Remarks on Lin-Nakamura-Wang's paper

TL;DR

This paper critically examines a previous theorem related to cavity reconstruction, asserting that the theorem in question is invalid, thereby questioning the validity of the original results.

Contribution

The paper provides a critical analysis showing the invalidity of a key theorem in prior work on cavity reconstruction from a single measurement.

Findings

Theorem 1.2 is not valid.

The critique impacts the validity of the original reconstruction method.

Highlights the need for revised theoretical foundations.

Abstract

Theorem 1.2 in their paper arXiv:1904.00999v1 [math.AP] 30 Mar 2019 "Reconstruction of unknown cavity by single measurement" is not valid.

Equations

$$\left\{\displaystyle\begin{array}[]{ll}\displaystyle\Delta u=0&\mbox{in $\Omega\setminus\overline{D}$,}\\ \\ \displaystyle\frac{\partial u}{\partial\nu}=0&\mbox{on $\partial D$,}\\ \\ \displaystyle u(R\cos\theta,R\sin\theta)=\left(R+\frac{1}{R}\right)\cos\theta,&\mbox{$\theta\in\,[0,\,2\pi[$}.\end{array}\right.$$

$$u(r\cos \theta ,r\sin \theta )=\left(r+\frac{1}{r}\right)\cos \theta .$$

$$\left\{\begin{array}[]{ll}\displaystyle\Delta z_{g}=0&\mbox{in $\Omega$,}\\ \\ \displaystyle z_{g}=g&\mbox{on $\partial\Omega$}\end{array}\right.$$

$$\Vert z_g{\Vert }_{H^1(G)}<ϵ.$$

$${\int }_{\partial \Omega }{\partial }_{\nu }w\cdot gds=-{\int }_{\partial D}u\cdot {\partial }_{\nu }{z}_{g}ds,$$

$$\left\{\begin{array}[]{ll}\displaystyle\Delta v=0&\mbox{in $\Omega$,}\\ \\ \displaystyle v=u&\mbox{on $\partial\Omega$.}\end{array}\right.$$

$$\tilde{u}(r\cos \theta ,r\sin \theta )=\left(r+\frac{1}{r}\right)\cos \theta .$$

$$\begin{array}[]{l}\displaystyle\,\,\,\,\,\,-\int_{\partial D}u\cdot\partial_{\nu}z_{g}\,ds\\ \\ \displaystyle=\int_{\partial D}\left(\partial_{\nu}u\cdot z_{g}-u\cdot\partial_{\nu}z_{g}\right)\,ds\\ \\ \displaystyle=\int_{\partial D}\left(\partial_{\nu}\tilde{u}\cdot z_{g}-\tilde{u}\cdot\partial_{\nu}z_{g}\right)\,ds.\end{array}$$

$${\int }_{\partial D}\left({\partial }_{\nu }\tilde{u}\cdot z_g-\tilde{u}\cdot {\partial }_{\nu }{z}_g\right)ds={\int }_C\left({\partial }_{\nu }\tilde{u}\cdot z_g-\tilde{u}\cdot {\partial }_{\nu }{z}_g\right)ds.$$

$${\int }_{\partial \Omega }{\partial }_{\nu }w\cdot gds={\int }_C\left({\partial }_{\nu }\tilde{u}\cdot z_g-\tilde{u}\cdot {\partial }_{\nu }{z}_g\right)ds.$$

$$\left|{\int }_{\partial \Omega }{\partial }_{\nu }w\cdot gds\right|\le Cϵ,$$

$${\int }_{\partial \Omega }{\partial }_{\nu }w\cdot gds={\int }_S\left({\partial }_{\nu }\tilde{u}\cdot z_g-\tilde{u}\cdot {\partial }_{\nu }{z}_g\right)ds.$$

$$\displaystyle\int_{\partial\Omega}\partial_{\nu}w\cdot g\,ds=-2\pi\nabla z_{g}(0,0)\cdot\mbox{\boldmath$e$}_{1},$$

$$\displaystyle E_{t}(x)=\log|x-t\mbox{\boldmath$e$}_{1}|$$

$$\underset{j\to \infty }{\lim }\Vert z_{g_j}-E_t{\Vert }_{H^1(G\cup B_{t/2})}=0.$$

$$\underset{j\to \infty }{\lim }{\int }_{\partial \Omega }{\partial }_{\nu }w\cdot g_{j}ds=\frac{2\pi }{t}.$$

$$\underset{j\to \infty }{\lim }\Vert z_{g_j}{\Vert }_{H^1(G)}=\Vert E_t{\Vert }_{H^1(G)}.$$

$${\tilde{g}}_j=\frac{ϵ}{2\Vert E_t{\Vert }_{H^1(G)}}{g}_j.$$

$$\Vert z_{{\tilde{g}}_j}{\Vert }_{H^1(G)}=\frac{ϵ}{2\Vert E_t{\Vert }_{H^1(G)}}\Vert z_{g_j}{\Vert }_{H^1(G)}<ϵ$$

$$\underset{j\to \infty }{\lim }{\int }_{\partial \Omega }{\partial }_{\nu }w\cdot {\tilde{g}}_{j}ds=\frac{2\pi }{t}\cdot \frac{ϵ}{2\Vert E_t{\Vert }_{H^1(G)}}$$

$${\int }_{\partial \Omega }{\partial }_{\nu }w\cdot gds={\int }_{S_{\eta }}\left({\partial }_{\nu }\tilde{u}\cdot z_g-\tilde{u}\cdot {\partial }_{\nu }{z}_g\right)ds.$$

$$\begin{array}[]{l}\displaystyle\,\,\,\,\,\,\int_{S_{\eta}}\partial_{\nu}\tilde{u}\cdot z_{g}\,ds\\ \\ \displaystyle=\left(1-\frac{1}{\eta^{2}}\right)\eta\int_{0}^{2\pi}\cos\theta\cdot z_{g}(\eta\cos\theta,\eta\sin\theta)d\theta\\ \\ \displaystyle=-\left(1-\frac{1}{\eta^{2}}\right)\eta\int_{0}^{2\pi}\sin\theta\cdot\frac{d}{d\theta}\left\{z_{g}(\eta\cos\theta,\eta\sin\theta)\right\}d\theta\\ \\ \displaystyle=-\left(1-\frac{1}{\eta^{2}}\right)\eta^{2}\int_{0}^{2\pi}\sin\theta\cdot\nabla z_{g}(\eta\cos\theta,\eta\sin\theta)\cdot(-\sin\theta,\cos\theta)^{T}\,d\theta\\ \\ \displaystyle\rightarrow\int_{0}^{2\pi}\sin\theta\cdot\nabla z_{g}(0,0)\cdot(-\sin\theta,\cos\theta)^{T}\,d\theta\\ \\ \displaystyle=-\pi\nabla z_{g}(0,0)\cdot\mbox{\boldmath$e$}_{1}.\end{array}$$

$$\begin{array}[]{l}\displaystyle\,\,\,\,\,\,\int_{S_{\eta}}\tilde{u}\cdot\partial_{\nu}z_{g}\,ds\\ \\ \displaystyle=(\eta^{2}+1)\int_{0}^{2\pi}\cos\theta\cdot\nabla z_{g}(\eta\cos\theta,\eta\sin\theta)\cdot(\cos\theta,\sin\theta)^{T}\,d\theta\\ \\ \displaystyle\rightarrow\pi\nabla z_{g}(0,0)\cdot\mbox{\boldmath$e$}_{1}.\end{array}$$

$${\partial }_{3}E(x)=-\frac{x_3}{|x{|}^3},$$

$${\partial }_3^{2}E(x)=\frac{1}{|x{|}^5}(3x_3^2-|x{|}^2).$$

$$\displaystyle\partial_{\nu_{x}}F_{\mbox{\boldmath$a$}}(x,y)=-\partial_{3}^{2}E(x-y)$$

$$\displaystyle\partial_{\nu_{x}}F_{\mbox{\boldmath$a$}}(x,y)=-\frac{1}{|x-y|^{5}}(2y_{3}^{2}-x_{1}^{2}-x_{2}^{2}).$$

$$\displaystyle\partial_{\nu_{x}}F_{\mbox{\boldmath$a$}}(x,y)\sim-\partial_{\nu_{x_{0}}}^{2}E(x-y).$$

$${\partial }_{{\nu }_{x_0}}^{2}E(x-y)=-({\partial }_{x_1}^2+{\partial }_{x_2}^2)E(x-y),$$

$$\displaystyle\partial_{\nu_{x}}F_{\mbox{\boldmath$a$}}(x,y)\sim(\partial_{x_{1}}^{2}+\partial_{x_{2}}^{2})E(x-y).$$

Full text

AMS: 35R30

KEY WORDS: No response test, enclosure method, probe method

1 A counter example

In [6] they state222Please refer to their paper [6] for the symbols used in this note without explanation. if $\overline{D}\not\subset\overline{G}$, then $I(G)=\infty$. However, in this note we give a simple example that $\displaystyle\overline{D}\not\subset\overline{G}$, however $I(G)=0$.

Let $\Omega=\{x\in{\rm\bf R}^{2}|\,|x|<R\}$ with $R>1$ and $D=\{x\in{\rm\bf R}^{2}\,|\,|x|<1\}$. Let $u$ solve

[TABLE]

Note that the solution has the explict form

[TABLE]

The key point of this note is the following trivial fact: $u$ has an extension to the domain $\tilde{\Omega}=\{x\in{\rm\bf R}^{2}|\,0<|x|<R\}=\Omega\setminus\{0\}$ as a solution of the Laplace equation.

Let $0<\delta<1$ and choose $G=\{x\in{\rm\bf R}^{2}\,|\,|x|<1-\delta\}$. We have $\overline{G}\subset D$ and thus $\overline{D}\not\subset\overline{G}$.

Given $\epsilon>0$ let $g\in H^{1/2}(\partial\Omega)$ be an arbitrary function such that the solution $z_{g}$ of

[TABLE]

satisfies

[TABLE]

By Lemma 2.1 in [6] we have

[TABLE]

where $w=u-v$ and $v$ solves

[TABLE]

Let $\tilde{u}$ denote the harmonic extension of $u$ into $\tilde{\Omega}$, that is

[TABLE]

Let $C=\{x\in{\rm\bf R}^{2}\,|\,|x|=1-\delta^{\prime}\}$ with $\delta<\delta^{\prime}<1$. We have $C\subset G$.

Write

[TABLE]

Since $\tilde{u}$ and $z_{g}$ are harmonic in $1-\delta^{\prime}<|x|<1$, one has the expression

[TABLE]

Thus (1.2) becomes

[TABLE]

It is easy to see that this right-hand side has the bound $O(\|z_{g}\|_{H^{1}(G)})$. Thus the condition (1.1) yields

[TABLE]

where $C$ is independent of $g$. Hence $I_{\epsilon}(G)\leq C\epsilon$ and $I(G)=\lim_{\epsilon\downarrow 0}I_{\epsilon}(G)=0$.

2 Looking at the example in Section 1 a little more

Let $u$ be the solution of (1.0) and $\tilde{u}$ its harmonic extension to $\tilde{\Omega}$. In this section $G$ denotes an arbitrary open subset of $\Omega$ such that $\overline{G}\subset\Omega$ and $\Omega\setminus\overline{G}$ is connected. In this section we prove

Proposition 2.1.**

(a) If $(0,0)\in G$, then $I(G)=0$.

(b) If $(0,0)\not\in\overline{G}$, then, for all $\epsilon$ $I_{\epsilon}(G)=\infty$.

Proof. First we prove (a). In this case one can find a cirecle $S$ centered at $(0,0)$ such that $S\subset G$. At this time, the following equation is obtained as in the previous section:

[TABLE]

Note that $z_{g}$ is the same as before. Thus this together with (1.2) yield $I_{\epsilon}(G)\leq C\epsilon$ with a positive constant $C$ independent of $g$. And hence $I(G)=\lim_{\epsilon\downarrow 0}I_{\epsilon}(G)=0$.

Next we prove (b). For this we claim the identity:

[TABLE]

where \mbox{\boldmathe}_{1}=(1,0)^{T}.

First of all admit equation (2.1) and move on. Consider the case $(0,0)\not\in\overline{G}$. One can find an open disc $B$ centered at $(0,0)$ and radius $t_{0}$ such that $\overline{B}\subset\Omega\setminus\overline{G}$. Let $B_{t}=\{x\in{\rm\bf R}^{2}\,|\,|x|<t\}$ with $0<t<t_{0}$. Since the function

[TABLE]

is harmonic in a neighbourhood of $\overline{G}\cup\overline{B_{t/2}}$, the Runge approximation property yields: there exists a sequence $\{g_{j}\}$ such that

[TABLE]

Then an interior regulerity estimate yields $z_{g_{j}}$ together with its all derivatives converges to $E_{t}$ and the corresponding derivatives compact uniformly in $B_{t/2}$. Thus (2.1) yields

[TABLE]

Note also that we have

[TABLE]

Given $\epsilon>0$ define

[TABLE]

Since the map $g\mapsto z_{g}$ is linear, we have

[TABLE]

for all $j>>1$.

And (2.3) gives

[TABLE]

Since $\overline{B}\cap\overline{G}=\emptyset$, Lebesgue’s dominated convergence theorem gives $\lim_{t\downarrow 0}\|E_{t}\|_{H^{1}(G)}=\|E_{0}\|_{H^{2}(G)}<\infty$. Thus the right-hand side on (2.4) blows up as $t\downarrow 0$. This yields $I_{\epsilon}(G)=\infty$.

$\Box$

Remarks.

(i) The case $(0,0)\in\partial G$ seems delicate (at the present time).

(ii) This type of sequence satisfying (2.2) has been used in the probe method [2] which aims at reconstructing unknown discontinuities such as cavities, inclusions and cracks. However, the probe method employs the Dirichlet-to-Neumann map, i.e., infinitely many pairs of the Cauchy data of the governing equation. Instead in the proof of (b) a single pair of Cauchy data is fixed and sequences $z_{g_{j}}$ produced by infinitely many $g_{j}$ are used as test functions.

(iii) The choices of $\{g_{j}\}$ in two cases (a) and (b) are different. Since we do not know the position of $\{(0,0)\}$ in advance, we have the question: what is the good choice of $\{g_{j}\}$ common to two cases. This is also a problem about the no response test.

2.1 Proof of (2.1)

Same as before, we have, for all circles $S_{\eta}$ centered at $(0,0)$ with radius $\eta\in\,]0,\,1[$

[TABLE]

We compute the limt of this right-hand side as $\eta\downarrow 0$.

First we have

[TABLE]

Second we have

[TABLE]

This completes the proof.

3 One can not apply Fatou’ s lemma

The key point of their argument on page 5 is the definiteness of the signature of \partial_{\nu_{x}}F_{\mbox{\boldmatha}}(x,y) for $x\in N_{y_{0}}\cap\partial D$ and $y\rightarrow y_{0}$ along the axis of the cylinder $N_{y_{0}}$. Here we give an example of $D$ that does not ensure this property.

Let $D$ be a bounded domain and in $x_{3}<0$. We assume that $y_{0}=(0,0,0)\in\partial D$ and $N_{y_{0}}\cap\partial D$ is flat and included in the plane $x_{3}=0$. Thus \nu_{x}=\nu_{y_{0}}=\mbox{\boldmathe}_{3}.

Let $E(x)=\frac{1}{|x|}$. We have

[TABLE]

and

[TABLE]

Since \mbox{\boldmatha}=\nu_{y_{0}}=\mbox{\boldmathe}_{3}, we have, for all $x\in N_{y_{0}}\cap\partial D$ and $y=(0,0,y_{3})$ with $0<y_{3}<<1$

[TABLE]

and thus

[TABLE]

Therefore we have

(i) if $x_{1}^{2}+x_{2}^{2}<2y_{3}^{2}$, then \partial_{\nu_{x}}F_{\mbox{\boldmatha}}(x,y)<0;

(ii) if $x_{1}^{2}+x_{2}^{2}>2y_{3}^{2}$, then \partial_{\nu_{x}}F_{\mbox{\boldmatha}}(x,y)>0.

Thus as $y_{3}\downarrow 0$ the sign of the function \partial_{\nu_{x}}F_{\mbox{\boldmatha}}(x,y) of $x\in N_{y_{0}}\cap\partial D$ can not have a definite sign.

This implies, one can not apply Fatou’s lemma as done (3.4) in this simplest case.

4 Another reason of invalidness of (3.5) on page 5: A heuristic explanation

Even general case one can not obtain (3.5). Its heuristic explanation is the following.

Since \mbox{\boldmatha}=\nu_{y_{0}}, if $x\in N_{y_{0}}\cap\partial D$ we expect

[TABLE]

However, $E$ satisfies the Laplace equation we have

[TABLE]

where $x_{1}$ and $x_{2}$ are tangential directions at $y_{0}$. Thus we can expect

[TABLE]

Then the integral

[TABLE]

may become

[TABLE]

Then applying integration by parts to this right-hand, one can reduce the singularity of integrand twice and gets an integral and additional terms which are bounded as $y\rightarrow y_{0}$.

5 Some comments on references

In [3] (1999!) using a single set of the Cauchy data, we have already given the reconstruction formula of the convex hull of unknown polygonal cavity $D$ and done its numerical testing in [5]. The method developed in this paper is called the enclosure method and based on the asymptotic behaviour of the integral with respect to a large parameter $\tau$

[TABLE]

where $g=e^{\tau x\cdot(\omega+i\omega^{\perp})}$ with two unit vectors $\omega$ and $\omega^{\perp}$ perpendicular each other. Note that in this case $z_{g}(x)=e^{\tau x\cdot(\omega+i\omega^{\perp})}$.

Besides, in the case when $\Omega$ is an ellipse, even though the homogeneous background is unknown, the enclosure method works and yields a reconstruction formula of the convex hull of the union of the polygonal cavity and the focal points of $\Omega$ by using a single flux corresponding to a band-limited surface potential [4].

These informations are missed in [6].

6 Extendability

The point is the extendability of the potential $u$ from $\Omega\setminus\overline{D}$ across $\partial D$ into $D$, for example, if $\partial D$ is a real analytic surface, then by applying the Cauchy-Kovalevskaya theorem one has such an extension locally. In this case, we can prove that, by doing the procedure above locally around $y_{0}\in\partial D\setminus\overline{G}$ on page 5 in [6], (3.5) in [6] is not valid. The enclosure method in [3] catches a corner where one can not have an extention of the potential (due to Friedman-Isakov’s extension argument [1] under the condition $\mbox{diam}\,D<\mbox{dist}\,(D,\partial\Omega)$).

So at least we have to find an argument that employs explicitly the impossibility of applying the Cauchy-Kovalevskaya theorem on $\partial D$.

7 Conclusion

The problem is not simple and still unsolved! I guess the complete version of the no response test with a single measurement tells us the limt of the extension of the soultion (continuation as a solution of the governing equation). Proposition 2.1 is an evidence of this belief.

[TABLE]

Acknowledgments

The author was partially supported by Grant-in-Aid for Scientific Research (C)(No. 17K05331) and (B)(No. 18H01126) of Japan Society for the Promotion of Science.

[TABLE]