Asymptotics of the Bosonic Ground State

The goal of this section is to study the properties of the ground state \(u_\mu \) of (7) in the limits \(\mu \rightarrow 0\) and \(\mu \rightarrow \infty \). Parts of the proofs follow the lines of [24, 28].

Recall that \(u_\mu \) is the unique radial decreasing positive solution of the equation

$$\begin - \Delta u = g_\mu (u), \quad } \quad g_\mu (t) = - t F_1'(t^2) - \mu t. \end$$

(23)

We also recall that we set

$$ \Lambda (\mu ):= \Vert u_\mu \Vert _^2, \quad } \quad }_1(u):= \int _^d} | \nabla u |^2 + \int _^d} F_1(u^2). $$

Finally, we recall the definition of \(F_1\), given in (2) by

$$ F_1(t):= - t^q & \quad t \le 1 \\ - + t - t^ & \quad t \ge 1. \end\right. }, \quad } \quad 1< r< \frac < q. $$

The coefficients \(, \) and \(\) are positive. We made this choice so that \(F_1\) is of class \(C^2\), concave, behaves as \(-t^q\) around 0, and as \(-t^r\) at infinity.

1.1 The Limit \(\mu \rightarrow 0\)

We first focus on the limit \(\mu \rightarrow 0\). It is useful to distinguish the case \(q < \frac\) (subcritical case) from the case \(q \ge \frac\) (the critical or supercritical case). Note that the critical and the supercritical cases may only occur in dimension \(d\ge 3\).

Lemma A.1

(Limit \(\mu \rightarrow 0\), subcritical case).  Assume \(q < \frac\). Let \(v_0\) be the (unique) ground state to the NLS equation

$$\begin - \Delta v = - v + q v^, \end$$

(24)

and set \(\mu _c:= v_0(\textbf)^\). Then, for all \(0< \mu < \mu _c\), we have

$$ u_\mu (}) = \mu ^} v_0(\sqrt }). $$

In particular, for \(0< \mu < \mu _c\), we have

$$\begin \Lambda (\mu ) \mu ^(q \frac)} \Vert v_0 \Vert _2^2 \quad \xrightarrow [\mu \rightarrow 0] \infty , \quad } \quad }_1(u_\mu ) = \mu ^ - \frac} }_1(v_0) > 0. \end$$

Proof

The existence of a ground state solution for (24) was first proved by Kwong [18], and follows from arguments similar to the one used in Proposition 3.7. Now, we perform the scaling \(u(}) = \alpha ^ v(\beta })\), and find that u solves (7) if and only if v solves

$$\begin - \Delta v + \frac v = q \beta ^ \alpha ^ v^ & \quad } \quad v(}) \le \alpha \\ - \frac} v + r \beta ^ \alpha ^ v^ & \quad } \quad v(}) > \alpha . \end\right. } \end$$

(25)

Choosing \(\beta ^2 = \alpha ^=\mu \) gives

$$\begin - \Delta v + v = q v^ & \quad } \quad v(}) \le \mu ^} \\ - \frac} v + r \mu ^} v^ & \quad } \quad v(}) > \mu ^} . \end\right. } \end$$

(26)

Since the function \(v_0\) is radial decreasing, we have \(v_0(}) \le v_0(\textbf)\). In the case \(\mu < \mu _c\), this implies that \(v_0(}) \le \mu ^} \) pointwise, and the second condition is never satisfied. We deduce that \(v_0\) is a ground state solution of (26). By uniqueness of the ground state, this implies that \(u_\mu (}) = \mu ^} v_0(\sqrt })\). The scaling for \(\Lambda (\mu )\) and \(}_1(u_\mu )\) are straightforward (note that \(F_1(u_\mu ^2) = - u_\mu ^\) for \(\mu < \mu _c\)). It remains to prove that \(}_1(v_0) > 0\).

We multiply (24) by \(v_0\) and integrate, and by \(x \cdot \nabla v_0\) and integrate to obtain respectively

$$ \int _^d} | \nabla v_0 |^2 = - \Vert v_0 \Vert ^2_2 + q \int _^d} | v_0 |^, \quad } \quad \frac \int _^d} | \nabla v_0 |^2 = - \Vert v_0 \Vert ^2_2 + \int _^d} | v_0 |^. $$

The second equality is the Pohozaev’s identity. Taking the difference gives

$$ \frac \int _^d} | \nabla v_0 |^2 = (q-1) \int _^d} | v_0 |^. $$

This leads to \(}_1(v_0) = \Vert \nabla v_0 \Vert ^2_2 \frac\left( q - \frac\right) > 0\). \(\square \)

The critical and supercritical case are more involved. We will use results by Berestycki and Lions from [3, 4] following [7]. We introduce the functions

$$ G_\mu (t):=G_*(t) - \frac \mu t^2, \quad } \quad G_*(t):= - \frac F_1(t^2) = \frac t^ & \quad t \le 1 \\ \frac - \frac} t^2 + \frac} t^ & \quad t \ge 1, \end\right. } $$

so that \(g_\mu = G_\mu '\). We also consider the corresponding optimization problems \(M(G_\mu )\), where we define

$$\begin M(G) := \sup \left\^d} G(v) , \quad v \in \dot^1(\mathbb ^d), \ \Vert \nabla v \Vert _^2 \le 1 \right\} . \end$$

(27)

We recall that \(\dot^1(\mathbb ^d)\) is the homogenous Sobolev space of \(\mathbb ^d\). It is the completion of \(C^\infty _0(\mathbb ^d)\) for the norm \(u \mapsto \Vert \nabla u \Vert _2^2\). Recall the Sobolev embedding that for \(d \ge 3\), we have \(\dot^1(\mathbb ^d) \hookrightarrow L^}(\mathbb ^d)\). Problem (27) is a slight modification of the dual version of the problem studied in [3, 4], see [3, Remark 3.2]. We prefer to work with this version, as the optimization space is independent of G. This will be more convenient when comparing the problems \(M(G_\mu )\) for different values of \(\mu \).

We will use results from [3, 4]. For the sake of clarity, let us recall the main results that we will use. The first main result of [3] is the following one.

Lemma A.2

(From [3], Theorem 2). Assume \(d \ge 3\), and that \(g:= G'\) satisfies the three following conditions:

$$\begin - \infty< \lim _ \frac < 0, \quad \lim _ \frac} \le 0, \quad \exists \zeta> 0 \quad } \quad G(\zeta ) > 0. \end$$

(28)

Then the problem M(G) has an optimizer \(v \in H^1(\mathbb ^d)\), which is positive radial decreasing. In addition, v satisfies the Euler–Lagrange equation

$$ - \Delta v = \theta g(v), \quad } \quad \theta := \left( \frac \right) \frac > 0. $$

All such optimizers satisfy \(\Vert \nabla v \Vert _ = 1\) and are of class \(C^2\). Finally, all positive radial decreasing maximizing sequences are pre-compact in \(H^1(\mathbb ^d)\), and converge to such an optimizer, up to a subsequence.

We note that the condition \(d \ge 3\) is always satisfied in the (super)critical case. The expression of \(\theta \) comes from the Pohozaev’s identity. Finally, setting \(u(}):= v (\theta ^ })\), we can check that u is a solution to

From [3, Section 5], we also recall the result corresponding to the \(\ll \)zero-mass\(\gg \) case.

Lemma A.3

(From [3], Theorem 4). Assume \(d \ge 3\), and that \(g:= G'\) satisfies the three conditions

$$\begin g(0) = 0, \ } \ \lim _ \frac}} \le 0, \quad \lim _ \frac} \le 0, \quad \exists \zeta> 0 \quad } \quad G(\zeta ) > 0. \end$$

(29)

Then the results of Lemma A.2 holds, upon replacing \(H^1(\mathbb ^d)\) by \(\dot^1(\mathbb ^d)\).

The differences with the previous case Lemma A.2 is that g(t)/t is now allowed to vanish at 0 (hence the term \(\ll \)zero mass\(\gg \)). However, the corresponding solutions may not belong to \(L^2(\mathbb ^d)\).

In our case with \(g = g_\mu \), we can check that the conditions (28) are satisfied, as \(\lim _ g_\mu (t)/t = - \mu < 0\), \(\lim _ g_\mu (t)/t^} = 0\) and \(G_\mu (t) > 0\) for t large enough. Finally, if \(v_\mu \) an optimizer of \(M(G_\mu )\), then \(u(}):= v_\mu (\theta _\mu ^ })\) is a positive solution of \(- \Delta u = g_\mu (u)\). However, we proved in Proposition 3.7 that this last PDE has a unique positive solution \(u_\mu \). This proves that \(u = u_\mu \), and that the problem \(M(G_\mu )\) has a unique radial non-increasing optimizer \(v_\mu \). The functions \(v_\mu \) and \(u_\mu \) are linked by the relations

$$\begin u_\mu (}) := v_\mu (\theta _\mu ^ }), \qquad \theta _\mu := \left( \frac \right) \frac > 0. \end$$

(30)

We are now in position to prove the result in the supercritical case.

Lemma A.4

(Limit \(\mu \rightarrow 0\), supercritical case). Assume \(q > \frac\). Consider the “zero-mass” equation

$$\begin -\Delta u = g_*(u), \qquad g_*(t) = G_*'(t) = q t^ & \quad t \le 1\\ -t + r t^ & \quad t \ge 1. \end\right. } \end$$

(31)

Then \(u_\mu \) converges to the unique positive radial decreasing solution \(u_*\) of (31), strongly in \(\dot^1\cap L^}(}^d)\). In addition, we have

$$\begin }_1(u_\mu ) \xrightarrow [\mu \rightarrow 0] \frac \left( \frac\frac\right) ^} > 0. \end$$

(32)

Proof

Note that the function \(g_*\) satisfies the hypothesis of Lemma A.3 (since \(q > \frac\)), but does not satisfy the ones of Lemma A.2, as \(\lim g(t)/t = 0\). Lemma A.3 then shows that \(u_*\) is of class \(C^2\). In addition, using [11, Theorem 5], we have that \(\lim _|x|^u_*(x)\) exists and is a positive finite number. As a consequence, we can use [31, Theorem 2] to prove the uniqueness of positive solution of (31). So the problem \(M(G_*)\) has a unique optimizer, that we denote by \(v_*\). The functions \(u_*\) and \(v_*\) are linked by a relation similar to (30). Note that the decay of \(u_*\) (hence \(v_*\)) implies that \(u_*\) and \(v_*\) are in \(L^2(\mathbb ^d)\) only in dimensions \(d \ge 5\).

For any \(\mu >0\), let \(v_\mu \in H^1(\mathbb ^d)\) be the unique positive radial decreasing optimizer of \(M(G_\mu )\), so that

$$ \int _^d}G_\mu (v_\mu )=M(G_\mu )>0, \quad \Vert \nabla v_\mu \Vert _^2 = 1. $$

Our goal is to show that, for any sequence \((\mu _n)_n\) such that \(\lim _\mu _n=0\), \((v_)_\) is a maximizing sequence for \(M(G_*)\). Let us first show that

$$\begin \lim _M(G_)=M(G_*). \end$$

(33)

First, since \(v_\in H^1(\mathbb ^d)\), we have

$$\begin M(G_*)\ge \int _^d} G_(v_)= \int _^d} G_(v_)+\frac\mu _n \int _^d} |v_|^2\ge M(G_). \end$$

(34)

For the converse inequality, let us first consider the case \(d \ge 5\). In this case, we have \(v_*(})\simeq _}\rightarrow } \frac}\right| ^}\), so that \(v_*\in L^2(\mathbb ^d)\), and

$$\begin M(G_)\ge \int _^d} G_(v_)= \int _^d} G_(v_)-\frac\mu _n \int _^d} |v_|^2= M(G_)(1+o(1)). \end$$

This proves (33) in dimensions \(d \ge 5\). For \(d\in \left\ \), the function \(v_*(})\) is not in \(L^2(}^d)\). We use a cutoff function and proceed similarly. Let \(R>0\) and let \(0\le \chi _R\le 1\) a \(C^_c\) a cutoff function such that \(\chi (s )=1\) for \(\left| s \right| \le R\), \(\chi (s )=0\) for \(\left| s \right| \ge 2R\) and \(\left| \chi '_R\right| \le 2/R\). Denote by \(v_R=\chi _Rv_*\). A straightforward computation gives, for R large enough,

$$\begin&\int _^d} |\nabla v_R|^2=\int _^d} |\nabla v_*|^2 + O(R^)\\&\int _^d} G_*(v_R)=\int _^d} G_*(v_*) + O(R^)\\&\int _^d} |v_R|^2= O(R) & d=3\\ O(\log (R)) & d=4. \end\right. } \end$$

Let \(\delta _R:=\Vert \nabla v_R\Vert _^2\) and define \(\tilde_R(})=v_R(\delta _R^})\), so that \(\Vert \nabla \tilde_R\Vert _^2=1\) and \(\tilde_R(})\) is admissible for the optimization problem \(M(G_)\) for any \(n\in \mathbb \). We have

$$\begin M(G_)\ge \int _^d} G_(\tilde_R)=\delta _R^}\left( \int _^d} G_(v_R)-\frac\int _^d}v_R^2\right) \end$$

Choosing \(R = R_n=\mu _n^\) gives

$$\begin \int _^d} G_(v_)-\frac\int _^d}v_^2&=\int _^d} G_*(v_*)\left( 1+O(\mu _n^) + O(\mu _n^) & d=3\\ O(\mu _n\log (\mu _n)) & d=4 \end\right. }\right) \\&= M(G_*)(1+o(1)) \end$$

and

$$\begin \delta _^}=\Vert \nabla v_*\Vert _^}(1+O(\mu _n^))^}\ge (1+o(1)). \end$$

As a conclusion, \(M(G_)\ge M(G_*)(1+o(1))\), which concludes the proof of (33).

Now, (33), together with (34), implies

$$\begin \lim _\mu _n\Vert v_\Vert _^2=0. \end$$

(35)

As a consequence,

$$\begin \int _^d} G_*(v_)=M(G_)+\frac\int _^d}v_^2\rightarrow M(G_*) \end$$

and \((v_)_n\) is a maximizing sequence for \(M(G_*)\). Together with Lemma A.3, \((v_)_n\) converges in \(\dot^1(\mathbb ^d)\) to v, an optimizer of \(M(G_*)\), up to a subsequence. Since the optimizer is unique, we deduce that the full sequence \((v_)_n\) converges to \(v_*\) in \(\dot^1\). To conclude the proof, we recall that \(u_\mu (}) = v_\mu (\theta _\mu ^ })\) with \(\theta _\mu = \left( \frac \right) \frac\) and \(u_*(}) = v_*(\theta _*^ })\) with \(\theta _* = \left( \frac \right) \frac\). So the convergence \(M(G_\mu ) \rightarrow M(G_*)\) implies the convergence \(u_\mu \) to \(u_*\) in \(\dot^1(\mathbb ^d)\).

It remains to prove (32), namely that \(}_1(u_\mu ) > 0\) for \(\mu \) small enough. Recall the scaling (30), which implies

$$ \int _^d} G_\mu (u_\mu ) = \theta _\mu ^} M(G_\mu ) \xrightarrow [\mu \rightarrow 0] \theta _*^\fracM(G_*), \quad } \quad \mu \int _^d} | u_\mu |^2 = \mu \theta _\mu ^} \int _^d} | v_\mu |^2 \xrightarrow [\mu \rightarrow 0] 0, $$

where we used (35) in the last limit. On the other hand, since \(- \Delta u_\mu = G_\mu '(u_\mu )\), we have the Pohozaev’s identity

$$ \Vert \nabla u_\mu \Vert _^2 = \frac \int _^d} G_\mu (u_\mu ). $$

Together with the fact that \(F_1(u^2) = - 2 G_\mu (u) - \mu u^2\), we get

$$\begin }_1(u_\mu )&= \Vert \nabla u_\mu \Vert ^2 - 2 \int _^d} G_\mu (u_\mu ) - \mu \int _^d} | u_\mu |^2 \xrightarrow [\mu \rightarrow 0] \frac \left( \frac \frac \right) ^}, \end$$

which concludes the proof. \(\square \)

It remains to prove the result in the critical case \(q = \frac\).

Lemma A.5

(Limit \(\mu \rightarrow 0\), critical case). Assume \(q = \frac\). There exists a sequence \((\lambda _\mu )_\), \(\lambda _\mu \in (0,\infty )\) such that the rescaled function

$$\begin \lambda _^}u_(\lambda _\mu \cdot ) \end$$

converges strongly in \(\dot^1\cap L^}(}^d)\), to the Sobolev optimizer

$$\begin S(})=\left( 1+\frac}|^2}\right) ^}. \end$$

(36)

In addition, for \(\mu \) small enough, we have \(}_1(u_\mu ) > 0\).

The function \(S(\cdot )\) is also the unique positive radial decreasing solution (up to dilations) the “Emden–Fowler” equation

$$\begin -\Delta S =\frac S^}. \end$$

(37)

Since \(S\le 1\) pointwise, it is also a solution to the zero-mass equation (31) when \(q = \frac\).

Proof

Let \(q = \frac\), we introduce the modified function \(\widetilde_*(t):=\fract^ = \frac t^}\) for all \(t\ge 0\), and we consider the corresponding optimization problem \(M(\widetilde_*)\) defined by (27). Note that the function \(\widetilde_*\) does not satisfy the hypothesis of Lemma A.2 nor the ones of Lemma A.3, since \(\widetilde_*:= \widetilde_*'\) satisfies

$$ \lim _ \dfrac_*(t)}}} = \frac > 0. $$

In this case, we resort to the results of [22] (which deals with the critical case). It follows from [22, Theorem I.1] that all radial decreasing maximizing sequences for \(M(\widetilde_*)\) are pre-compact in \(\dot^1(\mathbb ^d)\) up to a dilation. As a consequence, \(M(\widetilde_*)\) admits an optimizer \(\widetilde^*\) which satisfies the equation

$$\begin -\Delta \widetilde_*=\widetilde_* \frac\tilde_*^}\quad }\quad \widetilde_*=\frac\frac_*)}. \end$$

More precisely, any optimizer of \(M(\widetilde_*)\) is of the form \(\widetilde_(})=\lambda ^}\widetilde_(\lambda })\) with \(\widetilde_(})= S(_*}^})\). Here S is the Aubin–Talenti function defined by (36). Since \(S\le 1\) pointwise, the function S is a solution to the zero-mass equation (31). This is also true for \(\tilde_\) whenever \(\lambda \le 1\).

We claim that \(M(\widetilde_*)=M(G_*)\). Indeed, on the one hand, since \(G_*\le \widetilde_*\), we have \(M(\widetilde_*)\ge M(G_*)\). On the other hand, let \(\widetilde_*\) be an optimizer of \(M(\widetilde_*)\). Using the dilation described above, \(\widetilde_*\) can be chosen such that \(|\widetilde_*|\le 1\). As a consequence, \(\widetilde_*(\widetilde_*)=G_*(\widetilde_*)\) and

$$\begin M(G_*)\ge \int _^d} G_*(\widetilde_*)= \int _^d} \widetilde_*(\widetilde_*)= M(\widetilde_*). \end$$

For any \(\mu >0\), let \(v_\mu \) be the unique positive radial decreasing optimizer of \(M(G_\mu )\). We have \(v_\mu \in H^1(\mathbb ^d)\), \(\int _^d}G_\mu (v_\mu )=M(G_\mu )>0\) and \(\Vert \nabla v_\mu \Vert _^2 = 1\). Our goal is to show that, for any sequence \((\mu _n)_n\) such that \(\lim _\mu _n=0\), \((v_)_\) is a maximizing sequence for \(M(\widetilde_*)\).

As \(\tilde_*\) is an optimizer of \(M(G_*)\), we can reason as in the proof of (33) to obtain

$$\begin \lim _M(G_)=M( G_*)= M(\tilde_*). \end$$

(38)

Moreover, for any \(n\in \mathbb \),

$$\begin&0\le \frac\int _^d}v_^2=\int _^d} G_*(v_)-\int _^d} G_(v_)\\&\le \int _^d} \widetilde_*(v_)-M(G_)\le M(\widetilde_*)-M(G_), \end$$

As a consequence,

$$\begin \lim _\mu _n\Vert v_\Vert _^2=0. \end$$

(39)

and

$$\begin&0\le M(\widetilde_*)-\int _^d}\widetilde_*(v_)\le M(\widetilde_*)-\int _^d} G_*(v_)\\&=M(\widetilde_*)-M(G_)-\frac\int _^d}v_^2\xrightarrow [n \rightarrow \infty ] 0. \end$$

Hence, \((v_)_n\) is a maximizing sequence for \(M(G_*)\). It follows from [22, Theorem I.1] that there exists a sequence \((\lambda _n)_n\) such that the rescaled sequence \((\lambda _n^}\tilde_(\lambda _n\cdot ))_n\) converges strongly in \(\dot^1(\mathbb ^d)\) to \(\tilde_\), up to an undisplayed subsequence. Since \(u_\mu (}) = v_\mu (\theta _\mu ^ })\) with \(\theta _\mu = \left( \frac \right) \frac\) and \(S(}) = \tilde_(\tilde_*^ })\), we conclude that the rescaled sequence

$$\begin \lambda _n^}u_(\lambda _n\cdot ) \end$$

converges to S strongly in \(\dot^1(\mathbb ^d)\). The proof \(}_1(u_\mu ) > 0\) for \(\mu \) small enough follows the same lines as the proof of (32). \(\square \)

1.2 The Limit \(\mu \rightarrow \infty \)

The behaviour of \(u_\mu \) in the limit \(\mu \rightarrow \infty \) is more involved. One reason is that the functions

$$\begin g_\mu (t)=-\mu t+ qt^& \quad t\le 1\\ - t + r t^ & \quad t\ge 1 \end\right. } \end$$

diverge pointwise to \(- \infty \) as \(\mu \rightarrow \infty \). One way to handle this divergence is to make a change of variable. We introduce

$$ \widetilde_\mu (t):= A g_\mu ( B t), $$

where the positive parameters \(A = A(\mu )\) and \(B = B(\mu )\) will be chosen below. It is easy to check that \(u_\mu \) is a solution to \(-\Delta u=g_\mu (u)\) if and only if \(\widetilde_\mu \) is a solution to \(-\Delta u= \widetilde_\mu (u)\), with

$$ \widetilde_\mu (}) = B^ u_\mu \left( \sqrt }\right) . $$

Let us now choose the parameters A and B that will serve our purposes. We have

$$\begin \widetilde_\mu (t)&=-AB \mu t+ A B^q t^& Bt\le 1\\ - ABt + AB^r t^ & Bt\ge 1 \end\right. }\\&=-AB (\mu +) t+ AB^r t^+ ABt+ A B^q t^- AB^r t^& Bt\le 1\\ 0 & Bt\ge 1. \end\right. } \end$$

We make the following choice

$$ A & = (+ \mu )^}\\ B & = (+ \mu )^} \end\right. }, \quad } \quad AB \left( } + \mu \right) & = 1 \\ A B^ & = 1 \end\right. } $$

In particular, we have \(B \rightarrow \infty \) as \(\mu \rightarrow \infty \), so the condition \(Bt \le 1\) is satisfied on a smaller and smaller interval as \(\mu \) increases. Actually, with these values, the function \(\widetilde_\mu \) simplifies into

$$\begin \widetilde_\mu (t)=- t+ r t^+ \frac}+\mu } t+ (+\mu )^}q t^-r t^& t\le (+ \mu )^}\\ 0 & t\ge (+ \mu )^} \end\right. } \end$$

We have the following result.

Lemma A.6

Let Q be the (unique) ground state to the equation

$$\begin - \Delta w = - w + r w^. \end$$

(40)

For \(\mu > 0\) (large), we introduce the rescaled function \(\widetilde_\mu (}):= (\mu +)^} u_\mu ( }/\sqrt})\). Then, we have, in the limit \(\mu \rightarrow \infty \),

$$ \left\| \widetilde_\mu - Q \right\| _ = o(1). $$

In particular, we have

$$ \Lambda (\mu ) = \Vert u_\mu \Vert _2^2 = (\mu +)^( \frac - r)} \Vert Q \Vert _2^2 (1 + o(1)) \xrightarrow [\mu \rightarrow \infty ] \infty . $$

For any \(\mu >0\), let \(\alpha = \alpha _\mu =(+ \mu )^}\). As explained above, the rescaled function \(\widetilde_\mu \) is the (unique) ground state of

$$\begin - \Delta \widetilde + \widetilde= r \widetilde^ + E(\alpha ,\widetilde), \end$$

(41)

with the function

$$ E(\alpha ,t):= }\alpha ^ t +q \alpha ^ t^- r t^ & \quad } \quad t < \alpha ; \\ 0 & \quad } \quad t \ge \alpha . \end\right. } $$

Note that \(\alpha _\mu \rightarrow 0\) as \(\mu \rightarrow \infty \) and

$$ \Lambda (\mu )=\int \left| u_\mu \right| ^2=\left( +\mu \right) ^\left( \frac-r\right) }\left\| \widetilde_\mu \right\| ^2_2= \frac-r\right) }}\left\| \widetilde_\mu \right\| ^2_2. $$

Let us record some properties of the function E.

Proposition A.7

For all \(\alpha > 0\), the function \(E(\alpha ,\cdot )\) is \(C^1\), positive on \((0,\alpha )\), and satisfies the inequality

$$\begin&0\le \le (+q)\alpha ^t\quad }t\in (0,\alpha ),\\&\sup _t\left| \partial _tE(\alpha ,t)\right| \le \left( +q(2q-1)+r\right) \alpha ^. \end$$

Proof

Differentiating E w.r.t. to t for \(0< t < \alpha \), we have

$$\begin \partial _t E(\alpha ,t)&= }\alpha ^+ q(2q-1) \alpha ^t^ - r (2r-1)t^ \\ \partial _t^2 E(\alpha ,t)&= 2t^ \left( q(2q-1)(q-1) \alpha ^ t^ - r (2r-1)(r-1) \right) . \end$$

For the continuity of \(E(\alpha ,\cdot )\) and its derivative, we note that for \(t = \alpha \), we have

$$ E(\alpha ,\alpha ^-)= \left( +q-r\right) \alpha ^ = 0 $$

and

$$ \partial _t E(\alpha ,\alpha ^-) (+ q(2q-1)-r (2r-1))\alpha ^2(q(q-1)-r (r-1))\alpha ^0. $$

Next, we remark that \(\partial _t ^2 E(\alpha ,\cdot )\) vanishes only once in \((0,\alpha )\). As \(\partial _t E(\alpha ,0)>0\) and \(\partial _t E(\alpha ,\alpha )=0\), then \(\partial _t E(\alpha ,\cdot )\) is positive, then negative. It follows that \(E(\alpha ,\cdot )\) is increasing then decreasing, but since \(E(\alpha , 0)= E(\alpha , \alpha ) = 0\), E is positive on \((0, \alpha )\). The bounds are now easy to obtain. \(\square \)

Now, we show that \(\widetilde_\mu \) converges, in \(H^2(}^d)\), to Q, the unique positive solution to

$$ -\Delta w+ w= r w^. $$

We use a fixed point procedure. We introduce the linearized operator \(}_Q: W^\cap H^2_\textrm (\mathbb ^d) \rightarrow L^p \cap L^2_\textrm(\mathbb ^d)\) defined by

$$ }_Q: v \mapsto - \Delta v + v - r(2r-1) Q^ v. $$

As Q is non-degenerate [18], then \(}_Q\) is invertible for any \(p\ge 2\). For any \(\alpha >0\), we write \(\widetilde_\alpha = Q + w_\alpha \), so that \(\widetilde_\alpha \) satisfies (41) iff \(w_\alpha \) satisfies the fixed point equation \(w_\alpha = G_\alpha (w_\alpha )\) with

$$ G_\alpha (w):= }_Q^ \left\r \left[ (Q + w)^ - Q^ - (2r-1) Q^ w \right] \right\} . $$

Proposition A.8

There exists \(\alpha _0>0\) such that, for all \(\alpha \in (0,\alpha _0)\), there exists \(\eta =\eta (\alpha )>0\) so that the map \(G_\alpha \) is a contraction on \(}(\eta )\), where

$$ }( \eta ):= \left\ \cap H^2_\textrm(\mathbb ^d), \ \Vert w \Vert _} + \Vert w \Vert _ \le \eta \right\} . $$

In particular, the map \(G_\alpha \) admits a unique fixed point in \(}( \eta )\).

Our choice of the space is such that \(W^(}^d)\subset C^(}^d)\), \(\forall \gamma \in (0,1)\).

Proof

Let us first prove that for \(\alpha \) and \(\eta \) small enough, the map \(G_\alpha \) leaves \(}( \eta )\) invariant. Since \(}_Q^: L^p\cap L^2_\textrm (}^d)\rightarrow W^\cap H^2_\textrm (}^d)\) is bounded, it is enough to control, for all \(w\in }(0, \eta )\)

$$ \left\| E(\alpha , Q + w) + r \left[ (Q + w)^ - Q^ - (2r-1) Q^ w \right] \right\| _$$

for \(p\in \left\ \). For \(w \in }(\eta )\), we have from Proposition A.7 that

$$ \left\| E(\alpha , Q + w ) \right\| _p \le \left( + q \right) \alpha ^ \Vert Q + w \Vert _p \le \left( + q \right) \alpha ^ \left( \Vert Q \Vert _p + \eta \right) . $$

On the other hand, we have, with \(G(w):= |Q + w|^ - Q^ - (2r-1) Q^} w\)

$$ \left| G(w)\right| \le C w^ \quad & } \quad w \ge Q, \\ C Q^ w^2 = C (w/Q)^ w^ \quad & } \quad | w | \le Q. \end\right. } $$

If \(2r-3\ge 0\), then

$$ \left\| G(w) \right\| _\le C \left( \left\| Q \right\| _^+\left\| w \right\| _^\right) \left\| w \right\| _p^2\le C\left( \left\| Q \right\| _^+\eta ^\right) \eta ^2. $$

If \(2r-3<0\), when \(\left| w\right| \le Q\), we also have \(\left| G(w)\right| \le C \left| w\right| ^\), hence

$$ \left\| G(w) \right\| _p \le C \eta ^. $$

It follows that

$$\begin \left\| G_\alpha (w) \right\| _\cap H^2}\le C \left\| }_Q^ \right\| \left( \alpha ^\left( \left\| Q \right\| _p+\eta \right) + \eta ^2\left( \left\| Q \right\| _^+ \eta ^\right) \right) . \end$$

(42)

For any \(\alpha \) small enough, we can thus choose \(\eta \), which depends on \(\alpha \), such that the R.H.S. of (42) is smaller than \(\eta \), thus \(G_\alpha \) leaving the set \(}(\eta )\) invariant.

We now prove that \(G_\alpha \) is a contraction. First, we have, using again Proposition A.7,

$$ \left| E(\alpha , w) - E(\alpha , v) \right| \le \Vert E'(\alpha ;.) \Vert _\infty | w - v | \le C \alpha ^ | w - v |, $$

so

$$ \left\| E(\alpha , w) - E(\alpha , v) \right\| _p \le C \alpha ^ \Vert w - v \Vert _p. $$

Next, we have

$$\begin&\left| G(w) G(v) \right| \le (2r-1) \left( \int _0^1 \left| (Q tw (1 - t)v )^ - Q^ \right| }t \right) | w - v |. \end$$

Setting \(z = tw + (1 - t)v\), we have

$$ \left| |Q + z|^ - Q^\right| \le C z^ \quad & } \quad z \ge Q, \\ C (2r-2) Q^ z = C (2r-2) (z/Q)^ z^ \quad & } \quad | z | \le Q. \end\right. } $$

Again, in both cases, we get an upper bound of the form

$$\begin \left\| G(w) - G(v) \right\| _p \le C \left( \left\| Q \right\| _^ + \eta ^ \right) \eta \left\| w - v \right\| _p. \end$$

Altogether, we have proved that

$$ \left\| G_\alpha (w) - G_\alpha (v) \right\| _} \le C \left( \alpha ^ + \eta ^ + \left\| Q \right\| _\eta \right) \Vert w - v \Vert _}. $$

So, for \(\eta \) and \(\alpha \) small enough, we can apply the fixed point theorem, and deduce that \(G_\alpha \) has a unique fixed point in \(}(\eta )\). \(\square \)

Remark A.9

In the proof of Proposition A.8, given \(\alpha \) small enough, \(\eta \) has to be choosen such that

$$\begin C \left\| }_Q^ \right\| \left( \alpha ^\left( \left\| Q \right\| _p+\eta \right) + \eta ^2\left( \left\| Q \right\| _^+ \eta ^\right) \right) \le \eta \\ C \left( \alpha ^ + \eta ^ + \left\| Q \right\| _\eta \right) <\lambda \end$$

for some \(\lambda \in (0,1)\). In particular, one can choose \(\eta =\alpha ^\) with \(0<\beta <2(r-1)\).

Since \(\widetilde_\mu \) is the (unique) ground state of (41) with \(\alpha =\alpha _\mu \), we deduce that \(\widetilde_\mu =Q+w_\) with \(w_\) the unique fixed point of \(G_\). As a consequence,

$$ \left\| \widetilde_\mu - Q \right\| _= \left\| w_ \right\| _ = \left\| G_(w_) \right\| _\le \alpha _\mu ^=\left( \frac+\mu }\right) ^} \xrightarrow [\mu \rightarrow \infty ] 0. $$

This concludes the proof of Lemma A.6.