Weyl formula for the negative dissipative eigenvalues of Maxwell's equations

TL;DR

This paper derives a Weyl formula for counting negative eigenvalues of the Maxwell operator with dissipative boundary conditions in an exterior domain, providing insight into the spectral properties of such systems.

Contribution

It establishes a Weyl asymptotic formula for the negative eigenvalues of Maxwell's equations with dissipative boundary conditions in an exterior domain, a novel spectral analysis result.

Findings

Weyl formula for negative eigenvalues derived

Spectral properties of Maxwell's equations with dissipation analyzed

Asymptotic behavior of eigenvalues characterized

Abstract

Let $V(t) = e^{tG_b},\: t \geq 0,$ be the semigroup generated by Maxwell's equations in an exterior domain $\Omega \subset {\mathbb R}^3$ with dissipative boundary condition $E_{tan}- \gamma(x) (\nu \wedge B_{tan}) = 0, \gamma(x) > 0, \forall x \in \Gamma = \partial \Omega.$ We study the case when $\Omega = \{x \in {\mathbb R^3}:\: |x| > 1\}$ and $\gamma \neq 1$ is a constant. We establish a Weyl formula for the counting function of the negative real eigenvalues of $G_b.$

Equations

$${\partial }_{t}E=\mathrm{curl}B,{\partial }_{t}B=-\mathrm{curl}E\mathrm{in}{\mathbb{R}}_t^{+}\times \Omega ,$$

$$E_{tan}-\gamma (x)(\nu \wedge B_{tan})=0\mathrm{on}{\mathbb{R}}_t^{+}\times \Gamma ,$$

$$E(0,x)=E_0(x),B(0,x)=B_0(x).$$

$$(E,B)=V(t)f=e^{t{G}_b}f,t\ge 0,$$

$${\lambda }_m\le -\frac{1}{\max \{({\gamma }_0-1),\sqrt{{\gamma }_0-1}\}}=-c_0,$$

$$N(r)=\#\{\lambda \in {\sigma }_p(G_b)\cap {\mathbb{R}}^{-}:|\lambda |\le r\},r>r_0(\gamma ),$$

$$\mathrm{mult}({\lambda }_m)=\mathrm{rank}\frac{1}{2\pi \mathbf{i}}{\int }_{|{\lambda }_n-z|=ϵ}(z-G_b{)}^{-1}dz,$$

$$N(r)=({\gamma }_0^2-1)r^2+{\mathcal{O}}_{\gamma }(r),r\ge r_0(\gamma )>c_0.$$

$$N(r)=\frac{1}{4\pi}\Bigl{(}\int_{\Gamma}(\gamma^{2}(y)-1)dS_{y}\Bigr{)}r^{2}+{\mathcal{O}}_{\gamma}(r),\>r\geq r_{0}(\gamma_{0}).$$

$$\mathrm{curl}E=-\mathbf{i}\mu B,\mathrm{curl}B=\mathbf{i}\mu E.$$

$$\{\begin{array}{l}\mathrm{curl}E=\mathbf{i}\mu H,\mathrm{curl}H=-\mathbf{i}\mu E,\mathrm{for}x\in B_3,\\ E_{tan}+\gamma (\nu \wedge H_{tan})=0,\mathrm{for}x\in {\mathbb{S}}^2.\end{array}$$

$$h_n^{(1)}(z):=\frac{H_{n+1/2}^{(1)}(z)}{\sqrt{z}},n\ge 1$$

$$\displaystyle E(x)=\sum_{n=1}^{\infty}\sum_{|m|\leq n}\Bigl{[}\alpha_{n}^{m}\sqrt{n(n+1)}\frac{h^{(1)}_{n}(\mu r)}{r}Y_{n}^{m}(\omega)\omega$$

$$\displaystyle+\frac{\alpha_{n}^{m}}{r}(rh^{(1)}_{n}(\mu r))^{\prime}U_{n}^{m}(\omega)+\beta_{n}^{m}h^{(1)}_{n}(\mu)V_{n}^{m}(\omega)\Bigr{]},$$

$$\displaystyle H(x)=-\frac{1}{{\bf i}\mu}\sum_{n=1}^{\infty}\sum_{|m|\leq n}\Bigl{[}\beta_{n}^{m}\sqrt{n(n+1)}\frac{h^{(1)}_{n}(\mu r)}{r}Y_{n}^{m}(\omega)\omega$$

$$\displaystyle+\frac{\beta_{n}^{m}}{r}(rh^{(1)}_{n}(\mu r))^{\prime}U_{n}^{m}(\omega)+\mu^{2}\alpha_{n}^{m}h^{(1)}_{n}(\mu)V_{n}^{m}(\omega)\Bigr{]}.$$

$$L_t^2({\mathbb{S}}^2)=\{u(\omega )\in (L^2({\mathbb{S}}^2){)}^3:⟨\omega ,u(\omega )⟩=0\mathrm{on}{\mathbb{S}}^2\}.$$

$$(\nu\wedge H_{tan})(\omega)=-\frac{1}{{\bf i}\mu}\sum_{n=1}^{\infty}\sum_{|m|\leq n}\Bigl{[}\beta_{n}^{m}\Bigl{(}h^{(1)}_{n}(\mu)+\frac{d}{dr}h^{(1)}_{n}(\mu r)|_{r=1}\Bigr{)}V_{n}^{m}(\omega)$$

$$-\mu^{2}\alpha_{n}^{m}h^{(1)}_{n}(\mu)U_{n}^{m}(\omega)\Bigr{]}$$

$$\alpha_{n}^{m}\Bigl{[}h^{(1)}_{n}(\mu)+\frac{d}{dr}(h^{(1)}_{n}(\mu r))|_{r=1}-\gamma{\bf i}\mu h^{(1)}_{n}(\mu)\Bigr{]}=0,\>\forall n\in{\mathbb{N}},\>|m|\leq n,$$

$$-\frac{\beta_{n}^{m}\gamma}{{\bf i}\mu}\Bigl{[}h^{(1)}_{n}(\mu)+\frac{d}{dr}(h^{(1)}_{n}(\mu r))|_{r=1}-\frac{{\bf i}\mu}{\gamma}h^{(1)}_{n}(\mu)\Bigr{]}=0,\>\forall n\in{\mathbb{N}},\>|m|\leq n.$$

$$1+\frac{d}{dr}h^{(1)}_{n}(-{\bf i}\lambda r)\Bigl{|}_{r=1}(h^{(1)}_{n}(-{\bf i}\lambda))^{-1}-\lambda\gamma=0$$

$$g_{n}(\lambda):=\frac{1}{\lambda}+\frac{d}{d\lambda}\Bigl{(}h^{(1)}_{n}(-{\bf i}\lambda)\Bigr{)}(h^{(1)}_{n}(-{\bf i}\lambda))^{-1}-\gamma=0.$$

$$h^{(1)}_{n}(-{\bf i}\lambda)=(-{\bf i})^{n+1}\frac{e^{\lambda}}{-{\bf i}\lambda}R_{n}\Bigl{(}\frac{{\bf i}}{-2{\bf i}\lambda}\Bigr{)}=(-{\bf i})^{n}\frac{e^{\lambda}}{\lambda}R_{n}\Bigl{(}-\frac{1}{2\lambda}\Bigr{)}$$

$$R_n(z):=\sum _{m=0}^n{a}_{m,n}{z}^m,a_{m,n}=\frac{(n+m)!}{m!(n-m)!}>0.$$

$$G_{n,n+1}(\lambda )=\frac{\frac{d}{d\lambda }{h}_{n+1}^{(1)}(-\mathbf{i}\lambda )}{h_{n+1}^{(1)}(-\mathbf{i}\lambda )}-\frac{\frac{d}{d\lambda }{h}_n^{(1)}(-\mathbf{i}\lambda )}{h_n^{(1)}(-\mathbf{i}\lambda )}>0.$$

$$\Bigl{(}h^{(1)}_{n}(-{\bf i}\lambda)\frac{d}{d\lambda}h^{(1)}_{n+1}(-{\bf i}\lambda)-h^{(1)}_{n+1}(-{\bf i}\lambda)\frac{d}{d\lambda}h^{(1)}_{n}(-{\bf i}\lambda)\Bigr{)}\Bigl{(}h^{(1)}_{n+1}(-{\bf i}\lambda)h^{(1)}_{n}(-{\bf i}\lambda)\Bigr{)}^{-1}>0.$$

$$\xi_{n}(\lambda):=\frac{e^{\lambda}}{\lambda}R_{n}\Bigl{(}-\frac{1}{2\lambda}\Bigr{)},\>\eta_{n}(\lambda):=\lambda\xi_{n}(\lambda).$$

$$\Bigl{(}\xi_{n}(\lambda)\frac{d}{d\lambda}\xi_{n+1}(\lambda)-\xi_{n+1}(\lambda)\frac{d}{d\lambda}\xi_{n}(\lambda)\Bigr{)}\Bigl{(}\xi_{n+1}(\lambda)\xi_{n}(\lambda)\Bigr{)}^{-1}$$

$$=\Bigl{(}\eta_{n}(\lambda)\frac{d}{d\lambda}\eta_{n+1}(\lambda)-\eta_{n+1}(\lambda)\frac{d}{d\lambda}\eta_{n}(\lambda)\Bigr{)}\Bigl{(}\eta_{n+1}(\lambda)\eta_{n}(\lambda)\Bigr{)}^{-1}>0.$$

$$F(\lambda )={\eta }_n(\lambda )\frac{d}{d\lambda }{\eta }_{n+1}(\lambda )-{\eta }_{n+1}(\lambda )\frac{d}{d\lambda }{\eta }_n(\lambda )$$

$$F^{\prime }(\lambda )={\eta }_n(\lambda )\frac{d^2}{d{\lambda }^2}{\eta }_{n+1}(\lambda )-{\eta }_{n+1}(\lambda )\frac{d^2}{d{\lambda }^2}{\eta }_n(\lambda ).$$

Full text

1. Introduction

Let $K\subset\{x\in{\mathbb{R}}^{3}:\>|x|\leq a\}$ be an open connected domain and let $\Omega={\mathbb{R}}^{3}\setminus\bar{K}$ be connected domain with $C^{\infty}$ smooth boundary $\Gamma$. Consider the boundary problem

[TABLE]

with initial data $f=(E_{0},B_{0})\in(L^{2}(\Omega))^{6}={\mathcal{H}}.$ Here $\nu(x)$ is the unit outward normal to $\partial\Omega$ at $x\in\Gamma$ pointing into $\Omega$, $\langle\>,\rangle$ denotes the scalar product in ${\mathbb{C}}^{3}$, $u_{tan}:=u-\langle u,\nu\rangle\nu$, and $\gamma(x)\in C^{\infty}(\Gamma)$ satisfies $\gamma(x)>0$ for all $x\in\Gamma.$ The solution of the problem (1.1) is described by a contraction semigroup

[TABLE]

where the generator $G_{b}$ has domain $D(G_{b})$ which is the closure in the graph norm of functions $u=(v,w)\in(C_{(0)}^{\infty}({\mathbb{R}}^{3}))^{3}\times(C_{(0)}^{\infty}({\mathbb{R}}^{3}))^{3}$ satisfying the boundary condition $v_{tan}-\gamma(\nu\wedge w_{tan})=0$ on $\Gamma.$

In [1] it was proved that the spectrum of $G_{b}$ in the open half plan $\{z\in{\mathbb{C}}:\>\operatorname{Re}z<0\}$ is formed by isolated eigenvalues with finite multiplicities. Note that if $G_{b}f=\lambda f$ with $\operatorname{Re}\lambda<0$, the solution $u(t,x)=V(t)f=e^{\lambda t}f(x)$ of (1.1) has exponentially decreasing global energy. Such solutions are called asymptotically disappearing and they are very important for the inverse scattering problems (see [1]). In particular, the eigenvalues $\lambda$ with $\operatorname{Re}\lambda\to-\infty$ imply a very fast decay of the corresponding solutions. In [2] the existence of eigenvalues of $G_{b}$ has been studied for the ball $B_{3}=\{x\in{\mathbb{R}}^{3},\>|x|<1\}$ assuming $\gamma$ constant. It was proved for $\gamma=1$ there are no eigenvalues in $\{z\in{\mathbb{C}}:\operatorname{Re}z<0\}$, while for $\gamma\neq 1$ there is always an infinite number of real eigenvalues $\lambda_{m}$ and with exception of one they satisfy the estimate

[TABLE]

where $\gamma_{0}=\max\{\gamma,\frac{1}{\gamma}\}.$

In this Note we study the distribution of the negative eigenvalues and our purpose is to obtain a Weyl formula for the counting function

[TABLE]

where every eigenvalues $\lambda_{m}$ is counted with its algebraic multiplicity given by

[TABLE]

where $0<\epsilon\ll 1$. Our main result is the following

Theorem 1.1**.**

Let $\gamma\neq 1$ be a constant and let $\gamma_{0}=\max\{\gamma,\frac{1}{\gamma}\}.$ Then the counting function $N(r)$ for the ball $B_{3}$ has the asymptotic

[TABLE]

The proof of Theorem 1.1 is based on a precise analysis of the roots of the equation (3.1) involving spherical Hankel functions $h^{(1)}_{n}(\lambda)$ of first kind. We show in Section 3 that for $\gamma>1$ this equation has only one real root $\lambda_{n}<0$. Moreover, we have $\lambda_{n+1}<\lambda_{n},\>\forall n\in{\mathbb{N}},$ so we have a decreasing sequence of eigenvalues. The geometric multiplicity of $\lambda_{n}$ is $2n+1$. Since $C_{b}$ is not a self-adjoint operator the geometric multiplicity could be less than the algebraic one. In our case these multiplicities coincide and the proof is based on a representation of $(G_{b}-z)^{-1}.$ To estimate $\lambda_{n}$ as $n\to\infty$, we apply an approximation of the exterior semiclassical Dirichlet to Neumann map for the operator $(h^{2}\Delta+z)$ established in [6] (see also [8]) combined with an application of Rouché theorem.

We conjecture that in the general case of strictly convex obstacles and $\min_{y\in\Gamma}\gamma(y)=\gamma_{1}>1$ we have the asymptotic

[TABLE]

For the ball $B_{3}$ this agrees with (1.3).

2. Boundary problem for Maxwell system

Our purpose is to study the eigenvalues of $G_{b}$ in case the obstacle is equal to the ball $B_{3}=\{x\in{\mathbb{R}}^{3}:|x|\leq 1\}$. Setting $\lambda={\bf i}\mu$, $\operatorname{Im}\mu>0,$ an eigenfunction $(E,B)\neq 0$ of $G_{b}$ satisfies

[TABLE]

Replacing $B$ by $H=-B$ yields for $(E,H)\in(H^{2}(|x|\leq 1))^{6}$,

[TABLE]

Expand $E(x),H(x)$ in the spherical functions $Y_{n}^{m}(\omega),\>n=0,1,2,...,\>|m|\leq n,\>\omega\in{\mathbb{S}}^{2}$ and the spherical Hankel functions of first kind

[TABLE]

An application of Theorem 2.50 in [3] (in the notation of [3] it is necessary to replace $\omega$ by $\mu\in{\mathbb{C}}\setminus\{0\}$) says that the solution of the system (2.2) for $x=|x|\omega,r=|x|>0,\omega=\frac{x}{r}$ has the form

[TABLE]

[TABLE]

Here $U_{n}^{m}(\omega)=\frac{1}{\sqrt{n(n+1)}}{\rm grad}\,_{{\mathbb{S}}^{2}}Y_{n}^{m}(\omega)$ and $V_{n}^{m}(\omega)=\nu\wedge U_{n}^{m}(\omega)$ for $n\in{\mathbb{N}},-n\leq m\leq n$ form a complete orthonormal basis in

[TABLE]

To find a representation of $\nu\wedge H_{tan}$, observe that $\nu\wedge(\nu\wedge U_{n}^{m})=-U_{n}^{m},$ so for $r=1$ one has

[TABLE]

[TABLE]

and the boundary condition in (2.2) is satisfied if

[TABLE]

[TABLE]

3. Roots of the equation $g_{n}(\lambda)=0$

To examine the eigenvalues of $G_{b}$ it is necessary to find the roots of the equations (2) and (2). Since $h^{(1)}_{n}(\mu)\neq 0$ for $\operatorname{Im}\mu>0$, the problem is reduced to study the roots $\lambda\in{\mathbb{R}}^{-}$ of the equation

[TABLE]

and the same equation with $\gamma$ replaced by $\frac{1}{\gamma}$. Clearly, if $\mu=-{\bf i}\lambda$ is such that the expressions in the brackets $[...]$ in (2.5) and (2.6) are non-vanishing for every $n\geq 1$, we must have $\alpha_{n}^{m}=\beta_{n}^{m}=0$ which implies $E_{tan}=B_{tan}=0$. Hence $(E,B)=0$ because the boundary problem with $\gamma=0$ has no eigenvalues in $\{z\in{\mathbb{C}}:\>\operatorname{Re}z<0\}.$ In this section we suppose that $\gamma\neq 1$ and examine the equation

[TABLE]

It is well known that (see [5])

[TABLE]

with

[TABLE]

We will prove the following

Proposition 3.1**.**

For $\lambda<0$ we have

[TABLE]

Proof.

The purpose is to show that

[TABLE]

Introduce the functions

[TABLE]

Then $h^{(1)}_{n}(-{\bf i}\lambda)=(-{\bf i})^{n}\xi_{n}(\lambda)$ and the above inequality is equivalent to

[TABLE]

[TABLE]

Since $\eta_{n}(\lambda)\eta_{n+1}(\lambda)>0$ for $\lambda<0$, it suffices to show that the function

[TABLE]

has positive values for $\lambda\in(-\infty,0).$ Consider the derivative

[TABLE]

We have

[TABLE]

The function $\Xi_{n}(z)=zh^{(1)}_{n}(z)$ satisfies the equation

[TABLE]

and

[TABLE]

[TABLE]

Consequently,

[TABLE]

[TABLE]

On the other hand,

[TABLE]

[TABLE]

and

[TABLE]

since

[TABLE]

Finally, the function $F(\lambda)$ in the interval $(-\infty,0]$ is increasing from 0 to $+\infty$ and this completes the proof. ∎

Now if $\lambda_{n}<0$ is a solution the equation

[TABLE]

one has

[TABLE]

so $\lambda_{n}$ is not a root of the equation

[TABLE]

In the following we assume that $\gamma>1.$ Then for $\lambda\to-\infty$ we have $g_{n+1}(\lambda)\to 1-\gamma<0$, and since $g_{n+1}(\lambda_{n})>0$ the equation $g_{n+1}(\lambda)=0$ has at least one root $-\infty<\lambda_{n+1}<\lambda_{n}.$

Lemma 3.1**.**

Let $\gamma>1$. For every $n\geq 1$ the equation $g_{n}(\lambda)=0$ in the interval $(-\infty,0)$ has exactly one root $\lambda_{n}<0.$

Proof.

Setting $w=-\frac{1}{2\lambda}$, we write the equation (3.2) as ${\mathcal{R}}_{n}(w):=w^{2}R_{n}^{\prime}(w)+\alpha R_{n}(w)=0$, where $\alpha=\frac{1-\gamma}{2}<0.$ We will show that this equation has exactly one positive root. Since

[TABLE]

the polynomial ${\mathcal{R}}_{n}(w)$ has the representation

[TABLE]

with

[TABLE]

Taking into account the form of $a_{k,n}$, we deduce

[TABLE]

Thus the sign of $b_{k,n}$ depends on the sign of the function

[TABLE]

which for $k\geq 1$ is increasing since

[TABLE]

Clearly, $b_{0,n}=\alpha<0$ and $b_{n+1,n}>0.$ There are two cases:

(i) $b_{1,n}\leq 0.$ Then there is only one change of sing in the Descartes’ sequence $\{b_{n+1,n},b_{n,n},...,b_{1,n},b_{0,n}\}.$

(ii) $b_{1,n}>0$. Then $b_{k,n}>0$ for $1\leq k\leq n+1$ and in the Descartes’ sequence $\{b_{n+1,n},b_{n,n},...,b_{1,n},b_{0,n}\}$ one has again only one change of sign.

Applying the Descartes’ rule of signs, we conclude that the number of the positive roots of ${\mathcal{R}}_{n}(w)=0$ is exactly one.

∎

Combining Proposition 3.1 and Lemma 3.1, one obtain the following

Corollary 3.1**.**

Let $\gamma>1$. Then the generator $G_{b}$ has an infinite sequence of real eigenvalues

[TABLE]

and $\lambda_{n}$ has geometric multiplicity $2n+1.$

The last statement concerns the geometric multiplicity since the functions $\{Y_{m,n}(\omega)\}_{m=-n}^{m}$ are linearly independent. The algebraic multiplicity of $\lambda_{m}$ will be discussed in Section 5.

4. Estimation of the roots

Throughout this section we assume $\gamma>1$. Set $\lambda=\frac{{\bf i}\sqrt{z}}{h},\>0<h\ll 1$ with $z=-1+{\bf i}\eta,\>0\leq|\eta|\leq h^{1/2},\eta\in{\mathbb{R}}.$ Consider the Dirichlet problem

[TABLE]

and note that $\Delta+\frac{z}{h^{2}}=\Delta-\lambda^{2}.$ The solution of (4.1) has the form

[TABLE]

where

[TABLE]

The semiclassical Dirichlet-to-Neumann operator ${\mathcal{N}}_{ext}(h,z)=\frac{h}{{\bf i}}\frac{d}{dr}w|_{r=1}$ related to (4.1) becomes

[TABLE]

[TABLE]

By using the approximation of ${\mathcal{N}}_{ext}(h,z)$ established in [8],[6] for $z=-1+{\bf i}\eta$, one deduces

[TABLE]

with $\rho=\sqrt{z-r_{0}(x^{\prime},\xi^{\prime})}$ and a constant $C>0$ independent of $z,\lambda$ and $f$. Here $r_{0}(x^{\prime},\xi^{\prime})$ is the principal symbol of the semiclasssical Laplace-Beltrami operator $-h^{2}\Delta_{{\mathbb{S}}^{2}}=\frac{z}{\lambda^{2}}\Delta_{{\mathbb{S}}^{2}}$. Moreover, $\sqrt{z}={\bf i}\sqrt{1-{\bf i}\eta}={\bf i}(1-\frac{{\bf i}\eta}{2}+{\mathcal{O}}(\eta^{2}))$ and

[TABLE]

Hence, for $0<h\leq h_{0}$ we get

[TABLE]

On the other hand,

[TABLE]

Applying the spectral theorem, one deduces

[TABLE]

and

[TABLE]

[TABLE]

This implies

[TABLE]

which we write as

[TABLE]

Remark 4.1**.**

For bounded $1\leq n\leq N_{0}$ and sufficiently large $|\lambda|$ the estimate $(4.2)$ follows easily from the fact that $\frac{R_{n}^{\prime}(w)}{R_{n}(w)}=n(n+1)+{\mathcal{O}}(|w|)$ as $|w|\to 0.$

Remark 4.2**.**

The estimate $(4.2)$ is similar to that in Proposition 2.1 in [7], where the function $\frac{J_{\nu}^{\prime}(\lambda)}{J_{\nu}(\lambda)}$ for $\nu\geq 0$ and $0<C\leq|\operatorname{Im}\lambda|\leq\delta|\operatorname{Re}\lambda|,\>\operatorname{Re}\lambda>C_{1}$ has been approximated. Here $J_{\nu}(z)$ is the Bessel function, while the boundary problem examined in [7] is in the domain $|x|<1.$

Put $z=\lambda$ and for $z\in\Lambda_{0}$ consider the function

[TABLE]

with zeros

[TABLE]

In the following we set $z_{n}=-\sqrt{\frac{n(n+1)}{\gamma^{2}-1}}.$ Clearly,

[TABLE]

and $\frac{n(n+1)}{z_{n}^{2}}=\gamma^{2}-1,\>f_{n}^{\prime}(z_{n})=-\frac{\gamma^{2}-1}{\gamma z_{n}}.$ A calculus yields the second derivative

[TABLE]

[TABLE]

For $n$ large enough and $a>0$ to be fixed below introduce the contour

[TABLE]

Our purpose is to choose $a$ so that

[TABLE]

We have

[TABLE]

and

[TABLE]

On the other hand,

[TABLE]

Clearly, one has the estimate

[TABLE]

Set $C_{\gamma}=\frac{\gamma^{2}-1}{\gamma}>0$ and choose $a>0$ so that $C_{\gamma}a>4C_{0}$. We fix $a$ and obtain

[TABLE]

taking $n$ large enough to satisfy the inequality

[TABLE]

Next we arrange the inequality

[TABLE]

It is clear that

[TABLE]

where

[TABLE]

Note that for $z\in C_{n}(a)$ and $n$ large enough according to (4.4), the function $|G(\frac{n(n+1)}{z^{2}})|$ is bounded by a constant $B_{\gamma,a}$ depending on $\gamma$ and $a$ . Thus for large $n$ we get

[TABLE]

and the proof of (4.7) is reduced to

[TABLE]

which is satisfied taking again $n$ large. Finally, we proved the estimate (4.3) and we can apply Rouché theorem for the functions $g_{n}(z)$ and $f_{n}(z)$ and conclude that the function $g_{n}(z)$ has exactly one simple zero $\lambda_{n}$ in $C_{n}(a)$. Since $g_{n}(z)$ has only real zeros (see Appendix in [2]), this implies the following

Lemma 4.1**.**

There exist $n_{0}(\gamma)$ and $a(\gamma)>0$ depending on $\gamma$ such that for $n\geq n_{0}(\gamma)$ the negative root $\lambda_{n}$ of the equation (3.2) satisfies the estimate

[TABLE]

Remark 4.3**.**

According to Proposition 2.1, $n_{0}(\gamma)$ must satisfy the inequality

[TABLE]

5. Weyl asymptotics

We start with the analysis of the multiplicity of $\lambda_{n}.$

Lemma 5.1**.**

For $n\geq n_{0}(\gamma)$ we have ${\rm mult}(\lambda_{n})=2n+1.$

Proof.

Since the geometric multiplicity of $\lambda_{n}$ is $2n+1$, it is sufficient to show that

[TABLE]

Let $\lambda\in\Lambda_{0},$ where $\Lambda_{0}$ is the set introduced in the previous section and let $\lambda\notin\sigma(G_{b})$. If $0\neq(f,g)\in{\rm Image}\>G_{b}\cap L^{2}(\Omega)$, one has ${\rm div}\,f={\rm div}\,g=0$ and for $(u,v)=(G_{b}-\lambda)^{-1}(f,g)$ we get ${\rm div}\,u={\rm div}\,v=0.$ Consider the skew self-adjoint operator

[TABLE]

with boundary condition $\nu\wedge u=0$ on ${\mathbb{S}}^{2}.$ Then $\sigma(A)\subset{\bf i}{\mathbb{R}}$ and let $(u_{0}(x;\lambda),v_{0}(x;\lambda))=(A-\lambda)^{-1}(f,g),$ that is

[TABLE]

Since ${\rm div}\,u_{0}={\rm div}\,v_{0}=0$, the well known coercive estimates yield $(u_{0},v_{0})\in H^{1}(\Omega)$. Moreover the resolvent $(A-\lambda)^{-1}$ is analytic in $\{z\in{\mathbb{C}}:\>\mbox{\rm Re}\>z<0\}$ and $u_{0}(x;\lambda),\>v_{0}(x;\lambda)$ depend analytically on $\lambda$. We write $(u,v)=(u_{0},v_{0})+(u_{1},v_{1})$, where $(u_{1}(x;\lambda),v_{1}(x;\lambda))$ is the solution of the problem

[TABLE]

To solve (5.3), note that $-\gamma(\nu\wedge(v_{0})_{tan}(\omega;z))=F(\omega;\lambda)\in L^{2}({\mathbb{S}}^{2})$ with $F(\omega;\lambda)$ analytical in $\lambda$ for $\lambda\in\Lambda_{0}$. Thus we may write

[TABLE]

with analytical coefficients $\tilde{\alpha}_{n}^{m}(\lambda),\tilde{\beta}_{n}^{m}(\lambda)$. Now we can solve (2.5), (2.6) with right hand part $(\tilde{\alpha}_{n}^{m}(\lambda),\>\tilde{\beta}_{n}^{m}(\lambda))$. Finally, we obtain a representation of the solution of (5.3) with meromorphic coefficients

[TABLE]

[TABLE]

If $\gamma>1$ the analysis in the previous section shows that for $\lambda\in\Lambda_{0}$ the meromorphic function $\alpha_{n}^{m}(\lambda)$ has a simple pole at $\lambda_{n}<0$, while $\beta_{n}^{m}(\lambda)$ is analytic in $\Lambda_{0}.$ For $0<\gamma<1$ the function $\alpha_{n}^{m}(\lambda)$ is analytic in $\Lambda_{0}$ and $\beta_{n}^{m}(\lambda)$ is meromorphic. Next we integrate $(u(x;\lambda),v(x;\lambda))$ over the circle $|\lambda_{n}-\lambda|=\epsilon$, where $\epsilon$ is sufficiently small. The integral of $(u_{0}(x;\lambda),v_{0}(x;\lambda))$ vanish, while for the integral of $(u_{1}(x;\lambda),v_{1}(x;\lambda))$, taking into account the representation of the solution of (5.3), we will obtain a sum

[TABLE]

This completes the proof of (5.1). ∎

Passing to the analysis of $N(r)$, consider first the case $\gamma>1.$ The root $\lambda_{n}$ has algebraic multiplicity $2n+1$ and to find a lower bound of $N(r)$ we apply the estimate

[TABLE]

for $r\geq a(\gamma)+\frac{n_{0}(\gamma)+1}{\sqrt{\gamma^{2}-1}}.$ Then

[TABLE]

To get a upper bound for $N(r),$ we use the estimate

[TABLE]

for

[TABLE]

hence

[TABLE]

If $0<\gamma<1$, we have $\frac{1}{\gamma}>1$ and one applies our argument to the the equation (2.6). This completes the proof of theorem 1.1