1 Introduction

A liquid crystal is a state of matter between isotropic liquid and crystalline solid. Based on the molecular positional and orientational order of liquid crystals, there are three main types: sematic, cholesterics and nematic. The nematic liquid crystal is the most common type in which the general phases are uniaxial and biaxial. In 1971, de Gennes [11] used Q-tensor order parameters to formulate the elastic energy of liquid crystals with Landau’s bulk energy. The Landau-de Gennes theory has been verified in physics as a successful theory describing both uniaxial and biaxial phases in nematic liquid crystals. Indeed, Pierre-Gilles de Gennes was awarded a Nobel prize for physics in 1991 for his discoveries in liquid crystals and polymers.

In the Landau-de Gennes framework, the space of Q-tensors in the Landau-de Gennes theory is a space of symmetric, traceless \(3\times 3\) matrices defined by

$$\begin{aligned} S_0:=\left\{ Q\in {\mathbb {M}}^{3\times 3}:\quad Q^T=Q, \, \text{ tr } Q =0\right\} , \end{aligned}$$

where \({\mathbb {M}}^{3\times 3}\) denotes the space of \(3\times 3\) matrices. When \(Q\in S_0\) has two equal non-zero eigenvalues, a nematic liquid crystal is said to be uniaxial. When Q has three unequal non-zero eigenvalues, a nematic liquid crystal is said to be biaxial. For material constants abc, we define the constant order parameter

$$\begin{aligned}s_+:=\frac{b+\sqrt{b^2+24ac}}{4c} \end{aligned}$$

and denote the identity matrix by I. The subspace of uniaxial Q-tensors is given by

$$\begin{aligned}S_*:=\left\{ Q \in S_0:\quad Q =s_+ \left( u\otimes u-\frac{1}{3} I\right) ,\quad u\in S^2 \right\} .\end{aligned}$$

In this paper, we only consider the case of positive constants a, b, c, which corresponds to a lower temperature regime in liquid crystals (the constant a could also be negative; see [34, 35]).

Let \(\Omega \) be a domain in \({\mathbb {R}}^3\). For a tensor \(Q\in W^{1,2}(\Omega ; S_0)\), the Landau-de Gennes energy is defined by

$$\begin{aligned} E_{LG}(Q; \Omega )=\int _{\Omega }f_{LG}(Q,\nabla Q)\,dx:=\int _{\Omega }\left( f_E(Q,\nabla Q) + f_B(Q)\right) \,dx, \end{aligned}$$
(1.1)

where \( f_E\) is the elastic energy density with elastic constants \(L_1,\ldots ,L_4\) of the form

$$\begin{aligned} f_E(Q,\nabla Q):=\frac{L_1}{2}|\nabla Q|^2+\frac{L_2}{2} \frac{\partial Q_{ij}}{\partial x_j} \frac{\partial Q_{ik}}{\partial x_k} +\frac{L_3}{2}\frac{\partial Q_{ik}}{\partial x_j}\frac{\partial Q_{ij}}{\partial x_k}+\frac{L_4}{2}Q_{lk}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k} \end{aligned}$$
(1.2)

and \(f_B(Q)\) is the bulk energy density defined by

$$\begin{aligned} f_B(Q):=-\frac{ a}{2}{{\,\textrm{tr}\,}}( Q^2)-\frac{ b}{3}{{\,\textrm{tr}\,}}( Q^3)+\frac{ c}{4}\left[ {{\,\textrm{tr}\,}}( Q^2)\right] ^2 \end{aligned}$$
(1.3)

with positive material constants a, b and c. Here and in the sequel, we adopt the Einstein summation convention for repeated indices.

In [11], de Gennes discovered the first two terms of the elastic energy density in (1.2) with \(L_3=L_4=0\). Since both the Oseen-Frank theory and the Landau-de Gennes theory should unify for modeling uniaxial liquid crystals, Schiele and Trimper [38] pointed out that the early attempt of de Gennes’ work [11] was incomplete since it would require the splay and bend Frank constants to be equal (i.e. \(k_1=k_3\)) in the Oseen-Frank density (as defined below in (1.5)) of uniaxial tensors \(Q =s_+ (u\otimes u-\frac{1}{3} I)\). However, some experiments on liquid crystals showed that \(k_3>k_1\), so they added a third order term to original de Gennes’ elastic energy density by

$$\begin{aligned} \frac{L_1}{2}|\nabla Q|^2+\frac{L_2}{2} \frac{\partial Q_{ij}}{\partial x_j} \frac{\partial Q_{ik}}{\partial x_k} +\frac{L_4}{2} Q_{lk}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k} \end{aligned}$$

with \(L_4=\frac{1}{2s_+^3}(k_3-k_1)>0\). Later, Berreman and Meiboom [6] observed that the above two groups discarded the surface energy density in the Oseen-Frank density, which correlates the blue phase theory for liquid crystals, so they proposed to recover a second order term \(\frac{L_3}{2}\frac{\partial Q_{ik}}{\partial x_j}\frac{\partial Q_{ij}}{\partial x_k}\) with four third order terms, but their density is over-determined with the Oseen-Frank density. Later, Longa et al. [31] gave an extension of the Landau-de Gennes density with 22 independent parameters. Finally, combining the work of Schiele and Trimper [38] with Berreman and Meiboom [2], Dickmann [13] found the full density (1.2), which is consistent with the Oseen-Frank density in (1.5) for uniaxial nematic liquid crystals. Since then, the general form (1.2) of the Landau-de Gennes energy density has been widely used in the study of nematic liquid crystals (e.g. [1, 33, 35]). Under the relation that

$$\begin{aligned}&s_+^2L_1 =-\frac{1}{6}k_1+\frac{1}{2}k_2+\frac{1}{6}k_3,\, s_+^2L_2 =k_1-k_2-k_4,\, s_+^2L_3 =k_4,\, s_+^{3}L_4 =\frac{k_3-k_1}{2}, \end{aligned}$$

one can show (c.f. [33]) that for a uniaxial tensor \(Q =s_+ (u\otimes u-\frac{1}{3} I)\) with \(u\in S^2\),

$$\begin{aligned} f_E(Q,\nabla Q)=W(u,\nabla u), \end{aligned}$$
(1.4)

where the Oseen-Frank density \(W(u,\nabla u)\) is defined by

$$\begin{aligned} W(u,\nabla u) =&\frac{k_1}{2}({{\,\textrm{div}\,}}u)^2+ \frac{k_2 }{2}(u\cdot {{\,\textrm{curl}\,}}u)^2+\frac{k_3}{2}|u\times {{\,\textrm{curl}\,}}u|^2\nonumber \\&+\frac{k_2+k_4 }{2}({{\,\textrm{tr}\,}}(\nabla u)^2-({{\,\textrm{div}\,}}u)^2) \end{aligned}$$
(1.5)

for a unit director \(u\in W^{1,2} (\Omega ; S^2)\). In (1.5), \(k_1\), \(k_2\), \(k_3\) are the Frank constants for molecular distortion of splay, twist and bend respectively and \(k_4\) is the Frank constant for the surface energy (c.f. [12]). In 1937, Zvetkov established numerical values for p-azoxyanisole (PAA) at \(120^\circ C\) (with the unit \(10^{-12} \,m/J\)) as follows:

$$\begin{aligned}&k_1=5,\;k_2=3.8,\;k_3=10.1. \end{aligned}$$

Therefore, according to physical experiments on nematic liquid crystals, the elastic constant \(L_4=\frac{1}{2s_+^3}(k_3-k_1)\) is not equal to zero in general (c.f. [12]).

A fundamental problem in mathematics on the Landau-de Gennes theory is to establish existence of a minimizer of the energy functional \(E_{LG}(Q,\Omega )\) in \(W^{1,2}_{Q_0}(\Omega ; S_0)\) with \(L_4\ne 0\). If the functional density \(f_{LG}(Q,\nabla Q)\) satisfies the coercivity condition, one can prove existence of a minimizer of the functional \(E_{LG}(Q,\Omega )\) in \(W^{1,2}(\Omega ; S_0)\). In 2010, Ball and Majumdar [2] found an example where for \(Q\in S_0\), the general Landau-de Gennes energy density (1.2) with \(L_4\ne 0\) does not satisfy the coercivity condition. Very recently, Golovaty et al. [22] emphasized that “From the standpoint of energy minimization, unfortunately, such a version of Landau-de Gennes becomes problematic, since the inclusion of the cubic term leads to an energy which is unbounded from below”. Therefore, the Landau-de Gennes density (1.2) causes a knowledge gap between mathematical and physical theories on liquid crystals, since the energy functional \(E_{LG}(Q,\Omega )\) in \(W^{1,2}(\Omega ; S_0)\) does not satisfy the coercivity condition and violates the existence theorem of minimizers (e.g. [1, 18]). In physics, concerning the third order term \(\frac{L_4}{2} Q_{\alpha \beta }\frac{\partial Q_{ij}}{\partial x_\alpha }\frac{\partial Q_{ij}}{\partial x_\beta }\) with \(L_4\ne 0\) in (1.2), Longa et al. [31] questioned that “In the presence of biaxial fluctuations the general third order theory in \(Q_{\alpha \beta }\) becomes unstable and thus is thermodynamically incorrect". In order to overcome the difficulty, they extended Landau-de Gennes densities through 22 independent second, third, fourth order terms, to preserve the stability of the free energy. Although their result is very interesting, all energy densities in [31] are complicated and have not addressed the above coercivity problem for all Q-tensors. In 2020, following similar spirit in [31], Golovaty et al. [22] proposed a new physical interpretation of the density through fourth order terms to address the above coercivity problem for all Q-tensors. We would like to point out that the new density form in [22] is completely different from the original Landau-de Gennes density (1.2) although the density in [22] can recover the Oseen-Frank density for uniaxial Q-tensors.

In this paper, we will propose a new Landau-de Gennes energy density to solve the above coercivity problem with \(L_4\ne 0\). At first, we observe in Lemma 2.1 that for uniaxial tensors \(Q\in S_*\), the original third order term on \(L_4\) in (1.2), proposed by Schiele and Trimper [38, p. 268] in physics, is a linear combination of a fourth order term and a second order term in the following:

$$\begin{aligned} Q_{lk}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}=\frac{3}{s_+}\left( Q_{ln}\frac{\partial Q_{ij}}{\partial x_l}\right) \left( Q_{kn}\frac{\partial Q_{ij}}{\partial x_k}\right) -\frac{2s_+}{3}|\nabla Q|^2. \end{aligned}$$
(1.6)

In the case of \(L_4\ge 0\), we introduce a new elastic energy density

$$\begin{aligned} f_{E,1}(Q,\nabla Q)=&\left( \frac{L_1}{2}-\frac{s_+L_4}{3}\right) |\nabla Q|^2 +\frac{L_2}{2} \frac{\partial Q_{ij}}{\partial x_j}\frac{\partial Q_{ik}}{\partial x_k} \nonumber \\&+\frac{L_3}{2} \frac{\partial Q_{ik}}{\partial x_j}\frac{\partial Q_{ij}}{\partial x_k}+ \frac{3L_4}{2s_+} Q_{ln}Q_{kn}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k} \end{aligned}$$
(1.7)

for all \(Q\in S_0\). We should point out that the fourth order term \(Q_{ln}Q_{kn}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}\) in (1.7) is a non-negative square term, so our new Landau-de Gennes density (1.7) for \(Q\in S_0\) satisfies the coercivity condition in mathematics under suitable conditions on \(L_1,\cdots , L_4\). The first three terms in (1.7) keep the original form (1.2) for \(Q\in S_0\) and the new Landau-de Gennes density (1.7) is equivalent to the original density (1.2) for \(Q\in S_*\). We also remark that our fourth order term \(Q_{ln}Q_{kn}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}\) is a linear combination of three fourth order terms \(L^{(4)}_5, L^{(4)}_6, L_7^{(4)}\) in [31]; i.e., we verify in Lemma 2.2 that

$$\begin{aligned} Q_{ln}Q_{kn}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}=\frac{8}{5} L^{(4)}_5- \frac{2}{5}L^{(4)}_6 +\frac{2}{5}L_7^{(4)}. \end{aligned}$$

For a \(Q\in W^{1,2}(\Omega , S_0)\), we introduce a new Landau-de Gennes energy functional

$$\begin{aligned} E_{L}(Q; \Omega )= \int _{\Omega }f_{LG}(Q,\nabla Q)\,dx =\int _{\Omega }\left( f_{E,1}(Q, \nabla Q) +\frac{1}{L} \tilde{f}_B(Q)\right) \,dx, \end{aligned}$$
(1.8)

where \(f_{E,1}(Q, \nabla Q)\) has the form (1.7), \(\tilde{f}_B(Q):= f_B(Q)-\min _{Q\in S_0} f_B(Q)\ge 0\) and \(L>0\) is a parameter to drive all elastic constants to zero [5, 17, 35].

Although there are many differences between the Oseen-Frank theory and the Landau-de Gennes theory, it is of great interest in mathematics whether minimizers of the Landau-de Gennes energy functional can approach a minimizer of the Oseen-Frank energy functional. When \(L_2=L_3=L_4=0\) in (1.7), Majumdar and Zarnescu [34] first proved that as \(L\rightarrow 0\), minimizers \(Q_{L}\) of \(E_{L}\) converges to \(Q_*=s_+ (u^*\otimes u^*-\frac{1}{3} I)\), where \(Q_*\) is a minimizer of the Dirichlet energy functional in \(W^{1,2}_{Q_0}(\Omega ; S_*)\). Sine then, there exist many developments on the one-constant approximation (c.f. [1]) and some special cases of unequal constants \(L_2\), \(L_3\), \(L_4\) ( [5, 29]). In theory of liquid crystals, the general expectation on the elastic constants is that \(L_4\) is not always zero (c.f. [38, p. 268], [2]). For the case of \(L_4\ne 0\), we first prove

Theorem 1

Let \(L_1\), \(L_2\), \(L_3\) and \(L_4\) be elastic constants satisfying

$$\begin{aligned}&L_1-\frac{s_+L_4}{6}>0, \quad - L_1-\frac{s_+L_4}{6}<L_3<2 L_1-\frac{s_+L_4}{3},\nonumber \\&\, L_1-\frac{s_+L_4}{6}+\frac{5}{3}L_2+\frac{1}{6}L_3>0,\quad L_4\ge 0. \end{aligned}$$
(1.9)

Then, for each \(L>0\), \(f_{LG}(Q,\nabla Q)\) in (1.8) satisfies the coercivity condition so that there exists a minimizer \(Q_L\) of the functional (1.8) in \(W^{1,2}_{Q_0}(\Omega ; S_0 )\) with boundary value \(Q_0\in W^{1,2}(\Omega ; S_* )\). As \(L\rightarrow 0\), the minimizers \(Q_L\) of \(E_L(Q;\Omega )\) converge (up to a subsequence) strongly to \(Q_*\) in \(W^{1,2}_{Q_0}(\Omega ; S_0)\) and satisfies

$$\begin{aligned} \lim _{L\rightarrow 0}\frac{1}{L}\int _{\Omega } {{\tilde{f}}}_B(Q_L) \,dx=0. \end{aligned}$$

Furthermore, \(Q_*\) is a minimizer of the functional \(E(Q; \Omega ):=\int _{\Omega } f_{E,1}(Q, \nabla Q)\,dx\) for all uniaxial Q-tensors in \(W^{1,2}_{Q_0}(\Omega ; S_* )\).

Remark 1

In the case of \(L_4<0\), for each \(Q\in W^{1,2}(\Omega , S_0)\), we introduce an elastic energy density by

$$\begin{aligned} f_{E,-}(Q,\nabla Q):=&\left( \frac{L_1}{2}+\frac{s_+L_4}{3}\right) |\nabla Q|^2+\frac{L_2}{2} \frac{\partial Q_{ij}}{\partial x_j} \frac{\partial Q_{ik}}{\partial x_k}+ \frac{L_3}{2} \frac{\partial Q_{ik}}{\partial x_j}\frac{\partial Q_{ij}}{\partial x_k} \nonumber \\&-\frac{3}{s_+}L_4\left( |Q|^2|\nabla Q|^2-Q_{ln}Q_{kn}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}\right) . \end{aligned}$$
(1.10)

It is clear that \(|Q|^2|\nabla Q|^2-Q_{ln}\frac{\partial Q_{ij}}{\partial x_l}Q_{kn}\frac{\partial Q_{ij}}{\partial x_k}\ge 0\) for each \(Q\in W^{1,2}(\Omega , S_0)\), and that the elastic energy density \(f_E(Q,\nabla Q)\) in (1.2) is equal to \(f_{E,-}(Q,\nabla Q)\) for uniaxial tensors \(Q\in S_*\).

Next, we discuss critical points of the Landau-de Gennes energy functional (1.7) in \(W^{1,2}_{Q_0}(\Omega ; S_0)\). One can write \(f_{E}(Q,\nabla Q):=\frac{\alpha }{2} |\nabla Q|^2+V(Q, \nabla Q)\) for some \(\alpha >0\) so that \(V(Q, \nabla Q)\ge 0\) for all \(Q\in W^{1,2}_{Q_0}(\Omega ; S_0)\). Then, the Euler-Lagrange equation for the Landau-de Gennes energy functional (1.8) in \(W^{1,2}_{Q_0}(\Omega ; S_0)\cap L^{\infty }(\Omega ; S_0)\) is

$$\begin{aligned}&\alpha \Delta Q + \frac{1}{2} {\nabla _k\left( V_{Q_{x_k}}+ V^T_{Q_{x_ k}}\right) }-\frac{1}{3} I \text{ tr } \left( \nabla _k V_{Q_{x_k}}\right) -\frac{1}{2} \left( V_{Q} + V^T_{Q}\right) + \frac{1}{3} I \text{ tr } \left( V_{Q}\right) \nonumber \\&\quad = \frac{1}{L} \left( -aQ - b \left( Q Q -\frac{1}{3}I {{\,\textrm{tr}\,}}(Q^2)\right) +cQ {{\,\textrm{tr}\,}}(Q^2)\right) \end{aligned}$$
(1.11)

in the weak sense, where \(A^T\) denotes the transpose of A,

$$\begin{aligned} V_{Q}:=\frac{\partial V(Q,\nabla Q)}{\partial Q} \text{ and } V_{Q_{x_k}}:=\frac{\partial V(Q,\nabla Q)}{\partial Q_{x_k}}. \end{aligned}$$

For general elastic constants \(L_1, \cdots \), \(L_4\), we cannot find any reference having an explicit form of the Euler-Lagrange equation of \(E(Q; \Omega )\) for \(Q=s_+ (u\otimes u-\frac{1}{3} I)\in S_*\) with \(u\in S^2\), so we give an explicit form of the Euler-Lagrange equation in the following:

$$\begin{aligned}&\alpha \left( s_+ \Delta Q -2 \nabla _kQ\nabla _kQ +2 s_+^{-1}\left( Q+\frac{s_+}{3}I\right) |\nabla Q|^2\right) \nonumber \\&\qquad +\nabla _k \left( V_{Q_{x_k}} \left( Q +\frac{s_+}{3} I\right) +\left( Q +\frac{s_+}{3} I\right) V^T_{Q_{x_k}} \nonumber \right. \\&\left. \qquad -2s_+^{-1} \left( Q +\frac{s_+}{3} I\right) \left\langle Q+\frac{s_+}{3} I,\, V_{Q_{x_k}} \right\rangle \right) \nonumber \\&\qquad - V_{Q_{x_k}}\nabla _kQ -\nabla _k Q V^T_{Q_k}\nonumber \\&\qquad +2s_+^{-1}\left[ \left\langle V_{Q_k}, \nabla _k Q\right\rangle \left( Q +\frac{s_+}{3} I\right) + \left\langle V_{Q_{x_k}}, \left( Q+\frac{s_+}{3} I\right) \right\rangle \nabla _k Q\right] \nonumber \\&\qquad - \left( Q +\frac{s_+}{3} I\right) V^T_{Q} - V_{Q} \left( Q +\frac{s_+}{3} I\right) \nonumber \\&\qquad +2s_+^{-1}\left\langle V_{Q}, Q +\frac{s_+}{3} I\right\rangle \left( Q +\frac{s_+}{3} I\right) =0, \end{aligned}$$
(1.12)

which is equivalent to the Oseen-Frank system for \(u\in S^2\), where \(\langle A, B\rangle = {{\,\textrm{tr}\,}}(B^TA)\) is the standard inner product of two matrices A and B. In the case of \(L_2=L_3=L_4=0\), the Euler-Lagrange Eq. (1.12) reduces to

$$\begin{aligned} s_+\Delta Q-2 \frac{\partial Q}{\partial x_k} \frac{\partial Q}{\partial x_k}+2s_+^{-1}\left( Q+\frac{s_+}{3}I\right) |\nabla Q|^2=0, \end{aligned}$$

which is equivalent to the harmonic map equation of u (c.f. [36]).

Since the Landau-de Gennes theory has been successfully used for modeling both uniaxial and biaxial states of nematic liquid crystals, it is of great interest whether the Q-tensor type of the Oseen-Frank system can be approximated by the Landau-de Gennes system (1.11) without using minimizers. In general, the problem of the convergence of solutions of the Landau-de Gennes Eq. (1.11) without using minimizers is open. Indeed, Gartland [17] pointed out that the convergence of solutions of the Landau-de Gennes Eq. (1.11) is similar to the convergence of solutions of the Ginzburg-Landau approximate equation from superconductivity theory. The Ginzburg-Landau functional was introduced in [20] to study the phase transition in superconductivity. For a parameter \(\varepsilon >0\), the Ginzburg-Landau functional of \(u:\Omega \rightarrow {\mathbb {R}}^3\) is defined by

$$\begin{aligned} E_{\varepsilon }(u; \Omega ):=\int _{\Omega }\left( \frac{1}{2} |\nabla u|^2+\frac{1}{4\varepsilon ^2}(1-|u|^2)^2\right) \, dx. \end{aligned}$$
(1.13)

The Euler-Lagrange equation for the Ginzburg-Landau functional is

$$\begin{aligned} \Delta u_{\varepsilon }+\frac{1}{\varepsilon ^2} u_{\varepsilon }\left( 1-|u_{\varepsilon }|^2\right) =0. \end{aligned}$$
(1.14)

Using the cross product of vectors, Chen [7] first proved that as \(\varepsilon \rightarrow 0\), solutions \(u_{\varepsilon }\) of the Ginzburg-Landau system (1.14) weakly converge to a harmonic map in \(W^{1,2}(\Omega ; {\mathbb {R}}^3)\). Chen and Struwe [9] proved global existence of partial regular solutions to the heat flow of harmonic maps using the Ginzburg-Landau approximation. In [3, 4], Bethuel, Brezis and Hélein obtained many results on asymptotic behavior for minimizers of \(E_{\varepsilon }\) in two dimensions as \(\varepsilon \rightarrow 0\) (see also [39]). Recently, many works ( [15, 25,26,27]) have examined the convergence of the Ginzburg-Landau approximation for the Ericksen-Leslie system with unequal Frank’s constants \(k_1, k_2, k_3\). Motivated by the above results on the Ginzburg-Landau approximation, it is natural to investigate the converging problem on solutions of the Landau-de Gennes system (1.11) as \(L\rightarrow 0\). By comparing with the result of Chen [7] (see also [8]) on the weak convergence of solutions of the Ginzburg-Landau equations, it is interesting to study whether solutions \(Q_L\) of the Landau-de Gennes equations (1.11) with uniform bound of the energy converge weakly to a solution \(Q_*\) of (1.12) in \(W^{1,2}_{Q_0}(\Omega ; S_0 )\). However, it seems that the problem is not clear when \(L_2\), \(L_3\), \(L_4\) are not zero. Under a condition, we solve this problem and prove:

Theorem 2

Let \(Q_L\) be a weak solution to the equation (1.11) with a uniform bound in L. Assume that the solution \(Q_L\) converges strongly to \(Q_*\) in \(W^{1,2}_{Q_0}(\Omega ; S_0 )\) as \(L\rightarrow 0\) and satisfies

$$\begin{aligned} \lim _{L\rightarrow 0}\frac{1}{L}\int _{\Omega } {{\tilde{f}}}_B(Q_L) \,dx=0. \end{aligned}$$
(1.15)

Then, \(Q_*\) is a weak solution to (1.12).

For the proof of Theorem 2, we use a concept of a projection \(\pi \) in a neighborhood \(S_{\delta }\) of the space \(S_*\), where

$$\begin{aligned} S_\delta :=\left\{ Q\in S_0:\quad {{\,\textrm{dist}\,}}(Q;S_*)\le \delta \right\} . \end{aligned}$$

For a sufficiently small \(\delta >0\), there exists a smooth projection \(\pi : S_{\delta }\rightarrow S_*\) so that for any \(Q\in S_{\delta }\), \(\pi (Q)\in S_*\) (c.f. [9]). By the projection \(\pi \), we consider the modified bulk energy density

$$\begin{aligned}F(Q):=\tilde{f}_B(Q)+|Q-\pi (Q)|^2 \end{aligned}$$

so that the Hessian of F(Q) is positive definite for each any \(Q\in S_{\delta }\) with a sufficiently small \(\delta >0\). Then, choosing a suitable test function and using (1.15), we employ Taylor’s expansion of F(Q) to cancel the limit term involving \(\frac{1}{L}\nabla _{Q_{ij}} f_B(Q_L)\). With these results, we divide the domain into three parts and then employ Egoroff’s theorem to prove Theorem 2.

When \(L_2=L_3=L_4=0\), Majumdar and Zarnescu [34] first proved that minimizers \(Q_L\) of \(E_{LG}(Q; \Omega )\) uniformly converge to \(Q_*\) away from the singular set of \(Q_*\) since there exists a monotonicity formula for minimizers \(Q_L\) of \(E_{LG}(Q; \Omega )\) in \(W^{1,2}(\Omega , S_0)\). Later, Nguyen and Zarnescu [36] improved the result by proving local smooth convergence of minimizers \(Q_L\) away from the singular set of \(Q_*\). However, in the general case of non-zero elastic constants \(L_2\), \(L_3\) and \(L_4\), there exists no such monotonicity formula for minimizers \(Q_L\) of \(E_{LG}(Q; \Omega )\) in \(W^{1,2}(\Omega , S_0)\), so it is a very interesting question whether one can improve the convergence of \(Q_L\) for general cases. For this question, we generalize Nguyen and Zarnescu’s result in [36] to the case of non-zero elastic constants \(L_2, L_3, L_4\) as follows:

Theorem 3

For each \(L>0\), let \(Q_L\) be a weak solution to the equation (1.11) and let \(Q_*\) be the limiting map of \(Q_L\) in Theorem 2. Assume that \(Q_L\) is smooth and converges to \(Q_*\) uniformly inside \(\Omega \backslash \Sigma \), where \(\Sigma \) is the singular set of \(Q_*\). Then, as \(L\rightarrow 0\) (up to a subsequence), we have

$$\begin{aligned} Q_L\rightarrow Q_*\text { in }C^k_\textrm{loc}(\Omega \backslash \Sigma ) \end{aligned}$$
(1.16)

for any positive integer \(k\ge 0\).

We would like to point out that our proof of Theorem 3 is new and different from one in [36]. We outline main steps as follows:

Step I. For each \(Q \in S_\delta \), there exists a rotation \(R(Q)\in SO(3)\) such that \({{\tilde{Q}}}=R^T(Q)QR(Q)\) is diagonal. For any \(\xi \in S_0\), we prove

$$\begin{aligned} \sum _{i,j=1}^3\partial _{{{\tilde{Q}}}_{ii}}\partial _{{{\tilde{Q}}}_{jj}} f_B(\tilde{Q})\xi _{ii}\xi _{jj}\ge \frac{\lambda }{2} \left( \xi _{11}^2 + \xi _{22}^2+ \xi _{33}^2\right) , \end{aligned}$$
(1.17)

where \(\lambda =\min \{3a,s_+ b \}>0\). For each smooth \(Q(x)\in S_0\), \(R^T(Q(x))Q(x)R(Q(x))\) is diagonal. Then there exists a measure zero set \(\Omega _0\) such that Q(x) has a constant multiplicity of eigenvalues inside subdomains of \(\Omega \backslash \Omega _0\) and R(Q(x)) is almost differentiable in \(\Omega \). Using the geometric identity

$$\begin{aligned} \nabla \left( R^T(Q) Q R (Q)\right) _{ii} =\left( R^T(Q)\nabla Q R (Q)\right) _{ii}\end{aligned}$$

with \(i=1, 2, 3\), we apply (1.17) to obtain

$$\begin{aligned}&\int _{B_{r_0}(x_0)} |\nabla ^2 Q_L|^2\phi ^2\, dx \le C\int _{B_{r_0}(x_0)}|\nabla Q_L|^2 |\nabla \phi |^2\,dx, \end{aligned}$$

for a uniform constant C in L, where \(\phi \) is a cutoff function in \(B_{r_0}(x_0)\subset \Omega \backslash \Sigma \).

Step II. Using Step I with the technique of ‘filling hole’ on elliptic systems [18], we establish a uniform Caccioppoli inequality for solutions \(Q_L\) of (2.10) in L; i.e., there exists a uniform constant C independent of L such that

$$\begin{aligned} \int _{B_{r/2}(x_0)} |\nabla Q_L|^2\,dx\le \frac{C}{r^2}\int _{B_{r}(x_0)} | Q_L- Q_{L,x_0, r}|^2\,dx \end{aligned}$$
(1.18)

for any \(x_0\) with \(B_{r_0}(x_0)\subset \Omega \backslash \Sigma \) and any \(r\le r_0\), where .

Step III. Based on the uniform Caccioppoli inequality (1.18), we apply the well-known Gagliardo-Nirenberg interpolation (c.f. [28] or [15]) to obtain a control on the local \(L^3\)-estimate; i.e., there exists a uniform constant \(r_0>0\) such that

$$\begin{aligned} \int _{B_{r_0}(x_0)} |\nabla Q_L|^3\,dx\le \varepsilon _0 \end{aligned}$$
(1.19)

for some small \(\varepsilon _0>0\). Then, combining (1.19) with (1.17), we apply an induction method to obtain uniform estimates on higher derivatives \(\nabla ^kQ_L\) in L and prove Theorem 2.

Remark 2

When \(L_4\) is sufficiently small, the \(L^p\)-theory on the constant elliptic system (c.f. [18]) can assure that the weak solution \(Q_L\) to the equation (1.11) is smooth in \(\Omega \). In the case of \(L_4=0\), Contreras and Lamy [10] proved that the minimizers \(Q_L\) uniformly converge to \(Q_*\) in \(\Omega \backslash \Sigma \) by assuming that \(Q_L\) is uniformly bounded.

Remark 3

In a recent work [16] with Yu Mei, we can expand ideas on proofs of Theorems 2-3 to show that solutions for the Beris-Edwards system for biaxial Q-tensors converge smoothly to the solution of the Beris-Edwards system for uniaxial Q-tensors up to its maximal existence time.

Finally, we make some remarks about new forms of the Landau-de Gennes energy density through a strong Ericksen’s condition on the Oseen-Frank density. Recently, Golovaty et al. [22] proposed a novel form of the Landau-de Gennes energy density through the Oseen-Frank density. In addition to Ericksen’s inequalities that \(k_2\ge |k_4|, k_3\ge 0, 2k_1\ge k_2+k_4\), they also assumed that

$$\begin{aligned} k_1 \ge k_2 + k_4,\, k_3 \ge k_2 + k_4,\,k_4 \le 0. \end{aligned}$$
(1.20)

However, their result is not optimal. In fact, Golovaty et al. [22] made a further remark that their assumption (1.20) could be relaxed if one includes more cubic terms in [31], but they did not do it. In Sect. 5, we improve their result to derive an implicit form of \(f_E(Q,\nabla Q)\). Assuming the strong Ericksen inequality with a weaker assumption that \(2k_3> k_2+k_4 \) instead of the condition (1.20), we write the Oseen-Frank density into a new form, which satisfies the coercivity condition for all \(u\in {\mathbb {R}}^3\). In [12], de Gennes and Prost remarked that the bending constant \(k_3\) is much larger than others \(k_1\) and \(k_2\). Therefore, the assumption \(2k_3> k_2+k_4 \) is satisfied through a strong Ericksen’s condition. We give an explicit form of \(f_E(Q,\nabla Q)\) in Proposition 5.1, which satisfies the coercivity condition for all \(Q\in S_0\).

The paper is organized as follows. In Sect. 2, we prove Theorem 1. In Sect. 3, we prove Theorem 2. In Sect. 4, we prove Theorem 3. In Sect. 5, we obtain new forms of the Landau-de Gennes energy density through a strong Ericksen’s condition.

2 The coercivity condition and existence of minimizers

At first, we note

Lemma 2.1

For a uniaxial \(Q= s_+(u\otimes u-\frac{1}{3} I)\in S_*\) with \(u\in S^2\), we have

$$\begin{aligned} Q_{lk}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k} = \frac{3}{s_+}\left( Q_{ln}\frac{\partial Q_{ij}}{\partial x_l}\right) \left( Q_{kn}\frac{\partial Q_{ij}}{\partial x_k}\right) -\frac{2s_+}{3}|\nabla Q|^2. \end{aligned}$$
(2.1)

Proof

Using the fact that \(|u|=1\), we have

$$\begin{aligned} Q_{ln}Q_{kn}&= s_+^2\left( u_ku_n-\frac{1}{3} \delta _{kn}\right) \left( u_lu_n-\frac{1}{3} \delta _{ln}\right) \nonumber \\&= s_+^2\left( u_ku_lu_nu_n-\frac{1}{3} \delta _{kn}u_lu_n-\frac{1}{3} \delta _{ln}u_ku_n+\frac{1}{9} \delta _{kn}\delta _{ln}\right) \nonumber \\&=s^2\left( \frac{1}{3}u_ku_l+\frac{1}{9} \delta _{kl}\right) =\frac{s_+}{3} s_+\left( u_ku_l-\frac{1}{3} \delta _{lk}\right) +\frac{2s_+^2}{9} \delta _{kl} \nonumber \\&=\frac{s_+}{3} Q_{kl}+\frac{2s_+^2}{9} \delta _{kl}. \end{aligned}$$
(2.2)

Through the identity (2.2), we obtain

$$\begin{aligned} Q_{lk}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}&= \left( \frac{3}{s_+}Q_{ln}Q_{kn}-\frac{2s_+}{3} \delta _{kl}\right) \frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}\\&=\frac{3}{s_+}\left( Q_{ln}\frac{\partial Q_{ij}}{\partial x_l}\right) \left( Q_{kn}\frac{\partial Q_{ij}}{\partial x_k}\right) -\frac{2s_+}{3}|\nabla Q|^2. \end{aligned}$$

\(\square \)

Recall from Longa et al. [31] that

$$\begin{aligned} L^{(4)}_5:=&Q_{\alpha \rho }Q_{\rho \beta }\frac{\partial Q_{\alpha \mu }}{\partial x_ \beta }\frac{\partial Q_{\mu \nu }}{\partial x_\nu }, \quad L^{(4)}_6:=Q_{\alpha \rho }Q_{\rho \beta } \frac{\partial Q_{\alpha \mu }}{\partial x_ \mu } \frac{\partial Q_{\beta \nu }}{\partial x_ \nu }, \\ L^{(4)}_7:=&Q_{\alpha \rho }Q_{\rho \beta } \frac{\partial Q_{\alpha \mu }}{\partial x_ \nu }\frac{\partial Q_{\beta \mu }}{\partial x_\nu }. \end{aligned}$$

Then we have

Lemma 2.2

For a uniaxial \(Q\in S_*\), we obtain

$$\begin{aligned}&Q_{ln}Q_{kn}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}=\frac{8}{5} L^{(4)}_5- \frac{2}{5}L^{(4)}_6 +\frac{2}{5}L_7^{(4)}. \end{aligned}$$
(2.3)

Proof

Let \(Q= s_+(u\otimes u-\frac{1}{3} I)\) for \(u\in S^2\). Noting that \(u_i\nabla u_i =0\), we calculate

$$\begin{aligned} (u\times {{\,\textrm{curl}\,}}u)_1^2=&[u_2(\nabla _1u_2-\nabla _2u_1)-u_3(\nabla _3u_1-\nabla _1u_3)]^2 \\ =&(-u_1\nabla _1u_1-u_2\nabla _2u_1-u_3\nabla _3u_1)^2=[(u\cdot \nabla )u_1]^2. \end{aligned}$$

Similarly, we can calculate other terms to obtain

$$\begin{aligned} \sum _i[(u\cdot \nabla )u_i]^2=\sum _i(u\times {{\,\textrm{curl}\,}}u)_i^2=|u\times {{\,\textrm{curl}\,}}u|^2. \end{aligned}$$

Moreover, we calculate

$$\begin{aligned} Q_{lk}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}&=s_+^{3} \left( u_lu_k-\frac{1}{3}\delta _{lk}\right) \nabla _l(u_iu_j)\nabla _k(u_iu_j) \nonumber \\&= s_+^{3} \left( u_lu_k-\frac{1}{3}\delta _{lk}\right) (u_j\nabla _lu_i+u_i\nabla _lu_j)(u_j\nabla _ku_i+u_i\nabla _ku_j) \nonumber \\&= s_+^{3} \left( u_lu_k-\frac{1}{3}\delta _{lk}\right) (\nabla _lu_i\nabla _ku_i+\nabla _lu_j\nabla _ku_j) \nonumber \\&=2s_+^{3} \sum _i[( u\cdot \nabla ) u_i]^2-\frac{2}{3}s_+^{3} |\nabla u|^2=2s_+^{3} |u\times {{\,\textrm{curl}\,}}u|^2-\frac{2}{3}s_+^{3} |\nabla u|^2. \end{aligned}$$
(2.4)

It follows from using (2.1) and (2.4) that

$$\begin{aligned} Q_{ln}Q_{kn}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}&=\frac{1}{3}s_+ Q_{lk}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}+ \frac{2s^2_+}{9}|\nabla Q|^2\nonumber \\&=\frac{2}{3}s_+^{4}|u\times {{\,\textrm{curl}\,}}u|^2+\frac{2}{9}s_+^4|\nabla u|^2. \end{aligned}$$
(2.5)

We can verify from [31] that

$$\begin{aligned} 4L_5^{(4)}-L_6^{(4)}=\frac{5s_+^4}{3}|u\times {{\,\textrm{curl}\,}}u|^2,\quad L_7^{(4)}=\frac{5}{9}s_+^4|\nabla u|^2. \end{aligned}$$
(2.6)

Substituting (2.6) into (2.5), we have

$$\begin{aligned} Q_{ln}Q_{kn}\frac{\partial Q_{ij}}{\partial x_l}\frac{\partial Q_{ij}}{\partial x_k}=\frac{2}{5}(4L^{(4)}_5-L^{(4)}_6)+\frac{2}{5}L_7^{(4)}. \end{aligned}$$

\(\square \)

Under the condition (1.9), one can verify from Lemma 1.2 in [30] that there are two uniform constants \(\alpha >0\) and \(C>0\) such that the form \(f_{E,1}(Q,p)\) also satisfies

$$\begin{aligned} \frac{\alpha }{2} |p|^2&\le f_{E,1}(Q,p)\le C(1+|Q|^2)|p|^2, \quad |\partial _Qf_{E,1}(Q,p)|\le C(1+|Q|)|p|^2 \end{aligned}$$
(2.7)

for any \(Q\in {\mathbb {M}}^{3\times 3}\) and \(p \in {\mathbb {M}}^{3\times 3}\times {\mathbb {R}}^3\). Since \( f_{E,1}(Q,p)\) is quadratic in p and satisfies (2.7), it can be checked (c.f. [28]) that \( f_{E,1}(Q,p)\) is uniformly convex in p; that is

$$\begin{aligned} \partial ^2_{p^i_{kl}p^j_{mn}} f_{E,1}(Q,p)\xi ^i_{kl}\xi ^j_{mn}\ge \frac{\alpha }{2} |\xi |^2, \quad \forall \xi \in {\mathbb {M}}^{3\times 3}\times {\mathbb {R}}^3. \end{aligned}$$
(2.8)

Now we give a proof of Theorem 1.

Proof

Under the condition on \(L_1, \cdots , L_4\) in Theorem 1, it is clear from using Lemma 1.2 in [30] that

$$\begin{aligned} f_{E,1}(Q,\nabla Q)\ge \frac{\alpha }{2} |\nabla Q|^2, \quad \forall Q\in S_0. \end{aligned}$$

By the standard theory in the calculus of variations (c.f. [17]), there exists a minimizer \(Q_{L}\) of \(E_L\) in \(W_{Q_0}^{1,2}(\Omega ; S_0)\). For each \(Q\in W^{1,2}_{Q_0}(\Omega ; S_0)\), we set

$$\begin{aligned} E(Q ;\Omega ):=\int _{\Omega } f_{E,1}(Q,\nabla Q)\,dx. \end{aligned}$$

It implies that

$$\begin{aligned} E(Q_L ;\Omega )+ \int _{\Omega }(f_B(Q_L)- \inf _{S_0} f_B)\,dx \le E(Q ;\Omega ) \end{aligned}$$

for any \(Q\in W^{1,2}_{Q_0}(\Omega ; S_*)\) with the fact that \({{\tilde{f}}}_B(Q)=f_B(Q)- \inf _{S_0} f_B=0\).

As \(L\rightarrow 0\), minimizers \(Q_{L}\) converge (possible passing subsequence) weakly to a tensor \(Q_*\in W^{1,2}(\Omega ; S_0)\) with that \({{\tilde{f}}}_B( Q_*)=0\), which implies that \(Q_*\in S_*\) a.e. in \(\Omega \). Then, for any \(Q\in W_{Q_0}^{1,2}(\Omega ; S_*)\), we have

$$\begin{aligned} E (Q_* ;\Omega )\le \liminf _{L\rightarrow 0}E(Q_{L} ;\Omega ) \le {{\mathrm{lim\,sup}}}_{L\rightarrow 0}E(Q_{L} ;\Omega )\le E(Q;\Omega ). \end{aligned}$$

Therefore \(Q_*\) is also a minimizer of E in \(W^{1,2}_{Q_0}(\Omega ; S_*)\). Choosing \(Q=Q_*\) in the above inequality, it implies that

$$\begin{aligned} E (Q_* ;\Omega )=\lim _{L\rightarrow 0}E_L(Q_{L} ;\Omega ),\quad \lim _{L\rightarrow 0}\frac{1}{L}\int _{\Omega } {{\tilde{f}}}_B(Q_{L}) \,dx=0. \end{aligned}$$

Moreover, it is known that

$$\begin{aligned} \int _{\Omega } |\nabla Q_*|^2\,dx\le&\liminf _{L\rightarrow 0}\int _{\Omega } |\nabla Q_L|^2\,dx ,\\ \int _{\Omega } V(Q_*, \nabla Q_*)\,dx\le&\liminf _{L\rightarrow 0}\int _{\Omega } V(Q_{L},\nabla Q_L )\,dx. \end{aligned}$$

It implies that \(\int _{\Omega } |\nabla Q_*|^2\,dx= \liminf _{L\rightarrow 0}\int _{\Omega } |\nabla Q_L|^2\,dx\). Otherwise, there is a subsequence \(L_k\rightarrow 0\) such that

$$\begin{aligned} \int _{\Omega } |\nabla Q_*|^2\,dx< \lim _{L_k\rightarrow 0}\int _{\Omega } |\nabla Q_{L_k}|^2\,dx. \end{aligned}$$

Then

$$\begin{aligned} E (Q_* ;\Omega )&=\lim _{L_k\rightarrow 0}E_{L_k}(Q_{L_k} ;\Omega ),\\&= \frac{\alpha }{2} \lim _{L_k\rightarrow 0}\int _{\Omega } |\nabla Q_{L_k}|^2\,dx+ \lim _{L_k\rightarrow 0}\int _{\Omega } V(Q_{L_k}, \nabla Q_{L_k})\,dx\\&< E (Q_* ;\Omega ). \end{aligned}$$

This is impossible. Therefore, minimizers \(Q_{L_k}\) converge strongly, up to a subsequence, to a minimizer \(Q_*=s_+(u_*\otimes u_*-\frac{1}{3} I)\) of E in \(W^{1,2}_{Q_0}(\Omega ; S_0)\). Following from Lemma 3.1, \(Q_*\) satisfies (1.12) and \(u_*\) is a minimizer of the Oseen-Frank energy in \(W^{1,2}(\Omega ; S^2)\). Due to the well-known result of Hardt, Kinderlehrer and Lin [23], \(u_*\) is partially regular in \(\Omega \) (see also [24]). Thus \(Q_*\) is partially regular. \(\square \)

Lemma 2.3

If Q is a minimizer of \(E_{L}(Q;\Omega )\) from (1.8) in \(W^{1,2}_{Q_0}(\Omega ; S_0)\), it satisfies

$$\begin{aligned}&- \alpha \Delta Q_{ij}-\frac{1}{2} \nabla _k (V_{p^k_{ij}}+ V_{p^k_{ji}})+\frac{1}{3} \delta _{ij}\sum _{l=1}^3 \nabla _k V_{p^k_{ll}} \\&\qquad +\frac{1}{2} (V_{Q_{ij}} +V_{Q_{ji}})- \frac{1}{3} \delta _{ij}\sum _{l=1}^3 V_{Q_{ll}} \\&\qquad + \frac{1}{L} \left( -aQ_{ij}- b \left( Q_{ik}Q_{kj}-\frac{1}{3}\delta _{ij}{{\,\textrm{tr}\,}}(Q^2) \right) +cQ_{ij}{{\,\textrm{tr}\,}}(Q^2)\right) =0 \end{aligned}$$

in the weak sense, where \(V_{p^k_{ij}}:=V_{p^k_{ij}}(Q,p)\) with \(p=(\nabla _k Q_{ij})\).

Proof

For any test function \(\phi \in C^\infty _0(\Omega ; S_0)\), consider \(Q_t:=Q+t\phi \) for \(t\in {\mathbb {R}}\). Then for all \(\phi \in C^\infty _0(\Omega ; S_0)\), we calculate

$$\begin{aligned}&\left. \int _\Omega \frac{d}{d t}\left( {{\tilde{f}}}_{E,1}(Q_t,\nabla Q_t)+\frac{1}{L}{{\tilde{f}}}_B(Q_t)\right) \right| _{t=0}\,dx \\&\quad =\int _\Omega \alpha \frac{\partial Q_{ij}}{\partial x_k}\frac{\partial \phi _{ij}}{\partial x_k}+ V_{p^k_{ij}} \frac{\partial \phi _{ij}}{\partial x_k}+V_{Q_{ij}} \phi _{ij}\,dx \\&\qquad + \frac{1}{L}\int _\Omega -aQ_{ij}\phi _{ij}- b Q_{ik}Q_{kj}\phi _{ij} +c(Q_{ij}{{\,\textrm{tr}\,}}(Q^2)\phi _{ij})\,dx \\&\quad =\int _\Omega \left( -\alpha \Delta Q_{ij} -\frac{1}{2}\frac{\partial }{\partial x_k}(V_{p^k_{ij}} + V_{p^k_{ji}}) +\frac{1}{2} (V_{Q_{ij}} +V_{Q_{ji}}) \right) \phi _{ij}\,dx \\&\qquad + \frac{1}{L}\int _\Omega \left( -aQ_{ij}- b Q_{ik}Q_{kj} +cQ_{ij}{{\,\textrm{tr}\,}}(Q^2)\right) \phi _{ij}\,dx\\&\quad = \int _\Omega \left( -\alpha \Delta Q_{ij} -\frac{1}{2} \nabla _k(V_{p^k_{ij}}+ V_{p^k_{ji}}) -\frac{1}{3} \delta _{ij}\sum _l \nabla _k V_{p^k_{ll}} \right) \phi _{ij}\,dx \\&\qquad +\int _\Omega \left( \frac{1}{2} (V_{Q_{ij}} +V_{Q_{ji}}) - \frac{1}{3} \delta _{ij}\sum _l V_{Q_{ll}} \right) \phi _{ij}\,dx \\&\qquad + \frac{1}{L}\int _\Omega \left( -aQ_{ij}- b \left( Q_{ik}Q_{kj}-\frac{1}{3} \delta _{ij}{{\,\textrm{tr}\,}}(Q^2) \right) +cQ_{ij}{{\,\textrm{tr}\,}}(Q^2)\right) \phi _{ij}\,dx = 0, \end{aligned}$$

where we used the fact that \(\phi \) is traceless. This proves our claim. \(\square \)

In the case of \(L_2=L_3=L_4=0\), Majumdar and Zarnescu [34] proved that the weak solution of (1.11) is bounded by using a maximum principle. However, when \(L_2\), \(L_3\), \(L_4\) are non-zero, the system (1.11) is a nonlinear elliptic system, so there exists no such maximum principle for it (e.g. [17, 21]). Therefore, it is not clear whether each minimizer \(Q_L\) of \(E_{L}(Q; \Omega )\) in \(W_{Q_0}^{1,2}(\Omega , S_0)\) is bounded and the energy density \(f_{E,1}(Q, \nabla Q)\) in (1.8) can be bounded above by \(C|\nabla Q|^2 +C\). Without this above growth condition on the density, it is a well-known fact that a minimizer \(Q_L\) of the Landau-de Gennes energy functional in \(W^{1,2}_{Q_0}(\Omega ; S_0 )\) may not satisfy the Euler-Lagrange equation in \(W^{1,2}(\Omega , S_0)\). To overcome this difficulty, we can introduce a smooth cutoff function \(\eta (r)\) in \([0, \infty )\) so that \(\eta (r)=1\) for \(r\le M\) with a very large constant \(M>0\) and \(\eta (r)=0\) for \(r\ge M+1\). Then for each \(Q\in W^{1,2}(\Omega , S_0)\), one can modify the Landau-de Gennes density by

$$\begin{aligned} {{\widetilde{f}}}_E(Q, \nabla Q):= \frac{\alpha }{2} |\nabla Q|^2 +\tilde{V}(Q, \nabla Q)= \frac{\alpha }{2}|\nabla Q|^2 +\eta (|Q|) V(Q, \nabla Q) \end{aligned}$$
(2.9)

with the property that

$$\begin{aligned} \frac{\alpha }{2} |\nabla Q|^2\le {{\widetilde{f}}}_E(Q, \nabla Q)\le C|\nabla Q|^2. \end{aligned}$$

For a large \(M>0\) in (2.9), we consider a modified Landau-de Gennes functional

$$\begin{aligned} {{\tilde{E}}}_{L}(Q; \Omega )=\int _{\Omega }\left( {{\widetilde{f}}}_E(Q, \nabla Q) +\frac{1}{L} {{\tilde{f}}}_B(Q)\right) \,dx. \end{aligned}$$
(2.10)

Then we obtain

Lemma 2.4

Let \(Q_L\) be a weak solution to the equation (1.11) with the boundary value \(Q_0\in W^{1,2}(\Omega ; S_*)\) associated to the functional \({{\tilde{f}}}_E(Q,\nabla Q)\) in (2.9). Then, \(|Q_L|\le M+1\) for a sufficient large M.

Proof

Recall from the definition of \({{\tilde{f}}}_E(Q,\nabla Q)\) in (2.9) that for a \(Q\in S_0\) with \(|Q|\ge M+1\),

$$\begin{aligned} {{\tilde{f}}}_E(Q, \nabla Q)=\frac{\alpha }{2}|\nabla Q|^2. \end{aligned}$$

Similarly to one in [8], choose a test function \(\phi =Q (1-\min \{1, \frac{M+1}{|Q|}\})\). Multiplying (1.11) by the test function \(\phi \), we have

$$\begin{aligned}&\alpha \int _{|Q|\ge M+1}|\nabla Q|^2 \left( 1- \frac{M+1}{|Q|}\right) -(M+1) Q_{ij} \nabla _k Q_{ij} \nabla _k \frac{1}{|Q|})\,dx\\&\qquad + \frac{1}{L} \int _{|Q|\ge M+1} \left( -a|Q|^2- b Q_{ik}Q_{kj} Q_{ij} +c|Q|^4\right) \left( 1- \frac{M+1}{|Q|}\right) \,dx=0. \end{aligned}$$

Note the fact that \( \nabla _k |Q|^2=2Q_{ij} \nabla _k Q_{ij} \). The above second term is non-negative. For a sufficiently large \(M>0\), the third term is positive. This implies that the set \(\{|Q|\ge M+1\}\) is empty; i.e., \(|Q|\le M+1\) a.e. in \(\Omega \). \(\square \)

The following result is a variant result of Giaquinta-Giusti [19] (see more details in page 206 of [18]):

Proposition 2.1

For each \(L>0\), let \(Q_L\) be a bounded minimizer of (2.10) in \(W^{1,2}_{Q_L}(\Omega ; S_0 )\). Then there exists an open set \(\Omega _L\subset \Omega \) such that \(Q_L\in C_{loc}^{1, \alpha }(\Omega \backslash \Omega _L)\) for each \(\alpha <1\). Moreover, there is a small constant \(\varepsilon _0\) independent of \(Q_L\) such that

$$\begin{aligned} \Sigma _L:=\Omega \backslash \Omega _L=\left\{ x_0\in \Omega : \, \liminf _{R\rightarrow 0}\int _{B_r(x_0)}|\nabla Q_L|^2\,dx >\varepsilon _0\right\} \end{aligned}$$

and the Hausdorff measure \({\mathcal {H}}^{q}(\Sigma _L )=0\) with \(0<q<1\).

3 Proof of Theorem 2

At first, let us recall that for a uniaxial tensor \(Q\in W^{1,2}_{Q_0}(\Omega ; S_* )\), its energy is given by

$$\begin{aligned} E(Q; \Omega ):=\int _{\Omega } f_{E}(Q, \nabla Q)\,dx, \end{aligned}$$

where \(f_{E}(Q, \nabla Q)=\frac{\alpha }{2} |\nabla Q|^2+V(Q, \nabla Q)\). Then we have

Lemma 3.1

If Q is a minimizer of \(E(Q; \Omega )\) in \(W^{1,2}_{Q_0}(\Omega ; S_*)\), it satisfies

$$\begin{aligned}&\alpha \left( - s_+\Delta Q_{ij} +2 \nabla _kQ_{il}\nabla _kQ_{jl} -2 s_+^{-1}\left( Q_{ij}+\frac{s_+}{3}\delta _{ij}\right) |\nabla Q|^2\right) \\&\qquad -\nabla _k \left( \left( Q_{jl}+\frac{s_+}{3} \delta _{jl}\right) V_{p^k_{il}} +\left( Q_{il}+\frac{s_+}{3} \delta _{il}\right) V_{p^k_{jl}}\right. \\&\left. \qquad -2s_+^{-1} \left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) V_{p^k_{lm}} \right) \\&\qquad + V_{p^k_{il}}\nabla _kQ_{jl} + V_{p^k_{jl}}\nabla _k Q_{il}\\&\qquad -2s_+^{-1}V_{p^k_{lm}}\left( \nabla _k Q_{lm} \left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) +\left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) \nabla _k Q_{ij}\right) \\&\qquad +V_{Q_{il}}\left( Q_{jl}+\frac{s_+}{3} \delta _{jl}\right) +V_{Q_{jl}}\left( Q_{il}+\frac{s_+}{3} \delta _{il}\right) \\&\qquad -2s_+^{-1}V_{Q_{lm}}\left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) \left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \\&\quad =0 \end{aligned}$$

in the weak sense.

Proof

Let \(\phi \in C^\infty _0(\Omega ; {\mathbb {R}}^3)\) be a test function. For each \(u_t = \frac{u+t\phi }{|u+t\phi |}\) with \(t\in {\mathbb {R}}\), we define

$$\begin{aligned} Q_t(x):=Q(u_t(x)) = s_+\left( u_t(x)\otimes u_t(x)-\frac{1}{3} I\right) \in S_*. \end{aligned}$$
(3.1)

For any \(\eta \in C^\infty _0(\Omega ; S_0)\), we choose a test function \(\phi \) such that \(\phi _{i}:=u_k\eta _{ik} \). If Q is a minimizer, the first variation of the energy of Q is zero; that is

$$\begin{aligned} \left. \frac{\,d}{\,dt}\int _\Omega f_E(Q_t,\nabla Q_t)\,dx\right| _{t=0}=\int _\Omega \left. f_{Q_{ij}}\frac{\,dQ_{t; ij}}{\,dt} +f_{p_{ij}^k}\frac{\,d}{\,dt}\frac{\partial Q_{ij}}{\partial x^k}\,dx \right| _{t=0}=0. \end{aligned}$$

Note that

$$\begin{aligned} \frac{\,dQ_{t; ij}}{\,dt} =&\frac{\Big ( Q_{jl}+\frac{s_+}{3} \delta _{jl} +t(Q_{lm}+\frac{s_+}{3} \delta _{lm})\Big )\eta _{il} +\Big (Q_{il}+\frac{s_+}{3} \delta _{il}+t(Q_{lm}+\frac{s_+}{3} \delta _{lm})\Big )\eta _{jl}}{1+2ts_+^{-1}Q_{il}\eta _{il}+t^2s_+^{-1}(Q_{lm}+\frac{s_+}{3} \delta _{lm})\eta _{il}\eta _{im}} \\&-\frac{2s_+^{-1}\Big (Q_{ij}+\frac{s_+}{3} \delta _{ij}+t(Q_{lm}+\frac{s_+}{3} \delta _{lm})\eta _{il}\eta _{im}\Big ) (Q_{lm}+\frac{s_+}{3} \delta _{lm})\eta _{lm}}{|1+2ts_+^{-1}Q_{il}\eta _{il}+t^2s_+^{-1}(Q_{lm}+\frac{s_+}{3} \delta _{lm})\eta _{il}\eta _{im}|^2}, \end{aligned}$$

where we used the fact that \(|u|=1\) and \(\phi _{i}=u_l\eta _{il} \). Then we have

$$\begin{aligned} \left. \frac{\,dQ_{t; i,j}}{\,dt} \right| _{t=0}&=s_+\big ( u_j\phi _i+u_i\phi _j -2(u\cdot \phi )(u_i u_j)\big ) \nonumber \\&=\left( Q_{jl}+\frac{s_+}{3} \delta _{jl}\right) \eta _{il}+(Q_{il}+\frac{s_+}{3} \delta _{il})\eta _{jl} \nonumber \\&\quad -2s_+^{-1}\left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) \eta _{lm}. \end{aligned}$$
(3.2)

Using the fact that \(\nabla _k |u+t\phi |^2 =0\) at \(t=0\) and substituting \(\phi _{i}:=u_l\eta _{il} \), a simple calculation shows

$$\begin{aligned} \left. \frac{\,d}{\,dt}\frac{\partial Q_{t; ij}}{\partial x_k}\right| _{t=0}&=\left. \left( \frac{\partial }{\partial x_k}\frac{\,d}{\,dt} Q_{t; ij}\right) \right| _{t=0} \nonumber \\&=\frac{\partial Q_{jl}}{\partial x_k} \eta _{il}+\frac{\partial Q_{il}}{\partial x_k}\eta _{jl} \nonumber \\&\quad -2s_+^{-1}\left( \frac{\partial Q_{ij}}{\partial x_k}Q_{lm}+\frac{\partial Q_{lm}}{\partial x_k}\left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \right) \eta _{lm} \nonumber \\&\quad +\left( Q_{jl}+\frac{s_+}{3} \delta _{jl}\right) \frac{\partial \eta _{il}}{\partial x_k}+\left( Q_{il}+\frac{s_+}{3} \delta _{il}\right) \frac{\partial \eta _{jl}}{\partial x_k} \nonumber \\&\quad -2s_+^{-1}\left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) \frac{\partial \eta _{lm}}{\partial x_k}. \end{aligned}$$
(3.3)

In the special case of \(\frac{1}{2} \int _{\Omega } {|\nabla Q|^2}\,dx\), it follows from using (3.3) and \(\langle Q, \nabla Q\rangle =0\) that

$$\begin{aligned} \frac{d}{d t}\int _{\Omega }\left. \frac{|\nabla Q_t|^2}{2}\,dx\right| _{t=0}&=\int _\Omega \nabla _k Q_{t; ij} \left. \frac{ d \nabla _k Q_{t; ij}}{dt} \right| _{t=0}\,dx \nonumber \\&=\int _{\Omega }2 \nabla _kQ_{il}\nabla _kQ_{jl}\eta _{ij}-2 \left( s_+^{-1}Q_{ij}+\frac{1}{3}\delta _{ij}\right) |\nabla Q|^2 \eta _{ij}\,dx \nonumber \\&\quad +\frac{1}{2}s_+ \int _{\Omega }(\nabla _k Q_{il} \nabla _k\eta _{il} + \nabla _k Q_{lj}\nabla _k\eta _{jl})\,dx\nonumber \\&=\int _{\Omega }\left( -s_+ \Delta Q_{ij} +2 \nabla _kQ_{il}\nabla _kQ_{jl} -2 \left( s_+^{-1}Q_{ij}+\frac{1}{3}\delta _{ij}\right) |\nabla Q|^2\right) \eta _{ij}\,dx \end{aligned}$$
(3.4)

for all \(\eta \in C^\infty _0(\Omega ; S_0)\).

For the term \(V(Q,\nabla Q)\), using (3.2)-(3.3) and integrating by parts, we have

$$\begin{aligned}&\int _\Omega \left. \frac{\,d}{\,dt}V(Q_t,\nabla Q_t)\right| _{t=0}\,dx\nonumber \\&\quad =\int _\Omega \left. [V_{p^k_{ij}} \frac{ d \nabla _k Q_{t; ij}}{dt}+V_{Q_{ij}}\frac{d Q_{t; ij}}{dt} ]\right| _{t=0}\,dx \nonumber \\&\quad =\int _\Omega V_{p^k_{ij}}\left( (Q_{jl}+\frac{s_+}{3} \delta _{jl})\frac{\partial \eta _{il}}{\partial x_k}+\left( Q_{il}+\frac{s_+}{3} \delta _{il}\right) \frac{\partial \eta _{jl}}{\partial x_k}+\frac{\partial Q_{jl}}{\partial x_k} \eta _{il}+\frac{\partial Q_{il}}{\partial x_k}\eta _{jl}\right) \,dx \nonumber \\&\qquad -2s_+^{-1}\int _\Omega V_{p^k_{ij}}\left( \frac{\partial Q_{ij}}{\partial x_k} \left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) +\frac{\partial Q_{lm}}{\partial x_k}\left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \right) \eta _{lm}\,dx \nonumber \\&\qquad +\int _\Omega -2s_+^{-1}V_{p^k_{ij}}\left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) \frac{\partial \eta _{lm}}{\partial x_k}+V_{Q_{ij}} \left( Q_{jl}+\frac{s_+}{3} \delta _{jl}\right) \eta _{il}\,dx \nonumber \\&\qquad +\int _\Omega V_{Q_{ij}}\left( \left( Q_{il}+\frac{s_+}{3} \delta _{il}\right) \eta _{jl} -2s_+^{-1}\left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) \eta _{lm} \right) \,dx \nonumber \\&\quad =-\int _\Omega \frac{\partial }{\partial x_k}\left( \left( Q_{jl}+\frac{s_+}{3} \delta _{jl}\right) V_{p^k_{il}} +\left( Q_{il}+\frac{s_+}{3} \delta _{il}\right) V_{p^k_{jl}}\right) \eta _{ij} \,dx \nonumber \\&\qquad +\int _\Omega \frac{\partial }{\partial x_k}\left( 2s_+^{-1} \left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) V_{p^k_{lm}} \right) \eta _{ij}+V_{p^k_{il}}\frac{\partial Q_{jl}}{\partial x_k}\eta _{ij}\,dx \nonumber \\&\qquad +\int _\Omega \left( V_{p^k_{jl}}\frac{\partial Q_{il}}{\partial x_k} -2s_+^{-1}V_{p^k_{lm}}\left( \frac{\partial Q_{lm}}{\partial x_k}\left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) +\left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) \frac{\partial Q_{ij}}{\partial x_k}\right) \right) \eta _{ij}\,dx \nonumber \\&\qquad +\int _\Omega \left( V_{Q_{il}}\left( Q_{jl}+\frac{s_+}{3} \delta _{jl}\right) +V_{Q_{jl}}\left( Q_{il}+\frac{s_+}{3} \delta _{il}\right) \right) \eta _{ij}\,dx \nonumber \\&\qquad -2s_+^{-1}\int _\Omega V_{Q_{lm}}\left( Q_{lm}+\frac{s_+}{3} \delta _{lm}\right) \left( Q_{ij}+\frac{s_+}{3} \delta _{ij}\right) \eta _{ij}\,dx. \end{aligned}$$
(3.5)

Combining above two identities (3.4) and (3.5), we prove Lemma 3.1. \(\square \)

Corollary 1

Assume that \(Q=s_+(u\otimes u-\frac{1}{3} I)\). Then \(Q=(Q_{ij})\) is a solution of equation

$$\begin{aligned} \Delta Q_{ij}-2s_+^{-1}\nabla _kQ_{il}\nabla _kQ_{jl}+2s_+^{-1}\left( s_+^{-1}Q_{ij}+\frac{1}{3}\delta _{ij}\right) |\nabla Q|^2=0 \end{aligned}$$
(3.6)

if and only if u is a harmonic map from \(\Omega \) into \(S^2\); i.e., \(-\Delta u=|\nabla u|^2 u\).

Now we give a proof of Theorem 2.

Proof

For each \(L>0\), let \(Q_L\) be a weak solution to the equation (1.11) with boundary value \(Q_0\in W^{1,2}(\Omega , S_*)\) and assume that \(Q_L\) is uniformly bounded in \(\Omega \).

For each \(\delta >0\), define a set

$$\begin{aligned} \Sigma _{\delta }=S_0\backslash S_{\delta }=\{Q\in S_0: \text{ dist }(Q, S_*)\ge \delta \}. \end{aligned}$$

For each \(Q\in \Sigma _{\delta }\), we have \(\pi (Q)\in S_*\); i.e., \( \pi (Q) =s_+\left( u\otimes u-\frac{1}{3} I\right) \) with \(u\in S^2\).

For a \( \pi (Q_L) =s_+\left( u_L\otimes u_L-\frac{1}{3} I\right) \) with \(u_L\in S^2\) and a test function \(\phi \in C^\infty _0(\Omega ; {\mathbb {R}}^3)\) with a small \(t\in {\mathbb {R}}\), we set \(u_{L, t}:= \frac{u_L+t\phi }{|u_L+t\phi |}\). Then we have

$$\begin{aligned} \big (\pi (Q_L)\big )_t := \ s_+\left( u_{L, t}\otimes u_{L, t}-\frac{1}{3} I\right) \in S_*. \end{aligned}$$
(3.7)

For any \(Q\in S_{\delta }\), set

$$\begin{aligned} F(Q):={{\tilde{f}}}_B(Q)+|Q-\pi (Q)|^2. \end{aligned}$$

Using the Taylor expansion of \(F\left( (\pi (Q_L))_t\right) \) at \(Q_L\in S_{\delta }\), we derive

$$\begin{aligned} \frac{F\left( (\pi (Q_L))_t\right) }{L}&= \frac{F(Q_L)}{L} +\frac{(\nabla F(Q_L) )_{ij}}{L} \left( (\pi (Q_L))_t -Q_L\right) _{ij} \nonumber \\&\quad +\frac{1}{2L}\nabla ^2_{Q_{ij}Q_{kl}}F\left( Q_{\tau _1}\right) \left( (\pi (Q_L))_t-Q_L\right) _{ij}\left( (\pi (Q_L))_t-Q_L\right) _{kl}, \end{aligned}$$
(3.8)

where \(Q_{\tau _1}:=(1-\tau _1) (\pi (Q_L))_t+\tau _1 Q_L\) for some \(\tau _1\in [0,1]\). Note that

$$\begin{aligned} |Q_{\tau _1}-(\pi (Q_L))_t|\le |Q_L-\pi (Q_L)|+ |\pi (Q_L)-(\pi (Q_L))_t|. \end{aligned}$$

Since \((\pi (Q_L))_t\in S_*\), it implies that \(F((\pi (Q_L))_t)=0\). Note that the function F(Q) is smooth in Q. For sufficiently small t, we have \(|Q_{\tau _1}-(\pi (Q_L))_t|\le 3\delta \) for \(Q_L\in S_{2\delta }\). For each \(Q\in S^*\), it is known that the Hessian of \({{\tilde{f}}}_B(Q)\) is semi-positive definite at \(Q=Q^*\). Therefore each \(Q\in S_{\delta }\), the Hessian of \(F_B(Q)\) is positive definite with sufficiently small \(\delta >0\); i.e., for any \(Q_L\in S_{2\delta }\), we have

$$\begin{aligned}&\nabla ^2_{Q_{ij}Q_{kl}} F\big (Q_\tau \big )( (\pi (Q_L))_t-Q_L)_{ij}( (\pi (Q_L))_t-Q_L)_{kl}\ge \frac{1}{2} |(\pi (Q_L))_t-Q_L|^2 \end{aligned}$$
(3.9)

with sufficiently small t and \(\delta \). Then it follows from (3.8)-(3.9) that

$$\begin{aligned} \begin{aligned}&\int _{\Omega _{L,2\delta }} \frac{(\nabla F(Q_L) )_{ij}}{L} ((\pi (Q_L))_t -Q_L)_{ij}\,dx\\&\quad =\int _{\Omega _{L,2\delta }} \frac{\nabla _{Q_{ij}} f_B(Q_L)}{L}((\pi (Q_L))_t -Q_L)_{ij}\,dx \\&\qquad +2\int _{\Omega _{L,2\delta }}\partial _{Q_{ij}}(Q_L-\pi (Q_L))_{mn}\frac{(Q_L-\pi (Q_L))_{mn}}{L} ((\pi (Q_L))_t -Q_L)_{ij}\,dx \\&\quad \le -\frac{1}{2}\int _{\Omega _{L,2\delta }}\frac{|(\pi (Q_L))_t-Q_L|^2}{L}\,dx \end{aligned} \end{aligned}$$
(3.10)

provided \(\Omega _{L,2\delta }=\{x\in \Omega : Q_L(x)\in S_{2\delta }\}\) for \(\delta >0\).

By using Young’s inequality, we have

$$\begin{aligned} \begin{aligned}&\int _{\Omega _{L,2\delta }} \frac{1}{L}\nabla _{Q_{ij}} f_B(Q_L)((\pi (Q_L))_t -Q_L)_{ij} \,dx +\frac{1}{4}\int _{\Omega _{L,2\delta }}\frac{|(\pi (Q_L))_t-Q_L|^2}{L}\,dx \\&\quad \le C\int _{\Omega _{L,2\delta }}\frac{|Q_L-\pi (Q_L)|^2}{L} \,dx\le C \int _{\Omega _{L,2\delta }} \frac{1}{L} {{\tilde{f}}}_B(Q_L)\,dx, \end{aligned} \end{aligned}$$
(3.11)

where we used a result in [36] or Corollary 2 in Sect. 3.

In order to extend (3.11) to \(\Omega \), we define

$$\begin{aligned} {\hat{Q}}_{L,t}:={\left\{ \begin{array}{ll} \pi (Q_L)_t,&{} \text{ for } Q_L\in S_{\delta } \\ \frac{|Q_L-\pi (Q_L)|^2}{\delta ^2}\pi (Q_L)_t+\frac{\delta ^2-|Q_L-\pi (Q_L)|^2}{\delta ^2} Q_{*,t},&{} \text{ for } Q_L\in \Sigma _{\delta }\backslash \Sigma _{2\delta } \\ Q_{*,t},&{} \text{ for } Q_L\in \Sigma _{2\delta }. \end{array}\right. } \end{aligned}$$
(3.12)

It can be checked that \({\hat{Q}}_{L,t}\in W^{1,2}_{Q_0}(\Omega ; S_0)\). Then

$$\begin{aligned} {\hat{Q}}_{L,t}-Q_{*,t}={\left\{ \begin{array}{ll} \pi (Q_L)_t- Q_{*,t},&{} \text{ for } Q_L\in S_{\delta } \\ \frac{|Q_L-\pi (Q_L)|^2}{\delta ^2} (\pi (Q_L)_t- Q_{*,t}),&{} \text{ for } Q_L\in \Sigma _{\delta }\backslash \Sigma _{2\delta } \\ 0,&{} \text{ for } Q_L\in \Sigma _{2\delta }. \end{array}\right. } \end{aligned}$$
(3.13)

On the other hand, there exists a uniform bound \(C(\delta )>0\) such that for all \(x\in \Omega \backslash \Omega _{L,\delta }\), \(\tilde{f}_B(Q_L(x)) \ge C(\delta )\). Using Lemma 2.4, we observe that

$$\begin{aligned} \begin{aligned}&\int _{\Omega \backslash \Omega _{L,\delta } }\frac{1}{L}\nabla _{Q_{ij}} f_B(Q_L)({{\hat{Q}}}_{L,t}-Q_L)_{ij}\,dx \\&\quad =\int _{\Omega _{L,2\delta } \backslash \Omega _{L,\delta } }\frac{1}{L}\nabla _{Q_{ij}} f_B(Q_L)\left[ \frac{|Q_L-\pi (Q_L)|^2}{\delta ^2}(\pi (Q_L)_t-Q_{*,t})+(Q_{*,t}-Q_L)\right] _{ij}\,dx \\&\qquad +\int _{\Omega \backslash \Omega _{L,2\delta } }\frac{1}{L}\nabla _{Q_{ij}} f_B(Q_L)(Q_{*,t}-Q_L)_{ij}\,dx \\&\quad \le C \frac{| \Omega \backslash \Omega _{L,\delta }|}{L} \le \frac{C}{C(\delta )}\int _{\Omega \backslash \Omega _{L,\delta } }\frac{{{\tilde{f}}}_B(Q_L) }{L} \,dx. \end{aligned} \end{aligned}$$
(3.14)

By the assumption (1.15) in Theorem 2, we deduce from (3.11) and (3.14) that

$$\begin{aligned} \lim _{L\rightarrow 0}\int _{\Omega }\frac{1}{L}\nabla _{Q_{ij}} f_B(Q_L)({{\hat{Q}}}_{L,t}-Q_L)_{ij}\,dx\le 0. \end{aligned}$$
(3.15)

Multiplying (1.11) by \(({{\hat{Q}}}_{L,t}-Q_L)\) and using (3.15) yield

$$\begin{aligned}&\lim _{L\rightarrow 0}\int _{\Omega }\left( \alpha \nabla _kQ_{L,ij} +\tilde{V}_{p^k_{ij}}(Q_L,\nabla Q_L)-{{\tilde{V}}}_{Q_{ij}}(Q_L,\nabla Q_L)\right) \nabla _k ({{\hat{Q}}}_{L,t}-Q_L)_{ij}\,dx\ge 0. \end{aligned}$$
(3.16)

Here we used the fact that \({{\hat{Q}}}_{L,t}-Q_L\) is symmetric and traceless.

In order to pass a limit, we claim that \({\hat{Q}}_{L,t}\rightarrow Q_{*,t}\) strongly in \(W^{1,2}_{Q_0}(\Omega ; S_0)\).

In fact, it follows from (3.13) that

$$\begin{aligned} \begin{aligned}&\int _{\Omega } |\nabla ({\hat{Q}}_{L,t}-Q_{*,t})|^2 \,dx= \int _{ \Omega _{L,2\delta }} |\nabla ({\hat{Q}}_{L,t}-Q_{*,t})|^2 \,dx \\&\quad = \int _{\Omega _{L,\delta }} |\nabla ( {{\hat{Q}}}_{L,t}-Q_{*,t})|^2 \,dx +\int _{ \Omega _{L,2\delta } \backslash \Omega _{L,\delta } } \left| \nabla \left( \frac{|Q_L-\pi (Q_L)|^2}{\delta ^2} (\pi (Q_L)_t- Q_{*,t})\right) \right| ^2 \,dx \\&\quad \le \int _{\Omega _{L,\delta }} |\nabla ( (\pi (Q_L))_t-\pi (Q_{*})_t)|^2 \,dx + C \int _{ \Omega _{L,2\delta } \backslash \Omega _{L,\delta } }|\nabla ( (\pi (Q_L))_t-\pi (Q_{*})_t)|^2 \,dx \\&\qquad +C \int _{ \Omega _{L,2\delta } \backslash \Omega _{L,\delta } }\frac{|(\pi (Q_L))_t-Q_{*,t}|^2}{\delta ^4}\left( |\nabla ( Q_L-Q_*)|^2+ |\nabla ( \pi (Q_*)-\pi (Q_L))|^2\right) \,dx. \end{aligned} \end{aligned}$$
(3.17)

Note that

$$\begin{aligned}&\pi (Q_L)-\pi (Q_{*})=\nabla _{Q}\pi (Q_{\xi }) (Q_L-Q_{*}),\\&(\pi (Q_L))_t-\pi (Q_{*})_t=\nabla _{Q}\pi (Q_{\xi })_t (Q_L-Q_{*}). \end{aligned}$$

When \(Q_L\) approaches to \(Q_*\), \(\nabla _{Q}\pi (Q_{\xi })\) is close to the identity map I and \(\nabla _{Q}\pi (Q_{\xi })_t\) for small t. Therefore

$$\begin{aligned}&|\nabla (\pi (Q_L) -\pi (Q_{*})) |\le C|\nabla (Q_L-Q_{*})| +C |\nabla Q_{\xi }| |Q_L-Q_{*}|. \end{aligned}$$

As \(Q_L\rightarrow Q_*\), the term \(\pi (Q_L)_t\) is close to \(\pi (Q_*)_t\) and \(\nabla _Q\pi (Q_\xi )_t\) is close to the identity map for small t. Note that \(\nabla ^2_{QQ}\pi (Q_\xi )_t\) is bounded. Then

$$\begin{aligned} |\nabla ((\pi (Q_L))_t-\pi (Q_{*})_t)|&\le |\nabla _Q\pi (Q_\xi )_t\nabla (Q_L -Q_{*})|+|\nabla ^2_{QQ}\pi (Q_\xi )_t||\nabla Q_\xi ||Q_L -Q_{*}| \\&\le C|\nabla (Q_L-Q_{*})| + C|\nabla Q_{\xi }| |Q_L-Q_{*}|. \end{aligned}$$

Then the inequality (3.17) reads as

$$\begin{aligned}&\int _{\Omega } |\nabla ({\hat{Q}}_{L,t}-Q_{*,t})|^2 \,dx \\&\quad \le C\int _{ \Omega _{L,2\delta } }|\nabla (Q_L-Q_{*})|^2 +(|\nabla Q_L|^2 +|\nabla Q_*|^2)|Q_L-Q_{*}|^2\,dx \\&\quad \le C\int _{ \Omega }|\nabla (Q_L-Q_{*})|^2\,dx +C\left( \int _{ \Omega \backslash \Sigma _\varepsilon }+\int _{\Sigma _\varepsilon }\right) |\nabla Q_*|^2|Q_L-Q_{*}|^2\,dx. \end{aligned}$$

Here we employ the Egoroff theorem; i.e., for all \(\varepsilon >0\), there exists a measurable subset \(\Sigma _{\varepsilon } \subset \Omega \) such that

$$\begin{aligned} | \Sigma _\varepsilon |\le \varepsilon \text{ and } Q_L\rightarrow Q_* \text{ uniformly } \text{ on } \Omega \backslash \Sigma _{\varepsilon } \text{ as } L\rightarrow 0. \end{aligned}$$
(3.18)

As \(\varepsilon \rightarrow 0\) and \(L\rightarrow 0\), we prove the claim that \({{\hat{Q}}}_{L,t}\rightarrow Q_{*,t}\) strongly in \(W^{1,2}_{Q_0}(\Omega ; S_0)\).

We observe that

$$\begin{aligned}&\int _{\Omega }|{{\tilde{V}}}_{p^k_{ij}}(Q_L,\nabla Q_L)\nabla _k({{\hat{Q}}}_{L,t}-Q_L)_{ij}-{{\tilde{V}}}_{p^k_{ij}}(Q_*,\nabla Q_*)\nabla _k(Q_{*,t}-Q_*)_{ij}|\,dx \\&\quad \le \int _{\Omega }| {{\tilde{V}}}_{p^k_{ij}}(Q_L,\nabla Q_L)||(\nabla _kQ_{L,t} -\nabla _kQ_{*,t})_{ij}+(\nabla _kQ_*-\nabla _kQ_L)_{ij}|\,dx \\&\quad \quad +\left( \int _{ \Omega \backslash \Sigma _\varepsilon }+\int _{\Sigma _\varepsilon }\right) |\tilde{V}_{p^k_{ij}}(Q_L,\nabla Q_L)\nabla _k(Q_{*,t}-Q_*)_{ij}-\tilde{V}_{p^k_{ij}}(Q_*,\nabla Q_*)\nabla _k(Q_{*,t}-Q_*)_{ij}|\,dx \end{aligned}$$

and

$$\begin{aligned}&\int _{\Omega }|{{\tilde{V}}}_{Q_{ij}}(Q_L,\nabla Q_L) ({{\hat{Q}}}_{L,t}-Q_L)_{ij} -{{\tilde{V}}}_{Q_{ij}}(Q_*,\nabla Q_*) (Q_{*,t}-Q_*)_{ij}|\,dx \\&\quad \le \left( \int _{ \Omega \backslash \Sigma _\varepsilon }+\int _{\Sigma _\varepsilon }\right) |\tilde{V}_{Q_{ij}}(Q_L,\nabla Q_L) ({{\hat{Q}}}_{L,t}-Q_L)_{ij} -\tilde{V}_{Q_{ij}}(Q_*,\nabla Q_L) ({{\hat{Q}}}_{L,t}-Q_L)_{ij}|\,dx \\&\qquad +\int _{\Omega } |{{\tilde{V}}}_{Q_{ij}}(Q_*,\nabla Q_L) ({{\hat{Q}}}_{L,t}-Q_L)_{ij}-{{\tilde{V}}}_{Q_{ij}}(Q_*,\nabla Q_*) ({{\hat{Q}}}_{*,t}-Q_*)_{ij}|\,dx. \end{aligned}$$

Using the uniform convergence of \(Q_L\) in \(\Omega \backslash \Sigma _\varepsilon \) and strong convergence of \({{\hat{Q}}}_{L,t}, Q_L\) in \(W^{1,2}_{Q_0}(\Omega ,S_0)\), we derive

$$\begin{aligned}&\lim _{L\rightarrow 0}\int _{\Omega }|{{\tilde{V}}}_{Q_{ij}}(Q_L,\nabla Q_L) ({{\hat{Q}}}_{L,t}-Q_L)_{ij} -{{\tilde{V}}}_{Q_{ij}}(Q_*,\nabla Q_*) (Q_{*,t}-Q_*)_{ij}|\,dx=0, \\&\lim _{L\rightarrow 0}\int _{\Omega }| {{\tilde{V}}}_{p^k_{ij}}(Q_L,\nabla Q_L)\nabla _k({{\hat{Q}}}_{L,t}-Q_L)_{ij} -{{\tilde{V}}}_{p^k_{ij}}(Q_*,\nabla Q_*)\nabla _k(Q_{*,t}-Q_*)_{ij}|\,dx=0. \end{aligned}$$

As \(L\rightarrow 0\), the estimate (3.16) yields

$$\begin{aligned} \begin{aligned}&\int _{\Omega }\left( \alpha \nabla _kQ_{*,ij} +\tilde{V}_{p^k_{ij}}(Q_*,\nabla Q_*)\right) \nabla _k(Q_{*,t}-Q_*)_{ij}\,dx \\&\qquad +\int _{\Omega }{{\tilde{V}}}_{ij}(Q_*,\nabla Q_*) (Q_{*,t}-Q_*)_{ij}\,dx\ge 0. \end{aligned} \end{aligned}$$
(3.19)

For each \(\eta \in C_0^{\infty }(\Omega ,S_0)\), we define

$$\begin{aligned} \begin{aligned} \varphi _{ij} (Q,\eta ):=&\left( s_+^{-1}Q_{jl}+\frac{1}{3} \delta _{jl}\right) \eta _{il}+\left( s_+^{-1}Q_{il}+\frac{1}{3} \delta _{il}\right) \eta _{jl} \\&-2\left( s_+^{-1}Q_{ij}+\frac{1}{3} \delta _{ij}\right) \left( s_+^{-1}Q_{lm}+\frac{1}{3} \delta _{lm}\right) \eta _{lm}. \end{aligned} \end{aligned}$$
(3.20)

In view of (3.2) and (3.3), we have

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{(Q_t-Q_*)}{t}=\varphi (Q_*,\eta ),\quad \lim _{t\rightarrow 0}\nabla \frac{(Q_t-Q_*)}{t}=\nabla \varphi (Q_*,\eta ). \end{aligned}$$

For the estimate (3.19), the limit in t exists. Dividing (3.19) by t then as \(t\rightarrow 0^+\) and \(t\rightarrow 0^-\), we have

$$\begin{aligned} \int _{\Omega } \left( \alpha \nabla _kQ_{*,ij}+V_{p^k_{ij}}(Q_*,\nabla Q_*) \right) \nabla _k \varphi _{ij}(Q_*,\eta )+V_{Q_{ij}} (Q_*,\nabla Q_*)\varphi _{ij}(Q_*,\eta )\,dx= 0. \end{aligned}$$

Repeating the same steps in (3.4) and (3.5), we prove that \(Q_*\) satisfies (1.12). \(\square \)

4 Smooth convergence of solutions

In this section, we will prove Theorem 3. At first, we derive some key lemmas.

For any tensor \(Q \in S_0\), there exists a rotation \(R(Q)\in SO(3)\) such that \({{\tilde{Q}}}:=R^T(Q) Q R(Q)\) is diagonal. Moreover, the space \(S^*\) has only three diagonal tensors so for each \(Q\in S_*\), we assume that

$$\begin{aligned} R^T(Q)QR(Q)= \begin{pmatrix} \frac{-s_+}{3}&{}0&{}0\\ 0&{}\frac{-s_+}{3}&{}0\\ 0&{}0&{}\frac{2s_+}{3} \end{pmatrix}:=Q^+. \end{aligned}$$
(4.1)

Lemma 4.1

For any \(Q\in S_\delta \) and \(\xi \in S_0\) with a sufficiently small \(\delta >0\), the Hessian of the bulk density \(f_B(Q)\) satisfies the following estimate

$$\begin{aligned} \sum _{i,j=1}^3\partial _{{{\tilde{Q}}}_{ii}}\partial _{{{\tilde{Q}}}_{jj}} f_B(\tilde{Q})\xi _{ii}\xi _{jj}\ge \frac{\lambda }{2} ( \xi _{11}^2 + \xi _{22}^2+ \xi _{33}^2), \end{aligned}$$
(4.2)

where \(\lambda =\min \{3a,s_+ b \}>0\) and \({{\tilde{Q}}}=R^T(Q) Q R(Q)\) is diagonal.

Proof

For a fixed \(\pi (Q_0)\in S_*\), there exists a rotation \(R(\pi (Q_0))\in SO(3)\) in (4.1) such that \(R^T(\pi (Q_0))\pi (Q_0) R(\pi (Q_0))=Q^+\). For \(i=1, 2, 3\), we calculate the first derivative of \(f_B( {\tilde{Q}})\) by

$$\begin{aligned} \partial _{ {\tilde{Q}}_{ii}}f_B({\tilde{Q}}) =\left( -a {\tilde{Q}}_{ii}-b {\tilde{Q}}_{ik}{\tilde{Q}}_{ki}+c {\tilde{Q}}_{ii}| {\tilde{Q}}|^2\right) . \end{aligned}$$

Then the second derivative of \(f_B({\tilde{Q}})\) with \(i,j=1, 2, 3\) is

$$\begin{aligned}&\partial _{{\tilde{Q}}_{ii}}\partial _{{\tilde{Q}}_{jj}}f_B({\tilde{Q}})= -a\delta _{ij} -2b\delta _{ij}{\tilde{Q}}_{ii}+c(\delta _{ij}|\tilde{Q}|^2+2{\tilde{Q}}_{ii}{\tilde{Q}}_{jj}) . \end{aligned}$$
(4.3)

For the case of \(i=j\) at \(Q=Q_0\), \({\tilde{Q}}=Q^+\), From the equality \(\frac{2}{3} cs^2_+=\frac{1}{3}bs_++a \) (c.f. [34]), we find

$$\begin{aligned} \partial _{{\tilde{Q}}_{ii}}\partial _{{\tilde{Q}}_{ii}}f_B({\tilde{Q}}) =&-a -2{\tilde{Q}}_{ii}b+( |{\tilde{Q}}|^2+2{\tilde{Q}}_{ii}^2)c=-(2\tilde{Q}_{ii}-\frac{s_+}{3})b+2{\tilde{Q}}_{ii}^2c. \end{aligned}$$

Then, at \({\tilde{Q}}=Q^+\), we have

$$\begin{aligned} \partial _{{\tilde{Q}}_{11}}\partial _{{\tilde{Q}}_{11}}f_B({\tilde{Q}}) =&\left( s_+b+\frac{2s_+^2}{9}c\right) =\frac{1}{3}a+\frac{10s_+}{9}b, \end{aligned}$$
(4.4)
$$\begin{aligned} \partial _{{\tilde{Q}}_{22}}\partial _{{\tilde{Q}}_{22}}f_B({\tilde{Q}}) =&\frac{1}{3}a+\frac{10s_+}{9}b, \end{aligned}$$
(4.5)
$$\begin{aligned} \partial _{{\tilde{Q}}_{33}}\partial _{{\tilde{Q}}_{33}}f_B({\tilde{Q}})=&-s_+ b+\frac{8s_+}{9}c=\frac{4}{3}a-\frac{5s_+}{9}b. \end{aligned}$$
(4.6)

For the case of \(i\ne j\), at \({\tilde{Q}}=Q^+\), we observe that

$$\begin{aligned} 2\partial _{{\tilde{Q}}_{11}}\partial _{{\tilde{Q}}_{22}}f_B({\tilde{Q}}) =&4{\tilde{Q}}_{11}\tilde{Q}_{22}c=\frac{4s_+^2}{9}c=\frac{2}{3}a+\frac{2s_+}{9}b, \end{aligned}$$
(4.7)
$$\begin{aligned} 2\partial _{{\tilde{Q}}_{11}}\partial _{{\tilde{Q}}_{33}}f_B({\tilde{Q}}) =&4{\tilde{Q}}_{11}\tilde{Q}_{33}c=-\frac{8s_+^2}{9}c=-\left( \frac{4}{3}a+\frac{4s_+}{9}b\right) , \end{aligned}$$
(4.8)
$$\begin{aligned} 2\partial _{{\tilde{Q}}_{22}}\partial _{{\tilde{Q}}_{33}}f_B({\tilde{Q}}) =&4{\tilde{Q}}_{22}\tilde{Q}_{33}c=-\frac{8s_+^2}{9}c=-\left( \frac{4}{3}a+\frac{4s_+}{9}b\right) . \end{aligned}$$
(4.9)

In conclusion, using the fact that \(\xi _{33}=-(\xi _{11}+\xi _{22})\), we have at \({\tilde{Q}}=Q^+\)

$$\begin{aligned} \partial _{{{\tilde{Q}}}_{ii}}\partial _{{{\tilde{Q}}}_{jj}} f_B(\tilde{Q})\xi _{ii}\xi _{jj} =&\left( \frac{1}{3}a+\frac{10s_+}{9}b\right) (\xi _{11}^2+\xi _{22}^2)+\left( \frac{2}{3}a+\frac{2s_+}{9}b\right) \xi _{11}\xi _{22} \\&+\left( \frac{4}{3}a-\frac{5s_+}{9}b \right) \xi _{33}^2-\left( \frac{4}{3}a+\frac{4s_+}{9}b\right) \xi _{33}(\xi _{11}+\xi _{22}) \\ =&bs_+(\xi _{11}^2+\xi _{22}^2) + 3a\xi _{33}^2\ge \lambda (\xi _{11}^2 + \xi _{22}^2+\xi _{33}^2) \end{aligned}$$

with \(\lambda =\min \{3a,s_+b\}>0\). Then

$$\begin{aligned}&\partial _{{{\tilde{Q}}}_{ii}}\partial _{{{\tilde{Q}}}_{jj}} f_B(\tilde{Q})\xi _{ii}\xi _{jj} \ge \partial _{{{\tilde{Q}}}_{ii}}\partial _{{{\tilde{Q}}}_{jj}} f_B( Q^+)\xi _{ii}\xi _{jj}-C|\tilde{Q}-Q^+|\left| \sum _{i=1}^3\xi _{ii}\right| ^2. \end{aligned}$$

Due to the fact that \(|{{\tilde{Q}}}-Q^+| = |Q-\pi (Q)|\), we prove (4.2) for a sufficiently small \(\delta >0\). \(\square \)

Corollary 2

For any \(Q\in S_{\delta }\) with a sufficiently small \(\delta >0\), there exists constants \(C_1,C_2,C_3>0\) such that

$$\begin{aligned} C_1 {{\tilde{f}}}_B(Q)\le&|Q-\pi (Q)|^2\le C_2 {{\tilde{f}}}_B(Q), \end{aligned}$$
(4.10)
$$\begin{aligned} |Q-\pi (Q)|\le&C_3 |g_B(Q)|. \end{aligned}$$
(4.11)

Proof

It follows from the Taylor expansion of \({{\tilde{f}}}_B(Q)\) at \(\pi (Q)\) that

$$\begin{aligned} {{\tilde{f}}}_B(Q)=&{{\tilde{f}}}_B(\pi (Q))+\nabla _{Q_{ij}}f_B(\pi (Q))(Q-\pi (Q))_{ij} \nonumber \\&+\partial _{Q_{ij}}\partial _{Q_{kl}} f_B(Q_\tau )(Q-\pi (Q))_{ij}(Q-\pi (Q))_{kl}, \end{aligned}$$
(4.12)

where \(Q_\tau \) is an intermediate point between \(Q_L\) and \(\pi (Q)\).

Since Q commutes with \(\pi (Q)\) (c.f. [36]), they can be simultaneously diagonalized. Note that \({{\tilde{f}}}_B(\pi (Q))=0\), \(\nabla _{Q_{ij}}f_B(\pi (Q))=0\) and \(Q_\tau \) is sufficiently close to \(\pi (Q)\). Using Lemma 4.1 with the fact that

$$\begin{aligned}&\sum _{i,j=1}^3 \partial _{{\tilde{Q}}_{ii}}\partial _{\tilde{Q}_{jj}}f_B({{\tilde{Q}}})({{\tilde{Q}}}-{{\tilde{\pi }}}(Q))_{ii}({{\tilde{Q}}}-{\tilde{\pi }}(Q))_{jj} \\&\quad =\partial _{Q_{ij}}\partial _{Q_{kl}} f_B(Q)(Q-\pi (Q))_{ij}(Q-\pi (Q))_{kl}, \end{aligned}$$

we have

$$\begin{aligned} \partial _{Q_{ij}}\partial _{Q_{kl}} f_B(Q_\tau )(Q-\pi (Q))_{ij}(Q-\pi (Q))_{kl}\ge \frac{\lambda }{2}\sum _{i=1}^3({{\tilde{Q}}}-Q^+)_{ii}^2. \end{aligned}$$
(4.13)

Then we obtain

$$\begin{aligned} {{\tilde{f}}}_B(Q)\ge \frac{\lambda }{2}\sum _{i=1}^3({{\tilde{Q}}}-Q^+)_{ii}^2= \frac{\lambda }{2} |Q-\pi (Q)|^2. \end{aligned}$$

The left-hand side of (4.10) is a direct consequence of (4.12) by using Young’s inequality. Taking Taylor expansion of \( g_B(Q)\) at \(\pi (Q)\) yields

$$\begin{aligned} (g_B(Q))_{ij}=\partial _{Q_{ij}}\partial _{Q_{kl}} f_B(Q_{\tau _1})(Q-\pi (Q))_{kl}. \end{aligned}$$

Multiplying both side by \((Q-\pi (Q))_{ij}\) and using (4.13), we obtain (4.11). \(\square \)

From now on, for each \(L>0\), let \(Q_L\) be a solution to the equation (1.11) and assume that \(Q_L\) is smooth and converges to \(Q_*\) uniformly inside \(\Omega \backslash \Sigma \), where \(\Sigma \) is the singular set of \(Q_*\). For a sufficiently small L, \({{\,\textrm{dist}\,}}(Q_L;S_*)\le \delta \) inside \(\Omega \backslash \Sigma \).

Set

$$\begin{aligned} H(Q, \nabla Q):=&\alpha \Delta Q + \frac{1}{2} {\nabla _k(V_{Q_{x_k}}+ V^T_{Q_{x_ k}})}-\frac{1}{3} I \text{ tr } (\nabla _k V_{Q_{x_k}})\nonumber \\&-\frac{1}{2} ( V_{Q} + V^T_{Q})+ \frac{1}{3} I \text{ tr } (V_{Q}) \end{aligned}$$
(4.14)

and

$$\begin{aligned} g_B(Q)= \left( -aQ -b \left( Q Q -\frac{1}{3}I {{\,\textrm{tr}\,}}(Q^2)\right) -cQ {{\,\textrm{tr}\,}}(Q^2)\right) . \end{aligned}$$
(4.15)

Due to the fact that \({{\tilde{Q}}}:=R^T(Q) Q R(Q)\) is diagonal, \(g_B({{\tilde{Q}}})=R^T(Q)\, g_B(Q)R(Q)\) is also diagonal for a rotation \(R(Q)\in SO(3)\).

Let Q be differentiable in \(\Omega \). Then there exists a set \(\Sigma _Q\), which has measure zero, such that R(Q) is differentiable in \(\Omega \backslash \Sigma _Q\) (c.f. Corollary 2 [34, 37]). Therefore, we have the following geometric identity of rotations:

Lemma 4.2

Assume that for any \(x\in \Omega \backslash \Sigma _Q\), there exists a differentiable rotation R(Q) such that both \(R^T(Q)QR(Q)\) and \(R^T(Q)h(Q)R(Q)\) are diagonal. Then, for each i, we have

$$\begin{aligned} \nabla \left( R^T(Q) h(Q) R (Q)\right) _{ii} =\left( R^T(Q)\nabla h(Q) R (Q)\right) _{ii}. \end{aligned}$$
(4.16)

Proof

Let \(x_0\) be a fixed point in \(\Omega \backslash \Sigma _Q\) and fix \(i=1,2,3\). For \(Q_0=Q(x_0)\in S_0\), there exists \(R_0:=R(Q_0)\in SO(3)\) such that \(R_0^TQ_0R_0\) is diagonal. Denote \(\tilde{R}(Q)=R_0^TR(Q)\) with \({{\tilde{R}}}(Q_0)=I\). Fix \(Q_0\in S_0\), there is \(R_0:=R(Q_0)\in SO(3)\) such that \(R_0^TQ_0R_0\) and \(R_0^Th(Q_0)R_0\) diagonal. Denote \({{\tilde{R}}}(Q)=R_0^TR(Q)\), so \({{\tilde{R}}}(Q_0)=I\). Since \(R(Q)\in SO(3)\), \(R_{ki}(Q)R_{kj}(Q)=\delta _{ij}\). Then, for each i, we have at \(x_0\)

$$\begin{aligned}&\nabla \left( R^T(Q) h(Q) R (Q)\right) _{ii} \nonumber \\&\quad =\nabla \left( R^T(Q) R_0R^T_0 h(Q) R_0 R^T_0R (Q)\right) _{ii} \nonumber \\&\quad = \sum _{k,l=1}^3 \nabla {{\tilde{R}}}_{ki}(Q)(R^T_0 h(Q) R_0)_{kl}{{\tilde{R}}}_{li}(Q) \nonumber \\&\quad \quad + \sum _{k,l=1}^3 {{\tilde{R}}}_{ki}(Q)(R^T_0 h(Q) R_0)_{kl} \nabla _Q {{\tilde{R}}}_{li}(Q) \nonumber \\&\qquad + \sum _{k,l=1}^3 {{\tilde{R}}}_{ki}(Q) \nabla (R^T_0 h(Q) R_0)_{kl}{{\tilde{R}}}_{li}(Q). \end{aligned}$$
(4.17)

Note that \(R^T_0 h(Q_0) R_0\) is diagonal, \(\tilde{R}_{ik}(Q_0)=\delta _{ik}\) and \(\nabla {{\tilde{R}}}_{ii}(Q_0)=0\). It can be seen that the term \(\nabla {{\tilde{R}}}_{ki}(Q)(R^T_0Q_0 R_0)_{kl}{{\tilde{R}}}_{li}(Q)\) at \(Q=Q_0\) is zero. Therefore

$$\begin{aligned}\left. \nabla \left( R^T(Q) h(Q) R (Q)\right) _{ii}\right| _{Q=Q_0}=\left( R^T(Q_0)\nabla h(Q)|_{Q=Q_0} R(Q_0)\right) _{ii.}\end{aligned}$$

Since \(x_0\) is any point, we prove (4.16). \(\square \)

Denote the inner product by \(\langle A,B\rangle =A_{ij}B_{ij}\) for \(A,B\in {\mathbb {M}}^{3\times 3}\). Using the above geometric identity, we have

Lemma 4.3

Let k and l be two integers, for \(x\in \Omega \backslash \Sigma _Q\), \(Q= Q(x)\) and any smooth scalar function \(\phi \), we have

$$\begin{aligned}&\langle \nabla ^k (g_B({{\tilde{Q}}}) \phi ^2),R^T(Q)\nabla ^{l+1} QR(Q)\rangle \nonumber \\&\quad =\nabla \langle \nabla ^k (g_B({{\tilde{Q}}}) \phi ^2),R^T\nabla ^l QR\rangle \nonumber \\&\qquad -\langle \nabla ^{k+1} (g_B({{\tilde{Q}}}) \phi ^2) ,R^T(Q)\nabla ^l QR(Q)\rangle . \end{aligned}$$
(4.18)

Proof

For \(x_0\in \Omega \backslash \Sigma _Q\), R(x) is differentiable in the neighborhood of \(x_0\). Fixing \(Q_0=Q(x_0)\in S_0\), there exists \(R_0=R(Q(x_0))\) such that \(R_0^TQ_0R_0\) is diagonal. Recall that \({{\tilde{R}}}(Q)=R_0^TR(Q)\), \({{\tilde{R}}}_{ik}(Q_0)=\delta _{ik}\) and \(\nabla {{\tilde{R}}}_{ii}(Q_0)=0\). For any matrix A, let \(A_D\) be the diagonal part of A and \(A_N\) the non-diagonal part of A such that \(A=A_D+A_N\). Recall that \(\nabla ^k g_B({{\tilde{Q}}})\) and \(\nabla ^{k+1} g_B({{\tilde{Q}}})\) are diagonal. By employing an analogous argument in the proof of Lemma 4.2, we obtain

$$\begin{aligned} \begin{aligned}&\nabla \langle \nabla ^k (g_B({{\tilde{Q}}}) \phi ^2),{{\tilde{R}}}^T(Q)(R_0^T\nabla ^{l} QR_0)_D{{\tilde{R}}}(Q)\rangle \\&\quad =\langle \nabla ^{k+1} (g_B({{\tilde{Q}}}) \phi ^2) ,{{\tilde{R}}}^T(Q)(R_0^T\nabla ^{l} QR_0)_D{{\tilde{R}}}(Q)\rangle \\&\qquad +\langle \nabla ^{k} (g_B({{\tilde{Q}}}) \phi ^2) ,{{\tilde{R}}}^T(Q)(R_0^T\nabla ^{l+1} QR_0)_D{{\tilde{R}}}(Q)\rangle \\&\quad =\langle \nabla ^{k+1} (g_B({{\tilde{Q}}}) \phi ^2),R^T(Q)\nabla ^{l} QR(Q)\rangle \\&\quad \quad +\langle \nabla ^{k} (g_B({{\tilde{Q}}}) \phi ^2) ,R^T(Q)\nabla ^{l+1} QR(Q)\rangle \\&\quad =\nabla \langle \nabla ^k (g_B({{\tilde{Q}}}) \phi ^2),R^T(Q)\nabla ^l QR(Q)\rangle . \end{aligned} \end{aligned}$$
(4.19)

Here we used that

$$\begin{aligned} \langle \nabla ^{k+1} (g_B({{\tilde{Q}}}) \phi ^2) ,{{\tilde{R}}}^T(Q)(R_0^T\nabla ^{l} QR_0)_N{{\tilde{R}}}(Q)\rangle =0. \end{aligned}$$

Similarly, using (4.19), at \(Q=Q_0\), we find

$$\begin{aligned} \begin{aligned}&\langle \nabla ^k (g_B({{\tilde{Q}}}) \phi ^2),R^T(Q)\nabla ^{l+1} QR(Q)\rangle \\&\quad =\left\langle \nabla ^k (g_B({{\tilde{Q}}}) \phi ^2) ,\left( {{\tilde{R}}}^T(Q)\nabla (R_0^T\nabla ^{l} QR_0)_D{{\tilde{R}}}(Q)\right) _D\right\rangle \\&\quad =\left\langle \nabla ^k (g_B({{\tilde{Q}}}) \phi ^2) ,\nabla \left( \tilde{R}^T(Q)(R_0^T\nabla ^{l} QR_0)_D{{\tilde{R}}}(Q)\right) _D\right\rangle \\&\quad =\nabla \langle \nabla ^k (g_B({{\tilde{Q}}}) \phi ^2) ,\tilde{R}^T(Q)(R_0^T\nabla ^{l} QR_0)_D{{\tilde{R}}}(Q)\rangle \\&\qquad -\langle \nabla ^{k+1} (g_B({{\tilde{Q}}}) \phi ^2) ,\tilde{R}^T(Q)(R_0^T\nabla ^{l} QR_0)_D{{\tilde{R}}}(Q)\rangle \\&\quad =\nabla \langle \nabla ^k (g_B({{\tilde{Q}}}) \phi ^2) ,R^T(Q)\nabla ^l QR(Q)\rangle -\langle \nabla ^{k+1} (g_B({{\tilde{Q}}}) \phi ^2) ,R^T(Q)\nabla ^l QR(Q)\rangle . \end{aligned} \end{aligned}$$
(4.20)

Since \(x_0\in \Omega \backslash \Sigma _Q\) is arbitrary, this completes the proof. \(\square \)

Let \(R_L=R(Q_L)\) be a rotation such that \(R^T_LQ_LR\) is diagonal. Then we have

Lemma 4.4

Let \(x_0\in \Omega \) with some \(B_{r_0}(x_0)\subset \Omega \backslash \Sigma \) for a sufficiently small \(r_0\). Then, for any \(\phi \in C^\infty _0 ( B_{r_0}(x_0))\) and \(Q_L\in S_\delta \) with sufficiently small \(\delta \), we have

$$\begin{aligned} \int _{\Omega }\left( |\nabla ^2 Q_L|^2+ \frac{|(R^T_L\nabla Q_LR_L)_D|^2}{L} \right) \phi ^2\,dx\le C\int _{\Omega }|\nabla Q_L|^2 |\nabla \phi |^2\,dx, \end{aligned}$$
(4.21)

where C is a constant independent of L.

Proof

Let \(\varphi _\varepsilon \) be a cutoff function such that \(\varphi _\varepsilon (x)=0\) for \({{\,\textrm{dist}\,}}(x,\Sigma _{Q_L})\le \varepsilon \) and \(\varphi _\varepsilon (x)=1\) for \({{\,\textrm{dist}\,}}(x,\Sigma _{Q_L})\ge 2\varepsilon \). Multiplying (1.11) by \(\nabla (\phi ^2 \varphi _{\varepsilon }^2 \nabla Q_L)\) yields

$$\begin{aligned} \int _{\Omega } \langle H(Q_L, \nabla Q_L) ,\, \nabla (\phi ^2\varphi _{\varepsilon }^2 \nabla Q_L)\rangle \,dx =\int _{\Omega } \langle \frac{1}{L} g_B(Q_L) ,\, \nabla (\phi ^2\varphi _{\varepsilon }^2 \nabla Q_L)\rangle \,dx. \end{aligned}$$
(4.22)

Utilizing Lemma 4.1 with a sufficiently small \(\delta >0\), we derive

$$\begin{aligned}&\langle \nabla g_B({{\tilde{Q}}}_L), \nabla {{\tilde{Q}}}_L \rangle = \nabla _{{\alpha }}[\partial _{{{\tilde{Q}}}_{ii}}f_B({{\tilde{Q}}}_L)]\nabla _{\alpha } ({{\tilde{Q}}}_L)_{ii} \nonumber \\&\quad = \sum _{i,j}\partial ^2_{{{\tilde{Q}}}_{ii} \tilde{Q}_{jj}}f_B({{\tilde{Q}}}_L) \nabla _{\alpha } ({{\tilde{Q}}}_L)_{ii} \nabla _{\alpha }( {{\tilde{Q}}}_L)_{jj}\ge \frac{\lambda }{2}\sum _{i=1}^3|\nabla _{\alpha } ({{\tilde{Q}}}_L)_{ii}|^2=\frac{\lambda }{2}|\nabla _{\alpha } {{\tilde{Q}}}_L|^2. \end{aligned}$$
(4.23)

Using Lemma 4.2 and (4.23), we have

$$\begin{aligned}&\frac{1}{L}\int _{\Omega } \langle g_B(Q_L)\,, \nabla (\phi ^2\varphi _{\varepsilon }^2 \nabla Q_L)\rangle \,dx\nonumber \\&\quad =\frac{1}{L}\int _{\Omega } \langle g_B({{\tilde{Q}}}_L)\, , R^T_L\nabla (\phi ^2\varphi _{\varepsilon }^2 \nabla Q_L) R_L\rangle \,dx \nonumber \\&\quad = \frac{1}{L}\int _{\Omega } \langle g_B({{\tilde{Q}}}_L)\, , \nabla (\phi ^2\varphi _{\varepsilon }^2 R^T_L\nabla Q_L R_L) \rangle \,dx\nonumber \\&\quad = -\frac{1}{L}\int _{\Omega } \langle \nabla g_B({{\tilde{Q}}}_L)\, , \phi ^2\varphi _{\varepsilon }^2 R^T_L\nabla Q_L R_L\rangle \,dx\nonumber \\&\quad =-\frac{1}{L}\int _{\Omega } \langle \nabla g_B({{\tilde{Q}}}_L)\, , \phi ^2 \varphi _{\varepsilon }^2 \nabla {{\tilde{Q}}}_L) \rangle \,dx \le -\frac{\lambda }{2}\int _{\Omega }\frac{|\nabla {{\tilde{Q}}}_L|^2}{L}\phi ^2 \varphi _{\varepsilon }^2 \,dx \nonumber \\&\quad \le -\frac{\lambda }{2}\int _{\Omega }\frac{|(R^T_L\nabla Q_L R_L)_D|^2}{L}\phi ^2 \varphi _{\varepsilon }^2 \,dx. \end{aligned}$$
(4.24)

As \(\varepsilon \) tends to zero, we observe that

$$\begin{aligned}&\lim _{\varepsilon \rightarrow 0}\int _{\Omega } \langle H(Q_L, \nabla Q_L) ,\, \nabla (\phi ^2\varphi _{\varepsilon }^2 \nabla Q_L)\rangle \,dx \\&\quad = -\lim _{\varepsilon \rightarrow 0}\int _{\Omega } \langle \nabla ( H(Q_L, \nabla Q_L) ),\, \phi ^2\varphi _{\varepsilon }^2 \nabla Q_L \rangle \,dx\\&\quad =-\int _{\Omega } \langle \nabla _iH(Q_L, \nabla Q_L) ,\, \nabla _i Q_L\rangle \phi ^2\,dx. \end{aligned}$$

It follows from using integrating by parts, (2.8) and Young’s inequality that

$$\begin{aligned} \begin{aligned}&-\int _{\Omega } \langle \nabla _iH(Q_L, \nabla Q_L) ,\, \nabla _i Q_L\rangle \phi ^2\,dx \\&\quad =-\int _{\Omega }\nabla _i\left( \nabla _j\frac{\partial f_{E,1}(Q_L,\nabla Q_L)}{\partial {(Q_L)_{kl,j}}}-\frac{\partial f_{E,1}(Q_L,\nabla Q_L)}{\partial {(Q_L)_{kl}}}\right) \nabla _i (Q_L)_{kl}\phi ^2\,dx \\&\quad \ge \int _{\Omega }\frac{\partial ^2f_{E,1}(Q_L,\nabla Q_L)}{\partial ( Q_L)_{kl,j}\partial (Q_L)_{mn,r}}\nabla ^2_{ir}(Q_L)_{mn}\nabla ^2_{ij}(Q_L)_{kl}\phi ^2\,dx \\&\qquad -C\int _{\Omega }(|\nabla ^2 Q_L|+|\nabla Q_L|^2)|\nabla Q_L|^2\phi ^2\\&\qquad +(|\nabla ^2 Q_L||\nabla Q_L|+|\nabla Q_L|^3)|\nabla \phi ||\phi |\,dx \\&\quad \ge \frac{\alpha }{4} \int _{\Omega }|\nabla ^2 Q_L|^2\phi ^2\,dx-C\int _{\Omega }|\nabla Q_L|^4 \phi ^2+|\nabla Q_L|^2 |\nabla \phi |^2\,dx. \end{aligned} \end{aligned}$$
(4.25)

Combining (4.24) with (4.25) yields

$$\begin{aligned}&\int _{\Omega }\left( \frac{\alpha }{4}|\nabla ^2 Q_L|^2+\frac{\lambda }{2} \frac{|(R^T_L\nabla Q_L R_L)_D|^2}{L} \right) \phi ^2\,dx\le C \int _{\Omega }|\nabla Q_L|^4\phi ^2 + |\nabla Q_L|^2|\nabla \phi |^2 \,dx. \end{aligned}$$

Integrating by parts and using Young’s inequality, we deduce

$$\begin{aligned} \int _{\Omega }|\nabla Q_L|^4 \phi ^2\,dx&= \int _{\Omega }\langle \nabla Q_L, |\nabla Q_L|^2\nabla Q_L \rangle \phi ^2\,dx \\&=-\int _{\Omega }\langle Q_L-Q_{L; x_0, r}, \nabla (|\nabla Q_L|^2\nabla Q_L) \rangle \phi ^2\,dx \\&\quad +2\int _{\Omega }\langle Q_L-Q_{L; x_0, r}, |\nabla Q_L|^2\nabla Q_L\rangle \phi \nabla \phi \,dx\\&\quad \le \frac{1}{2} \int _{\Omega }|\nabla Q_L|^4 \phi ^2\,dx\\&\quad + C\int _{\Omega }|Q_L-Q_{L; x_0, r}|^2 |\nabla ^2 Q_L|^2\phi ^2+ |\nabla Q_L|^2 |\nabla \phi |^2\,dx. \end{aligned}$$

Here . Note that

$$\begin{aligned} |Q_L(x)-Q_{L; x_0, r}|\le |Q_L(x)-Q_*(x)|+|Q_{L; x_0, r}- {Q_*}_{x_0, R}|+|Q_*(x)-{Q_*}_{x_0, R}| \end{aligned}$$

and for \(x\in B_r(x_0)\subset \Omega \backslash \Sigma \), \(Q_L(x)\) uniformly converges to \(Q_*(x)\). For a sufficiently small \(r_0\) and L, we see that

$$\begin{aligned} \int _{\Omega }|\nabla Q_L|^4 \phi ^2\,dx\le&C\int _{\Omega }|Q_L-Q_{L; x_0, r}|^2 |\nabla ^2 Q_L|^2\phi ^2\,dx +C\int _{\Omega } |\nabla Q_L|^2 |\nabla \phi |^2\,dx \\ \le&\frac{\alpha }{4}\int _{\Omega }|\nabla ^2 Q_L|^2\phi ^2\,dx +C\int _{\Omega } |\nabla Q_L|^2 |\nabla \phi |^2\,dx. \end{aligned}$$

Then we conclude that

$$\begin{aligned}&\int _{\Omega } \left( |\nabla ^2 Q_L|^2+ \frac{|(R^T_L\nabla Q_LR_L)_D|^2}{L} \right) \phi ^2\, dx \le C\int _{\Omega }|\nabla Q_L|^2 |\nabla \phi |^2\,dx. \end{aligned}$$

\(\square \)

As an application of Lemma 4.4, we obtain a uniform Caccioppoli inequality for minimizer \(Q_L\) as follows.

Lemma 4.5

Let \(x_0\in \Omega \) with \(B_{r_0}(x_0)\subset \Omega \backslash \Sigma \) for a sufficiently small \(r_0>0\). Then for any \(r\le r_0\), we have

$$\begin{aligned} \int _{B_{r/2}(x_0)} |\nabla Q_L|^2\,dx\le \frac{C}{r^2}\int _{B_{r}(x_0)} | Q_L- Q_{L; x_0, r}|^2\,dx, \end{aligned}$$
(4.26)

where and C is a constant independent of L.

Proof

For two st such that \(\frac{r}{2}\le t< s\le r\), choose a cutoff function \(\phi \in C_0^{\infty } (B_s(x_0))\) such that \(0\le \phi \le 1\), \(\phi =1\) on \(B_t\) and \(|\nabla \phi |\le C/(s-t)\).

Integrating by parts and using Young’s inequality, we have

$$\begin{aligned} \int _{\Omega } |\nabla Q_L(x)|^2 \phi ^2\,dx=&- \int _{\Omega } \langle \Delta Q_L, Q_L(x)-Q_{L; x_0, r}\rangle \phi ^2\,dx\\&- \int _{\Omega } \langle \nabla Q, (Q(x)-Q_{x_0, r})\nabla \phi ^2 \rangle \,dx\\ \le&\frac{1}{2}\int _{\Omega } |\nabla Q_L(x)|^2 \phi ^2 \,dx- \int _{\Omega } \langle \Delta Q_L, Q_L(x)-Q_{L; x_0, r}\rangle \phi ^2\,dx\\&+C \int _{\Omega }|Q_L(x)-Q_{L; x_0, r}|^2 |\nabla \phi |^2\,dx. \end{aligned}$$

Then, by Young’s inequality and Lemma 4.4, we obtain

$$\begin{aligned} \int _{B_t} |\nabla Q_L|^2 \,dx\le&(s-t)^2\int _{\Omega } |\nabla ^2 Q_L|^2 \,dx +\frac{C}{(s-t)^2} \int _{B_r(x_0)}|Q_L(x)-Q_{L; x_0, r}|^2 \,dx \\ \le&C_1\int _{B_s\backslash B_t}|\nabla Q_L|^2 \,dx+ \frac{C}{(s-t)^2} \int _{B_r(x_0)}|Q_L(x)-Q_{L; {x_0, r}}|^2 \,dx. \end{aligned}$$

Through the standard technique of ’filling hole’, we have

$$\begin{aligned} \int _{B_t} |\nabla Q_L|^2 \phi ^2\,dx\le \theta \int _{B_s}|\nabla Q_L|^2 \,dx+ \frac{C}{(s-t)^2} \int _{B_r(x_0)}|Q_L(x)-Q_{L; x_0, r}|^2 \,dx \end{aligned}$$

for \(\theta = \frac{C_1}{1+C_1} <1\) and two st such that \(r/2\le t< s\le r\). In view of Lemma 3.1 in Chapter V of [18], the relation (4.21) follows. \(\square \)

Using Lemma 4.5, we have local uniform estimates on higher derivatives.

Lemma 4.6

Let \(x_0\in \Omega \) with \(B_{r_0}(x_0)\subset \Omega \backslash \Sigma \). Assume that there exists a constant \(\varepsilon _0>0\) such that

$$\begin{aligned} \int _{B_{r_0}(x_0)}|\nabla Q_L|^3\,dx\le \varepsilon _0^3. \end{aligned}$$
(4.27)

Then, for any integer \(k\ge 1\), there exist a constant \(r_k\ge r_0/2\) and a positive constant \(C_k\) independent of L such that

$$\begin{aligned}&\int _{B_{r_k}(x_0)} |\nabla ^{k+1}Q_L|^2+\frac{1}{L} |(R^T_L\nabla ^k Q_LR_L)_D|^2 \,dx\le C_k. \end{aligned}$$
(4.28)

Proof

For simplicity of notations, we denote \(Q=Q_L\) and \(R=R_L\). The claim (4.28) is true for \(k=1\). At first, we show the case of \(k=2\).

Assume that there exists a constant \(C_1>0\) such that

$$\begin{aligned} \int _{B_{r_1}(x_0)}\left( |\nabla Q|^4 +|\nabla ^2 Q|^2+ \frac{1}{L} |(R^T\nabla QR)_D|^2\right) \,dx\le C_1. \end{aligned}$$

Let \(\phi \) be a cutoff function in \(C_0^{\infty }(B_{r_1}(x_0))\), where \(r_2\) satisfies \(\frac{r_0}{2}<r_2<r_1<r_0\) and \(\phi =1\) in \(B_{r_2}(x_0)\). Let \(\varphi _\varepsilon \) be another cutoff function such that \(\varphi _\varepsilon (x)=0\) for \({{\,\textrm{dist}\,}}(x,\Sigma _Q)\le \varepsilon \) and \(\varphi _\varepsilon (x)=1\) for \({{\,\textrm{dist}\,}}(x,\Sigma _Q)\ge 2\varepsilon \). We differentiate (1.11) twice and multiply by \(\nabla ^2 Q\phi ^2\varphi _\varepsilon ^2\) to get

$$\begin{aligned}&\int _{B_{r_0}(x_0)}\left\langle \nabla ^2_{\beta \gamma }H(Q, \nabla Q) , \nabla ^2_{\beta \gamma } Q\phi ^2\varphi _\varepsilon ^2 \right\rangle \,dx =\int _{B_{r_0}(x_0)}\left\langle \frac{1}{L} \nabla ^2_{\beta \gamma } g_B( Q),\nabla ^2_{\beta \gamma } Q\phi ^2\varphi _\varepsilon ^2\right\rangle \,dx. \end{aligned}$$
(4.29)

Applying Lemma 4.3 and Lemma 4.1 to the right-hand side of (4.29), we find

$$\begin{aligned} \begin{aligned}&\frac{1}{L}\int _{B_{r_0}(x_0)}\langle \nabla ^2_{\beta \gamma } g_B(Q) , \nabla ^2_{\beta \gamma } Q \phi ^2\varphi _\varepsilon ^2 \rangle \,dx\\&\quad =\frac{1}{L}\int _{B_{r_0}(x_0)} \langle g_B(Q) , \nabla ^2_{\beta \gamma } (\nabla ^2_{\beta \gamma } Q \phi ^2\varphi _\varepsilon ^2) \rangle \,dx \\&\quad =\frac{1}{L}\int _{B_{r_0}(x_0)}\langle g_B({{\tilde{Q}}}) , R^T\nabla ^2_{\beta \gamma } (\nabla ^2_{\beta \gamma } Q \phi ^2\varphi _\varepsilon ^2) R\rangle \,dx \\&\quad =\frac{1}{L}\int _{B_{r_0}(x_0)}\langle \nabla ^2_{\beta \gamma }(g_B({{\tilde{Q}}}) ), R^T\nabla ^2_{\beta \gamma } Q R \phi ^2\varphi _\varepsilon ^2 \rangle \,dx\\&\quad =\frac{1}{L}\int _{B_{r_0}(x_0)}\langle \nabla ^2_{\beta \gamma }g_B({{\tilde{Q}}}) ,\nabla ^2_{\beta \gamma }{{\tilde{Q}}}\rangle \phi ^2\varphi _\varepsilon ^2\,dx \\&\quad = \frac{1}{L} \int _{B_{r_0}(x_0)}\sum _{i,j} \partial ^2_{\tilde{Q}_{ii}{{\tilde{Q}}}_{jj}}f_B({{\tilde{Q}}}) \nabla ^2_{\gamma \beta } \tilde{Q}_{ii}\nabla ^2_{\gamma \beta } {{\tilde{Q}}}_{jj} \phi ^2\varphi _\varepsilon ^2\,dx \\&\qquad + \frac{1}{L} \int _{B_{r_0}(x_0)}\sum _{i,j,m}\partial ^3_{\tilde{Q}_{ii} {{\tilde{Q}}}_{jj}{{\tilde{Q}}}_{mm}}f_B({\tilde{Q}}) \nabla _{\gamma } \tilde{Q}_{ii} \nabla _\beta {{\tilde{Q}}} _{jj} \nabla ^2_{\gamma \beta } \tilde{Q}_{mm} \phi ^2\varphi _\varepsilon ^2 \,dx \\&\quad \ge \frac{\lambda }{4L } \int _{B_{r_0}(x_0)} |\nabla ^2 {{\tilde{Q}}}|^2 \phi ^2\varphi _\varepsilon ^2\,dx-\frac{ C}{L} \int _{B_{r_0}(x_0)}|\nabla {\tilde{Q}}|^4 \phi ^2 \varphi _\varepsilon ^2 \,dx. \\&\quad \ge \frac{\lambda }{4L } \int _{B_{r_0}(x_0)} |(R^T\nabla ^2 QR)_D|^2 \phi ^2\varphi _\varepsilon ^2\,dx\\&\quad \quad - C \int _{B_{r_0}(x_0)}|\nabla Q|^2\frac{|(R^T\nabla Q R)_D|^2}{L} \phi ^2 \varphi _\varepsilon ^2 \,dx, \end{aligned} \end{aligned}$$
(4.30)

where we used that

$$\begin{aligned} \int _{B_{r_0}(x_0)} |\nabla ^2 {{\tilde{Q}}}|^2 \phi ^2\varphi _\varepsilon ^2\,dx&=-\int _{B_{r_0}(x_0)} \langle \nabla (R^T QR)_D,\nabla (\nabla ^2{{\tilde{Q}}} \phi ^2\varphi _\varepsilon ^2)\rangle \,dx \nonumber \\&=\int _{B_{r_0}(x_0)} \langle (R^T \nabla ^2QR)_D \phi ^2\varphi _\varepsilon ^2,\nabla ^2{{\tilde{Q}}}\rangle \,dx\nonumber \\&=\int _{B_{r_0}(x_0)}|(R^T\nabla ^2 QR)_D|^2 \phi ^2\varphi _\varepsilon ^2\,dx. \end{aligned}$$
(4.31)

Observer that, for any fixed \(\varepsilon >0\), R(x) in (4.30) is differentiable. Then it follows from (4.1) that \(Q^+=R^T\pi (Q) R\) and

$$\begin{aligned}&C \int _{B_{r_0}(x_0)}|\nabla Q|^2\frac{|R^T\nabla Q R|^2}{L} \phi ^2 \varphi _\varepsilon ^2 \,dx \nonumber \\&\quad =\frac{C}{L}\int _{B_{r_0}(x_0)}\langle (R^T\nabla Q R)_D,(R^T\nabla Q R)_D\rangle |\nabla Q|^2\phi ^2 \varphi _\varepsilon ^2 \,dx \nonumber \\&\quad =\frac{C}{L}\int _{B_{r_0}(x_0)}\langle \nabla ( R^T Q R-Q^+)_D,\nabla (R^T Q R-Q^+)_D\rangle |\nabla Q|^2\phi ^2 \varphi _\varepsilon ^2 \,dx \nonumber \\&\quad =C \int _{B_{r_0}(x_0)}|\nabla Q|^2\frac{|(R^T\nabla (Q-\pi (Q)) R)_D|^2}{L} \phi ^2 \varphi _\varepsilon ^2 \,dx \nonumber \\&\quad \le C \int _{B_{r_0}(x_0)}|\nabla Q|^2\frac{|\nabla (Q-\pi (Q))|^2}{L} \phi ^2 \varphi _\varepsilon ^2 \,dx. \end{aligned}$$
(4.32)

Letting \(\varepsilon \rightarrow 0\) in (4.32), using (1.11) and the fact that \(|Q-\pi (Q)|\le C| g_B(Q)|\) in (4.11), we find

$$\begin{aligned}&C \int _{B_{r_0}(x_0)}|\nabla Q|^2\frac{|\nabla (Q-\pi (Q))|^2}{L} \phi ^2\,dx \nonumber \\&\quad =-\frac{C}{L} \int _{B_{r_0}(x_0)}\langle (Q-\pi (Q),\nabla _\beta \big (\nabla _\beta (Q-\pi (Q))|\nabla Q|^2 \phi ^2\big )\rangle \,dx \nonumber \\&\quad \le C\int _{B_{r_0}(x_0)}\frac{|Q-\pi (Q)|}{L}|\nabla ^2 Q||\nabla Q|^2 \phi ^2\,dx +C\int _{B_{r_0}(x_0)}\frac{|Q-\pi (Q)|}{L}|\nabla Q|^3 |\nabla \phi ||\phi |\,dx \nonumber \\&\quad \le C\int _{B_{r_0}(x_0)}|H(Q,\nabla Q)||\nabla ^2 Q||\nabla Q|^2 \phi ^2\,dx +C\int _{B_{r_0}(x_0)}|H(Q,\nabla Q)||\nabla Q|^3 |\nabla \phi ||\phi |\,dx \nonumber \\&\quad \le C\int _{B_{r_0}(x_0)}( |\nabla Q|^6 +|\nabla Q|^2|\nabla ^2 Q|^2)\phi ^2\,dx+C\int _{B_{r_0}(x_0)}(|\nabla Q|^4 +|\nabla ^2 Q|^2)|\nabla \phi |^2\,dx. \end{aligned}$$
(4.33)

In view of (4.32)-(4.32), we deduce (4.30) to

$$\begin{aligned}&\int _{B_{r_0}(x_0)}\left\langle \frac{1}{L} \nabla ^2_{\beta \gamma } g_B( Q),\nabla ^2 Q\phi ^2\right\rangle \,dx \nonumber \\&\quad \ge \frac{\lambda }{4L } \int _{B_{r_0}(x_0)} |(R^T\nabla ^2 QR)_D|^2 \phi ^2\,dx- C\int _{B_{r_0}(x_0)}( |\nabla Q|^6 +|\nabla Q|^2|\nabla ^2 Q|^2)\phi ^2\,dx \nonumber \\&\qquad -C\int _{B_{r_0}(x_0)}(|\nabla Q|^4 +|\nabla ^2 Q|^2)|\nabla \phi |^2\,dx. \end{aligned}$$
(4.34)

Applying (2.8) and Young’s inequality to the left-hand side of (4.29), we obtain

$$\begin{aligned}{} & {} \int _{B_{r_0}(x_0)}\left\langle \nabla ^2_{\beta \gamma }H(Q, \nabla Q) , \nabla ^2_{\beta \gamma } Q\phi ^2 \right\rangle \,dx \nonumber \\{} & {} \quad =\int _{B_{r_0}(x_0)} \nabla ^2_{\beta \gamma }\left( \nabla _k\frac{\partial f_{E,1}(Q,\nabla Q)}{\partial {Q_{ij,k}}}-\frac{\partial f_{E,1}(Q,\nabla Q)}{\partial {Q_{ij}}}\right) \nabla ^2_{\beta \gamma } Q_{ij} \phi ^2\,dx \nonumber \\{} & {} \quad \le -\int _{B_{r_0}(x_0)} \frac{\partial ^2 f_E}{\partial {Q_{ij,\gamma }}\partial Q_{mn,l}} \nabla ^3_{\alpha \beta l}Q_{mn} \nabla ^3_{\alpha \beta \gamma } Q_{ij}\phi ^2 \,dx \nonumber \\{} & {} \qquad +C \int _{B_{r_0}(x_0)} \left| \nabla \left( \frac{\partial ^2 f_{E,1}(Q,\nabla Q)}{\partial p\partial Q} \nabla Q\right) \right| |\nabla ^3 Q|\phi ^2\,dx \nonumber \\{} & {} \qquad +C \int _{B_{r_0}(x_0)}\left| \nabla \frac{\partial ^2 f_{E,1}(Q,\nabla Q)}{\partial p\partial p} \right| |\nabla ^2 Q||\nabla ^3Q|\phi ^2\,dx \nonumber \\{} & {} \qquad +C \int _{B_{r_0}(x_0)} \left| \nabla ^2\frac{\partial f_{E,1}(Q,\nabla Q)}{\partial p}\right| |\nabla ^2 Q||\nabla \phi ||\phi |+\left| \nabla ^2\frac{\partial f_{E,1}(Q,\nabla Q)}{\partial Q}\right| |\nabla ^2 Q|\phi ^2\,dx \nonumber \\{} & {} \quad \le \int _{B_{r_0}(x_0)}\frac{\alpha }{4}|\nabla ^3 Q|^2\phi ^2\,dx +C\int _{B_{r_0}(x_0)}(|\nabla Q|^4 +|\nabla ^2 Q|^2)|\nabla \phi |^2\,dx \nonumber \\{} & {} \qquad +C\int _{B_{r_0}(x_0)}( |\nabla Q|^6 +|\nabla Q|^2|\nabla ^2 Q|^2)\phi ^2\,dx. \end{aligned}$$
(4.35)

Using Hölder’s inequality, we have

$$\begin{aligned} \int _{B_{r_0}(x_0)}(|\nabla ^2Q|^2|\nabla Q|^2+|\nabla Q|^6)\phi ^2\,dx\le C\varepsilon _0 \int _{B_{r_0}(x_0)}|\nabla ^3Q|^2 \phi ^2+|\nabla ^2 Q|^2|\nabla \phi |^2\,dx. \end{aligned}$$
(4.36)

Combining (4.34) with (4.36) and choosing \(\varepsilon _0\) sufficiently small, we obtain

$$\begin{aligned} \int _{B_{r_2}(x_0)}|\nabla ^3Q|^2 + \frac{1}{L} |(R^T\nabla ^2 QR)_D|^2\,dx\le C_2. \end{aligned}$$

Set \(r_k:=(1-\sum _{i=1}^k2^{-(i+1)})r_0>\frac{r_0}{2}\). For any \(k\le l\), we can assume that there is a constant \(C_k\) such that

$$\begin{aligned} \int _{B_{r_k}(x_0)} (|\nabla ^{k+1}Q|^2+\frac{1}{L} |(R^T\nabla ^k QR)_D|^2)\,dx\le C_k. \end{aligned}$$
(4.37)

As a consequence of the Sobolev inequality, we have

$$\begin{aligned} \Vert \nabla ^{k-1}Q\Vert _{L^{\infty }(B_{r_k}(x_0))} \le C_k \end{aligned}$$
(4.38)

for any \(k\le l-1\).

Next, we prove it for \(k=l+1\). Let \(r_{l+1}\) be the constant satisfying

$$\begin{aligned} \frac{r_0}{2}<r_{l+1}<r_{l}<\cdots<r_1 <r_0. \end{aligned}$$

Let \(\phi \) be a cutoff function in \(C_0^{\infty }(B_{r_{l}}(x_0))\) with \(\phi =1\) in \(B_{r_{l+1}}(x_0)\). We apply \(\nabla ^{l+1}\) to (1.11) and multiply by \( \nabla ^{l+1} Q\phi ^2\varphi _\varepsilon ^2 \) to have

$$\begin{aligned}&\int _{B_{r_0}(x_0)}\langle \nabla ^{l+1}H(Q, \nabla Q),\nabla ^{l+1} Q\phi ^2\varphi _\varepsilon ^2\rangle \,dx \nonumber \\&\quad =\int _{B_{r_0}(x_0)} \langle \frac{1}{L} \nabla ^{l+1}g_B( Q),\nabla ^{l+1} Q\phi ^2\varphi _\varepsilon ^2\rangle \,dx. \end{aligned}$$
(4.39)

It follows from Lemma 4.3 that

$$\begin{aligned} \begin{aligned}&\frac{1}{L}\int _{B_{r_0}(x_0)}\langle \nabla ^{l+1}g_B( Q),\nabla ^{l+1} Q\phi ^2\varphi _\varepsilon ^2\rangle \,dx \\&\quad = (-1)^{l+1}\frac{1}{L}\int _{B_{r_0}(x_0)}\langle g_B( \tilde{Q}),R^T\nabla ^{l+1}(\nabla ^{l+1} Q\phi ^2\varphi _\varepsilon ^2)R\rangle \,dx \\&\quad = \frac{1}{L}\int _{B_{r_0}(x_0)}\langle \nabla ^{l+1}g_B( \tilde{Q})\phi ^2\varphi _\varepsilon ^2,R^T\nabla ^{l+1} QR\rangle \,dx \\&\quad =(-1)^{l}\frac{1}{L}\int _{B_{r_0}(x_0)}\langle \nabla (\nabla ^{l+1}g_B( {{\tilde{Q}}})\phi ^2\varphi _\varepsilon ^2),\nabla {\tilde{Q}}\rangle \,dx \\&\quad =\frac{1}{L}\int _{B_{r_0}(x_0)}\langle \nabla ^{l+1}g_B( \tilde{Q}),\nabla ^{l+1}{\tilde{Q}}\rangle \phi ^2\varphi _\varepsilon ^2\,dx \\&\quad =\frac{1}{L}\int _{B_{r_0}(x_0)}\sum _{i,j}\nabla ^{l}\big (\partial ^2_{\tilde{Q}_{ii}{{\tilde{Q}}}_{jj}} f_B({{\tilde{Q}}})\nabla \tilde{Q}_{jj}\big )\nabla ^{l+1}{\tilde{Q}}_{ii} \phi ^2\varphi _\varepsilon ^2\,dx. \end{aligned} \end{aligned}$$
(4.40)

Using a similar argument in (4.31), for \(i\ge 3\), one can check that

$$\begin{aligned} \int _{B_{r_0}(x_0)} |\nabla ^{i} {{\tilde{Q}}}|^2Z(x) \varphi _\varepsilon ^2\,dx&=(-1)^{i-1}\int _{B_{r_0}(x_0)} \langle \nabla {{\tilde{Q}}},\nabla ^{i-1}(\nabla ^{i} {{\tilde{Q}}}Z(x) \varphi _\varepsilon ^2)\rangle \,dx \nonumber \\&=\int _{B_{r_0}(x_0)} \langle \big (R^T\nabla ^i QR)_DZ(x) \varphi _\varepsilon ^2\big ),\nabla ^{i} {{\tilde{Q}}}\rangle \,dx\nonumber \\&=\int _{B_{r_0}(x_0)} |(R^T\nabla ^{i} QR)_D|^2Z(x) \varphi _\varepsilon ^2\,dx \end{aligned}$$
(4.41)

for some scalar function Z(x). Observe that \(\partial ^j_{\tilde{Q}}f_B({{\tilde{Q}}})=0\) for \(j\ge 5\). Applying Lemma 4.1 with a sufficiently small \(\delta \) and (4.41) to (4.40), we obtain

$$\begin{aligned}{} & {} \frac{1}{L}\int _{B_{r_0}(x_0)}\langle \nabla ^{l+1}g_B( Q),\nabla ^{l+1} Q\phi ^2\varphi _\varepsilon ^2\rangle \,dx \nonumber \\{} & {} \quad \ge \frac{1}{2 L} \int _{B_{r_0}(x_0)} |\nabla ^{l+1} {{\tilde{Q}}}|^2\phi ^2 \varphi _\varepsilon ^2\,dx \nonumber \\{} & {} \qquad -\frac{C}{ L} \int _{B_{r_0}(x_0)}\sum _{\begin{array}{c} \mu _1\le \mu _2\le l\\ \mu _1+\mu _2=l+1 \end{array}}|\partial ^3_{{{\tilde{Q}}}} f_B({{\tilde{Q}}})|^2|\nabla ^{\mu _1} {{\tilde{Q}}}|^2|\nabla ^{\mu _2} {{\tilde{Q}}}|^2\phi ^2 \varphi _\varepsilon ^2\,dx \nonumber \\{} & {} \qquad -\frac{C}{ L} \int _{B_{r_0}(x_0)}\sum _{\begin{array}{c} \mu _1\le \mu _2\le \mu _3\le l-1\\ \mu _1+\mu _2+\mu _3=l+1 \end{array}}|\partial ^4_{{{\tilde{Q}}}} f_B({{\tilde{Q}}})|^2|\nabla ^{\mu _1} {{\tilde{Q}}}|^2|\nabla ^{\mu _2} {{\tilde{Q}}}|^2|\nabla ^{\mu _3} {{\tilde{Q}}}|^2\phi ^2 \varphi _\varepsilon ^2\,dx \nonumber \\{} & {} \quad \ge \frac{1}{2 L} \int _{B_{r_0}(x_0)} |(R^T\nabla ^{l+1} QR)_D|^2\phi ^2 \varphi _\varepsilon ^2\,dx \nonumber \\{} & {} \qquad -\frac{C}{ L} \int _{B_{r_0}(x_0)}\sum _{\begin{array}{c} \mu _1\le \mu _2\le l\\ \mu _1+\mu _2=l+1 \end{array}}|\partial ^3_{{{\tilde{Q}}}} f_B({{\tilde{Q}}})|^2|(R^T\nabla ^{\mu _1} QR)_D|^2|(R^T\nabla ^{\mu _2} QR)_D|^2\phi ^2 \varphi _\varepsilon ^2\,dx \nonumber \\{} & {} \qquad -\frac{C}{ L} \int _{B_{r_0}(x_0)}\sum _{\begin{array}{c} \mu _1\le \mu _2\le \mu _3\le l-1\\ \mu _1+\mu _2+\mu _3=l+1 \end{array}}|\partial ^4_{{{\tilde{Q}}}} f_B({{\tilde{Q}}})|^2|(R^T(\nabla ^{\mu _1} Q)R)_D|^2 \nonumber \\{} & {} \qquad \times |(R^T\nabla ^{\mu _2} QR)_D|^2|(R^T\nabla ^{\mu _3} QR)_D|^2\phi ^2 \varphi _\varepsilon ^2\,dx. \end{aligned}$$
(4.42)

As \(\varepsilon \) tends to zero, we obtain from (4.38) and (4.42) that

$$\begin{aligned} \begin{aligned}&\lim _{\varepsilon \rightarrow 0}\frac{1}{L}\int _{B_{r_0}(x_0)}\langle \nabla ^{l+1}g_B( Q),\nabla ^{l+1} Q\phi ^2\varphi _\varepsilon ^2\rangle \,dx \\&\quad \ge \frac{1}{4 L} \int _{B_{r_0}(x_0)} |(R^T\nabla ^{l+1} Q R)_D|^2\phi ^2 \,dx -\frac{C}{L} \int _{B_{r_0}(x_0)} |(R^T\nabla Q R)_D|^2|(R^T\nabla ^{l} Q R)_D|^2\phi ^2 \,dx \\&\qquad -\frac{C}{L} \int _{B_{r_0}(x_0)} |(R^T\nabla ^2 Q R)_D|^2|(R^T\nabla ^{l-1} Q R)_D|^2\phi ^2 \,dx-C_k \\&\quad \ge \frac{1}{4 L} \int _{B_{r_0}(x_0)} |(R^T\nabla ^{l+1} Q R)_D|^2\phi ^2 \,dx-C_k. \end{aligned} \end{aligned}$$
(4.43)

Using Young’s inequality and integration by parts, we have

$$\begin{aligned} \begin{aligned}&\int _{B_{r_0}(x_0)}\langle \nabla ^{l+1}H(Q, \nabla Q),\nabla ^{l+1} Q\phi ^2\rangle \,dx \\&\quad =-\int _{{\mathbb {R}}^3} \frac{\partial ^2 f_E}{\partial {Q_{ij,\gamma }}\partial Q_{mn,\beta }} \nabla ^{l+1}\nabla _{ \beta }Q_{mn} \nabla ^{l+1}\nabla _{ \gamma } Q_{ij}\phi ^2 \,dx \\&\qquad +C \int _{B_{r_0}(x_0)} \left| \nabla ^{l+1}\frac{\partial f_{E,1}(Q,\nabla Q)}{\partial p}\right| |\nabla ^{l+1} Q||\nabla \phi ||\phi |\,dx \\&\qquad +C \int _{B_{r_0}(x_0)} \left| \nabla ^{l+1}\frac{\partial f_{E,1}(Q,\nabla Q)}{\partial Q}\right| |\nabla ^{l+1} Q|\phi ^2\,dx\\&\qquad +C \int _{B_{r_0}(x_0)} \left| \nabla ^l\left( \frac{\partial ^2 f_{E,1}(Q,\nabla Q)}{\partial p\partial Q} \nabla Q\right) \right| |\nabla ^{l+2} Q|\phi ^2\,dx \\&\qquad +C \int _{B_{r_0}(x_0)}\left| \nabla ^{l-1}\left( \nabla \frac{\partial ^2 f_{E,1}(Q,\nabla Q)}{\partial p\partial p} \nabla ^2Q\right) \right| |\nabla ^{l+2} Q|\phi ^2\,dx \\&\quad \le -\int _{B_{r_0}(x_0)}\frac{\alpha }{4}|\nabla ^{l+2} Q|^2\phi ^2\,dx \\&\qquad + C\int _{B_{r_0}(x_0)} \sum _{\begin{array}{c} \mu _1+\cdots + \mu _i=l+2\\ \mu _1\le \cdots \le \mu _{i-1} <l+2 \end{array} }|\nabla ^{\mu _1}Q|^2|\nabla ^{\mu _2} Q|^2\cdots |\nabla ^{\mu _{i-1}}Q|^2 |\nabla ^{\mu _i}\phi |^2 \,dx. \end{aligned} \end{aligned}$$
(4.44)

In view of (4.37)-(4.38), we have

$$\begin{aligned}&C\int _{B_{r_0}(x_0)} \sum _{\begin{array}{c} \mu _1+\cdots + \mu _i=l+2\\ \mu _1\le \cdots \le \mu _{i-1} <l+2 \end{array} }|\nabla ^{\mu _1}Q|^2|\nabla ^{\mu _2} Q|^2\cdots |\nabla ^{\mu _{i-1}}Q|^2 |\nabla ^{\mu _i}\phi |^2 \le C_l. \end{aligned}$$
(4.45)

Combining (4.44) with (4.45), our claim (4.28) follows for \(k=l+1\). \(\square \)

Now we give a proof of Theorem 3.

Proof

For any \(x_0\in \Omega \backslash \Sigma \), let \(B_{2R_0}(x_0)\) be a ball such that \(B_{2r_0}(x_0)\subset \Omega \backslash \Sigma \). From Lemma 4.5, we deduce the following estimate

(4.46)

It follows from (4.21) that

$$\begin{aligned} r_0\int _{B_{r_0}(x_0)} |\nabla ^2 Q_L|^2\,dx\le \frac{C}{r_0}\int _{B_{\frac{4}{3} r_0}(x_0)} |\nabla Q_L|^2 \,dx. \end{aligned}$$

As a consequence of the Gagliardo-Nirenberg interpolation (c.f. [15]), we have

$$\begin{aligned} \int _{B_{r_0}(x_0)} |\nabla Q_L|^3\,dx\le&C\left( \int _{B_{r_0}(x_0)}r_0|\nabla ^2 Q_L|^2+\frac{1}{r_0}|\nabla Q_L|^2 \,dx\right) ^{3/2}\\&+C\left( r_0^{-1}\int _{B_{r_0}(x_0)}|\nabla Q_L|^2\,dx\right) ^{3/2}\le \varepsilon _0^3. \end{aligned}$$

Using Lemma 4.5 with any \(k\ge 1\), we obtain

$$\begin{aligned} \Vert Q_L\Vert _{W^{k,2}_\textrm{loc}(\Omega \backslash \Sigma )}<C_k. \end{aligned}$$
(4.47)

Then, \(Q_L\) converges smoothly to \(Q_*\) in \(\Omega \backslash \Sigma \). \(\square \)

5 The Landau-de Gennes density through the Oseen-Frank density

In this section, we will obtain a new form of the Landau-de Gennes energy density through the Oseen-Frank density. Under the condition (1.9), it was shown in [33] that for each \(Q =s_+ (u\otimes u-\frac{1}{3} I)\in S_*\), one has

$$\begin{aligned} W(u,\nabla u) =f_E(Q,\nabla Q). \end{aligned}$$

Assuming the strong Ericksen condition

$$\begin{aligned} k_2>|k_4|,\quad k_3>0,\quad 2k_1>k_2+k_4, \end{aligned}$$
(5.1)

it was pointed out in [28] (see also [1, 14, 15]) that there are positive constants \(\lambda \) and C such that the density \(W(u,\nabla u)\) is equivalent to a new form that \({{\widetilde{W}}}(u,p)\) satisfies

$$\begin{aligned} \lambda |p|^2\le {{\widetilde{W}}}(u,p)\le C(1+|u|^2) |p|^2 \end{aligned}$$

for any \(u\in {\mathbb {R}}^3\) and any \(p\in {\mathbb {M}}^{3\times 3}\). However, there seems no reference for an explicit form of \(\widetilde{W}(u,\nabla u)\), so we give an explicit form \({{\widetilde{W}}}(u,\nabla u)\) here. For \(u\in {\mathbb {R}}^3\), it can be checked that

$$\begin{aligned}&2|\nabla u|^2-|{{\,\textrm{curl}\,}}u|^2-({{\,\textrm{div}\,}}u)^2 \nonumber \\&\quad =(\nabla _1 u_1+\nabla _2 u_2)^2+(\nabla _2 u_2+\nabla _3 u_3)^2+(\nabla _1 u_1+\nabla _3 u_3)^2 \nonumber \\&\qquad +(\nabla _2u_3+\nabla _3u_2)^2+(\nabla _1u_3+\nabla _3u_1)^2+(\nabla _1u_2+\nabla _2u_1)^2. \end{aligned}$$
(5.2)

Lemma 5.1

Assume the Frank constants \(k_1, \cdots ,k_4\) satisfy

$$\begin{aligned} k_2>|k_4|,\quad \min \{k_1, k_3\}>\frac{1}{2} (k_2+k_4). \end{aligned}$$
(5.3)

Then the density \(W(u, \nabla u)\) of the form (1.5) for each \(u\in S^2\) is equivalent to the new form

$$\begin{aligned} \begin{aligned} {{\widetilde{W}}}(u,\nabla u)=&\frac{ {{\tilde{\alpha }}}}{2} |\nabla u|^2+\frac{2k_1- k_2-k_4- \alpha }{4} ({{\,\textrm{div}\,}}u)^2 \\&+\frac{k_2-k_4- \alpha }{4} (u\cdot {{\,\textrm{curl}\,}}u)^2 +\frac{2k_3-k_2-k_4- \alpha }{4}|u\times {{\,\textrm{curl}\,}}u|^2 \\&+\frac{k_2+k_4- \alpha }{4} \sum _{i\ne j}\left( (\nabla _i u_i+\nabla _j u_j)^2+(\nabla _iu_j+\nabla _ju_i)^2\right) \end{aligned} \end{aligned}$$
(5.4)

where \( \alpha =\min \{k_2-|k_4|,2k_1- k_2-k_4,2k_3- k_2-k_4\}>0\).

Proof

Note that \(W(u, \nabla u)\) is rotational invariant (c.f. [24]); i.e., for each \(R\in SO(3)\), \({{\tilde{x}}}=R(x-x_0)\) and \({{\tilde{u}}} = R u(x)=R u\). Then we have

$$\begin{aligned} W({{\tilde{u}}}, {{\tilde{\nabla }}} {\tilde{u}})= W(Ru, R\nabla uR^T)=W(u,\nabla u) . \end{aligned}$$

Then for any \(u\in S^2\), we can find some \(R=R(u(x_0))\in SO(3)\) at each point \(x_0\in \Omega \) such that

$$\begin{aligned} {{\tilde{u}}}(0):=R u(x_0) = (0,0,1)^T. \end{aligned}$$

Using the relation

$$\begin{aligned} \frac{\partial {{\tilde{u}}}_3}{\partial {{\tilde{x}}}_i} = -({{\tilde{u}}}_1 \frac{\partial {{\tilde{u}}}_1}{\partial {{\tilde{x}}}_i}+ {{\tilde{u}}}_2 \frac{\partial \tilde{u}_2}{\partial {{\tilde{x}}}_i})=0 \end{aligned}$$

for all \(i=1,2,3\), we evaluate four terms of the Oseen-Frank energy density at \({{\tilde{x}}}_0\)

$$\begin{aligned} \begin{aligned} ({{\tilde{\nabla }}} \cdot {{\tilde{u}}})^2=&({{\tilde{\nabla }}}_1 {{\tilde{u}}}_1+{{\tilde{\nabla }}}_2 {{\tilde{u}}}_2)^2,&({{\tilde{u}}}\cdot {{\,\textrm{curl}\,}}{{\tilde{u}}})^2=&({\tilde{\nabla }}_1{{\tilde{u}}}_2-{{\tilde{\nabla }}}_2{{\tilde{u}}}_1)^2 , \\ |{{\tilde{u}}}\times {{\,\textrm{curl}\,}}{{\tilde{u}}}|^2 =&|{{\tilde{\nabla }}}_3{{\tilde{u}}}_1|^2+|{{\tilde{\nabla }}}_3{{\tilde{u}}}_2|^2,&({{\,\textrm{tr}\,}}({{\tilde{\nabla }}} {{\tilde{u}}})^2-({{\tilde{\nabla }}} \cdot {{\tilde{u}}})^2)=&2{\tilde{\nabla }}_1{{\tilde{u}}}_2{{\tilde{\nabla }}}_2{{\tilde{u}}}_1-2{{\tilde{\nabla }}}_1 {{\tilde{u}}}_1{\tilde{\nabla }}_2 {{\tilde{u}}}_2. \end{aligned} \end{aligned}$$
(5.5)

Substituting above identities into the density, we have

$$\begin{aligned} \begin{aligned} W({{\tilde{u}}},{{\tilde{\nabla }}} {{\tilde{u}}}) =&\frac{k_1}{2}({{\tilde{\nabla }}}_1 \tilde{u}_1+{{\tilde{\nabla }}}_2 {{\tilde{u}}}_2)^2 +\frac{k_2}{2} (|{{\tilde{\nabla }}}_1\tilde{u}_2|^2+|{{\tilde{\nabla }}}_2{{\tilde{u}}}_1|^2) \\&+\frac{k_3}{2}(|{{\tilde{\nabla }}}_3{{\tilde{u}}}_1|^2+|{{\tilde{\nabla }}}_3\tilde{u}_2|^2)+ {k_4} {{\tilde{\nabla }}}_1{{\tilde{u}}}_2{{\tilde{\nabla }}}_2\tilde{u}_1-(k_2+k_4)({{\tilde{\nabla }}}_1 {{\tilde{u}}}_1{{\tilde{\nabla }}}_2 {{\tilde{u}}}_2) \\ =&\frac{ {{\tilde{\alpha }}}}{2} |{{\tilde{\nabla }}} {{\tilde{u}}}|^2+\frac{2k_1-k_2-k_4- {{\tilde{\alpha }}}}{4}({{\tilde{\nabla }}}_1 {{\tilde{u}}}_1+{{\tilde{\nabla }}}_2 {{\tilde{u}}}_2)^2 \\&+\frac{k_2+k_4- {{\tilde{\alpha }}}}{4}({{\tilde{\nabla }}}_1 {{\tilde{u}}}_1-{\tilde{\nabla }}_2 {{\tilde{u}}}_2)^2+\frac{k_2-k_4- {{\tilde{\alpha }}}}{2}(|{\tilde{\nabla }}_1{{\tilde{u}}}_2|^2+|{{\tilde{\nabla }}}_2{{\tilde{u}}}_1|^2) \\&+\frac{(k_3- {{\tilde{\alpha }}})}{2}(|{{\tilde{\nabla }}}_3{{\tilde{u}}}_1|^2+|{\tilde{\nabla }}_3{{\tilde{u}}}_2|^2)+\frac{k_4}{2}({{\tilde{\nabla }}}_1{{\tilde{u}}}_2+{\tilde{\nabla }}_2{{\tilde{u}}}_1)^2 \\ =&\frac{ {{\tilde{\alpha }}}}{2}| \nabla u|^2+\frac{2k_1- k_2-k_4- {{\tilde{\alpha }}}}{4} ({{\,\textrm{div}\,}}u)^2+\frac{k_2-k_4- {{\tilde{\alpha }}}}{4} (u\cdot {{\,\textrm{curl}\,}}u)^2 \\&+\frac{k_3- {{\tilde{\alpha }}}}{2}|u\times {{\,\textrm{curl}\,}}u|^2+\frac{k_2+k_4- {{\tilde{\alpha }}}}{4}\left( |{{\tilde{\nabla }}}_1 {{\tilde{u}}}_1-{{\tilde{\nabla }}}_2 \tilde{u}_2|^2+|{{\tilde{\nabla }}}_1{{\tilde{u}}}_2+{{\tilde{\nabla }}}_2{{\tilde{u}}}_1|^2\right) , \end{aligned} \end{aligned}$$
(5.6)

where \( {{\tilde{\alpha }}}\) is a positive constant due to the strong Ericksen condition (5.1). Using (5.5), we find

$$\begin{aligned}&|{{\tilde{\nabla }}}_1 {{\tilde{u}}}_1-{{\tilde{\nabla }}}_2 {{\tilde{u}}}_2|^2+|{\tilde{\nabla }}_1{{\tilde{u}}}_2+{{\tilde{\nabla }}}_2{{\tilde{u}}}_1|^2+|{{\tilde{\nabla }}}_3\tilde{u}_1|^2+|{{\tilde{\nabla }}}_3{{\tilde{u}}}_2|^2 \\&\quad = |{{\tilde{\nabla }}} {{\tilde{u}}}|^2+{{\,\textrm{tr}\,}}({{\tilde{\nabla }}} \tilde{u})^2-({{\tilde{\nabla }}}\cdot {{\tilde{u}}})^2=2|\nabla u|^2-|{{\,\textrm{curl}\,}}u|^2-({{\,\textrm{div}\,}}u)^2. \end{aligned}$$

If we further assume that \(2k_3>k_2+k_4\), then we can rewrite (5.6) into

$$\begin{aligned} {{\widetilde{W}}}( u, \nabla u)=&\frac{{{\tilde{\alpha }}}}{2}| \nabla u|^2+\frac{2k_1- k_2-k_4- \alpha }{4} ({{\,\textrm{div}\,}}u)^2 \nonumber \\&+\frac{k_2-k_4- \alpha }{4} (u\cdot {{\,\textrm{curl}\,}}u)^2+\frac{2k_3-k_2-k_4- \alpha }{4}|u\times {{\,\textrm{curl}\,}}u|^2 \nonumber \\&+\frac{k_2+k_4- \alpha }{4}\left( 2|\nabla u|^2-|{{\,\textrm{curl}\,}}u|^2-({{\,\textrm{div}\,}}u)^2\right) . \end{aligned}$$
(5.7)

From (5.2), we prove (5.4). \(\square \)

It is clear that the new form \({{\widetilde{W}}}(u,p)\) in (5.4) with \(p=\nabla u\) satisfies

$$\begin{aligned} \frac{{{\tilde{\alpha }}}}{2} |p|^2\le {{\widetilde{W}}}(u,p)\le C(1+|u|^2)|p|^2 \end{aligned}$$

for all \(u\in {\mathbb {R}}^3\) and \(p\in {\mathbb {M}}^{3\times 3}\).

Through the relation (5.4), we can have the new Landau-de Gennes energy density satisfying the coercivity in the following:

Proposition 5.1

Assume that \({{\hat{L}}}_1\), \({{\hat{L}}}_2\), \({{\hat{L}}}_3\) and \({{\hat{L}}}_4\) satisfy the condition

$$\begin{aligned} {{\hat{L}}}_1:=&s_+^{-2}\frac{2k_1- k_2-k_4}{2},&{{\hat{L}}}_2:=&s_+^{-2}\frac{2k_3-k_2-k_4}{2} \nonumber \\ {{\hat{L}}}_3:=&s_+^{-2}\frac{k_2-k_4}{2},&{{\hat{L}}}_4:=&s_+^{-2}\frac{k_2+k_4}{2}. \end{aligned}$$
(5.8)

Then for each \(Q\in S_*\), we obtain

$$\begin{aligned} f_{E,2}(Q,\nabla Q):=&\frac{ \alpha }{2}|\nabla Q|^2 + \frac{{{\hat{L}}}_1-{{\tilde{\alpha }}}}{2} \sum _{i=1}^3 \Big ((s_+^{-1}Q +\frac{1}{3}I)_{ij}(\nabla \cdot Q_j)\Big )^2 \\&+\frac{{{\hat{L}}}_2-{{\tilde{\alpha }}}}{2} \left| (s_+^{-1}Q+\frac{1}{3}I)_i\times {{\,\textrm{curl}\,}}Q_{i}\right| ^2 \\&+ \frac{{{\hat{L}}}_3-{{\tilde{\alpha }}}}{2} \sum _{i=1}^3\left( (s_+^{-1}Q +\frac{1}{3}I)_{ij}({{\,\textrm{curl}\,}}Q_j)_i\right) ^2 \\&+\frac{{{\hat{L}}}_4-{{\tilde{\alpha }}}}{2}\sum _{i\ne j}\sum _{k=1}^3 \Big ( (\nabla _iQ_{ik} +\nabla _jQ_{jk})^2+(\nabla _iQ_{jk}+\nabla _jQ_{ik})^2 \Big ), \end{aligned}$$

where \(Q_i\) is the i-th column of the Q matrix and \({\tilde{\alpha }}\) is given by

$$\begin{aligned}&\alpha =\min \{{{\hat{L}}}_1,{{\hat{L}}}_2,{{\hat{L}}}_3,{{\hat{L}}}_4 \} >0 . \end{aligned}$$
(5.9)

Proof

Due to the fact that \(|u|^2=1\), a direct calculation yields

$$\begin{aligned} \nabla _ku_i =u_j\nabla _k (u_i u_j)=s_+^{-1}(u_1\nabla _kQ_{i1}+u_2\nabla _kQ_{i2}+u_3\nabla _kQ_{i3})=s_+^{-1}u_j\nabla _kQ_{ij}. \end{aligned}$$

One can verify that

$$\begin{aligned} u_l\nabla _ku_i =s_+^{-1}(s_+^{-1}Q +\frac{1}{3}I)_{lj}\nabla _kQ_{ij}. \end{aligned}$$
(5.10)

Here we used the fact that \(|Q|=\sqrt{\frac{2}{3}}s_+\). Then we can derive \(f_E(Q,\nabla Q)\) from (5.4) that

$$\begin{aligned}&s_+^2({{\,\textrm{div}\,}}u)^2=s_+^2\sum _i (u_i{{\,\textrm{div}\,}}u)^2=\sum _i \Big ((s_+^{-1}Q +\frac{1}{3}I)_{ij}(\nabla \cdot Q_j)\Big )^2, \end{aligned}$$
(5.11)
$$\begin{aligned}&s_+^2|u\times {{\,\textrm{curl}\,}}u|^2=s_+^2|u\times (s_+^{-1}u_j({{\,\textrm{curl}\,}}Q_{j}))|^2=\left| (s_+^{-1}Q+\frac{1}{3}I)_i\times {{\,\textrm{curl}\,}}Q_{i}\right| ^2, \end{aligned}$$
(5.12)
$$\begin{aligned}&s_+^2(u\cdot {{\,\textrm{curl}\,}}u)^2=s_+^2(s_+^{-1}u_iu_j({{\,\textrm{curl}\,}}Q_j)_i)^2=\left( (s_+^{-1}Q +\frac{1}{3}I)_{ij}({{\,\textrm{curl}\,}}Q_j)_i\right) ^2. \end{aligned}$$
(5.13)

It then follows from (5.2) and (5.10) that

$$\begin{aligned}&s_+^2\left( 2|\nabla u|^2-|{{\,\textrm{curl}\,}}u|^2-({{\,\textrm{div}\,}}u)^2\right) \nonumber \\&\quad =\frac{1}{2} \sum _{l,k=1}^3\sum _{i\ne j}((s_+^{-1}Q +\frac{1}{3}I)_{lk})^2\left( (\nabla _iQ_{ik} +\nabla _jQ_{jk})^2+(\nabla _iQ_{jk}+\nabla _jQ_{ik})^2 \right) . \end{aligned}$$
(5.14)

Substituting the identities (5.11)-(5.14) into the equation (5.4), we complete a proof. \(\square \)