On the p-Laplacian with Robin boundary conditions and boundary trace theorems

Abstract

Let $\Omega \subset \mathbb {R}^\nu$, $\nu \ge 2$, be a $C^{1,1}$ domain whose boundary $\partial \Omega$ is either compact or behaves suitably at infinity. For $p\in (1,\infty )$ and $\alpha >0$, define

$$\begin{aligned} \Lambda (\Omega ,p,\alpha ):=\inf _{\begin{array}{c} u\in W^{1,p}(\Omega )\\ u\not \equiv 0 \end{array}}\dfrac{\displaystyle \int _\Omega |\nabla u|^p \mathrm {d} x - \alpha \displaystyle \int _{\partial \Omega } |u|^p\mathrm {d}\sigma }{\displaystyle \int _\Omega |u|^p\mathrm {d} x}, \end{aligned}$$

where $\mathrm {d}\sigma$ is the surface measure on $\partial \Omega$. We show the asymptotics

$$\begin{aligned} \Lambda (\Omega ,p,\alpha )=-(p-1)\alpha ^{\frac{p}{p-1}} - (\nu -1)H_\mathrm {max}\, \alpha + o(\alpha ), \quad \alpha \rightarrow +\infty , \end{aligned}$$

where $H_\mathrm {max}$ is the maximum mean curvature of $\partial \Omega$. The asymptotic behavior of the associated minimizers is discussed as well. The estimate is then applied to the study of the best constant in a boundary trace theorem for expanding domains, to the norm estimate for extension operators and to related isoperimetric inequalities.

Appendices

Appendix 1: Solvable cases

For the sake of completeness, let us mention some cases in which $\Lambda (\Omega ,p,\alpha )$ can be computed explicitly.

Proposition 6.7

For any $p\in (1,\infty )$ and $\alpha >0$ there holds $\Lambda (\mathbb {R}_+,p,\alpha )=(1-p)\alpha ^\frac{p}{p-1}$, and the minimizer $u_*$ for Eq. (1) is given by $u_*(t)=\exp \big (-\alpha ^{\frac{1}{p-1}} t\big )$.

Proof

By computing the right-hand side of (1) for $u=u_*$ we obtain the inequality $\Lambda (\mathbb {R}_+,p,\alpha )\le (1-p)\alpha ^\frac{p}{p-1}$. For the reverse inequality, we remark that $\lim _{x\rightarrow +\infty } u(x)=0$ for any $u\in W^{1,p}(\mathbb {R}_+)$. Indeed, since $|u|^{p-1} | u|'\in L^1(\mathbb {R}_+)$, by the the Hölder inequality, it follows that the limit

$$\begin{aligned} \lim _{x\rightarrow \infty } \big |u(x)\big |^p= \lim _{x\rightarrow \infty } p\int _0^{x} \big | u\big |^{p-1} | u|' \,\mathrm {d}t + \big |u(0)\big |^p \end{aligned}$$

exists and therefore must be equal to zero. Hence

$$\begin{aligned} \big |u(0)\big |^p= & {} -p\int _0^{+\infty } \big | u\big |^{p-1} | u|' \,\mathrm {d}t \le p\int _0^{+\infty } \big | u\big |^{p-1} \big || u|'\big | \,\mathrm {d}t\le p\Vert u\Vert _p^{p-1} \big \Vert |u|' \big \Vert _p\\\le & {} p\Vert u\Vert ^{p-1}_p \Vert u'\Vert _p, \end{aligned}$$

and we have

$$\begin{aligned}&\inf _{\begin{array}{c} u\in W^{1,p}(\mathbb {R}_+)\\ u\not \equiv 0 \end{array}}\dfrac{\displaystyle \int _0^{+\infty } \big |u'(t)\big |^p \mathrm {d}t - \alpha \big |u(0)\big |^p}{\displaystyle \int _0^{+\infty } |u(t)|^p\mathrm {d}t} = \inf _{\begin{array}{c} v\in W^{1,p}(\mathbb {R}_+)\\ \Vert v\Vert _p=1 \end{array}} \Big ( \Vert v'\Vert ^p_p - \alpha \big |v(0)\big |^p\Big )\\&\quad \ge \inf _{\begin{array}{c} v\in W^{1,p}(\mathbb {R}_+)\\ \Vert v\Vert _p=1 \end{array}} \Big ( \Vert v'\Vert ^p_p - \alpha p \Vert v'\Vert _p\Big ) \ge \inf _{x\in \mathbb {R}_+} (x^p-p\alpha x)=(1-p)\alpha ^\frac{p}{p-1}, \end{aligned}$$

which gives the sought result. $\square$

As observed in [25], the one-dimensional result can be used to study the infinite planar sectors

$$\begin{aligned} U_\theta := \big \{ (x_1,x_2)\in \mathbb {R}^2: \big |\arg (x_1+i x_2)\big |<\theta \big \}, \quad 0<\theta <\pi . \end{aligned}$$

Proceeding literally as in Lemma 2.6 and Lemma 2.8 of [25] one arrives at the following result:

Proposition 6.8

Let $\alpha >0$ and $p\in (1,+\infty )$, then for $\theta \ge \dfrac{\pi }{2}$ there holds $\Lambda (U_\theta ,p,\alpha )=(1-p)\alpha ^\frac{p}{p-1}$. On the other hand, for $\theta <\dfrac{\pi }{2}$ the infimum is attained on

$$\begin{aligned} u(x_1,x_2)=\exp \Big (- \Big (\dfrac{\alpha }{\sin \theta }\Big )^{\frac{1}{p-1}}x_1\Big ), \end{aligned}$$

and satisfies

$$\begin{aligned} \Lambda (U_\theta ,p,\alpha )=(1-p)\Big (\dfrac{\alpha }{\sin \theta }\Big )^\frac{p}{p-1}<(1-p)\alpha ^\frac{p}{p-1}. \end{aligned}$$

Appendix 2: An auxiliary inequality

Proposition 6.9

Let $p>1$, then for any $\varepsilon \in (0,1]$ and for all $a,b\ge 0$ there holds

$$\begin{aligned} (a+b)^p\le (1+\varepsilon ) a^p +\big (1-2^{\frac{1}{1-p}}\big )^{1-p}\, \dfrac{b^p}{\varepsilon ^{p-1}}\, . \end{aligned}$$

Proof

By homogenity, it is sufficient to show that $(1+t)^p\le (1+\varepsilon ) t^p+ c\varepsilon ^{1-p}$ for all $t\ge 0$. Denote $f_\varepsilon (t):=(1+t)^p-(1+\varepsilon ) t^p$, then one simply needs an upper estimate for $C(\varepsilon ):=\varepsilon ^{p-1}\sup _{t\in \mathbb {R}_+}f_\varepsilon (t)$. We have $f_\varepsilon (0)=1$ and $f_\varepsilon (+\infty )=-\infty$. The equation $f'_\varepsilon (t)=0$ has a unique solution

$$\begin{aligned} t=t_\varepsilon =\dfrac{1}{(1+\varepsilon )^{\frac{1}{p-1}}-1}, \quad f(t_\varepsilon )=\dfrac{1+\varepsilon }{\big ((1+\varepsilon )^{\frac{1}{p-1}}-1\big )^{p-1}}, \end{aligned}$$

implying

$$\begin{aligned} C(\varepsilon )=\max \Big \{\dfrac{\varepsilon ^{p-1}(1+\varepsilon )}{\big ((1+\varepsilon )^{\frac{1}{p-1}}-1\big )^{p-1}},1\Big \}. \end{aligned}$$

Denote $\sigma :=(1+\varepsilon )^{\frac{1}{p-1}}-1$, then

$$\begin{aligned} \sup _{\varepsilon \in (0,1)} \dfrac{\varepsilon ^{p-1}(1+\varepsilon )}{\big ((1+\varepsilon )^{\frac{1}{p-1}}-1\big )^{p-1}} =\sup _{\sigma \in (0,\sigma _1)}\Big (\dfrac{(1+\sigma )^p-(1+\sigma )}{\sigma }\Big )^{p-1}, \quad \sigma _1:= 2^{\frac{1}{p-1}}-1. \end{aligned}$$

Using the convexity of $\varphi (\sigma )=(1+\sigma )^p-(1+\sigma )$ we have

$$\begin{aligned} \varphi (\sigma )\le \varphi (0) + \dfrac{\varphi (\sigma _1)-\varphi (0)}{\sigma _1}\, \sigma \equiv \dfrac{2^{\frac{1}{p-1}}}{2^{\frac{1}{p-1}}-1}\, \sigma , \quad \sigma \in (0,\sigma _1). \end{aligned}$$

Since $\big (1-2^{\frac{1}{1-p}}\big )^{1-p} > 1$, this gives the sought inequality. $\square$

Appendix 3: Remainder estimates for more regular domains

If a stronger regularity of $\Omega$ is imposed, the remainder in Theorem 1.1 can be made more explicit.

Proposition 6.10

In Theorem 1.1 assume additionally that the boundary of $\Omega$ it $C^3$ smooth and that the mean curvature attains the maximum value $H_\mathrm {max}$, then the remainder estimate in (5) can be improved to $\mathcal {O}(\alpha ^{1-\kappa })$ with

$$\begin{aligned} \kappa ={\left\{ \begin{array}{ll} \dfrac{1}{p+1}, &{} p\in (1,2],\\ \dfrac{1}{3(p-1)}, &{} p \in (2,\infty ). \end{array}\right. } \end{aligned}$$

(48)

If, in addition, $\Omega$ is $C^4$ smooth, then one can take

$$\begin{aligned} \kappa ={\left\{ \begin{array}{ll} \dfrac{2}{p+2}, &{} p\in (1,2],\\ \dfrac{1}{2(p-1)}, &{} p \in (2,\infty ). \end{array}\right. } \end{aligned}$$

(49)

Proof

Remark first that the result of Sect. 5 imply

$$\begin{aligned} \Lambda (\Omega ,p,\alpha )\ge (1-p)\alpha ^{\frac{p}{p-1}}-(\nu -1) H_\mathrm {max}\alpha +\mathcal {O}(\alpha ^{\frac{p}{p-1}}\log \alpha ), \end{aligned}$$

and $\alpha ^{\frac{p}{p-1}}\log \alpha =o(\alpha ^{1-\kappa })$ for $\kappa$ given by (48) or (49). Therefore, it is sufficient to show the upper bound, which will be done by taking another test function in the computations of Sect. 4.

Let $s_0\in S$ be such that $M(s_0)=M_\mathrm {max}$. As $\partial \Omega$ is $C^3$ smooth, then M is $C^1$, and for some $m\ge 0$ we have $-M_\mathrm {max}\le -M(s)\le -M_\mathrm {max}+ m d(s,s_0)$ with $d(\cdot ,\cdot )$ standing for the geodesic distance on S and for s sufficiently close to $s_0$.

Let us choose a function $f\in C_c^\infty (\mathbb {R})$ which equals 1 in a neighborhood of the origin and consider the functions $v\in W^{1,p}(S)$ given by

$$\begin{aligned} v(s)= f\Big (\dfrac{d(s,s_0)}{\mu }\Big ), \end{aligned}$$

(50)

where $\mu$ is a positive parameter which tends to 0 as $\beta \rightarrow +\infty$ and will be chosen later. One can estimate, for large $\beta$,

$$\begin{aligned} \begin{aligned} A:=\int _S \big |v(s)\big |^p \mathrm {d}\sigma (s)&=a \mu ^{\nu -1}+\mathcal {O}(\mu ^\nu ),\quad A^{-1}=\mathcal {O}(\mu ^{1-\nu }),\\ \int _S \big (-M(s)\big ) \big |v(s)\big |^p \mathrm {d}\sigma (s)&= -M_\mathrm {max}A + \mathcal {O}(\mu ^\nu ),\\ \int _S \big |\nabla v(s)\big |^p\mathrm {d}\sigma (s)&=\mathcal {O}(\mu ^{\nu -p-1}). \end{aligned} \end{aligned}$$

(51)

Consider first the case $p\in (1,2]$. The substitution into (18) gives

$$\begin{aligned}&\Lambda ^+(p,\alpha )\le \Bigg \{ \mathcal {O}(\mu ^{\nu -p-1}\beta ^{-1}) +\,\dfrac{1}{p}\beta ^{p-1} A - \dfrac{1}{p^2} \beta ^{p-2}M_\mathrm {max}A+\mathcal {O}(\mu ^\nu \beta ^{p-2})\\&\qquad +\mathcal {O}(\mu ^{\nu -1}\beta ^{p-3}) - \beta ^{p-1} A\Bigg \}\\&\qquad \times \Big \{ \dfrac{1}{p\beta } A - \dfrac{M_\mathrm {max}}{p^2\beta ^2} A + \mathcal {O}(\mu ^\nu \beta ^{-2})+\mathcal {O}(\mu ^{\nu -1}\beta ^{-3}) \Big \}^{-1}\\&\quad =\Big \{ \dfrac{1-p}{p} \beta ^{p-1} A - \dfrac{1}{p^2} M_\mathrm {max}\beta ^{p-2} A + \mathcal {O}(\mu ^{\nu -p-1}\beta ^{-1}+\mu ^\nu \beta ^{p-2}+\mu ^{\nu -1}\beta ^{p-3}) \Big \}\\&\qquad \times \Big \{ \dfrac{A}{p\beta } \Big (1 - \dfrac{M_\mathrm {max}}{p\beta }+ \mathcal {O}(\mu \beta ^{-1}+\beta ^{-2}) \Big ) \Big \}^{-1}\\&\quad =\Big ( (1-p)\beta ^p - \dfrac{1}{p} M _\mathrm {max}\beta ^{p-1}+ \mathcal {O}\big (\mu ^{-p}+\mu \beta ^{p-1}+\beta ^{p-2}\big ) \Big )\\&\qquad \times \Big (1 + \dfrac{M_\mathrm {max}}{p\beta }+ \mathcal {O}(\mu \beta ^{-1}+\beta ^{-2}) \Big )\\&\quad =(1-p)\beta ^p - M_\mathrm {max}\beta ^{p-1} +\mathcal {O}\big ( \mu ^{-p}+\mu \beta ^{p-1}+\beta ^{p-2}\big ). \end{aligned}$$

The remainder is optimized by $\mu =\beta ^{-\frac{p-1}{p+1}}$, and we arrive

$$\begin{aligned} \Lambda ^+(p,\alpha )&\le (1-p)\beta ^p - M_\mathrm {max}\beta ^{p-1} + \mathcal {O}(\beta ^{p-1 - \frac{p-1}{p+1}})\\&= (1-p)\alpha ^\frac{p}{p-1} - M_\mathrm {max}\alpha + \mathcal {O}\big (\alpha ^{1- \frac{1}{p+1}}\big ). \end{aligned}$$

If, in addition, $\Omega$ is $C^4$ smooth, then M is $C^2$ smooth, and for some $m\ge 0$ we have $-M_\mathrm {max}\le -M(s)\le -M_\mathrm {max}+ m d(s,s_0)^2$ as s is sufficiently close to $s_0$, which allows one to replace the second estimate in (51) by

$$\begin{aligned} \int _S \big (-M(s)\big ) \big |v(s)\big |^p \mathrm {d}\sigma (s)= -M_\mathrm {max}A + \mathcal {O}(\mu ^{\nu +1}), \end{aligned}$$

(52)

and a similar computation gives

$$\begin{aligned} \Lambda ^+(p,\alpha )\le (1-p)\beta ^p - M_\mathrm {max}\beta ^{p-1} +\mathcal {O}\big ( \mu ^{-p}+\mu ^2 \beta ^{p-1}+\beta ^{p-2}\big ). \end{aligned}$$

Hence, taking $\mu =\beta ^{-\frac{p-1}{p+2}}$ we arrive at

$$\begin{aligned} \Lambda ^+(p,\alpha )&\le (1-p)\beta ^p - M_\mathrm {max}\beta ^{p-1} + \mathcal {O}(\beta ^{p-1 - \frac{2(p-1)}{p+2}})\\&= (1-p)\alpha ^\frac{p}{p-1} - M_\mathrm {max}\alpha + \mathcal {O}\big (\alpha ^{1- \frac{2}{p+2}}\big ). \end{aligned}$$

For the case $p\in (2,+\infty )$, the substitution of (51) into (23) gives

$$\begin{aligned} \Lambda ^+(p,\alpha )&\le \Big \{ \dfrac{1-p}{p} \beta ^{p-1} A - \dfrac{M_\mathrm {max}}{p^2} \beta ^{p-2} A + \mathcal {O}\big ( \varepsilon _0^\frac{2-p}{2}\mu ^{\nu -p-1}\beta ^{-1} + \mu ^\nu \beta ^{p-2}\\&\quad {}+ \mu ^{\nu -1}\beta ^{p-3} +\varepsilon _0 \mu ^{\nu -1}\beta ^{p-1} \big )\Big \} \times \Big \{ \dfrac{A}{p\beta } \Big (1 - \dfrac{M_\mathrm {max}}{p\beta }+ \mathcal {O}(\mu \beta ^{-1}+\beta ^{-2}) \Big ) \Big \}^{-1}\\&=\Big ( (1-p)\beta ^p - \dfrac{1}{p} M _\mathrm {max}\beta ^{p-1}+ \mathcal {O}\big (\varepsilon _0^\frac{2-p}{2}\mu ^{-p}+\mu \beta ^{p-1}+\beta ^{p-2} +\varepsilon _0 \beta ^p\big ) \Big )\\&\quad \times \Big (1 + \dfrac{M_\mathrm {max}}{p\beta }+ \mathcal {O}(\mu \beta ^{-1}+\beta ^{-2}) \Big )\\&=(1-p)\beta ^p - M _\mathrm {max}\beta ^{p-1}+ \mathcal {O}\big (\varepsilon _0^\frac{2-p}{2}\mu ^{-p}+\mu \beta ^{p-1}+\beta ^{p-2} +\varepsilon _0 \beta ^p\big ). \end{aligned}$$

In order to optimize we solve $\varepsilon _0^\frac{2-p}{2}\mu ^{-p}=\mu \beta ^{p-1}=\varepsilon _0 \beta ^p$, which gives

$$\begin{aligned} \mu =\beta ^{-1/3}, \quad \varepsilon _0=\beta ^{-4/3}, \quad \mathcal {O}\big (\varepsilon _0^\frac{2-p}{2}\mu ^{-p}+\mu \beta ^{p-1}+\beta ^{p-2} +\varepsilon _0 \beta ^p\big )=\mathcal {O}(\beta ^{p-\frac{4}{3}}), \end{aligned}$$

and, therefore,

$$\begin{aligned} \Lambda ^+(p,\alpha )\le & {} (1-p)\beta ^p - M_\mathrm {max}\beta ^{p-1} + \mathcal {O}(\beta ^{p-\frac{4}{3}}) = (1-p)\alpha ^\frac{p}{p-1}\\&- M_\mathrm {max}\alpha + \mathcal {O}\big (\alpha ^{1- \frac{1}{3(p-1)}}\big ). \end{aligned}$$

For $C^4$ domains, a similar computation using (52) gives

$$\begin{aligned} \Lambda ^+(p,\alpha )\le (1-p)\beta ^p - M _\mathrm {max}\beta ^{p-1}+ \mathcal {O}\big (\varepsilon _0^\frac{2-p}{2}\mu ^{-p}+\mu ^2 \beta ^{p-1}+\beta ^{p-2} +\varepsilon _0 \beta ^p\big ). \end{aligned}$$

The remainder is optimized by

$$\begin{aligned} \mu =\beta ^{-1/4}, \quad \varepsilon _0=\beta ^{-3/2}, \quad \mathcal {O}\big (\varepsilon _0^\frac{2-p}{2}\mu ^{-p}+\mu \beta ^{p-1}+\beta ^{p-2} +\varepsilon _0 \beta ^p\big )=\mathcal {O}(\beta ^{p-\frac{3}{2}}), \end{aligned}$$

which gives the sought result. We remark that for $p=2$ in (49) we obtain $\kappa =1/2$, which is optimal, see [23, 32]. $\square$

Finally, an easy revision of the proof of Theorem 6.6 gives the following result:

Corollary 6.11

Under the assumptions of Proposition 6.10, the right-hand-side of (45) can be improved to $\mathcal {O}(\alpha ^{-\kappa })$.