A note on Wang’s conjecture for harmonic functions with nonlinear boundary condition

Abstract.

We obtain some Liouville type theorems for positive harmonic functions on compact Riemannian manifolds with nonnegative Ricci curvature and strictly convex boundary and partially verifies Wang’s conjecture (J. Geom. Anal. 31 (2021)).

For the specific manifold $\mathbb{B}^{n}$, we present a new proof of this conjecture, which has been resolved by Gu-Li (Math. Ann. 391 (2025)). Our proof is based on a general principle of applying the P-function method to such Liouville type results. As a further application of this method, we obtain some classification results for nonnegative solutions of some semilinear elliptic equations with a nonlinear boundary condition.

1. Introduction

Wang proposed a conjecture for Liouville type result for harmonic functions with some specific nonlinear boundary condition [Wan21, Conjecture 1].

Conjecture 1 (Wang, [Wan21]).

Let $(M^{n},g)\ (n\geq 3)$ be a compact Riemannian manifold with $\mathrm{Ric}\geq 0$ and the second fundamental form $\Pi\geq 1$ on $\partial M$. If $u\in C^{\infty}(M)$ is a positive solution of the following equation

(1.1) $$\begin{cases}\Delta u=0&\text{in }M,\\ \frac{\partial u}{\partial\nu}+\lambda u=u^{q}&\text{on }\partial M,\end{cases}$$

where $1<q\leq\frac{n}{n-2}$ and $0<\lambda\leq\frac{1}{q-1}$ are constants. Then either $u$ is a constant function, or $q=\frac{n}{n-2},\,\lambda=\frac{n-2}{2}$, $(M^{n},g)$ is isometric to $\mathbb{B}^{n}\subset\mathbb{R}^{n}$ and $u$ corresponds to

$$u(x)=\left(\frac{n-2}{2}\frac{1-|a|^{2}}{|a|^{2}|x|^{2}-2\langle x,a\rangle+1}\right)^{\frac{n-2}{2}},$$

where $a\in\mathbb{B}^{n}$.

This conjecture, if proved to be true, have several interesting geometric consequences, such as a sharp upper bound of the area of the boundary and a sharp lower bound of Steklov eigenvalue on such manifolds. See [Wan21, Section 2] and also [GHW21, Section 5] for detailed discussions.

Inspired by Xia-Xiong’s work on Steklov eigenvalue estimate [XX24], Guo-Hang-Wang [GHW21, Theorem 2] verified conjecture 1 for some special cases under nonnegative sectional curvature condition:

Theorem 1 (Guo-Hang-Wang, [GHW21]).

Let $(M^{n},g)\ (n\geq 3)$ be a compact Riemannian manifold with $\mathrm{Sec}\geq 0$ and $\Pi\geq 1$ on $\partial M$. Then the only positive solution to 1.1 is constant provided $3\leq n\leq 8$ and $1<q\leq\frac{4n}{5n-9}$, $0<\lambda\leq\frac{1}{q-1}$.

In the first part of this note, we shall partially confirm conjecture 1 under Ricci curvature condition via integration by parts and meticulously choosing the parameters.

Theorem 2.

Let $(M^{n},g)\ (n\geq 3)$ be a compact Riemannian manifold with $\mathrm{Ric}\geq 0$ and $\Pi\geq 1$ on $\partial M$. Then the only positive solution to 1.1 is constant provided one of the following two conditions holds:

1. $3\leq n\leq 7$, $1<q\leq\frac{3n}{4(n-2)}$, and $0<\lambda\leq\min\{\frac{1}{2(q-1)},\frac{3(n-1)}{2q}\}$
   

2. $3\leq n\leq 9$, $1<q\leq\frac{n+1+\sqrt{(5n-1)(n-1)}}{4(n-2)}$ and $0<\lambda\leq\min\{\frac{1}{2(q-1)},\frac{6q+1}{2q+1}\frac{n-1}{2q}\}$

Remark 1.1.

Let us compare the ranges of $q$ and $\lambda$ in theorem 2 with those in theorem 1.

• Condition (2) in theorem 2 allows one more dimension $n=9$ compared with theorem 1.

• theorem 2 yields a larger range of $q$ than theorem 1 in some dimensions. Explicitly, we have $\frac{n+1+\sqrt{(5n-1)(n-1)}}{4(n-2)}\geq\frac{3n}{4(n-1)}\geq\frac{4n}{5n-9}$ if $3\leq n\leq 5$, and $\frac{n+1+\sqrt{(5n-1)(n-1)}}{4(n-2)}\geq\frac{4n}{5n-9}\geq\frac{3n}{4(n-1)}$ if $6\leq n\leq 9$.

• The range of $\lambda$ in theorem 2 is not sharp. Moreover, condition (2) doesn’t cover condition (1) since $\frac{6q+1}{2q+1}\frac{n-1}{2q}<\frac{3(n-1)}{2q}$.

Except for studying it within some special ranges of parameters, another way toward conjecture 1 is to confine ourselves to some specific manifolds. Guo and Wang proposed such an individual conjecture on the model space $\mathbb{B}^{n}$ [GW20, Conjecture 1]:

Conjecture 2 (Guo-Wang, [GW20]).

If $u\in C^{\infty}(\mathbb{B}^{n})$ is a positive solution of the following equation

(1.2) $$\begin{cases}\Delta u=0&\mathrm{in}\ \mathbb{B}^{n},\\ \frac{\partial u}{\partial\nu}+\lambda u=u^{q}&\mathrm{on}\ \mathbb{S}^{n-1},\end{cases}$$

where $1<q\leq\frac{n}{n-2}$ and $0<\lambda\leq\frac{1}{q-1}$ are constants. Then either $u$ is a constant function, or $q=\frac{n}{n-2},\,\lambda=\frac{n-2}{2}$ and $u$ corresponds to

(1.3) $$u(x)=\left(\frac{n-2}{2}\frac{1-|a|^{2}}{|a|^{2}|x|^{2}-2\langle x,a\rangle+1}\right)^{\frac{n-2}{2}},$$

where $a\in\mathbb{B}^{n}$.

Historically, Escobar [Esc90, Theorem 2.1] (see also [Esc88]) classified all solution of 1.2 by Obata’s method [Oba71] when $q=\frac{n}{n-2}$ and $\lambda=\frac{n-2}{2}$. After several works in this field [GW20, GHW21, LO23, Ou24], Gu-Li [GL25, Theorem 1.1] finally give an affirmative answer to conjecture 2.

Theorem 3 (Gu-Li, [GL25]).

If $u\in C^{\infty}(\mathbb{B}^{n})$ is a positive solution of the equation 1.2 for some constants $1<q<\frac{n}{n-2}$ and $0<\lambda\leq\frac{1}{q-1}$, then $u$ is a constant function.

Gu-Li’s method is based on sophisticated integration by parts, with several delicately chosen parameters, and the computation therein is more or less formidable.

In the second part of this note, we shall provide a simplified proof of theorem 3. We have to admit that our proof is essentially equivalent to Gu-Li’s original proof. However, our argument is based on a general principle for classifying solutions of such semilinear elliptic equations, which is a continuation and development of that in [Wan22].

The basic idea is to start with the critical power case (i.e. $q=\frac{n}{n-2}$) and study the model solution 1.3 and come up with an appropriate function, known as the P-function in literature (in honer of L. Payne), whose constancy implies the rigidity of the solution $u$. For the subcritical power case (i.e. $1<q<\frac{n}{n-2}$), we regard the equation 1.2 as the critical one in a larger dimension space (see 3.3), then a modified argument as in the critical power case implies the conclusion. Readers interested in the P-function method are invited to [Pay68, Wei71, Dan11, CFP24] for more research in this realm.

Specific to conjecture 2, one merit of our argument is that the choice of the parameters is naturally indicated from the viewpoint of the P-function method. As an advantage, the computation is streamlined. Another contribution is that our calculation is in the spirit of Escobar’s work [Esc90] and clarifies the role of the weight function as providing a closed conformal vector field (see Lemma 3.2 and 3.10). This could shed light on some key difficulties for resolving conjecture 1.

Finally, we mention that Escobar [Esc90, Theorem 2.1] also classified all conformal metrics on $\mathbb{B}^{n}$ with nonzero constant scalar curvature and constant boundary mean curvature by studying the solution of the following semilinear elliptic equation with a nonlinear boundary condition:

$$\begin{cases}-\Delta u=\frac{n-2}{4(n-1)}R\,u^{\frac{n+2}{n-2}}&\mathrm{in}\ \mathbb{B}^{n},\\ \frac{\partial u}{\partial\nu}+\frac{n-2}{2}u=\frac{n-2}{2(n-1)}Hu^{\frac{n}{n-2}}&\mathrm{on}\ \mathbb{S}^{n-1},\end{cases}$$

where $R,\,H$ are constants. The same strategy as our proof of theorem 3 could be applied to derive a slightly more general classification result for such kind of semilinear elliptic equations.

Theorem 4.

If $u\in C^{\infty}(\mathbb{B}^{n})$ is a nonnegative solution of the following equation

(1.4) $$\begin{cases}-\Delta u=\frac{n-2}{4(n-1)}R\,u^{p}&\mathrm{in}\ \mathbb{B}^{n},\\ \frac{\partial u}{\partial\nu}+\lambda u=\frac{n-2}{2(n-1)}Hu^{\frac{p+1}{2}}&\mathrm{on}\ \mathbb{S}^{n-1},\end{cases}$$

where $R>0,H\geq 0$ are constants, $1<p\leq\frac{n+2}{n-2}$ and $0\leq\lambda\leq\frac{2}{p-1}$.

1. If $1<p\leq\frac{n+2}{n-2}$ and $\lambda=0$, then $u\equiv 0$.

2. If $p=\frac{n+2}{n-2}$ and $0<\lambda<\frac{2}{p-1}=\frac{n-2}{2}$, then

   $$u(x)=\left(\frac{n(n-1)}{R}\right)^{\frac{n-2}{4}}\left(\frac{1}{2\epsilon}|x|^{2}+\frac{\epsilon}{2}\right)^{-\frac{n-2}{2}},$$

   where $0<\epsilon=\sqrt{\frac{n(n-1)}{R}}\left(\frac{n-2}{2\lambda}\frac{H}{n-1}+\sqrt{(\frac{n-2}{2\lambda})^{2}(\frac{H}{n-1})^{2}+\frac{R}{n(n-1)}(\frac{n-2}{\lambda}-1)}\right)$.

3. If $p=\frac{n+2}{n-2}$ and $\lambda=\frac{2}{p-1}=\frac{n-2}{2}$, then

   $$u(x)=\left(\frac{n(n-1)}{R}\right)^{\frac{n-2}{4}}\left(\frac{(1+\epsilon^{2}|a|^{2})|x|^{2}+2(1+\epsilon^{2})\langle x,a\rangle+(\epsilon^{2}+|a|^{2})}{2\epsilon(1-|a|^{2})}\right)^{-\frac{n-2}{2}},$$

   where $a\in\mathbb{B}^{n}$, $0<\epsilon=\sqrt{\frac{n(n-1)}{R}}\left(\frac{H}{n-1}+\sqrt{(\frac{H}{n-1})^{2}+\frac{R}{n(n-1)}}\right)$.

Remark 1.2.

The case $R=0$ is completely solved by theorem 3, so we don’t include it in the above theorem.

Remark 1.3.

The case $R>0,\,H=0$ has been obtained by Dou-Hu-Xu [DHX25, Theorem 1.1]. Our arguments present a simplified proof of their result. We also invite readers to [DHX25] for some consequences of such a Liouville type result.

We also mention that our strategy for theorem 3 could be applied to give another proof of [BV91, Theorem 6.1] and hence answer a question raised by Wang [Wan22, Section 5]. This is not the main theme of this note, so we put it in the appendix. We hope that this strategy and general principle may also be useful in some other situations.

This note is organized as follows. In Section 2, we prove theorem 2. In Section 3, we present our new proof of theorem 3. In Section 4, we prove theorem 4. Finally, in appendix A, we include our proof of [BV91, Theorem 6.1].

Acknowledgement: This work was completed during a research visit to Michigan State University, supported by the Zhiyuan Scholarship of Shanghai Jiao Tong University. The author wishes to thank the university for its hospitality and is especially grateful to Prof. Xiaodong Wang and Dr. Zhixin Wang for their helpful discussions.

2. Proof of theorem 2

Our proof exploits two main ingredients in the method of integration by parts: Bochner formula (Step 1) and the equation itself (Step 2).

Proof.

Step 1: Let $v:=u^{-\frac{1}{a}}$, where $a>0$ is to be determined. We follow the notations in [GHW21] and define $\chi:=\frac{\partial v}{\partial\nu},\,f:=v|_{\partial M}$. Then it’s straightforward to see that

(2.1) $$\begin{cases}\Delta v=(a+1)v^{-1}|\nabla v|^{2}&\text{in }M,\\ \chi=\frac{1}{a}(\lambda f-f^{a+1-aq})&\text{on }\partial M.\end{cases}$$

By Bochner formula, there holds

$$\frac{1}{2}\Delta|\nabla v|^{2}=\left|\nabla^{2}v-\frac{\Delta}{n}g\right|^{2}+\frac{1}{n}(\Delta v)^{2}+\langle\nabla\Delta v,\nabla v\rangle+\mathrm{Ric}(\nabla v,\nabla v).$$

Multiply both sides by $v^{b}$ and integrate it over $M$, where $b$ is a constant to be determined, we have

(2.2) $$\frac{1}{2}\int_{M}v^{b}\Delta|\nabla v|^{2}=\int_{M}v^{b}\left(\left|\nabla^{2}v-\frac{\Delta}{n}g\right|^{2}+\mathrm{Ric}(\nabla v,\nabla v)\right)+\frac{1}{n}\int_{M}v^{b}(\Delta v)^{2}+\int_{M}v^{b}\langle\nabla\Delta v,\nabla v\rangle.$$

It follows from 2.1, divergence theorem and the boundary curvature assumption $\Pi\geq 1,H\geq n-1$ that

$$\frac{1}{2}\int_{M}v^{b}\Delta|\nabla v|^{2}=\frac{1}{2}\int_{M}\mathrm{div}(v^{b}\nabla|\nabla v|^{2})-\langle\nabla v^{b},\nabla|\nabla v|^{2}\rangle$$

$$= \int_{\partial M}f^{b}\langle\nabla_{\nabla f+\chi\nu}\nabla v,\nu\rangle-b\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle$$

$$= \int_{\partial M}f^{b}\left(\langle\nabla f,\nabla\chi\rangle-\langle\nabla v,\nabla_{\nabla f}\nu\rangle+\chi\nabla^{2}v(\nu,\nu)\right)-b\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle$$

$$= \int_{\partial M}f^{b}\left(\langle\nabla f,\nabla\chi\rangle-\Pi(\nabla f,\nabla f)+\chi(\Delta v-\Delta f-H\chi)\right)-b\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle$$

$$= \int_{\partial M}f^{b}\left(2\langle\nabla f,\nabla\chi\rangle-\Pi(\nabla f,\nabla f)+(a+1)\chi f^{-1}(|\nabla f|^{2}+\chi^{2})+bf^{-1}\chi|\nabla f|^{2}-H\chi^{2}\right)$$

$$-b\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle$$

$$\leq (2(q-1)\lambda-1)\int_{\partial M}f^{b}|\nabla f|^{2}+\left(a+1+b+2(a+1-aq)\right)\int_{\partial M}f^{b}\frac{\chi}{f}|\nabla f|^{2}+(a+1)\int_{\partial M}f^{b}\frac{\chi}{f}\chi^{2}$$

$$-(n-1)\int_{\partial M}f^{b}\chi^{2}-b\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle$$

$$= (2(q-1)\lambda-1)\int_{\partial M}f^{b}|\nabla f|^{2}+\frac{a+1+b+2(a+1-aq)}{a}\lambda\int_{\partial M}f^{b}|\nabla f|^{2}$$

$$-\frac{a+1+b+2(a+1-aq)}{a}\int_{\partial M}f^{b+a-aq}|\nabla f|^{2}+\frac{a+1}{a}\lambda\int_{\partial M}f^{b}\chi^{2}-\frac{a+1}{a}\int_{\partial M}f^{b+a-aq}\chi^{2}$$

$$-(n-1)\int_{\partial M}f^{b}\chi^{2}-b\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle$$

$$= \left(\frac{a+b+3}{a}\lambda-1\right)\int_{\partial M}f^{b}|\nabla f|^{2}+\left(\frac{a+1}{a}\lambda-(n-1)\right)\int_{\partial M}f^{b}\chi^{2}$$

(2.3) $$+\left(2(q-1)-\frac{a+b+3}{a}\right)\int_{\partial M}f^{b+a-aq}|\nabla f|^{2}-\frac{a+1}{a}\int_{\partial M}f^{b+a-aq}\chi^{2}-b\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle.$$

On the other hand, the right hand side of 2.2 could be written as

$$\int_{M}v^{b}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\mathrm{Ric}(\nabla v,\nabla v)\right)+\frac{(a+1)^{2}}{n}\int_{M}wv^{b-2}|\nabla v|^{4}-(a+1)\int_{M}v^{b-2}|\nabla v|^{4}$$

(2.4) $$+2(a+1)\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle.$$

Therefore, by 2.2, Section 2 and Section 2 we have

$$\left(\frac{a+b+3}{a}\lambda-1\right)\int_{\partial M}f^{b}|\nabla f|^{2}+\left(\frac{a+1}{a}\lambda-(n-1)\right)\int_{\partial M}f^{b}\chi^{2}$$

$$+\left(2(q-1)-\frac{a+b+3}{a}\right)\int_{\partial M}f^{b+a-aq}|\nabla f|^{2}-\frac{a+1}{a}\int_{\partial M}f^{b+a-aq}\chi^{2}$$

$$\geq \int_{M}v^{b}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\mathrm{Ric}(\nabla v,\nabla v)\right)+\left(\frac{(a+1)^{2}}{n}-(a+1)\right)\int_{M}v^{b-2}|\nabla v|^{4}$$

(2.5) $$+(2(a+1)+b)\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle.$$

Step 2: Multiply both sides of 2.1 by $v^{b}$ and integrate it over $M$, we have

(2.6) $$\int_{M}v^{b}(\Delta v)^{2}=(a+1)\int_{M}v^{b-1}|\nabla v|^{2}\Delta v.$$

It follows from 2.1 that the left hand side of 2.6 could be written as

$$\int_{M}v^{b}(\Delta v)^{2}=\int_{M}\mathrm{div}(v^{b}\Delta v\nabla v)-\langle\nabla v^{b},\nabla v\rangle\Delta v-\langle\nabla\Delta v,\nabla v\rangle v^{b}$$

$$= (a+1)\int_{\partial M}f^{b-1}(|\nabla f|^{2}+\chi^{2})\chi-(a+1)b\int_{M}v^{b-2}|\nabla v|^{4}$$

$$+(a+1)\int_{M}v^{b-2}|\nabla v|^{4}-2(a+1)\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle$$

$$= \frac{a+1}{a}\lambda\int_{\partial M}f^{b}|\nabla f|^{2}+\frac{a+1}{a}\lambda\int_{\partial M}f^{b}\chi^{2}-\frac{a+1}{a}\int_{\partial M}f^{b+a-aq}|\nabla f|^{2}-\frac{a+1}{a}\int_{\partial M}f^{b+a-aq}\chi^{2}$$

(2.7) $$+(a+1)(1-b)\int_{M}v^{b-2}|\nabla v|^{4}-2(a+1)\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle.$$

Therefore, by 2.6 and Section 2 we have

$$\frac{a+1}{a}\lambda\int_{\partial M}f^{b}|\nabla f|^{2}+\frac{a+1}{a}\lambda\int_{\partial M}f^{b}\chi^{2}-\frac{a+1}{a}\int_{\partial M}f^{b+a-aq}|\nabla f|^{2}-\frac{a+1}{a}\int_{\partial M}f^{b+a-aq}\chi^{2}$$

(2.8) $$= \left((a+1)^{2}+(a+1)(b-1)\right)\int_{M}v^{b-2}|\nabla v|^{4}+2(a+1)\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle.$$

Step 3: Now consider $\lx@cref{creftype~refnum}{eq. equation 1}+c\lx@cref{creftype~refnum}{eq. equation 2}$:

$$\left(\frac{a+b+3}{a}\lambda-1+\frac{a+1}{a}\lambda c\right)\int_{\partial M}f^{b}|\nabla f|^{2}+\left(\frac{a+1}{a}\lambda-(n-1)+\frac{a+1}{a}\lambda c\right)\int_{\partial M}f^{b}\chi^{2}$$

$$+\left(2(q-1)-\frac{a+b+3}{a}-\frac{a+1}{a}c\right)\int_{\partial M}f^{b+a-aq}|\nabla f|^{2}-(\frac{a+1}{a}+\frac{a+1}{a}c)\int_{\partial M}f^{b+a-aq}\chi^{2}$$

$$\geq \int_{M}v^{b}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\mathrm{Ric}(\nabla v,\nabla v)\right)+\left((\frac{1}{n}+c)(a+1)^{2}+(cb-c-1)(a+1)\right)\int_{M}v^{b-2}|\nabla v|^{4}$$

$$+\left(2(c+1)(a+1)+b\right)\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle.$$

Define $\beta$ by setting $b=-\beta(a+1)$, $x:=\frac{1}{a}$ and choose $c=-1+\frac{\beta}{2}$ to eliminate the term $\int_{M}v^{b-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle$, we obtain

$$\left(\frac{4-\beta}{2}\lambda x-(1+\frac{\beta\lambda}{2})\right)\int_{\partial M}f^{b}|\nabla f|^{2}+\left(\frac{\beta\lambda}{2}(x+1)-(n-1)\right)\int_{\partial M}f^{b}\chi^{2}$$

$$+\left(-\frac{4-\beta}{2}x+2(q-1)+\frac{1}{2}\beta\right)\int_{\partial M}f^{b+a-aq}|\nabla f|^{2}-\frac{1}{2}\beta(x+1)\int_{\partial M}f^{b+a-aq}\chi^{2}$$

$$\geq \int_{M}v^{b}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\mathrm{Ric}(\nabla v,\nabla v)\right)$$

(2.9) $$+\frac{1}{x^{2}}(x+1)\left(-(\frac{1}{2}\beta^{2}-\beta+\frac{n-1}{n})(x+1)+\frac{1}{2}\beta\right)\int_{M}v^{b-2}|\nabla v|^{4}.$$

Step 4: Now we verify the condition (1) in theorem 2: Take $\beta=1,x=\frac{\beta+4(q-1)}{4-\beta}=\frac{1+4(q-1)}{3}$ such that $-\frac{4-\beta}{2}x+2(q-1)+\frac{1}{2}\beta=0$. Equivalently, $a=\frac{3}{1+4(q-1)}$ and $b=-(a+1)=-\frac{4q}{4q-3}$. Then Section 2 turns out to be

$$\left(2(q-1)\lambda-1\right)\int_{\partial M}f^{b}|\nabla f|^{2}+\left(\frac{2}{3}q\lambda-(n-1)\right)\int_{\partial M}f^{b}\chi^{2}-\frac{2}{3}q\int_{\partial M}f^{b+a-aq}\chi^{2}$$

(2.10) $$\geq \int_{M}v^{b}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\mathrm{Ric}(\nabla v,\nabla v)\right)+\left(\frac{3}{1+4(q-1)}\right)^{2}\frac{4}{3}q\left(\frac{1}{2}-\frac{2(n-2)}{3n}q\right)\int_{M}v^{b-2}|\nabla v|^{4}.$$

Then the condition (1) in theorem 2 implies that the left hand side of Section 2 is less or equal to zero, while the right hand side of Section 2 is larger or equal to zero. It follows that $0\equiv\chi=\frac{\partial v}{\partial\nu}$. Hence by 2.1 we deduce that $v|_{\partial M}=f$ is constant, and $u$ is a harmonic function in $M$ with constant boundary value on $\partial M$. Therefore $u$ is constant in $M$.

Step 5: Now we verify the condition (2) in theorem 2: By analyzing the range of $x$ such that the left hand side of Section 2 is less or equal to zero and the right hand side of Section 2 is larger or equal to zero, we obtain the optimal value of $\beta$ as $\beta=\frac{4q+2}{4q+1}$, $x=\frac{\beta+4(q-1)}{4-\beta}=\frac{8q^{2}-4q-1}{6q+1}$. Equivalently, we choose $a=\frac{6q+1}{8q^{2}-4q-1}$ and $b=-\frac{4q+2}{4q+1}(a+1)=-\frac{4q(2q+1)}{8q^{2}-4q-1}$. Then Section 2 turns out to be

$$\left(2(q-1)\lambda-1\right)\int_{\partial M}f^{b}|\nabla f|^{2}+\left(\frac{2q(2q+1)}{6q+1}\lambda-(n-1)\right)\int_{\partial M}f^{b}\chi^{2}-\frac{2q(2q+1)}{6q+1}\int_{\partial M}f^{b+a-aq}\chi^{2}$$

$$\geq \int_{M}v^{b}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\mathrm{Ric}(\nabla v,\nabla v)\right)$$

(2.11) $$+\left(\frac{6q+1}{8q^{2}-4q-1}\right)^{2}\left(\frac{2q(4q+1)}{6q+1}\right)\left(\frac{-4(n-2)q^{2}+2(n+1)q+n}{(6q+1)n}\right)\int_{M}v^{b-2}|\nabla v|^{4}.$$

Then the condition (2) in theorem 2 implies that the left hand side of Section 2 is less or equal to zero, while the right hand side of Section 2 is larger or equal to zero. As before, we could conclude that $u$ is a constant function on $M$. ∎

3. Proof of theorem 3

We shall first establish some identities that holds on general Riemannian manifolds.

Lemma 3.1.

Let $(M^{n},g)$ be a Riemannian manifold. For any $v\in C^{\infty}(M)$ and constant $d>0$, there holds

(3.1) $$\mathrm{div}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right)=\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}+\left(1-\frac{1}{d}\right)\langle\nabla\Delta v,\nabla v\rangle+\mathrm{Ric}(\nabla v,\nabla v).$$

Proof.

This is a straightforward corollary of Bochner formula. ∎

Lemma 3.2.

Let $(M^{n},g)$ be a Riemannian manifold admitting a smooth function $w$ with $\nabla^{2}w=\frac{\Delta w}{n}g$, then for any $v\in C^{\infty}(M)$ and constant $d>0$ there holds

(3.2) $$\mathrm{div}\left(\nabla_{\nabla w}\nabla v-\frac{\Delta v}{d}\nabla w\right)=\left(\frac{1}{n}-\frac{1}{d}\right)\Delta v\Delta w-\left(1-\frac{1}{n}\right)\langle\nabla\Delta w,\nabla v\rangle+\left(1-\frac{1}{d}\right)\langle\nabla\Delta v,\nabla w\rangle.$$

Proof.

Notice that

$$\mathrm{div}(\nabla_{\nabla w}\nabla v) =\frac{1}{2}\langle\nabla^{2}v,\mathcal{L}_{\nabla w}g\rangle+\langle\nabla\Delta v,\nabla w\rangle+Ric(\nabla v,\nabla w)$$

$$=\frac{1}{n}\Delta v\Delta w+\langle\nabla\Delta v,\nabla w\rangle+Ric(\nabla v,\nabla w),$$

where we used Ricci’s identity

$$\mathrm{div}(\nabla^{2}v)=\nabla\Delta v+\mathrm{Ric}(\nabla v,\cdot).$$

The hypothesis on $w$ implies that the curvature operator $R$ satisfies

$$R(X,Y)\nabla w=\left\langle X,\frac{\nabla\Delta w}{n}\right\rangle Y-\left\langle Y,\frac{\nabla\Delta w}{n}\right\rangle X,\quad\forall X,Y\in T_{p}(M).$$

It follows that

$$Ric(\nabla w,\cdot)=-\frac{n-1}{n}\nabla\Delta w.$$

Therefore, we derive

$$\mathrm{div}(\nabla_{\nabla w}\nabla v)=\frac{1}{n}\Delta v\Delta w+\langle\nabla\Delta v,\nabla w\rangle-\frac{n-1}{n}\langle\nabla\Delta w,\nabla v\rangle.$$

Hence the desired identity follows. ∎

Now we are ready to present our proof of theorem 3. Roughly speaking, we regard the equation 1.2 as the critical power case in a $d-$dimensional space, where

(3.3) $$d:=\frac{2q}{q-1}\geq n.$$

Then we modify Escobar’s argument [Esc90] to fit in this ”critical case”, and use the boundary condition to tackle the emerging terms in this case.

At the end of our proof, we shall review and compare our choice of parameters with those in Gu-Li [GL25].

Proof of theorem 3:.

Denote $g$ to be the Euclidean metric on $\mathbb{B}^{n}$ and $\nu$ to be the unit outer normal vector of $\partial\mathbb{B}^{n}=\mathbb{S}^{n-1}$.

Step 1: Consider the intrinsic dimension $d:=\frac{2q}{q-1}$ so that $q=\frac{d}{d-2}$. Then $d\geq n\Leftrightarrow q\leq\frac{n}{n-2}$. Let $v:=u^{-\frac{2}{d-2}}$, $\chi:=\frac{\partial v}{\partial\nu},f:=v|_{\mathbb{S}^{n-1}}$, then $v$ satisfies

(3.4) $$\begin{cases}\Delta v=\frac{d}{2}v^{-1}|\nabla v|^{2}&\mathrm{in}\ \mathbb{B}^{n},\\ \chi=\frac{2}{d-2}(\lambda f-1)&\mathrm{on}\ \mathbb{S}^{n-1}.\end{cases}$$

We shall consider the P-function $P:=v^{-1}|\nabla v|^{2}=\frac{2}{d}\Delta v$, which satisfies $P\equiv constant$ if $u$ is the model solution 1.3, and derive the equation satisfied by $P$ (i.e. 3.6).

It’s straightforward to see from 3.4 that

(3.5) $$\nabla\Delta v=dv^{-1}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right).$$

Hence by Lemma 3.1 we have

(3.6) $$\mathrm{div}\left(v^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right)\right)=v^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right).$$

Integrate this equation over $\mathbb{B}^{n}$ and use the boundary value condition 3.4, we get

$$\int_{\mathbb{B}^{n}}v^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right)$$

$$= \int_{\mathbb{S}^{n-1}}f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nabla v,\nu)=\int_{\mathbb{S}^{n-1}}f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nabla f+\chi\nu,\nu)$$

$$= \int_{\mathbb{S}^{n-1}}f^{1-d}(\langle\nabla f,\nabla\chi\rangle-|\nabla f|^{2})+\int_{\mathbb{S}^{n-1}}\chi f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu)$$

$$= \left(\frac{2}{d-2}\lambda-1\right)\int_{\mathbb{S}^{n-1}}f^{1-d}|\nabla f|^{2}+\frac{2}{d-2}\lambda\int_{\mathbb{S}^{n-1}}f^{2-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu)$$

(3.7) $$-\frac{2}{d-2}\int_{\mathbb{S}^{n-1}}f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu).$$

Step 2: Take $w:=\frac{1-|x|^{2}}{2}$. Then $w$ satisfies: $w|_{\mathbb{S}^{n-1}}=0,\ \nabla w|_{\mathbb{S}^{n-1}}=-\nu,\ \nabla^{2}w=\frac{\Delta w}{n}g,\ \Delta w\equiv-n$. We shall use $w$ to tackle the last two terms in the right hand side of Section 3.

Now use Lemma 3.2, 3.5 and $\Delta w\equiv-n$, we derive

(3.8) $$\mathrm{div}\left(v^{2-d}\left(\nabla_{\nabla w}\nabla v-\frac{\Delta v}{d}\nabla w\right)\right)=\left(\frac{n}{d}-1\right)v^{2-d}\Delta v+v^{1-d}\left\langle\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v,\nabla w\right\rangle.$$

It follows from 3.6 and 3.8 that

$$wv^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right)=w\mathrm{div}\left(v^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right)\right)$$

$$= \mathrm{div}\left(wv^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right)\right)-v^{1-d}\left\langle\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v,\nabla w\right\rangle$$

$$= \mathrm{div}\left(wv^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right)\right)-\mathrm{div}\left(v^{2-d}\left(\nabla_{\nabla w}\nabla v-\frac{\Delta v}{d}\nabla w\right)\right)+\left(\frac{n}{d}-1\right)v^{2-d}\Delta v.$$

Integrate this equation over $\mathbb{B}^{n}$ and notice that $\nabla w|_{\mathbb{S}^{n-1}}=-\nu$, we get

$$\int_{\mathbb{B}^{n}}wv^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right)$$

(3.9) $$= \int_{\mathbb{S}^{n-1}}f^{2-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu)+\left(\frac{n}{d}-1\right)\int_{\mathbb{B}^{n}}v^{2-d}\Delta v.$$

Once again,3.5 and Lemma 3.2 implies

$$\mathrm{div}\left(v^{1-d}\left(\nabla_{\nabla w}\nabla v-\frac{\Delta v}{d}\nabla w\right)\right)=\left(\frac{n}{d}-1\right)v^{1-d}\Delta v.$$

Integrate this equation over $\mathbb{B}^{n}$ and notice that $\nabla w|_{\mathbb{S}^{n-1}}=-\nu$, we get

(3.10) $$\int_{\mathbb{S}^{n-1}}f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu)=\left(1-\frac{n}{d}\right)\int_{\mathbb{B}^{n}}v^{1-d}\Delta v.$$

Note that 3.10 is exactly the Pohozaev identity [Sch88, Proposition 1.4] used by Escobar [Esc90], whose validity comes from the fact that $\nabla w$ is a closed conformal vector field.

Plug Section 3 and 3.10 into Section 3, we derive the key integral identity

$$\int_{\mathbb{B}^{n}}\left(1-\frac{2}{d-2}\lambda w\right)v^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right)$$

$$= \left(\frac{2}{d-2}\lambda-1\right)\int_{\mathbb{S}^{n-1}}f^{1-d}|\nabla f|^{2}+\frac{2}{d-2}\left(\frac{n}{d}-1\right)\left(\int_{\mathbb{B}^{n}}v^{1-d}\Delta v-\lambda\int_{\mathbb{B}^{n}}v^{2-d}\Delta v\right).$$

Notice that $1-\frac{2}{d-2}\lambda w\geq 1-\frac{1}{d-2}\lambda\geq 0$ , $\frac{2}{d-2}\lambda-1\leq 0$ and $\frac{n}{d}-1\leq 0$, our proof would be complete once we show that

(3.11) $$\int_{\mathbb{B}^{n}}v^{1-d}\Delta v-\lambda\int_{\mathbb{B}^{n}}v^{2-d}\Delta v\geq 0.$$

Step 3: We shall use the boundary condition 3.4 to make 3.11 homogeneous.

The boundary condition in 3.4 implies

$$\int_{\mathbb{S}^{n-1}}f^{2-d}\chi=\int_{\mathbb{B}^{n}}\mathrm{div}(v^{2-d}\nabla v)=\int_{\mathbb{B}^{n}}(v^{2-d}\Delta v+(2-d)v^{-d}|\nabla v|^{2})=\frac{4-d}{d}\int_{\mathbb{B}^{n}}v^{2-d}\Delta v,$$

$$\int_{\mathbb{S}^{n-1}}f^{1-d}\chi=\int_{\mathbb{B}^{n}}\mathrm{div}(v^{1-d}\nabla v)=\int_{\mathbb{B}^{n}}(v^{1-d}\Delta v+(1-d)v^{-d}|\nabla v|^{2})=\frac{2-d}{d}\int_{\mathbb{B}^{n}}v^{1-d}\Delta v.$$

Therefore we have

$$\int_{\mathbb{S}^{n-1}}f^{1-d}\chi^{2}=\frac{2}{d-2}\lambda\int_{\mathbb{S}^{n-1}}f^{2-d}\chi-\frac{2}{d-2}\int_{\mathbb{S}^{n-1}}f^{1-d}\chi=\frac{2(4-d)}{d(d-2)}\lambda\int_{\mathbb{B}^{n}}v^{2-d}\Delta v+\frac{2}{d}\int_{\mathbb{B}^{n}}v^{1-d}\Delta v.$$

Equivalently,

$$\int_{\mathbb{B}^{n}}v^{1-d}\Delta v=\frac{d}{2}\int_{\mathbb{S}^{n-1}}f^{1-d}\chi^{2}+\frac{d-4}{d-2}\lambda\int_{\mathbb{B}^{n}}v^{2-d}\Delta v.$$

Hence we obtain

$$\int_{\mathbb{B}^{n}}v^{1-d}\Delta v-\lambda\int_{\mathbb{B}^{n}}v^{2-d}\Delta v=\frac{d}{2}\int_{\mathbb{S}^{n-1}}f^{1-d}\chi^{2}-\frac{2}{d-2}\lambda\int_{\mathbb{B}^{n}}v^{2-d}\Delta v.$$

Therefore, it suffices to show that $\frac{d}{2}\int_{\mathbb{S}^{n-1}}f^{1-d}\chi^{2}-\frac{2}{d-2}\lambda\int_{\mathbb{B}^{n}}v^{2-d}\Delta v\geq 0$.

Step 4: Using $\nabla^{2}w=-g$ and 3.4, we could express $\int_{\mathbb{S}^{n-1}}f^{1-d}\chi^{2}$ as

$$\int_{\mathbb{S}^{n-1}}f^{1-d}\chi^{2}=-\int_{\mathbb{B}^{n}}\mathrm{div}\left(v^{1-d}\langle\nabla v,\nabla w\rangle\nabla v\right)$$

$$= -\int_{\mathbb{B}^{n}}\left\{v^{1-d}\langle\nabla v,\nabla w\rangle\Delta v+(1-d)v^{-d}\langle\nabla v,\nabla w\rangle|\nabla v|^{2}+v^{1-d}\langle\nabla_{\nabla v}\nabla v,\nabla w\rangle+v^{1-d}\nabla^{2}w(\nabla v,\nabla v)\right\}$$

$$= -\int_{\mathbb{B}^{n}}v^{1-d}\left\langle\nabla_{\nabla v}\nabla v-\frac{d-2}{d}\Delta v\nabla v,\nabla w\right\rangle+\frac{2}{d}\int_{\mathbb{B}^{n}}v^{2-d}\Delta v$$

$$= \int_{\mathbb{B}^{n}}\left\{w\mathrm{div}\left(v^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{d-2}{d}\Delta v\nabla v\right)\right)-\mathrm{div}\left(wv^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{d-2}{d}\Delta v\nabla v\right)\right)\right\}$$

$$+\frac{2}{d}\int_{\mathbb{B}^{n}}v^{2-d}\Delta v.$$

Therefore,

$$\frac{d}{2}\int_{\mathbb{S}^{n-1}}f^{1-d}\chi^{2}-\frac{2}{d-2}\lambda\int_{\mathbb{B}^{n}}v^{2-d}\Delta v$$

(3.12) $$= \left(1-\frac{2}{d-2}\lambda\right)\int_{\mathbb{B}^{n}}v^{2-d}\Delta v+\frac{d}{2}\int_{\mathbb{B}^{n}}w\mathrm{div}\left(v^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{d-2}{d}\Delta v\nabla v\right)\right).$$

Finally, we use Lemma 3.1 and 3.5 to calculate the last term in Section 3 as follows:

$$w\mathrm{div}\left(v^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{d-2}{d}\Delta v\nabla v\right)\right)$$

$$= wv^{1-d}\left(\mathrm{div}\left(\nabla_{\nabla v}\nabla v-\frac{d-2}{d}\Delta v\nabla v\right)+(1-d)v^{-1}\left\langle\nabla_{\nabla v}\nabla v-\frac{d-2}{d}\Delta v\nabla v,\nabla v\right\rangle\right)$$

$$= wv^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{d-2}{d}\right)(\Delta v)^{2}+2v^{-1}\left\langle\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v,\nabla v\right\rangle\right.$$

$$\quad\quad\quad\left.+(1-d)v^{-1}\left\langle\nabla_{\nabla v}\nabla v-\frac{d-2}{d}\Delta v\nabla v,\nabla v\right\rangle\right)$$

$$= wv^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+(3-d)v^{-1}\langle\nabla_{\nabla v}\nabla v,\nabla v\rangle+\left(\frac{1}{n}-\frac{d-2}{d}+\frac{2(d-3)}{d}\right)(\Delta v)^{2}\right)$$

$$= wv^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g+\frac{3-d}{2}\frac{\mathrm{d}v\otimes\mathrm{d}v}{v}\right|^{2}+\frac{(d-3)(d+2n)+3(d-n)}{nd^{2}}(\Delta v)^{2}\right)\geq 0.$$

This finishes the proof of theorem 3. ∎

Remark 3.1.

Gu-Li chose the power of $v$ as $a=-\frac{q+1}{q-1}$ (see [GL25, Section 3.1]), which is exactly $1-d$ in our 3.6. They chose the combination coefficient of the vector field as $b=\frac{q-1}{2q}$ (see [GL25, Section 3.2]), which is exactly $\frac{1}{d}$ in our 3.5. They finally choose the combination coefficient of the weight function as $\frac{2}{1+c}=\lambda(q-1)$ (see [GL25, Section 3.3]), which equals $\frac{2}{d-2}\lambda$ and appears naturally in our Section 3.

4. Proof of theorem 4

Proof of theorem 4:.

Denote $g$ to be the Euclidean metric on $\mathbb{B}^{n}$ and $\nu$ to be the unit outer normal vector of $\partial\mathbb{B}^{n}=\mathbb{S}^{n-1}$.

Assume $u$ is not identically zero, then $u$ is superharmonic by 1.4. Hence the maximum principle and the Hopf lemma imply that $u>0$ on $\overline{\mathbb{B}^{n}}$.

The remaining calculation is a minor modification of our proof of theorem 3 in Section 3. For completeness, we repeat the calculation and provide all details.

Step 1: Define $d:=\frac{2(p+1)}{p-1}$. Then $d\geq n\Leftrightarrow p\leq\frac{n+2}{n-2}$. Let $v:=u^{-\frac{2}{d-2}}$, $\chi:=\frac{\partial v}{\partial\nu},f:=v|_{\mathbb{S}^{n-1}}$, then $v$ satisfies

(4.1) $$\begin{cases}\Delta v=\frac{d}{2}v^{-1}|\nabla v|^{2}+\frac{n-2}{2(d-2)}\frac{R}{n-1}v^{-1}&\mathrm{in}\ \mathbb{B}^{n},\\ \chi=\frac{2}{d-2}\lambda f-\frac{n-2}{d-2}\frac{H}{n-1}&\mathrm{on}\ \mathbb{S}^{n-1}.\end{cases}$$

The key point is that 3.5 still holds in this case for a solution $v$ in 4.1:

(4.2) $$\nabla\Delta v=dv^{-1}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right).$$

Hence by Lemma 3.1 we get 3.6. Integrate this over $\mathbb{B}^{n}$ and use the boundary condition 4.1, we get

$$\int_{\mathbb{B}^{n}}v^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right)$$

$$= \int_{\mathbb{S}^{n-1}}f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nabla v,\nu)=\int_{\mathbb{S}^{n-1}}f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nabla f+\chi\nu,\nu)$$

$$= \int_{\mathbb{S}^{n-1}}f^{1-d}(\langle\nabla f,\nabla\chi\rangle-|\nabla f|^{2})+\int_{\mathbb{S}^{n-1}}\chi f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu)$$

$$= \left(\frac{2}{d-2}\lambda-1\right)\int_{\mathbb{S}^{n-1}}f^{1-d}|\nabla f|^{2}+\frac{2}{d-2}\lambda\int_{\mathbb{S}^{n-1}}f^{2-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu)$$

(4.3) $$-\frac{n-2}{d-2}\frac{H}{n-1}\int_{\mathbb{S}^{n-1}}f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu).$$

Step 2: Take $w:=\frac{1-|x|^{2}}{2}$. Then $w$ satisfies: $w|_{\mathbb{S}^{n-1}}=0,\ \nabla w|_{\mathbb{S}^{n-1}}=-\nu,\ \nabla^{2}w=\frac{\Delta w}{n}g,\ \Delta w\equiv-n$. We shall use $w$ to tackle the last two terms in the right hand side of Section 4

Now use Lemma 3.2, 4.2 and $\Delta w\equiv-n$, we derive 3.8. It follows from 3.6 and 3.8 that

$$wv^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right)=w\mathrm{div}\left(v^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right)\right)$$

$$= \mathrm{div}\left(wv^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right)\right)-v^{1-d}\left\langle\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v,\nabla w\right\rangle$$

$$= \mathrm{div}\left(wv^{1-d}\left(\nabla_{\nabla v}\nabla v-\frac{\Delta v}{d}\nabla v\right)\right)-\mathrm{div}\left(v^{2-d}\left(\nabla_{\nabla w}\nabla v-\frac{\Delta v}{d}\nabla w\right)\right)+\left(\frac{n}{d}-1\right)v^{2-d}\Delta v.$$

Integrate this equation over $\mathbb{B}^{n}$ and notice that $\nabla w|_{\mathbb{S}^{n-1}}=-\nu$, we get Section 3:

$$\int_{\mathbb{B}^{n}}wv^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right)$$

$$= \int_{\mathbb{S}^{n-1}}f^{2-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu)+\left(\frac{n}{d}-1\right)\int_{\mathbb{B}^{n}}v^{2-d}\Delta v.$$

Once again, 4.2 and Lemma 3.2 implies

$$\mathrm{div}\left(v^{1-d}\left(\nabla_{\nabla w}\nabla v-\frac{\Delta v}{d}\nabla w\right)\right)=\left(\frac{n}{d}-1\right)v^{1-d}\Delta v.$$

Integrate this equation over $\mathbb{B}^{n}$ and notice that $\nabla w|_{\mathbb{S}^{n-1}}=-\nu$, we get 3.10:

$$\int_{\mathbb{S}^{n-1}}f^{1-d}\left(\nabla^{2}v-\frac{\Delta v}{d}g\right)(\nu,\nu)=\left(1-\frac{n}{d}\right)\int_{\mathbb{B}^{n}}v^{1-d}\Delta v.$$

Plug Section 3 and 3.10 into Section 4, we derive the following key integral identity:

$$0\leq \int_{\mathbb{B}^{n}}\left(1-\frac{2}{d-2}\lambda w\right)v^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right)$$

(4.4) $$= \left(\frac{2}{d-2}\lambda-1\right)\int_{\mathbb{S}^{n-1}}f^{1-d}|\nabla f|^{2}+\frac{2}{d-2}\left(\frac{n}{d}-1\right)\left(\frac{n-2}{2}\frac{H}{n-1}\int_{\mathbb{B}^{n}}v^{1-d}\Delta v-\lambda\int_{\mathbb{B}^{n}}v^{2-d}\Delta v\right).$$

Step 3: Now we verify the first case in theorem 4.

If $1<p\leq\frac{n+2}{n-2}$ and $\lambda=0$, then Section 4 reduces to

$$0\leq \int_{\mathbb{B}^{n}}v^{1-d}\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\left(\frac{1}{n}-\frac{1}{d}\right)(\Delta v)^{2}\right)$$

$$= -\int_{\mathbb{S}^{n-1}}f^{1-d}|\nabla f|^{2}+\frac{n-2}{d-2}\frac{H}{n-1}\left(\frac{n}{d}-1\right)\int_{\mathbb{B}^{n}}v^{1-d}\Delta v$$

$$\leq 0.$$

This forces

$$\nabla^{2}v=\frac{\Delta v}{n}g\quad\mathrm{in}\ \mathbb{B}^{n},$$

and $v|_{\mathbb{S}^{n-1}}=constant$. It follows from Ricci’s identity that

$$\nabla\Delta v=\mathrm{div}(\nabla^{2}v)=\mathrm{div}\left(\frac{\Delta v}{n}g\right)=\frac{\nabla\Delta v}{n}.$$

Therefore, $\Delta v$ is a constant in $\mathbb{B}^{n}$. Combining with $v|_{\mathbb{S}^{n-1}}=constant$, we could set $v(x)=r|x|^{2}+s$ for some constants $r,s$. The equation 4.1 reduces to

$$\begin{cases}2nr^{2}|x|^{2}+2nrs=2dr^{2}|x|^{2}+\frac{n-2}{2(d-2)}\frac{R}{n-1}&\mathrm{in}\ \mathbb{B}^{n},\\ 2r=-\frac{n-2}{d-2}\frac{H}{n-1}&\mathrm{on}\ \mathbb{S}^{n-1}.\end{cases}$$

If follows that $d=n$ and

$$\begin{cases}rs=\frac{1}{4n}\frac{R}{n-1},\\ r=-\frac{1}{2}\frac{H}{n-1}.\end{cases}$$

If $H=0$, then $r=0$ and $R=0$. It’s a contradiction.

If $H>0$, then $r=-\frac{1}{2}\frac{H}{n-1}<0,s=-\frac{1}{2n}\frac{R}{H}<0$. This contradicts with the fact that $v>0$ on $\mathbb{B}^{n}$.

In conclusion, there is no positive solution in this case, and the only nonnegative solution of 1.4 is $u\equiv 0$.

Step 4: If $p=\frac{n+2}{n-2}$ and $0<\lambda<\frac{2}{p-1}=\frac{n-2}{2}$, then Section 4 reduces to

$$0\leq \int_{\mathbb{B}^{n}}\left(1-\frac{2}{n-2}\lambda w\right)v^{1-n}\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}=\left(\frac{2}{n-2}\lambda-1\right)\int_{\mathbb{S}^{n-1}}f^{1-n}|\nabla f|^{2}\leq 0.$$

This forces

$$\nabla^{2}v=\frac{\Delta v}{n}g\quad\mathrm{in}\ \mathbb{B}^{n},$$

and $v|_{\mathbb{S}^{n-1}}=constant$. As before, one could derive that $\Delta v$ is a constant in $\mathbb{B}^{n}$ and set $v(x)=r|x|^{2}+s$ for some constants $r,s$. Then it follows from the equation 4.1 that

$$\begin{cases}4rs=\frac{R}{n(n-1)},\\ r=\frac{\lambda}{n-2}(r+s)-\frac{H}{2(n-1)}.\end{cases}$$

Since $v$ is positive, we have $s>0$ and we solve

$$r=\frac{1}{2\epsilon}\sqrt{\frac{R}{n(n-1)}},\quad s=\frac{\epsilon}{2}\sqrt{\frac{R}{n(n-1)}},$$

where $0<\epsilon=\sqrt{\frac{n(n-1)}{R}}\left(\frac{n-2}{2\lambda}\frac{H}{n-1}+\sqrt{(\frac{n-2}{2\lambda})^{2}(\frac{H}{n-1})^{2}+\frac{R}{n(n-1)}(\frac{n-2}{\lambda}-1)}\right)$.

In conclusion, $v(x)=\sqrt{\frac{R}{n(n-1)}}\left(\frac{1}{2\epsilon}|x|^{2}+\frac{\epsilon}{2}\right)$ and $u=v^{-\frac{n-2}{2}}$ is the desired solution.

Step 5: If $p=\frac{n+2}{n-2}$ and $\lambda=\frac{n-2}{2}$, then Section 4 reduces to

$$0\leq \int_{\mathbb{B}^{n}}\left(1-\frac{2}{n-2}\lambda w\right)v^{1-n}\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}=\left(\frac{2}{n-2}\lambda-1\right)\int_{\mathbb{S}^{n-1}}f^{1-n}|\nabla f|^{2}=0.$$

As before, we could derive that $\Delta v$ is a constant in $\mathbb{B}^{n}$ (but $v|_{\mathbb{S}^{n-1}}$ is not necessarily constant) and set $v(x)=r|x|^{2}+\langle\xi,x\rangle+s$, where $r,s$ are constants and $\xi\in\mathbb{R}^{n}$. It follows from the equation 4.1 that

$$\begin{cases}4rs=|\xi|^{2}+\frac{R}{n(n-1)},\\ r=s-\frac{H}{n-1}.\end{cases}$$

Since $v$ is positive, we have $s>0$ and we solve

$$\begin{cases}r=\frac{1}{2}\left(\sqrt{(\frac{H}{n-1})^{2}+\frac{R}{n(n-1)}+|\xi|^{2}}-\frac{H}{n-1}\right),\\ s=\frac{1}{2}\left(\sqrt{(\frac{H}{n-1})^{2}+\frac{R}{n(n-1)}+|\xi|^{2}}+\frac{H}{n-1}\right).\end{cases}$$

For the fixed $\xi\in\mathbb{R}^{n}$, there exists a unique $a\in\mathbb{B}^{n}$ such that

$$\xi=2\sqrt{\left(\frac{H}{n-1}\right)^{2}+\frac{R}{n(n-1)}}\,\frac{a}{1-|a|^{2}}.$$

Hence we derive that

$$\begin{cases}r=\frac{1}{2}\left(\frac{1+|a|^{2}}{1-|a|^{2}}\sqrt{(\frac{H}{n-1})^{2}+\frac{R}{n(n-1)}}-\frac{H}{n-1}\right)=\sqrt{\frac{R}{n(n-1)}}\,\frac{1+\epsilon^{2}|a|^{2}}{2\epsilon(1-|a|^{2})},\\ s=\frac{1}{2}\left(\frac{1+|a|^{2}}{1-|a|^{2}}\sqrt{(\frac{H}{n-1})^{2}+\frac{R}{n(n-1)}}+\frac{H}{n-1}\right)=\sqrt{\frac{R}{n(n-1)}}\,\frac{\epsilon^{2}+|a|^{2}}{2\epsilon(1-|a|^{2})},\\ \xi=\sqrt{\frac{R}{n(n-1)}}\frac{(1+\epsilon^{2})a}{\epsilon(1-|a|^{2})},\end{cases}$$

where $0<\epsilon=\sqrt{\frac{n(n-1)}{R}}\left(\frac{H}{n-1}+\sqrt{(\frac{H}{n-1})^{2}+\frac{R}{n(n-1)}}\right)$.

In conclusion, $v(x)=\sqrt{\frac{R}{n(n-1)}}\left(\frac{1+\epsilon^{2}|a|^{2}}{2\epsilon(1-|a|^{2})}|x|^{2}+\frac{1+\epsilon^{2}}{\epsilon(1-|a|^{2})}\langle x,a\rangle+\frac{\epsilon^{2}+|a|^{2}}{2\epsilon(1-|a|^{2})}\right)$ and $u=v^{-\frac{n-2}{2}}$ is the desired solution. ∎

Appendix A

Bidaut-Véron and Véron [BV91, Theorem 6.1] established the following uniqueness result:

Theorem 5 ([BV91]).

Let $(M^{n},g),n\geq 3$ be a complete, compact Riemannian manifold without boundary. Assume $Ric\geq n-1$, and $1<q\leq\frac{n+2}{n-2}$, $0<\lambda\leq\frac{n}{q-1}$ are constants. Let $u\in C^{2}(M)$ be a positive solution of

(A.1) $$-\Delta u+\lambda u=u^{q}\quad\mathrm{in~}M,$$

Then either $u$ is constant on $M$ or $q=\frac{n+2}{n-2},\,\lambda=\frac{n(n-2)}{4}$ and $(M^{n},g)$ is isometric to $(\mathbb{S}^{n},g_{0})$. In the latter case, there holds

$$u(x)=\frac{c_{n}}{(\cosh t+(\sinh t)x\cdot\xi)^{\frac{n-2}{2}}}$$

for some $\xi\in\mathbb{S}^{n}$ and some constant $t\geq 0$.

For the critical power case (i.e. $q=\frac{n+2}{n-2}$), the solution could be classified via a conceptually simple strategy as follows. By carefully analyzing the model solution, Wang [Wan22, Section 2] come up with an appropriate function $\phi$, known as a P-function, which is constant if and only if the solution $u$ is given by the model case. Then the Bochner formula implies that, up to a first order term, $\phi$ is a subharmonic function. Hence the maximum principle could be applied to show that $\phi$ must be constant and the proof finishes.

In the subcritical power case (i.e. $1<q<\frac{n+2}{n-2}$), as pointed out by Wang [Wan22, Section 5], the choice of parameters (and the P-function) is delicate if one wishes to use the same strategy. In the following, we shall utilize our principle used in our proof of theorem 3 to present the appropriate P-function and prove theorem 5 in its full generality.

The idea of the proof is as follows. As in our proof in Section 3, to handle the subcritical power case, we introduce the intrinsic dimension $d$ such that $q=\frac{d+2}{d-2}$. Then we mimic Wang’s proof of the critical power case [Wan22] and finally use an integral inequality (Proposition A.1) to treat the emerging terms in the subcritical case and conclude the results.

We note that Proposition A.1 is crucial for the sharp range of $\lambda$ in the subcritical power case. Moreover, it also has its independent interest.

Proposition A.1.

Let $(M^{n},g)$ be a compact manifold without boundary, satisfying $\mathrm{Ric}\geq(n-1)k\geq 0$, then for any constant $s\in(-\infty,0]\cup[\frac{4(n+2)}{4n-1},+\infty)$, there holds

(A.2) $$\int_{M}\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}v^{s}+\frac{1}{n}\int_{M}(\Delta v)^{2}v^{s}\geq k\int_{M}|\nabla v|^{2}v^{s},\quad\forall\ 0<v\in C^{\infty}(M).$$

Moreover, if A.2 holds for some $s\in(-\infty,0)\cup(\frac{4(n+2)}{4n-1},+\infty)$ and some $0<v\in C^{\infty}(M)$, then $v$ is a constant function.

We shall postpone the proof of Proposition A.1 and prove theorem 5 right now.

Proof of theorem 5.

Let $d:=\frac{2(q+1)}{q-1}$. Then $q\leq\frac{n+2}{n-2}\Leftrightarrow d\geq n$, and $\lambda\leq\frac{n}{q-1}\Leftrightarrow\lambda\leq\frac{n(d-2)}{4}$. Define $v:=u^{-\frac{q-1}{2}}=u^{-\frac{2}{d-2}}$. A straightforward calculation gives

$$\Delta v=\frac{d}{2}v^{-1}|\nabla v|^{2}+\frac{2}{d-2}v^{-1}-\frac{2\lambda}{d-2}v.$$

Consider the P-function

$$\phi:=\Delta v+\frac{4\lambda}{d-2}v=v^{-1}\left(\frac{d}{2}|\nabla v|^{2}+\frac{2}{d-2}+\frac{2\lambda}{d-2}v^{2}\right).$$

On the one hand,

$$\Delta(v\phi)=v\Delta\phi+2\langle\nabla v,\nabla\phi\rangle+(\Delta v)^{2}+\frac{4\lambda}{d-2}v\Delta v.$$

On the other hand, by Bochner formula, there holds

$$\Delta(v\phi) =d\left(|\nabla^{2}v|^{2}+\left\langle\nabla v,\nabla\left(\phi-\frac{4\lambda}{d-2}v\right)\right\rangle+Ric(\nabla v,\nabla v)\right)+\frac{2\lambda}{d-2}(2|\nabla v|^{2}+2v\Delta v)$$

$$=d\left(\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+\frac{(\Delta v)^{2}}{n}+\langle\nabla v,\nabla\phi\rangle+Ric(\nabla v,\nabla v)\right)-\frac{4(d-1)\lambda}{d-2}|\nabla v|^{2}+\frac{4\lambda}{d-2}v\Delta v.$$

We deduce that $\phi$ satisfies the following equation

$$v\Delta\phi+(2-d)\langle\nabla v,\nabla\phi\rangle=d\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}+(\frac{d}{n}-1)(\Delta v)^{2}+d\ \mathrm{Ric}(\nabla v,\nabla v)-\frac{4(d-1)\lambda}{d-2}|\nabla v|^{2}.$$

Multiply both sides of this equation by $v^{1-d}$ and use $Ric\geq n-1$, we derive the following key inequality:

$$\mathrm{div}(v^{2-d}\nabla\phi)\geq (d-n)\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}v^{1-d}+\left(\frac{d}{n}-1\right)(\Delta v)^{2}v^{1-d}$$

(A.3) $$+\left(d(n-1)-\frac{4(d-1)\lambda}{d-2}\right)|\nabla v|^{2}v^{1-d}.$$

Now integrate appendix A over $M$ and use Proposition A.1 (with parameter $s=1-d$), we get

(A.4) $$0=\int_{M}div(v^{2-d}\nabla\phi)\geq\left(d-n+d(n-1)-\frac{4(d-1)\lambda}{d-2}\right)\int_{M}|\nabla v|^{2}v^{1-d}\geq 0,$$

since $\lambda\leq\frac{n(d-2)}{4}$. Therefore, equality holds in A.4.

For $1<q<\frac{n+2}{n-2}$, this forces that the equality holds in Proposition A.1. Hence $v$ is a constant.

For $0<\lambda<\frac{n}{q-1}$, A.4 implies that $v$ is a constant.

For $q=\frac{n+2}{n-2}$ and $\lambda=\frac{n(n-2)}{4}$, we have $\nabla^{2}v-\frac{\Delta v}{n}g\equiv 0$ and $\mathrm{Ric}(\nabla v,\nabla v)\equiv n-1$. It follows that $(M^{n},g)$ is isometric to the round sphere $\mathbb{S}^{n}$. ∎

Finally we present the proof of Proposition A.1 with the help of two auxiliary lemmas.

Lemma A.1.

Let $(M^{n},g)$ be a compact manifold without boundary. Then for any $u\in C^{\infty}(M)$, there holds

$$\int_{M}(\Delta u)^{2}= \frac{n}{n-1}\int_{M}\left|\nabla^{2}u-\frac{\Delta u}{n}g\right|^{2}+\frac{n}{n-1}\int_{M}\mathrm{Ric}(\nabla u,\nabla u).$$

Proof.

By Bochner formula,

$$\frac{1}{2}\Delta|\nabla u|^{2}=\left|\nabla^{2}u-\frac{\Delta u}{n}g\right|^{2}+\frac{1}{n}(\Delta u)^{2}+\langle\nabla\Delta u,\nabla u\rangle+\mathrm{Ric}(\nabla u,\nabla u).$$

Integrate it over M, we get

$$0=\int_{M}\left(\left|\nabla^{2}u-\frac{\Delta u}{n}g\right|^{2}+\mathrm{Ric}(\nabla u,\nabla u)\right)+\frac{1}{n}\int_{M}(\Delta u)^{2}+\int_{M}(\mathrm{div}(\Delta u\nabla u)-(\Delta u)^{2}).$$

Rearrange it and the proof finishes. ∎

Lemma A.2.

Let $(M^{n},g)$ be a compact manifold without boundary. If $u\in C^{\infty}(M)$ is a positive function, then

$$\int_{M}u^{-1}|\nabla u|^{2}\Delta u= \frac{n}{n+2}\int_{M}u^{-2}|\nabla u|^{4}-\frac{2n}{n+2}\int_{M}\left\langle\nabla^{2}u-\frac{\Delta u}{n}g,\frac{du\otimes du}{u}-\frac{1}{n}\frac{|\nabla u|^{2}}{u}g\right\rangle.$$

Proof.

By divergence theorem,

$$\int_{M}u^{-1}|\nabla u|^{2}\Delta u= \int_{M}\mathrm{div}(u^{-1}|\nabla u|^{2}\nabla u)-(2u^{-1}\langle\nabla_{\nabla u}\nabla u,\nabla u\rangle-u^{-2}|\nabla u|^{4})$$

$$= \int_{M}u^{-2}|\nabla u|^{4}-2\int_{M}\left\langle\nabla^{2}u,\frac{du\otimes du}{u}-\frac{1}{n}\frac{|\nabla u|^{2}}{u}g\right\rangle-\frac{2}{n}\int_{M}u^{-1}|\nabla u|^{2}\Delta u.$$

Rearrange it and the proof finishes. ∎

Proof of Proposition A.1.

First assume that $s\in(-\infty,-2)\cup(-2,0]\cup[\frac{4(n+2)}{4n-1},+\infty)$. Let $t:=s+2\in(-\infty,0)\cup(0,2]\cup[\frac{6(2n+1)}{4n-1},+\infty)$ and $u:=v^{\frac{t}{2}}$. Then we have

$$v^{\frac{s}{2}} =u^{-\frac{2}{t}+1},$$

$$v^{\frac{s}{2}}(\nabla^{2}v-\frac{\Delta v}{n}g) =\frac{2}{t}\left(\nabla^{2}u-\frac{\Delta u}{n}g\right)+\frac{2}{t}\left(\frac{2}{t}-1\right)\left(\frac{du\otimes du}{u}-\frac{1}{n}\frac{|\nabla u|^{2}}{u}g\right),$$

$$v^{\frac{s}{2}}\Delta v =\frac{2}{t}\left(\frac{2}{t}-1\right)u^{-1}|\nabla u|^{2}+\frac{2}{t}\Delta u,$$

$$v^{\frac{s}{2}}\nabla v =\frac{2}{t}\nabla u.$$

By Lemma A.1 , Lemma A.2 and the fact that $|\frac{du\otimes du}{u}-\frac{1}{n}\frac{|\nabla u|^{2}}{u}g|^{2}=\frac{n-1}{n}u^{-2}|\nabla u|^{4}$, we have

$$\int_{M}\left|\nabla^{2}v-\frac{\Delta v}{n}g\right|^{2}v^{s}+\frac{1}{n}\int_{M}(\Delta v)^{2}v^{s}$$

$$= \left(\frac{2}{t}\right)^{2}\int_{M}|\nabla^{2}u-\frac{\Delta u}{n}g|^{2}+\left(\frac{2}{t}\right)^{2}\left(\frac{2}{t}-1\right)^{2}\frac{n-1}{n}\int_{M}u^{-2}|\nabla u|^{4}$$

$$+2\left(\frac{2}{t}\right)^{2}\left(\frac{2}{t}-1\right)\int_{M}\left\langle\nabla^{2}u-\frac{\Delta u}{n}g,\frac{du\otimes du}{u}-\frac{1}{n}\frac{|\nabla u|^{2}}{u}g\right\rangle$$

$$+\frac{1}{n}\left(\frac{2}{t}\right)^{2}\left(\frac{2}{t}-1\right)^{2}\int_{M}u^{-2}|\nabla u|^{4}+\frac{1}{n}\left(\frac{2}{t}\right)^{2}\frac{n}{n-1}\int_{M}\left(\left|\nabla^{2}u-\frac{\Delta u}{n}g\right|^{2}+\mathrm{Ric}(\nabla u,\nabla u)\right)$$

$$+\frac{2}{n}\left(\frac{2}{t}\right)^{2}\left(\frac{2}{t}-1\right)\frac{n}{n+2}\int_{M}u^{-2}|\nabla u|^{4}$$

$$-\frac{2}{n}\left(\frac{2}{t}\right)^{2}\left(\frac{2}{t}-1\right)\frac{2n}{n+2}\int_{M}\left\langle\nabla^{2}u-\frac{\Delta u}{n}g,\frac{du\otimes du}{u}-\frac{1}{n}\frac{|\nabla u|^{2}}{u}g\right\rangle$$

$$= \frac{n}{n-1}\left(\frac{2}{t}\right)^{2}\int_{M}\left|\nabla^{2}u-\frac{\Delta u}{n}g\right|^{2}+\frac{2n}{n+2}\left(\frac{2}{t}\right)^{2}\left(\frac{2}{t}-1\right)\int_{M}\left\langle\nabla^{2}u-\frac{\Delta u}{n}g,\frac{du\otimes du}{u}-\frac{1}{n}\frac{|\nabla u|^{2}}{u}g\right\rangle$$

$$+\left(\left(\frac{2}{t}\right)^{2}\left(\frac{2}{t}-1\right)^{2}+\frac{2}{n+2}\left(\frac{2}{t}\right)^{2}\left(\frac{2}{t}-1\right)\right)\int_{M}u^{-2}|\nabla u|^{4}+\frac{1}{n-1}\left(\frac{2}{t}\right)^{2}\int_{M}\mathrm{Ric}(\nabla u,\nabla u)$$

$$= \frac{n}{n-1}\left(\frac{2}{t}\right)^{2}\int_{M}\left|\nabla^{2}u-\frac{\Delta u}{n}g+\left(\frac{2}{t}-1\right)\frac{n-1}{n+2}\left(\frac{du\otimes du}{u}-\frac{1}{n}\frac{|\nabla u|^{2}}{u}g\right)\right|^{2}$$

$$+\frac{3(2n+1)}{(n+2)^{2}}\left(\frac{2}{t}\right)^{2}\left(\frac{2}{t}-1\right)\left(\frac{2}{t}-\frac{4n-1}{3(2n+1)}\right)\int_{M}u^{-2}|\nabla u|^{4}+\frac{1}{n-1}\left(\frac{2}{t}\right)^{2}\int_{M}\mathrm{Ric}(\nabla u,\nabla u)$$

$$= \frac{n}{n-1}\left(\frac{2}{t}\right)^{2}\int_{M}\left|\nabla^{2}u-\frac{\Delta u}{n}g+\left(\frac{2}{t}-1\right)\frac{n-1}{n+2}\left(\frac{du\otimes du}{u}-\frac{1}{n}\frac{|\nabla u|^{2}}{u}g\right)\right|^{2}$$

$$+\frac{4n-1}{4(n+2)^{2}}s\left(s-\frac{4(n+2)}{4n-1}\right)\int_{M}v^{s-2}|\nabla v|^{4}+\frac{1}{n-1}\int_{M}v^{s}\mathrm{Ric}(\nabla v,\nabla v)$$

$$\geq \frac{4n-1}{4(n+2)^{2}}s\left(s-\frac{4(n+2)}{4n-1}\right)\int_{M}v^{s-2}|\nabla v|^{4}+k\int_{M}|\nabla v|^{2}v^{s}\geq k\int_{M}|\nabla v|^{2}v^{s}.$$

Hence we have proved A.2 for $s\in(-\infty,-2)\cup(-2,0]\cup[\frac{4(n+2)}{4n-1},+\infty)$.

The case $s=-2$ could be obtained by taking the limit $s\to-2$ in A.2. ∎