Uniqueness for the inverse boundary value problem with singular potentials in 2D

Abstract

In this paper we consider the inverse boundary value problem for the Schrödinger equation with potential in $L^p$ class, $p>4/3$. We show that the potential is uniquely determined by the boundary measurements.

Appendices

Appendix 1: Cauchy operator and integration by parts

We define the two fundamental tools for solving the two-dimensional inverse problem of the Schrödinger operator in this section: the Cauchy operators and an integration by parts formula for the Cauchy operator conjugated by an exponential. These were used by Bukhgeim [7] for solving the problem.

Definition 6.1

Let $u \in \mathscr {E}'(\mathbb {R}^2)$ be a compactly supported distribution. Then we define the Cauchy operators by

$$\begin{aligned} {\overline{\partial }^{-1}}u = \frac{1}{\pi z} *u, \qquad {\partial ^{-1}}u = \frac{1}{\pi \overline{z}} *u. \end{aligned}$$

Remark 6.2

The notations $\overline{\partial }^{-1}$ and $\partial ^{-1}$ cause no problems because $1/(\pi z)$ and $1/(\pi \overline{z})$ are the fundamental solutions to the operators $\overline{\partial } = (\partial _1 + i\partial _2)/2$ and $\partial = (\partial _1 - i\partial _2)/2$.

Lemma 6.3

Let $\tau > 0$, $z_0 \in \mathbb {C}$ and $\Phi (z) = (z-z_0)^2$. Let $\psi \in C^\infty _0(\mathbb {R}^2)$ with $\psi \equiv 1$ in a neighbourhood of 0, and write

$$\begin{aligned} \psi _\tau (z) = \psi ( \tau ^{1/2}(z-z_0) ), \qquad h(z) = \frac{1 - \psi _\tau (z) }{\overline{z}-\overline{z_0}}. \end{aligned}$$

Then for $a \in C^\infty _0(\mathbb {R}^2)$ we have the integration by parts formula

$$\begin{aligned}&{\overline{\partial }^{-1}}(e^{-i\tau (\Phi + \overline{\Phi })}a) = {\overline{\partial }^{-1}}\big (e^{-i\tau (\Phi + \overline{\Phi })} \psi _\tau a\big ) \\&\qquad - \frac{1}{2i\tau } \big ( e^{-i\tau (\Phi + \overline{\Phi })} h a - {\overline{\partial }^{-1}}(e^{-i\tau (\Phi + \overline{\Phi })} \overline{\partial }h a) - {\overline{\partial }^{-1}}( e^{-i\tau (\Phi + \overline{\Phi })} h \overline{\partial }a) \big ). \end{aligned}$$

If we had set $h(z) = (1-\psi _\tau (z))/(z-z_0)$ instead then

$$\begin{aligned}&{\partial ^{-1}}(e^{-i\tau (\Phi + \overline{\Phi })} a) = {\partial ^{-1}}\big (e^{-i\tau (\Phi + \overline{\Phi })} \psi a\big )\\&\qquad - \frac{1}{2i\tau }\big ( e^{-i\tau (\Phi + \overline{\Phi })} h a - {\partial ^{-1}}(e^{-i\tau (\Phi + \overline{\Phi })} \partial h a) - {\partial ^{-1}}(e^{-i\tau (\Phi + \overline{\Phi })} h \partial a)\big ). \end{aligned}$$

Proof

The proof follows by differentiating $e^{-i\tau (\Phi +\overline{\Phi })} h a$ and noting that by Remark 6.2 the operators ${\overline{\partial }^{-1}}\overline{\partial }$ and ${\partial ^{-1}}\partial$ are the identity on compactly supported distributions.

Lemma 6.4

Let $X \subset \mathbb {R}^2$ be a bounded domain and $1<p<\infty$. Then the Cauchy operators ${\overline{\partial }^{-1}}$ and ${\partial ^{-1}}$ are bounded $L^p(X) \rightarrow W^{1,p}(X)$.

Proof

If $f\in L^p(X)$ we extend it by zero to $\mathbb {R}^2{\setminus } X$ to create a compactly supported distribution and thus ${\overline{\partial }^{-1}}f$ is well defined by Definition 6.1. The convolution kernel $1/(\pi z)$ is locally integrable, so by Young’s inequality

$$\begin{aligned} \left\Vert {\overline{\partial }^{-1}}f \right\Vert _{L^p(X)} \le C \left\Vert f \right\Vert _{L^p(X)}, \end{aligned}$$

because in essence ${\overline{\partial }^{-1}}f$ has the same values in X as the convolution of f with the kernel $\chi _{X-X}(z)/(\pi z)$, where $X-X = \{ z \in \mathbb {R}^2 \mid z = z_1-z_2, z_j \in \mathbb {R}^2\}$.

For the derivatives note that by Remark 6.2 we have $\overline{\partial }{\overline{\partial }^{-1}}f = f$. On the other hand $\partial {\overline{\partial }^{-1}}f = \Pi f$ which is the Beurling transform, and hence bounded $L^p(X) \rightarrow L^p(X)$. For reference see for example Section 4.5.2 in [2] or [24] for a more classical approach. $\square$

Appendix 2: Cut-off function estimates

This section contains all the technical cut-off function construction and norm estimates used in the paper.

Lemma 7.1

Let $\psi \in C^\infty _0(\mathbb {R}^2)$. For $z_0\in \mathbb {R}^2$ and $\tau >0$ write $\psi _\tau (z) = \psi ( \tau ^{1/2}(z-z_0) )$. Then, given any vector $v\in \mathbb {C}^2$, we have

$$\begin{aligned} \left\Vert \psi _\tau \right\Vert _{L^p(\mathbb {R}^2)} = \left\Vert \psi \right\Vert _{L^p(\mathbb {R}^2)} \tau ^{-1/p}, \qquad \left\Vert v\cdot \nabla \psi _\tau \right\Vert _{L^p(\mathbb {R}^2)} = \left\Vert v\cdot \nabla \psi \right\Vert _{L^p(\mathbb {R}^2)} \tau ^{1/2-1/p} \end{aligned}$$

for $1\le p \le \infty$.

Proof

This follows directly from the scaling properties and translation invariance of $L^p$-norms in $\mathbb {R}^2$. $\square$

Lemma 7.2

Let $\tau >0$ and set $\mathbb {R}^2_\tau = \mathbb {R}^2 {\setminus } B(0,\tau ^{-1/2})$. Then

$$\begin{aligned} \left\Vert z^{-a} \right\Vert _{L^p(\mathbb {R}^2_\tau )} = \left( \frac{2\pi }{ap-2} \right) ^{1/p} \tau ^{a/2-1/p} \end{aligned}$$

for $a>0$ and $2/a<p\le \infty$.

Proof

This is a direct computation using the polar coordinates integral transform $\int _{\mathbb {R}^2_\tau } \ldots dz = \int _{\tau ^{-1/2}}^\infty \int _{\mathbb S^1} \ldots d\sigma (\theta ) r dr$, with $z = r\theta$. $\square$

Lemma 7.3

Let $\psi \in C^\infty _0(\mathbb {R}^2)$ be a test function supported in B(0, 2) with $0\le \psi \le 1$ and $\psi \equiv 1$ in B(0, 1). For $\tau >0$ and $z_0\in \mathbb {R}^2$ write $\psi _\tau (z) = \psi ( \tau ^{1/2}(z-z_0) )$. Let $h(z) = (1 - \psi _\tau (z)) / (\overline{z} - \overline{z_0})$. Then

$$\begin{aligned} \left\Vert h \right\Vert _{L^p(\mathbb {R}^2)} \le C_p \tau ^{1/2-1/p} \end{aligned}$$

for $C_p<\infty$ when $2<p\le \infty$ and for any complex vector $v\in \mathbb {C}^2$ we have

$$\begin{aligned} \left\Vert v\cdot \nabla h \right\Vert _{L^p(\mathbb {R}^2)}\le C_{\psi ,p,v} \tau ^{1-1/p} \end{aligned}$$

for $C_{\psi ,p,v}<\infty$ when $1\le p \le \infty$. The same conclusions hold if we had defined h by dividing $1-\psi _\tau$ by $z-z_0$ instead of its complex conjugate.

Proof

For the first claim note that $\left|h(z) \right| \le \left|z-z_0 \right|^{-1}$ and ${{\,\mathrm{supp}\,}}h \subset \mathbb {R}^2_\tau + z_0 = \mathbb {R}^2 {\setminus } B(z_0,\tau ^{-1/2})$. Hence $\left\Vert h \right\Vert _{L^p(\mathbb {R}^2)} \le \left\Vert z^{-1} \right\Vert _{L^p(\mathbb {R}^2_\tau )}$ and Lemma 7.2 takes care of the first estimate.

For the second estimate

$$\begin{aligned} v\cdot \nabla h(z) = \frac{v\cdot \nabla \psi _\tau (z)}{\overline{z}-\overline{z_0}} - \frac{1-\psi _\tau (z)}{(\overline{z}-\overline{z_0})^2}. \end{aligned}$$

The $L^p$-norm of the first term is bounded by $\left\Vert v\cdot \nabla \psi _\tau \right\Vert _{L^p} \left\Vert z^{-1} \right\Vert _{L^\infty (\mathbb {R}^2_\tau )}$ which is at most $C_{\psi ,p,v} \tau ^{1-1/p}$ according to Lemmas 7.1 and 7.2. The second term is supported in $\mathbb {R}^2{\setminus } B(z_0,\tau ^{-1/2})$ and bounded pointwise by $\left|z-z_0 \right|^{-2}$. Hence, as in the first paragraph, it has the required bound.