Critical SQG in Bounded Domains

Abstract

We consider the critical dissipative SQG equation in bounded domains, with the square root of the Dirichlet Laplacian dissipation. We prove global a priori interior $C^{\alpha }$ and Lipschitz bounds for large data.

Introduction

The Surface Quasigeostrophic equation (SQG) of geophysical origin [18] was proposed as a two dimensional model for the study of inviscid incompressible formation of singularities [5, 9]. While the global regularity of all solutions of SQG whose initial data are smooth is still unknown, the original blow-up scenario of [9] has been ruled out analytically [13] and numerically [11], and nontrivial examples of global smooth solutions have been constructed [4]. Solutions of SQG and related equations without dissipation and with non-smooth (piece-wise constant) initial data give rise to interface dynamics [3, 17] with potential finite time blow up [15].

The addition of fractional Laplacian dissipation produces globally regular solutions if the power of the Laplacian is larger or equal than one half. When the linear dissipative operator is precisely the square root of the Laplacian, the equation is commonly referred to as the “critical dissipative SQG”, or “critical SQG”. This active scalar equation [5] has been the object of intensive study in the past decade. The solutions are transported by divergence-free velocities they create, and are smoothed out and decay due to nonlocal diffusion. Transport and diffusion do not add size to a solution: the solution remains bounded, if it starts so [22]. The space $L^{\infty }({\mathbb R}^2)$ is not a natural phase space for the nonlinear evolution: the nonlinearity involves Riesz transforms and these are not well behaved in $L^{\infty }$. Unfortunately, for the purposes of studies of global in time behavior of solutions, $L^{\infty }$ is unavoidable: it quantifies the most important information freely available. The equation is quasilinear and $L^{\infty }$-critical, and there is no “ wiggle room”, nor a known better (smaller) space which is invariant for the evolution. One must work in order to obtain better information. A pleasant aspect of criticality is that solutions with small initial $L^{\infty }$ norm are smooth [10]. The global regularity of large solutions was obtained independently in [2] and [20] by very different methods: using harmonic extension and the De Georgi methodology of zooming in, and passing from $L^2$ to $L^{\infty }$ and from $L^{\infty }$ to $C^{\alpha }$ in [2], and constructing a family of time-invariant moduli of continuity in [20]. Several subsequent proofs were obtained (please see [12] and references therein). All the proofs are dimension-independent, but are in either ${\mathbb R}^d$ or on the torus ${\mathbb {T}}^d$. The proofs of [7] and [12] were based on an extension of the Córdoba–Córdoba inequality [14]. This inequality states that

$$\begin{aligned} \Phi '(f)\Lambda f - \Lambda \Phi (f)\ge 0 \end{aligned}$$

(1)

pointwise. Here $\Lambda = \sqrt{-\Delta }$ is the square root of the Laplacian in the whole space ${\mathbb R}^d$, $\Phi$ is a real valued convex function of one variable, normalized so that $\Phi (0) = 0$ and f is a smooth function. The fractional Laplacian in the whole space has a (very) singular integral representation, and this can be used to obtain (1). In [7] specific nonlinear maximum principle lower bounds were obtained and used to prove the global regularity. A typical example is

$$\begin{aligned} D(f) =f\Lambda f- \frac{1}{2}\Lambda \left( {f^2}\right) \ge c\left( \Vert \theta \Vert _{L^{\infty }}\right) ^{-1} {f^3} \end{aligned}$$

(2)

pointwise, for $f=\partial _i\theta$ a component of the gradient of a bounded function $\theta$. This is a useful cubic lower bound for a quadratic expression, when $\Vert \theta \Vert _{L^{\infty }}\le \Vert \theta _0\Vert _{L^{\infty }}$ is known to be bounded above. The critical SQG equation in ${\mathbb R}^2$ is

$$\begin{aligned} \partial _t \theta + u\cdot \nabla \theta + \Lambda \theta = 0 \end{aligned}$$

(3)

where

$$\begin{aligned} u = \nabla ^{\perp }\Lambda ^{-1}\theta = R^{\perp }{\theta } \end{aligned}$$

(4)

and $\nabla ^{\perp } = (-\partial _2, \partial _1)$ is the gradient rotated by $\frac{\pi }{2}$. Because of the conservative nature of transport and the good dissipative properties of $\Lambda$ following from (1), all $L^p$ norms of $\theta$ are nonincreasing in time. Moreover, because of properties of Riesz transforms, u is essentially of the same order of magnitude as $\theta$. Differentiating the equation we obtain the stretching equation

$$\begin{aligned} \left( \partial _t + u\cdot \nabla + \Lambda \right) \nabla ^{\perp }\theta = (\nabla u)\nabla ^{\perp }\theta . \end{aligned}$$

(5)

(In the absence of $\Lambda$ this is the same as the stretching equation for three dimensional vorticity in incompressible Euler equations, one of the main reasons SQG was considered in [5, 9] in the first place.) Taking the scalar product with $\nabla ^{\perp }\theta$ we obtain

$$\begin{aligned} \frac{1}{2}(\partial _t + u\cdot \nabla + \Lambda )q^2 + D(q) = Q \end{aligned}$$

(6)

for $q^2 = |\nabla ^{\perp }\theta |^2$, with

$$\begin{aligned} Q = (\nabla u)\nabla ^{\perp }\theta \cdot \nabla ^{\perp }\theta \le |\nabla u| q^2. \end{aligned}$$

The operator $\partial _t + u\cdot \nabla + \Lambda$ is an operator of advection and fractional diffusion: it does not add size. Using the pointwise bound (2) we already see that the dissipative lower bound is potentially capable of dominating the cubic term Q, but there are two obstacles. The first obstacle is that constants matter: the two expressions are cubic, but the useful dissipative cubic lower bound $D(q)\ge K |q|^{3}$ has perhaps too small a prefactor K if the $L^{\infty }$ norm of $\theta _0$ is too large. The second obstacle is that although

$$\begin{aligned} \nabla u = R^{\perp }(\nabla \theta ) \end{aligned}$$

has the same size as $\nabla ^{\perp }\theta$ (modulo constants) in all $L^p$ spaces $1<p<\infty$, it fails to be bounded in $L^{\infty }$ by the $L^{\infty }$ norm of $\nabla ^{\perp }\theta$. In order to overcome these obstacles, in [7] and [12], instead of estimating directly gradients, the proof proceeds by estimating finite differences, with the aim of obtaining bounds for $C^{\alpha }$ norms first. In fact, in critical SQG, once the solution is bounded in any $C^{\alpha }$ with $\alpha >0$, it follows that it is $C^{\infty }$. More generally, if the equation has a dissipation of order s, i.e., $\Lambda$ is replaced by $\Lambda ^s$ with $0<s\le 1$ then if $\theta$ is bounded in $C^{\alpha }$ with $\alpha >1-s$, then the solution is smooth [8]. (This condition is sharp, if one considers general linear advection diffusion equations, [23]. In [12] the smallness of $\alpha$ is used to show that the term corresponding to Q in the finite difference version of the argument is dominated by the term corresponding to D(q).

In this paper we consider the critical SQG equation in bounded domains. We take a bounded open domain $\Omega \subset {\mathbb R}^d$ with smooth (at least $C^{2,\alpha }$) boundary and denote by $\Delta$ the Laplacian operator with homogeneous Dirichlet boundary conditions and by $\Lambda _D$ its square root defined in terms of eigenfunction expansions. Because no explicit kernel for the fractional Laplacian is available in general, our approach, initiated in [6] is based on bounds on the heat kernel.

The critical SQG equation is

$$\begin{aligned} \partial _t \theta + u\cdot \nabla \theta + \Lambda _D\theta =0 \end{aligned}$$

(7)

with

$$\begin{aligned} u = \nabla ^{\perp }\Lambda _D^{-1}\theta = R_D^{\perp }\theta \end{aligned}$$

(8)

and smooth initial data. We obtain global regularity results, in the spirit of the ones in the whole space. There are quite significant differences between the two cases. First of all, the fact that no explicit formulas are available for kernels requires a new approach; this yields as a byproduct new proofs even in the whole space. The main difference and additional difficulty in the bounded domain case is due to the lack of translation invariance. The fractional Laplacian is not translation invariant, and from the very start, differentiating the equation (or taking finite differences) requires understanding the respective commutators. For the same reason, the Riesz transforms $R_D$ are not spectral operators, i.e., they do not commute with functions of the Laplacian, and so velocity bounds need a different treatment. In [6] we proved using the heat kernel approach the existence of global weak solutions of (7) in $L^2(\Omega )$. A proof of local existence of smooth solutions is provided in the present paper in $d=2$. The local existence is obtained in Sobolev spaces based on $L^2$ and uses Sobolev embeddings. Because of this, the proof is dimension dependent. A proof in higher dimensions is also possible but we do not pursue this here. We note that for regular enough solutions (e.g. $\theta \in H_0^1(\Omega )$) the normal component of the velocity vanishes at the boundary $\left( R_D^{\perp }\theta \cdot N\right) _{\left| \right. \partial \Omega }=0$ because the stream function $\psi = \Lambda _D^{-1}\theta$ vanishes at the boundary and its gradient is normal to the boundary. Let us remark here that even in the case of a half-space and $\theta \in C_0^{\infty }(\Omega )$, the tangential component of the velocity need not vanish: there is tangential slip.

In order to state our main results, let

$$\begin{aligned} d(x) = dist(x,\partial \Omega ) \end{aligned}$$

(9)

denote the distance from x to the boundary of $\Omega$. We introduce the $C^{\alpha }(\Omega )$ space for interior estimates:

Definition 1

Let $\Omega$ be a bounded domain and let $0<\alpha <1$ be fixed. We say that $\theta \in C^{\alpha }(\Omega )$ if $\theta \in L^{\infty }(\Omega )$ and

$$\begin{aligned}{}[f]_{\alpha } = \sup _{x\in \Omega }(d(x))^{\alpha }\left( \sup _{h\ne 0,|h|<d(x)}\frac{|f(x+h)-f(x)|}{|h|^{\alpha }}\right) <\infty . \end{aligned}$$

(10)

The norm in $C^{\alpha }(\Omega )$ is

$$\begin{aligned} \Vert f\Vert _{C^{\alpha }} = \Vert f\Vert _{L^{\infty }(\Omega )} + [f]_{\alpha }. \end{aligned}$$

(11)

Our main results are the following:

Theorem 1

Let $\theta (x,t)$ be a smooth solution of (7) on a time interval [0, T), with $T\le \infty$, with initial data $\theta (x,0)= \theta _0(x)$. Then the solution is uniformly bounded,

$$\begin{aligned} \sup _{0\le t< T}\Vert \theta (t)\Vert _{L^{\infty }(\Omega )}\le \Vert \theta _0\Vert _{L^{\infty }(\Omega )}. \end{aligned}$$

(12)

There exists $\alpha$ depending only on $\Vert \theta _0\Vert _{L^{\infty }(\Omega )}$ and $\Omega$, and a constant $\Gamma$ depending only on the domain $\Omega$ (and in particular, independent of T) such that

$$\begin{aligned} \sup _{0\le t<T}\Vert \theta (t)\Vert _{C^{\alpha }(\Omega )} \le \Gamma \Vert \theta _0\Vert _{C^{\alpha }(\Omega )} \end{aligned}$$

(13)

holds.

The second theorem is about global interior gradient bounds:

Theorem 2

Let $\theta (x,t)$ be a smooth solution of (7) on a time interval [0, T), with $T\le \infty$, with initial data $\theta (x,0)= \theta _0(x)$. There exists a constant $\Gamma _1$ depending only on $\Omega$ such that

$$\begin{aligned} \sup _{x\in \Omega , 0\le t<T}d(x)|\nabla _x\theta (x,t)|\le \Gamma _1\left[ \sup _{x\in \Omega }d(x)|\nabla _x\theta _0(x)| + \left( 1+\Vert \theta _0\Vert _{L^{\infty }(\Omega )}\right) ^4\right] \end{aligned}$$

(14)

holds.

Remark 1

Higher interior regularity can be proved also. In fact, once global interior $C^{\alpha }$ bounds are obtained for any $\alpha >0$, the interior regularity problem becomes subcritical, meaning that “there is room to spare”. This is already the case for Theorem 2 and justifies thinking that the equation is $L^{\infty }$ interior-critical. However, we were not able to obtain global uniform $C^{\alpha }(\bar{\Omega })$ bounds. Moreover, we do not know the implication $C^{\alpha }(\bar{\Omega }) \Rightarrow C^{\infty }(\bar{\Omega })$ uniformly, and thus the equation is not $L^{\infty }$ critical up to the boundary. This is due to the fact that the commutator between normal derivatives and the fractional Dirichlet Laplacian is not controlled uniformly up to the boundary. The example of half-space is instructive because explicit kernels and calculations are available. In this example odd reflection across the boundary permits the construction of global smooth solutions, if the initial data are smooth and compactly supported away from the boundary. The support of the solution remains compact and cannot reach the boundary in finite time, but the gradient of the solution might grow in time at an exponential rate.

The proofs of our main results use the following elements. First, the inequality (1) which has been proved in [6] for the Dirichlet $\Lambda _D$ is shown to have a lower bound

$$\begin{aligned} D(f)(x) = \left( f\Lambda _Df - \frac{1}{2}\Lambda _D\left( {f^2}\right) \right) (x) \ge c\frac{f^2(x)}{d(x)} \end{aligned}$$

(15)

with $c>0$ depending only on $\Omega$. Note that in ${\mathbb R}^d$, $d(x)=\infty$, which is consistent with (1). This lower bound (valid for general $\Phi$ convex, with c independent of $\Phi$, see (46)) provides a strong damping boundary repulsive term, which is essential to overcome boundary effects coming from the lack of translation invariance.

The second element of proofs consists of nonlinear lower bounds in the spirit of [7]. A version for derivatives in bounded domains, proved in [6] is modified for finite differences. In order to make sense of finite differences near the boundary in a manner suitable for transport, we introduce a family of good cutoff functions depending on a scale $\ell$ in Lemma 3. The finite difference nonlinear lower bound is

$$\begin{aligned} D(f)(x)\ge c\left( |h|\Vert \theta \Vert _{L^{\infty }(\Omega )}\right) ^{-1}|f(x)|^3+ c\frac{|f(x)|^2}{d(x)} \end{aligned}$$

(16)

when $f=\chi \delta _h\theta$ is large (see (48)), where $\chi$ belongs to the family of good cutoff functions.

Once global interior $C^{\alpha }(\Omega )$ bounds are obtained, in order to obtain global interior bounds for the gradient, we use a different nonlinear lower bound,

$$\begin{aligned} D(f) = \left( f\Lambda _Df -\frac{1}{2}(\Lambda _Df^2)\right) (x) \ge c \frac{|f(x)|^{3+\frac{\alpha }{1-\alpha }}}{\Vert \theta \Vert _{C^{\alpha }(\Omega )}^{\frac{1}{1-\alpha }}}(d(x))^{\frac{\alpha }{1-\alpha }} + c\frac{f^2(x)}{d(x)} \end{aligned}$$

(17)

for large $f=\chi \nabla \theta$ (see (61)). This is a super-cubic bound, and makes the gradient equation look subcritical. Similar bounds were obtained in the whole space in [7]. Proving the bounds (16) and (17) requires a different approach and new ideas because of the absence of explicit formulas and lack of translation invariance.

The third element of proofs are bounds for $R_D^{\perp }\theta$ based only on global a priori information on $\Vert \theta \Vert _{L^{\infty }}$ and the nonlinear lower bounds on D(f) for appropriate f. Such an approach was initiated in [7] and [12]. In the bounded domain case, again, the method of proof is different because the kernels are not explicit, and reference is made to the heat kernels. The boundaries introduce additional error terms. The bound for finite differences is

$$\begin{aligned} |\delta _h R^{\perp }_D\theta (x)| \le C\left( \sqrt{\rho D(f)(x)} + \Vert \theta \Vert _{L^{\infty }}\left( \frac{|h|}{d(x)}+ \frac{|h|}{\rho }\right) + |\delta _h\theta (x)|\right) \end{aligned}$$

(18)

for $\rho \le cd(x)$, with $f=\chi \delta _h \theta$ and with C a constant depending on $\Omega$ (see 90). The bound for gradient is

$$\begin{aligned} |\nabla R^{\perp }_D\theta (x)| \le C\left( \sqrt{\rho D(f)(x)} + \Vert \theta \Vert _{L^{\infty }(\Omega )}\left( \frac{1}{d(x)} + \frac{1}{\rho }\right) + |\nabla \theta (x)|\right) \end{aligned}$$

(19)

for $\rho \le cd(x)$ with $f=\chi \nabla \theta$ with a constant C depending on $\Omega$ (see (107)). These are remarkable pointwise bounds (clearly not valid for the case of the Laplacian even in the whole space, where $D(f)(x) = |\nabla f(x)|^2$).

The fourth element of the proof are bounds for commutators. These bounds

$$\begin{aligned} \left| \left[ \chi \delta _h, \Lambda _D\right] \theta (x)\right| \le C\frac{|h|}{d(x)^2}\Vert \theta \Vert _{L^{\infty }(\Omega )}, \end{aligned}$$

(20)

for $\ell \le d(x)$, (see (112)), and

$$\begin{aligned} \left| \left[ \chi \nabla , \Lambda _D\right] \theta (x)\right| \le \frac{C}{d(x)^2}\Vert \theta \Vert _{L^{\infty }(\Omega )}, \end{aligned}$$

(21)

for $\ell \le d(x)$, (see (115)), reflect the difficulties due to the boundaries. They are remarkable though in that the only price to pay for a second order commutator in $L^{\infty }$ is $d(x)^{-2}$. Note that in the whole space this commutator vanishes ($\chi =1$). This nontrivial situation in bounded domains is due to cancellations and bounds on the heat kernel representing translation invariance effects away from boundaries (see (37, 38)). Although the heat kernel in bounded domains has been extensively studied, and the proofs of (37) and (38) are elementary, we have included them in the paper because we have not found them readily available in the literature and for the sake of completeness.

The paper is organized as follows: after preliminary background, we prove the nonlinear lower bounds. We have separate sections for bounds for the Riesz transforms and the commutators. The proof of the main results are then provided, using nonlinear maximum principles. We give some of the explicit calculations in the example of a half-space and conclude the paper by proving the translation invariance bounds for the heat kernel (37), (38), and a local well-posedness result in two appendices.

Preliminaries

The $L^2(\Omega )$ - normalized eigenfunctions of $-\Delta$ are denoted $w_j$, and its eigenvalues counted with their multiplicities are denoted $\lambda _j$:

$$\begin{aligned} -\Delta w_j = \lambda _j w_j. \end{aligned}$$

(22)

It is well known that $0<\lambda _1\le \cdots \le \lambda _j\rightarrow \infty$ and that $-\Delta$ is a positive selfadjoint operator in $L^2(\Omega )$ with domain ${\mathcal {D}}\left( -\Delta \right) = H^2(\Omega )\cap H_0^1(\Omega )$. The ground state $w_1$ is positive and

$$\begin{aligned} c_0d(x) \le w_1(x)\le C_0d(x) \end{aligned}$$

(23)

holds for all $x\in \Omega$, where $c_0, \, C_0$ are positive constants depending on $\Omega$. Functional calculus can be defined using the eigenfunction expansion. In particular

$$\begin{aligned} \left( -\Delta \right) ^{\beta }f = \sum _{j=1}^{\infty }\lambda _j^{\beta } f_j w_j \end{aligned}$$

(24)

with

$$\begin{aligned} f_j =\int _{\Omega }f(y)w_j(y)dy \end{aligned}$$

for $f\in {\mathcal {D}}\left( \left( -\Delta \right) ^{\beta }\right) = \{f\left| \right. \; (\lambda _j^{\beta }f_j)\in \ell ^2(\mathbb N)\}$. We will denote by

$$\begin{aligned} \Lambda _D^s = \left( -\Delta \right) ^{\frac{s}{2}}, \end{aligned}$$

(25)

the fractional powers of the Dirichlet Laplacian, with $0\le s \le 2$ and with $\Vert f\Vert _{s,D}$ the norm in ${\mathcal {D}}\left( \Lambda _D^s\right)$:

$$\begin{aligned} \Vert f\Vert _{s,D}^2 = \sum _{j=1}^{\infty }\lambda _j^{s}f_j^2. \end{aligned}$$

(26)

It is well-known and easy to show that

$$\begin{aligned} {\mathcal {D}}\left( \Lambda _D\right) = H_0^1(\Omega ). \end{aligned}$$

Indeed, for $f\in {\mathcal {D}}\left( -\Delta \right)$ we have

$$\begin{aligned} \Vert \nabla f\Vert ^2_{L^2(\Omega )} = \int _{\Omega }f\left( -\Delta \right) fdx = \Vert \Lambda _Df\Vert _{L^2(\Omega )}^2 = \Vert f\Vert ^2_{1,D}. \end{aligned}$$

(27)

We recall that the Poincaré inequality implies that the Dirichlet integral on the left-hand side above is equivalent to the norm in $H_0^1(\Omega )$ and therefore the identity map from the dense subset ${\mathcal {D}}\left( -\Delta \right)$ of $H_0^1(\Omega )$ to ${\mathcal D}\left( \Lambda _D\right)$ is an isometry, and thus $H_0^1(\Omega )\subset {\mathcal {D}}\left( \Lambda _D\right)$. But ${\mathcal {D}}\left( -\Delta \right)$ is dense in ${\mathcal D}\left( \Lambda _D\right)$ as well, because finite linear combinations of eigenfunctions are dense in ${\mathcal D}\left( \Lambda _D\right)$. Thus the opposite inclusion is also true, by the same isometry argument. Note that in view of the identity

$$\begin{aligned} \lambda ^{\frac{s}{2}} = c_{s}\int _0^{\infty }(1-e^{-t\lambda })t^{-1-\frac{s}{2}}dt, \end{aligned}$$

(28)

with

$$\begin{aligned} 1 = c_{s} \int _0^{\infty }(1-e^{-\tau })\tau ^{-1-\frac{s}{2}}d\tau , \end{aligned}$$

valid for $0\le s <2$, we have the representation

$$\begin{aligned} \left( \left( \Lambda _D\right) ^{s}f\right) (x) = c_{s}\int _0^{\infty }\left[ f(x)-e^{t\Delta }f(x)\right] t^{-1-\frac{s}{2}}dt \end{aligned}$$

(29)

for $f\in {\mathcal {D}}\left( \left( -\Lambda _D\right) ^{s}\right)$. We use precise upper and lower bounds for the kernel $H_D(t,x,y)$ of the heat operator,

$$\begin{aligned} (e^{t\Delta }f)(x) = \int _{\Omega }H_D(t,x,y)f(y)dy. \end{aligned}$$

(30)

These are as follows [16, 24, 25]. There exists a time $T>0$ depending on the domain $\Omega$ and constants c, C, k, K, depending on T and $\Omega$ such that

$$\begin{aligned}&c\min \left( \frac{w_1(x)}{|x-y|}, 1\right) \min \left( \frac{w_1(y)}{|x-y|}, 1\right) t^{-\frac{d}{2}}e^{-\frac{|x-y|^2}{kt}}\nonumber \\&\le H_D(t,x,y)\le C \min \left( \frac{w_1(x)}{|x-y|}, 1\right) \min \left( \frac{w_1(y)}{|x-y|}, 1\right) t^{-\frac{d}{2}}e^{-\frac{|x-y|^2}{Kt}} \end{aligned}$$

(31)

holds for all $0\le t\le T$. Moreover

$$\begin{aligned} \frac{\left| \nabla _x H_D(t,x,y)\right| }{H_D(t,x,y)}\le C\left\{ \begin{array}{ll} \frac{1}{d(x)}, &{} \quad {\hbox {if}}\; \sqrt{t}\ge d(x),\\ \frac{1}{\sqrt{t}}\left( 1 + \frac{|x-y|}{\sqrt{t}}\right) ,&{} \quad {\hbox {if}}\; \sqrt{t}\le d(x) \end{array} \right. \end{aligned}$$

(32)

holds for all $0\le t\le T$. Note that, in view of

$$\begin{aligned} H_D(t,x,y) = \sum _{j=1}^{\infty }e^{-t\lambda _j}w_j(x)w_j(y), \end{aligned}$$

(33)

elliptic regularity estimates and Sobolev embedding which imply uniform absolute convergence of the series (if $\partial \Omega$ is smooth enough), we have that

$$\begin{aligned} \partial _1^{\beta }H_D(t,y,x) = \partial _2^{\beta }H_D(t,x,y) = \sum _{j=1}^{\infty }e^{-t\lambda _j}\partial _y^{\beta }w_j(y)w_j(x) \end{aligned}$$

(34)

for positive t, where we denoted by $\partial _1^{\beta }$ and $\partial _2^{\beta }$ derivatives with respect to the first spatial variables and the second spatial variables, respectively.

Therefore, the gradient bounds (32) result in

$$\begin{aligned} \frac{\left| \nabla _y H_D(t,x,y)\right| }{H_D(t,x,y)}\le C\left\{ \begin{array}{ll} \frac{1}{d(y)}, &{} \quad {\hbox {if}}\; \sqrt{t}\ge d(y),\\ \frac{1}{\sqrt{t}}\left( 1 + \frac{|x-y|}{\sqrt{t}}\right) ,&{} \quad {\hbox {if}}\;\sqrt{t}\le d(y). \end{array} \right. \end{aligned}$$

(35)

We also use a bound

$$\begin{aligned} \nabla _x\nabla _x H_D(x,y,t) \le Ct^{-1-\frac{d}{2}}e^{-\frac{|x-y|^2}{\tilde{K}t}} \end{aligned}$$

(36)

valid for $t\le cd(x)^2$ and $0<t\le T$, which follows from the upper bounds (31), (32).

Important additional bounds we need are

$$\begin{aligned} \int _{\Omega }\left| (\nabla _x +\nabla _y)H_D(x,y,t)\right| dy \le Ct^{-\frac{1}{2}}e^{-\frac{d(x)^2}{\tilde{K}t}} \end{aligned}$$

(37)

and

$$\begin{aligned} \int _{\Omega }\left| \nabla _x(\nabla _x +\nabla _y)H_D(x,y,t)\right| dy \le Ct^{-1}e^{-\frac{d(x)^2}{\tilde{K}t}} \end{aligned}$$

(38)

valid for $t\le cd(x)^2$ and $0<t\le T$. These bounds reflect the fact that translation invariance is remembered in the solution of the heat equation with Dirichlet boundary data for short time, away from the boundary. We sketch the proofs of (36), (37) and (38) in Appendix 1.

Nonlinear Lower Bounds

We prove bounds in the spirit of [7]. The proofs below are based on the method of [6], but they concern different objects (finite differences, properly localized) or different assumptions ($C^{\alpha }$). Nonlinear lower bounds are an essential ingredient in proofs of global regularity for drift-diffusion equations with nonlocal dissipation.

We start with a couple lemmas. In what follows we denote by c and C generic positive constants that depend on $\Omega$. When the logic demands it, we temporarily manipulate them and number them to show that the arguments are not circular. There is no attempt to optimize constants, and their numbering is local in the proof, meaning that, if for instance $C_2$ appears in two proofs, it need not be the same constant. However, when emphasis is necessary we single out constants, but then we avoid the letters c, C with or without subscripts.

Lemma 1

The solution of the heat equation with initial datum equal to 1 and zero boundary conditions,

$$\begin{aligned} \Theta (x,t) = \int _{\Omega }H_D(x,y,t)dy \end{aligned}$$

(39)

obeys $0\le \Theta (x,t)\le 1$, because of the maximum principle. There exist constants T, c, C depending only on $\Omega$ such that the following inequalities hold:

$$\begin{aligned} \Theta (x,t)\ge c\min \left\{ 1, \left( \frac{d(x)}{\sqrt{t}}\right) ^d\right\} \end{aligned}$$

(40)

for all $0\le t\le T$, and

$$\begin{aligned} \Theta (x,t) \le C \frac{d(x)}{\sqrt{t}} \end{aligned}$$

(41)

for all $0\le t\le T$. Let $0<s<2$. There exists a constant c depending on $\Omega$ and s such that

$$\begin{aligned} \int _0^{\infty }t^{-1-\frac{s}{2}}(1-\Theta (x,t))dt \ge c d(x)^{-s} \end{aligned}$$

(42)

holds.

Remark 2

$\Lambda _D^{s} 1$ is defined by duality by the left hand side of (42) and belongs to $H^{-1}(\Omega )$.

Proof

Indeed,

$$\begin{aligned} \Theta (x,t) = \int _{\Omega }H_D(t,x,y)dy\ge \int _{|x-y|\le \frac{d(x)}{2}} H_D(t,x,y)dy \end{aligned}$$

because $H_D$ is positive. Using the lower bound in (23) we have that $|x-y|\le \frac{d(x)}{2}$ implies

$$\begin{aligned} \frac{w_1(x)}{|x-y|}\ge 2c_0,\quad \frac{w_1(y)}{|x-y|}\ge c_0, \end{aligned}$$

and then, using the lower bound in (31) we obtain

$$\begin{aligned} H_D(t,x,y) \ge 2cc_0^2t^{-\frac{d}{2}}e^{-\frac{|x-y|^2}{kt}}. \end{aligned}$$

Integrating it follows that

$$\begin{aligned} \Theta (x,t) \ge 2 cc_0^2\omega _{d-1}k^{\frac{d}{2}}\int _0^{\frac{d(x)}{2\sqrt{kt}}}\rho ^{d-1}e^{-\rho ^2}d\rho . \end{aligned}$$

If $\frac{d(x)}{2\sqrt{kt}}\ge 1$ then the integral is bounded below by $\int _0^1\rho ^{d-1}e^{-\rho ^2}d\rho$. If $\frac{d(x)}{2\sqrt{kt}}\le 1$ then $\rho \le 1$ implies that the exponential is bounded below by $e^{-1}$ and so (40) holds. $\square$

Now (41) holds immediately from (23) and the upper bound in (31) because the integral

$$\begin{aligned} \int _{{\mathbb R}^d}|\xi |^{-1}e^{-\frac{|\xi |^2}{K}}d\xi <\infty \end{aligned}$$

if $d\ge 2$.

Regarding (42) we use

$$\begin{aligned} \int _0^{\infty }t^{-1-\frac{s}{2}}(1-\Theta (x,t))dt\ge \int _\tau ^{T}t^{-1-\frac{s}{2}}(1-\Theta (x,t))dt \end{aligned}$$

and choose appropriately $\tau$. In view of (41), if

$$\begin{aligned} \frac{d(x)}{\sqrt{\tau }}\le \frac{1}{2C} \end{aligned}$$

then, when $\tau \le t\le T$ we have

$$\begin{aligned} 1-\Theta (x,t)\ge \frac{1}{2}, \end{aligned}$$

and therefore

$$\begin{aligned} \int _{\tau }^{T} t^{-1-\frac{s}{2}}\left( 1-\Theta (x,t)\right) dt \ge \frac{1}{s} \tau ^{-\frac{s}{2}}\left( 1- \left( \frac{\tau }{T}\right) ^{\frac{s}{2}}\right) \end{aligned}$$

holds. The choice

$$\begin{aligned} \frac{d(x)}{\sqrt{\tau }} =\frac{1}{2C} \end{aligned}$$

implies (42) provided $2\tau \le T$ which is the same as $d(x)\le \frac{\sqrt{T}}{2C\sqrt{2}}$. On the other hand, $\Theta$ is exponentially small if t is large enough, so the contribution to the integral in (42) is bounded below by a nonzero constant. This ends the proof of the lemma.

Lemma 2

Let $0\le \alpha <1$. There exists constant C depending on $\Omega$ and $\alpha$ such that

$$\begin{aligned} \int _{\Omega }|\nabla _y H_D(t,x,y)| |x-y|^{\alpha }dy \le C t^{-\frac{1-\alpha }{2}} \end{aligned}$$

(43)

holds for $0\le t\le T$.

Indeed, the upper bounds (31) and (35) yield

$$\begin{aligned} \begin{array}{l} \int \nolimits _{d(y)\ge \sqrt{t}}|\nabla _y H_D(t,x,y)||x-y|^{\alpha }dy \\ \le C_2t^{-\frac{1}{2}} \int _{{\mathbb R}^d}\left( 1 + \frac{|x-y|}{\sqrt{t}}\right) t^{-\frac{d}{2}}e^{-\frac{|x-y|^2}{Kt}}|x-y|^{\alpha }dy\\ = C_3t^{-\frac{1-\alpha }{2}} \end{array} \end{aligned}$$

and, in view of the upper bound in (23), $\frac{1}{d(y)}w_1(y)\le C_0$ and the upper bound in (31), we have

$$\begin{aligned} \begin{array}{l} \int _{d(y)\le \sqrt{t}}|\nabla _y H_D(t,x,y)||x-y|^{\alpha }dy \\ \le C_4\int _{{\mathbb R}^d}\frac{1}{|x-y|}t^{-\frac{d}{2}}e^{-\frac{|x-y|^2}{Kt}}|x-y|^{\alpha }dy = C_5t^{-\frac{1-\alpha }{2}}. \end{array} \end{aligned}$$

This proves (43). We introduce now a good family of cutoff functions $\chi$ depending on a length scale $\ell$.

Lemma 3

Let $\Omega$ be a bounded domain with $C^2$ boundary. For $\ell >0$ small enough (depending on $\Omega$) there exist cutoff functions $\chi$ with the properties: $0\le \chi \le 1$, $\chi (y)=0$ if $d(y)\le \frac{\ell }{4}$, $\chi (y)= 1$ for $d(y)\ge \frac{\ell }{2}$, $|\nabla ^k\chi |\le C\ell ^{-k}$ with C independent of $\ell$ and

$$\begin{aligned} \int _{\Omega }\frac{(1-\chi (y))}{|x-y|^{d+j}}dy \le C\frac{1}{d(x)^{j}} \end{aligned}$$

(44)

and

$$\begin{aligned} \int _{\Omega }|\nabla \chi (y)|\frac{1}{|x-y|^{d-\alpha }}\le Cd(x)^{-(1-\alpha )} \end{aligned}$$

(45)

hold for $j>-d$, $\alpha <d$ and $d(x)\ge \ell$. We will refer to such $\chi$ as a “good cutoff”.

Proof

There exists a length $\ell _0$ such that if P is a point of the boundary $\partial \Omega$, and if $|P-y|\le 2\ell _0$, then $y\in \Omega$ if and only if (after a rotation and a translation) $y_d>F(y')$, where $y'=(y_1, \ldots , y_{d-1})$ and F is a $C^2$ function with $F(0)=0$, $\nabla F(0)=0$, $|\nabla F|\le \frac{1}{10}$. We took thus without loss of generality coordinates such that $P= (0,0)$ and the normal to $\partial \Omega$ at P is $(0,\ldots , 0, 1)$. Now if $\ell <\ell _0$ and $d(x)\ge \ell$ and $|y-P|\le \frac{\ell _0}{2}$ satisfies $d(y)\le \frac{\ell }{2}$, then there exists a point $Q\in B(P,\ell _0)$ such that

$$\begin{aligned} |x-y|^2\ge \frac{1}{16}(|y-Q|^2 + d(x)^2)\ge \frac{1}{16}(|y'-Q'|^2 + d(x)^2) \end{aligned}$$

Indeed, if $|x-P|\ge \ell _0$ we take $Q=P$ because then $|x-y| =|x-P+P-y|\ge \ell _0-\frac{\ell _0}{2}$, so $|x-y|\ge \frac{|y-Q|}{2}$. But also $|x-y|\ge \frac{d(x)}{2}$ because there exists a point $P_1 =(p, F(p))\in \partial \Omega$ such that $|y-P_1| = d(y)\le \frac{\ell }{2}$ while obviously $|x-P_1|\ge d(x)\ge \ell$. If, on the other hand $|x-P|< \ell _0$, then x is in the neighborhood of P and we take $Q=x$. Because $y-P_1= (y'-p, y_d-F(p))$ we have

$$\begin{aligned} d(y)\le |y_d-F(y')| \le \frac{11}{10}d(y) \end{aligned}$$

for $y\in B(P,\ell _0)$. We take a partition of unity of the form $1= \psi _0 +\sum _{j=1}^N\psi _j$ with $\psi _k\in C_0^{\infty }({\mathbb R}^d)$, subordinated to the cover of the boundary with neighborhoods as above, and with $\psi _0$ supported in $d(x)\ge \frac{\ell _0}{4}$, identically 1 for $d(x)\ge \frac{\ell _0}{2}$, $\psi _j$ supported near the boundary $\partial \Omega$ in balls of size $2\ell _0$ and identically 1 on balls of radius $\ell _0$. $\square$

The cutoff will be taken of the form $\chi = \alpha _0 +\sum _{j=1}^N \chi _j(\frac{y_d-F(y')}{\ell })\alpha _j(y)$, where of course the meaning of y changes in each neighborhood. The smooth functions $\chi _j(z)$, are identically zero for $|z|\le \frac{11}{40}$ and identically 1 for $|z| \ge \frac{10}{22}$. The integrals in (44) and (45) reduce to integrals of the type

$$\begin{aligned} \begin{array}{l} \int _{y_d>F(y'), |y'|\le \ell _0}\frac{\left( 1-\chi _1\left( \frac{y_d-F(y')}{\ell }\right) \right) }{|x-y|^{d+j}}dy\\ \le C\left( \int _{0}^{\infty }\left( 1-\chi _1\left( \frac{u}{\ell }\right) \right) du\right) \left( \int _{{\mathbb R}^{d-1}}\frac{dy'}{\left( {|y'-Q'|^2 +d(x)^2}\right) ^{\frac{d+j}{2}}}\right) \\ \le C\ell d(x)^{-1-j}\le Cd(x)^{-j} \end{array} \end{aligned}$$

and

$$\begin{aligned} \begin{array}{l} \int _{y_d>F(y'), |y'|\le \ell _0}\frac{\left| \nabla _y\chi _1\left( \frac{y_d-F(y')}{\ell }\right) \right| }{|x-y|^{d-\alpha }}dy\\ \le C\left( \int _{-{\infty }}^{\infty }|\nabla \chi _1(z)|dz\right) \left( \int _{{\mathbb R}^{d-1}}\frac{dy'}{\left( {|y'-Q'|^2 +d(x)^2}\right) ^{\frac{d-\alpha }{2}}}\right) \\ \le C d(x)^{-(1-\alpha )}. \end{array} \end{aligned}$$

This completes the proof.

We recall from [6] that the Córdoba-Córdoba inequality [14] holds in bounded domains. In fact, more is true: there is a lower bound that provides a strong boundary repulsive term:

Proposition 1

Let $\Omega$ be a bounded domain with smooth boundary. Let $0\le s<2$. There exists a constant $c>0$ depending only on the domain $\Omega$ and on s, such that, for any $\Phi$, a $C^2$ convex function satisfying $\Phi (0)= 0$, and any $f\in C_0^{\infty }(\Omega )$, the inequality

$$\begin{aligned} \Phi '(f)\Lambda _D^s f - \Lambda _D^s(\Phi (f))\ge \frac{c}{d(x)^s}\left( f\Phi '(f)-\Phi (f)\right) \end{aligned}$$

(46)

holds pointwise in $\Omega$.

The proof follows in a straightforward manner from the proof of [6] using convexity, approximation, and the lower bound (42). We prove below two nonlinear lower bounds for the case $\Phi (f)= \frac{f^2}{2}$, one when f is a localized finite difference, and one when f is a localized first derivative. The proof of Proposition 1 can be left as an exercise, following the same pattern as below.

Theorem 3

Let $f\in L^{\infty }(\Omega )$ be smooth enough ($C^2$, e.g.) and vanish at the boundary, $f\in {\mathcal {D}}(\Lambda _D^{s})$ with $0\le s<2$. Then

$$\begin{aligned} \begin{array}{l} D(f) = f\Lambda _D^{s} f -\frac{1}{2}\Lambda _D^{s} f^2\\ =\gamma _0\smallint _0^{\infty }t^{-1-\frac{s}{2}}dt\smallint _{\Omega }H_D(x,y,t)(f(x)-f(y))^2dy\\ \quad +\, \gamma _0 f^2(x)\smallint _0^{\infty }t^{-1-\frac{s}{2}}\left[ 1-e^{t\Delta }1\right] (x)dt\\ = \gamma _0\smallint _0^{\infty }t^{-1-\frac{s}{2}}dt\smallint _{\Omega }H_D(x,y,t)(f(x)-f(y))^2dy\\ \quad + \, f^2(x)\frac{1}{2}\Lambda _D^{s} 1. \end{array} \end{aligned}$$

(47)

holds for all $x\in \Omega$. Here $\gamma _0 =\frac{c_{s}}{2}$ with $c_s$ of (29). Let $\ell >0$ be a small number and let $\chi \in C_0^{\infty }(\Omega )$, $0\le \chi \le 1$ be a good cutoff function, with $\chi (y)=1$ for $d(y)\ge \frac{\ell }{2}$, $\chi (y) =0$ for $d(y)\le \frac{\ell }{4}$ and with $|\nabla \chi (y)|\le \frac{C}{\ell }$. There exist constants $\gamma _1>0$ and $M>0$ depending on $\Omega$ such that, if q(x) is a smooth function in $L^{\infty }(\Omega )$ then if

$$\begin{aligned} f(x)=\chi (x)(\delta _h q(x)) =\chi (x)(q(x+h)-q(x)) \end{aligned}$$

then

$$\begin{aligned} D(f) = (f\Lambda _D^{s} f)(x) -\frac{1}{2}(\Lambda _D^{s} f^2)(x) \ge \gamma _1 |h|^{-s}\frac{|f_d(x)|^{2+s}}{\Vert q\Vert _{L^{\infty }}^{s}} + \gamma _1\frac{f^2(x)}{d(x)^{s}} \end{aligned}$$

(48)

holds pointwise in $\Omega$ when $|h|\le \frac{\ell }{16}$, and $d(x)\ge \ell$ with

$$\begin{aligned} |f_d(x)| = \left\{ \begin{array}{ll} |f(x)|, &{} \quad {\hbox {if}}\;\; |f(x)| \ge M\Vert q\Vert _{L^{\infty }(\Omega )}\frac{|h|}{d(x)},\\ 0, &{} \quad {\hbox {if}}\;\; |f(x)| \le M\Vert q\Vert _{L^{\infty }(\Omega )} \frac{|h|}{d(x)}. \end{array} \right. \end{aligned}$$

(49)

Proof

We start by proving (47):

$$\begin{aligned} \begin{array}{l} f(x)\Lambda _D^{s}f(x) - \frac{1}{2}\Lambda _D^{s} f^2(x) \\ \quad = c_{s}\int _0^{\infty }t^{-1-\frac{s}{2}}\int _{\Omega }\left\{ \left[ \frac{1}{|\Omega |} f(x)^2 - f(x)H_D(t,x,y)f(y)\right] \right. \\ \qquad \left. - \frac{1}{2|\Omega |}f^{2}(x) + \frac{1}{2}H_D(t,x,y)f^2(y)\right\} dy\\ \quad =c_{s}\int _0^{\infty }t^{-1-\frac{s}{2}}dt\int _{\Omega }\left\{ \frac{1}{2}\left[ H_D(t,x,y)(f(x)-f(y))^2\right] \right. \\ \qquad \left. + \frac{1}{2}f^2(x)\left[ \frac{1}{|\Omega |} -H_D(t,x,y)\right] \right\} dy \\ \quad = c_{s}\int _0^{\infty }t^{-1-\frac{s}{2}}dt\int _{\Omega }\left\{ \frac{1}{2}\left[ H_D(t,x,y)(f(x){-}f(y))^2\right] dy {+} \frac{1}{2}f^2(x)\left[ 1-e^{t\Delta }1\right] (x)\right\} \\ \quad = c_{s}\int _0^{\infty }t^{-1-\frac{s}{2}}dt\int _{\Omega }\frac{1}{2}\left[ H_D(t,x,y)(f(x)-f(y))^2\right] dy + \frac{1}{2}f^2(x)\Lambda _D^{s}1. \end{array} \end{aligned}$$

It follows that

$$\begin{aligned} \left( f\Lambda _D^{s}f - \frac{1}{2}\Lambda _D^{s} f^2\right) (x)\ge & {} \frac{1}{2}c_{s}\smallint _0^{\infty }\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint _{\Omega }H_D(t,x,y)(f(x)\nonumber \\&-f(y))^2dy + \frac{1}{2}f^2(x)\Lambda _D^{s}1 \end{aligned}$$

(50)

where $\tau >0$ is arbitrary and $0\le \psi (s)\le 1$ is a smooth function, vanishing identically for $0\le s\le 1$ and equal identically to 1 for $s\ge 2$. We restrict to $t\le T$,

$$\begin{aligned} \begin{array}{l} \left( f\Lambda _D^{s}f - \frac{1}{2}\Lambda _D^{s} f^2\right) (x)\\ \ge \frac{1}{2}c_{s}\smallint _0^{T}\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint _{\Omega }H_D(t,x,y)\left( f(x)-f(y)\right) ^2dy +\frac{1}{2}f^2(x)\Lambda _D^{s}1 \end{array} \end{aligned}$$

(51)

$\square$

and open brackets in (51):

$$\begin{aligned} \begin{array}{l} \left( f\Lambda _D^{s}f - \frac{1}{2}\Lambda _D^{s} f^2\right) (x) \ge \frac{1}{2}f^2(x)c_{s}\int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\int _{\Omega }H_D(t,x,y)dy\\ \qquad - f(x)c_{s}\int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\int _{\Omega }H_D(t,x,y)f(y)dy + \frac{1}{2}f^2(x)\Lambda _D^{s}1\\ \quad \ge |f(x)|\left[ \frac{1}{2}|f(x)| I(x) - J(x)\right] +\frac{1}{2}f^2(x)\Lambda _D^{s}1 \end{array} \end{aligned}$$

(52)

with

$$\begin{aligned} I(x) = c_{s}\int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\int _{\Omega }H_D(t,x,y)dy, \end{aligned}$$

(53)

and

$$\begin{aligned} J(x)= & {} c_{s}\left| \smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }H_D(t,x,y)f(y)dy\right| \nonumber \\= & {} c_{s}\left| \smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }H_D(t,x,y)\chi (y)\delta _hq(y)dy\right| . \end{aligned}$$

(54)

We proceed with a lower bound on I and an upper bound on J. For the lower bound on I we note that, in view of (40) and the fact that

$$\begin{aligned} I(x) = c_{s}\int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}\Theta (x,t)dt \end{aligned}$$

we have

$$\begin{aligned} I(x)\ge & {} c_1\int _0^{\min (T, d^2(x))}\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\\= & {} c_1\tau ^{-\frac{s}{2}}\int _1^{\tau ^{-1}(\min (T, d^2(x)))}\psi (u)u^{-1-\frac{s}{2}}du. \end{aligned}$$

Therefore we have that

$$\begin{aligned} I(x)\ge c_2 \tau ^{-\frac{s}{2}} \end{aligned}$$

(55)

with $c_2 = c_1\int _1^2\psi (u)u^{-1-\frac{s}{2}}du$, a positive constant depending only on $\Omega$ and s, provided $\tau$ is small enough,

$$\begin{aligned} \tau \le \frac{1}{2}\min (T, d^2(x)). \end{aligned}$$

(56)

In order to bound J from above we use (43) with $\alpha =0$. Now

$$\begin{aligned} J\le & {} c_{s}\left| \smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }\delta _{-h}H_D(t,x,y)\chi (y)q(y)dy\right| \\&+\,c_{s}\left| \smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }H_D(t,x,y-h)(\delta _{-h}\chi (y))q(y)dy\right| \end{aligned}$$

We have that

$$\begin{aligned} J_2= & {} c_{s}\left| \int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\int _{\Omega }H_D(t,x,y-h)(\delta _{-h}\chi (y))q(y)dy\right| \\\le & {} C_{6}\frac{|h|}{d(x)}\Vert q\Vert _{L^{\infty }}\tau ^{-\frac{s}{2}}. \end{aligned}$$

Indeed,

$$\begin{aligned} t^{-{\frac{d}{2}}}e^{-\frac{|x-y|^2}{Kt}}\le C_K |x-y|^{-d} \end{aligned}$$

so the bound follows from (31) and (45). On the other hand,

$$\begin{aligned} J_1= & {} c_{s}\left| \smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }\delta _{-h}H_D(t,x,y)\chi (y)q(y)dy\right| \\\le & {} \Vert q\Vert _{L^{\infty }(\Omega )}|h|\smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }|\nabla _y H_D(t,x,y)|dy \end{aligned}$$

and therefore, in view of (43)

$$\begin{aligned} J_1\le C_1 |h|\Vert q\Vert _{L^{\infty }(\Omega )}\int _0^T\psi \left( \frac{t}{\tau }\right) t^{-\frac{3}{2}-\frac{s}{2}}dt \end{aligned}$$

and therefore

$$\begin{aligned} J_1 \le C_7|h|\Vert q\Vert _{L^{\infty }(\Omega )}\tau ^{-\frac{1}{2}-\frac{s}{2}} \end{aligned}$$

(57)

with

$$\begin{aligned} C_7 = C_1\int _1^{\infty }\psi (u)u^{-\frac{3}{2}-\frac{s}{2}}du \end{aligned}$$

a constant depending only on $\Omega$ and s. In conclusion

$$\begin{aligned} |J| \le C_8\tau ^{-\frac{s}{2}}|h| (\tau ^{-\frac{1}{2}} + d(x)^{-1})\Vert q\Vert _{L^{\infty }}. \end{aligned}$$

(58)

Now, because of the lower bound (52), if we can choose $\tau$ so that

$$\begin{aligned} J(x) \le \frac{1}{4} |f(x)|I(x) \end{aligned}$$

then it follows that

$$\begin{aligned} \left[ f\Lambda _D^{s}f - \frac{1}{2}\Lambda _D^{s} f^2\right] (x) \ge \frac{1}{4}f^2(x)I(x) + \frac{1}{2}f^2(x)\Lambda _D^{s}1. \end{aligned}$$

(59)

Because of the bounds (55), (58), if

$$\begin{aligned} |f(x)|\ge \frac{8C_8}{c_2}\frac{|h|}{d(x)}\Vert q\Vert _{L^{\infty }}, \end{aligned}$$

then a choice

$$\begin{aligned} \tau (x)^{-\frac{1}{2}} = C_9{\Vert q\Vert _{L^{\infty }}^{-1}}|f(x)||h|^{-1} \end{aligned}$$

(60)

with $C_9 = c_2 (8C_8)^{-1}$ achieves the desired bound. This concludes the proof.

We are providing now a lower bound for D(f) for a different situation.

Theorem 4

Let $\ell >0$ be a small number and let $\chi \in C_0^{\infty }(\Omega )$, $0\le \chi \le 1$ be a good cutoff function, with $\chi (y)=1$ for $d(y)\ge \frac{\ell }{2}$, $\chi (y) =0$ for $d(y)\le \frac{\ell }{4}$ and with $|\nabla \chi (y)|\le \frac{C}{\ell }$. There exist constants $\gamma _2>0$ and $M>0$ depending on $\Omega$ such that, if q(x) is a smooth function in $C^{\alpha }(\Omega )$ with $0<\alpha <1$ and

$$\begin{aligned} f(x)=\chi (x)\nabla q(x), \end{aligned}$$

then

$$\begin{aligned} D(f) = (f\Lambda _D^{s} f)(x) -\frac{1}{2}(\Lambda _D^{s} f^2)(x) \ge \gamma _2 \frac{|f_d(x)|^{2+\frac{s}{1-\alpha }}}{\Vert q\Vert _{C^{\alpha }(\Omega )}^{\frac{s}{1-\alpha }}}(d(x))^{\frac{s\alpha }{1-\alpha }} + \gamma _1\frac{f^2(x)}{d(x)^{s}} \end{aligned}$$

(61)

holds pointwise in $\Omega$ when $d(x)\ge \ell$, with

$$\begin{aligned} |f_d(x)| = \left\{ \begin{array}{ll} |f(x)|, &{} \quad {\hbox {if}}\;\; |f(x)| \ge M\Vert q\Vert _{L^{\infty }(\Omega )}(d(x))^{-1},\\ 0, &{} \quad {\hbox {if}}\;\; |f(x)| \le M\Vert q\Vert _{L^{\infty }(\Omega )}(d(x))^{-1}. \end{array} \right. \end{aligned}$$

(62)

Proof

We follow exactly the proof of Theorem 3 up to, and including the definition of I(x) given in (53). In particular, the lower bound (55) is still valid, provided $\tau$ is small enough (56). The term J starts out the same, but is treated slightly differently,

$$\begin{aligned} J(x)= & {} c_{s}\left| \smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }H_D(t,x,y)f(y)dy\right| \nonumber \\= & {} c_{s}\left| \smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }H_D(t,x,y)\chi (y)\nabla _y (q(y)-q(x))dy\right| . \end{aligned}$$

(63)

In order to bound J we use (45) and (43).

$$\begin{aligned} |J(x)|\le & {} c_{s}\left| \smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }\partial _yH_D(t,x,y)\chi (y)(q(y)-q(x))dy\right| \\&+\, c_{s}\left| \smallint \nolimits _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }H_D(t,x,y)(\nabla \chi (y))(q(y)-q(x))dy\right| \\= & {} J_1(x) + J_2(x) \end{aligned}$$

We have from (31) and (45), as before,

$$\begin{aligned} J_2(x)= & {} c_{s}\left| \int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\int _{\Omega }H_D(t,x,y)(\nabla \chi (y))(q(y)-q(x))dy\right| \\\le & {} Cd(x)^{-1}\Vert q\Vert _{L^{\infty }}\tau ^{-\frac{s}{2}}. \end{aligned}$$

On the other hand,

$$\begin{aligned} J_1(x)= & {} c_{s}\left| \int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\int _{\Omega }\nabla _yH_D(t,x,y)\chi (y)(q(y)-q(x))dy\right| \\\le & {} c_s(d(x))^{-\alpha }\Vert q\Vert _{C^{\alpha }(\Omega )}\int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\\&\times \, \int _{\Omega \cap |x-y|\le d(x)}|\nabla _y H_D(t,x,y)||x-y|^{\alpha }dy \\&+\, c_s\Vert q\Vert _{L^{\infty }}\int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}dt\int _{\Omega \cap |x-y|\ge d(x) }|\nabla _y H_D(t,x,y)|dy \\= & {} J_{11}(x) + J_{12}(x). \end{aligned}$$

In view of (43)

$$\begin{aligned} J_{11}(x)\le C_1 d(x)^{-\alpha } \Vert q\Vert _{C^{\alpha }(\Omega )}\int _0^T\psi \left( \frac{t}{\tau }\right) t^{-\frac{3-\alpha }{2}-\frac{s}{2}}dt \end{aligned}$$

and so

$$\begin{aligned} J_{11}(x) \le C_2(d(x))^{-\alpha }\Vert q\Vert _{C^{\alpha }(\Omega )}\tau ^{-\frac{1-\alpha }{2}-\frac{s}{2}} \end{aligned}$$

(64)

with

$$\begin{aligned} C_2 = C_1\int _1^{\infty }\psi (z)z^{-\frac{3-\alpha }{2}-\frac{s}{2}}dz \end{aligned}$$

a constant depending only on $\Omega$ and s. Regarding $J_{12}(x)$ we have in view of (35)

$$\begin{aligned} J_{12}(x)\le & {} C\Vert q\Vert _{L^{\infty }(\Omega )}\int _0^T\psi \left( \frac{t}{\tau }\right) t^{-1-\frac{s}{2}}\left( \frac{1}{\sqrt{t}} + \frac{1}{d(x)}\right) e^{-\frac{d(x)^2}{2Kt}}dt\\\le & {} C_3\tau ^{-\frac{s}{2}}d(x)^{-1}\Vert q\Vert _{L^{\infty }(\Omega )} \end{aligned}$$

because, in view of (23)

$$\begin{aligned} \frac{w_1(y)}{|x-y|}\le C_0\frac{d(y)}{|x-y|} \le C_0\frac{d(y)}{d(x)} \end{aligned}$$

on the domain of integration. $\square$

In conclusion

$$\begin{aligned} |J(x)| \le C_3\tau ^{-\frac{s}{2}}\left[ \tau ^{-\frac{1-\alpha }{2}}(d(x))^{-{\alpha }}\Vert q\Vert _{C^{\alpha }(\Omega )} + d(x)^{-1}\Vert q\Vert _{L^{\infty }(\Omega )}\right] . \end{aligned}$$

(65)

The rest is the same as in the proof of Theorem 3: If $|f(x)|\ge Md(x)^{-1}\Vert q\Vert _{L^{\infty }(\Omega )}$ for suitable M, ($M= 8C_3c_2^{-1}$) then we choose $\tau$ such that

$$\begin{aligned} \frac{|f(x)|}{\Vert q\Vert _{C^{\alpha }(\Omega )}} = M\tau ^{-\frac{1-\alpha }{2}}(d(x))^{-\alpha }, \end{aligned}$$

and this yields $|f(x)| I\ge 4 |J(x)|$, and consequently, in view of (59) which is then valid, the result (61) is proved.

We specialize from now on to $s=1$.

Bounds for Riesz Transforms

We consider u given in (8),

$$\begin{aligned} u = \nabla ^{\perp }\Lambda _D^{-1}\theta . \end{aligned}$$

We are interested in estimates of u in terms of $\theta$, and in particular estimates of finite differences and the gradient. We fix a length scale $\ell$ and take a good cutoff function $\chi \in C_0^{\infty }(\Omega )$ which satisfies $\chi (x) =1$ if $d(x)\ge \frac{\ell }{2}$, $\chi (x) = 0$ if $d(x)\le \frac{\ell }{4}$, $|\nabla \chi (x)|\le C\ell ^{-1}$, (44) and (45). We take $|h|\le \frac{\ell }{14}$. In view of the representation

$$\begin{aligned} \Lambda _D^{-1} = c\int _0^{\infty }t^{-\frac{1}{2}}e^{t\Delta }dt \end{aligned}$$

(66)

we have on the support of $\chi$

$$\begin{aligned} \delta _h u(x) = c\int _0^{\infty }t^{-\frac{1}{2}}dt\int _{\Omega }\delta _h^x\nabla _x^{\perp }H_D(x,y,t)\theta (y)dy. \end{aligned}$$

(67)

We split

$$\begin{aligned} \delta _h u = \delta _h u^{in} + \delta _h u^{out} \end{aligned}$$

(68)

with

$$\begin{aligned} \delta _h u(x)^{in} = c\int _0^{\rho ^2}t^{-\frac{1}{2}}dt\int _{\Omega }\delta _h^x\nabla _x^{\perp }H_D(x,y,t)\theta (y)dy \end{aligned}$$

(69)

and $\rho =\rho (x,h)>0$ a length scale to be chosen later but it will be smaller than the distance from x to the boundary of $\Omega$:

$$\begin{aligned} \rho \le c d(x). \end{aligned}$$

(70)

We represent

$$\begin{aligned} \delta _h u^{in}(x) = u_h(x) + v_h(x) \end{aligned}$$

(71)

where

$$\begin{aligned} u_h(x) = c\int _0^{\rho ^2}t^{-\frac{1}{2}}dt\int _{\Omega }\nabla _x^{\perp }H(x,y,t)(\chi (y)\delta _h\theta (y)-\chi (x)\delta _h\theta (x))dy \end{aligned}$$

(72)

and where

$$\begin{aligned} v_h(x) = e_1(x) + e_2(x) + e_3(x) + \chi (x)\delta _h\theta (x)e_4(x) \end{aligned}$$

(73)

with

$$\begin{aligned} e_1(x)= & {} c\int _0^{\rho ^2}t^{-\frac{1}{2}}dt\int _{\Omega }\nabla _x^{\perp }(H_D(x+h,y,t)-H_D(x,y,t))(1-\chi (y))\theta (y)dy, \nonumber \\ \end{aligned}$$

(74)

$$\begin{aligned} e_2(x)= & {} c\int _0^{\rho ^2}t^{-\frac{1}{2}}dt\int _{\Omega }\nabla _x^{\perp }(H_D(x+h,y,t)-H_D(x,y-h,t))\chi (y)\theta (y)dy, \nonumber \\\end{aligned}$$

(75)

$$\begin{aligned} e_3(x)= & {} c\int _0^{\rho ^2}t^{-\frac{1}{2}}dt\int _{\Omega }\nabla _x^{\perp }H_D(x,y,t)(\chi (y+h)-\chi (y))\theta (y+h)dy, \end{aligned}$$

(76)

and

$$\begin{aligned} e_4(x) =c\int _0^{\rho ^2}t^{-\frac{1}{2}}dt\int _{\Omega }\nabla _x^{\perp }H_D(x,y,t)dy. \end{aligned}$$

(77)

We used here the fact that $(\chi \theta )(\cdot )$ and $(\chi \theta )(\cdot + h)$ are compactly supported in $\Omega$ and hence

$$\begin{aligned} \int _{\Omega }\nabla _x^{\perp }H_D(x,y-h,t)\chi (y)\theta (y)dy = \int _{\Omega }\nabla _x^{\perp }H_D(x,y,t)\chi (y+h)\theta (y+h)dy. \end{aligned}$$

The following elementary lemma will be used in several instances:

Lemma 4

Let $\rho >0$, $p>0$. Then

$$\begin{aligned} \int _0^{\rho ^2} t^{-1-\frac{m}{2}}\left( \frac{p}{\sqrt{t}}\right) ^je^{-\frac{p^2}{Kt}}dt \le C_{K,m,j}p^{-m} \end{aligned}$$

(78)

if $m\ge 0$, $j\ge 0$, $m+j>0$, and

$$\begin{aligned} \int _0^{\rho ^2} t^{-1}e^{-\frac{p^2}{Kt}}dt \le C_K\left( 1+ 2\log _{+}\left( \frac{\sqrt{K}\rho }{p}\right) \right) \end{aligned}$$

(79)

if $m=0$ and $j=0$, with constants $C_{K,m,j}$ and $C_K$ independent of $\rho$ and p. Note that when $m+j>0$, $\rho =\infty$ is allowed.

We start estimating the terms in (73). For $e_1$ we use the inequality (36), and it then follows from Lemma 4 with $m= {d} +1$ that

$$\begin{aligned} |e_1(x)|\le C|h|\Vert \theta \Vert _{L^{\infty }}\int _0^1d\lambda \int _{\Omega }\frac{1}{|x +\lambda h -y |^{d+1}}(1-\chi (y))dy \end{aligned}$$

and therefore we have from (44) that

$$\begin{aligned} |e_1(x)| \le C \Vert \theta \Vert _{L^{\infty }} \frac{|h|}{d(x)} \end{aligned}$$

(80)

holds for $d(x)\ge \ell$. Concerning $e_3$ we use Lemma (4) with $m= d$ and $j=0,1$ in conjunction with (32) and obtain

$$\begin{aligned} |e_3(x)|\le C{|h|}\Vert \theta \Vert _{L^{\infty }}\int _{\Omega }|\nabla \chi (y)|\frac{1}{|x-y|^d}dy \end{aligned}$$

and therefore we obtain from (45)

$$\begin{aligned} |e_3(x)| \le C \Vert \theta \Vert _{L^{\infty }}\frac{|h|}{d(x)} \end{aligned}$$

(81)

holds for $d(x)\ge \ell$. Regarding $e_4$ we can split it into

$$\begin{aligned} e_4 (x) = e_5(x) + e_6(x) \end{aligned}$$

with

$$\begin{aligned} e_{5}(x) = \int _0^{\rho ^2}t^{-\frac{1}{2}}\int _{\Omega }\nabla _x^{\perp }H_D(x,y,t)\chi (y)dy \end{aligned}$$

and

$$\begin{aligned} e_{6}(x) = \int _0^{\rho ^2}t^{-\frac{1}{2}}\int _{\Omega }\nabla _x^{\perp }H_D(x,y,t)(1-\chi (y))dy. \end{aligned}$$

Now $e_6$ is bounded using the Lemma (4) with $m= d$ and $j=0,1$ in conjunction with (32) and (44) and obtain

$$\begin{aligned} |e_6(x)|\le C\int _{\Omega }\frac{(1-\chi (y))}{|x-y|^d}dy \le C \end{aligned}$$

(82)

for $d(x)\ge \ell$, with a constant independent of $\ell$. For $e_5$ we use the fact that $\chi$ is a fixed smooth function which vanishes at the boundary.

In order to bound the terms $e_2$ and $e_5$ we need to use additional information, namely the inequalities (37) and (38). For $e_5$ we write

$$\begin{aligned} e_5(x)= & {} \smallint \nolimits _0^{\rho ^2}t^{-\frac{1}{2}}dt\smallint \nolimits _{\Omega }\left( \nabla _x^{\perp }H_D(x,y,t) + \nabla _y^{\perp } H_D(x,y,t)\right) \chi (y)dy \\&+\, \smallint \nolimits _0^{\rho ^2}t^{-\frac{1}{2}}dt\smallint \nolimits _{\Omega }H_D(x,y,t)\nabla _y^{\perp }\chi (y)dy, \end{aligned}$$

and using (37) and Lemma 4 with $m=0$, $j=0$ and (45) we obtain the bound

$$\begin{aligned} |e_5(x)|\le C\left( 1+ \log _{+}\left( \frac{\rho }{d(x)}\right) \right) + C\rho \int _{\Omega }\frac{|\nabla \chi (y)|}{|x-y|^d}dy \end{aligned}$$

and therefore, in view of (45) and $\rho \le d(x)$ we have

$$\begin{aligned} |e_5(x)|\le C \end{aligned}$$

(83)

for $d(x)\ge \ell$, with C depending on $\Omega$ but not on $\ell$. Consequently, we have

$$\begin{aligned} |e_4(x)|\le C \end{aligned}$$

(84)

for $d(x)\le \ell$, with a constant C depending on $\Omega$ only. In order to estimate $e_2$ we write

$$\begin{aligned} H_D(x+h,y,t)- H_D(x,y-h,t) = h\cdot \int _0^1 (\nabla _x+\nabla _y)H_D(x+\lambda h, y + (\lambda -1)h,t)d\lambda \end{aligned}$$

(85)

and use (38) and Lemma 4 with $m=1$, $j=0$ to obtain

$$\begin{aligned} |e_2(x)|\le & {} |h|\Vert \theta \Vert _{L^{\infty }}\smallint \nolimits _0^1d\lambda \smallint \nolimits _0^{\rho ^2}t^{-\frac{1}{2}}dt\smallint \nolimits _{\Omega }|\nabla _x^{\perp }(\nabla _x +\nabla _y)H_D\\&\times (x+\lambda h, y+(\lambda -1)h)||\chi (y)|dy\\\le & {} C|h|\Vert \theta \Vert _{L^{\infty }}\smallint \nolimits _0^1d\lambda \smallint \nolimits _0^{\rho ^2}t^{-\frac{3}{2}}e^{-\frac{d(x)^2}{4Kt}}dt \end{aligned}$$

and thus

$$\begin{aligned} |e_2(x)|\le C\Vert \theta \Vert _{L^{\infty }}\frac{|h|}{d(x)} \end{aligned}$$

(86)

holds for $d(x)\ge \ell$. Summarizing, we have that

$$\begin{aligned} |v_h(x)| \le C\Vert \theta \Vert _{L^{\infty }}\frac{|h|}{d(x)} + C|\delta _h\theta (x)| \end{aligned}$$

(87)

for $d(x)\ge \ell$. We now estimate $u_h$ using (32) and a Schwartz inequality

$$\begin{aligned} |u_h(x)|\le & {} c\smallint \nolimits _0^{\rho ^2}t^{-1}\smallint \nolimits _{\Omega }\left( 1+\frac{|x-y|}{\sqrt{t}}\right) H_D(x,y,t)(\chi (\delta _h\theta )(y)-\chi \delta _h\theta (x))dy\\\le & {} \sqrt{\rho }\left\{ \smallint \nolimits _0^{\rho ^2}t^{-\frac{3}{2}}dt\smallint \nolimits _{\Omega }H_D(x,y,t)(\chi (\delta _h\theta )(y)-\chi \delta _h\theta )^2dy\right\} ^{\frac{1}{2}} \end{aligned}$$

We have therefore

$$\begin{aligned} |u_h(x)|\le C\sqrt{\rho D(f)(x)}. \end{aligned}$$

(88)

where $f=\chi \delta _h\theta$ and D(f) is given in Theorem 3. Regarding $\delta _h u^{out}$ we have

$$\begin{aligned} |\delta _h u^{out}(x)| \le C\Vert \theta \Vert _{L^{\infty }}\frac{|h|}{\rho } \end{aligned}$$

(89)

in view of (36). Putting together the estimates (87), (88) and (89) we have

Proposition 2

Let $\chi$ be a good cutoff, and let u be defined by (8). Then

$$\begin{aligned} |\delta _h u(x)| \le C\left( \sqrt{\rho D(f)(x)} + \Vert \theta \Vert _{L^{\infty }}\left( \frac{|h|}{d(x)}+ \frac{|h|}{\rho }\right) + |\delta _h\theta (x)|\right) \end{aligned}$$

(90)

holds for $d(x)\ge \ell$, $\rho \le cd(x)$, $f=\chi \delta _h \theta$ and with C a constant depending on $\Omega$.

Now we will obtain similar estimates for $\nabla u$. We start with the representation

$$\begin{aligned} \nabla u(x) = \nabla u^{in}(x) + \nabla u^{out}(x) \end{aligned}$$

(91)

where

$$\begin{aligned} \nabla u^{in}(x) = c\int _0^{\rho ^2}t^{-\frac{1}{2}}\int _{\Omega }\nabla _x\nabla _x^{\perp }H_D(x,y,t)\theta (y)dy \end{aligned}$$

(92)

and $\rho = \rho (x) \le c d(x)$. In view of (36) we have

$$\begin{aligned} |\nabla u^{out}(x)|\le \frac{C}{\rho }\Vert \theta \Vert _{L^{\infty }(\Omega )} \end{aligned}$$

(93)

We split now

$$\begin{aligned} \nabla u^{in}(x) = g(x) + g_1(x) + g_2(x) + g_3(x) + g_4(x)f(x) \end{aligned}$$

(94)

where

$$\begin{aligned} f(x) = \chi (x)\nabla \theta (x) \end{aligned}$$

(95)

and with

$$\begin{aligned} g(x) = c\int _0^{\rho ^2}t^{-\frac{1}{2}}\int _{\Omega }\nabla _x^{\perp }H_D(x,y,t)(f(y)-f(x))dy, \end{aligned}$$

(96)

and

$$\begin{aligned} g_1(x)= & {} c\int _0^{\rho ^2}t^{-\frac{1}{2}}\int _{\Omega }\nabla _x\nabla _x^{\perp }(H_D(x,y,t)(1-\chi (y))\theta (y)dy, \end{aligned}$$

(97)

$$\begin{aligned} g_2(x)= & {} c\int _0^{\rho ^2}t^{-\frac{1}{2}}\int _{\Omega }\nabla _x^{\perp }(\nabla _x+ \nabla _y)H_D(x,y,t)\chi (y)\theta (y)dy, \end{aligned}$$

(98)

$$\begin{aligned} g_3(x)= & {} c\int _0^{\rho ^2}t^{-\frac{1}{2}}\int _{\Omega }\nabla _x^{\perp }H_D(x,y,t)(\nabla _y\chi (y))\theta (y)dy, \end{aligned}$$

(99)

and

$$\begin{aligned} g_4(x) =c\int _0^{\rho ^2}t^{-\frac{1}{2}}\int _{\Omega }\nabla _x^{\perp }H_D(x,y,t)dy. \end{aligned}$$

(100)

Now

$$\begin{aligned} |g_1(x)| \le \frac{C}{d(x)}\Vert \theta \Vert _{L^{\infty }(\Omega )} \end{aligned}$$

(101)

holds for $d(x)\ge \ell$ because of (36), time integration using Lemma 4 and then use of (44). For $g_2(x)$ we use (38) and then Lemma 4 to obtain

$$\begin{aligned} |g_2(x)| \le \frac{C}{d(x)}\Vert \theta \Vert _{L^{\infty }(\Omega )} \end{aligned}$$

(102)

for $d(x)\ge \ell$. Now

$$\begin{aligned} |g_3(x)| \le \frac{C}{d(x)}\Vert \theta \Vert _{L^{\infty }(\Omega )} \end{aligned}$$

(103)

holds because of (32), Lemma 4 and then use of (45). Regarding $g_4$, in view of

$$\begin{aligned} \int _{\Omega }\nabla _y^{\perp }H_D(x,y,t)dy = 0 \end{aligned}$$

(104)

we have

$$\begin{aligned} g_4(x) = c\int _0^{\rho ^2}t^{-\frac{1}{2}}\int _{\Omega }\left( \nabla _x^{\perp } + \nabla _y^{\perp }\right) H_D(x,y,t)dy \end{aligned}$$

and, we thus obtain from (37) and from Lemma 4 with $m=j=0$

$$\begin{aligned} |g_4(x)| \le C \end{aligned}$$

(105)

because $\rho \le cd(x)$.

Finally we have using a Schwartz inequality like for (88)

$$\begin{aligned} |g(x)| \le C\sqrt{\rho D(f)}. \end{aligned}$$

(106)

Gathering the bounds we have proved

Proposition 3

Let $\chi$ be a good cutoff with scale $\ell$ and let u be given by (8). Then

$$\begin{aligned} |\nabla u(x)| \le C\left( \sqrt{\rho D(f)} + \Vert \theta \Vert _{L^{\infty }(\Omega )}\left( \frac{1}{d(x)} + \frac{1}{\rho }\right) + |\nabla \theta (x)|\right) \end{aligned}$$

(107)

holds for $d(x)\ge \ell$, $\rho \le cd(x)$ and $f=\chi \nabla \theta$ with a constant C depending on $\Omega$.

Commutators

We consider the finite difference

$$\begin{aligned} (\delta _h\Lambda _D\theta )(x) = \Lambda _D\theta (x+h)-\Lambda _D\theta (x) \end{aligned}$$

(108)

with $d(x)\ge \ell$ and $|h|\le \frac{\ell }{16}$. We use a good cutoff $\chi$ again and denote

$$\begin{aligned} f(x) = \chi (x)\delta _h\theta (x). \end{aligned}$$

(109)

We start by computing

$$\begin{aligned} (\delta _h\Lambda _D\theta )(x)= & {} (\Lambda _Df)(x) + c\int _0^{\infty }t^{-\frac{3}{2}}dt\int _{\Omega }(H_D(x,y,t)\nonumber \\&-\,H_D(x+h,y,t))(1-\chi (y))\theta (y)dy\nonumber \\&-\,C\int _0^{\infty }t^{-\frac{3}{2}}dt\int _{\Omega }(H_D(x+h,y,t)-H_D(x,y-h,t))\chi (y)\theta (y)dy\nonumber \\&-\,c\int _0^{\infty }t^{-\frac{3}{2}}dt\int _{\Omega }H_D(x,y,t)(\delta _h\chi )(y)\theta (y+h)dy\nonumber \\= & {} (\Lambda _Df)(x) + E_1(x) + E_2(x) + E_3(x). \end{aligned}$$

(110)

Lemma 5

There exists a constant $\Gamma _0$ such that the commutator

$$\begin{aligned} C_h(\theta ) = \delta _h\Lambda _D\theta -\Lambda _D(\chi \delta _h\theta ) \end{aligned}$$

(111)

obeys

$$\begin{aligned} \left| C_h(\theta )(x)\right| \le \Gamma _0\frac{|h|}{d(x)^2}\Vert \theta \Vert _{L^{\infty }(\Omega )} \end{aligned}$$

(112)

for $d(x)\ge \ell$, $|h|\le \frac{\ell }{16}$, $f=\chi \delta _h\theta$ and $\theta \in H_0^1(\Omega )\cap L^{\infty }(\Omega )$.

Proof

We use (110). For $E_1(x)$ we use a similar argument as for $e_1$ leading to (80), namely the inequality (31) and Lemma 4 with $m=d+2$, $j=0$, and (44) to obtain

$$\begin{aligned} |E_1(x)|\le C\frac{|h|}{d(x)^2}\Vert \theta \Vert _{L^{\infty }}. \end{aligned}$$

For $E_2$ we proceed in a manner analogous to the one leading to the bound (86), by using (85), (37), Lemma 4 with $m=d+2$, $j=0$, and (45) to obtain

$$\begin{aligned} |E_2(x)|\le C\frac{|h|}{d(x)^2}\Vert \theta \Vert _{L^{\infty }}. \end{aligned}$$

For $E_3$ we use

$$\begin{aligned} |E_3(x)|\le |h|\Vert \theta \Vert _{L^{\infty }}\int _0^{\infty }t^{-\frac{3}{2}}dt\int _{\Omega }H_D(x,y,t)|\nabla (\chi )(y)|dy \end{aligned}$$

and using Lemma 4 with $m=d+1$, $j=0$ and (45) we obtain

$$\begin{aligned} |E_3(x)|\le C\frac{|h|}{d(x)^2}\Vert \theta \Vert _{L^{\infty }}, \end{aligned}$$

concluding the proof. $\square$

We consider now the commutator $[\nabla , \Lambda _D]$.

Lemma 6

There exists a constant $\Gamma _3$ depending on $\Omega$ such that for any smooth function f vanishing at $\partial \Omega$ and any $x\in \Omega$ we have

$$\begin{aligned} \left| [\nabla , \Lambda _D]f(x)\right| \le \frac{\Gamma _3}{d(x)^2}\Vert f\Vert _{L^{\infty }(\Omega )}. \end{aligned}$$

(113)

If $\chi$ is a good cutoff with scale $\ell$ and if $\theta$ is a smooth bounded function in ${\mathcal {D}}\left( \Lambda _D\right)$, then

$$\begin{aligned} C_{\chi }(\theta ) = \nabla \Lambda _D\theta -\Lambda _D\chi \nabla \theta \end{aligned}$$

(114)

obeys

$$\begin{aligned} |C_{\chi }(\theta )(x)| = \left| (\nabla \Lambda _D\theta - \Lambda _D(\chi \nabla \theta ))(x)\right| \le \frac{\Gamma _3}{d(x)^2}\Vert \theta \Vert _{L^{\infty }(\Omega )} \end{aligned}$$

(115)

for $d(x)\ge \ell$, with a constant $\Gamma _3$ independent of $\ell$.

Proof

We note that

$$\begin{aligned} {[}\nabla ,\Lambda _D]f(x) = -c_1\int _0^{\infty }t^{-\frac{3}{2}}\int _{\Omega }\left( \nabla _xH_D(x,y,t)f(y) - H_D(x,y,t)\nabla _y f(y)\right) dy \end{aligned}$$

(116)

and therefore

$$\begin{aligned} {[}\nabla ,\Lambda _D]f(x) = -c_1\int _0^{\infty }t^{-\frac{3}{2}}\int _{\Omega }\left( \nabla _x+\nabla _y\right) H_D(x,y,t)f(y)dy. \end{aligned}$$

(117)

The inequality (113) follows from (37) and Lemma 4. For the inequality (115) we need also to estimate

$$\begin{aligned} C(x) = c_s\left| \int _0^{\infty }t^{-\frac{3}{2}}\int _{\Omega }H_D(x,y,t)(\nabla \chi (y))\theta (y)dy\right| \end{aligned}$$

by the right hand side of (115), and this follows from (45) in view of (31). $\square$

SQG: Hölder Bounds

We consider the equation (7) with u given by (8) and with smooth initial data $\theta _0$ compactly supported in $\Omega$. We note that by the Córdoba-Córdoba inequality we have

$$\begin{aligned} \Vert \theta (t)\Vert _{L^{\infty }}\le \Vert \theta _0\Vert _{L^{\infty }}. \end{aligned}$$

(118)

We prove the following uniform interior Hölder bound:

Theorem 5

Let $\theta (x,t)$ be a smooth solution of (7) in the smooth bounded domain $\Omega$. There exists a constant $0<\alpha <1$ depending only on $\Vert \theta _0\Vert _{L^{\infty }(\Omega )}$, and a constant $\Gamma >0$ depending on the domain $\Omega$ such that, for any $\ell >0$ sufficiently small

$$\begin{aligned} \sup _{d(x)\ge \ell , \; |h|\le \frac{\ell }{16},\; t\ge 0}\frac{|\theta (x+h,t)-\theta (x,t)|}{|h|^{\alpha }} \le \Vert \theta _0\Vert _{C^{\alpha }} + \Gamma \ell ^{-\alpha }\Vert \theta _0\Vert _{L^{\infty }(\Omega )} \end{aligned}$$

(119)

holds.

Proof

We take a good cutoff $\chi$ used above, $|h|\le \frac{\ell }{16}$ and observe that, from the SQG equation we obtain the equation

$$\begin{aligned} (\partial _t + u\cdot \nabla + (\delta _h u)\cdot \nabla _h)(\delta _h\theta )+ \Lambda _D(\chi \delta _h \theta ) + C_h(\theta ) = 0 \end{aligned}$$

(120)

where $C_h(\theta )$ is the commutator given above in (111). Denoting (as before in (109)) $f=\chi \delta _h\theta$ we have after multiplying by $\delta _h\theta$ and using the fact that $\chi (x)=1$ for $d(x)\ge \ell$,

$$\begin{aligned} \frac{1}{2}L_{\chi }\left( \delta _h\theta \right) ^2 + D(f) + (\delta _h\theta ) C_h(\theta ) = 0 \end{aligned}$$

(121)

where

$$\begin{aligned} L_{\chi }g = \partial _t g + u\cdot \nabla _x g + \delta _h u\cdot \nabla _h g + \Lambda _D(\chi ^2 g) \end{aligned}$$

(122)

and D(f) is given in Theorem 3. $\square$

Multiplying by $|h|^{-2\alpha }$ where $\alpha >0$ will be chosen small to be small enough we obtain

$$\begin{aligned} \frac{1}{2}L_{\chi }\left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }}\right) + |h|^{-2\alpha } D(f) \le 2\alpha \frac{|\delta _h u|}{|h|}\left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }}\right) + |C_h(\theta )| |\delta _h\theta ||h|^{-2\alpha }. \end{aligned}$$

(123)

The factor $2\alpha$ comes from the differentiation $\delta _h u\cdot \nabla _h (|h|^{-2\alpha })$ and its smallness will be crucial below. Let us record here the inequality (47) in the present case:

$$\begin{aligned} D(f) \ge \gamma _1 |h|^{-1}\Vert \theta \Vert _{L^{\infty }}^{-1}|(\delta _h\theta )_d|^3 + \gamma _1 (d(x))^{-1} |\delta _h\theta |^2, \end{aligned}$$

(124)

valid pointwise, when $|h|\le \frac{\ell }{16}$ and $d(x)\ge \ell$, where

$$\begin{aligned} |(\delta _h\theta )_d| = |\delta _h\theta |, \quad {\hbox {if}}\; |\delta _h\theta (x)|\ge M\Vert \theta \Vert _{L^{\infty }}\frac{|h|}{d(x)}, \end{aligned}$$

and $|(\delta _h\theta )_d|=0$ otherwise.

We use now the estimates (90), (112) and a Young inequality for the term involving $\sqrt{\rho D(f)}$ to obtain

$$\begin{aligned}&\frac{1}{2}L_{\chi }\left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }}\right) + \frac{1}{2}|h|^{-2\alpha } D(f) \le C_1\alpha ^2|h|^{-2-2\alpha }\rho |\delta _h\theta |^4 \nonumber \\&\quad +\, C_1\alpha \Vert \theta \Vert _{L^{\infty }}\left( \frac{1}{d(x)} + \frac{1}{\rho }\right) |h|^{-2\alpha }|\delta _h\theta |^2 + C_1\alpha |\delta _h\theta ||h|^{-1-2\alpha }|\delta _h\theta |^2 \nonumber \\&\quad +\, \Gamma _0\frac{|h|}{d(x)^2}\Vert \theta \Vert _{L^{\infty }} |\delta _h\theta ||h|^{-2\alpha } \end{aligned}$$

(125)

for $d(x)\ge \ell$, $|h|\le \frac{\ell }{16}$. Let us choose $\rho$ now. We set

$$\begin{aligned} \rho = \left\{ \begin{array}{ll} |\delta _h\theta (x)|^{-1}|h|\Vert \theta \Vert _{L^{\infty }}, &{} \quad {\hbox {if}}\; |\delta _h\theta (x)|\ge M_1\Vert \theta \Vert _{L^{\infty }}\frac{|h|}{d(x)},\\ d(x), &{} \quad {\hbox {if}}\quad |\delta _h\theta (x)|\le M_1\Vert \theta \Vert _{L^{\infty }}\frac{|h|}{d(x)}, \end{array} \right. \end{aligned}$$

(126)

where we put

$$\begin{aligned} M_1 = M + \sqrt{\frac{8\Gamma _0}{\gamma _1}} + 1, \end{aligned}$$

(127)

where M is the constant from Theorem 3, $\Gamma _0$ is the constant from (112) and $\gamma _1$ is the constant from (124). This choice was made in order to use the lower bound on D(f) to estimate the contribution due to the inner piece $u_h$ (see (72)) of $\delta _h u$ and the contribution from the commutator $C_h(\theta )$. We distinguish two cases. The first case is when $|\delta _h\theta (x)|\ge M_1\Vert \theta \Vert _{L^{\infty }}\frac{|h|}{d(x)}$. Then we have

$$\begin{aligned}&\frac{1}{2}L_{\chi }\left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }}\right) + \frac{1}{2}|h|^{-2\alpha }D(f) \nonumber \\&\quad \le C_1\left[ (\alpha \Vert \theta \Vert _{L^{\infty }})^2 +(2+\frac{1}{M_1})\alpha \Vert \theta \Vert _{L^{\infty }}\right] \,|\delta _h\theta |^3|h|^{-1-2\alpha }\Vert \theta \Vert _{L^{\infty }}^{-1} \nonumber \\&\qquad +\, \Gamma _0\frac{|h|}{d(x)^2}\Vert \theta \Vert _{L^{\infty }} |\delta _h\theta ||h|^{-2\alpha }. \end{aligned}$$

(128)

The choice of $M_1$ was such that, in this case

$$\begin{aligned} \Gamma _0\frac{|h|}{d(x)^2}\Vert \theta \Vert _{L^{\infty }} |\delta _h\theta (x)||h|^{-2\alpha } \le \frac{\gamma _1}{8} |\delta _h\theta (x)|^3|h|^{-1-2\alpha }\Vert \theta \Vert _{L^{\infty }}^{-1}. \end{aligned}$$

We choose now $\alpha$ by requiring

$$\begin{aligned} \epsilon = \alpha \Vert \theta \Vert _{L^{\infty }} \end{aligned}$$

(129)

to satisfy

$$\begin{aligned} C_1M_1^2(\epsilon ^2 + (2+M_1^{-1})\epsilon ) \le \frac{\gamma _1}{8} \end{aligned}$$

(130)

and obtain from (128)

$$\begin{aligned} \frac{1}{2}L_{\chi }\left( \frac{|\delta _h\theta (x)|^2}{|h|^{2\alpha }}\right) + \frac{1}{4}|h|^{-2\alpha }D(f) \le 0 \end{aligned}$$

(131)

for $d(x)\ge \ell$, $|h|\le \frac{\ell }{16}$, in the case $|f|\ge M_1\Vert \theta \Vert _{L^{\infty }}\frac{|h|}{d(x)}$.

The second case is when the opposite inequality holds, i.e, when $|\delta _h\theta (x)|\le M_1\Vert \theta \Vert _{L^{\infty }}\frac{|h|}{d(x)}$. Then, using $\rho = d(x)$ we obtain from (125)

$$\begin{aligned} \frac{1}{2}L_{\chi }\left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }}\right) + \frac{1}{2}|h|^{-2\alpha }D(f)\le & {} C_1(M_1^2\epsilon ^2 + (M_1+2)\epsilon )\frac{1}{d(x)} (\delta _h\theta (x))^2|h|^{-2\alpha }\nonumber \\&+\, \Gamma _0d(x)^{-2}\Vert \theta \Vert _{L^{\infty }}|\delta _h\theta ||h|^{1-2\alpha }\nonumber \\\le & {} \frac{\gamma _1}{8d(x)}\left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }}\right) + 2\Gamma _0M_1\Vert \theta \Vert _{L^{\infty }}^2d(x)^{-3}|h|^{2-2\alpha }.\nonumber \\ \end{aligned}$$

(132)

Summarizing, in view of the inequalities (131) and (132), the damping term $\frac{\gamma _1}{d(x)}|\delta _h\theta (x)|^2$ in (124) and the choice of small $\epsilon$ in (130), we have that

$$\begin{aligned} L_{\chi }\left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }}\right) + \frac{\gamma _1}{4d(x)}\left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }}\right) \le B \end{aligned}$$

(133)

holds for $d(x)\ge \ell$ and $|h|\le \frac{\ell }{16}$ where

$$\begin{aligned} B= 2(16)^{-2+2\alpha }\Gamma _0M_1\Vert \theta \Vert _{L^{\infty }}^2d(x)^{-1-2\alpha } = \Gamma _1\frac{\gamma _1}{4}\Vert \theta \Vert _{L^{\infty }}^2d(x)^{-1-2\alpha } \end{aligned}$$

(134)

with $\Gamma _1$ depending on $\Omega$. Without loss of generality we may take $\Gamma _1> 4(16)^{2\alpha }$ so that

$$\begin{aligned} \frac{|\delta _h\theta |^2}{|h|^{2\alpha }} <\Gamma _1\ell ^{-2\alpha }\Vert \theta _0\Vert _{L^{\infty }}^2 \end{aligned}$$

when $|h|\ge \frac{\ell }{16}$. We note that

$$\begin{aligned} L_\chi \left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }}\right) + \frac{\gamma _1}{4d(x)}\left( \frac{\delta _h\theta (x)^2}{|h|^{2\alpha }} - \Gamma _1\ell ^{-2\alpha }\Vert \theta \Vert _{L^{\infty }}^2\right) \le 0 \end{aligned}$$

(135)

holds for any t, $x\in \Omega$ with $d(x)\ge \ell$ and $|h|\le \frac{\ell }{16}$.

We take $\delta >0$, $T>0$. We claim that, for any $\delta >0$ and any $T>0$

$$\begin{aligned} \sup _{d(x)\ge \ell , |h|\le \frac{\ell }{16}, 0\le t \le T}\frac{|\delta _h\theta (x)|^2}{|h|^{2\alpha }} \le (1+\delta )\left[ \Vert \theta _0\Vert _{C^{\alpha }}^2 + \Gamma _1 \ell ^{-2\alpha }\Vert \theta _0\Vert _{L^{\infty }}^2\right] \end{aligned}$$

holds.

The rest of the proof is done by contradiction. Indeed, assume by contradiction that there exists $\tilde{t}\le T$, $\tilde{x}$ and $\tilde{h}$ with $d({\tilde{x}})\ge \ell$ and $|{\tilde{h}}|\le {\frac{\ell }{16}}$ such that

$$\begin{aligned} \frac{|\theta (\tilde{x} +\tilde{h}, \tilde{t})- \theta (\tilde{x},\tilde{t}) |^2}{|h|^{2\alpha }} >(1+\delta )\left[ \Vert \theta _0\Vert _{C^{\alpha }}^2 + \Gamma _1 \ell ^{-2\alpha }\Vert \theta _0\Vert _{L^{\infty }}^2\right] = R \end{aligned}$$

holds. Because the solution is smooth, we have

$$\begin{aligned} \frac{|\delta _h\theta (x,t)|^2}{|h|^{2\alpha }} \le (1+\delta )\Vert \theta _0\Vert _{C^{\alpha }}^2 \end{aligned}$$

for a short time $0\le t\le t_1$. (Note that this is not a statement about well-posedness in this norm: $t_1$ may depend on higher norms.) Also, because the solution is smooth, it is bounded in $C^1$, and

$$\begin{aligned} \sup _{d(x)\ge \ell , |h|\le \frac{\ell }{16}}\frac{|\delta _h\theta (x)|^2}{|h|^2}\le C \end{aligned}$$

on the time interval [0, T]. It follows that there exists $\delta _1>0$ such that

$$\begin{aligned} \sup _{d(x)\ge \ell , |h|\le \delta _1}\frac{|\delta _h\theta (x)|^2}{|h|^{2\alpha }} \le C\delta _1^{2-2\alpha }\le \frac{R}{2}. \end{aligned}$$

In view of these considerations, we must have $\tilde{t} >t_1$, $|\tilde{h}|\ge \delta _1$. Moreover, the supremum is attained: there exists $\bar{x}\in \Omega$ with $d(\bar{x})\ge \ell$ and $\bar{h} \ne 0$ such that $\delta _1\le |\bar{h}|\le \frac{\ell }{16}$ such that

$$\begin{aligned} \frac{|\theta (\bar{x}+\bar{h}, \tilde{t})-\theta (\bar{x},\tilde{t})|^2}{|\bar{h}|^{2\alpha }} = s(\tilde{t}) =\sup _{d(x)\ge \ell , |h|\le \frac{\ell }{16}}\frac{|\delta _h\theta (\tilde{t})|^2}{|h|^{2\alpha }} > R. \end{aligned}$$

Because of (135) we have that

$$\begin{aligned} \frac{d}{dt}\frac{|\theta (\bar{x}+\bar{h},t)-\theta (\bar{x},t)|^2}{|\bar{h}|^{2\alpha }}_{\left| \right. t=\tilde{t}} <0 \end{aligned}$$

and therefore there exists $t'<\tilde{t}$ such that $s(t')>s(\tilde{t})$. This implies that $\inf \{t>t_1\left| \right. s(t)>R\} = t_1$ which is absurd because we made sure that $s(t_1)< R$. Now $\delta$ and T are arbitrary, so we have proved

$$\begin{aligned} \sup _{d(x)\ge \ell , |h|\le \frac{\ell }{16}, t\ge 0}\frac{|\delta _h\theta (x)|^2}{|h|^{2\alpha }} \le \left[ \Vert \theta _0\Vert _{C^{\alpha }}^2 + \Gamma _1 \ell ^{-2\alpha }\Vert \theta _0\Vert _{L^{\infty }}^2\right] \end{aligned}$$

(136)

which finishes the proof of the theorem.

Proof of Theorem 1

The proof follows from (136) because $\Gamma _1$ does not depend on $\ell$. For any fixed $x\in \Omega$ we may take $\ell$ such that $\ell \le d(x)\le 2 \ell$. Then (136) implies

$$\begin{aligned} d(x)^{2\alpha }\frac{|\delta _h\theta (x,t)|^2}{|h|^{2\alpha }}\le \left[ \Vert \theta _0\Vert _{C^{\alpha }}^2 + \Gamma _1 2^{2\alpha }\Vert \theta _0\Vert _{L^{\infty }}^2\right] . \end{aligned}$$

(137)

Gradient Bounds

We take the gradient of (7). We obtain

$$\begin{aligned} (\partial _t + u\cdot \nabla )\nabla \theta + (\nabla u)^*\nabla \theta +\nabla \Lambda _D\theta =0 \end{aligned}$$

where $(\nabla u)^*$ is the transposed matrix. Let us take a good cutoff $\chi$. Then $g=\nabla \theta$ obeys everywhere

$$\begin{aligned} \partial _t g + u\cdot \nabla g +\Lambda _D(\chi g) + C_{\chi }(\theta ) + (\nabla u)^*g = 0 \end{aligned}$$

(138)

with $C_{\chi }$ given in (114). We multiply by g and, using the fact that $\chi (x)=1$ when $d(x)\ge \ell$ we obtain

$$\begin{aligned} \frac{1}{2}L_{\chi }g^2 + D(f) + gC_{\chi }(\theta ) + g(\nabla u)^*g = 0 \end{aligned}$$

(139)

when $d(x)\ge \ell$, where $L_{\chi }$ is similar to the one defined in (122):

$$\begin{aligned} L_{\chi }(\phi ) = \partial _t\phi + u\cdot \nabla \phi + \Lambda _D(\chi ^2\phi ) \end{aligned}$$

(140)

and $f=\chi g$. Recall that $D(f) = f\Lambda _Df-\Lambda _D\left( \frac{f^2}{2}\right)$. Then, using (115) and (107) we deduce

$$\begin{aligned} \frac{1}{2}L_{\chi }g^2 + D(f) \le \frac{\Gamma _3}{d(x)^2}|g|\Vert \theta \Vert _{L^{\infty }(\Omega )} + C\left( \sqrt{\rho D(f)} + \Vert \theta \Vert _{L^{\infty }(\Omega )}\left( \frac{1}{d(x)} + \frac{1}{\rho }\right) + |\nabla \theta (x)|\right) g^2 \nonumber \\ \end{aligned}$$

(141)

for $d(x)\ge \ell$. Using a Young inequality we deduce

$$\begin{aligned} L_{\chi }g^2 + D(f) \le \frac{2\Gamma _3}{d(x)^2}\Vert \theta \Vert _{L^{\infty }(\Omega )}|g| + C_4\rho g^4 + C_4\Vert \theta \Vert _{L^{\infty }(\Omega )}\left( \frac{1}{d(x)} + \frac{1}{\rho }\right) g^2 + C_4|g|^3\nonumber \\ \end{aligned}$$

(142)

for $d(x)\ge \ell$. Now $|g| = |f|$ when $d(x)\ge \ell$. If $|g(x)|\ge M\Vert \theta \Vert _{L^{\infty }(\Omega )}d(x)^{-1}$ then, in view of (61)

$$\begin{aligned} D(f)\ge \gamma _2\Vert \theta \Vert _{C^{\alpha }(\Omega )}^{-\frac{1}{1-\alpha }}|g|^{3+\frac{\alpha }{1-\alpha }}(d(x))^{\frac{\alpha }{1-\alpha }} + \frac{\gamma _1}{d(x)}g^2 \end{aligned}$$

(143)

which is a super-cubic lower bound. We choose in this case

$$\begin{aligned} \rho ^{-1} = C_5 |g(x)|, \end{aligned}$$

(144)

and the right hand side of (142) becomes at most cubic in g:

$$\begin{aligned} L_{\chi }g^2 + D(f) \le |g|^3\left[ \frac{2\Gamma _3}{M^2\Vert \theta \Vert _{L^{\infty }(\Omega )}} + C_4\left( \frac{1}{C_5} + \frac{1}{M} + C_5\Vert \theta \Vert _{L^{\infty }(\Omega )} + 1\right) \right] = K|g|^3.\nonumber \\ \end{aligned}$$

(145)

In view of (143) we see that

$$\begin{aligned} L_{\chi }g^2 + |g|^3\left( \gamma _2\left( \Vert \theta \Vert _{C^{\alpha }(\Omega )}^{-\frac{1}{\alpha }}|g(x)|d(x)\right) ^{\frac{\alpha }{1-\alpha }} - K\right) \le 0 \end{aligned}$$

(146)

holds for $d(x)\ge \ell$, if $|g|\ge M\Vert \theta \Vert _{L^{\infty }}d(x)^{-1}$. In the opposite case, $|g(x)|\le M\Vert \theta \Vert _{L^{\infty }}d(x)^{-1}$ we choose

$$\begin{aligned} \rho (x) = d(x) \end{aligned}$$

(147)

and obtain from (142)

$$\begin{aligned} \begin{array}{l} L_{\chi }g^2 + D(f) \\ \le \frac{1}{d(x)^3}\left[ C_4M^4\Vert \theta \Vert _{L^{\infty }(\Omega )}^4 + C_4M^3\Vert \theta \Vert _{L^{\infty }(\Omega )}^3\right. \\ \quad \left. + 2C_4M^2\Vert \theta \Vert _{L^{\infty }(\Omega )}^3+ 2M\Gamma _3\Vert \theta \Vert _{L^{\infty }(\Omega )}^2\right] = \frac{K_1}{d(x)^3} \end{array} \end{aligned}$$

(148)

and using the convex damping inequality (61)

$$\begin{aligned} D(f)\ge \gamma _1\frac{g^2}{d(x)} \end{aligned}$$

we obtain in this case

$$\begin{aligned} L_{\chi }g^2 + \frac{1}{d(x)}\left( \gamma _1g^2(x) - \frac{K_1}{d(x)^2}\right) \le 0. \end{aligned}$$

(149)

Putting together (146) and (149) and 119 we obtain

Theorem 6

Let $\theta$ be a smooth solution of (7). Then

$$\begin{aligned} \sup _{d(x)\ge \ell }|\nabla \theta (x,t)| \le C\left[ \Vert \nabla \theta _0\Vert _{L^{\infty }(\Omega )} + \frac{P(\Vert \theta \Vert _{L^{\infty }(\Omega )})}{\ell }\right] \end{aligned}$$

(150)

where $P(\Vert \theta \Vert _{L^{\infty }(\Omega )})$ is a polynome of degree four.

Proof of Theorem 2

The proof follows by choosing $\ell$ depending on x, because the constants in (150) do not depend on $\ell$.

Example: Half Space

The case of the half space is interesting because global smooth solutions of (7) are easily obtained by reflection: If the initial data $\theta _0$ is smooth and compactly supported in $\Omega = {\mathbb R}^d_+$ and if we consider its odd reflection

$$\begin{aligned} \widetilde{\theta _0}(x) = \left\{ \begin{array}{ll} \theta _0(x_1,\ldots x_d), &{} \quad {\hbox {if}}\; x_d>0,\\ -\theta _0(x_1,\ldots , -x_d) &{} \quad {\hbox {if}}\; x_d<0 \end{array} \right. \end{aligned}$$

(151)

then the solution of the critical SQG equation in the whole space, with intitial data $\widetilde{\theta _0}$ is globally smooth and its restriction to $\Omega$ solves (7) there. This follows because of reflection properties of the heat kernel and of the Dirichlet Laplacian.

The heat kernel with Dirichlet boundary conditions in $\Omega ={\mathbb R}^d_{+}$ is

$$\begin{aligned} H(x,y,t) = ct^{-\frac{d}{2}}\left( e^{-\frac{|x-y|^2}{4t}} - e^{-\frac{|x-{\widetilde{y}}|^2}{4t}}\right) \end{aligned}$$

where $\widetilde{y} = (y_1,\ldots , y_{d-1}, -y_d)$. More precisely,

$$\begin{aligned} H(x,y,t) = G^{(d-1)}_t(x'-y')\left[ G_t(x_d-y_d)-G_t(x_d+y_d)\right] \end{aligned}$$

(152)

with $x'= (x_1,\ldots , x_{d-1})$,

$$\begin{aligned} G_t^{(d-1)}(x') = \left( \frac{1}{4\pi t}\right) ^{\frac{d-1}{2}}e^{-\frac{|x'|^2}{4t}} \end{aligned}$$

(153)

and

$$\begin{aligned} G_t(\xi ) = \left( \frac{1}{4\pi t}\right) ^{\frac{1}{2}} e^{-\frac{\xi ^2}{4t}} \end{aligned}$$

(154)

Let us note that

$$\begin{aligned} \nabla _x H = H\left( \begin{array}{c} -\frac{x'-y'}{2t}\\ -\frac{x_d -y_d}{2t} + \frac{y_d}{t}\frac{e^{-\frac{x_dy_d}{t}}}{1-e^{-\frac{x_dy_d}{t}}} \end{array} \right) \end{aligned}$$

(155)

We check that (32) is obeyed. Indeed, because $1-e^{-p}\ge \frac{p}{2}$ when $0\le p\le 1$ it follows that

$$\begin{aligned} \frac{y_d}{t}e^{-\frac{x_dy_d}{t}}(1-e^{-\frac{x_dy_d}{t}})^{-1}\le \frac{y_d}{t}\frac{2t}{x_dy_d} \end{aligned}$$

if $\frac{x_dy_d}{t}\le 1$, and if $p=\frac{x_dy_d}{t} \ge 1$ then

$$\begin{aligned} \frac{y_d}{t}e^{-\frac{x_dy_d}{t}}(1-e^{-\frac{x_dy_d}{t}})^{-1}\le \frac{e}{e-1}\frac{y_d}{t}e^{-\frac{x_dy_d}{t}}. \end{aligned}$$

In this case, if $\frac{x_d}{\sqrt{t}} \ge 1$ then $\frac{y_d}{t}\le t^{-\frac{1}{2}}p$ and $pe^{-p}$ is bounded; if $\frac{x_d}{\sqrt{t}}\le 1$ we write $\frac{y_d}{t} = t^{-\frac{1}{2}}(\frac{y_d-x_d}{\sqrt{t}} + \frac{x_d}{\sqrt{t}})$ and thus we obtain:

$$\begin{aligned} \left| \nabla _x H\right| \le C H\left[ \frac{1}{\sqrt{t}}(1+ \frac{|x-y|}{\sqrt{t}}) + \frac{1}{x_d}\right] \end{aligned}$$

(156)

We check (37): First we have

$$\begin{aligned} (\nabla _x + \nabla _y)H = \left( \begin{array}{c} 0\\ \frac{x_d+y_d}{t}G_t(x_d+y_d)G^{(d-1)}_t(x'-y') \end{array} \right) \end{aligned}$$

(157)

and then

$$\begin{aligned} \int _{\Omega }\left| (\nabla _x+ \nabla _y)H(x,y,t)\right| dy \le Ct^{-\frac{1}{2}}e^{-\frac{x_d^{2}}{4t}}. \end{aligned}$$

(158)

Indeed, the only nonzero component occurs when the differentiation is with respect to the normal direction, and then

$$\begin{aligned} \left| (\partial _{x_d} + \partial _{y_d})H(x,y,t)\right| = ct^{-\frac{d}{2}}e^{-\frac{|x'-y'|^2}{4t}}\left( \frac{x_d+y_d}{t}\right) e^{-\frac{(x_d+y_d)^2}{4t}} \end{aligned}$$

(159)

Therefore

$$\begin{aligned} \smallint \nolimits _{\Omega } \left| (\nabla _x+ \nabla _y)H(x,y,t)\right| dy\le & {} Ct^{-\frac{1}{2}}\smallint \nolimits _0^{\infty }\left( \frac{x_d+y_d}{t}\right) e^{-\frac{(x_d+y_d)^2}{4t}}dy_d\nonumber \\= & {} Ct^{-\frac{1}{2}}\smallint \nolimits _{\frac{x_d}{\sqrt{t}}}^{\infty }\xi e^{-\frac{\xi ^2}{4}}d\xi \nonumber \\= & {} Ct^{-\frac{1}{2}}e^{-\frac{x_d^2}{4t}}. \end{aligned}$$

(160)

We check (38): first

$$\begin{aligned} \partial _{x'}(\nabla _x + \nabla _y)H= & {} -\frac{x_d+y_d}{t}G_t(x_d+y_d)\frac{(x'-y')}{2t}G_t^{(d-1)}(x'-y') \nonumber \\ \partial _{x_d}(\nabla _x + \nabla _y)H= & {} \left( \frac{1}{t} - \frac{(x_d+y_d)^2}{2t^2}\right) G_t(x_d+y_d)G_t^{(d-1)}(x'-y') \end{aligned}$$

(161)

Consequently

$$\begin{aligned} |\nabla _x(\nabla _x + \nabla _y)H(x,y,t)|\le Ct^{-\frac{d}{2} -1}\left( 1+\frac{|x'-y'|}{\sqrt{t}}\right) \left( 1+ \frac{(x_d+y_d)^2}{t}\right) e^{-\frac{|x'-y'|^2}{4t}}e^{-\frac{(x_d+y_d)^2}{4t}} \end{aligned}$$

(162)

and (38) follows:

$$\begin{aligned} \int _{\Omega }|\nabla _x(\nabla _x + \nabla _y)H(x,y,t)|dy \le Ct^{-1}\int _{\frac{x_d}{\sqrt{2t}}}^{\infty }(1+z^2)e^{-\frac{z^2}{2}}dz. \end{aligned}$$

We compute $\Theta$ and $\Lambda _D1$:

$$\begin{aligned} \Theta (x,t)= (e^{t\Delta }1)(x) = \int _{\Omega }H(x,y,t)dy = \frac{1}{\sqrt{2\pi }}\int _{-\frac{x_d}{\sqrt{2t}}}^{\frac{x_d}{\sqrt{2t}}}e^{-\frac{\xi ^2}{2}}d\xi \end{aligned}$$

(163)

and therefore

$$\begin{aligned} \int _0^{\infty }t^{-\frac{3}{2}}(1-e^{t\Delta }1)dt = \frac{2}{\sqrt{2\pi }}\int _0^{\infty }t^{-\frac{3}{2}}dt\int _{\frac{x_d}{\sqrt{2t}}}e^{-\frac{\xi ^2}{2}}d\xi = \frac{4}{x_d\sqrt{\pi }}. \end{aligned}$$

Remark 3

We note here that $\Lambda _D^{s} 1 = C_sy_d^{-s}$ is calculated by duality:

$$\begin{aligned} \left( \Lambda _D^{s}1, \phi \right)= & {} \left( 1, \Lambda _D^{s}\phi \right) \\= & {} c_{s}\smallint \nolimits _{\Omega }dx\smallint \nolimits _0^{\infty }t^{-1-\frac{s}{2}}dt\left[ \phi (x)-\smallint \nolimits _{\Omega }H(x,y,t)\phi (y)dy\right] \\= & {} c_{s}\smallint \nolimits _0^{\infty }t^{-1-\frac{s}{2}}dt\left[ \smallint \nolimits _{\Omega }\phi (x)dx -\smallint \nolimits _{\Omega }\Theta (y_d,t)\phi (y)dy\right] \\= & {} c_{s}\smallint \nolimits _0^{\infty }t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\Omega }\left( 1-\Theta (y_d,t)\right) \phi (y)dy\\= & {} \frac{2c_{s}}{\sqrt{2\pi }}\smallint \nolimits _{\Omega }\phi (y)\smallint \nolimits _0^{\infty }t^{-1-\frac{s}{2}}dt\smallint \nolimits _{\frac{y_d}{\sqrt{2t}}}^{\infty }e^{-\frac{\xi ^2}{2}}d\xi \\= & {} C_{s}\smallint \nolimits _{\Omega }y_d^{-s}\phi (y)dy \end{aligned}$$

where we used the symmetry of the kernel H and (163).

We observe that if we consider horizontal finite differences, i.e. $h_d= 0$ then $C_h(\theta )$ vanishes, and we deduce that

$$\begin{aligned} \sup _{x,h',t}|h'|^{-\alpha }|\theta (x'+ h', x_d, t)-\theta (x', x_d,t)|\le C_{1,\alpha } \end{aligned}$$

(164)

with $C_{1,\alpha }$ the partial $C^{\alpha }$ norm of the initial data. This inequality can be used to prove that $u_2$ is bounded when $d=2$. Indeed

$$\begin{aligned} u_2(x,t) = c\int _{\Omega }\left( \frac{1}{|x-y|^3} -\frac{1}{|x-\widetilde{y}|^3}\right) (x_1-y_1)\theta (y,t)dy \end{aligned}$$

(165)

and the bound is obtained using the partial Hölder bound on $\theta$ (164) and the uniform bounds $\Vert \theta \Vert _{L^p}$ for $p=1, \infty$. The outline of the proof is as follows: we split the integral

$$\begin{aligned} u_2 = u_{2}^{in} + u_2^{out} \end{aligned}$$

(166)

with

$$\begin{aligned} u_2^{in}(x)= & {} c\int _{|x_1-y_1|\le \delta , |x_2-y_2|\le \delta }\left( \frac{1}{|x-y|^3} -\frac{1}{|x-\widetilde{y}|^3}\right) (x_1-y_1)\left( \theta (y_1, y_2, t)\right. \nonumber \\&\left. -\, \theta ( x_1, y_2,t)\right) dy \end{aligned}$$

(167)

and

$$\begin{aligned} u_2^{out}(x) = c\int _{\max \{|x_1-y_1|, |x_2-y_2|\}\ge \delta }\left( \frac{1}{|x-y|^3} -\frac{1}{|x-\widetilde{y}|^3}\right) (x_1-y_1)\theta (y_1,y_2,t)dy \end{aligned}$$

(168)

where in (167) we used the fact that the kernel is odd in the first variable. Then, for $u^{in}$ we use the bound (164) to derive

$$\begin{aligned} |u_2^{in}(x)|\le C_{1,\alpha }C\int _0^{\sqrt{2}\delta }\rho ^{-1+\alpha }d\rho = CC_{1,\alpha }\delta ^{\alpha } \end{aligned}$$

(169)

and for $u^{out}$, if we have no other information on $\theta$ we just bound

$$\begin{aligned} |u_2^{out}(x)|\le C\log \left( \frac{L}{\delta }\right) \Vert \theta _0\Vert _{L^{\infty }} + CL^{-2}\Vert \theta _0\Vert _{L^1} \end{aligned}$$

(170)

with some $L\ge \delta$. Both $\delta$ and L are arbitrary.

Finally, let us note that even if $\theta \in C_0^{\infty }(\Omega )$, the tangential component of the velocity need not vanish at the boundary because it is given by the integral

$$\begin{aligned} u_1(x_1, 0, t) = -c\int _{{\mathbb R}^2_{+}}\frac{2y_2}{\left( (x_1-y_1)^2 +y_2^2\right) ^{\frac{3}{2}}} \theta (y,t)dy. \end{aligned}$$

Appendices

Appendix 1

We sketch here the proofs of (36) (37) and (38). We take a point $\bar{x}\in \Omega$, a point $y\in \Omega$ and distinguish between two cases, if $d(\bar{x})< \frac{|\bar{x}-y|}{4}$ and if $d(\bar{x})\ge \frac{|\bar{x}-y|}{4}$. In the first case we take a ball B of radius $\delta =\frac{d(\bar{x})}{8}$ centered at $\bar{x}$ and in the second case we take also a ball B centered at $\bar{x}$ but with radius $\delta =\frac{d(\bar{x})}{2}$. We note that in both cases the radius $\delta$ is proportional to $d(\bar{x})$. We take $x\in B(\bar{x}, \frac{\delta }{2})$, we fix $y\in \Omega$, take the function $h(z,t) = H_D(z,y,t)$, and apply Green’s identity in the domain $U = B\times (0,t)$. We obtain

$$\begin{aligned} 0= & {} \smallint \nolimits _U\left[ (\partial _s-\Delta _z)h(z,s)G_{t-s}(x-z) + h(z,s)(\partial _s+\Delta _z)G_{t-s}(x-z)\right] dzds \\= & {} h(x,t)-G_t(x-y) + \smallint \nolimits _0^t\smallint \nolimits _{\partial B}\left[ \frac{\partial G_{t-s}(x-z)}{\partial n}h(z,s)-\frac{\partial h(z,s)}{\partial n}G_{t-s}(x-z)\right] \end{aligned}$$

and thus

$$\begin{aligned} H_D(x,y,t) = G_t(x-y) - \int _0^t\int _{\partial B}\left[ \frac{\partial G_{t-s}(x-z)}{\partial n}h(z,s)-\frac{\partial h(z,s)}{\partial n}G_{t-s}(x-z)\right] \end{aligned}$$

We note that the x dependence is only via G, and $x-z$ is bounded away from zero. We differentiate twice under the integral sign, and use the upper bounds (31), (32). We have

$$\begin{aligned}&|\nabla _x\nabla _x H_D(x,y,t)-\nabla _x\nabla _x G_{t}(x-y)| \\&\quad \le C\int _0^t\int _{\partial B}(t-s)^{-\frac{d+3}{2}}p_3(\frac{|x-z|}{\sqrt{t-s}})e^{-\frac{|x-z|^2}{4(t-s)}}s^{-\frac{d}{2}}e^{-\frac{|y-z|^2}{Ks}}dzds\\&\qquad +\int _0^{\min \{t; d^2(y)\}}\int _{\partial B} (t-s)^{-\frac{d+2}{2}}p_2(\frac{|x-z|}{\sqrt{t-s}})e^{-\frac{|x-y|^2}{4(t-s)}}s^{-\frac{d+1}{2}}p_1(\frac{|y-z|}{\sqrt{s}})e^{-\frac{|y-z|^2}{Ks}}dzds\\&\qquad + \int _{\min \{t; d^2(y)\}}^t\int _{\partial B} (t-s)^{-\frac{d+2}{2}}p_2(\frac{|x-z|}{\sqrt{t-s}})e^{-\frac{|x-y|^2}{4(t-s)}}s^{-\frac{d}{2}}\frac{1}{d(y)}\frac{w_1(y)}{|y-z|}e^{-\frac{|y-z|^2}{Ks}}dzds \\ \end{aligned}$$

where $p_k(\xi )$ are polynomials of degree k. The integrals are not singular. In both cases $|x-z|\ge \frac{\delta }{2}$, and any negative power $(t-s)^{-\frac{k}{2}}$ can be absorbed by $e^{-\frac{|x-z|^2}{8(t-s)}}$ at the price $|x-z|^{-k}\le C\delta ^{-k}$, still leaving $e^{-\frac{|x-z|^2}{8(t-s)}}$ available. Similarly, in the first case $|y-z|\ge |\bar{x}-y| -\delta \ge \delta$ and in the second case $|y-z|\ge |\bar{x}-z| - |\bar{x}-y| \ge \frac{\delta }{2}$. Any power $s^{-\frac{k}{2}}$ can be absorbed by $e^{-\frac{|y-z|^2}{2Ks}}$ at the price $|y-z|^{-k}\le C\delta ^{-k}$ still leaving $e^{-\frac{|y-z|^2}{2Ks}}$ available. We note that if $d(y)<d(x)$ so that $d(y)^2<t$ is possible, then, in view of (23) we have $\frac{w_1(y)}{|y-z|d(y)}\le C\delta ^{-1}$. We also note that view of the fact that

$$\begin{aligned} |x-y|^2t^{-1}\le 2\left( \frac{t-s}{t}\left( \frac{|x-z|^2}{t-s}\right) + \frac{s}{t}\left( \frac{|y-z|^2}{s}\right) \right) \end{aligned}$$

we have a bound

$$\begin{aligned} e^{-\frac{|x-z|^2}{8(t-s)}-\frac{|y-z|^2}{2Ks}}\le e^{-\frac{|x-y|^2}{\tilde{K}t}} \end{aligned}$$

with $\tilde{K} = 16 + 4K$. Pulling this exponential out and estimating all the rest in terms of $\delta$ we obtain, in both cases, all the integrals bounded by $Ct\delta ^{-d-4}$ and therefore we have, in both cases,

$$\begin{aligned} |\nabla _x\nabla _x H_D(x,y,t)-\nabla _x\nabla _x G_{t}(x-y)|\le Ce^{-\frac{|x-y|^2}{\tilde{K}t}}t\delta ^{-d-4}\le Ct^{-1-\frac{d}{2}}e^{-\frac{|x-y|^2}{\tilde{K}t}} \end{aligned}$$

because $t\le c\delta ^2$. This proves (36).

For (37) and (38) we start by noticing that it is enough to prove the estimates

$$\begin{aligned} \int _{B(x, \frac{d(x)}{14})}|(\nabla _x+\nabla _y)H_D(x,y,t)|dy \le C t^{-\frac{1}{2}}e^{-\frac{d(x)^2}{Kt}} \end{aligned}$$

(171)

and

$$\begin{aligned} \int _{B(x, \frac{d(x)}{14})}|\nabla _x(\nabla _x+\nabla _y)H_D(x,y,t)|dy \le C t^{-1}e^{-\frac{d(x)^2}{Kt}} \end{aligned}$$

(172)

for $t<cd^2(x)$. Indeed, if $|x-y|\ge \frac{d(x)}{14}$, individual Gaussian upper bounds for up to two derivatives of $H_D$ suffice (there is no need for cancellations). In order to prove (171) and (172) we use a good cutoff $\chi$ with a scale $\ell = \frac{d(x)}{100}$. We take $y\in B(x, \frac{d(x)}{14})$. Both x and y are fixed for now. We note that the function

$$\begin{aligned} z\mapsto h(z) = \chi (z)G_t(z-y) \end{aligned}$$

solves

$$\begin{aligned} (\partial _t-\Delta )h(z,t) = -\left[ (\Delta \chi (z))G_t(z-y) +2(\nabla \chi (z))\cdot \nabla G_t(z-y)\right] = F(z,y,t), \end{aligned}$$

vanishes for $z\in \partial \Omega$, and has initial datum $h_0= \chi (z)\delta (z-y)$, so, by Duhamel

$$\begin{aligned} h(z,t) = e^{t\Delta }h_0 + \int _0^t e^{(t-s)\Delta }F(s)ds, \end{aligned}$$

which, in view of $(e^{t\Delta }f)(z) = \int _{\Omega }H_D(z,w,t)f(w)dw$ yields

$$\begin{aligned} \chi (z)G_t(z-y) = \chi (y)H_D(z,y,t) + \int _0^t\int _{\Omega }H_D(z,w,t-s)F(w,s)dwds \end{aligned}$$

for all z, and recalling that $\chi (x)=\chi (y) =1$, and reading at $z=x$ we have

$$\begin{aligned} H_D(x,y,t)= & {} G_t(x-y) + \int _0^t\int _{\Omega }H_D(x,w,t-s)\left[ \Delta \chi (w)G_s(w-y)\right. \nonumber \\&\left. +\,2\nabla \chi (w)\cdot \nabla G_s(w-y)\right] dwds. \end{aligned}$$

(173)

The right hand side integral is not singular and can be differentiated because the support of $\nabla \chi$ is far from the ball $B(x,\frac{d(x)}{14})$. Differentiation $\nabla _x+\nabla _y$ cancels the Gaussian $G_t(x-y)$. The estimates of the right hand side

$$\begin{aligned} \left| (\nabla _x+\nabla _y) \int _0^t \int _{\Omega }H_D(x,w,t-s)F(w,y,s)dwds\right| \le Ct^{-\frac{d+1}{2}}e^{-\frac{d(x)^2}{Kt}} \end{aligned}$$

and

$$\begin{aligned} \left| \nabla _x(\nabla _x+\nabla _y) \int _0^t \int _{\Omega }H_D(x,w,t-s)F(w,y,s)dwds\right| \le Ct^{-{\frac{d+2}{2}}}e^{-\frac{d(x)^2}{Kt}} \end{aligned}$$

for $t<cd^2(x)$ follow from Gaussian upper bounds. Integration dy on the ball $B(\frac{d(x)}{14})$ picks up the volume of the ball, and thus (171) and (172) are verified.

Appendix 2

We sketch here the proof of local wellposedness of the equation (7). We start by defining a Galerkin approximation. We consider the projectors $P_n$

$$\begin{aligned} P_n f= \sum _{j=1}^nf_jw_j \end{aligned}$$

(174)

with $f_j= \int _{\Omega }f(x)w_j(x)dx$. We consider for fixed n the approximate system

$$\begin{aligned} \partial _t \theta _n + P_n\left( u_n\cdot \nabla \theta _n\right) + \Lambda _D\theta _n = 0 \end{aligned}$$

(175)

where

$$\begin{aligned} u_n = \nabla ^{\perp }\Lambda _D^{-1}\theta _n = R_D^{\perp }\theta _n \end{aligned}$$

(176)

with

$$\begin{aligned} (P_n\theta _n)(x,t) = \theta _n(x,t) = \sum _{j=1}^n\theta _{n,j}(t)w_j(x) \end{aligned}$$

(177)

and with initial data $\theta _n(0) = P_n\theta _0$ where $\theta _0$ is a fixed smooth function belonging to $H_0^1(\Omega )\cap H^2(\Omega )$. Although it was written as a PDE, the system (175) is a system of ODEs for the coefficients $\theta _{n,j}(t)= \int _{\Omega }\theta _nw_jdx$. Let us note that $P_n$ does not commute with $\nabla$ but does commute with $-\Delta$ and functions of it. The function $u_n$ is divergence-free and it is a finite sum of divergence-free functions,

$$\begin{aligned} u_n(x) = \sum _{j=1}^n\lambda _j^{-\frac{1}{2}}\theta _{n,j}(t)\nabla ^{\perp }w_j(x). \end{aligned}$$

(178)

Note however that $u_n\notin P_nL^{2}(\Omega )$. The normal component of $u_n$ vanishes at the boundary because $\nabla ^{\perp }w_j\cdot \nu _{\left| \right. \Omega } =0$. Moreover, because

$$\begin{aligned} \int _{\Omega }P_n(u_n\cdot \nabla \theta _n)\theta _n dx = \int _{\Omega }(u_n\cdot \nabla \theta _n)\theta _ndx = 0 \end{aligned}$$

it follows that $\Vert \theta _n(t)\Vert _{L^2(\Omega )}$ is bounded in time and therefore the solution exists for all time. The following upper bound for higher norms is uniform only for short time, and it is the bound that is used for local existence of smooth solutions. We apply $\Lambda _D^2=-\Delta$ to (175) and use the fact that it is a local operator, it commutes with $P_n$ and with derivatives:

$$\begin{aligned} \partial _t\Lambda _D^2\theta _n + P_n\left( u_n\cdot \nabla \Lambda _D^2\theta _n -2\nabla u_n\nabla \nabla \theta _n + (\Lambda _D^2 u_n)\cdot \nabla \theta _n\right) + \Lambda _D^3\theta _n = 0 \end{aligned}$$

(179)

We take the scalar product with $\Lambda _D^2\theta _n$. Because this is finite linear combinations of eigenfunctions, it vansihes at $\partial \Omega$ and integration by parts is allowed. We obtain

$$\begin{aligned} \frac{d}{2dt}\Vert \Lambda _D^2\theta _n\Vert ^2_{L^2(\Omega )} + \Vert \Lambda _D^{\frac{5}{2}}\theta _n\Vert ^2_{L^2(\Omega )}\le & {} \Vert \Lambda _D^2u_n\Vert ^2_{L^2(\Omega )}\Vert \Lambda _D^2\theta _n\Vert _{L^2(\Omega )}\Vert \nabla \theta _n\Vert _{L^{\infty }(\Omega )}\nonumber \\&+\, 2\Vert \nabla u_n\Vert _{L^{\infty }(\Omega )}\Vert \nabla \nabla \theta \Vert _{L^2(\Omega )}\Vert \Lambda _D^2\theta \Vert _{L^2(\Omega )}\nonumber \\ \end{aligned}$$

(180)

We note now that

$$\begin{aligned} \Lambda _D^2u_n = \sum _{j=1}^n\theta _{n,j}(-\Delta )\lambda _j^{-\frac{1}{2}}\nabla ^{\perp }w_j= \nabla ^{\perp }\Lambda _D^{-1}(\Lambda _D^2\theta _n) = R_D^{\perp }(\Lambda _D^2\theta _n). \end{aligned}$$

(181)

Now $R_D$ is bounded in $L^2(\Omega )$ (It is in fact an isometry on components; this follows from (27)), therefore

$$\begin{aligned} \Vert \Lambda _D^2 u_n\Vert _{L^2{(\Omega )}} \le \Vert \Lambda _D^2\theta _n\Vert _{L^2(\Omega )}. \end{aligned}$$

(182)

The fact that $R_D$ is bounded in $L^4(\Omega )$ is also true [19]. Then

$$\begin{aligned} \Vert \Lambda _D^2 u_n\Vert _{L^4{(\Omega )}} \le \Vert \Lambda _D^{2}\theta _n\Vert _{L^4(\Omega )}. \end{aligned}$$

(183)

Moreover, it is known (see for instance [1]) that in $d=2$ we have

$$\begin{aligned} \Vert f\Vert _{L^4(\Omega )} \le C\Vert \Lambda _D^{\frac{1}{2}}f\Vert _{L^2(\Omega )} \end{aligned}$$

and therefore

$$\begin{aligned} \Vert \Delta \theta _n\Vert _{L^{4}(\Omega )}\le \Vert \Lambda _D^{\frac{5}{2}} \theta _n\Vert _{L^2{(\Omega )}}. \end{aligned}$$

(184)

and

$$\begin{aligned} \Vert \Delta u_n\Vert _{L^{4}(\Omega )}\le C\Vert \Lambda _D^{\frac{5}{2}} \theta _n\Vert _{L^2{(\Omega )}}. \end{aligned}$$

(185)

Now we use the Sobolev embedding

$$\begin{aligned} \Vert \nabla \phi \Vert _{L^{\infty }(\Omega )}\le C\left( \Vert \Delta \phi \Vert _{L^{4}(\Omega )} + \Vert \phi \Vert _{L^2(\Omega )}\right) \end{aligned}$$

(186)

and deduce, using also a Poincaré inequality

$$\begin{aligned} \frac{d}{dt}\Vert \Lambda _D^2\theta _n\Vert _{L^2(\Omega )}^2 + \Vert \Lambda _D^{\frac{5}{2}}\theta _n\Vert ^2_{L^2(\Omega )} \le C\Vert \Lambda _D^2\theta _n\Vert _{L^2(\Omega )}^2\Vert \Lambda _D^{\frac{5}{2}}\theta \Vert _{L^2(\Omega )}. \end{aligned}$$

(187)

Thus, after a Young inequality we deduce that

$$\begin{aligned} \sup _{t\le T} \Vert \Lambda _D^2\theta _n\Vert _{L^2(\Omega )}^2 + \int _0^T\Vert \Lambda _D^{\frac{5}{2}}\theta _n\Vert ^2_{L^2(\Omega )}dt \le C\Vert \Lambda _D^2\theta _0\Vert _{L^2(\Omega )}^2 \end{aligned}$$

(188)

holds for T depending only on $\Vert \Lambda _D^2\theta _0\Vert _{L^2(\Omega )}$, with a constant independent of n. The following result can now be obtained by assing to the limit in a subsequence and using a Aubin-Lions lemma [21]:

Proposition 4

Let $\theta _0\in H_0^1(\Omega )\cap H^{2}(\Omega )$ in $d=2$. There exists $T>0$ a unique solution of (7) with initial datum $\theta _0$ satisfying

$$\begin{aligned} \theta \in L^{\infty }(0,T; H_0^1(\Omega )\cap H^{2}(\Omega ))\cap L^2\left( 0,T; {\mathcal {D}}\left( \Lambda _D^{2.5}\right) \right) . \end{aligned}$$

(189)

Higher regularity can be obtained as well. Because the proof uses $L^2$- based Sobolev spaces and Sobolev embedding, it is dimension dependent. A proof in higher dimensions is also possible, but it requires using higher powers of $\Delta$, and will not be pursued here.