DN spectrum of a concentric radial conductivity inclusion

Let $n=\mathrm{\mid }k\mathrm{\mid }$, $s=r\mathrm{/}a$, and $\sigma =1+\beta$. In polar coordinates $\rho =\mathrm{\mid }x-x_0\mathrm{\mid }$, seek a solution with angular mode $e^{ik\theta }$, using $u_i=A{\rho }^n{e}^{ik\theta }$ for $\rho <r$ and $u_o=(B{\rho }^n+C{\rho }^{-n})e^{ik\theta }$ for $r<\rho <a$. Continuity of $u$ and continuity of $\gamma {\mathrm{\partial }}_{\rho }u$ at $\rho =r$ give $C{r}^{-2n}=-\eta B$ because $(\sigma -1)\mathrm{/}(\sigma +1)=\eta$. The boundary condition $u_o(a,\theta )=e_k(\theta )$ then gives $B{a}^n=(1-\eta s^{2n}{)}^{-1}$.

Each piece is harmonic on its subdomain, and its value and conductivity-weighted normal derivative agree at $\rho =r$ by step 1.1. Both pieces and their classical gradients are bounded up to the interface because $r>0$. Testing a distributional first derivative against $\phi \in C_c^{\mathrm{\infty }}(D)$ and integrating by parts on the inner disk and annulus produces interface terms equal to the jump in $u$ times $\phi$; the matching values cancel these terms, leaving the bounded piecewise classical derivatives. By [F7] the glued function belongs to $H^1(D)$. For any $z\in C_c^{\mathrm{\infty }}(D)$, integration by parts on the inner disk and outer annulus cancels the interface flux terms and proves the weak equation. Since $C_c^{\mathrm{\infty }}(D)$ is dense in $H_0^1(D)$ by [F1] and the energy form is continuous, the equation holds for every $z\in H_0^1(D)$. By [F1] the function is the unique solution for trace $e_k$. For any smooth representative $V$ of a boundary trace, the same piecewise integration by parts gives ${\int }_D{\gamma }_r\mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }V}dx={\int }_{\mathrm{\partial }D}{\mathrm{\partial }}_{\rho }{u}_o(a,\theta )\bar{V}dS,$ because the two interface flux terms cancel. The identity extends to every $V\in H^1(D)$ by the density of smooth ambient restrictions [F5], continuity of the energy form, and continuity of the trace. At $\rho =a$, the conductivity is $1$, so the outward flux is ${\mathrm{\partial }}_{\rho }{u}_o(a,\theta )=\frac{n}{a}\frac{1+\eta s^{2n}}{1-\eta s^{2n}}{e}_k(\theta )\mathrm{.}$ Thus this is the weak DN functional, with the displayed eigenvalue under the $L^2(\mathrm{\partial }D)$ pairing. For ${\gamma }_1\equiv 1$, the harmonic extension $(\rho \mathrm{/}a{)}^n{e}_k(\theta )$ gives flux $(n\mathrm{/}a)e_k$, and the constant solution has zero flux.

For each angle $\theta$, define $(R_{\theta }U)(x)=U(x_0+Q_{\theta }(x-x_0))$ on $D$, where $Q_{\theta }$ is the planar rotation matrix, and define $P_{k}U=(2\pi {)}^{-1}{\int }_0^{2\pi }{e}^{-ik\theta }{R}_{\theta }Ud\theta$. By density [F5] and the rotation isometries, $\theta \mapsto R_{\theta }U$ is strongly continuous in $H^1(D)$ for every $U$, so this integral is the $H^1$ limit of its Riemann sums. Rotations preserve the $H^1(D)$ inner product; Minkowski's inequality makes $P_k$ a contraction, and character orthogonality [F2] makes it a self-adjoint projection with pairwise orthogonal ranges. For $w\in C_c^{\mathrm{\infty }}(D)$, the map $\theta \mapsto R_{\theta }w$ is continuous in $H^1$ and every Riemann sum of the displayed integral is in $C_c^{\mathrm{\infty }}(D)$, so $P_{k}w\in H_0^1(D)$; contraction and density in the definition of $H_0^1$ extend this invariance to all of $H_0^1(D)$. For a smooth restriction $U$, the polar-coordinate chain rule gives the $H^1$ norm as the integral of $\mathrm{\mid }U{\mathrm{\mid }}^2+\mathrm{\mid }{\mathrm{\partial }}_{\rho }U{\mathrm{\mid }}^2+{\rho }^{-2}\mathrm{\mid }{\mathrm{\partial }}_{\theta }U{\mathrm{\mid }}^2$ against $\rho d\rho d\theta$. Parseval [F2] for these three terms and monotone convergence [F6] in the radial integral show that the finite orthogonal sums $S_N:={\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}{P}_k$ converge to $U$ in $H^1(D)$. For a general $U\in H^1(D)$, approximate by smooth restrictions using [F5]; on each annulus $ϵ<\rho <a$ the polar map is a $C^1$ diffeomorphism, so the polar norm and Fourier identities pass to the limit there, and monotone convergence [F6] as $ϵ\downarrow 0$ gives the full-disk identities. Each $S_N$ is an orthogonal projection and hence a contraction; density [F5] therefore extends $S_{N}U\to U$ to every $U\in H^1(D)$. The induced quotient modes are thus a complete orthogonal family in $\mathcal{D}(D)$, and the constant mode is orthogonal to every nonzero mode. Since the trace is bounded and rotation-equivariant on smooth functions, [F4] and [F5] extend commutation to all of $H^1(D)$; the trace-kernel identity in [F4] then shows that each projected boundary class is a scalar multiple of one mode. For any $H^1$ representative $U$ of the boundary mode $e_k$, let $v(\rho )$ be its $k$th angular Fourier coefficient. Parseval gives $\mathrm{\parallel }U{\mathrm{\parallel }}_{H^1(D)}^2\ge 2\pi {\int }_0^a(\mathrm{\mid }{v}^{\mathrm{\prime }}(\rho ){\mathrm{\mid }}^2\rho +\frac{n^2\mathrm{\mid }v(\rho ){\mathrm{\mid }}^2}{\rho })d\rho \mathrm{.}$ The trace condition gives $v(a)=1$. On every interval $[ϵ,a]$, the one-dimensional Sobolev representative of $v$ is absolutely continuous, and expansion of the square yields ${\int }_{ϵ}^a(\mathrm{\mid }{v}^{\mathrm{\prime }}{\mathrm{\mid }}^2\rho +\frac{n^2\mathrm{\mid }v{\mathrm{\mid }}^2}{\rho })d\rho ={\int }_{ϵ}^a{\mid v^{\mathrm{\prime }}-\frac{n}{\rho }v\mid }^2\rho d\rho +n\mathrm{\mid }v(a){\mathrm{\mid }}^2-n\mathrm{\mid }v(ϵ){\mathrm{\mid }}^2\mathrm{.}$ The full energy being finite implies ${\int }_0^a\mathrm{\mid }v(\rho ){\mathrm{\mid }}^2{\rho }^{-1}d\rho <\mathrm{\infty }$, so there is a sequence ${ϵ}_j\downarrow 0$ with $v({ϵ}_j)\to 0$; otherwise $\mathrm{\mid }v\mathrm{\mid }$ would be bounded below near zero and this integral would diverge. Letting $ϵ={ϵ}_j$ and using nonnegativity of the square proves that the radial energy is at least $n$. Thus every representative has squared $H^1$ norm at least $2\pi n$, so [F4] gives $Q_n^{\mathcal{D}}:=\mathrm{\parallel }{e}_k{\mathrm{\parallel }}_{\mathcal{D}(D)}^2\ge 2\pi n$. The harmonic lift $(\rho \mathrm{/}a{)}^n{e}_k$ has squared $H^1$ norm $2\pi n+\pi a^2\mathrm{/}(n+1)$ by direct radial integration, so [F4] gives $Q_n^{\mathcal{D}}\le 2\pi n+\pi a^2\mathrm{/}(n+1)$. Orthogonality of the constant mode to $e_k$ gives $\mathrm{\parallel }{e}_k+c\mathbf{1}{\mathrm{\parallel }}_{\mathcal{D}}^2=Q_n^{\mathcal{D}}+\mathrm{\mid }c{\mathrm{\mid }}^2\mathrm{\parallel }\mathbf{1}{\mathrm{\parallel }}_{\mathcal{D}}^2$; the constant class has positive norm by the trace isomorphism [F4]. Taking the infimum over $c$ shows that the statement's further quotient norm satisfies $Q_n=Q_n^{\mathcal{D}}$.

The weak DN map is diagonal on the complete orthogonal quotient modes from step 3.1; the constant mode is orthogonal to all nonzero modes, so quotienting by constants leaves $Q_n$ unchanged: angular integration makes the pairing of mode $e_k$ with any distinct mode zero, and step 2.1 gives its diagonal coefficient. For finite sums, the dual norm of the difference applied to $\sum a_k{e}_k$ is $({\sum }_{k\mathrm{\ne }0}\mathrm{\mid }{a}_k{\mathrm{\mid }}^2(2\pi a\mathrm{\mid }{\lambda }_n(r)-n\mathrm{/}a\mathrm{\mid }{)}^2\mathrm{/}{Q}_n{)}^{1\mathrm{/}2}$; boundedness [F1] and density extend this formula to all inputs. Therefore $\mathrm{\parallel }{\dot{\mathrm{\Lambda }}}_{{\gamma }_r}-{\dot{\mathrm{\Lambda }}}_{{\gamma }_1}\mathrm{\parallel }={\sup }_{k\mathrm{\ne }0}\frac{2\pi a\mathrm{\mid }{\lambda }_n(r)-n\mathrm{/}a\mathrm{\mid }}{Q_n},{\lambda }_n(r)-\frac{n}{a}=\frac{2n}{a}\frac{\eta s^{2n}}{1-\eta s^{2n}}\mathrm{.}$ Using $Q_n\ge 2\pi n$, $0<s\le 1\mathrm{/}2$, and $0<\eta <1$, this is at most ${\sup }_{n\ge 1}\frac{2\eta s^{2n}}{1-\eta s^{2n}}\le \frac{2\eta }{1-\eta }{s}^2\mathrm{.}$ This proves all asserted eigenvalue and operator-norm formulas.