Abstract
As has been outlined in the introduction, the eigenvalues \(\lambda <\delta _1^2\) of \(L=-\mathcal{T}^2-\mathcal{K}\mathcal{T}\) from (1.16) are in one-to-one correspondence with the eigenvalues 1 of a certain Birman-Schwinger type operator \(\mathcal{Q}_\lambda \) that acts on functions \(\Psi =\Psi (r)\).
Access provided by Autonomous University of Puebla. Download chapter PDF
As has been outlined in the introduction, the eigenvalues \(\lambda <\delta _1^2\) of \(L=-\mathcal{T}^2-\mathcal{K}\mathcal{T}\) from (1.16) are in one-to-one correspondence with the eigenvalues 1 of a certain Birman-Schwinger type operator \(\mathcal{Q}_\lambda \) that acts on functions \(\Psi =\Psi (r)\).
4.1 The Operator \(\mathcal{Q}_z\)
Let \(L^2_r\) denote the \(L^2\)-Lebesgue space of radially symmetric functions \(\Psi (x)=\Psi (r)\) on \(\mathbb R^3\), where we take
as the inner product of \(\Psi , \Phi \in L^2_r\). Unless otherwise stated, a generic constant (denoted by C) is allowed to depend only upon Q.
Definition 4.1
For \(z\in \Omega =\mathbb C\setminus [\delta _1^2, \infty [\), we introduce
where \(r_{\pm }(e, \ell )\) and \(\theta (r, e, \ell )\) are as in Appendix I, Sect. A.1, and D is given by (3.1). Along with \(\mathcal{Q}_z\), we also introduce the integral kernels
Remark 4.2
-
(a)
If \(z=a+ib\in \mathbb C\setminus \mathbb R\), then \(|k^2\omega _1^2(e, \ell )-z|\ge |b|>0\). More precisely,
$$\begin{aligned} |k|\ge \Big [\frac{\sqrt{2|a|}}{\delta _1}\Big ]+1\,\,\Longrightarrow \,\, |k^2\omega _1^2(e, \ell )-z|^2= & {} (k^2\omega _1^2(e, \ell )-a)^2+b^2 \nonumber \\\ge & {} (k^2\delta _1^2-|a|)^2+b^2 \nonumber \\\ge & {} \frac{1}{4} k^4\delta _1^4+b^2. \end{aligned}$$(4.3)
On the other hand, if \(z=\lambda \in ]-\infty , \delta _1^2[\), then
and hence
In particular, \(\frac{1}{k^2\omega _1^2(e, \ell )-z}\) in (4.1) and (4.2) is well-defined for \(z\in \Omega \).
(b) In the definitions, we understand the factor \(|Q'(e)|\) to be zero outside of K, the support of Q, instead of carrying around another characteristic function all the time. In particular, always \(r_+(e, \ell )\le r_Q\) holds, which means the following: in (4.1), \(\int _0^\infty d\tilde{r}\,\Psi (\tilde{r})\) can be replaced by \(\int _0^{r_Q} d\tilde{r}\,\Psi (\tilde{r})\); \((\mathcal{Q}_z\Psi )(r)\) can be replaced by \((\mathcal{Q}_z\Psi )(r)\,\mathbf{1}_{\{0\le r\le r_Q\}}\) and \(K_z(r, \tilde{r})\) can be replaced by \(K_z(r, \tilde{r})\,\mathbf{1}_{\{0\le r, \tilde{r}\le r_Q\}}\). \(\diamondsuit \)
Lemma 4.3
[Properties of \({\mathcal Q}_z\)] The following assertions hold.
-
(a)
For every \(z\in \Omega \), we have \(\mathcal{Q}_z\in \mathcal{B}(L^2_r)\), the space of linear and bounded operators on \(L^2_r\). In addition, the map
$$\begin{aligned} \Omega \ni z\mapsto \mathcal{Q}_z\in \mathcal{B}(L^2_r) \end{aligned}$$(4.5)is analytic, and for the derivatives
$$\begin{aligned} (\mathcal{Q}^{(j)}_z\Psi )(r)= & {} \frac{16\pi j!}{r^2}\sum _{k\ne 0}\int _0^\infty d\tilde{r}\,\Psi (\tilde{r}) \iint \limits _D d\ell \,\ell \,de\,\mathbf{1}_{\{r_-(e,\,\ell )\le r,\,\tilde{r}\le r_+(e,\,\ell )\}}\\&\qquad \qquad \qquad \times \frac{\omega _1(e, \ell )\,|Q'(e)|}{(k^2\omega _1^2(e, \ell )-z)^{j+1}} \sin (k\theta (r, e, \ell ))\sin (k\theta (\tilde{r}, e, \ell )) \end{aligned}$$for \(\Psi \in L^2_r\).
-
(b)
If \(z\in \Omega \), then
$$\begin{aligned} (\mathcal{Q}_z\Psi )(r)=\langle K_{\bar{z}}(r, \cdot ), \Psi \rangle \end{aligned}$$for \(\Psi \in L^2_r\). In particular,
$$\begin{aligned} \langle \mathcal{Q}_z\Psi , \Phi \rangle =\langle \Psi , \mathcal{Q}_{\bar{z}}\Phi \rangle \end{aligned}$$for \(\Psi , \Phi \in L^2_r\), so that \(Q_z^*=Q_{\bar{z}}\). Thus, if \(\lambda \in ]-\infty , \delta _1^2[\), then \(\mathcal{Q}_\lambda \) is symmetric.
-
(c)
If \(z\in \Omega \), then \(\mathcal{Q}_z\) is a Hilbert-Schmidt operator on \(L^2_r\).
-
(d)
If \(z\in \Omega \), then
$$\begin{aligned}&{\langle \mathcal{Q}_z\Psi , \Psi \rangle } \\= & {} 64\pi ^2\sum _{k\ne 0} \iint \limits _D d\ell \,\ell \,de\,\frac{\omega _1(e, \ell )\,|Q'(e)|}{k^2\omega _1^2(e, \ell )-\bar{z}} \,\bigg |\int _{r_-(e,\,\ell )}^{r_+(e,\,\ell )}\Psi (r)\sin (k\theta (r, e, \ell ))\,dr\bigg |^2 \end{aligned}$$for \(\Psi \in L^2_r\). In particular, if \(\lambda \in ]-\infty , \delta _1^2[\), then \(\langle \mathcal{Q}_\lambda \Psi , \Psi \rangle \ge 0\) for \(\Psi \in L^2_r\), i.e., \(\mathcal{Q}_\lambda \) is positive. In addition, for the derivatives
$$\begin{aligned} {\langle \mathcal{Q}^{(j)}_z\Psi , \Psi \rangle }= & {} 64\pi ^2 j!\sum _{k\ne 0} \iint \limits _D d\ell \,\ell \,de\,\frac{\omega _1(e, \ell )\,|Q'(e)|}{(k^2\omega _1^2(e, \ell )-\bar{z})^{j+1}} \nonumber \\&\qquad \qquad \qquad \qquad \times \,\bigg |\int _{r_-(e,\,\ell )}^{r_+(e,\,\ell )}\Psi (r)\sin (k\theta (r, e, \ell ))\,dr\bigg |^2 \end{aligned}$$(4.6)for \(\Psi \in L^2_r\).
-
(e)
There is a constant \(C>0\) such that for \(\lambda , \tilde{\lambda }\in ]-\infty , \delta _1^2[\),
$$ {\Vert \mathcal{Q}_\lambda -\mathcal{Q}_{\tilde{\lambda }}\Vert }_{\mathrm{HS}} \le C\,\Big (1+\frac{1}{(\delta _1^2-\lambda )(\delta _1^2-\tilde{\lambda })}\Big ) \,|\lambda -\tilde{\lambda }|, $$where \({\Vert \cdot \Vert }_{\mathrm{HS}}\) denotes the Hilbert-Schmidt norm.
-
(f)
If \(\lambda \in ]-\infty , \delta _1^2[\), then the spectrum of \(\mathcal{Q}_\lambda \) consists of \(\mu _1(\lambda )\ge \mu _2(\lambda )\ge \ldots \rightarrow 0\) (the eigenvalues are listed according to their multiplicities). In addition,
$$\begin{aligned} \mu _1(\lambda )=\Vert \mathcal{Q}_\lambda \Vert =\sup \,\{\langle \mathcal{Q}_\lambda \Psi , \Psi \rangle : {\Vert \Psi \Vert }_{L^2_r}\le 1\}, \end{aligned}$$(4.7)where \(\Vert \cdot \Vert ={\Vert \cdot \Vert }_{\mathcal{B}(L^2_r)}\), and every function
$$\begin{aligned} \mu _k(\cdot ):\,\,]-\infty , \delta _1^2[\,\rightarrow \,]0, \infty [ \end{aligned}$$for \(k\in \mathbb N\) is monotone increasing and locally Lipschitz continuous (and hence differentiable a.e. by Rademacher’s Theorem).
Proof
(a) Let \(z\in \Omega \) be fixed. By Remark 4.2(a), there is \(\alpha _0>0\) such that \(|k^2\omega _1^2\) \((e, \ell )-z|\ge \alpha _0\) for \(|k|\ge 1\) and \((e, \ell )\in D\). In addition, according to (4.3) and (4.4), there is \(k_0\in \mathbb N\) so that \(|k^2\omega _1^2(e, \ell )-z|\ge \frac{1}{2} k^2\delta _1^2\) for \(|k|\ge k_0\) and \((e, \ell )\in D\); if \(k_0\) is taken to be large enough, we can also make sure that \(\frac{1}{2} k^2\delta _1^2\ge k^{3/2}\). First, we observe that
To establish this claim, we recall from (3.7) that \(\ell ^2=2r_-^2(e-U_Q(r_-))\) holds, where \(r_{\pm }=r_{\pm }(e, \ell )\). Since \(U_Q\) is increasing and \(e\le e_0\), we get \(\ell ^2\le 2r_-^2(e_0-U_Q(0))\le 2r^2(e_0-U_Q(0))\).
For \(1\le |k|\le k_0\) and \(i\in \mathbb N_0\), we now apply (4.8) to r and \(\tilde{r}\) in order to estimate
as
Analogously, for \(|k|\ge k_0\) and \(i\in \mathbb N_0\), we deduce
It follows that
for
this constant depends upon z and Q, but \(k_0\) is independent of i. Therefore,
Next, note that
Thus, using Hölder’s inequality,
and this in turn leads to
To prove the analyticity of (4.5), we recall that it suffices to show weak analyticity, in the sense that all maps \(\Omega \ni z\mapsto \langle \Psi , Q_z\Phi \rangle \in \mathbb C\) for \(\Psi , \Phi \in L^2_r\) are analytic; see [85, Thm. 3.1.12]. Fix \(z_0\in \Omega \). If \(|z-z_0|\) is sufficiently small, then \(z\in \Omega \) and we have the series expansion
for every \(k\ne 0\) and \((e, l)\in D\), which suggests that
for
We are going to show that the series (4.13) converges near \(z_0\). For this, due to (4.10) and (4.12), we deduce that
If we write the constant \(C_{1, i}\) from (4.11) as \(C_{1, i}=\tilde{C}_1\alpha _0^{-(i+1)}+\hat{C}_1\), with \(\alpha _0\) depending only on \(z_0\), then \(|z-z_0|<\min \{\frac{\alpha _0}{2}, \frac{1}{2}\}\) ensures that
which has a finite \(\sum _{i=0}^\infty \). It follows that (4.13) converges for \(z\in \Omega \) such that \(|z-z_0|<\min \{\frac{\alpha _0}{2}, \frac{1}{2}\}\), i.e., on a sufficiently small ball about \(z_0\). The formula for the derivative is gotten from \(a_1\) and those for the higher order derivatives follow from this one inductively.
-
(b)
By the definition of \(K_z\) in (4.2), we have
$$\begin{aligned} K_z(r, \tilde{r})=\frac{4}{r^2\tilde{r}^2}\sum _{k\ne 0} s_{k, 0}(r, \tilde{r}, z). \end{aligned}$$(4.14)
Hence,
observing that \(\overline{K_z}=K_{\bar{z}}\). Due to \(K_z(r, \tilde{r})=K_z(\tilde{r}, r)\), we hence obtain
-
(c)
According to (b), the operator \(\mathcal{Q}_z\) on \(L^2_r\) has the integral kernel \(K_{\bar{z}}\). Hence, in order to verify that \(\mathcal{Q}_z\) is Hilbert-Schmidt, we need to verify that
for every \(z\in \Omega \), where \(K_z\) is viewed both as a function of \((x, \bar{x})\) and a function of \((r, \tilde{r})\) and we used Remark 4.2(b); see [35, Prop. 6.36]. From (4.14), (4.10) and (4.12), we get
Note that from \(\mathcal{Q}_z\) being Hilbert-Schmidt it follows that \(\mathcal{Q}_z\) is bounded and \(\Vert \mathcal{Q}_z\Vert \le {\Vert \mathcal{Q}_z\Vert }_{\mathrm{HS}}\), i.e., once again we see that (a) holds. However, since the key of the argument is (4.10) and (4.12), it needs very little additional work to derive both bounds. (d) Here, we calculate
The proof of (4.6) is analogous. (e) For \(\lambda , \tilde{\lambda }<\delta _1^2\), we have, cf. (4.16),
Using (4.8) and (4.12), we may continue this estimate for suitable constants \(C, \hat{C}>0\) as
For \(k\ge 2\), we know from Remark 4.2(a) that \(k^2\omega _1^2(e, \beta )-\lambda \ge k^2\delta _1^2/2\) and \(k^2\omega _1^2(e, \beta )-\tilde{\lambda }\ge k^2\delta _1^2/2\) are verified. If \(k=1\), then always \(\omega _1^2(e, \beta )-\lambda \ge \delta _1^2-\lambda \) and \(\omega _1^2(e, \beta )-\tilde{\lambda }\ge \delta _1^2-\tilde{\lambda }\) hold. Thus, we arrive at
and this yields the claim.
(f) According to (b–d), \(\mathcal{Q}_\lambda \) is a symmetric and positive Hilbert-Schmidt operator, which is in particular compact. Thus, the assertions up to and including (4.7) are a consequence of the spectral theory for compact positive self-adjoint operators; see [35, Section 6]. Concerning the \(\mu _k(\lambda )\), we have the characterization
according to the Courant max-min principle. In the present situation, this follows from the spectral decomposition theorem for symmetric and compact operators. By (d), we obtain for \(\tilde{\lambda }\ge \lambda \), both in \(]-\infty , \delta _1^2[\) and \(\Psi \in L^2_r\),
where \(r_{\pm }=r_{\pm }(e, \ell )\) and \(\theta =\theta (r, e, \ell )\). Hence, (4.18) implies that \(\mu _k(\tilde{\lambda })\ge \mu _k(\lambda )\) for all \(k\in \mathbb N\). To establish the local Lipschitz continuity of \(\mu _k(\cdot )\), note that
whence we deduce from (e) and \(\Vert \cdot \Vert \le {\Vert \cdot \Vert }_{\mathrm{HS}}\) that for \(\Psi \in L^2_r\) satisfying \({\Vert \Psi \Vert }_{L^2_r}\le 1\), one has
Applying (4.18) once more, we arrive at
which completes the proof. \(\square \)
In the following, we are going to derive some more specific properties of the \(\mathcal{Q}_z\). See Appendix II, Sect. B.1 below for the function spaces that are being used. Once again, we understand that \(|Q'(e_Q)|\) vanishes outside of K.
Lemma 4.4
If \(z\in \Omega \) and \(\psi (r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi (r)\) for \(\Psi \in L^2_r\), then \(\psi \in X^0_{\mathrm{odd}}\),
and
In particular,
Moreover, if also \(\tilde{\psi }(r, p_r, \ell )=|Q'(e_Q)|\,p_r\tilde{\Psi }(r)\) for some \(\tilde{\Psi }\in L^2_r\), then
Proof
First, note that \(\psi \) is odd in v and has its support in K. Furthermore, due to Remark B.2(a), Lemma 2.5 and (A.32),
Thus, \(\psi \in X^0_{\mathrm{odd}}\subset X^0_0\), and accordingly Corollary B.14 yields
On the other hand,
by Lemma B.5. Therefore, we arrive at
and this completes the proof of (4.21), by the definition of \(\mathcal{Q}_z\). Concerning (4.22), the first part follows from \(\mathcal{K} g=|Q'(e_Q)|\,p_r\,U'_g(r)\), see (B.37), and for the second part, one just has to use Lemma 2.4. Lastly, (4.23) is a direct consequence of (4.20) and the fact that \((\tilde{\psi }-\psi )(r, p_r, \ell ) =|Q'(e_Q)|\,p_r (\tilde{\Psi }-\Psi )(r)\). \(\square \)
Now, we can make the connection from eigenvalues \(\lambda <\delta _1^2\) of the self-adjoint operator
cf. (1.16) and Corollary B.19, to eigenvalues 1 of \(Q_\lambda \).
Theorem 4.5
Let \(\lambda <\delta _1^2\). Then \(\lambda \) is an eigenvalue of L if and only if 1 is an eigenvalue of \(\mathcal{Q}_\lambda \). More precisely,
-
(a)
if \(u\in X^2_{\mathrm{odd}}\) is an eigenfunction of L for the eigenvalue \(\lambda \), then \(\Psi =U'_{\mathcal{T}u}\in L^2_r\) for \(r\in [0, r_Q]\) is an eigenfunction of \(\mathcal{Q}_\lambda \) for the eigenvalue 1;
-
(b)
if \(\Psi \in L^2_r\) is an eigenfunction of \(\mathcal{Q}_\lambda \) for the eigenvalue 1, then \(u=(-\mathcal{T}^2-\lambda )^{-1}(|Q'(e_Q)|\,p_r\Psi )\in X^2_{\mathrm{odd}}\) is an eigenfunction of L for the eigenvalue \(\lambda \).
Proof
First, suppose that \(Lu=\lambda u\) for some \(u\in X^2_{\mathrm{odd}}\) and \(u\ne 0\). Then \((-\mathcal{T}^2-\lambda )u=\mathcal{K}\mathcal{T}u\). Defining \(\psi =(-\mathcal{T}^2-\lambda )u\in X^0_{\mathrm{odd}}\), Remark B.18(a) implies that \(\psi =\mathcal{K}\mathcal{T}(-\mathcal{T}^2-\lambda )^{-1}\psi \). Since \(\mathcal{K} g=|Q'(e_Q)|\,p_r\,U'_g(r)\) by (B.37), we can put
for \(r\in [0, r_Q]\) to obtain \(\psi =|Q'(e_Q)|\,p_r\Psi (r)\). Then \(\Psi \ne 0\), as otherwise \(\psi =0\) and \(u=0\). Next, we are going to verify that \(\Psi \in L^2_r\). Using (B.40) from Lemma B.15 and Lemma B.8(c), we get
In particular, Lemma B.8(a) implies \({\Vert \Psi \Vert }^2_{L^2_r}\le 4\pi {\Vert \mathcal{T}u\Vert }^2_{X^0} \le 4\pi \Delta _1^2\,{\Vert u\Vert }^2_{X^1}<\infty \), so that indeed \(\Psi \in L^2_r\). Thus, we deduce from Lemma 4.4 that
and consequently \(\mathcal{Q}_\lambda \Psi =\Psi \).
Conversely, suppose that \(\mathcal{Q}_\lambda \Psi =\Psi \) is verified for some \(\Psi \in L^2_r\) and \(\Psi \ne 0\). According to Remark 4.2(b), \(\Psi \) has its support in \([0, r_Q]\). Defining \(\psi =|Q'(e_Q)\) \(|\,p_r\Psi (r)\), we obtain \(\psi \in X^0_{\mathrm{odd}}\) from Lemma 4.4. As a consequence, \(u=(-\mathcal{T}^2-\lambda )^{-1}\psi \in X^2_{\mathrm{odd}}\). Also \(u\ne 0\), since otherwise \(\psi =0\) and \(\Psi =0\). From Lemma 4.4, we finally get
so that \(Lu=-\mathcal{T}^2 u-\mathcal{K}\mathcal{T}u=\lambda u\). \(\square \)
Lemma 4.6
The following assertions hold.
-
(a)
To \(\Psi \in L^2_r\) we associate the function \(\psi (r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi (r)\). If \(z\in \Omega \), then
$$\begin{aligned} \langle \mathcal{Q}_z\Psi , \Psi \rangle =64\pi ^4\sum _{k\ne 0}\iint \limits _D d\ell \,\ell \,de\,\frac{1}{\omega _1(e, \ell )\,|Q'(e)|} \,\frac{1}{k^2\omega _1^2(e, \ell )-\bar{z}} \,|\psi _k(I, \ell )|^2. \end{aligned}$$(4.25) -
(b)
Let \(\Psi \in L^2_r\) be given and suppose that \(F(r)=F(0)+\int _0^r\Psi (s)\,ds\) for \(r\in [0, r_Q]\) as well as \(g=-|Q'(e_Q)|(F-F_0)\), where \(F_0\) is the zero’th Fourier coefficient of F. Then \(\mathcal{Q}_0\Psi =U'_g\) and furthermore
$$\begin{aligned} \langle \mathcal{Q}_0\Psi , \Psi \rangle =4\pi \iint \limits _K\frac{dx\,dv}{|Q'(e_Q)|}\,|g|^2 =4\pi \iint \limits _K |Q'(e_Q)|\,(F-F_0)^2\,dx\,dv. \end{aligned}$$(4.26) -
(c)
Let \(\Psi \in L^2_r\) be given and suppose that \(F(r)=F(0)+\int _0^r\Psi (s)\,ds\) for \(r\in [0, r_Q]\). Define \(u=-\mathcal{T}^{-1}(|Q'(e_Q)|(F-F_0))\). Then \(u\in X^2_{\mathrm{odd}}\) and
$$\begin{aligned} {(Lu, u)}_{X^0} =\frac{1}{4\pi }\,\Big (\langle \mathcal{Q}_0\Psi , \Psi \rangle -{\Vert \mathcal{Q}_0\Psi \Vert }^2_{L^2_r}\Big ). \end{aligned}$$(4.27)
Proof
- (a)
-
(b)
Owing to Lemma B.9, we have \(g\in X^1_{\mathrm{even}}\) as well as \(\mathcal{T}g=-\psi \) for \(\psi \) as in (a). In addition, \(g_0=0\) by (B.24), so that \(g\in X^1_0\). Thus, Lemma B.13(c) yields \(-\mathcal{T}^{-1}\psi =g-g_0=g\).
Next, recall that \(\psi \) is odd in v and \({\Vert \psi \Vert }_{X^0} \le \rho _Q(0)^{1/2}\,{\Vert \Psi \Vert }_{L^2_r}<\infty \) by (4.20), which means that \(\psi \in X^0_{\mathrm{odd}}\subset X^0_0\). As a consequence, \(\mathcal{T}{(-\mathcal{T}^2)}^{-1}\psi =-\mathcal{T}^{-1}\psi =g\) by Lemma B.13(e). Hence, if we take \(z=0\in \Omega \) in (4.22) of Lemma 4.4, then we get
To verify (4.26), note first that \(ik\omega _1 g_k=-\psi _k\) for \(k\in \mathbb Z\). Applying (B.4) from Remark B.2(a), we obtain
where we have used that \(\frac{\partial e}{\partial I}=\omega _1\) owing to (A.18). Thus, the claim follow from (a) for \(z=0\). (c) We continue to use the notation and the observations from (b). Since \(g\in X_0^1\), we have \(u=\mathcal{T}^{-1}g\in X_0^2\). As also \(g\in X^1_{\mathrm{even}}\) and \(\mathcal{T}^{-1}\) reverses the parity by Remark B.18, we get \(u\in X^2_{\mathrm{odd}}\). Accordingly, we deduce from (B.44) in Corollary B.19 that
Now \(\mathcal{T}u=\mathcal{T}\mathcal{T}^{-1}g=g\) due to Lemma B.13(d), so that
by Remark B.2(a) and (4.26). Furthermore, using (B.40) from Lemma B.15 in conjunction with (b), it follows that
Altogether, this yields (4.27). \(\square \)
Lemma 4.7
Let \(\mu _1:\,]-\infty , \delta _1^2[\,\rightarrow \,]0, \infty [\) be defined as in Lemma 4.3(f). Then
-
(a)
\(0<\mu _1(0)<1\).
-
(b)
If \(\lambda _*<\delta _1^2\) and \(\lambda \in [0, \lambda _*]\), or \(\lambda _*=\delta _1^2\) and \(\lambda \in [0, \lambda _*[=[0, \delta _1^2[\), then \(\mu _1(\lambda )\le 1\).
-
(c)
For \(\lambda \in [0, \delta _1^2[\), let \(\Psi _\lambda \in L^2_r\) denote a normalized eigenfunction of \(\mathcal{Q}_\lambda \) for \(\mu _1(\lambda )\). Define \(\psi _\lambda (r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi _\lambda (r)\in X^0_{\mathrm{odd}}\) and \(g_\lambda =(-\mathcal{T}^2-\lambda )^{-1}\psi _\lambda \in X^2_{\mathrm{odd}}\). Then
$$\begin{aligned} \mu _1(\lambda )=4\pi \,{(\psi _\lambda , g_\lambda )}_{X^0} \end{aligned}$$and
$$\begin{aligned} Lg_\lambda =(1-\mu _1(\lambda ))\psi _\lambda +\lambda g_\lambda , \end{aligned}$$as well as
$$ {(Lg_\lambda , g_\lambda )}_Q =\frac{1}{4\pi }\,\mu _1(\lambda )(1-\mu _1(\lambda )) +\lambda {\Vert g_\lambda \Vert }^2_{X^0}. $$ -
(d)
The function \(\mu _1:\,]-\infty , \delta _1^2[\,\rightarrow \,]0, \infty [\) is convex.
-
(e)
We have
$$\begin{aligned} \mu _1(\lambda )\le & {} 16\pi \bigg (\int _0^{r_Q}\int _0^{r_Q} \frac{dr}{r^2}\,\frac{d\tilde{r}}{\tilde{r}^2} \bigg |\sum _{k\ne 0}\iint \limits _D d\ell \,\ell \,de\,\mathbf{1}_{\{r_-(e,\,\ell )\le r,\,\tilde{r}\le r_+(e,\,\ell )\}} \\&\qquad \qquad \qquad \quad \times \,\frac{\omega _1(e, \ell )\,|Q'(e)|}{k^2\omega _1^2(e, \ell )-\lambda } \sin (k\theta (r, e, \ell ))\sin (k\theta (\tilde{r}, e, \ell ))\,\bigg |^2\bigg )^{1/2}. \end{aligned}$$
Proof
(a) Clearly \(\mu _1(0)>0\), since otherwise \(\Vert \mathcal{Q}_0\Vert =0\), and thus \(\mathcal{Q}_0=0\). To show that \(\mu _1(0)<1\), let \(\Psi \in L^2_r\) be given. Define \(F(r)=\int _0^r\Psi (s)\,ds\) as well as \(u=-\mathcal{T}^{-1}(|Q'(e_Q)|(F-F_0))\). Then \(u\in X^2_{\mathrm{odd}}\) and
by (1.20) and (4.27) from Lemma 4.6. As a consequence,
implies that \(\mu _1(0)=\Vert \mathcal{Q}_0\Vert \le 1\). Lastly, suppose that \(\mu _1(0)=1\). Since \(\mu _1(0)\) is an eigenvalue, we have \(\mathcal{Q}_0\Psi =\Psi \) for some \(\Psi =\Psi (r)\ne 0\) such that \(\Psi \in L^2_r\); Remark 4.2(b) implies that \(\Psi \) has its support in \([0, r_Q]\). For the corresponding u, we deduce \(u=0\) from (4.28). Therefore, (B.24), Lemma B.13(d) and Lemma B.9(b) lead to
which is impossible. (b) Recall from Lemma 3.18 that \(\lambda _*\le \delta _1^2\). Thus, if we fix \(\lambda \) in one of the two cases: (i) \(\lambda _*<\delta _1^2\) and \(\lambda \in [0, \lambda _*]\); or (ii) \(\lambda _*=\delta _1^2\) and \(\lambda \in [0, \lambda _*[=[0, \delta _1^2[\), then \(\lambda \in [0, \delta _1^2[\). Let \(\Psi _\lambda \in L^2_r\) denote a normalized eigenfunction for \(\mu _1(\lambda )\), i.e., we have \(\mathcal{Q}_\lambda \Psi _\lambda =\mu _1(\lambda )\Psi _\lambda \) and \({\Vert \Psi _\lambda \Vert }_{L^2_r}=1\). For \(\psi _\lambda (r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi _\lambda (r)\), we get \(\psi _\lambda \in X^0_{\mathrm{odd}}\), cf. the proof of Lemma 4.6(a). Thus, \(g_\lambda =(-\mathcal{T}^2-\lambda )^{-1}\psi _\lambda \in X^2_{\mathrm{odd}}\). Using (4.21) from Lemma 4.4, we calculate
In addition,
This yields
hence in particular
Thus, by the Antonov stability estimate, Theorem 1.2,
so that
Now, (B.26) in Corollary B.10 yields
and in particular \({(\psi _\lambda , g_\lambda )}_{X^0}>0\), as otherwise \(\psi _\lambda =0\) and consequently \(\Psi _\lambda =0\), which is impossible. Hence, (4.31) shows that \(\mu _1(\lambda )\le 1\).
(c) Note that due to Lemma 4.6(a),
and therefore the first claim follows by comparing to (4.32). The other relations are due to (4.29) and (4.30). (d) If \(\lambda \in ]-\infty , \delta _1^2[\) and \(\Psi \in L^2_r\), then
by (4.6) from Lemma 4.3(d). Thus, every function \(]-\infty , \delta _1^2[\,\ni \lambda \mapsto \langle \mathcal{Q}_\lambda \Psi , \Psi \rangle \) is convex. As a consequence of (4.7), also \(\mu _1(\lambda )=\sup \,\{\langle \mathcal{Q}_\lambda \Psi , \Psi \rangle : {\Vert \Psi \Vert }_{L^2_r}\le 1\}\) is convex. (e) Here, we use
and the fact that
cf. [35, Prop. 6.36] and (4.16). \(\square \)
According to Lemma 4.3(f), the monotone limits
do exist. Of particular importance to us will be the number
Remark 4.8
If \(\lambda _*=\delta _1^2\), then \(\mu _*\le 1\). This follows from Lemma 4.7(b). \(\diamondsuit \)
The next result will use assumption (\(\omega _1\)-3). If \(\omega _1\) attains its minimum at an interior point \((\hat{e}, \hat{\beta })\in \mathring{D}\), then we are in the situation of (\(\omega _1\)-2), and Corollary 4.16 below applies. Otherwise, since \(\omega _1\) is continuous on D, its minimum is attained on the boundary \(\partial D\), which consists of three parts: the left side
the lower boundary curve
and the upper line
Corollary 3.16 shows that the minimum can only be attained on this upper line (4.34) at a point \((e_0, \hat{\beta })\), and (\(\omega _1\)-3) roughly concerns the case where both \(\frac{\partial \omega _1}{\partial e}(e_0, \hat{\beta })\ne 0\) and \(\frac{\partial \omega _1}{\partial \beta }(e_0, \hat{\beta })\ne 0\), which is reasonable to expect for a minimum on the boundary.
Lemma 4.9
Suppose that (\(\omega _1\)-3) is satisfied. Then
does exist in the Hilbert-Schmidt norm \({\Vert \cdot \Vert }_{\mathrm{HS}}\) of \(L^2_r\). In particular, the kernel of the symmetric and positive Hilbert-Schmidt operator \(\mathcal{Q}_{\delta _1^2}\) is given by
and \(\mu _*=\Vert \mathcal{Q}_{\delta _1^2}\Vert <\infty \). More generally, the k’th eigenvalue of \(\mathcal{Q}_{\delta _1^2}\) is \(\mu _{*, k}\). For \(k\in \mathbb N\), the functions
are monotone increasing, locally Lipschitz continuous on \(]-\infty , \delta _1^2[\) and continuous on \(]-\infty , \delta _1^2]\), if we set \(\mu _k(\delta _1^2)=\mu _{*, k}\). In particular, the \(\mu _k\) are differentiable a.e. Furthermore, \(\mu _1:\,\,]-\infty , \delta _1^2]\,\rightarrow \,]0, \infty [\) is a convex function.
Proof
We need to refine (4.17), from where we know that
for \(\lambda , \tilde{\lambda }<\delta _1^2\) and a suitable constant \(\hat{C}>0\). Thus
where
We assert that
and to establish this claim, we are going to use Lebesgue’s dominated convergence in \(\int _0^{r_Q}\int _0^{r_Q} dr\,d\bar{r}\) together with the condition
from (\(\omega _1\)-3), where \((e_0, \hat{\beta })\in D\) satisfies \(\omega _1(e_0, \hat{\beta })=\delta _1\). Let \(r, \bar{r}>0\) be fixed and define
If \((e, \beta )\in D\) are such that \(\beta \le \hat{C}r\bar{r}\) and \((e, \beta )\ne (\hat{e}, \hat{\beta })\), then \(\omega _1(e, \beta )-\delta _1\ge \alpha >0\) for \(\alpha =\alpha (e, \beta )\) by (4.38), and accordingly
for this \((e, \beta )\). On the other hand,
by (4.38). Next, we are going to bound
Case 1: \(\hat{\beta }>0\). If \(\beta \le R\le \hat{\beta }/2\), then \(|(e, \beta )-(e_0, \hat{\beta })|\ge |\beta -\hat{\beta }|\ge \hat{\beta }/2\) and hence
If \(R\ge \hat{\beta }/2\), then we always have
Case 2: \(\hat{\beta }=0\). Then
Thus, if we summarize (4.39) and (4.42)–(4.44) for \(R=\hat{C}r\bar{r}\), it follows that \(\tau (r, \bar{r})\rightarrow 0\) as \(\lambda \rightarrow \delta _1^2-\) for all \(r, \bar{r}>0\). Hence, to complete the proof of (4.37), we need to obtain an integrable majorant. For, using (4.40), we can bound
Case 1: \(\hat{\beta }>0\). Let \(\hat{\varepsilon } =\min \{r_Q, \frac{\hat{\beta }}{2\hat{C}r_Q}\}\). If \(r\le \hat{\varepsilon }\) or \(\hat{r}\le \hat{\varepsilon }\), then \(\hat{C}r\bar{r}\le \hat{C}\hat{\varepsilon } r_Q\le \hat{\beta }/2\), and thus we can apply (4.42) in this case, as well as (4.43) in the opposite case. Therefore, we split the integral to obtain
which shows that a suitably large constant provides an integrable majorant. Case 2: \(\hat{\beta }=0\). By (4.44), we get
Since \(1+|\ln r|^2+|\ln \bar{r}|^2\) is integrable on \([0, r_Q]\times [0, r_Q]\), we have found an integrable majorant also in this case. Altogether, we have shown that (4.37) is verified. At the same time, this yields \(\lim _{\tilde{\lambda }\rightarrow \delta _1^2-} T(\tilde{\lambda })=0\), and going back to (4.36), we deduce that (4.35) holds for an appropriate Hilbert-Schmidt operator \(\mathcal{Q}_{\delta _1^2}\) on \(L^2_r\). Since \({\Vert \cdot \Vert }_{\mathcal{B}(L^2_r)}\le {\Vert \cdot \Vert }_{\mathrm{HS}}\), (4.35) in particular implies that \(\mathcal{Q}_{\delta _1^2}=\lim _{\lambda \rightarrow \delta _1^2-} \mathcal{Q}_\lambda \) in \(\mathcal{B}(L^2_r)\). Recalling from (4.7) that \(\mu _1(\lambda )={\Vert \mathcal{Q}_\lambda \Vert }_{\mathcal{B}(L^2_r)}\), we can use (4.33) to get
as claimed.
Let \(\kappa _1\ge \kappa _2\ge \ldots \rightarrow 0\) denote the eigenvalues (listed according to their multiplicities) of the symmetric and positive Hilbert-Schmidt operator \(\mathcal{Q}_{\delta _1^2}\). Then
by the Courant max-min principle. Passing to the limit \(\lim _{\tilde{\lambda }\rightarrow \delta _1^2-}\) in (4.36), we derive
where \(\lim _{\lambda \rightarrow \delta _1^2-} T(\lambda )=0\). Thus, if \(\Psi \in L^2_r\) is such that \({\Vert \Psi \Vert }_{L^2_r}=1\), then we have
Since the \(\mu _k(\lambda )\) are also characterized by the Courant max-min principle, see (4.18), it follows that
and accordingly \(\mu _{*, k}=\lim _{\lambda \rightarrow \delta _1^2-}\mu _k(\lambda )=\kappa _k\).
The next assertion is due to the definition of \(\mu _{*, k}\) and Lemma 4.3(f), whereas the convexity of \(\mu _1\) on \(]-\infty , \delta _1^2]\) is a consequence of Lemma 4.7(d). \(\square \)
Corollary 4.10
Suppose that (\(\omega _1\)-3) is satisfied.
(a) There is a constant \(C>0\) such that for every \(\lambda \in [0, \delta _1^2]\) and \(r, \tilde{r}\in ]0, r_Q]\), we have
(b) For \(\lambda \in [0, \delta _1^2[\), let \(\Psi _\lambda \in L^2_r\) denote a normalized eigenfunction of \(\mathcal{Q}_\lambda \) for \(\mu _1(\lambda )\). Then there is a constant \(C>0\) such that for every \(\lambda \in [0, \delta _1^2[\) and \(r\in ]0, r_Q]\), we have
(c) For \(\Psi _\lambda \) as in (b), define \(\psi _\lambda (r, p_r, \ell ) =|Q'(e_Q)|\,p_r\Psi _\lambda (r)\in X^0_{\mathrm{odd}}\). Then there is a constant \(C>0\) such that for every \(\lambda \in [0, \delta _1^2[\) and \(k\in \mathbb Z\), we have
where \({(\psi _\lambda )}_k\) are the Fourier coefficients of \(\psi _\lambda \).
Proof
(a) From (4.14) and similar to the argument following (4.9), we obtain with \(\min \{r^2, \tilde{r}^2\}\le r^2\) and using (\(\omega _1\)-3)
By means of the function I from (4.41), this can be expressed as
for certain constants \(C, \hat{C}>0\) that only depend on Q. Once again, we distinguish two cases. Case 1: \(\hat{\beta }>0\). If \(r^2\le \frac{\hat{\beta }}{2\hat{C}}\), then we can apply (4.42) to get
On the other hand, if \(r^2\ge \frac{\hat{\beta }}{2\hat{C}}\), then (4.43) leads to
Case 2: \(\hat{\beta }=0\). Here, we invoke (4.44) to deduce that
Hence, in any case, we arrive at the bound
as desired. (b) Using (a), we obtain from (4.15) and Remark 4.2(b)
so that
for a certain constant \(C_*>0\) that only depends on Q. Fix \(a_*\in ]0, r_Q[\) such that \(\int _0^{a_*}(1+|\ln r|)\,dr\le \frac{1}{2C_*}\). Then
entails \(\int _0^{a_*} |\Psi _\lambda (\tilde{r})|\,d\tilde{r} \le \int _{a_*}^{r_Q} |\Psi _\lambda (\tilde{r})|\,d\tilde{r}\). Going back to (4.45), it follows by means of Hölder’s inequality that
from where a suitable \(C>0\) can be read off. (c) Owing to (4.24), Theorem 3.5 and (b), we have
which completes the proof. \(\square \)
Corollary 4.11
Suppose that (\(\omega _1\)-3) is satisfied. Let \((\lambda _j)\subset [0, \delta _1^2[\) be such that \(\lim _{j\rightarrow \infty }\lambda _j=\delta _1^2\). For \(j\in \mathbb N\), let \(\Psi _j\in L^2_r\) denote a normalized eigenfunction of \(Q_{\lambda _j}\) for \(\mu _1(\lambda _j)\). Furthermore, define \(\psi _j(r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi _j(r)\in X^0_{\mathrm{odd}}\). Then there is a subsequence \(j'\rightarrow \infty \) so that
does exist in \(L^2_r\) and
does exist in \(X^0\), where \(\psi _*(r, p_r, \ell ) =|Q'(e_Q)|\,p_r\Psi _*(r)\). In addition, \({\Vert \Psi _*\Vert }_{L^2_r}=1\) and \(\mathcal{Q}_{\delta _1^2}\Psi _*=\mu _*\Psi _*\) as well as \(\mu _*=\Vert \mathcal{Q}_{\delta _1^2}\Vert \).
Proof
Recall from (4.33) and Lemma 4.9 that \(\mu _*\in [\mu _1(0), \infty [\). For \(j, k\in \mathbb N\), we can estimate
According to Lemma 4.9, we have \(\lim _{\lambda \rightarrow \delta _1^2-} {\Vert \mathcal{Q}_{\delta _1^2}-\mathcal{Q}_\lambda \Vert }_{\mathrm{HS}}=0\) and \(\mathcal{Q}_{\delta _1^2}: L^2_r\rightarrow L^2_r\) is a Hilbert-Schmidt operator, and hence compact. Thus, since \({\Vert \Psi _j\Vert }_{L^2_r}=1\), the set \(\{\mathcal{Q}_{\delta _1^2}\Psi _j: j\in \mathbb N\}\subset L^2_r\) is relatively compact. Therefore, there is a subsequence \(j'\rightarrow \infty \) and a function \(\hat{\Psi }\in L^2_r\) so that \(\lim _{j'\rightarrow \infty } \mathcal{Q}_{\delta _1^2}\Psi _{j'} =\hat{\Psi }\) in \(L^2_r\). From (4.46), we deduce that along the subsequence
As a consequence, \(\Psi _*=\lim _{j'\rightarrow \infty }\Psi _{j'}\) does exist in \(L^2_r\). Since
by (4.23), also \(\psi _*=\lim _{j'\rightarrow \infty }\psi _{j'}\) does exist in \(X^0\), where \(\psi _*(r, p_r, \ell ) =|Q'(e_Q)|\) \(p_r\Psi _*(r)\) a.e. Lastly,
implies that \(\mathcal{Q}_{\delta _1^2}\Psi _*=\mu _*\Psi _*\). \(\square \)
The following criterion is useful for proving that \(\delta _1^2\) is an eigenvalue of L in the case where \(\mu _*=1\).
Lemma 4.12
Suppose that (\(\omega _1\)-3) is satisfied and that \(\mu _*=1\). Let \((\lambda _j)\subset [0, \delta _1^2[\) be such that \(\lim _{j\rightarrow \infty }\lambda _j=\delta _1^2\). For \(j\in \mathbb N\), let \(\Psi _j\in L^2_r\) denote a normalized eigenfunction of \(Q_{\lambda _j}\) for \(\mu _1(\lambda _j)\). Furthermore, define \(\psi _j(r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi _j(r)\in X^0_{\mathrm{odd}}\) and \(g_j=(-\mathcal{T}^2-\lambda _j)^{-1}\psi _j\in X^2_{\mathrm{odd}}\). If \((g_j)\subset X^0=L^2_{\mathrm{sph},\,\frac{1}{|Q'|}}(K)\) is bounded, then \(\delta _1^2\) is an eigenvalue of L.
Proof
From (4.21), we deduce
Since \(-\mathcal{T}^2 g_j=\psi _j+\lambda _j g_j\), using Corollary B.19, this implies that for every odd function \(h\in X^{00}\), we have
Next, from (4.20), we get \({\Vert \psi _j\Vert }_{X^0}\le \rho _Q(0)^{1/2}\,{\Vert \Psi _j\Vert }_{L^2_r}\le \rho _Q(0)^{1/2}\). Since \(\lim _{j\rightarrow \infty }\mu _1(\lambda _j)=\mu _*=1\), this yields in particular that
By assumption, \((g_j)\subset X^0\) is bounded. Hence, passing to a subsequence (that is not relabeled), we may assume that \(g_j\rightharpoonup g_*\) weakly in \(X^0\) as \(j\rightarrow \infty \) for some function \(g_*\in X^0_{\mathrm{odd}}\). Suppose that \(g_*=0\). Then \(g_j\rightharpoonup 0\) weakly in \(X^0\) implies that \(\mathcal{K}\mathcal{T}g_j\rightharpoonup 0\) weakly in \(X^0\) as \(j\rightarrow \infty \), by Lemma B.15(d). Due to (4.47), this yields \(\psi _j\rightharpoonup 0\) weakly in \(X^0\) as \(j\rightarrow \infty \). On the other hand, by Corollary 4.11, we may pass to a subsequence \(j'\rightarrow \infty \) so that \(\Psi _*=\lim _{j'\rightarrow \infty }\Psi _{j'}\) does exist in \(L^2_r\) and \(\psi _*=\lim _{j'\rightarrow \infty }\psi _{j'}\) does exist in \(X^0\) as strong limits; the functions are linked via \(\psi _*(r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi _*(r)\). But then we must have \(\psi _*=0\) and accordingly \(\Psi _*=0\), which however contradicts \({\Vert \Psi _*\Vert }_{L^2_r}=1\), cf. Corollary 4.11. As a consequence, it follows that \(g_*\in X^0_{\mathrm{odd}}\) satisfies \(g_*\ne 0\). Passing to the limit \(j\rightarrow \infty \) in (4.48) and using (4.49), we moreover infer that \({(g_*, Lh)}_{X^0} =\delta _1^2 {(g_*, h)}_{X^0}\) for every odd function \(h\in X^{00}\). From Lemma C.11, we conclude that \(g_*\in X^2_{\mathrm{odd}}\) and \(Lg_*=\delta _1^2 g_*\), which completes the proof. \(\square \)
4.2 Relating \(\mu _*\) to the Fact That \(\lambda _*\) is an Eigenvalue of L
Theorem 4.13
We have
In this case, \(\mu _1(\lambda _*)=1\) and \(\lambda _*\) is an eigenvalue of L.
Proof
If \(\mu _*>1\), then \(\lambda _*=\delta _1^2\) is impossible by Remark 4.8, so that we must have \(\lambda _*<\delta _1^2\). Conversely, suppose that \(\lambda _*<\delta _1^2\) holds. Then, according to Theorem C.8, \(\lambda _*\) is an eigenvalue of L. Let \(u_*\in X^2_{\mathrm{odd}}\) be an eigenfunction of L for the eigenvalue \(\lambda _*\). Using Theorem 4.5(a), it follows that \(\Psi _*=U'_{\mathcal{T}u_*}\in L^2_r\) for \(r\in [0, r_Q]\) is an eigenfunction of \(\mathcal{Q}_{\lambda _*}\) for the eigenvalue 1. Since \(\mu _1(\lambda _*)\) is the largest eigenvalue of \(\mathcal{Q}_{\lambda _*}\), we get \(\mu _1(\lambda _*)\ge 1\). On the other hand, \(\mu _1(\lambda _*)\le 1\) by Lemma 4.7(b), and hence \(\mu _1(\lambda _*)=1\). It remains to show that \(\mu _*>1\). Suppose that on the contrary \(\mu _*\le 1\) is satisfied. For \(\lambda \in [\lambda _*, \delta _1^2[\), the monotonicity of \(\mu _1\) then yields \(1=\mu _1(\lambda _*)\le \mu _1(\lambda )\le \mu _*\le 1\), which means that \(\mu _1(\lambda )=1\) is constant for \(\lambda \in [\lambda _*, \delta _1^2[\). Take \(\lambda _*\le \tilde{\lambda }<\lambda <\delta _1^2\). and let \(\Psi _{\tilde{\lambda }}\) denote a normalized eigenfunction for \(\mu _1(\tilde{\lambda })\). Then, by (4.19) and (4.7),
which means that \(\langle \mathcal{Q}_\lambda \Psi _{\tilde{\lambda }}, \Psi _{\tilde{\lambda }}\rangle =1\) for all \(\lambda _*\le \tilde{\lambda }<\lambda <\delta _1^2\). Differentiating this relation w.r. to \(\lambda \) at a fixed \(\lambda _0\in ]\tilde{\lambda }, \delta _1^2[\), it follows from (4.6) that
for all \(\tilde{\lambda }\in [\lambda _*, \lambda _0[\). Defining \(\psi _{\tilde{\lambda }}(r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi _{\tilde{\lambda }}(r)\in X^0_{\mathrm{odd}}\), then (4.24) implies that \({(\psi _{\tilde{\lambda }})}_k=0\) for \(k\in \mathbb Z\), so that \(\psi _{\tilde{\lambda }}=0\) and in turn \(\Psi _{\tilde{\lambda }}=0\), which however is impossible. \(\square \)
Theorem 4.14
Suppose that (\(\omega _1\)-1) is satisfied. If \(\mu _*<1\), then \(\lambda _*=\delta _1^2\) and this is not an eigenvalue of L.
Proof
The approach is inspired by [20, Section 2]. Since \(\lambda _*\le \delta _1^2\) by Lemma 3.18, \(\mu _*<1\) together with Theorem 4.13 implies \(\lambda _*=\delta _1^2\). Now suppose on the contrary that there is a function \(u_*\in X^2_{\mathrm{odd}}\) such that \({\Vert u_*\Vert }_{X^0}=1\) and \(Lu_*=\delta _1^2 u_*\). If we define \(\Psi _*(r)=U'_{\mathcal{T}u_*}(r)\) for \(r\in [0, r_Q]\), then \(\Psi _*\in L^2_r\) and (B.37) yields \(\mathcal{K}\mathcal{T}u_*=|Q'(e_Q)|\,p_r\,U'_{\mathcal{T}u_*}(r) =|Q'(e_Q)|\,p_r\,\Psi _*(r)\). Hence, for \(a>0\) and \(b\in \mathbb R\), we get
Since \(z=\delta _1^2-a+ib\in \Omega \), it follows from (4.21) that
and therefore,
We claim that if \(a=a(\varepsilon )\rightarrow 0^+\) and \(b=b(\varepsilon )\rightarrow 0\) as \(\varepsilon \rightarrow 0\), then
in \(L^2_r\). For, we can invoke Corollary B.16 as well as (B.25) to deduce
If \(|k|\ge 2\), then \(k^2\omega _1^2(I, \ell )-\delta _1^2+a\ge (k^2-1)\delta _1^2\ge 3\delta _1^2\). Thus,
for \(\phi _1(I, \ell )=\frac{|{(u_*)}_1(I, \ell )|^2}{|Q'(e)|}\in L^1(D)\). For almost all \((I, \ell )\in D\), we know from hypothesis (\(\omega _1\)-1) that \(\omega _1(I, \ell )\ne \delta _1\), i.e., \(\omega _1(I, \ell )>\delta _1\). For such an \((I, \ell )\), we have
Since always
it follows by using Lebesgue’s dominated convergence theorem that indeed (4.51) is verified. Going back to (4.50), this entails that
Next, we are going to compare \(\mathcal{Q}_{\delta _1^2-a+ib}\) to \(\mathcal{Q}_{\delta _1^2-a}\). Here, we find
cf. Lemma 4.3(d) and the definition of \(\mathcal{Q}_z\). Using (4.8), (4.12) and similar arguments as in the proof of Lemma 4.3(a), we obtain
Now
so that
So if we take for instance \(b(\varepsilon )=\varepsilon ^3\) and \(a(\varepsilon )=\varepsilon \), it follows that
Using also (4.52), we conclude that
As a consequence,
Since \(\mu _*<1\), this enforces \(\Psi _*=0\) and hence \(\mathcal{K}\mathcal{T}u_*=0\). Therefore, \(-\mathcal{T}^2 u_*=-\mathcal{T}^2 u_*-\mathcal{K}\mathcal{T}u_*=Lu_*=\delta _1^2 u_*\), i.e., \(\delta _1^2\) is an eigenvalue of \(-\mathcal{T}^2\) with eigenfunction \(u_*\). However, this contradicts Lemma B.12. \(\square \)
The next result clarifies the case where \(\mu _*=1\).
Theorem 4.15
Suppose that (\(\omega _1\)-3) is satisfied and that \(\mu _*=1\). Then \(\lambda _*=\delta _1^2\), and this is an eigenvalue of L if and only if
holds.
Proof
Since \(\lambda _*\le \delta _1^2\) by Lemma 3.18, \(\mu _*=1\) together with Theorem 4.13 imply \(\lambda _*=\delta _1^2\). For the actual proof, recall from Lemma 4.3(f) that \(\mu _1(\cdot ):\,\,]-\infty , \delta _1^2[\,\rightarrow \,]0, \infty [\) is differentiable a.e., so (4.53) makes sense.
First, we consider the case where \(\delta _1^2\) is an eigenvalue of L. Let \(u_*\in X^2_{\mathrm{odd}}\) be such that \({\Vert u_*\Vert }_{X^0}=1\) and \(Lu_*=\delta _1^2 u_*\). If we define \(\Psi _*(r)=U'_{\mathcal{T}u_*}(r)\) for \(r\in [0, r_Q]\), then \(\Psi _*\in L^2_r\) and (B.37) implies that \(\mathcal{K}\mathcal{T}u_*=|Q'(e_Q)|\,p_r\,U'_{\mathcal{T}u_*}(r) =|Q'(e_Q)|\,p_r\,\Psi _*(r)=:\psi _*\in X^0_{\mathrm{odd}}\). For \(\lambda <\delta _1^2\), we have
and hence
for the Fourier coefficients. Since
by (B.40) from Lemma B.15(b), taking the inner product in \(X^0\) of (4.54) with \(u_*\), we deduce
Next, due to (4.25) from Lemma 4.6, we have
Thus, by (B.4), (A.18), Lemma B.8(b) and (4.55) applied twice,
Comparing to (4.56), this yields
If we had \(\Psi _*=0\), then also \(\psi _*=0\) and consequently \((k^2\omega _1^2-\delta _1^2){(u_*)}_k=0\) in D for \(k\ne 0\) by (4.55). This implies that \({(u_*)}_k=0\) for \(|k|\ge 2\) and \((\omega _1-\delta _1){(u_*)}_1=0\) in D. Owing to (\(\omega _1\)-1), this enforces \({(u_*)}_1=0\) a.e. and therefore \(u_*=0\), which is a contradiction. In other words, we do know that \(\Psi _*\ne 0\). Hence, by (4.7) and (4.57),
Thus,
and upon using (4.55) one more time, we conclude that
for all \(\lambda <\delta _1^2\). Since \(\mu _1\) is convex on \(]-\infty , \delta _1^2]\) by Lemma 4.9(d), the difference quotients
for \(h>0\) are monotone increasing in \(\lambda \) (and also in h); see [14, p. 13/14]. Let \(\lambda _0\in ]-\infty , \delta _1^2[\) be a point where \(\mu _1\) is differentiable and let \(h>0\). For \(\lambda _1=\lambda _0-h\) and \(\lambda _2=\delta _1^2-h\), we have \(\lambda _1<\lambda _2\), whence \(\mu _1(\delta _1^2)=\mu _*=1\) in conjunction with (4.58) for \(\lambda =\delta _1^2-h\) leads to
It follows that \({\Vert \mu '_1\Vert }_{L^\infty (]-\infty , \delta _1^2[)} \le \frac{4\pi }{{\Vert \Psi _*\Vert }_{L^2_r}^2}\,{\Vert u_*\Vert }_{X^0}^2\), which proves (4.53).
To establish the converse, we assume (4.53) to hold, and we are going to verify that \(\delta _1^2\) is an eigenvalue of L. For this, we are going to use Lemma 4.12. The operator family \(Q_z\) for \(z\in \Omega =\mathbb C\setminus [\delta _1^2, \infty [\) satisfies the assumptions of Lemma D.1 with \(\lambda _0=\delta _1^2\) and \(H=L^2_r\), by Lemmas 4.3 and 4.9. Hence, there are sequences \(\lambda _j\nearrow \delta _1^2\), \(\varepsilon _j>0\) and \(\Phi _{j, \lambda }\in L^2_r\) for \(\lambda \in ]\lambda _j-\varepsilon _j, \lambda _j+\varepsilon _j[\) such that \({\Vert \Phi _{j, \lambda }\Vert }_{L^2_r}=1\),
is real analytic for \(j\in \mathbb N\), and \(Q_\lambda \Phi _{j, \lambda }=\mu _1(\lambda )\Phi _{j, \lambda }\) for \(j\in \mathbb N\) and \(\lambda \in ]\lambda _j-\varepsilon _j, \lambda _j+\varepsilon _j[\). Furthermore, \(\mu _1\) is real analytic in \(]\lambda _j-\varepsilon _j, \lambda _j+\varepsilon _j[\) and satisfies
for \(\lambda \in ]\lambda _j-\varepsilon _j, \lambda _j+\varepsilon _j[\). By decreasing \(\varepsilon _j\) further, if necessary, we may assume that \(\varepsilon _j\rightarrow 0\) as \(j\rightarrow \infty \). Due to (4.53), there exists a set \(N\subset ]-\infty , \delta _1^2[\) of measure zero such that \(S=\sup _{\lambda \in ]-\infty , \delta _1^2[\setminus N} |\mu '_1(\lambda )|<\infty \). For each \(j\in \mathbb N\), pick \(\hat{\lambda }_j\in ]\lambda _j-\varepsilon _j, \lambda _j+\varepsilon _j[\setminus N\) and define \(\Psi _j=\Phi _{j, \hat{\lambda }_j}\). It follows that \(\lim _{j\rightarrow \infty }\hat{\lambda }_j=\delta _1^2\) and \({\Vert \Psi _j\Vert }_{L^2_r}=1\). In addition, \(Q_{\hat{\lambda }_j}\Psi _j=Q_{\hat{\lambda }_j}\Phi _{j, \hat{\lambda }_j} =\mu _1(\hat{\lambda }_j)\Phi _{j, \hat{\lambda }_j}=\mu _1(\hat{\lambda }_j)\Psi _j\), i.e., \(\Psi _j\) is a normalized eigenfunction for the eigenvalue \(\mu _1(\hat{\lambda }_j)\) of \(Q_{\hat{\lambda }_j}\) such that
the latter due (4.59); recall that generally \(\langle Q'_\lambda \Psi , \Psi \rangle \ge 0\) by (4.6). Now define \(\psi _j(r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi _j(r)\in X^0_{\mathrm{odd}}\) and \(g_j=(-\mathcal{T}^2-\hat{\lambda }_j)^{-1}\psi _j\in X^2_{\mathrm{odd}}\). To complete the proof, we need to show that \((g_j)\subset X^0\) is bounded. From (B.4), (A.18), (B.25), (4.24) and (4.6), we obtain
Thus, the claim follows from (4.60). \(\square \)
4.3 Some Further Results
The following observation corresponds to the situation where \(\omega _1\) is differentiable and attains its minimum at an interior point \((\hat{e}, \hat{\beta })\) of D; cf. assumption (\(\omega _1\)-2).
Corollary 4.16
Suppose that (\(\omega _1\)-2) is satisfied. Then \(\mu _*=\infty \), \(\lambda _*<\delta _1^2\), \(\mu _1(\lambda _*)=1\) and \(\lambda _*\) is an eigenvalue of L.
Proof
We only need to show that \(\mu _*=\infty \), then the remaining assertions do follow from Theorem 4.13. The lower boundary curve \({(\partial D)}_3=\{(e, \beta )\in D: e=e_{\mathrm{min}}(\beta )\}\) of D characterizes the \((e, \beta )\) where \(r_-(e, \beta )=r_0(\beta )=r_+(e, \beta )\). Since \((\hat{e}, \hat{\beta })\in \mathrm{int}\,D =\{(e, \beta ): \beta \in ]0, \beta _*[, e\in ]e_{\mathrm{min}}(\beta ), e_0[\} \subset D\setminus {(\partial D)}_3\) by hypothesis, we have that \(r_+(\hat{e}, \hat{\beta })-r_-(\hat{e}, \hat{\beta })=6\eta >0\). The functions \(r_{\pm }\) are known to be continuous (even \(C^1\)) on \(\mathrm{int}\,D\); see [30, 50] and [88, Def./Thm. 2.4(b)]. Thus, by shrinking the neighborhood U of \((\hat{e}, \hat{\beta })\) if necessary, we may assume that
is verified, along with
from (1.31). Next, we have \(\theta (r_-(\hat{e}, \hat{\beta }), \hat{e}, \hat{\beta })=0\) and \(\theta (r_+(\hat{e}, \hat{\beta }), \hat{e}, \hat{\beta })=\pi \). Since \(\frac{\partial \theta }{\partial r} =\frac{\omega _1}{p_r}\) due to (A.21) and \(p_r>0\) along the half-orbit, \(\theta (\cdot , \hat{e}, \hat{\beta })\) is strictly increasing. In particular, we obtain
As also
is continuous, there is \(\varepsilon \in ]0, \eta ]\) such that \(\sin \theta (r, e, \beta )\ge \sigma \) for \((e, \beta )\in U\) so that \(|e-\hat{e}|\le \varepsilon \), \(|\beta -\hat{\beta }|\le \varepsilon \) and \(r\in [\hat{r}_m-\varepsilon , \hat{r}_m+\varepsilon ]\cap ]r_-(e, \beta ), r_+(e, \beta )[ =[\hat{r}_m-\varepsilon , \hat{r}_m+\varepsilon ]\). If \(\varepsilon >0\) is small enough, we may assume that \([\hat{e}-\varepsilon , \hat{e}+\varepsilon ] \times [\hat{\beta }-\varepsilon , \hat{\beta }+\varepsilon ]\subset U\subset \mathrm{int}\,D\) as well as \([\hat{r}_m-\varepsilon , \hat{r}_m+\varepsilon ]\subset [0, r_Q]\). Furthermore, note that in general \(\sin \theta (r, e, \beta )\ge 0\) for \((e, \beta )\in \mathrm{int}\,D\) and \(r_-(e, \beta )<r<r_+(e, \beta )\). Next, owing to \((\hat{e}, \hat{\beta })\in \mathrm{int}\,D\), we have \(e\in ]U_Q(0), e_0[\). Using (Q2), we can thus make sure that \(\inf \{|Q'(e)|: e\in [\hat{e}-\varepsilon , \hat{e}+\varepsilon ]\}=\alpha >0\). Now, we consider the function
for which \({\Vert \Psi _0\Vert }_{L^2_r}=1\). Hence, for \(\lambda <\delta _1^2\) by Lemma 4.3(d),
where \(a=\delta _1^2-\lambda >0\). From Theorem 3.5 and (4.61), we deduce that
As a consequence,
Thus, if we pass to the limit \(\lambda \rightarrow \delta _1^2-\), i.e., \(a\rightarrow 0^+\), in (4.62), it follows that \(\mu _*=\infty \). \(\square \)
Regarding Theorem 4.15, if (\(\omega _1\)-3) holds and if \(\mu _*=1\), then one can show that \(\lambda =\delta _1^2\) is an eigenvalue of L, provided one is able to gain a little bit from the term \(|Q'(e)|\), in the sense that \(Q'(e_0)=0\) in a controlled way, as expressed by (Q5); then the inherent logarithmic singularity can be dealt with. To simplify the presentation, we additionally assume that \(\mu _*\) is simple as an eigenvalue of \(Q_{\delta _1^2}\), but with some more technical efforts, this assumption could be disposed of.
Corollary 4.17
Suppose that (\(\omega _1\)-3) and (Q5) are satisfied, and assume that \(\mu _*=1\) is a simple eigenvalue of \(Q_{\delta _1^2}\). Then \(\lambda _*=\delta _1^2\), and this is an eigenvalue of L.
Proof
We already know that \(\lambda _*=\delta _1^2\); see the proof of Theorem 4.15. To verify that \(\delta _1^2\) is an eigenvalue of L, we are going to use Theorem 4.15. According to Lemma D.2, there is \(\varepsilon >0\) such that \(]\delta _1^2-\varepsilon , \delta _1^2[\ni \lambda \mapsto \mu _1(\lambda )\) is real analytic. In addition, there are \(\Psi _\lambda \in L^2_r\) satisfying \({\Vert \Psi _\lambda \Vert }_{L^2_r}=1\), \(Q_\lambda \Psi _\lambda =\mu _1(\lambda )\Psi _\lambda \), and \(]\delta _1^2-\varepsilon , \delta _1^2[\ni \lambda \mapsto \Psi _\lambda \) is real analytic. Also \(\mu '_1(\lambda )=\langle Q'_\lambda \Psi _\lambda , \Psi _\lambda \rangle \) holds for \(\lambda \in ]\delta _1^2-\varepsilon , \delta _1^2[\). By Lemma 4.9, the function \(\mu _1\) is convex, so that \(\mu ''_1\ge 0\) and \(\mu '_1\) is increasing. In other words,
does exist in \(]0, \infty ]\), and the issue is to show that \(\mu '_*<\infty \). Defining \(\psi _\lambda (r, p_r, \ell )=|Q'(e_Q)|\,p_r\Psi _\lambda (r)\in X^0_{\mathrm{odd}}\) as before, we get, from Lemma 4.3(d), (4.24) and Corollary 4.10(c),
Thus, using (\(\omega _1\)-3) and (Q5),
where \(x=(x_1, x_2)\). Therefore \(\mu '_*\le C\) and the proof is complete. \(\square \)
Author information
Authors and Affiliations
Corresponding author
Rights and permissions
Copyright information
© 2021 The Author(s), under exclusive license to Springer Nature Switzerland AG
About this chapter
Cite this chapter
Kunze, M. (2021). A Birman-Schwinger Type Operator. In: A Birman-Schwinger Principle in Galactic Dynamics. Progress in Mathematical Physics, vol 77. Birkhäuser, Cham. https://doi.org/10.1007/978-3-030-75186-9_4
Download citation
DOI: https://doi.org/10.1007/978-3-030-75186-9_4
Published:
Publisher Name: Birkhäuser, Cham
Print ISBN: 978-3-030-75185-2
Online ISBN: 978-3-030-75186-9
eBook Packages: Mathematics and StatisticsMathematics and Statistics (R0)