A Capillary Surface with No Radial Limits

TL;DR

This paper constructs a capillary surface in a convex domain with a corner that exhibits no radial limits at the boundary point, extending previous examples to cases where the contact angle remains bounded away from zero and pi.

Contribution

It generalizes prior work by demonstrating a capillary surface with no radial limits in a convex corner domain with bounded contact angles.

Findings

Existence of a capillary surface with no radial limits at a convex corner.

The contact angle remains bounded away from zero and pi.

Extension of previous non-radial limit examples to convex corners.

Abstract

In 1996, Kirk Lancaster and David Siegel investigated the existence and behavior of radial limits at a corner of the boundary of the domain of solutions of capillary and other prescribed mean curvature problems with contact angle boundary data. In Theorem 3, they provide an example of a capillary surface in a unit disk $D$ which has no radial limits at $(0,0)\in\partial D.$ In their example, the contact angle ($\gamma$) cannot be bounded away from zero and $\pi.$ Here we consider a domain $\Omega$ with a convex corner at $(0,0)$ and find a capillary surface $z=f(x,y)$ in $\Omega\times\mathbb{R}$ which has no radial limits at $(0,0)\in\partial\Omega$ such that $\gamma$ is bounded away from $0$ and $\pi.$

Equations

$${\partial }^{+}{\Omega }^{\ast }=\{(r\cos (\alpha ),r\sin (\alpha )):0\le r\le {\delta }^{\ast }\}$$

$${\partial }^{-}{\Omega }^{\ast }=\{(r\cos (-\alpha ),r\sin (-\alpha )):0\le r\le {\delta }^{\ast }\}.$$

$$\mathrm{div}(Tf)=\frac{1}{2}f\mathrm{in}\Omega$$

$$Tf\cdot \nu =\cos \left(\gamma \right)\mathrm{on}\partial \Omega \setminus \{\mathcal{O}\},$$

$$Rf(\theta )=\underset{r\to 0^{+}}{\lim }f(r\cos \theta ,r\sin \theta ),-\alpha <\theta <\alpha$$

$$\pi -2\alpha <{\underline{\gamma }}^{+}+{\underline{\gamma }}^{-}\le {\overline{\gamma }}^{+}+{\overline{\gamma }}^{-}<\pi +2\alpha$$

$$(-1{)}^{j}f\left(r_j,0\right)>1\mathrm{for}\mathrm{each}j\in \mathrm{IN}.$$

$$Rf(\theta )\stackrel{\text{makebox}[0.0pt]\text{mbox}def}{=}\underset{r\downarrow 0}{\lim }f(r\cos (\theta ),r\sin (\theta ))$$

$$\underset{{\Omega }_{\delta }}{\sup }|g-h|<\eta \mathrm{and}\left|{\gamma }_g-\frac{\pi }{2}\right|\le \alpha +ϵ,$$

$$\gamma_{\beta}=\left\{\begin{array}[]{ccc}\frac{\pi}{2}-\alpha-\epsilon&{\rm on}&\overline{B_{\beta}\left({\cal O}\right)}\\ \gamma_{0}&{\rm on}&\overline{\Omega}\setminus B_{\beta}\left({\cal O}\right).\\ \end{array}\right.$$

$$w(r\cos \theta ,r\sin \theta )=\frac{\cos \theta -\sqrt{k^2-{\sin }^2\theta }}{k\kappa r},$$

$$M=\underset{\Omega \cap \partial B_{{\delta }_1}(\mathcal{O})}{\sup }|w-g_{\beta }|\mathrm{and}{w}_{\beta }=w-M.$$

$$(-1{)}^j{f}_n\left(r_j,0\right)>1\mathrm{for}\mathrm{each}j=1,\dots ,n.$$

$$f_1\left(r_1,0\right)+1<f_2\left(r_1,0\right)-f_1\left(r_1,0\right)<-\left(f_1\left(r_1,0\right)+1\right)$$

$$f_2\left(r_1,0\right)+1<f_3\left(r_1,0\right)-f_2\left(r_1,0\right)<-\left(f_2\left(r_1,0\right)+1\right)$$

$$-\left(f_2\left(r_2,0\right)-1\right)<f_3\left(r_2,0\right)-f_2\left(r_2,0\right)<f_2\left(r_2,0\right)-1;$$

$${\eta }_{n+1}=\underset{1\le j\le n}{\min }|f_n(r_j,0)-(-1{)}^j|\mathrm{and}{\delta }_{n+1}=\min \left\{r_n,\frac{1}{n}\right\}.$$

$$-\left(f_n\left(r_j,0\right)-1\right)<f_{n+1}\left(r_j,0\right)-f_n\left(r_j,0\right)<f_n\left(r_j,0\right)-1$$

$$f_n\left(r_j,0\right)+1<f_{n+1}\left(r_j,0\right)-f_n\left(r_j,0\right)<-\left(f_n\left(r_j,0\right)+1\right)$$

$$\underset{r\to 0^{+}}{\lim }f(r,0)\mathrm{does}\mathrm{not}\mathrm{exist},\mathrm{even}\mathrm{as}\mathrm{an}\mathrm{infinite}\mathrm{limit},$$

Full text

1 Introduction

Let $\Omega$ be a domain in ${{\rm IR}}^{2}$ with locally Lipschitz boundary and ${\cal O}=(0,0)\in\partial\Omega$ such that $\partial\Omega\setminus\{{\cal O}\}$ is a $C^{4}$ curve and $\Omega\subset B_{1}\left(0,1\right),$ where $B_{\delta}\left({\cal N}\right)$ is the open ball in ${\rm IR}^{2}$ of radius $\delta$ about ${\cal N}\in{{\rm IR}}^{2}.$ Denote the unit exterior normal to $\Omega$ at $(x,y)\in\partial\Omega$ by $\nu(x,y)$ and let polar coordinates relative to ${\cal O}$ be denoted by $r$ and $\theta.$ We shall assume there exists a $\delta^{*}\in(0,2)$ and $\alpha\in\left(0,\frac{\pi}{2}\right)$ such that $\partial\Omega\cap B_{\delta^{*}}({\cal O})$ consists of the line segments

[TABLE]

and

[TABLE]

Set $\Omega^{*}=\Omega\cap B_{\delta^{*}}({\cal O}).$ Let $\gamma:\partial\Omega\to[0,\pi]$ be given. Let $\left(x^{\pm}(s),y^{\pm}(s)\right)$ be arclength parametrizations of $\partial^{\pm}\Omega$ with $\left(x^{+}(0),y^{+}(0)\right)=\left(x^{-}(0),y^{-}(0)\right)=(0,0)$ and set $\gamma^{\pm}(s)=\gamma\left(x^{\pm}(s),y^{\pm}(s)\right).$

Consider the capillary problem of finding a function $f\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\})$ satisfying

[TABLE]

and

[TABLE]

where $Tf=\frac{\nabla f}{\sqrt{1+|\nabla f|^{2}}}.$ We are interested in the existence of the radial limits $Rf(\cdot)$ of a solution $f$ of (1)–(2), where

[TABLE]

and $Rf(\pm\alpha)=\lim_{\partial^{\pm}\Omega^{*}\ni{\bf x}\rightarrow{\cal O}}f({\bf x}),{\bf x}=(x,y)$, which are the limits of the boundary values of $f$ on the two sides of the corner if these exist. In [2], the following is proven:

Proposition 1**.**

Let $f$ be a bounded solution to (1) satisfying (2) on $\partial^{\pm}\Omega^{*}\setminus\{{\cal O}\}$ which is discontinuous at ${\cal O}.$ If $\alpha>\pi/2$ then $Rf(\theta)$ exists for all $\theta\in(-\alpha,\alpha).$ If $\alpha\leq\pi/2$ and there exist constants $\underline{\gamma}^{\,\pm},\overline{\gamma}^{\,\pm},0\leq\underline{\gamma}^{\,\pm}\leq\overline{\gamma}^{\,\pm}\leq\pi,$ satisfying

[TABLE]

so that $\underline{\gamma}^{\pm}\leq\gamma^{\pm}(s)\leq\overline{\gamma}^{\,\pm}$ for all $s,0<s<s_{0},$ for some $s_{0}$, then again $Rf(\theta)$ exists for all $\theta\in(-\alpha,\alpha)$.

In [5], Lancaster and Siegel proved this theorem with the additional restriction that $\gamma$ be bounded away from [math] and $\pi;$ Figure 1 illustrates these cases.

In Theorem 3 of [5], Lancaster and Siegel also proved

Proposition 2**.**

Let $\Omega$ be the disk of radius $1$ centered at $(1,0).$ Then there exists a solution to $Nf=\frac{1}{2}f$ in $\Omega,|f|\leq 2,f\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus O),O=(0,0)$ so that no radial limits $Rf(\theta)$ exist ($\theta\in[-\pi/2,\pi/2]$).

In this case, $\alpha=\frac{\pi}{2};$ if $\gamma$ is bounded away from [math] and $\pi,$ then Proposition 1 would imply that $Rf(\theta)$ exists for each $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and therefore the contact angle $\gamma=\cos^{-1}\left(Tf\cdot\nu\right)$ in Proposition 2 is not bounded away from [math] and $\pi.$

In our case, the domain $\Omega$ has a convex corner of size $2\alpha$ at ${\cal O}$ and we wish to investigate the question of whether an example like that in Proposition 2 exists in this case when $\gamma$ is bounded away from [math] and $\pi.$ In terms of the Concus-Finn rectangle, the question is whether, given $\epsilon>0,$ there is a $f\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\})$ of (1)–(2) such that no radial limits $Rf(\theta)$ exist ($\theta\in[-\alpha,\alpha]$) and $|\gamma-\frac{\pi}{2}|\leq\alpha+\epsilon;$ this is illustrated in Figure 2.

Theorem 1**.**

For each $\epsilon>0$, there is a domain $\Omega$ as described above and a solution $f\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\})$ of (1) such that the contact angle $\gamma=\cos^{-1}\left(Tf\cdot\nu\right):\partial\Omega\setminus\{{\cal O}\}\to[0,\pi]$ satisfies $|\gamma-\frac{\pi}{2}|\leq\alpha+\epsilon$ and there exist a sequence $\{r_{j}\}$ in $(0,1)$ with $\lim_{j\to\infty}r_{j}=0$ such that

[TABLE]

Assuming $\Omega$ and $\gamma$ are symmetric with respect to the line $\{(x,0):x\in{\rm IR}\},$ this implies that no radial limit

[TABLE]

exists for any $\theta\in[-\alpha,\alpha].$

We note that our Theorem is an extension of Theorem 3 of [5] to contact angle data in a domain with a convex corner. As in [4, 5], we first state and prove a localization lemma; this is analogous to the Lemma in [4] and Lemma 2 of [5].

Lemma 1**.**

Let $\Omega\subseteq{{\rm IR}}^{2}$ be as above, $\epsilon>0,$ $\eta>0$ and $\gamma_{0}:\partial\Omega\setminus\{{\cal O}\}\to[0,\pi]$ such that $|\gamma_{0}-\frac{\pi}{2}|\leq\alpha+\epsilon.$ For each $\delta\in(0,1)$ and $h\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\})$ which satisfies (1) and (2) with $\gamma=\gamma_{0},$ there exists a solution $g\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\})$ of (1) such that $\lim_{\overline{\Omega}\ni(x,y)\to(0,0)}g(x,y)=+\infty,$

[TABLE]

where $\Omega_{\delta}=\overline{\Omega}\setminus B_{\delta}\left({\cal O}\right)$ and $\gamma_{g}=\cos^{-1}\left(Tg\cdot\nu\right):\partial\Omega\setminus\{{\cal O}\}\to[0,\pi]$ is the contact angle which the graph of $g$ makes with $\partial\Omega\times{\rm IR}.$

Proof.

Let $\epsilon,\eta,\delta,\Omega,h$ and $\gamma_{0}$ be given. For $\beta\in(0,\delta)$, let $g_{\beta}\in C^{2}\left(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\}\right)$ satisfy (1) and (2) with $\gamma=\gamma_{\beta},$ where

[TABLE]

As in the proof of Theorem 3 of [5], $g_{\beta}$ converges to $h,$ pointwise and uniformly in the $C^{1}$ norm on $\overline{\Omega_{\delta}}$ as $\beta$ tends to zero. Fix $\beta>0$ small enough that $\sup_{\Omega_{\delta}}|g-h|<\eta.$

Set $\Sigma=\{(r\cos(\theta),r\sin(\theta)):r>0,-\alpha\leq\theta\leq\alpha\}.$ Now define $w:\Sigma\to{\rm IR}$ by

[TABLE]

where $k=\sin\alpha\sec\left(\frac{\pi}{2}-\alpha-\epsilon\right)=\sin\alpha\csc(\alpha+\epsilon).$ As in [1], there exists a $\delta_{1}>0$ such that ${\rm div}(Tw)-\frac{1}{2}w\geq 0$ on $\Sigma\cap B_{\delta_{1}}({\cal O}),$ $Tw\cdot\nu=\cos\left(\frac{\pi}{2}-\alpha-\epsilon\right)$ on $\partial\Sigma\cap B_{\delta_{1}}({\cal O}),$ and $\lim_{r\to 0^{+}}w(r\cos\theta,r\sin\theta)=\infty$ for each $\theta\in[-\alpha,\alpha].$ We may assume $\delta_{1}\leq\delta^{*}.$ Let

[TABLE]

Since ${\rm div}(Tw_{\beta})-\frac{1}{2}w_{\beta}\geq\frac{M}{2}\geq 0={\rm div}(Tg_{\beta})-\frac{1}{2}g_{\beta}$ in $\Omega\cap B_{\delta_{1}}({\cal O}),$ $w_{\beta}\leq g_{\beta}$ on $\Omega\cap\partial B_{\delta_{1}}({\cal O})$ and $Tg_{\beta}\cdot\nu\geq Tw_{\beta}\cdot\nu$ on $\partial\Omega\cap B_{\delta_{1}}({\cal O}),$ we see that $g_{\beta}\geq w_{\beta}$ on $\Omega\cap\partial B_{\delta_{1}}({\cal O}).$ ∎

We may now prove Theorem 1.

Proof.

We shall construct a sequence ${f_{n}}$ of solutions of (1) and a sequence $\{r_{n}\}$ of positive real numbers such that $\lim_{n\to\infty}r_{n}=0,$ $f_{n}(x,y)$ is even in $y$ and

[TABLE]

Let $\gamma_{0}=\frac{\pi}{2}$ and $f_{0}=0.$ Set $\eta_{1}=1$ and $\delta_{1}=\delta_{0}.$ From Lemma 1, there exists a $f_{1}\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\})$ which satisfies (1) such that $\sup_{\Omega_{\delta_{1}}}|f_{1}-f_{0}|<\eta_{1},$ $\left|\gamma_{1}-\frac{\pi}{2}\right|\leq\alpha+\epsilon$ and $\lim_{\Omega\ni(x,y)\to{\cal O}}f_{1}(x,y)=-\infty,$ where $\gamma_{1}=\cos^{-1}\left(Tf_{1}\cdot\nu\right).$ Then there exists $r_{1}\in\left(0,\delta_{1}\right)$ such that $f_{1}\left(r_{1},0\right)<-1.$

Now set $\eta_{2}=-\left(f_{1}\left(r_{1},0\right)+1\right)>0$ and $\delta_{2}=r_{1}.$ From Lemma 1, there exists a $f_{2}\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\})$ which satisfies (1) such that $\sup_{\Omega_{\delta_{2}}}|f_{2}-f_{1}|<\eta_{2},$ $\left|\gamma_{2}-\frac{\pi}{2}\right|\leq\alpha+\epsilon$ and $\lim_{\Omega\ni(x,y)\to{\cal O}}f_{2}(x,y)=\infty,$ where $\gamma_{2}=\cos^{-1}\left(Tf_{2}\cdot\nu\right).$ Then there exists $r_{2}\in\left(0,\delta_{2}\right)$ such that $f_{2}(r_{2},0)>1.$ Since $(r_{1},0)\in\Omega_{\delta_{2}},$

[TABLE]

and so $f_{2}\left(r_{1},0\right)<-1.$

Next set $\eta_{3}=\min\left\{-\left(f_{2}\left(r_{1},0\right)+1\right),f_{2}\left(r_{2},0\right)-1\right\}>0$ and $\delta_{3}=r_{2}.$ From Lemma 1, there exists a $f_{3}\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\})$ which satisfies (1) such that $\sup_{\Omega_{\delta_{3}}}|f_{3}-f_{2}|<\eta_{3},$ $\left|\gamma_{3}-\frac{\pi}{2}\right|\leq\alpha+\epsilon$ and $\lim_{\Omega\ni(x,y)\to{\cal O}}f_{3}(x,y)=-\infty,$ where $\gamma_{3}=\cos^{-1}\left(Tf_{3}\cdot\nu\right).$ Then there exists $r_{3}\in\left(0,\delta_{3}\right)$ such that $f_{3}(r_{3},0)<-1.$ Since $(r_{1},0),(r_{2},0)\in\Omega_{\delta_{2}},$ we have

[TABLE]

and

[TABLE]

hence $f_{3}\left(r_{1},0\right)<-1$ and $1<f_{3}\left(r_{2},0\right).$

Continuing to define $f_{n}$ and $r_{n}$ inductively, we set

[TABLE]

From Lemma 1, there exists $f_{n+1}\in C^{2}(\Omega)\cap C^{1}(\overline{\Omega}\setminus\{{\cal O}\})$ which satisfies (1) such that $\sup_{\Omega_{\delta_{n+1}}}|f_{n+1}-f_{n}|<\eta_{n+1},$ $\left|\gamma_{n+1}-\frac{\pi}{2}\right|\leq\alpha+\epsilon$ and $\lim_{\Omega\ni(x,y)\to{\cal O}}f_{n+1}(x,y)=(-1)^{n+1}\infty,$ where $\gamma_{n+1}=\cos^{-1}\left(Tf_{n+1}\cdot\nu\right).$ Then there exists $r_{n+1}\in\left(0,\delta_{n+1}\right)$ such that $(-1)^{n+1}f_{n+1}(r_{n+1},0)>1.$ For each $j\in\{1,\dots,n\}$ which is an even number, we have

[TABLE]

and so $1<f_{n+1}\left(r_{j},0\right).$ For each $j\in\{1,\dots,n\}$ which is an odd number, we have

[TABLE]

and so $f_{n+1}\left(r_{j},0\right)<-1.$

As in [5, 6], there is a subsequence of $\{f_{n}\},$ still denoted $\{f_{n}\},$ which converges pointwise and uniformly in the $C^{1}$ norm on $\overline{\Omega_{\delta}}$ for each $\delta>0$ as $n\to\infty$ to a solution $f\in C^{2}(\Omega)\cap C^{1}\left(\overline{\Omega}\setminus{\cal O}\right)$ of (1). For each $j\in{\rm IN}$ which is even, $f_{n}\left(r_{j},0\right)>1$ for each $n\in{\rm IN}$ and so $f\left(r_{j},0\right)\geq 1.$ For each $j\in{\rm IN}$ which is odd, $f_{n}\left(r_{j},0\right)<-1$ for each $n\in{\rm IN}$ and so $f\left(r_{j},0\right)\leq-1.$ Therefore

[TABLE]

and so $Rf(0)$ does not exist.

Since $\Omega$ is symmetric with respect to the $x-$axis and $\gamma_{n}(x,y)$ is an even function of $y,$ $f(x,y)$ is an even function of $y.$ Now suppose that there exists $\theta_{0}\in[-\alpha,\alpha]$ such that $Rf(\theta_{0})$ exists; then $\theta_{0}\neq 0.$ From the symmetry of $f,$ $Rf(-\theta_{0})$ must also exist and $Rf(-\theta_{0})=Rf(\theta_{0})$. Set $\Omega^{\prime}=\{(r\cos\theta,r\sin\theta):0<r<\delta_{0},-\theta_{0}<\theta<\theta_{0}\}\subset\Omega.$ Since $f$ has continuous boundary values on $\partial\Omega^{\prime},$ $f\in C^{0}\left(\overline{\Omega^{\prime}}\right)$ and so $Rf(0)$ does exist, which is a contradiction. Thus $Rf(\theta)$ does not exist for any $\theta\in[-\alpha,\alpha].$ ∎