1 Introduction and Statement of the Results

The baby Skyrme model is a two dimensional version of the original Skyrme model for baryons and mesons (see [16, 17]); in the static case, the two dimensional Skyrme field is a map \(u:{\mathbb {R}}^2\rightarrow S^2\) which goes to a constant at infinity, so it can be identified with a map from \(S^2\) to \(S^2\) with a given topological degree Q(u), where

$$\begin{aligned} Q(u)= \frac{1}{4\pi } \int _{{\mathbb {R}}^2} u\cdot \partial _1 u\times \partial _2 u\, dx, \end{aligned}$$

and the problem is to minimize the energy functional

$$\begin{aligned} E(u)=\int _{{\mathbb {R}}^2}\Big ( \frac{1}{2} |\nabla u|^2+ |\partial _1 u\times \partial _2 u|^2+V(u)\Big )\, dx \end{aligned}$$

over the maps with a fixed degree \(Q(u)\ne 0\); these energy-minimizing fields are called baby skyrmions. The first two terms in the energy density are, respectively, the sigma model term and the two dimensional Skyrme term, quartic in derivatives. The third term is a non negative potential which is mandatory in two dimensions, otherwise E(u) can be decreased by a simple scaling argument (see [6]), and the minimum is not attained.

Notice that the convergence of a minimizing sequence is not trivial because of the lack of compactness of E(u). The existence of baby skyrmions is proved in [10, 11] for the potential \(V(u)=|u-e_3|^4\), where \(e_3=(0,0,1)\) is the north pole of the sphere \(S^2\), and in [9] for more general potentials.

Gauged versions of the baby Skyrme (and related) models have been recently studied with numerical methods by many authors (see [1,2,3, 5, 8, 12, 13, 15] and their bibliographies), in general by using the Skyrme ansatz in order to reduce the problem to a system of ordinary differential equations.

In this paper, we consider the gauged version of E(u) introduced in [8], namely the functional

$$\begin{aligned} F(u,A)= & {} \frac{1}{2} \int _{{\mathbb {R}}^2}\big (|D_1 u|^2+|D_2 u|^2+ |D_1 u\times D_2 u|^2+2(1-e_3\cdot u)\\&\quad +\gamma (\partial _1 A_2-\partial _2 A_1)^2\big )\, dx \end{aligned}$$

where \(A:{\mathbb {R}}^2\rightarrow {\mathbb {R}}^2\) is the gauge field, with \(A(x)=(A_1(x),A_2(x))\), \(D_i u=\partial _i u+A_i e_3\times u\) are the covariant derivatives, and \(\partial _1 A_2-\partial _2 A_1\equiv F_{1,2}\) is the magnetic component of the field strength. We have, for simplicity, fixed the coefficient of the Skyrme term and the potential, while \(\gamma >0\) represents the coupling strength of the gauge field.

Since \(2(1-e_3\cdot u)\ge (1-e_3\cdot u)^2\), using the well known topological lower bound (see [14]):

$$\begin{aligned} \frac{1}{2} \int _{{\mathbb {R}}^2}\big (|D_ 1 u|^2+|D_ 2 u|^2+(1-e_ 3\cdot u)^2+(\partial _1 A_ 2-\partial _2 A_ 1)^2\big )\, dx\ge 4\pi |Q(u)|, \end{aligned}$$

we get

$$\begin{aligned} F(u,A)\ge 4\pi \min (1,\gamma ) |Q(u)|\end{aligned}$$

(provided that u(x) and A(x) have an appropriate behavior at infinity) so that F(u, A) is bounded away from zero on every non trivial topological sector. Using the Skyrme ansatz, namely functions of the form

$$\begin{aligned} u_\theta (x)= & {} (\sin (\theta (r))\cos (k\varphi ),\sin (\theta (r)) \sin (k\varphi ),\cos (\theta (r))),\nonumber \\ A(x)= & {} k \frac{a(r)}{r^2} (-x_2,x_1), \end{aligned}$$
(1)

where \(k\in {\mathbb {Z}}\), \(x=(x_1,x_2)\), \(r=|x|\), and \(\theta (r)\), a(r) are functions defined for \(r>0\), the functional F(u, A) becomes

$$\begin{aligned} F(\theta ,a)= & {} \pi \int _0^{+\infty }\Big (k^2 (1+a(r))^2(1+\theta '(r)^2) \frac{\sin ^2(\theta (r))}{r^2} \\&\quad +\theta '(r)^2+\gamma \frac{k^2 a'(r)^2}{r^2} +2 \big (1-\cos (\theta (r))\big )\Big ) r\, dr. \end{aligned}$$

We aim to prove the existence of topologically non trivial minimum points \((\theta ,a)\) of \(F(\theta ,a)\), and to study the behavior of \((\theta ,a)\) as the electromagnetic coupling constant \(\gamma \) tends to infinity or to zero, namely in the regime of weak or strong coupling respectively. More precisely, we consider the set X of the couples \((\theta ,a)\) of continuous functions \(\theta ,a:[0,+\infty [\rightarrow {\mathbb {R}}\) which are absolutely continuous on every compact subinterval of \(]0,+\infty [\), such that \(F(\theta ,a)<+\infty \), and the usual boundary conditions \(\theta (0)=\pi \), \(\theta (\infty )=0\), \(a(0)=0\) are satisfied. If \((\theta ,a)\in X\), for the function \(u_{\theta }(x)\) in (1) we have \(Q(u_{\theta })=-k\) (see Remark 1). From now on we assume \(k\ne 0\). The following theorem holds.

Theorem 1.1

There exists \((\theta ,a)\in X\) such that \(F(\theta ,a)=\inf \{F(\theta ,a)\mid (\theta ,a)\in X \}\), where \(\theta \in C([0,+\infty [)\cap C^\infty (]0,+\infty [)\), \(a\in C^2([0,+\infty [)\cap C^\infty (]0,+\infty [)\). Moreover \(0<\theta (r)<\pi \) and \(-1<a(r)<0\) for every \(r>0\). Finally, a(r) is strictly decreasing on \([0,+\infty [\) from zero to a value \(a(\infty )\in [-1,0[\), and \(\theta (r)\) is strictly decreasing for large values of r.

In Theorem 1.1 we prove the existence of a minimizer of \(F(\theta ,a)\) by using analytical methods rather than numerical simulations. In the three dimensional case (without potential) an existence result has been proven in [18], always using Skyrme anzatz.

Moreover we recall that the existence of a minimizer of the ungauged functional \(F(\theta ,0)\) is well known (see [4] or [11]).

We want now to study what happens when \(\gamma \rightarrow +\infty \) or \(\gamma \rightarrow 0\). To highlight the dependence of the minimizer \((\theta ,a)\) on the coupling constant \(\gamma \), sometimes we will denote it by \((\theta _{\gamma },a_{\gamma })\). Then we can state the following theorems.

Theorem 1.2

Let \((\gamma _n)_n\) be a sequence of coupling constants such that \(\gamma _n\rightarrow +\infty \); then we have \(\lim _{n\rightarrow +\infty } a_{\gamma _n}(\infty )=0\) and \(\lim _{n\rightarrow +\infty } a_{\gamma _n}(r)=0\) uniformly on \([0,+\infty [\); we have moreover, up subsequences, \(\lim _{n\rightarrow +\infty }\theta _{\gamma _n}(r)={\hat{\theta }}(r)\) uniformly on every compact subinterval of \(]0,+\infty [\), where \({\hat{\theta }}\) is a minimizer of the ungauged functional \(F(\theta ,0)\).

Theorem 1.3

Let \((\gamma _n)_n\) be a sequence of coupling constants such that \(\gamma _n\rightarrow 0\); then we have \(\lim _{n\rightarrow +\infty } a_n(\infty )=-1\); moreover, for every \(R>0\), we have, up subsequences, \(\lim \limits _{n\rightarrow +\infty } a_{\gamma _n}(r)=-1\) and \(\lim \limits _{n\rightarrow +\infty }\theta _{\gamma _n}(r)=0\) uniformly on \([R,+\infty [\).

Notice that, since

$$\begin{aligned} \int _{{\mathbb {R}}^2}(\partial _1 A_2-\partial _2 A_1)\, dx=2\pi \int _0^{+\infty } k a_\gamma '(r)\, dr=2\pi k \,a_{\gamma }(\infty ), \end{aligned}$$

\(a_\gamma (\infty )\) is proportional to the magnetic flux, and Theorem 1.3 says that, even if the magnetic flux can be any value in the interval \([-2\pi k,0[\), in the strong coupling regime it is quantized, and the skyrmion profile as well as the magnetic field are localized at the origin.

The behavior of \((\theta _\gamma , a_\gamma )\) described in the theorems above was observed, by means of numerical simulations, in [8], and also, for different gauged baby Skyrme models, in [12, 13, 15].

2 The Functional Framework

The natural domain of the functional \(F(\theta ,a)\) in the Introduction is the set Y of the functions \((\theta ,a)\) continuous on \(]0,+\infty [\), absolutely continuous on every compact subinterval of \(]0,+\infty [\), such that \(F(\theta ,a)<+\infty \). Clearly this implies

$$\begin{aligned} \int _0^{+\infty } \frac{a'(r)^2}{r} \, dr<+\infty , \end{aligned}$$

so that \(a'(r)= \frac{a'(r)}{\sqrt{r}} \sqrt{r}\in L^1([0,R])\) for every \(R>0\), and a(r) is absolutely continuous on [0, R] for every \(R>0\). In particular, \(a(0)\in {\mathbb {R}}\). Moreover, since

$$\begin{aligned} \frac{|a(r)-a(0)|}{r} \le \frac{1}{r} \int _0^r \frac{|a'(s)|}{\sqrt{s}} \sqrt{s}\, ds\le \sqrt{ \frac{1}{2} \int _0^r \frac{a'(s)^2}{s} \, ds}, \end{aligned}$$

we have \(a'(0)=0\). We have now the following simple lemmas.

Lemma 2.1

If \((\theta ,a)\in Y\) and \(a(0)\ne -1\), then there exists \(p\in {\mathbb {Z}}\) such that \(\lim \limits _{r\rightarrow 0} \theta (r)=p\pi \).

Proof

Since \(a(0)\ne -1\) and \(F(\theta ,a)<+\infty \), we have

$$\begin{aligned} 2\int _0^{\delta }|\sin (\theta (r))||\theta '(r)|\, dr\le \int _0^{\delta }\Big ( \frac{\sin ^2(\theta (r))}{r} +\theta '(r)^2 r\Big )\, dr<+\infty \end{aligned}$$
(2)

for some \(\delta >0\), so that \(\partial _r\sin ^2(\theta (r))\in L^1([0,\delta ])\), and, in particular, \(\sin ^2(\theta (r))\) tends to some \(\ell \in [0,1]\) for \(r\rightarrow 0\); but (2) implies \(\ell =0\), and since the set \(\sin ^2x<\varepsilon \) is disconnected for small \(\varepsilon \), we get the lemma. \(\square \)

Lemma 2.2

If \((\theta ,a)\in Y\), then there exists \(q\in {\mathbb {Z}}\) such that \(\lim \limits _{r\rightarrow +\infty }\theta (r)=2\pi q\).

Proof

\((\theta ,a)\in Y\) implies

$$\begin{aligned} \int _0^{+\infty }\big (\theta '(r)^2 r+\big (1-\cos (\theta (r))\big ) r\big )\, dr<+\infty , \end{aligned}$$

so that \(\liminf _{r\rightarrow +\infty }(1-\cos (\theta (r)))=0\). On the other hand, since for \(1<R<r_1<r_2\) we have

$$\begin{aligned} \big |\big (1-\cos (\theta (r_2))\big )^2-\big (1-\cos (\theta (r_1))\big )^2\big |\\ \qquad \qquad \le 2\sqrt{\int _R^{+\infty } \big (1-\cos (\theta (r))\big )^2 r\, dr}\sqrt{\int _R^{+\infty }\theta '(r)^2 r\, dr}, \end{aligned}$$

and \((1-\cos x)^2\le 2(1-\cos x)\), the function \(1-\cos (\theta (r)\) verifies the Cauchy condition at infinity, so \(\lim _{r\rightarrow +\infty } (1-\cos (\theta (r)))=0\), and we can conclude as in the previous lemma. \(\square \)

Remark 1

If \((\theta ,a)\in Y\) and \(a(0)\ne -1\), the topological degree Q(u) of the function \(u(x)=(\sin (\theta (r))\cos (k\varphi ),\sin (\theta (r))\sin (k\varphi ),\cos (\theta (r)))\) is well defined and \(Q(u)=0\) or \(Q(u)=-k\), in fact

$$\begin{aligned} Q(u)= \frac{k}{2} \int _0^{+\infty }\sin (\theta (r))\theta '(r)\, dr =-\frac{k}{2} \big (1-\cos (\theta (0))\big )= \frac{k}{2} \big ((-1)^p-1\big ) \end{aligned}$$

since \(\theta (0)=p\pi \) (Lemma 2.1). Clearly \(Q(u)\ne 0\) implies \(k\ne 0\), p odd, and \(Q(u)=-k\).

Let us consider now the set X defined in Sect. 1, namely the functions \( (\theta ,a)\in Y\) which satisfy the boundary conditions \(\theta (0)=\pi \), \(\theta (\infty )=0\), \(a(0)=0\). As we shall see soon, the functional \(F(\theta ,a)\) can be studied in the set \(X_0=\{ (\theta ,a)\in X\mid 0\le \theta (r)\le \pi , -1\le a (r)\le 0\}\). In fact we have the following result.

Lemma 2.3

If \((\theta ,a)\in X\), then there exists \(({\tilde{\theta }},{\tilde{a}})\in X_0\) such that \(F({\tilde{\theta }},{\tilde{a}})\le F(\theta ,a)\).

Proof

Set \(h(s)= \arccos (\cos (s))\) and \({\tilde{\theta }}(r)=h(\theta (r))\). Clearly \(0\le {\tilde{\theta }}(r)\le \pi \), \(\cos ({\tilde{\theta }}(r))=\cos (\theta (r))\), \(\sin ^2({\tilde{\theta }}(r))=\sin ^2(\theta (r))\), and \({\tilde{\theta }}(r)\) fulfills the boundary conditions \({\tilde{\theta }}(0)=\pi \), \({\tilde{\theta }}(\infty )=0\). Since h(s) is Lipschitz continuous, \({\tilde{\theta }}(r)\) is continuous on \([0,+\infty [\) and absolutely continuous on every compact subinterval of \(]0,+\infty [\); moreover the chain rule holds true, namely \({\tilde{\theta }}'(r)=h'(\theta (r))\theta '(r)\) if h is derivable at \(\theta (r)\), and \({\tilde{\theta }}'(r)=0\) if \(\theta (r)\) is a corner point of h (see [7], Theorem 7.8). Since \(h'(s)=\pm 1\) for \(s\ne n\pi \), we have clearly \(|{\tilde{\theta }}'(r)|\le |\theta '(r)|\) a.e. on \([0,+\infty [\), and the inequality \(F({\tilde{\theta }},a)\le F(\theta ,a)\) follows immediately.

In a similar way, if we set now \(h(s)=(|1+s|-1-||1+s|-1|)/2\) and \({\tilde{a}}(r)=h(a(r))\), it is easy to check that \({\tilde{a}}(0)=0\) and \(-1\le {\tilde{a}}(r)\le 0\). Moreover, since \((1+h(s))^2\le (1+s)^2\), we have \((1+{\tilde{a}}(r))^2\le (1+a(r))^2\) for every \(r\ge 0\). Finally \(|{\tilde{a}}'(r)|\le |a'(r)|\) a.e. on \([0,+\infty [\), so that \(F({\tilde{\theta }},{\tilde{a}})\le F(\theta ,a)\), and the lemma is proved. \(\square \)

We notice the following property of \((\theta ,a)\in X_0\).

Lemma 2.4

If \( (\theta ,a)\in X_ 0\), then, for every \(r\ge 1\), we have:

$$\begin{aligned} \theta (r)\le \sqrt{\pi ^2+2F(\theta ,a)} \frac{1}{\sqrt{r}}. \end{aligned}$$

Proof

Since \(0\le \theta (r)\le \pi \), and \(x^2\le \pi ^2 (1-\cos x)/2\) on \([0,\pi ]\), we have

$$\begin{aligned} \int _0^{+\infty }\big (\theta '(r)^2+\theta (r)^2\big ) r\, dr\le F(\theta ,a); \end{aligned}$$

therefore, for every \(r \ge 1\) we get:

$$\begin{aligned} r\theta (r)^2-\theta (1)^2= & {} \int _1^r\big (\theta (s)^2+2 s\theta (s)\theta '(s)\big )\, ds\le \int _1^r\big (\theta (s)^2 s+2 s\theta (s)\theta '(s)\big )\, ds \\\le & {} \int _1^r\theta (s)^2 s\, ds+2\sqrt{\int _1^r\theta (s)^2 s\, ds}\sqrt{\int _1^r\theta '(s)^2 s\, ds} \\\le & {} \int _0^{+\infty }\big (\theta '(r)^2+2\theta (r)^2\big ) r\, dr \le 2F(\theta ,a), \end{aligned}$$

so that \(r \theta (r)^2\le \pi ^2+2 F(\theta ,a)\), and we get the lemma. \(\square \)

We conclude this section by proving that \(F(\theta ,a)\) is bounded away from zero on X.

Lemma 2.5

For every \((\theta ,a)\in X\) we have \(F(\theta ,a)\ge 4\pi \min (1,\gamma )|k|\).

Proof

Clearly we can assume \((\theta ,a)\in X_0\); from the decomposition (see [14]):

$$\begin{aligned}&\frac{1}{2} \Big (|D_1 u|^2+|D_2 u|^2+(1-e_3\cdot u)^2+(\partial _1 A_2-\partial _2 A_1)^2\Big ) \\&\quad =\pm u\cdot D_1 u\times D_2 u\pm (\partial _1 A_2-\partial _2 A_1)(1-e_3\cdot u) \\&\qquad + \frac{1}{2} |D_1 u\pm u\times D_2 u|^2+ \frac{1}{2} \Big (\partial _1 A_2-\partial _2 A_1\mp (1-e_3\cdot u)\Big )^2 \end{aligned}$$

we get

$$\begin{aligned}&F(\theta ,a) \ge \pm 2\pi \min (1,\gamma )\int _0^{+\infty }\Big ( \frac{k\sin (\theta (r))\theta '(r)}{r} + \frac{k\partial _r ((1-\cos (\theta (r))) a(r))}{r} \Big ) r\, dr \\&\qquad =\pm 2\pi \min (1,\gamma )\Big (-2k+k \int _0^{+\infty }\partial _r ((1-\cos (\theta (r))) a(r))\, dr\Big ); \end{aligned}$$

since \(a(0)=0\), and \(-1\le a(r)\le 0\), the last integral is equal to zero, and the lemma follows. \(\square \)

3 Existence of Gauged Baby Skyrmions

In this Section we want to show that the infimum of \(F(\theta ,a)\) on X is attained.

Proposition 3.1

There exists \((\theta ,a)\in X_0\) such that \(F(\theta ,a)= \min \limits _X F(\theta ,a)\).

Proof

Let us consider a minimizing sequence \((\theta _n,a_n)_n\) for \(F(\theta ,a)\); from Lemma 2.3 we can assume \((\theta _n,a_n)_n\subset X_0\); moreover \(F(\theta _n,a_n)\le M\) for some \(M>0\) that does not depend on n. Since

$$\begin{aligned} \int _0^{+\infty }\big (\theta _n'(r)^2+\theta _n(r)^2\big ) r\, dr \le F(\theta _n,a_n), \end{aligned}$$

the sequence \((\theta _n)_n\) is bounded in \(W^{1,2}([r_1,r_2])\) for every \([r_1,r_2]\subset ]0,+\infty [\), so that, by using a standard diagonal subsequence argument (see for instance [4]), we get a function \(\theta :]0,+\infty [\rightarrow {\mathbb {R}}\) such that (modulo subsequences) \(\theta _n\rightarrow \theta \) weakly in \(W^{1,2}([r_1,r_2])\) and strongly in \(C([r_1,r_2])\) for every \([r_1,r_2]\subset ]0,+\infty [\). We observe now that since \(-1\le a_n(r)\le 0\) and

$$\begin{aligned} \int _0^R a_n'(r)^2\, dr\le R\int _0^R \frac{a_n'(r)^2}{r} \, dr\le \frac{R}{\pi k^2\gamma } F(\theta _n,a_n), \end{aligned}$$

the sequence \((a_n)_n\) is bounded in \(W^{1,2}([0,R])\) for every \(R>0\), so that, arguing as above, there exists a function \(a:[0,+\infty [\rightarrow {\mathbb {R}}\) such that \(a_n\rightarrow a\) weakly in \(W^{1,2}([0,R])\) and strongly in C([0, R]) for every \(R>0\). Clearly \(0\le \theta (r)\le \pi \), \(-1\le a(r)\le 0\), and moreover \(a(0)=0\).

We have also \(F(\theta ,a)<+\infty \); in fact, let us denote by \(F(\theta ,a ;[r_1,r_2])\) the integral of the energy density on the interval \([r_1,r_2]\subset ]0,+\infty [\). From the weak lower semicontinuity of \(F(\theta ,a ; [r_1,r_2])\), we get

$$\begin{aligned} F(\theta ,a ; [r_1,r_2])\le \liminf _{n\rightarrow +\infty }F(\theta _n,a_n ; [r_1,r_2])\le \liminf _{n\rightarrow +\infty }F(\theta _n,a_n)\le M \end{aligned}$$

and since \([r_1,r_2]\) was arbitrary, the claim is proved.

From \(F(\theta ,a)<+\infty \) and Lemma 2.2 we get \(\theta (+\infty )=0\). To conclude the proof, it remains to show the crucial fact that \(\theta (0)=\pi \), and therefore the minimum is not topologically trivial.

In fact, since \(a(0)=0\) and \(a_n\rightarrow a\) uniformly on every interval [0, R], the sequence \((1+a_n(r))_n\) is bounded away from zero on \([0,\delta ]\) for some \(\delta >0\), so that the bound \(F(\theta _n,a_n)\le M\) implies

$$\begin{aligned} \int _0^{\delta } \frac{\sin ^2(\theta _n(r))}{r} \theta _n'(r)^2\, dr\le c, \end{aligned}$$

where \(c>0\) does not depend on n. Then, since \(\theta _n(0)=\pi \), for every \(r\in ]0,\delta [\) we have

$$\begin{aligned} \cos (\theta _n(r))+1\le & {} \int _0^r \frac{|\sin (\theta _n(s))||\theta _n'(s)|}{\sqrt{s}} \sqrt{s}\, ds \\\le & {} \sqrt{\int _0^r \frac{\sin ^2(\theta _n(s))}{s} \theta _n'(s)^2\, ds} \frac{r}{\sqrt{2}} \le \frac{\sqrt{c}}{\sqrt{2}} r. \end{aligned}$$

For \(n\rightarrow +\infty \) we get \(\cos (\theta (r))+1\le \frac{\sqrt{c}}{\sqrt{2}}r\), an so \(\theta (0)=\pi \). \(\square \)

Let \( (\theta ,a)\in X_ 0\) be a minimum point for the functional \(F(\theta ,a)\); by using variations of the form \((\theta +\varepsilon _1\varphi _1,a+\varepsilon _2\varphi _2)\), where \(\varphi _i\), \(i=1,2\) are smooth functions with compact support in \(]0,+\infty [\), we obtain the Euler–Lagrange equations for \(F(\theta ,a)\); namely (see also [8]):

$$\begin{aligned}&\Big (1+k^2 \frac{(1+a(r))^2}{r^2} \sin ^2(\theta (r))\Big ) \theta ''(r) =-k^2(1+a(r))^2 \frac{\sin (2\theta (r))}{2 r^2} \theta '(r)^2 \nonumber \\&\quad -\frac{1}{r} \Big (1-k^2 (1+a(r))(1+a(r)-2 r a'(r)\Big ) \frac{\sin ^2(\theta (r))}{r^2}\theta '(r) \nonumber \\&\quad +k^2 (1+a(r))^2 \frac{\sin (2\theta (r))}{2 r^2} +\sin (\theta (r)) \end{aligned}$$
(3)

and

$$\begin{aligned} \partial _r \frac{a'(r)}{r} = \frac{1}{\gamma } (1+a(r))(1+\theta '(r)^2) \frac{\sin ^2(\theta (r))}{r}. \end{aligned}$$
(4)

To shorten notations we will write in the following the first Euler–Lagrange equation as

$$\begin{aligned} A(r)\theta ''(r)=B(r)\theta '(r)^2+C(r)\theta '(r)+D(r). \end{aligned}$$
(5)

From (3) we get \(\theta '\in W^{1,1}([r_1,r_2])\) for every \([r_1,r_2]\subset ]0,+\infty [\), so that \(\theta \in C^1(]0,+\infty [)\), and, from (4), \(a\in C^2(]0,+\infty [)\). We observe now that the right hand side of (4) is summable on every interval [0, R]; in fact, since \(1+a(r)\) is bounded away from zero on some \([0,\delta ]\), from \(F(\theta ,a)<+\infty \) we get the summability on \([0,\delta ]\); moreover

$$\begin{aligned}&\int _{\delta }^R \frac{1}{\gamma } (1+a(r))(1+\theta '(r)^2) \frac{\sin ^2(\theta (r))}{r} \, dr\le \int _{\delta }^R \frac{1}{\gamma } (1+\theta '(r)^2) \frac{1}{r} \, dr \\&\quad \le \frac{1}{\delta ^2} \int _{\delta }^R \frac{1}{\gamma } (1+\theta '(r)^2) r\, dr<+\infty \end{aligned}$$

because of \(\int _0^{+\infty }\theta '(r)^2 r\, dr<+\infty \). Then \(r\rightarrow a'(r)/r\) is uniformly continuous on ]0, R] for every \(R>0\); in particular, \(\lim _{r\rightarrow 0}a'(r)/r=a''(0)\in {\mathbb {R}}\), and \(a\in C^2([0,+\infty [)\). Proceeding in the same way, we get the regularity stated in Theorem 1.1.

We can prove now Theorem 1.1.

Proof of Theorem 1.1

The existence of a minimizer \((\theta ,a)\) and the regularity of \(\theta (r)\) and a(r) have been proved above. Moreover we know that \(0\le \theta (r)\le \pi \) and \(-1\le a (r)\le 0\); if \(\theta (r_0)=\pi \) for some \(r_ 0>0\) then \(\theta '(r_ 0)=0\), and, from the Euler–Lagrange equations we get \(\theta (r)=\pi \) for every \(r>0\), which is impossible, so that \(\theta (r)<\pi \). Arguing in the same way we have \(0<\theta (r)<\pi \) and \(-1<a(r)<0\) for every \(r>0\).

Moreover the function \(a'(r)/r\) is strictly increasing because of (4), so that \(a'(r)<0\) (for if not, we would have \(a'(r_0)/r_0>0\) for some \(r_0>0\), and then \(a'(r)>a'(r_0)r/r_0\) for \(r>r_0\), so that \(a(+\infty )=+\infty \)); then a(r) is strictly decreasing on \([0,+\infty [\).

It remain to prove that \(\theta (r)\) is strictly decreasing for large value of r; more precisely, since \(\theta (+\infty )=0\), there exists \(R>0\) such that \(\theta (r)<\pi /2\) for \(r>R\). We claim that \(\theta (r)\) is strictly decreasing on \([R,+\infty [\). In fact, let us suppose that \(\theta '(r_1)>0\) for some \(r_1\ge R\); then there exists \(r_2>r_1\) such that \(\theta (r_1)=\theta (r_2)\equiv t\) and \(\theta (r)>t\) on \(]r_1,r_2[\); let us consider now the function \({\tilde{\theta }}(r)= t\) for \(r\in [r_1,r_2]\), and \({\tilde{\theta }}(r)=\theta (r)\) for \(r\notin [r_1,r_2]\); since \(\sin ^2 x\) and \(2 (1-\cos x)\) are increasing on \([0,\pi /2]\), we get \(F[{\tilde{\theta }},a]<F(\theta ,a)\), which is a contradiction, so \(\theta '(r)\le 0\) on \([R,+\infty [\). On the other hand, from Eqs. (3), (4) we see that \(\theta (r)\) can not be constant on a subinterval of \(]0,+\infty [\), and the claim is proved. \(\square \)

Remark 2

Let \((\theta ,a)\) be a minimizer of \(F(\theta ,a)\) on X as in Theorem 1.1; for future references we point out that

$$\begin{aligned} \lim _{r\rightarrow +\infty } r a'(r)=0. \end{aligned}$$
(6)

and

$$\begin{aligned} -\frac{2}{r}\le a'(r)<0 \end{aligned}$$
(7)

for every \(r>0\).

In fact, since \(\sin ^2 x\le \pi ^2(1-\cos x)/2\) for \(x\in [0,\pi ]\) and \(F(\theta ,a)<+\infty \), we have \(\sin ^2(\theta (r))r\in L^1(]0,+\infty [)\). Then, from (4) we get

$$\begin{aligned} 0<r^2\partial _r \frac{a'(r)}{r}< & {} \frac{1}{\gamma } (1+\theta '(r)^2)\sin ^2(\theta (r)) r \\\le & {} \frac{1}{\gamma } (\sin ^2(\theta (r)) r+\theta '(r)^2 r)\in L^1(]0,\infty [); \end{aligned}$$

but \(\partial _r(r a'(r))=2 a'(r)+r^2\partial _r(a'(r)/r)\), so that \(\partial _r(r a'(r))\in L^1(]0,+\infty [)\), and the limit (6) exists and it is \(\le 0\). Clearly \(\lim _{r\rightarrow +\infty } r a'(r)<0\) implies \(r a'(r)<K<0\) on some interval \([R,+\infty [\), and this gives \(a(+\infty )=-\infty \), whereas \(a(+\infty )\ge -1\), so that (6) is proved.

Moreover, for every s, \(r>0\), with \(s\le r\), we have \(a'(s)/s\le a'(r)/r\), so that \(a'(s)\le (a'(r)/r)s\); integrating over [0, r] we get (7).

4 The Weak Coupling Regime

In this section we have to prove Theorem 1.2. We start by proving that \((F(\theta _{\gamma },a_{\gamma }))_{\gamma }\) is bounded from above.

Lemma 4.1

There exists \(C>0\) such that, for every \(\gamma >0\), we have \(F(\theta _{\gamma },a_{\gamma })\le C\).

Proof

Let \(\theta (r)\) be such that

$$\begin{aligned} \pi \int _0^{+\infty }\left( k^2(1+\theta '(r)^2) \frac{\sin ^2(\theta (r))}{r^2} +\theta '(r)^2 \!+\! 2 \big (1-\cos (\theta (r))\big )\right) r\, dr\equiv C_1< \!+\!\infty , \end{aligned}$$

and let

$$\begin{aligned} {\tilde{a}}(r)= {\left\{ \begin{array}{ll} - \frac{1}{\gamma } r^2 &{} \hbox { if}\ 0\le r\le \sqrt{\gamma } \\ -1 &{} \hbox { if}\ r> \sqrt{\gamma } \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} F(\theta ,{\tilde{a}})\le C_1+\pi \int _0^{+\infty }\gamma \frac{k^2{\tilde{a}}'(r)^2}{r} \, dr=C_1+2 k^2\pi , \end{aligned}$$

and since \(F(\theta _{\gamma },a_{\gamma })\le F(\theta ,{\tilde{a}})\), the lemma is proved. \(\square \)

The following lemma show that \(F(\theta ,a)\) can be written in a simpler form at a minimizer \((\theta _{\gamma },a_{\gamma })\).

Lemma 4.2

For every \(\gamma >0\) we have:

$$\begin{aligned} F(\theta _{\gamma },a_{\gamma })= & {} \pi \int _0^{+\infty }\Big ( k^2(1+a_{\gamma }(r))(1+\theta _{\gamma }'(r)^2) \frac{\sin ^2(\theta _{\gamma }(r))}{r^2}\\&\quad +\theta _{\gamma }'(r)^2+2\big (1-\cos (\theta _{\gamma }(r))\big )\Big ) r\, dr. \end{aligned}$$

Proof

By multiplying the Eq. (4) by \(a_\gamma (r)\) and integrating, we have

$$\begin{aligned} \int _0^{+\infty } \frac{a_{\gamma }'(r)^2}{r} \, dr= -\int _0^{+\infty } \frac{1}{\gamma } (1+a_{\gamma }(r)) a_{\gamma }(r)(1+\theta _{\gamma }'(r)^2) \frac{\sin ^2(\theta _{\gamma }(r))}{r} \, dr, \end{aligned}$$

so, inserting the right hand side of this equation in the expression of \(F(\theta _{\gamma },a_{\gamma })\) we get the lemma. \(\square \)

Proof of Theorem 1.2

Let \((\gamma _n)_n\) be a sequence of coupling constants such that \(\gamma _n\rightarrow +\infty \), and set for brevity, \((\theta _{\gamma _n},a_{\gamma _n})=(\theta _n,a_n)\). By multiplying the Eq. (4) by \(r^2\) and integrating we have (see also (6)):

$$\begin{aligned} -a_n(\infty )=\int _0^{+\infty } \frac{1}{2\gamma _n} (1+a_n(r))(1+\theta _n'(r)^2)\sin ^2(\theta _n(r)) r\, dr. \end{aligned}$$
(8)

But, since \(\sin ^2 x\le \pi ^2(1-\cos x)/2\) on \([0,\pi ]\), and by using the Lemma 4.1:

$$\begin{aligned}&\int _0^{+\infty }(1+a_n(r))(1+\theta _n'(r)^2)\sin ^2(\theta _n(r)) r\, dr \\&\quad \le \int _0^{+\infty }\Big (\frac{\pi ^2}{2}\big (1-\cos (\theta _n(r))\big )+\theta _n'(r)^2\Big ) r\, dr\le F(\theta _n,a_n)\le C, \end{aligned}$$

so that \(-a_n(\infty )\le C/2\gamma _n\); passing to the limit for \(n\rightarrow +\infty \), we get the first claim of Theorem 1.2.

To complete the proof, let us denote by \({\bar{X}}\) the set of continuous functions \(\theta :[0,+\infty [ \rightarrow {\mathbb {R}}\) which are absolutely continuous on every compact subinterval of \(]0,+\infty [\), satisfies the boundary conditions \(\theta (0)=\pi \), \(\theta (\infty )=0\) and, moreover, \(F(\theta ,0)<+\infty \). Clearly \((\theta _n)_n\subset {\bar{X}}\); we claim that \((\theta _n)_n\) is a minimizing sequence for the ungauged functional \(F(\theta ,0)\). In fact, we have

$$\begin{aligned} F(\theta _n,a_n)\le \inf _{\theta \in {\bar{X}}}F(\theta ,0)\le F(\theta _n,0); \end{aligned}$$

moreover, since \(a_n(\infty )\rightarrow 0\), we can assume \(1+a_n(r)>\frac{1}{2}\), so that

$$\begin{aligned}&\pi \int _0^{+\infty } k^2(1+\theta _n'(r)^2) \frac{\sin ^2(\theta _n(r))}{r} \, dr\\&\quad \le 4\pi \int _0^{+\infty } k^2(1+a_n(r))^2(1+\theta _n'(r)^2) \frac{\sin ^2(\theta _n(r))}{r} \, dr\\&\quad \le 4 F(\theta _n,a_n)\le 4 C, \end{aligned}$$

where \(C>0\) is the constant of Lemma 4.1. Then, from Lemma 4.2 we have

$$\begin{aligned} 0<F(\theta _n,0)-F(\theta _n,a_n)= & {} \pi \int _0^{+\infty } k^2(-a_n(r))(1+\theta _n'(r)^2) \frac{\sin ^2(\theta _n(r))}{r} \, dr\\&<-4 a_n(\infty ) C \rightarrow 0, \end{aligned}$$

therefore \(F(\theta _n,0)\rightarrow \inf _{\theta \in {\bar{X}}} F(\theta ,0)\) and the claim is proved. But, for the ungauged functional \(F(\theta ,0)\), it is well known (from [4], or by using the arguments of Theorem 1.1) that there exists \({\hat{\theta }}\in {\bar{X}}\), such that \(F({\hat{\theta }},0)=\inf _{\theta \in {\bar{X}}} F(\theta ,0)\), and (up a subsequence) \(\theta _n\rightarrow {\hat{\theta }}\) weakly in \(W^{1,2}([r_1,r_2])\) and uniformly on every compact subinterval \([r_1,r_2]\subset ]0,+\infty [\), and the proof is complete. \(\square \)

5 The Strong Coupling Regime

Let us consider a sequence \((\gamma _n)_n\) such that \(\gamma _n\rightarrow 0\), and set again \((\theta _{\gamma _n},a_{\gamma _n})=(\theta _n,a_n)\). From (7) and the fact that \(-1\le a_n(r)\le 0\), we get that for every \([r_1,r_2]\subset ]0,+\infty [\), \((a_n)_n\) is bounded in \(W^{1,2}([r_1,r_2])\). Then, there exists a continuous function \({\bar{a}}:]0,+\infty [\rightarrow {\mathbb {R}}\) such that, up subsequences, \(a_n\rightarrow {\bar{a}}\) weakly in \(W^{1,2}([r_1,r_2])\) and uniformly on \([r_1,r_2]\) for every \([r_1,r_2]\subset ]0,+\infty [\); of course we have \(-1\le {\bar{a}}(r)\le 0\) and \({\bar{a}}(r)\) is decreasing.

Since \(F(\theta _n,a_n)\le C\) (see Lemma 4.1) implies that the sequence \(\big (\int _0^{+\infty }(\theta _n(r)^2+\theta _n'(r)^2) r\, dr\big )_n\) is bounded, we get in the same way a continuous function \({\bar{\theta }}:]0,+\infty [\rightarrow {\mathbb {R}}\) such that, up subsequences, \(\theta _n\rightarrow {\bar{\theta }}\) weakly in \(W^{1,2}([r_1,r_2])\) and uniformly on \([r_1,r_2]\) for every \([r_1,r_2]\subset ]0,+\infty [\); clearly \(0\le {\bar{\theta }}(r)\le \pi \), and moreover \(\int _0^{+\infty }({\bar{\theta }}(r)^2+{\bar{\theta }}'(r)^2) r\, dr<+\infty \), so that we have \({\bar{\theta }}(\infty )=0\).

We aim to show that \({\bar{a}}(r)\equiv -1\) and \({\bar{\theta }}(r)\equiv 0\). We start by proving the following lemma.

Lemma 5.1

There exists \(R>0\) such that \({\bar{a}}(R)=-1\).

Proof

Let us suppose, by contradiction, that \({\bar{a}}(R)>-1\) for every \(R>0\), and fix \(R>0\); since \(a_n(r)\) is strictly decreasing, from (8) we have

$$\begin{aligned} \int _0^R\sin ^2(\theta _n(r)) r\, dr\le & {} \frac{1}{1+a_n(R)} \int _0^R(1+a_n(r))(1+\theta _n'(r)^2)\sin ^2(\theta _n(r)) r\, dr\\\le & {} -\frac{2\gamma _n a_n(\infty )}{1+a_n(R)} \le \frac{2\gamma _n}{1+a_n(R)}\rightarrow 0, \end{aligned}$$

and since for every r with \(0<r<R\) we have \(\theta _n\rightarrow {\bar{\theta }}\) uniformly on [r, R], we obtain \(\int _r^R\sin ^2({\bar{\theta }}(r)) r\, dr=0\), so that \({\bar{\theta }}(r)\equiv 0\) or \({\bar{\theta }}(r)\equiv \pi \) on ]0, R]. On the other hand, from Lemma 4.1 we have

$$\begin{aligned} \int _0^R\theta _n'(r)^2 \frac{\sin ^2(\theta _n(r))}{r} \, dr\le & {} \int _0^R(1+\theta _n'(r)^2) \frac{\sin ^2(\theta _n(r))}{r} \, dr\\\le & {} \frac{1}{\pi k^2(1+a_n(R))^2} \pi \int _0^R k^2(1+a_n(r))^2(1+\theta _n'(r)^2)\\&\frac{\sin ^2(\theta _n(r))}{r} \,dr\\\le & {} \frac{C}{\pi k^2(1+a_n(R))^2}, \end{aligned}$$

so that, for every \(r\in [0,R]\):

$$\begin{aligned} \cos (\theta _n(r))+1\le & {} \int _0^r \frac{|\sin (\theta _n(s))||\theta _n'(s)|}{\sqrt{s}} \sqrt{s}\, ds\\\le & {} \sqrt{\int _0^R\theta _n'(s)^2 \frac{\sin ^2(\theta _n(s))}{s} \, ds}\, \frac{r}{\sqrt{2}} \le \frac{\sqrt{C}}{\sqrt{2\pi }|k|(1+a_n(R))} r. \end{aligned}$$

Passing to the limit we get \(\cos ({\bar{\theta }}(r))+1\le (\sqrt{C}/\sqrt{2\pi }|k|(1+{\bar{a}}(R)))r\) for every \(r\in [0,R]\), and then we must have \({\bar{\theta }}(r)\equiv \pi \) on ]0, R]. Since R is arbitrary, we have \({\bar{\theta }}(r)\equiv \pi \) on \(]0,+\infty [\), and this is impossible, since \({\bar{\theta }}(\infty )=0\). \(\square \)

From now on we set \(R_0=\inf \{ r>0\mid {\bar{a}}(R)=-1\}\). Of course \(R_0\ge 0\) and \({\bar{a}}(r)\equiv -1\) on \([R_0,+\infty [\); moreover, if \(R_0>0\), we have, from the proof of the lemma above, \({\bar{\theta }}(r)\equiv \pi \) on \(]0,R_0]\).

Lemma 5.2

For every \([r_1,r_2]\subset ]R_0,+\infty [\) we have \(\lim _{n\rightarrow +\infty }a_n'(r)=0\) uniformly on \([r_1,r_2]\).

Proof

Let \(r>R_0\) be fixed; for every \(s\in [R_0,r]\) we have \(a_n'(s)<(a_n'(r)/r)s\); then, by integration, \(a_n(r)-a_n(R_0)<a_n'(r)(r^2-R_0^2)/2r<0\), so that, since \(a_n(r)-a_n(R_0)\rightarrow 0\), we have \(a_n'(r)\rightarrow 0\) for every \(r>R_0\). Therefore \(a_n'(r)/r\) goes to zero uniformly on every \([r_1,+\infty [\subset ]R_0,+\infty [\), and so \(a_n'(r)\rightarrow 0\) on every \([r_1,r_2]\subset ]R_0,+\infty [\). \(\square \)

Lemma 5.3

For every \([r_1,r_2]\subset ]0,+\infty [\), the sequences \((\theta _n')_n\) and \((\theta _n'')_n\) are bounded in \(L^{\infty }([r_1,r_2])\).

Proof

Let \([r_1,r_2]\subset ]0,+\infty [\); for every n there exists \(R_n\in [r_1,r_2]\) such that \(\theta _n(r_2)-\theta _n(r_1)=\theta _n'(R_n)(r_2-r_1)\), and since \(\theta _n(r_2)-\theta _n(r_1)\rightarrow {\bar{\theta }}(r_2)-{\bar{\theta }}(r_1)\), the sequence \((\theta _n'(R_n))_n\) is bounded. Moreover \(\theta _n(r)\) satisfies the Eq. (5), namely

$$\begin{aligned} A_n(r)\theta _n''(r)=B_n(r)\theta _n'(r)^2+C_n(r)\theta _n'(r)+D_n(r), \end{aligned}$$

where the coefficients depend on n. Recalling that \(-1<a_n(r)<0\) and \(-2\le r a_n'(r)<0\) because of (7), it is easy to check that \(B_n(r)\), \(C_n(r)\) and \(D_n(r)\) are bounded in \(L^{\infty }([r_1,r_2])\), so that

$$\begin{aligned} |\theta _n''(r)|\le C_1\theta _n'(r)^2+C_2 |\theta _n'(r)|+C_3 \end{aligned}$$
(9)

on \([r_1,r_2]\), where \(C_i\), \(i=1,2,3\), does not depend on n; Lemma 4.1 implies that \((\theta _n')_n\) is bounded in \(L^2([r_1,r_2])\), so that \((\theta _n'')_n\) is bounded in \(L^1([r_1,r_2])\). Since clearly \(|\theta _n'(r)-\theta _n'(R_n)|\le \int _{R_n}^r |\theta _n''(r)|\, dr\), we get the boundness of \((\theta _n')_n\) in \(L^{\infty }([r_1,r_2])\) and, by (9), also \((\theta _n'')_n\) is bounded in \(L^{\infty }([r_1,r_2])\) as claimed.

\(\square \)

Lemma 5.4

The function \({\bar{\theta }}(r)\) is differentiable on \(]0,+\infty [\), twice differentiable on \(]R_0,+\infty [\), and \(\lim _{n\rightarrow +\infty }\theta _n'(r)={\bar{\theta }}'(r)\) uniformly on every \([r_1,r_2]\subset ]0,+\infty [\), \(\lim _{n\rightarrow +\infty }\theta _n''(r)={\bar{\theta }}''(r)\) uniformly on every \([r_1,r_2]\subset ]R_0,+\infty [\). Moreover \({\bar{\theta }}(r)\) satisfies the equation \(\partial _r(r{\bar{\theta }}'(r))=\sin ({\bar{\theta }}(r))r\) on \(]R_0,+\infty [\).

Proof

Let us consider the sequence \((\theta _n')_n\); since \((\theta _n'')_n\) is bounded over the compact subsets of \(]0,+\infty [\), by the Ascoli-Arzelà theorem there exists a continuous function \(\eta :]0,+\infty [\rightarrow {\mathbb {R}}\) such that, up subsequences, \(\theta _n'\rightarrow \eta \) uniformly on every \([r_1,r_2]\subset ]0,+\infty [\). But we know that \(\theta _n\rightarrow {\bar{\theta }}\), so that \({\bar{\theta }}\) is differentiable and \({\bar{\theta }}'=\eta \), and the first claim is proved.

We observe now that, since \(\theta _n\rightarrow {\bar{\theta }}\), \(a_n\rightarrow -1\) and \(a_n'\rightarrow 0\) uniformly on every \([r_1,r_2]\subset ]R_0,+\infty [\), by the Eq. (3) the same holds true for the sequence \((\theta _n'')_n\); but \(\theta _n'\rightarrow {\bar{\theta }}'\), so that \({\bar{\theta }}'\) is differentiable on \(]R_0,+\infty [\), and \(\theta _n''(r)\rightarrow {\bar{\theta }}''(r)\) uniformly on every \([r_1,r_2]\subset ]R_0,+\infty [\). Moreover, from (3) we get also the equation \(\partial _r(r{\bar{\theta }}'(r))=\sin ({\bar{\theta }}(r))r\). \(\square \)

We can prove now Theorem 1.3.

Proof of Theorem 1.3

First of all, we claim that \(R_0=0\); in fact, let us suppose \(R_0>0\); then \({\bar{\theta }}(r)\equiv \pi \) on \(]0,R_0]\); since \({\bar{\theta }}(r)\) is differentiable at \(R_0\), we must have \({\bar{\theta }}'(R_0)=0\), and, from the equation \(\partial _r(r{\bar{\theta }}'(r))=\sin ({\bar{\theta }}(r))r\) we get \({\bar{\theta }}(r)\equiv \pi \); but this is impossible since \({\bar{\theta }}(\infty )= 0\). Therefore \(R_0=0\), so that \({\bar{a}}(r)\equiv -1\), and, since \(a_n(r)\) is strictly decreasing, \(a_n(r)\rightarrow -1\) uniformly on \([R,+\infty [\) for every \(R>0\); in particular, \(a_n(\infty )\rightarrow -1\).

We want to show now that \({\bar{\theta }}(r)\equiv 0\); in fact, since \(0\le {\bar{\theta }}(r)\le \pi \), from equation \(\partial _r(r{\bar{\theta }}'(r))=\sin ({\bar{\theta }}(r))r\), we deduce that \(r{\bar{\theta }}'(r)\) is not decreasing; clearly we must have \({\bar{\theta }}'(r)\le 0\); for if not \(0<r_0{\bar{\theta }}'(r_0)/r\le {\bar{\theta }}'(r)\) on some interval \([r_0,+\infty [\), and, by integration, \({\bar{\theta }}(\infty )= +\infty \). Therefore the function \(r{\bar{\theta }}'(r)\) is \(\le 0\) and not decreasing. In particular, there exists the limit

$$\begin{aligned} \lim _{r\rightarrow 0}r{\bar{\theta }}'(r)\le 0. \end{aligned}$$

We claim that the above limit is equal to zero. For if not, there exists \(K\in {\mathbb {R}}\) with \(K<0\), and a neighbourhood \(]0,r_0[\) of zero such that \({\bar{\theta }}'(r)<K/r\) on \(]0,r_0[\) and, by integration, \({\bar{\theta }}(r_0)-{\bar{\theta }}(r)<K \log (r_0/r)\) for every \(r\in ]0,r_0[\), so that \({\bar{\theta }}(r)\rightarrow +\infty \) as \(r\rightarrow 0\), whereas \(0\le {\bar{\theta }}(r)\le \pi \). Then \({\bar{\theta }}'(r)\equiv 0\), and so \({\bar{\theta }}(r)\equiv 0\) as claimed, and therefore \(\theta _n(r)\rightarrow 0\) on \(]0,+\infty [\).

Finally, let \(R>0\) be fixed; for every n there exists \(R_n>0\) such that \(\theta _n(R_n)=\pi /2\), and \(\theta _n(r)\) is strictly decreasing on \([R_n,+\infty [\) (see the proof of Theorem 1.1). From Lemmas 2.4 and 4.1 we have

$$\begin{aligned} R_n\le \max \Big (1,\frac{4}{\pi ^2}\big (\pi ^2+2 C\big )\Big ); \end{aligned}$$

on the other hand, \(\theta _n(r)\rightarrow 0\) uniformly on every \([r_1,r_2]\subset ]0,+\infty [\), so that we must have \(R_n\rightarrow 0\). Then, for n large enough, we have \(R_n<R\), so \(\theta _n(r)\) is strictly decreasing on \([R,+\infty [\), and, since \(\theta _n(R)\rightarrow 0\), we get \(\theta _n\rightarrow 0\) uniformly on \([R,+\infty [\). \(\square \)