Higher-order boundary regularity estimates for nonlocal parabolic equations

Abstract

We establish sharp higher-order Hölder regularity estimates up to the boundary for solutions to equations of the form $\partial _tu-Lu=f(t,x)$ in $I\times \Omega$ where $I\subset \mathbb {R}$, $\Omega \subset \mathbb {R}^n$ and f is Hölder continuous. The nonlocal operators L that we consider are those arising in stochastic processes with jumps, such as the fractional Laplacian $(-\Delta )^s$, $s\in (0,1)$. Our main result establishes that, if f is $C^\gamma$ is space and $C^{\gamma /2s}$ in time, and $\Omega$ is a $C^{2,\gamma }$ domain, then $u/d^s$ is $C^{s+\gamma }$ up to the boundary in space and u is $C^{1+\gamma /2s}$ up the boundary in time, where d is the distance to $\partial \Omega$. This is the first higher order boundary regularity estimate for nonlocal parabolic equations, and is new even for the fractional Laplacian in $C^\infty$ domains.

Appendix

In this Appendix we prove Proposition 4.1. The proof will follow the same compactness argument as the boundary regularity. We start with the following Liouville type result:

Lemma 5.1

Let $s\in (0,1)$, $\beta \in (0,1)$ and $\alpha \in (0,s)$ satisfying $\alpha +\beta <2s$. Let w satisfy

$$\begin{aligned} (\partial _t-L)(w(\cdot +\tau ,\cdot +h)-w(\cdot ,\cdot )) = 0 \,\text { in }\, (-\infty ,0)\times \mathbb {R}^n \end{aligned}$$

for all $h\in \mathbb {R}^n$ and $\tau <0$. Assume also that

$$\begin{aligned}{}[w]_{C_{t,x}^{\frac{\beta }{2s},\beta }(Q_R)}\le CR^{\alpha +s} \end{aligned}$$

(5.1)

for all $R\ge 1$. Then

$$\begin{aligned} w(t,x)=at+x^TAx+p\cdot x+q \end{aligned}$$

for some $p\in \mathbb {R}^n$ and $a,q\in \mathbb {R}$. Moreover, if $\alpha +\beta <s$ then $a=0$, if $\alpha +\beta +s<2$ then $A=0$, and if $\alpha +\beta +s<1$ then $p=0$.

Proof

Let $\rho >0$ and define

$$\begin{aligned} v(t,x)=\frac{w(\rho ^{2s}(t+\tau ),\rho (x+h))-w(\rho ^{2s}t,\rho x)}{\rho ^{\alpha +s}(|\rho \tau |^{\beta /2s}+|\rho h|^{\beta })}. \end{aligned}$$

Note that

$$\begin{aligned} \partial _tv-Lv=0 \end{aligned}$$

in $Q_1$ and, by (5.1),

$$\begin{aligned} \Vert v\Vert _{L^{\infty }(Q_R)}\le CR^{\alpha +s} \end{aligned}$$

for all $R\ge 1$ and in particular

$$\begin{aligned} \Vert v\Vert _{L^{\infty }(Q_1)}\le C. \end{aligned}$$

Then, by the interior estimates of Theorem 1.3 in [13] with can get, for $\theta >\alpha +s$

$$\begin{aligned} \Vert v\Vert _{C_{t,x}^{\theta /2s,\theta }(Q_{1/2})}\le C. \end{aligned}$$

Hence, the incremental quotients of order $\beta$ of w are uniformly bounded in $C_{t,x}^{\theta /2s,\theta }$. This implies (see [8], Lemma 5.6) $w\in C_{t,x}^{\frac{\theta +\beta }{2s},\theta +\beta }$ and scaling back we get

$$\begin{aligned}{}[w]_{C_{t,x}^{\frac{\alpha +\beta +s}{2s},\alpha +\beta +s}(Q_\rho )}\le [w]_{C_{t,x}^{\frac{\theta +\beta }{2s},\theta +\beta }(Q_\rho )}\le \rho ^{\alpha +s-\theta } \end{aligned}$$

and when we let $\rho \rightarrow \infty$ we get

$$\begin{aligned}{}[w]_{C_{t,x}^{\frac{\alpha +\beta +s}{2s},\alpha +\beta +s}((-\infty ,0)\times \mathbb {R}^n)}=0. \end{aligned}$$

This gives the desired result. $\square$

The next proposition is the main tool to prove Proposition 4.1.

Proposition 5.2

Let $s\in (0,1)$, $\gamma \in (0,s)$, $\beta \in (\gamma ,1)$ and $\alpha =s+\gamma -\beta \in (0,s)$. Let w be a solution of

$$\begin{aligned} \partial _tw-Lw=f\quad \text { in } Q_1 \end{aligned}$$

with L an operator of the form (1.2). Assume $w\in C^{\frac{\beta }{2s},\beta }_{t,x}((-\infty ,0)\times \mathbb {R}^n)$, $a\in C^{1,\gamma }(S^{n-1})$ and $f\in C^{\frac{\gamma }{2s},\gamma }_{t,x}(Q_1)$. Then

$$\begin{aligned} \sup _{r>0} r^{-\alpha -s}[w-(a_rt+x^TA_r x+p_r\cdot x+q_r)]_{C_{t,x}^{\frac{\beta }{2s},\beta }(Q_r)}\le C [w]_{C_{t,x}^{\frac{\beta }{2s},\beta }((-\infty ,0)\times \mathbb {R}^n)} \end{aligned}$$

(5.2)

where

$$\begin{aligned} a_rt+x^TA_r x+p_r\cdot x+q_r=\arg \min _{\mathcal {P}}\int _{-r^{2s}}^0\int _{B_r}(w-(at+x^TAx+p\cdot x+q))^2\,dxdt. \end{aligned}$$

(5.3)

where $\mathcal {P}$ is the space of all polynomials of degree at most $\lfloor (\alpha +\beta +s)/2s\rfloor$ in t and $\lfloor \alpha +\beta +s\rfloor$ in x (i.e., and $a_r=0$ if $\alpha +\beta <s$, $A_r=0$ if $\alpha +\beta +s<2$, and $p_r=0$ if $\alpha +\beta +s<1$).

In particular

$$\begin{aligned}{}[w]_{C_{t,x}^{\frac{\alpha +\beta +s}{2s},\alpha +\beta +s}(Q_{1/2})}\le C[w]_{C_{t,x}^{\frac{\beta }{2s},\beta }((-\infty ,0)\times \mathbb {R}^n)}. \end{aligned}$$

(5.4)

Proof

The proof is similar to that of Proposition 3.3. Assume (5.2) does not hold. Then there are sequences $w_k$, $L_k$ and $f_k$ satisfying:

• $L_k$ is of the form (1.2) and $a_k\in C^{1,\gamma }(S^{n-1})$

• $w_k\in C_{t,x}^{\beta /2s,\beta }((-\infty ,0)\times \mathbb {R}^n)$ and $f_k\in C_{t,x}^{\gamma /2s,\gamma }(Q_1)$

• $w_k$ is a solution of

  $$\begin{aligned} \partial _tw_k-L_kw_k=f_k\quad \text { in } Q_1 \end{aligned}$$

  (5.5)

such that

$$\begin{aligned} \sup _k\,\sup _{r>0}\,r^{-\alpha -s}[w_k-(a_{r,k}t+x^TA_{r,k}x+p_{r,k}\cdot x+q_{r,k})]_{C_{t,x}^{\beta /2s,\beta }(Q_r)}=\infty . \end{aligned}$$

with

$$\begin{aligned} a_{r,k}t+x^TA_{r,k}x+p_{r,k}\cdot x+q_{r,k}=\arg \min _{\mathcal {P}}\int _{-r^{2s}}^0\int _{B_r}(w_k-(at+x^TAx+p\cdot x+q))^2\,dxdt. \end{aligned}$$

Define

$$\begin{aligned} \theta (r):=\sup _k\,\sup _{r'>r} (r')^{-\alpha -s}[w_k-(a_{r',k}t+x^TA_{r',k}x+p_{r',k}\cdot x+q_{r',k})]_{C_{t,x}^{\frac{\beta }{2s},\beta }(Q_{r'})}. \end{aligned}$$

As before, $\theta$ is a nondecreasing function of r that goes to $\infty$ as r goes to 0. Moreover, since for any fixed r we have $\theta (r)<\infty$, if we take the sequence 1 / m there are $r'_m$ and $k_m$ such that $r'_m\ge 1/m$ and

$$\begin{aligned} (r_m')^{-\alpha -s}[w_{k_m}-(a_{r_m',k_m}t+x^TA_{r_m',k_m}x+p_{r_m',k_m}\cdot x+q_{r_m',k_m})]_{C_{t,x}^{\beta /2s,\beta }(Q_{r_m'})}\ge \frac{1}{2}\theta (1/m)\ge \frac{1}{2}\theta (r'_m). \end{aligned}$$

Let’s denote

$$\begin{aligned} w_m=w_{k_m} \quad A_m=A_{r_m',k_m} \quad p_m=p_{r_m',k_m} \quad q_m=q_{r_m',k_m} \quad a_m=a_{r_m',k_m} \end{aligned}$$

to make the notation cleaner, and further denote

$$\begin{aligned} P_{r,m}(t,x)=a_mr^{2s}t+x^TA_mxr^2+p_m\cdot rx+q_m. \end{aligned}$$

With this notation we define the blow-up sequence

$$\begin{aligned} v_m(t,x)=\frac{w_m((r'_m)^{2s}t,r'_mx)-P_{r'_m,m}(t,x)}{(r'_m)^{\alpha +\beta +s}\theta (r'_m)}. \end{aligned}$$

Notice that

$$\begin{aligned}{}[v_m]_{C_{t,x}^{\beta /2s,\beta }(Q_1)}\ge 1/2 \end{aligned}$$

(5.6)

and that

$$\begin{aligned} \int _{-1}^0\int _{B_1}v_m(at+x^TAx+p\cdot x+q)\,dx\,dt=0 \end{aligned}$$

(5.7)

for all $A\in \mathbb {R}^{n\times n}$, $p\in \mathbb {R}^n$ and $a,q\in \mathbb {R}$ because of the minimization condition (5.3).

Next we show that

$$\begin{aligned}{}[v_m]_{C^{\beta /2s,\beta }(Q_R)}\le CR^{\alpha +s} \end{aligned}$$

(5.8)

for any $R\ge 1$. As before, this requires an estimate of the form

$$\begin{aligned}{}[P_{Rr'_m,m}-P_{r'_m,m}]_{C_{t,x}^{\beta /2s,\beta }(Q_{Rr'_m})}\le \theta (r'_m)(Rr_m')^{\alpha +s}. \end{aligned}$$

(5.9)

The only difference between the proof (5.9) and (3.10) is the presence of a quadratic term so, to keep the presentation clear, let us assume for simplicity that the linear terms vanish and prove the bound for the quadratic term.

Hence, proceeding dyadically as in Proposition 3.3, we have to show that

$$\begin{aligned}{}[x^TA_{2^kr'_m,m}x-x^TA_{r'_m,m}x]_{C_{t,x}^{\beta /2s,\beta }(Q_{Rr'_m})}\le C\theta (r'_m)(Rr_m')^{\alpha +s}. \end{aligned}$$

Let us start with $k=1$ and noticing that

$$\begin{aligned}&\frac{1}{(r'_m)^{\alpha +s}\theta (r'_m)}|x^T(A_{2r'_m,m}-A_{r'_m,m})x|(r'_m)^{2-\beta }\\&\quad \le \frac{1}{(r'_m)^{\alpha +s}\theta (r'_m)}[x^TA_{2r'_m,m}x-x^TA_{r'_m,m}x]_{C_{t,x}^{\beta /2s,\beta }(Q_{r'_m})}\\&\quad \le \frac{2^{\alpha +s}\theta (2r'_m)}{\theta (r'_m)}\frac{[w_m-x^TA_{2r'_m,m}x]_{C_{t,x}^{\beta /2s,\beta }(Q_{2r'_m})}}{(2r'_m)^{\alpha +s}\theta (2r'_m)}\\&\qquad \frac{1}{(r'_m)^{\alpha +s}\theta (r'_m)}[w_m-x^TA_{r'_m,m}x]_{C_{t,x}^{\beta /2s,\beta }(Q_{r'_m})} \\&\quad \le C \end{aligned}$$

so that, denoting by $\Vert \cdot \Vert$ is the $L^2$ matrix norm, i.e. $\Vert M\Vert =\sup _{|x|=1}|x^TMx|$ and taking supremum over x

$$\begin{aligned} \Vert A_{2r'_m,m}-A_{r'_m,m}\Vert \le C(r'_m)^{\alpha +\beta +s-2}\theta (r'_m). \end{aligned}$$

But on the other hand,

$$\begin{aligned}{}[x^TA_{2r'_m,m}x-x^TA_{r'_m,m}x]_{C_{t,x}^{\beta /2s,\beta }(Q_{2r'_m})}\le C\Vert A_{2r'_m,m}-A_{r'_m,m}\Vert (r'_m)^{2-\beta } \end{aligned}$$

so the result follows in this case. For $R=2^k$ just use a telescopic sum as in Proposition 3.3.

Using (5.9) we have

$$\begin{aligned}{}[v_m]_{C_{t,x}^{\beta /2s,\beta }(Q_R)}= & {} \frac{1}{(r'_m)^{\alpha +s}\theta (r'_m)}[w_m-P_{r'_m,m}]_{C_{t,x}^{\beta /2s, \beta }(Q_{Rr'_m})}\\= & {} \frac{R^{\alpha +s}}{(Rr'_m)^{\alpha +s}\theta (r'_m)} [w_m-P_{r'_m,m}]_{C_{t,x}^{\beta /2s,\beta }(Q_{Rr'_m})} \\\le & {} \frac{R^{\alpha +s}}{(Rr'_m)^{\alpha +s}\theta (r'_m)} [w_m-P_{Rr'_m,m}]_{C_{t,x}^{\beta /2s,\beta }(Q_{Rr'_m})}\\&+\,\frac{R^{\alpha +s}}{(Rr'_m)^{\alpha +s}\theta (r'_m)}[P_{Rr'_m,m} -P_{r'_m,m}]_{C_{t,x}^{\beta /2s,\beta }(Q_{Rr'_m})} \\\le & {} \frac{R^{\alpha +s}\theta (Rr'_m)}{\theta (r'_m)}+C\frac{R^{\alpha +s}}{(Rr'_m)^{\alpha +s}\theta (r'_m)}(Rr'_m)^{\alpha +s}\theta (r'_m)\\\le & {} CR^{\alpha +s} \end{aligned}$$

and we get (5.8). From (5.8) the bound

$$\begin{aligned} \Vert v_m\Vert _{L^\infty {(Q_R)}}\le CR^{\alpha +\beta +s} \end{aligned}$$

(5.10)

follows as in Proposition 3.3.

By the Arzelà–Ascoli theorem, (5.8) and (5.10) imply that a subsequence of $\{v_m\}_m$ converges uniformly on compact subsets of $(-\infty ,0)\times \mathbb {R}^n$ to a uniformly continuous function v. Let’s check that v satisfies the hypotheses of Lemma 5.1. As in Proposition 3.3 we have

$$\begin{aligned} (\partial _t-L_{k_m})(v_m(t+\tau ,x+h)-v_m(t,x))\longrightarrow 0 \end{aligned}$$

as $m\rightarrow \infty$ (notice that the quadratic term can only appear when $s>1/2$ and in this case the operator is well defined on linear functions and vanishes identically). So, again using Lemma 3.1

$$\begin{aligned} (\partial _t-L)(v(t+\tau ,x+h)-v(t,x))=0 \end{aligned}$$

for some L of the form (1.2). Finally, because of (5.8), v also satisfies the growth condition of Lemma 5.1. Hence

$$\begin{aligned} v(t,x)=at+x^TAx+p\cdot x+q, \end{aligned}$$

and by (5.7) we obtain $v\equiv 0$. But passing to the limit in (5.6) we get a contradiction, so (5.2) holds. The fact that (5.2) implies (5.4) is quite standard. $\square$

Proof of Proposition 4.1

Fix $r>0$ and consider the following rescaling of w:

$$\begin{aligned} \tilde{w}(t,x)=\frac{1}{r^{\alpha +\beta +s}}w(r^{2s}t,rx). \end{aligned}$$

Then $\tilde{w}$ will satisfy

$$\begin{aligned} \Vert \tilde{w}\Vert _{L^\infty ((-\infty ,0)\times \mathbb {R}^n)}<\infty \quad \text { and } \Vert \tilde{w}\Vert _{L^\infty (Q_R)}\le CR^{\alpha +\beta +s}. \end{aligned}$$

Let now $\eta$ be a cut-off function in space that vanishes outside $B_1(e_n)$. More precisely, let $\eta \in C^\infty _c(B_1(e_n))$ with $\eta \equiv 1$ in $B_{5/6}(e_n)$. Notice that $\tilde{w}\eta$ satisfies an equation like the one in Proposition 5.2. Indeed, because $a\in C^{1,\gamma }(S^{n-1})$ we have that,

$$\begin{aligned} \partial _t\tilde{w\eta }-L\tilde{w}\eta =\tilde{f}\quad \text { in }\quad (-1,0)\times B_{3/4}(e_n) \end{aligned}$$

and hence (5.4) gives (recall $\tilde{w}=\tilde{w}\eta$ in $B_{1/2}$)

$$\begin{aligned}{}[\tilde{w}]_{C_{t,x}^{\frac{\alpha +\beta +s}{2s},\alpha +\beta +s}((-1,0)\times B_{1/2}(e_n))}\le C[\tilde{w}]_{C_{t,x}^{\frac{\beta }{2s},\beta }((-1,0)\times B_1(e_n))}, \end{aligned}$$

and in particular also

$$\begin{aligned}{}[\tilde{w}]_{C_{t,x}^{\frac{\alpha +\beta }{2s},\alpha +\beta }((-1,0)\times B_{1/2}(e_n))}\le C[\tilde{w}]_{C_{t,x}^{\frac{\beta }{2s},\beta }((-1,0)\times B_1(e_n))}. \end{aligned}$$

Rescaling this estimate back we get

$$\begin{aligned}{}[w]_{C_{t,x}^{\frac{\alpha +\beta +s}{2s},\alpha +\beta +s}((-r^{2s},0)\times B_{r/2}(e_n))}\le r^{-\alpha -s}C[w]_{C_{t,x}^{\frac{\beta }{2s},\beta }((-1,0)\times B_r(e_n))}\le Cr^{-\alpha -s}r^{s+\alpha }=C \end{aligned}$$

and

$$\begin{aligned}{}[w]_{C_{t,x}^{\frac{\alpha +\beta }{2s},\alpha +\beta }((-r^{2s},0)\times B_{r/2}(e_n))}\le r^{-\alpha }C[w]_{C_{t,x}^{\frac{\beta }{2s},\beta }((-1,0)\times B_r(e_n))}\le Cr^{-\alpha }r^{s+\alpha }=Cr^s, \end{aligned}$$

as wanted.$\square$