Calderón’s Problem for Some Classes of Conductivities in Circularly Symmetric Domains

Abstract

In this note, we study Calderón’s problem for certain classes of conductivities in domains with circular symmetry in two and three dimensions. Explicit formulas are obtained for the reconstruction of the conductivity from the Dirichlet-to-Neumann map. As a consequence, we show that the reconstruction is Lipschitz stable.

1 Introduction

Consider a conductor in a domain ${\varOmega }\subset \mathbb {R}^{n}$ with conductivity γ(x). When a voltage potential $f\in {H^{\frac {1}{2}}}(\partial {\varOmega } )$ is applied at the boundary ∂Ω, the induced potential u in Ω is the unique weak solution in H¹(Ω) of

$$ \begin{cases} \nabla \cdot (\gamma \nabla u) = 0 & \text{ in }{\varOmega},\\ u = f & \text{ on }\partial {\varOmega}. \end{cases}$$

The Dirichlet-to-Neumann map is given by ${\varLambda }_{\gamma }(f)= \gamma {\partial _{\nu } }u|_{\partial {\varOmega }}\in {H^{- \frac {1}{2}}}(\partial {\varOmega } )$. Here, ν denotes the exterior unit normal to ∂Ω. The problem studied by Calderón in [8] is to determine the conductivity γ from Λ_γ.

For n ≥ 3 and γ ∈ C², that Λ_γ uniquely determines γ was proved by Sylvester and Uhlmann in [16]. Recently, based on the breakthrough work by Haberman and Tataru [13], Caro and Rogers [9] proved uniqueness for Lipschitz conductivities. There is also a related work by Haberman [12].

In two dimensions, and C² conductivities, the uniqueness was proved by Nachman [15]. Later, Astala and Päivärinta [4] proved uniqueness for bounded measurable conductivities.

After the uniqueness has been established, it is natural to study the stability of the reconstruction, i.e., we would like to estimate γ₁ − γ₂ in certain norm by

$$||{\varLambda}_{1}-{\varLambda}_{2}||_{\star}=\sup\limits_{f\in H^{\frac{1}{2}}(\partial{\varOmega})\atop f\not=0}\frac{||({\varLambda}_{1}-{\varLambda}_{2})f||_{H^{-\frac{1}{2}}(\partial{\varOmega})}}{||f||_{H^{\frac{1}{2}}(\partial {\varOmega})}}.$$

In [1], Alessandrini proved that the following log-stability estimate holds:

$$||\gamma_{1}-\gamma_{2}||_{L^{\infty}({\varOmega})}\le C\big(\log(1+||{\varLambda}_{1}-{\varLambda}_{2}||_{\star}^{-1})\big)^{-\sigma},$$

where C,σ are positive constants and γ_j ∈ H^s+ 2(Ω),s > n/2. Later, Mandache [14] showed that such estimate is optimal.

To improve the stability estimate, Alessandrini and Vessella [3] considered special classes of piecewise constant conductivities, for n ≥ 3. The Lipschitz stability obtained therein has been generalized to other classes of conductivities in [2, 7] and [11].

The analog of the result of [3] was proven for the two-dimensional case in [5]. Subsequent generalizations of this result were obtained in [6] and [10].

In this paper, we prove Lipschitz stability estimate for two special cases of domains with circular symmetry. In the first case, we consider ${\varOmega }=B(0,1)\subset \mathbb {R}^{2}$ with conductivities of the form

$$ {\gamma_{\alpha} (x)=}\begin{cases} \alpha_{1}+\alpha_{2}(a-r)& \text{ if } 0 \le r<a, \\ \alpha_{0} &\text{ if } a\le r <1, \end{cases}$$

where r = |x| and ε₀ ≤ α₀,α₁ ≤ M, 0 ≤ α₂ ≤ N. We denote this set of conductivities by μ(a,ε₀,M,N). In the second case, we consider ${\varOmega }=B(0,1)\times (0,+\infty )\subset \mathbb {R}^{3}$ with conductivities of the form

$$ {\gamma_{\alpha} (z)=}\begin{cases} 1+{\alpha_{1}} & \text{ if }h \le z < \infty,\\ 1+ {\alpha_{2}} &\text{ if } 0 \le z < h , \end{cases}$$

where ${\alpha _{j}} \in \left [ {0,M} \right ],j = 1,2,M > 0,h>0$. We denote this set of conductivities by μ(h,M). We give a formula for the Dirichlet-to-Neumann map in each case, together with a formula to recover the conductivity from the Dirichlet-to-Neumann map. As a consequence, we show that the map Λ_γ↦γ is Lipschitz. More precisely, our main results are as follows.

Theorem 1.1

Let Ω = B(0,1) and a ∈ (0,1),ε₀,M > 0,N ≥ 0. There exists a positive constant C = C(a,ε₀,M,N) such that

$${\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_ \star } \ge C\left( {\left| {{\alpha_{0}} - {{\upbeta}_{0}}} \right| + \left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right| + \left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|} \right), \forall {\gamma_{\alpha} },{\gamma_{\upbeta} } \in \mu (a,{\varepsilon_{0}},M,N).$$

Theorem 1.2

Let ${\varOmega }=B(0, 1)\times (0, \infty )$ and $h \in (0,\infty ), M>0$. There exists a positive constant C = C(h,M) such that

$$\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_{{H_{rad}^{\frac{1}{2}}(B)} \to {H_{rad}^{- \frac{1}{2}}(B)}} \ge C\left( {\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right| + \left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right|} \right),\forall \gamma_{\alpha} ,\gamma_{\upbeta}\in \mu(h,M).$$

2 Proof of Theorem 1.1

Consider the Dirichlet problem in the unit disc B = B(0,1) on the plane

$$ \begin{array}{@{}rcl@{}} \begin{cases} \nabla \cdot (\gamma_{\alpha} \nabla u) = 0 & \text{ in }B, \\ u = f &\text{ on }\partial B, \end{cases} \end{array} $$

(2.1)

where the conductivity γ_α ∈ μ(a,ε₀,M,N). In the polar coordinate, if $u(x)={\sum }_{n\in \mathbb {Z}} {u_{n}(r)e^{in\theta }} \in H^{1}(B)$, then the equation in (2.1) is

$$ \begin{cases} (\gamma_{\alpha} u^{\prime}_{n})^{\prime}+ \frac{\gamma_{\alpha} }{r}u^{\prime}_{n} - \frac{{{n^{2}}\gamma_{\alpha} }}{r^{2}}{u_{n}} = 0, \forall n\in\mathbb Z, \\ \lim\limits_{r \to {a^ - }} {u_{n}}(r) = \lim\limits_{r \to {a^ + }} {u_{n}}(r), \\ \lim\limits_{r \to {a^ - }} \left( {\gamma {u^{\prime}_{n}}} \right)(r) = \lim\limits_{r \to {a^ + }} \left( {\gamma {u^{\prime}_{n}}} \right)(r). \end{cases} $$

Solving these systems, we obtain

$$u_{0}(r)=c_{0}, \quad 0\le r<1,$$

and for n≠ 0,

$$ \begin{array}{@{}rcl@{}} {u_{n}(r)=}\begin{cases} {b_{n}}{r^{\left| n \right|}} + {c_{n}}{r^{- \left| n \right|}}& \text{ if }a\le r <1, \\ \sum\limits_{k\ge |n|} {{a_{k}}{r^{k}}} &\text{ if }0< r <a, \end{cases} \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} {a_{|n| + m}} &=& \frac{{{\alpha_{2}}}}{{{\alpha_{1}} + a{\alpha_{2}}}}\frac{{(2m - 1)\left| n \right| + m(m - 1)}}{{2m\left| n \right| + {m^{2}}}}{a_{\left| n \right| + m - 1}}\\ &=&{\left( {\frac{{{\alpha_{2}}}}{{{\alpha_{1}} + a{\alpha_{2}}}}} \right)^{m}}\prod\limits_{j = 1}^{m} {\frac{{(2j - 1)\left| n \right| + j(j - 1)}}{{2j\left| n \right| + {j^{2}}}}} {a_{\left| n \right|}}, m=1,2,\cdots. \end{array} $$

Note that α₂ ≥ 0,α₁ ≥ 𝜖₀ > 0, the power series ${\sum }_{k\ge |n|}a_{k}r^{k}$ is uniformly convergent on [0,a]. From that, we get

$$ \frac{{{c_{n}}}}{{{b_{n}}}} = \frac{{{a^{2\left| n \right|}}\left[ {\left| n \right|{u_{n}}({a^ - }) - a\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{u^{\prime}_{n}}({a^ - })} \right]}}{{\left| n \right|{u_{n}}({a^ - }) + a\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{u^{\prime}_{n}}({a^ - })}}.$$

The Dirichlet-to-Neumann map ${\varLambda }_{\alpha }:{H^{\frac {1}{2}}}(\partial B) \to {H^{- \frac {1}{2}}}(\partial B)$ is determined by

$$ {\varLambda}_{\alpha} f(\theta) =\sum\limits_{n \in \mathbb{Z}} {\widehat {{{\varLambda}_{{\alpha }}}f}(n){e^{in\theta }}}, $$

where $f(\theta )={\sum }_{n\in \mathbb Z}\hat {f}(n)e^{in\theta }\in H^{\frac {1}{2}}(\partial B)$ and

$$ \begin{array}{@{}rcl@{}} \widehat {{{\varLambda}_{{\alpha }}}f}(n) &=& \lim\limits_{r \to {1^ - }} {\gamma_{\alpha} }(r){u^{\prime}_{n}}(r) = \alpha_{0} \left| n \right|\widehat f(n)\frac{{{b_{n}} - {c_{n}}}}{{{b_{n}} + {c_{n}}}}\\ &=& {\alpha_{0}}\left| n \right|\widehat f(n)\frac{{1 - {a^{2\left| n \right|}} + (1 + {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)}}{{1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)}},\forall n \in \mathbb{Z},\\ B_{n}(b)&=& 1 + \frac{b}{{2\left| n \right| + 1}}\times \frac{{1 + \sum\limits_{m = 2}^{\infty} {m{b^{m - 1}}{h_{m,n}}} }}{{1 + \frac{{\left| n \right|}}{{2\left| n \right| + 1}}b\left( {1 + \sum\limits_{m = 2}^{\infty} {{b^{m - 1}}{h_{m,n}}} } \right)}},\\ {h_{m,n}} &=& \prod\limits_{j = 2}^{m} {\frac{{(2j - 1)\left| n \right| + j(j - 1)}}{{2j\left| n \right| + {j^{2}}}}},\quad b = \frac{{a{\alpha_{2}}}}{{{\alpha_{1}} + a{\alpha_{2}}}}. \end{array} $$

Note that $0\le b\le b_{0}=\frac {{aN}}{{{\varepsilon _{0}} + aN}}<1$. To obtain some properties of B_n(b), we need the following technical lemma.

Lemma 2.1

(i) ${\sum }_{m = 2}^{\infty } {m{b^{m - 1}}{\prod }_{j = 2}^{m} {\frac {{2j - 1}}{{2j}}} }={(1 - b)^{- \frac {3}{2}}} - 1$. (ii) ${\sum }_{m = 2}^{\infty } {{b^{m - 1}}{\prod }_{j = 2}^{m} {\frac {{2j - 1}}{{2j}}} }=\frac {2}{{1 - b + \sqrt {1 - b} }} - 1$. (iii) $\lim _{n \to \infty } h_{mn}= {\prod }_{j=2}^m \frac {2j-1}{2j}$.

We have the following proposition

Proposition 2.2

B_n’s satisfy the following properties. (i) 1 ≤ B_n(b) ≤ d₀, where $d_{0}=1 + \frac {{{b_{0}}}}{{{{(1 - {b_{0}})}^{\frac {3}{2}}}}}$. (ii) $\lim _{n \to \infty } {B_{n}}(b) = 1$. (iii) $\lim _{n \to \infty } (2\left | n \right | + 1)({B_{n}}(b) - 1) = \frac {b}{{1 - b}}$. (iv) $\lim _{n \to \infty } \frac {{\frac {{{\alpha _{1}}}}{{{\alpha _{0}}}}{B_{n}}(b) - 1}}{{\frac {{{\alpha _{1}}}}{{{\alpha _{0}}}}{B_{n + 1}}(b) - 1}} = 1, b\not =0$. (v) $\frac {{1 - {b_{0}}}}{{2\left | n \right | + 1}} \le {B^{\prime }_{n}}(b) \le \frac {A}{{2\left | n \right | + 1}}$, where A = A(a,ε₀,N) is a constant.

Proof

We rewrite B_n(b) as follows

$$ {B_{n}}(b) = 1 + \frac{b}{{2\left| n \right| + 1}}\times \frac{{1 + \sum\limits_{m = 2}^{\infty} {m{b^{m - 1}}{h_{m,n}}} }}{{1 + \frac{{\left| n \right|}}{{2\left| n \right| + 1}}b\left( {1 + \sum\limits_{m = 2}^{\infty} {{b^{m - 1}}{h_{m,n}}} } \right)}}. $$

(2.2)

(i) From (2.2), it is easy to see that B_n(b) ≥ 1. We now show that B_n(b) ≤ d₀. Indeed, using (i) in Lemma 2.2, we have

$$ B_{n}(b)=\frac{{1 + \frac{{\left| n \right| + 1}}{{2\left| n \right| + 1}}b + \sum\limits_{m = 2}^{\infty} {\frac{{\left| n \right| + m}}{{2\left| n \right| + 1}}{b^{m}}{h_{m,n}}} }}{{1 + \frac{{\left| n \right|}}{{2\left| n \right| + 1}}b + \sum\limits_{m = 2}^{\infty} {\frac{{\left| n \right|}}{{2\left| n \right| + 1}}{b^{m}}{h_{m,n}}} }}\le 1 + b_{0} + \sum\limits_{m = 2}^{\infty} {m{{b^{m}_{0}}}}\prod\limits_{j = 2}^{m} {\frac{{2j - 1}}{{2j}}} = d_{0}. $$

(ii) From (2.2), it is not difficult to get $\lim _{n \to \infty } {B_{n}}(b) = 1$.

(iii) We have

$$ \left( {2\left| n \right| + 1} \right)\left( {{B_{n}}(b) - 1} \right) = \frac{{b\left( {1 + \sum\limits_{m = 2}^{\infty} {m{b^{m - 1}}{h_{m,n}}} } \right)}}{{1 + \frac{{\left| n \right|}}{{2\left| n \right| + 1}}b\left( {1 + \sum\limits_{m = 2}^{\infty} {{b^{m - 1}}{h_{m,n}}} } \right)}}. $$

Hence, from Lemma 2.1, we obtain

$$\lim\limits_{n \to \infty } \left( {2\left| n \right| + 1} \right)({B_{n}}(b) - 1) = \frac{{b{{(1 - b)}^{- \frac{3}{2}}}}}{{{{(1 - b)}^{- \frac{1}{2}}}}} = \frac{b}{{1 - b}}.$$

(iv) We consider two casesCase 1 α₀≠α₁. From (ii), we have

$$ \lim\limits_{n \to \infty } \frac{{\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b) - 1}}{{\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n + 1}}(b) - 1}} = \frac{{\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}} - 1}}{{\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}} - 1}}=1.$$

Case 2 α₀ = α₁. We need to prove $\lim _{n \to \infty } \frac {{{B_{n}}(b) - 1}}{{{B_{n + 1}}\left (b \right ) - 1}} = 1$. From (iii), we get

$$ \lim\limits_{n \to \infty } \frac{{{B_{n}}(b) - 1}}{{{B_{n + 1}}(b) - 1}} =\lim\limits_{n \to \infty } \frac{{(2\left| n \right| + 1)\left( {{B_{n}}(b) - 1} \right)}}{{(2\left| n \right| + 3)\left( {{B_{n + 1}}(b) - 1} \right)}} = 1.$$

(v) We denote by M_n(b) and N_n(b) the numerator and denominator of $B^{\prime }_{n}(b)$, respectively. Direct computation gives

$$ {M_{n}}(b) = \frac{1}{{2\left| n \right| + 1}} + \sum\limits_{m = 2}^{\infty} {\frac{{\left| n \right|{{(m - 1)}^{2}}}}{{{{(2\left| n \right| + 1)}^{2}}}}} {b^{m}}{h_{m,n}} + \sum\limits_{m = 2}^{\infty} {\frac{{{m^{2}}}}{{2\left| n \right| + 1}}} {b^{m - 1}}{h_{m,n}}+ {I_{n}}(b), $$

where

$$ \begin{array}{@{}rcl@{}} I_{n}(b)&=& \left( {\sum\limits_{l = 2}^{\infty} {\frac{{{l^{2}}{b^{l - 1}}}}{{2\left| n \right| + 1}}{h_{l,n}}} } \right)\left( {\sum\limits_{k = 2}^{\infty} {\frac{{\left| n \right|{b^{k}}}}{{2\left| n \right| + 1}}{h_{k,n}}} } \right) - \left( {\sum\limits_{l = 2}^{\infty} {\frac{{l\left| n \right|{b^{l - 1}}}}{{2\left| n \right| + 1}}{h_{l,n}}} } \right)\\ &&\times\left( {\sum\limits_{k = 2}^{\infty} {\frac{{k{b^{k}}}}{{2\left| n \right| + 1}}{h_{k,n}}} } \right). \end{array} $$

The coefficient of b^m in I_n(b) is

$$ \begin{array}{@{}rcl@{}} \sum\limits_{k + l -1= m} {\left( {\frac{{{l^{2}}\left| n \right|}}{{{{\left( {2\left| n \right| + 1} \right)}^{2}}}} - \frac{{l\left| n \right|k}}{{{{\left( {2\left| n \right| + 1} \right)}^{2}}}}} \right){h_{l,n}}{h_{k,n}}} & =& \sum\limits_{k + l-1 = m} {\frac{{l\left| n \right|(l - k)}}{{{{\left( {2\left| n \right| + 1} \right)}^{2}}}}{h_{l,n}}{h_{k,n}}}\\ &=&\frac{1}{2}\sum\limits_{k + l-1 = m} {\frac{{\left| n \right|{{(k - l)}^{2}}}}{{{{\left( {2\left| n \right| + 1} \right)}^{2}}}}{h_{l,n}}{h_{k,n}}}. \end{array} $$

From this, we obtain

$$ {M_{n}}(b) \ge \frac{1}{{2\left| n \right| + 1}}. $$

(2.3)

Moreover, we have

$$ \begin{array}{@{}rcl@{}} \sum\limits_{m = 2}^{\infty} {\frac{{\left| n \right|{{(m - 1)}^{2}}}}{{{{(2\left| n \right| + 1)}^{2}}}}} {b^{m}}{h_{m,n}} &\le& \frac{1}{{2(2\left| n \right| + 1)}}\sum\limits_{m = 0}^{\infty} {{m^{2}}{b^{m}}} \\ &=& \frac{1}{{2(2\left| n \right| + 1)}} \frac{{b(b + 1)}}{{{{(1 - b)}^{3}}}} \le \frac{1}{{2(2\left| n \right| + 1)}} \frac{{{b_{0}}({b_{0}} + 1)}}{{{{(1 - {b_{0}})}^{3}}}}. \end{array} $$

(2.4)

Next, we have

$$ \begin{array}{@{}rcl@{}} \sum\limits_{m = 2}^{\infty} {\frac{{{m^{2}}}}{{2\left| n \right| + 1}}} {b^{m - 1}}{h_{m,n}} \le \frac{1}{2({2\left| n \right| + 1})}\sum\limits_{m = 0}^{\infty} {{m^{2}}} {b^{m - 1}}\le \frac{1}{{2(2\left| n \right| + 1)}} \frac{{{b_{0}} + 1}}{{{{(1 - {b_{0}})}^{3}}}}. \end{array} $$

(2.5)

We see that

$$\frac{1}{2}\sum\limits_{k + l - 1 = m} {\frac{{\left| n \right|{{(k - l)}^{2}}}}{{{{\left( {2\left| n \right| + 1} \right)}^{2}}}}{h_{l,n}}{h_{k,n}}} \le \frac{1}{{4(2\left| n \right| + 1)}}{(m + 1)^{3}},m = 3,4, {\ldots} $$

It follows that

$$ \begin{array}{@{}rcl@{}} {I_{n}}(b) &\le& \frac{1}{{4(2\left| n \right| + 1)}}\sum\limits_{m = 3}^{\infty} {{{(m + 1)}^{3}}} {b^{m}} \le \frac{1}{{4(2\left| n \right| + 1)}}\sum\limits_{m = 0}^{\infty} {{{(m + 1)}^{3}}} {b^{m}}\\ & \le &\frac{{{b_{0}}}}{{4\left( {2\left| n \right| + 1} \right)}}\left( {\frac{{2{b_{0}} + 1}}{{{{(1 - {b_{0}})}^{3}}}} + \frac{{3{b_{0}}({b_{0}} + 1)}}{{{{(1\! -\! {b_{0}})}^{4}}}}} \right). \end{array} $$

(2.6)

From (2.3), (2.4), (2.5), and (2.6), we deduce that

$$ \frac{1}{{2\left| n \right| + 1}} \le {M_{n}}(b) \le \frac{A}{{2\left| n \right| + 1}},\quad A\text{ is a constant depending on }a, \varepsilon_{0}, N. $$

(2.7)

On the other hand, we have

$$ 1 \le {N_{n}}(b) = {\left( {1 + \frac{{\left| n \right|}}{{2\left| n \right| + 1}}b + \sum\limits_{m = 2}^{\infty} {\frac{{\left| n \right|}}{{2\left| n \right| + 1}}{b^{m}}{h_{m,n}}} } \right)^{2}} \le \frac{1}{{1 - b}} \le \frac{1}{{1 - {b_{0}}}}. $$

(2.8)

From (2.7) and (2.8), we have

$$ \frac{{1 - {b_{0}}}}{{2\left| n \right| + 1}} \le {B^{\prime}_{n}}(b) \le \frac{A}{{2\left| n \right| + 1}},$$

where A is a constant depending on a,ε₀,N. □

We now give an explicit formula to reconstruct the parameters a and α from the Dirichlet-to-Neumann map. We define

$${C_{n}}{ = \frac{{\varLambda}_{\alpha} ({e^{in\theta }})}{{\left| n \right|{e^{in\theta }}}} = \alpha_{0}}\frac{{1 - {a^{2\left| n \right|}} + (1 + {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)}}{1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)}.$$

If there is a strictly increasing sequence of positive integers $\{n_{k}\}_{k=1}^{\infty }$ such that $C_{n_{k}}=\alpha _{0}$, it is easy to obtain α₀ = α₁,α₂ = 0; i.e., the conductor is homogeneous. Otherwise, we have the following proposition.

Proposition 2.3

The following formulas hold: (i) $\alpha _{0}=\lim _{n \to \infty } {C_{n}}$. (ii) $a^{-2}=\lim _{n \to \infty } \frac {{{C_{n}} - {\alpha _{0}}}}{{{C_{n + 1}} - {\alpha _{0}}}}$. (iii) α₁ = α₀D, where

$$D = \frac{{\lim\limits_{n \to \infty } \frac{{{C_{n}} - {\alpha_{0}}}}{{2{a^{2\left| n \right|}}{\alpha_{0}}}} + 1}}{{1 - \lim\limits_{n \to \infty } \frac{{{C_{n}} - {\alpha_{0}}}}{{2{a^{2\left| n \right|}}{\alpha_{0}}}}}}.$$

(iv) aα₂ = α₁E, where

$$ E = \lim\limits_{n \to \infty } (2\left| n \right| + 1)\left[ {\frac{{{\alpha_{0}}\left( {2{\alpha_{0}}{a^{2\left| n \right|}} + ({C_{n}} - {\alpha_{0}})(1 + {a^{2\left| n \right|}})} \right)}}{{{\alpha_{1}}\left( {2{\alpha_{0}}{a^{2\left| n \right|}} - ({C_{n}} - {\alpha_{0}})(1 - {a^{2\left| n \right|}})} \right)}} - 1} \right]. $$

Proof

(i) From (ii) in Proposition 2.2

$$ \lim\limits_{n \to \infty } {C_{n}} = {\alpha_{0}}. $$

(ii) Next, we have

$$ \begin{array}{@{}rcl@{}} &&\lim\limits_{n \to \infty } \frac{{{C_{n}} - {\alpha_{0}}}}{{{C_{n + 1}} - {\alpha_{0}}}} \\ &=& \lim\limits_{n \to \infty }\frac{{2{a^{2\left| n \right|}}\left( {\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b) - 1} \right)}}{{2{a^{2\left| n \right| + 2}}\left( {\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n + 1}}(b) - 1} \right)}}\frac{{1 + {a^{2\left| n \right| + 2}} + (1 - {a^{2\left| n \right| + 2}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n + 1}}(b)}}{{1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)}}. \end{array} $$

Using (ii) and (iv) in Proposition 2.2, we obtain

$$ \lim\limits_{n \to \infty } \frac{{{C_{n}} - {\alpha_{0}}}}{{{C_{n + 1}} - {\alpha_{0}}}} = \frac{1}{{{a^{2}}}}.$$

(iii) Using (ii) in Proposition 2.2, we have

$$ \lim\limits_{n \to \infty } \frac{{{C_{n}} - {\alpha_{0}}}}{{2{a^{2\left| n \right|}}{\alpha_{0}}}} = \lim\limits_{n \to \infty }\frac{ {\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b) - 1} }{{1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)}}=\frac{{\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}} - 1}}{{\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}} + 1}}. $$

This leads to

$${\alpha_{1}} = \frac{{{\alpha_{0}}\left( {\lim\limits_{n \to \infty } \frac{{{C_{n}} - {\alpha_{0}}}}{{2{a^{2\left| n \right|}}{\alpha_{0}}}} + 1} \right)}}{{1 - \lim\limits_{n \to \infty } \frac{{{C_{n}} - {\alpha_{0}}}}{{2{a^{2\left| n \right|}}{\alpha_{0}}}}}}.$$

(iv) We now calculate α₂. From

$${C_{n}} - {\alpha_{0}} = {\alpha_{0}}\frac{{2{a^{2\left| n \right|}}\left( {\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b) - 1} \right)}}{{1 + {a^{2}{\left| n \right|}} + (1 - {a^{2}{\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)}}$$

we calculate

$${B_{n}}(b) = \frac{{{\alpha_{0}}\left( {2{\alpha_{0}}{a^{2\left| n \right|}} + ({C_{n}} - {\alpha_{0}})(1 + {a^{2\left| n \right|}})} \right)}}{{{\alpha_{1}}\left( {2{\alpha_{0}}{a^{2\left| n \right|}} - ({C_{n}} - {\alpha_{0}})(1 - {a^{2\left| n \right|}})} \right)}}.$$

From that and (iii) in Proposition 2.2, we get

$$ \begin{array}{@{}rcl@{}} E &=& \lim\limits_{n \to \infty } (2\left| n \right| + 1)\left[ {\frac{{{\alpha_{0}}\left( {2{\alpha_{0}}{a^{2\left| n \right|}} + ({C_{n}} - {\alpha_{0}})(1 + {a^{2\left| n \right|}})} \right)}}{{{\alpha_{1}}\left( {2{\alpha_{0}}{a^{2\left| n \right|}} - ({C_{n}} - {\alpha_{0}})(1 - {a^{2\left| n \right|}})} \right)}} - 1} \right] \\ &=& \lim\limits_{n \to \infty } (2\left| n \right| + 1)({B_{n}}(b) - 1) = \frac{b}{{1 - b}}. \end{array} $$

From $b={\frac {{a{\alpha _{2}}}}{{{\alpha _{1}} + a{\alpha _{2}}}}}$, we obtain aα₂ = α₁E. □

We now prove Theorem 1.1.

Proof Proof of Theorem 1.1

For $\gamma _{\alpha }, \gamma _{\upbeta } \in \mu (a,\varepsilon _{0},M,N), f\in H^{\frac {1}{2}}(\partial B)$, we have

$$\left\| {\left( {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right)f} \right\|_{{H^{- \frac{1}{2}}}(\partial B)}^{2} = \sum\limits_{n \in \mathbb{Z}} {\frac{{{n^{2}}}}{{{{(1 + {n^{2}})}^{\frac{1}{2}}}}}{{({A_{n}} - {B_{n}})}^{2}}{{\left| {\widehat f(n)} \right|}^{2}}} ,$$

where b = aα₂/(α₁ + aα₂),c = aβ₂/(β₁ + aβ₂), and

$${A_{n}} = \alpha_{0}\frac{{1 - {a^{2\left| n \right|}} + (1 + {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)}}{{1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)}},{B_{n}} = {\upbeta}_{0}\frac{{1 - {a^{2\left| n \right|}} + (1 + {a^{2\left| n \right|}})\frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(c)}}{{1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(c)}}.$$

By direct computation, we obtain

$$ \begin{array}{@{}rcl@{}} &&{A_{n}} -{B_{n}} \\ &= &\frac{{({\alpha_{0}} + {{\upbeta}_{0}})\left( {\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b) - \frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(c)} \right)2{a^{2\left| n \right|}}}}{{\left( {1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)} \right)\left( {1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(c)} \right)}}\\ &&+\frac{{({\alpha_{0}} - {{\upbeta}_{0}})\left[ {(1 - {a^{4\left| n \right|}})\left( {1 + \frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}\frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(b){B_{n}}(c)} \right) + (1 + {a^{4\left| n \right|}})\left( {\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b) + \frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(c)} \right)} \right]}}{{\left( {1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b)} \right)\left( {1 + {a^{2\left| n \right|}} + (1 - {a^{2\left| n \right|}})\frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(c)} \right)}}. \end{array} $$

We denote by K_n and H_n the numerator and denominator of A_n − B_n, respectively. We have ${H_{n}} \le {({2 + \frac {M}{{{\varepsilon _{0}}}}{d_{0}}} )^{2}}$ and

$$ \begin{array}{@{}rcl@{}} {\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_{\star}} &=& \sup\limits_{{f \in {H^{\frac{1}{2}}}(\partial B)}\atop f \ne 0} \frac{{{{\left\| {\left( {{{\varLambda}_{{\alpha }}} - {{\varLambda}_{{\upbeta }}}} \right)f} \right\|}_{{H^{- \frac{1}{2}}}(\partial B)}}}}{{{{\left\| f \right\|}_{{H^{\frac{1}{2}}}(\partial B)}}}} \ge \sup\limits_{n\not=0} \frac{{\left| {{K_{n}}} \right|}}{{\left| {{2H_{n}}} \right|}} \ge \sup\limits_{n\not=0} \frac{{\left| {{K_{n}}} \right|}}{{{2\left( {2 + \frac{M}{{{\varepsilon_{0}}}}{d_{0}}} \right)^{2}}}},\\ \left| {{K_{n}}} \right| &\ge& \frac{{2{\varepsilon_{0}}}}{M}\left| {{\alpha_{0}} - {{\upbeta}_{0}}} \right| - \frac{{8{M^{2}}{d_{0}}}}{{{\varepsilon_{0}}}}{a^{2\left| n \right|}}. \end{array} $$

When α₀≠β₀, for n big enough, we obtain

$$\frac{{8{M^{2}}{d_{0}}}}{{{\varepsilon_{0}}}}{a^{2\left| n \right|}} \le \frac{{{\varepsilon_{0}}}}{M}\left| {{\alpha_{0}} - {{\upbeta}_{0}}} \right|.$$

Hence,

$$ {\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_ \star } \ge \frac{{{\varepsilon_{0}}}}{2M}{\left( {2 + \frac{M}{{{\varepsilon_{0}}}}{d_{0}}} \right)^{- 2}}\left| {{\alpha_{0}} - {{\upbeta}_{0}}} \right|. $$

(2.9)

For α₀ = β₀, we also have (2.9).

Next, we have

$$\left| {{K_{n}}} \right| \ge 4{\varepsilon_{0}}{a^{2\left| n \right|}}\left| {\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b) - \frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(c)} \right| - \left| {{\alpha_{0}} - {{\upbeta}_{0}}} \right|\left| {1 + \frac{{{M^{2}}}}{{{\varepsilon_{0}^{2}}}}{d_{0}} + 4\frac{M}{{{\varepsilon_{0}}}}} \right|.$$

From (2.9), we have

$$ {\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_{\star}} \ge {C_{1}}4{\varepsilon_{0}}{a^{2\left| n \right|}}\left| {\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b) - \frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(c)} \right|, $$

(2.10)

where C₁ = C₁(a,ε₀,M) is a constant. We now consider

$$ \begin{array}{@{}rcl@{}} &&\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}{B_{n}}(b) - \frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}{B_{n}}(c)\\ & =& \frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}\left( {{B_{n}}(b) - {B_{n}}(c)} \right) + \left( {\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}} - \frac{{{{\upbeta}_{1}}}}{{{{\upbeta}_{0}}}}} \right){B_{n}}(c)\\ &=&\frac{{{\alpha_{1}}}}{{{\alpha_{0}}}}(b - c){B^{\prime}_{n}}(\xi ) +\frac{{{{\upbeta}_{0}}({\alpha_{1}} - {{\upbeta}_{1}}) + {{\upbeta}_{1}}({{\upbeta}_{0}} - {\alpha_{0}})}}{{{\alpha_{0}}{{\upbeta}_{0}}}}{B_{n}}(c) \quad(\text{for } \xi \in (b,c)) \\ &=& - \frac{{{{\upbeta}_{1}}{B_{n}}(c)}}{{{\alpha_{0}}{{\upbeta}_{0}}}}({\alpha_{0}} - {{\upbeta}_{0}}) + \left[ {\frac{{{B_{n}}(c)}}{{{\alpha_{0}}}} - \frac{{{\alpha_{1}}{{\upbeta}_{2}}a{B^{\prime}_{n}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}}} \right]({\alpha_{1}} - {{\upbeta}_{1}})\\ &&+\frac{{{\alpha_{1}}{{\upbeta}_{1}}a{B^{\prime}_{n}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}}({\alpha_{2}} - {{\upbeta}_{2}}). \end{array} $$

So from (2.9) and (2.10), we have

$$ {\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_{\star}} \ge {C_{2}}{a^{2\left| n \right|}}{D_{n}}, $$

(2.11)

where C₂ = C₂(a,ε₀,M) and

$$ \begin{array}{@{}rcl@{}} {D_{n}} &=& \frac{1}{\alpha_{0}}\left| \left[ B_{n}(c) - \frac{{{\alpha_{1}}{{\upbeta}_{2}}a{B^{\prime}_{n}}(\xi )}}{{({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}} \right]({\alpha_{1}} - {{\upbeta}_{1}}) \right.\\ &&\left.+ \frac{{{\alpha_{1}}{{\upbeta}_{1}}a{B^{\prime}_{n}}(\xi )}}{{({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}}({\alpha_{2}} - {{\upbeta}_{2}}) \right|. \end{array} $$

Using (i) and (v) in Proposition 2.2, we get

$$ \begin{array}{@{}rcl@{}} &&\frac{1}{M}\left( {1 - \frac{A}{{2\left| n \right| + 1}}} \right) \le \frac{{{B_{n}}(c)}}{{{\alpha_{0}}}} - \frac{{{\alpha_{1}}{{\upbeta}_{2}}a{B^{\prime}_{n}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}} \le \frac{{{d_{0}}}}{{{\varepsilon_{0}}}},\\ && 0\le \frac{{{\varepsilon_{0}^{2}}a(1 - {b_{0}})}}{{M{{({\varepsilon_{0}} + Na)}^{2}}\left( {2\left| n \right| + 1} \right)}} \le \frac{{{\alpha_{1}}{{\upbeta}_{1}}a{B^{\prime}_{n}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}} \le \frac{A}{{{\varepsilon_{0}}\left( {2\left| n \right| + 1} \right)}}. \end{array} $$

There exists an n₀ = n₀(a,ε₀,N) such that for every n ≥ n₀ then

$$0\le \frac{1}{{2M}} \le \frac{1}{M}\left( {1 - \frac{A}{{2\left| n \right| + 1}}} \right).$$

We now show that

$$ {\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_ \star } \ge C(a,{\varepsilon_{0}},M,N){ }\left( {\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right| + \left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|} \right). $$

(2.12)

We consider three cases.Case 1 (α₁ −β₁)(α₂ −β₂) ≥ 0. We have

$$ \begin{array}{@{}rcl@{}} {D_{{n_{0}}}}& \ge& \frac{1}{{2M}}\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right| + \frac{{{\varepsilon_{0}^{2}}a(1 - {b_{0}})}}{{M{{({\varepsilon_{0}} + Na)}^{2}}\left( {2\left| {{n_{0}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|\\ &\ge& \min \left\{ {\frac{1}{{2M}},\frac{{{\varepsilon_{0}^{2}}a(1 - {b_{0}})}}{{M{{({\varepsilon_{0}} + Na)}^{2}}\left( {2\left| {{n_{0}}} \right| + 1} \right)}}} \right\}\left( {\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right| + \left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|} \right). \end{array} $$

(2.13)

From (2.11) and (2.13), we obtain (2.12).Case 2 ${\left ({{\alpha _{1}} - {\upbeta }_{1} } \right )}\left ({{\alpha _{2}} - {{\upbeta }_{2}}} \right ) < 0$ and

$$ \begin{array}{@{}rcl@{}} {D_{{n_{0}}}} &=& \left[ {\frac{{{B_{{n_{0}}}}(c)}}{{{\alpha_{0}}}} - \frac{{{\alpha_{1}}{{\upbeta}_{2}}a{B^{\prime}_{n}}_{_{0}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}}} \right]\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right| \\ &&- \frac{{{\alpha_{1}}{{\upbeta}_{1}}a{B^{\prime}_{n}}_{_{0}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|. \end{array} $$

From that, we have

$$ \frac{d_{0}}{\varepsilon_{0}}\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right| - \frac{{{\varepsilon_{0}^{2}}a(1 - {b_{0}})}}{{M{{({\varepsilon_{0}} + Na)}^{2}}\left( {2\left| {{n_{0}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right| \ge {D_{{n_{0}}}} \ge 0. $$

Then there exists an n₁ = n₁(a,ε₀,M,N) > n₀ such that

$$ \frac{{\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right|}}{{4M}} \ge \frac{{d_{0}AM{{({\varepsilon_{0}} + Na)}^{2}}}}{{{\varepsilon_{0}^{2}}a(1 - {b_{0}})}}\frac{{\left( {2\left| {{n_{0}}} \right| + 1} \right)}}{{\left( {2\left| {{n_{1}}} \right| + 1} \right)}}\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right| \ge \frac{A}{{{\varepsilon_{0}}\left( {2\left| {{n_{1}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|.$$

We get

$$ {D_{{n_{1}}}} \ge \frac{{\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right|}}{{2M}} - \frac{A}{{{\varepsilon_{0}}\left( {2\left| {{n_{1}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right| \ge \frac{{\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right|}}{{4M}}>\frac{A}{{{\varepsilon_{0}}\left( {2\left| {{n_{1}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|. $$

(2.14)

From (2.11) and (2.14), we have (2.12)Case 3 ${\left ({{\alpha _{1}} - {\upbeta }_{1} } \right )}\left ({{\alpha _{2}} - {{\upbeta }_{2}}} \right ) < 0$ and

$$ \begin{array}{@{}rcl@{}} {D_{{n_{0}}}} = \frac{{{\alpha_{1}}{{\upbeta}_{1}}a{B^{\prime}_{{n_{0}}}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right| - \left[ {\frac{{{B_{{n_{0}}}}(c)}}{{{\alpha_{0}}}} - \frac{{{\alpha_{1}}{{\upbeta}_{2}}a{B^{\prime}_{{n_{0}}}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}}} \right]\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right|. \end{array} $$

There exists an n₂ = n₂(a,ε₀,M,N) > n₀ such that

$$ \frac{{{\varepsilon_{0}^{2}}a(1 - {b_{0}})}}{{2M{{({\varepsilon_{0}} + Na)}^{2}}\left( {2\left| {{n_{0}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right| \ge \frac{{2d_{0}MA}}{{{\varepsilon_{0}^{2}}(2\left| {{n_{2}}} \right| + 1)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|. $$

(2.15)

If

$$ \begin{array}{@{}rcl@{}} {D_{{n_{2}}}} = \left[ {\frac{{{B_{{n_{2}}}}(c)}}{{{\alpha_{0}}}} - \frac{{{\alpha_{1}}{{\upbeta}_{2}}aB^{\prime}_{n_{2}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}}} \right]\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right| - \frac{{{\alpha_{1}}{{\upbeta}_{1}}B^{\prime}_{n_{2}}(\xi )}}{{{\alpha_{0}}({\alpha_{1}} + {\alpha_{2}}a)({{\upbeta}_{1}} + {{\upbeta}_{2}}a)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|,\end{array} $$

we return to Case 2. Otherwise,

$$ \frac{A}{{{\varepsilon_{0}}\left( {2\left| {{n_{2}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right| - \frac{1}{{2M}}\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right| \ge {D_{{n_{2}}}} \ge 0. $$

(2.16)

From (2.15) and (2.16), we obtain

$$\frac{{{\varepsilon_{0}^{2}}a(1 - {b_{0}})}}{{2M{{({\varepsilon_{0}} + Na)}^{2}}\left( {2\left| {{n_{0}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right| \ge \frac{d_{0}}{{{\varepsilon_{0}}}}\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right|.$$

Moreover, we have

$$ \begin{array}{@{}rcl@{}} {D_{{n_{0}}}}& \ge& \frac{{{\varepsilon_{0}^{2}}a(1 - {b_{0}})}}{{M{{({\varepsilon_{0}} + Na)}^{2}}\left( {2\left| {{n_{0}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right| - \frac{d_{0}}{{{\varepsilon_{0}}}}\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right|\\ & \ge& \frac{{{\varepsilon_{0}^{2}}a(1 - {b_{0}})}}{{2M{{({\varepsilon_{0}} + Na)}^{2}}\left( {2\left| {{n_{0}}} \right| + 1} \right)}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|\ge \frac{d_{0}}{{{\varepsilon_{0}}}}\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right|. \end{array} $$

(2.17)

From (2.11) and (2.17), we have (2.12). From (2.9) and (2.12), the conclusion follows. □

3 Proof of Theorem 1.2

Consider the Dirichlet problem

$$ \begin{array}{@{}rcl@{}} \begin{cases} \nabla \cdot (\gamma_{\alpha} \nabla u) = 0 & \text{ in }B\times (0,+\infty), \\ u = 0 &\text{ on }\partial B \times (0,+\infty),\\ u = f &\text{ on } B\times \{0\}, \end{cases} \end{array} $$

(3.1)

where the conductivity γ_α ∈ μ(h,M).

Definition 3.1

(i) We denote

$$ \begin{array}{@{}rcl@{}} &&L_{rad}^{2}(B) = \left\{ u \in {L^{2}}(B),u(x, y)=f\left( \sqrt{x^{2}+y^{2}}\right) \right\}.\\ &&L_{rad}^{2}(B\times (0,+\infty)) = \left\{ {u \in {L^{2}}(B\times (0,+\infty))}, u(x, y, z)=f\left( \sqrt{x^{2}+y^{2}}, z\right) \right\}. \end{array} $$

(ii) Let

$$H_{rad}^{\frac{1}{2}}(B) = \left\{ {f \in L_{rad}^{2}(B),\sum\limits_{n = 1}^{\infty} {{{(1 + {{\left| {{\lambda_{n}}} \right|}^{2}})}^{\frac{1}{2}}}{{\left| {\widehat f(n)} \right|}^{2}}{J_{1}^{2}}(} {\lambda_{n}}) < \infty } \right\}, $$

where

$$\widehat f(n) = \frac{{2{{\int}_{0}^{1}} {f(} r){J_{0}}({\lambda_{n}}r)rdr}}{{{{({J_{1}}({\lambda_{n}}))}^{2}}}},$$

J₀(λ_nr) is Bessel function of order zero, λ_n is positive zero of function J₀

$${\lambda_{1}} < {\lambda_{2}} < {\ldots} {\lambda_{n}} {\ldots} ,{\lambda_{n}} \sim \left( {n - \frac{1}{4}} \right)\pi , \text{ when }n\to \infty.$$

J₁(λ_n) is Bessel function of order 1 and

$${J_{1}}({\lambda_{n}} ) = \sum\limits_{m = 0}^{\infty} {\frac{{{{(- 1)}^{m}}{\lambda_{n}^{2m + 1}}}}{{{2^{2m + 1}}(m + 1)!m!}}},\quad {J_{1}}({\lambda_{n}})=-J_{0}^{\prime}({\lambda_{n}})$$

with ${J_{1}}({\lambda _{n}}) \sim \sqrt {\frac {2}{{\pi {\lambda _{n}}}}} {\cos \limits } ({{\lambda _{n}} - \frac {{3\pi }}{4}} ) + \mathrm {O}({\frac {1}{{\lambda _{n}^{3/2}}}} )$ when $n\to \infty$. The norm of $f \in H_{rad}^{\frac {1}{2}}(B)$ is given by

$${\left\| f \right\|_{H_{rad}^{\frac{1}{2}}(B)}} = {\left( {\sum\limits_{n = 1}^{\infty} {{{(1 + {{\left| {{\lambda_{n}}} \right|}^{2}})}^{\frac{1}{2}}}{{\left| {\widehat f(n)} \right|}^{2}}{J_{1}^{2}}(} {\lambda_{n}})} \right)^{\frac{1}{2}}}.$$

(iii) The dual space of $H_{rad}^{\frac {1}{2}}(B)$ is defined by

$$H_{rad}^{- \frac{1}{2}}(B)={\left( {H_{rad}^{\frac{1}{2}}(B)} \right)^{*}} = \left\{ {f:H_{rad}^{\frac{1}{2}}(B) \to \mathbb{C}} \text{ bounded linear functional}\right\}$$

with the norm

$${\left\| f \right\|_{H_{rad}^{- \frac{1}{2}}(B)}} = {\left( {\sum\limits_{n = 1}^{\infty} {{{(1 + {{\left| {{\lambda_{n}}} \right|}^{2}})}^{- \frac{1}{2}}}{{\left| {\widehat f(n)} \right|}^{2}}{J_{1}^{2}}(} {\lambda_{n}})} \right)^{\frac{1}{2}}}.$$

(iv) We denote

$$H_{rad}^{1}(B \times (0, + \infty )) = \left\{ {u \in L_{rad}^{2}(B \times (0, + \infty )):|\nabla u| \in L_{rad}^{2}(B \times (0, \infty ))} \right\}.$$

In the cylindrical coordinates, if $u(r, z)={\sum }_{n=1}^{\infty } u_{n}(z)J_{0}(\lambda _{n} r)$ we have

$${\left\| u \right\|_{{H^{1}_{rad}}(B \times (0, + \infty ))}} = \pi \sum\limits_{n = 1}^{\infty} {{J_{1}^{2}}({\lambda_{n}})} {\int}_{0}^{\infty} [(1 + {\lambda_{n}^{2}}){{\left| {{u_{n}}(z)} \right|}^{2}} + {{\left| {{u^{\prime}_{n}}(z)} \right|}^{2}} ]dz.$$

For $f \in {H_{rad}^{\frac {1}{2}}(B)}$, the Dirichlet problem (3.1) in cylindrical coordinates is

$$ \begin{array}{@{}rcl@{}} \begin{cases} \gamma_{\alpha}u_{rr} + \frac{\gamma_{\alpha} }{r }{u_{r} } + \partial_z(\gamma_{\alpha}u_z) = 0,&B \times (0,\infty ), \\ u(1,z) = 0,& 0<z<\infty,\\ u(r,0)=f,& 0\le r<1, \end{cases} \end{array} $$

and have a unique solution $u \in H^{1}_{rad} \left ({B \times (0,\infty )} \right )$.

We expand $u = {\sum }_{n = 1}^{\infty } {{u_{n}}(z){J_{0}}({\lambda _{n}}r )}$. By direct computation, we have

$$ \begin{array}{@{}rcl@{}} {u_{n}(z)=}\begin{cases} {a_{n}}{e^{- {\lambda_{n}}z}}& \text{ if }h \le z < \infty,\\ {b_{n}}{e^{- {\lambda_{n}}z}} + {c_{n}}{e^{{\lambda_{n}}z}}&\text{ if }0 \le z < h. \end{cases} \end{array} $$

At z = h, we have

$$ \begin{array}{@{}rcl@{}} \begin{cases} \lim\limits_{z \to {h^ + }} {u_{n}}(z) = \lim\limits_{z \to {h^ - }} {u_{n}}(z),\\ \lim\limits_{z \to {h^ + }} \left( \gamma_{\alpha} {u^{\prime}_{n}} \right)(z) = \lim\limits_{z \to {h^ - }} \left( \gamma_{\alpha}u^{\prime}_{n} \right)(z). \end{cases} \end{array} $$

It follows that

$$ \begin{array}{@{}rcl@{}} \frac{c_{n}}{b_{n}}=\frac{\alpha_{2}-\alpha_{1}}{(2+\alpha_{1}+\alpha_{2})e^{2\lambda_{n}h}}. \end{array} $$

The Dirichlet-to-Neumann map ${\varLambda }_{\alpha }:{H_{rad}^{\frac {1}{2}}(B)} \to {H_{rad}^{- \frac {1}{2}}(B)}$ is determined by

$$ {\varLambda}_{\alpha} f(r) = - \sum\limits_{n = 1}^{\infty} {(1 + {\alpha_{2}})\frac{{({\alpha_{2}} - {\alpha_{1}}){e^{- 2{\lambda_{n}}h}} - \left( {2 + {\alpha_{1}} + {\alpha_{2}}} \right)}}{{({\alpha_{2}} - {\alpha_{1}}){e^{- 2{\lambda_{n}}h}} + 2 + {\alpha_{1}} + {\alpha_{2}}}}} {\lambda_{n}}\hat f(n){J_{0}}({\lambda_{n}}r ). $$

We now give an explicit formula to reconstruct the parameters h, α from the Dirichlet-to-Nemann map. Define

$$ {A_{n}} = -\frac{{{{\varLambda}_{\alpha} }({J_{0}}({\lambda_{n}}r))}}{{{\lambda_{n}}{J_{0}}({\lambda_{n}}r)}} = (1 + {\alpha_{2}})\frac{{2 + {\alpha_{1}} + {\alpha_{2}} - ({\alpha_{2}} - {\alpha_{1}}){e^{- 2{\lambda_{n}}h}}}}{{2 + {\alpha_{1}} + {\alpha_{2}} + ({\alpha_{2}} - {\alpha_{1}}){e^{- 2{\lambda_{n}}h}}}}. $$

(3.2)

If A₁ = 1 + α₂ then α₁ = α₂; i.e., the conductor is homogeneous. Otherwise, $A_{n}\not =1+\alpha _{2}, \forall n\in \mathbb N$ and we have the following proposition.

Proposition 3.2

We reconstruct h,α_j as follows: (i) ${\alpha _{2}} = \lim _{n \to \infty } {A_{n}} - 1.$ (ii) $h = \frac {1}{2\pi }\ln (\lim _{n \to \infty } \frac {{{A_{n}} - 1-\alpha _{2}}}{{{A_{n + 1}} - 1-\alpha _{2}}})$. (iii) ${\alpha _{1}} = \frac {{2A + (A + 2){\alpha _{2}}}}{{2 - A}},$ where

$$A=\lim\limits_{n \to \infty } \frac{{\left( {{A_{n}} - 1-\alpha_{2}} \right){e^{2{\lambda_{n}}h}}}}{{1 + {\alpha_{2}}}}.$$

Proof

(i) It is easy to show that ${\alpha _{2}} = \lim _{n \to \infty } {A_{n}} - 1$.

(ii) We have

$$\frac{{{A_{n}} - 1-\alpha_{2}}}{{{A_{n + 1}} - 1-\alpha_{2}}} = \frac{{{e^{- 2{\lambda_{n}}h}}}}{{{e^{- 2{\lambda_{n + 1}}h}}}}\frac{{2 + {\alpha_{1}} + {\alpha_{2}} + ({\alpha_{2}} - {\alpha_{1}}){e^{- 2{\lambda_{n + 1}}h}}}}{{2 + {\alpha_{1}} + {\alpha_{2}} + ({\alpha_{2}} - {\alpha_{1}}){e^{- 2{\lambda_{n}}h}}}}.$$

Note that ${\lambda _{n}} \sim \left ({n - \frac {1}{4}} \right )\pi , \text { when }n\to \infty$. We obtain

$$\lim\limits_{n \to \infty } \frac{{{A_{n}} - 1-\alpha_{2}}}{{{A_{n + 1}} - 1-\alpha_{2}}} = {e^{2\pi h}}.$$

Hence,

$$h = \frac{1}{2\pi }\ln \left( {\lim\limits_{n \to \infty } \frac{{{A_{n}} - 1-\alpha_{2}}}{{{A_{n + 1}} - 1-\alpha_{2}}}} \right).$$

(iii) Since

$$A={\lim\limits_{n \to \infty } \frac{{\left( {{A_{n}} - 1-\alpha_{2}} \right){e^{2{\lambda_{n}}h}}}}{{1 + {\alpha_{2}}}} = \frac{{2({\alpha_{1}} - {\alpha_{2}})}}{{2 + {\alpha_{1}} + {\alpha_{2}}}}},$$

so α₁ = (2A + (A + 2)α₂)/(2 − A). □

Remark 3.3

We can reconstruct h,α₁ from α₂,A₁,A₂ as follows

$$ \begin{array}{@{}rcl@{}} h& =& \frac{1}{2(\lambda_{1}-\lambda_{2})}\ln\left( \frac{(A_{1}+1+\alpha_{2})(A_{2}-1-\alpha_{2})}{(A_{1}-1-\alpha_{1})(A_{2}+1+\alpha_{2})}\right)\\ \alpha_{1}&=& \frac{A_{1}(2+\alpha_{2}(1+e^{-2\lambda_{1}h}))-(1+\alpha_{2})(2+\alpha_{2}(1-e^{-2\lambda_{2}h}))}{(1+\alpha_{2})(1+e^{-2\lambda_{1}h})-A_{1}(1-e^{-2\lambda_{1}h})}. \end{array} $$

We now prove Theorem 1.2.

Proof Proof of Theorem 1.2

Firstly, for each $\gamma _{\alpha }, \gamma _{\upbeta }\in \mu (h,M), f\in H^{\frac {1}{2}}_{rad}(B)$, we have

$$\left\| {\left( {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right)f} \right\|_{H_{rad}^{- \frac{1}{2}}(B)}^{2}=\sum\limits_{n = 1}^{\infty} {\frac{{{\lambda_{n}^{2}}}}{{{{(1 + {\lambda_{n}^{2}})}^{\frac{1}{2}}}}}{\left( A_{n} - B_{n} \right)^{2}}{{|\hat f(n)|}^{2}}{{\left( {{J_{1}}({\lambda_{n}})} \right)}^{2}}},$$

where

$$ \begin{array}{@{}rcl@{}} &&{A_{n}} = -(1 + {\alpha_{2}}) \frac{{({\alpha_{2}} - {\alpha_{1}}){e^{- 2{\lambda_{n}}h}} - \left( {2 + {\alpha_{1}} + {\alpha_{2}}} \right)}}{{({\alpha_{2}} - {\alpha_{1}}){e^{^{- 2{\lambda_{n}}h}}} + 2 + {\alpha_{1}} + {\alpha_{2}}}},\\ &&{B_{n}} = - (1 + {{\upbeta}_{2}})\frac{{({{\upbeta}_{2}} - {{\upbeta}_{1}}){e^{- 2{\lambda_{n}}h}} - \left( {2 + {{\upbeta}_{1}} + {{\upbeta}_{2}}} \right)}}{{({{\upbeta}_{2}} - {{\upbeta}_{1}}){e^{^{- 2{\lambda_{n}}h}}} + 2 + {{\upbeta}_{1}} + {{\upbeta}_{2}}}}. \end{array} $$

By direct computation we obtain

$$ {A_{n}} - {B_{n}} = \frac{{(A - B{e^{- 2{\lambda_{n}}h}} - C{e^{- 4{\lambda_{n}}h}})({\alpha_{2}} - {{\upbeta}_{2}}) + D{e^{- 2{\lambda_{n}}h}}({\alpha_{1}} - {{\upbeta}_{1}})}}{{(2 + {\alpha_{1}} + {\alpha_{2}} + ({\alpha_{2}} - {\alpha_{1}}){e^{- 2{\lambda_{n}}h}})(2 + {{\upbeta}_{1}} + {{\upbeta}_{2}} + ({{\upbeta}_{2}} - {{\upbeta}_{1}}){e^{- 2{\lambda_{n}}h}})}}, $$

where

$$ \begin{array}{@{}rcl@{}} A &=& (2 + {\alpha_{1}} + {\alpha_{2}})(2 + {{\upbeta}_{1}} + {{\upbeta}_{2}}) \in [4,4{(M + 1)^{2}}],\\ B &=& (2 + {\alpha_{2}} + {{\upbeta}_{2}})(2 + {{\upbeta}_{1}})\in [4,4{(M + 1)^{2}}],\\ C &=& ({\alpha_{2}} - {\alpha_{1}})({{\upbeta}_{2}} - {{\upbeta}_{1}})\in [ - {M^{2}},{M^{2}}],\\ D &=& (2 + {\alpha_{2}} + {{\upbeta}_{2}})(2 + {{\upbeta}_{2}})\in [4,4{(M + 1)^{2}}]. \end{array} $$

We denote by K_n and H_n the numerator and denominator of (A_n − B_n), respectively. We have H_n ≤ (2 + 3M)² and

$$ \begin{array}{@{}rcl@{}} {\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_{H_{rad}^{\frac{1}{2}}(B) \to H_{rad}^{- \frac{1}{2}}(B)}}& = \sup\limits_{f\in H_{rad}^{\frac{1}{2}}(B)\atop f \ne 0} \frac{{{{\left\| {\left( {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right)f} \right\|}_{H_{rad}^{- \frac{1}{2}}(B)}}}}{{{{\left\| f \right\|}_{H_{rad}^{\frac{1}{2}}(B)}}}}\\ &\ge \sup\limits_{n\not=0} \frac{{\left| {{K_{n}}} \right|}}{{\left| {{2H_{n}}} \right|}} \ge \sup\limits_{n\not=0} \frac{{\left| {{K_{n}}} \right|}}{{{2\left( {2 + 3M} \right)^{2}}}}. \end{array} $$

(3.3)

Hence,

$$ \sup\limits_{n} \left| {{K_{n}}} \right| \ge (A + Be^{- 2\lambda_{n} h} + |C|e^{- 4\lambda_{n} h})\left|\alpha_{2} - {\upbeta}_{2} \right| - D e^{- 2\lambda_{n} h}\left|\alpha_{1} - {\upbeta}_{1} \right|.$$

For α₂≠β₂, we choose n big enough so that

$$ \sup\limits_{n} \left| K_{n} \right| \ge 2\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|. $$

(3.4)

From (3.4), (3.3) becomes

$$ {\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_{H_{rad}^{\frac{1}{2}}(B) \to H_{rad}^{- \frac{1}{2}}(B)}} \ge \frac{1}{{{{(2 + 3M)}^{2}}}}\left| {{\alpha_{2}} - {{\upbeta}_{2}}} \right|. $$

(3.5)

For α₂ = β₂, we also have (3.5). It is easy to get

$$ |K_{1}|\ge 4e^{-2\lambda_{1}h}|\alpha_{1}-{\upbeta}_{1}|-4(M+1)^{2}(1+e^{-2\lambda_{1}h})^{2}|\alpha_{2}-{\upbeta}_{2}|. $$

Therefore, from (3.3) and (3.5), we have

$$ {\left\| {\varLambda}_{\alpha} - {\varLambda}_{\upbeta} \right\|_{H_{rad}^{\frac{1}{2}}(B) \to H_{rad}^{- \frac{1}{2}}(B)}} \ge \frac{e^{-2\lambda_{1}h}}{{{{2(2 + 3M)}^{2}}(M+1)^{2}(1+e^{-2\lambda_{1}h})^{2}}}\left| {{\alpha_{1}} - {{\upbeta}_{1}}} \right|. $$

(3.6)

From (3.5) and (3.6), we are done. □