On the <Emphasis Type="Italic">p</Emphasis>-Laplacian with Robin boundary conditions and boundary trace theorems

Hynek Kovařík; Konstantin Pankrashkin

doi:10.1007/s00526-017-1138-4

Calculus of Variations and Partial Differential Equations

April 2017, 56:49 | Cite as

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

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Hynek KovaříkEmail author
Konstantin Pankrashkin

Article

First Online: 21 March 2017

Received: 28 September 2016

Accepted: 20 February 2017

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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.

Mathematics Subject Classification

49R05 35P30 58C40

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Notes

Acknowledgements

We thank the referee for a detailed reading of our paper and for many valuable comments, in particular, on the computations of Remark 6.3. H. K. has been partially supported by Gruppo Nazionale per Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM). The support of MIUR-PRIN2010-11 grant for the project “Calcolo delle variazioni” (H. K.), is also gratefully acknowledged. K.P. has been partially supported by CNRS GDR 2279 DynQua. The authors thank Carlo Nitsch and Cristina Trombetti for useful comments on a preliminary version of the work.

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 $\mathcal {O}(\alpha ^{-\kappa })$.

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Authors and Affiliations

Hynek Kovařík
- 1
Email author
Konstantin Pankrashkin
- 2

1.DICATAM, Sezione di MatematicaUniversità degli studi di BresciaBresciaItaly
2.Laboratoire de Mathématiques d’OrsayUniv.Paris-Sud, CNRS, Université Paris-SaclayOrsayFrance

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

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