Phase transition of an anisotropic Ginzburg–Landau equation

Abstract

We study the effective geometric motions of an anisotropic Ginzburg–Landau equation with a small parameter $\varepsilon >0$ which characterizes the width of the transition layer. For well-prepared initial datum, we show that as $\varepsilon$ tends to zero the solutions will develop a sharp interface limit which evolves under mean curvature flow. The bulk limits of the solutions correspond to a vector field ${\textbf{u}}(x,t)$ which is of unit length on one side of the interface, and is zero on the other side. The proof combines the modulated energy method and weak convergence methods. In particular, by a (boundary) blow-up argument we show that ${\textbf{u}}$ must be tangent to the sharp interface. Moreover, it solves a geometric evolution equation for the Oseen–Frank model in liquid crystals.

Appendices

Appendix A: Proof of Proposition 1.1

Proof of Proposition 1.1

We first recall that $\sigma =1$ (cf. (2.5)), $I_0\subset \Omega$ is the initial interface and $\eta _0$ is the cut-off function in (2.12). Then we define

$$\begin{aligned} s_\varepsilon (x):= \eta _0\left( x \right) \theta \left( \frac{d_{I_0}(x)}{\varepsilon }\right) +\Big (1-\eta _0\left( x \right) \Big ){\textbf{1}}_{\Omega ^+_0}, \end{aligned}$$

(A.1)

where $\theta (z)$ is the solution of the ODE

$$\begin{aligned} -\theta ''(z)+f'(\theta )=0,\quad \theta (-\infty )=0,~\theta (+\infty )=1. \end{aligned}$$

(A.2)

We note that $d_{I_0}$ is Lipschitz continuous in $\Omega$, and thus by Rademacher’s theorem we have $|\nabla d_{I_0}|\leqslant 1$ a.e. in $\Omega$. Recalling (1.19), we define

$$\begin{aligned} {\textbf{u}}_\varepsilon ^{in} (x):=s_\varepsilon (x) {\textbf{u}}^{in}(x). \end{aligned}$$

(A.3)

One can verify that ${\textbf{u}}_\varepsilon ^{in}\in W^{1,2}_0(\Omega )\cap L^\infty (\Omega )$, $\Vert {\textbf{u}}_\varepsilon ^{in}\Vert _{L^\infty (\Omega )}\leqslant 1$, and

$$\begin{aligned} {\textbf{u}}_\varepsilon ^{in}=\left\{ \begin{array}{ll} {\textbf{u}}^{in} &{} \quad \text {if}~ x\in \Omega ^+_0\backslash B_{2\delta _0}(I_0),\\ \theta \left( \frac{d_{I_0}}{\varepsilon }\right) {\textbf{u}}^{in} &{} \quad \text {if}~ x\in B_{\delta _0}(I_0),\\ 0&{}\quad \text {if}~ x\in \Omega ^-_0\backslash B_{2\delta _0}(I_0). \end{array} \right. \end{aligned}$$

(A.4)

So the condition (1.14a) is verified. To verify the others, we first compute the modulated energy in (1.7) for the initial datum ${\textbf{u}}_\varepsilon ^{in}$. We write (A.1) as

$$\begin{aligned} s_\varepsilon (x)= \theta \left( \frac{d_{I_0}(x)}{\varepsilon }\right) +{\hat{s}}_\varepsilon (x), \end{aligned}$$

(A.5)

where ${\hat{s}}_\varepsilon (x):=\left( 1-\eta _0\left( x \right) \right) \left( {\textbf{1}}_{\Omega ^+_0} -\theta \left( \frac{d_{I_0}(x)}{\varepsilon }\right) \right)$. Invoking (2.12) and the exponential convergence of $\theta (z)$ as $z\rightarrow \pm \infty$ (cf. (A.2)), we deduce that

$$\begin{aligned} \Vert {\hat{s}}_\varepsilon \Vert _{L^\infty (\Omega )}+\Vert \nabla {\hat{s}}_\varepsilon \Vert _{L^\infty (\Omega )}\leqslant Ce^{-\frac{C}{\varepsilon }}, \end{aligned}$$

(A.6)

for some constant $C>0$ that only depends on $I_0$. By a Taylor’s expansion, we find

$$\begin{aligned} F({\textbf{u}}_\varepsilon ^{in} )=f(\theta +{\hat{s}}_\varepsilon )=f(\theta )+ O(e^{-C/\varepsilon }). \end{aligned}$$

Combining (A.3), (A.5) with (A.6), we obtain

$$\begin{aligned} |\nabla {\textbf{u}}_\varepsilon ^{in} |^2= \varepsilon ^{-2}\theta '^2 +\theta ^2 |\nabla {\textbf{u}}^{in}|^2+O(e^{-C/\varepsilon })(|\nabla {\textbf{u}}^{in}|^2+1). \end{aligned}$$

Note that we have also employed the identities $\partial _{x_i} {\textbf{u}}^{in}\cdot {\textbf{u}}^{in}=0$ a.e. in $\Omega$. Recalling (1.8), we have

$$\begin{aligned} \psi _\varepsilon \Big |_{t=0}=\int _0^{\theta +{\hat{s}}_\varepsilon }\sqrt{2f(s)}\, ds: \Omega \mapsto [0,1]. \end{aligned}$$

(A.7)

So we can compute

$$\begin{aligned}&\frac{\varepsilon }{2} \left| \nabla {\textbf{u}}_\varepsilon ^{in}\right| ^2+\frac{1}{\varepsilon } F({\textbf{u}}_\varepsilon ^{in})-\varvec{\xi }\cdot \nabla \psi _\varepsilon \Big |_{t=0}\nonumber \\ =&\frac{1}{2\varepsilon } \theta '^2 +\frac{1}{\varepsilon } f(\theta )-\varepsilon ^{-1} \varvec{\xi }\cdot {\textbf{n}}_{I_0} \theta ' \sqrt{2f(\theta )}+\frac{\varepsilon }{2}\theta ^2 |\nabla {\textbf{u}}^{in}|^2+O(e^{-C/\varepsilon })(|\nabla {\textbf{u}}^{in}|^2+1). \end{aligned}$$

(A.8)

It follows from (2.10) that $1-\varvec{\xi }\cdot {\textbf{n}}_{I_0}=O(d_I^2)$. So we have

$$\begin{aligned} \varepsilon ^{-1} \varvec{\xi }\cdot {\textbf{n}}_{I_0} \theta ' \sqrt{2f(\theta )}=\varepsilon ^{-1} \theta ' \sqrt{2f(\theta )}+O(e^{-C/\varepsilon })+ \varepsilon ^{-1} O(d_{I_0}^2) \theta ' \sqrt{2f(\theta )}. \end{aligned}$$

Note that the last term can be written as

$$\begin{aligned} \varepsilon ^{-1} O(d_{I_0}^2) \theta ' \sqrt{2f(\theta )}= O(\varepsilon )z^2 \theta '(z) \sqrt{2f(\theta (z))}\big |_{z= \frac{d_{I_0}(x)}{\varepsilon } }. \end{aligned}$$

Substituting the above two equations into (A.8), we find

$$\begin{aligned}&\int _\Omega \left( \frac{\varepsilon }{2} \left| \nabla {\textbf{u}}_\varepsilon ^{in}\right| ^2+\frac{1}{\varepsilon } F({\textbf{u}}_\varepsilon ^{in})-\varvec{\xi }\cdot \nabla \psi _\varepsilon \right) \, dx\nonumber \\&\quad = \int _\Omega \underbrace{\left( \frac{1}{2\varepsilon } \theta '^2 +\frac{1}{\varepsilon } f(\theta )-\frac{1}{\varepsilon } \theta ' \sqrt{2f(\theta ) }\right) }_{=0}\,dx\nonumber \\&\quad \quad +\int _\Omega \frac{\varepsilon }{2}\theta ^2 |\nabla {\textbf{u}}^{in}|^2\,dx+O(e^{-C/\varepsilon })\int _\Omega (|\nabla {\textbf{u}}^{in}|^2+1)\,dx +O(\varepsilon )\qquad \text { at } t=0. \end{aligned}$$

(A.9)

Note that the integrand of the first integral on the right-hand side of (A.9) vanishes due to the identity $\theta '^2(z)=2f(\theta (z))$, which follows from (A.2). Now we turn to the first term in (1.7). Using (A.6) we can estimate

$$\begin{aligned} |{\text {div}}{\textbf{u}}_\varepsilon ^{in}|^2\leqslant 2|\nabla \theta \cdot {\textbf{u}}^{in}|^2+2\theta ^2|{\text {div}}{\textbf{u}}^{in}|^2+O(e^{-C/\varepsilon })(1+|{\text {div}}{\textbf{u}}^{in}|^2). \end{aligned}$$

(A.10)

By the exponential decay of $\theta '(z)$ as $z\rightarrow \pm \infty$, we deduce that

$$\begin{aligned} |\nabla \theta \cdot {\textbf{u}}^{in}|={\left\{ \begin{array}{ll} \Big |\frac{d_{I_0} }{\varepsilon }\theta '\left( \frac{d_{I_0} }{\varepsilon }\right) \frac{{\textbf{u}}^{in}\cdot {\textbf{n}}_{I_0}}{d_{I_0} }\Big |\leqslant C\Big | \frac{{\textbf{u}}^{in}\cdot {\textbf{n}}_{I_0}}{d_{I_0}}\Big |\quad \text { in } B_{\delta _0}(I_0)\backslash I_0,\\ \\ \Big |\frac{1}{\varepsilon }\theta '\left( \frac{d_{I_0} }{\varepsilon }\right) {\textbf{u}}^{in}\cdot {\textbf{n}}_{I_0}\Big |\leqslant e^{-\frac{C}{\varepsilon }}\quad \qquad \text { in } \Omega \backslash B_{\delta _0}(I_0). \end{array}\right. } \end{aligned}$$

(A.11)

Using this, (1.19) and Hardy’s inequality (cf. [7]), we find

$$\begin{aligned} \int _\Omega |\nabla \theta \cdot {\textbf{u}}^{in}|^2\, dx&\leqslant C \int _{I_0}\int _{-\delta _0}^{\delta _0} \Big | \frac{{\textbf{u}}^{in}\cdot {\textbf{n}}_{I_0}}{d_{I_0}}\Big |^2\, dr\,d{\mathcal {H}}^{d-1}+C\nonumber \\&\leqslant C\left( \int _\Omega |\nabla {\textbf{u}}^{in}|^2\, dx+1\right) . \end{aligned}$$

(A.12)

Combining this with (A.10) and (A.9) we derive $E_\varepsilon [{\textbf{u}}_\varepsilon ^{in} | I_0]\leqslant C\varepsilon$. Recalling (1.21), we have also obtained (1.14b). To verify (1.14c), we shall compute (1.12) at $t=0$. By (A.7), we see that

$$\begin{aligned} B [{\textbf{u}}_\varepsilon ^{in} | I_0]= 2\int _\Omega \Big ( \tfrac{\chi +1}{2}-\psi _\varepsilon \Big )\eta \circ d_I \, dx. \end{aligned}$$

We shall only give the estimate in $B_{\delta _0}(I_0)\cap \Omega _0^+$ because the one in $B_{\delta _0}(I_0)\cap \Omega _0^-$ follows in the same way, and the one in $\Omega \backslash B_{\delta _0}(I_0)$ is due to (A.6) and the exponential convergence of $\theta (z)$ at $\pm \infty$.

$$\begin{aligned}&\int _{B_{\delta _0}(I_0)\cap \Omega _0^+} |\psi _\varepsilon -1| d_{I}(x)\, dx\Big |_{t=0}\nonumber \\&\overset{(A.7)}{=} \int _{B_{\delta _0}(I_0)\cap \Omega _0^+}\left( \int _{s_\varepsilon (x)}^1 \sqrt{2f(s)}\, ds\right) d_{I}(x)\, dx\Big |_{t=0}\nonumber \\&\overset{ (A.6)}{=}\varepsilon \int _{B_{\delta _0}(I_0)\cap \Omega _0^+}\left( \int _{\theta (\frac{d_{I}(x)}{\varepsilon })}^1 \sqrt{2f(s)}\, ds\right) \frac{d_{I}(x)}{\varepsilon }\, dx\Big |_{t=0}+O(e^{-C/\varepsilon })\leqslant C \varepsilon ^2, \end{aligned}$$

(A.13)

where the last step is due to the exponential decay of $Q(z):=z\int _{\theta (z)}^1\sqrt{2f(s)}\, ds$ as $z\uparrow \infty$. $\square$

Appendix B: Proof of Proposition 2.1

Lemma B.1

The following identity holds:

$$\begin{aligned}&\int \nabla {\textbf{H}}: (\varvec{\xi }\otimes {\textbf{n}}_\varepsilon )\left| \nabla \psi _\varepsilon \right| \, d x -\int (\nabla \cdot {\textbf{H}}) \, \varvec{\xi }\cdot \nabla \psi _\varepsilon \, d x\nonumber \\ =&\,\int \nabla {\textbf{H}}: (\varvec{\xi }-{\textbf{n}}_\varepsilon ) \otimes {\textbf{n}}_\varepsilon \left| \nabla \psi _\varepsilon \right| \, d x+\int {\textbf{H}}_\varepsilon \cdot {\textbf{H}}|\nabla {\textbf{u}}_\varepsilon |\, d x \nonumber \\&+\int \nabla \cdot {\textbf{H}}\left( \frac{\varepsilon }{2} |\nabla {\textbf{u}}_\varepsilon |^2 +\frac{1}{\varepsilon }F ({\textbf{u}}_\varepsilon ) -|\nabla \psi _\varepsilon | \right) \, d x +\int \nabla \cdot {\textbf{H}}( |\nabla \psi _\varepsilon |-\varvec{\xi }\cdot \nabla \psi _\varepsilon )\, d x\nonumber \\&-\int (\nabla {\textbf{H}})_{ij} \,\varepsilon \left( \partial _i {\textbf{u}}_\varepsilon \cdot \partial _j {\textbf{u}}_\varepsilon \right) \, d x +\int \nabla {\textbf{H}}: ({\textbf{n}}_\varepsilon \otimes {\textbf{n}}_\varepsilon )\left| \nabla \psi _\varepsilon \right| \, d x. \end{aligned}$$

(B.1)

Proof

We introduce the stress tensor $(T_\varepsilon )_{ij}:= \left( \frac{\varepsilon }{2} |\nabla {\textbf{u}}_\varepsilon |^2 +\frac{1}{\varepsilon } F ({\textbf{u}}_\varepsilon ) \right) \delta _{ij} - \varepsilon \partial _i {\textbf{u}}_\varepsilon \cdot \partial _j {\textbf{u}}_\varepsilon .$ By (2.20b), we have the identity $\nabla \cdot T_\varepsilon ={\textbf{H}}_\varepsilon |\nabla {\textbf{u}}_\varepsilon |.$ Testing this identity with ${\textbf{H}}$, integrating by parts and using (2.14c), we obtain

$$\begin{aligned} \begin{aligned}&\int {\textbf{H}}_\varepsilon \cdot {\textbf{H}}|\nabla {\textbf{u}}_\varepsilon |\,d x =- \int \nabla {\textbf{H}}:T_\varepsilon \,d x\\&=- \int \nabla \cdot {\textbf{H}}\left( \frac{\varepsilon }{2} |\nabla {\textbf{u}}_\varepsilon |^2 +\frac{1}{\varepsilon } F ({\textbf{u}}_\varepsilon ) \right) \, dx+ \int (\nabla {\textbf{H}})_{ij} \, \varepsilon \left( \partial _i {\textbf{u}}_\varepsilon \cdot \partial _j {\textbf{u}}_\varepsilon \right) d x. \end{aligned} \end{aligned}$$

So adding zero leads to

$$\begin{aligned} \begin{aligned}&\int \nabla {\textbf{H}}: {\textbf{n}}_\varepsilon \otimes {\textbf{n}}_\varepsilon \left| \nabla \psi _\varepsilon \right| d x\\&=\int {\textbf{H}}_\varepsilon \cdot {\textbf{H}}|\nabla {\textbf{u}}_\varepsilon |\, dx+\int \nabla \cdot {\textbf{H}}\left( \frac{\varepsilon }{2} |\nabla {\textbf{u}}_\varepsilon |^2 +\frac{1}{\varepsilon } F ({\textbf{u}}_\varepsilon ) -|\nabla \psi _\varepsilon | \right) \,d x+\int \nabla \cdot {\textbf{H}}|\nabla \psi _\varepsilon |\,d x\\&-\int (\nabla {\textbf{H}})_{ij}\,\varepsilon \left( \partial _i {\textbf{u}}_\varepsilon \cdot \partial _j {\textbf{u}}_\varepsilon \right) d x+\int (\nabla {\textbf{H}}): ({\textbf{n}}_\varepsilon \otimes {\textbf{n}}_\varepsilon )\left| \nabla \psi _\varepsilon \right| d x, \end{aligned} \end{aligned}$$

which yields (B.1). $\square$

Lemma B.2

Under the assumptions of Theorem 1.1, the following identity holds:

$$\begin{aligned} \frac{d}{d t} E&\left[ {\textbf{u}}_\varepsilon | I\right] +\frac{1}{2\varepsilon }\int \left( \varepsilon ^2 \left| \partial _t {\textbf{u}}_\varepsilon \right| ^2-|{\textbf{H}}_\varepsilon |^2\right) \,dx\nonumber \\&+\frac{1}{2\varepsilon }\int \Big | \varepsilon \partial _t {\textbf{u}}_\varepsilon -(\nabla \cdot \varvec{\xi }) D \textrm{d}^F ({\textbf{u}}_\varepsilon ) \Big |^2d x +\frac{1}{2\varepsilon }\int \Big | {\textbf{H}}_\varepsilon -\varepsilon |\nabla {\textbf{u}}_\varepsilon | {\textbf{H}}\Big |^2\,d x\nonumber \\ =&\,\frac{1}{2\varepsilon } \int \Big | (\nabla \cdot \varvec{\xi }) | D \textrm{d}^F ({\textbf{u}}_\varepsilon )|{\textbf{n}}_\varepsilon +\varepsilon |\Pi _{{\textbf{u}}_\varepsilon } \nabla {\textbf{u}}_\varepsilon | {\textbf{H}}\Big |^2\,d x \end{aligned}$$

(B.2a)

$$\begin{aligned}&+\frac{\varepsilon }{2} \int |{\textbf{H}}|^2\left( |\nabla {\textbf{u}}_\varepsilon |^2-|\Pi _{{\textbf{u}}_\varepsilon }\nabla {\textbf{u}}_\varepsilon |^2\right) \,d x -\int \nabla {\textbf{H}}\cdot (\varvec{\xi }-{\textbf{n}}_\varepsilon )^{\otimes 2}\left| \nabla \psi _\varepsilon \right| \,d x \end{aligned}$$

(B.2b)

$$\begin{aligned}&+\int \left( \nabla \cdot {\textbf{H}}\right) \left( \frac{\varepsilon }{2} |\nabla {\textbf{u}}_\varepsilon |^2 +\frac{1}{\varepsilon }F ({\textbf{u}}_\varepsilon ) -|\nabla \psi _\varepsilon | \right) \,d x \end{aligned}$$

(B.2c)

$$\begin{aligned}&+\int \left( \nabla \cdot {\textbf{H}}\right) \left( 1-\varvec{\xi }\cdot {\textbf{n}}_\varepsilon \right) |\nabla \psi _\varepsilon |\, d x+ \int (J_\varepsilon ^1+J_\varepsilon ^2)\, d x, \end{aligned}$$

(B.2d)

where

$$\begin{aligned} J_\varepsilon ^1 :=&\, \nabla {\textbf{H}}: {\textbf{n}}_\varepsilon \otimes {\textbf{n}}_\varepsilon \left( |\nabla \psi _\varepsilon |-\varepsilon |\nabla {\textbf{u}}_\varepsilon |^2\right) \nonumber \\&+\varepsilon \nabla {\textbf{H}}:({\textbf{n}}_\varepsilon \otimes {\textbf{n}}_\varepsilon )\left( |\nabla {\textbf{u}}_\varepsilon |^2-|\Pi _{{\textbf{u}}_\varepsilon } \nabla {\textbf{u}}_\varepsilon |^2\right) \nonumber \\&-\sum _{ij}\varepsilon (\nabla {\textbf{H}})_{ij} \Big ((\partial _i {\textbf{u}}_\varepsilon -\Pi _{{\textbf{u}}_\varepsilon } \partial _i {\textbf{u}}_\varepsilon )\cdot (\partial _j {\textbf{u}}_\varepsilon -\Pi _{{\textbf{u}}_\varepsilon } \partial _j {\textbf{u}}_\varepsilon )\Big ), \end{aligned}$$

(B.3)

$$\begin{aligned} J_\varepsilon ^2:=&- \left( \partial _t \varvec{\xi }+\left( {\textbf{H}}\cdot \nabla \right) \varvec{\xi }+\left( \nabla {\textbf{H}}\right) ^{{\textsf{T}}} \varvec{\xi }\right) \cdot \nabla \psi _\varepsilon . \end{aligned}$$

(B.4)

Proof

We shall employ the Einstein summation convention by summing over repeated indices. Using the energy dissipation law in (2.6) and adding zero, we find

$$\begin{aligned}&\frac{d}{d t} E_\varepsilon [ {\textbf{u}}_\varepsilon | I] +\varepsilon \int |\partial _t {\textbf{u}}_\varepsilon |^2\,d x-\int (\nabla \cdot \varvec{\xi }) {D \textrm{d}^F } ({\textbf{u}}_\varepsilon )\cdot \partial _t {\textbf{u}}_\varepsilon \,d x\nonumber \\ =&\int \left( {\textbf{H}}\cdot \nabla \right) \varvec{\xi }\cdot \nabla \psi _\varepsilon \,d x +\int \left( \nabla {\textbf{H}}\right) ^{{\textsf{T}}} \varvec{\xi }\cdot \nabla \psi _\varepsilon \,d x+ \int J_\varepsilon ^2\, d x. \end{aligned}$$

(B.5)

By the symmetry of $\nabla ^2\psi _\varepsilon$ and the boundary conditions in (2.14c), we have

$$\begin{aligned} \int \nabla \cdot (\varvec{\xi }\otimes {\textbf{H}}) \cdot \nabla \psi _\varepsilon \, d x = \int \nabla \cdot ({\textbf{H}}\otimes \varvec{\xi }) \cdot \nabla \psi _\varepsilon \, d x. \end{aligned}$$

Hence, the first integral on the right-hand side of (B.5) can be rewritten as

$$\begin{aligned}&\int \left( {\textbf{H}}\cdot \nabla \right) \,\varvec{\xi }\cdot \nabla \psi _\varepsilon \, d x\nonumber \\&=\int \nabla \cdot (\varvec{\xi }\otimes {\textbf{H}}) \cdot \nabla \psi _\varepsilon \, d x -\int (\nabla \cdot {\textbf{H}}) \,\varvec{\xi }\cdot \nabla \psi _\varepsilon \, d x\nonumber \\&= \int (\nabla \cdot \varvec{\xi }) \,{\textbf{H}}\cdot \nabla \psi _\varepsilon \, d x +\int (\varvec{\xi }\cdot \nabla ) \,{\textbf{H}}\cdot \nabla \psi _\varepsilon \, d x -\int (\nabla \cdot {\textbf{H}}) \,\varvec{\xi }\cdot \nabla \psi _\varepsilon \,d x. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned}&\frac{d}{d t} E_\varepsilon [ {\textbf{u}}_\varepsilon | I] +\varepsilon \int |\partial _t {\textbf{u}}_\varepsilon |^2\,d x-\int (\nabla \cdot \varvec{\xi }) {D \textrm{d}^F } ({\textbf{u}}_\varepsilon )\cdot \partial _t {\textbf{u}}_\varepsilon \,d x \\ =&\int (\nabla \cdot \varvec{\xi })\, {\textbf{H}}\cdot \nabla \psi _\varepsilon d x+\int (\varvec{\xi }\cdot \nabla ) \,{\textbf{H}}\cdot \nabla \psi _\varepsilon \,d x -\int (\nabla \cdot {\textbf{H}})\, \varvec{\xi }\cdot \nabla \psi _\varepsilon \,d x\\&+\int \nabla {\textbf{H}}: \left( \varvec{\xi }\otimes {\textbf{n}}_\varepsilon \right) \left| \nabla \psi _\varepsilon \right| d x+ \int J_\varepsilon ^2\, d x. \end{aligned} \end{aligned}$$

Now using (B.1) to replace the third and the fourth integrals on the right-hand side of the above equation, we find

$$\begin{aligned}&\frac{d}{d t} E_\varepsilon [ {\textbf{u}}_\varepsilon | I] +\varepsilon \int |\partial _t {\textbf{u}}_\varepsilon |^2\,d x-\int (\nabla \cdot \varvec{\xi }) {D \textrm{d}^F } ({\textbf{u}}_\varepsilon )\cdot \partial _t {\textbf{u}}_\varepsilon \,d x \nonumber \\ =&\int (\nabla \cdot \varvec{\xi }) \,{\textbf{H}}\cdot \nabla \psi _\varepsilon \,d x +\int (\varvec{\xi }\cdot \nabla )\, {\textbf{H}}\cdot \nabla \psi _\varepsilon \,d x +\int \nabla {\textbf{H}}: (\varvec{\xi }-{\textbf{n}}_\varepsilon ) \otimes {\textbf{n}}_\varepsilon \left| \nabla \psi _\varepsilon \right| \,d x\nonumber \\&+\int {\textbf{H}}_\varepsilon \cdot {\textbf{H}}|\nabla {\textbf{u}}_\varepsilon |\,d x + \int \nabla \cdot {\textbf{H}}\left( \frac{\varepsilon }{2} |\nabla {\textbf{u}}_\varepsilon |^2 +\frac{1}{\varepsilon }F ({\textbf{u}}_\varepsilon ) -|\nabla \psi _\varepsilon | \right) \,d x\nonumber \\&+\int \nabla \cdot {\textbf{H}}\left( |\nabla \psi _\varepsilon |-\varvec{\xi }\cdot \nabla \psi _\varepsilon \right) \,d x-\int (\nabla {\textbf{H}})_{ij} \,\varepsilon \left( \partial _i {\textbf{u}}_\varepsilon \cdot \partial _j {\textbf{u}}_\varepsilon \right) \, d x\nonumber \\&+\int \nabla {\textbf{H}}: {\textbf{n}}_\varepsilon \otimes {\textbf{n}}_\varepsilon \left| \nabla \psi _\varepsilon \right| \,d x +\int J_\varepsilon ^2\, dx. \end{aligned}$$

(B.6)

We shall show that $J_\varepsilon ^1$ arises from the second and the third to last integrals by proving the following identity:

$$\begin{aligned} \Pi _{{\textbf{u}}_\varepsilon } \partial _i {\textbf{u}}_\varepsilon \cdot \Pi _{{\textbf{u}}_\varepsilon } \partial _j {\textbf{u}}_\varepsilon =n_\varepsilon ^i n_\varepsilon ^j |\Pi _{{\textbf{u}}_\varepsilon } \nabla {\textbf{u}}_\varepsilon |^2 ~\text { a.e. in }\Omega , \end{aligned}$$

(B.7)

where $(n_\varepsilon ^\ell )_{1\leqslant \ell \leqslant 3}={\textbf{n}}_\varepsilon$. Such an identity holds obviously on the set $\{ x\mid {\textbf{u}}_\varepsilon =0\}$ by (2.22). It also holds on $\{ x \mid g(|{\textbf{u}}_\varepsilon |)>0\}$ due to the following identity which follows from (2.22) and (2.23a):

$$\begin{aligned} \Pi _{{\textbf{u}}_\varepsilon } \partial _i {\textbf{u}}_\varepsilon \cdot \Pi _{{\textbf{u}}_\varepsilon } \partial _j {\textbf{u}}_\varepsilon | {D \textrm{d}^F }({\textbf{u}}_\varepsilon )|^2=\partial _i \psi _\varepsilon ~ \partial _j \psi _\varepsilon =n_\varepsilon ^i ~ n_\varepsilon ^j |\Pi _{{\textbf{u}}_\varepsilon } \nabla {\textbf{u}}_\varepsilon |^2| {D \textrm{d}^F }({\textbf{u}}_\varepsilon )|^2. \end{aligned}$$

On the open set $\{ x\mid |{\textbf{u}}_\varepsilon |>0\}$ which includes $\{ x\mid |{\textbf{u}}_\varepsilon |=1\}$, we deduce from (2.22) and (2.19a) that $\Pi _{{\textbf{u}}_\varepsilon } \partial _j {\textbf{u}}_\varepsilon =(\partial _j |{\textbf{u}}_\varepsilon |) ~{\textbf{u}}_\varepsilon$. By [18, Theorem 4.4] we have $\partial _j |{\textbf{u}}_\varepsilon |=0$ a.e. on $\{ x\mid |{\textbf{u}}_\varepsilon |=1\}$. We thus complete the proof of (B.7).

Now by (B.7) and adding zero, we find

$$\begin{aligned} \begin{aligned}&\nabla {\textbf{H}}: {\textbf{n}}_\varepsilon \otimes {\textbf{n}}_\varepsilon \left| \nabla \psi _\varepsilon \right| - (\nabla {\textbf{H}})_{ij} \,\varepsilon \left( \partial _i {\textbf{u}}_\varepsilon \cdot \partial _j {\textbf{u}}_\varepsilon \right) \\ \overset{(2.22)}{=}&\nabla {\textbf{H}}: {\textbf{n}}_\varepsilon \otimes {\textbf{n}}_\varepsilon \left| \nabla \psi _\varepsilon \right| - \varepsilon (\nabla {\textbf{H}})_{ij}(\Pi _{{\textbf{u}}_\varepsilon } \partial _i {\textbf{u}}_\varepsilon \cdot \Pi _{{\textbf{u}}_\varepsilon } \partial _j {\textbf{u}}_\varepsilon ) \\&- (\nabla {\textbf{H}})_{ij} \,\varepsilon \Big ((\partial _i {\textbf{u}}_\varepsilon -\Pi _{{\textbf{u}}_\varepsilon } \partial _i {\textbf{u}}_\varepsilon )\cdot (\partial _j {\textbf{u}}_\varepsilon -\Pi _{{\textbf{u}}_\varepsilon } \partial _j {\textbf{u}}_\varepsilon )\Big ) \overset{(B.3)}{=} J_\varepsilon ^1~\text { a.e. in }\Omega . \end{aligned} \end{aligned}$$

Using the identities $\nabla \psi _\varepsilon ={\textbf{n}}_\varepsilon |\nabla \psi _\varepsilon |$ and $\nabla {\textbf{H}}:(\varvec{\xi }\otimes \varvec{\xi })=0$ (due to (2.14b)), we merge the second and the third integrals on the right-hand side of (B.6):

$$\begin{aligned} \frac{d}{d t} E_\varepsilon [ {\textbf{u}}_\varepsilon | I]&=-\varepsilon \int |\partial _t {\textbf{u}}_\varepsilon |^2\, dx+\int (\nabla \cdot \varvec{\xi }) {D \textrm{d}^F } ({\textbf{u}}_\varepsilon )\cdot \partial _t {\textbf{u}}_\varepsilon \, dx\nonumber \\&\quad + \int (\nabla \cdot \varvec{\xi }) \,{\textbf{H}}\cdot \nabla \psi _\varepsilon \, dx+\int {\textbf{H}}_\varepsilon \cdot {\textbf{H}}|\nabla {\textbf{u}}_\varepsilon | \, dx -\int \nabla {\textbf{H}}: (\varvec{\xi }-{\textbf{n}}_\varepsilon )^{\otimes 2}\left| \nabla \psi _\varepsilon \right| \, dx\nonumber \\&\quad +\int \left( \nabla \cdot {\textbf{H}}\right) \Big ( \frac{\varepsilon }{2} |\nabla {\textbf{u}}_\varepsilon |^2 +\frac{1}{\varepsilon }F ({\textbf{u}}_\varepsilon ) -|\nabla \psi _\varepsilon | \Big )\, dx\nonumber \\&\quad +\int (\nabla \cdot {\textbf{H}}) \left( 1-\varvec{\xi }\cdot {\textbf{n}}_\varepsilon \right) |\nabla \psi _\varepsilon |\, dx+ \int (J_\varepsilon ^1+J_\varepsilon ^2) \, dx. \end{aligned}$$

(B.8)

Now we complete squares for the first four terms on the right-hand side of (B.8). Reordering terms, we have

$$\begin{aligned} -&\varepsilon |\partial _t {\textbf{u}}_\varepsilon |^2+ (\nabla \cdot \varvec{\xi }) D \textrm{d}^F ({\textbf{u}}_\varepsilon )\cdot \partial _t {\textbf{u}}_\varepsilon + (\nabla \cdot \varvec{\xi }) {\textbf{H}}\cdot \nabla \psi _\varepsilon + {\textbf{H}}_\varepsilon \cdot {\textbf{H}}|\nabla {\textbf{u}}_\varepsilon |\\&= -\,\frac{1}{2\varepsilon } \Big ( |\varepsilon \partial _t {\textbf{u}}_\varepsilon |^2 -2(\nabla \cdot \varvec{\xi }) D \textrm{d}^F ({\textbf{u}}_\varepsilon )\cdot \varepsilon \partial _t {\textbf{u}}_\varepsilon +(\nabla \cdot \varvec{\xi })^2 | D \textrm{d}^F ({\textbf{u}}_\varepsilon )|^2 \Big )\\&\quad - \frac{1}{2\varepsilon } |\varepsilon \partial _t {\textbf{u}}_\varepsilon |^2 + \frac{1}{2\varepsilon }(\nabla \cdot \varvec{\xi })^2 | D \textrm{d}^F ({\textbf{u}}_\varepsilon )|^2 + (\nabla \cdot \varvec{\xi }) {\textbf{H}}\cdot \nabla \psi _\varepsilon \\&\quad - \frac{1}{2\varepsilon } \Big ( |{\textbf{H}}_\varepsilon |^2 - 2 \varepsilon |\nabla {\textbf{u}}_\varepsilon | {\textbf{H}}_\varepsilon \cdot {\textbf{H}}+ \varepsilon ^2 |\nabla {\textbf{u}}_\varepsilon |^2 |{\textbf{H}}|^2\Big ) + \frac{1}{2\varepsilon } \Big ( |{\textbf{H}}_\varepsilon |^2 + \varepsilon ^2 |\nabla {\textbf{u}}_\varepsilon |^2 |{\textbf{H}}|^2\Big )\\&= -\,\frac{1}{2\varepsilon } \Big |\varepsilon \partial _t {\textbf{u}}_\varepsilon - (\nabla \cdot \varvec{\xi }) D \textrm{d}^F ({\textbf{u}}_\varepsilon ) \Big |^2 - \frac{1}{2\varepsilon } \Big |{\textbf{H}}_\varepsilon - \varepsilon |\nabla {\textbf{u}}_\varepsilon | {\textbf{H}}\Big |^2 - \frac{1}{2\varepsilon } |\varepsilon \partial _t {\textbf{u}}_\varepsilon |^2 +\frac{1}{2\varepsilon } |{\textbf{H}}_\varepsilon |^2\\&\quad + \frac{1}{2\varepsilon } \Big ( (\nabla \cdot \varvec{\xi })^2 | D \textrm{d}^F ({\textbf{u}}_\varepsilon )|^2 + 2\varepsilon (\nabla \cdot \varvec{\xi }) \nabla \psi _\varepsilon \cdot {\textbf{H}}+ |\varepsilon \Pi _{{\textbf{u}}_\varepsilon }\nabla {\textbf{u}}_\varepsilon |^2 |{\textbf{H}}|^2 \Big )\\&\quad +\frac{\varepsilon }{2} \left( |\nabla {\textbf{u}}_\varepsilon |^2- | \Pi _{{\textbf{u}}_\varepsilon }\nabla {\textbf{u}}_\varepsilon |^2\right) |{\textbf{H}}|^2. \end{aligned}$$

Substituting this identity into (B.8), we arrive at (B.2). $\square$

Proof of Proposition 2.1

The proof here is the same as the case $\mu =0$, done in [40, Lemma 4.4]. This is because the form of the energy dissipation law (2.6) remains unchanged in the presence of the divergence term in (1.2a).

We first estimate the right-hand side of (B.2) by $E_\varepsilon [{\textbf{u}}_\varepsilon | I]$ up to a constant that only depends on $I_t$. Concerning (B.2a), it follows from the triangle inequality that

$$\begin{aligned} \begin{aligned} \int&\left| \frac{1}{\sqrt{\varepsilon }} (\nabla \cdot \varvec{\xi }) | D \textrm{d}^F ({\textbf{u}}_\varepsilon )|{\textbf{n}}_\varepsilon +\sqrt{\varepsilon } |\Pi _{{\textbf{u}}_\varepsilon } \nabla {\textbf{u}}_\varepsilon | {\textbf{H}}\right| ^2d x \\ \leqslant&\int \left| (\nabla \cdot \varvec{\xi }) \left( \frac{1}{\sqrt{\varepsilon }} | D \textrm{d}^F ({\textbf{u}}_\varepsilon )|-\sqrt{\varepsilon } |\Pi _{{\textbf{u}}_\varepsilon } \nabla {\textbf{u}}_\varepsilon | \right) {\textbf{n}}_\varepsilon \right| ^2\,d x\\ {}&+\int \Big |(\nabla \cdot \varvec{\xi })\sqrt{\varepsilon } |\Pi _{{\textbf{u}}_\varepsilon } \nabla {\textbf{u}}_\varepsilon |({\textbf{n}}_\varepsilon -\varvec{\xi })\Big |^2\, d x \\&+ \int \left| \big ((\nabla \cdot \varvec{\xi }) \varvec{\xi }+{\textbf{H}}\big ) \sqrt{\varepsilon } |\Pi _{{\textbf{u}}_\varepsilon } \nabla {\textbf{u}}_\varepsilon | \right| ^2\,d x. \end{aligned} \end{aligned}$$

The first integral on the right-hand side of the above inequality is controlled using (2.26c). Due to the elementary inequality $|\varvec{\xi }- {\textbf{n}}_\varepsilon |^2 \leqslant 2 (1-{\textbf{n}}_\varepsilon \cdot \varvec{\xi })$, the second integral is controlled by (2.26d). The third integral can be treated using the relation ${\textbf{H}}=({\textbf{H}}\cdot \varvec{\xi }) \varvec{\xi }+O(d_I(x,t))$ and (2.15a). So it can be controlled by (2.26e).

The integrals in (B.2b) can be controlled using (2.26c) and (2.26d). The one in (B.2c) is controlled by (2.26a). The first term in (B.2d) can be controlled using (2.26d). It remains to estimate (B.3) and (B.4). The integrals of the last two terms defining $J_\varepsilon ^1$ can be controlled by (2.26b). Therefore,

$$\begin{aligned} \int J_\varepsilon ^1\, dx \overset{(2.26b)}{\leqslant }&\int \nabla {\textbf{H}}: \left( {\textbf{n}}_\varepsilon \otimes ({\textbf{n}}_\varepsilon -\varvec{\xi })\right) \left( |\nabla \psi _\varepsilon |-\varepsilon |\nabla {\textbf{u}}_\varepsilon |^2\right) \, dx\nonumber \\&+\int (\varvec{\xi }\cdot \nabla ) {\textbf{H}}\cdot {\textbf{n}}_\varepsilon \left( |\nabla \psi _\varepsilon |-\varepsilon |\nabla {\textbf{u}}_\varepsilon |^2\right) \, dx+ C E_\varepsilon [{\textbf{u}}_\varepsilon | I]\nonumber \\ \overset{(2.14b)}{\leqslant }&C\Big ( \int |{\textbf{n}}_\varepsilon -\varvec{\xi }| \Big (\varepsilon |\nabla {\textbf{u}}_\varepsilon |^2-\varepsilon |\Pi _{{\textbf{u}}_\varepsilon }\nabla {\textbf{u}}_\varepsilon |^2\Big )\, dx\\&+\int |{\textbf{n}}_\varepsilon -\varvec{\xi }| \left| \varepsilon |\Pi _{{\textbf{u}}_\varepsilon }\nabla {\textbf{u}}_\varepsilon |^2-|\nabla \psi _\varepsilon |\right| \, dx\nonumber \\&+ \int \min \left( d^2_I ,1\right) \left( |\nabla \psi _\varepsilon |+\varepsilon |\nabla {\textbf{u}}_\varepsilon |^2\right) \, dx+ E_\varepsilon [{\textbf{u}}_\varepsilon | I]\Big ). \end{aligned}$$

The first and the third integrals in the last display can be estimated using (2.26b) and (2.26e) respectively. Then we employ (2.23a) to find

$$\begin{aligned} \int J_\varepsilon ^1\, dx \leqslant&\,C \Big ( \int |{\textbf{n}}_\varepsilon -\varvec{\xi }| \left| \varepsilon |\Pi _{{\textbf{u}}_\varepsilon }\nabla {\textbf{u}}_\varepsilon |^2-|\nabla \psi _\varepsilon |\right| \, dx+ E_\varepsilon [{\textbf{u}}_\varepsilon | I]\Big )\nonumber \\ \overset{(2.23a)}{=}&\, C\Big ( \int |{\textbf{n}}_\varepsilon -\varvec{\xi }| \sqrt{\varepsilon } |\Pi _{{\textbf{u}}_\varepsilon }\nabla {\textbf{u}}_\varepsilon | \left| \sqrt{\varepsilon } |\Pi _{{\textbf{u}}_\varepsilon }\nabla {\textbf{u}}_\varepsilon |-\frac{1}{\sqrt{\varepsilon }}| D \textrm{d}^F ({\textbf{u}}_\varepsilon )|\right| \, dx+ E_\varepsilon [{\textbf{u}}_\varepsilon | I]\Big ). \end{aligned}$$

Finally applying the Cauchy-Schwarz inequality and then (2.26c) and (2.26d), we obtain $\int J_\varepsilon ^1 \,dx\leqslant C E_\varepsilon [{\textbf{u}}_\varepsilon | I].$ As for $J_\varepsilon ^2$ (B.4), we employ (2.15c) and (2.26e) to obtain $\int J_\varepsilon ^2\,dx \leqslant C E_\varepsilon [{\textbf{u}}_\varepsilon | I].$ All in all, we have proved that the right-hand side of (B.2) is bounded by $E_\varepsilon [{\textbf{u}}_\varepsilon | I]$ up to a multiplicative constant which only depends on $I_t$. $\square$