Absorbing boundary conditions for the two-dimensional Schrödinger equation with an exterior potential

Abstract

Four families of ABCs where built in Antoine et al. (Math Models Methods Appl Sci, 22(10), 2012) for the two-dimensional linear Schrödinger equation with time and space dependent potentials and for general smooth convex fictitious surfaces. The aim of this paper is to propose some suitable discretization schemes of these ABCs and to prove some semi-discrete stability results. Furthermore, the full numerical discretization of the corresponding initial boundary value problems is considered and simulations are provided to compare the accuracy of the different ABCs.

Appendix: \(\mathcal Z \)-transform: technical annex

For the sake of clarity, we precise some notations and results about the $\mathcal Z$-transform of a discrete signal [14].

Definition 2

Let $(f_n)_{n\in \mathbb{N }}$ be a discrete signal. We call $\mathcal Z$-transform of $(f_n)$, and we denote by $\mathcal Z (f_n)$ or $\hat{f}$, the function of the $z$ variable defined by

$$\begin{aligned} \hat{f}(z) = \mathcal Z (f_n)(z) = \sum _{n=0}^{+\infty } f_n z^{-n}, \qquad for\,|z| > \hat{R_f}, \end{aligned}$$

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where $\hat{R_f}$ denotes the convergence radius of the series $\hat{f}$ which is defined by

$$\begin{aligned} \hat{R_f} = \inf \left\{ R>0 \ ; \ \sum _n f_n R^{-n} < +\infty \right\} . \end{aligned}$$

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Thereby, $\hat{R_f}$ is the inverse of the convergence radius of the power series $\sum f_n z^n$.

We denote by $\star$ the usual convolution product

$$\begin{aligned} {\displaystyle }a_n \star b^n = (a\star b )_n = \sum _{k=0}^n a_k b^{n-k}. \end{aligned}$$

Let us recall some classical properties of the $\mathcal Z$ transform.

Proposition 3

Let $(f_n)_{n\in \mathbb{N }}$ and $(g_n)_{n\in \mathbb{N }}$ be two discrete signals with convergence radius $\hat{R}_f$ and $\hat{R}_g$, respectively. Then, the following results hold

• 1. $\mathcal Z (f_{n+1}) = z\hat{f} - zf(0)$,

• 2. $\mathcal Z (f_{n+1} \pm f_n) = (z\pm 1)\hat{f}(z) - zf(0)$,

• 3. $\mathcal Z (f_n\star g_n) = \hat{f}(z) \hat{g}(z)$,    for $|z|> \max \, (\hat{R_f}, \hat{R_g})$.

We also have the following lemma used in the stability proofs of the paper.

Lemma 1

Let $(u_p)_{p\in \mathbb{N }}$ and $(h_p)_{p\in \mathbb{N }}$ be two sequences. We define the sequence $(y_p)_{p\in \mathbb{N }}$ by

$$\begin{aligned} y_p=\sum _{k=0}^p h_k u_{p-k}, \end{aligned}$$

and by $\hat{h}$ the $\mathcal Z$-transform of $(h_p)_{p\in \mathbb{N }}$, for which we assume that $R_{\hat{h}} \ge 1$. Let $\mathbb{H }(\mathbb{E })$ be the Hardy space on $\mathbb{E }=\left\{ z\in \mathbb{C }, |z|>1\right\}$

$$\begin{aligned} \mathbb{H }(\mathbb{E }) = \left\{ \mathcal H \, holomorphic\; on \, \mathbb{E }\, s.t. \, \sup _{r>1}\int _{-\pi }^{\pi } \left| \mathcal H (re^{i\omega })\right| \, d\omega < +\infty \right\} . \end{aligned}$$

If $\hat{h} \in \mathbb{H }(\mathbb{E })$, then one has

$$\begin{aligned} \sum _{p=0}^n \overline{u_p}y_p=\frac{1}{2\pi } \int _{-\pi }^{\pi } \hat{h}(e^{i\omega }) \left| \sum _{p=0}^n u_p e^{-i\omega p} \right| ^2 \, d\omega . \end{aligned}$$

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Proof

Let us define, for $\rho \ge 1,\,y_p(\rho )=\sum _{k=0}^p h_k \rho ^{-k}u_{p-k}$. We fix $n<\infty$ and consider the Laurent polynomials $\hat{y}_\rho (z):=\sum _{p=0}^n y_p(\rho )z^{-p}$ and $\hat{u}(z):=\sum _{p=0}^n u_pz^{-p}$. By using the Cauchy product, one has for all $z$ s.t. $|z|>\rho$

$$\begin{aligned} \hat{h}(\rho z) \cdot \hat{u}(z) = \hat{h}_{\rho }(z) + \sum _{p=n+1}^\infty \left( \sum _{k=p-n}^p h_k \rho ^{-k} u_{p-k}\right) z^{-p}. \end{aligned}$$

In particular, this is true for the unit circle. We compute the $L^2$ scalar product on the unit circle for the measure $\frac{1}{2\pi } d\omega$. The orthogonality of $z^p$ implies that

$$\begin{aligned} \left\langle \hat{u}, \hat{h}\cdot \hat{u}\right\rangle = \left\langle \hat{u},\hat{y}_\rho \right\rangle = \sum _{p=0}^n \overline{u_p}y_p(\rho ). \end{aligned}$$

The left hand side of this equality is reduced to

$$\begin{aligned} \frac{1}{2\pi } \int _{-\pi }^\pi \hat{h}(\rho e^{i\omega }) \left| \sum _{p=0}^n u_p e^{-i\omega p}\right| ^2 \, d\omega . \end{aligned}$$

But, $\hat{h}(\rho e^{i\omega })$ converges to $\hat{h}(e^{i\omega })$ in $L^1$ when $\rho \rightarrow 1^+$. Therefore, since $\lim _{\rho \rightarrow 1^+}y_p(\rho )=y_p$, this ends the proof of Lemma 1. $\square$

This lemma is mainly used in the following result.

Lemma 2

Let $(\alpha _n)_n,\,(\beta _n)_n$ and $(\gamma _n)_n$ be the sequences given by (31), and $(\varphi ^k)_{k\in \mathbb{N }}$ a sequence of complex numbers. We have the following properties:

$$\begin{aligned} Q_{\alpha }&= \sum _{p=0}^n \overline{\varphi ^p} \sum _{k=0}^p \alpha _{p-k} \varphi ^k \quad \in e^{i\pi /4}\mathbb{R }^+ \cup e^{-i\pi /4}\mathbb{R }^+, \end{aligned}$$

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$$\begin{aligned} Q_{\beta }&= \sum _{p=0}^n \overline{\varphi ^p} \sum _{k=0}^p \beta _{p-k} \varphi ^k \quad \in e^{i\pi /4}\mathbb{R }^+ \cup e^{-i\pi /4}\mathbb{R }^+, \end{aligned}$$

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$$\begin{aligned} Q_{\gamma }&= \sum _{p=0}^{n} \overline{\varphi ^{p}} \sum _{k=0}^{p} \gamma _{p-k} \varphi ^k \quad \in \{{}\mathrm{Re}(z)\ge 0\}. \end{aligned}$$

(82)

Proof

The proof of the result for the terms $Q_{\alpha }$ and $Q_{\beta }$ mainly relies on Lemma 1 (see Annex 5). Let us consider here $Q_{\alpha }$ (the proof is similar for $Q_{\beta }$). The $\mathcal Z$-transform of the sequence $(\alpha _n)_n$ evaluated on the unit circle for $\omega \in (-\pi ,\pi )$ is given by $\hat{\alpha }(e^{i\omega })=\sqrt{\frac{e^{i\omega }+1}{e^{i\omega }-1}} \in L^1(-\pi ,\pi )$. It is easy to see that $\hat{\alpha }\in \mathbb H (\mathbb E )$. Therefore, Lemma 1 holds and we have

$$\begin{aligned} Q_{\alpha } = \frac{1}{2\pi } \int _{-\pi }^{\pi } \sqrt{\frac{e^{i\omega }+1}{e^{i\omega }-1}} \, \left| \sum _{n=0}^P v^{n}e^{-i\omega n}\right| ^2 d\omega . \end{aligned}$$

But for $\omega \in (-\pi ;\pi )$, one has $\sqrt{\frac{e^{i\omega }+1}{e^{i\omega }-1}} = \sqrt{i\tan \left( \frac{\omega }{2}\right) }$. Hence, the application $z\mapsto \sqrt{\frac{z+1}{z-1}}$ maps the unit circle onto $e^{i\pi /4}\mathbb{R }^+\cup e^{-i\pi /4}\mathbb{R }^+$. This implies that

$$\begin{aligned} Q_{\alpha } \in e^{i\pi /4}\mathbb{R }^+\cup e^{-i\pi /4}\mathbb{R }^+. \end{aligned}$$

This proof cannot be extended to $Q_{\gamma }$ since the $\mathcal Z$-transform of the sequence $(\gamma _n)_n$ evaluated on the unit circle for $\omega \in (-\pi ,\pi )$ does not belong to $\mathbb{H }(\mathbb{E })$. We therefore proceed in a different way. The term $Q_{\gamma }$ can be interpreted as an hermitian form

$$\begin{aligned} Q_{\gamma } = \sum _{p=0}^{n} \overline{\varphi ^{p}} \left( \gamma _p\star \varphi ^p\right) = {}^t\overline{\varvec{\varphi }} A \varvec{\varphi } = \langle \varvec{\varphi }, A \varvec{\varphi } \rangle \end{aligned}$$

where $\varvec{\varphi }$ is the vector with size $n+1$ and complex coefficients $\varvec{\varphi } = \left( \varphi _0, \cdots , \varphi _n\right) ^T$ and $A$ designates the real coefficients matrix of size $(n+1)\times (n+1)$ defined by

$$\begin{aligned} A = \begin{pmatrix} 1 &{} 0 &{} \dots &{} \dots &{} 0 \\ 2 &{} 1 &{} \ddots &{} &{} \vdots \\ 2 &{} 2 &{} 1 &{} \ddots &{} \vdots \\ \vdots &{} &{} \ddots &{} \ddots &{} 0 \\ 2 &{} \dots &{} \dots &{} 2 &{} 1 \end{pmatrix}. \end{aligned}$$

Since $A$ is positive, for any real valued vector $\mathbf x$ we have

$$\begin{aligned} \langle \mathbf x , A\mathbf x \rangle \ge 0. \end{aligned}$$

We now decompose the complex valued vector $\varvec{\varphi }$ as $\varvec{\varphi }=\mathbf x +i\mathbf y$, with $\mathbf x$ and $\mathbf y$ two real valued vectors. We compute the hermitian product

$$\begin{aligned} Q_{\gamma } = \langle \varvec{\varphi },A\varvec{\varphi } \rangle = {}&\langle \mathbf x ,A\mathbf x \rangle + \langle \mathbf y ,A\mathbf y \rangle + i \big [ \langle \mathbf x ,A\mathbf y \rangle - \langle \mathbf y ,A\mathbf x \rangle \big ]. \end{aligned}$$

Then we have

$$\begin{aligned} \text{ Re }(Q_{\gamma }) = \langle \mathbf x ,A\mathbf x \rangle + \langle \mathbf y ,A\mathbf y \rangle \ge 0, \end{aligned}$$

and

$$\begin{aligned} \text{ Im }(Q_{\gamma }) = \langle \mathbf x ,A\mathbf y \rangle - \langle \mathbf y ,A\mathbf x \rangle , \end{aligned}$$

this term being non null if $\mathbf x$ or $\mathbf y$ are not equal to zero since $A$ is not symmetric. Consequently, we have

$$\begin{aligned} Q_{\gamma } \in \{{}\text{ Re }(z)\ge 0\}. \end{aligned}$$