Remarks on the Vanishing Viscosity Process of State-Constraint Hamilton–Jacobi Equations

Abstract

We investigate the convergence rate in the vanishing viscosity process of the solutions to the subquadratic state-constraint Hamilton–Jacobi equations. We give two different proofs of the fact that, for non-negative Lipschitz data that vanish on the boundary, the rate of convergence is ${\mathcal {O}}(\sqrt{\varepsilon })$ in the interior. Moreover, the one-sided rate can be improved to ${\mathcal {O}}(\varepsilon )$ for non-negative compactly supported data and ${\mathcal {O}}(\varepsilon ^{1/p})$ (where $1<p<2$ is the exponent of the gradient term) for non-negative data $f\in \mathrm {C}^2({\overline{\varOmega }})$ such that $f = 0$ and $Df = 0$ on the boundary. Our approach relies on deep understanding of the blow-up behavior near the boundary and semiconcavity of the solutions.

Appendix

1.1 A.1 Estimates on Solutions

We present here a proof for the gradient bound of the solution to (PDE$_\varepsilon$) using Bernstein’s method (see also [21, 23]). Another proof using Berstein’s method inside a doubling variable argument is given in [1].

Proof of Theorem 4

Let $\theta \in (0,1)$ be chosen later, $\varphi \in \mathrm {C}_c^\infty (\varOmega )$, $0\le \varphi \le 1$, $\mathrm {supp}\;\varphi \subset \varOmega$ and $\varphi = 1$ on $\varOmega _\delta$ such that

$$\begin{aligned} \left|\varDelta \varphi (x)\right|\le C\varphi ^\theta \qquad \text {and}\qquad \left|D \varphi (x)\right|^2 \le C\varphi ^{1+\theta },\quad \forall x\in \varOmega , \end{aligned}$$

(66)

where $C = C(\delta ,\theta )$ is a constant depending on $\delta ,\theta$.

Define $w(x) := \left|Du^\varepsilon (x)\right|^2$ for $x \in \varOmega$. The equation for w is given by

$$\begin{aligned} -\varepsilon \varDelta w + 2 p\left|D u^\varepsilon \right|^{p-2}D u^\varepsilon \cdot D w + 2 w - 2 D f\cdot D u^\varepsilon + 2 \varepsilon \left|D^2u^\varepsilon \right|^2 = 0 \qquad \text {in}\;\varOmega . \end{aligned}$$

Then an equation for $(\varphi w)$ can be derived as follows.

$$\begin{aligned}&-\varepsilon \varDelta (\varphi w) + 2 p\left|D u^\varepsilon \right|^{p-2}D u^\varepsilon \cdot D (\varphi w) + 2 (\varphi w) + 2 \varepsilon \varphi \left|D^2u^\varepsilon \right|^2 + 2\varepsilon \frac{D \varphi }{\varphi }\cdot D (\varphi w) \\&\quad = \varphi (D f\cdot D u^\varepsilon ) + 2 p\left|D u^\varepsilon \right|^{p-2}(D u^\varepsilon \cdot D \varphi )w -\varepsilon w \varDelta \varphi + 2\varepsilon \frac{\left|D \varphi \right|^2}{\varphi }w \qquad \text {in}\;\mathrm {supp}\;\varphi . \end{aligned}$$

Assume that $\varphi w$ achieves its maximum over ${\overline{\varOmega }}$ at $x_0\in \varOmega$. And we can further assume that $x_0\in \mathrm {supp}\;\varphi$, since otherwise the maximum of $\varphi w$ over ${\overline{\varOmega }}$ is zero. By the classical maximum principle,

$$\begin{aligned} -\varepsilon \varDelta (\varphi w)(x_0)\ge 0 \qquad \text {and}\qquad \left|D(\varphi w)(x_0)\right|= 0. \end{aligned}$$

Use this in the equation of $\varphi w$ above to obtain

$$\begin{aligned} \varepsilon \varphi \left|D^2u^\varepsilon \right|^2 \le \varphi (Df\cdot Du^\varepsilon )+ 2 p\left|Du^\varepsilon \right|^{p-1} \left|D\varphi \right|w + \varepsilon w \left|\varDelta \varphi \right|+ 2\varepsilon w\frac{\left|D\varphi \right|^2}{\varphi }, \end{aligned}$$

where all terms are evaluated at $x_0$. From (66), we have

$$\begin{aligned} \varepsilon \varphi \left|D^2u^\varepsilon \right|^2 \le \varphi \left|Df\right|w^{\frac{1}{2}}+ 2 Cp w^{\frac{p-1}{2}+1} \varphi ^{\frac{1+\theta }{2}} + C\varepsilon w \varphi ^{\theta } + 2C\varepsilon w\varphi ^\theta . \end{aligned}$$

(67)

By Cauchy–Schwartz inequality, $n\left|D^2u^\varepsilon \right|^2\ge (\varDelta u^\varepsilon )^2$. Thus, if $n\varepsilon < 1$, then

$$\begin{aligned} \begin{aligned} \varepsilon \left|D^2u^\varepsilon \right|^2&\ge \frac{(\varepsilon \varDelta u^\varepsilon )^2}{n\varepsilon } \ge (\varepsilon \varDelta u^\varepsilon )^2 = \left( u^\varepsilon + \left|Du^\varepsilon \right|^p - f\right) ^2\\&\ge \left|Du^\varepsilon \right|^{2p} - 2C\left|Du^\varepsilon \right|^p \ge \frac{\left|Du^\varepsilon \right|^{2p}}{2} - 2C, \end{aligned} \end{aligned}$$

(68)

where C depends on $\max _{{\overline{\varOmega }}}f$ only. Using (68) in (67), we obtain that

$$\begin{aligned} \varphi \left( \frac{1}{2}w^p - 2C\right) \le \varphi \left|Df\right|w^{\frac{1}{2}}+ 2 Cp w^{\frac{p-1}{2}+1} \varphi ^{\frac{1+\theta }{2}} + 3C\varepsilon w \varphi ^{\theta }. \end{aligned}$$

Multiply both sides by $\varphi ^{p-1}$ to deduce that

$$\begin{aligned} (\varphi w)^p \le 4C\varphi ^{p-1} + 2\Vert Df\Vert _{L^\infty }\varphi ^p w^{\frac{1}{2}} + 4Cp \varphi ^{\frac{2p+\theta - 1}{2}}w^{\frac{p+1}{2}} + 6C\varepsilon \varphi ^{p+\theta - 1}w. \end{aligned}$$

Choose $2p+\theta -1 \ge p+1$, i.e., $p+\theta \ge 2$. This is always possible with the requirement $\theta \in (0,1)$, as $1<p <\infty$. Then we get

$$\begin{aligned} (\varphi w)^p \le C\left( 1+ (\varphi w)^\frac{1}{2} + (\varphi w)^\frac{p+1}{2} +(\varphi w)\right) . \end{aligned}$$

(69)

As a polynomial in $z = (\varphi w)(x_0)$, this implies that $(\varphi w)(x_0)\le C$ where C depends on coefficients of the right hand side of (69), which gives our desired gradient bound since $w(x)=(\varphi w)(x) \le (\varphi w)(x_0)$ for $x \in {\overline{\varOmega }}_\delta \subset \mathrm {supp}\;\varphi$. $\square$

1.2 A.2 Well-Posedness of (PDE$_\varepsilon$)

Proof of Theorem 5

If $p\in (1,2)$, we use the ansatz $u(x) = C_\varepsilon d(x)^{-\alpha }$ to find a solution to (PDE$_\varepsilon$). Plug the ansatz into (PDE$_\varepsilon$) and compute

$$\begin{aligned} \begin{aligned} \left|Du (x)\right|^p&= \frac{(\alpha C_\varepsilon )^p }{d(x)^{p(\alpha +1)}}\left|D d(x)\right|^p,\\ \varepsilon \varDelta u(x)&= \frac{\varepsilon C_\varepsilon \alpha (\alpha +1)}{d(x)^{\alpha +2}}\left|D d(x)\right|^2 - \frac{\varepsilon C_\varepsilon \alpha }{d(x)^{\alpha +1}}\varDelta d(x). \end{aligned} \end{aligned}$$

Since $\left|D d(x)\right|= 1$ for x near $\partial \varOmega$, as $x\rightarrow \partial \varOmega$, the explosive terms of the highest order are

$$\begin{aligned} C_\varepsilon ^p \alpha ^p d^{-(\alpha +1)p} -\varepsilon C_\varepsilon \alpha (\alpha +1)d^{-(\alpha +2)}. \end{aligned}$$

Set the above to be zero to obtain that

$$\begin{aligned} \displaystyle \alpha = \frac{2-p}{p-1} \qquad \text {and}\qquad C_\varepsilon = \left( \frac{1}{\alpha }(\alpha +1)^\frac{1}{p-1}\right) \varepsilon ^{\frac{1}{p-1}} = \frac{1}{\alpha }(\alpha +1)^{\alpha +1}\varepsilon ^{\alpha +1}. \end{aligned}$$

(70)

For $0<\delta < \frac{1}{2}\delta _0$ and $\eta$ small, define

$$\begin{aligned} \begin{aligned} {\overline{w}}_{\eta ,\delta }(x)&:= \frac{(C_\alpha +\eta )\varepsilon ^{\alpha +1}}{(d(x)-\delta )^\alpha } + M_\eta , \qquad x\in \varOmega _\delta ,\\ {\underline{w}}_{\eta ,\delta }(x)&:= \frac{(C_\alpha -\eta )\varepsilon ^{\alpha +1}}{(d(x)+\delta )^\alpha } - M_\eta , \qquad x\in \varOmega ^\delta , \end{aligned} \end{aligned}$$

where $C_\alpha := \frac{1}{\alpha } (\alpha +1)^{\alpha +1}$, $M_\eta$ to be chosen. Next, we show that ${\overline{w}}_{\eta ,\delta }$ is a supersolution of (PDE$_\varepsilon$) in $\varOmega _\delta$, while ${\underline{w}}_{\eta ,\delta }$ is a subsolution of (PDE$_\varepsilon$) in $\varOmega ^\delta$. Compute

$$\begin{aligned} {\mathcal {L}}^\varepsilon \left[ {\overline{w}}_{\eta ,\delta }\right] (x) =&\frac{ (C_\alpha + \eta )\varepsilon ^{\alpha +1}}{(d(x)-\delta )^\alpha } + M_\eta + \frac{(C_\alpha +\eta )^p \alpha ^p\varepsilon ^{\alpha +2}}{(d(x)-\delta )^{\alpha +2}}\left|D d(x)\right|^p - f(x) \\&- \frac{(C_\alpha +\eta )\alpha (\alpha +1)\varepsilon ^{\alpha +2}}{(d(x)-\delta )^{\alpha +2}}\left|D d(x)\right|^2 + \frac{(C_\alpha +\eta )\alpha \varepsilon ^{\alpha +2}}{(d(x)-\delta )^{\alpha +1}}\varDelta d(x)\\ \ge&M_\eta - f(x) \\&+ \underbrace{\frac{\nu C_\alpha \alpha (\alpha +1)\varepsilon ^{\alpha +2}}{(d(x)-\delta )^{\alpha +2}}\left[ \nu ^{p-1}\left|Dd(x)\right|^{p} - \left|Dd(x)\right|^2 + \frac{(d(x)-\delta )\varDelta d(x)}{\alpha +1}\right] }_{I}, \end{aligned}$$

where we use $(C_\alpha \alpha )^p = C_\alpha \alpha (\alpha +1)$ and $\nu = \frac{C_\alpha +\eta }{C_\alpha } \in (1,2)$ for small $\eta$. Let

$$\begin{aligned} \delta _\eta : = \frac{\alpha +1}{K_2}\left[ \nu ^{p-1}-1\right] \end{aligned}$$

and $\delta _\eta \rightarrow 0$ as $\eta \rightarrow 0$. To get ${\mathcal {L}}^\varepsilon \left[ {\overline{w}}_{\eta ,\delta }\right] \ge 0$, there are two cases to consider, depending on how large $d(x)-\delta$ is.

• If $0< d(x)-\delta<\delta _\eta < \delta _0$ for $\eta$ small and fixed, then $\left|Dd(x)\right|= 1$, and thus $I\ge 0$. Hence, ${\mathcal {L}}^\varepsilon \left[ {\overline{w}}_{\eta ,\delta }\right] \ge 0$ if we choose $M_\eta \ge \max _{{\overline{\varOmega }}} f$.

• If $d(x)-\delta \ge \delta _\eta$, then

  $$\begin{aligned} I \le \left( \frac{1}{\delta _\eta }\right) ^{\alpha +2}\nu C_\alpha \alpha (\alpha +1)\left[ \nu ^{p-1}K_1^{p}+K_1^2+K_2K_0\right] \varepsilon ^{\alpha +2}. \end{aligned}$$

  Thus, we can choose $M_\eta = \max _{{\overline{\varOmega }}} f + C\varepsilon ^{\alpha +2}$ for C large enough (depending on $\eta$) so that ${\mathcal {L}}^\varepsilon \left[ {\overline{w}}_{\eta ,\delta }\right] \ge 0$.

Therefore, ${\overline{w}}_{\eta ,\delta }$ is a supersolution in $\varOmega _\delta$.

Similarly, we have

$$\begin{aligned}&{\mathcal {L}}_\varepsilon \left[ {\underline{w}}_{\eta ,\delta }\right] (x) \\&\quad = \frac{ (C_\alpha - \eta )\varepsilon ^{\alpha +1}}{(d(x)+\delta )^\alpha } - M_\eta + \frac{(C_\alpha -\eta )^p \alpha ^p\varepsilon ^{\alpha +2}}{(d(x)+\delta )^{\alpha +2}}\left|D d(x)\right|^p - f(x) \\&\qquad - \frac{(C_\alpha -\eta )\alpha (\alpha +1)\varepsilon ^{\alpha +2}}{(d(x)+\delta )^{\alpha +2}}\left|D d(x)\right|^2 + \frac{(C_\alpha -\eta )\alpha \varepsilon ^{\alpha +2}}{(d(x)+\delta )^{\alpha +1}}\varDelta d(x)\\&\quad = - M_\eta - f(x) \\&\qquad +\underbrace{\frac{\nu C_\alpha \alpha (\alpha +1)\varepsilon ^{\alpha +2}}{(d(x)+\delta )^{\alpha +2}}\left[ \nu ^{p-1}\left|Dd(x)\right|^{p} - \left|Dd(x)\right|^2 + \frac{(d(x)+\delta )\varDelta d(x)}{\alpha +1}+\frac{(d(x)+\delta )^2}{\alpha (\alpha +1)\varepsilon }\right] }_{J}, \end{aligned}$$

where $\nu = \frac{C_\alpha -\eta }{C_\alpha }\in (0,1)$ for small $\eta$. Let

$$\begin{aligned} \delta _\eta := \left( 1-\nu ^{p-1}\right) \left( \frac{\alpha (\alpha +1)\varepsilon }{1+K_2\alpha \varepsilon }\right) \end{aligned}$$

and $\delta _\eta \rightarrow 0$ as $\eta \rightarrow 0$. To obtain ${\mathcal {L}}^\varepsilon \left[ {\underline{w}}_{\eta ,\delta }\right] \le 0$, there are two cases to consider depending on how large $d(x)+\delta$ is.

• If $0<d(x)+\delta<\delta _\eta < \delta _0$ for $\eta$ small and fixed, then $\left|Dd(x)\right|= 1$, and thus $J\le 0$. Hence, ${\mathcal {L}}^\varepsilon \left[ {\underline{w}}_{\eta ,\delta }\right] \le 0$ if we choose $M_\eta \ge -\max _{\varOmega }f$.

• If $d(x)+\delta \ge \delta _\eta$, then

  $$\begin{aligned} \left|J\right|\le \left( \frac{1}{\delta _\eta }\right) ^{\alpha +2} \nu C_\alpha \alpha (\alpha +1)\left[ \nu ^{p-1}K_1^{p}+K_1^2 + \frac{(K_0+1)K_2}{\alpha +1} + \frac{(K_0+1)^2}{\alpha (\alpha +1)\varepsilon }\right] \varepsilon ^{\alpha +2} \end{aligned}$$

  Thus, we can choose $M_\eta = -\max _{{\overline{\varOmega }}} f - C\varepsilon ^{\alpha +2}$ for C large enough (depending on $\eta$) so that ${\mathcal {L}}^\varepsilon \left[ {\underline{w}}_{\eta ,\delta }\right] \le 0$.

Therefore, ${\underline{w}}_{\eta ,\delta }$ is a subsolution in $\varOmega ^\delta$.

For $p=2$, we use the ansatz $u(x) = -C_\varepsilon \log (d(x))$ instead. Similar to the previous case, one can find $u(x) = -\varepsilon \log (d(x))$. For $0<\delta <\frac{1}{2}\delta _0$, define

$$\begin{aligned} \begin{aligned}&{\overline{w}}_{\eta ,\delta }(x) = -(1+\eta )\varepsilon \log (d(x)-\delta ) + M_\eta , \qquad x\in \varOmega _\delta ,\\&{\underline{w}}_{\eta ,\delta }(x) = -(1-\eta )\varepsilon \log (d(x)+\delta ) - M_\eta , \qquad x\in \varOmega ^\delta , \end{aligned} \end{aligned}$$

where $M_\eta$ is to be chosen so that ${\overline{w}}_{\eta ,\delta }(x)$ is a supersolution in $\varOmega _\delta$ and ${\underline{w}}_{\eta ,\delta }$ is a subsolution in $\varOmega ^\delta$. The computations are omitted here, as they are similar to the previous case.

We divide the rest of the proof into 3 steps. We first construct a minimal solution, then a maximal solution to (PDE$_\varepsilon$), and finally show that they are equal to conclude the existence and the uniqueness of the solution to (PDE$_\varepsilon$).

Step 1 There exists a minimal solution ${\underline{u}}\in \mathrm {C}^2(\varOmega )$ of (PDE$_\varepsilon$) such that $v\ge {\underline{u}}$ for any other solution $v\in \mathrm {C}^2(\varOmega )$ solving (PDE$_\varepsilon$).

Proof

Let $w_{\eta ,\delta }\in \mathrm {C}^2(\varOmega )$ solve

$$\begin{aligned} {\left\{ \begin{array}{ll} {\mathcal {L}}^\varepsilon \left[ w_{\eta ,\delta }\right] = 0 &{}\qquad \text {in}\;\varOmega ,\\ w_{\eta ,\delta } = {\underline{w}}_{\eta ,\delta } &{}\text {on}\;\partial \varOmega . \end{array}\right. } \end{aligned}$$

(71)

• Fix $\eta >0$. As $\delta \rightarrow 0^+$, the value of ${\underline{w}}_{\eta ,\delta }$ blows up on the boundary. Therefore, by the standard comparison principle for the second-order elliptic equation with the Dirichlet boundary, $\delta _1 \le \delta _2$ implies $w_{\eta ,\delta _1}\ge w_{\eta ,\delta _2}$ on ${\overline{\varOmega }}$.

• For $\delta '>0$, since ${\underline{w}}_{\eta ,\delta '}$ is a subsolution in ${\overline{\varOmega }}$ with finite boundary,

  $$\begin{aligned} 0<\delta \le \delta '\qquad \Longrightarrow \qquad {\underline{w}}_{\eta ,\delta '} \le w_{\eta _,\delta '}\le w_{\eta ,\delta } \qquad \text {on}\;{\overline{\varOmega }}. \end{aligned}$$

  (72)

• Similarly, since ${\overline{w}}_{\eta ,\delta '}$ is a supersolution on $\varOmega _{\delta '}$ with infinity value on the boundary $\partial \varOmega _{\delta '}$, by the comparison principle,

  $$\begin{aligned} w_{\eta ,\delta } \le {\overline{w}}_{\eta , \delta '} \qquad \text {in}\;\varOmega _{\delta '} \qquad \Longrightarrow \qquad w_{\eta ,\delta } \le {\overline{w}}_{\eta ,0} \qquad \text {in}\;\varOmega . \end{aligned}$$

  (73)

From (72) and (73), we have

$$\begin{aligned} 0<\delta \le \delta '\qquad \Longrightarrow \qquad {\underline{w}}_{\eta ,\delta '} \le w_{\eta _,\delta '}\le w_{\eta ,\delta } \le {\overline{w}}_{\eta ,0} \qquad \text {in}\;\varOmega . \end{aligned}$$

(74)

Thus, $\{w_{\eta ,\delta }\}_{\delta >0}$ is locally bounded in $L^{\infty }_{\mathrm {loc}}(\varOmega )$ ($\{w_{\eta ,\delta }\}_{\delta >0}$ is uniformly bounded from below). Using the local gradient estimate for $w_{\eta ,\delta }$ solving (71), we deduce that $\{w_{\eta ,\delta }\}_{\delta >0}$ is locally bounded in $W^{1,\infty }_{\mathrm {loc}}(\varOmega )$. Since $w_{\eta ,\delta }$ solves (71), we further have that $\{w_{\eta ,\delta }\}_{\delta >0}$ is locally bounded in $W^{2,r}_{\mathrm {loc}}(\varOmega )$ for all $r<\infty$ by Calderon–Zygmund estimates.

Local boundedness of $\{w_{\eta ,\delta }\}_{\delta >0}$ in $W^{2,r}_{\mathrm {loc}}(\varOmega )$ implies weak$^*$ compactness, that is, there exists a function $u\in W^{2,r}_{\mathrm {loc}}(\varOmega )$ such that (via subsequence and monotonicity)

$$\begin{aligned} w_{\eta ,\delta } \rightharpoonup u \qquad \text {weakly in}\;W^{2,r}_{\mathrm {loc}}(\varOmega ),\qquad \text {and}\qquad w_{\eta ,\delta } \rightarrow u \qquad \text {strongly in}\;W^{1,r}_{\mathrm {loc}}(\varOmega ). \end{aligned}$$

In particular, $w_{\eta ,\delta }\rightarrow u$ in $\mathrm {C}^1_{\mathrm {loc}}(\varOmega )$ thanks to Sobolev compact embedding. Let us rewrite the equation ${\mathcal {L}}^\varepsilon \left[ w_{\eta ,\delta }\right] = 0$ as $\varepsilon \varDelta w_{\eta ,\delta }(x) = F[w_{\eta ,\delta }](x)$ for $x \in U\subset \subset \varOmega$, where

$$\begin{aligned} F[w_{\eta ,\delta }](x) = w_{\eta ,\delta }(x) + H(x,Dw_{\eta ,\delta }(x)). \end{aligned}$$

Since $w_{\eta ,\delta }\rightarrow u$ in $\mathrm {C}^1(U)$ as $\delta \rightarrow 0$, we have $F[w_{\eta ,\delta }](x) \rightarrow F(x)$ uniformly in U as $\delta \rightarrow 0$, where

$$\begin{aligned} F(x) = u(x) + H(x,Du(x)). \end{aligned}$$

In the limit, we obtain that $u\in L^2(U)$ is a weak solution of $\varepsilon \varDelta u = F$ in U where F is continuous. Thus, $u\in \mathrm {C}^2(\varOmega )$ and by stability, u solves ${\mathcal {L}}^\varepsilon [u] = 0$ in $\varOmega$. From (74), we also have

$$\begin{aligned} {\underline{w}}_{\eta ,0} \le u \le {\overline{w}}_{\eta ,0} \qquad \text {in}\;\varOmega . \end{aligned}$$

Moreover, $u(x)\rightarrow \infty$ as $\mathrm {dist}(x,\partial \varOmega )\rightarrow 0$ with the precise rate like (9) or (10). Note that by construction, u may depend on $\eta$. But next, we will show that u is independent of $\eta$, by proving u is the unique minimal solution of ${\mathcal {L}}^\varepsilon [u] = 0$ in $\varOmega$ with $u = +\infty$ on $\partial \varOmega$.

Let $v\in \mathrm {C}^{2}(\varOmega )$ be a solution to (PDE$_\varepsilon$). Fix $\delta >0$. Since $v(x)\rightarrow \infty$ as $x\rightarrow \partial \varOmega$ while $w_{\eta ,\delta }$ remains bounded on $\partial \varOmega$, the comparison principle yields

$$\begin{aligned} v\ge w_{\eta ,\delta } \qquad \text {in} \; \varOmega . \end{aligned}$$

Let $\delta \rightarrow 0$ and we deduce that $v\ge u$ in $\varOmega$. This concludes that u is the minimal solution in $\mathrm {C}^2(\varOmega )(\forall \,r<\infty )$ and thus u is independent of $\eta$. $\square$

Step 2 There exists a maximal solution ${\overline{u}}\in \mathrm {C}^2(\varOmega )$ of (PDE$_\varepsilon$) such that $v\le {\overline{u}}$ for any other solution $v\in \mathrm {C}^2(\varOmega )$ solving (PDE$_\varepsilon$).

Proof

For each $\delta >0$, let $u_\delta \in \mathrm {C}^2(\varOmega _\delta )$ be the minimal solution to ${\mathcal {L}}^\varepsilon [u_\delta ] = 0$ in $\varOmega _\delta$ with $u_\delta = +\infty$ on $\partial \varOmega _\delta$. By the comparison principle, for every $\eta >0$, there holds

$$\begin{aligned} {\underline{w}}_{\eta ,\delta } \le u_\delta \le {\overline{w}}_{\eta ,\delta } \qquad \text {in}\;\varOmega _\delta , \end{aligned}$$

and

$$\begin{aligned} 0<\delta <\delta ' \qquad \Longrightarrow \qquad u_\delta \le u_\delta ' \qquad \text {in}\;\varOmega _{\delta '}\,. \end{aligned}$$

The monotoniciy, together with the local boundedness of $\{u_\delta \}_{\delta >0}$ in $W^{2,r}_{\mathrm {loc}}(\varOmega )$, implies that there exists $u\in W^{2,r}_{\mathrm {loc}}(\varOmega )$ for all $r<\infty$ such that $u_\delta \rightarrow u$ strongly in $\mathrm {C}^1_{\mathrm {loc}}(\varOmega )$. Using the equation ${\mathcal {L}}^\varepsilon [u_\delta ] = 0$ in $\varOmega _\delta$ and the regularity of Laplace’s equation, we can further deduce that $u\in \mathrm {C}^2(\varOmega )$ solves (PDE$_\varepsilon$) and

$$\begin{aligned} {\underline{w}}_{\eta ,0} \le u\le {\overline{w}}_{\eta ,0} \qquad \text {in}\;\varOmega \end{aligned}$$

for all $\eta >0$. As $u_\delta$ is independent of $\eta$ by the previous argument in Step 1, it is clear that u is also independent of $\eta$. Now we show that u is the maximal solution of (PDE$_\varepsilon$). Let $v\in \mathrm {C}^2(\varOmega )$ solve (PDE$_\varepsilon$). Clearly $v\le u_\delta$ on $\varOmega _\delta$. Therefore, as $\delta \rightarrow 0$, we have $v\le u$. $\square$

In conclusion, we have found a minimal solution ${\underline{u}}$ and a maximal solution ${\overline{u}}$ in $\mathrm {C}^2(\varOmega )$ such that

$$\begin{aligned} {\underline{w}}_{\eta ,0} \le {\underline{u}}\le {\overline{u}}\le {\overline{w}}_{\eta ,0} \qquad \text {in}\;\varOmega \end{aligned}$$

(75)

for any $\eta >0$. This extra parameter $\eta$ now enables us to show that ${\overline{u}} = {\underline{u}}$ in $\varOmega$. The key ingredient here is the convexity in the gradient slot of the operator.

Step 3 We have ${\overline{u}}\equiv {\underline{u}}$ in $\varOmega$. Therefore, the solution to (PDE$_\varepsilon$) in $\mathrm {C}^2(\varOmega )$ is unique.

Proof

Let $\theta \in (0,1)$. Define $w_\theta = \theta {\overline{u}} + (1-\theta ) \inf _{\varOmega } f$. It can be verified that $w_\theta$ is a subsolution to (PDE$_\varepsilon$). Then one may argue that by the comparison principle,

$$\begin{aligned} w_\theta = \theta {\overline{u}} + (1-\theta )\inf _{\varOmega } f\le {\underline{u}} \qquad \text {in}\;\varOmega , \end{aligned}$$

and conclude that ${\overline{u}} \le {\underline{u}}$ by letting $\theta \rightarrow 1$. But we have to be careful here. As they are both explosive solutions, to use the comparison principle, we need to show that $w_\theta \le {\underline{u}}$ in a neighborhood of $\partial \varOmega$. From (75), we see that

$$\begin{aligned}&1\le \frac{{\overline{u}}(x)}{{\underline{u}}(x)} \le \frac{{\overline{w}}_{\eta ,0}(x)}{{\underline{w}}_{\eta ,0}(x)} = \frac{(C_\alpha +\eta )+ M_\eta d(x)^\alpha }{(C_\alpha -\eta )- M_\eta d(x)^\alpha },&1<p<2,\\&1 \le \frac{{\overline{u}}(x)}{{\underline{u}}(x)} \le \frac{{\overline{w}}_{\eta ,0}(x)}{{\underline{w}}_{\eta ,0}(x)} = \frac{-(1+\eta )\log (d(x)) + M_\eta }{-(1-\eta )\log (d(x))-M_\eta },&p=2, \end{aligned}$$

for $x\in \varOmega$. Hence,

$$\begin{aligned}&1\le \lim _{d(x)\rightarrow 0} \left( \frac{{\overline{u}}(x)}{{\underline{u}}(x)}\right) \le \frac{C_\alpha +\eta }{C_\alpha -\eta },&1< p < 2,\\&1\le \lim _{d(x)\rightarrow 0} \left( \frac{{\overline{u}}(x)}{{\underline{u}}(x)}\right) \le \frac{-(1+\eta )}{-(1-\eta )},&p = 2. \end{aligned}$$

Since $\eta >0$ is chosen arbitrary, we obtain

$$\begin{aligned} \lim _{d(x)\rightarrow 0} \left( \frac{{\overline{u}}(x)}{{\underline{u}}(x)}\right) = 1. \end{aligned}$$

This means for any $\varsigma \in (0,1)$, there exists ${\delta _1}(\varsigma )>0$ small such that

$$\begin{aligned} \frac{{\overline{u}}(x)}{{\underline{u}}(x)}\le (1+\varsigma )\Longrightarrow \left( \frac{1}{1+\varsigma }\right) {\overline{u}}(x) \le {\underline{u}}(x) \qquad \text {in}\; \varOmega \backslash \varOmega _{\delta _1}. \end{aligned}$$

For a fixed $\theta \in (0,1)$, one can always choose $\varsigma$ small enough so that $\displaystyle (1+\varsigma )^{-1} \ge \frac{1+\theta }{2}$. Since ${\overline{u}}(x) \rightarrow +\infty$ as $d(x) \rightarrow 0$, there exists $\delta _2 > 0$ such that ${\overline{u}}(x) \ge 2 \inf _\varOmega f$ for all $x \in \varOmega \setminus \varOmega _{\delta _2}$. Now we have

$$\begin{aligned} {\underline{u}}(x)\ge \left( \frac{1}{1+\varsigma }\right) {\overline{u}}(x) \ge \theta {\overline{u}}(x)+ \left( \frac{1-\theta }{2}\right) {\overline{u}}(x)\ge \theta {\overline{u}}(x) + (1-\theta )\left( \inf _\varOmega f\right) \end{aligned}$$

for all $x \in \varOmega \setminus \varOmega _\delta$ where $\delta : = \min \{\delta _1, \delta _2\}$. This implies for any fixed $\theta \in (0,1)$, $w_\theta \le {\underline{u}}$ in a neighborhood of $\partial \varOmega$. Hence, by the comparison principle,

$$\begin{aligned} w_\theta = \theta {\overline{u}} + (1-\theta )\inf _{\varOmega } f\le {\underline{u}} \qquad \text {in}\;\varOmega , \end{aligned}$$

for any $\theta \in (0,1)$. Then let $\theta \rightarrow 1$ to get the conclusion.

This finishes the proof of the well-posedness of (PDE$_\varepsilon$) for $1<p\le 2$. $\square$

Proof of Lemma 6

The proof is a variation of Perron’s method (see [10]) and we proceed by contradiction. Let $\varphi \in \mathrm {C}({\overline{\varOmega }})$ and $x_0\in {\overline{\varOmega }}$ such that $u(x_0) = \varphi (x_0)$ and $u-\varphi$ has a global strict minimum over ${\overline{\varOmega }}$ at $x_0$ with

$$\begin{aligned} \varphi (x_0) + H(x_0,D\varphi (x_0)) < 0. \end{aligned}$$

(76)

Let $\varphi ^\varepsilon (x) = \varphi (x) - \left|x-x_0\right|^2 + \varepsilon$ for $x\in {\overline{\varOmega }}$. Let $\delta > 0$. We see that for $x\in \partial B(x_0,\delta )\cap {\overline{\varOmega }}$,

$$\begin{aligned} \varphi ^\varepsilon (x) = \varphi (x) - \delta ^2 +\varepsilon \le \varphi (x) - \varepsilon \end{aligned}$$

if $2\varepsilon \le \delta ^2$. We observe that

$$\begin{aligned} \begin{aligned} \varphi ^\varepsilon (x) - \varphi (x_0)&= \varphi (x)-\varphi (x_0) + \varepsilon - \left|x-x_0\right|^2 \\ D\varphi ^\varepsilon (x) - D\varphi (x_0)&= D\varphi (x) - D\varphi (x_0) - 2(x-x_0) \end{aligned} \end{aligned}$$

for $x\in B(x_0,\delta )\cap {\overline{\varOmega }}$. By the continuity of H(x, p) near $(x_0,D\varphi (x_0))$ and the fact that $\varphi \in \mathrm {C}^1({\overline{\varOmega }})$, we can deduce from (76) that if $\delta$ is small enough and $0<2\varepsilon < \delta ^2$, then

$$\begin{aligned} \varphi ^\varepsilon (x)+H(x,D\varphi ^\varepsilon (x)) < 0 \qquad \text {for}\;x\in B(x_0,\delta )\cap {\overline{\varOmega }}. \end{aligned}$$

(77)

We have found $\varphi ^\varepsilon \in \mathrm {C}^1({\overline{\varOmega }})$ such that $\varphi ^\varepsilon (x_0)>u(x_0)$, $\varphi ^\varepsilon <u$ on $\partial B(x_0,\delta )\cap {\overline{\varOmega }}$ and (77). Let

$$\begin{aligned} {\tilde{u}}(x) = {\left\{ \begin{array}{ll} \max \big \lbrace u(x),\varphi ^\varepsilon (x) \big \rbrace &{}x\in B(x_0,\delta )\cap {\overline{\varOmega }},\\ u(x)&{}x\notin B(x_0,\delta )\cap {\overline{\varOmega }}.\\ \end{array}\right. } \end{aligned}$$

We see that ${\tilde{u}}\in \mathrm {C}({\overline{\varOmega }})$ is a subsolution of (PDE$_0$) in $\varOmega$ with ${\tilde{u}}(x_0) > u(x_0)$, which is a contradiction. Thus, u is a supersolution of (PDE$_0$) on ${\overline{\varOmega }}$. $\square$

1.3 A.3 Semiconcavity

We present a proof for the semiconcavity of solution to first-order Hamilton–Jacobi equation using the doubling variable method (see also [8]).

Theorem 19

Let $H(x,p) = G(p)-f(x)$ where $G\ge 0$ with $G(0) = 0$ is a convex function from ${\mathbb {R}}^n\rightarrow {\mathbb {R}}^n$ and $f\in \mathrm {C}^2_c({\mathbb {R}}^n)$. Let $u\in \mathrm {C}_c({\mathbb {R}}^n)$ be a viscosity solution to $u+H(x,Du) = 0$ in ${\mathbb {R}}^n$. Then u is semiconcave, i.e., u is a viscosity solution of $-D^2u \ge -c\;{\mathbb {I}}_n$ in ${\mathbb {R}}^n$ where

$$\begin{aligned} c = \max \big \lbrace D_{\xi \xi }f(x): \left|\xi \right|=1, x\in {\mathbb {R}}^n \big \rbrace \ge 0. \end{aligned}$$

Proof

Consider the auxiliary functional

$$\begin{aligned} \varPhi (x,y,z) = u(x)-2u(y)+u(z) - \frac{\alpha }{2}\left|x-2y+z\right|^2 - \frac{c}{2}\left|y-x\right|^2-\frac{c}{2}\left|y-z\right|^2 \end{aligned}$$

for $(x,y,z)\in {\mathbb {R}}^n\times {\mathbb {R}}^n\times {\mathbb {R}}^n$. By the a priori estimate, u is bounded and Lipschitz. Thus, we can assume $\varPhi$ achieves its maximum over ${\mathbb {R}}^n\times {\mathbb {R}}^n\times {\mathbb {R}}^n$ at $(x_\alpha ,y_\alpha ,z_\alpha )$. The viscosity solution tests give us

$$\begin{aligned}&u(x_\alpha ) + G\big (p_\alpha +c(x_\alpha -y_\alpha )\big ) \le f(x_\alpha )\\&u(z_\alpha ) + G\big (p_\alpha +c(z_\alpha -y_\alpha )\big ) \le f(z_\alpha )\\&u(y_\alpha ) + G\left( p_\alpha +\frac{c}{2}(x_\alpha -y_\alpha ) + \frac{c}{2}(z_\alpha -y_\alpha )\right) \ge f(y_\alpha ), \end{aligned}$$

where $p_\alpha = \alpha (x_\alpha -2y_\alpha +z_\alpha )$. By the convexity of G, we have

$$\begin{aligned} 2G\left( p_\alpha \!+\! \frac{c}{2}(x_\alpha -y_\alpha ) \!+\! \frac{c}{2}(z_\alpha -y_\alpha )\right) \!\le \! G\big (p_\alpha +c(x_\alpha -y_\alpha )\big ) \!+\! G\big (p_\alpha +c(z_\alpha -y_\alpha )\big ) \end{aligned}$$

Therefore,

$$\begin{aligned} u(x_\alpha ) - 2u(y_\alpha ) + u(z_\alpha ) \le f(x_\alpha ) - 2f(y_\alpha ) + f(z_\alpha ). \end{aligned}$$

• $\varPhi (x_\alpha ,y_\alpha ,z_\alpha )\ge \varPhi (0,0,0)$ gives us

  $$\begin{aligned} \frac{\alpha }{2}\left|x_\alpha -2y_\alpha +z_\alpha \right|^2 + \frac{c}{2}\left|y_\alpha -x_\alpha \right|^2+\frac{c}{2}\left|y_\alpha -z_\alpha \right|^2 \le C. \end{aligned}$$

  Thus, $(x_\alpha -y_\alpha )\rightarrow h_0$ and $(y_\alpha -z_\alpha )\rightarrow h_0$ as $\alpha \rightarrow \infty$ for some $h_0 \in {\mathbb {R}}^n$.

• $\varPhi (x_\alpha ,y_\alpha ,z_\alpha )\ge \varPhi (y_\alpha +h_0,y_\alpha ,y_\alpha -h_0)$ gives us

  $$\begin{aligned} u(x_\alpha )-2u(y_\alpha ) + u(z_\alpha ) - \frac{\alpha }{2}\left|x_\alpha - 2y_\alpha +z_\alpha \right|^2 - \frac{c}{2}\left|x_\alpha -y_\alpha \right|^2 - \frac{c}{2}\left|y_\alpha -z_\alpha \right|^2 \\ \ge u(y_\alpha +h_0) - 2u(y_\alpha ) + u(y_\alpha -h_0) - c\left|h_0\right|^2. \end{aligned}$$

  Therefore, by the fact that u is Lipschitz, we have

  $$\begin{aligned} \begin{aligned} \frac{\alpha }{2}\left|x_\alpha - 2y_\alpha +z_\alpha \right|^2 \le&c\left( \frac{2\left|h_0\right|^2 - \left|x_\alpha -y_\alpha \right|^2 - \left|y_\alpha -z_\alpha \right|^2}{2}\right) \\&+ C\Big (\left|(x_\alpha -y_\alpha ) - h_0\right|+ \left|(z_\alpha - y_\alpha ) + h_0\right|\Big ) \rightarrow 0 \end{aligned} \end{aligned}$$

  as $\alpha \rightarrow \infty$.

For any $x\in {\mathbb {R}}^n$, we have $\varPhi (x_\alpha ,y_\alpha ,z_\alpha ) \ge \varPhi (x+h,x,x-h)$, i.e.,

$$\begin{aligned}&u(x+h)-2u(x)+u(x-h)-c\left|h\right|^2\\&\quad \le f(x_\alpha )-2f(y_\alpha ) + f(z_\alpha ) \\&\qquad - \frac{\alpha }{2}\left|x_\alpha - 2y_\alpha +z_\alpha \right|^2-\frac{c}{2}\left|y_\alpha -x_\alpha \right|^2-\frac{c}{2}\left|y_\alpha -z_\alpha \right|^2. \end{aligned}$$

If $\{y_\alpha \}$ is unbounded, then since $f\in \mathrm {C}_c^2({\mathbb {R}}^n)$, we have $f(y_\alpha )\rightarrow 0$ as $\alpha \rightarrow \infty$. As a consequence, $x_\alpha ,z_\alpha \rightarrow \infty$ as well and thus $f(x_\alpha )-2f(y_\alpha ) + f(z_\alpha )\rightarrow 0$ as $\alpha \rightarrow \infty$. Therefore,

$$\begin{aligned} u(x+h)-2u(x)+u(x-h)-c\left|h\right|^2 \le 0. \end{aligned}$$

If $\{y_\alpha \}$ is bounded, then $y_\alpha \rightarrow y_0$ for some $y_0 \in {\mathbb {R}}^n$ as $\alpha \rightarrow \infty$. Thus,

$$\begin{aligned}&u(x+h)-2u(x)+u(x-h)-c\left|h\right|^2\\&\quad \le f(y_0+h_0)- 2f(y_0) + f(y_0-h_0) -c\left|h_0\right|^2. \end{aligned}$$

Let $\xi = h_0$ and we have

$$\begin{aligned} {\left\{ \begin{array}{ll} f(y_0+h_0) - f(y_0) = \displaystyle \int _0^1 D_x f(y_0 + t\xi )\cdot \xi dt, \\ f(y_0) - f(y_0-h_0) = \displaystyle \int _0^1 D_x f(y_0 - \xi + t\xi )\cdot \xi dt. \end{array}\right. } \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} f(y_0+h_0) - 2f(y_0) + f(y_0-h_0)&= \int _0^1 \Big (D_x f(y_0 + t\xi ) - D_x f(y_0 - \xi + t\xi ) \Big )\cdot \xi dt\\&= \int _0^1 \int _0^1 \xi ^{\mathsf {T}} D^2 f(y_0-\xi +t\xi +s \xi ) \xi \;dsdt. \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \left|f(y_0+h_0) - 2f(y_0) + f(y_0-h_0)\right|\le \left( \max _{\left|\xi \right|=1}D_{\xi \xi } f\right) \left|\xi \right|^2. \end{aligned}$$

Hence,

$$\begin{aligned} u(x+h)-2u(x)+u(x-h)-c\left|h\right|^2 \le 0 \end{aligned}$$

and thus u is semiconcave. It is easy to see that if $\varphi$ is smooth and $u-\varphi$ has a local min at x, then $D^2\varphi (x) \le c\;{\mathbb {I}}$, i.e., $-D^2\varphi (x)\ge -c\;{\mathbb {I}}$. $\square$