2 Preliminaries

Besides (1.7), (1.8), and (1.11), we recall several results of the case $p \neq 2$ .

Assume $u_{\varepsilon }$ is a minimizer of $E_{\varepsilon }(u,G)$ in W. Denote the disc with x as center and r as radius by $B(x,r)$ or $B_r(x)$ . Let $\lambda , \mu $ be two positive constants which are independent of $\varepsilon $ . If

$$ \begin{align*}\int_{G\cap B(x^{\varepsilon}, 2\lambda \varepsilon)}(1-|u_{\varepsilon}|^2)^2 \leq \mu\varepsilon^2, \end{align*} $$

then $B(x^{\varepsilon },\lambda \varepsilon )$ is called a good disc. Otherwise $B(x^{\varepsilon },\lambda \varepsilon )$ is called a bad disc. When $p>2$ , [Reference Lei17, Remark 2.7] shows that the zeros of $u_{\varepsilon }$ are included in finite nonintersecting bad discs $B(x_i^{\varepsilon },h\varepsilon )$ ( $i=1,2,...,N_1$ ), where $N_1$ and $h>0$ are independent of $\varepsilon $ . As $\varepsilon \to 0$ , there exists a subsequence $x_i^{\varepsilon _k}$ of the center $x_i^{\varepsilon }$ of bad discs such that $x_i^{\varepsilon _k} \to a_i \in \overline {G} (i=1,2,...,N_1).$ Since there may be at least two subsequences that converge to the same point, we denote the limit points by $a_1,a_2,...,a_N (N \leq N_1)$ . Write $\Lambda _j:=\{i; x_i^{\varepsilon _k} \to a_j$ when $k \to \infty \}$ for each j. We can choose $\sigma>0$ suitably small such that $B_\sigma (a_j) \subset G$ when $a_j \in G$ , and $B_\sigma (a_j) \cap B_\sigma (a_m) = \emptyset $ for $j \neq m$ . When $p \in (1,2)$ , [Reference Lei16, Section 3] also presents the same conclusions above. Clearly, these results still hold for the regularized minimizers.

2.1 Case of $p>2$

We now introduce several results in [Reference Lei17] and [Reference Lei18] which will be used later.

Proposition 2.1. [Reference Lei17, Proposition 2.1]

Let $u_{\varepsilon }$ be a minimizer of $E_{\varepsilon }(u,G)$ in W. Then, there exists a constant $C>0$ which is independent of $\varepsilon \in (0,1)$ , such that $E_{\varepsilon }(u_{\varepsilon },G) \leq C\varepsilon ^{2-p}$ .

Proposition 2.2. [Reference Lei17, Proposition 2.2]

There exists a constant $C_0>0$ which is independent of $\varepsilon \in (0,1)$ , such that for any $x,y \in \overline {G}$ ,

$$ \begin{align*}|u_{\varepsilon}(x)-u_{\varepsilon}(y)| \leq C_0\varepsilon^{(2-p)/p}|x-y|^{(p-2)/p}. \end{align*} $$

Proposition 2.3. [Reference Lei17, Theorem 2.6]

All the zeros of $u_{\varepsilon }$ are contained in finite, disjointed bad discs $\{B(x_j^{\varepsilon },h\varepsilon );j=1,2,\ldots ,N_1\}$ . In addition,

$$ \begin{align*}|u_{\varepsilon}(x)|\geq \frac{1}{2}, \quad \forall x \in \overline{G}\setminus \cup_{j=1}^{N_1}B(x_j^{\varepsilon},h\varepsilon). \end{align*} $$

Proposition 2.4. [Reference Lei18, Proposition 2.4]

Write $d_i:= \deg (u_\varepsilon , \partial B(x_i^{\varepsilon },h\varepsilon ))$ . Then for each i, there exists a subsequence $\varepsilon _k$ of $\varepsilon $ such that $d_i$ is independent of $\varepsilon _k$ as long as k is sufficiently large.

Proposition 2.5. [Reference Lei17, Theorem 3.1]

Let $u_\varepsilon $ be a minimizer of $E_{\varepsilon }(u,G)$ in W. Then for any compact subset K of $G\setminus \{a_1,a_2,...,a_N\}$ , there exists a constant $C>0$ which does not depend on $\varepsilon $ , such that $E_{\varepsilon }(u_{\varepsilon },K)\leq C$ .

2.2 Case of $p \in (1,2)$

In [Reference Lei16], the following free energy functional was studied

$$ \begin{align*}E_{\varepsilon,\varrho}(u)= \frac{1}{2}\int_{B_1}|\nabla u|^2 +\frac{1}{4\varepsilon^2}\int_{B_1\setminus B_{\varrho}}(1-|u|^2)^2 +\frac{1}{2\varepsilon^2}\int_{B_{\varrho}}|u|^2, \end{align*} $$

where $B_r=\{x \in \mathbb {R}^2;|x|<r\}$ , $\varepsilon $ and $\varrho $ are small positive parameters. It is associated with the model of superconductivity with normal impurity inclusion such as superconducting-normal junctions. There are two major differences between $E_{\varepsilon ,\varrho }(u)$ and $E_\varepsilon (u,G)$ . The former models an heterogenous superconductor and the latter models the homogenous case. In addition, the domain considered in [Reference Lei16] is the unit disk.

In fact, the domain considered in [Reference Lei16] can be replaced by the more general G and the purpose of using $B_1$ there is for convenience (see also [Reference Ding, Liu and Yu10]). In addition, if $\varrho =0$ , then $E_{\varepsilon ,\varrho }(u)$ becomes $E_\varepsilon (u,G)$ , and we can see the results from the corresponding conclusions of Case I (i.e., $\varrho =O(\varepsilon )$ ) in [Reference Lei16], as long as we replace $B_1$ by G and p approaches the dimension $2$ .

First, by [Reference Lei16, (1–6) in Theorem 1.2], we have

Proposition 2.6. Let $p\in(1,2),\; u_\varepsilon $ be a minimizer of $E_\varepsilon (u,G)$ in W. Then there exists a constant $C>0$ which is independent of $\varepsilon $ , such that $E_\varepsilon (u_\varepsilon ,G)\leq C.$

Next, by [Reference Lei16, Theorem 3.1 and Proposition 2.2], we have

Proposition 2.7. Assume that $p\in(1,2),\;u_{\varepsilon }\in W$ satisfies the (1.4) in the weak sense. Then there is a constant $\rho _0>0$ , such that for $\rho \in (0,\rho _0)$ ,

(2.1) $$ \begin{align} |u_\varepsilon(x)|\geq \frac{1}{2},\quad as \quad x\in G \setminus G^{2\rho\varepsilon}. \end{align} $$

Here, $G^{\rho \varepsilon }:=\{x \in G; dist(x,\partial G)> \rho \varepsilon \}$ . In addition, for any $\rho>0$ , there exists a positive constant $C_1$ which is independent of $\varepsilon $ , such that

(2.2) $$ \begin{align} \|\nabla u_{\varepsilon}(x)\|_{L^{\infty} (B(x,\rho\varepsilon))} \leq C_1\varepsilon^{-1},\quad as \quad x \in G^{\rho\varepsilon}. \end{align} $$

By [Reference Lei16, (3-1) in Proposition 3.2], we have

Proposition 2.8. Let $p\in(1,2),\;u_\varepsilon $ be a minimizer of $E_\varepsilon (u,G)$ in W. Then, there exists a constant $C>0$ which is independent of $\varepsilon \in (0,\varepsilon _0)$ with $\varepsilon _0$ sufficiently small, such that,

$$ \begin{align*}\frac{1}{\varepsilon^2} \int_{G}(1-|u_\varepsilon|^2)^2 \leq C. \end{align*} $$

By [Reference Lei16, (3-10)], we have

Proposition 2.9. Let $p\in(1,2),\;u_\varepsilon $ be a minimizer of $E_\varepsilon (u,G)$ in W. For any given $\sigma>0$ , and $0<\varepsilon \ll \sigma $ , there holds

$$ \begin{align*}|u_{\varepsilon}(x)|\geq \frac{1}{2}, \quad \forall x \in \overline{G}\setminus (\cup_{j=1}^{Card J}B(a_j,\sigma)). \end{align*} $$

The next proposition shows the reversed Hölder inequality, which is crucial in the proofs of (1.14) and (1.15). We now use the same ideas in [Reference Giaquinta11, Chapter 5] to prove this reversed Hölder inequality.

Proposition 2.10. Assume $p \in [3/2,2)$ , $u_{\varepsilon }$ is a minimizer of $E_{\varepsilon }(u,G)$ in W. Then there exists a constant $R_0 \in (0,1/2)$ which is independent of $\varepsilon $ , such that for any $B_R \subset G (2R<R_0)$ , we have

$$ \begin{align*}\left(\int_{B_R}|\nabla u_{\varepsilon}|^{\tilde{q}}\right)^{1/\tilde{q}} \leq C\left(\int_{B_{2R}}(|\nabla u_{\varepsilon}|^2+1) ^{p/2}\right)^{1/p}, \quad \forall \tilde{q} \in [p,p(1+t)), \end{align*} $$

where $C>0$ depends on $R_0,p,\tilde {q}$ , and $t \in (0,1/2)$ only depends on $R_0$ .

Proof . Step 1. Let $Q \subset \mathbb {R}^2$ be a square and $m \in [3/2,2)$ . Suppose

(2.3) $$ \begin{align} \frac{1}{|Q_R|}\int_{Q_R(x_0)}g^m \leq b \left( \frac{1}{|Q_{2R}|} \int_{Q_{2R}(x_0)}g \right)^m \end{align} $$

for each $x_0 \in Q$ and each $0<R<\frac {1}{3}\min \{dist(x_0,\partial Q), R_0\}$ , where $b>1$ and $R_0>0$ are absolute constants. By the same argument of [Reference Giaquinta11, proof of Theorem 1.2], we claim that $g \in L_{loc}^q(Q)$ for $q \in [m,m(1+t))$ and

(2.4) $$ \begin{align} \left(\frac{1}{|Q_R|}\int_{Q_R}g^q \right)^{\frac{1}{q}} \leq C\left(\frac{1}{|Q_{2R}|}\int_{Q_{2R}}g^m\right)^{\frac{1}{m}} \end{align} $$

for $Q_{2R} \subset Q$ , $0<R<R_0$ , where t is a positive constant only depending on b, and $C>0$ depends on $b,m,q$ . In particular, C is blowing up when q approaches $m(1+t)$ .

In fact, when we check [Reference Giaquinta and Modica12, Proposition 5.1], (2.3) implies $\theta =0$ and $f(x) \equiv 0$ . Set

$$ \begin{align*}E(h,s)=\{x \in Q; h(x)>s\}, \quad \alpha_k=(3^2\cdot 4^k)^{1/m}, \quad G(x)=g(x)\|g\|_{L^m(Q)}^{-1}, \end{align*} $$

$$ \begin{align*}\Phi(x)=\alpha_k^{-1}G(x) \ in \ C_k:=\{x \in Q;2^{-k}<dist(x, \partial Q) \leq 2^{-k+1}\}. \end{align*} $$

As in [Reference Giaquinta and Modica12, proof of (5.4)], by the Calderon–Zygmund subdivision argument and an iteration, we still obtain

(2.5) $$ \begin{align} \int_{E(\Phi,T)}\Phi^m \leq aT^{m-1}\int_{E(\Phi,T)}\Phi \end{align} $$

for $T \geq 1$ . According to the result of [Reference Giaquinta and Modica12] (see line 7 in page 167), a in (2.5) can be chosen as

$$ \begin{align*} a=b\left(\frac{5m}{m-1}\right)^{m-1}(30^2\cdot 5m+2^2). \end{align*} $$

In view of $m \in [3/2,2)$ , a can be bounded by an absolute constant $\tilde {a}$ (which is independent of m). Now, (2.5) with $a=\tilde {a}$ is (1.6) in Chapter 5 of [Reference Giaquinta11], which implies that the conditions of [Reference Giaquinta11, Lemma 1.2 in Chapter 5] are satisfied if we write $h(T)=\int _{E(\Phi ,T)}\Phi $ and $H(T) \equiv 0$ . By applying Lemma 1.2 in Chapter 5 of [Reference Giaquinta11], we can deduce (2.4) for $q \in [m,\frac {\tilde {a}}{\tilde {a}-1}m)$ . Set $\frac {\tilde {a}}{\tilde {a}-1}=1+t$ , then

$$ \begin{align*}t=\frac{1}{\tilde{a}-1}. \end{align*} $$

This implies that t is independent of m. In addition, C in (2.4) depends on $m[\tilde {a}m- (\tilde {a}-1)q]^{-1}$ , which implies that C is blowing up when q approaches $m(1+t)$ .

Step 2. Let $y=x\varepsilon ^{-1}$ in $E_{\varepsilon }(u,G)$ and denote $ v_{\varepsilon }(y)=u_{\varepsilon }(x)$ , $G_{\varepsilon }=\{y=x\varepsilon ^{-1};x \in G\}$ . Then

$$ \begin{align*}\begin{array}{ll} E_{\varepsilon}(u_{\varepsilon},G) &\displaystyle =\varepsilon^{2-p}\left(\frac{1}{p} \int_{G_{\varepsilon}} |\nabla v_{\varepsilon}|^p dy+ \frac{1}{4} \int_{G_{\varepsilon}}(1-|v_{\varepsilon}|^2)^2dy \right)\\[3mm] & :=\varepsilon^{2-p}E(v_{\varepsilon},G_{\varepsilon}). \end{array} \end{align*} $$

It is clear that $v_{\varepsilon }$ is also a minimizer of $E(v,G_{\varepsilon })$ .

Let $m=(p+2)/2$ , then $m \in [3/2,2)$ when $p \in [3/2,2)$ . Clearly, $(A+B)^m \leq (2\max \{A,B\})^m \leq 2^m(A^m+B^m)$ as long as $A,B$ are positive. Checking the proof of Theorem 3.1 in Chapter 5 of [Reference Giaquinta11], we find that $c(m)$ in lines 13–14 of page 160 satisfies $c(m) = 2^m \leq 4$ for $p \in [3/2,2)$ . Therefore, $c_4$ in (3.5) (see line 5 of page 161) is independent of p after an iteration (by Lemma 3.1). Next, the Sobolev–Poincare inequality shows that $c_5$ can be chosen as a suitably large absolute constant in view of $p \in [3/2,2)$ . Thus, we also derive a condition which satisfies (2.3) with $b=c_5$ . Using the reversed Hölder inequality (2.4) with $g=(\varepsilon +|\nabla v_\varepsilon |)^{\frac {2p}{2+p}}$ , we know that there exist constants $t, R_0 \in (0,1/2)$ and $C>0$ , such that for any $B_R \subset B_{R_0/2} \subset G$ and $q \in [m,m(1+t))$ , the inequality

$$ \begin{align*}&\displaystyle \left(\frac{1}{|B_{R/\varepsilon}|} \int_{B_{R/\varepsilon}} |\nabla v_{\varepsilon}|^{\frac{2pq}{2+p}}dy \right)^{1/q}\\&\quad\displaystyle \leq \left(\frac{1}{|B_{R/\varepsilon}|} \int_{B_{R/\varepsilon}} (|\nabla v_{\varepsilon}|+\varepsilon)^{\frac{2pq}{2+p}}dy\right)^{1/q}\\&\quad\displaystyle \leq C\left(\frac{1}{|B_{2R/\varepsilon}|} \int_{B_{2R/\varepsilon}} (|\nabla v_{\varepsilon}|+ \varepsilon)^{p}dy\right)^{\frac{2}{p+2}}\end{align*} $$

holds, where $B_{R/\varepsilon }=\{y=x\varepsilon ^{-1};x \in B_R\}$ . Letting $x=y\varepsilon $ and multiplying by $\varepsilon ^{-\frac {2p}{p+2}}$ , we obtain

$$ \begin{align*}\left(\int_{B_R}|\nabla u_{\varepsilon}|^{\frac{2pq}{2+p}}dx\right)^{\frac{p+2}{2pq}} \leq C\left(\int_{B_{2R}} (|\nabla u_{\varepsilon}|^2+1)^{p}dx\right)^{1/p}. \end{align*} $$

Let $\tilde {q}=\frac {2pq}{p+2}$ . Noticing $q \in [m,m(1+t))$ , we can see $\tilde {q} \in [p,p(1+t))$ . And hence the proposition holds.

Clearly, the results above (Propositions 2.6–2.10) still hold for the regularized minimizers.

By [Reference Lei16, Proposition 5.3], we have

Proposition 2.11. Assume $p \in (1, 2)$ approaches the dimension $2$ , and $\tilde {u}_\varepsilon $ is a regularized minimizer. Then, for any compact subset $K \subset G \setminus (\cup _{j=1}^N\{a_j\})$ , there exists a positive constant $C=C(K)$ which is independent of $\varepsilon $ , such that

$$ \begin{align*}\|\frac{1}{\varepsilon^p} (1-|\tilde{u}_\varepsilon|^2)\|_{L^{\infty}(K)} \leq C. \end{align*} $$

3 Proof of Theorem 1.1

In this section, we assume $p> 2$ , and $u_\varepsilon $ is the minimizer of $E_\varepsilon (u,G)$ in W.

Proof of (1.10).

Noting $\alpha \geq 2-\frac {4}{p}$ , using (1.5), the Hölder inequality and Proposition 2.1, we deduce that

$$ \begin{align*}&\displaystyle\int_{G} |\nabla u_\varepsilon|^2(1-|u_\varepsilon|^2)^{\alpha}\\&\quad\leq C\displaystyle\int_{G} |\nabla u_\varepsilon|^2(1-|u_\varepsilon|^2)^{2-\frac{4}{p}}\\&\quad\leq C \left(\displaystyle\int_{G} (1-|u_\varepsilon|^2)^2\right)^{1-\frac{2}{p}} \left(\displaystyle\int_{G} |\nabla u_\varepsilon|^p\right)^{\frac{2}{p}}\\&\quad\leq C\varepsilon^{2(1-\frac{2}{p})+\frac{2}{p}(2-p)} = C.\end{align*} $$

Namely, $|\nabla u_\varepsilon |^2(1-|u_\varepsilon |^2)^{\alpha }$ is bounded in $L^1(G)$ . Thus, there exists a Radon measure $\omega _1$ such that

$$ \begin{align*}\lim_{\varepsilon_k \to 0}|\nabla u_{\varepsilon_k}|^2(1-|u_{\varepsilon_k}|^2)^{\alpha} =\omega_1, \quad \mbox{weakly star in} \quad C(\overline{G}). \end{align*} $$

Next, using the Hölder inequality and Proposition 2.5, we see that as $\varepsilon \to 0$ ,

(3.1) $$ \begin{align}\begin{split}&\displaystyle \int_{G\setminus \cup_{j=1}^N B(a_j,\sigma)}|\nabla u_\varepsilon|^2(1-|u_\varepsilon|^2)^{\alpha}\\&\quad \leq \left[\displaystyle\int_{G\setminus \cup_j B(a_j,\sigma)}|\nabla u_\varepsilon|^p\right]^{\frac{2}{p}} \\&\qquad \cdot\left[\displaystyle\int_{G\setminus \cup_jB(a_j,\sigma)} (1-|u_\varepsilon|^2)^{\frac{p\alpha}{p-2}} \right]^{\frac{p-2}{p}} \to 0.\end{split}\end{align} $$

Thus, $supp(\omega _1) \subset \{a_j\}_{j=1}^N$ . Therefore, we can find $L_j\geq 0$ such that

$$ \begin{align*}\omega_1=\sum_{j=1}^N L_j\delta_{a_j}. \end{align*} $$

We claim $L_j>0$ for each j (and hence $supp(\omega _1) = \{a_j\}_{j=1}^N$ ). For convenience, we here drop k from $\varepsilon _k$ .

Noting $B_{h\varepsilon }(x_i)$ contains zeros of $u_\varepsilon $ , by Propositions 2.3 we have

$$ \begin{align*}\frac{1}{2} \leq |u_\varepsilon| \leq \frac{3}{4}, \quad on \ \partial B_{h\varepsilon}(x_i) \end{align*} $$

as long as $h\varepsilon $ is suitably small. For each $x_1 \in \partial B_{h\varepsilon }(x_i)$ , by Proposition 2.2, we get

$$ \begin{align*}\frac{3}{8} \leq |u_\varepsilon(y)| \leq \frac{7}{8}, \quad \forall y \in \overline{B_{\hat{l}\varepsilon}(x_1)} \cap B_{h\varepsilon}(x_i). \end{align*} $$

Here, $\hat {l}:=\min \{(8C_0)^{\frac {p}{2-p}},h\}$ . Therefore,

(3.2) $$ \begin{align} \frac{3}{8} \leq |u_\varepsilon(y)| \leq \frac{7}{8}, \quad \forall y \in B_{h\varepsilon}(x_i) \setminus B_{(h-\hat{l}/2)\varepsilon}(x_i). \end{align} $$

As in [Reference Hong15, Theorem 3.9], by (3.2) we can write

$$ \begin{align*}u_\varepsilon(x)=|u_\varepsilon(x)|\phi(r,\tau), \ r=|x|, \ \tau=\frac{x}{|x|}, \end{align*} $$

and hence

$$ \begin{align*}|\nabla u_\varepsilon|^2 \geq |u_\varepsilon|^2r^{-2}|\nabla_{\tau}\phi(r,\tau)|^2. \end{align*} $$

Thus, by the Hölder inequality, there holds

$$ \begin{align*}&\displaystyle \int_{B_{h\varepsilon}(x_i) \setminus B_{(h-\hat{l}/2)\varepsilon}(x_i)} (1-|u_\varepsilon|^2)^{\alpha}|\nabla u_\varepsilon|^2 \\&\quad\displaystyle \geq \left(1-\frac{7}{8}\right)^{\alpha} \int_{B_{h\varepsilon}(x_i) \setminus B_{(h-\hat{l}/2)\varepsilon}(x_i)} |u_\varepsilon|^2 r^{-2}|\nabla _{\tau}\phi(r,\tau)|^2 \\&\quad\displaystyle \geq \left(\frac{1}{8}\right)^\alpha \left(\frac{3}{8}\right)^2 \int_{(h-\hat{l}/2)\varepsilon}^{h\varepsilon} \left(\int_{S^1}|\nabla_{\tau}\phi|^2ds\right)\frac{dr}{r}\\&\quad\displaystyle \geq \left(\frac{1}{8}\right)^\alpha \left(\frac{3}{8}\right)^2 \int_{(h-\hat{l}/2)\varepsilon}^{h\varepsilon}\frac{1}{2\pi} \left(\int_{S^1}|\nabla_{\tau}\phi|ds\right)^2\frac{dr}{r}.\end{align*} $$

Theorem 8.2 in [Reference Brezis, Coron and Lieb4] shows that

$$ \begin{align*}\int_{S^1}|\nabla_{\tau}\phi|ds \geq 2\pi|d_i|, \end{align*} $$

where $d_i=\deg (u_\varepsilon ,\partial B_{h\varepsilon }(x_i))$ is independent of $\varepsilon $ (see Proposition 2.4). Therefore,

$$ \begin{align*}\int_{B_{h\varepsilon}(x_i) \setminus B_{(h-\hat{l}/2)\varepsilon}(x_i)} (1-|u_\varepsilon|^2)^{\alpha}|\nabla u_\varepsilon|^2 \geq \left(\frac{1}{8}\right)^\alpha \left(\frac{3}{8}\right)^2 (2\pi d_i^2) \cdot \left(\log \frac{h}{h-\hat{l}/2}\right). \end{align*} $$

This implies $L_j>0$ , and hence (1.10) is proved.

Proof of (1.9).

Denote $u_\varepsilon $ by u. Clearly, by the Cauchy inequality,

(3.3) $$ \begin{align} |\det(\nabla u)|=|u_{1x_1}u_{2x_2}-u_{1x_2}u_{2x_1}| \leq \frac{1}{2}|\nabla u|^2. \end{align} $$

By Proposition 2.1, we obtain that for each j,

(3.4) $$ \begin{align} \varepsilon^{p-2}\int_{G\cap B(a_j,\sigma)}|\det(\nabla u)|^{p/2} \leq \frac{\varepsilon^{p-2}}{2}\int_{G}|\nabla u|^p \leq C. \end{align} $$

Using Proposition 2.5, we get

(3.5) $$ \begin{align} \varepsilon^{p-2}\int_{G \setminus \cup_j B(a_j,\sigma)}|\det(\nabla u)|^{p/2} \leq \frac{\varepsilon^{p-2}}{2}\int_{G \setminus \cup_j B(a_j,\sigma)}|\nabla u|^p \leq C\varepsilon^{p-2}. \end{align} $$

Combining (3.4) and (3.5), we see that $\varepsilon ^{p-2}|\det (\nabla u)|^{p/2}$ is bounded in $L^1(G)$ . Thus, there exists a Radon measure $\omega _2$ such that

$$ \begin{align*}\lim_{\varepsilon_k \to 0}\varepsilon_k^{p-2} |\det(\nabla u_{\varepsilon_k})|^{p/2} =\omega_2, \quad \mbox{weakly star in} \quad C(\overline{G}). \end{align*} $$

By virtue of (3.5), $supp(\omega _2) \subset \{a_j\}_{j=1}^N$ . Thus, we can find $P_j\geq 0$ such that

$$ \begin{align*}\omega_2=\sum_{j=1}^N P_j\delta_{a_j}. \end{align*} $$

Equation (1.9) is proved.

4 Convergence rate of $|u_\varepsilon |-1$ when $p \in (1,2)$

In this section, we consider the case of $p \in (1,2)$ . Proposition 2.8 implies that $|{u}_{\varepsilon }|-1$ tends to zero and the decay rate is presented. Here, we will give a decay rate of the gradient of $|{u}_{\varepsilon }|-1$ .

Proof of Theorem 1.3

Proof. Step 1. Let $R>0$ be a small constant such that $B(x,2R) \subset \subset G\setminus \cup _{i=1}^N\{a_i\}$ . Applying (1.5) and Proposition 2.9, we have

(4.1) $$ \begin{align} \frac{1}{2}\leq |{u}_{\varepsilon}|\leq 1 \quad in \ B(x,2R). \end{align} $$

By the integral mean value theorem and Proposition 2.6, there is $r \in [R,2R]$ such that

(4.2) $$ \begin{align} \int_{\partial B(x,r)}|\nabla |{u}_\varepsilon||^pd\xi +\frac{1}{\varepsilon^p} \int_{\partial B(x,r)}(1-|{u}_{\varepsilon}|^2)^2d\xi \leq C \end{align} $$

with $C=C(r)>0$ independent of $\varepsilon $ . Denote $B(x,r)$ by B. If $\rho _1$ is a minimizer of the functional

$$ \begin{align*}E(\rho,B)=\frac{1}{p} \int_B(|\nabla \rho|^2+1)^{p/2} +\frac{1}{2\varepsilon^p} \int_B(1-\rho)^2 \end{align*} $$

in $W_{|{u}_{\varepsilon }|}^{1,p}(B,\mathbb {R}^+\cup \{0\})$ , then it solves

(4.3) $$ \begin{align} -div[(|\nabla\rho|^2+1)^{(p-2)/2}\nabla\rho] =\frac{1}{\varepsilon^p}(1-\rho), \end{align} $$

(4.4) $$ \begin{align} \rho|_{\partial B}=|{u}_{\varepsilon}|, \end{align} $$

By the maximum principle, it follows from (4.1) that

(4.5) $$ \begin{align} \frac{1}{2}\leq \rho_1\leq 1 \quad on \ \overline{B}. \end{align} $$

Applying Proposition 2.6 we see easily that

(4.6) $$ \begin{align} E(\rho_1,B)\leq E(|{u}_{\varepsilon}|,B) \leq C(E_{\varepsilon}({u}_{\varepsilon},B)+1) \leq C. \end{align} $$

Step 2. Multiplying (4.3) with $\rho =\rho _1$ by $(\nu \cdot \nabla \rho _1)$ , we have

(4.7) $$ \begin{align} \begin{array}{ll} -\displaystyle\int_{\partial B}v^{(p-2)/2}|\partial_\nu\rho_1|^2d\xi &+\displaystyle\int_Bv^{(p-2)/2}\nabla \rho_1 \cdot \nabla (\nu \cdot \nabla \rho_1)\\[3mm] &=\displaystyle\frac{1}{\varepsilon^p} \int_B(1-\rho_1)(\nu \cdot \nabla \rho_1), \end{array} \end{align} $$

where $\nu $ denotes the unit outside norm vector on $\partial B$ and $v=|\nabla \rho _1|^2+1$ . Integrating by parts yields

(4.8) $$ \begin{align} \begin{split}& \displaystyle\int_Bv^{(p-2)/2}\nabla \rho_1 \cdot \nabla (\nu \cdot \nabla \rho_1)\\&\quad= \displaystyle\int_Bv^{(p-2)/2}|\nabla \rho_1|^2 +\frac{1}{p}\int_B \nu \cdot \nabla (v^{p/2})\\&\quad= \displaystyle\int_Bv^{(p-2)/2}|\nabla \rho_1|^2 +\frac{1}{p}\int_{\partial B}v^{p/2}d\xi -\frac{1}{p}\int_B v^{p/2} ({\textrm{div}} \nu), \end{split} \end{align} $$

and

$$ \begin{align*}\displaystyle\frac{1}{\varepsilon^p} \int_B(1-\rho_1)(\nu \cdot \nabla \rho_1) = \frac{1}{2\varepsilon^p} \int_B(1-\rho_1)^2({\textrm{div}}\nu) -\frac{1}{2\varepsilon^p} \int_{\partial B}(1-\rho_1)^2d\xi. \end{align*} $$

Substitute this result and (4.8) into (4.7). Noting div $\nu =r^{-1}>0$ , we can use (4.5), (4.6), (4.4), and (4.2) to obtain

(4.9) $$ \begin{align} \int_{\partial B}v^{(p-2)/2}|\partial_\nu\rho_1|^2d\xi \leq C+\frac{1}{p}\int_{\partial B}v^{p/2}d\xi. \end{align} $$

Step 3. By the Jensen inequality $(1+a^2+b^2)^{1/2} \leq 1+|a|+|b|$ and (4.4), there holds

$$ \begin{align*}\int_{\partial B} v^{p/2}d\xi \leq \int_{\partial B}v^{(p-1)/2} [(1+|\partial_\tau |u_\varepsilon||) +|\partial_\nu \rho_1|]d\xi, \end{align*} $$

where $\tau $ denotes the unit tangent vector on $\partial B$ . Using the Hölder inequality and (4.2), we deduce from the result above that

$$ \begin{align*}\int_{\partial B} v^{p/2}d\xi \leq C\left(\int_{\partial B} v^{\frac{p}{2}}d\xi\right)^{\frac{p-1}{p}} +\left(\int_{\partial B}v^{\frac{p-2}{2}}|\partial_\nu \rho_1|^2d\xi\right)^{\frac{1}{2}} \left(\int_{\partial B} v^{\frac{p}{2}}d\xi\right)^{\frac{1}{2}}. \end{align*} $$

Inserting (4.9) into this result and using the Young inequality, we obtain that for any $\delta \in (0,1)$ ,

$$ \begin{align*}\int_{\partial B} v^{p/2}d\xi \leq C(\delta)+\left(\delta+\displaystyle\frac{1}{2p} +\frac{1}{2}\right)\int_{\partial B}v^{p/2}d\xi, \end{align*} $$

Therefore, by choosing $\delta>0$ sufficiently small we get

(4.10) $$ \begin{align} \int_{\partial B}(|\nabla \rho_1|^2+1)^{p/2}d\xi \leq C. \end{align} $$

Multiply (4.3) by $(1-\rho _1)$ . In view of (4.4), applying the Hölder inequality, and using (4.2) and (4.10), we get

(4.11) $$ \begin{align} \begin{split}&\displaystyle\int_B[(|\nabla\rho_1|^2+1)^{(p-2)/2}|\nabla\rho_1|^2 +\frac{1}{\varepsilon^p}(1-\rho_1)^2]\\&\quad\leq \left\vert\displaystyle\int_{\partial B}(1-\rho_1)(|\nabla\rho_1|^2+1)^{(p-2)/2} (\nu\cdot\nabla\rho_1)d\xi\right\vert \leq C\varepsilon. \end{split} \end{align} $$

Step 4. Set $U=\rho _1w$ on B; $U={u}_{\varepsilon }$ on $G\setminus B$ , where $w={u}_{\varepsilon }/|{u}_{\varepsilon }|$ . Then $U \in W$ . Since ${u}_{\varepsilon }$ is a minimizer of $E_{\varepsilon }(u,G)$ , we have

(4.12) $$ \begin{align} E_{\varepsilon}(u_{\varepsilon},G) \leq E_{\varepsilon}(U,G) = E_{\varepsilon}(\rho_1 w,B) +E_{\varepsilon}(u_{\varepsilon},G\setminus B). \end{align} $$

In view of $p \in (1,2)$ , we get

$$ \begin{align*}&\displaystyle\int_B(|\nabla \rho_1|^2 +\rho_1^2|\nabla w|^2)^{p/2} -\int_B(\rho_1^2|\nabla w|^2)^{p/2}\\&\quad=\displaystyle\frac{p}{2}\int_B \int_0^1(s|\nabla \rho_1|^2 +\rho_1^2|\nabla w|^2)^{(p-2)/2} |\nabla \rho_1|^2dsdx\\&\quad\leq \displaystyle\frac{p}{2}\int_B|\nabla \rho_1|^pdx \cdot \int_0^1s^{(p-2)/2}ds =\int_B|\nabla \rho_1|^p.\end{align*} $$

Here, we use the fact $\int _0^1s^{(p-2)/2}ds=2/p$ . Combining this with (4.12) leads to

$$ \begin{align*}E_{\varepsilon}(u_{\varepsilon},B) \leq E_{\varepsilon}(\rho_1 w,B) \leq \frac{1}{p} \int_B(\rho_1^2|\nabla w|^2)^{p/2} +E_\varepsilon(\rho_1,B). \end{align*} $$

Using (4.5) and (4.11), we can see from the result above that

(4.13) $$ \begin{align} E_{\varepsilon}(u_{\varepsilon},B) \leq \frac{1}{p} \int_B |\nabla w|^p+C\varepsilon. \end{align} $$

Step 5. Here, we denote $|u_\varepsilon |$ by $h_\varepsilon $ .

Clearly, $(sa^2+b^2)^{(p-2)/2} \geq (a^2+b^2)^{(p-2)/2}$ when $s \in (0,1)$ and $p \in (1,2)$ . Therefore, according to the mean value theorem, it follows that

$$ \begin{align*}&\displaystyle(|\nabla h_\varepsilon|^2 +h_\varepsilon^2|\nabla w|^2)^{p/2} -(h_\varepsilon|\nabla w|^2)^{p/2}\\[3mm]&\quad=\displaystyle\frac{p}{2}\left(\int_0^1[s(|\nabla h_\varepsilon|^2 +h_\varepsilon^2|\nabla w|^2) +(1-s)(h_\varepsilon^2|\nabla w|^2)]^{\frac{p-2}{2}}ds\right) \cdot |\nabla h_\varepsilon|^2\\[3mm]&\quad\geq \displaystyle\frac{p}{2}|\nabla u_\varepsilon|^{p-2}|\nabla h_\varepsilon|^2. \end{align*} $$

This and (4.13) imply

(4.14) $$ \begin{align}&\displaystyle\frac{1}{2}\int_B|\nabla u_\varepsilon|^{p-2}|\nabla h_\varepsilon|^2 +\frac{1}{p}\int_B(h_\varepsilon^p-1)|\nabla w|^p +\frac{1}{4\varepsilon^p}\int_B(1-h_\varepsilon^2)^2\notag\\[3mm]&\quad\leq E_{\varepsilon}(u_{\varepsilon},B) -\displaystyle\frac{1}{p}\int_B|\nabla w|^p \leq C\varepsilon. \end{align} $$

In view of (4.1), there holds

$$ \begin{align*}\frac{1}{p} \int_B(1-h_\varepsilon^p)|\nabla w|^p \leq \frac{2^p}{p}\int_B(1-h_\varepsilon^2)|\nabla u_\varepsilon|^p. \end{align*} $$

Applying the Hölder inequality and Propositions 2.10 and 2.6, we can obtain

$$ \begin{align*}\frac{1}{p} \int_B(1-h_\varepsilon^p)|\nabla w|^p \leq C\varepsilon^{2t_0/(1+t_0)} \end{align*} $$

with $t_0 \in (0,t)$ is suitably small. Combining this with (4.14), we can see

(4.15) $$ \begin{align} \frac{1}{2}\int_B|\nabla u_\varepsilon|^{p-2}|\nabla h_\varepsilon|^2 +\frac{1}{4\varepsilon^p}\int_B(1-h_\varepsilon^2)^2 \leq C\varepsilon^{2t_0/(1+t_0)}. \end{align} $$

Using the Hölder inequality, by (4.15) and Proposition 2.6, we see that

$$ \begin{align*}\int_B|\nabla h|^p \leq \left(\int_B|\nabla u_\varepsilon|^{p-2}|\nabla h_\varepsilon|^2\right)^{\frac{p}{2}} \left(\int_B|\nabla u_\varepsilon|^p\right)^{\frac{2-p}{2}} \leq C\varepsilon^{pt_0/(1+t_0)}. \end{align*} $$

Combining with (4.15) and by an argument of finite coverings, we can see (1.15).

5 Proof of Theorem 1.2

Proof of (1.12).

By (3.3) and Proposition 2.6, we get

(5.1) $$ \begin{align} \int_{G}|\det(\nabla u_\varepsilon)|^{p/2} \leq \int_G|\nabla u_\varepsilon|^p \leq C. \end{align} $$

Therefore, we can find a Radon measure $\omega _3$ and a subsequence $\varepsilon _k$ of $\varepsilon $ such that

(5.2) $$ \begin{align} \lim_{k \to \infty}|\det(\nabla u_{\varepsilon_k})|^{p/2}=\omega_3, \quad \mbox{weakly star in} \quad C(\overline{G}). \end{align} $$

Proposition 2.9 implies that there exists $\phi _\varepsilon $ such that when $x \in G \setminus \cup _jB_\sigma (a_j)$ ,

$$ \begin{align*}u_\varepsilon(x)=h_\varepsilon(x)(\cos\phi_\varepsilon(x), \sin\phi_\varepsilon(x)), \end{align*} $$

where $h_\varepsilon (x)=|u_\varepsilon (x)|$ . Thus, on $G \setminus \cup _jB_\sigma (a_j)$ ,

$$ \begin{align*}|\det(\nabla u_\varepsilon)|=h_\varepsilon |\partial_\nu h_\varepsilon \partial_{\tau}\phi_\varepsilon -\partial_{\tau}h_\varepsilon \partial_{\nu}\phi_\varepsilon|. \end{align*} $$

Therefore, by the Hölder inequality, there holds

(5.3) $$ \begin{align}\begin{split}&\displaystyle\int_{G \setminus \cup_jB_\sigma(a_j)}|\det(\nabla u_\varepsilon)|^{p/2}\\&\quad \leq \left(\int_{G \setminus \cup_jB_\sigma(a_j)}|\nabla h_\varepsilon|^p\right)^{\frac{1}{2}} \left(\int_{G \setminus \cup_jB_\sigma(a_j)}|h_\varepsilon\nabla \phi_\varepsilon|^p\right)^{\frac{1}{2}}. \end{split} \end{align} $$

In view of Proposition 2.6 and (1.15),

$$ \begin{align*}\int_{G \setminus \cup_jB_\sigma(a_j)}|\det(\nabla u_\varepsilon)|^{p/2} \leq C\left(\int_{G \setminus \cup_jB_\sigma(a_j)}|\nabla h_\varepsilon|^p\right)^{\frac{1}{2}} \to 0 \end{align*} $$

when $\varepsilon \to 0$ . Therefore, $supp(\omega _3) \subset \{a_j\}_{j=1}^N$ . Thus, there exists constants $Q_j \geq 0$ such that

$$ \begin{align*}\omega_3=\sum_{j=1}^N Q_j\delta_{a_j}. \end{align*} $$

Equation (1.12) is proved.

Proof of (1.14).

First we observe that $2-p<pt$ since p is sufficiently close to $2$ , where t is the constant in Proposition 2.10 (which implies that t is independent of p). Thus we can choose $\gamma \in (2-p,pt)$ . By (1.5), the Hölder inequality and Propositions 2.10 and 2.6, we have

(5.4) $$ \begin{align} \int_{G\setminus G^{2\rho\varepsilon}} (1-|\tilde{u}_\varepsilon|^2)^{\alpha} |\nabla \tilde{u}_\varepsilon|^2 \leq \left(\int_{G\setminus G^{2\rho\varepsilon}} |\nabla \tilde{u}_\varepsilon|^{p+\gamma}\right)^{\frac{2}{p+\gamma}} |G\setminus G^{2\rho\varepsilon}|^{1-\frac{2}{p+\gamma}} \to 0 \end{align} $$

when $\varepsilon \to 0$ .

In view of (1.5), the right hand side of (1.4) with $u=\tilde {u}_\varepsilon $ is bounded by $\varepsilon ^{-p}$ . Thus, checking the proof of Proposition 5.1 in [Reference Tolksdorf22], and applying Proposition 2.6 we get

$$ \begin{align*}\|\nabla \tilde{u}_\varepsilon\|_{L^{\infty}(G^{\rho\varepsilon} \cap B_\sigma(a_j))}^p \leq C\sigma^{-2}\varepsilon^{-p} \int_G(1+|\nabla \tilde{u}_\varepsilon|^p) \leq C\varepsilon^{-p}. \end{align*} $$

Therefore, by the Hölder inequality, we obtain that for any $\gamma>0$ ,

(5.5) $$ \begin{align}\begin{split}&\displaystyle\int_{G^{\rho\varepsilon}\cap B(a_j,\sigma)} (1-|\tilde{u}_\varepsilon|^2)^{\alpha} |\nabla \tilde{u}_\varepsilon|^2 \\&\quad\leq C\varepsilon^{p-2}\displaystyle\int_{G^{\rho\varepsilon}\cap B(a_j,\sigma)} (1-|\tilde{u}_\varepsilon|^2)^{\alpha} |\nabla \tilde{u}_\varepsilon|^p \\&\quad\leq C\varepsilon^{p-2} \left(\displaystyle\int_{G^{\rho\varepsilon}\cap B(a_j,\sigma)} |\nabla \tilde{u}_\varepsilon|^{p+\gamma}\right)^{\frac{p}{p+\gamma}}\\&\qquad \cdot\left(\displaystyle\int_{G^{\rho\varepsilon}\cap B(a_j,\sigma)} (1-|\tilde{u}_\varepsilon|^2)^{(p+\gamma)\alpha/\gamma}\right)^{\frac{\gamma}{p+\gamma}}. \end{split}\end{align} $$

We set

$$ \begin{align*}\gamma_0=2-p. \end{align*} $$

Then for any $\alpha \geq 2-p$ ,

(5.6) $$ \begin{align} \frac{(p+\gamma_0)\alpha}{\gamma_0} \geq 2. \end{align} $$

In addition, $\gamma _0 \in (0,pt)$ since p is sufficiently close to $2$ . Here, t is the constant in Proposition 2.10.

According to Proposition 2.10, and by Proposition 2.6, it follows that

$$ \begin{align*}\left(\int_{G^{\rho\varepsilon}\cap B(a_j,\sigma)} |\nabla \tilde{u}_\varepsilon|^{p+\gamma_0}\right)^{\frac{p}{p+\gamma_0}} \leq C\int_G(|\nabla \tilde{u}_\varepsilon|^2+1)^{p/2} \leq C. \end{align*} $$

Inserting this result into (5.5) with $\gamma =\gamma _0$ yields

(5.7) $$ \begin{align}\begin{split}&\displaystyle\int_{G^{\rho\varepsilon}\cap B(a_j,\sigma)} (1-|u_\varepsilon|^2)^{\alpha} |\nabla \tilde{u}_\varepsilon|^2\\&\quad\leq C\varepsilon^{p-2} \displaystyle \left(\int_{G^{\rho\varepsilon}\cap B(a_j,\sigma)} (1-|\tilde{u}_\varepsilon|^2)^{(p+\gamma_0)\alpha/\gamma_0} \right)^{\frac{\gamma_0}{p+\gamma_0}}.\end{split}\end{align} $$

Using (5.6), Proposition 2.8, we can deduce from (5.7) that

(5.8) $$ \begin{align} \int_{G^{\rho\varepsilon}\cap B(a_j,\sigma)} (1-|\tilde{u}_\varepsilon|^2)^{\alpha} |\nabla \tilde{u}_\varepsilon|^2 \leq C\varepsilon^{p-2+\frac{2\gamma_0}{p+\gamma_0}}=C. \end{align} $$

Next, using

$$ \begin{align*}\mathop{\mathrm{sup}}\limits_{G\setminus \cup_jB(a_j,\sigma)}|\nabla \tilde{u}_\varepsilon| \leq C \end{align*} $$

and

$$ \begin{align*}\mathop{\mathrm{sup}}\limits_{G\setminus \cup_jB(a_j,\sigma)}(1-|\tilde{u}_\varepsilon|^2) \to 0 \end{align*} $$

which are implied by (1.11), we derive that when $\varepsilon \to 0$ ,

(5.9) $$ \begin{align}\begin{split}&\displaystyle\int_{G^{\rho\varepsilon}\setminus \cup_jB(a_j,\sigma)}(1-|\tilde{u}_\varepsilon|^2)^{\alpha} |\nabla \tilde{u}_\varepsilon|^2 \\&\quad\leq \displaystyle\left(\mathop{\mathrm{sup}}\limits_{G\setminus \cup_jB(a_j,\sigma)}(1-|\tilde{u}_\varepsilon|^2)^{\alpha}\right) \left(\mathop{\mathrm{sup}}\limits_{G\setminus \cup_jB(a_j,\sigma)}|\nabla \tilde{u}_\varepsilon|^2\right) |G\setminus \cup_jB_\sigma(a_j)|\\&\quad \to 0. \end{split} \end{align} $$

Combining (5.4), (5.8) with (5.9), we obtain that $(1-|\tilde {u}_\varepsilon |^2)^{\alpha } |\nabla \tilde {u}_\varepsilon |^2$ is bounded in $L^1(G)$ , and hence

$$ \begin{align*}\lim_{\varepsilon_k \to 0}(1-|\tilde{u}_{\varepsilon_k}|^2)^{\alpha} |\nabla \tilde{u}_{\varepsilon_k}|^2 =\omega_4, \quad \mbox{weakly star in} \quad C(\overline{G}), \end{align*} $$

where $\omega _4$ is a Radon measure. In addition, (5.9) implies $supp(\omega _4) \subset \{a_j\}_{j=1}^N$ . Thus, there exists constants $M_j \geq 0$ such that

$$ \begin{align*}\omega_4=\sum_{j=1}^N M_j\delta_{a_j}. \end{align*} $$

By (5.4), we see that $M_j=0$ when $a_j \in \partial G$ .

For the other $a_j$ which are contained in G, using (2.2) instead of Proposition 2.2 in the proof of $L_j>0$ , we also get $M_j>0$ . Equation (1.14) is proved.

Proof of (1.13).

By Proposition 2.8, $\frac {(1-|\tilde {u}_\varepsilon |^2)^2}{\varepsilon ^2}$ is bounded in $L^1(G)$ . Thus, there exist a subsequence $\varepsilon _k$ of $\varepsilon $ and a Radon measure $\omega _5$ , such that

$$ \begin{align*}\lim_{k \to 0}\frac{(1-|\tilde{u}_{\varepsilon_k}|^2)^2}{\varepsilon_k^2} =\omega_5, \quad \mbox{weakly star in}\quad C(\overline{G}). \end{align*} $$

In addition, according to Proposition 2.11,

$$ \begin{align*}\frac{1}{\varepsilon_k^2}\int_{G \setminus \cup_j B_\sigma(a_j)} (1-|\tilde{u}_{\varepsilon_k}|^2)^2 \leq C\varepsilon_k^{2(p-1)} \to 0 \end{align*} $$

when $\varepsilon _k \to 0$ . Therefore, we can see that $supp(\omega _5) \subset \{a_j\}_{j=1}^N$ , and hence

$$ \begin{align*}\omega_5=\sum_{j=1}^N K_j\delta_{a_j}, \end{align*} $$

where $K_j \geq 0$ .

We claim $K_j>0$ for each j. In fact, there exists $x_j^\varepsilon $ which is the center of the bad disc tends to $a_j$ when $\varepsilon \to 0$ . Recalling the definition of the bad disc in Section 2, we have

$$ \begin{align*}\frac{1}{\varepsilon^2}\int_{B(x_j^\varepsilon,h\varepsilon)}(1-|\tilde{u}_\varepsilon|^2)^2 \geq \mu>0. \end{align*} $$

for each j. This implies $K_j>0$ (and hence $supp(\omega _5) = \{a_j\}_{j=1}^N$ ). Equation (1.13) is proved.

6 Proof of Theorem 1.4

Step 1. First we claim

(6.1) $$ \begin{align} |u| \leq 1 \quad a.e. \ on \ \mathbb{R}^2. \end{align} $$

For convenience, sometimes we denote $B_R(0)$ by $B_R$ .

From (1.17), it follows that

$$ \begin{align*} 0 & =\lim_{R \to \infty} \int_{B_{2R}\setminus B_R} |\nabla u|^p dx \\ & =\lim_{R \to \infty} \int_R^{2R}\left[r\int_{\partial B_r} |\nabla u|^p d\xi\right] \frac{dr}{r} \\ & \geq \lim_{R \to \infty} \inf_{r \in [R,2R]}\left[r\int_{\partial B_r} |\nabla u|^p d\xi\right] \cdot (\log 2). \end{align*} $$

Therefore, we can find a subsequence $R_k$ of R such that

(6.2) $$ \begin{align} \lim_{R_k \to \infty}R_k\int_{\partial B_{R_k}(0)}|\nabla u|^pd\xi=0. \end{align} $$

When $p \in (1,2)$ , the Sobolev inequality implies $u \in L^{\frac {2p}{2-p}}(\mathbb {R}^2)$ . By the same proof of (6.2), there also holds that

(6.3) $$ \begin{align} \lim_{R_k \to \infty}R_k\int_{\partial B_{R_k}(0)}|u|^{\frac{2p}{2-p}}d\xi=0. \end{align} $$

Here, $R_k$ is also a subsequence of R.

Set $\Phi =u-u\min \{1,|u|\}/|u|$ and $B_+=\{x \in \mathbb {R}^2; |u(x)|>1\}$ , then

$$ \begin{align*}\left\{\!\begin{array}{ll}\nabla \Phi=\nabla u-|u|^{-1}\nabla u+(u\cdot\nabla u)u/|u|^3, \quad &on \ B_+;\\[3mm]\nabla \Phi=0 &on \ \mathbb{R}^2 \setminus B_+. \end{array} \right. \end{align*} $$

Multiplying (1.16) by $\Phi $ and integrating on $B_{R_k}$ and then letting $R_k \to \infty $ , we get

(6.4) $$ \begin{align} \begin{array}{ll} &-\displaystyle\lim_{R_k \to \infty}\int_{\partial B_{R_k}(0)} u|\nabla u|^{p-2}\partial_{\nu}ud\xi\\[3mm] &+\displaystyle\lim_{R_k \to \infty}\int_{\partial B_{R_k}(0)} \frac{u}{|u|}\min\{1,|u|\}|\nabla u|^{p-2}\partial_{\nu}u d\xi\\[3mm] &+\displaystyle\int_{B_+}(1-1/|u|)|\nabla u|^p +\int_{B_+}|\nabla u|^{p-2}(u \cdot \nabla u)^2/|u|^3\\[3mm] &+\displaystyle\int_{B_+}|u|(|u|+1)(|u|-1)^2=0. \end{array} \end{align} $$

When $R_k \to \infty $ , by the Hölder inequality and (6.2) and (6.3), there holds

(6.5) $$ \begin{align} \begin{split}&\displaystyle \int_{\partial B_{R_k}} |u||\nabla u|^{p-1} d\xi\\&\quad\leq \displaystyle R_k^{-\frac{1}{2}} \left(R_k\int_{\partial B_{R_k}}|\nabla u|^pd\xi\right)^{1-\frac{1}{p}} \left(R_k\int_{\partial B_{R_k}}|u|^{\frac{2p}{2-p}}d\xi\right)^{\frac{2-p}{2p}} |\partial B_{R_k}|^{\frac{1}{2}}\\&\quad\to 0, \end{split}\end{align} $$

which implies both the first and the second terms of the left hand side of (6.4) are equal to zero. This shows $|B_+|=0$ and hence (6.1) is proved when $p \in (1.2)$ .

When $p>2$ , set $\Phi :=(|u|-1)_+$ . Then,

$$ \begin{align*}\left\{\!\begin{array}{l}\nabla\Phi=0,\quad on \quad \mathbb{R}^2\setminus B_+;\\[3mm]\nabla\Phi=\frac{u \cdot \nabla u}{|u|},\quad on \quad B_+. \end{array}\right. \end{align*} $$

Obviously, (1.17) implies

(6.6) $$ \begin{align} \|\nabla \Phi\|_{L^p(\mathbb{R}^2)}<\infty. \end{align} $$

Let $\zeta \in C^{\infty }(\mathbb {R}^2,[0,1])$ be a cut-off function satisfying $\zeta (y)=1$ for $|y|\leq 1$ , and $\zeta (y)=0$ for $|y|\geq 2$ . Set $\zeta _t(y)=\zeta (\frac {y}{t})$ . Multiply (1.16) by $\xi $ , where

$$ \begin{align*}\left\{\!\begin{array}{l} \xi=0,\quad on \quad \mathbb{R}^2\setminus B_+;\\[3mm]\xi=\frac{u}{|u|}\zeta_t,\quad on \quad B_+. \end{array}\right. \end{align*} $$

Then,

$$ \begin{align*}\int_{B_+}|\nabla u|^{p-2}\nabla u\nabla(\frac{u}{|u|}\zeta_t) =-\int_{B_+}|u|(1+|u|)\Phi\zeta_t. \end{align*} $$

By calculating the left hand side, we can obtain that

(6.7) $$ \begin{align} \begin{split}&\displaystyle\int_{B_+}|u|^{-1}|\nabla u|^p\zeta_t -\int_{B_+}|u|^{-3}|\nabla u|^{p-2}(u \cdot \nabla u)^2\zeta_t\\&\quad+\displaystyle\int_{B_+}|u|(1+|u|)\Phi\zeta_t +\int_{B_+}|\nabla u|^{p-2}\nabla \Phi\nabla\zeta_t=0. \end{split} \end{align} $$

In view of $|\nabla u|^2 \geq (u \cdot \nabla u)^2/|u|^2$ , the first and the second terms in the left hand side of (6.7) is nonnegative. Therefore, using (1.17) and (6.6), we can deduce that

$$ \begin{align*}&\displaystyle\int_{B_+}|u|(1+|u|)\Phi\zeta_t \leq |\int_{B_+}|\nabla u|^{p-2}\nabla \Phi\nabla\zeta_t|\\&\quad\leq \displaystyle\frac{1}{t}\int_{{B_+}\cap\{y;t\leq|y|\leq2t\}} |\nabla u|^{p-2}|\nabla \Phi| \leq \frac{1}{t}\|\nabla u\|_p^{p-2}\|\nabla\Phi\|_p t^{2/p}.\end{align*} $$

When $t \to \infty $ , the right hand side converges to zero by virtue of $p>2$ . And hence so is the left hand side. This means $|B_+|=0$ or $\Phi =0$ a.e. on $\mathbb {R}^2$ . Thus, (6.1) is proved when $p>2$ .

Step 2. Multiplying (1.16) by u and integrating on $B_R(0)$ yield

(6.8) $$ \begin{align} \int_{B_R(0)}|u|^2(1-|u|^2) =\int_{B_R(0)} |\nabla u|^p - \int_{\partial B_R(0)}u|\nabla u|^{p-2}\partial_{\nu}ud\xi. \end{align} $$

When $p \in (1,2)$ , we use (6.5) and (1.17) to get

(6.9) $$ \begin{align} \int_{\mathbb{R}^2}|u|^2(1-|u|^2) =\int_{\mathbb{R}^2} |\nabla u|^p <\infty. \end{align} $$

When $p>2$ , by (6.1) and (6.2), there holds

$$ \begin{align*}\left\vert\int_{\partial B_{R_k}}u|\nabla u|^{p-2}\partial_{\nu}ud\xi\right\vert \leq CR_k^{\frac{1}{p}-1}\left(R_k\int_{\partial B_{R_k}}|\nabla u|^pd\xi\right)^{1-\frac{1}{p}} |\partial B_{R_k}|^{\frac{1}{p}} \to 0 \end{align*} $$

when $R_k \to \infty $ . From (6.8) with $R=R_k$ , we also deduce (6.9).

Multiply (1.16) by $(x \cdot \nabla u)$ and integrate on $B_R(0)$ . Integrating by parts, we can see the left hand side

$$ \begin{align*}&-\displaystyle\int_{B_R}{\textrm{div}}(|\nabla u|^{p-2}\nabla u)(x \cdot \nabla u)\\&\quad=-R\displaystyle\int_{\partial B_R}|\nabla u|^{p-2}|\partial_{\nu}u|^2d\xi +\int_{B_R}|\nabla u|^p+\frac{1}{p}\int_{B_R} x\cdot \nabla (|\nabla u|^p)\\&\quad=-R\displaystyle\int_{\partial B_R}|\nabla u|^{p-2}|\partial_{\nu}u|^2d\xi +(1-\frac{2}{p})\int_{B_R}|\nabla u|^p+\frac{R}{p}\int_{\partial B_R} |\nabla u|^pd\xi.\end{align*} $$

Therefore,

(6.10) $$ \begin{align} \begin{split}&\displaystyle\left(\frac{2}{p}-1\right)\int_{B_R}|\nabla u|^p +\frac{1}{2}\int_{B_R}(1-|u|^2)^2\\&\quad=\displaystyle\frac{R}{4}\int_{\partial B_R}(1-|u|^2)^2 d\xi -R\int_{\partial B_R}|\nabla u|^{p-2}|\partial_{\nu}u|^2d\xi\\&\qquad +\displaystyle\frac{R}{p}\int_{\partial B_R} |\nabla u|^pd\xi, \end{split} \end{align} $$

and

(6.11) $$ \begin{align} \begin{split}&\displaystyle \left(1-\frac{2}{p}\right)\int_{B_R}|\nabla u|^p +\int_{B_R}(|u|^2-\frac{1}{2}|u|^4)\\&\quad=\displaystyle\frac{R}{2}\int_{\partial B_R}(|u|^2-\frac{1}{2}|u|^4) d\xi +R\int_{\partial B_R}|\nabla u|^{p-2}|\partial_{\nu}u|^2d\xi\\&\qquad -\displaystyle\frac{R}{p}\int_{\partial B_R} |\nabla u|^pd\xi. \end{split} \end{align} $$

Step 3. We claim that there exists $R_0>0$ such that either $|u|<\frac {1}{4}$ or $|u|>T$ on $\mathbb {R}^2 \setminus B_{R_0}$ , where

$$ \begin{align*}T:=\left \{\! \begin{array}{ll} 3/4, \quad & when \ p \in (1,2);\\ \sqrt{p/(3p-4)}, \quad & when \ p>2. \end{array}\right. \end{align*} $$

In fact, the following set is bounded

$$ \begin{align*}S:=\left\{x \in \mathbb{R}^2; \frac{1}{4} \leq |u(x)| \leq T\right\}. \end{align*} $$

Otherwise, there exists a sequence $x_m \to \infty $ (when $m \to \infty $ ) such that $\frac {1}{4} \leq |u(x_m)| \leq T$ . In view of (1.17) and (6.1), by the Tolksdorf theorem (cf. [Reference Tolksdorf22]), we have $|\nabla u(x)| \leq C$ for each $x \in \mathbb {R}^2$ . Therefore, for $\sigma \in (0,1-T)$ , we can find $\delta \in (0,1)$ which is independent of m such that

$$ \begin{align*}\frac{1}{8} \leq |u(x)| \leq T+\sigma, \quad when \ x \in B(x_m,\delta). \end{align*} $$

Therefore,

(6.12) $$ \begin{align} \int_{B(x_m,\delta)}|u|^2(1-|u|^2) \geq \left(\frac{1}{8}\right)^2 (1-(T+\sigma)^2)\pi\delta^2:=M_*. \end{align} $$

Clearly, $M_*$ is independent of m. On the other hand, by (6.9), we can find $R_*>0$ such that

(6.13) $$ \begin{align} \int_{|x|>R_*}|u|^2(1-|u|^2)<M_*. \end{align} $$

Noting $B(x_m,\delta ) \subset \mathbb {R}^2 \setminus B_{R_*}$ for sufficiently large m, (6.13) contradicts with (6.12). This implies S is bounded. Namely, there exists $R_0>0$ such that $S \subset B_{R_0}$ . Since $\mathbb {R}^2 \setminus B_{R_0}$ is connected and u is continuous, then either $|u|<\frac {1}{4}$ or $|u|>T$ on $\mathbb {R}^2 \setminus B_{R_0}$ in view of the definition of S.

Step 4. When $p \in (1,2)$ , we will prove Theorem 1.4.

When $|u|<\frac {1}{4}$ on $\mathbb {R}^2 \setminus B_{R_0}$ , (6.9) and (6.1) lead to

(6.14) $$ \begin{align} u \in L^2(\mathbb{R}^2) \cap L^4(\mathbb{R}^2). \end{align} $$

This implies

$$ \begin{align*}\frac{R_k}{2}\int_{\partial B_{R_k}}(|u|^2-\frac{1}{2}|u|^4) d\xi \to 0 \end{align*} $$

when $R_k \to \infty $ . Inserting this and (6.2) into (6.11) with $R=R_k$ , we get

(6.15) $$ \begin{align} \left(1-\frac{2}{p}\right)\int_{\mathbb{R}^2}|\nabla u|^p +\int_{\mathbb{R}^2}(|u|^2-\frac{1}{2}|u|^4)=0. \end{align} $$

Inserting (6.9) into (6.15) yields

$$ \begin{align*}\left(\frac{3}{2}-\frac{2}{p}\right)\int_{\mathbb{R}^2}|u|^4=\left(2-\frac{2}{p}\right)\int_{\mathbb{R}^2}|u|^2 \geq \left(2-\frac{2}{p}\right)\int_{\mathbb{R}^2}|u|^4. \end{align*} $$

This implies $|u| \equiv 0$ .

When $|u|>3/4$ on $\mathbb {R}^2 \setminus B_{R_0}$ , (6.9) and (6.1) lead to

(6.16) $$ \begin{align} 1-|u|^2 \in L^2(\mathbb{R}^2). \end{align} $$

This implies

$$ \begin{align*}\frac{R_k}{4}\int_{\partial B_{R_k}}(1-|u|^2)^2 d\xi \to 0 \end{align*} $$

when $R_k \to \infty $ . Inserting this and (6.2) into (6.10), we get

(6.17) $$ \begin{align} \left(\frac{2}{p}-1\right)\int_{\mathbb{R}^2}|\nabla u|^p +\frac{1}{2}\int_{\mathbb{R}^2}(1-|u|^2)^2 =0. \end{align} $$

In view of $p \in (1,2)$ , (6.17) implies $|\nabla u|=1-|u|^2=0$ on $\mathbb {R}^2$ . Therefore, $u \equiv C$ with $|C|=1$ .

Step 5. When $p>2$ , we will prove Theorem 1.4.

When $|u|<\frac {1}{4}$ on $\mathbb {R}^2 \setminus B_{R_0}$ , by (6.1) and (6.15) we get $u \equiv 0$ .

When $|u|>T$ on $\mathbb {R}^2 \setminus B_{R_0}$ , (6.17) still holds. Combining with (6.9) leads to

$$ \begin{align*}\left(2-\frac{4}{p}\right)\int_{\mathbb{R}^2}|u|^2(1-|u|^2)=\int_{\mathbb{R}^2}(1-|u|^2)^2. \end{align*} $$

Namely,

$$ \begin{align*}\int_{\mathbb{R}^2}(1-|u|^2)[(3-\frac{4}{p})|u|^2-1]=0. \end{align*} $$

By (6.1) and $|u|>T=\sqrt {p/(3p-4)}$ , we have $|u| \equiv 1$ on $\mathbb {R}^2$ . Inserting this into (6.9) we see that $u \equiv C$ with $|C|=1$ .