Proof. Like [16], [17], by using the moving planes technique together with the Kelvin transform as in [26], we can prove the existence of positive constants \(\delta\) and \(\gamma\) depending only on \(\Omega\), such that the following holds:

• For each \(x\in\Omega_{2\delta}:=\{y\in\Omega: dist(y,\partial\Omega)<2\delta\}\), there exists a measurable set \(I_x\) such that

  • \(I_x\subset \Omega\setminus\Omega_{\delta}=\{y\in\Omega: dist(y,\partial\Omega)\geq\delta\}\),

  • \(|I_x|\geq \gamma\), where \(|I_x|\) denotes the Lebesgue measure of \(I_x\),

  • \(u_{k,p}(x)\leq u_{k,p}(\xi)\) for all \(\xi\in I_x\) and \(k=1,2\).

For instance, when \(\Omega\) is convex, \(I_x\) can be taken to be a part of a cone with a vertex at \(x\), while non-convex domains can be treated by the Kelvin inversion of neighborhoods of boundary points where the boundary is not convex. In particular, this implies the assertion (1).

To prove the assertion (2), we denote \(\psi\) to be the first eigenfunction of \(-\Delta\) in \(H_0^1(\Omega)\) with the first eigenvalue \(\lambda_1=\lambda_1(\Omega)\): \[\begin{align} \label{fc-lambda1} \begin{cases} -\Delta \psi=\lambda_1 \psi \quad\text{in}\quad\Omega,\\ \psi>0\quad\text{in }\Omega,\quad \psi=0\quad\text{on}\;\partial\Omega, \end{cases} \text{normalized such that}\;\|\psi\|_{\infty}=1. \end{align}\tag{9}\] It follows that for \(k\in\{1,2\}\), \[\begin{align} \label{fc2-2} \lambda_1 \int_{\Omega} u_{k,p} \psi ~dx=\int_{\Omega}( \mu_k u_{k,p}^{p}+\beta u_{k,p}^{\frac{p-1}{2}}u_{3-k,p}^{\frac{p+1}{2}})\psi ~dx\geq \mu_k\int_{\Omega} u_{k,p}^{p} \psi dx. \end{align}\tag{10}\] Since \[\mu_k u_{k,p}^{p}\geq 2\lambda_1 u_{k,p}\quad \text{if}\quad u_{k,p}\geq t_{0}:=\sup_{p\geq p_0}\left(\frac{2\lambda_1}{\min\{\mu_1,\mu_2\}}\right)^{\frac{1}{p-1}},\] we have \[\begin{align} &2\lambda_1 \int_{\Omega} (u_{1,p}+u_{2,p})\psi ~dx\nonumber\\ =&\sum_{k=1}^2 2\lambda_1\int_{\{u_{k,p}>t_{0}\}}u_{k,p}\psi ~dx+ \sum_{k=1}^2 2\lambda_1\int_{\{u_{k,p}\leq t_{0}\}}u_{k,p}\psi ~dx\\ \leq &\int_{\Omega} (\mu_1 u_{1,p}^p+\mu_2 u_{2,p}^p)\psi~ dx+ 4\lambda_1 t_0|\Omega|\nonumber\\ \leq & \lambda_1 \int_{\Omega} (u_{1,p}+u_{2,p})\psi ~dx+ 4\lambda_1 t_0|\Omega|,\nonumber \end{align}\] which implies \[\int_{\Omega} (u_{1,p}+u_{2,p})\psi ~dx\leq C.\] From here and 10 , we obtain \[\begin{align} \label{eq-5-17} \int_{\Omega} u_{k,p}^p \psi~dx\leq \frac{\lambda_1}{\mu_k}\int_{\Omega} u_{k,p}\psi ~dx\leq C. \end{align}\tag{11}\] Note \[c_0:= \inf_{\Omega\setminus\Omega_{\delta}} \psi>0.\] For any \(y\in\Omega_{2\delta}\), it follows from the properties (1)-(3) of \(I_y\subset \Omega\setminus\Omega_{\delta}\) that \[\begin{align} \int_{\Omega} u_{k,p}^p \psi~dx\geq \int_{I_y} u_{k,p}^p \psi~dx\geq \gamma u_{k,p}^p(y) \inf_{I_y} \psi \geq \gamma c_0 u_{k,p}^p(y). \end{align}\] Therefore, \[\begin{align} \int_{\Omega} u_{k,p}^p dx &\leq \int_{\Omega_{2\delta}} u_{k,p}^p dx +\int_{\Omega\setminus\Omega_{2\delta}} u_{k,p}^p dx\\ &\leq \frac{|\Omega|}{\gamma c_0} \int_{\Omega} u_{k,p}^p \psi~dx+\frac{1}{c_0}\int_{\Omega\setminus\Omega_{2\delta}} u_{k,p}^p\psi dx\leq C\int_{\Omega} u_{k,p}^p \psi~dx. \end{align}\] From here and 11 , we obtain \(\int_{\Omega} u_{k,p}^p dx \leq C\) and \(\int_{\Omega} u_{k,p}^p dx \leq C\int_{\Omega} u_{k,p} dx\). ◻

Proof. Without loss of generality, we may assume \[\begin{align} \mathcal{M}_p:=u_{1,p}(x_{1,p})=\max_{\Omega}u_{1,p}\geq \max_{\Omega}u_{2,p}. \end{align}\] Then Lemma 6 implies \(dist(x_{1,p},\partial\Omega)\geq\delta\). Without loss of generality, we may assume \(x_{1,p}=0\) and so \(B_\delta\subset\Omega\). Furthermore, by a scaling argument as [16], we may also assume \(\Omega\subset B_1\).

Let \(G(y)\) be the Green function for \(-\Delta\) in \(\Omega\) with the Dirichlet boundary condition and the pole at the origin. Then \[G(y)=\frac{1}{2\pi}\log\frac{1}{|y|}-g(y),\] where \(g\) is harmonic in \(\Omega\) with \(g(y)=\frac{1}{2\pi}\log\frac{1}{|y|}\in [0, \frac{1}{2\pi}\log \frac{1}{\delta}]\) for \(y\in \partial\Omega\), so \(0\leq g(y)\leq \frac{1}{2\pi}\log \frac{1}{\delta}\) for all \(y\in \Omega\). Noting from Lemma 6 and Hölder inequality that \(\int_{\Omega} u_{1,p}^{\frac{p-1}{2}} u_{2,p}^{\frac{p+1}{2}} dx \leq C\), we see from the Green’s representation formula that \[\begin{align} \label{eq-5-25} \mathcal{M}_p= u_{1,p}(0)=&\frac{1}{2\pi}\int_{\Omega}\log(\frac{1}{|y|}) ( \mu_1 u_{1,p}^{p}+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}})dy\\ &-\int_{\Omega}g(y)( \mu_1 u_{1,p}^{p}+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}})dy\nonumber\\ \geq &\frac{\mu_1}{2\pi}\int_{\Omega}\log(\frac{1}{|y|}) u_{1,p}^{p} dy-C.\nonumber \end{align}\tag{12}\] Denote \[\begin{align} v_{k,p}(x):=1-\frac{u_{k,p}(\mathcal{M}_p^{-\frac{p-1}{2}}x)}{\mathcal{M}_p}, \end{align}\] then \[\begin{align} \begin{cases} \Delta v_{1,p}=\mu_1(1-v_{1,p})^p+\beta (1-v_{1,p})^{\frac{p-1}{2}}(1-v_{2,p})^{\frac{p+1}{2}}\;\text{in}\; \widetilde{\Omega}:=\mathcal{M}_p^{\frac{p-1}{2}}\Omega\subset B_{\mathcal{M}_p^{(p-1)/2}},\\ v_{1,p}(0)=0, \quad 0\leq v_{1,p}\leq 1, \quad 0\leq v_{2,p}\leq 1. \end{cases} \end{align}\] Note from 12 that \[\int_{\widetilde{\Omega}}\log(\mathcal{M}_p^{\frac{p-1}{2}}/|x|)(1-v_{1,p})^pdx\leq \frac{2\pi}{\mu_1}+\frac{2\pi C}{\mu_1 \mathcal{M}_p}.\]

If \(\mathcal{M}_p\leq 1\), we get the desired conclusion. So we assume that \(\mathcal{M}_p> 1\), which implies \(B_{\delta}\subset\Omega\subset\widetilde{\Omega}\). Since \[\begin{align} 0\leq \mu_1(1-v_{1,p})^p+\beta (1-v_{1,p})^{\frac{p-1}{2}}(1-v_{2,p})^{\frac{p+1}{2}}\leq \mu_1+\beta, \end{align}\] it follows from \(v_{1,p}(0)=0\) and the inhomogeneous Harnack inequality [27] that \[\|v_{1,p}\|_{L^\infty(B_r)}\leq Cr^2,\quad \forall r\in (0,\delta),\] and so \[\begin{align} v_{1,p}\leq \frac{C}{p},\quad\text{in}\quad B_{\frac{\delta}{\sqrt{p}}}, \end{align}\]which implies \[\begin{align} &(1-v_{1,p})^p \geq\left(1-\frac{C}{p}\right)^p\geq e^{-C},\quad\text{in}\quad B_{\frac{\delta}{\sqrt{p}}}. \end{align}\] It follows that \[\begin{align} \label{eq-5-31} \frac{2\pi}{\mu_1}+\frac{2\pi C}{\mu_1} \geq &\frac{2\pi}{\mu_1}+\frac{2\pi C}{\mu_1\mathcal{M}_p}\geq\int_{B_{\frac{\delta}{\sqrt{p}}}}\log(\mathcal{M}_p^{\frac{p-1}{2}}/|x|)(1-v_{1,p})^pdx\nonumber\\ \geq &C\int^{\frac{\delta}{\sqrt{p}}}_0 \log\left({\mathcal{M}_p^{\frac{p-1}{2}}}/{r}\right) r dr \geq \frac{C}{p}\log\left({\sqrt{p}\mathcal{M}_p^{\frac{p-1}{2}}}/{\delta}\right) \nonumber\\ \geq &C\left(\frac{\log p}{p}+\frac{p-1}{p}\log{\mathcal{M}_p}\right)\geq C\log{\mathcal{M}_p}\nonumber, \end{align}\tag{13}\] where we have used that for any \(\alpha,\rho>0\), \[\begin{align} \int_0^\rho \log(\alpha/r)r dr\geq \frac{1}{2} \log(\alpha/\rho)\rho^2. \end{align}\] This implies \({\mathcal{M}_p}\leq C\). ◻

Proof. By direct computations, we have for \(k=1,2\), \[\begin{align} &\langle x-y, \nabla u_{k,p} \rangle \Delta u_{k,p} =\operatorname{div}\left(\langle x-y, \nabla u_{k,p} \rangle \nabla u_{k,p}-\frac{|\nabla u_{k,p}|^2}{2}(x-y)\right),\\ &\langle x-y, \nabla u_{1,p} \rangle \left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)+\langle x-y, \nabla u_{2,p} \rangle \left(\mu_2 u_{2,p}^p +\beta u_{2,p}^{\frac{p-1}{2}}u_{1,p}^{\frac{p+1}{2}}\right)\\ =&\frac{1}{p+1} \operatorname{div}\left( \left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)(x-y)\right)\nonumber\\ &-\frac{2}{p+1}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right). \nonumber \end{align}\] Then multiplying the \(k\)-th equation of 1 with \(\langle x-y, \nabla u_{k,p} \rangle\) and integrating over \(\Omega'\), we obtain ?? . Similarly, we have \[\begin{align} &\partial_i u_{k,p} \Delta u_{k,p}=\operatorname{div}(\partial_i u_{k,p} \nabla u_{k,p})-\frac{1}{2} \partial_i(|\nabla u_{k,p}|^2),\\ &\partial_i u_{1,p} \left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)+\partial_i u_{2,p} \left(\mu_2 u_{2,p}^p +\beta u_{2,p}^{\frac{p-1}{2}}u_{1,p}^{\frac{p+1}{2}}\right)\\ =&\frac{1}{p+1}\partial_i\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right).\nonumber \end{align}\] Then multiplying the \(k\)-th equation of 1 with \(\partial_i u_{k,p}\) and integrating over \(\Omega'\), we obtain ?? . ◻

Proof of Theorem 1. Assume in addition that the smooth bounded domain \(\Omega\) is star-shaped. Without loss of generality, we may assume that \(\Omega\) is star-shaped with respect to the origin, i.e. \[\begin{align} \label{eq-5-34} \langle x,\Vec{n}(x)\rangle\geq 0,\quad\forall x\in\partial\Omega. \end{align}\tag{14}\] Applying ?? with \(\Omega'=\Omega\) and \(y=0\), it follows from \(\nabla u_{k,p}=-|\nabla u_{k,p}|\vec{n}\) on \(\partial\Omega\) that \[\begin{align} \label{eq-5-35} &\frac{4}{p+1}\int_{\Omega}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dx\nonumber\\ =&\int_{\partial\Omega}\left(|\nabla u_{1,p}|^2 \langle x, \Vec{n}(x)\rangle+|\nabla u_{2,p}|^2 \langle x, \Vec{n}(x)\rangle\right)dS_x. \end{align}\tag{15}\] Using Theorem 9 with \(t=1\), we get that for any \(x\in\partial\Omega\), \[|\nabla u_{k,p}(x)|=-\langle \nabla u_{k,p}(x),\vec{n}(x)\rangle\geq \inf_{\Omega}\frac{u_{k,p}(\cdot)}{dist(\cdot,\partial\Omega)}\geq C\|u_{k,p}\|_{L^1(\Omega)}.\] From here, 14 and Lemma 6, we get \[\begin{align} \int_{\partial\Omega}|\nabla u_{k,p}|^2 \langle x, \Vec{n}(x)\rangle dS_x&\geq C^2\|u_{k,p}\|_{L^1(\Omega)}^2\int_{\partial\Omega}\langle x, \Vec{n}(x)\rangle dS_x\\ &=C^2\|u_{k,p}\|_{L^1(\Omega)}^2\int_{\Omega}{\rm div}(x)dx=C_1\|u_{k,p}\|_{L^1(\Omega)}^2\\ &\geq C_2\left(\int_{\Omega}u_{k,p}^{p}dx\right)^2. \end{align}\] This, together with Lemma 7 and 15 , implies \[\begin{align} \frac{C_2}{2}\left(\sum_{k=1}^2\int_{\Omega}u_{k,p}^{p}dx\right)^2&\leq C_2\sum_{k=1}^2\left(\int_{\Omega}u_{k,p}^{p}dx\right)^2\leq \sum_{k=1}^2 \int_{\partial\Omega}|\nabla u_{k,p}|^2 \langle x, \Vec{n}(x)\rangle dS_x\\ & =\frac{4}{p+1}\int_{\Omega}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dx\\ &\leq \frac{C_3}{p}\sum_{k=1}^2\int_{\Omega}u_{k,p}^{p}dx, \end{align}\] so \[\begin{align} \label{eq-5-43} p\int_{\Omega} \left( u_{1,p}^p+ u_{2,p}^p \right)dx\leq C. \end{align}\tag{16}\] Consequently, \[\begin{align} &p\int_{\Omega} |\nabla u_{1,p}|^2+|\nabla u_{2,p}|^2 dx\nonumber\\ =&p\int_{\Omega} \left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} \right) dx\\ \leq & C p \int_{\Omega} \left( u_{1,p}^p+ u_{2,p}^p \right)dx \leq C.\nonumber \end{align}\] The proof is complete. ◻

Proof. Recalling 9 that \(\lambda_1>0\) denotes the first eigenvalue of the operator \(-\Delta\) in \(H_0^1(\Omega)\), we see from 21 that for \(p>3\), \[\begin{align} \lambda_1\int_{\Omega} u_{k,p}^2 dx\leq &\int_{\Omega} |\nabla u_{k,p}|^2 dx= \mu_{k}\int_{\Omega}u_{k,p}^{p+1}dx+\beta\int_{\Omega}u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} dx\nonumber\\ \leq& (\mu_{k}+\beta)\Vert M_{p}\Vert_{\infty}^{p-1}\int_{\Omega}u_{k,p}^2 dx, \end{align}\] so \[\begin{align} \Vert M_{p}\Vert_{\infty}^{p-1}\geq \frac{\lambda_1}{\mu_{k}+\beta}, \end{align}\] which implies ?? .

By 18 , it follows that \[\begin{align} \label{eq-4462} p\int_{\Omega}u_{k,p}^{p+1}dx =\frac{1}{\mu_{k}}p\int_{\Omega} |\nabla u_{k,p}|^2 dx-\frac{\beta}{\mu_{k}} p\int_{\Omega}u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} dx\leq C. \end{align}\tag{22}\] This, together with Hölder inequality, shows that \[\begin{align} p\int_{\Omega}u_{k,p}^{p} dx\leq \left(p\int_{\Omega}u_{k,p}^{p+1} dx \right)^{\frac{p}{p+1}}\left(p|\Omega|\right)^{\frac{1}{p+1}}\leq C. \end{align}\] On the other hand, by Lemma 10, we have \[\begin{align} \label{eq-44618} \left(\int_{\Omega}u_{k,p}^{p+1}dx\right)^{\frac{2}{p+1}}\leq S_p^2 (p+1)\int_{\Omega} |\nabla u_{k,p}|^2 dx, \end{align}\tag{23}\] where \(S_p^{-2}=8\pi e+o(1)\) as \(p\to+\infty\). This, together with 22 and Hölder inequality, gives that for \(p\) large, \[\begin{align} &(8\pi e+o(1))\left(\int_{\Omega}u_{k,p}^{p+1}dx\right)^{\frac{2}{p+1}}\\ \leq&(p+1)\int_{\Omega} |\nabla u_{k,p}|^2 dx=\mu_{k}(p+1)\int_{\Omega}u_{k,p}^{p+1}dx+\beta(p+1)\int_{\Omega}u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} dx\nonumber\\ \leq &(p+1)\left(\mu_{k}\left(\int_{\Omega}u_{k,p}^{p+1}dx\right)^{\frac{1}{2}}+\beta\left(\int_{\Omega}u_{3-k,p}^{p+1}dx\right)^{\frac{1}{2}}\right)\left(\int_{\Omega}u_{k,p}^{p+1}dx\right)^{\frac{1}{2}}\nonumber\\ \leq & C(p+1)^\frac{1}{2}\left(\int_{\Omega}u_{k,p}^{p+1}dx\right)^{\frac{1}{2}},\nonumber \end{align}\] and so \[(p+1)\left(\int_{\Omega}u_{k,p}^{p+1}dx\right)^{1-\frac{4}{p+1}}\geq C.\] Thus ?? holds and \[C_1\leq p\int_{\Omega}u_{k,p}^{p+1} dx \leq Cp\int_{\Omega}u_{k,p}^{p} dx,\] i.e. ?? holds.

Finally, assume by contradiction that ?? does not hold for some \(k\in\{1,2\}\), namely up to a subsequence, \(p\|u_{k,p}\|_{\infty}\leq C\). Then \[C_1\leq p\int_{\Omega}u_{k,p}^{p+1} dx\leq C\frac{C^{p+1}}{p^p}\to 0,\quad\text{as }p\to\infty,\] a contradiction. Thus ?? holds. This completes the proof. ◻

Proof. By direct computations, we have \[\begin{align} \label{eq-3-50} \begin{cases} -\Delta v_{1,p,1}=\mu_1\left(1+\frac{v_{1,p,1}}{p}\right)^p+\beta\left(1+\frac{v_{1,p,1}}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{2,p,1}}{p}\right)^{\frac{p+1}{2}},\\ -\Delta v_{2,p,1}=\mu_2\left(1+\frac{v_{2,p,1}}{p}\right)^p+\beta\left(1+\frac{v_{2,p,1}}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{1,p,1}}{p}\right)^{\frac{p+1}{2}},\\ v_{k,p,1}\leq 0,\quad \max\{v_{1,p,1}(0), v_{2,p,1}(0)\}=0,\\ 0\leq 1+\frac{v_{k,p,1}(z)}{p}=\frac{u_{k,p}(\varepsilon_{1,p}z+x_{p,1})}{M_p(x_{p,1})}\leq 1,\quad k=1,2. \end{cases} \end{align}\tag{28}\] For any \(R>1\), we have \(B_R(0)\subset\widetilde{\Omega}_{p,1}\) for \(p\) large by 27 . For \(k\in\{1,2\}\), we let \[\begin{align} v_{k,p,1}=\phi_{k,p,1}+\psi_{k,p,1} \quad\text{ in }\,\,B_R(0), \end{align}\] where \[\begin{cases} -\Delta \phi_{k,p,1}=-\Delta v_{k,p,1} \quad\text{in }\; B_R(0),\\ \phi_{k,p,1} =0\quad\text{on }\;\partial B_R(0), \end{cases} \, \begin{cases} -\Delta \psi_{k,p,1}=0 \quad\text{in }\; B_R(0),\\ \psi_{k,p,1}=v_{k,p,1} \quad\text{on }\;\partial B_R(0). \end{cases}\] It follows from 28 that \(|\Delta \phi_{k,p,1}|\leq \mu_k+\beta\). By the standard elliptic theory, \(\phi_{k,p,1}\) is uniformly bounded in \(B_R(0)\) for \(p\) large. From here and \(v_{k,p,1}\leq 0\), we see that \(\psi_{k,p,1}\) is uniformly bounded from above in \(B_R(0)\). Then by the Harnack inequality, there are two possibilities:

• \(\psi_{k,p,1}\) is uniformly bounded in \(B_{R/2}(0)\), and so is \(v_{k,p,1}\).

• \(\psi_{k,p,1}\rightarrow -\infty\) uniformly in \(B_{R/2}(0)\), and so is \(v_{k,p,1}\).

Note that once (1) (resp. (2)) holds for some \(R>1\), then (1) (resp. (2)) holds for all \(R>1\). Since \(\max\{v_{1,p,1}(0),v_{2,p,1}(0)\}=0\), there is \(k\in \{1,2\}\) such that \(v_{k,p,1}(0)=0\). Then the above argument implies that there are two possibilities:

• Both \(v_{1,p,1}\) and \(v_{2,p,1}\) are locally uniformly bounded in \(\mathbb{R}^2\) (this case happens if \(v_{3-k,p,1}(0)\) is uniformly bounded).

• \(v_{k,p,1}\) is locally uniformly bounded in \(\mathbb{R}^2\) and \(\psi_{3-k,p,1}\rightarrow -\infty\) uniformly in any compact subset of \(\mathbb{R}^2\) (this case happens if \(v_{3-k,p,1}(0)\to-\infty\)).

From here and the standard elliptic theory, up to a subsequence, we obtain that

• either \((v_{1,p,1},v_{2,p,1})\rightarrow (V_{1},V_{2})\) in \(C^{2}_{loc}(\mathbb{R}^2)\times C^{2}_{loc}(\mathbb{R}^2)\), where \[\begin{align} \label{fc-3-26} \begin{cases} -\Delta V_{1} = \mu_1 e^{V_{1}} +\beta e^{\frac{V_{1}+V_{2}}{2}} \quad\text{in} \, \mathbb{R}^2,\\ -\Delta V_{2}= \mu_2 e^{V_{2}} +\beta e^{\frac{V_{1}+V_{2}}{2}} \quad\text{in} \, \mathbb{R}^2,\\ V_{1}\leq 0,\quad V_{2}\leq 0,\quad \max\{V_{1}(0), V_{2}(0)\}=0, \end{cases} \end{align}\tag{29}\]

• or \(v_{k,p,1}\rightarrow \widetilde{V}_{k}\) in \(C^{2}_{loc}(\mathbb{R}^2)\), where \[\begin{align} \label{fc-3-27} \begin{cases} -\Delta \widetilde{V}_{k}=\mu_{k} e^{\widetilde{V}_{k}} \quad\text{in} \, \mathbb{R}^2,\\ V_{k,1}\leq 0=V_{k,1}(0), \end{cases} \end{align}\tag{30}\] and \(v_{3-k,p,1}\rightarrow -\infty\) uniformly in any compact subset of \(\mathbb{R}^2\).

For any \(R>1\), by the dominated convergence theorem, Proposition 12, 20 and 26 , we have that for the first case 29 , \[\begin{align} \label{fc-3-29} \int_{B_R(0)} e^{V_{k}} dz =&\lim_{p\to+\infty}\int_{B_R(0)}\left(1+\frac{v_{k,p,1}}{p}\right)^pdz\\ =& \lim_{p\to+\infty}\int_{B_R(0)} \frac{u_{k,p}(\varepsilon_{p,1}z+x_{p,1})^p}{M_p(x_{p,1})^p} dz\nonumber\\ = &\lim_{p\to+\infty}\frac{p}{M_p(x_{p,1})}\int_{B_{R\varepsilon_{p,1}}(x_{p,1})} u_{k,p}^p dx\leq C,\nonumber \end{align}\tag{31}\] so \[\int_{\mathbb{R}^2} e^{V_{1}} dz<+\infty,\quad \int_{\mathbb{R}^2} e^{V_{2}} dz<+\infty,\] and then the assertion ?? follows directly from Theorem 11 (Note that we assume \(\mu_1\geq \mu_2\), from which and Theorem 11 we must have \(V_{1}(0)\leq V_{2}(0)=0\)).

Similarly, for the second case 30 , we have \(\int_{\mathbb{R}^2} e^{\widetilde{V}_{k}} dz<+\infty\), and then the assertion ?? follows directly from Chen-Li’s famous result [29].

The proof is complete. ◻

Proof. In this proof, we adopt some ideas from [23] and [24]. We divide the proof into several steps.

Step 1. We prove that for some \(n\geq 1\), if \((\mathcal{P}_1^n)\)-\((\mathcal{P}_2^n)\) hold, then either \((\mathcal{P}_1^{n+1})\)-\((\mathcal{P}_2^{n+1})\) hold or \((\mathcal{P}_3^{n+1})\) holds.

Assume that \((\mathcal{P}_1^n)\)-\((\mathcal{P}_2^n)\) hold but \((\mathcal{P}_3^{n+1})\) fails, then up to a subsequence, there exists \(x_{p,n+1}\in\Omega\) such that \[\begin{align} \label{eq-3-64} p R_{p,n}(x_{p,n+1})^2M_p(x_{p,n+1})^{p-1}:=\max_{x\in\Omega} p R_{p,n}(x)^2M_p(x)^{p-1}\to+\infty. \end{align}\tag{36}\] Define \[\begin{align} \label{fc-vpn431} \varepsilon_{p,n+1}^{-2}:=p M_p(x_{p,n+1})^{p-1}\to+\infty, \end{align}\tag{37}\] and \[\begin{align} v_{k,p,n+1}(z):=\frac{p}{M_p(x_{p,n+1})}\left(u_{k,p}(\varepsilon_{p,n+1}z+x_{p,n+1})-M_p(x_{p,n+1})\right),\; k=1,2, \end{align}\] for \(z\in \widetilde{\Omega}_{p,n+1}:=\left(\Omega-x_{p,n+1}\right)/\varepsilon_{p,n+1}\). Similarly as 28 , we have \[\begin{align} \label{fc-3-41} \begin{cases} -\Delta v_{1,p,n+1}=\mu_1\left(1+\frac{v_{1,p,n+1}}{p}\right)^p+\beta\left(1+\frac{v_{1,p,n+1}}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{2,p,n+1}}{p}\right)^{\frac{p+1}{2}},\\ -\Delta v_{2,p,n+1}=\mu_1\left(1+\frac{v_{2,p,n+1}}{p}\right)^p+\beta\left(1+\frac{v_{2,p,n+1}}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{1,p,n+1}}{p}\right)^{\frac{p+1}{2}},\\ 1+\frac{v_{k,p,n+1}(z)}{p}=\frac{u_{k,p}(\varepsilon_{p,n+1}z+x_{p,n+1})}{M_p(x_{p,n+1})}\geq 0, \; \max\{v_{1,p,n+1}(0), v_{2,p,n+1}(0)\}=0. \end{cases} \end{align}\tag{38}\]

Step 1-1. We prove that \((\mathcal{P}_1^{n+1})\) holds.

First, we prove that for any \(i\in\{1,\ldots,n\}\), \[\begin{align} \label{eq-3-67} \lim\limits_{p\rightarrow\infty}\frac{|x_{p,i}-x_{p,n+1}|}{\varepsilon_{p,i}}=+\infty. \end{align}\tag{39}\] Assume by contradiction that up to a subsequence, there exists \(y_0\in\mathbb{R}^2\) such that \[\begin{align} \frac{x_{p,n+1}-x_{p,i}}{\varepsilon_{p,i}}\to y_0,\quad\text{for some }\;i\in\{1,\ldots,n\}. \end{align}\] We may also assume \(M_p(x_{p,n+1})=u_{k_0,p}(x_{p,n+1})\) for some \(k_0\in \{1,2\}\). Then by \((\mathcal{P}_2^n)\) and \[1+\frac{v_{k,p,i}(z)}{p}=\frac{u_{k,p}(\varepsilon_{p,i}z+x_{p,i})}{M_p(x_{p,i})}\geq 0,\] we have \[\begin{align} &p|x_{p,i}-x_{p,n+1}|^2 M_p(x_{p,n+1})^{p-1}\nonumber\\ =&p|x_{p,i}-x_{p,n+1}|^2 u_{k_0,p}(x_{p,n+1})^{p-1}\nonumber\\ =&\left(\frac{|x_{p,n+1}-x_{p,i}|}{\varepsilon_{p,i}}\right)^2 \left(1+\frac{v_{k_0,p,i}(\frac{x_{p,n+1}-x_{p,i}}{\varepsilon_{p,i}})}{p}\right)^{p-1}\\ \leq&|y_0|^2e^{\max\{V_{1}(y_0), V_{2}(y_0), \widetilde{V}_1(y_0), \widetilde{V}_2(y_0)\}}+o(1)\leq C\nonumber, \end{align}\] for \(p\) large, which is a contradiction with 36 . This proves 39 . Furthermore, it follows from 36 that \[\begin{align} \label{eq-3-70} \frac{|x_{p,i}-x_{p,n+1}|}{\varepsilon_{p,n+1}}\geq \left(p R_{p,n}(x_{p,n+1})^2M_p(x_{p,n+1})^{p-1}\right)^{\frac{1}{2}}\rightarrow+\infty, \quad\forall i\leq n. \end{align}\tag{40}\] Thus, \((\mathcal{P}_1^{n+1})\) follows from 39 , 40 and \((\mathcal{P}_1^n)\).

Step 1-2. We prove the convergence of \(v_{k,p,n+1}\) such that \((\mathcal{P}_2^{n+1})\) holds.

Notice that comparing to 20 and Lemma 13, the different thing here is that \(x_{p,n+1}\) is no longer a local maximum point of \(M_{p}\), so we need to control \(M_p\) by \(M_p(x_{p,n+1})\) near \(x_{p+1}\). Fix any \(\eta\in (0,1)\) and any large \(R>1\). By 40 , we have that for \(p\) large and for any \(z\in B_R(0)\cap\widetilde{\Omega}_{p,n+1}\), \[\begin{align} \frac{|x_{p,i}-x_{p,n+1}|}{\varepsilon_{p,n+1}}\geq \frac{R}{\eta} \geq \frac{|z|}{\eta},\quad\forall i\leq n, \end{align}\] \[\begin{align} |x_{p,i}-x_{p,n+1}-\varepsilon_{p,n+1}z| \geq (1-\eta)|x_{p,i}-x_{p,n+1}|,\;\forall i\leq n, \end{align}\] and so \[\begin{align} 0<\frac{R_{p,n}(x_{p,n+1})}{R_{p,n}(\varepsilon_{p,n+1}z+x_{p,n+1})}\leq \frac{1}{1-\eta}. \end{align}\] From here and 36 , we obtain that for \(p\) large and for any \(z\in B_R(0)\cap\widetilde{\Omega}_{p,n+1}\), \[\begin{align} \label{eq-3-74} \left(\frac{M_p(\varepsilon_{p,n+1}z+x_{p,n+1})}{M_p(x_{p,n+1})}\right)^{p-1}\leq &\frac{R_{p,n}(x_{p,n+1})^2}{R_{p,n}(\varepsilon_{p,n+1}z+x_{p,n+1})^2} \leq \frac{1}{(1-\eta)^2}. \end{align}\tag{41}\] This, together with 38 , implies that \(|\Delta v_{1,p,n+1}|\) and \(|\Delta v_{2,p,n+1}|\) are both uniformly bounded in \(B_R(0)\cap\widetilde{\Omega}_{p,n+1}\) for \(p\) large. Recall from 37 that \(\varepsilon_{p,n+1}\rightarrow 0\) as \(p\rightarrow\infty\) and \[\begin{align} \label{fc-3-43} \liminf_{p\rightarrow\infty}M_p(x_{p,n+1})\geq 1. \end{align}\tag{42}\] Since \(t^{p}-1-p(t-1)\geq 0\) for all \(t\geq 0\), we see from 41 that for \(p\) large and for any \(z\in B_R(0)\cap\widetilde{\Omega}_{p,n+1}\), \[\begin{align} v_{k,p,n+1}(z)\leq & p\left(\frac{M_p(\varepsilon_{p,n+1}z+x_{p,n+1})}{M_p(x_{p,n+1})}-1\right)\\ \leq &\left(\frac{M_p(\varepsilon_{p,n+1}z+x_{p,n+1})}{M_p(x_{p,n+1})}\right)^{p}-1\leq C.\nonumber \end{align}\] On the other hand, we will prove \[\label{fc-xpn} \frac{dist(x_{p,n+1},\partial\Omega)}{\varepsilon_{p,n+1}}\to+\infty,\quad\text{and so }\;\widetilde{\Omega}_{p,n+1}\to\mathbb{R}^2.\tag{43}\] Once 43 is proved, thanks to 38 and 41 43 , and since \(R>1\) is arbitrary, we can repeat the proof of Lemma 13 to obtain the convergence of \(v_{k,p,n+1}\). For example, if both \(v_{1,p,n+1}(0)\) and \(v_{2,p,n+1}(0)\) are uniformly bounded, then up to a subsequence, \((v_{1,p,n+1},v_{2,p,n+1})\rightarrow (V_{1,n+1},V_{2,n+1})\) in \(C^2_{loc}(\mathbb{R}^2)\times C^2_{loc}(\mathbb{R}^2)\), where \((V_{1,n+1},V_{2,n+1})\) is a solution of the Liouville-type system 17 . Since 41 gives that for each fixed \(z\), \[\left(1+\frac{v_{k,p,n+1}(z)}{p}\right)^{p-1}=\left(\frac{u_{k,p}(\varepsilon_{p,n+1}z+x_{p,n+1})}{M_p(x_{p,n+1})}\right)^{p-1}\leq \frac{1}{(1-\eta)^2},\] letting \(p\to\infty\), we obtain \(e^{V_{k,n+1}(z)}\leq \frac{1}{(1-\eta)^2}\). Since \(\eta\in (0,1)\) is arbitrary, we have \(e^{V_{k,n+1}(z)}\leq 1\) and so \(V_{k,n+1}(z)\leq 0\). Thus \[V_{1,n+1}\leq 0,\quad V_{2,n+1}\leq 0,\quad \max\{V_{1,n+1}(0), V_{2,n+1}(0)\}=0.\] Then the same proof of Lemma 13 implies that \((V_{1,n+1}, V_{2,n+1})=(V_1, V_2)\), where \((V_1, V_2)\) is given by 33 . The other case that \(v_{k,p,n+1}(0)\) is uniformly bounded and \(v_{3-k,p,n+1}(0)\to-\infty\) for some \(k\in\{1,2\}\) can be discussed similarly. This proves that \((\mathcal{P}_2^{n+1})\) holds.

Now we turn to prove 43 , and in the proof we omit the subscript \(n+1\) in \(v_{k,p,n+1}, \varepsilon_{p,n+1}, x_{p,n+1}, \widetilde{\Omega}_{p,n+1}\) just for convenience. Assume by contradiction that up to a subsequence, \(dist(x_{p},\partial\Omega)=O(\varepsilon_{p})\), then up to a rotation, we may assume \(\widetilde{\Omega}_{p}\rightarrow (-\infty,t_0)\times\mathbb{R}\) for some \(0\leq t_0<\infty\). Recalling 38 , without loss of generality, we may assume \[v_{1,p}(0)=0.\] Fix any \(R_0\geq t_0+1\). Then for any \(z\in B_{R_0}(0)\cap\widetilde{\Omega}_{p}\), by the Green’s representation formula and 4 , we have \[\begin{align} \label{eq-4-23} |\nabla v_{1,p}(z)|=&\frac{p\varepsilon_{p}}{M_p(x_{p})}|\nabla u_{1,p}(x_{p}+\varepsilon_{p}z)|\nonumber\\ =&\frac{p\varepsilon_{p}}{M_p(x_{p})}\left|\int_{\Omega}\nabla_x G(x_{p}+\varepsilon_{p}z,y)\left(\mu_1 u_{1,p}^{p}+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)dy\right|\\ \leq & C \frac{ p\varepsilon_{p}}{M_p(x_{p})}\int_{\Omega}\frac{1}{|x_{p}+\varepsilon_{p}z-y|}\left(\mu_1 u_{1,p}^{p}+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)dy.\nonumber \end{align}\tag{44}\] Note that for \(y\in \Omega\setminus B_{2R_0\varepsilon_{p}}(x_{p})\), \[|x_{p}+\varepsilon_{p}z-y|\geq |x_{p}-y|-\varepsilon_{p}|z|\geq \varepsilon_{p}R_0,\] so it follows from ?? and 42 that \[\begin{align} \label{eq-44624} \frac{ p\varepsilon_{p}}{M_p(x_{p})}&\int_{\Omega\setminus B_{2R_0\varepsilon_{p}}(x_{p})}\frac{1}{|x_{p}+\varepsilon_{p}z-y|}\left(\mu_1 u_{1,p}^{p}+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)dy\\ \leq \frac{p}{M_p(x_{p})R_0}&\int_{\Omega}\mu_1 u_{1,p}^{p}+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}} dy\leq C.\nonumber \end{align}\tag{45}\] For \(y\in B_{2R_0\varepsilon_{p}}(x_{p})\), using 41 with \(R=2R_0\), we have \[\max\{u_{1,p}(y), u_{2,p}(y)\}\leq (1-\eta)^{-\frac{2}{p-1}}M_p(x_p),\] This together with 37 implies that (write \(y=x_{p}+\varepsilon_{p}\tau\)) \[\begin{align} \label{eq-44625} &\frac{ p\varepsilon_{p}}{M_p(x_{p})}\int_{B_{2R_0\varepsilon_{p}}(x_{p})}\frac{1}{|x_{p}+\varepsilon_{p}z-y|}\left(\mu_1 u_{1,p}^{p}+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)dy\\ \leq &C p\varepsilon_{p} M_p(x_{p})^{p-1} \int_{B_{2R_0\varepsilon_{p}}(x_{p})}\frac{1}{|x_{p}+\varepsilon_{p}z-y|} dy\nonumber\\ = & C\int_{B_{2R_0}(0)}\frac{1}{|\tau-z|}d\tau\leq C.\nonumber \end{align}\tag{46}\] Inserting 45 and 46 into 44 , we get \(|\nabla v_{1,p}(z)|\leq C\) for any \(z\in B_{R_0}(0)\cap\widetilde{\Omega}_{p}\), so \[|v_{1,p}(z)|=|v_{1,p}(z)-v_{1,p}(0)|\leq C,\quad\forall z\in\overline{B_{R_0}(0)\cap\widetilde{\Omega}_{p}}.\] From here and \(R_0\geq t_0+1\), for \(p\) large there exists \(z_p\in\partial\widetilde{\Omega}_{p}\cap B_{R_0}(0)\) such that \(p=|v_{1,p}(z_p)|\leq C\), a contradiction with \(p\to+\infty\). This proves 43 .

Step 2. We prove there exists \(n_0\geq 1\) such that \((\mathcal{P}_1^{n_0}),\;(\mathcal{P}_2^{n_0})\) and \((\mathcal{P}_3^{n_0})\) hold.

Assume by contradiction that such \(n_0\) does not exist, then it follows from Step 1 that \((\mathcal{P}_1^{n})\) and \((\mathcal{P}_2^{n})\) hold for any \(n\geq 1\). Recalling 33 34 and letting \(b_0=\frac{4\pi}{\mu_1(1+\frac{\beta}{\sqrt{\mu_{1}\mu_{2}}})}\), then there is \(R_1>1\) such that \[\min\Big\{\int_{B_{R_1}(0)}e^{V_{k}}dz,\; \int_{B_{R_1}(0)}e^{\widetilde{V}_{k}}dz\Big\}\geq b_0,\quad k=1,2.\] Recall ?? that \[p\int_{\Omega}u_{k,p}^{p} dx \leq C_2,\quad k=1,2.\] We fix \(n>\frac{2C_2}{b_0}\). Then by \((\mathcal{P}_1^{n})\), for \(p\) large we have that for any \(i,j\in\{1,\ldots,n\}\) with \(i\neq j\), \[\begin{align} B_{\varepsilon_{p,i} R_1}(x_{p,i})\cap B_{\varepsilon_{p,j} R_1}(x_{p,j})=\emptyset. \end{align}\] Clearly those estimates in Remark 14 also hold for any \(j\), so we obtain \[\begin{align} 2C_2&\geq \liminf_{p\to+\infty}p\int_{\Omega}u_{1,p}^{p} +u_{2,p}^{p} dx\\ &\geq \liminf_{p\to+\infty}\sum_{j=1}^np\int_{B_{\varepsilon_{p,j} R_1}(x_{p,j})}u_{1,p}^{p} +u_{2,p}^{p} dx\\ &\geq n \min\Big\{\int_{B_{R_1}(0)}e^{V_{k}}dz, \int_{B_{R_1}(0)}e^{\widetilde{V}_{k}}dz\Big\}\geq nb_0>2C_2, \end{align}\] a contradiction.

Therefore, there exists \(n_0\geq 1\) such that \((\mathcal{P}_1^{n_0}),\;(\mathcal{P}_2^{n_0})\) and \((\mathcal{P}_3^{n_0})\) hold. Consequently, ?? follows directly from the definition of \(\mathcal{S}\) and \((\mathcal{P}_3^{n_0})\), and ?? follows directly from ?? and \((\mathcal{P}_3^{n_0})\).

Step 3. To prove ?? , we only need to prove that for any \(x\in\Omega\setminus\{x_{p,1},\cdots, x_{p,{n_0}}\}\) and \(k=1,2\), \[\begin{align} \label{eq-3-80} p\left|\nabla u_{k,p}(x)\right|=O\left(\sum\limits_{j=1}^{n_0}\frac{1}{|x-x_{p,j}|}\right). \end{align}\tag{47}\]

Let us take \(k=1\) for example. Define \[\begin{align} \Omega_{p,j}:=\Big\{x\in\Omega\;:\;R_{p,n_0}(x)=\min_{1\leq i\leq n_0}|x_{p,i}-x|=|x_{p,j}-x|\Big\},\;\forall j, \end{align}\] so \(\cup_j\Omega_{p,j}=\Omega\). Then by the Green’s representation formula and 4 , \[\begin{align} \label{eq-3-81} \left|p\nabla u_{1,p}(x)\right|=&\left|p\int_{\Omega}\nabla_x G(x,y) \left(\mu_1 u_{1,p}^p+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dy \right|\\ \leq & C p\int_{\Omega}\frac{1}{|x-y|} \left(\mu_1 u_{1,p}^p+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dy\nonumber\\ \leq & C p \sum\limits_{j=1}^{n_0}\left(\int_{\Omega_{p,j}}\frac{1}{|x-y|} u_{1,p}^p dy +\int_{\Omega_{p,j}}\frac{1}{|x-y|} u_{2,p}^p dy \right).\nonumber \end{align}\tag{48}\] For \(y\in\Omega_{p,j}^1:=\Omega_{p,j}\cap B_{{|x_{p,j}-x|}/{2}}(x_{p,j})\), we have \[|x-y|\geq |x-x_{p,j}|-|y-x_{p,j}|\geq {|x_{p,j}-x|}/{2},\] which together with ?? implies \[\begin{align} \label{eq-3-84} p\int_{\Omega_{p,j}^1}\frac{1}{|x-y|} u_{k,p}^p dy\leq \frac{C}{|x_{p,j}-x|} p\int_{\Omega_{p,j}} u_{k,p}^p dy\leq \frac{C}{|x_{p,j}-x|}. \end{align}\tag{49}\] For \(y\in\Omega_{p,j}^2:=\Omega_{p,j}\setminus B_{{2|x_{p,j}-x|}}(x_{p,j})\), we have \[|x-y|\geq |y-x_{p,j}|-|x-x_{p,j}|\geq {|x_{p,j}-x|},\] which also implies \[\begin{align} \label{eq-3-85} p\int_{\Omega_{p,j}^2}\frac{1}{|x-y|} u_{k,p}^p dy\leq \frac{C}{|x_{p,j}-x|} p\int_{\Omega_{p,j}} u_{k,p}^p dy\leq \frac{C}{|x_{p,j}-x|}. \end{align}\tag{50}\] For \(y\in\Omega_{p,j}^3:=\Omega_{p,j}\cap( B_{{2|x_{p,j}-x|}}(x_{p,j})\setminus B_{{|x_{p,j}-x|}/{2}}(x_{p,j}) )\), by \(R_{p,n_0}(y)=|y-x_{p,j}|\) and \((\mathcal{P}_3^{n_0})\), we have \[\begin{align} p|y-x_{p,j}|^2 u_{k,p}(y)^{p-1}\leq p R_{p,n}(y)^2M_p(y)^{p-1}\leq C, \end{align}\] which together with \(\|u_{k,p}\|_{\infty}\leq C\) implies \[\begin{align} \label{eq-3-87} &p\int_{\Omega_{p,j}^3}\frac{1}{|x-y|} u_{k,p}^pdy \leq Cp\int_{\Omega_{p,j}^3}\frac{1}{|x-y|} u_{k,p}^{p-1}dy\nonumber\\ \leq & C\int_{B_{{2|x_{p,j}-x|}}(x_{p,j})\setminus B_{{|x_{p,j}-x|}/{2}}(x_{p,j})}\frac{dy}{|x-y||y-x_{p,j}|^2} \\ \leq &\frac{C}{|x_{p,j}-x|^2}\int_{B_{{2|x_{p,j}-x|}}(0)\setminus B_{{|x_{p,j}-x|}/{2}}(0)}\frac{dy}{|y-(x-x_{p,j})|}\nonumber\\ \leq &\frac{C}{|x_{p,j}-x|^2}\int_{B_{{3|x_{p,j}-x|}}(0)}\frac{dy}{|y|}\leq \frac{C}{|x_{p,j}-x|}.\nonumber \end{align}\tag{51}\] Inserting 49 51 into 48 , we obtain 47 with \(k=1\).

The proof is complete. ◻

Proof. The proof is standard and we sketch it here for later usage. Let \(\mathscr{K}\Subset\overline{\Omega}\backslash\mathcal{S}\) be any compact subset and take \(r\in (0,r_0)\) small such that \[\begin{align} \label{new3467} \mathscr{K}\subset\overline{\Omega}\Big\backslash\left(\bigcup\limits_{i=1}^N B_{2r}(x_i)\right). \end{align}\tag{53}\] For \(x\in\mathscr{K}\) and \(m\in \{0,1\}\), by the Green’s representation formula, we have that for \(\rho\in (0,r)\), \[\begin{align} \label{3-22} p\nabla^m u_{1,p}(x)&=p \int_{\Omega}\nabla_x^m G(x,y)\left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)dy\nonumber\\ &=\sum_{i=1}^{N} \nabla_x^m G(x,x_i) p \int_{\Omega\cap B_\rho(x_i)}\left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dy\\ &+\sum_{i=1}^{N} p \int_{\Omega\cap B_\rho(x_i)}\nabla_x^m (G(x,y)- G(x,x_i))\left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dy\nonumber\\ &+p \int_{\Omega\backslash \bigcup_{i=1}^{N} B_\rho(x_i)} \nabla_x^m G(x,y)\left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dy.\nonumber \end{align}\tag{54}\] Since ?? implies \[\begin{align} \label{new3468} \Vert u_{k,p}\Vert_{L^{\infty}(\Omega\backslash \bigcup_{i=1}^{N} B_{\rho}(x_i))}\leq \frac{C_\rho}{p},\quad k=1,2, \end{align}\tag{55}\] and 4 implies \[\int_{\Omega\backslash \bigcup_{i=1}^{N} B_\rho(x_i)} |\nabla_x^m G(x,y)|dy\leq \int_{\Omega} |\nabla_x^m G(x,y)|dy \leq C,\] we obtain \[\begin{align} \label{995} \left|p \int_{\Omega\backslash \bigcup_{i=1}^{N} B_\rho(x_i)} \nabla_x^m G(x,y)\left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dy\right|\leq C\frac{C_\rho^{p}}{p^{p-1}}. \end{align}\tag{56}\] Furthermore, it follows from the fact (note 53 ) \[\begin{align} \label{new34611} |\nabla_y\nabla_x^mG(x,y)|\leq C,\quad\forall x\in\mathscr{K},\; y\in \cup_{i=1}^N B_{r}(x_i), \end{align}\tag{57}\] and ?? that \[\begin{align} &p\left|\int_{\Omega\cap B_\rho(x_i)}\nabla_x^m (G(x,y)- G(x,x_i))\left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)dy\right|\label{new34612}\\ \leq& p\int_{\Omega\cap B_\rho(x_i)}\rho|\nabla_y\nabla_x^m G(x,\lambda_y y+(1-\lambda_y)x_i)| \left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dy\leq C\rho,\nonumber \end{align}\tag{58}\] where \(\lambda_y\in (0,1)\) depends on \(y\). Inserting 56 and 58 into 54 , and letting \(p\to\infty\) first and then \(\rho\to 0\), we get \(p \nabla^m u_{1,p}(x) \to\sum_{i=1}^{N}\gamma_{1,i} \nabla_x^mG(x,x_i)\) uniformly in \(\mathcal{K}\). Thus \[pu_{1,p}(x) \to \sum_{i=1}^{N}\gamma_{1,i} G(x,x_i) \quad\text{in} \quad C_{loc}^1(\overline{\Omega}\backslash\mathcal{S}).\] Similarly, we have \(pu_{2,p}(x) \rightarrow\sum_{i=1}^{N}\gamma_{2,i} G(x,x_i)\) in \(C_{loc}^1(\overline{\Omega}\backslash\mathcal{S})\). Then by standard elliptic estimates, the convergence is also in \(C_{loc}^2(\overline{\Omega}\backslash\mathcal{S})\). ◻

Proof. The proof follows the idea from [30]. Assume by contradiction that \(\mathcal{S}\cap\partial\Omega\neq\emptyset\), say \(x_{N}\in \mathcal{S}\cap \partial\Omega\) for example. Note from 52 that \(B_{2r_0}(x_N)\cap\mathcal{S}=\{x_N\}\), and recall that \(x_N=\lim_{p\rightarrow\infty}x_{p,j}\) for some \(2\leq j\leq n_0\).

Recalling \((\mathcal{P}_{2}^{n_0})\), if Type \(\mathcal{A}\) happens for \(x_{p,j}\), similarly as Remark 14, we have that for any \(r\in(0,r_0)\), \[\begin{align} &\liminf_{p\to\infty}p\int_{B_{r}(x_N)\cap\Omega} \left(\mu_{k} u_{k,p}^{p+1} +\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dx\nonumber\\ \geq & \int_{\mathbb{R}^2}\left(\mu_{k} e^{V_{k}} +\beta e^{\frac{V_{1}+V_{2}}{2}}\right)dz=8\pi,\quad k=1,2.\nonumber \end{align}\] If Type \(\mathcal{B}\) or Type \(\mathcal{C}\) happens for \(x_{p,j}\), similarly we have that for any \(r\in(0,r_0)\), \[\begin{align} \liminf_{p\to\infty}p\int_{B_{r}(x_N)\cap\Omega} \mu_{k} u_{k,p}^{p+1}dx \geq \int_{\mathbb{R}^2}\mu_{k} e^{\widetilde{V}_{k}} dz= 8\pi\quad\text{for some }\,k\in\{1,2\}. \end{align}\] Therefore, we always have that for any \(r\in(0,r_0)\), \[\begin{align} \label{eq-3-93} \liminf_{p\to\infty}p\int_{B_{r}(x_N)\cap\Omega} \left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1} +2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dx\geq 8\pi. \end{align}\tag{59}\]

Let \(\Vec{n}(x)\) be the outer normal vector of \((\partial B_{r}(x_N)\cap\Omega)\cup(\partial\Omega\cap B_{r}(x_N))\) and define \(y_p:=x_N+\rho_{r,p}\Vec{n}(x_N)\), where \[\begin{align} \rho_{r,p}:=\frac{\int_{\partial\Omega\cap B_{r}(x_N)}\langle x-x_N, \Vec{n}(x)\rangle (|\nabla u_{1,p}|^2+|\nabla u_{2,p}|^2) dS_x}{\int_{\partial\Omega\cap B_{r}(x_N)} \langle \Vec{n}(x_N),\Vec{n}(x)\rangle (|\nabla u_{1,p}|^2+|\nabla u_{2,p}|^2) dS_x}, \end{align}\] which implies that \[\begin{align} \label{eq-3-95} \sum_{k=1}^2\int_{\partial\Omega\cap B_{r}(x_N)}|\nabla u_{k,p}|^2 \langle x-y_p, \Vec{n}(x)\rangle dS_x=0. \end{align}\tag{60}\] We can take \(r >0\) small enough such that \(\langle \Vec{n}(x_1),\Vec{n}(x)\rangle\in [\frac{1}{2},1]\) for any \(x\in\partial\Omega\cap B_{r}(x_N)\). Then \(|\rho_{r,p}|\leq 2r\). Note from \(u_{k,p}=0\) on \(\partial\Omega\) that \[\begin{align} \nabla u_{k,p}(x)=-|\nabla u_{k,p}(x)|\Vec{n}(x)\quad\text{ for }\,\,\forall x\in\partial\Omega\cap B_{r}(x_N). \end{align}\] which shows that \[\begin{align} \label{eq-3-97} &\sum_{k=1}^2\int_{\partial\Omega\cap B_{r}(x_N)}\langle \nabla u_{k,p},x-y_p\rangle \langle \nabla u_{k,p}, \Vec{n}(x)\rangle dS_x\\ =&\sum_{k=1}^2\int_{\partial\Omega\cap B_{r}(x_N)}|\nabla u_{k,p}|^2 \langle x-y_p, \Vec{n}(x)\rangle dS_x=0.\nonumber \end{align}\tag{61}\] Applying the Pohozaev identity ?? with \(\Omega'=B_{r}(x_N)\cap\Omega\) and \(y=y_p\), and using 60 and 61 , we obtain \[\begin{align} \label{eq-3-98} &\frac{2p^2}{p+1}\int_{B_{r}(x_N)\cap\Omega}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dx\\ =&\sum_{k=1}^2\int_{\partial B_{r}(x_N)\cap\Omega}\langle p\nabla u_{k,p},x-y_p\rangle \langle p\nabla u_{k,p}, \Vec{n}(x)\rangle dS_x\nonumber\\ &-\frac{1}{2}\sum_{k=1}^2\int_{\partial B_{r}(x_N)\cap\Omega}|p\nabla u_{k,p}|^2 \langle x-y_p, \Vec{n}(x)\rangle dS_x\nonumber\\ &+\frac{p^2}{p+1}\int_{\partial B_{r}(x_N)\cap\Omega}\langle x-y_p, \Vec{n}(x)\rangle\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)dS_x\nonumber\\ =&I_p+II_p+III_p.\nonumber \end{align}\tag{62}\]

Noting from \(|\rho_{r,p}|\leq 2r\) that \(|x-y_p|\leq 3r\) for any \(x\in \partial B_r(x_N)\), we see from ?? that \[\big|III_p\big|\leq C\frac{C_r^{p+1}}{p^p}\to 0\quad\text{as }\;p\to\infty.\] Up to a subsequence, we may assume \(y_p\to \tilde{y}_r\) with \(|\tilde{y}_r-x_N|\leq 2r\). Recall Lemma 16 that \[p u_{k,p}\rightarrow F_k:=\sum\limits_{i=1}^N \gamma_{k,i} G(\cdot, x_i)\quad\text{ in }\,\,C^{2}_{loc}(\overline{\Omega\cap B_{r}(x_N)}\setminus\{x_N\}).\] Since \(x_N\in\partial\Omega\) implies \(G(x,x_N)\equiv 0\) for \(x\neq x_N\), we have (see e.g. [11]) \[F_k(x)=O(1),\quad \nabla F_k(x)=O(1),\quad \text{\it for }x\in \overline{\Omega\cap B_{r_0}(x_N)}\setminus\{x_N\}.\] From here and \(0<r<r_0\), we obtain \[\begin{align} \lim_{p\to\infty}I_p =&\sum_{k=1}^2\int_{\partial B_{r}(x_N)\cap\Omega}\langle \nabla F_k,x-\tilde{y}_r\rangle \langle \nabla F_k, \Vec{n}(x)\rangle dS_x\\ =& O(1) \int_{\partial B_{r}(x_N)\cap\Omega}|x-\tilde{y}_r| d S_x=O(r^2), \end{align}\] and similarly \[\lim_{p\to\infty}II_p=O(r^2).\] Inserting these estimates into 62 , we finally obtain \[\begin{align} \lim\limits_{r\rightarrow 0}\lim\limits_{p\rightarrow\infty}\frac{2p^2}{p+1}\int_{B_{r}(x_N)\cap\Omega}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dx=0, \end{align}\] which is a contradiction with 59 . ◻

Proof. Since 64 implies \(M_p(y_{p,i})\geq M_p(x_{p,m})>0\), we have \[\begin{align} 0<\varepsilon_{p,i}^*=\left(p M_p(y_{p,i})^{p-1}\right)^{-\frac{1}{2}}\leq \left(p M_p(x_{p,m})^{p-1}\right)^{-\frac{1}{2}}=\varepsilon_{p,m}\to 0. \end{align}\] This, together with \((\mathcal{P}_3^{n_0})\), shows that \[\begin{align} \label{fc-3-75} \frac{|x_{p,m}-y_{p,i}|}{\varepsilon_{p,m}}=&\frac{|x_{p,m}-y_{p,i}|}{\varepsilon_{p,i}^*}\frac{\varepsilon_{p,i}^*}{\varepsilon_{p,m}}\leq \frac{|x_{p,m}-y_{p,i}|}{\varepsilon_{p,i}^*}\\ = &\left(p R_{p,n_0} (y_{p,i})^2 M_p(y_{p,i})^{p-1}\right)^{\frac{1}{2}}\leq C.\nonumber \end{align}\tag{67}\] Thus \[\begin{align} y_{p,i}\rightarrow x_i\;\text{and}\; \frac{B_{r_0}(x_i)-y_{p,i}}{\varepsilon_{p,i}^*}\rightarrow \mathbb{R}^2,\quad\text{as }p\to\infty. \end{align}\] Furthermore, up to a subsequence, we may assume \[\frac{y_{p,i}-x_{p,m}}{\varepsilon_{p,m}}\to z_\infty\in\mathbb{R}^2.\]

Case 1. Assuming \((v_{1,p,m},v_{2,p,m})\) satisfies type \(\mathcal{A}\), we prove that \((v_{1,p,i}^*,v_{2,p,i}^*)\) also satisfies type \(\mathcal{A}\).

By \((\mathcal{P}_2^{n_0})\), we have \[\begin{align} v_{k,p,m}\left(\frac{y_{p,i}-x_{p,m}}{\varepsilon_{p,m}}\right)\rightarrow V_{k}(z_{\infty})\leq 0,\quad k=1,2, \end{align}\] which shows that \[\begin{align} \label{eq-3-113} 1\leq & \left(\frac{\varepsilon_{p,m}}{\varepsilon_{p,i}^*}\right)^2 = \left(\frac{M_p(y_{p,i})}{M_p(x_{p,m})}\right)^{p-1} = \max\left\{\left(1+\frac{ v_{1,p,m}\left(\frac{y_{p,i}-x_{p,m}}{\varepsilon_{p,m}}\right)}{p}\right)^{p-1},\right.\\ &\left.\left(1+\frac{ v_{2,p,m}\left(\frac{y_{p,i}-x_{p,m}}{\varepsilon_{p,m}}\right)}{p}\right)^{p-1}\right\}\rightarrow \max \{e^{V_{1}(z_{\infty})},e^{V_{2}(z_{\infty})}\}\leq 1.\nonumber \end{align}\tag{68}\] This implies that \(z_\infty=0\), \[\begin{align} \frac{y_{p,i}-x_{p,m}}{\varepsilon_{p,m}}\to 0,\quad \frac{\varepsilon_{p,i}^*}{\varepsilon_{p,m}}\nearrow 1,\quad\frac{M_p(x_{p,m})}{M_p(y_{p,i})}\nearrow 1.\label{eq-3-114} \end{align}\tag{69}\] We claim that \[\begin{align} p\left(\frac{M_p(x_{p,m})}{M_p(y_{p,i})}-1\right)\to 0.\label{eq-3-1149442} \end{align}\tag{70}\] If not, then there exists \(C_1< 0\) such that up to a subsequence, \[\begin{align} p\left(\frac{M_p(x_{p,m})}{M_p(y_{p,i})}-1\right)\leq C_1. \end{align}\] It follows that \[\begin{align} 1\leftarrow \left(\frac{\varepsilon_{p,i}^*}{\varepsilon_{p,m}}\right)^2=\left(\frac{M_p(x_{p,m})}{M_p(y_{p,i})}\right)^{p-1}\leq\left(1+\frac{C_1}{p}\right)^{p-1}\to e^{C_1}<1, \end{align}\] which is a contradiction.

Recalling the definition 65 of \(v_{k,p,i}^*\) and the definition 32 of \(v_{k,p,m}\), a direct computation gives \[\begin{align} \label{fc-3-81} v_{k,p,i}^*(z)=\frac{M_p(x_{p,m})}{M_p(y_{p,i})}v_{k,p,i}\left(\frac{\varepsilon_{p,i}^*}{\varepsilon_{p,m}}z+\frac{y_{p,i}-x_{p,m}}{\varepsilon_{p,m}}\right)+ p\left(\frac{M_p(x_{p,m})}{M_p(y_{p,i})}-1\right). \end{align}\tag{71}\] From here, 69 70 , and \((\mathcal{P}_2^{n_0})\) which says that \((v_{1,p,m},v_{2,p,m})\rightarrow ({V}_{1},{V}_{2})\) in \(C^2_{loc}(\mathbb{R}^2)\times C^2_{loc}(\mathbb{R}^2)\), we conclude that \((v_{1,p,i}^*,v_{2,p,i}^*)\rightarrow ({V}_{1},{V}_{2})\) in \(C^2_{loc}(\mathbb{R}^2)\times C^2_{loc}(\mathbb{R}^2)\).

Case 2. Assuming \((v_{1,p,m},v_{2,p,m})\) satisfies type \(\mathcal{B}\) (the case type \(\mathcal{C}\) is similar), we prove that \((v_{1,p,i}^*,v_{2,p,i}^*)\) satisfies the same type \(\mathcal{B}\).

By \((\mathcal{P}_2^{n_0})\), we have \[\begin{align} v_{1,p,m}\left(\frac{y_{p,i}-x_{p,m}}{\varepsilon_{p,m}}\right)\rightarrow \widetilde{V}_{1}(z_{\infty})\leq 0,\quad v_{2,p,m}\left(\frac{y_{p,i}-x_{p,m}}{\varepsilon_{p,m}}\right)\rightarrow -\infty. \end{align}\] Then the same argument as 68 implies that 69 70 also holds. Therefore, it follows from 71 and \((\mathcal{P}_2^{n_0})\) that \(v_{1,p,i}^*\rightarrow \widetilde{V}_{1}\) in \(C^{2}_{loc}(\mathbb{R}^2)\) and and \(v_{2,p,i}^*\rightarrow -\infty\) uniformly in any compact subsets of \(\mathbb{R}^2\). This completes the proof. ◻

Proof of Theorem 3. Clearly Theorem 3 follows from Proposition 15 and Lemma 18. ◻

Proof. First, for any \(0<\rho_1<\rho_2<r_0\), it follows from ?? that \[p\int_{B_{\rho_2}(x_i)\setminus B_{\rho_1}(x_i)}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} \right)dx\leq C\frac{C^{p+1}}{p^p}\to 0,\] so the left hand side of ?? is independent of the choice of \(\rho\), namely \[\begin{align} \label{fc-4-01}\lim_{\rho\to 0} \lim\limits_{p\rightarrow \infty} p\int_{B_{\rho}(x_i)}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} \right)dx\\ =\lim\limits_{p\rightarrow \infty} p\int_{B_{\rho}(x_i)}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} \right)dx.\nonumber \end{align}\tag{72}\]

Applying the Pohozaev identity ?? with \(\Omega'=B_{\rho}(x_i)\) and \(y=x_i\), we have \[\begin{align} \label{eq-4-2-1} &\frac{2 p^2}{p+1}\int_{B_{\rho}(x_i)}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dx\nonumber\\ =&\rho\sum_{k=1}^2\int_{\partial B_{\rho}(x_i)} \langle p\nabla u_{k,p}, \Vec{n}(x)\rangle ^2 dS_x -\frac{\rho}{2}\sum_{k=1}^2\int_{\partial B_{\rho}(x_i)}|p\nabla u_{k,p}|^2dS_x\\ &+\frac{\rho p^2}{p+1}\int_{\partial B_{\rho}(x_i)}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)dS_x\nonumber, \end{align}\tag{73}\] Again by ?? , we get \[\begin{align} \frac{\rho p^2}{p+1}\int_{\partial B_{\rho}(x_i)}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right)dS_x\to 0.\label{eq-4-3-1} \end{align}\tag{74}\] By Lemma 16, we have that for \(x\in\partial B_{\rho}(x_i)\) and \(k=1,2\), \[\begin{align} p\nabla u_{k,p}=-\frac{\gamma_{k,i}}{2\pi}\frac{x-x_i}{|x-x_i|^2}+O(1), \end{align}\] which implies that \[\begin{align} &\rho\int_{\partial B_{\rho}(x_i)}\langle p\nabla u_{k,p}, \Vec{n}(x)\rangle ^2 dS_x-\frac{\rho}{2}\int_{\partial B_{\rho}(x_i)}|p\nabla u_{k,p}|^2 dS_x=\frac{\gamma_{k,i}^2}{4\pi}+O(\rho).\label{eq-4-6-1} \end{align}\tag{75}\] Inserting 74 and 75 into 73 , letting \(p\to\infty\) first and then \(\rho\to 0\), we obtain ?? . ◻

Proof. Since \(M_p(y_{p,i})=\max\limits_{x\in\overline{B_{r_0}(x_i)}}M_p(x)\), we have \[\begin{align} &p\int_{B_{\rho}(x_i)}\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} \,dx\nonumber\\ \leq & M_p(y_{p,i}) \sum_{k=1}^2p\int_{B_{\rho}(x_i)} \mu_k u_{k,p}^p +\beta u_{k,p}^{\frac{p-1}{2}}u_{3-k,p}^{\frac{p+1}{2}} dx,\nonumber \end{align}\] which together with ?? shows that \[\begin{align} \label{eq-4-59442} &\frac{1}{8\pi}(\gamma_{1,i}^2+\gamma_{2,i}^2) \leq (\gamma_{1,i}+\gamma_{2,i})\lim_{p\rightarrow\infty}M_p(y_{p,i}) . \end{align}\tag{79}\] Inserting \(\gamma_{k,i}=\sigma_{k,i}\lim\limits_{p\rightarrow\infty}M_p(y_{p,i})\) into 79 and using \(1\leq \lim\limits_{p\rightarrow\infty}M_p(y_{p,i})<+\infty\), we obtain 78 . ◻

Proof. Similarly as Remark 14, we have \[\begin{align} \sigma_{k,i}= &\lim\limits_{p\rightarrow \infty} \frac{p}{M_p(y_{p,i})} \int_{B_{\rho}(x_{i})} \left(\mu_k u_{k,p}^p +\beta u_{k,p}^{\frac{p-1}{2}}u_{3-k,p}^{\frac{p+1}{2}}\right) dx\nonumber\\ \geq & \int_{\mathbb{R}^2}\left(\mu_{k} e^{V_{k}} +\beta e^{\frac{V_{1}+V_{2}}{2}}\right)dz=8\pi,\quad k=1,2.\nonumber \end{align}\] From here and Corollary 1, we obtain \(\sigma_{1,i}=\sigma_{2,i}=8\pi\). ◻

Proof. Take \(k=1\) for example. There exists large \(R_{\eta}>1\) such that \[\begin{align} \int_{B_{R_{\eta}}(0)} \mu_1 e^{V_{1}} +\beta e^{\frac{{V}_{1}+{V}_{2}}{2}} dz\geq 8\pi-\frac{\eta}{2}. \end{align}\] Denote \[\begin{align} \label{fc-gp} g_p(z):=\mu_1\left(1+\frac{v_{1,p,i}^*(z)}{p}\right)^p+\beta\left(1+\frac{v_{1,p,i}^*(z)}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{2,p,i}^*(z)}{p}\right)^{\frac{p+1}{2}}. \end{align}\tag{80}\] Then by \((v_{1,p,i}^*, v_{2,p,i}^*)\to (V_1, V_2)\) in \(C_{loc}^2(\mathbb{R}^2)\times C_{loc}^2(\mathbb{R}^2)\), there exists large \(p_{\eta}>1\) such that for all \(p>p_{\eta}\), \[\begin{align} \label{eq-4-18} &\int_{B_{R_{\eta}}(0)} g_p(z) dz\geq \int_{B_{R_{\eta}}(0)} \mu_1 e^{V_{1}} +\beta e^{\frac{{V}_{1}+{V}_{2}}{2}} dz-\frac{\eta}{2}\geq 8\pi-\eta. \end{align}\tag{81}\] Furthermore, letting \(\delta=r_0/3\), it follows from ?? and 76 that \[\begin{align} \lim_{p\to\infty}\int_{B_{\frac{2\delta}{\varepsilon_{p,i}^*}}(0)}g_p(w)dw=\lim_{p\to\infty}\frac{p}{M_p(y_{p,i})}\int_{B_{2\delta}(y_{p,i})} \mu_{1}u_{1,p}^{p}+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}} dx=\sigma_{1,i}, \end{align}\] so by taking \(p_{\eta}\) larger if necessary, we have that for all \(p>p_{\eta}\), \[\begin{align} \label{eq-4-18-0}\int_{B_{\frac{2\delta}{\varepsilon_{p,i}^*}}(0)}g_p(w)dw\leq \sigma_{1,i}+\eta. \end{align}\tag{82}\]

By the Green’s representation formula, we have \[\begin{align} u_{1,p}(\varepsilon_{p,i}^* z+y_{p,i})=\int_{\Omega} G(\varepsilon_{p,i}^* z+y_{p,i},y) \left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}} u_{2,p}^{\frac{p+1}{2}}\right) dy. \end{align}\] Denoting \(\widetilde{\Omega}_{p,i}:=\frac{\Omega-y_{p,i}}{\varepsilon_{p,i}^*}\), it follows from 65 that \[\begin{align} v_{1,p,i}^*(z)=-p+\int_{\widetilde{\Omega}_{p,i}} G(\varepsilon_{p,i}^* z+y_{p,i},\varepsilon_{p,i}^* w+y_{p,i}) g_p(w)dw. \end{align}\] From here and 5 , we have \[\begin{align} \label{eq-4-22} &v_{1,p,i}^*(z)-v_{1,p,i}^*(0)\nonumber\\ =&\int_{\widetilde{\Omega}_{p,i}} \left[G(\varepsilon_{p,i}^* z+y_{p,i},\varepsilon_{p,i}^* w+y_{p,i})-G(y_{p,i},\varepsilon_{p,i}^* w+y_{p,i})\right] g_p(w) dw\\ =&\frac{1}{2\pi}\int_{B_{\frac{2\delta}{\varepsilon_{p,i}^*}}(0)} \log{\frac{|w|}{|w-z|}} g_p(w) dw+\frac{1}{2\pi}\int_{\widetilde{\Omega}_{p,i}\setminus B_{\frac{2\delta}{\varepsilon_{p,i}^*}}(0)} \log{\frac{|w|}{|w-z|}} g_p(w) dw\nonumber \\ &-\int_{\widetilde{\Omega}_{p,i}} \left[H(\varepsilon_{p,i}^* z+y_{p,i},\varepsilon_{p,i}^* w+y_{p,i})-H(y_{p,i},\varepsilon_{p,i}^* w+y_{p,i})\right] g_p(w) dw\nonumber\\ =&I(z)+II(z)+III(z).\nonumber \end{align}\tag{83}\]

Now we let \(2R_{\eta}\leq |z|\leq \frac{\delta}{\varepsilon_{p,i}^*}\) and \(p>p_{\eta}\). Since \(H(\cdot,\cdot)\) is bounded in \(B_{r_0}(x_i)\times\Omega\), it follows that \[\begin{align} III(z)&=O(1)\int_{\widetilde{\Omega}_{p,i}}g_p(w) dw\\ &=O(1)\frac{p}{M_p(y_{p,i})}\int_{\Omega}\mu_{1}u_{1,p}^{p}+\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}} dx=O(1).\nonumber \end{align}\] For \(|z|\leq {\delta}/{\varepsilon_{p,i}^*}\) and \(|w|\geq {2\delta}/{\varepsilon_{p,i}^*}\), we have \(\frac{2}{3}\leq \frac{|w|}{|w-z|}\leq 2\), which shows that \[\begin{align} II(z)=O(1)\int_{\widetilde{\Omega}_{p,i}\setminus B_{{2\delta}/{\varepsilon_{p,i}^*}}(0)}g_p(w) dw=O(1). \end{align}\] Next, we divide \(I(z)\) into the following parts: \[\begin{align} I(z)=&\frac{1}{2\pi}\int_{B_{R_{\eta}}(0)} \log{\frac{|w|}{|w-z|}} g_p(w) dw\\ &+\frac{1}{2\pi}\int_{B_{{2\delta}/{\varepsilon_{p,i}^*}}(0)\setminus B_{R_{\eta}}(0) \cap\{|w|\leq 2|z-w|\}} \log{\frac{|w|}{|w-z|}} g_p(w) dw\nonumber\\ &+\frac{1}{2\pi}\int_{B_{{2\delta}/{\varepsilon_{p,i}^*}}(0)\setminus B_{R_{\eta}}(0)\cap\{|w|\geq 2|z-w|\}} \log{|w|} g_p(w) dw\nonumber\\ &+\frac{1}{2\pi}\int_{B_{{2\delta}/{\varepsilon_{p,i}^*}}(0)\setminus B_{R_{\eta}}(0) \cap\{|w|\geq 2|z-w|\}} \log{\frac{1}{|w-z|}} g_p(w) dw\nonumber\\ =&I_1(z)+I_2(z)+I_3(z)+I_4(z).\nonumber \end{align}\] For \(|w|\leq R_{\eta}\) and \(|z|\geq 2R_{\eta}\), we have \(\frac{|w|}{|w-z|}\leq \frac{2R_{\eta}}{|z|}\leq 1\). This, together with 81 , implies that \[\begin{align} I_1(z)\leq& \frac{1}{2\pi} \log{\frac{2R_{\eta}}{|z|}}\int_{B_{R_{\eta}}(0)} g_p(w) dw \leq -\frac{1}{2\pi}\left(8\pi-\eta \right)\log|z|+C. \end{align}\] Also, \[\begin{align} I_2(z)\leq \frac{\log{2}}{2\pi} \int_{B_{{2\delta}/{\varepsilon_{p,i}^*}}(0)\setminus B_{R_{\eta}}(0)} g_p(z) dz\leq C. \end{align}\] For \(|w|\geq 2|z-w|\), we have \(|w|\leq 2|z|\). So we see from 81 82 that \[\begin{align} I_3(z)\leq &\frac{1}{2\pi}\log(2|z|)\int_{B_{{2\delta}/{\varepsilon_{p,i}^*}}(0)\setminus B_{R_{\eta}}(0)} g_p(w) dw\\ \leq & \frac{1}{2\pi}(\sigma_{1,i}-8\pi+2\eta)\log|z|+C.\nonumber \end{align}\] Since \(1+\frac{v_{k,p,i}^*(w)}{p}=\frac{u_{k,p}(\varepsilon_{p,i}^*w+y_{p,i})}{M_p(y_{p,i})}\leq 1\) and so \(0\leq g_p(w)\leq \mu_k+\beta\) for \(|z|\leq \frac{2\delta}{\varepsilon_{p,i}^*}\), we have \[\begin{align} I_4(z)= &\frac{1}{2\pi}\int_{B_{{2\delta}/{\varepsilon_{p,i}^*}}(0)\setminus B_{R_{\eta}}(0) \cap\{2\leq 2|z-w|\leq|w|\}} \log{\frac{1}{|w-z|}} g_p(w) dw\\ &+\frac{1}{2\pi}\int_{B_{{2\delta}/{\varepsilon_{p,i}^*}}(0)\setminus B_{R_{\eta}}(0) \cap\{|z-w|\leq 1\}} \log{\frac{1}{|w-z|}} g_p(w) dw\nonumber\\ \leq & C \int_{\{|z-w|\leq 1\}} \log{\frac{1}{|w-z|}} dw\leq C.\nonumber \end{align}\] Inserting above estimates into 83 and using \(v_{1,p,i}^*(0)\to V_1(0)\), we obtain \[v_{1,p,i}^*(z)\leq -\left(8-\frac{\sigma_{1,i}}{2\pi}-\frac{3\eta}{2\pi}\right)\log|z|+C_\eta\] for any \(2R_{\eta}\leq |z|\leq \frac{\delta}{\varepsilon_{p,i}^*}\) and \(p>p_{\eta}\). ◻

Proof. Note that \(v_{k,p,i}^*(0)=V_k(0)+o(1)=O(1)\), which, together with \(u_{k,p}(y_{p,i})=M_p(y_p)(1+\frac{v_{k,p,i}*(0)}{p})\), implies that \[\begin{align} \label{eq-4-3543-} u_{k,p}(y_{p,i})=M_p(y_{p,i})+O\left(\frac{1}{p}\right)\geq \frac{1}{2},\quad\text{for p large.} \end{align}\tag{84}\] By the Green’s representation formula, we have \[\begin{align} u_{1,p}(y_{p,i})=&\int_{\Omega}G(y_{p,i},x)\left(\mu_1 u_{1,p}^p+\beta u_{1,p}^{\frac{p-1}{2}} u_{2,p}^{\frac{p+1}{2}} \right) dx\\ =&\int_{B_{r_0/3}(y_{p,i})}G(y_{p,i},x)\left(\mu_1 u_{1,p}^p+\beta u_{1,p}^{\frac{p-1}{2}} u_{2,p}^{\frac{p+1}{2}} \right) dx+O\left(\frac{1}{p}\right)\nonumber, \end{align}\] where we have used that \[\begin{align} &\int_{\Omega\setminus B_{r_0/3}(y_{p,i})}G(y_{p,i},x)\left(\mu_1 u_{1,p}^p+\beta u_{1,p}^{\frac{p-1}{2}} u_{2,p}^{\frac{p+1}{2}} \right) dx\\ = &O(1)\int_{\Omega} \mu_1 u_{1,p}^p+\beta u_{1,p}^{\frac{p-1}{2}} u_{2,p}^{\frac{p+1}{2}} dx=O\left(\frac{1}{p}\right). \end{align}\] Recalling \(g_p\) defined in 80 , it follows from 5 that \[\begin{align} \label{eq-4-3743-} u_{1,p}(y_{p,i})=&\frac{M_p(y_{p,i})}{p}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}G(y_{p,i},\varepsilon_{p,i}^*z+y_{p,i})g_p(z) dz+O\left(\frac{1}{p}\right)\\ =&-\frac{M_p(y_{p,i})}{p}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}H(y_{p,i},\varepsilon_{p,i}^*z+y_{p,i})g_p(z) dz\nonumber\\ &-\frac{M_p(y_{p,i})}{2\pi p}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}(\log|z|) g_p(z) dz\nonumber\\ &-\frac{M_p(y_{p,i})}{2\pi p}\log{\varepsilon_{p,i}^*}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}g_p(z) dz+O\left(\frac{1}{p}\right).\nonumber \end{align}\tag{85}\] From Remark 22 and the dominated convergence theorem, we obtain \[\begin{align} &\lim\limits_{p\rightarrow+\infty}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}g_p(z) dz=\int_{\mathbb{R}^2} \mu_1 e^{V_1} +\beta e^{\frac{V_1+V_2}{2}} dz=8\pi, \end{align}\] \[\begin{align} \lim\limits_{p\rightarrow+\infty}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}(\log|z|) g_p(z) dz=\int_{\mathbb{R}^2}(\log|z|)\left( \mu_1 e^{V_1} +\beta e^{\frac{V_1+V_2}{2}} \right) dz=O(1). \end{align}\] Furthermore, \[\begin{align} &\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}H(y_{p,i},\varepsilon_{p,i}^*z+y_{p,i}) g_p(z) dz=O(1)\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}g_p(z) dz=O(1). \end{align}\] These, together with 84 and 85 , imply \[\begin{align} \frac{\log{\varepsilon_{p,i}^*}}{p}=-\frac{1}{4}+o(1). \end{align}\] Since \[\log{p}+2\log{\varepsilon_{p,i}^*}+(p-1)\log{M_p(y_{p,i})}=0,\] we obtain \(M_p(y_{p,i})\to \sqrt{e}\) as \(p\to\infty\). Consequently, ?? follows from 84 , and \(\gamma_{k,i}=\sigma_{k,i}\lim M_p(y_{p,i})=8\pi\sqrt{e}\). Finally, ?? follows from ?? and \[M_p(y_{p,i})=\max\limits_{x\in\overline{B_{r_0}(x_i)}}M_p(x).\] The proof is complete. ◻

Proof. Similarly as Remark 14, we have \[\begin{align} \sigma_{1,i}= &\lim\limits_{p\rightarrow \infty} \frac{p}{M_p(y_{p,i})} \int_{B_{\rho}(x_{i})} \left(\mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) dx\nonumber\\ \geq & \int_{\mathbb{R}^2}\mu_{1} e^{\widetilde{V}_{1}} dz=8\pi.\nonumber \end{align}\] Then Corollary 1 implies \(8\pi\leq\sigma_{1,i}\leq 4(1+\sqrt{2})\pi\). Consequently, the same proof as Lemma 21 implies that for any \(\eta\in(0,0.1]\), there exist large \(R_{\eta}>1\), \(p_{\eta}>1\) and \(C_{\eta}>0\) such that \[\begin{align} \label{eq-4-324343-} v_{1,p,i}^*(z)&\leq -\left(8-\frac{\sigma_{1,i}}{2\pi}-\eta\right)\log|z|+C_{\eta}\\ &\leq -(6-2\sqrt{2}-\eta)\log|z|+C_{\eta}\leq -3\log|z|+C_{\eta},\nonumber \end{align}\tag{86}\] for any \(2R_{\eta}\leq |z|\leq \frac{\delta}{\varepsilon_{p,i}^*}\) and \(p>p_{\eta}\), where \(\delta=r_0/3\). Then similarly as Remark 22, we have that for \(m\in \{-1,0,1\}\) and \(p\) large, \[\begin{align} \label{eq-4-324343} \left(1+\frac{v_{1,p,i}^*}{p}\right)^{p+m}\leq \frac{C}{(1+|z|)^3},\quad\forall |z|\leq \frac{\delta}{\varepsilon_{p,i}^*}. \end{align}\tag{87}\]

Fix any large \(R>1\). By type \(\mathcal{B}\) in \((\mathcal{P}_2^{n_0})\) and the dominated convergence theorem, we have for any \(m=-1,0,1\) and \(p\) large enough, \[\begin{align} &\int_{B_{R}(0)} \left(1+\frac{v_{2,p,i}^*}{p}\right)^{p+m} dz=o(1),\\ &\int_{B_{R}(0)} \left(1+\frac{v_{1,p,i}^*}{p}\right)^{p+m} dz= \int_{B_{R}(0)} e^{\widetilde{V}_{1}} dz+o(1),\\ &\int_{B_{\frac{\delta}{\varepsilon_{p,i}^*}}(0)} \left(1+\frac{v_{1,p,i}^*}{p}\right)^{p+m} dz=\int_{\mathbb{R}^2} e^{\widetilde{V}_{1}} dz+o(1), \end{align}\] and so \[\begin{align} \label{fc-4-29} 0<&\int_{B_{\frac{\delta}{\varepsilon_{p,i}^*}}(0)} \left(1+\frac{v_{1,p,i}^*}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{2,p,i}^*}{p}\right)^{\frac{p+1}{2}} dz\\ \leq & \left(\int_{B_{R}(0)} \left(1+\frac{v_{1,p,i}^*}{p}\right)^{p-1} dz\right)^{\frac{1}{2}} \left(\int_{B_{R}(0)} \left(1+\frac{v_{2,p,i}^*}{p}\right)^{p+1} dz\right)^{\frac{1}{2}} \nonumber\\ &+\left(\int_{B_{\frac{\delta}{\varepsilon_{p,i}^*}}(0)\setminus B_{R}(0)} \left(1+\frac{v_{1,p,i}^*}{p}\right)^{p-1} dz\right)^{\frac{1}{2}} \left(\frac{p}{M_p(y_{p,i})^2}\int_{\Omega} u_{2,p}^{p+1} dx\right)^{\frac{1}{2}} \nonumber\\ \leq & o(1)+O\left(\int_{\mathbb{R}^2\setminus B_{R}(0)} e^{\widetilde{V}_{1}} dz+o(1)\right).\nonumber \end{align}\tag{88}\] Letting \(p\to\infty\) first and then \(R\to \infty\), we obtain \[\begin{align} \label{fc-4-29-0} \lim_{p\to\infty}\int_{B_{\frac{\delta}{\varepsilon_{p,i}^*}}(0)} \left(1+\frac{v_{1,p,i}^*}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{2,p,i}^*}{p}\right)^{\frac{p+1}{2}} dz=0. \end{align}\tag{89}\] Consequently, \[\begin{align} \label{fc-4-30} \sigma_{1,i}=&\lim_{p\to\infty}\frac{p}{M_p(y_{p,i})} \int_{B_{\delta}(y_{p,i})} \mu_1 u_{1,p}^p +\beta u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}} dx\nonumber\\ =& \lim_{p\to\infty}\int_{B_{\frac{\delta}{\varepsilon_{p,i}^*}}(0)} \mu_1\left(1+\frac{v_{1,p,i}^*}{p}\right)^p+\beta\left(1+\frac{v_{1,p,i}^*}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{2,p,i}^*}{p}\right)^{\frac{p+1}{2}} dz\\ =&\int_{\mathbb{R}^2} \mu_1 e^{\widetilde{V}_{1}} dz =8\pi.\nonumber \end{align}\tag{90}\] This completes the proof. ◻

Proof. By type \(\mathcal{B}\), we have \(v_{2,p,i}^*(0)\rightarrow-\infty\) and so \[\begin{align} p(u_{2,p}(y_{p,i})-M_p(y_{p,i}))\rightarrow-\infty, \end{align}\] which implies that \[u_{1,p}(y_{p,i})=M_p(y_{p,i})=\|u_{1,p}\|_{L^\infty(B_{r_0}(x_i))}.\] Then it follows from 85 that \[\begin{align} \label{eq-4-5543-} 1 =&\frac{1}{p}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}H(y_{p,i},\varepsilon_{p,i}^*z+y_{p,i})g_p(z) dz\\ &-\frac{1}{2\pi p}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}(\log|z|) g_p(z) dz\nonumber\\ &-\frac{1}{2\pi p}\log{\varepsilon_{p,i}^*}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}g_p(z) dz+O\left(\frac{1}{p}\right).\nonumber \end{align}\tag{91}\] Recall 88 89 that \[\begin{align} \lim_{p\to\infty} \int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)} \left(1+\frac{v_{1,p,i}^*}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{2,p,i}^*}{p}\right)^{\frac{p+1}{2}} dz =0. \end{align}\] Similarly, we can obtain \[\begin{align} &\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)} \left|\log|z|\right|\left(1+\frac{v_{1,p,i}^*}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{2,p,i}^*}{p}\right)^{\frac{p+1}{2}} dz\\ \leq & o(1)+ C \left(\int_{\mathbb{R}^2\setminus B_R(0)}|\log|z||^2 e^{\widetilde{V}_{1}} dz+o(1)\right),\quad\forall R>1, \end{align}\] and so \[\lim_{p\to\infty}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)} \left|\log|z|\right|\left(1+\frac{v_{1,p,i}^*}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{2,p,i}^*}{p}\right)^{\frac{p+1}{2}} dz=0.\] Then a similar argument as 90 leads to \[\begin{align} &\lim\limits_{p\rightarrow+\infty}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}g_p(z) dz=\int_{\mathbb{R}^2} \mu_1 e^{\widetilde{V}_{1}} dz=8\pi,\\ &\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}H(y_{p,i},\varepsilon_{p,i}^*z+y_{p,i}) g_p(z) dz=O(1)\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}g_p(z) dz=O(1),\\ &\lim\limits_{p\rightarrow+\infty}\int_{B_{{r_0}/{3\varepsilon_{p,i}^*}}(0)}(\log|z|) g_p(z) dz= \int_{\mathbb{R}^2}\mu_1(\log|z|)e^{\widetilde{V}_{1}} dz=O(1). \end{align}\] Inserting these estimates into 91 , we can prove as Lemma 23 that \(M_p(y_{p,i})\to \sqrt{e}\). This completes the proof. ◻

Proof. Similarly as 89 90 , we have \[\begin{align} &\lim_{p\to\infty}p\int_{B_{\rho}(x_i)}u_{k,p}^{\frac{p-1}{2}}u_{3-k,p}^{\frac{p+1}{2}} dx\\ =& \lim_{p\to\infty} M_p(y_{p,i})\int_{B_{{\rho}/{\varepsilon_{p,i}^*}}(0)} \left(1+\frac{v_{k,p,i}^*}{p}\right)^{\frac{p-1}{2}}\left(1+\frac{v_{3-k,p,i}^*}{p}\right)^{\frac{p+1}{2}} dz =0,\;k=1,2,\\ &\lim_{p\to\infty}p\int_{B_{\rho}(x_i)}\mu_1 u_{1,p}^{p+m} dx\\ =&\lim_{p\to\infty} M_p(y_{p,i})^{1+m}\int_{B_{{\rho}/{\varepsilon_{p,i}^*}}(0)} \left(1+\frac{v_{1,p,i}^*}{p}\right)^{p+m}dz=8\pi e^{\frac{1+m}{2}},\quad m=0,1, \end{align}\] which implies that \[\begin{align} &\gamma_{2,i}=\lim_{p\rightarrow\infty}p\int_{B_{\rho}(x_i)}\mu_2 u_{2,p}^{p} +\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p-1}{2}} dx=\lim\limits_{p\rightarrow \infty} p\int_{B_{\rho}(x_i)}\mu_2 u_{2,p}^{p} dx, \end{align}\] and (using 21 ) \[\begin{align} p\int_{B_{\rho}(x_i)}u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} dx=O(1)p\int_{B_{\rho}(x_i)}u_{1,p}^{\frac{p-1}{2}}u_{2,p}^{\frac{p+1}{2}} dx\to 0. \end{align}\] This completes the proof. ◻

Proof. Recalling 21 that we can take \(c_0>1\) such that \(\|u_{k,p}\|_{L^\infty(\Omega)}\leq c_0\) for \(k=1,2\) and all \(p\geq p_0\). Denote \[\bar u_p:=c_0pu_{2,p} \quad\text{and}\quad f_p:=c_0p\left(\mu_2 u_{2,p}^p +\beta u_{2,p}^{\frac{p-1}{2}}u_{1,p}^{\frac{p+1}{2}}\right)\geq 0,\] then \[\begin{cases} -\Delta \bar u_p=f_p,\quad \bar u_p>0,\quad\text{in}~\Omega,\\ \bar u_p=0,\quad\text{on}~\partial\Omega. \end{cases}\] Note from ?? that \(\|f_p\|_{L^1(\Omega)}\leq C\) for all \(p\geq p_0\). Since \(\log t\le\frac{1}{e}t\) for any \(t\in(0,+\infty)\), we obtain \[\log (p^{\frac{2}{p-1}}c_0u_{2,p})\le \frac{1}{e}p^{\frac{2}{p-1}}c_0u_{2,p},\quad \forall x\in\Omega.\] Then we have that for any \(x\in\Omega\), \[\label{fc-fp1} \begin{align} f_p&\leq Cp(c_0u_{2,p})^{\frac{p-1}{2}}=Ce^{\frac{p-1}{2}\log(p^{\frac{2}{p-1}}c_0u_{2,p})}\leq Ce^{\frac{p-1}{2e} p^{\frac{2}{p-1}}c_0u_{2,p}}\\ &\leq Ce^{\frac{1}{e} \bar u_p},\quad\text{for p large.} \end{align}\tag{93}\] Besides, we see from \(\gamma_{2,i}=0\) and Corollary 2 that \[\label{fc-fp2}\lim_{p\to\infty}\int_{B_{r_0/3}(x_i)}f_pdx=0.\tag{94}\] Thanks to 93 and 94 , the same proof as [31] implies that \[\|c_0pu_{2,p}\|_{L^\infty(B_{r_0/12}(x_i))}=\|\bar u_p\|_{L^\infty(B_{r_0/12}(x_i))}\leq C,\quad\text{ for p large}.\] Together with ?? , we finally obtain \(\|pu_{2,p}\|_{L^\infty(B_{r_0}(x_i))}\leq C\) for \(p\) large. ◻

Proof. By Corollary 1 and Lemma 24, we have \(0<\sigma_{2,i}\leq 8\pi\). Also recall from Lemma 25 that \[0<\gamma_{2,i}=\sigma_{2,i}\lim_{p\rightarrow+\infty}M_p(y_{p,i})=\sigma_{2,i}\sqrt{e}\leq 8\pi\sqrt{e}.\] From Lemma 19, Corollary 2 and \(\gamma_{1,i}=8\pi\sqrt{e}\), we get that for any \(\rho\in (0, r_0/3]\), \[\begin{align} \label{eq-4-634343} \left(\lim\limits_{p\rightarrow \infty} p\int_{B_{\rho}(x_i)}\mu_2 u_{2,p}^{p} dx\right)^2=\gamma_{2,i}^2=8\pi \lim\limits_{p\rightarrow \infty} p\int_{B_{\rho}(x_i)}\mu_2 u_{2,p}^{p+1} dx. \end{align}\tag{95}\] Now we claim that \[\begin{align} \label{fc-4-37} \lim_{p\rightarrow \infty} p\int_{B_{\rho}(x_i)}\mu_2 u_{2,p}^{p+1} dx\geq 8\pi e. \end{align}\tag{96}\]

Take \(\chi_i \in C_0^{\infty}(B_{2\rho}(x_i))\) such that \(0\leq\chi_i\leq1\) and \[\chi_i(x)= \begin{cases} 1, & if|x-x_i|\leq \rho \\ 0, & if|x-x_i|\in[\frac{4\rho}{3},2\rho). \end{cases}\notag\] Define \[\widetilde{u}_{2,p}:=u_{2,p} \chi_i \in H_0^1(B_{2\rho}(x_i)).\] Then by ?? ?? and Corollary 2, we obtain \[\begin{align} p\int_{B_{2\rho}(x_i)} \widetilde{u}_{2,p}^{p+1} dx&=p\int_{B_{\rho}(x_i)} {u}_{2,p}^{p+1} dx+o(1),\\ p\int_{B_{2\rho}(x_i)}\left|\nabla \widetilde{u}_{2,p}\right|^2 dx &= p\int_{B_{\rho}(x_i)}\left|\nabla u_{2,p}\right|^2 dx+o(1),\tag{97}\\ p\int_{B_{\rho}(x_i)}\left|\nabla u_{2,p}\right|^2 dx= p\int_{B_{\rho}(x_i)} &\mu_2 u_{2,p}^{p+1}+\beta u_{2,p}^{\frac{p+1}{2}} u_{1,p}^{\frac{p+1}{2}}dx -p \int_{\partial B_{\rho}(x_i)}u_{2,p}\frac{\partial u_{2,p}}{\partial \Vec{n}}dS_x\notag\\ = p\int_{B_{\rho}(x_i)}&\mu_2 u_{2,p}^{p+1} dx +o(1)\tag{98}. \end{align}\] Note from 95 that for \(p\) large, \[\begin{align} p\int_{B_{\rho}(x_i)} u_{2,p}^{p+1} dx &\geq \frac{\gamma_{2,i}^2}{10\pi\mu_2}>0.\nonumber \end{align}\] Then, by applying Lemma 10 with \(D=B_{2\rho}(x_i)\) to \(\widetilde{u}_{2,p}\), we have that for \(p\) large, \[\begin{align} p\int_{B_{2\rho}(x_i)}\left|\nabla \widetilde{u}_{2,p}\right|^2 dx &\geq \frac{p^{\frac{p-1}{p+1}}}{(p+1) S_p^2} \left[ p\int_{B_{2\rho}(x_i)} \widetilde{u}_{2,p}^{p+1} dx \right]^{\frac{2}{p+1}}\notag\\ &= (8\pi e+o(1))\left[ p\int_{B_{\rho}(x_i)}u_{2,p}^{p+1} dx +o(1)\right]^{\frac{2}{p+1}}\geq 8\pi e+o(1) .\label{44610} \end{align}\tag{99}\] This, together with 97 and 98 , proves 96 .

Clearly 95 96 imply \(\gamma_{2,i}\geq 8\pi \sqrt{e}\) and so \(\gamma_{2,i}=8\pi \sqrt{e}\), which gives \(\sigma_{2,i}=8\pi\). Furthermore, 96 becomes \[\begin{align} \label{fc-4-001}8\pi e&= \lim_{p\rightarrow \infty} p\int_{B_{\rho}(x_i)}\mu_2 u_{2,p}^{p+1} dx\leq \lim_{p\rightarrow \infty}\Vert u_{2,p}\Vert_{L^\infty(B_\rho(x_i))} p\int_{B_{\rho}(x_i)}\mu_2 u_{2,p}^{p} dx\\&=\gamma_{2,i}\lim_{p\rightarrow \infty}\Vert u_{2,p}\Vert_{L^\infty(B_\rho(x_i))}=8\pi\sqrt{e}\lim_{p\rightarrow \infty}\Vert u_{2,p}\Vert_{L^\infty(B_\rho(x_i))},\nonumber \end{align}\tag{100}\] so \[\begin{align} \lim_{p\rightarrow \infty}\Vert u_{2,p}\Vert_{L^\infty(B_\rho(x_i))}\geq\sqrt{e}. \end{align}\] From here and \(\Vert u_{2,p}\Vert_{L^\infty(B_{r_0}(x_i))}\leq M_p(y_{p,i})\to \sqrt{e}\), we obtain ?? . Finally, ?? follows from the first equality of 100 and ?? . This completes the proof. ◻

Proof of Theorem 4. Theorem 4 follows from Lemmas 20, 24 and 28. ◻

Proof. In Section 4, we have proved that \(\gamma_{k,i}=8\pi \sqrt{e}\) for \(x_i\in \mathcal{S}_k\) and \(\gamma_{k,i}=0\) for \(x_i\in\mathcal{S}\setminus\mathcal{S}_k\). Then Lemma 16 imlies \[p u_{k,p}\rightarrow 8\pi \sqrt{e} \sum_{x_i\in\mathcal{S}_k} G(x_i,\cdot)\quad\text{ in }\,\,C^{2}_{loc}(\overline{\Omega}\setminus\mathcal{S}).\] On the other hand, Lemma 27 shows that \[p\Vert u_{k,p}\Vert_{L^{\infty}(\cup_{x_i\in \mathcal{S}\setminus\mathcal{S}_k}B_{r_0}(x_i))}=O(1),\] then the same proof as Lemma 16 actually implies that the convergence holds in \(C^{2}_{loc}(\overline{\Omega}\setminus\mathcal{S}_k)\), namely ?? holds.

If \(\mathcal{S}_k=\emptyset\) for some \(k\), then Lemma 27 and ?? together imply \(p\|u_{k,p}\|_{L^\infty(\Omega)}\leq C\), a contradiction with Proposition 12. This proves \(\mathcal{S}_k\neq\emptyset\) for \(k=1,2\). ◻

Proof. The proof is standard and let us take \(i=1\) for example. Fix any \(r\in (0, r_0/3)\). By applying the Pohozaev identity ?? with \(\Omega'=B_{r}(x_1)\), we obtain \[\begin{align} \label{fc-ppp} &\sum_{k=1}^2\int_{\partial B_{r}(x_1)}\frac{1}{2}|p\nabla u_{k,p}|^2 n_i -p \partial_i u_{k,p} \langle p\nabla u_{k,p}, \Vec{n}(x)\rangle dS_x\\ =&\frac{p^2}{p+1}\int_{\partial B_{r}(x_1)}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) n_i dS_x,\;i=1,2.\nonumber \end{align}\tag{101}\] Again by ?? , we have \[\begin{align} \label{eq-4-114} \frac{p^2}{p+1}\int_{\partial B_{r}(x_1)}\left(\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}}\right) n_i dS_x=o(1). \end{align}\tag{102}\] By Proposition 29, we get that for \(x\in \partial B_{r}(x_1)\), \[\begin{align} \frac{1}{8\pi \sqrt{e}}p \nabla u_{k,p} (x)=- \frac{m_{k,1}}{2\pi r^2}(x-x_1) -m_{k,1} \nabla_x H(x,x_1) +\sum_{l=2 }^N m_{k,l} \nabla_x G(x,x_l)+o(1),\nonumber \end{align}\] which implies that \[\begin{align} &\frac{1}{2}|p\nabla u_{k,p}|^2 n_i -p \partial_i u_{k,p} \langle p\nabla u_{k,p}, \Vec{n}(x)\rangle\\ =&-\frac{8 em_{k,1}^2}{r^3}(x-x_1)_i-\frac{32\pi em_{k,1}}{r}\left(m_{k,1} \nabla_x H(x,x_1) -\sum_{l=2 }^N m_{k,l} \nabla_x G(x,x_l)\right)_i+O(1).\nonumber \end{align}\] Thus \[\begin{align} &\sum_{k=1}^2\int_{\partial B_{r}(x_1)}\frac{1}{2}|p\nabla u_{k,p}|^2 n_i -p \partial_i u_{k,p} \langle p\nabla u_{k,p}, \Vec{n}(x)\rangle dS_x \nonumber\\ =& -64\pi^2 e \sum_{k=1}^2m_{k,1}\left(m_{k,1} \nabla_x H(x_1,x_1) -\sum_{l=2 }^N m_{k,l} \nabla_x G(x_1,x_l)\right)_i+O(r).\nonumber \end{align}\] From here and 102 , and letting first \(p\to\infty\) and then \(r\to0\) in 101 , we obtain \[\begin{align} \sum_{k=1}^2 m_{k,1}\left( m_{k,1} \nabla_x H(x_1,x_1) -\sum_{l=2 }^N m_{k,l} \nabla_x G(x_1,x_l)\right)=0. \end{align}\] The proof is complete by using \(R(x)=H(x,x)\). ◻

Proof. By system 1 and ?? , we have that for any fixed \(\rho\in (0, r_0)\), \[\begin{align} &p\int_{\Omega}|\nabla u_{1,p}|^2+|\nabla u_{2,p}|^2 dx\\ =&p\int_{\Omega} \mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} dx\\ =&\sum_{i=1}^N p\int_{B_{\rho}(x_i)}\mu_1 u_{1,p}^{p+1}+\mu_2 u_{2,p}^{p+1}+2\beta u_{1,p}^{\frac{p+1}{2}}u_{2,p}^{\frac{p+1}{2}} dx+o(1). \end{align}\] Since we have proved in Section 4 that \[\mathcal{S}_k=\{x_i\in\mathcal{S}\; : \; \gamma_{k,i}\neq 0\}=\{x_i\in\mathcal{S}\; : \; \gamma_{k,i}=8\pi\sqrt{e}\}, \quad k=1,2\] and \[m_{k,i}= \begin{cases} 0 \quad x_i\not\in\mathcal{S}_k,\\ 1 \quad x_i\in\mathcal{S}_k, \end{cases}\] we have \((\gamma_{1,i},\gamma_{2,i})=(8\pi\sqrt{e}m_{1,i}, 8\pi\sqrt{e}m_{2,i})\) for any \(i\). From here and ?? , we obtain the desired assertion ?? . ◻

Proof of Theorem 2. Clearly Theorem 2 follows from Propositions 29-31. ◻