Rigidity versus symmetry breaking via nonlinear flows on cylinders and Euclidean spaces

Abstract

This paper is motivated by the characterization of the optimal symmetry breaking region in Caffarelli-Kohn-Nirenberg inequalities. As a consequence, optimal functions and sharp constants are computed in the symmetry region. The result solves a longstanding conjecture on the optimal symmetry range. As a byproduct of our method we obtain sharp estimates for the principal eigenvalue of Schrödinger operators on some non-flat non-compact manifolds, which to the best of our knowledge are new. The method relies on generalized entropy functionals for nonlinear diffusion equations. It opens a new area of research for approaches related to carré du champ methods on non-compact manifolds. However, key estimates depend as much on curvature properties as on purely nonlinear effects. The method is well adapted to functional inequalities involving simple weights and also applies to general cylinders. Beyond results on symmetry and symmetry breaking, and on optimal constants in functional inequalities, rigidity theorems for nonlinear elliptic equations can be deduced in rather general settings.

Appendix: Regularity and decay estimates

We denote by $'$ and $\nabla$ the differentiation with respect to s and $\omega$ respectively. We work in the general setting and do not assume that $\mathfrak M={\mathbb {S}}^{d-1}$.

Proposition 8.1

Any positive solution $\varphi \in {\mathrm {H}}^1({\mathcal {C}})$ of (1.7) with $p\in (2, 2^*)$ is uniformly bounded and smooth. Moreover there are two positive constants, $C_1$ and $C_2$ such that, for all $(s,\omega )\in {\mathcal {C}}$,

$$\begin{aligned}&C_1\,e^{-\sqrt{\Lambda }\,|s|}\le \varphi (s,\omega )\le C_2\,e^{-\sqrt{\Lambda }\,|s|},\\&|\varphi '(s,\omega )|, |\varphi ''(s,\omega )|, |\nabla \varphi (s,\omega )|, |\nabla \varphi '(s,\omega )|, |\Delta \,\varphi (s,\omega )|\le C_2\,e^{-\sqrt{\Lambda }\,|s|}. \end{aligned}$$

Proof

A similar result was proved in [13]. Here we work in a more general setting when $\mathfrak M\ne {\mathbb {S}}^{d-1}$. For sake of completeness, we sketch the main steps of the proof.

Step 1. The solution is bounded, smooth and $\lim _{|s|\rightarrow +\infty }\varphi (s,\omega )=0$ for any $\omega \in \mathfrak M$. Boundedness is obtained by a Moser iteration scheme. The $C^\infty$ regularity follows by a localized boot-strap argument based on, e.g., [32, Corollary 7.11, Theorem 8.10, and Corollary 8.11]. If $s\mapsto \chi (s)$ is a smooth truncation function such that $0\le \chi \le 1$, $\chi \equiv 1$ if $|s|\le 1$ and $\chi \equiv 0$ if $|s|\ge 2$, then $\varphi _\varepsilon (s,\omega ):=\varphi (s,\omega )\,\big (1-\chi (\varepsilon \,s)\big )$ has an arbitrary small norm in ${\mathrm {H}}^1({\mathcal {C}})$ and $\lim _{\varepsilon \rightarrow 0_+}\Vert {\varphi _\varepsilon }\Vert _{{\mathrm {L}}^{\infty }({\mathcal {C}})}=0$, again by a Moser iteration scheme.

Step 2. Exponential decay of $\varphi$ in |s|. For any $\mu \in (0,\sqrt{\Lambda })$, let $h(s):=e^{-\mu \,|s|}$ and define

$$\begin{aligned} s_\mu :=\inf \left\{ s>0\,:\,|\varphi (\sigma ,\omega )|^{p-2}<\Lambda -\mu ^2\quad \forall \,(\sigma ,\omega )\in {\mathcal {C}}\cap \{|\sigma |>s\}\right\} . \end{aligned}$$

By the Strong Maximum Principle applied to the function $(h-\varphi )$ which solves

$$\begin{aligned} -\,\partial ^2_s\,(h-\varphi )-\Delta \,(h-\varphi )+\mu ^2\,(h-\varphi )\ge \left( \Lambda -\mu ^2-|\varphi |^{p-2}\right) \,\varphi \ge 0 \end{aligned}$$

for $|s| \ge s_\mu$, we get the estimate

$$\begin{aligned} 0<\varphi \le \Vert {\varphi }\Vert _{{\mathrm {L}}^{\infty }({\mathcal {C}})}\,e^{-\mu \,(|s|-s_\mu )}\quad \forall \,(s,\omega )\in {\mathcal {C}}\cap \{|s|>s_\mu \}. \end{aligned}$$

Step 3. Optimal exponential decay of $\varphi$ in |s|. The function $h_1(s,\omega ):=e^{-\sqrt{\Lambda }\,|s|}$ satisfies the equation $-\,\Delta \,h_1+\Lambda \,h_1=0$ on ${\mathcal {C}}\cap \{|s|>1\}$. Hence, by the Strong Maximum Principle, we have

$$\begin{aligned} \varphi (s,\omega )\ge \left( \min _{{\mathcal {C}}\cap \{|s|\le 1\}}\varphi \right) \,e^{-\sqrt{\Lambda }\,(|s|-1)}. \end{aligned}$$

From Step 2 we know that for some positive M and $\bar{s}$, we have

$$\begin{aligned}\textstyle -\,\partial ^2_s\,\varphi -\Delta \,\varphi +\left( \Lambda -\frac{M}{s^2}\right) \varphi \le 0\quad \text{ in }\quad {\mathcal {C}}\cap \{|s|>\bar{s}\}, \end{aligned}$$

while the function $h_2(s,\omega ):=e^{-\sqrt{\Lambda }\,|s|}\,e^{\frac{\lambda }{|s|}}$ satisfies

$$\begin{aligned} -\,\partial ^2_s\,h_2-\Delta \,h_2+\left( \Lambda -\frac{M}{s^2}\right) h_2=-\frac{1}{s^2}\left( M+2\,\lambda \,\sqrt{\Lambda }+\frac{2\,\lambda }{s}+\frac{\lambda }{s^2}\right) h_2\quad \text{ in }\quad {\mathcal {C}}\cap \{|s|>\bar{s}\}. \end{aligned}$$

By taking $\lambda <-\frac{M}{2\sqrt{\Lambda }}$ and applying the Strong Maximum Principle for $S>0$ large enough, we obtain

$$\begin{aligned} 0<\varphi \le \Vert \varphi \Vert _{L^\infty ({\mathcal {C}})}\,e^{-\frac{\lambda }{S}}\,e^{-\sqrt{\Lambda }\,(|s|-S)}\quad \text{ in }\quad {\mathcal {C}}\cap \{|s|>S\}. \end{aligned}$$

Step 4. Optimal exponential decay in |s| for $\nabla \varphi$, $\Delta \,\varphi$. Using local charts and [32, Theorem 8.32, p. 210] on local $C^{1,\alpha }$ estimates, all first derivatives of $\varphi$ converge to 0 with rate $e^{-\sqrt{\Lambda }\,|s|}$ as $|s|\rightarrow +\infty$. [32, Theorem 8.10, p. 186] provides local ${\mathrm {W}}^{k+2,2}$ estimates of the order $e^{-\sqrt{\Lambda }\,|s|}$ for |s| large enough. The result follows from [32, Corollary 7.11, Theorem 8.10, and Corollary 8.11] if k is taken large enough. $\square$

Next we rephrase the results of Proposition 8.1 in the language of the pressure function ${\mathsf {p}}$ of Sect. 4 using (6.1) and establish the estimates needed in Lemmas 4.3 and 6.1.

Proposition 8.2

Let $m=1-1/n$ and $\varphi \in {\mathrm {H}}^1({\mathcal {C}})$ be a positive solution of (1.9) with $p\in (2, 2^*)$. Then the functions ${\mathsf {p}}$ associated with $\varphi$ according to (6.1) are such that ${\mathsf {p}}''$, ${\mathsf {p}}'/r$, ${\mathsf {p}}/r^2$, $\nabla {\mathsf {p}}'/r$, $\nabla {\mathsf {p}}/r^2$ and $\Delta {\mathsf {p}}/r^2$ are bounded as $r\rightarrow +\infty$ and of class $C^\infty$ on $(0,\infty )\times \mathfrak M$. Moreover, if $\alpha \le \alpha _\mathrm{FS}$, as $r\rightarrow 0_+$, we have

• (i) $\int _{\mathfrak M}{|{\mathsf {p}}'(r,\omega )|^2}\,dv_g\le O(1)$,

• (ii) $\int _{\mathfrak M}{|\nabla {\mathsf {p}}(r,\omega )|^2}\,dv_g\le O(r^2)$,

• (iii) $\int _{\mathfrak M}{|{\mathsf {p}}''(r,\omega )|^2}\,dv_g\le O(1/r^2)$,

• (iv) $\int _{\mathfrak M}{\left| \nabla {\mathsf {p}}'(r,\omega )-\tfrac{1}{r}\,\nabla {\mathsf {p}}(r,\omega )\right| ^2}\,dv_g\le O(1)$,

• (v) $\int _{\mathfrak M}{\left| \Delta {\mathsf {p}}(r,\omega )\right| ^2}\,dv_g\le O(1/r^2)$.

Moreover, with the notations defined by (4.8) and (6.3),

$$\begin{aligned} \lim _{r\rightarrow 0_+}{\mathsf {b}}(r)=0=\lim _{r\rightarrow +\infty }{\mathsf {b}}(r) \end{aligned}$$

and

$$\begin{aligned} \lim _{r\rightarrow 0_+}{\mathsf {c}}(r)=0=\lim _{r\rightarrow +\infty }{\mathsf {c}}(r). \end{aligned}$$

Proof

We say that $f(s,\omega )\sim g(s,\omega )$ as $s\rightarrow +\infty$ (resp. $s\rightarrow -\infty$) if the ratio f / g is bounded from above and from below by positive constants, independent of $\omega$, and for s (resp. $-s$) large enough.

There are some easy consequences of the change of variables (6.1) and of Proposition 8.1: since $\varphi (s,\omega )\sim e^{\sqrt{\Lambda }\,s}$ as $s\rightarrow -\infty$, $\varphi (-\log r/\alpha ,\omega )\sim r^{-2/(p-2)}$ as $r\rightarrow +\infty$ and it is straightforward to check that ${\mathsf {p}}''$, ${\mathsf {p}}'/r$, ${\mathsf {p}}/r^2$, $\nabla {\mathsf {p}}'/r$ and $\nabla {\mathsf {p}}/r^2$ are bounded as $r\rightarrow +\infty$. As a consequence, we obtain that

$$\begin{aligned} |{\mathsf {b}}(r)|, \quad {\mathsf {c}}(r)\le O\left( r^{2-n}\right) \rightarrow 0\quad \text{ as }\quad r\rightarrow +\infty \end{aligned}$$

because, by assumption, we know that $n>d\ge 2$.

To complete the proof, one has to establish that $\lim _{r\rightarrow 0_+}{\mathsf {b}}(r)=\lim _{r\rightarrow 0_+}{\mathsf {c}}(r)=0$. A convenient method for that relies on the Kelvin transformation. Let

$$\begin{aligned} u(r,\omega )=r^{-2n}\,\widetilde{u}(R,\omega )\quad \text{ and }\quad {\mathsf {p}}(r,\omega )=r^2\,\widetilde{{\mathsf {p}}}(R,\omega ) \end{aligned}$$

with $R=1/r$. It is a remarkable fact to observe that $\widetilde{u}$ solves the same equation as u, which can be easily seen after applying the Emden-Fowler transformation $w(r,\omega )=r^{2-n}\,\widetilde{w}(R,\omega )$ to the function w such that $u(r,\omega )=|w(r,\omega )|^\frac{2\,n}{n-2}$. With evident notations if $\varphi$ and $\widetilde{\varphi }$ are given in terms of w and $\widetilde{w}$ by (1.5), then $\widetilde{\varphi }(s,\omega )=\varphi (-s,\omega )$ for any $(s,\omega )\in {\mathbb {R}}\times \mathfrak M$ and it is clear that Eq. (1.9) is invariant under the transformation $s\mapsto -\,s$.

According to Proposition 8.1, ${\mathsf {p}}(r,\omega )=r^2\,\widetilde{{\mathsf {p}}}(1/r,\omega )$ is bounded away from 0 and from infinity, and, uniformly in $\omega$,

$$\begin{aligned}&|{\mathsf {p}}'(r,\omega )|=|2\,r\,\widetilde{{\mathsf {p}}}\left( \tfrac{1}{r},\omega \right) -\widetilde{{\mathsf {p}}}'\left( \tfrac{1}{r},\omega \right) |\le O\left( \frac{1}{r}\left( \sqrt{\Lambda }-\frac{\widetilde{\varphi }'(s,\omega )}{\widetilde{\varphi }(s,\omega )}\right) \right) ,\\&\tfrac{1}{r}\,|\nabla {\mathsf {p}}(r,\omega )|=r\,|\nabla \,\widetilde{{\mathsf {p}}}\left( \tfrac{1}{r},\omega \right) |\le O\left( \frac{1}{r}\,\frac{\nabla \widetilde{\varphi }(s,\omega )}{\widetilde{\varphi }(s,\omega )}\right) , \end{aligned}$$

which are of order at most 1 / r. Moreover, also uniformly in $\omega$,

$$\begin{aligned}&|{\mathsf {p}}''(r,\omega )| = \left| 2\,\widetilde{{\mathsf {p}}}\left( \tfrac{1}{r},\omega \right) -\frac{2}{r}\,\widetilde{{\mathsf {p}}}'\left( \tfrac{1}{r},\omega \right) +\frac{1}{r^2}\,\widetilde{{\mathsf {p}}}''\left( \tfrac{1}{r},\omega \right) \right| \\&\qquad \qquad \quad \;\le O\left( \frac{1}{r^2}\left( \frac{\widetilde{\varphi }''(s,\omega )}{\widetilde{\varphi }(s,\omega )}-\,\frac{p}{2}\,\frac{|\widetilde{\varphi }'(s,\omega )|^2}{|\widetilde{\varphi }(s,\omega )|^2}+\,\alpha \,\frac{\widetilde{\varphi }'(s,\omega )}{\widetilde{\varphi }(s,\omega )}\right) \right) ,\\&\left| \tfrac{1}{r}\,\nabla {\mathsf {p}}'(r,\omega )-\tfrac{1}{r^2}\,\nabla {\mathsf {p}}(r,\omega )|=|\nabla \,\widetilde{{\mathsf {p}}}\left( \tfrac{1}{r},\omega \right) -\tfrac{1}{r}\,\nabla \,\widetilde{{\mathsf {p}}}'\left( \tfrac{1}{r},\omega \right) \right| \\&\quad \le O\left( \frac{1}{r^2}\left( \frac{p}{2}\,\frac{\widetilde{\varphi }'(s,\omega )\,\nabla \widetilde{\varphi }(s,\omega )}{|\widetilde{\varphi }(s,\omega )|^2}-\frac{\nabla \widetilde{\varphi }'(s,\omega )}{\widetilde{\varphi }(s,\omega )}\right) \right) ,\\&\frac{1}{r^2}\,|\Delta \,{\mathsf {p}}(r,\omega )|=\left| \Delta \,\widetilde{{\mathsf {p}}}\left( \tfrac{1}{r},\omega \right) \right| \le O\left( \frac{1}{r^2}\left( \frac{\Delta \widetilde{\varphi }(s,\omega )}{\widetilde{\varphi }(s,\omega )}-\,\frac{p}{2}\,\frac{|\nabla \widetilde{\varphi }(s,\omega )|^2}{|\widetilde{\varphi }(s,\omega )|^2}\right) \right) , \\ \end{aligned}$$

which are of order at most $1/r^2$. This shows that $|{\mathsf {b}}(r)|$, ${\mathsf {c}}(r)\le O(r^{n-4})$ and concludes the proof if $4\le d<n$. When $d=2$ or 3 and $p>4$, i.e., $n<4$, more detailed estimates are needed. We will actually prove Properties (i)–(v) as $r\rightarrow 0_+$. Using the fact that $\widetilde{\varphi }$ and $\varphi$ solve the same equation, this amounts to prove that

• (i) $\int _{\mathfrak M}{\left| \frac{\varphi '(s,\omega )}{\varphi (s,\omega )}-\sqrt{\Lambda }\right| ^2}\,dv_g\le O(e^{2\alpha s})$,

• (ii) $\int _{\mathfrak M}{\left| \frac{\nabla \varphi (s,\omega )}{\varphi (s,\omega )}\right| ^2}\,dv_g\le O(e^{2\alpha s})$,

• (iii) $\int _{\mathfrak M}{\left| \frac{\varphi ''(s,\omega )}{\varphi (s,\omega )}-\,\frac{p}{2}\,\frac{|\varphi '(s,\omega )|^2}{|\varphi (s,\omega )|^2}+\,\alpha \,\frac{\varphi '(s,\omega )}{\varphi (s,\omega )}\right| ^2}\,dv_g\le O(e^{2\alpha s})$,

• (iv) $\int _{\mathfrak M}{\left| \frac{p}{2}\,\frac{\varphi '(s,\omega )\,\nabla \varphi (s,\omega )}{|\varphi (s,\omega )|^2}-\frac{\nabla \varphi '(s,\omega )}{\varphi (s,\omega )}\right| ^2}\,dv_g\le O(e^{2\alpha s})$,

• (v) $\int _{\mathfrak M}{\left| \frac{\Delta \varphi (s,\omega )}{\varphi (s,\omega )}-\,\frac{p}{2}\,\frac{|\nabla \varphi (s,\omega )|^2}{|\varphi (s,\omega )|^2}\right| ^2}\,dv_g\le O(e^{2\alpha s})$,

as $s\rightarrow -\infty$.

Proof of (i). Let us consider a positive solution $\varphi$ to (1.9) and define on ${\mathbb {R}}$ the function

$$\begin{aligned} \varphi _0(s)=\int _{\mathfrak M}{\varphi (s,\omega )}\,dv_g. \end{aligned}$$

By integrating (1.9) on $\mathfrak M$, we know that $\varphi _0$ solves

$$\begin{aligned} -\,\varphi _0''+\Lambda \,\varphi _0=\int _{\mathfrak M}{\varphi ^{p-1}}\,dv_g=:h_0(s)\sim e^{-(p-1)\sqrt{\Lambda }\,|s|}\quad \text{ in }\quad {\mathbb {R}}. \end{aligned}$$

From the integral representation

$$\begin{aligned} \varphi _0(s)=\frac{e^{-\sqrt{\Lambda }s}}{2\,\sqrt{\Lambda }}\int _{-\infty }^se^{\sqrt{\Lambda }t}\,h_0(t)\,dt+\frac{e^{\sqrt{\Lambda }s}}{2\,\sqrt{\Lambda }}\int _s^\infty e^{-\sqrt{\Lambda }t}\,h_0(t)\,dt, \end{aligned}$$

we deduce that $\varphi _0(s)\sim e^{\sqrt{\Lambda }s}\sim \varphi (s,\omega )$ as $s\rightarrow -\infty$ and

$$\begin{aligned} \frac{\varphi _0'(s)-\sqrt{\Lambda }\,\varphi _0(s)}{\varphi (s,\omega )}\sim & {} -\,e^{-2\sqrt{\Lambda }s}\int _{-\infty }^se^{\sqrt{\Lambda }t}\,h_0(t)\,dt\\= & {} O(e^{2\alpha s})\quad \text{ as }\quad s\rightarrow -\infty . \end{aligned}$$

If we define $\psi (s,\omega ):=e^{\sqrt{\Lambda }\,|s|}\,\big (\varphi (s,\omega )-\varphi _0(s)\big )$, we may observe that it is bounded and solves the equation

$$\begin{aligned} -\,\partial ^2_s\psi -\,\Delta \,\psi -\,2\,\sqrt{\Lambda }\,\partial _s\psi= & {} e^{\sqrt{\Lambda }\,|s|}\,\big (\varphi ^{p-1}-\varphi _0^{p-1}\big )\nonumber \\=: & {} H\le O\left( e^{-2\alpha |s|}\right) \end{aligned}$$

(8.1)

and

$$\begin{aligned} \frac{\partial _s\varphi (s,\omega )}{\varphi (s,\omega )}-\sqrt{\Lambda }=O\left( e^{2\alpha s}\right) +\frac{\partial _s\psi (s,\omega )}{e^{-\sqrt{\Lambda }\,s}\,\varphi (s,\omega )}\quad \text{ as }\quad s\rightarrow -\infty . \end{aligned}$$

We recall that $e^{-\sqrt{\Lambda }\,s}\,\varphi (s,\omega )$ is bounded from above and from below by positive constants as $s\rightarrow -\infty$, and $|e^{-\sqrt{\Lambda }\,s}\,\partial _s\varphi (s,\omega )|$ is bounded above. As a consequence, we know that $\partial _s H=O(e^{2\alpha s})$ as $s\rightarrow -\infty$. Hence we know that

$$\begin{aligned} \Big |\frac{\partial _s\varphi (s,\omega )}{\varphi (s,\omega )}-\sqrt{\Lambda }\Big |\le C\,|\partial _s\psi (s,\omega )|+O(e^{2\alpha s}), \end{aligned}$$

where C is a constant.

We differentiate (8.1) with respect to s. The function $\partial _s\psi$ solves

$$\begin{aligned} -\,\partial ^2_s(\partial _s\psi )-\,\Delta \,(\partial _s\psi )-\,2\,\sqrt{\Lambda }\,\partial _s(\partial _s\psi )=\partial _sH, \end{aligned}$$

(8.2)

with

$$\begin{aligned} |\partial _sH(s,\omega )|\le O(e^{2\alpha s})\quad \text{ as }\quad s\rightarrow -\infty . \end{aligned}$$

Let us define

$$\begin{aligned} \chi _1(s):=\frac{1}{2}\int _{\mathfrak M}{|\partial _s\psi |^2}\,dv_g, \end{aligned}$$

multiply (8.2) by $\partial _s\psi$ and integrate over $\mathfrak M$. Using

$$\begin{aligned} \chi _1'=\int _{\mathfrak M}{\partial _s\psi \,\partial _s^2\psi }\,dv_g \end{aligned}$$

and

$$\begin{aligned} \chi _1''=\int _{\mathfrak M}{\partial _s\psi \,\partial ^2_s(\partial _s\psi )}\,dv_g+\int _{\mathfrak M}{|\partial _s^2\psi |^2}\,dv_g, \end{aligned}$$

we obtain that the nonnegative function $\chi _1$ solves

$$\begin{aligned} -\,\chi _1''+\int _{\mathfrak M}{|\partial _s^2\psi |^2}\,dv_g+\underbrace{\int _{\mathfrak M}{\left( |\nabla (\partial _s\psi )|^2-\lambda _1\,|\partial _s\psi |^2\right) }\,dv_g}_{\ge 0} +\,2\,\lambda _1\,\chi _1-\,2\,\sqrt{\Lambda }\,\chi _1'=h_1,\nonumber \\ \end{aligned}$$

(8.3)

where the Poincaré inequality

$$\begin{aligned} \int _{\mathfrak M}{|\nabla (\partial _s\psi )|^2}\,dv_g\ge \lambda _1\int _{\mathfrak M}{|\partial _s\psi |^2}\,dv_g \end{aligned}$$

holds because $\int _{\mathfrak M}{\partial _s\psi }\,dv_g=0$ for any $s\in {\mathbb {R}}$, by definition of $\psi$, and where

$$\begin{aligned} h_1:=\int _{\mathfrak M}{\partial _sH\,\partial _s\psi }\,dv_g. \end{aligned}$$

From the Cauchy-Schwarz inequality, we deduce that

$$\begin{aligned} |\chi _1'(s)|^2= & {} \left( \int _{\mathfrak M}{\partial _s\psi \,\partial _s^2\psi }\,dv_g\right) ^2\\\le & {} \int _{\mathfrak M}{|\partial _s\psi |^2}\,dv_g\int _{\mathfrak M}{|\partial _s^2\psi |^2}\,dv_g=2\,\chi _1(s)\int _{\mathfrak M}{|\partial _s^2\psi |^2}\,dv_g, \end{aligned}$$

that is,

$$\begin{aligned} \int _{\mathfrak M}{|\partial _s^2\psi |^2}\,dv_g\ge \frac{|\chi _1'|^2}{2\,\chi _1}, \end{aligned}$$

and inserting this estimate in (8.3) we obtain

$$\begin{aligned} -\,\chi _1''+\frac{|\chi _1'|^2}{2\,\chi _1}+\,2\,\lambda _1\,\chi _1-\,2\,\sqrt{\Lambda }\,\chi _1'\le h_1. \end{aligned}$$

Let $\zeta _1=\sqrt{\chi _1}$ and observe that it solves

$$\begin{aligned} -\,\zeta _1''+\,\lambda _1\,\zeta _1-\,2\,\sqrt{\Lambda }\,\zeta _1'\le \frac{h_1}{2\,\zeta _1}. \end{aligned}$$

By Cauchy-Schwarz inequality, for $s\rightarrow -\infty$,

$$\begin{aligned} |h_1 (s)| \le \sqrt{2}\, \left( \int _{\mathfrak M}{|\partial _sH|^2}\,dv_g\right) ^{1/2}\zeta _1(s)\le C\, e^{2\alpha s}\, \zeta _1(s). \end{aligned}$$

Using once more an integral representation of the solution, with $\mu :=\sqrt{\Lambda +\lambda _1}$, it is easy to check that

$$\begin{aligned} e^{\sqrt{\Lambda } s}\,\zeta _1(s)\le \frac{e^{-\mu s}}{4\,\mu }\int _{-\infty }^se^{(\mu +\sqrt{\Lambda })t}\,\frac{h_1(t)}{\zeta _1(t)}\,dt+\frac{e^{\mu s}}{4\,\mu }\int _s^\infty e^{(\sqrt{\Lambda }-\mu )t}\,\frac{h_1(t)}{\zeta _1(t)}\,dt, \end{aligned}$$

which is enough to deduce that $\zeta _1(s)\le O\big (e^{\alpha s}\big )$ as $s\rightarrow -\infty$. Note that the condition that

$$\begin{aligned} \mu - \sqrt{\Lambda }= \sqrt{\Lambda +\lambda _1} - \sqrt{\Lambda }\ge \alpha \end{aligned}$$

is equivalent to the inequality $\alpha \le \alpha _\mathrm{FS}$. Hence we have shown that for $\alpha \le \alpha _\mathrm{FS}$,

$$\begin{aligned} \chi _1(s)\le O\left( e^{2\alpha s}\right) \quad \text{ as }\quad s\rightarrow -\infty . \end{aligned}$$

(8.4)

This ends the proof of (i).

Proof of (ii). By differentiating (1.9) with respect to $\omega$, we obtain

$$\begin{aligned} -\,\partial ^2_s\,\nabla \varphi -\,\nabla \,\Delta \,\varphi +\Lambda \,\nabla \varphi =(p-1)\,\varphi ^{p-2}\,\nabla \varphi \quad \text{ in }\quad {\mathcal {C}}. \end{aligned}$$

We proceed as in case (i). With similar notations, by defining

$$\begin{aligned} \chi _2(s):=\frac{1}{2}\int _{\mathfrak M}{|\nabla \varphi |^2}\,dv_g, \end{aligned}$$

after multiplying the equation by $\nabla \varphi$ and using the fact that

$$\begin{aligned} \int _{\mathfrak M}{(\Delta \varphi )^2}\,dv_g\ge \lambda _1\int _{\mathfrak M}{|\nabla \varphi |^2}\,dv_g \end{aligned}$$

as, e.g., in [25, Lemma 7] and the Cauchy-Schwarz inequality, we obtain

$$\begin{aligned} -\,\chi _2''+\frac{|\chi _2'|^2}{2\,\chi _2}+\,2\,(\Lambda +\lambda _1)\,\chi _2\le h_2 \, \end{aligned}$$

with $h_2 := (p-1)\,\int _{\mathfrak M}{\varphi ^{p-2} |\nabla \varphi |^2}\,dv_g = O\big (e^{-\,p\,\sqrt{\Lambda }\,|s|}\big )$. The function $\zeta _2=\sqrt{\chi _2}$ satisfies

$$\begin{aligned} -\,\zeta _2''+\,(\Lambda +\lambda _1)\,\zeta _2\le \frac{h_2}{2\,\zeta _2}. \end{aligned}$$

By the Cauchy-Schwarz inequality, $h_2/\zeta _2 = O\big (e^{-\,(p-1)\,\sqrt{\Lambda }\,|s|}\big )$. We easily deduce that

$$\begin{aligned} \chi _2(s)\le O\big (e^{2(\alpha +\sqrt{\Lambda })s}\big )\quad \text{ as }\quad s\rightarrow -\infty . \end{aligned}$$

Indeed, we observe that $\sqrt{\Lambda +\lambda _1}-\sqrt{\Lambda }\ge \alpha$ for any $\Lambda \in (0,\Lambda _{\,{\mathrm{FS}}})$ and $\varphi (s,\omega )\sim e^{\sqrt{\Lambda }\,s}$ as $s\rightarrow -\infty$, which ends the proof of (ii).

Proof of (iii). With $\psi =e^{-\sqrt{\Lambda }s}\,(\varphi -\varphi _0)$ and $\varphi _0(s)=\int _{\mathfrak M}{\varphi (s,\omega )}\,dv_g$ as in case (i), we can check that

$$\begin{aligned} \frac{\varphi ''}{\varphi }-\,\frac{p}{2}\,\frac{|\varphi '|^2}{|\varphi |^2}+\,\alpha \,\frac{\varphi '}{\varphi }= & {} O(e^{2\alpha s})+\frac{e^{\sqrt{\Lambda }s}}{\varphi }\,\partial _s^2\psi \nonumber \\&+\,\left( (2-p)\,\sqrt{\Lambda }+\alpha -p\,\frac{\varphi '_0-\sqrt{\Lambda }\varphi _0}{\varphi }\right) \,\frac{e^{\sqrt{\Lambda }s}}{\varphi }\,\partial _s\psi \nonumber \\&-\, \frac{p}{2}\, \Big ( \frac{e^{\sqrt{\Lambda }s}}{\varphi } \Big )^2 |\partial _s\psi |^2. \end{aligned}$$

(8.5)

Because according to Proposition 8.1 $\partial _s\psi$ and $\frac{\partial _s\varphi }{\varphi }$ are bounded as $s\rightarrow -\infty$, and taking into account (8.4), it remains to prove that

$$\begin{aligned} \chi _3(s):=\frac{1}{2}\int _{\mathfrak M}{|\partial _s^2\psi |^2}\,dv_g \end{aligned}$$

is of order $O(e^{2\alpha s})$. We differentiate (8.1) twice with respect to s. After multiplying the equation by $\partial _s^2\psi$ and using the fact that

$$\begin{aligned} \int _{\mathfrak M}{|\nabla (\partial _s^2\psi )|^2}\,dv_g\ge \lambda _1\int _{\mathfrak M}{|\partial _s^2\psi |^2}\,dv_g \end{aligned}$$

because $\int _{\mathfrak M}{\partial _s^2\psi }\,dv_g=0$, we obtain

$$\begin{aligned} -\,\chi _3''+\frac{|\chi _3'|^2}{2\,\chi _3}+\,2\,\lambda _1\,\chi _3-\,2\,\sqrt{\Lambda }\,\chi _3'\le h_3, \end{aligned}$$

with $h_3:=\int _{\mathfrak M}{\partial ^2_sH\,\partial ^2_s\psi }\,dv_g.$ With the same arguments as in case (i), we deduce that

$$\begin{aligned} \chi _3(s)\le O\left( e^{2\alpha s}\right) \quad \text{ as }\quad s\rightarrow -\infty . \end{aligned}$$

This ends the proof of (iii).

Proof of (iv). The term $\frac{\varphi '(s,\omega )\,\nabla \varphi (s,\omega )}{|\varphi (s,\omega )|^2}$ is easily bounded after integrating with respect to $\omega$ because $\frac{\partial _s\,\varphi }{\varphi }$ is bounded according to Proposition 8.1 and by (ii). As for the term $\frac{\nabla \varphi '(s,\omega )}{\varphi (s,\omega )}$, we proceed like in case (ii). By applying the operator $\nabla \partial _s$ to (1.9), we obtain

$$\begin{aligned}&-\,\partial ^2_s\,(\nabla \partial _s\varphi )-\,\nabla \Delta (\partial _s\varphi )+\Lambda \,\nabla \partial _s\varphi =\partial _s\nabla H\\&\quad =(p-1)\,\varphi ^{p-2}\left( \nabla \partial _s\varphi +(p-2)\,\frac{\partial _s\varphi \,\nabla \varphi }{\varphi }\right) \;\text{ in }\;{\mathcal {C}}. \end{aligned}$$

With similar notations, by defining

$$\begin{aligned} \chi _4(s):=\frac{1}{2}\int _{\mathfrak M}{|\nabla \partial _s\varphi |^2}\,dv_g, \end{aligned}$$

after multiplying the equation by $\nabla \partial _s\varphi$ and using the Poincaré inequality

$$\begin{aligned} \int _{\mathfrak M}{|\Delta (\partial _s\varphi )|^2}\,dv_g\ge \lambda _1\int _{\mathfrak M}{|\nabla (\partial _s\varphi )|^2}\,dv_g \end{aligned}$$

we obtain

$$\begin{aligned} -\,\chi _4''+\frac{|\chi _4'|^2}{2\,\chi _4}+\,2\,(\Lambda +\lambda _1)\,\chi _4\le h_4, \end{aligned}$$

with $h_4:=\int _{\mathfrak M}{\partial _s \nabla H\,\partial _s \nabla \varphi }\,dv_g$. With the same arguments, we deduce that

$$\begin{aligned} \chi _4(s)\le O\big (e^{2(\alpha +\sqrt{\Lambda })\,s}\big )\quad \text{ as }\quad s\rightarrow -\infty . \end{aligned}$$

We end the proof of (iv) by observing that $\varphi (s,\omega )\sim e^{\sqrt{\Lambda }\,s}$ as $s\rightarrow -\infty$.

Proof of (v). By applying the Laplace-Beltrami operator to (1.9), we obtain

$$\begin{aligned}&-\,\partial ^2_s\,(\Delta \varphi )-\,\Delta ^2\varphi +\Lambda \,\Delta \varphi =\Delta H\\&\quad =(p-1)\,\varphi ^{p-2}\left( \Delta \varphi +(p-2)\,\frac{|\nabla \varphi |^2}{\varphi }\right) \quad \text{ in }\quad {\mathcal {C}}. \end{aligned}$$

We proceed as in case (ii). With similar notations, by defining

$$\begin{aligned} \chi _5(s):=\frac{1}{2}\int _{\mathfrak M}{|\Delta \varphi |^2}\,dv_g, \end{aligned}$$

after multiplying the equation by $\Delta \varphi$ and using the fact that

$$\begin{aligned} -\int _{\mathfrak M}{\Delta \varphi \,\Delta ^2\varphi }\,dv_g=\int _{\mathfrak M}{|\nabla \Delta \varphi |^2}\,dv_g\ge \lambda _1\int _{\mathfrak M}{|\Delta \varphi |^2}\,dv_g, \end{aligned}$$

we obtain

$$\begin{aligned} -\,\chi _5''+\frac{|\chi _5'|^2}{2\,\chi _5}+\,2\,(\Lambda +\lambda _1)\,\chi _5\le h_5 \end{aligned}$$

with $h_5:= \int _{\mathfrak M}{\Delta H \, \Delta \varphi }\,dv_g$. With the same arguments, we deduce that

$$\begin{aligned} \chi _5(s)\le O\big (e^{2(\alpha +\sqrt{\Lambda })\,s}\big )\quad \text{ as }\quad s\rightarrow -\infty , \end{aligned}$$

using again the fact that $\sqrt{\Lambda +\lambda _1}-\sqrt{\Lambda }\ge \alpha$ for any $\Lambda \in (0,\Lambda _{\,{\mathrm{FS}}})$ and $\varphi (s,\omega )\sim e^{\sqrt{\Lambda }\,s}$ as $s\rightarrow -\infty$. The estimate for the other term follows from (ii).

This ends the proof of (v). $\square$