_{L^2(S^{N-1)}}. \end{aligned}$$

The last inequality implies

$$\begin{aligned} \Vert S_{ij}\phi \Vert ^2_{L^2(S^{N-1)}}\le \sqrt{\lambda _n}\Vert \phi \Vert _{L^2(S^{N-1)}} \end{aligned}$$

which, combined with (32), yields (30). Finally, applying (28) and (30), it follows for every $\omega \in S^{N-1}$

$$\begin{aligned} |\nabla _\tau \phi (\omega )|^2&=\sum _{i<j}|S_{ij}\phi (\omega )|^2\le \frac{N(N-1)}{2}\lambda _n \mathrm{{dim}}\mathcal H_n \,\Vert \phi \Vert _{L^2(S^{N-1)}}, \end{aligned}$$

which proves (31). $\square $

Corollary 4.7

The tangential derivative of the zonal harmonics $\mathbb {Z}^{(n)}$ satisfies

$$\begin{aligned} \Vert \nabla _{\tau }\mathbb {Z}^{(n)}\Vert _{\infty }\le \sqrt{\frac{1}{|S^{N-1}|}\frac{N(N-1)}{2}\lambda _n }\left( \mathrm{{dim}}\mathcal H_n\right) ^\frac{3}{2}. \end{aligned}$$

In particular, for a constant $C=C(N)$ we have

$$\begin{aligned} \Vert \nabla _{\tau }\mathbb {Z}^{(n)}\Vert _{\infty }\le C n^{\frac{3n-4}{2}}. \end{aligned}$$

Proof

The first property is an immediate consequence of (31) and of the estimate

$$\begin{aligned} \Vert \mathbb {Z}^{(n)}\Vert _{L^2(S^{N-1})}\le \sqrt{|S^{N-1}|}\,\Vert \mathbb {Z}^{(n)}\Vert _\infty \le \sqrt{|S^{N-1}|}\,\frac{\mathrm{{dim}}\mathcal H_n}{|S^{N-1}|}. \end{aligned}$$

To prove the second inequality it is sufficient to recall the asymptotic behaviours $\mathrm{{dim}}\mathcal H_n\approx n^{N-2}$, $\lambda _n\approx n^2$ for $n\rightarrow \infty $. $\square $

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G. Metafune
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L. Negro
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C. Spina
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1.Dipartimento di Matematica “Ennio De Giorgi”Università del SalentoLecceItaly

Gradient Estimates for Elliptic Operators with Second-Order Discontinuous Coefficients

Abstract

Keywords

Mathematics Subject Classification

Notes

References

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