Infinitely Many Solutions for the Nonlinear Schrödinger–Poisson System

Abstract

In this paper, we consider the following nonlinear Schrödinger–Poisson system in $\mathbb {R}^{3}$:

$$ \left\{\begin{array}{ll} -{\Delta} u+u+K(|y|){\Phi}(y)u=Q(|y|)|u|^{p-1}u& \text{in}\hspace{1.14mm} \mathbb{R}^{3},\vspace{0.2cm}\\ -{\Delta} {\Phi}=K(|y|)u^{2}& \text{in}\hspace{1.14mm} \mathbb{R}^{3}, \end{array} \right. $$

(0.1)

where 1 < p < 5, K(|y|) and Q(|y|) are two radial, positive, and locally Hölder continuous functions satisfying that

$$ K(r)=\frac{a}{r^{m}}+O\Big(\frac{1}{r^{m+\theta}}\Big),~~Q(r)=1+\frac{b}{r^{n}} +O\Big(\frac{1}{r^{n+\varepsilon}}\Big)~~\text{as}~r\rightarrow+\infty, $$

where $b\in \mathbb {R}$, a,m,n,𝜃 and ε are positive constants. Simulated by the work of Wang and Zhao [31], either a > 0, b < 0 or a > 0, b > 0, m > 2n holds, we use the Lyapunov–Schmidt reduction method to construct infinitely many nonradial positive solutions of (0.1) with arbitrary large energy.

Appendix. Some useful estimates

In this section, we present some technical computations. We first give the following basic estimate.

Lemma A.1

Let r ∈Λ_k with k sufficiently large, then

$$ \sum\limits_{j=2}^{k}U(l|x_{j}-x_{1}|)=2(1+O(e^{-\frac{l\rho}{2}}))U(l \rho),\quad \sum\limits_{j=2}^{k}U^{l}(|x_{j}-x_{1}|)=2(1+O(e^{-\frac{l\rho}{2}}))U^{l}(\rho), $$

where 0 < l ≤ 1, $\rho =2r\sin \limits \frac {\pi }{k}$.

Proof

For k large enough, we have

$$\aligned \sum\limits_{j=3}^{[\ln k]+1}U(l|x_{j}-x_{1}|)&\leq \sum\limits_{j=3}^{[\ln k]+1}e^{-2lr\sin\frac{(j-1)\pi}{k}}\leq \sum\limits_{j=3}^{[\ln k]+1}e^{-2(j-1)l(1+o(1))\pi\frac{r}{k}}\\ &<\sum\limits_{j=3}^{\infty}e^{-\frac{7l\pi(j-1)}{4}\frac{r}{k}}\leq 2e^{-\frac{7l\pi r}{2k}}\leq 2e^{-\frac{l\rho}{2}}U(l\rho). \endaligned$$

Then it follows from symmetry that

$$ \aligned \sum\limits_{j=2}^{k}U(l|x_{j}-x_{1}|)&=2\sum\limits_{j=2}^{[\ln k]+1}U(l|x_{j}-x_{1}|)+O(ke^{-2lr\sin\frac{[\ln k]\pi}{k}})\\ &=2U(l|x_{j}-x_{1}|)+O(e^{-\frac{l\rho}{2}}U(l\rho))=2(1+O(e^{-\frac{l\rho}{2}}))U(l \rho). \endaligned $$

So the first estimation holds. By a similar argument, we see easily that the second estimation also holds. □

Lemma A.2

Let r ∈Λ_k with k sufficiently large, then for any $y\in {\Omega }_{j_{0}}$, j₀ = 1, 2,⋯ ,k,

$$ \begin{array}{@{}rcl@{}} \mathbb{U}_{r}(y)=U(y-x_{j_{0}})+O\Big(U\Big(\frac{\rho}{2}\Big)\Big), \end{array} $$

(A.1)

$$ \begin{array}{@{}rcl@{}} \mathbb{U}_{r}(y)=U(y-x_{j_{0}})+O\big(e^{-\frac{\rho}{4}}U^{\frac{1}{2}}(y-x_{j_{0}})\big). \end{array} $$

(A.2)

Proof

For any y ∈Ω₁, it is easy to see that |y − x_j|≥|y − x₁| with j≠ 1. If $|y-x_{1}|\geq \frac {|x_{j}-x_{1}|}{2}$, then we have $|y-x_{j}|\geq \frac {|x_{j}-x_{1}|}{2}$. If $|y-x_{1}|<\frac {|x_{j}-x_{1}|}{2}$, then we have

$$ |y-x_{j}|\geq |x_{j}-x_{1}|-|y-x_{1}|\geq\frac{|x_{j}-x_{1}|}{2}. $$

Thus it follows that $|y-x_{j}|\geq \frac {|x_{j}-x_{1}|}{2}$ for any y ∈Ω₁. By Lemma A.1, we get

$$ \aligned \sum\limits_{j=1}^{k} U(y-x_{j})&=U(y-x_{1})+\sum\limits_{j=2}^{k} U(y-x_{j})\\ &=U(y-x_{1})+O\Big(\sum\limits_{j=2}^{k} U\Big(\frac{|x_{j}-x_{1}|}{2}\Big)\Big)\\ &=U(y-x_{1})+O\Big(U\Big(\frac{\rho}{2}\Big)\Big). \endaligned$$

By symmetry, we see that (A.1) holds.

Similarly, Lemma A.1 gives that for any y ∈Ω₁,

$$\aligned \sum\limits_{j=1}^{k} U(y-x_{j})&=U(y-x_{1})+\sum\limits_{j=2}^{k} U(y-x_{j})\\ &=U(y-x_{1})+O\Big(\sum\limits_{j=2}^{k} U^{\frac{1}{2}}\Big(\frac{|x_{j}-x_{1}|}{2}\Big)U^{\frac{1}{2}}(y-x_{1})\Big)\\ &=U(y-x_{1})+O\big(e^{-\frac{\rho}{4}}U^{\frac{1}{2}}(y-x_{1})\big). \endaligned$$

By symmetry, (A.2) also follows. □

Lemma A.3

For any α > 0 and 0 < μ ≤ 1, $\mathbb {U}_{r}$, $\mathbb {V}_{r,\mu }$ and $\frac {\mathbb {U}_{r}^{\alpha }}{\mathbb {V}_{r,\mu }}$ are bounded in $\mathbb {R}^{3}$.

Proof

By (A.2) in Lemma A.2, there exists a C > 0 independent of k such that $0<\mathbb {U}_{r}\leq C$ for any y ∈Ω_j with j = 1, 2,⋯ ,k. Note that $\mathbb {R}^{3}=\cup _{j=1}^{k}{\Omega }_{j}$, then $0<\mathbb {U}_{r}\leq C$ for any $y\in \mathbb {R}^{3}$. So $\mathbb {U}_{r}$ is bounded in $\mathbb {R}^{3}$.

The asymptotical behavior of V_μ at infinity yields that for any y ∈Ω₁ and some C > 0,

$$\aligned 0&<\mathbb{V}_{r,\mu}=\sum\limits_{j=1}^{k}V_{\mu}(y-x_{j})\leq V_{\mu}(y-x_{1})+ \sum\limits_{j=2}^{k}\frac{1}{|y-x_{j}|}{\int}_{\mathbb{R}^{3}}U^{2\mu}(y)dy\\ &\leq \sum\limits_{j=2}^{k}\frac{2}{|x_{1}-x_{j}|}{\int}_{\mathbb{R}^{3}}U^{2\mu}(y)dy+C\leq \frac{Ck}{r} \sum\limits_{j=2}^{k}\frac{1}{j-1}\leq \frac{Ck\ln k}{r}\leq C. \endaligned $$

Thus, we see that $\sum \limits _{j=1}^{k}V_{\mu }(y-x_{j})$ is bounded by symmetry.

For any y ∈Ω₁, we have

$$\aligned \frac{\mathbb{U}_{r}^{\alpha}}{\mathbb{V}_{r,\mu}}&\leq\sum\limits_{j=1}^{k}\frac{U^{\alpha}(y-x_{j})}{V_{\mu}(y-x_{j})} =\frac{U^{\alpha}(y-x_{1})}{V_{\mu}(y-x_{1})}+\sum\limits_{j=2}^{k}\frac{U^{\alpha}(y-x_{j})}{V_{\mu}(y-x_{j})}\\ &=\frac{U^{\alpha}(y-x_{1})}{V_{\mu}(y-x_{1})}+O\Big(\sum\limits_{j=2}^{k}|y-x_{j}|^{1-\min\{\alpha,1\}} e^{-\min\{\alpha,1\}|y-x_{j}|}\Big)\\ &=\frac{U^{\alpha}(y-x_{1})}{V_{\mu}(y-x_{1})}+O\big(e^{-\frac{\min\{\alpha,1\}\rho}{3}}\big). \endaligned$$

Note that $\frac {U^{\alpha }(y-x_{1})}{V_{\mu }(y-x_{1})}\to 0$ as $|y-x_{1}|\to \infty$, then $\frac {U^{\alpha }(y-x_{1})}{V_{\mu }(y-x_{1})}$ is bounded. This implies that $\frac {\mathbb {U}_{r}^{\alpha }}{\mathbb {V}_{r,\mu }}$ is bounded in Ω₁. By symmetry, $\frac {\mathbb {U}_{r}^{\alpha }}{\mathbb {V}_{r,\mu }}$ is bounded in $\mathbb {R}^{3}$. □

Lemma A.4

For any 0 < μ ≤ 1, there exists a 0 < 𝜃_μ < 1 such that

$$ \begin{array}{@{}rcl@{}} {\Phi}_{\mathbb{U}_{r}^{\mu}}(y)=\Big[\frac{a}{r^{m}}+O\Big(\frac{1}{r^{m+\theta_{\mu}}}\Big)\Big]\sum\limits_{j=1}^{k} V_{\mu}(y-x_{j}). \end{array} $$

(A.3)

Proof

Direct computation shows that

$$\aligned &{\int}_{{\Omega}_{j}}\frac{K(|x|)}{|y-x|}\mathbb{U}_{r}^{2\mu}(x)dx\\ =&{\int}_{{\Omega}_{j}}\frac{K(|x|)}{|y-x|}\Big(U(y-x_{j})+O\Big(e^{-\frac{\rho}{4}}e^{-\frac{|x-x_{j}|}{2}}\Big)\Big) ^{2\mu}dx\\ =&\Big({\int}_{{\Omega}_{j}\cap B_{\frac{r}{2}}(x_{j})}+{\int}_{{\Omega}_{j}\setminus B_{\frac{r}{2}}(x_{j})}\Big) \frac{K(|x|)}{|y-x|}\Big[U^{2\mu}(x-x_{j})+O\Big(e^{-\frac{\mu\rho}{4}}e^{-\frac{3\mu|x-x_{j}|}{2}} +e^{-\frac{\mu\rho}{2}}e^{-\mu|x-x_{j}|}\Big)\Big]dx. \endaligned$$

From the fact that for any α > 0,

$$ \frac{1}{|x+x_{j}|^{\alpha}}=\frac{1}{|x_{j}|^{\alpha}}\left[1+O\left( \frac{|x|}{|x_{j}|}\right)\right], ~~~x\in B_{\frac{r}{2}}(0), $$

we have

$$\aligned &{\int}_{{\Omega}_{j}\cap B_{\frac{r}{2}}(x_{j})}\frac{K(|x|)}{|y-x|}U^{2\mu}(x-x_{j})dx\\ =&{\int}_{({\Omega}_{j}-x_{j})\cap B_{\frac{r}{2}}(0)}\frac{K(|x+x_{j}|)}{|y-x_{j}-x|}U^{2\mu}(x)dx\\ =&{\int}_{({\Omega}_{j}-x_{j})\cap B_{\frac{r}{2}}(0)}\Big[\frac{a}{|x+x_{j}|^{m}} +O\Big(\frac{1}{|x+x_{j}|^{m+\theta}}\Big)\Big]\frac{U^{2\mu}(x)}{|y-x_{j}-x|}dx\\ =&\frac{a}{r^{m}}{\int}_{({\Omega}_{j}-x_{j})\cap B_{\frac{r}{2}}(0)} \Big[1+O\Big(\frac{|x|}{|x_{j}|}\Big)\Big]\frac{U^{2\mu}(x)}{|y-x_{j}-x|}dx +O\Big(\frac{V_{\mu}(y-x_{j})}{r^{m+\theta}}\Big)\\ =&\frac{a}{r^{m}}{\int}_{\mathbb{R}^{3}}\Big[1+O\Big(\frac{|x|}{|x_{j}|}\Big)\Big]\frac{U^{2\mu}(x)}{|y-x_{j}-x|}dx +O\Big(\frac{e^{-\frac{\mu r}{2}}V_{\frac{\mu}{2}}(y-x_{j})}{r^{m}}+\frac{V_{\mu}(y-x_{j})}{r^{m+\theta}}\Big)\\ =&\frac{a}{r^{m}}V_{\mu}(y-x_{j})+O\Big(\frac{1}{r^{m+1}}{\int}_{\mathbb{R}^{3}}\frac{|x|U^{2\mu}(x)}{|y-x_{j}-x|}dx +\frac{V_{\mu}(y-x_{j})}{r^{m+\theta}}\Big)\\ =&\Big[\frac{a}{r^{m}}+O\Big(\frac{1}{r^{m+\theta}}\Big)\Big]V_{\mu}(y-x_{j}). \endaligned$$

A similar argument implies that

$$ {\int}_{{\Omega}_{j}\setminus B_{\frac{r}{2}}(x_{j})}\frac{K(|x|)}{|y-x|}U^{2\mu}(x-x_{j})dx\leq Ce^{-\frac{\mu r}{2}}V_{\frac{\mu}{2}}(y-x_{j})\leq \frac{C}{r^{m+\theta}}V_{\mu}(y-x_{j}), $$

$$ {\int}_{{\Omega}_{j}}\frac{K(|x|)}{|y-x|}\Big(e^{-\frac{\mu\rho}{4}}e^{-\frac{3\mu|x-x_{j}|}{2}} +e^{-\frac{\mu\rho}{2}}e^{-\mu|x-x_{j}|}\Big)\leq\frac{C}{r^{m+\theta(\mu)}}V_{\mu}(y-x_{j}),~~~ 0<\theta(\mu)<1. $$

Then there exists a 0 < 𝜃_μ < 1 such that

$$ {\int}_{{\Omega}_{j}}\frac{K(|x|)}{|y-x|}\mathbb{U}_{r}^{2\mu}(x)dx =\Big[\frac{a}{r^{m}}+O\Big(\frac{1}{r^{m+\theta_{\mu}}}\Big)\Big]V_{\mu}(y-x_{j}). $$

It follows that

$$ {\int}_{\mathbb{R}^{3}}\frac{K(|x|)}{|y-x|}\mathbb{U}_{r}^{2\mu}(x)dx =\sum\limits_{j=1}^{k}{\int}_{{\Omega}_{j}}\frac{K(|x|)}{|y-x|}\mathbb{U}_{r}^{2\mu}(x)dx =\Big[\frac{a}{r^{m}}+O\Big(\frac{1}{r^{m+\theta_{\mu}}}\Big)\Big]\sum\limits_{j=1}^{k}V_{\mu}(y-x_{j}). $$

Thus (A.3) holds and we finish the proof of this lemma. □

Remark A.5

By Lemma A.4, there exists a constant C_μ > 0 such that

$$ |{\Phi}_{\mathbb{U}^{\mu}_{r}}|\leq \frac{C_{\mu}}{r^{m}}. $$

Proposition A.6

There exists a small positive number σ > 0 such that

$$ \begin{array}{@{}rcl@{}} I(\mathbb{U}_{r}) =k\left( A+\frac{a^{2}(B_{1}+B_{k}(r))}{r^{2m}}-\frac{B_{2}}{r^{n}}-B_{3}\left( \frac{r}{k}\right)^{-1}e^{-\frac{2\pi r}{k}}+O\left( \frac{1}{r^{2m+\sigma}}+\frac{1}{r^{n+\sigma}}\right)\right), \end{array} $$

where I is defined as in (2.1),

$$ A = \left( \frac{1}{2} - \frac{1}{p+1}\right)\!{\int}_{\mathbb{R}^{3}}U^{p+1},~~~B_{1} = \frac{1}{4}{\int}_{\mathbb{R}^{3}}{\int}_{\mathbb{R}^{3}}\frac{U^{2}(x)U^{2}(y)}{|y-x|}dxdy,~~B_{2} = \frac{b}{p+1}{\int}_{\mathbb{R}^{3}}U^{p+1}, $$

B₃ > 0 is a positive constant and B_k(r) is a function satisfying

$$ 0<B_{k}(r)<\frac{Ck\ln k}{r}{\int}_{\mathbb{R}^{3}}U^{2}. $$

Proof

Clearly,

$$\aligned I(\mathbb{U}_{r})&=\frac 12 {\int}_{\mathbb{R}^{3}} (|\nabla\mathbb{U}_{r}|^{2} +\mathbb{U}_{r}^{2})-\frac{1}{p+1}{\int}_{\mathbb{R}^{3}}Q(|y|)\mathbb{U}_{r}^{p+1}+\frac{1}{4}{\int}_{\mathbb{R}^{3}} K(|y|){\Phi}_{\mathbb{U}_{r}}\mathbb{U}_{r}^{2}\\ &=\frac 12 {\int}_{\mathbb{R}^{3}} (|\nabla\mathbb{U}_{r}|^{2} +\mathbb{U}_{r}^{2})-\frac{1}{p+1}{\int}_{\mathbb{R}^{3}}\mathbb{U}_{r}^{p+1}\\ &\quad+\frac{1}{p+1}{\int}_{\mathbb{R}^{3}}\left( 1-Q(|y|)\right)\mathbb{U}_{r}^{p+1} +\frac{1}{4}{\int}_{\mathbb{R}^{3}} K(|y|){\Phi}_{\mathbb{U}_{r}}\mathbb{U}_{r}^{2}\\ :&=I_{1}+I_{2}+I_{3}. \endaligned $$

For I₁, the symmetry gives that

$$ \aligned I_{1}&=\frac 12 {\int}_{\mathbb{R}^{3}}\Big[\sum\limits_{j=1}^{k}U^{p}(y-x_{j})\Big]\mathbb{U}_{r} -\frac{1}{p+1}{\int}_{\mathbb{R}^{3}}\mathbb{U}_{r}^{p+1}\\ &=\frac{k}{2} {\int}_{\mathbb{R}^{3}}U^{p+1}+\frac{k}{2}\sum\limits_{i=2}^{k}{\int}_{\mathbb{R}^{3}}U^{p}(y-x_{1})U(y-x_{i}) -\frac{k}{p+1}{\int}_{{\Omega}_{1}}\mathbb{U}_{r}^{p+1}\\ &=k\left( \frac{1}{2}{\int}_{\mathbb{R}^{3}}U^{p+1}-\frac{1}{2}\sum\limits_{i=2}^{k}{\int}_{\mathbb{R}^{3}}U^{p}(y-x_{1}) U(y-x_{i})-\frac{1}{p+1}{\int}_{\mathbb{R}^{3}}U^{p+1}+O\left( e^{-(1+\sigma)\rho}\right)\right)\\ &=k\Big(A-\Big(\frac{1}{2}+o(1)\Big)\sum\limits_{i=2}^{k}U(|x_{1}-x_{i}|)+O\left( e^{-(1+\sigma)\rho}\right)\Big), \endaligned $$

(A.4)

where 0 < σ < 1. Here we use the estimation:

$$ \begin{array}{@{}rcl@{}} \frac{1}{p+1}{\int}_{{\Omega}_{1}}\mathbb{U}^{p+1}_{r}&=&\frac{1}{p+1}{\int}_{{\Omega}_{1}}U^{p+1}(y-x_{1})dy +{\sum}_{j=2}^{k}{\int}_{{\Omega}_{1}}U^{p}(y-x_{1})U(y-x_{j})dy\\ &&+O\Big({\int}_{{\Omega}_{1}}U^{p-1}(y - x_{1})\Big({\sum}_{j=2}^{k}U(y - x_{j})\Big)^{2}dy + {\int}_{{\Omega}_{1}} \Big({\sum}_{j=2}^{k}U(y-x_{j})\Big)^{p+1}dy\Big)\\ &=&\frac{1}{p+1}{\int}_{\mathbb{R}^{3}}U^{p+1}(y-x_{1})dy +{\sum}_{j=2}^{k}{\int}_{\mathbb{R}^{3}}U^{p}(y-x_{1})U(y-x_{j})dy+O(e^{-(1+\sigma)\rho}). \end{array} $$

Next, we estimate I₂:

$$ \begin{array}{@{}rcl@{}} &&{\int}_{\mathbb{R}^{3}} \big(1-Q(|y|)\big)\mathbb{U}_{r}^{p+1}\\ &&\quad=k\left( {\int}_{\mathbb{R}^{3}}\big(1-Q(|y|)\big)U^{p+1}(y-x_{1})+(p+1){\sum}_{i=2}^{k}{\int}_{\mathbb{R}^{3}} \big(1-Q(|y|)\big)U^{p}(y-x_{1})U(y-x_{i})+O\left( e^{-(1+\sigma)\rho}\right)\right)\\ &&\quad=k\left( -\frac{b}{r^{n}}{\int}_{\mathbb{R}^{3}}U^{p+1}+O\left( e^{-(1+\sigma)\rho}+\frac{1}{r^{n+\kappa}}\right) \right)\\ &&\quad=k\left( -\frac{(p+1)B_{2}}{r^{n}}+O\left( e^{-(1+\sigma)\rho}+\frac{1}{r^{n+\kappa}}\right)\right), \end{array} $$

where 0 < κ ≤ ε. It is easy to see that

$$ I_{2}=\frac{1}{p+1}{\int}_{\mathbb{R}^{3}}\left( 1-Q(|y|)\right)\mathbb{U}_{r}^{p+1}=k\left( -\frac{B_{2}}{r^{n}} +O\left( e^{-(1+\sigma)\rho}+\frac{1}{r^{n+\kappa}}\right)\right). $$

(A.5)

Finally, we claim that

$$ \aligned I_{3}=\frac{k}{4} \left( \frac{a^{2}}{r^{2m}}\left( 4B_{1}+\sum\limits_{j=2}^{k}V_{1}(x_{j}-x_{1}){\int}_{\mathbb{R}^{3}}U^{2}\right) +O\left( \frac{1}{r^{2m+\sigma}}\right)\right). \endaligned $$

(A.6)

Note that

$$ \begin{array}{@{}rcl@{}} &&{\int}_{\mathbb{R}^{3}} K(|y|){\Phi}_{\mathbb{U}_{r}}\mathbb{U}_{r}^{2}=k{\int}_{{\Omega}_{1}} K(|y|){\Phi}_{\mathbb{U}_{r}}\mathbb{U}_{r}^{2}\\ &&\quad=k{\int}_{{\Omega}_{1}}K(|y|)\left( \frac{a}{r^{m}}{\sum}_{j=1}^{k}V_{1}(y-x_{j})+O\left( \frac{1}{r^{m+\theta}} {\sum}_{j=1}^{k}\frac{1}{1+|y-x_{j}|}\right)\right)\left( U_{1}+O\left( \frac{1}{r^{\sigma}} e^{-\frac{1}{2}|y-x_{1}|}\right)\right)^{2}\\ &&\quad=k\left[\frac{a}{r^{m}}{\int}_{{\Omega}_{1}}K(|y|){\sum}_{j=1}^{k}V_{1}(y-x_{j}){U_{1}^{2}}+O\left( \frac{1} {r^{m+\sigma}}{\int}_{\mathbb{R}^{3}}K(|y|){\sum}_{j=1}^{k}\frac{e^{-|y-x_{1}|}}{1+|y-x_{j}|}\right)\right]\\ &&\qquad+kO\left( \frac{1}{r^{m+\theta}}{\int}_{{\Omega}_{1}}K(|y|){\sum}_{j=1}^{k}\frac{{U_{1}^{2}}}{1+|y-x_{j}|} \right)\\ &&\quad=k\left( \frac{a^{2}}{r^{2m}}\left( {\int}_{\mathbb{R}^{3}}U^{2}V_{1}+{\sum}_{j=2}^{k}V_{1}(x_{j}-x_{1}) {\int}_{\mathbb{R}^{3}}U^{2}\right)+O\left( \frac{1}{r^{2m+\sigma}} \left( 1+{\sum}_{j=2}^{k}\frac{1}{|x_{j}-x_{1}|}\right)\right)\right). \end{array} $$

Since

$$ {\int}_{\mathbb{R}^{3}}U^{2}V_{1}=4B_{1} $$

and

$$ \sum\limits_{j=2}^{k}V_{1}(x_{j}-x_{1})\leq C\sum\limits_{j=2}^{k}\frac{1}{|x_{j}-x_{1}|}\leq C\frac{k}{r}\sum\limits_{j=2}^{k}\frac{1}{j}\leq \frac{Ck\ln k}{r}\leq C^{\prime}. $$

Combining (A.4), (A.5) and (A.6), we yield that

$$ \begin{array}{@{}rcl@{}} I(\mathbb{U}_{r}) &=k\left( A+\frac{a^{2}}{r^{2m}}\left( B_{1}+\frac{1}{4}\sum\limits_{j=2}^{k}V_{1}(x_{j}-x_{1}){\int}_{\mathbb{R}^{3}}U^{2}\right)-\frac{B_{2}}{r^{n}}-B_{3}\left( \frac{r}{k}\right)^{-1}e^{-\frac{2\pi r}{k}}+O\left( \frac{1}{r^{2m+\sigma}}+\frac{1}{r^{n+\sigma}}\right)\right). \end{array} $$

Thus, we finish the proof. □