Existence of $C^\alpha$ solutions to integro-PDEs

Abstract

This paper is concerned with existence of a $C^{\alpha }$ viscosity solution of a second order non-translation invariant integro-PDE. We first obtain a weak Harnack inequality for such integro-PDE. We then use the weak Harnack inequality to prove Hölder regularity and existence of solutions of the integro-PDEs.

Appendix

In this “Appendix”, we give the proof of Theorem 3.4.

Lemma A.1

Let $\Omega$ be a bounded domain and let $\{\Omega _j\}_{j=1}^{\infty }$ be a set of domains such that $\Omega _j\subset \Omega _{j+1}$ and $\cup _{j=1}^{\infty }\Omega _j=\Omega$. Let $u_j$ be a continuous function defined on ${\mathbb {R}}^d$ such that $u_j$ converges uniformly to a continuous function u in ${\widetilde{\Omega }}_{\mathrm{diam}\Omega }$. Then

$$\begin{aligned} \limsup _{j\rightarrow \infty }\Gamma _{\Omega _j,r}^{n,+}(u_j)\subset \Gamma _{\Omega ,r}^{n,+}(u) \end{aligned}$$

where

$$\begin{aligned} \Gamma _{\Omega ,r}^{n,+}(u):=\{x\in \Omega :\,\, \exists p,|p|\le r,\, \text {such that}\,\, u(y)\le u(x)+ \langle p, y-x \rangle \,\,\text {for}\,\, y\in {{\tilde{\Omega }}}_{\mathrm{diam}\Omega }\}. \end{aligned}$$

Proof

The proof is very similar to Lemma A.1 in [3]. $\square$

For any $\epsilon >0$ and $u:{\mathbb {R}}^d\rightarrow {\mathbb {R}}$, we define the sup-convolution of u by

$$\begin{aligned} u^\epsilon (x)=\sup _{y\in {\mathbb {R}}^d}\left\{ u(y)-\frac{|x-y|^2}{2\epsilon }\right\} . \end{aligned}$$

The following Lemma A.2 can be found in [3] and [11].

Lemma A.2

Let u be a bounded continuous function in ${\mathbb {R}}^d$. Then

• (i) $u^\epsilon \rightarrow u$ as $\epsilon \rightarrow 0$ uniformly in any compact set of ${\mathbb {R}}^d$;

• (ii) $u^\epsilon$ has the Taylor expansion up to second order at a.e. $x\in {\mathbb {R}}^d$, i.e.

  $$\begin{aligned} u^\epsilon (y)= & {} u^\epsilon (x)+Du^\epsilon (x)\cdot (y-x)+\frac{1}{2}D^2u^\epsilon (x)(y-x)\cdot (y-x)\\&+o(|x-y|^2)\quad \text {a.e. }x\in {\mathbb {R}}^d; \end{aligned}$$

• (iii) $D^2u^\epsilon (x)\ge -\frac{1}{\epsilon }I$ a.e. in ${\mathbb {R}}^d$.

• (iv) If $u_\delta ^\epsilon$ is a standard modification of $u^\epsilon$, then $D^2u_\delta ^\epsilon \ge -\frac{1}{\epsilon }I$ and

  $$\begin{aligned} D^2u_\delta ^\epsilon (x)\rightarrow D^2u^\epsilon (x)\quad \text {a.e. in }{\mathbb {R}}^d\text { as }\delta \rightarrow 0. \end{aligned}$$

Lemma A.3

Let $\Omega$ be a bounded domain. Let u be a bounded continuous function and solve

$$\begin{aligned} -{\mathcal {P}}^+(D^2u)(x)-{\mathcal {P}}_{K}^+(u)(x)-C_0|Du(x)|\le f(x)\quad \text {in }\Omega \end{aligned}$$

(A.1)

in the viscosity sense, then

$$\begin{aligned} -{\mathcal {P}}^+(D^2u^{\epsilon })(x)-{\mathcal {P}}_{K}^+(u^\epsilon )(x)-C_0|Du^\epsilon (x)|\le f(x^\epsilon ),\quad \text {a.e in }\Omega _{2\left( \epsilon \Vert u\Vert _{L^\infty ({\mathbb {R}}^d)}\right) ^{\frac{1}{2}}}\nonumber \\ \end{aligned}$$

(A.2)

where $x^\epsilon \in \Omega$ is any point such that

$$\begin{aligned} u^{\epsilon }(x):=\sup _{y\in {\mathbb {R}}^d}\left\{ u(y)-\frac{|y-x|^2}{2\epsilon } \right\} =u(x^\epsilon )-\frac{|x^\epsilon -x|^2}{2\epsilon }. \end{aligned}$$

(A.3)

Proof

Suppose that x is any point in $\Omega _{2\left( \epsilon \Vert u\Vert _{L^\infty ({\mathbb {R}}^d)}\right) ^{\frac{1}{2}}}$ at which $u^\epsilon$ has the taylor expansion up to second order. For each $\delta >0$, there exists $\varphi _\delta \in C_b^2({\mathbb {R}}^d)$ such that $\varphi _\delta$ touches $u^\epsilon$ from above at x,

$$\begin{aligned} \varphi _\delta (y)= & {} u^\epsilon (x)+Du^\epsilon (x)\cdot (y-x)+\frac{1}{2}\left( D^2u^\epsilon (x)+\delta I\right) (y-x)\cdot (y-x)\\&+o(|x-y|^2)\quad \text {a.e. }x\in {\mathbb {R}}^d \end{aligned}$$

and $\varphi _\delta \rightarrow u^\epsilon$ as $\delta \rightarrow 0$ a.e. in ${\mathbb {R}}^d$. Let $x^\epsilon$ be the one in (A.3). It is standard to obtain that $x^\epsilon \in \Omega$ and u is touched from above at $x^\epsilon$ by $\varphi _\delta (\cdot -x^\epsilon +x)$. Therefore

$$\begin{aligned}&-{\mathcal {P}}^+\left( D^2\varphi _\delta (\cdot -x^\epsilon +x)\right) (x^\epsilon ) -{\mathcal {P}}_{K}^+\left( \varphi _\delta (\cdot -x^\epsilon +x)\right) (x^\epsilon )\nonumber \\&\quad -C_0|D\varphi _\delta (\cdot -x^\epsilon +x)(x)|\le f(x^\epsilon ). \end{aligned}$$

(A.4)

(A.2) follows from letting $\delta \rightarrow 0$ in (A.4). $\square$

Theorem A.4

Let $\Omega$ be a bounded domain in ${\mathbb {R}}^d$ and $f\in L^d(\Omega )\cap C(\Omega )$. Then there exists a constant C such that, if u solves (A.1) in the viscosity sense, then

$$\begin{aligned} \sup _{\Omega } u\le \sup _{\Omega ^c}u+C\mathrm{diam}(\Omega )\Vert f^+\Vert _{L^d(\Gamma _\Omega ^{n,+}(u^+))}. \end{aligned}$$

(A.5)

Proof

By Theorem 3.1 in [38], we know that (A.5) holds if $u\in C^2(\Omega )\cap C_b({\mathbb {R}}^d)$. Using Lemma A.3, $u^\epsilon$ satisfies

$$\begin{aligned} -{\mathcal {P}}^+(D^2u^{\epsilon })(x)-{\mathcal {P}}_{K}^+(u^\epsilon )(x)-C_0|Du^\epsilon (x)|\le f_\epsilon (x),\quad \text {a.e in }\Omega _{2\left( \epsilon \Vert u\Vert _{L^\infty ({\mathbb {R}}^d)}\right) ^{\frac{1}{2}}} \end{aligned}$$

where

$$\begin{aligned} f_\epsilon (x)=\sup _{B_{2\left( \epsilon \Vert u\Vert _{L^\infty ({\mathbb {R}}^d)} \right) ^{\frac{1}{2}}}(x)}f(y). \end{aligned}$$

Because $u^\epsilon \rightarrow u$ as $\epsilon \rightarrow 0$ uniformly in any compact set of ${\mathbb {R}}^d$, if $r<r_0(u)$ and $\epsilon$ is sufficiently small where

$$\begin{aligned} r_0(u):=\frac{\sup _\Omega u-\sup _{\Omega ^c}u}{2d}, \end{aligned}$$

then $r<r_0(u^\epsilon )$ and $\Gamma _{\Omega ,r}^{n,+}(u^\epsilon )$ remains in a fixed compact subset of $\Omega$. Let $u_\delta ^\epsilon$ be a standard mollification of $u^\epsilon$. It follows from the proof of Theorem 3.1 in [38], for $\kappa \ge 0$ and small $\delta$, we have

$$\begin{aligned} \int _{B_r}\left( |p|^{\frac{d}{d-1}}+\kappa ^{\frac{d}{d-1}}\right) ^{1-d}dp\le \int _{\Gamma _{\Omega ,r}^{n,+}(u_\delta ^\epsilon )}\left( |Du_\delta ^\epsilon |^{\frac{d}{d-1}}+\kappa ^{\frac{d}{d-1}}\right) ^{1-d}\left( \frac{-\mathrm{Tr}(D^2u_\delta ^\epsilon )}{d}\right) ^ddx.\nonumber \\ \end{aligned}$$

(A.6)

Since $-\frac{1}{\epsilon }I\le D^2u_\delta ^\epsilon \le 0$ in $\Gamma _{\Omega ,r}^{n,+}(u_\delta ^\epsilon )$, the bounded convergence theorem combining with Lemma A.1, Lemma A.2(iv) implies (A.6) holds with $u^\epsilon$ in place of $u_\delta ^\epsilon$ by taking $\delta \rightarrow 0$. Then the arguments in Theorem 3.1 in [38] remain unchanged to obtain

$$\begin{aligned} r\le & {} \left( \mathrm{exp}\left( \frac{2^{d-2}}{|B_1|d^{d}}\left( 1+\int _{\Gamma _{\Omega ,r}^{n,+}(u^\epsilon )}\frac{[\gamma +C_2(1+\mathrm{diam}(\Omega )^{-1})]^d}{\lambda ^d}dx\right) \right) -1\right) ^{\frac{1}{d}}\nonumber \\&\frac{\Vert f_\epsilon ^+\Vert _{L^d(\Gamma _{\Omega ,r}^{n,+}(u^\epsilon ))}}{\lambda } \end{aligned}$$

(A.7)

where $C_2\ge 0$ depends on $\mathrm{diam}(\Omega )$ and K. Then the result follows from letting $\epsilon \rightarrow 0$ in (A.7). $\square$

Proof of Theorem 3.4

The result follows from applying Theorem A.4 to $-u$. $\square$