Clustering of boundary interfaces for an inhomogeneous Allen–Cahn equation on a smooth bounded domain

Abstract

We consider the inhomogeneous Allen–Cahn equation

$$\begin{aligned} \epsilon ^2\Delta u\,+\,V(y)(1-u^2)\,u\,=\,0\quad \text{ in }\ \Omega , \qquad \frac{\partial u}{\partial \nu }\,=\,0\quad \text{ on }\ \partial \Omega , \end{aligned}$$

where $\Omega$ is a bounded domain in ${\mathbb {R}}^2$ with smooth boundary $\partial \Omega$ and V(x) is a positive smooth function, $\epsilon >0$ is a small parameter, $\nu$ denotes the unit outward normal of $\partial \Omega$. For any fixed integer $N\ge 2$, we will show the existence of a clustered solution $u_{\epsilon }$ with N-transition layers near $\partial \Omega$ with mutual distance $O(\epsilon |\ln \epsilon |)$, provided that the generalized mean curvature $\mathcal {H}$ of $\partial \Omega$ is positive and $\epsilon$ stays away from a discrete set of values at which resonance occurs. Our result is an extension of those (with dimension two) by Malchiodi et al. (Pac. J. Math. 229(2):447–468, 2007) and Malchiodi et al. (J. Fixed Point Theory Appl. 1(2):305–336, 2007).

Appendices

Appendices

The computations of (3.30)

The main objective in this section is to compute the quantities in (3.30). For the cases $n=3,\ldots ,N$, from the expressions of ${\mathbf {b}}_{1n}, {\mathbf {b}}_{2n}$ as in (3.20)–(3.21), we can obtain that

$$\begin{aligned} {\mathbf {b}}_{1n}&\,=\,- 2 e^{- \sqrt{2} x_n} e^{- \sqrt{2} \beta (f_n-f_{n-1})} +O(e^{- 2\sqrt{2} |x_n+\beta (f_n-f_{n-1})|}) \nonumber \\&\,=\,-2\epsilon (N-n+1) e^{- \sqrt{2} x_n} e^{- \sqrt{2} \beta (\mathfrak {f}_{n}-\mathfrak {f}_{n-1}) } +O(e^{- 2\sqrt{2} |x_n+\beta (f_n-f_{n-1})|}), \end{aligned}$$

(A.1)

$$\begin{aligned} {\mathbf {b}}_{2n}&\,=\,- 2 e^{- \sqrt{2} x_n} e^{- \sqrt{2} \beta (f_n-f_{n-2})} +O(e^{- 2\sqrt{2} |x_n+\beta (f_n-f_{n-2})|}) \nonumber \\&\,=\,-2 \epsilon ^2 (N-n+1)(N-n+2) e^{- \sqrt{2} x_n} e^{- \sqrt{2} \beta (\mathfrak {f}_{n}-\mathfrak {f}_{n-2})} +O(e^{- 2\sqrt{2} |x_n+\beta (f_n-f_{n-2})|}). \end{aligned}$$

(A.2)

Specially, when $n=1, 2$, we obtain

$$\begin{aligned} {\mathbf {b}}_{11} \,&= \, H\big (\beta (s+ f_{1} ) \big ) -1\\&\,=\,- 2 e^{- \sqrt{2} \,x_1} e^{-2 \sqrt{2} \,\beta \, f_1 } +O(e^{- 2\sqrt{2} \,|x_1+2 \,\beta \, f_1 |}) \\&\,=\,-2\epsilon N \, e^{- \sqrt{2} \,x_1} e^{\,- 2 \sqrt{2} \,\beta \, \mathfrak {f}_{1} \,} +O(e^{- 2\sqrt{2} \,|x_1+2\,\beta \, f_1 |}),\\ {\mathbf {b}}_{21}&\,=\, H\big (\beta (s+ f_{2} ) \big ) -1\\&\,=\, - 2 e^{- \sqrt{2} \,x_1} e^{- \sqrt{2} \,\beta \, (f_{2}+ f_{1}) } +O(e^{- 2\sqrt{2} \,|x_1+ \,\beta \, (f_{2}+ f_{1}) |})\\&\,= - 2 \epsilon ^2N(N-1) e^{- \sqrt{2} \,x_1} e^{- \sqrt{2} \,\beta \, ( \mathfrak {f}_{2}+ \mathfrak {f}_{1}) } +O(e^{- 2\sqrt{2} \,|x_1+ \,\beta \, (f_{2}+ f_{1}) |}), \end{aligned}$$

and

$$\begin{aligned} {\mathbf {b}}_{12} \,&= \, H\big (\beta (s- f_{1} ) \big ) -1\\&\,=\,- 2 e^{- \sqrt{2} \,x_2} e^{- \sqrt{2} \,\beta \, (f_{2}-f_{1}) } +O(e^{- 2\sqrt{2} \,|x_2+ \,\beta \,(f_{2}-f_{1}) |}) \\&\,=\,-2 ( N -1) \epsilon \, e^{- \sqrt{2} \,x_2} e^{- \sqrt{2} \,\beta \, (\mathfrak {f}_{2}-\mathfrak {f}_{1}) } +O(e^{- 2\sqrt{2} \,|x_2+ \,\beta \,(f_{2}-f_{1}) |}),\\ {\mathbf {b}}_{22}&\,=\, H\big (\beta (s+ f_{1} ) \big ) -1 \\&\,=\, - 2 e^{- \sqrt{2} \,x_2} e^{- \sqrt{2} \,\beta \, (f_{2}+f_{1}) } +O(e^{- 2\sqrt{2} \,|x_2+ \,\beta \, (f_{2}+f_{1}) |})\\&\,= - 2 \epsilon ^2 N(N-1)^2 e^{- \sqrt{2} \,x_2} e^{- \sqrt{2} \,\beta \, (f_{2}+f_{1}) } +O(e^{- 2\sqrt{2} \,|x_2+ 2 \,\beta \, f_{2} |}). \end{aligned}$$

Similarly, for the cases $n =1, \ldots , N-2$, we can also obtain

$$\begin{aligned} {\mathbf {b}}_{3n}&\,=\,2 e^{ \sqrt{2} x_n} e^{- \sqrt{2} \beta (f_{n+1}-f_{n})} +O(e^{- 2\sqrt{2} |x_n+\beta ( f_{n}-f_{n+1})|}) \nonumber \\&\,=\, 2\epsilon (N-n) e^{ \sqrt{2} x_n} e^{- \sqrt{2} \beta (\mathfrak {f}_{n+1}-\mathfrak {f}_n)}+O(e^{- 2\sqrt{2} |x_n+\beta (f_n-f_{n+ 1})|}), \end{aligned}$$

(A.3)

$$\begin{aligned} {\mathbf {b}}_{4n}&\,=\,2 e^{ \sqrt{2} x_n} e^{- \sqrt{2} \beta (f_{n+2}-f_{n})} +O(e^{- 2\sqrt{2} |x_n+\beta (f_{n}-f_{n+2})|})\nonumber \\&\,=\, 2 \epsilon ^2(N-n)(N-n-1) e^{ \sqrt{2} x_n} e^{- \sqrt{2} \beta (\mathfrak {f}_{n+2}-\mathfrak {f}_{n}) } +O \big ( e^{- 2\sqrt{2} |x_n+\beta (f_n-f_{n+2})|} \big ). \end{aligned}$$

(A.4)

And specially, when $n=N-1, N$, recalling the notation $f_{N+1} = + \infty$, we have

$$\begin{aligned} {\mathbf {b}}_{3\, N-1} \,&=\, H\big ( \beta (s-f_N) \big ) +1\\&\, = 2 e^{- \sqrt{2} \,x_{N-1}} e^{- \sqrt{2} \,\beta \,(f_{N}-f_{N-1})} +O(e^{- 2\sqrt{2} \,|x_{N-1}+\,\beta (\,f_{N-1} -f_{N-3}\,)|}) \\ \,&= \, 2 \epsilon \, e^{ \sqrt{2} \,x_{N-1}} e^{- \sqrt{2} \,\beta \,(\mathfrak {f}_{N}-\mathfrak {f}_{N-1})} +O(e^{- 2\sqrt{2} \,|x_{N}+\,\beta (\,f_{N}-f_{N-1}\,)|}),\\ {\mathbf {b}}_{4\, N-1} \,&=\, 0, \end{aligned}$$

and

$$\begin{aligned} {\mathbf {b}}_{3\, N} = {\mathbf {b}}_{4\, N} = 0. \end{aligned}$$

By combining the above formulas, we obtain the following:

Case 1: When $n=3,\ldots ,N-2$, (3.30) holds.

Case 2: When $n=1, 2$, we obtain that

$$\begin{aligned}&{\mathbf {b}}_{11}\,-\,{\mathbf {b}}_{21}\,+\,{\mathbf {b}}_{31}\,-\,{\mathbf {b}}_{41}\\&\quad \,=\, -2\epsilon N \, e^{- \sqrt{2} \,x_1} e^{\,- 2 \sqrt{2} \,\beta \, \mathfrak {f}_{1} \,} +O(e^{- 2\sqrt{2} \,|x_1+2\,\beta \, f_1 |})\\&\,\qquad +\, 2\epsilon (N-1) \, e^{ \sqrt{2} \,x_1} e^{- \sqrt{2} \,\beta \,(\mathfrak {f}_{2}-\mathfrak {f}_1)}+O(e^{- 2\sqrt{2} \,|x_1+\,\beta \,(f_1-f_{2}\,)|})\\&\,\qquad + 2\epsilon ^2 N(N-1) e^{- \sqrt{2} \,x_1} e^{- \sqrt{2} \,\beta \, ( \mathfrak {f}_{2}+ \mathfrak {f}_{1}) } +O(e^{- 2\sqrt{2} \,|x_1+ \,\beta \, (f_{2}+ f_{1}) |})\\&\,\qquad \,- 2 \epsilon ^2 (N-2)(N-1)\, e^{ \sqrt{2} \,x_1} e^{- \sqrt{2} \,\beta \,(\mathfrak {f}_{3} -\mathfrak {f}_{1})} +O(e^{- 2\sqrt{2} \,|x_1+\,\beta \,(f_1-f_{3}\,)|}), \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&{\mathbf {b}}_{12}\,-\,{\mathbf {b}}_{22}\,+\,{\mathbf {b}}_{32}\,-\,{\mathbf {b}}_{42}\\&\quad \,=\, -2 \epsilon ( N -1) \, e^{- \sqrt{2} \,x_2} e^{- \sqrt{2} \,\beta \, (\mathfrak {f}_{2}-\mathfrak {f}_{1}) } +O(e^{- 2\sqrt{2} \,|x_2+ \,\beta \,(f_{2}-f_{1}) |}) \\&\,\qquad +\, 2 \epsilon (N-2) \, e^{ \sqrt{2} \,x_2} e^{- \sqrt{2} \,\beta \,(\mathfrak {f}_{3}-\mathfrak {f}_2)}+O(e^{- 2\sqrt{2} \,|x_2+\,\beta \,(f_2-f_{3}\,)|})\\&\,\qquad + 2 \epsilon ^2N(N-1)^2 e^{- \sqrt{2} \,x_2} e^{- \sqrt{2} \,\beta \, (\mathfrak {f}_{2} +\mathfrak {f}_{1} ) } +O(e^{- 2\sqrt{2} \,|x_2+ 2 \,\beta \, f_{2} |})\\&\,\qquad \,- 2 \epsilon ^2(N-2)(N-3) \, e^{ \sqrt{2} \,x_2} e^{- \sqrt{2} \,\beta \,(\mathfrak {f}_{4} -\mathfrak {f}_{2})} +O(e^{- 2\sqrt{2} \,|x_2+\,\beta \,(f_2-f_{4}\,)|}). \end{aligned} \end{aligned}$$

Case 3: When $n=N-1, N$, we get that

$$\begin{aligned} \begin{aligned}&{\mathbf {b}}_{1 N-1}\,-\,{\mathbf {b}}_{2 N-1}\,+\,{\mathbf {b}}_{3 N-1}\,-\,{\mathbf {b}}_{4N-1}\\&\quad \,=\,-4\epsilon e^{- \sqrt{2} x_{N-1} } e^{- \sqrt{2} \beta (\mathfrak {f}_{N-1}-\mathfrak {f}_{N-2}) } +O(e^{- 2\sqrt{2} |x_{N-1} +\beta (f_{N-1} -f_{N-2})|}) \\&\,\qquad + 2 \epsilon \, e^{ \sqrt{2} \,x_{N-1}} e^{- \sqrt{2} \,\beta \,(\mathfrak {f}_{N}-\mathfrak {f}_{N-1})} +O(e^{- 2\sqrt{2} \,|x_{N}+\,\beta (\,f_{N}-f_{N-1}\,)|})\\&\,\qquad + 12 \epsilon ^2 e^{- \sqrt{2} x_{N-1}} e^{- \sqrt{2} \beta (\mathfrak {f}_{N-1}-\mathfrak {f}_{N-3})} +O(e^{- 2\sqrt{2} |x_{N-1}+\beta (f_{N-1}-f_{N-3})|}), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&{\mathbf {b}}_{1N}\,-\,{\mathbf {b}}_{2N}\,+\,{\mathbf {b}}_{3N}\,-\,{\mathbf {b}}_{4N}\\&\quad \,=\,-2 \epsilon \, e^{- \sqrt{2} \,x_N} e^{- \sqrt{2} \,\beta \,(\mathfrak {f}_{N}-\mathfrak {f}_{N-1})} +O(e^{- 2\sqrt{2} \,|x_N+\,\beta (\,f_N-f_{N-1}\,)|}) \\&\,\qquad + \, 4 \epsilon ^2\, e^{- \sqrt{2} \,x_N} e^{- \sqrt{2} \,\beta \,(\mathfrak {f}_{N}-\mathfrak {f}_{N-2})}+O(e^{- 2\sqrt{2} \,|x_N+\,\beta (\,f_N-f_{N-2}\,)|}). \end{aligned} \end{aligned}$$

Linear problem

We first set

$$\begin{aligned} {\mathcal {S}}=\mathbb {R}\times (0, \ell /\epsilon ), \end{aligned}$$

and provide the following lemma in [13].

Lemma B.1

(Lemma 4.2 in [13]) For a given function $\Phi _*(x, z)\in L^2({\mathcal {S}})$ with

$$\begin{aligned} \int _{\mathbb {R}} \Phi _*(x,z) H_x \, {\mathrm {d}}x\,=\, 0,\quad 0<z<\frac{\ell }{\epsilon }, \end{aligned}$$

let us consider the following problem

$$\begin{aligned} \phi _{*,xx} \, +\, \phi _{*,zz} \, +\, (1-3H^2)\, \phi _{*}\,=\, \Phi _* \quad \text{ in } {\mathcal {S}}, \end{aligned}$$

(B.1)

with the conditions

$$\begin{aligned}&\phi _{*}(x, 0)\,=\, \phi _{*}(x, \ell /{\epsilon }), \quad \phi _{*, z}(x, 0)\,=\, \phi _{*, z}(x, \ell /{\epsilon }), \quad x\in \mathbb {R}, \end{aligned}$$

(B.2)

$$\begin{aligned}&\int _\mathbb {R}{\phi _{*}(x,z) H_x}\,{\mathrm {d}}x =0,\quad 0<z<\frac{\ell }{\epsilon }. \end{aligned}$$

(B.3)

The problem (B.1)–(B.3) has a unique solution $\phi _{*} \in H^2({\mathcal {S}})$. $\square$

Then, by using the above lemma, we can obtain following result:

Lemma B.2

For a given function $\Phi ^*(x, z)\in L^2({\mathcal {S}})$ with

$$\begin{aligned} \int _{\mathbb {R}} \Phi ^*(x,z) H_x \, {\mathrm {d}}x\,=\, 0,\quad 0<z<\frac{\ell }{\epsilon }, \end{aligned}$$

consider the following problem

$$\begin{aligned} \phi ^*_{zz}+ \beta ^{2}\,\Big [\phi ^{*}_{xx}\, +\, (1-3H^2)\phi ^{*}\Big ]=\, \Phi ^* \quad \text{ in } {\mathcal {S}}, \end{aligned}$$

(B.4)

with the conditions

$$\begin{aligned}&\phi ^{*}(x, 0)\,=\, \phi ^{*}(x, \ell /{\epsilon }), \quad \phi ^{*}_z (x, 0)\,=\, \phi ^*_{ z}(x, \ell /{\epsilon }), \quad x\in \mathbb {R}, \end{aligned}$$

(B.5)

$$\begin{aligned}&\int _\mathbb {R}{\phi ^{*}(x,z) H_x}\,{\mathrm {d}}x =0,\quad 0<z<\frac{\ell }{\epsilon }. \end{aligned}$$

(B.6)

There exists a unique solution $\phi ^{*} \in H^2({\mathcal {S}})$ to problem (B.4)–(B.6), which satisfies

$$\begin{aligned} \Vert \phi _{*}\Vert _{H^2( {\mathcal {S}})} \le C \Vert \Phi ^* \Vert _{L^2({\mathcal {S}})}. \end{aligned}$$

(B.7)

Proof

Let

$$\begin{aligned} \phi ^{*}(x,z)= \tilde{\phi }^{*} (x, \iota (z)), \quad \iota (z) = \epsilon ^{-1} \int _0^{\epsilon z} \beta (r) {\mathrm {d}}r. \end{aligned}$$

Here, the map

$$\begin{aligned} \iota : \Big [ 0, \frac{\ell }{\epsilon }\Big )\rightarrow \Big [ 0, \frac{\hat{\ell }}{\epsilon }\Big ), \quad {\tilde{z}} = \iota (z) \end{aligned}$$

is a diffeomorphism, where $\hat{\ell } = \int _0^\ell \beta (r) {\mathrm {d}}r.$

It is easy to derive that

$$\begin{aligned} \phi ^{*}_{z} (x,z) \,=\, \beta \,\tilde{\phi }^{*}_{{\tilde{z}}}(x,{\tilde{z}}), \qquad \phi ^{*}_{zz} (x,z)\,=\, \beta ^2\, \tilde{\phi }^{*}_{{\tilde{z}}{\tilde{z}}}(x,{\tilde{z}}) \,+\, \epsilon \, \beta '\,\tilde{\phi }^{*}_{{\tilde{z}}}(x,{\tilde{z}}), \end{aligned}$$

while differentiation in x does not change. Therefore, problem (B.4)–(B.6) can be rewritten as

$$\begin{aligned} \tilde{\phi }^{*}_{{\tilde{z}}{\tilde{z}}} \,+\, \Big [ \tilde{\phi }^{*}_{xx}\, +\, (1-3H^2) \tilde{\phi }^{*}\Big ]&=\, \tilde{ \Phi }^*\,-\, \epsilon \,\beta ^{-2}\, \beta '\,\tilde{\phi }^{*}_{{\tilde{z}}}\quad \text{ in } \mathbb {R}\times \Big [0,\frac{\hat{\ell }}{\epsilon }\Big ), \end{aligned}$$

(B.8)

$$\begin{aligned} \int _{\mathbb {R}} \tilde{\Phi }^*(x, {\tilde{z}}) H_x \, {\mathrm {d}}x\,=\,&0,\quad 0<{\tilde{z}}< \frac{\hat{\ell }}{\epsilon }, \end{aligned}$$

(B.9)

with the conditions

$$\begin{aligned}&\tilde{\phi }^{*}(x, 0)\,=\, \phi ^{*}(x, \hat{\ell } /{\epsilon }), \quad \tilde{\phi }^{*}_{{\tilde{z}}} (x, 0)\,=\, \tilde{\phi }^*_{ {\tilde{z}} }(x, \hat{\ell }/{\epsilon }), \quad x\in \mathbb {R}, \end{aligned}$$

(B.10)

$$\begin{aligned}&\int _\mathbb {R} {\tilde{\phi }}^*(x,{\tilde{z}}) H_x\,{\mathrm {d}}x=0,\quad 0<{\tilde{z}}<\frac{\hat{\ell }}{\epsilon }. \end{aligned}$$

(B.11)

From Lemma B.1, we can know that problem (B.8)–(B.11) has a unique solution $\tilde{\phi }^{*}(x, {\tilde{z}})$. The result follows by transforming $\tilde{\phi }^{*}(x, \iota (z) )$ into $\phi ^{*}(x, z)$ via change of variables. By using the method of sub-supersolutions, we can get the estimate (B.7). This concludes the proof of the lemma. $\square$