Reply to Comment on ‘Revisiting the probe and enclosure methods’

What should you do when you re-read the paper you intended to cite and found an error in the proof? In section 4 of Ikehata (2022 Inverse Problems 38 075009) the author pointed out that the proof of lemma 3.5 in Sini and Yoshida (2012 Inverse Problems 28 055013; 2013 Inverse Problems 29 039501) does not work for a Lipschitz obstacle case, by applying their approach to an essentially same lemma (lemma 3.2 in Ikehata (2022 Inverse Problems 38 075009)). For this, Sini (2023 Inverse Problems accepted) gave a comment. This is a reply to their comment.


The Lipschitz obstacle case
They seem to have no objection to the explanation I provided in section 4 of my paper [2] regarding the 'proof' of lemma 3.5 in their paper [4,5] for the Lipschitz obstacle case.The common conclusion in their notation is as follows.

Conclusion.
In the Lipschitz obstacle case, given an arbitrary direction ρ it is not always possible for us to choose an orthogonal coordinate system (y 1 , y 2 , y 3 ) centered at an arbitrary point x 0 with x 0 • ρ = h D (ρ) in such a way that the both of (a) and (b) described below are satisfied: Original Content from this work may be used under the terms of the Creative Commons Attribution 4.0 licence.Any further distribution of this work must maintain attribution to the author(s) and the title of the work, journal citation and DOI.
(a) the vector ρ is parallel to the y 3 -axis and other axes parpendicular to each other are directed to two lines on the plane x • ρ = h D (ρ) passing through x 0 ; (b) D and ∂D in a neighborhood of x 0 = (0, 0, 0) in the new coordinates (y 1 , y 2 , y 3 ) has the expression y 3 > l(y 1 , y 2 ) and y 3 = l(y 1 , y 2 ), respectively with a Lipschiz function l(y 1 , y 2 ) ⩾ 0 with l(0, 0) = 0.
Therefore, one cannot have the estimates (1) and ( 2) of [3] in the Lipschitz obstacle case.However, it seems, they do not read the following sentence placed at the end of section 3.1 of [2] (Note that the notation is taken from [2]).'If D is open and ∂D is Lipschitz, then D is 3-regular with respect to all unit vectors ω.Thus lemma 3.2 is valid also for such D without any restriction on ω.The advantage of the proof above is: we do not make use of any local coordinate system of D in a neighborhood of ∂D.'And the footnote 12 on page 31 of [2]: 'However, using the argument in the proof of lemma 3.2, one could prove the validity of (6.6) for a class of obstacle including Lipschitz one.'Note that, as explained in Introduction of [2], their main important contribution is the lower estimate in the system (1.3) in [2] together with the idea of the proof.It is independent of lemma 3.5 in [4] and covers the Lipschitz obstacle case.Therefore one can say that their final result in [4,5] that combines lemma 3.5 and estimates (1.3) itself covers the Lipschitz obstacle case, however, provided the proof of lemma 3.5 in [4] is replaced with a proof using the argument in that of lemma 3.2 in [2].

C 1 -regular case
Their geometric description of the proof in [3] that the C 1 -regularity is enough to ensure the existence of the orthogonal coordinate system (y 1 , y 2 , y 3 ) satisfying (a) and (b) at the same time, is not clear in the following sense.
The argument starts under the Lipschitz obstacle assumption (just saying), however, suddenly restricts the argument to the C 1 -case.It is not clear what is a 'gap' between Lipschitz and C 1 in choosing the desired coordinate system.
In the following appendix we give a justification of their approach presented in section 4 of [2] in the case when ∂D is C 1 .It suffices to show that the condition (4.2) in [2] is satisfied in that case.By showing it, the reader shall know the difficulty of doing the same argument in the Lipschitz boundary case starting from the definition of Lipschitz domain [1].
Lemma A.1.Let x 0 ∈ ∂D and assume that there exist a positive number δ, orthogonal matrix A and a function l ∈ C 0 (R 2 ) with l(0, 0) = 0 such that ) .

Thus one gets
) .
Here assume that ) .