On a Fractional Nirenberg Problem Involving the Square Root of the Laplacian on ${\mathbb {S}}^{3}$

Abstract

In this paper, we are devoted to establishing the compactness and existence results of the solutions to the fractional Nirenberg problem for $n=3,$ $\sigma =1/2,$ when the prescribing $\sigma$-curvature function satisfies the $(n-2\sigma )$-flatness condition near its critical points. The compactness results are new and optimal. In addition, we obtain a degree-counting formula of all solutions. From our results, we can know where blow up occur. Moreover, for any finite distinct points, the sequence of solutions that blow up precisely at these points can be constructed. We extend the results of Li (Commun Pure Appl Math 49:541–597, 1996) from the local problem to nonlocal cases.

A Appendix

In this appendix, we provide some elementary calculations which have been used in the proof of Theorem 1.3.

Lemma A.1

Let $\alpha \ge 2,$ there exists a positive constant C depending only on $\alpha$ such that, for any $a \ge 0,$ $b \in {\mathbb {R}},$

$$\begin{aligned} \Big | | a+b|^{\alpha -1}(a+b)-a^{\alpha }-\alpha a^{\alpha -1} b-\frac{\alpha (\alpha -1)}{2} a^{\alpha -2} b^{2} \Big | \le C\left( |b|^{\alpha }+a^{\gamma }|b|^{\alpha -\gamma }\right) , \end{aligned}$$

where $\gamma =\max \{0,\alpha -3\}.$

Lemma A.2

Let $1<\beta <2,$ there exists a universal positive constant C such that, for any $a>0,$ $b\in {\mathbb {R}},$

$$\begin{aligned} \left| |a+b|^{\beta -1}(a+b)-a^{\beta }-\beta a^{\beta -1}b\right| \le C|b|^{\beta }. \end{aligned}$$

Lemma A.3

Let $\beta >1$ and $k\in {\mathbb {N}}_{+},$ there exists a constant C, such that for any $(a_{1},\ldots , a_{k}) \in {\mathbb {R}}^{k},$

$$\begin{aligned} \Big |\Big (\sum _{i=1}^{k} a_i\Big )^{\beta }-\sum _{i=1}^{k} a_{i}^{\beta } \Big | \le C\sum _{i \ne j}|a_{i}|^{\beta -1} |a_{j}|. \end{aligned}$$

Lemma A.4

Let $\varepsilon _{0}, \tau >0$ be suitably small and $A>0$ be suitably large. Let $A^{-1} \tau ^{-1 / 2}<t_{1}, t_{2}<A \tau ^{-1 / 2},$ $P_{1},P_{2}\in {\mathbb {S}}^3,$ $|P_{1}-P_{2}|\ge \varepsilon _{0},$ $\delta _{P_{i},t_{i}}$ be as in (4.3) and $G_{P_1}(P_2)$ be as in (1.12) $(|P_{1}-P_{2}|$ represents the distance between two points $P_{1}$ and $P_{2}$ after through a stereographic projection). Then, we have,

$$\begin{aligned}&\int _{{\mathbb {S}}^3}\delta _{P_1, t_1}^2\delta _{P_2,t_2} =4\pi |{\mathbb {S}}^2|\frac{G_{P_1}(P_2)}{t_1t_2} +O(\tau ^{3/2}), \end{aligned}$$

(A.1)

$$\begin{aligned}&\int _{{\mathbb {S}}^3}\delta _{P_1,t_1}^{2-\tau }\delta _{P_2,t_2}=O(\tau ), \end{aligned}$$

(A.2)

$$\begin{aligned}&\frac{\partial }{\partial t_{1}} \int _{{\mathbb {S}}^3}\delta _{P_1,t_{1}}^2\delta _{P_2, t_{2}}= -4\pi |{\mathbb {S}}^2|\frac{G_{P_1}(P_2)}{t_1^2t_2}+O(\tau ^2), \end{aligned}$$

(A.3)

$$\begin{aligned}&\frac{\partial }{\partial t_{1}} \int _{{\mathbb {S}}^3}\delta _{P_2, t_{2}}\delta _{P_1,t_{1}}^{2-\tau } =\int _{{\mathbb {S}}^3}\delta _{P_2,t_2}\delta _{P_1,t_1}^{2}+O(\tau ^{5/2}|\log \tau |), \end{aligned}$$

(A.4)

$$\begin{aligned}&\frac{\partial }{\partial t_1} \int _{{\mathbb {S}}^3} \delta _{P_2, t_2}^{2-\tau }\delta _{P_1, t_1}= \frac{\partial }{\partial t_1}\int _{{\mathbb {S}}^3} \delta _{P_2, t_2}^{2}\delta _{P_1, t_1}+O(\tau ^{5/2}|\log \tau |), \end{aligned}$$

(A.5)

$$\begin{aligned}&\int _{{\mathbb {S}}^3}|P-P_1|^2\delta _{P_1,t_1}^{3-\tau } =\frac{1}{t_{1}^2}\frac{3\pi }{2}|{\mathbb {S}}^2| +O(\tau ^{2}| \log \tau |), \end{aligned}$$

(A.6)

$$\begin{aligned}&\frac{\partial }{\partial t_1}\int _{{\mathbb {S}}^3} \delta ^{3-\tau }_{P_1,t_1}= -\frac{\tau }{t_1}\frac{\pi }{2}|{\mathbb {S}}^2| +O(\tau ^{5/2}|\log \tau |), \end{aligned}$$

(A.7)

$$\begin{aligned}&\frac{\partial }{\partial t_1} \int _{{\mathbb {S}}^3} |P-P_1|^2\delta _{P_1,t_1}^{3-\tau }= - \frac{3\pi }{t_{1}^3}|{\mathbb {S}}^2|+O(\tau ^{5/2}|\log \tau |). \end{aligned}$$

(A.8)

Proof

Because the computation is elementary and routine, we only take (A.2) as an example to prove.

By the stereographic projection, we have

$$\begin{aligned} \int _{{\mathbb {S}}^3}\delta _{P_1, t_1}^{2-\tau }\delta _{P_2,t_2}&= \int _{{\mathbb {R}}^3} \left( \frac{2t_{1}}{(1+t_1^2|y-y_1|^2)}\right) ^{2-\tau } \left( \frac{2t_{2}}{(1+t_2^2|y-y_2|^2)}\right) \left( \frac{2}{1+|y|^2}\right) ^{\tau }\,dy\\&\le C_1\int _{{\mathbb {R}}^3} \frac{t_1^{2-\tau }t_2}{(1+t_1^2|y-y_1|^2)^{2-\tau } (1+t_2^2|y-y_2|^2)}\,dy=:C_1\widetilde{{\mathcal {I}}}_{2-\tau }. \end{aligned}$$

Let

$$\begin{aligned} a_{12}=\frac{y_2-y_1}{2};\quad z =y-\frac{y_1+y_2}{2}. \end{aligned}$$

Then we have

$$\begin{aligned} \widetilde{{\mathcal {I}}}_{2-\tau }=\frac{1}{t_1^{2-\tau }t_2}\int _{{\mathbb {R}}^3} \frac{1}{(\frac{1}{t_1^2}+|z+a_{12}|^2)^{2-\tau } (\frac{1}{t_2^2}+|z-a_{12}|^2)}\,dy. \end{aligned}$$

By symmetry arguments, we may assume

$$\begin{aligned} t_2\le t_1. \end{aligned}$$

Because we have

$$\begin{aligned} t_1t_2|a_{12}|^2\ge C\frac{t_1}{t_2}, \quad \text{ with } C \text{ a } \text{ large } \text{ constant }. \end{aligned}$$

Thus,

$$\begin{aligned} t_2^2|a_{12}|^2\ge C;\quad t_1^2|a_{12}|^2\ge C, \end{aligned}$$

and it follows that if $|z+a_{12}|\le \frac{1}{t_1} \le |a_{12}|$, then $|z-a_{12}|\ge |a_{12}|$; and if $|z-a_{12}|\le \frac{1}{t_2}\le |a_{12}|$, then $|z+a_{12}|\ge |a_{12}|$.

Then

$$\begin{aligned} \widetilde{{\mathcal {I}}}_{2-\tau }&\le \frac{1}{t_1^{2-\tau }t_2} \Big \{ \int \limits _{|z+a_{12}|\le \frac{1}{t_1}}\frac{C_1 t_1^{2(2-\tau )}}{(\frac{1}{t_2^2}+|a_{12}|^2)} +\int \limits _{|z-a_{12}|\le \frac{1}{t_2}}\frac{C_1 t_2^2}{(\frac{1}{t_1^2}+|a_{12}|^2)^{2-\tau }} \Big \}\nonumber \\&\quad +\frac{2^3}{t_1^{2-\tau }t_2} \Big \{ \int _{\begin{array}{c} |z+a_{12}|>\frac{1}{t_1} \\ |z-a_{12}|>\frac{1}{t_2} \end{array}} \frac{C_1}{|z+a_{12}|^{2(2-\tau )}|z-a_{12}|^2}\,dz \Big \}, \end{aligned}$$

(A.9)

where $C_1$ is a constant.

Since

$$\begin{aligned}&\int _{\begin{array}{c} |z+a_{12}|>\frac{1}{t_1} \\ |z-a_{12}|>\frac{1}{t_2} \end{array}} \frac{1}{|z+a_{12}|^{2(2-\tau )}|z-a_{12}|^2}\,dz\nonumber \\&\quad \le \int \limits _{\frac{1}{t_1}<|z+a_{12}|\le |a_{12}|}\frac{dz}{|z+a_{12}|^{2(2-\tau )}|a_{12}|^2} +\int \limits _{\frac{1}{t_2}<|z-a_{12}|\le |a_{12}|}\frac{dz}{|a_{12}|^{2(2-\tau )}|z-a_{12}|^2}\nonumber \\&\qquad +\int _{\begin{array}{c} |z+a_{12}|>|a_{12}| \\ |z-a_{12}|>|a_{12}| \end{array}} \frac{dz}{|z+a_{12}|^{2(2-\tau )}|z-a_{12}|^2}. \end{aligned}$$

(A.10)

It is easy to see that if $|z\pm a_{12}|\le |a_{12}|$, then $|z\mp a_{12}| \ge |a_{12}|$ again, and if $|z+a_{12}|\ge |a_{12}|$ and $|z-a_{12}|\ge |a_{12}|$, then $|z-a_{12}|\ge \frac{1}{C_2^{\frac{1}{3}}}|z+a_{12}|$ with $C_2$ large enough. if $|z-a_{12}|>|a_{12}|$, then $|z-a_{12}|>|a_{12}|>|z+a_{12}|$, we have

$$\begin{aligned}&\int _{\begin{array}{c} |z+a_{12}|>|a_{12}| \\ |z-a_{12}|>|a_{12}| \end{array}} \frac{dz}{|z+a_{12}|^{2(2-\tau )}|z-a_{12}|^2}\\&\le C_2 \int _{|z+a_{12}|\ge |a_{12}|}\frac{dz}{|z+a_{12}|^{2(2-\tau )+2}} =\frac{C_2|{\mathbb {S}}^2|}{(3-2\tau )}\frac{1}{|a_{12}|^{3-2\tau }}. \end{aligned}$$

On the other hand,

$$\begin{aligned}&\frac{1}{|a_{12}|^2}\int \limits _{\frac{1}{t_1}<|z+a_{12}|\le |a_{12}|}\frac{dz}{|z+a_{12}|^{2(2-\tau )}} \nonumber \\&=\frac{|{\mathbb {S}}^2|}{|a_{12}|^2}\int _{\frac{1}{t_1}}^{|a_{12}|} \frac{1}{r^{2-2\tau }}dr =\frac{|{\mathbb {S}}^2|}{|a_{12}|^2}\frac{1}{2\tau -1}(|a_{12}|^{2\tau -1}-\frac{1}{t_1^{2\tau -1}}), \end{aligned}$$

(A.11)

and

$$\begin{aligned}&\frac{1}{|a_{12}|^{2(2-\tau )}}\int \limits _{\frac{1}{t_2}<|z-a_{12}|\le |a_{12}|}\frac{dz}{|z-a_{12}|^2} \nonumber \\&=\frac{|{\mathbb {S}}|^2}{|a_{12}|^{2(2-\tau )}}\int \limits _{\frac{1}{t_2}<|z-a_{12}| \le |a_{12}|}\frac{1}{r}dr =\frac{|{\mathbb {S}}|^2}{|a_{12}|^{2(2-\tau )}}(|a_{12}|-\frac{1}{t_2}). \end{aligned}$$

(A.12)

By (A.10), (A.11) and (A.12), we have

$$\begin{aligned}&\frac{2^3}{t_1^{2-\tau }t_2}\int _{\begin{array}{c} |z+a_{12}|>\frac{1}{t_1} \\ |z-a_{12}|>\frac{1}{t_2} \end{array}} \frac{1}{|z+a_{12}|^{2(2-\tau )}|z-a_{12}|^2}\,dz\nonumber \\&\le C\left( \frac{1}{\mu ^{\frac{3}{2}-\tau }t_2^{\tau }} +\frac{1}{\mu t_1^{\tau }} +\frac{1}{\mu ^{\frac{3}{2}-\tau }t_2^{\tau }} +\frac{1}{\mu ^{2-\tau }t_2^{\tau }} +\frac{1}{\mu ^{\frac{3}{2}-\tau }t_2^{\tau }} \right) \nonumber \\&\le C(\tau ^{\frac{3}{2}}+\tau +\tau ^{2})=O(\tau ), \end{aligned}$$

(A.13)

recalling that (A.9), one has

$$\begin{aligned}&\frac{2^3}{t_1^{2-\tau }t_2} \int \limits _{|z+a_{12}|\le \frac{1}{t_1}}\frac{C_1 t_1^{2(2-\tau )}}{(\frac{1}{t_2^2}+|a_{12}|^2)}\nonumber \\&\le C\frac{1}{t_1^{\tau }(\frac{t_1}{t_2}+\mu )} =O\left( \mu ^{-1}\right) =O( \tau ), \end{aligned}$$

(A.14)

similarly,

$$\begin{aligned} \frac{2^3}{t_1^{2-\tau }t_2} \int \limits _{|z-a_{12}|\le \frac{1}{t_2}}\frac{C_1 t_2^2}{(\frac{1}{t_1^2}+|a_{12}|^2)^{2-\tau }} =O\left( \mu ^{-(2-\tau )}\right) =O( \tau ^{2}). \end{aligned}$$

(A.15)

Thus, by (A.13), (A.14) and (A.15), we obtain (A.2). $\square$

Lemma A.5

Under the hypotheses of Lemma A.4, in addition that $\Gamma _{1},\Gamma _{2}$ are positive constants independent of $\tau .$ Then, we have,

$$\begin{aligned}&\langle \delta _{P_{1}, t_{1}}, \delta _{P_{1}, t_{1}}\rangle = |{\mathbb {S}}^3|,\quad \Bigg \langle \frac{\partial \delta _{P_1,t_1}}{\partial P_1^{(\ell )}}, \frac{\partial {\delta _{P_1,t_1}}}{\partial P_1^{(\ell )}} \Bigg \rangle =\Gamma _{1}t_1^2, \end{aligned}$$

(A.16)

$$\begin{aligned}&\Bigg \langle \frac{\partial \delta _{P_1, t_1}}{\partial t_{1}}, \frac{\partial \delta _{P_1, t_1}}{\partial t_{1}} \Bigg \rangle =\Gamma _2 t_1^{-2}, \end{aligned}$$

(A.17)

$$\begin{aligned}&\langle \delta _{P_{1}, t_{1}}, \delta _{P_{2}, t_{2}}\rangle =O(\tau ), \end{aligned}$$

(A.18)

$$\begin{aligned}&\Vert \delta _{P_1,t_1}^{1-\tau }\delta _{P_2, t_2}\Vert _{L^{3/2}({\mathbb {S}}^3)}=O(\tau |\log \tau |), \end{aligned}$$

(A.19)

$$\begin{aligned}&\Big \Vert \delta _{P_2,t_2}^{1-\tau } \frac{\partial \delta _{P_1,t_1}}{\partial t_1}\Big \Vert _{L^{3/2}({\mathbb {S}}^3)} = O(\tau ^{3/2}|\log \tau |), \end{aligned}$$

(A.20)

$$\begin{aligned}&\Big \Vert \delta _{P_1,t_1}^{1-\tau }\delta _{P_{2}, t_{2}}\frac{\partial \delta _{P_1, t_1}}{\partial t_1}\Big \Vert _{L^1({\mathbb {S}}^3)}=O(\tau ^{3/2}|\log \tau |), \end{aligned}$$

(A.21)

$$\begin{aligned} \begin{aligned}&\Vert \delta _{P_1,t_1}^{2-\tau }-\delta _{P_1, t_1}^2 \Vert _{L^{3/2}({\mathbb {S}}^3)}= O(\tau |\log \tau |),\\&\Vert \delta _{P_1,t_1}^{1-\tau }-\delta _{P_1, t_1} \Vert _{L^{3}({\mathbb {S}}^3)} =O(\tau |\log \tau |), \end{aligned} \end{aligned}$$

(A.22)

$$\begin{aligned} \Vert \delta _{P_1,t_1}^{3-\tau }-\delta _{P_1,t_1}^3\Vert _{L^{1}({\mathbb {S}}^3)} =O(\tau |\log \tau |), \end{aligned}$$

(A.23)

$$\begin{aligned} \begin{aligned}&\big \Vert |P-P_1|\delta _{P_1, t_1}^{2}\big \Vert _{L^{3/2}({\mathbb {S}}^3)}=O( \tau ^{1/2}),\\&\big \Vert |P-P_1|^2 \delta ^2_{P_{1}, t_1}\big \Vert _{L^{3/2}({\mathbb {S}}^3)} =O(\tau ), \end{aligned} \end{aligned}$$

(A.24)

$$\begin{aligned} \int _{{\mathbb {S}}^3}|P-P_1|^3\delta _{P_1,t_1} \frac{\partial \delta _{P_1,t_1}}{\partial t_1}=O(\tau ^2). \end{aligned}$$

(A.25)

Proof

Because the computation is elementary and routine, we only take (A.16) and (A.17) as an example to prove.

Proof of:(A.16)

$$\begin{aligned} \left\langle \delta _{P_{1}, t_{1}}, \delta _{P_{1}, t_{1}}\right\rangle =\int _{{\mathbb {S}}^3}\delta _{P_{1},t_{1}}^{3}&=2^3\int _{{\mathbb {R}}^3} \frac{t_1^3}{(1+t_1^2|y|^2)^3}\,dy\\&=2^3|{\mathbb {S}}^2|\int _{0}^{\infty }\frac{r^2}{(1+r^2)^3}\,dr =|{\mathbb {S}}^3|. \end{aligned}$$

Since $P_{\sigma }\delta _{P_{1},t_{1}}=\delta _{P_1,t_{1}}^2$, then $P_{\sigma }\big (\frac{\partial \delta _{P_{1},t_1}}{\partial P_1}\big ) =2\delta _{P_1,t_1}\big (\frac{\partial \delta _{P_{1},t_1}}{\partial P_1}\big ).$ It follows that

$$\begin{aligned} \Bigg \langle \frac{\partial \delta _{P_1, t_1}}{\partial P_1^{(\ell )}}, \frac{\partial \delta _{P_1, t_1}}{\partial P_{1}^{(\ell )}} \Bigg \rangle&=\int _{{\mathbb {S}}^3}\Big (P_{\sigma }\frac{\partial \delta _{P_{1},t_1}}{\partial P_1^{(\ell )}}\Big ) \frac{\partial \delta _{P_{1},t_1}}{\partial P_1^{(\ell )}}\\&=\int _{{\mathbb {S}}^3} 2\delta _{P_1,t_1}\Big (\frac{\partial \delta _{P_{1},t_1}}{\partial P_1^{(\ell )}}\Big )^2\\&=\int _{{\mathbb {R}}^3}\frac{2t_1}{1+t_1^2|y|^2} \Big [\frac{4t_1^3(y)^{(\ell )}}{(1+t_1^2|y|^2)^2} \Big ]^2 dy\\&=2^5\int _{{\mathbb {R}}^3}\frac{t_1^{2}(x^{(\ell )})^2}{(1+|x|^2)^5}=\Gamma _{1}t_1^{2}. \end{aligned}$$

where $\Gamma _{1}$ is a constant.

Proof of:(A.17) Since $P_{\sigma }\delta _{P_{1},t_{1}}=\delta _{P_1,t_{1}}^2,$ then $P_{\sigma }\big (\frac{\partial \delta _{P_{1},t_1}}{\partial t_1}\big ) =2\delta _{P_1,t_1}\big (\frac{\partial \delta _{P_{1},t_1}}{\partial t_1}\big ).$ It follows that

$$\begin{aligned}&\int _{{\mathbb {S}}^3}2\delta _{P_1,t_1} \Big (\frac{\partial \delta _{P_1,t_1}}{\partial t_1}\Big )^2\\&\quad =2\int _{{\mathbb {R}}^3} \frac{2t_1}{1+t_1^2|y|^2} \Big ( \frac{2}{1+t_1^2|y|^2}-\frac{2^2t_1^2|y|^2}{(1+t_1^2|y|^2)^2}\Big )^2\,dy\\&\quad =2^4t_1^{-2} \int _{{\mathbb {R}}^3}\frac{1}{(1+|x|^2)^3} \Big ( 1-\frac{2|x|^2}{1+|x|^2}\Big )^2\,dy:=\Gamma _2t_1^{-2} \end{aligned}$$

where $\Gamma _2$ is a constant. $\square$

Lemma A.6

Let $\varepsilon _0, \tau , A$ be as in Lemma A.4, $P_{1},P_2,P_3 \in {\mathbb {S}}^3$ satisfy $|P_i-P_j|\ge \varepsilon _0,$ $i\ne j,$ and $A^{-1}\tau ^{-1/2}<t_1,t_2,t_3\le A\tau ^{-1/2}.$ Then, we have,

$$\begin{aligned} \Big |\frac{\partial }{\partial P_{1}}\int _{{\mathbb {S}}^3} \delta _{P_{1},t_{1}}^{3-\tau }\Big |=O(\tau ^{1/2}),\quad \Big \Vert \delta _{P_2,t_2} \frac{\partial \delta _{P_1,t_1}}{\partial P_1} \Big \Vert _{L^{3/2}({\mathbb {S}}^3)}=O( \tau ^{1/2}|\log \tau |), \end{aligned}$$

(A.26)

$$\begin{aligned} \Big |\frac{\partial }{\partial P_1} \int _{{\mathbb {S}}^3}\delta _{P_2, t_2}^{2-\tau }\delta _{P_1, t_1}\Big |=O(\tau ^{1/2}),\quad \Big |\frac{\partial }{\partial P_1} \int _{{\mathbb {S}}^3}\delta _{P_2, t_2}^2\delta _{P_1, t_1}\Big |=O(\tau ^{1/2}), \end{aligned}$$

(A.27)

$$\begin{aligned}&\Big \Vert \delta _{P_2,t_2}^{1-\tau }\delta _{P_{3}, t_{3}} \frac{\partial \delta _{P_1,t_1}}{\partial P_1} \Big \Vert _{L^1({\mathbb {S}}^3)} =O(\tau ^{1/2}|\log \tau |). \end{aligned}$$

(A.28)

Lemma A.7

In addition to the hypotheses of Lemma A.4, we assume that $K\in C^{1}({\mathbb {S}}^3)$. Then

$$\begin{aligned} \frac{\partial }{\partial t_1}\int _{{\mathbb {S}}^3}(K(P)-K(P_2))\delta _{P_2, t_2}^{2-\tau } \delta _{P_1, t_1}= O(\tau ^2). \end{aligned}$$

(A.29)

Lemma A.8

In addition to the hypotheses of Lemma A.7, we assume that $v\in E_{P_1,t_1}$. Then

$$\begin{aligned}&\int _{{\mathbb {S}}^3} (K(P)-K(P_1))\delta _{P_1,t_1}^{1-\tau } \frac{\partial \delta _{P_1,t_1}}{\partial P_{1}}v =O(\tau ^{1/2}|\log \tau | \Vert v\Vert _{\sigma }). \end{aligned}$$

(A.30)