Regularity of minimizers of a tensor-valued variational obstacle problem in three dimensions

Abstract

Motivated by Ball and Majumdar’s modification of Landau-de Gennes model for nematic liquid crystals, we study energy-minimizer Q of a tensor-valued variational obstacle problem in a bounded 3-D domain with prescribed boundary data. The energy functional is designed to blow up as Q approaches the obstacle. Under certain assumptions, especially on blow-up profile of the singular bulk potential, we prove higher interior regularity of Q, and show that the contact set of Q is either empty, or small with characterization of its Hausdorff dimension. We also prove boundary partial regularity of the energy-minimizer.

Appendices

Proof of (2.6)

In order to prove (2.6), we first prove the following lemma.

Lemma A.1

Let $M,\,N,\,P$ be $3\times 3$-symmetric traceless matrices. Then

$$\begin{aligned} \Vert M\Vert _2^2\le \frac{2}{3}|M|^2, \end{aligned}$$

(A.1)

and

$$\begin{aligned} \sum \limits _{i=1}^3(M_{i1}+N_{i2}+P_{i3})^2\le \frac{5}{3}(|M|^2+|N|^2+|P|^2). \end{aligned}$$

(A.2)

Proof

To prove (A.1), we can only consider the case when M is diagonal, due to the fact that M is symmetric and both Frobenius norm and matrix 2-norm are unitarily invariant. Without loss of generality we assume $M=\mathrm {diag}\{\lambda _1,\lambda _2,-\lambda _1-\lambda _2\}$ with $\lambda _1\lambda _2\ge 0$, we deduce that

$$\begin{aligned} 2|M|^2-3\Vert M\Vert _2^2 =2(\lambda _1^2+\lambda _2^2+(\lambda _1+\lambda _2)^2)-3(\lambda _1+\lambda _2)^2=2(\lambda _1-\lambda _2)^2\ge 0. \end{aligned}$$

Then (A.1) follows. The equality holds if and only if M has two equal eigenvalues.

For (A.2), we compute

$$\begin{aligned} \begin{aligned}&5(|M|^2+|N|^2+|P|^2)-3\sum \limits _{i=1}^3(M_{i1}+N_{i2}+P_{i3})^2\\&\quad \ge \left( 5(M_{11}^2+M_{22}^2+M_{33}^2+2N_{12}^2+2P_{13}^2)-3(M_{11}+N_{12}+P_{13})^2 \right) \\&\qquad +\left( 5(N_{11}^2+N_{22}^2+N_{33}^2+2M_{21}^2+2P_{23}^2)-3(M_{21}+N_{22}+P_{23})^2 \right) \\&\qquad +\left( 5(P_{11}^2+P_{22}^2+P_{33}^2+2M_{31}^2+2N_{32}^2)-3(M_{31}+N_{32}+P_{33})^2 \right) . \end{aligned} \end{aligned}$$

(A.3)

The equality holds if and only if $M_{23}=N_{13}=P_{12}=0$. It suffices to prove non-negativity of each term on the right hand side of (A.3). We only show this for the first term; the others can be handled similarly.

$$\begin{aligned}&5(M_{11}^2+M_{22}^2+M_{33}^2+2N_{12}^2+2P_{13}^2)-3(M_{11}+N_{12}+P_{13})^2\\&\quad \ge \left( 5+\frac{5}{2}-3\right) M_{11}^2+7N_{12}^2+7P_{13}^2-6M_{11}N_{12}-6M_{11}P_{13}-6N_{12}P_{13}\\&\quad = \left( \frac{3}{2}M_{11}-2N_{12}\right) ^2+\left( \frac{3}{2} M_{11}-2P_{13}\right) ^2+3(N_{12}-P_{13})^2\ge 0. \end{aligned}$$

Here we used $\sum _{i=1}^3 M_{ii}=0$ in the second line. The equality holds if and only if

$$\begin{aligned} N_{12}=P_{13}=\frac{3}{4}M_{11}, \quad M_{22}=M_{33}=-\frac{1}{2} M_{11}. \end{aligned}$$

This completes the proof of (A.2). $\square$

To this end, (2.6) follows immediately from the lemma if we take $M=D_k^h Q$ in (A.1), and take $M=D_k^h \partial _1 Q,\, N=D_k^h \partial _2 Q,\, P=D_k^h \partial _3 Q$ in (A.3).

Formula of p(A)

Lemma B.1

For $A>-\frac{3}{5}$, let

$$\begin{aligned} p(A) {:}{=} \sup _{\omega \in [0,1],\,\omega +\frac{5}{3}A\ge 0}p(A,\omega ), \end{aligned}$$

where $p(A,\omega )$ is defined in (2.7). Then p(A) is given by (1.10).

Proof

We rewrite (2.7) as

$$\begin{aligned} p(A,\omega ) = 1+\frac{\frac{9}{5}\left( \omega +\frac{5}{3} A\right) +\sqrt{\frac{9}{5}\left( \omega +\frac{5}{3} A\right) \left[ \left( \frac{9}{5}-2A^2\right) \left( \omega +\frac{5}{3} A\right) +2A^2\left( 1+\frac{5}{3}A\right) \right] }}{2A^2}. \end{aligned}$$

(B.1)

It is easy to see that if $\frac{9}{5}-2A^2 \ge 0$, $p(A,\omega )$ achieves its supremum at $\omega = 1$, which gives

$$\begin{aligned} p(A) = 1+\frac{3}{A}+\frac{9}{5A^2}. \end{aligned}$$

It suffices to consider $2A^2 > \frac{9}{5}$, i.e., $A> \frac{3\sqrt{10}}{10}$. Define

$$\begin{aligned} g(y) {:}{=} \frac{9}{5}y+\sqrt{\frac{9}{5}y(B_1y+B_2)}, \end{aligned}$$

where

$$\begin{aligned} B_1 = \frac{9}{5}-2A^2,\quad B_2 = 2A^2\left( 1+\frac{5}{3}A\right) . \end{aligned}$$

Then

$$\begin{aligned} p(A,\omega ) = 1+(2A^2)^{-1}g\left( \omega +\frac{5}{3} A\right) . \end{aligned}$$

Since

$$\begin{aligned} g'(y) = \frac{9}{5}+\sqrt{\frac{9}{5}}\cdot \frac{2B_1 y+ B_2}{2\sqrt{y(B_1y+B_2)}}, \end{aligned}$$

we find that $g'(y)<0$ if and only if

$$\begin{aligned} B_1 y +\frac{B_2}{2}< -\sqrt{\frac{9}{5} y(B_1y+B_2)}, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \left( B_1 y +\frac{B_2}{2}\right) ^2>\frac{9}{5} y(B_1y+B_2)\quad \text{ and } \quad B_1y +\frac{B_2}{2}<0. \end{aligned}$$

Solving these inequalities under the assumption $A>\frac{3\sqrt{10}}{10}$, we find that

$$\begin{aligned} y>\frac{A\left( 1+\frac{5}{3}A\right) }{2A- \sqrt{\frac{18}{5}}}=:y_{+}. \end{aligned}$$

This implies that within the domain of g(y), i.e.,

$$\begin{aligned} y = \omega +\frac{5}{3}A \in \left[ \frac{5}{3} A, 1+\frac{5}{3}A\right] . \end{aligned}$$

g(y) is decreasing if and only if $y\ge y_+$.

• 1. When

  $$\begin{aligned} 1+\frac{5}{3} A \le y_+\quad \Leftrightarrow \quad A\in \left( \frac{3\sqrt{10}}{10},\sqrt{\frac{18}{5}}\right] , \end{aligned}$$

  $p(A,\cdot )$ is increasing on [0, 1]. Hence,

  $$\begin{aligned} p(A) = p(A,1) = 1+\frac{3}{A}+\frac{9}{5A^2}. \end{aligned}$$

• 2. When

  $$\begin{aligned} \frac{5}{3} A< y_+< 1+\frac{5}{3} A\quad \Leftrightarrow \quad A\in \left[ \sqrt{\frac{18}{5}}, \frac{3}{5}+\sqrt{\frac{18}{5}}\right] , \end{aligned}$$

  then supremum of $p(A,\cdot )$ is achieved at $\omega _*$ such that $\omega _*+\frac{5}{3} A = y_+$. Combining this with (B.1) yields that

  $$\begin{aligned} p(A) = 1+\frac{3+5A}{2\sqrt{10}A- 6}. \end{aligned}$$

• 3. When

  $$\begin{aligned} \frac{5}{3} A \ge y_+\quad \Leftrightarrow \quad A\ge \frac{3}{5}+\sqrt{\frac{18}{5}}, \end{aligned}$$

  $p(A,\cdot )$ is decreasing on [0, 1]. Hence,

  $$\begin{aligned} p(A) = p(A,0) = 1+\frac{3+\sqrt{9+6A}}{2A}. \end{aligned}$$

This completes the derivation. $\square$

Study of d(Q)

We study the properties of d(Q) in this section. It is known that every $Q\in {\mathcal {Q}}_{phy}$ can be represented by

$$\begin{aligned} Q=\lambda _1 n\otimes n+\lambda _2 m\otimes m+\lambda _3 p\otimes p, \end{aligned}$$

(C.1)

where

$$\begin{aligned} \lambda _1+\lambda _2+\lambda _3=0,\, \lambda _i\in \left[ -\frac{1}{3},\frac{2}{3}\right] ,\,\lambda _1\le \lambda _2\le \lambda _3, \end{aligned}$$

(C.2)

and where (n, m, p) forms an orthonormal frame in ${\mathbb {R}}^3$. Then we have the following characterization of d(Q).

Lemma C.1

Let $Q\in {\mathcal {Q}}_{phy}$ be given by (C.1) and (C.2). Then $d(Q)=|Q-Q'|$, where

$$\begin{aligned} Q'=-\frac{1}{3} n\otimes n+\left( \lambda _2+\frac{\lambda _1+\frac{1}{3}}{2}\right) m\otimes m+\left( \lambda _3+\frac{\lambda _1+\frac{1}{3}}{2}\right) p\otimes p. \end{aligned}$$

As a result,

$$\begin{aligned} d(Q) = \frac{\sqrt{6}}{2}\left( \lambda _1+\frac{1}{3}\right) . \end{aligned}$$

(C.3)

Proof

Since the distance between two matrices is invariant under orthogonal transforms, without loss of generality, we may assume $n = (1,0,0)$, $m = (0,1,0)$ and $p = (0,0,1)$. Let $s=2\lambda _1+\lambda _2$ and $r=2\lambda _2+\lambda _1$. Then (C.1) becomes

$$\begin{aligned} Q=s\left( n\otimes n-\frac{1}{3}I\right) +r\left( m\otimes m-\frac{1}{3} I\right) , \quad \frac{r-1}{2}\le s\le r\le 0. \end{aligned}$$

Now we are going to look for $Q'\in \partial {\mathcal {Q}}_{phy}$ such that $|Q-Q'|$ is minimized. Assume

$$\begin{aligned} Q'=s'\left( n'\otimes n'-\frac{1}{3}I\right) +r'\left( m'\otimes m'-\frac{1}{3} I\right) , \end{aligned}$$

with

$$\begin{aligned} n' = (a,b,c),\quad m'=(u,v,w),\quad \frac{r'-1}{2}\le s'\le r'\le 0, \end{aligned}$$

and $(n',m')$ being an orthonormal pair.

First we are going to show that when $s',r'$ is fixed, $|Q-Q'|$ is minimized when $n'=n,\,m'=m$. We calculate that

$$\begin{aligned} \begin{aligned}&|Q-Q'|^2\\&\quad =|Q|^2+|Q'|^2\\&\qquad -2\left[ s\left( n\otimes n-\frac{1}{3}I\right) +r\left( m\otimes m-\frac{1}{3} I\right) \right] :\left[ s'\left( n'\otimes n'-\frac{1}{3}I\right) +r'\left( m'\otimes m'-\frac{1}{3} I\right) \right] \\&\quad =C(s,r,s',r')-2\left[ ss'\left( ( n,n')^2-\frac{1}{3}\right) +sr'\left( ( n,m')^2-\frac{1}{3}\right) \right. \\&\qquad \left. +rs'\left( (m,n')^2-\frac{1}{3}\right) +rr'\left( ( m,m')^2-\frac{1}{3}\right) \right] \\&\quad = C_(s,r,s',r')-2(ss'a^2+rs'b^2+sr'u^2+rr'v^2). \end{aligned} \end{aligned}$$

Here $C(s,r,s',r')$ represents some constant depending only on s, r, $s'$ and $r'$, whose definition changes from line to line.

Then it suffices to show that

$$\begin{aligned} ss'a^2+rs'b^2+sr'u^2+rr'v^2\le ss'+rr'. \end{aligned}$$

Recall that

$$\begin{aligned} a^2+b^2+c^2=1,\quad u^2+v^2+w^2=1,\quad au+bv+cw=0. \end{aligned}$$

(C.4)

We claim that $u^2\le b^2+c^2$. Indeed, by (C.4),

$$\begin{aligned} a^2u^2=|bv+cw|^2\le (b^2+c^2)(v^2+w^2)=(1-a^2)(1-u^2), \end{aligned}$$

which implies that $u^2\le 1-a^2=b^2+c^2$. Then we deduce that

$$\begin{aligned} \begin{aligned}&ss'+rr'-(ss'a^2+rs'b^2+sr'u^2+rr'v^2)\\&\quad = (s-r)s' (b^2+c^2)-(s-r)r'u^2+rr'w^2+rs'c^2\\&\quad \ge (s-r)(s'-r')u^2\ge 0. \end{aligned} \end{aligned}$$

Here we used (C.4) and the facts that $s\le r\le 0$ and $s'\le r' \le 0$.

To this end, we have showed that if Q is given by (C.1) and if $Q'\in \partial {\mathcal {Q}}_{phy}$ minimizes $|Q-Q'|$, $Q'$ should be represented by

$$\begin{aligned} Q'=\mu _1 n\otimes n+\mu _2 m\otimes m+\mu _3 p\otimes p, \end{aligned}$$

for some $-1/3= \mu _1\le \mu _2\le \mu _3\le 2/3$ such that $\mu _1+\mu _2+\mu _3 =0$. The constraints on $\mu _i$ are due to the characterization of $\partial {\mathcal {Q}}_{phy}$ in (1.2). Moreover,

$$\begin{aligned} |Q-Q'|=\sqrt{(\lambda _1-\mu _1)^2+(\lambda _2-\mu _2)^2+(\lambda _3-\mu _3)^2}. \end{aligned}$$

Therefore, $|Q-Q'|$ achieves its minimum if

$$\begin{aligned} \mu _1=-\frac{1}{3},\quad \mu _2=\lambda _2+\frac{\lambda _1+\frac{1}{3}}{2},\quad \mu _3=\lambda _3+\frac{\lambda _1+\frac{1}{3}}{2}. \end{aligned}$$

(C.3) follows immediately. This completes the proof. $\square$

An immediate consequence of Lemma C.1 is

Lemma C.2

d(Q) is Lipschitz continuous in ${\mathcal {Q}}_{phy}$.

Proof

The difference between the smallest eigenvalues of two matrice in ${\mathcal {Q}}_{phy}$ can be bounded by their distance. Combining this fact with Lemma C.1, we complete the proof of the Lemma. $\square$

A construction of $\{f_b^\varepsilon \}$

In this section, we provide a construction of $\{f_b^\varepsilon \}_{0<\varepsilon \ll 1}$ used in Sect. 4. For convenience, we recall the conditions on $\{f_b^\varepsilon \}_{0<\varepsilon \ll 1}$:

(i$'$):

For all $0<\varepsilon \ll 1$, $f_b^\varepsilon (Q)\in [0, \infty )$ for all $Q\in {\mathcal {Q}}$;

(ii$'$):

$f_b^\varepsilon$ are convex and smooth in ${\mathcal {Q}}$;

(iii$'$):

$f_b^\varepsilon (Q)\le f_b(Q)$ for all $Q\in {\mathcal {Q}}$.

(iv$'$):

Moreover,

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0^+} f_b^\varepsilon (Q) = f_b(Q),\quad \lim _{\varepsilon \rightarrow 0^+} Df_b^\varepsilon (Q) = Df_b(Q) \end{aligned}$$

locally uniformly in ${\mathcal {Q}}^{\mathrm {o}}_{phy}$.

Here $Df_b^\varepsilon (Q)$ denotes the gradient of $f_b^\varepsilon$ with respect to Q.

Proof of Lemma 4.1

Define

$$\begin{aligned} {\mathcal {Q}}_{phy}^\varepsilon = \{Q\in {\mathcal {Q}}^{\mathrm {o}}_{phy}:\, f_b(Q)< \varepsilon ^{-1}\}. \end{aligned}$$

Take $\varepsilon \ll 1$, such that ${\mathcal {Q}}_{phy}^\varepsilon$ is a non-empty open subset of ${\mathcal {Q}}_{phy}$. Then we define on the entire ${\mathcal {Q}}$ that

$$\begin{aligned} F_b^\varepsilon (Q) = \sup _{Q'\in {\mathcal {Q}}_{phy}^\varepsilon } f_b(Q')+Df_b(Q')(Q-Q'). \end{aligned}$$

It is not difficult to show that $\{F_b^\varepsilon \}_{0<\varepsilon \ll 1}$ satisfies all the conditions above except for the smoothness issue. Indeed, $F_b^\varepsilon \equiv f_b$ on ${\mathcal {Q}}_{phy}^\varepsilon$, while outside ${\mathcal {Q}}_{phy}^\varepsilon$, $F_b^\varepsilon$ is only Lipschitz continuous and $DF_b^\varepsilon$ exists in the $L^\infty$-sense but may not be well-defined pointwise. In particular, for all $Q_1,Q_2\in {\mathcal {Q}}$,

$$\begin{aligned} |F_b^\varepsilon (Q_1)-F_b^\varepsilon (Q_2)|\le |Q_1-Q_2|\sup _{{\mathcal {Q}}_{phy}^\varepsilon }|Df_b| =: |Q_1-Q_2|\omega _\varepsilon . \end{aligned}$$

(D.1)

Note that $\omega _\varepsilon \rightarrow +\infty$ as $\varepsilon \rightarrow 0^+.$

We shall make a little modification of $\{F_b^\varepsilon \}$ to construct smooth $\{f_b^\varepsilon \}$. Let $\phi$ be a non-negative $C_0^\infty$-mollifier in ${\mathcal {Q}}$ supported on the unit ball, such that $\int _{\mathcal {Q}}\phi (Q)\,dQ = 1$. Then we define

$$\begin{aligned} f_b^\varepsilon (Q) = \int _{{\mathcal {Q}}} \phi (Q')F_b^\varepsilon (Q-\varepsilon \omega _\varepsilon ^{-1} Q')\,dQ'-\varepsilon . \end{aligned}$$

We derive that for arbitrary $Q\in {\mathcal {Q}}$,

$$\begin{aligned} \begin{aligned} |f_b^\varepsilon (Q)+\varepsilon -F_b^\varepsilon (Q)| \le&\; \int _{{\mathcal {Q}}}\phi (Q')|F_b^\varepsilon (Q-\varepsilon \omega _\varepsilon ^{-1} Q')-F_b^\varepsilon (Q)|\,dQ'\\ \le&\; \int _{{\mathcal {Q}}}\phi (Q')\cdot \varepsilon \omega _\varepsilon ^{-1}\cdot \omega _\varepsilon \,dQ'= \varepsilon . \end{aligned} \end{aligned}$$

(D.2)

In the first inequality, we used the fact that $\phi$ is non-negative and normalized; in the second inequality, we applied (D.1) as well as that $\phi$ is supported on the unit ball in ${\mathcal {Q}}$. (D.2) implies that $F_b^\varepsilon (Q)-2\varepsilon \le f_b^\varepsilon (Q)\le F_b^\varepsilon (Q)\le f_b(Q)$.

It is then easy to verify that $\{f_b^\varepsilon \}_{0<\varepsilon \ll 1}$ satisfies all the conditions we need. $\square$