Free boundary theory for singular/degenerate nonlinear equations with right hand side: a non-variational approach

Abstract

We use non-variational type arguments to prove optimal regularity and smoothness of the free boundary for one-phase solutions to inhomogeneous nonlinear free boundary problems (FBP) governed by singular/degenerate elliptic PDEs with a nonzero right hand side (RHS). In a precise way, we show that viscosity solutions to FBP, as previously mentioned, are locally Lipschitz continuous and under certain conditions, flat or Lipschitz free boundaries, are $C^{1, \alpha }.$

Appendix: Relationship between types of solutions to \(\Delta _g\)

In this appendix, we answered the issue in Remark 1.5. Let G as in (2.1) and g satisfying (1.3). Consider u a solution to

$$\begin{aligned} \Delta _g u = f \ \ \ \text { in } \ \Omega , \end{aligned}$$

(10.1)

where $f \in L^q(\Omega ), \ \ q > n$.

Definition 10.1

A function $u \in W_{loc}^{1,G}(\Omega )$ is a weak supersolution(subsolution) to (10.1) if for any $0 \le \phi \in C^{\infty }_0(\Omega )$ holds

$$\begin{aligned} \int _{\Omega } \frac{g(\vert \nabla u \vert )}{\vert \nabla u \vert } \nabla u \nabla \phi dx \ge - \int _{\Omega } f \phi dx \ \ \ (\text {resp. } \ \le ). \end{aligned}$$

The function u will be a weak solution to (10.1) if

$$\begin{aligned} \int _{\Omega } \frac{g(\vert \nabla u \vert )}{\vert \nabla u \vert } \nabla u \nabla \phi dx = - \int _{\Omega } f \phi dx, \end{aligned}$$

for any $\phi \in C_0^{\infty }(\Omega )$.

Definition 10.2

A function $u: \Omega \longrightarrow (-\infty , +\infty ]$ is called a G-superharmonic function, if u satisfies

• (1) u is lower semicontinuous,

• (2) u is bounded almost everywhere,

• (3) the comparison principle holds: if v is a weak solution of $\Delta _g v = f$ in $D \subset \Omega$, u is continuous in $\overline{D}$ and $u \ge v$ on $\partial D$, then $u \ge v$ in D.

A function $u:\Omega \longrightarrow [ - \infty , + \infty )$ is G-subharmonic, if $-u$ is G-superharmonic.

Definition 10.3

A function $u: \Omega \longrightarrow (-\infty , +\infty ]$ is called a $L^q$-viscosity supersolution(subsolution) to (10.1) if u satisfies

• (1) u is lower semicontinuous;

• (2) u is bounded almost everywhere;

• (3) If there exists $\phi \in W^{2,q}_{loc}(\Omega )$ touching u by below(above) strictly at $x_0 \in \Omega$ and $\nabla \phi (x_0) \ne 0$¹, then

  $$\begin{aligned} ess\liminf \limits _{x \rightarrow x_{0}} \Big (\Delta _g \phi (x) - f(x)\Big ) \le 0 \ \ \ \left( ess\limsup \limits _{x \rightarrow x_{0}} \Big (\Delta _g \phi (x) - f(x)\Big ) \ge 0 \right) . \end{aligned}$$

We have the following results.

Lemma 10.4

Let $u \in C^0(\Omega )$. Are equivalents:

• (A) u is a weak supersolution to (10.1);

• (B) u is a G-superharmonic function in $\Omega$;

Similar consequences holds for solutions and subsolutions.

Proof

The proof follows the same guidelines of Theorems 2.4 and 2.5 of [18].

Proposition 10.5

Let u be a weak supersolution to (10.1). Then, u is a $L^q$-viscosity supersolution solution to (10.1).

Proof

By Lemma 10.4 it is sufficient to prove that G-superharmonic functions in $\Omega$ are $L^q$-viscosity supersolutions. Assume, by contradiction, that u is G-superharmonic but u is a not a $L^q$-viscosity supersolution of (10.1). In this case, there exists $\phi \in W^{2,q}_{loc}(\Omega )$ touching u by below strictly at $x_0 \in \Omega$ with $\nabla \phi (x_0) \ne 0$ and $\mu > 0$ such that

$$\begin{aligned} ess\liminf \limits _{x \rightarrow x_{0}} \Big (\Delta _g \phi (x) - f(x)\Big ) \ge \mu > 0. \end{aligned}$$

Now, we know that there exists $r_0 > 0$ such that in $B_{r_0}(x_0)$,

$$\begin{aligned} \nabla u(x) \ne 0 \ \ \ \text { and } \ \ \ \ \Delta _g u(x) > f(x) \ \ a.e. \end{aligned}$$

Let $m = \inf _{\partial B_{r_0}(x_0)} \vert u - \phi \vert > 0$ and $\psi = \phi + m$. Since $A) \Leftrightarrow B)$ we have that $\psi$ is a weak subsolution of (10.1) and $u \ge \psi$ on $\partial B_{r_0}(x_0)$. By comparison principle (see Proposition 5.2 of [5]), $\psi \le u$ in $B_{r_0}(x_0)$. However,

$$\begin{aligned} \psi (x_0) = \phi (x_0) + m > u(x_0), \end{aligned}$$

which is a contradiction.