Porous Medium Equation with a Drift: Free Boundary Regularity

Abstract

We study regularity properties of the free boundary for solutions of the porous medium equation with the presence of drift. We show the $C^{1,\alpha }$ regularity of the free boundary when the solution is directionally monotone in space variable in a local neighborhood. The main challenge lies in establishing a local non-degeneracy estimate (Theorem 1.3 and Proposition 1.5), which appears new even for the zero drift case.

Appendices

Appendix A: Proof of Lemma 2.6

Let us only consider the case when $U={\mathbb {R}}^d$. The case of $U=B_1$ follows similarly.

Fix one non-negative $\phi \in C_c^\infty ({\mathbb {R}}^d\times [0,T))$. Denote

$$\begin{aligned} U_0:=\{\phi>0\}\cap \{\psi >0\}. \end{aligned}$$

For any $\varepsilon >0$, take finitely many space time balls $U_i, i=1,\ldots ,n$ such that

• 1. for each $i\geqq 1$, $|U_i|\leqq \varepsilon ^{d+1}$ and $U_i$ is in the $\varepsilon$-neighbourhood of $\Gamma (\psi )$,

• 2. $\{ U_i\}_{i=1,\ldots ,n}$ is an open cover of $\Gamma (\psi ) \cap \{\phi >0\}.$

Since $\Gamma (\psi )$ is of dimension d, we can assume

$$\begin{aligned} n\lesssim \varepsilon ^{-d}. \end{aligned}$$

(A.1)

Take a partition of unity $\{\rho _i, i=0,\ldots ,n\}$ which is subordinate to the open cover $\{U_i\}_{i\geqq 0}$. Then for $i\geqq 1$,

$$\begin{aligned} |\nabla \rho _i|+|\partial _t\rho _i|\lesssim 1/\varepsilon . \end{aligned}$$

(A.2)

By the assumption, $\psi$ is a supersolution in the interior of its positive set. And since $\varepsilon$ can be arbitrarily small, to show (2.4) we only need to show

$$\begin{aligned} I_\varepsilon :=\sum _{i=1}^{n(\varepsilon )}\left( \int _0^T\int _{{\mathbb {R}}^d} \psi \,(\phi \rho _i)_t-(\nabla \psi ^m+\psi \,\vec {b})\nabla (\phi \rho _i)\;\mathrm{d}x\mathrm{d}t - \int _{{\mathbb {R}}^d} \psi (0,x)\phi (0,x)\rho _i\mathrm{d}x\right) \rightarrow 0 \end{aligned}$$

as $\varepsilon \rightarrow 0$.

By property 1 of $U_i$ and the regularity assumption on $\psi$, in all $U_i, i\geqq 1$ we have

$$\begin{aligned} \psi \leqq C\varepsilon ^{\frac{1}{\alpha }},\quad |\nabla \psi ^m|\leqq C\psi ^{m-\alpha }|\nabla \psi ^\alpha |\leqq C\varepsilon ^{\frac{m-\alpha }{\alpha }}. \end{aligned}$$

Now from (A.1), (A.2) and $\alpha <m$, it follows that

$$\begin{aligned} |I_\varepsilon |&\leqq C\varepsilon ^{-d}\left( \iint _{U_i} \frac{1}{\varepsilon } (\psi +|\nabla \psi ^m|) \;\mathrm{d}x\mathrm{d}t +\int _{U_i\cap \{t=0\}} \psi (0,x)\mathrm{d}x\right) \\&\leqq C(\varepsilon ^{\frac{1}{\alpha }}+\varepsilon ^{\frac{m-\alpha }{\alpha }}+\varepsilon ) \end{aligned}$$

which indeed converges to 0 as $\varepsilon \rightarrow 0$.

Appendix B: Sketch of the proof of Lemma 5.3

We follow the idea of Lemma 9 [8] and compute

$$\begin{aligned} \Delta f(0)={\overline{\lim }}_{r\rightarrow 0}\left( \oint _{B_r}f(x)-f(0)\mathrm{d}x\right) . \end{aligned}$$

Without loss of generality, suppose locally near the origin that

$$\begin{aligned} f(x)=\inf _{|\nu |=1}h\left( x+\psi (x)\nu \right) , \end{aligned}$$

because otherwise $\Delta f(0)=0$. Choosing an appropriate system of coordinates, we can have

$$\begin{aligned}&f(0)=h(\psi (0)e_n);\\&\nabla \psi (0)=\alpha e_1+\beta e_n. \end{aligned}$$

We will evaluate f(x) by above by choosing $\nu (x)=\frac{\nu _*(x)}{|\nu _*(x)|}$ where

$$\begin{aligned} \nu _*(x):=e_n+\frac{\beta x_1-\alpha x_n}{\psi (0)}e_1 +\frac{\gamma }{\psi (0)}\left( \Sigma _{i=2}^{d-1}\,x_i\,e_i\right) \end{aligned}$$

where $\gamma$ satisfies

$$\begin{aligned} (1+\gamma )^2=(1+\beta )^2+\alpha ^2. \end{aligned}$$

With this choice of $\nu$, we define $y:=x+\psi (x)\nu (x)$ and so $y(0)=\psi (0)e_n$. After direct computations (also see [8]), we can write

$$\begin{aligned} y=Y_*(x)+\psi (0)e_n+o(|x|^2) \end{aligned}$$

such that the first-order term, except the translation $\varphi (0)e_n$, satisfies

$$\begin{aligned} Y_*(x):=x+(\alpha x_1+\beta x_n)e_n+(\beta x_1-\alpha )e_1+\gamma \Sigma _{i=1}^d x_ie_i. \end{aligned}$$

Hence $Y_*(x)$ is a rigid rotation plus a dilation and we have

$$\begin{aligned} \left| \frac{D (Y_*-x)}{Dx}\right| \leqq \sigma \Vert \nabla \psi \Vert _\infty . \end{aligned}$$

(B.1)

Then

$$\begin{aligned} \oint _{B_r}f(x)-f(0)\mathrm{d}x&\leqq \oint _{B_r}h(y(x))-h(y(0))\mathrm{d}x\\&\leqq \oint _{B_r}h(y(x))\\&\quad -h(Y_*(x)+y(0))\mathrm{d}x+ \oint _{B_r}h(Y_*(x)+y(0))-h(y(0))\mathrm{d}x. \end{aligned}$$

By the condition on $\psi$ and the computations done in Lemma 9 [8], the first term is non-positive.

Since h is smooth, the second term converges to

$$\begin{aligned} \left( \left| \frac{D Y_*}{D x}\right| _{x=0}\right) ^2(\Delta h)(y(0))\quad \text { as }r \rightarrow 0. \end{aligned}$$

Now, using (B.1) and the assumption that $\Delta h\geqq -C$ and $\Vert \nabla \psi \Vert _\infty \leqq 1$, we get

$$\begin{aligned} \oint _{B_r}f(x)-f(0)\mathrm{d}x&\leqq \oint _{B_r}h(Y_*(x)+y(0))-h(y(0))\mathrm{d}x\\&\leqq (1+\sigma \Vert \nabla \psi \Vert _\infty )(\Delta h) (y(0))+\sigma \Vert \nabla \psi \Vert _\infty C. \end{aligned}$$

Thus we have finished the proof.

Appendix C: Proof of Lemma 5.4

Let us suppose $x=0$ and $f(0)=h(y)$ for a unique y. We only compute $\partial _1 f(0)=\partial _{x_1}f(0)$. If $\nabla h(y)=0$, it is not hard to see

$$\begin{aligned} \partial _1 f(0)=\partial _1 h(y)=0. \end{aligned}$$

Next suppose $\nabla h(y)\ne 0$. We know that h obtains its minimum over $B(0,\psi (0))$ at point $y\in \partial B(0,{\psi (0)})$. Let us assume

$$\begin{aligned} y=(y_1,y_2,0,\ldots ,0),\quad \text { and thus }|y_1|^2+|y_2|^2=(\psi (0))^2. \end{aligned}$$

For smooth h, it is not hard to see that

$$\begin{aligned} \nabla h(y)=-k y\quad \text {with }k=\frac{|\nabla h|}{\psi (0)}. \end{aligned}$$

Near point y

$$\begin{aligned} h(x)-h(y)=-ky_1(x_1-y_1)-ky_2(x_2-y_2)+o(|x-y|). \end{aligned}$$

To estimate $w((\delta ,0,\ldots ,0))$, consider the leading terms:

$$\begin{aligned}&A(\delta ):=-ky_1(x_1-y_1)-ky_2(x_2-y_2)\\&=-ky_1(x_1-\delta )-ky_2x_2+ky^2_1 +ky^2_2-ky_1\delta . \end{aligned}$$

By a standard argument, under the constrain

$$\begin{aligned} |x_1-\delta |^2+|x_2|^2+|x_3|^2+\cdots +|x_n|^2\leqq \psi (\delta ,0,\ldots ,0)^2, \end{aligned}$$

$A(\delta )$ achieves its minimum at

$$\begin{aligned} x_1=y_1\psi (\delta ,0,\ldots ,0)/(y^2_1+y^2_2)^{\frac{1}{2}}+\delta ,\,x_2 =y_2\psi (\delta ,0, \ldots ,0)/(y^2_1+y^2_2)^{\frac{1}{2}} \end{aligned}$$

with value

$$\begin{aligned}&-k\psi (\delta ,0, \ldots ,0)(y^2_1+y^2_2)^{\frac{1}{2}}+ky^2_1+ky^2_2-ky_1\delta \\&=-k\psi (\delta ,0, \ldots ,0)\psi (0)+k\psi (0)^2-ky_1\delta . \end{aligned}$$

Thus

$$\begin{aligned} \partial _1 f(0)=\lim _{\delta \rightarrow 0}A(\delta )/\delta =-k\psi (0)\,\partial _1 \psi (0)-ky_1. \end{aligned}$$

Notice that $\partial _1 h(y)=-ky_1$. So we find

$$\begin{aligned} \partial _1 f(0)-\partial _1 h(y)=-k\psi (0)\,\partial _1 \psi (0) =-|\nabla h|\,\partial _1 \psi (0). \end{aligned}$$

This leads to the conclusion.