Local minimizer and De Giorgi’s type conjecture for the isotropic–nematic interface problem

Abstract

In this paper, we investigate the structure of local minimizers for the isotropic–nematic interface based on the Landau-de Gennes energy. In the absence of the anisotropic energy, the uniaxial solution is the only local minimizer in 1-D. In 3-D, we propose a De Giorgi’s type conjecture and give an affirmative answer under a mild assumption. In the presence of the anisotropic energy with $L_2>-\,1$ and homeotropic anchoring, the uniaxial solution is also the only local minimizer in a class of diagonal form in 1-D.

1 Introduction

Liquid crystal is a state of matter between liquid and solid, in which molecules tend to align a preferred direction. One of the most common phases is the nematic phase, in which the molecules tend to have the same alignment but their positions are not correlated. In physics, the different order parameters are used to describe the anisotropic behavior of liquid crystals.

The most simple one is the vector theory, which uses a unit vector field $\mathbf {n}(x)$ to describe the locally preferred alignment of liquid crystal molecules near the material point x. Onsager introduced the molecular theory, in which the orientational distribution function $f(x,\mathbf {m})$ is introduced to describe the number density of molecules whose orientation is parallel to $\mathbf {m}$ at material point x. The Q-tensor theory uses a symmetric traceless $3\times 3$ matrix $\mathbf {Q}$ to describe the alignment behaviour of liquid crystals. Physically, $\mathbf {Q}$ could be understood as the second momentum of f:

$$\begin{aligned} \mathbf {Q}(x)=\int _{{\mathbb {S}^2}}\left( \mathbf {m}\mathbf {m}-\frac{1}{3}\mathbf {I}\right) f(x,\mathbf {m})\mathrm {d}\mathbf {m}. \end{aligned}$$

We may classify a liquid crystal by the tensor $\mathbf {Q}$: uniaxial if $\mathbf {Q}$ has only two distinct eigenvalues;biaxial if $\mathbf {Q}$ has three distinct eigenvalues; isotropic if all eigenvalues are zero.

Since the order tensor $\mathbf {Q}$ vanishes when f is the probability density $\frac{1}{4\pi }$ for the isotropic phase, the tensor $\mathbf {Q}$ measures how the second moments of a given probability density deviates from the isotropic value. Thus, it is convenient to use the Q-tensor theory to model the isotropic–nematic phase transition problem, which is based on the so-called Landau-de Gennes energy:

$$\begin{aligned} \mathcal {F}(\mathbf {Q},\nabla \mathbf {Q})&=\int _{\Omega }\left\{ \underbrace{\frac{a}{2}\mathrm {Tr}\mathbf {Q}^2 -\frac{b}{3}\mathrm {Tr}\mathbf {Q}^3+\frac{c}{4}(\mathrm {Tr}\mathbf {Q}^2)^2}_{F_b:\text {bulk energy}}\right. \\&\left. \quad +\,\underbrace{\frac{1}{2}\left( L_1|\nabla \mathbf {Q}|^2+L_2Q_{ij,j}Q_{ik,k} +L_3Q_{ij,k}Q_{ik,j}+L_4Q_{ij}Q_{kl,i}Q_{kl,j}\right) }_{F_e:\text {elastic energy}} \right\} \mathrm {d}x. \end{aligned}$$

Here a, b, c are material and temperature dependent nonnegative constants and $L_i(i=1,2,3,4)$ are material dependent elastic constants. We refer the reader to [3] for more details. As in [2, 10], we take $L_3=L_4=0$ in the elastic energy $F_e$. Then the energy functional is reduced to

$$\begin{aligned} \mathcal F(\mathbf {Q},\nabla \mathbf {Q})=\int _{\Omega }\Big \{\frac{1}{2}\left( L_1|\nabla \mathbf {Q}|^2+L_2Q_{ij,j}Q_{ik,k} \right) +F_b(\mathbf {Q}) \Big \}\mathrm {d}x. \end{aligned}$$

(1.1)

The bulk energy $F_b$ can characterize the isotropic–nematic phase transition for liquid crystals. The critical points of the bulk energy are

$$\begin{aligned} \mathbf {Q}=0\quad \text { or }\quad s^{\pm }\left( \mathbf {n}\mathbf {n}-\frac{1}{3}\mathbf {I}\right) , \end{aligned}$$

(1.2)

where $s^{\pm }$ are the solutions of $3a-bs+2cs^2=0,$ and $\mathbf {n}\in {\mathbb {S}^2}$. In addition, if $0<a<\frac{b^2}{24c}$, then $\mathbf {Q}=0$ and $\mathbf {Q}=s^+(\mathbf {n}\mathbf {n}-\frac{1}{3}\mathbf {I})$ are stable critical points, which correspond to isotropic phase and nematic phase respectively, and $\mathbf {Q}=s^-(\mathbf {n}\mathbf {n}-\frac{1}{3}\mathbf {I})$ is unstable. See [14] for the details. Since we are interested in the stable interface between the two co-existence phases, we impose the condition $b^2=27ac$, which means that the bulk energy at each phase is equal.

In this paper, we are concerned with the structure of the molecular directional field near the phase transition and the shape of the interface. To this end, we assume that the transition between 0 and $s^{+}\left( \mathbf {n}\otimes \mathbf {n}-\frac{1}{3}\right)$ appears in a thin region of width $\sqrt{L_1}$. By a rescaling and limiting process as in [10], the energy functional (1.1) can be reduced to

$$\begin{aligned} \mathcal F(\mathbf {Q},\nabla \mathbf {Q})=\int _{\mathbf {R}^3}\left\{ \frac{1}{6}|\nabla \mathbf {Q}|^2+\frac{L}{4} Q_{ij,j}Q_{ik,k}+\frac{1}{3} F_{b}(\mathbf {Q}) \right\} \,\mathrm {d}x. \end{aligned}$$

(1.3)

Furthermore, after scaling if necessary, we may take

$$\begin{aligned} a=1, \quad b=9,\quad c=3 \end{aligned}$$

so that $s^+=1$ and $s^-=1/2$ and $F_b(\mathbf {Q})=0 \text { if } \mathbf {Q}=0 \text { or }\mathbf {n}\otimes \mathbf {n}-\frac{1}{3}\mathbf I$. The energy functional (1.3) becomes

$$\begin{aligned} \mathcal {F}(\mathbf {Q},\nabla \mathbf {Q})=&\int _{\mathbf {R}^3} \left\{ \frac{1}{6}|\nabla \mathbf {Q}|^2+\frac{L}{4} Q_{ij,j}Q_{ik,k} + \frac{1}{6}\mathrm {Tr}\mathbf {Q}^2 -\mathrm {Tr}\mathbf {Q}^3+\frac{1}{4}(\mathrm {Tr}\mathbf {Q}^2)^2\right\} \mathrm {d}x. \end{aligned}$$

(1.4)

Now the isotropic–nematic interface problem is reduced to study the minimizers of (1.4) in the class so that $\mathcal {F}(\mathbf {Q},\nabla \mathbf {Q})$ is finite and satisfying

$$\begin{aligned} \mathbf {Q}(x_1,x_2, +\,\infty )=\mathbf {n}\mathbf {n}-\frac{1}{3}\mathbf {I},\quad \mathbf {Q}(x_1,x_2,-\,\infty )=0. \end{aligned}$$

(1.5)

Let us refer to [6, 8, 9, 11] and references therein for some numerical results on the isotropic–nematic interface problem based on the Landau-de Gennes’s framework. Numerical results show that the ratio of the coefficients in the isotropic and anisotropic energies and different anchoring condition make an important effect on the structure and stability of minimizers. To our knowledge, there are few rigorous results on this problem. In [10], the authors made a first effort to 1-D problem, which will be introduced in next section.

Let us conclude the introduction by introducing the following notations. We define

$$\begin{aligned} \mathcal {A}= \left\rbrace \mathbf {Q}\in W^{1,2}(\mathbf {R}^N, S_0) : \mathbf {Q}(x_1,\ldots ,x_{N-1}, \,+\infty ) = \mathbf {n}\mathbf {n}- \frac{1}{3} \mathbf {I}, \ \mathbf {Q}(x_1,\ldots ,x_{N-1}, -\,\infty ) = 0 \right\lbrace \end{aligned}$$

so that $\mathcal F(\mathbf {Q}, \nabla \mathbf {Q})$ is finite for $\mathbf {Q}\in \mathcal {A}$, where $S_0$ denotes the set of all symmetric traceless $3 \times 3$ matrices. We take $N=3$ or 1 throughout this paper.

Definition 1.1

$\mathbf {Q}^* \in \mathcal {A}$ is called a global minimizer of $\mathcal F(\mathbf {Q}, \nabla \mathbf {Q})$ if it satisfies $\mathcal F(\mathbf {Q}^*, \nabla \mathbf {Q}^*) \le \mathcal F(\mathbf {Q}, \nabla \mathbf {Q})$ for all $\mathbf {Q}\in \mathcal {A}$; $\mathbf {Q}^* \in \mathcal {A}$ is called a local minimizer if it satisfies $\mathcal F(\mathbf {Q}^*, \nabla \mathbf {Q}^*) \le \mathcal F(\mathbf {Q}, \nabla \mathbf {Q})$ for $\mathbf {Q}\in \mathcal {A}$ in some open neighbourhood of $\mathbf {Q}^*$; $\mathbf {Q}^* \in \mathcal {A}$ is called a stable solution of the Euler-Lagrange equation associated with $\mathcal F$, e.g. (2.1), if it admits a local minimizer of $\mathcal F$.

2 De Giorgi’s type conjecture

Let us first consider the case of $L=0$. The associated Euler-Lagrange equation takes

$$\begin{aligned} -\,\Delta \mathbf {Q}+ \mathbf {Q}- 9 \mathbf {Q}^2 + 3|\mathbf {Q}|^2 \mathbf {Q}+ 3|\mathbf {Q}|^2 \mathbf {I}= 0 \end{aligned}$$

(2.1)

with boundary condition

$$\begin{aligned} \mathbf {Q}(x_1,x_2, +\,\infty ) = \mathbf {n}\mathbf {n}- \frac{1}{3} \mathbf {I}, \quad \mathbf {Q}(x_1,x_2,-\,\infty ) = 0. \end{aligned}$$

(2.2)

This system is similar to the Allen-Cahn equation

$$\begin{aligned} \Delta u-(1-u^2)u=0\quad \text {in}\quad \mathbf {R}^N, \end{aligned}$$

(2.3)

which also arises from the phase transition problem. In 1978, De Giorgi made the following well-known conjecture:

Let u be a bounded solution of (2.3) such that $\partial _{x_N}u > 0$. Then the level sets of u are all hyperplanes, at least for dimension $N\le 8$.

This conjecture has been solved by Ghoussoub-Gui [7] for $N=2$, Ambrosio-Cabré [1] for $N=4$ and Savin [12] for $4\le N\le 8$. The conjecture is not true for $N\ge 9$ [4].

Motivated by De Giorgi’s conjecture, we propose a similar conjecture:

(GDC):  Let $\mathbf {Q}$ be symmetric, traceless and a bounded solution of (2.1)–(2.2). Let $\lambda _3$ be the largest eigenvalue of $\mathbf {Q}$. If $\partial _{x_3} \lambda _3 >0$, then all level sets $\{x\in \mathbf {R}^3: Q_{ij}(x) =s \}$ are hyperplanes.

Compared with (2.3), this conjecture seems more difficult even in 1-D since (2.1) is a system with five independent components. In [5], Fazly and Ghoussoub also considered the De Giorgi type conjecture for the elliptic system: $\Delta u = \nabla H(u)$ in $\mathbf {R}^N$, where $u : \mathbf {R}^N \rightarrow \mathbf {R}^m$ and $H \in C^2(\mathbf {R}^m)$, and proved that the solution $u = \{ u_i \}_{i=1}^m$ is necessarily one-dimensional under various conditions on the nonlinearity H and some monotonicity assumption on u. The elliptic system we considered in this paper does not fall into this system due to the traceless constraint on $\mathbf {Q}$. Even for the results in 1-D to be presented, there are no additional assumptions on the solution. Let us mention [13], which may be the first paper on the vectorial Allen-Cahn system.

In [10], Park et al. consider the global minimizer of the 1-D total energy functional

$$\begin{aligned} \mathcal F(\mathbf {Q},\mathbf {Q}')=\int _{\mathbb {R}}\left\{ \frac{1}{6}|\mathbf {Q}^{\prime }{}|^2+ \frac{1}{6}\mathrm {Tr}\mathbf {Q}^2 -\mathrm {Tr}\mathbf {Q}^3+\frac{1}{4}(\mathrm {Tr}\mathbf {Q}^2)^2 \right\} \,\mathrm {d}s, \end{aligned}$$

(2.4)

where ${}^{^{\prime }{}}$ denotes $\frac{d}{ds}=\frac{d}{dz}$. The associated Euler-Lagrange equation becomes

$$\begin{aligned} -\,\mathbf {Q}''+\mathbf {Q}-9\mathbf {Q}^2+3|\mathbf {Q}|^2\mathbf {Q}+3|\mathbf {Q}|^2\mathbf {I}=0 \end{aligned}$$

(2.5)

with the boundary condition

$$\begin{aligned} \mathbf {Q}( +\,\infty )=\mathbf {n}\mathbf {n}-\frac{1}{3}\mathbf {I},\quad \mathbf {Q}(-\,\infty )=0. \end{aligned}$$

(2.6)

Theorem 2.1

[10] The global minimizer of (2.4) subject to (2.6) must take the form

$$\begin{aligned} \mathbf {Q}(s)=\frac{1}{2}\left( 1+\tanh \frac{1}{2}(s-t)\right) \left( \mathbf {n}\mathbf {n}-\frac{1}{3}\mathbf {I}\right) , \end{aligned}$$

(2.7)

where t is an arbitrary constant due to translation symmetry.

To solve (GDC), a key step is to study the structure of local minimizers in 1-D. Thanks to Theorem 2.1, it is natural to ask the following question:

Whether the local minimizers of (2.4) subject to (2.6) must take the form (2.7)?

In Sect. 3, we will give an affirmative answer to this question. For the global minimizer, one can reduce $\mathbf {Q}$ to a diagonal form (2.8). For the local minimizer, we have to consider general $\mathbf {Q}$. To reduce the problem, we used many elegant structures hidden in the system. In Sect. 4, we will prove the conjecture (GDC) under the mild assumption that the eigenvector of $\mathbf {Q}$ corresponding to the largest eigenvalue is a constant vector.

For the case of $L\ne 0$, the problem becomes more complex. In this case, the direction vector $\mathbf {n}$ on the anchoring condition at $+\,\infty$ could make a significant effect on the behavior for the minimizers. There are three different types of the alignment director $\mathbf {n}$ on the boundary as below:

• (1) Homeotropic anchoring: $\mathbf {n}\cdot (0,0,1)=1;$

• (2) Planar anchoring: $\mathbf {n}\cdot (0,0,1)=0;$

• (3) Tilt anchoring: $0<\mathbf {n}\cdot (0,0,1)<1.$

For simplicity, we will first seek minimizers of the diagonal form

$$\begin{aligned} \mathbf {Q}=\left( \begin{array}{ccc}-\frac{1}{3}(S+T)&{}0&{}0\\ 0&{} -\,\frac{1}{3}(S-T)&{}0\\ 0&{}0&{} \frac{2}{3}S\end{array} \right) , \end{aligned}$$

(2.8)

which is meaningful due to the rotation invariant of the bulk energy. Then in 1-D, the energy functional is reduced to

$$\begin{aligned} \mathcal F_L(S,T)=\frac{2}{9}\int _{\mathbb {R}}\left( \frac{1+L}{2}(S')^2+\frac{1}{6}(T')^2+ \frac{1}{6}(3S^2+T^2)-S(S^2-T^2)+\frac{1}{18}(3S^2+T^2)^2\right) \mathrm {d}s. \end{aligned}$$

(2.9)

The associated Euler–Lagrange equation of (2.9) takes as follows

$$\begin{aligned} \left\{ \begin{array}{l} -\,\frac{1+L}{2}S''+\frac{S}{2}-\frac{3S^2}{2}+\frac{T^2}{2}+\frac{S(3S^2+T^2)}{3}=0,\\ -\,\frac{1}{6}T''+\frac{T}{6}+{ST}+\frac{T(3S^2+T^2)}{9}=0. \end{array}\right. \end{aligned}$$

(2.10)

Here we consider the homeotropic anchoring condition, which leads to the following boundary conditions for (S, T):

$$\begin{aligned} S(+\,\infty )=1,\quad T(+\,\infty )=S(-\,\infty )=T(-\,\infty )=0. \end{aligned}$$

(2.11)

It is obvious that (2.10)–(2.11) has a uniaxial solution with $T=0$ and S(s) solving

$$\begin{aligned} -\,(1+L)S''+{S}-{3S^2}+2S^3=0. \end{aligned}$$

That is, an uniaxial equilibrium state takes

$$\begin{aligned} \mathbf {Q}_0(s)=S(s)\text {diag}\left\{ -\,\frac{1}{3},-\,\frac{1}{3}, \frac{2}{3}\right\} ,\quad S(s)=S^*(s/\sqrt{1+L}), \end{aligned}$$

(2.12)

where $S^*$ solves

$$\begin{aligned} -\, S^{\prime \prime } + S - 3S^2 + 2S^3 = 0, \quad S(-\,\infty )= 0,\,\, S(+\,\infty ) = 1. \end{aligned}$$

(2.13)

In [10], the authors investigate the stability of this solution.

Theorem 2.2

The uniaxial equilibrium state $\mathbf {Q}_0$ is stable for the energy functional (2.4) when $L\le 0$ and unstable when $L>0$.

In Sect. 5, we will prove that all the solutions of (2.10)–(2.11) must take $\mathbf {Q}_0(s)$ when $L>-1$. This in particular implies that the stable equilibrium state for the energy functional (2.4) cannot be of diagonal form (2.8) when $L>0$ and under the homeotropic anchoring condition.

For $L\ne 0$, the structure of equilibrium solutions with planar and tilt anchoring boundary conditions is still an open question. In this case, the anisotropic term should play a key role in the study of the behavior for minimizers near the isotropic–nematic phase transition. See [8, 9] for numerical results and [10] for more discussions and open questions.

3 Local minimizer in 1-D for $L=0$

In this section, we study the local minimizer of (2.4) with $L=0$.

Theorem 3.1

All the local minimizers of (2.4) subject to (2.6) must take the form

$$\begin{aligned} \mathbf {Q}(s)=\frac{1}{2}\left( 1+\tanh \frac{1}{2}(s-t)\right) \left( \mathbf {n}\mathbf {n}-\frac{1}{3}\mathbf {I}\right) , \end{aligned}$$

(3.1)

where t is an arbitrary parameter. In fact, (3.1) gives all solutions of (2.5)–(2.6).

For the global minimizer, the proof relies on the fact that we may assume that $\mathbf {Q}$ is the diagonal form (2.8). Thus, it suffices to consider an ODE system with two components. For the local minimizer, we have to consider an ODE system with five components. To reduce the problem, we need the following key lemmas. In what follows, we always assume that $\mathbf {Q}$ is a solution of (2.5)–(2.6). We will often use the following spectral decomposition of $\mathbf {Q}$:

$$\begin{aligned} \mathbf {Q}= \lambda _1 \mathbf {n}_1 \mathbf {n}_1 + \lambda _2 \mathbf {n}_2 \mathbf {n}_2 + \lambda _3 \mathbf {n}_3 \mathbf {n}_3 . \end{aligned}$$

(3.2)

Lemma 3.2

It holds that

$$\begin{aligned} \mathrm {Tr}(\mathbf {Q}^{\prime })^2= \sum _{i=1}^3 |\lambda ^{\prime }|^2 + 2 \sum _{i<j} (\lambda _i - \lambda _j)^2 (\mathbf {n}_i^{\prime } \cdot \mathbf {n}_j)^2 =\mathrm {Tr}(\mathbf {Q}^2) - 6\mathrm {Tr}(\mathbf {Q}^3 ) +\frac{3}{2} ( |\mathbf {Q}|^2 )^2. \end{aligned}$$

Moreover, $\mathbf {Q}(x^*) \ne 0$ for all finite $x^*$.

Proof

Using (3.2) and the identities like $|\mathbf {n}_1^{\prime }|^2 = |\mathbf {n}_1^{\prime } \cdot \mathbf {n}_2 |^2 + |\mathbf {n}_1^{\prime }\cdot \mathbf {n}_3|^2$, we obtain

$$\begin{aligned} \begin{aligned} | \mathbf {Q}^{\prime } |^2&= \sum _{i=1}^{3} | \lambda ^{\prime }_i|^2 +2 \sum _{i=1}^3 \lambda _i^2 | \mathbf {n}_i^{\prime }|^2 + 4 \sum _{i<j } \lambda _i \lambda _j (\mathbf {n}_i \cdot \mathbf {n}_j^{\prime }) (\mathbf {n}_j \cdot \mathbf {n}_i^{\prime }) \\&= \sum _{i=1}^{3} | \lambda ^{\prime }_i|^2 +2 \sum _{i<j } \left( \lambda _i (\mathbf {n}_j\cdot \mathbf {n}_i ^{\prime }) + \lambda _j (\mathbf {n}_i \cdot \mathbf {n}_j^{\prime }) \right) ^2\\&= \sum _{i=1}^{3} | \lambda ^{\prime }_i |^2 +2 \sum _{i<j } ( \lambda _i-\lambda _j)^2 (\mathbf {n}_j\cdot \mathbf {n}_i ^{\prime })^2, \end{aligned} \end{aligned}$$

which gives the first equality.

Using the Eq. (2.5), we have

$$\begin{aligned} 0&= \big \langle -\,\mathbf {Q}^{\prime \prime } +\mathbf {Q}- 9\mathbf {Q}^2 + 3|\mathbf {Q}|^2 \mathbf {Q}+ 3 |\mathbf {Q}|^2 I , 2\mathbf {Q}^{\prime }\big \rangle \\&= \left( \mathrm {Tr}\left( -\,\mathbf {Q}^{\prime 2} +\mathbf {Q}^2 - 6 \mathbf {Q}^3 \right) + \frac{3}{2}( |\mathbf {Q}|^2)^2 \right) ^{\prime }, \end{aligned}$$

where $\big \langle A, B\big \rangle \triangleq Tr(A^TB)$ denote the matrix inner product. Therefore, there exists some constant $C_0$ so that

$$\begin{aligned} | \mathbf {Q}^{\prime }|^2 = |\mathbf {Q}|^2 - 6\mathrm {Tr}(\mathbf {Q}^3) +\frac{3}{2}( |\mathbf {Q}^2|)^2 + C_0, \end{aligned}$$

(3.3)

which implies $C_0\ge 0$ due to $\mathbf {Q}(-\,\infty )=0$. Moreover, due to that $\mathbf {Q}$ is a continuous function and the boundary condition (2.6), $\mathbf {Q}$ is bounded, and thus there exists some constant $C_1>0$ so that

$$\begin{aligned} |\mathbf {Q}^{\prime }|^2 \le C_1. \end{aligned}$$

(3.4)

Using (2.5) again, we get by integration by parts and (3.3) that

$$\begin{aligned} \begin{aligned} 0&= \int _{x_1}^{x_2} \big \langle - \mathbf {Q}^{\prime \prime } + \mathbf {Q}- 9 \mathbf {Q}^2 + 3|\mathbf {Q}|^2 \mathbf {Q}+ 3 |\mathbf {Q}|^2 I , \mathbf {Q}\big \rangle \mathrm {d}x \\&= -\,\mathrm {Tr}( \mathbf {Q}^{\prime } \mathbf {Q}) \bigg |_{x_1}^{x_2} + \int _{x_1}^{x_2} ( |\mathbf {Q}^{\prime }|^2 +|\mathbf {Q}|^2 - 9\mathrm {Tr}(\mathbf {Q}^3) + 3 ( |\mathbf {Q}|^2)^2 ) \mathrm {d}x \\&= -\, \mathrm {Tr}( \mathbf {Q}^{\prime } \mathbf {Q}) \bigg |_{x_1}^{x_2} + \int _{x_1}^{x_2} ( C_0+2|\mathbf {Q}|^2 - 15\mathrm {Tr}(\mathbf {Q}^3) + \frac{9}{2} ( |\mathbf {Q}|^2)^2 )\mathrm {d}x. \end{aligned} \end{aligned}$$

(3.5)

If $C_0 > 0$, due to $\mathbf {Q}(-\,\infty ) = 0$, there exists $M<0$ sufficiently small so that

$$\begin{aligned} C_0+2|\mathbf {Q}|^2 - 15\mathrm {Tr}(\mathbf {Q}^3) + \frac{9}{2} ( |\mathbf {Q}|^2)^2 \ge \frac{C_0}{2} ,\quad \ |\mathbf {Q}|^2 \le 1 \quad \forall x < M. \end{aligned}$$

(3.6)

Choosing $x_1<x_2 < M$, and combining (3.4), (3.5) and (3.6), we deduce that

$$\begin{aligned} \begin{aligned} 2 \sqrt{C_1}&\ge 2 \sqrt{\mathrm {Tr}( \mathbf {Q}^{\prime 2} ) \mathrm {Tr}(\mathbf {Q}^2) } \ge \mathrm {Tr}(\mathbf {Q}^{\prime }\mathbf {Q}) \bigg |_{x_1}^{x_2} \\&=\int _{x_1}^{x_2} (C_0+2|\mathbf {Q}|^2 - 15\mathrm {Tr}(\mathbf {Q}^3) + \frac{9}{2} ( |\mathbf {Q}|^2)^2 )\mathrm {d}x \ge \frac{C_0}{2}\ (x_2 -x_1) . \end{aligned} \end{aligned}$$

Letting $x_1 \rightarrow -\,\infty$, we get a contradiction, and thus $C_0 = 0$. This proves the second equality.

If $\mathbf {Q}(x^*) = 0$ for some finite $x^*$, then we have

$$\begin{aligned} \mathrm {Tr}(\mathbf {Q}^{\prime 2}(x^*)) = 0 \Longrightarrow \mathbf {Q}^{\prime } (x^*) = 0. \end{aligned}$$

Due to $\mathbf {Q}(x^*) =\mathbf {Q}^{\prime }(x^*) = 0$, the uniqueness theorem of ODE gives $\mathbf {Q}(x) \equiv 0$, which contradicts the boundary condition $\mathbf {Q}(+\,\infty ) \ne 0$ in (2.6). $\square$

We introduce two important quantities

$$\begin{aligned} A(x)=|\mathbf {Q}(x)| ^2,\quad B(x)=|\mathbf {Q}^{\prime }(x)|^2. \end{aligned}$$

By Lemma 3.2, we have

$$\begin{aligned} \mathrm {Tr}(\mathbf {Q}^3) = \frac{A-B}{6} + \frac{A^2}{4} , \quad \mathbf {Q}^{\prime }(\pm \, \infty ) = 0. \end{aligned}$$

By (2.5), we have

$$\begin{aligned} \begin{aligned} 0&=\big \langle - \,\mathbf {Q}^{\prime \prime } + \mathbf {Q}- 9 \mathbf {Q}^2 + 3|\mathbf {Q}|^2 \mathbf {Q}+ 3 |\mathbf {Q}|^2 I ,\mathbf {Q}\big \rangle \\&= -\, \frac{ ( \mathrm {Tr}(\mathbf {Q}^2) )^{\prime \prime } - 2\mathrm {Tr}(\mathbf {Q}^{\prime 2} ) }{2} + |\mathbf {Q}|^2 - 9\mathrm {Tr}(\mathbf {Q}^3) + 3( |\mathbf {Q}|^2)^2 \\&=-\, \frac{ (\mathrm {Tr}(\mathbf {Q}^2) )^{\prime \prime } - 2\mathrm {Tr}(\mathbf {Q}^{\prime 2} ) }{2} + |\mathbf {Q}|^2 - 9 \left( \frac{A-B}{6} + \frac{A^2}{4}\right) + 3( |\mathbf {Q}|^2)^2 \\&= -\,\frac{A^{\prime \prime } - 2 B }{2} + A - 9\left( \frac{A-B}{6} + \frac{A^2}{4}\right) + 3 A^2. \end{aligned} \end{aligned}$$

This gives the following useful differential equation

$$\begin{aligned} A^{\prime \prime } = -\, A + 5B + \frac{3}{2}A^2. \end{aligned}$$

(3.7)

Moreover,

$$\begin{aligned} B(\pm \, \infty ) = 0, \quad A(-\,\infty ) = 0, \quad A(+\,\infty )= \frac{2}{3}, \quad A^{\prime }(\pm \, \infty ) = 0. \end{aligned}$$

(3.8)

Lemma 3.3

It holds that

$$\begin{aligned} B\ge A - \sqrt{6} A^{3/2} + \frac{3}{2} A^2. \end{aligned}$$

The equality holds if and only if $\lambda _i = 2a >0, \lambda _j = -\,a , j \ne i$ for some i and $a > 0$.

Proof

By Lemma 3.2, it suffices to prove

$$\begin{aligned} \mathrm {Tr}(\mathbf {Q}^3) \le \frac{1}{ \sqrt{6} } A^{3/2}. \end{aligned}$$

(3.9)

Thanks to $\sum _{i=1}^{3} \lambda _i = 0$, $\mathrm {Tr}(\mathbf {Q}^3) = \sum _{i=1}^3 \lambda _i^3 = 3 \lambda _1 \lambda _2 \lambda _3$. If $\lambda _1 \lambda _2 \lambda _3 \le 0$, then we have $\mathrm {Tr}(\mathbf {Q}^3) \le 0 < A^{3/2}/\sqrt{6}$. Otherwise, we assume $\lambda _1, \lambda _2 <0$ without loss of generality. Obviously,

$$\begin{aligned} (\mathrm {Tr}(\mathbf {Q}^3) )^2&= 9 \lambda _1^2 \lambda _2^2 \lambda _3^2 = 9 ( \lambda _1\lambda _2 ) ( \lambda _1\lambda _2 ) (\lambda _1 +\lambda _2)^2 \\&= 36 ( \lambda _1\lambda _2 ) ( \lambda _1\lambda _2 ) \frac{( \lambda _1 +\lambda _2)^2}{4} \le 36\left( \frac{ \lambda _1\lambda _2 + \lambda _1\lambda _2 + ( \lambda _1 +\lambda _2)^2/4 }{3} \right) ^3, \end{aligned}$$

from which and the following fact

$$\begin{aligned} \frac{ \lambda _1^2 + \lambda _2^2 + \lambda _1 \lambda _2}{3} - \frac{ \lambda _1\lambda _2 + \lambda _1\lambda _2 + ( \lambda _1 +\lambda _2)^2/4 }{3} = \frac{1}{4} (\lambda _1 - \lambda _2)^2 \ge 0, \end{aligned}$$

we infer that

$$\begin{aligned} (\mathrm {Tr}(\mathbf {Q}^3))^2 \le 36 \left( \frac{ \lambda _1^2 + \lambda _2^2 + \lambda _1 \lambda _2}{3} \right) ^3 =36 \left( \frac{A}{6} \right) ^3 = \frac{A^3}{6}, \end{aligned}$$

which gives (3.9). The above arguments also show that the equality holds if and only if $\lambda _1 =\lambda _2 =-\,a , \lambda _3 = 2a$ for some $a>0$. $\square$

Lemma 3.4

For any $x \in \mathbf {R},\, 0< A(x) < \frac{2}{3}$ and $A^{\prime }(x) > 0$.

Proof

First of all, $A(x)>0$ follows from Lemma 3.2. Next we show that $A(x) \le \frac{2}{3}$. Otherwise, thanks to $A(-\,\infty ) = -\,\infty$ and $A(+\,\infty )=\frac{2}{3}$, A(x) achieves its global maximum at a finite point $x^*$ and $A(x^*) > 2/3$. Then we infer that

$$\begin{aligned} A^{\prime \prime }(x^*) = -\,A(x^*) +5B(x^*) + \frac{3}{2} A^{2} (x^*)>0, \end{aligned}$$

which contradicts with the fact that $A(x^*)$ is the global maximum. Hence, we have $A(x)\le 2/3$.

To rule out $A(x) = 2/3$ for some x, we assume $A(x^*) = 2/3$. Again, $A(x^*)$ is the global maximum due to $A(x) \le 2/3$ and it leads to $A^{\prime \prime } \le 0$. Then we have

$$\begin{aligned} 0 \ge A^{\prime \prime }(x^*) = -\,A(x^*) +5B(x^*) + \frac{3}{2} A^{2}(x^*) = 5B(x^*). \end{aligned}$$

As $B(x^*)$ is non-negative, we deduce that $B(x^*)=\mathrm {Tr}(\mathbf {Q}^{\prime 2}(x^*))=0$, hence, $\mathbf {Q}^{\prime }(x^*) = 0$. Then it follows from Lemma 3.3 that

$$\begin{aligned} 0 = B(x^*) \ge \left( A - \sqrt{6} A^{3/2} + \frac{3}{2} A^2\right) (x^*) = \frac{2}{3} -\, \frac{4}{3} + \frac{2}{3} = 0. \end{aligned}$$

The equality implies that

$$\begin{aligned} \lambda _i(x^*) = 2a >0,\quad \lambda _j (x^*)= -a <0\,\,\text {for}\,\, j \ne i \end{aligned}$$

for some i and $a> 0$. Hence, $2/3 = A(x^*) =\mathrm {Tr}(\mathbf {Q}^2) (x^*)=6a^2$, then $a = \frac{1}{3}$. Therefore,

$$\begin{aligned} Q(x^*) = \sum _{k=1}^3 \lambda _k \mathbf {n}_k \mathbf {n}_k\big |_{x=x^*}&= -\,\frac{1}{3} \sum _{j\ne i} \mathbf {n_j n_j} +\frac{2}{3} \mathbf {n_i n_i}\big |_{x=x^*}\\&= -\,\frac{1}{3} (\mathbf {I} - \mathbf {n_i n_i})+\frac{2}{3} \mathbf {n_i n_i}\big |_{x=x^*} = \mathbf {n_i n_i} - \frac{1}{3} \mathbf {I}\big |_{x=x^*}. \end{aligned}$$

Let $\widetilde{\mathbf {Q}}(x) \equiv \mathbf {n_i(x^*) n_i(x^*)} - \frac{1}{3} \mathbf {I}$. Then $\widetilde{\mathbf {Q}}(x)$ is a solution of (2.5) with

$$\begin{aligned} \widetilde{\mathbf {Q}}(x^*) =\mathbf {n_i(x^*) n_i(x^*)} - \,\frac{1}{3} \mathbf {I} = \mathbf {Q}(x^*),\quad \widetilde{\mathbf {Q}}^{\prime }(x^*) = 0 =\mathbf {Q}^{\prime }(x^*). \end{aligned}$$

The uniqueness theorem of ODE implies that $\widetilde{\mathbf {Q}}(x) \equiv \mathbf {Q}(x)$. However, $\widetilde{\mathbf {Q}}(-\,\infty ) \ne 0 =\mathbf {Q}(-\,\infty )$ and it leads to a contradiction. This shows that

$$\begin{aligned} 0< A(x) < \frac{2}{3}\quad \forall \,\, x \in \mathbf {R}. \end{aligned}$$

Finally, we show that A(x) is strictly monotonic. Using (3.7) and Lemma 3.3, we get

$$\begin{aligned} A^{\prime \prime }(x) = -\,A + B + \frac{3}{2} A^2 \ge -A + \frac{3}{2} A^2 + 5\left( A - \sqrt{6} A^{3/2} + \frac{3}{2} A^2 \right) = 4 A - 5 \sqrt{6} A^{3/2} + 9 A^2. \end{aligned}$$

(3.10)

Thanks to $A(-\,\infty ) = A^{\prime }(-\,\infty ) = 0$, we have

$$\begin{aligned} A^{\prime \prime }(x) \ge 4 A - 5 \sqrt{6} A^{3/2} + 9 A^2 = A(4 - 5\sqrt{6A} + 9A ) >0 \quad \ \forall x <-\,M \end{aligned}$$

for some large $M>0$. Consequently, A(x) is strictly convex and monotonic for $x <-M$. We define

$$\begin{aligned} x_1=\sup \big \{ x: A^{\prime }(t) >0 , \ \forall t \in (-\,\infty , x)\big \} ,\quad J= (-\,\infty , x_1). \end{aligned}$$

We aim to prove $x_1 = +\,\infty$. Otherwise, we have $A^{\prime }(x_1) = 0$. Multiplying $2 A^{\prime }(y)$ on both side of (3.10) and then integrating from $-\,\infty$ to some $x \in J$, we obtain

$$\begin{aligned} \begin{aligned} 0&\le \int _{-\,\infty }^x \left( A^{\prime \prime }(y) - \left( 4 A - 5 \sqrt{6} A^{3/2} + 9 A^2\right) (y) \right) \cdot 2 A^{\prime }(y) dy \\&= ( A^{\prime 2} - ( 4A^2 - 4\sqrt{6}A^{3/2} + 6 A^3 )) \bigg | _{-\,\infty }^{x} = A^{\prime 2}(x)-( 4A^2 - 4\sqrt{6}A^{3/2} + 6 A^3 )(x), \end{aligned} \end{aligned}$$

which gives

$$\begin{aligned} A^{\prime }(x)^2 \ge ( 4A^2 - 4\sqrt{6}A^{3/2} + 6 A^3 )(x) = A^2 (2 - \sqrt{6A} )^2(x). \end{aligned}$$

Recall that $0<A(x) < 2/3$ and $A^{\prime }(x) > 0 \ (x\in J)$. Thus,

$$\begin{aligned} A^{\prime }(x) \ge A(2-\sqrt{6A})(x). \end{aligned}$$

Letting $x \rightarrow x_1$ gives

$$\begin{aligned} 0 = A^{\prime }(x_1) \ge A(2-\sqrt{6A})(x_1) > 0. \end{aligned}$$

This is a contradiction. Therefore, $x_1 = +\,\infty$ and $A^{\prime }(x) > 0$ for $x \in \mathbf {R}$. $\square$

Now we are in a position to prove Theorem 3.1.

Proof of Theorem 3.1

We infer from Lemma 3.4 and (3.10) that

$$\begin{aligned} 0&\le \int _{-\,\infty }^x \left( A^{\prime \prime }(y) - \left( 4 A - 5 \sqrt{6} A^{3/2} + 9 A^2\right) (y) \right) \cdot 2 A^{\prime }(y) dy\\&\quad = A^{\prime }(x)^2 - A^2(2 - \sqrt{6A})^2(x). \end{aligned}$$

Due to $A(+\,\infty )=A'(+\,\infty )=0$, letting $x \rightarrow +\, \infty$ gives

$$\begin{aligned} 0 \le \int _{-\,\infty }^{+\,\infty } \left( A^{\prime \prime }(y) - \left( 4 A - 5 \sqrt{6} A^{3/2} + 9 A^2\right) (y) \right) \cdot 2 A^{\prime }(y) dy= 0, \end{aligned}$$

which implies that for $x\in \mathbf {R}$,

$$\begin{aligned} A^{\prime \prime }(x) - (4 A - 5 \sqrt{6} A^{3/2} + 9 A^2 )(x ) = 0. \end{aligned}$$

(3.11)

Consequently,

$$\begin{aligned} A^{\prime }(x)^2 - A^2(2 - \sqrt{6A})^2(x)=0,\quad \text {i.e.}\quad A^{\prime }(x) = A (2-\sqrt{6A})(x). \end{aligned}$$

(3.12)

Using (3.7) and (3.11), we find

$$\begin{aligned} -\,A + 5B + \frac{3}{2}A^2 = A^{\prime \prime } =4 A - 5 \sqrt{6} A^{3/2} + 9 A^2 \Longrightarrow B = A - \sqrt{6} A^{3/2} + \frac{3}{2} A^2.\quad \end{aligned}$$

(3.13)

While, by Lemma 3.3, the equality on B holds if and only if two eigenvalues of $\mathbf {Q}(x)$ are equal and negative for any $x\in \mathbf {R}$. As $\mathbf {Q}(x) \ne 0$ due to Lemma 3.2, it must hold that

$$\begin{aligned} \lambda _i(x) = \lambda _j(x) <0 \ \forall x \in \mathbf {R}, \end{aligned}$$

and $i\ne j$ are fixed for any x. Therefore, without loss of generality, we assume $\lambda _1(x) = \lambda _2(x) < 0$ and denote

$$\begin{aligned} \lambda _1(x) =\lambda _2(x ) = -\,\frac{S(x)}{3}, \quad \lambda _3(x) = \frac{2S(x)}{3} ,\quad S(x) > 0. \end{aligned}$$

The boundary condition (2.6) implies that $S(-\,\infty )= 0 ,\ S(+\,\infty ) = 1$. Consequently, we get $A= \mathrm {Tr}(\mathbf {Q}^2) = \frac{2S^2}{3}$, and then we see from (3.12) that

$$\begin{aligned} \left( \frac{2S^2}{3}\right) ^{\prime } = \frac{2S^2}{3} (2-2S) \iff S^{\prime } = S(S-1). \end{aligned}$$

(3.14)

With the boundary condition of S, we derive the explicit formula of $S, A,\lambda _i$ as follows

$$\begin{aligned} S(x ) = \frac{\exp (x- x_0)}{1+\exp (x-x_0)}, \quad (\lambda _1,\lambda _2,\lambda _3) = \left( -\,\frac{S}{3}, -\,\frac{S}{3} ,\frac{2S}{3} \right) , \quad A =\frac{2S^2}{3}, \end{aligned}$$

(3.15)

where $x_0$ is a parameter due to translation.

By Lemma 3.3, we have

$$\begin{aligned} B= \sum _{i=1}^3 |\lambda ^{\prime }_i|^2 + 2 \sum _{i<j} (\lambda _i - \lambda _j)^2 (\mathbf {n}_i^{\prime } \cdot \mathbf {n}_j )^2 = \mathrm {Tr}(\mathbf {Q}^2) - 6 \mathrm {Tr}(\mathbf {Q}^3 ) +\frac{3}{2} ( |\mathbf {Q}|^2 )^2. \end{aligned}$$

(3.16)

We infer from (3.14) and (3.15) that

$$\begin{aligned} B=\frac{2}{3} ( S^{2} - 2 S^3 + S^{4}), \end{aligned}$$

(3.17)

and

$$\begin{aligned} \sum _{i=1}^3 |\lambda ^{\prime }_i|^2 =\left( \frac{1}{9} + \frac{1}{9}+\frac{4}{9 } \right) S^{\prime 2} = \frac{2}{3} S^{\prime 2} = \frac{2}{3} S^{2}(S-1)^2. \end{aligned}$$

(3.18)

Combining (3.16), (3.17) and (3.18), we obtain

$$\begin{aligned} 2 \sum _{i<j} (\lambda _i - \lambda _j)^2 (\mathbf {n}_i^{\prime }\cdot \mathbf {n}_j)^2 = B - \sum _{i=1}^3 |\lambda ^{\prime }_i|^2 = \frac{2}{3} ( S^{2} - 2 S^{3} + S^{4}) - \frac{2}{3} S^{2}(S-1)^2 = 0. \end{aligned}$$

As $\lambda _1 ,\lambda _2 \ne \lambda _3$, we deduce

$$\begin{aligned} (\mathbf {n}_1^{\prime } \cdot \mathbf {n}_3)^2 = (\mathbf {n}_2^{\prime }\cdot \mathbf {n}_3)^2 = 0 \Longrightarrow |\mathbf {n}_3^{\prime }|^2 = ( \mathbf {n}_1^{\prime } \cdot \mathbf {n}_3)^2 + (\mathbf {n}_2^{\prime }\cdot \mathbf {n}_3)^2 = 0 \Longrightarrow \mathbf {n}_3^{\prime } = 0. \end{aligned}$$

Hence, $\mathbf {n}_3(x)$ is a constant vector $\mathbf {n}$. Finally, we obtain

$$\begin{aligned} \begin{aligned} \mathbf {Q}(x)&=\sum _{i=1}^3 \lambda _i \mathbf {n}_i \mathbf {n}_i = -\frac{S}{3} \left( \mathbf {n}_1\mathbf {n}_1 + \mathbf {n}_2\mathbf {n}_2 \right) + \frac{2S}{3}\mathbf {n}_3\mathbf {n}_3 = -\frac{S}{3} (\mathbf {I} -\mathbf {n}_3 \mathbf {n}_3) + \frac{2S}{3} \mathbf {n}_3\mathbf {n}_3 \\&= S\left( \mathbf {n}_3\mathbf {n}_3 - \mathbf { \frac{I}{3} } \right) = S \left( \mathbf {n n - \frac{I}{3}} \right) = \frac{\exp (x-x_0)}{1+ \exp (x-x_0)} \left( \mathbf {n n - \frac{I}{3}} \right) . \end{aligned} \end{aligned}$$

This proves Theorem 3.1. $\square$

4 Local minimizer in 3-D for $L=0$

The Landau-de Gennes energy (1.4) without anisotropic term in 3-D takes

$$\begin{aligned} \mathcal {F}(\mathbf {Q},\nabla \mathbf {Q}) = \int _{\mathbf {R}^3} \left\{ \frac{1}{6} | \nabla \mathbf {Q}|^2 + \frac{1}{6}\mathrm {Tr}\mathbf {Q}^2 -\mathrm {Tr}\mathbf {Q}^3 +\frac{1}{4} (\mathrm {Tr}\mathbf {Q}^2)^2\right\} dx, \end{aligned}$$

and the associated Euler–Lagrange equation takes

$$\begin{aligned} -\,\Delta \mathbf {Q}+\mathbf {Q}-9\mathbf {Q}^2+3|\mathbf {Q}|^2\mathbf {Q}+3|\mathbf {Q}|^2\mathbf {I}=0 \end{aligned}$$

(4.1)

with boundary condition

$$\begin{aligned} \mathbf {Q}(x_1,x_2, +\,\infty )=\mathbf {n}\mathbf {n}-\frac{1}{3}\mathbf {I},\quad \mathbf {Q}(x_1,x_2,-\,\infty )=0. \end{aligned}$$

(4.2)

Considering the spectral decomposition of $\mathbf {Q}$:

$$\begin{aligned} \mathbf {Q}=\mathbf {N}\mathbf {\Lambda }\mathbf {N}^T = \lambda _1 \mathbf {n}_1 \mathbf {n}_1 + \lambda _2 \mathbf {n}_2 \mathbf {n}_2 + \lambda _3 \mathbf {n}_3 \mathbf {n}_3, \quad \mathbf {N}= (\mathbf {n}_1,\mathbf {n}_2,\mathbf {n}_3) ,\,\,\mathbf {\Lambda }=\text {diag}(\lambda _1,\lambda _2,\lambda _3). \end{aligned}$$

Then we can reformulate (4.1) as follows

$$\begin{aligned} -\, \mathbf {N}^T ( \Delta \mathbf {Q}) \mathbf {N}+ \mathbf {\Lambda }- 9 \mathbf {\Lambda }^2 + 3 |\mathbf {Q}|^2 \mathbf {\Lambda }+ 3|\mathbf {Q}|^2 \mathbf {I}= 0, \end{aligned}$$

which gives that for the diagonal elements,

$$\begin{aligned} \begin{aligned}&-\,\Delta \lambda _1- 2 \Delta \mathbf {n}_1 \cdot \mathbf {n}_1 \lambda _1 - 2 \lambda _2 | \mathbf {n}_2 \cdot \nabla \mathbf {n}_1|^2 - 2\lambda _3 |\mathbf {n}_3 \cdot \nabla \mathbf {n}_1|^2 + \lambda _1 - 9\lambda _1^2 + 3|\mathbf {Q}|^2 \lambda _1 + 3|\mathbf {Q}|^2 = 0, \\&-\,\Delta \lambda _2- 2 \Delta \mathbf {n}_2 \cdot \mathbf {n}_2 \lambda _2 - 2 \lambda _1 | \mathbf {n}_1 \cdot \nabla \mathbf {n}_2|^2 - 2\lambda _3 |\mathbf {n}_3 \cdot \nabla \mathbf {n}_2|^2 + \lambda _2 - 9\lambda _2^2 + 3|\mathbf {Q}|^2 \lambda _2 + 3|\mathbf {Q}|^2 = 0, \\&-\,\Delta \lambda _3- 2 \Delta \mathbf {n}_3 \cdot \mathbf {n}_3 \lambda _3 - 2 \lambda _1 | \mathbf {n}_1 \cdot \nabla \mathbf {n}_3|^2 - 2\lambda _2 |\mathbf {n}_2 \cdot \nabla \mathbf {n}_3|^2 + \lambda _3 - 9\lambda _3^2 + 3|\mathbf {Q}|^2 \lambda _3 + 3|\mathbf {Q}|^2 = 0, \end{aligned}\nonumber \\ \end{aligned}$$

(4.3)

and for the off-diagonal elements,

$$\begin{aligned}&\lambda _1 \Delta \mathbf {n}_1 \cdot \mathbf {n}_2 + \lambda _2 \Delta \mathbf {n}_2 \cdot \mathbf {n}_1 + 2 \sum _{k=1}^3 \partial _k (\lambda _1-\lambda _2 ) \cdot (\partial _k \mathbf {n}_1 \cdot \mathbf {n}_2)\nonumber \\&+ 2 \lambda _3 \sum _{k=1}^3 (\partial _k \mathbf {n}_3 \cdot \mathbf {n}_1) ( \partial _k \mathbf {n}_3 \cdot \mathbf {n}_2) = 0, \nonumber \\&\lambda _2 \Delta \mathbf {n}_2 \cdot \mathbf {n}_3 + \lambda _3 \Delta \mathbf {n}_3 \cdot \mathbf {n}_2 + 2 \sum _{k=1}^3 \partial _k (\lambda _2-\lambda _3 ) \cdot (\partial _k \mathbf {n}_2 \cdot \mathbf {n}_3)\nonumber \\&+ 2 \lambda _1 \sum _{k=1}^3 (\partial _k \mathbf {n}_1 \cdot \mathbf {n}_2) ( \partial _k \mathbf {n}_1 \cdot \mathbf {n}_3) = 0, \nonumber \\&\lambda _3 \Delta \mathbf {n}_3 \cdot \mathbf {n}_1 + \lambda _1 \Delta \mathbf {n}_1 \cdot \mathbf {n}_3 + 2 \sum _{k=1}^3 \partial _k (\lambda _3-\lambda _1 ) \cdot (\partial _k \mathbf {n}_3 \cdot \mathbf {n}_1) \nonumber \\&+ 2 \lambda _2 \sum _{k=1}^3 (\partial _k \mathbf {n}_2 \cdot \mathbf {n}_3) ( \partial _k \mathbf {n}_2 \cdot \mathbf {n}_1) = 0. \end{aligned}$$

(4.4)

It follows from (4.2) that

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\lambda _i(x_1,x_2,+\,\infty ) = \frac{2}{3}, \ \mathbf {n}_i(x_1,x_2,+\,\infty ) = \mathbf {n}, \ \lambda _j(x_1,x_2,+\,\infty ) = -\frac{1}{3} \ \forall j \ne i;\\ &{}\lambda _k(x_1,x_2,-\,\infty ) = 0, \ k =1,2,3 \end{array}\right. }\quad \end{aligned}$$

(4.5)

for some $i \in \{1,2,3\}$. Without loss of generality, we assume $i=3$.

The system (4.3)–(4.5) are difficult to solve since it is a system with degree of freedom 5. For this, we assume that the eigenvector of $\mathbf {Q}$ corresponding to the largest eigenvalue is a constant vector, and then we can reduce (4.3)–(4.5) to a PDE system with degree of freedom 3. In this case, we give an affirmative answer to (GDC) proposed in Sect. 2.

Theorem 4.1

The level set of global solutions of (4.1)–(4.2) [or (4.3)–(4.5)] satisfying $\mathbf {n}_3(x_1,x_2,x_3) \equiv \mathbf {n}$ and $\partial _{x_3} \lambda _3>0$ are hyperplanes in $\mathbf {R}^3$. Moreover, $\mathbf {Q}(x_1,x_2,x_3) = S^*(x_3) (\mathbf {n}\mathbf {n}- \frac{1}{3} \mathbf {I})$, where $S^*(x_3)$ solves (2.13).

Using identities like $- \Delta \mathbf {n}_1 \cdot \mathbf {n}_1 = | \nabla \mathbf {n}_1|^2 = | \mathbf {n}_2 \cdot \nabla \mathbf {n}_1 |^2+ |\mathbf {n}_3 \cdot \nabla \mathbf {n}_1 |$, we have

$$\begin{aligned} -\, 2 \Delta \mathbf {n}_i \cdot \mathbf {n}_i \lambda _i -\sum _{j\ne i} 2 \lambda _j | \mathbf {n}_j \cdot \nabla \mathbf {n}_i|^2&= 2 | \nabla \mathbf {n}_i|^2 \lambda _i - \sum _{j\ne i} 2 \lambda _j | \mathbf {n}_j \cdot \nabla \mathbf {n}_i|^2\\&=2 \sum _{j\ne i} (\lambda _i -\lambda _j) | \mathbf {n}_j \cdot \nabla \mathbf {n}_i|^2. \end{aligned}$$

Thus, under the assumption that $\mathbf {n}_3$ is a constant vector, we can reduce (4.3) to

$$\begin{aligned} \Delta \lambda _1&= 2( \lambda _1 - \lambda _2) | \mathbf {n}_2 \cdot \nabla \mathbf {n}_1|^2 + \lambda _1 - 9\lambda _1^2 + 3|\mathbf {Q}|^2 \lambda _1 + 3|\mathbf {Q}|^2, \end{aligned}$$

(4.6)

$$\begin{aligned} \Delta \lambda _2&= 2( \lambda _2 - \lambda _1) | \mathbf {n}_1 \cdot \nabla \mathbf {n}_2|^2 + \lambda _2 - 9\lambda _2^2 + 3|\mathbf {Q}|^2 \lambda _2 + 3|\mathbf {Q}|^2, \end{aligned}$$

(4.7)

$$\begin{aligned} \Delta \lambda _3&= \lambda _3 - 9\lambda _3^2 + 3|\mathbf {Q}|^2 \lambda _3 + 3|\mathbf {Q}|^2. \end{aligned}$$

(4.8)

Yet, the above equations are not independent due to $\sum _{i=1}^3 \lambda _i = 0$. Denote

$$\begin{aligned} (\lambda _1,\lambda _2,\lambda _3) = \left( -\,\frac{S+T}{3}, -\,\frac{S-T}{3}, \frac{2S}{3}\right) . \end{aligned}$$

Then $|\mathbf {Q}|^2 = 2(3S^2+T^2)/9$, and (4.6)–(4.8) are equivalent to

$$\begin{aligned} \Delta S&= S -3S ^2 +T ^2 + \frac{2S(3S^2 +T^2)}{3}, \end{aligned}$$

(4.9)

$$\begin{aligned} \Delta T&= 4| \mathbf {n}_1 \cdot \nabla \mathbf {n}_2|^2 \cdot T +T+ 6ST +\frac{2T(3S^2 +T^2) }{3}, \end{aligned}$$

(4.10)

where (4.9) comes from (4.8), and (4.10) comes from subtracting (4.7) by (4.6). Due to (4.5), the boundary conditions of (S, T) take

$$\begin{aligned} \lim _{x_3 \rightarrow \pm \, \infty } T(x_1,x_2,x_3)= \lim _{x_3 \rightarrow -\,\infty } S(x_1,x_2,x_3) =0, \quad \lim _{x_3 \rightarrow +\,\infty } S(x_1,x_2,x_3) = 1. \end{aligned}$$

(4.11)

To prove Theorem 4.1, we need the following lemmas.

Lemma 4.2

If $S\ge 0$ and T is bounded, then $T\equiv 0$.

Proof

Firstly, we apply $S\ge 0$ and (4.10) to yield

$$\begin{aligned} \Delta T^2 = 2 | \nabla T|^2 + 2T^2\left( 4| \mathbf {n}_1 \cdot \nabla \mathbf {n}_2|^2+1+6S+2S^2 + \frac{2T^2}{3}\right) \ge 0. \end{aligned}$$

That is, $T^2$ is subharmonic. Define

$$\begin{aligned} \varphi (x,r)= {\mathchoice{{-\int }}{{-\int }}{{-\int }}{{-\int }}} _{\partial B(x,r)} { T(y)^2 dy }\quad \text {for}\,\, x \in \mathbf {R}^3,\,r >0\quad \text {and} \quad \varphi ( x,0 ) = T(x)^2. \end{aligned}$$

Taking r derivatives on $\varphi$ gives

$$\begin{aligned} \frac{\partial \varphi }{\partial r} = \frac{2}{4\pi r^2} \int _{B(x,r)} { |\nabla T |^2 + \Delta T \cdot T dy } \ge \frac{2}{4\pi r^2} \int _{B(x,r)}{\Delta T \cdot T dy} , \end{aligned}$$

which along with (4.10) gives

$$\begin{aligned} \frac{\partial \varphi }{\partial r} \ge \frac{2}{4\pi r^2} \int _{B(x,r)}{ T^2\left( 4| \mathbf {n}_1 \cdot \nabla \mathbf {n}_2|^2+ 1+6S+2S^2 + \frac{2T^2}{3} \right) dy} \ge \frac{2}{4\pi r^2} \int _{B(x,r)}{ T^2 \ dy}. \end{aligned}$$

As $T^2$ is subharmonic, we infer

$$\begin{aligned} \frac{\partial \varphi }{\partial r} \ge \frac{2}{4\pi r^2} \cdot \frac{4 \pi r^3}{3} T(x)^2 = \frac{2r}{3} T(x)^2. \end{aligned}$$

Since T is bounded and $| \varphi (x,r) | \le ||T||^2_{\infty }$, it must hold that $T(x) = 0$. $\square$

Lemma 4.3

If $0\le S \le M$ for some $M>0$, then $T\equiv 0$.

Proof

By Lemma 4.2, we only need to prove that T is bounded. Define

$$\begin{aligned} \psi (x,r)= {\mathchoice{{-\int }}{{-\int }}{{-\int }}{{-\int }}} _{\partial B(x,r)} {S(y) \ dy}\quad \text {for}\, x \in \mathbf {R}^3, r >0; \quad \psi (x,0) =S(x). \end{aligned}$$

We claim that $T^2(x) \le 2$. Otherwise, there exists $x^* \in \mathbf {R}^3$ so that $T(x^*)^2 >2$. Taking r derivatives on $\psi$ gives

$$\begin{aligned} \frac{\partial \psi }{ \partial r} = \frac{1}{4\pi r^2} \int _{B(x^*,r)}{\Delta S(y)} dy. \end{aligned}$$

Applying (4.9) and the fact that $T^2$ is subharmonic, we obtain

$$\begin{aligned} \begin{aligned} \frac{\partial \psi }{ \partial r}&= \frac{1}{4\pi r^2} \int _{B(x^*,r)}{\left( S -3S ^2 +T ^2 + \frac{2S(3S^2 +T^2)}{3} \right) } dy \\&\ge \frac{1}{4\pi r^2} \int _{B(x^*,r)}{ (T^2 - 1) dy } \ge \frac{1}{4\pi r^2} \cdot \frac{4\pi r^3}{3} (T(x^*)^2 -1) \ge \frac{r}{3}. \end{aligned} \end{aligned}$$

Here we used $S-3S^2 + 2S^3 \ge -1$ and $S\ge 0$. As $0 \le S \le M$ and $|\psi (x,r)| \le M$, we get

$$\begin{aligned} 2M \ge |\psi (x^*,6M+2) - \psi (x^*, 2)| = 6M| \psi _r(x^*,\xi ) | \ge 6M \cdot \frac{\xi }{3} > 2M \end{aligned}$$

for some $\xi \in [2, 6M+2]$. This is a contradiction. Hence, T is bounded. $\square$

Proof of Theorem 4.1

The monotonic assumption on $\lambda _3 = 2S/3$ implies $\partial _{x_3} S >0$, which along with the boundary condition

$$\begin{aligned} \lim _{x_3 \rightarrow -\,\infty } S(x_1,x_2,x_3) =0, \,\,\lim _{x_3 \rightarrow +\,\infty } S(x_1,x_2,x_3) = 1. \end{aligned}$$

gives $0\le S \le 1$. Lemma 4.3 yields $T\equiv 0$. Consequently, we can reduce (4.9)–(4.11) to

$$\begin{aligned} \Delta S = S -3S ^2 +2S^3 \end{aligned}$$

(4.12)

with the boundary conditions

$$\begin{aligned} \lim _{x_3 \rightarrow -\,\infty } S(x_1,x_2,x_3) =0, \,\,\lim _{x_3 \rightarrow +\,\infty } S(x_1,x_2,x_3) = 1, \end{aligned}$$

and yield

$$\begin{aligned} \mathbf {Q}= \sum _{i=1}^3 \lambda _i \mathbf {n}_i \mathbf {n}_i = -\frac{S}{3} \mathbf {n}_1\mathbf {n}_1 -\frac{S}{3} \mathbf {n}_2\mathbf {n}_2 + \frac{2S}{3} \mathbf {n}_3\mathbf {n}_3 = S\left( \mathbf {n}_3\mathbf {n}_3 - \frac{1}{3} \mathbf {I}\right) = S\left( \mathbf {n}\mathbf {n}- \frac{1}{3} \mathbf {I}\right) . \end{aligned}$$

Replacing S by $(u+1)/2$ and x by $x/\sqrt{2}$, we can reformulate (4.12) as

$$\begin{aligned} \Delta u = u^3 - u\ \end{aligned}$$

(4.13)

with

$$\begin{aligned} \partial _3 u >0,\quad \lim _{x_3 \rightarrow \pm \, \infty } u(x_1,x_2,x_3) = \pm \,1. \end{aligned}$$

By De Giorgi’s conjecture for (4.13), the level set of u are hyperplanes. As $S= (u+1)/2$, we deduce that the level set of S are hyperplanes. It further implies

$$\begin{aligned} S(x,y,z) = S^*(z), \quad \mathbf {Q}(x,y,z) = S^*(z) \left( \mathbf {n}\mathbf {n}- \frac{1}{3} \mathbf {I}\right) , \end{aligned}$$

where $S^*$ solves (2.13), and the level set of each component of $\mathbf {Q}$ are hyperplanes. $\square$

Finally, under the assumption that $\mathbf {n}_3$ is a constant vector $\mathbf {n}$ as Theorem 4.1, we give a criterion that $\mathbf {Q}$ is uniaxial, i.e. $\mathbf {Q}= S (\mathbf {n}\mathbf {n}-\frac{1}{3} \mathbf {I})$, which may be independent of interest.

Theorem 4.4

Let

$$\begin{aligned} \mathbf {Q}= -\,\frac{S+T}{3}\mathbf {n}_1 \mathbf {n}_1 - \frac{S-T}{3}\mathbf {n}_2 \mathbf {n}_2 + \frac{2S}{3} \mathbf {n}\mathbf {n}\end{aligned}$$

be the solution of (4.1)–(4.2) [ or equivalently (4.4), (4.9)–(4.11)], where $\mathbf {n}$ is a constant vector. Assume that $S\ge 0$ and there exist $0<\alpha \cdot \ln 4 < 1$, a constant C and $x^*\in \mathbf {R}^3$ so that

$$\begin{aligned} |A(r)| \le C e^{\alpha \cdot \ln ^2(r) } \quad \text { for }\, r\gg 1, \end{aligned}$$

(4.14)

where $A(r)= {\mathchoice{{-\int }}{{-\int }}{{-\int }}{{-\int }}} _{B(x^*,r)}{S(y)dy }$. Then $T \equiv 0$ and $\mathbf {Q}= S(\mathbf {n}\mathbf {n}-\frac{1}{3}\mathbf {I})$, where S solves (4.12) with boundary condition (4.11).

Remark 4.1

• (a) The non-negative assumption on $\lambda _3 =2S/3$ is natural since $\lambda _3 (x,y,z)$ is always the largest eigenvalue of $\mathbf {Q}(x,y,z)$.

• (b) If S is monotonic in $x_3$, then we have $0\le S \le 1$. The boundedness of S leads to that A(r) is bounded. Hence, we can apply Theorem 4.1 to yield $T\equiv 0$.

• (c) The condition (4.14) states that T is trivial unless the growth rate of the average of S in some ball B(x, r) is faster than any polynomial growth.

Proof

By Lemma 4.2, it suffices to prove that T is bounded. Without loss of generality, we take $x^*=0$. For $A> 0$ to be determined later, there exists a constant $B > 0$ so that

$$\begin{aligned} S - 3S^2 + 2S^3 \ge A\cdot S -B. \end{aligned}$$

(4.15)

If T is unbounded, then there exists $x_0$ depending on A such that $T(x_0)^2> B>0$. As S is non-negative, $T^2$ is subharmonic from the proof of Lemma 4.2.

Let $\psi (x,r)$ be as in Lemma 4.3 and $D = B(x_0, r)$. Then we get by (4.15) that

$$\begin{aligned} \begin{aligned} \frac{\partial \psi }{ \partial r} \Big | _{x=x_0}&\ge \frac{1}{4\pi r^2} \int _D{\left( S -3S ^2 +T ^2 + \frac{2S(3S^2 +T^2)}{3} \right) } dy\\&\ge \frac{1}{4\pi r^2} \int _D {\left( AS - B +T ^2 \left( 1+\frac{2S}{3}\right) \right) } dy \\&\ge \frac{1}{4\pi r^2}\left( \int _D\ AS\ dy + \int _{B(x_0,r)}{ (T^2 - B) dy}\right) \\&\ge \frac{A}{4\pi r^2} \int _D S dy + \frac{1}{4\pi r^2}\frac{4\pi r^3}{3}(T(x_0)^2 - B) \\&> \frac{A}{4\pi r^2} \int _DS dy \ge 0. \end{aligned} \end{aligned}$$

(4.16)

Hence, $\psi (x_0,r)$ is strictly monotonic. We drop the notation $x_0$ in $\psi$ for convenience. Note that

$$\begin{aligned} \int _{B(x_0,r) }{S(y)\ dy } = \int _0^r { 4\pi s^2 \psi (s)\ ds }, \end{aligned}$$

which along with (4.16) gives

$$\begin{aligned} \frac{\partial \psi }{ \partial r} \ge \frac{A}{r^2} \int _0^r { s^2 \psi (s)\ ds } \quad \text {for any}\,\, r> 0. \end{aligned}$$

(4.17)

For any $r > 0$, there exists $\xi \in [2r,4r]$ so that $\psi (4r)- \psi (2r) = 2r \cdot \psi _r(\xi )$. As $\psi$ is monotonic, we infer from (4.17) that

$$\begin{aligned} \psi (4r)\ge & {} \psi (4r)- \psi (2r) = 2r \cdot \psi _r(\xi ) \ge 2r \frac{A}{\xi ^2} \int _0^\xi { s^2 \psi (s)\ ds }\nonumber \\\ge & {} 2r \frac{A}{\xi ^2} \frac{\xi }{2} \left( \frac{\xi }{2}\right) ^2\psi \left( \frac{\xi }{2}\right) \ge \frac{A r^2}{2} \psi (r), \end{aligned}$$

(4.18)

which implies

$$\begin{aligned} \psi (4^n) \ge \left( \frac{A}{2}\right) ^n 4^{n(n-1)}\psi (1) = \left( \frac{A}{8}\right) ^n 4^{n^2}\psi (1). \end{aligned}$$

Define $M(x,r) := \int _ {B(x,r)}S(y) \ dy$. Then

$$\begin{aligned} M(x_0,2\cdot 4^{n}) = \int _0^{2\cdot 4^n} { 4\pi s^2 \psi (s)\ ds } > 4\pi (4^{n})^3 \psi (4^n) \ge 4\pi (8A)^n 4^{n^2} \psi (1). \end{aligned}$$

For sufficiently large n so that $|x_0| \le 2\cdot 4^n$, we have

$$\begin{aligned} M(0,4^{n+1} ) \ge M(x_0,2\cdot 4^{n})\ge 4\pi (8A)^n 4^{n^2} \psi (1). \end{aligned}$$

Now we choose $A> 1$ (therefore, $x_0$ can be determined) and combine (4.14) to obtain

$$\begin{aligned} e^{\ln 4 \cdot n^2}\psi (1) = 4^{n^2} \psi (1) < M(0,4^{n+1} ) \le C 4^{3n} e^{\alpha \ln ^2(4^{n+1})} = C4^{3n}e^{(\alpha \cdot \ln ^2 4) (n+1)^2}. \end{aligned}$$

Since $\alpha \cdot \ln 4 <1$, we reach a contradiction as $n\rightarrow +\,\infty$. Hence, T is bounded. $\square$

5 Local minimizer in 1-D for $L\ne 0$

In this section, we study the local minimizer of (2.4) with $L\ne 0$. However, we consider $\mathbf {Q}$ in the diagonal form (2.8) with the homeotropic anchoring. In this case, (S, T) satisfies the following Euler–Lagrange equation

$$\begin{aligned} \left\{ \begin{array}{l} (1+L)S^{\prime \prime }= S - 3S^2 + T^2 + \frac{2S(3S^2 +T^2)}{3} , \\ T^{\prime \prime }= T + 6ST + \frac{2T(3S^2 +T^2)}{3}, \end{array}\right. \end{aligned}$$

(5.1)

with the boundary condition

$$\begin{aligned} S(-\,\infty ) = T(\pm \, \infty ) = 0,\quad S(+\,\infty ) =1. \end{aligned}$$

(5.2)

Theorem 5.1

For all $L> -1$, the ODE system (5.1) and (5.2) has only one solution $T(x) \equiv 0,\,S(x) = S^*(\frac{x}{\sqrt{1+L}})$, where $S^*$ is the solution of the following equation

$$\begin{aligned} -\,S^{\prime \prime } + S - 3S^2 + 2S^3 = 0, \quad S(-\,\infty )= 0,\, S(+\,\infty ) = 1. \end{aligned}$$

(5.3)

Remark 5.1

For $L>0$, the solution (0, S) is unstable by Theorem 2.2. Thus, the stable equilibrium $\mathbf {Q}$ of (2.4) subject to the homeotropic anchoring can not be of the diagonal form (2.8) when $L>0$.

Lemma 5.2

Let (S, T) be the solution of (5.1)– (5.2). Then $S\le 1$ and $S^{\prime }(+\,\infty ) = 0$.

Proof

We first prove that $S\le 1$. Otherwise, the boundary condition $|S(\pm \, \infty )|\le 1$ and the fact that $S\in C^2$ implies that there exists a global maximum point $x^*$ of S. Consequently, $S(x^*) > 1$ and $S^{\prime \prime }(x^*) \le 0$. Using (5.1), we find

$$\begin{aligned} (1+L)S^{\prime \prime }(x^*) =\left( S - 3S^2 + 2 S^3 + T^2\left( 1+ \frac{2S}{3}\right) \right) (x^*)>0 \end{aligned}$$

due to $S(x^*) > 1$ and $S - 3S^2 + 2 S^3 = S(S-1)(2S-1 )$. This contradicts with $S^{\prime \prime }(x^*) \le 0$. Hence, $S\le 1$.

Next, we show that $S^{\prime }(+\,\infty ) = 0$. From the boundary condition $T(+ \infty ) = 0$ and $S(+\,\infty ) =1$, there exists $M> 0$ such that for $x > M$, $S(x)>2/3, | T(x) | < 1/10$, and $1+ 6S + 2(S^2 +T^2/3) >1$.

We claim that there does not exist $x_1,x_2\in [M, +\,\infty )$ so that $\text {sign}(T(x_1)) \ne \text {sign}(T(x_2))$, i.e $\text {sign}(T)$ is preserved in $[M,+\,\infty )$. Otherwise, assume that $\text {sign}(T(x_1)) \ne \text {sign}(T(x_2))$ for $M\le x_1< x_2$. Obviously, there exists $x_0 \in (x_1,x_2)$ such that $T(x_0) = 0$. As $T(x_0) = T(+\,\infty ) = 0$ and $T(x_2) \ne 0$, there exists a critical point $x_3$ of T in the interval $(x_0, +\,\infty )$ with nonzero value. Using (5.1), we have

$$\begin{aligned} T^{\prime \prime }(x_3) = T(x_3) \left( 1+ 6S(x_3) + \frac{2\big (3S(x_3) ^2+T(x_3)^2\big )}{3} \right) , \end{aligned}$$

which along with the fact $1+ 6S(x_3) + 2\big (S(x_3)^2 +T(x_3)^2/3\big ) >0$ gives

$$\begin{aligned} \text {sign}(T^{\prime \prime }(x_3) ) = \text {sign}( T(x_3) ). \end{aligned}$$

(5.4)

However, if $x_3$ is a maximum point, then $T(x_3) >0$ and $T^{\prime \prime }(x_3)\le 0$, which contradicts with (5.4). Similarly, if $x_3$ is a minimum point, then $T(x_3)<0$ and $T^{\prime \prime }(x_3)\ge 0$, which also contradicts with (5.4). Hence, $\text {sign}(T)$ is preserved in $[M,+\,\infty )$.

Without loss of generality, assume $T(x)\ge 0$ for $x \in [M, + \infty )$. Accordingly, T is a convex function in this interval, and thus $T(+\,\infty ) = 0$ gives $T^{\prime }(+ \infty ) =0$. Subtracting the first equation of (5.1) by $\frac{1}{3}\cdot$ the second equation of (5.1), we obtain

$$\begin{aligned} \left( (1+L)S - \frac{T}{3}\right) ^{\prime \prime } = \left( S- \frac{T}{3}\right) \left( 1 - 3S-3T+ 2\left( S^2 + \frac{T^2}{3}\right) \right) \le 0 \end{aligned}$$

due to $1/10 > T(x)\ge 0,\, S(x)\ge 2/3$ for $x>M$. That is, $(1+L)S - \frac{T}{3}$ is concave on $[M,+\,\infty )$. Consequently, the boundary condition $((1+L)S-\frac{T}{3})(+\,\infty ) = 1+ L$ shows that $( (1+L)S- \frac{T}{3} )^{\prime }(+\,\infty ) = 0$. Combining it with $T^{\prime }(+\,\infty ) = 0$ yields $S'(+\,\infty ) = 0$. $\square$

Lemma 5.3

If $T(x_0) =0, S(x_0) =0$ for some $-\,\infty \le x_0 <+\,\infty$ and $S(x) \ge 0$ for $x> x_0$, then $x_0=-\,\infty$, $T\equiv 0$ and $S(x) = S^*(x/ \sqrt{1+L})$. Here $S^*$ is a solution of (5.3).

Proof

Firstly, we reformulate the second equation of (5.1) as follows

$$\begin{aligned} T^{\prime \prime } = T \left( 1 + 6S+ \frac{2}{3} \big (3S^2 + T^2\big ) \right) . \end{aligned}$$

Due to $1 + 6S+ 2/3(3S^2 + T^2) >0$, we have

$$\begin{aligned} \text {sign}(T^{\prime \prime }) = \text {sign}(T) \quad \text {for}\,\,x > x_0 . \end{aligned}$$

We claim that $T(x) =0$ for $x \ge x_0$. Otherwise, there exists a local critical point of T with nonzero value. A similar argument as in (5.4) yields a contradiction. Hence, $T(x) = 0$ for $x \ge x_0$.

Next, we show that $x_0 = -\,\infty$. Actually, if $x_0$ is finite, we have $T^{\prime }(x_0) =0$, and (5.1) can be reduced to

$$\begin{aligned} (1+L)S^{\prime \prime } = S - 3S^2 + 2S^3 \quad \text {for}\,\, x \ge x_0 \end{aligned}$$

(5.5)

with $S(x_0) = 0, S(+\,\infty ) =1$. Multiplying $2S^{\prime }$ on both side of (5.5) and then integrating them on $[x_0, a]$ for some finite $a>x_0$, we obtain

$$\begin{aligned} (1+L) [ ( S^{\prime }( a) )^2 - (S^{\prime }(x_0)^2 ) ]&= ( S^2 - 2 S^3 + S^4)(a) - ( S^2 - 2 S^3 + S^4)(x_0)\\&= ( S^2 - 2 S^3 + S^4)(a). \end{aligned}$$

Recall that $S^{\prime }(+\,\infty ) =0$ due to Lemma 5.2 and $S(+\,\infty ) = 1$. Letting $a \rightarrow +\,\infty$ gives $- S^{\prime }(x_0)^2 = 0$. Thus, we have

$$\begin{aligned} S(x_0) = T(x_0) = S^{\prime }(x_0) = T^{\prime }(x_0) = 0, \end{aligned}$$

which implies that $S,T \equiv 0$, which contradicts with (5.2). Hence, $x_0=-\,\infty$.

Thus, $T \equiv 0$ and S is a solution of (5.5). By a scaling, we conclude $S(x) = S^*(x/\sqrt{1+L} )$.$\square$

Proof of Theorem 5.1

We only need to prove that $T\equiv 0$. If $S(x)\ge 0$ for $x \in \mathbf {R}$, we may take $x_0 = -\,\infty$ and apply Lemma 5.3 to yield that $T\equiv 0$.

Otherwise, there exists some $x_0$ such that $S(x_0) < 0$. We aim to derive a contraction. Due to $S(+\,\infty ) =1$, we can find some $x>x_0$ so that $S(x) = 0$. Consequently, we can define $x_1 := \sup \{ x : x> x_0, S(x) = 0 \}$. Obviously, $x_0< x_1 <+\,\infty$ and

$$\begin{aligned} S(x_1) = 0, \ S^{\prime }(x_1) \ge 0 , \ S(x ) \ge 0 \ \forall x \ge x_0. \end{aligned}$$

We claim $T(x_1) \ne 0$. Otherwise, Lemma 5.3 shows that $T \equiv 0$ and $x_1 = -\,\infty$, which contradicts that $x_1$ is finite.

On the other hand, from (5.1), we have

$$\begin{aligned} (1+L)S^{\prime \prime }(x_1) = T^2(x_1)> 0 \Longrightarrow S^{\prime \prime }(x_1) > 0, \end{aligned}$$

where we used $1+ L >0$. As a result, there exists $\delta > 0$ so that for $x \in (x_1,x_1+ \delta )$, we have $S^{\prime }(x) >S^{\prime }(x_1) \ge 0$.

Let $A \triangleq \{x>x_1 : S^{\prime }(x) = 0 \}$. We then define $x_2$ as follows

$$\begin{aligned} x_2\triangleq {\left\{ \begin{array}{ll} +\,\infty &{}\quad \ if \ A = \emptyset ; \\ \inf \{x \in A\} &{}\quad otherwise .\\ \end{array}\right. } \end{aligned}$$

Then we have $S^{\prime }(x_2) = 0$. By definition of $\delta$ and $x_2$, we know that $x_2 \ge x_1 + \delta > x_1$. Next, we reformulate the first equation of (5.1) as follows

$$\begin{aligned} (1+L)S^{\prime \prime } = S - 3S^2 + 2 S^3 + T^2\left( 1+ \frac{2S}{3}\right) . \end{aligned}$$

Multiplying $2S^{\prime }$ on both sides and then integrating them from $x_1$ to $x_2$ (if $x_2=+\,\infty$, we first integrate both side from $x_1$ to a and then let $a \rightarrow +\,\infty$), we obtain

$$\begin{aligned} \int _{x_1}^{x2} {2(1+L)S^{\prime } S^{\prime \prime } dx } = \int _{x_1}^{x2} { 2S^{\prime }( S - 3S^2 + 2 S^3 ) dx} + \int _{x_1}^{x2}{ 2S^{\prime }T^2\left( 1+ \frac{2S}{3}\right) dx } . \end{aligned}$$

Using $S^{\prime }(x_2) = 0$ and $S(x_1) = 0$, we obtain

$$\begin{aligned} -\,(1+L)S^{\prime 2}(x_1) = S^2(S-1)^2(x_2) + \int _{x_1}^{x2} { 2S^{\prime }T^2\left( 1+ \frac{2S}{3}\right) dx }. \end{aligned}$$

(5.6)

By definition of $x_2$, $S(x)^{\prime } > 0 \ \forall x \in (x_1,x_2)$, which further implies that $S(x) > S(x_1) = 0 \ \forall x \in (x_1,x_2]$. Now, the left hand side of (5.6) is non-positive, the first and second term on the right hand side are non-negative. Hence, we conclude

$$\begin{aligned} S^{\prime } (x_1) = 0, \, S(x_2) = 1,\ T(x)=0 \ \forall x \in (x_1 , x_2 ) . \end{aligned}$$

Hence $T^{\prime }(x) = 0$ for $x \in (x_1,x_2)$. The continuity of $T,T^{\prime }$ implies that $T^{\prime }(x_1) = T(x_1) = 0$. Therefore

$$\begin{aligned} S^{\prime }(x_1) = S(x_1) = T^{\prime }(x_1) = T(x_1) = 0 . \end{aligned}$$

Thus, $S \equiv T \equiv 0$, which contradicts with (5.2). $\square$