# Peeling Sequences

Adrian Dumitrescu Algoresearch L.L.C., Milwaukee, WI, USA. Email
<EMAIL_ADDRESS>
###### Abstract

Given an $n$-element point set in the plane, in how many ways can it be peeled
off until no point remains? Only one extreme point can be removed at a time.
The answer obviously depends on the point set. If the points are in convex
position, there are exactly $n!$ ways, which is the maximum number of ways for
$n$ points. But what is the minimum number?

Keywords: integer sequence, convexity, geometric construction.

##  1 Introduction

Given $n$ points on a line, in how many ways can they be removed, when we are
only allowed to remove from one end? We can remove from either end until one
point remains, when we have exactly one choice. Consequently, there are two
choices at each step except the last one, thus $2^{n-1}$ ways.

A set of points in the plane is said to be in _general position_ if no three
of them are collinear. Let $P$ be a set of $n$ points in the plane in general
position, labeled by $1,2,\ldots,n$. Consider removing points one by one until
no point remains, under the provision that exactly one extreme point (i.e., a
convex hull vertex) is removed in the current step. If the points are in
convex position, there are exactly $n!$ ways, which is clearly the maximum
number of ways for removing $n$ points. What is the minimum number?

A _peeling sequence_ for $P$ is any permutation of $[n]$ that can be obtained
by writing the labels of the points removed one by one. We are interested in
the _minimum_ number of such permutations that can be obtained, over all point
sets of size $n$.

### Definitions.

Given an $n$-element point set $P$ in general position the plane, let $g(P)$
count the number of peeling sequences for $P$; and $g(n)$ denote the minimum
of $g(P)$ over all $n$-element point sets $P$ in general position. It should
be clear that $g(n)$ is an increasing function of $n$.

### Our results.

As a warm-up we determine the values of $g(\cdot)$ for the first few small
values of $n$. Trivially, we have $g(1)=1$, and $g(2)=2$.

######  Proposition 1.

The following exact values can be observed.

$\displaystyle g(3)$ $\displaystyle=6,$ $\displaystyle g(4)$
$\displaystyle=18,$ $\displaystyle g(5)$ $\displaystyle=60,$ $\displaystyle
g(6)$ $\displaystyle=180.$

We then establish the following bounds (all logarithms are in base $2$).

######  Theorem 1.

Every $n$-element point set in general position admits $\Omega(3^{n})$ peeling
sequences. On the other hand, there are sets with only $2^{O(n\log\log{n})}$
peeling sequences. Equivalently, the following inequalities are in effect:
$g(n)=\Omega(3^{n})$ and $g(n)=2^{O(n\log\log{n})}$.

It will be evident from the proof of Theorem 1 that $g(n)$ is a multiple of
$6$ for every $n\geq 3$ and that $g(n)$ satisfies the following recurrence:

$g(n)\geq 3\cdot g(n-1),\text{ for }n\geq 3.$ (1)

Let $h(n)=\log{g(n)}$. Then obviously $h(n)=\Omega(n)$ and
$h(n)=O(n\log\log{n})$.

### Related work.

The concept of “peeling” a convex hull has been associated to dynamic convex
hull algorithms and convex hull determination [7]. For a planar point set $P$,
the _convex layers_ of $P$ are the convex polygons obtained by iterating the
following procedure: compute its convex hull and remove its vertices from $P$
[3]. The process of peeling a point set has been shown very useful in
obtaining robust estimators in statistics [3, 8]. A common estimator of a
(unidimensional) sample is its arithmetic mean, however this estimator can be
severely affected by outliers. A method that performs better is is to discard
the highest and lowest $\alpha$-fraction of the data and take the mean of the
remainder. This is the $\alpha$-_trimmed mean_. The median is the special case
$\alpha=1/2$. The higher dimensional analog of trimming, called “peeling” by
Tukey, consists of successively removing extreme points of the convex hull of
the data until a certain fixed fraction of the points remains.

A quadratic algorithm for peeling (i.e., for convex layer decomposition) was
initially proposed by Shamos [9]. A faster algorithm, running in
$O(n\log^{2}{n})$ is due to Overmars and Van Leeuwen [7]. Finally, an optimal
algorithm, running in $O(n\log{n})$ time was obtained by Chazelle [3]. In the
area of dynamic convex hull algorithms, the latest developments are due to
Chan [2]. Computing the convex layers and studying their structure in a random
setting have been studied by Dalal [4]. Har-Peled and Lidický [6] have studied
the number of steps needed for peeling the integer grid $G_{n}$ with $n$
points (with, say, $n=k^{2}$); note that $G_{n}$ is _not_ in general position,
however the peeling process can be executed on any point set.

It should be noted that while in the discussion above all the extreme points
in a layer are removed in parallel (i.e., at the same time), in our study — of
the function $g(n)$ — the extreme points are removed sequentially one by one.

##  2 Preliminaries

######  Observation 1.

Let $P$ be any $n$-element point set. Then $g(P)\leq n!$.

###### Proof.

By definition, any peeling sequence for $P$ is a permutation of $[n]$ and the
upper bound is implied. ∎

### Proof of Proposition 1.

Let $h$ denote the size of the convex hull of $P$. The case $n=3$ is clear:
there are $3!=6$ permutations and each of them is a valid peeling sequence.
The remaining cases are illustrated in Fig. 1.

Figure 1: Illustration for $n=4,5,6$.

Let now $n=4$. If the points are in convex position, there are $4!=24$
permutations. If the points are not in convex position, let $4$ be the
interior point. Then any permutation that starts with $4$ is invalid, however,
all remaining $4!-3!=18$ permutations are valid.

Let now $n=5$. If $h\geq 4$, there are at least $4\times g(4)=72$
permutations. The case $h=3$ yields the smallest number, $24+18+18=60$, of
permutations; indeed, removing one of the extreme points yields a convex
quadrilateral, while the other two removals can result in a triangle with a
point inside.

Finally, let $n=6$. If $h\geq 4$, there are at least $4\times g(5)=240$
permutations. The case $h=3$ yields the smallest number: $3\times g(5)=3\times
60=180$ permutations; indeed, removing each of the extreme points may yield
the minimizer for $n=5$ discussed above. ∎

##  3 Proof of Theorem 1

### Lower bound.

It is clear that $g(2)=2$. Assume now that $n\geq 3$. Let $P$ be any
$n$-element point set in general position. By the assumption, ${\rm conv}(P)$
has at least three extreme vertices, removal of each yields a set of $n-1$
points in general position. Any two peeling sequences resulting from the
removal of two different extreme vertices are clearly different. As such,
$g(\cdot)$ satisfies the recurrence $g(n)\geq 3\cdot g(n-1)$. Consequently,
$g(n)=\Omega(3^{n})$. By Proposition 1 we also have $g(n)\geq 180\cdot
3^{n-6}$ for every $n\geq 6$.

### Upper bound.

Here we prove the upper bound in Theorem 1. Our construction is remotely
reminiscent of that in [1] in that it uses long slightly concave chains of
points (vs. long slightly concave chains of unit disks). The long chains of
disks in [1] have the effect that the construction is hard to dismantle and a
significant segment of the initial moves are in a certain sense wasted.
Similarly here, the long concave chains of points are very restrictive with
respect to the points that can be removed in an initial phase of the removal
procedure.

Figure 2: Illustration for $k=5$; the figure is not to scale. Left: the point
set resembles a slightly deflated pentagon. Right: after peeling off some of
the points.

Consider a regular $k$-gon $K$ with vertex set $P_{0}$. Replace each side of
the $k$-gon by a circular arc of very large radius $R$ centered in the
exterior of $K$ that is incident to the two endpoints of the side. Distribute
evenly $a-1$ new points on each arc, where $a$ is a positive integer. Denote
by $P$ the resulting set of $n=k+k(a-1)=ka$ points, see Fig. 2. Note that
${\rm conv}(P)$ has exactly $k$ extreme points. The key property of the
constructed point set $P$ with respect to repeated deletion of points is that
this situation persists (apart from a factor of $2$) as long as about $n/k$
points remain. This is made precise in the following.

######  Lemma 1.

Let $P^{\prime}\subset P$ be a subset of points that remains after removing
$|P|-|P^{\prime}|$ points from $P$ (i.e., peeling $P$ according to the rules
for that many steps). If $|P^{\prime}|\geq a+2$ then ${\rm conv}(P^{\prime})$
has at most $2k$ extreme points.

###### Proof.

Let $C$ be any fixed chain in $P$ and let $C^{\prime}=C\cap P^{\prime}$.
Observe that $C^{\prime}$ consists of a contiguous subchain of the points in
$C$. It suffices to show that $C^{\prime}$ contributes at most two extreme
points in $P^{\prime}$. This is clear if $|C^{\prime}|\leq 2$, so assume that
$|C^{\prime}|\geq 3$. Each concave chain has initially $(a-1)+2=a+1$ points.
Since $|P^{\prime}|\geq a+2$, $P^{\prime}$ contains points from at least two
different initial concave chains. By construction (recall that $R$ is very
large), ${\rm conv}(P^{\prime})$ has an extreme vertex that does not belong to
$C$ and exactly two extreme vertices in $C^{\prime}\subset C$. In either case,
$C^{\prime}$ contributes at most two extreme points in $P^{\prime}$, as
claimed. ∎

We return to the upper bound proof. It remains to estimate the number of
resulting peeling sequences and set a suitable value for $k$ to minimize the
bound. Divide any peeling sequence (of length $n$) into two subsequences, a
prefix subsequence of length $n-n/k-1$ and a suffix subsequence of length
$n/k+1$. By Lemma 1, the number of prefix subsequences is bounded from above
by $(2k)^{n-n/k-1}$. By Observation 1, the number of suffix subsequences
corresponding to a fixed prefix subsequence is bounded from above by
$(n/k+1)!$. Consequently, the total number of peeling sequences is bounded
from above by the product rule:

$g(n)\leq(2k)^{n-n/k-1}\cdot\left(\frac{n}{k}+1\right)!.$

Using a trivial estimate for the factorial $\log{n!}\leq n\log{n}-n/2$ and
taking logarithms yields

$\displaystyle h(n)$
$\displaystyle\leq\left(n-\frac{n}{k}-1\right)\log{(2k)}+\log{\left(\left(\frac{n}{k}+1\right)!\right)}$
$\displaystyle\leq\left(n-\frac{n}{k}\right)\log{(2k)}+\frac{n}{k}\log{\frac{n}{k}}.$

Setting

$k=\left\lceil\frac{\log{n}}{\log\log{n}}\right\rceil$

balances the two terms and further yields $h(n)=O(n\log\log{n})$. This
completes the proof of Theorem 1. ∎

### A second proof of the upper bound.

Here we give another construction with a shorter proof. For every $k\geq 3$,
Erdős and Szekeres [5] constructed sets with $n=2^{k-1}$ points in general
position containing no subset of size $k+1$ in convex position. Let $P$ be
such a set. That is, the largest convex subset of $P$ has at most $k$ points.
This implies that once we start peeling $P$, the convex hull of the current
set has never more than $k$ vertices, and so the number of choices at each
step is at most $k$. Consequently, the total number of peeling sequences is
bounded from above by $k^{n}$ (this bound can be somewhat reduced due to fewer
choices at the end of the process). Taking logarithms as before yields
$h(n)\leq n\log{k}=O(n\log\log{n})$.

##  4 Concluding remarks

Observe that the convex layer decomposition of a point set $P$ —mentioned in
Section 1 — yields a set of peeling sequences naturally derived from it:
remove the points from one layer, one by one, before moving to the next layer.
Indeed, this is so, since any point of the layer under removal is still
extreme at that step. In general, if there are $m$ layers and their sizes are
$h_{1},h_{2},\ldots,h_{m}$ counting from outside, where
$h_{1},\ldots,h_{m-1}\geq 3$, $h_{m}\geq 1$, and $\sum_{i=1}^{m}h_{i}=n$, then
there are $h_{1}!h_{2}!\cdots h_{m}!$ peeling sequences given by the convex
layer decomposition.

Apart from the case of points in convex position, the set of peeling sequences
corresponding to layer by layer removal of the points is a strict subset of
the set of peeling sequences of $P$. It is worth noting that this subset can
be much smaller than the whole set. For example, consider a point set with
$n/3$ layers, where each layer is a triangle (and so the point set is the
vertex set of $n/3$ nested triangles). Then there are
$6^{n/3}=1.817\ldots^{n}$ peeling sequences given by the convex layer
decomposition, whereas the total number of peeling sequences is
$\Omega(3^{n})$ by Theorem 1.

Our estimates on the growth rate of $g(n)$ are rather loose; indeed, even
after taking logarithms of the lower and upper bounds there is still a gap.
According to the bounds we have obtained, it is perfectly possible that $g(n)$
grows faster than any exponential:

######  Problem 1.

Is there a constant $a>1$ such that $g(n)=O(a^{n})$?

Another interesting problem is to improve the trivial lower bound in Theorem 1
(say, increase the base by $1$):

######  Problem 2.

Prove (or disprove) that $g(n)=\Omega(4^{n})$.

## References

  * [1] Sergey Bereg, Adrian Dumitrescu, and János Pach, Sliding disks in the plane, _International Journal of Computational Geometry & Applications_, 15 (2008), 373–387.
  * [2] Timothy M. Chan, Dynamic planar convex hull operations in near-logarithmic amortized time, _Journal of the ACM_ 48(1) (2001), 1–12.
  * [3] Bernard Chazelle, On the convex layers of a planar set, _IEEE Transactions on Information Theory_ 31(4) (1985), 509–517.
  * [4] Ketan Dalal, Counting the onion, _Random Structures and Algorithms_ 24(2) (2004), 155–165.
  * [5] Paul Erdős and György Szekeres, On some extremum problems in elementary geometry, _Annales Univ. Sci. Budapest_ 3-4 (1960–61), 53–62.
  * [6] Sariel Har-Peled and Bernard Lidický, Peeling the grid, _SIAM Journal of Discrete Mathematics_ 27(2) (2013), 650–655.
  * [7] Mark H. Overmars and Jan van Leeuwen, Maintenance of configurations in the plane, _Journal of Computer and System Sciences_ 23(2) (1981), 166–204.
  * [8] Michael I. Shamos, Geometry and statistics: problems at the interface, in _Recent Results and New Directions in Algorithms and Complexity_ (Joseph F. Traub, editor), pp. 251–280, Academic Press, New York, 1976.
  * [9] Michael I. Shamos, _Problems in Computational Geometry_ , PhD Thesis, Yale University, 1978.


# Reconstruction of cracks in Calderón’s inverse conductivity problem using
energy comparisons

Henrik Garde  Department of Mathematics, Aarhus University, Aarhus, Denmark.
<EMAIL_ADDRESS>and  Michael S. Vogelius  Department of Mathematics, Rutgers
University, New Brunswick, NJ, USA<EMAIL_ADDRESS>
###### Abstract.

We derive exact reconstruction methods for cracks consisting of unions of
Lipschitz hypersurfaces in the context of Calderón’s inverse conductivity
problem. Our first method obtains upper bounds for the unknown cracks, bounds
that can be shrunk to obtain the exact crack locations upon verifying certain
operator inequalities for differences of the local Neumann-to-Dirichlet maps.
This method can simultaneously handle perfectly insulating and perfectly
conducting cracks, and it appears to be the first rigorous reconstruction
method capable of this. Our second method assumes that only perfectly
insulating cracks or only perfectly conducting cracks are present. Once more
using operator inequalities, this method generates approximate cracks that are
guaranteed to be subsets of the unknown cracks that are being reconstructed.

Keywords: Calderón’s problem, electrical impedance tomography, cracks,
inclusion detection.

2020 Mathematics Subject Classification: 35R30, 35R05, 47H05.

##  1\. Introduction

For an integer $d\geq 2$, let $\Omega$ be a bounded Lipschitz domain in
$\mathbb{R}^{d}$ with connected complement, and let
$\Gamma\subseteq\partial\Omega$ be a non-empty relatively open subset. In the
presence of a collection of cracks, $D$, and an imposed boundary current, $f$
on $\Gamma$, the steady state voltage potential, $u$, satisfies

$\displaystyle-\nabla\cdot(\gamma_{0}\nabla u)$ $\displaystyle=0\quad\text{in
}\Omega\setminus\overline{D},$ $\displaystyle\gamma_{0}\frac{\partial
u}{\partial\nu}$ $\displaystyle=\begin{dcases}f&\text{ on }\Gamma,\\\ 0&\text{
on }\partial\Omega\setminus\overline{\Gamma},\end{dcases}$
$\displaystyle\gamma_{0}\frac{\partial u}{\partial n}$
$\displaystyle=0\quad\text{on }D_{0},$ (1.1) $\displaystyle u\text{ is locally
constant on }D_{\infty},$
$\displaystyle\int_{D_{i}}\Bigl{[}\gamma_{0}\frac{\partial u}{\partial
n}\Bigr{]}\mathrm{d}S$ $\displaystyle=0\quad\text{for each component $D_{i}$
of $D_{\infty}$}.$

We “ground” $u$ by requiring that $\int_{\Gamma}u\,\mathrm{d}S=0$. $D_{0}$ are
those cracks in the collection $D$ that are perfectly insulating, and
$D_{\infty}$ are those that are perfectly conducting, $D=D_{0}\cup
D_{\infty}$. Note that the constants on different connected components of
$D_{\infty}$ will generally be different (and determined by the integral
conditions). $\nu$ denotes the outer unit normal on $\partial\Omega$ and $n$
denotes a unit normal on $D$. We use $[\,v\,]$ to denote the jump of a
function $v$ across $D$, relative to the normal vector $n$, with the
convention that $[\,v\,]=v_{-}-v_{+}$, where $v_{-}$ is the trace (on $D$)
taken from that part of $\Omega$, which $n$ points away from, and $v_{+}$ is
the trace (on $D$) taken from that part of $\Omega$, which $n$ points into.
The imposed boundary current satisfies

$f\in L^{2}_{\diamond}(\Gamma)=\\{g\in L^{2}(\Gamma)\mid\langle
1,g\rangle=0\\},$ (1.2)

where $\langle\,\cdot\,,\,\cdot\,\rangle$ is the usual $L^{2}(\Gamma)$ inner
product. $\gamma_{0}$ is a known background conductivity coefficient, which is
defined in all of $\Omega$.

For the major part of our analysis it suffices that $\gamma_{0}\in
L^{\infty}(\Omega)$ with $\mathop{\textup{ess\,inf}}(\gamma_{0})>0$. However,
at certain points we crucially assume that the conductivity equation
$\nabla\cdot(\gamma_{0}\nabla v)=0$ has a weak unique continuation property
(UCP) on connected open subsets of $\Omega$ and also from Cauchy data on
$\Gamma$, see e.g. [6, Definition 3.3]. For $d=2$ the UCP follows directly
[1], while in higher dimensions this requires some regularity of $\gamma_{0}$
such as $\gamma_{0}$ being Lipschitz continuous or piecewise analytic [24, 25,
28]. At one particular point (Lemma 5.3) we require additional $C^{2}$
regularity, namely to ensure that the given cracks cannot be invisible to all
possible boundary measurements. However, if it turns out that the visibility
can be shown for a lower regularity (still satisfying the UCP) then the proofs
of our main results adapt to this.

Formally the problem introduced in (1) can be considered a problem related to
the conductivity profile given by

$\begin{dcases}\gamma_{0}&\text{ in }\Omega\setminus D,\\\ 0&\text{ in
}D_{0},\\\ \infty&\text{ in }D_{\infty},\end{dcases}$

which also explains our use of the notation $D_{0}$ and $D_{\infty}$. The
problem of determining an internal conductivity profile from boundary data is
often referred to as Calderón’s inverse conductivity problem [5, 30], and our
emphasis here is therefore on a special case of this, namely: to reconstruct
the unknown cracks in $\Omega$ (the set $D$) based on boundary measurements on
$\Gamma\subseteq\partial\Omega$.

In order to state the main results of this paper, it is necessary to be
precise about the collection of cracks we reconstruct.

######  Definition 1.1.

A collection of cracks $\chi$ lies in the class $\mathcal{X}$ if for some
$N\in\mathbb{N}_{0}$

$\chi=\bigcup_{i=1}^{N}\sigma_{i},$

where the $\sigma_{i}\subset\Omega$ are $(d-1)$ dimensional orientable
Lipschitz surfaces with non-empty Lipschitz boundary $\partial\sigma_{i}$, and
with

$\mathop{\textup{dist}}(\sigma_{i},\sigma_{j})>0\text{ for }i\neq j\text{ and
}\mathop{\textup{dist}}(\sigma_{i},\partial\Omega)>0\text{ for all }i.$

Notice that $\mathop{\textup{dist}}(\chi,\partial\Omega)>0$ and
$\Omega\setminus\overline{\chi}$ is connected. In particular, we do not allow
open subsets of $\Omega$ to be completely encapsulated by a crack. We refer to
$D\in\mathcal{X}$ as “a $(D_{0},D_{\infty})$ collection of cracks” if
$D=D_{0}\cup D_{\infty}$ for $D_{0},D_{\infty}\in\mathcal{X}$,
$\mathop{\textup{dist}}(D_{0},D_{\infty})>0$, each crack in $D_{0}$ is
perfectly insulating, and each crack in $D_{\infty}$ is perfectly conducting.
Note that $D_{0}$ or $D_{\infty}$ may possibly be empty.

Our first result is an exact monotonicity-based reconstruction method via
shape approximations from above, based on shrinking test inclusions $C$
containing $D$. Whether $C$ contains $D$ or not is checked via operator
inequalities related to the boundary electric power. The set of admissible
test inclusions are

$\displaystyle\mathcal{A}$ $\displaystyle=\\{C\Subset\Omega\mid C\text{ is the
closure of an open set, has connected }$
$\displaystyle\hphantom{{}=\\{C\subset\Omega\mid{}}\text{complement, and has
Lipschitz boundary }\partial C\\}.$

For $C\in\mathcal{A}$ we define $\Lambda_{C}^{\emptyset}$ to be the Neumann-
to-Dirichlet map (the ND map) for the conductivity coefficient which equals
$0$ in $C$ and $\gamma_{0}$ elsewhere. We define $\Lambda_{\emptyset}^{C}$ to
be the ND map for the conductivity coefficient which equals $\infty$ in $C$
and $\gamma_{0}$ elsewhere. Similarly we define $\Lambda_{D_{0}}^{D_{\infty}}$
to be the ND map associated with the $(D_{0},D_{\infty})$ collection of
cracks. For the precise definitions see section 2 or e.g. [6]. We shall
actually only be concerned with a local version of these maps from
$L^{2}_{\diamond}(\Gamma)$ to $L^{2}_{\diamond}(\Gamma)$. For self-adjoint
operators $A,B\in\mathscr{L}(L^{2}_{\diamond}(\Gamma))$ we use the notation
$A\geq B$ if and only if $A-B$ is positive semi-definite, that is

$\langle(A-B)f,f\rangle\geq 0,\qquad\forall f\in L^{2}_{\diamond}(\Gamma).$

Based on knowledge of the local ND map $\Lambda_{D_{0}}^{D_{\infty}}:f\mapsto
u|_{\Gamma}$, and in particular its relation to $\Lambda_{C}^{\emptyset}$ and
$\Lambda_{\emptyset}^{C}$, we can now reconstruct the cracks $D$ as follows.

######  Theorem 1.2.

Let $D$ be a $(D_{0},D_{\infty})$ collection of cracks. Given any
$C\in\mathcal{A}$, then

$D\subset C\qquad\text{if and only
if}\qquad\Lambda_{C}^{\emptyset}\geq\Lambda_{D_{0}}^{D_{\infty}}\geq\Lambda_{\emptyset}^{C}.$

###### Proof.

The direction “$\Rightarrow$” is proved in Proposition 3.1 in section 3 with
$\Sigma_{0}=D_{0}$ and $\Sigma_{\infty}=D_{\infty}$. The other direction
“$\Leftarrow$” is proved in Proposition 6.1 in section 6. ∎

Our second result provides an exact monotonicity-based reconstruction method
based on generating approximate cracks $\chi$ inside $D$. By checking operator
inequalities we can determine if $\chi$ is contained in $D$ or not. This
result assumes that we either only have perfectly insulating cracks or only
have perfectly conducting cracks.

######  Theorem 1.3.

Let $D\in\mathcal{X}$.

  1. (i)

Given any $\chi\in\mathcal{X}$, then

$\chi\subseteq D\qquad\text{if and only
if}\qquad\Lambda_{D}^{\emptyset}\geq\Lambda_{\chi}^{\emptyset}.$

  2. (ii)

Given any $\chi\in\mathcal{X}$, then

$\chi\subseteq D\qquad\text{if and only
if}\qquad\Lambda_{\emptyset}^{\chi}\geq\Lambda_{\emptyset}^{D}.$

###### Proof.

The direction “$\Rightarrow$” of (i) is proved in Proposition 3.2(i) in
section 3 with $\Sigma_{0}=D$ and $\Sigma_{\infty}=\emptyset$, and the same
direction for (ii) is proved in Proposition 3.2(ii) with
$\Sigma_{0}=\emptyset$ and $\Sigma_{\infty}=D$. The other direction
“$\Leftarrow$” for both (i) and (ii) is proved in Proposition 6.2 in section
6. ∎

######  Remark 1.4.

  1. (i)

Theorem 1.2 implies that

$\overline{D}=\cap\\{C\in\mathcal{A}\mid\Lambda_{C}^{\emptyset}\geq\Lambda_{D_{0}}^{D_{\infty}}\geq\Lambda_{\emptyset}^{C}\\}.$

Moreover, as seen in the proof of Proposition 6.1, if $D_{0}=\emptyset$ we
only need to consider
$\Lambda_{\emptyset}^{D_{\infty}}\geq\Lambda_{\emptyset}^{C}$ and, likewise,
if instead we have $D_{\infty}=\emptyset$ we only need to consider
$\Lambda_{C}^{\emptyset}\geq\Lambda_{D_{0}}^{\emptyset}$.

  2. (ii)

Theorem 1.3 implies that, in the perfectly insulating case

$D=\cup\\{\chi\in\mathcal{X}\mid\Lambda_{D}^{\emptyset}\geq\Lambda_{\chi}^{\emptyset}\\},$

and in the perfectly conducting case

$D=\cup\\{\chi\in\mathcal{X}\mid\Lambda_{\emptyset}^{\chi}\geq\Lambda_{\emptyset}^{D}\\}.$

###  1.1. Some related results

For large classes of cracks there are optimal (non-constructive) unique
identifiability results using merely two boundary measurements [2, 4, 9].
However, the _reconstruction of general cracks_ from finitely many
measurements appears to be an open problem. For spatial dimension $d=2$ there
is a factorisation method [3] to reconstruct _either_ perfectly insulating
cracks _or_ perfectly conducting cracks from an ND map. The results of [3]
resemble the two-dimensional version of Theorem 1.3, in the sense that one
determines if an artificial crack is a subset of the unknown crack. For
perfectly insulating cracks, probe and enclosure methods are discussed in [22,
23]. These methods determine the location of “boundary singularities”
associated with the cracks or the convex hull of the cracks, respectively. It
should also be mentioned that these reconstruction methods assume that
$\gamma_{0}\equiv 1$.

Our results Theorem 1.2 and Theorem 1.3 are applicable in spatial dimension
$d\geq 2$ for very general cracks and also for non-homogeneous background
conductivities. It appears that Theorem 1.2 is the only proven method capable
of reconstructing cracks with both perfectly insulating and perfectly
conducting parts. Moreover, the test operators $\Lambda_{C}^{\emptyset}$ and
$\Lambda_{\emptyset}^{C}$ are precisely the same test operators that are used
in [6, 12] for reconstructing general inclusions of positive volume (the
support of conductivity perturbations on open sets). So numerical
implementations of this method directly applies to cracks without requiring
any modifications. Arguably Theorem 1.3 becomes demanding to implement
numerically for $d>2$ and is likely only suitable for computations in $d=2$,
while the numerical implementation of Theorem 1.2 very naturally generalises
to higher dimensions using a peeling approach as in [14].

For inclusions of positive volume, the monotonicity of ND mappings with
respect to the conductivity coefficient was used in [29] to give bounds on the
inclusions. In [19, 20] this approach was proven to give an exact
reconstruction method for finite perturbations to the background conductivity,
also in cases with both positive and negative perturbations if lower and upper
bounds are known for the perturbed conductivity. This was generalised in [6],
where the perturbed conductivity could now simultaneously have parts with
finite positive and negative perturbations as well as extreme parts that are
perfectly insulating and perfectly conducting. Moreover, the need for lower
and upper bounds was removed. Finally, in [12] the method was shown to also
handle degenerate and singular perturbations based on $A_{2}$-Muckenhoupt
weights, allowing for continuous decay to zero and growth to infinity. This is
currently the most general method for reconstructing inclusions of positive
volume in Calderón’s problem based on a local ND map. Rigorous connections
have also been made to practically relevant electrode models [13, 14, 17, 21].

The same monotonicity-based methodology has also been used to design a
reconstruction method for the partial data Calderón problem, for the case of
layered piecewise constant conductivities [10, 11]; note that this is not a
finite-dimensional setting, as the piecewise constant partitioning is also
reconstructed. In finite-dimensional settings this methodology has led to
Lipschitz stability results in Calderón’s problem with finitely many
measurements [17] and to reformulating the finite-dimensional Calderón problem
as a convex semidefinite optimisation problem [18]. The methods are based on
the unique continuation principle and its connection to the localised
potentials from [15]. For applications to other inverse problems, we refer to
the list of references in [17].

The proofs of Theorems 1.2 and 1.3 also rely on localised potentials. However,
the approach here is quite different since the known localisation results
apply to open sets, and there is no open set inside the cracks to localise in.
Moreover, the monotonicity inequalities in [6, Appendix A] become trivial in
the limit of approximating cracks by open sets. Additionally, it should be
noted that using approximations of cracks with open sets and applying the
results from [6] will not result in proofs of our main results. The key
ingredient for our proofs turns out to be the localisation on open sets
containing parts of the cracks, to give simultaneous blow-up of potentials for
conductivity profiles both with and without the cracks, and crucially also to
have blow-up for their difference. The latter is the most technically
difficult to obtain, and here the constructive version of localised potentials
[15, Lemma 2.8] turns out to be invaluable, since the existence results for
localised potentials would not necessarily give blow-up for the difference of
two localised sequences of potentials.

###  1.2. Outline of the paper

In section 2 we introduce the relevant forward problems and associated
function spaces. In section 3 we prove the direction “$\Rightarrow$” of
Theorems 1.2 and 1.3 (the easy direction). Sections 4 and 5 are dedicated to
results about localised potentials, that are needed in section 6, where we
prove the direction “$\Leftarrow$” of Theorems 1.2 and 1.3 (the difficult
direction).

##  2\. Forward problems

Let $\Sigma$ be a $(\Sigma_{0},\Sigma_{\infty})$ collection of cracks. Let

$\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}=\\{v\in
H_{\diamond}^{1}(\Omega\setminus\overline{\Sigma_{0}})\mid\text{$v$ is locally
constant on $\Sigma_{\infty}$}\\},$

where the $\diamond$-symbol refers to a mean-free condition for the Dirichlet
trace on $\Gamma$, as in (1.2). A Poincaré inequality holds on
$\Omega\setminus\overline{\Sigma_{0}}$, see for instance [26]. As $\Sigma_{0}$
has $d$-Lebesgue measure zero, the gradient of
$v\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$ (defined in
$\Omega\setminus\overline{\Sigma_{0}}$) is equivalent to a function in
$L^{2}(\Omega)^{d}$ on the entire domain $\Omega$. Hence a norm on
$\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$, equivalent to the $H^{1}$-norm,
is

$\lVert v\rVert_{*}=\Bigl{(}\int_{\Omega}\gamma_{0}\lvert\nabla
v\rvert^{2}\,\mathrm{d}x\Bigr{)}^{1/2},$

with associated inner product111Throughout this paper we assume, without loss
of generality, that all our solutions and function spaces are real-valued.

$\langle w,v\rangle_{*}=\int_{\Omega}\gamma_{0}\nabla w\cdot\nabla
v\,\mathrm{d}x.$

By the Lax-Milgram lemma, the interior electric potential $u$ in (1), with
$D_{0}=\Sigma_{0}$ and $D_{\infty}=\Sigma_{\infty}$, is the unique solution in
$\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$ of the weak problem

$\langle u,v\rangle_{*}=\langle f,v|_{\Gamma}\rangle,\qquad\forall
v\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}.$ (2.1)

We will sometimes use the notation $u=u_{\Sigma_{0},f}^{\Sigma_{\infty}}$ to
clarify which cracks (possibly empty) are present, and which current density,
$f$, is used. Define the functional
$J:\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}\to\mathbb{R}$ by

$J(v)=\lVert v\rVert_{*}^{2}-2\langle f,v|_{\Gamma}\rangle,\qquad
v\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}.$

With this definition, $u$ is also the unique minimiser of $J$ (cf. [16, Remark
12.23] and [7, Theorem 1.1.2]). The corresponding local ND map
$\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}:f\mapsto u|_{\Gamma}$ satisfies

$\langle\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}f,f\rangle=\langle
u|_{\Gamma},f\rangle=\langle f,u|_{\Gamma}\rangle=\lVert u\rVert_{*}^{2},$

and is a compact self-adjoint operator on $L^{2}_{\diamond}(\Gamma)$.

For $C\in\mathcal{A}$ and $f\in L^{2}_{\diamond}(\Gamma)$, let
$u_{C,f}^{\emptyset}$ be the potential coming from a conductivity with
perfectly insulating inclusions in $C$ and $\gamma_{0}$ elsewhere. Then
$u_{0}=u_{C,f}^{\emptyset}$ is the unique solution in

$\mathcal{H}_{C}^{\emptyset}=H_{\diamond}^{1}(\Omega\setminus C)$

of the variational problem

$\int_{\Omega\setminus C}\gamma_{0}\nabla u_{0}\cdot\nabla
v\,\mathrm{d}x=\langle f,v|_{\Gamma}\rangle,\qquad\forall
v\in\mathcal{H}_{C}^{\emptyset}.$

Likewise, let $u_{\emptyset,f}^{C}$ be the potential coming from a
conductivity with perfectly conducting inclusions in $C$ and $\gamma_{0}$
elsewhere. Then $u_{\infty}=u_{\emptyset,f}^{C}$ is the unique solution in

$\mathcal{H}_{\emptyset}^{C}=\\{w\in H_{\diamond}^{1}(\Omega)\mid w\text{ is
locally constant in }C\\}$

of the variational problem

$\int_{\Omega\setminus C}\gamma_{0}\nabla u_{\infty}\cdot\nabla
v\,\mathrm{d}x=\langle f,v|_{\Gamma}\rangle,\qquad\forall
v\in\mathcal{H}_{\emptyset}^{C}.$

In terms of the functional

$\widetilde{J}(v)=\int_{\Omega\setminus C}\gamma_{0}\lvert\nabla
v\rvert^{2}\,\mathrm{d}x-2\langle f,v|_{\Gamma}\rangle,$

$u_{0}$ is the unique minimizer of
$\widetilde{J}|_{\mathcal{H}_{C}^{\emptyset}}$ and $u_{\infty}$ is the unique
minimizer of $\widetilde{J}|_{\mathcal{H}_{\emptyset}^{C}}$. The corresponding
ND maps are compact self-adjoint operators on $L^{2}_{\diamond}(\Gamma)$ and
satisfy

$\displaystyle\langle\Lambda_{C}^{\emptyset}f,f\rangle$ $\displaystyle=\langle
u_{0}|_{\Gamma},f\rangle=\langle
f,u_{0}|_{\Gamma}\rangle=\int_{\Omega\setminus C}\gamma_{0}\lvert\nabla
u_{0}\rvert^{2}\,\mathrm{d}x,$
$\displaystyle\langle\Lambda_{\emptyset}^{C}f,f\rangle$ $\displaystyle=\langle
u_{\infty}|_{\Gamma},f\rangle=\langle
f,u_{\infty}|_{\Gamma}\rangle=\int_{\Omega\setminus C}\gamma_{0}\lvert\nabla
u_{\infty}\rvert^{2}\,\mathrm{d}x.$

##  3\. The direction “$\Rightarrow$” in Theorems 1.2 & 1.3

Below we prove the “easy direction” of the _if and only if_ statements of
Theorems 1.2 and 1.3.

######  Proposition 3.1.

Let $\Sigma$ be a $(\Sigma_{0},\Sigma_{\infty})$ collection of cracks. Given
any $C\in\mathcal{A}$, then

$\Sigma\subset
C\qquad\text{implies}\qquad\Lambda_{C}^{\emptyset}\geq\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}\geq\Lambda_{\emptyset}^{C}.$

###### Proof.

For $f\in L_{\diamond}^{2}(\Gamma)$ denote $u_{0}=u_{C,f}^{\emptyset}$,
$u_{\infty}=u_{\emptyset,f}^{C}$, and $u=u_{\Sigma_{0},f}^{\Sigma_{\infty}}$.
Since $\Sigma\subset C$ we have $u|_{\Omega\setminus
C}\in\mathcal{H}_{C}^{\emptyset}$ and
$u_{\infty}|_{\Omega\setminus\overline{\Sigma_{0}}}\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$.
We now compare the ND maps, while taking the related minimization properties
from section 2 into account

$\displaystyle-\langle\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}f,f\rangle$
$\displaystyle=\int_{\Omega}\gamma_{0}\lvert\nabla
u\rvert^{2}\,\mathrm{d}x-2\langle f,u|_{\Gamma}\rangle$
$\displaystyle\geq\int_{\Omega\setminus C}\gamma_{0}\lvert\nabla
u\rvert^{2}\,\mathrm{d}x-2\langle f,u|_{\Gamma}\rangle$
$\displaystyle\geq\int_{\Omega\setminus C}\gamma_{0}\lvert\nabla
u_{0}\rvert^{2}\,\mathrm{d}x-2\langle f,u_{0}|_{\Gamma}\rangle$
$\displaystyle=-\langle\Lambda_{C}^{\emptyset}f,f\rangle,$

and

$\displaystyle-\langle\Lambda_{\emptyset}^{C}f,f\rangle$
$\displaystyle=\int_{\Omega\setminus C}\gamma_{0}\lvert\nabla
u_{\infty}\rvert^{2}\,\mathrm{d}x-2\langle f,u_{\infty}|_{\Gamma}\rangle$
$\displaystyle=\int_{\Omega}\gamma_{0}\lvert\nabla
u_{\infty}\rvert^{2}\,\mathrm{d}x-2\langle f,u_{\infty}|_{\Gamma}\rangle$
$\displaystyle\geq\int_{\Omega}\gamma_{0}\lvert\nabla
u\rvert^{2}\,\mathrm{d}x-2\langle f,u|_{\Gamma}\rangle$
$\displaystyle=-\langle\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}f,f\rangle.$

Here we used that $\nabla u_{\infty}\equiv 0$ in the interior of $C$. ∎

######  Proposition 3.2.

Let $\Sigma$ be a $(\Sigma_{0},\Sigma_{\infty})$ collection of cracks.

  1. (i)

Given any $\chi\in\mathcal{X}$, then

$\chi\subseteq\Sigma_{0}\qquad\text{implies}\qquad\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}\geq\Lambda_{\chi}^{\Sigma_{\infty}}.$

  2. (ii)

Given any $\chi\in\mathcal{X}$, then

$\chi\subseteq\Sigma_{\infty}\qquad\text{implies}\qquad\Lambda_{\Sigma_{0}}^{\chi}\geq\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}.$

###### Proof.

For $f\in L_{\diamond}^{2}(\Gamma)$ denote
$u=u_{\Sigma_{0},f}^{\Sigma_{\infty}}$.

Proof of (i): We denote $u_{1}=u_{\chi,f}^{\Sigma_{\infty}}$. Since
$\chi\subseteq\Sigma_{0}$ we have
$u_{1}|_{\Omega\setminus\overline{\Sigma_{0}}}\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$.
Using the related minimization problems for $u$ and $u_{1}$, we have

$-\langle\Lambda_{\chi}^{\Sigma_{\infty}}f,f\rangle=\lVert
u_{1}\rVert_{*}^{2}-2\langle f,u_{1}|_{\Gamma}\rangle\geq\lVert
u\rVert_{*}^{2}-2\langle
f,u|_{\Gamma}\rangle=-\langle\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}f,f\rangle.$

Proof of (ii): We denote $u_{2}=u_{\Sigma_{0},f}^{\chi}$. Since
$\chi\subseteq\Sigma_{\infty}$ we have $u\in\mathcal{H}_{\Sigma_{0}}^{\chi}$.
Using the related minimization problems for $u$ and $u_{2}$, we have

$-\langle\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}f,f\rangle=\lVert
u\rVert_{*}^{2}-2\langle f,u|_{\Gamma}\rangle\geq\lVert
u_{2}\rVert_{*}^{2}-2\langle
f,u_{2}|_{\Gamma}\rangle=-\langle\Lambda_{\Sigma_{0}}^{\chi}f,f\rangle.\qed$

##  4\. Some lemmas in preparation for localised potentials

If $\Sigma$ is a $(\Sigma_{0},\Sigma_{\infty})$ collection of cracks then we
have

$\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}\subseteq\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}\subseteq\mathcal{H}_{\Sigma_{0}}^{\emptyset},$

in the sense that $v\in\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}$ satisfies
$v|_{\Omega\setminus\overline{\Sigma_{0}}}\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$.
Moreover, these are all Hilbert spaces with the same inner product
$\langle\,\cdot\,,\,\cdot\,\rangle_{*}$.

Let $P$ and $P^{\perp}$ be the orthogonal projections of
$\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$ onto
$\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}$ and
$(\mathcal{H}_{\emptyset}^{\Sigma_{\infty}})^{\perp}$, respectively, in the
$\langle\,\cdot\,,\,\cdot\,\rangle_{*}$ inner product. Likewise we let $Q$ and
$Q^{\perp}$ denote the orthogonal projections of
$\mathcal{H}_{\Sigma_{0}}^{\emptyset}$ onto
$\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$ and
$(\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}})^{\perp}$, respectively, in the
$\langle\,\cdot\,,\,\cdot\,\rangle_{*}$ inner product.

######  Lemma 4.1.

Let $\Sigma$ be a $(\Sigma_{0},\Sigma_{\infty})$ collection of cracks and let
the projections $P$, $P^{\perp}$, $Q$, and $Q^{\perp}$ be given as above. For
$f\in L^{2}_{\diamond}(\Gamma)$ we denote
$u_{0}=u_{\Sigma_{0},f}^{\emptyset}$,
$u_{\infty}=u_{\emptyset,f}^{\Sigma_{\infty}}$, and
$u=u_{\Sigma_{0},f}^{\Sigma_{\infty}}$.

  1. (i)

We have $u_{\infty}=Pu$ and

$\langle(\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}-\Lambda_{\emptyset}^{\Sigma_{\infty}})f,f\rangle=\lVert
P^{\perp}u\rVert_{*}^{2}.$

  2. (ii)

We have $u=Qu_{0}$ and

$\langle(\Lambda_{\Sigma_{0}}^{\emptyset}-\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}})f,f\rangle=\lVert
Q^{\perp}u_{0}\rVert_{*}^{2}.$

###### Proof.

Proof of (i): From the weak formulation for $u$ and for all
$v\in\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}\subseteq\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$,
we have

$\langle f,v|_{\Gamma}\rangle=\langle u,v\rangle_{*}=\langle Pu,v\rangle_{*}.$

However, by the weak formulation for $u_{\infty}$ and its unique solvability
in $\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}$, we have $u_{\infty}=Pu$. Thus
$u-u_{\infty}=P^{\perp}u$. Using the weak formulation for $u$ again, we have

$\langle(\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}-\Lambda_{\emptyset}^{\Sigma_{\infty}})f,f\rangle=\langle
f,(P^{\perp}u)|_{\Gamma}\rangle=\langle u,P^{\perp}u\rangle_{*}=\lVert
P^{\perp}u\rVert_{*}^{2}.$

Proof of (ii): From the weak formulation $u_{0}$ and for all
$v\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}\subseteq\mathcal{H}_{\Sigma_{0}}^{\emptyset}$,
we have

$\langle f,v|_{\Gamma}\rangle=\langle u_{0},v\rangle_{*}=\langle
Qu_{0},v\rangle_{*}.$

However, by the weak formulation for $u$ and its unique solvability in
$\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$, we have $u=Qu_{0}$. Thus
$u_{0}-u=Q^{\perp}u_{0}$. Using the weak formulation for $u_{0}$ again, we
have

$\langle(\Lambda_{\Sigma_{0}}^{\emptyset}-\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}})f,f\rangle=\langle
f,(Q^{\perp}u_{0})|_{\Gamma}\rangle=\langle
u_{0},Q^{\perp}u_{0}\rangle_{*}=\lVert Q^{\perp}u_{0}\rVert_{*}^{2}.\qed$

######  Remark 4.2.

While we do not use this fact, we note that, for sufficiently regular
$\gamma_{0}$ one can write the differences of the ND maps in Lemma 4.1 as
integrals on $\Sigma_{0}$ and $\Sigma_{\infty}$ via integration by parts:

$\displaystyle\langle(\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}-\Lambda_{\emptyset}^{\Sigma_{\infty}})f,f\rangle$
$\displaystyle=-\int_{\Sigma_{0}}\Bigl{(}\gamma_{0}\frac{\partial
u_{\infty}}{\partial n}\Bigr{)}{[u]}\,\mathrm{d}S,$
$\displaystyle\langle(\Lambda_{\Sigma_{0}}^{\emptyset}-\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}})f,f\rangle$
$\displaystyle=-\int_{\Sigma_{\infty}}\Bigl{[}\gamma_{0}\frac{\partial
u}{\partial n}\Bigr{]}{u_{0}}\,\mathrm{d}S.$

The integrals should be understood in the appropriate weak sense.

Next we state two general lemmas in functional analysis that are proven in
[15]. The first is a lemma relating the ranges of operators to bounds on their
adjoints.

######  Lemma 4.3 (Lemma 2.5 in [15]).

Let $H$, $K_{1}$, and $K_{2}$ be Hilbert spaces and let
$A_{j}\in\mathscr{L}(K_{j},H)$ for $j=1,2$. Then

$R(A_{1})\subseteq R(A_{2})\qquad\text{if and only if}\qquad\exists
C>0,\forall x\in H:\lVert A_{1}^{*}x\rVert_{K_{1}}\leq C\lVert
A_{2}^{*}x\rVert_{K_{2}}.$

The second lemma is the constructive version for the localised potentials from
[15].

######  Lemma 4.4 (Lemma 2.8 in [15]).

Let $H$, $K_{1}$, and $K_{2}$ be Hilbert spaces, let
$A_{j}\in\mathscr{L}(K_{j},H)$ for $j=1,2$, and assume that $A_{2}^{*}$ is
injective. Assume that there exists $y_{0}\in R(A_{1})$ such that
$y_{0}\not\in R(A_{2})$. For $n\in\mathbb{N}$ we define

$\xi_{n}=\bigl{(}A_{2}A_{2}^{*}+\tfrac{1}{n}I\bigr{)}^{-1}y_{0}$

and

$x_{n}=\frac{\xi_{n}}{\lVert A_{2}^{*}\xi_{n}\rVert^{3/2}_{K_{2}}}.$

Then

$\lim_{n\to\infty}\lVert
A_{1}^{*}x_{n}\rVert_{K_{1}}=\infty\qquad\text{and}\qquad\lim_{n\to\infty}\lVert
A_{2}^{*}x_{n}\rVert_{K_{2}}=0.$

##  5\. Localised potentials with cracks

Let $V\in\mathcal{A}$ and let $\Sigma$ be a $(\Sigma_{0},\Sigma_{\infty})$
collection of cracks. For $F\in L^{2}(V)^{d}$ we define
$w=w_{\Sigma_{0},F}^{\Sigma_{\infty}}\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$
as the unique solution of the following variational problem:

$\int_{\Omega}\gamma_{0}\nabla w\cdot\nabla
v\,\mathrm{d}x=\int_{V}F\cdot\nabla v\,\mathrm{d}x,\qquad\forall
v\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}.$ (5.1)

The dependence on $V$ is not explicitly given in the notation of
$w_{\Sigma_{0},F}^{\Sigma_{\infty}}$, but is indirectly given via $F$. We now
define an operator $L_{\Sigma_{0}}^{\Sigma_{\infty}}(V):L^{2}(V)^{d}\to
L_{\diamond}^{2}(\Gamma)$ as

$L_{\Sigma_{0}}^{\Sigma_{\infty}}(V)F=w^{\Sigma_{\infty}}_{\Sigma_{0},F}|_{\Gamma}.$

Let $u=u_{\Sigma_{0},f}^{\Sigma_{\infty}}$, then from the variational problems
of $u$ and $w$ we have

$\displaystyle\langle f,L_{\Sigma_{0}}^{\Sigma_{\infty}}(V)F\rangle$
$\displaystyle=\langle f,w|_{\Gamma}\rangle=\langle
u,w\rangle_{*}=\langle\nabla u|_{V},F\rangle_{L^{2}(V)^{d}}.$

Hence we have

$(L_{\Sigma_{0}}^{\Sigma_{\infty}}(V))^{*}f=\nabla
u_{\Sigma_{0},f}^{\Sigma_{\infty}}|_{V},\qquad f\in L_{\diamond}^{2}(\Gamma).$
(5.2)

Next we will prove results for the ranges of these variational operators, that
will be essential for the localised potentials in Proposition 5.4.

######  Proposition 5.1.

Let $V\in\mathcal{A}$ and let $\Sigma$ be a $(\Sigma_{0},\Sigma_{\infty})$
collection of cracks.

  1. (i)

If $\Sigma_{0}\Subset V$ then
$R(L_{\emptyset}^{\Sigma_{\infty}}(V))=R(L_{\Sigma_{0}}^{\Sigma_{\infty}}(V))$.

  2. (ii)

If $\Sigma_{\infty}\Subset V$ then
$R(L_{\Sigma_{0}}^{\emptyset}(V))=R(L_{\Sigma_{0}}^{\Sigma_{\infty}}(V))$.

###### Proof.

We give a proof of (i), and note that the proof of (ii) is almost the same
with obvious modifications.

Proof of “$\subseteq$”: Let $f\in R(L_{\emptyset}^{\Sigma_{\infty}}(V))$, then
there exists $F_{1}\in L^{2}(V)^{d}$ such that $f=w|_{\Gamma}$ with
$w=w_{\emptyset,F_{1}}^{\Sigma_{\infty}}$. We have

$\int_{\Omega}\gamma_{0}\nabla w\cdot\nabla
v^{\prime}\,\mathrm{d}x=\int_{V}F_{1}\cdot\nabla
v^{\prime}\,\mathrm{d}x,\qquad\forall
v^{\prime}\in\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}.$ (5.3)

Let $W_{1},W_{2}\in\mathcal{A}$ such that $\Sigma_{0}\Subset W_{1}\Subset
W_{2}\subseteq V$ and with $\mathop{\textup{dist}}(W_{2},\Sigma_{\infty})>0$.
Using the $C^{\infty}$-Urysohn lemma [8, Lemma 8.18], we may construct
$\varphi\in C_{\textup{c}}^{\infty}(W_{2})$ with $0\leq\varphi\leq 1$ and
$\varphi\equiv 1$ in $W_{1}$. In the following we will let
$v\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$ be arbitrary but fixed, and
let $c$ be the constant such that

$v_{0}=v-c$

has vanishing mean on $\partial V$. Moreover, we also define

$v^{\prime}=(1-\varphi)v_{0}.$

We note that $v^{\prime}$ (extended by zero to ${\overline{\Sigma_{0}}}$) lies
in $\mathcal{H}^{\Sigma_{\infty}}_{\emptyset}$. Therefore $\eqref{eq:weq1}$ is
applicable for $v^{\prime}$, and using the fact that
$v^{\prime}|_{\Omega\setminus V}=v_{0}|_{\Omega\setminus V}$ we now get

$\displaystyle\int_{\Omega}\gamma_{0}\nabla w\cdot\nabla v\,\mathrm{d}x$
$\displaystyle=\int_{V}\gamma_{0}\nabla{w}\cdot\nabla
v_{0}\,\mathrm{d}x+\int_{\Omega\setminus V}\gamma_{0}\nabla w\cdot\nabla
v^{\prime}\,\mathrm{d}x$ $\displaystyle=\int_{V}\gamma_{0}\nabla w\cdot\nabla
v_{0}\,\mathrm{d}x+\int_{\Omega}\gamma_{0}\nabla w\cdot\nabla
v^{\prime}\,\mathrm{d}x-\int_{V}\gamma_{0}\nabla w\cdot\nabla
v^{\prime}\,\mathrm{d}x$ $\displaystyle=\int_{V}\gamma_{0}\nabla w\cdot\nabla
v_{0}\,\mathrm{d}x+\int_{V}(F_{1}-\gamma_{0}\nabla w)\cdot\nabla
v^{\prime}\,\mathrm{d}x.$ (5.4)

Define $F_{2}=F_{1}-\gamma_{0}\nabla w|_{V}$. Using the definitions of
${v^{\prime}}$ and $\varphi$, and that $\Sigma_{0}\Subset W_{1}$ where
$1-\varphi$ vanishes, we have

$\int_{V}F_{2}\cdot{\nabla{v^{\prime}}}\,\mathrm{d}x=\int_{V}(1-\varphi)F_{2}\cdot\nabla
v_{0}\,\mathrm{d}x+\int_{V}(F_{2}\cdot\nabla(1-\varphi))v_{0}\,\mathrm{d}x.$
(5.5)

Define $F_{3}=\gamma_{0}\nabla{w}|_{V}+(1-\varphi)F_{2}$ and
$g=F_{2}\cdot\nabla(1-\varphi)$, then (5.4) and (5.5) collectively give

$\int_{\Omega}\gamma_{0}\nabla{w}\cdot{\nabla
v}\,\mathrm{d}x=\int_{V}F_{3}\cdot{\nabla
v_{0}}\,\mathrm{d}x+\int_{V}gv_{0}\,\mathrm{d}x.$ (5.6)

We now consider an auxiliary problem. Let $V^{\circ}$ be the interior of $V$
and let

$\mathcal{H}_{V}=\Bigl{\\{}\widehat{v}\in
H^{1}(V^{\circ}\setminus\overline{\Sigma_{0}})\mid\int_{\partial
V}\widehat{v}|_{\partial V}\,\mathrm{d}S=0\Bigr{\\}},$

then $\mathcal{H}_{V}$ is a Hilbert space with inner product (whose norm is
equivalent to the usual $H^{1}$-norm)

$\langle\widehat{v}_{1},\widehat{v}_{2}\rangle_{\mathcal{H}_{V}}=\int_{V}\nabla\widehat{v}_{1}\cdot\nabla\widehat{v}_{2}\,\mathrm{d}x.$

Using the Lax-Milgram lemma, we now define $\widehat{w}$ to be the unique
solution in $\mathcal{H}_{V}$ of the variational problem

$\int_{V}\nabla\widehat{w}\cdot{\nabla\widehat{v}}\,\mathrm{d}x=\int_{V}g\widehat{v}\,\mathrm{d}x,\qquad\forall\widehat{v}\in\mathcal{H}_{V}.$
(5.7)

Since $v_{0}$ has vanishing mean on $\partial V$ the restriction of $v_{0}$ to
$V$ belongs to $\mathcal{H}_{V}$. Let $F=F_{3}+\nabla\widehat{w}$, then (5.6)
and (5.7) lead to

$\int_{\Omega}\gamma_{0}\nabla w\cdot\nabla
v\,\mathrm{d}x=\int_{V}F\cdot\nabla v_{0}\,\mathrm{d}x=\int_{V}F\cdot\nabla
v\,\mathrm{d}x.$

As $v\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$ was arbitrary (and
$w\in\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}\subseteq\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$),
we have proven

$f=w|_{\Gamma}=L_{\Sigma_{0}}^{\Sigma_{\infty}}(V)F,$

and therefore $R(L_{\emptyset}^{\Sigma_{\infty}}(V))\subseteq
R(L_{\Sigma_{0}}^{\Sigma_{\infty}}(V))$.

Proof of “$\supseteq$”: Suppose now that $f\in
R(L_{\Sigma_{0}}^{\Sigma_{\infty}}(V))$, hence there exists $F_{1}\in
L^{2}(V)^{d}$ such that $f=w|_{\Gamma}$ with
$w=w_{\Sigma_{0},F_{1}}^{\Sigma_{\infty}}$. Thus we have

$\int_{\Omega}\gamma_{0}\nabla w\cdot\nabla
v\,\mathrm{d}x=\int_{V}F_{1}\cdot\nabla v\,\mathrm{d}x,\qquad\forall
v\in\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}.$ (5.8)

Let $\varphi$ be defined as before, and define
$\widetilde{w}\in\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}$ via the formula
$\widetilde{w}=(1-\varphi)w$ and extended by zero on $\overline{\Sigma_{0}}$.
Recall that
$\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}\subseteq\mathcal{H}_{\Sigma_{0}}^{\Sigma_{\infty}}$
and let $v\in\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}$ be arbitrary, then
(5.8) implies

$\displaystyle\int_{\Omega}\gamma_{0}\nabla\widetilde{w}\cdot\nabla
v\,\mathrm{d}x$ $\displaystyle=\int_{\Omega}\gamma_{0}\nabla w\cdot\nabla
v\,\mathrm{d}x-\int_{V}\gamma_{0}\nabla(\varphi w)\cdot\nabla v\,\mathrm{d}x$
$\displaystyle=\int_{V}(F_{1}-\gamma_{0}\nabla(\varphi w))\cdot\nabla
v\,\mathrm{d}x.$

Hence, with $F=F_{1}-\gamma_{0}\nabla(\varphi w)|_{V}$

$f=w|_{\Gamma}=\widetilde{w}|_{\Gamma}=L_{\emptyset}^{\Sigma_{\infty}}(V)F,$

and we have verified that $R(L_{\emptyset}^{\Sigma_{\infty}}(V))\supseteq
R(L_{\Sigma_{0}}^{\Sigma_{\infty}}(V))$. ∎

######  Remark 5.2.

If $\Sigma\Subset V$ in Proposition 5.1, then one can combine (i) and (ii) to
obtain:

$R(L_{\emptyset}^{\emptyset}(V))=R(L_{\Sigma_{0}}^{\emptyset}(V))=R(L_{\emptyset}^{\Sigma_{\infty}}(V))=R(L_{\Sigma_{0}}^{\Sigma_{\infty}}(V)).$

Indeed, after using (i) and (ii) as they are stated, we can use (i) again for
a $(\Sigma_{0}^{\prime},\Sigma_{\infty}^{\prime})$ collection of cracks with
$\Sigma_{0}^{\prime}=\Sigma_{0}$ and $\Sigma_{\infty}^{\prime}=\emptyset$.

The following result will be needed in the proof of Proposition 5.4. For
reasons of clarity we state it as a separate lemma with its own proof.222The
additional regularity required of $\gamma_{0}$ in this lemma guarantees the
validity of the well known (jump) relations for the (single and double) layer
potentials.

######  Lemma 5.3.

Let $\Sigma$ be a $(\Sigma_{0},\Sigma_{\infty})$ collection of cracks. Assume
that $\gamma_{0}\in C^{2}(\overline{\Omega})$ and is positive.

  1. (i)

If $\Sigma_{0}\neq\emptyset$ then there exists $f\in L^{2}_{\diamond}(\Gamma)$
such that $u_{\emptyset,f}^{\Sigma_{\infty}}\neq
u_{\Sigma_{0},f}^{\Sigma_{\infty}}$.

  2. (ii)

If $\Sigma_{\infty}\neq\emptyset$ then there exists $f\in
L^{2}_{\diamond}(\Gamma)$ such that $u_{\Sigma_{0},f}^{\emptyset}\neq
u_{\Sigma_{0},f}^{\Sigma_{\infty}}$.

###### Proof.

Proof of (i): It suffices to prove that there exists an $f\in
L^{2}_{\diamond}(\Gamma)$ such that

$\frac{\partial}{\partial n}u_{\emptyset,f}^{\Sigma_{\infty}}\text{ does not
vanish identically on }\Sigma_{0}.$

Let $\Phi(x,y)$ denote the (fundamental) solution to the problem

$\displaystyle-\nabla\cdot(\gamma_{0}\nabla\Phi(\,\cdot\,,y))$
$\displaystyle=\delta_{y}\quad\text{in
}\Omega\setminus\overline{\Sigma}_{\infty},$
$\displaystyle\gamma_{0}\frac{\partial}{\partial\nu}\Phi(\,\cdot\,,y)$
$\displaystyle=-\frac{1}{|\partial\Omega|}\quad\text{on }\partial\Omega,$
$\displaystyle\Phi(\,\cdot\,,y)\text{ is locally constant on
}\Sigma_{\infty},$
$\displaystyle\int_{\Sigma_{i}}\Bigl{[}\gamma_{0}\frac{\partial}{\partial
n}\Phi(\,\cdot\,,y)\Bigr{]}\mathrm{d}S$ $\displaystyle=0\quad\text{for each
component $\Sigma_{i}$ of $\Sigma_{\infty}$}.$

In terms of this $\Phi$ we have the representation formula

$u_{\emptyset,f}^{\Sigma_{\infty}}(y)=\int_{\Gamma}\Phi(x,y)f(x)\,\mathrm{d}S_{x},$

and so

$\frac{\partial}{\partial
n}u_{\emptyset,f}^{\Sigma_{\infty}}(y)=\int_{\Gamma}\frac{\partial}{\partial
n_{y}}\Phi(x,y)f(x)\,\mathrm{d}S_{x}\quad\text{on }\Sigma_{0}.$

Now suppose that $\frac{\partial}{\partial
n}u_{\emptyset,f}^{\Sigma_{\infty}}$ vanishes identically on $\Sigma_{0}$ for
all $f\in L^{2}_{\diamond}(\Gamma)$. Let $\psi$ be a non-zero continuous
function on $\Sigma_{0}$, supported away from $\partial\Sigma_{0}$, then it
would follow that

$0=\int_{\Sigma_{0}}\psi(y)\int_{\Gamma}\frac{\partial}{\partial
n_{y}}\Phi(x,y)f(x)\,\mathrm{d}S_{x}\,\mathrm{d}S_{y}=\int_{\Gamma}f(x)\int_{\Sigma_{0}}\frac{\partial}{\partial
n_{y}}\Phi(x,y)\psi(y)\,\mathrm{d}S_{y}\,\mathrm{d}S_{x}$

for all $f\in L^{2}_{\diamond}(\Gamma)$. In other words, there is a constant
$C$ such that

$\int_{\Sigma_{0}}\frac{\partial}{\partial
n_{y}}\Phi(x,y)\psi(y)\,\mathrm{d}S_{y}\equiv C\quad\text{on }\Gamma.$

We also have that

$\gamma_{0}(x)\frac{\partial}{\partial\nu}\int_{\Sigma_{0}}\frac{\partial}{\partial
n_{y}}\Phi(x,y)\psi(y)\,\mathrm{d}S_{y}=\int_{\Sigma_{0}}\frac{\partial}{\partial
n_{y}}\Bigl{(}\gamma_{0}(x)\frac{\partial}{\partial\nu_{x}}\Phi(x,y)\Bigr{)}\psi(y)\,\mathrm{d}S_{y}\equiv
0\quad\text{on }\Gamma.$

By unique continuation of the operator
$\nabla\cdot(\gamma_{0}\nabla(\,\cdot\,))$ it follows that

$\int_{\Sigma_{0}}\frac{\partial}{\partial
n_{y}}\Phi(x,y)\psi(y)\,\mathrm{d}S_{y}\equiv C\quad\text{in
}\Omega\setminus\overline{\Sigma_{0}}.$

However, from the jump relations for double layer potentials (see [31, 27]),
it now follows that $\psi$ must identically vanish on $\Sigma_{0}$, contrary
to the initial assumption that $\psi$ is non-zero.333The $C^{2}$ regularity of
$\gamma_{0}$ is sufficient to guarantee that the fundamental solution
$\Phi(x,y)$ has the form
$\Phi(x,y)=\gamma_{0}(y)^{-1}\Phi_{\Delta}(x,y)+R(x,y)$, where $\Phi_{\Delta}$
is the standard Green’s function for the Laplacian, and the “regular term” $R$
does not contribute to the jump relations. The references cited pertain to
Lipschitz surfaces.

Proof of (ii): It suffices to prove that there exists an $f\in
L^{2}_{\diamond}(\Gamma)$ such that

$u_{\Sigma_{0},f}^{\emptyset}\text{ is not locally constant on
}\Sigma_{\infty}.$

Let $\Psi(x,y)$ denote the (fundamental) solution to the problem

$\displaystyle-\nabla\cdot(\gamma_{0}\nabla\Psi(\,\cdot\,,y))$
$\displaystyle=\delta_{y}\quad\text{in }\Omega\setminus\overline{\Sigma}_{0},$
$\displaystyle\gamma_{0}\frac{\partial}{\partial\nu}\Psi(\,\cdot\,,y)$
$\displaystyle=-\frac{1}{|\partial\Omega|}\quad\text{on }\partial\Omega,$
$\displaystyle\gamma_{0}\frac{\partial}{\partial n}\Psi(\,\cdot\,,y)$
$\displaystyle=0\quad\text{on }\Sigma_{0}.$

We have the representation formula

$u_{\Sigma_{0},f}^{\emptyset}(y)=\int_{\Gamma}\Psi(x,y)f(x)\,\mathrm{d}S_{x}.$

Now suppose that $u_{\Sigma_{0},f}^{\emptyset}$ is locally constant on
$\Sigma_{\infty}$ for all $f\in L^{2}_{\diamond}(\Gamma)$. Let $\varphi$ be a
non-zero continuous function on $\Sigma_{\infty}$, supported away from
$\partial\Sigma_{\infty}$, and with vanishing mean on each component of
$\Sigma_{\infty}$ (i.e. orthogonal to $u_{\Sigma_{0},f}^{\emptyset}$ on
$\Sigma_{\infty}$ in the $L^{2}$ inner product), then it would follow that

$0=\int_{\Sigma_{\infty}}\varphi(y)\int_{\Gamma}\Psi(x,y)f(x)\,\mathrm{d}S_{x}\,\mathrm{d}S_{y}=\int_{\Gamma}f(x)\int_{\Sigma_{\infty}}\Psi(x,y)\varphi(y)\,\mathrm{d}S_{y}\,\mathrm{d}S_{x}$

for all $f\in L^{2}_{\diamond}(\Gamma)$. Thus, there is a constant $C$ such
that

$\int_{\Sigma_{\infty}}\Psi(x,y)\varphi(y)\,\mathrm{d}S_{y}\equiv
C\quad\text{on }\Gamma.$

Since $\varphi$ has vanishing mean on $\Sigma_{\infty}$, we also have that

$\gamma_{0}(x)\frac{\partial}{\partial\nu}\int_{\Sigma_{\infty}}\Psi(x,y)\varphi(y)\,\mathrm{d}S_{y}=\int_{\Sigma_{\infty}}\Bigl{(}\gamma_{0}(x)\frac{\partial}{\partial\nu_{x}}\Psi(x,y)\Bigr{)}\varphi(y)\,\mathrm{d}S_{y}\equiv
0\quad\text{on }\Gamma.$

By unique continuation, it follows that

$\int_{\Sigma_{\infty}}\Psi(x,y)\varphi(y)\,\mathrm{d}S_{y}\equiv
C\quad\text{in }\Omega\setminus\overline{\Sigma}.$

However, from the jump relations for the normal derivative of single layer
potentials, it now follows that $\varphi$ must identically vanish on
$\Sigma_{\infty}$; a contradiction. ∎

######  Proposition 5.4.

Let $\Sigma$ be a $(\Sigma_{0},\Sigma_{\infty})$ collection of cracks. Assume
that $\Sigma_{0}\Subset V$ and $\Sigma_{\infty}\Subset W$ for
$V,W\in\mathcal{A}$ with $\mathop{\textup{dist}}(V,W)>0$.

  1. (i)

If $\Sigma_{0}\neq\emptyset$ then there exists a sequence $(f_{n})$ in
$L^{2}_{\diamond}(\Gamma)$ such that

$\displaystyle\lim_{n\to\infty}\langle(\Lambda_{W}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})f_{n},f_{n}\rangle$
$\displaystyle=0,$
$\displaystyle\lim_{n\to\infty}\langle(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{W})f_{n},f_{n}\rangle$
$\displaystyle=0,$
$\displaystyle\lim_{n\to\infty}\langle(\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}-\Lambda_{\emptyset}^{\Sigma_{\infty}})f_{n},f_{n}\rangle$
$\displaystyle=\infty.$

  2. (ii)

If $\Sigma_{\infty}\neq\emptyset$ then there exists a sequence $(g_{n})$ in
$L^{2}_{\diamond}(\Gamma)$ such that

$\displaystyle\lim_{n\to\infty}\langle(\Lambda_{V}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})g_{n},g_{n}\rangle$
$\displaystyle=0,$
$\displaystyle\lim_{n\to\infty}\langle(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{V})g_{n},g_{n}\rangle$
$\displaystyle=0,$
$\displaystyle\lim_{n\to\infty}\langle(\Lambda_{\Sigma_{0}}^{\emptyset}-\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}})g_{n},g_{n}\rangle$
$\displaystyle=\infty.$

###### Proof.

We give a proof of (i), and note that the proof of (ii) is almost the same
with obvious modifications.

The first part of this proof, up to (5.9), is a slight modification of [20,
Proof of Theorem 3.6], and the arguments are essentially the same, however,
our setup includes perfectly conducting cracks. Due to the notational
differences we give the full details.

Let $Y\in\mathcal{A}$ such that $W\Subset Y$ and
$\mathop{\textup{dist}}(V,Y)>0$. We will construct a localisation for certain
electric potentials, such that their energy tends to zero on the slightly
larger set $Y$ instead of just on $W$. We make use of this fact to establish
(5.20) towards the end of the proof.

Let $U=\overline{\Omega}\setminus(V\cup Y)$ then $U$ is connected, $U$ is
relatively open in $\overline{\Omega}$, and $\partial\Omega\subset U$. For any

$h\in R(L_{\emptyset}^{\Sigma_{\infty}}(V))\cap
R(L_{\emptyset}^{\Sigma_{\infty}}(Y))$

there exist $F_{V}\in L^{2}(V)^{d}$ and $F_{Y}\in L^{2}(Y)^{d}$ such that
$h=L_{\emptyset}^{\Sigma_{\infty}}(V)F_{V}=L_{\emptyset}^{\Sigma_{\infty}}(Y)F_{Y}$.
Let $u_{V}=w_{\emptyset,F_{V}}^{\Sigma_{\infty}}$ and
$u_{Y}=w_{\emptyset,F_{Y}}^{\Sigma_{\infty}}$. Thus
$u_{V},u_{Y}\in\mathcal{H}_{\emptyset}^{\Sigma_{\infty}}$ and from the
variational formulations we obtain the equations

$\displaystyle-\nabla\cdot(\gamma_{0}\nabla u_{V})$
$\displaystyle=0\quad\text{in
}\Omega\setminus(V\cup\overline{\Sigma_{\infty}}),$
$\displaystyle-\nabla\cdot(\gamma_{0}\nabla u_{Y})$
$\displaystyle=0\quad\text{in }\Omega\setminus Y,$ $\displaystyle u_{V}=u_{Y}$
$\displaystyle=h\quad\text{on }\Gamma,$ $\displaystyle\gamma_{0}\frac{\partial
u_{V}}{\partial\nu}=\gamma_{0}\frac{\partial u_{Y}}{\partial\nu}$
$\displaystyle=0\quad\text{on }\partial\Omega.$

The variational formulation also gives that

$\int_{\Sigma_{i}}\Bigl{[}\gamma_{0}\frac{\partial u_{V}}{\partial
n}\Bigr{]}\mathrm{d}S=0\quad\text{for each component $\Sigma_{i}$ of
$\Sigma_{\infty}$.}$

Due to the unique continuation property (UCP) we get that
$u_{V}|_{U}=u_{Y}|_{U}$. By gluing $u_{V}$ and $u_{Y}$ together, we obtain the
function

$u=\begin{cases}u_{V}&\text{in }\Omega\setminus V,\\\ u_{Y}&\text{in
}V,\end{cases}$

which, due to the boundary value problem it solves, must equal
$u_{\emptyset,0}^{\Sigma_{\infty}}=0$. Consequently $h=u|_{\Gamma}=0$, and so
we have proven

$R(L_{\emptyset}^{\Sigma_{\infty}}(V))\cap
R(L_{\emptyset}^{\Sigma_{\infty}}(Y))=\\{0\\}.$ (5.9)

Since $\Sigma_{0}\Subset V$, we get from Proposition 5.1(i) that

$R(L_{\emptyset}^{\Sigma_{\infty}}(V))=R(L_{\Sigma_{0}}^{\Sigma_{\infty}}(V)).$
(5.10)

Define

$A=L_{\Sigma_{0}}^{\Sigma_{\infty}}(V)-L_{\emptyset}^{\Sigma_{\infty}}(V).$

Due to (5.2) we have

$A^{*}f=(L_{\Sigma_{0}}^{\Sigma_{\infty}}(V))^{*}f-(L_{\emptyset}^{\Sigma_{\infty}}(V))^{*}f=\nabla(u^{\Sigma_{\infty}}_{\Sigma_{0},f}-u^{\Sigma_{\infty}}_{\emptyset,f})|_{V}.$
(5.11)

Using unique continuation and the mean-free conditions on $\Gamma$ for
$u^{\Sigma_{\infty}}_{\Sigma_{0},f}$ and $u^{\Sigma_{\infty}}_{\emptyset,f}$,
we now conclude that

$A^{*}f=0\qquad\text{if and only if}\qquad
u^{\Sigma_{\infty}}_{\Sigma_{0},f}=u^{\Sigma_{\infty}}_{\emptyset,f}.$

Since $\Sigma_{0}\neq\emptyset$, lemma 5.3(i) implies $A^{*}\neq 0$ and
therefore $A\neq 0$. As a consequence there exists $g\in
R(A)\setminus\\{0\\}$, and because of (5.9) and (5.10) this $g$ satisfies
$g\in R(L_{\emptyset}^{\Sigma_{\infty}}(V))$ and $g\not\in
R(L_{\emptyset}^{\Sigma_{\infty}}(Y))$. Moreover,
$(L_{\emptyset}^{\Sigma_{\infty}}(Y))^{*}$ is injective by (5.2) and unique
continuation. We then apply the constructive result on localised potentials in
Lemma 4.4, and for $n\in\mathbb{N}$ define

$\xi_{n}=\bigl{(}L_{\emptyset}^{\Sigma_{\infty}}(Y)(L_{\emptyset}^{\Sigma_{\infty}}(Y))^{*}+\tfrac{1}{n}I\bigr{)}^{-1}g$

and

$f_{n}=\frac{\xi_{n}}{\lVert(L_{\emptyset}^{\Sigma_{\infty}}(Y))^{*}\xi_{n}\rVert_{L^{2}(Y)^{d}}^{3/2}}.$

Lemma 4.4 implies that

$\displaystyle\lim_{n\to\infty}\lVert(L_{\emptyset}^{\Sigma_{\infty}}(Y))^{*}f_{n}\rVert_{L^{2}(Y)^{d}}$
$\displaystyle=0,$ (5.12)
$\displaystyle\lim_{n\to\infty}\lVert(L_{\emptyset}^{\Sigma_{\infty}}(V))^{*}f_{n}\rVert_{L^{2}(V)^{d}}$
$\displaystyle=\infty,$ (5.13) $\displaystyle\text{and
}~{}\lim_{n\to\infty}\lVert A^{*}f_{n}\rVert_{L^{2}(V)^{d}}$
$\displaystyle=\infty,$ (5.14)

simultaneously. Since $\Sigma_{\infty}\Subset Y$, Proposition 5.1(ii) gives
that $R(L_{\emptyset}^{\Sigma_{\infty}}(Y))=R(L_{\emptyset}^{\emptyset}(Y))$.
This range equality, together with (5.12) and Lemma 4.3, implies

$\lim_{n\to\infty}\lVert(L_{\emptyset}^{\emptyset}(Y))^{*}f_{n}\rVert_{L^{2}(Y)^{d}}=0.$
(5.15)

From Proposition 5.1(ii) we have $R(L_{\emptyset}^{\Sigma_{\infty}}(V\cup
Y))=R(L_{\emptyset}^{\emptyset}(V\cup Y))$. So from (5.2) and Lemma 4.3 there
exists a $c>0$ such that

$\displaystyle\lVert(L_{\emptyset}^{\emptyset}(V))^{*}f_{n}\rVert_{L^{2}(V)^{d}}^{2}+\lVert(L_{\emptyset}^{\emptyset}(Y))^{*}f_{n}\rVert_{L^{2}(Y)^{d}}^{2}$
$\displaystyle=\lVert(L_{\emptyset}^{\emptyset}(V\cup
Y))^{*}f_{n}\rVert_{L^{2}(V\cup Y)^{d}}^{2}$ $\displaystyle\geq
c\lVert(L_{\emptyset}^{\Sigma_{\infty}}(V\cup Y))^{*}f_{n}\rVert_{L^{2}(V\cup
Y)^{d}}^{2}$ $\displaystyle\geq
c\lVert(L_{\emptyset}^{\Sigma_{\infty}}(V))^{*}f_{n}\rVert_{L^{2}(V)^{d}}^{2}.$
(5.16)

A combination of (5.13), (5.15), and (5.16) gives

$\lim_{n\to\infty}\lVert(L_{\emptyset}^{\emptyset}(V))^{*}f_{n}\rVert_{L^{2}(V)^{d}}=\infty.$
(5.17)

We are now ready to associate the different operators to differences of ND
mappings. Let $\widehat{u}_{n}=u_{\emptyset,f_{n}}^{\emptyset}$. Then (5.2),
(5.15), and (5.17) translate into

$\lim_{n\to\infty}\int_{V}\lvert\nabla\widehat{u}_{n}\rvert^{2}\,\mathrm{d}x=\infty\qquad\text{and}\qquad\lim_{n\to\infty}\int_{Y}\lvert\nabla\widehat{u}_{n}\rvert^{2}\,\mathrm{d}x=0.$
(5.18)

Using [6, Lemma A.1(ii)], we obtain the existence of a constant $K>0$,
independent of $f_{n}$, such that

$0\leq\langle(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{W})f_{n},f_{n}\rangle\leq
K\int_{W}\lvert\nabla\widehat{u}_{n}\rvert^{2}\,\mathrm{d}x.$ (5.19)

Let $\breve{u}_{n}$ equal $u_{W,f_{n}}^{\emptyset}$ in $\Omega\setminus W$ and
satisfy the following Dirichlet problem in $W$:

$\displaystyle-\nabla\cdot(\gamma_{0}\nabla\breve{u}_{n})$
$\displaystyle=0\quad\text{in interior of }W,$ $\displaystyle\breve{u}_{n}$
$\displaystyle=u_{W,f_{n}}^{\emptyset}\quad\text{on }\partial W.$

The result in [6, Lemma 5.3] ensures that if an extreme inclusion (either
perfectly insulating or perfectly conducting) is introduced, _compactly
contained_ in $Y$ where the energy of $\widehat{u}_{n}$ tends to zero, then
the corresponding electric potentials with this new conductivity profile will
have the same localisation as in (5.18). For a perfectly insulating inclusion,
$W\Subset Y$, the localisation applies to $\breve{u}_{n}$, using precisely the
extension in $W$ defined above. We thus have

$\lim_{n\to\infty}\int_{Y}\lvert\nabla\breve{u}_{n}\rvert^{2}\,\mathrm{d}x=0.$
(5.20)

Use of [6, Lemma A.1(iii)] yields

$0\leq\langle(\Lambda_{W}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})f_{n},f_{n}\rangle\leq\mathop{\textup{ess\,sup}}(\gamma_{0})\int_{W}\lvert\nabla\breve{u}_{n}\rvert^{2}\,\mathrm{d}x.$
(5.21)

From (5.18)–(5.21) we now conclude

$\lim_{n\to\infty}\langle(\Lambda_{W}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})f_{n},f_{n}\rangle=\lim_{n\to\infty}\langle(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{W})f_{n},f_{n}\rangle=0.$

It remains to prove that
$\langle(\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}-\Lambda_{\emptyset}^{\Sigma_{\infty}})f_{n},f_{n}\rangle\to\infty$
for $n\to\infty$. Let $u_{n}=u_{\Sigma_{0},f_{n}}^{\Sigma_{\infty}}$ and
$\widetilde{u}_{n}=u_{\emptyset,f_{n}}^{\Sigma_{\infty}}$, then from Lemma
4.1(i) we have

$\langle(\Lambda_{\Sigma_{0}}^{\Sigma_{\infty}}-\Lambda_{\emptyset}^{\Sigma_{\infty}})f_{n},f_{n}\rangle=\lVert
u_{n}-\widetilde{u}_{n}\rVert_{*}^{2}\geq\mathop{\textup{ess\,inf}}(\gamma_{0})\int_{V}\lvert\nabla(u_{n}-\widetilde{u}_{n})\rvert^{2}\,\mathrm{d}x.$

At the same time from (5.11)

$A^{*}f_{n}=\nabla(u_{n}-\widetilde{u}_{n})|_{V},$

so use of (5.14) concludes the proof. ∎

##  6\. The direction “$\Leftarrow$” in Theorems 1.2 & 1.3

Finally we show the “difficult direction” of the _if and only if_ statements
of Theorems 1.2 and 1.3, although the majority of the work has already been
done in Proposition 5.4.

######  Proposition 6.1.

Let $D$ be a $(D_{0},D_{\infty})$ collection of cracks. Given any
$C\in\mathcal{A}$, then

$\Lambda_{C}^{\emptyset}\geq\Lambda_{D_{0}}^{D_{\infty}}\geq\Lambda_{\emptyset}^{C}\qquad\text{implies}\qquad
D\subset C.$

###### Proof.

We will prove the contrapositive statement, i.e. assume $D\not\subset C$. As
$C$ is closed there is a relatively open part of $D$ that belongs to
$\Omega\setminus C$. Now we will show that either
$\Lambda_{C}^{\emptyset}\not\geq\Lambda_{D_{0}}^{D_{\infty}}$ or
$\Lambda_{D_{0}}^{D_{\infty}}\not\geq\Lambda_{\emptyset}^{C}$.

Since $C$ has connected complement we have either of two cases, which will
dictate which operator inequality that will fail to hold:

  1. (a)

There are $V,W\in\mathcal{A}$ with $\mathop{\textup{dist}}(V,W)>0$ and non-
empty $\chi\in\mathcal{X}$, such that

$\chi\subseteq D_{0},\qquad\chi\Subset V,\qquad C\subseteq
W,\qquad\text{and}\qquad D_{\infty}\Subset W.$

  2. (b)

There are $V,W\in\mathcal{A}$ with $\mathop{\textup{dist}}(V,W)>0$ and non-
empty $\chi\in\mathcal{X}$, such that

$\chi\subseteq D_{\infty},\qquad\chi\Subset W,\qquad C\subseteq
V,\qquad\text{and}\qquad D_{0}\Subset V.$

Case (a): Since $C\cup D_{\infty}\subseteq W$ then [6, Theorem 3.7] and (the
second inequality of) Proposition 3.1 give

$\Lambda_{C}^{\emptyset}\leq\Lambda_{W}^{\emptyset}\qquad\text{and}\qquad\Lambda_{\emptyset}^{D_{\infty}}\geq\Lambda_{\emptyset}^{W}.$

As $\chi\subseteq D_{0}$ then Proposition 3.2(i) gives

$\Lambda_{D_{0}}^{D_{\infty}}\geq\Lambda_{\chi}^{D_{\infty}}.$

Hence we have

$\displaystyle\Lambda_{C}^{\emptyset}-\Lambda_{D_{0}}^{D_{\infty}}$
$\displaystyle=(\Lambda_{C}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})+(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{D_{\infty}})+(\Lambda_{\emptyset}^{D_{\infty}}-\Lambda_{D_{0}}^{D_{\infty}})$
$\displaystyle\leq(\Lambda_{W}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})+(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{W})+(\Lambda_{\emptyset}^{D_{\infty}}-\Lambda_{\chi}^{D_{\infty}}).$

Now, since $\emptyset\neq\chi\Subset V$ and $D_{\infty}\Subset W$, Proposition
5.4(i) gives the existence of a sequence $(f_{n})$ in
$L^{2}_{\diamond}(\Gamma)$ such that

$\lim_{n\to\infty}\langle(\Lambda_{C}^{\emptyset}-\Lambda_{D_{0}}^{D_{\infty}})f_{n},f_{n}\rangle=-\infty,$

and in particular we conclude that
$\Lambda_{C}^{\emptyset}\not\geq\Lambda_{D_{0}}^{D_{\infty}}$.

Case (b): Since $C\cup D_{0}\subseteq V$ then [6, Theorem 3.7] and (the first
inequality of) Proposition 3.1 gives

$\Lambda_{\emptyset}^{C}\geq\Lambda_{\emptyset}^{V}\qquad\text{and}\qquad\Lambda_{D_{0}}^{\emptyset}\leq\Lambda_{V}^{\emptyset}.$

As $\chi\subseteq D_{\infty}$ then Proposition 3.2(ii) gives

$\Lambda_{D_{0}}^{D_{\infty}}\leq\Lambda_{D_{0}}^{\chi}.$

Hence we have

$\displaystyle\Lambda_{D_{0}}^{D_{\infty}}-\Lambda_{\emptyset}^{C}$
$\displaystyle=(\Lambda_{D_{0}}^{D_{\infty}}-\Lambda_{D_{0}}^{\emptyset})+(\Lambda_{D_{0}}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})+(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{C})$
$\displaystyle\leq(\Lambda_{D_{0}}^{\chi}-\Lambda_{D_{0}}^{\emptyset})+(\Lambda_{V}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})+(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{V}).$

Now, since $\emptyset\neq\chi\Subset W$ and $D_{0}\Subset V$, Proposition
5.4(ii) gives the existence of a sequence $(g_{n})$ in
$L^{2}_{\diamond}(\Gamma)$ such that

$\lim_{n\to\infty}\langle(\Lambda_{D_{0}}^{D_{\infty}}-\Lambda_{\emptyset}^{C})g_{n},g_{n}\rangle=-\infty,$

and in particular we conclude that
$\Lambda_{D_{0}}^{D_{\infty}}\not\geq\Lambda_{\emptyset}^{C}$. ∎

######  Proposition 6.2.

Let $D\in\mathcal{X}$.

  1. (i)

Given any $\chi\in\mathcal{X}$, then

$\Lambda_{D}^{\emptyset}\geq\Lambda_{\chi}^{\emptyset}\qquad\text{implies}\qquad\chi\subseteq
D.$

  2. (ii)

Given any $\chi\in\mathcal{X}$, then

$\Lambda_{\emptyset}^{\chi}\geq\Lambda_{\emptyset}^{D}\qquad\text{implies}\qquad\chi\subseteq
D.$

###### Proof.

We will prove the contrapositive statements, i.e., assume $\chi\not\subseteq
D$.

Proof of (i): We need to show that
$\Lambda_{D}^{\emptyset}\not\geq\Lambda_{\chi}^{\emptyset}$. There exist
$V,W\in\mathcal{A}$ with $\mathop{\textup{dist}}(V,W)>0$ and a non-empty
$\chi^{\prime}\in\mathcal{X}$, such that

$\chi^{\prime}\subseteq\chi,\qquad\chi^{\prime}\Subset
V,\qquad\text{and}\qquad D\subset W.$

Using Propositions 3.1 and 3.2(i) we have

$\Lambda_{D}^{\emptyset}\leq\Lambda_{W}^{\emptyset}\quad\text{and}\quad\Lambda_{\chi}^{\emptyset}\geq\Lambda_{\chi^{\prime}}^{\emptyset}.$

Hence

$\Lambda_{D}^{\emptyset}-\Lambda_{\chi}^{\emptyset}=(\Lambda_{D}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})+(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\chi}^{\emptyset})\leq(\Lambda_{W}^{\emptyset}-\Lambda_{\emptyset}^{\emptyset})+(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\chi^{\prime}}^{\emptyset}).$

Now, since $\emptyset\neq\chi^{\prime}\Subset V$, Proposition 5.4(i) gives the
existence of a sequence $(f_{n})$ in $L^{2}_{\diamond}(\Gamma)$ such that

$\lim_{n\to\infty}\langle(\Lambda_{D}^{\emptyset}-\Lambda_{\chi}^{\emptyset})f_{n},f_{n}\rangle=-\infty,$

and in particular we conclude that
$\Lambda_{D}^{\emptyset}\not\geq\Lambda_{\chi}^{\emptyset}$.

Proof of (ii): We need to show that
$\Lambda_{\emptyset}^{\chi}\not\geq\Lambda_{\emptyset}^{D}$. There exist
$V,W\in\mathcal{A}$ with $\mathop{\textup{dist}}(V,W)>0$ and a non-empty
$\chi^{\prime}\in\mathcal{X}$, such that

$\chi^{\prime}\subseteq\chi,\qquad\chi^{\prime}\Subset
W,\qquad\text{and}\qquad D\subset V.$

Using Propositions 3.1 and 3.2(ii) we have

$\Lambda_{\emptyset}^{D}\geq\Lambda_{\emptyset}^{V}\qquad\text{and}\qquad\Lambda_{\emptyset}^{\chi}\leq\Lambda_{\emptyset}^{\chi^{\prime}}.$

Hence

$\Lambda_{\emptyset}^{\chi}-\Lambda_{\emptyset}^{D}=(\Lambda_{\emptyset}^{\chi}-\Lambda_{\emptyset}^{\emptyset})+(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{D})\leq(\Lambda_{\emptyset}^{\chi^{\prime}}-\Lambda_{\emptyset}^{\emptyset})+(\Lambda_{\emptyset}^{\emptyset}-\Lambda_{\emptyset}^{V}).$

Now, since $\emptyset\neq\chi^{\prime}\Subset W$, Proposition 5.4(ii) gives
the existence of a sequence $(g_{n})$ in $L^{2}_{\diamond}(\Gamma)$ such that

$\lim_{n\to\infty}\langle(\Lambda_{\emptyset}^{\chi}-\Lambda_{\emptyset}^{D})g_{n},g_{n}\rangle=-\infty,$

and in particular we conclude that
$\Lambda_{\emptyset}^{\chi}\not\geq\Lambda_{\emptyset}^{D}$. ∎

### Acknowledgements

The research of HG was partially supported by grant 8021–00084B from Independent Research Fund Denmark | Natural Sciences. The research of MSV was partially supported by NSF grant DMS-22-05912.

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