Higher Order Conformally Invariant Equations in R^3 with Prescribed Volume

TL;DR

This paper investigates higher order conformally invariant equations in three-dimensional space, establishing existence of solutions with prescribed volume and characterizing the range of possible volumes for specific cases.

Contribution

It proves the existence of positive smooth radial solutions with prescribed volume for certain higher order conformally invariant equations in R^3, revealing new volume range properties.

Findings

For m=2, volume range is (0,Λ*], a bounded interval.

For m=3, volume range is (0,∞).

Contrasts with m=1 case where volume is fixed.

Abstract

In this paper we study the following conformally invariant poly-harmonic equation $$\Delta^mu=-u^\frac{3+2m}{3-2m}\quad\text{in }\mathbb{R}^3,\quad u>0,$$ with $m=2,3$. We prove the existence of positive smooth radial solutions with prescribed volume $\int_{\mathbb{R}^3} u^\frac{6}{3-2m}dx$. We show that the set of all possible values of the volume is a bounded interval $(0,\Lambda^*]$ for $m=2$, and it is $(0,\infty)$ for $m=3$. This is in sharp contrast to $m=1$ case in which the volume $\int_{\mathbb{R}^3} u^\frac{6}{3-2m}dx$ is a fixed value.

Equations

$${\Delta }^{m}u=-u^{\frac{3+2m}{3-2m}}\text{in }{\mathbb{R}}^3,u>0,$$

$${\int }_{{\mathbb{R}}^3}d{V}_{g_u}={\int }_{{\mathbb{R}}^3}\sqrt{|g_u|}dx={\int }_{{\mathbb{R}}^3}{u}^{\frac{6}{3-2m}}dx,$$

$$\underset{r\to \infty }{\lim }\frac{u(r)}{r}\in (0,\infty )\text{or }\underset{r\to \infty }{\lim }\frac{u(r)}{r^2}\in (0,\infty ).$$

$$U_0(r)=\sqrt{\sqrt{1/15}+r^2}.$$

$$\underset{r\to \infty }{\lim }\frac{u(r)}{r^3}\in (0,\infty )\text{or }\underset{r\to \infty }{\lim }\frac{u(r)}{r^4}\in (0,\infty ).$$

$$U_1(r)={\left(315^{-\frac{1}{3}}+r^2\right)}^{\frac{3}{2}}.$$

$$U(x)=c_m{\int }_{{\mathbb{R}}^3}|x-y{|}^p{U}^{\frac{3+2m}{3-2m}}(y)dy,$$

$${\Delta }^{2}u=-\frac{1}{u^7}\text{in }{\mathbb{R}}^3,u>0,{\Lambda }_u:={\int }_{{\mathbb{R}}^3}\frac{dx}{u^6(x)}$$

$${\Lambda }^{\ast }:={\int }_{{\mathbb{R}}^3}\frac{dx}{U_0^6(x)}={\int }_{{\mathbb{R}}^3}\frac{dx}{(\sqrt{1/15}+|x{|}^2{)}^3}.$$

$${\Delta }^{3}u=-\frac{1}{u^3}\text{in }{\mathbb{R}}^3$$

$${\int }_{{\mathbb{R}}^3}\frac{dx}{u^2(x)}=\Lambda .$$

$$(-\Delta {)}^{\frac{n}{2}}u=(n-1)!e^{nu}\text{in }{\mathbb{R}}^n,V:={\int }_{{\mathbb{R}}^n}{e}^{nu}dx<\infty .$$

$$u(x)=u_{\lambda ,x_0}(x):=\log \left(\frac{2\lambda }{1+{\lambda }^2|x-x_0{|}^2}\right).$$

$$\displaystyle\left\{\begin{array}[]{ll}\Delta^{k}u=f(u)&\quad\text{on }(0,R)\\ \Delta^{j}u_{1}(0)\geq\Delta^{j}u_{2}(0)&\quad\text{for every }j\in J\\ (\Delta^{j}u_{1})^{\prime}(0)=(\Delta^{j}u_{2})^{\prime}(0)=0&\quad\text{for every }j\in J,\end{array}\right.$$

$$\displaystyle\left\{\begin{array}[]{ll}\Delta^{2}u_{\rho}=-\frac{1}{u_{\rho}^{7}}\\ u_{\rho}(0)=U_{0}(0)+\rho\\ \Delta u_{\rho}(0)=\Delta U_{0}(0)\\ u_{\rho}^{\prime}(0)=(\Delta u_{\rho})^{\prime}(0)=0.\end{array}\right.$$

$$\underset{\rho \to \infty }{\lim }{\int }_{{\mathbb{R}}^3}\frac{dx}{u_{\rho }^6(x)}=0.$$

$$[0,\infty )\ni \rho \mapsto {\int }_{{\mathbb{R}}^3}\frac{dx}{u_{\rho }^6(x)}$$

$$\Delta u_{\rho }(r)\le \Delta U_0(r)-\epsilon ,r\ge 1,$$

$$w(r)=w(0)+\frac{1}{4\pi }{\int }_0^r\frac{1}{t^2}{\int }_{B_t}\Delta w(x)dxdt\text{for }w\in C_{rad}^2,$$

$$u_{\rho }(r)\le C-C_{\epsilon }{r}^2\text{ on }(0,\infty ),$$

$$\displaystyle\left\{\begin{array}[]{ll}\Delta^{3}u=-\frac{1}{u^{3}}\\ u(0)=k\\ \Delta u(0)=-\varepsilon\\ \Delta^{2}u(0)=1\\ u^{\prime}(0)=(\Delta u)^{\prime}(0)=(\Delta^{2}u)^{\prime}(0)=0.\end{array}\right.$$

$$k-\frac{\epsilon }{6}{r}^2\le u(r)\le k-\frac{\epsilon }{6}{r}^2+\frac{r^4}{120}\text{on }(0,\infty ).$$

$${\Delta }^{2}u\ge \frac{1}{2}\text{on }(0,\bar{\delta }).$$

$$\Delta u(r)$$

$$u(r)\ge k-\frac{\epsilon }{6}{r}^2+\frac{r^4}{240}\ge \frac{k}{2}+\frac{r^4}{250},$$

$$\frac{1}{4\pi }{\int }_0^{\infty }\frac{1}{t^2}{\int }_{B_t}\frac{dx}{{\left(\frac{k_0}{2}+\frac{|x{|}^4}{250}\right)}^3}dt\le \frac{1}{3}.$$

$${\Delta }^{2}u(r)$$

$$\ge 1-\frac{1}{4\pi }{\int }_0^{\infty }\frac{1}{t^2}{\int }_{B_t}\frac{dx}{{\left(\frac{k_0}{2}+\frac{|x{|}^4}{250}\right)}^3}dt$$

$$\ge \frac{2}{3}.$$

$$0\le {\Delta }^{2}u(\infty )\le {\Delta }^{2}u\le 1\text{ on }(0,\infty ).$$

Full text

[TABLE]

with $m=2,3$. We prove the existence of positive smooth radial solutions with prescribed volume $\int_{\mathbb{R}^{3}}u^{\frac{6}{3-2m}}dx$. We show that the set of all possible values of the volume is a bounded interval $(0,\Lambda^{*}]$ for $m=2$, and it is $(0,\infty)$ for $m=3$. This is in sharp contrast to $m=1$ case in which the volume $\int_{\mathbb{R}^{3}}u^{\frac{6}{3-2m}}dx$ is a fixed value.

1 Introduction to the problem

We consider the negative exponent problem

[TABLE]

where $m$ is either $2$ or $3$. Geometrically, if $u$ is a smooth solution to (1) then the conformal metric $g_{u}:=u^{\frac{4}{3-2m}}|dx|^{2}$ ($|dx|^{2}$ is the Euclidean metric on $\mathbb{R}^{3}$) has constant $Q$-curvature on $\mathbb{R}^{3}$, see [1, 2, 4, 6, 21, 22]. Moreover, the volume of the metric $g_{u}$ is

[TABLE]

which is invariant under the scaling $u_{\lambda}(x):=\lambda^{\frac{3-2m}{2}}u(\lambda x)$ with $\lambda>0$.

Equation (1) belongs to the class of conformally invariant equations. When $m=1$ this is called Yamabe equation; while for $m=2$ it is $Q$-curvature equation. In recent years Problem (1) has been extensively studied in [4, 5, 9, 12, 13, 17, 20] for $m=2$, in [6] for $m=3$ and in [8, 14, 18] for higher order case (but to an integral equation). We recall that radial solutions to (1) with $m=2$ has either exactly liner growth or exactly quadratic growth at infinity, that is,

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The solution with exactly linear growth is unique (up to a scaling) and is given by

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However, there are infinitely many (radial or nonradial) solutions with quadratic growth, see [5, 9, 12]. For $m=3$, radial solutions grow either cubically or quatrically at infinity, that is,

[TABLE]

In this case also we have an explicit solution which grows cubically at infinity, namely

[TABLE]

It is worth pointing out that both solutions $U_{0}$ and $U_{1}$ can be obtained by pulling back the round metric of $S^{3}$ via stereographic projection, and they satisfy an integral equation of the form

[TABLE]

where $p=1$ for $m=2$ and $p=3$ for $m=3$. Nevertheless, $U_{1}$ is not unique (up to scaling) among the radial solutions having exactly cubic growth at infinity.

We now state our main results concerning the existence of radial solutions to (1) with prescribed volume. For $m=2$ we prove:

Theorem 1.1

There exists a radial solution to

[TABLE]

if and only if $\Lambda_{u}\in(0,\Lambda^{*}]$, where $\Lambda^{*}$ is the volume of the metric $g_{U_{0}}$, that is,

[TABLE]

Moreover, if $\Lambda_{u}=\Lambda^{*}$ for some radial solution $u$ to (3) then up to a scaling we have $u=U_{0}$.

For $m=3$ we prove the existence of radial solution for every prescribed volume.

Theorem 1.2

For every $\Lambda>0$ there exists a positive radial solution to

[TABLE]

such that

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A similar phenomena has already been exhibited in a higher order Liouville equation, namely

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(Here $V$ is the volume of the conformal metric $g_{u}=e^{2u}|dx|^{2}$). More precisely, if $u$ is a solution to (7) with $n=4$ then necessarily $V\in(0,V^{*}]$, and $V=V^{*}$ if and only if $u$ is a spherical solution, that is, for some $\lambda>0$ and $x_{0}\in\mathbb{R}^{n}$ we have

[TABLE]

However, if $n\geq 5$ then for every $V\in(0,\infty)$ there exists a radial solution to (7). See [3, 10, 11, 15, 16, 19] and the references therein.

Finally, we remark that the upper bound of $V$ in (7) with $n=4$ comes from a Pohozaev type identity, and it holds for every solutions to (7) (radial and non-radial). However, from a similar Pohozaev type identity one does not get the same conclusion on the volume of the metric $g_{u}:=u^{\frac{4}{3-2m}}$, compare [12, Lemma 2.3].

Notations For a radially symmetric function $u$ we will write $u(|x|)$ to denote the same function $u(x)$.

2 Proof of the theorems

We shall use the following comparison lemma of two radial solutions to $\Delta^{n}u=f(u)$, whose proof follows from the ODE local uniqueness theorem, and a repeated use of the identity (13). See also Lemma 3.2 in [17] and Proposition A.2 in [7].

Lemma 2.1

Let $f$ be a locally Lipschitz continuous and monotone increasing function on $(0,\infty)$. Let $u_{1},u_{2}\in C^{2k}([0,R))$ be two positive solutions of

[TABLE]

where $J:=\{0,1,\dots,k-1\}$. Then $\Delta^{j}u_{1}\geq\Delta^{j}u_{2}$ and $(\Delta^{j}u_{1})^{\prime}\geq(\Delta^{j}u_{2})^{\prime}$ on $(0,R)$ for every $j\in J$. Moreover, if $\Delta^{j}u_{1}(0)>\Delta^{j}u_{2}(0)$ for some $j\in J$ then $\Delta^{j}u_{1}>\Delta^{j}u_{2}$ and $(\Delta^{j}u_{1})^{\prime}>(\Delta^{j}u_{2})^{\prime}$ on $(0,R)$ for every $j\in J$.

With the help of above comparison lemma and the fact that $\Delta U_{0}(\infty)=0$ we prove Theorem 1.1.

Proof of Theorem 1.1 For $\rho\in(-U_{0}(0),\infty)$ we consider the initial value problem

[TABLE]

Then by ODE local existence theorem $u_{\rho}$ exists in a neighborhood of the origin. Moreover, for every $\rho>0$ we have $u_{\rho}>U_{0}$ on $(0,\infty)$, thanks to Lemma 2.1. In fact, $u_{\rho}(r)\geq\rho+U_{0}(r)$ on $(0,\infty)$, which implies that

[TABLE]

Since $0<u_{\rho}^{-6}\leq U_{0}^{-6}$ for $\rho\geq 0$, by dominated convergence theorem, we have that the map

[TABLE]

is continuous. Hence, for every $\Lambda\in(0,\Lambda^{*}]$ there exists a solution $u$ to (3) with $\Lambda=\Lambda_{u}$.

To prove the converse we essentially follow [4, 9, 17]. Let $u$ be a solution of (3) for some $\Lambda_{u}>0$. Then we have $\Delta u>0$ in $\mathbb{R}^{3}$ (see e.g. [4, Lemma 2.2]). We set $\bar{u}(x):=\lambda^{\frac{-1}{2}}u(\lambda x)$ where $\lambda>0$ is such that $\Delta\bar{u}(0)=\Delta U_{0}(0)$. Then we have $\Lambda_{u}=\Lambda_{\bar{u}}$, and $\bar{u}=u_{\rho}$ for some $\rho\in\mathbb{R}$, where $u_{\rho}$ is the solution to (12). We claim that $\rho\geq 0$. In order to prove the claim we assume by contradiction that $\rho<0$. Then, it follows from Lemma 2.1 that

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for some $\varepsilon>0$. Therefore, as $\Delta U_{0}(\infty)=0$, we have $\Delta u_{\rho}(r)\leq-\frac{\varepsilon}{2}$ on $(R,\infty)$ for some $R>>1$. In particular, from the identity

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for some $C_{\varepsilon}>0$ we obtain

[TABLE]

a contradiction as $u_{\rho}>0$ on $\mathbb{R}^{3}$. Thus $\rho\geq 0$, and hence by Lemma 2.1 we have $\bar{u}\geq U_{0}$ on $(0,\infty)$. This in turn implies that $\Lambda_{\bar{u}}\leq\Lambda^{*}$, and $\Lambda_{\bar{u}}=\Lambda^{*}$ if and only if $\bar{u}=U_{0}$. $\square$

We now move to the proof of Theorem 1.2. We start with the following lemma.

Lemma 2.2

For $k$ large and $\varepsilon\in(0,1)$ there exists a positive entire radial solution to

[TABLE]

Moreover, if $u$ is a positive entire radial solution to (19) for some $\varepsilon\in\mathbb{R}$ then necessarily $\varepsilon\leq\sqrt{\frac{6k}{5}}$, and the solution $u$ satisfies

[TABLE]

Proof.

It follows from the ODE local existence theorem that for every $\varepsilon>0$ there exists a unique positive solution to (19) in a neighborhood of the origin. We let $(0,\delta)$ to be the maximum interval of existence.

From the identity (13) we see that $\Delta^{2}u$ is strictly monotone decreasing on $(0,\delta)$. Let $\bar{\delta}\in(0,\delta]$ be the largest number such that

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Using (21) in (13) with $w=\Delta u$ one obtains

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Again by (13) with $w=u$ we obtain for $r\in(0,\bar{\delta})$

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for $k\geq k_{0}$ for some $k_{0}$ sufficiently large and for every $\varepsilon\in(0,1)$. We can also choose $k_{0}$ large enough so that

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Now we use (22) in (13) with $w=\Delta^{2}u$ to obtain a lower bound of $\Delta^{2}u$. Indeed, for $k\geq k_{0}$ and $r\in(0,\bar{\delta})$ we have

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Thus, from the definition of $\bar{\delta}$ we get that $\bar{\delta}=\delta$. In particular, (22) holds on $(0,\delta)$. This shows that $\delta=\infty$, and we conclude the first part of the lemma.

To prove (20) we let $u$ be a positive entire radial solution to (19) for some $\varepsilon\in\mathbb{R}$. It follows from (13) that $\Delta^{2}u$ is strictly monotone decreasing on $(0,\infty)$. Therefore, as $\Delta^{2}u>0$ in $\mathbb{R}^{3}$ (see e.g. [6, Lemma 2.2]), we get

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This implies that $\Delta u$ is monotone increasing on $(0,\infty)$, and a repeated use of (13) gives (20). Finally, the upper bound of $u$ in (20) and the positivity of $u$ implies that $\varepsilon\leq\sqrt{\frac{6k}{5}}$.

We conclude the lemma. ∎

As a consequence of the above lemma the number $\varepsilon_{k}^{*}$ given by (for $k$ large)

[TABLE]

exists, and it satisfies the estimate $\varepsilon_{k}^{*}\leq\sqrt{\frac{6k}{5}}$. Moreover, for every $\varepsilon\in(-\infty,\varepsilon_{k}^{*})$ there exists a positive entire solution to (19), thanks to Lemma 2.1.

Lemma 2.3

For $k$ large (19) has a positive entire solution with $\varepsilon=\varepsilon_{k}^{*}$.

Proof.

For simplicity we ignore the subscript $k$ and we write $\varepsilon^{*}$ instead of $\varepsilon_{k}^{*}$. Let $u$ be the solution to (19) with $\varepsilon=\varepsilon^{*}$, and let $(0,R)$ be the maximum interval of existence. We assume by contradiction that $R<\infty$. Then necessarily we have

[TABLE]

It follows from the definition of $\varepsilon^{*}$ that there exists a sequence of positive entire solutions $(u_{n})$ to (19) with $\Delta u_{n}(0)\downarrow-\varepsilon^{*}$. Then, from the continuous dependence of the solutions on the initial data, we have that $u_{n}\to u$ locally uniformly in $[0,R)$. In particular, there exists $x_{n}\to R$ such that $u_{n}(x_{n})\to 0$. We claim that there exists $C>0$ such that

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Indeed, as $0<\Delta^{2}u_{n}\leq 1$ on $(0,\infty)$, by (13) we obtain

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This gives $|u_{n}^{\prime}|\leq C$ on $(0,R+3)$ for some $C>0$, and hence we have (24). Therefore, by (13) and together with (24) we get

[TABLE]

a contradiction as $\Delta^{2}u_{n}>0$ on $(0,\infty)$.

We conclude the lemma. ∎

Lemma 2.4

Let $u$ be a positive entire radial solution to (5). Assume that $\Delta^{2}u(\infty)>0$. Then there exists a positive entire radial solution $v$ to (5) such that

[TABLE]

Proof.

For $\rho>0$ small we consider the initial value problem

[TABLE]

Since $\Delta^{2}u(\infty)>0$, it follows that $u(r)\geq\delta r^{4}$ at infinity for some $\delta>0$. Therefore, we can choose $\rho_{0}>0$ small so that

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We fix $R_{1}>>1$ such that

[TABLE]

where $\varepsilon>0$ will be chosen later. By continuous dependence on the initial data we can choose $\rho\in(0,\rho_{0})$ sufficiently small such that the solution $v=v(\rho,u)$ to (30) exists on $(0,R_{1})$ and it satisfies

[TABLE]

We claim that for such $\rho>0$ the solution $v$ exists entirely.

In order to prove the claim we let $R_{2}>0$ (possibly the largest one) be such that $v\geq\frac{u}{2}$ on $(0,R_{2})$. (Note that $v\leq u$ on the common interval of existence, and for $\varepsilon>0$ small enough we have $R_{2}>R_{1}$). Then for $0<r<R_{2}$ we have

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The above estimate and a repeated use of (13) leads to

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Now we fix $\varepsilon>0$ sufficiently small so that $C_{3}\varepsilon r^{4}<\frac{1}{6}u(r)$ on $(0,\infty)$. Then we have

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Thus, $v\geq\frac{u}{2}$ on $(0,R_{2})$ implies that $v\geq\frac{2}{3}u$ on $(0,R_{2})$, and hence $R_{2}=\infty$.

This finishes the proof of the claim. ∎

Proof of Theorem 1.2 Let $(u_{k})$ be a sequence of positive entire radial solutions to (19) with $\varepsilon=\varepsilon_{k}^{*}$ as given by Lemma 2.3. We claim that

[TABLE]

First we note that $\Delta^{2}u_{k}(\infty)=0$, that is

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which is a consequence of Lemma 2.4 and the definition of $\varepsilon_{k}^{*}$. Moreover,

[TABLE]

thanks to (20) and the estimate $\varepsilon_{k}^{*}\leq\sqrt{\frac{6k}{5}}$. Now we consider the following two cases, and we show that (31) holds in each case.

Case 1 $\min_{(0,\infty)}u_{k}\to\infty$.

Since $u_{k}\to\infty$ locally uniformly in $\mathbb{R}^{3}$, from (32) we obtain

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which gives (31).

Case 2 $\min_{(0,\infty)}u_{k}=:u_{k}(x_{k})\leq C$.

Since $u_{k}\to\infty$ locally uniformly in $\mathbb{R}^{3}$, we have $x_{k}\to\infty$. We claim that

[TABLE]

In order to prove the claim we note that $u_{k}^{\prime}\geq 0$ on $[x_{k},\infty)$ and $u_{k}^{\prime}(x_{k})=0$. Moreover, as $\Delta^{2}u_{k}\leq\Delta^{2}u_{k}(0)=1$, by (13) we have

[TABLE]

Hence, $u_{k}^{\prime\prime}(x_{k}+r)\leq\frac{1}{6}(x_{k}+r)^{2}$, and by a Taylor expansion, we have our claim. Therefore, as $x_{k}\to\infty$, we get

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This proves (31).

Theorem 1.2 follows immediately as the integral in (31) depends continuously on the initial data, and for every fixed $k$ (large)

[TABLE]

where $u_{\rho,k}$ is the entire positive solution to (19) with $\Delta u_{\rho,k}(0)=\rho>-\varepsilon_{k}^{*}$ . $\square$