Uniqueness and nondegeneracy of ground states to nonlinear scalar field equations involving the Sobolev critical exponent in their nonlinearities for high frequencies

Abstract

The study of the uniqueness and nondegeneracy of ground state solutions to semilinear elliptic equations is of great importance because of the resulting energy landscape and its implications for the various dynamics. In Akahori et al. (Global dynamics above the ground state energy for the combined power-type nonlinear Schrödinger equation with energy-critical growth at low frequencies, preprint), semilinear elliptic equations with combined power-type nonlinearities involving the Sobolev critical exponent are studied. There, it is shown that if the dimension is four or higher, and the frequency is sufficiently small, then the positive radial ground state is unique and nondegenerate. In this paper, we extend these results to the case of high frequencies when the dimension is five and higher. After suitably rescaling the equation, we demonstrate that the main behavior of the solutions is given by the Sobolev critical part for which the ground states are explicit, and their degeneracy is well characterized. Our result is a key step towards the study of the different dynamics of solutions of the corresponding nonlinear Schrödinger and Klein–Gordon equations with energies above the energy of the ground state. Our restriction on the dimension is mainly due to the existence of resonances in dimension three and four.

Appendices

Existence of ground state

In this section, we sketch the proof of Proposition 1.1. Since we restrict nonlinearities to combined power-type ones, the proof is much simpler than the general case dealt with in [38]. In particular, we can use a positive functional ${\mathcal {I}}_{\omega }$ given by

$$\begin{aligned} {\mathcal {I}}_{\omega }(u)&:= {\mathcal {S}}_{\omega }(u)-\frac{1}{p+1}{\mathcal {N}}_{\omega }(u) \nonumber \\&= \frac{p-1}{2(p+1)} \Big \{ \Vert \nabla u \Vert _{L^{2}}^{2} + \omega \Vert u \Vert _{L^{2}}^{2} \Big \} + \frac{4-(d-2)(p-1)}{2d(p+1)} \Vert u \Vert _{L^{2^{*}}}^{2^{*}}. \end{aligned}$$

(A.1)

Moreover, we easily verify the following structures of ${\mathcal {S}}_{\omega }$ and ${\mathcal {N}}_{\omega }$ (cf. [37, Chapter 4]):

• For any $u \in H^{1}({\mathbb {R}}^{d})\setminus \{0\}$, there exists a unique $\lambda (u)>0$ such that

  $$\begin{aligned} {\mathcal {N}}_{\omega }(\lambda u) \left\{ \begin{array}{rcl}>0 &{} \text{ if }&{} 0< \lambda<\lambda (u), \\ =0 &{} \text{ if }&{} \lambda =\lambda (u), \\ <0 &{} \text{ if }&{} \lambda >\lambda (u). \end{array} \right. \end{aligned}$$

  (A.2)

• For any $u \in H^{1}({\mathbb {R}}^{d})\setminus \{0\}$,

  $$\begin{aligned} \text {the function }\lambda \mapsto {\mathcal {I}}_\omega (\lambda u)\text { is non-decreasing in }[0,\infty ). \end{aligned}$$

  (A.3)

Next, we introduce several variational values:

$$\begin{aligned} \sigma&:= \inf \{ \Vert \nabla u \Vert _{L^{2}}^{2} : u \in \dot{H}^{1}({\mathbb {R}}^{d}) \text{ with } \Vert u\Vert _{L^{2^{*}}}=1\}, \end{aligned}$$

(A.4)

$$\begin{aligned} m_{\omega }&:=\inf \{ {\mathcal {S}}_{\omega }(u) : u\in H^{1}({\mathbb {R}}^{d})\setminus \{0\} \text{ with } {\mathcal {N}}_{\omega }(u)=0 \}, \nonumber \\ {\widetilde{m}}_{\omega }&:=\inf \big \{{\mathcal {I}}_{\omega }(u) : u\in H^{1}({\mathbb {R}}^{d})\setminus \{0\} \text{ with } {\mathcal {N}}_{\omega }(u)\le 0 \big \} . \end{aligned}$$

(A.5)

By a standard argument (cf. [37, Chapter 4]), it is known that a minimizer for $m_\omega$ becomes a ground state to (1.1). Hence, in order to prove Proposition 1.1, it suffices to show the existence of minimizer for $m_\omega$.

We first state the relationship between $m_\omega$ and $\widetilde{m}_\omega$ (cf. [1, Proposition 1.2]):

Lemma A.1

Assume $d\ge 3$ and $1<p<\frac{d+2}{d-2}$. Then, for any $\omega >0$, we have the following:

(i) :

$m_{\omega }={\widetilde{m}}_{\omega }>0$

(ii) :

Any minimizer for ${\widetilde{m}}_{\omega }$ is also a minimizer for $m_{\omega }$, and vice versa.

Proof

We shall prove claim (i). Since ${\mathcal {I}}_\omega (u) = {\mathcal {S}}_\omega (u)$ for every $u \in H^1({\mathbb {R}}^d)$ with ${\mathcal {N}}_\omega (u) = 0$, it is clear that $\widetilde{m}_\omega \le m_\omega$. For the opposite inequality $m_\omega \le \widetilde{m}_\omega$, fix any $u \in H^1({\mathbb {R}}^d) \setminus \{0\}$ with ${\mathcal {N}}_\omega (u) \le 0$. By (A.2), there exists a $\lambda \in (0,1]$ such that ${\mathcal {N}}_\omega (\lambda u) = 0$. By (A.3),

$$\begin{aligned} m_\omega \le {\mathcal {S}}_\omega (\lambda u) = {\mathcal {I}}_\omega (\lambda u) \le {\mathcal {I}}_\omega (u), \end{aligned}$$

which yields $m_\omega \le \widetilde{m}_\omega$. Thus, $m_\omega = \widetilde{m}_\omega$. It remains to prove that $\widetilde{m}_\omega >0$. Let $u \in H^{1}({\mathbb {R}}^{d})\setminus \{0\}$ with ${\mathcal {N}}_{\omega }(u)\le 0$. Then, it follows from ${\mathcal {N}}_{\omega }(u)\le 0$ and Sobolev’s inequality that

$$\begin{aligned} \min \{1, \omega \} \Vert u\Vert _{H^{1}}^{2} \lesssim \Vert u\Vert _{H^{1}}^{p+1}+\Vert u\Vert _{H^{1}}^{2^{*}} . \end{aligned}$$

This implies that there exists a constant $c(\omega )>0$ such that $\Vert u\Vert _{H^{1}}^{2}\ge c(\omega )$ and therefore ${\mathcal {I}}_\omega (u) \gtrsim \min \{1,\omega \}c(\omega )$. Since u is arbitrary, we find that $\widetilde{m}_\omega >0$.

Next, we shall prove claim (ii). Since ${\mathcal {I}}_{\omega }={\mathcal {S}}_{\omega }-\frac{1}{p+1}{\mathcal {N}}_{\omega }$ and $m_{\omega }={\widetilde{m}}_{\omega }$, it suffices to prove that ${\mathcal {N}}_{\omega }({\widetilde{Q}}_{\omega })=0$ for all minimizer ${\widetilde{Q}}_{\omega }$ for ${\widetilde{m}}_{\omega }$. Suppose the contrary that there exists a minimizer ${\widetilde{Q}}_{\omega }$ for ${\widetilde{m}}_{\omega }$ such taht ${\mathcal {N}}_{\omega }({\widetilde{Q}}_{\omega })<0$. Then, it follows from (A.2) that there exists a unique $\lambda _{0}\in (0,1)$ such that ${\mathcal {N}}_{\omega }(\lambda _{0}{\widetilde{Q}}_{\omega })=0$. Furthermore, we have

$$\begin{aligned} {\widetilde{m}}_{\omega }\le {\mathcal {I}}_{\omega }(\lambda _{0}{\widetilde{Q}}_{\omega }) < {\mathcal {I}}_{\omega }({\widetilde{Q}}_{\omega })={\widetilde{m}}_{\omega }, \end{aligned}$$

which is a contradiction. Thus, ${\mathcal {N}}_{\omega }({\widetilde{Q}}_{\omega })=0$. $\square$

Next, we state a key inequality to show the existence of minimizer for $m_\omega$ (cf. [38, Lemma 2.2]):

Lemma A.2

Assume that $d\ge 3$ and $3< p<5$, or $d\ge 4$ and $1<p<\frac{d+2}{d-2}$. Then, the following estimate holds

$$\begin{aligned} m_{\omega }< \frac{1}{d}\sigma ^{\frac{d}{2}} = \frac{1}{2} \Vert \nabla W \Vert _{L^2}^2 - \frac{1}{2^*} \Vert W \Vert _{L^{2^*}}^{2^*} = \frac{1}{d} \Vert \nabla W \Vert _{L^2}^2 . \end{aligned}$$

(A.6)

Proof

Let $\chi$ be an even smooth function on ${\mathbb {R}}$ such that $\chi (r)=1$ for $0\le r\le 1$, $\chi (r)=0$ for $r\ge 2$, and $\chi$ is non-increasing on $[0,\infty )$. Then, we define

$$\begin{aligned} W_\varepsilon (x) := \varepsilon ^{-\frac{d-2}{4}}W \left( \frac{x}{\sqrt{\varepsilon }} \right) = \varepsilon ^{\frac{d-2}{4}}\left( \varepsilon +\frac{|x|^{2}}{d(d-2)}\right) ^{-\frac{d-2}{2}}, \quad V_{\varepsilon }(x) := \chi (|x|)W_{\varepsilon }(x). \end{aligned}$$

Then, we can verify that $\sigma ^{\frac{d}{2}} = \Vert \nabla W_\varepsilon \Vert _{L^2}^2 = \Vert W_\varepsilon \Vert _{L^{2^*}}^{2^*}$ and

$$\begin{aligned} \Vert \nabla V_{\varepsilon }\Vert _{L^{2}}^{2}&= \sigma ^{\frac{d}{2}}+O(\varepsilon ^{\frac{d-2}{2}}) , \end{aligned}$$

(A.7)

$$\begin{aligned} \Vert V_{\varepsilon } \Vert _{L^{2^{*}}}^{2^{*}}&= \sigma ^{\frac{d}{2}}+O(\varepsilon ^{\frac{d}{2}}) . \end{aligned}$$

(A.8)

Moreover, we find that

$$\begin{aligned} \Vert V_{\varepsilon }\Vert _{L^{q+1}}^{q+1} = \left\{ \begin{array}{rcl} O (\varepsilon ^{\frac{2d-(d-2)(q+1)}{4}}) &{}\quad \text{ if }&{} \frac{2}{d-2}< q, \\ O(\varepsilon ^{\frac{d}{4}}|\log {\varepsilon }|) &{}\quad \text{ if }&{} q=\frac{2}{d-2}, \\ O(\varepsilon ^{\frac{(d-2)(q+1)}{4}}) &{}\quad \text{ if }&{} 0< q< \frac{2}{d-2} . \end{array} \right. \end{aligned}$$

(A.9)

Next, for a given $\varepsilon \in (0,1)$, we introduce a function $y_{\varepsilon } :(0,\infty ) \rightarrow {\mathbb {R}}$ as

$$\begin{aligned} y_{\varepsilon }(t):=\frac{1}{2}t^{2} \big \{ \Vert \nabla V_{\varepsilon }\Vert _{L^{2}}^{2} + \omega \Vert V_{\varepsilon }\Vert _{L^{2}}^{2} \big \} -\frac{t^{2^{*}}}{2^{*}} \Vert V_{\varepsilon }\Vert _{L^{2^{*}}}^{2^{*}} . \end{aligned}$$

It is easy to verify that the function $y_{\varepsilon }$ attains its maximum only at the point

$$\begin{aligned} \tau _{\varepsilon , \max }:=\frac{\big \{ \Vert \nabla V_{\varepsilon }\Vert _{L^{2}}^{2} + \omega \Vert V_{\varepsilon }\Vert _{L^{2}}^{2} \big \}^{\frac{d-2}{4}} }{ \Vert V_{\varepsilon }\Vert _{L^{2^{*}}}^{\frac{d}{2}} }. \end{aligned}$$

It follows from the definition of $\sigma$ [see (A.4)], (A.7) and (A.8) that

$$\begin{aligned} \frac{\Vert \nabla V_{\varepsilon }\Vert _{L^{2}}^{2}}{\Vert V_{\varepsilon }\Vert _{L^{2^{*}}}^{2}} = \frac{\sigma ^{\frac{d}{2}} + O( \varepsilon ^{ \frac{d-2}{2}} )}{\sigma ^{\frac{d-2}{2}}+O(\varepsilon ^{\frac{d}{2}})} = \sigma + O(\varepsilon ^{\frac{d-2}{2}}). \end{aligned}$$

Moreover, we see from (A.7) and (A.9) that

$$\begin{aligned} \frac{\Vert V_{\varepsilon }\Vert _{L^{2}}^{2}}{\Vert \nabla V_{\varepsilon }\Vert _{L^{2}}^{2}} = \left\{ \begin{array}{lll} O(\varepsilon ^{\frac{1}{2}}) &{}\quad \text{ if }&{} d=3, \\ O( \varepsilon |\log {\varepsilon }| ) &{}\quad \text{ if }&{} d=4, \\ O( \varepsilon ) &{}\quad \text{ if }&{} d\ge 5. \end{array}\right. \end{aligned}$$

Hence, we find that

$$\begin{aligned} \begin{aligned} y(\tau _{\varepsilon ,\max })&= \frac{1}{d} \frac{\Vert \nabla V_{\varepsilon }\Vert _{L^{2}}^{d} }{\Vert V_{\varepsilon }\Vert _{L^{2^{*}}}^{d}} \bigg ( 1+ \omega \frac{\Vert V_{\varepsilon }\Vert _{L^{2}}^{2}}{ \Vert \nabla V_{\varepsilon }\Vert _{L^{2}}^{2}} \bigg )^{\frac{d}{2}} \\&= \left\{ \begin{array}{lll} \frac{1}{d}\sigma ^{\frac{d}{2}} + O(\varepsilon ^{\frac{1}{2}}) &{}\quad \text{ if }&{} d=3, \\ \frac{1}{d}\sigma ^{\frac{d}{2}} + O(\varepsilon |\log {\varepsilon }|) &{}\quad \text{ if }&{} d=4, \\ \frac{1}{d}\sigma ^{\frac{d}{2}} + O(\varepsilon ) &{}\quad \text{ if }&{} d\ge 5, \end{array} \right. \end{aligned} \end{aligned}$$

(A.10)

On the other hand, for each $\varepsilon \in (0,1)$, there exists $\tau _{\varepsilon ,0}>0$ such that ${\mathcal {N}}_{\omega }(\tau _{\varepsilon ,0}V_{\varepsilon })=0$.

Now, we assume that $d=3$ and $3< p< 5$. Then, it follows from (A.7), (A.8) and (A.9) that

$$\begin{aligned} \begin{aligned} 0&={\mathcal {N}}_{\omega }(\tau _{\varepsilon ,0}V_{\varepsilon }) \\&= \omega \tau _{\varepsilon ,0}^{2}O(\varepsilon ^{\frac{1}{2}}) + \tau _{\varepsilon ,0}^{2} \{\sigma ^{\frac{3}{2}}+O(\varepsilon ^{\frac{1}{2}})\} - \tau _{\varepsilon ,0}^{p+1} O(\varepsilon ^{\frac{5-p}{4}}) - \tau _{\varepsilon ,0}^{6} \{\sigma ^{\frac{3}{2}}+O(\varepsilon ^{\frac{3}{2}})\}. \end{aligned} \end{aligned}$$

Divide both sides above by $\tau _{\varepsilon ,0}^{2}\{\sigma ^{\frac{3}{2}}+O(\varepsilon ^{\frac{3}{2}})\}$. Then, we obtain

$$\begin{aligned} \tau _{\varepsilon ,0}^{4} = \omega O(\varepsilon ^{\frac{1}{2}}) + 1+O(\varepsilon ^{\frac{1}{2}}) - \tau _{\varepsilon ,0}^{p-1}O(\varepsilon ^{\frac{5-p}{4}}) . \end{aligned}$$

Since $p-1 < 4$, this implies that for any $\omega >0$,

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\tau _{\varepsilon ,0} \ge \frac{1}{2}. \end{aligned}$$

(A.11)

Furthermore, it follows from the definition of $m_{\omega }$ [see (A.5)], (A.9), (A.10), (A.11) and $2< p <2^{*}-1$ that

$$\begin{aligned} \begin{aligned} m_{\omega }&\le {\mathcal {S}}_{\omega }(\tau _{\varepsilon ,0} V_{\varepsilon }) = y_{\varepsilon }(\tau _{\varepsilon ,0}) - \frac{\tau _{\varepsilon ,0}^{p+1}}{p+1} \Vert V_{\varepsilon }\Vert _{L^{p+1}}^{p+1} \\&\le y_{\varepsilon }(\tau _{\varepsilon ,\max }) - \frac{\tau _{\varepsilon ,0}^{p+1}}{p+1} c_{1} \varepsilon ^{\frac{5-p}{4}} = \frac{1}{3}\sigma ^{\frac{3}{2}} + O(\varepsilon ^{\frac{1}{2}}) -c_{2} \varepsilon ^{\frac{5-p}{4}} \end{aligned} \end{aligned}$$

for some positive constants $c_{1}$ and $c_{2}$ depending only on p. Thus, we find that if $p>3$ and $\varepsilon$ is sufficiently small depending only on p and $\omega$, then

$$\begin{aligned} m_{\omega } < \frac{1}{3}\sigma ^{\frac{3}{2}}. \end{aligned}$$

Similarly, we can prove that if $d\ge 4$, then claim (A.6) is true. $\square$

Now, we are ready to prove Proposition 1.1.

Proof of Proposition 1.1

By Lemma A.1, it suffices to prove the existence of minimizer for ${\widetilde{m}}_{\omega }$. To this end, we consider a minimizing sequence $\{u_{n}\}$ for ${\widetilde{m}}_{\omega }$. We denote the Schwarz symmetrization of $u_{n}$ by $u_{n}^{*}$. Note that $\Vert \nabla u_n^*\Vert _{L^2} \le \Vert \nabla u_n \Vert _{L^2}$ and $\Vert u_n^*\Vert _{L^q} = \Vert u_n \Vert _{L^q}$ hold for each $q \in [2,2^*]$. For example, see [24]. From these properties, we have

$$\begin{aligned}&{\mathcal {N}}_{\omega }(u_{n}^{*})\le 0 \quad \text{ for } \text{ any } n \ge 1, \end{aligned}$$

(A.12)

$$\begin{aligned}&\lim _{n\rightarrow \infty }{\mathcal {I}}_{\omega }(u_{n}^{*})={\widetilde{m}}_{\omega }, \\ \nonumber&\left\| u_{n}^{*} \right\| _{H^{1}} < \infty \quad \text{ for } \text{ any } n \ge 1. \end{aligned}$$

(A.13)

Since $\{u_{n}^{*}\}$ is radially symmetric and bounded in $H^{1}({\mathbb {R}}^{d})$, there exists a radially symmetric function $Q \in H^{1}({\mathbb {R}}^{d})$ such that, passing to some subsequence,

$$\begin{aligned} \lim _{n\rightarrow \infty }u_{n}^{*}=Q \quad \text{ weakly } \text{ in } H^{1}({\mathbb {R}}^{d}) \text{ and } \text{ strongly } \text{ in } L^{p+1}({\mathbb {R}}^d). \end{aligned}$$

(A.14)

We shall show that Q becomes a minimizer for $m_{\omega }$.

We first show $Q \not \equiv 0$. Suppose the contrary that $Q \equiv 0$. Then, it follows from (A.12) and (A.14) that, passing to some subsequence,

$$\begin{aligned} 0\ge \lim _{n\rightarrow \infty }{\mathcal {N}}_{\omega }(u_{n}^{*}) \ge \lim _{n\rightarrow \infty }\left\{ \left\| \nabla u_{n}^{*} \right\| _{L^{2}}^{2} -\left\| u_{n}^{*} \right\| _{L^{2^{*}}}^{2^{*}} \right\} . \end{aligned}$$

(A.15)

If $\Vert \nabla u_n^* \Vert _{L^2} \rightarrow 0$, then $\Vert u_n^* \Vert _{L^q} \rightarrow 0$ for all $2<q \le 2^*$. By (A.12) and (A.13), one has $\Vert u_n^* \Vert _{L^2} \rightarrow 0$ and ${\widetilde{m}}_{\omega } = 0$. However, this contradicts ${\widetilde{m}}_{\omega } > 0$ (see Lemma A.1). Therefore, we may assume $\lim _{n\rightarrow \infty } \Vert \nabla u_n^* \Vert _{L^2} > 0$.

Now, (A.15) with the definition of $\sigma$ gives us

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\| \nabla u_{n}^{*} \right\| _{L^{2}}^{2} \ge \sigma \lim _{n\rightarrow \infty }\left\| u_{n}^{*} \right\| _{L^{2^{*}}}^{2} \ge \sigma \lim _{n\rightarrow \infty } \left\| \nabla u_{n}^{*} \right\| _{L^{2}}^{\frac{2(d-2)}{d}}. \end{aligned}$$

From $\lim _{n\rightarrow \infty } \Vert \nabla u_n^* \Vert _{L^2} >0$, it follows that

$$\begin{aligned} \sigma ^{\frac{d}{2}} \le \lim _{n\rightarrow \infty }\left\| \nabla u_{n}^{*} \right\| _{L^{2}}^2. \end{aligned}$$

(A.16)

Hence, we see from (A.1), (A.13), (A.15) and (A.16) that

$$\begin{aligned} \begin{aligned} {\widetilde{m}}_{\omega }&= \lim _{n\rightarrow \infty }{\mathcal {I}}_{\omega }(u_{n}^{*}) \\&\ge \lim _{n\rightarrow \infty } \left\{ \frac{p-1}{2(p+1)}\left\| \nabla u_{n}^{*} \right\| _{L^{2}}^{2} + \frac{4-(d-2)(p-1)}{2d(p+1)}\left\| u_{n}^{*} \right\| _{L^{2^{*}}}^{2^{*}} \right\} \\&\ge \frac{1}{d}\lim _{n\rightarrow \infty }\left\| \nabla u_{n}^{*} \right\| _{L^{2}}^{2} \ge \frac{1}{d}\sigma ^{\frac{d}{2}}. \end{aligned} \end{aligned}$$

However, this contradicts (A.6). Thus, $Q \not \equiv 0$.

Next, we shall show that ${\mathcal {N}}_{\omega }(Q)=0$. Using the Brezis-Lieb Lemma [7], we have

$$\begin{aligned} {\mathcal {I}}_{\omega }(u_{n}^{*}) - {\mathcal {I}}_{\omega }(u_{n}^{*}-Q) - {\mathcal {I}}_{\omega }(Q)&=o_{n}(1), \end{aligned}$$

(A.17)

$$\begin{aligned} {\mathcal {N}}_{\omega }(u_{n}^{*})-{\mathcal {N}}_{\omega }(u_{n}^{*}-Q) -{\mathcal {N}}_{\omega }(Q)&=o_{n}(1) . \end{aligned}$$

(A.18)

Furthermore, (A.17) together with (A.13) and the positivity of ${\mathcal {I}}_{\omega }$ implies that

$$\begin{aligned} {\mathcal {I}}_{\omega }(Q)\le {\widetilde{m}}_{\omega }. \end{aligned}$$

(A.19)

Let us suppose ${\mathcal {N}}_{\omega }(Q)<0$ and derive a contradiction. Note that (A.19) implies that ${\mathcal {I}}_{\omega }(Q)={\widetilde{m}}_{\omega }$. Moreover, it follows from (A.2) that there exists a unique $\lambda _{0}\in (0,1)$ such that ${\mathcal {N}}_{\omega }(\lambda _{0}Q)=0$. Hence, we have

$$\begin{aligned} {\widetilde{m}}_{\omega }\le {\mathcal {I}}_{\omega }(\lambda _{0}Q)<{\mathcal {I}}_{\omega }(Q)={\widetilde{m}}_{\omega }. \end{aligned}$$

This is a contradiction.

Next, suppose that ${\mathcal {N}}_{\omega }(Q)>0$. Then, it follows from (A.12) and (A.18) that ${\mathcal {N}}_{\omega }(u_{n}^{*}-Q)<0$ for any sufficiently large n. Hence, we can take $\lambda _{n}\in (0,1)$ such that ${\mathcal {N}}_{\omega }(\lambda _{n}(u_{n}^{*}-Q))=0$. Furthermore, we see from (A.13) and (A.17) that

$$\begin{aligned} \begin{aligned} {\widetilde{m}}_{\omega }&\le {\mathcal {I}}_{\omega }( \lambda _{n} (u_{n}^{*}-Q)) \le {\mathcal {I}}_{\omega }(u_{n}^{*}-Q) ={\mathcal {I}}_{\omega }(u_{n}^{*}) - {\mathcal {I}}_{\omega }(Q)+o_{n}(1) \\&= {\widetilde{m}}_{\omega } - {\mathcal {I}}_{\omega }(Q)+o_{n}(1). \end{aligned} \end{aligned}$$

Hence, we conclude that ${\mathcal {I}}_{\omega }(Q)=0$ and $Q \equiv 0$. However, this is a contradiction. Thus ${\mathcal {N}}_{\omega }(Q)=0$.

Since $Q \not \equiv 0$ and ${\mathcal {N}}_{\omega }(Q)=0$, we have

$$\begin{aligned} m_{\omega } \le {\mathcal {S}}_{\omega }(Q)={\mathcal {I}}_{\omega }(Q). \end{aligned}$$

(A.20)

Moreover, it follows from (A.17) and Proposition A.1 that

$$\begin{aligned} {\mathcal {I}}_{\omega }(Q) \le \liminf _{n\rightarrow \infty }{\mathcal {I}}_{\omega }(u_{n}^{*}) \le {\widetilde{m}}_{\omega }=m_{\omega }. \end{aligned}$$

(A.21)

Combining (A.20) and (A.21), we obtain ${\mathcal {S}}_{\omega }(Q)={\mathcal {I}}_{\omega }(Q)=m_{\omega }$. Thus, we have proved that Q is a minimizer for $m_{\omega }$. $\square$

The Moser iteration

Here we state a result used in Sects. 3 and 4 to obtain the uniform decay estimates.

Proposition B.1

Assume $d\ge 3$. Let a(x) and b(x) be functions on $B_{4}$, and let $u \in H^1(B_4)$ be a weak solution to

$$\begin{aligned} -\Delta u + a(x) u = b (x) u \quad \mathrm{in} \ B_4. \end{aligned}$$

Suppose that a(x) and u satisfy that

$$\begin{aligned} a(x) \ge 0 \quad \mathrm{for\ a.a.}\ x \in B_4, \quad \int _{B_4} a(x) |u(x)v(x)| d x < \infty \quad \mathrm{for\ each\ } v \in H^1_0(B_4). \end{aligned}$$

(i) :

Assume that for any $\varepsilon \in (0,1)$, there exists $t_{\varepsilon }>0$ such that

$$\begin{aligned} \left\| \chi _{[|b|>t_{\varepsilon }]}b \right\| _{L^{d/2}(B_4)} \le \varepsilon \end{aligned}$$

where $[|b|>t] := \left\{ x \in B_4 : |b(x)| > t \right\}$, and $\chi _A(x)$ denotes the characteristic function of $A \subset {\mathbb {R}}^d$. Then, for any $q \in (0,\infty )$, there exists a constant $C(d,q, t_{\varepsilon })$ such that

$$\begin{aligned} \Vert |u|^{q+1} \Vert _{H^1(B_1)} \le C(d,q,t_{\varepsilon }) \Vert u \Vert _{L^{2^*}(B_4)}. \end{aligned}$$

(ii) :

Let $s>d/2$ and assume that $b \in L^{s}(B_{4})$. Then, there exists a constant $C(d,s, \Vert b\Vert _{L^{s}(B_{4})})$ such that

$$\begin{aligned} \Vert u \Vert _{L^\infty (B_1)} \le C \left( d,s, \Vert b\Vert _{L^{s}(B_{4})} \right) \Vert u \Vert _{L^{2^*}(B_4)}. \end{aligned}$$

Here, the constants $C(d,q,t_{\varepsilon })$ and $C(d,s, \Vert b\Vert _{L^{s}(B_{4})})$ in (i) and (ii) remain bounded as long as q, $t_{\varepsilon }$ and $\Vert b\Vert _{L^{s}(B_{4})}$ are bounded.

By the assumption of Proposition B.1, notice that

$$\begin{aligned} \int _{B_4} a (x) \varphi u^2 dx \ge 0 \quad \text {for all } u \in H^1(B_4) \end{aligned}$$

where $\varphi \in C^\infty _0(B_4)$ with $\varphi \ge 0$. Using this fact and arguing as in [25, Proof of Proposition 2.2] and [16] (cf. [6]), we may prove Proposition B.1. Therefore, we omit the details of the proof.

The Pucci–Serrin condition

In this section, we give the range of space dimension d and the subcritical power p for which [35, Theorem 1] is applicable to the case of Eq. (1.1).

Proposition C.1

Let $3 \le d \le 6$ and assume $\frac{4}{d-2} \le p < \frac{d+2}{d-2}$ with $1<p$. Then, for any $\omega >0$, the Eq. (1.1) admits at most one positive radial solution.

Proof

In order to apply [35, Theorem 1], what we need to check is [35, (2.5)]. In our case, this condition becomes

$$\begin{aligned} \frac{d}{du} \left[ \frac{F(u)}{f(u)} \right] \ge \frac{d-2}{2d} \quad \mathrm{for\ } u > 0, \ u \ne a \end{aligned}$$

(C.1)

where

$$\begin{aligned} f(u) : = - \omega u + u^p + u^q, \quad F(u) := - \frac{\omega }{2} u^2 + \frac{u^{p+1}}{p+1} + \frac{u^{q+1}}{q+1}, \quad q := \frac{d+2}{d-2}, \quad f(a) = 0. \end{aligned}$$

We first rewrite (C.1). Since

$$\begin{aligned} \frac{d}{du}\left[ \frac{F(u)}{f(u)} \right] = 1 - \frac{F(u) f'(u) }{ (f(u))^2 }, \end{aligned}$$

(C.1) is equivalent to

$$\begin{aligned} 0 \le q (f(u))^2 - (q+1) F(u) f'(u) =:g(u) \quad \mathrm{for\ all}\ u > 0. \end{aligned}$$

(C.2)

Next, we expand g(u) as follows:

$$\begin{aligned} g(u) = A_{2} \omega ^{2} u^{2} + A_{p+1} \omega u^{p+1} + A_{q+1} \omega u^{q+1} + A_{2p} u^{2p} + A_{p+q} u^{p+q} \end{aligned}$$

where

$$\begin{aligned} A_{2}&:=\frac{q-1}{2}, \nonumber \\ A_{p+1}&:=\frac{(p^{2}-3p-2)q+p^{2}+p+2}{2(p+1)}, \nonumber \\ A_{q+1}&:=\frac{(q-1)(q-2)}{2}, \nonumber \\ A_{2p}&:=\frac{q-p}{p+1}, \nonumber \\ A_{p+q}&:=\frac{(q-p)(p+1-q)}{p+1}. \end{aligned}$$

(C.3)

We remark that our assumption yields $q - 1 \le p < q$ and $2 \le q$. Hence, it is easily seen that

$$\begin{aligned} A_2>0, \ A_{q+1} \ge 0, \ A_{2p} > 0, \ A_{p+q} \ge 0. \end{aligned}$$

(C.4)

To show (C.2), we divide the arguments into two cases:

Case 1. $d=3$.

In this case, we have $q = 5$ and $4 \le p < 5$, which implies

$$\begin{aligned} 2 (p+1) A_{p+1} = 6p^2 - 14 p - 8 > 0 \quad \text {for all } 4 \le p < 5. \end{aligned}$$

By (C.4), (C.2) holds.

Case 2. $d=4, 5, 6$.

In this case, we rewrite g(u) as follows:

$$\begin{aligned} g(u) = \omega ^2 u^2 \left\{ A_2 + A_{p+1} \frac{u^{p-1}}{\omega } + A_{2p} \left( \frac{u^{p-1}}{\omega } \right) ^2 \right\} + A_{q+1} \omega u^{q+1} + A_{p+q} u^{p+q}. \end{aligned}$$

By (C.4), it suffices to show

$$\begin{aligned} Q(r) := A_2 + A_{p+1} r + A_{2p} r^2 \ge 0 \quad \text {for each }r \ge 0 \text { and }q-1 \le p < q. \end{aligned}$$

When $d=4$, one has $q = 3$ and

$$\begin{aligned} Q(r) = 1 + \frac{2(p^2-2p-1)}{p+1} r + \frac{3-p}{p+1} r^2. \end{aligned}$$

If $p^{2}-2p-1 \ge 0$, then $Q(r)> 0$ for all $r\ge 0$. On the other hand, if $p^{2}-2p-1 < 0$, then we obtain $2 \le p < 1 + \sqrt{2} = :p_0$ and simple computations give

$$\begin{aligned} \min _{r\ge 0}{Q(r)}= 1 - \frac{(p^2-2p-1)^2}{(p+1)(3-p)}. \end{aligned}$$

(C.5)

Set

$$\begin{aligned} h(p) := (p+1)(3-p) - (p^2 - 2p - 1)^2 = -p^4 + 4p^3 - 3p^2 -2 p +2. \end{aligned}$$

Note that

$$\begin{aligned} h'(p) = -4p^3 + 12p^2 - 6p -2, \quad h''(p) = -12p^2 + 24 p - 6 < 0 \quad \mathrm{in} \ [2,p_0]. \end{aligned}$$

We also observe that

$$\begin{aligned} h'(2) = 2>0, \quad h'(p_0) = -2p_0 + 2 < 0, \quad h(2) = 2 = h(p_0) > 0. \end{aligned}$$

Hence, $h(p) > 0$ in $[2,p_0]$ and by (C.5), we have $\min _{r \ge 0} Q(r) \ge 0$ and (C.2).

When $d=5$, we see

$$\begin{aligned} Q(r) = \frac{1}{3} \left\{ 2 + \frac{5p^2 - 9p - 4}{p+1} r + \frac{7-3p}{p+1} r^2 \right\} . \end{aligned}$$

Remark that $5p^2 -9p - 4 < 0$ is equivalent to $9 - \sqrt{161}< 10 p < 9 + \sqrt{161}$ and also that $\frac{4}{3}< \frac{9+\sqrt{161}}{10} < \frac{7}{3}$. Hence, if $\frac{9+\sqrt{161}}{10} \le p < \frac{7}{3}$, then $Q(r) \ge 0$ for all $r \ge 0$.

When $\frac{4}{3} \le p < \frac{9+\sqrt{161}}{10}$, observe that

$$\begin{aligned} \min _{r \ge 0} Q(r) = \frac{1}{3} \left\{ 2 - \frac{(5p^2 - 9p - 4)^2}{4(7-3p)(p+1)} \right\} . \end{aligned}$$

Since

$$\begin{aligned} \begin{aligned} 8(7-3p)(p+1) - (5p^2-9p-4)^2&= - 25p^4 + 90 p^3 - 65p^2 - 40p + 40 \\&=(p-1)^2 ( -25p^2 +40p + 40 ) =: (p-1)^2 h(p), \end{aligned} \end{aligned}$$

by $\frac{9+\sqrt{161}}{10} < \frac{9+13}{10} = \frac{11}{5}$ and $h(\frac{11}{5}) = 7 > 0$, we obtain $h(p) \ge 0$ for every $p \in [ \frac{4}{3} , \frac{9+\sqrt{161}}{10} ]$. Thus, $\min _{r\ge 0} Q(r) \ge 0$ and (C.2) holds.

When $d=6$, we observe

$$\begin{aligned} Q(r) = \frac{1}{2} \left\{ 1 - \frac{(3p+1)(2-p)}{p+1} r + \frac{2(2-p)}{p+1} r^2 \right\} , \ \min _{r \ge 0} Q(r) = \frac{1}{2} \left\{ 1 - \frac{(2-p)(3p+1)^2}{8(p+1)} \right\} . \end{aligned}$$

Setting $h(p) := 8(p+1) - (2-p) (3p+1)^2$ for $1 \le p < 2$, we obtain

$$\begin{aligned} h(1) = 0, \quad h'(p) = 27p^2 - 24 p -3 \ge 0 \quad \text {for each } 1 \le p <2. \end{aligned}$$

Hence, $h(p) \ge 0$ for all $1 \le p < 2$ and (C.2) holds. $\square$

Remark C.1

When $d \ge 7$ and $1< p < \frac{d+2}{d-2} = q$, condition (C.1) is not satisfied for $\omega \gg 1$. In fact, we have $1< q < 2$ and $A_{q+1} < 0$ in (C.3). Fix an $\alpha \in ( \frac{1}{q-1}, \frac{1}{p-1} )$ and observe that

$$\begin{aligned} \alpha ( p + q)< 1 + \alpha ( q + 1), \quad 2 + 2\alpha < 1 + \alpha ( q + 1). \end{aligned}$$

Noting also $1+(p+1) \alpha < 1+(q+1)\alpha$ and $2 p \alpha < (p+q) \alpha$, we see that

$$\begin{aligned} g( \omega ^{\alpha } )= & {} A_2 \omega ^{2+2\alpha } + A_{p+1} \omega ^{1+ (p+1)\alpha } + A_{q+1} \omega ^{ 1 + (q+1) \alpha } + A_{2p} \omega ^{2\alpha p} + A_{p+q} \omega ^{(p+q)\alpha } \\= & {} \omega ^{1+(q+1)\alpha } \left( A_{q+1} + o(1) \right) . \end{aligned}$$

Since $A_{q+1} < 0$, we obtain $g(\omega ^\alpha ) < 0$ for $\omega \gg 1$ and (C.1) is not satisfied.

It is worth noting that for any $\Phi _\omega \in {\mathcal {G}}_\omega$ we have $\Phi _\omega (0) = \Vert \Phi _\omega \Vert _{L^\infty } \sim \omega ^{\frac{1}{p-1}}$ by Lemma 4.1. Hence, (C.1) breaks down even in the interval $[0,\Phi _\omega (0)]$.