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.
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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
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 Λγ.
For n ≥ 3 and γ ∈ C2, that Λγ 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
In [1], Alessandrini proved that the following log-stability estimate holds:
where C,σ are positive constants and γj ∈ Hs+ 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
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
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 Λγ↦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
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
2 Proof of Theorem 1.1
Consider the Dirichlet problem in the unit disc B = B(0,1) on the plane
where the conductivity γα ∈ 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
Solving these systems, we obtain
and for n≠ 0,
where
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
The Dirichlet-to-Neumann map \({\varLambda }_{\alpha }:{H^{\frac {1}{2}}}(\partial B) \to {H^{- \frac {1}{2}}}(\partial B)\) is determined by
where \(f(\theta )={\sum }_{n\in \mathbb Z}\hat {f}(n)e^{in\theta }\in H^{\frac {1}{2}}(\partial B)\) and
Note that \(0\le b\le b_{0}=\frac {{aN}}{{{\varepsilon _{0}} + aN}}<1\). To obtain some properties of Bn(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
Bn’s satisfy the following properties. (i) 1 ≤ Bn(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 Bn(b) as follows
(i) From (2.2), it is easy to see that Bn(b) ≥ 1. We now show that Bn(b) ≤ d0. Indeed, using (i) in Lemma 2.2, we have
(ii) From (2.2), it is not difficult to get \(\lim _{n \to \infty } {B_{n}}(b) = 1\).
(iii) We have
Hence, from Lemma 2.1, we obtain
(iv) We consider two casesCase 1 α0≠α1. From (ii), we have
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
(v) We denote by Mn(b) and Nn(b) the numerator and denominator of \(B^{\prime }_{n}(b)\), respectively. Direct computation gives
where
The coefficient of bm in In(b) is
From this, we obtain
Moreover, we have
Next, we have
We see that
It follows that
From (2.3), (2.4), (2.5), and (2.6), we deduce that
On the other hand, we have
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
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
(iv) aα2 = α1E, where
Proof
(i) From (ii) in Proposition 2.2
(ii) Next, we have
Using (ii) and (iv) in Proposition 2.2, we obtain
(iii) Using (ii) in Proposition 2.2, we have
This leads to
(iv) We now calculate α2. From
we calculate
From that and (iii) in Proposition 2.2, we get
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
where b = aα2/(α1 + aα2),c = aβ2/(β1 + aβ2), and
By direct computation, we obtain
We denote by Kn and Hn the numerator and denominator of An − Bn, respectively. We have \({H_{n}} \le {({2 + \frac {M}{{{\varepsilon _{0}}}}{d_{0}}} )^{2}}\) and
When α0≠β0, for n big enough, we obtain
Hence,
For α0 = β0, we also have (2.9).
Next, we have
From (2.9), we have
where C1 = C1(a,ε0,M) is a constant. We now consider
So from (2.9) and (2.10), we have
where C2 = C2(a,ε0,M) and
Using (i) and (v) in Proposition 2.2, we get
There exists an n0 = n0(a,ε0,N) such that for every n ≥ n0 then
We now show that
We consider three cases.Case 1 (α1 −β1)(α2 −β2) ≥ 0. We have
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
From that, we have
Then there exists an n1 = n1(a,ε0,M,N) > n0 such that
We get
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
There exists an n2 = n2(a,ε0,M,N) > n0 such that
If
we return to Case 2. Otherwise,
From (2.15) and (2.16), we obtain
Moreover, we have
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
where the conductivity γα ∈ μ(h,M).
Definition 3.1
(i) We denote
(ii) Let
where
J0(λnr) is Bessel function of order zero, λn is positive zero of function J0
J1(λn) is Bessel function of order 1 and
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
(iii) The dual space of \(H_{rad}^{\frac {1}{2}}(B) \) is defined by
with the norm
(iv) We denote
In the cylindrical coordinates, if \(u(r, z)={\sum }_{n=1}^{\infty } u_{n}(z)J_{0}(\lambda _{n} r)\) we have
For \(f \in {H_{rad}^{\frac {1}{2}}(B)}\), the Dirichlet problem (3.1) in cylindrical coordinates is
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
At z = h, we have
It follows that
The Dirichlet-to-Neumann map \({\varLambda }_{\alpha }:{H_{rad}^{\frac {1}{2}}(B)} \to {H_{rad}^{- \frac {1}{2}}(B)}\) is determined by
We now give an explicit formula to reconstruct the parameters h, α from the Dirichlet-to-Nemann map. Define
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,αj 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
Proof
(i) It is easy to show that \({\alpha _{2}} = \lim _{n \to \infty } {A_{n}} - 1\).
(ii) We have
Note that \({\lambda _{n}} \sim \left ({n - \frac {1}{4}} \right )\pi , \text { when }n\to \infty \). We obtain
Hence,
(iii) Since
so α1 = (2A + (A + 2)α2)/(2 − A). □
Remark 3.3
We can reconstruct h,α1 from α2,A1,A2 as follows
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
where
By direct computation we obtain
where
We denote by Kn and Hn the numerator and denominator of (An − Bn), respectively. We have Hn ≤ (2 + 3M)2 and
Hence,
For α2≠β2, we choose n big enough so that
For α2 = β2, we also have (3.5). It is easy to get
Therefore, from (3.3) and (3.5), we have
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Acknowledgments
Part of this work was done when the second author visited the Vietnam Institute for Advanced Study in Mathematics (VIASM) whom we thank for support and hospitality. We thank N. A. Tu for useful conversation and the referee for helpful comments.
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Dung, M.T.K., Tuan, D.A. Calderón’s Problem for Some Classes of Conductivities in Circularly Symmetric Domains. Acta Math Vietnam 45, 849–863 (2020). https://doi.org/10.1007/s40306-020-00374-2
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DOI: https://doi.org/10.1007/s40306-020-00374-2