Gaussian Beam Ansatz for Finite Difference Wave Equations

Abstract

This work is concerned with the construction of Gaussian Beam (GB) solutions for the numerical approximation of wave equations, semi-discretized in space by finite difference schemes. GB are high-frequency solutions whose propagation can be described, both at the continuous and at the semi-discrete levels, by microlocal tools along the bi-characteristics of the corresponding Hamiltonian. Their dynamics differ in the continuous and the semi-discrete setting, because of the high-frequency gap between the Hamiltonians. In particular, numerical high-frequency solutions can exhibit spurious pathological behaviors, such as lack of propagation in space, contrary to the classical space-time propagation properties of continuous waves. This gap between the behavior of continuous and numerical waves introduces also significant analytical difficulties, since classical GB constructions cannot be immediately extrapolated to the finite difference setting, and need to be properly tailored to accurately detect the propagation properties in discrete media. Our main objective in this paper is to present a general and rigorous construction of the GB ansatz for finite difference wave equations, and corroborate this construction through accurate numerical simulations.

Appendices

Appendix A. Proof of Theorem 4.1

We give here the proof of Theorem 4.1, concerning the construction of a GB ansatz for the wave equation (1.1). To this end, we shall first need the following technical result.

Proposition A.1

Let ${\varvec{x}}_0\in \mathbb {R}^d$, $N\in \mathbb {N}$ and $f\in L^\infty (\mathbb {R}^d)$ be a function satisfying

$$\begin{aligned} |{\varvec{x}}-{\varvec{x}}_0|^{-N} f({\varvec{x}})\in L^\infty (\mathbb {R}^d). \end{aligned}$$

(A.1)

Then, for any positive constant $0<\beta \in \mathbb {R}$, we have

$$\begin{aligned} \int _{\mathbb {R}^d} \left| f({\varvec{x}})e^{-k\beta |{\varvec{x}}-{\varvec{x}}_0|^2}\right| ^2\,\textrm{d}{\varvec{x}}\le \mathcal Ck^{-\frac{d}{2}-N} \end{aligned}$$

(A.2)

for some $\mathcal C= \mathcal C(d,N,\beta ) > 0$ that does not depend on k.

Proof

Using (A.1), we have that there exists a function $g\in L^\infty (\mathbb {R}^d)$ such that

$$\begin{aligned} f({\varvec{x}}) = |{\varvec{x}}-{\varvec{x}}_0|^N g({\varvec{x}}). \end{aligned}$$

In view of this, we can apply the Hölder inequality to estimate

$$\begin{aligned} \int _{\mathbb {R}^d} \left| f({\varvec{x}})e^{-k\beta |{\varvec{x}}-{\varvec{x}}_0|^2}\right| ^2\,\textrm{d}{\varvec{x}}&= \int _{\mathbb {R}^d} \left| |{\varvec{x}}-{\varvec{x}}_0|^Ng({\varvec{x}})e^{-k\beta |{\varvec{x}}-{\varvec{x}}_0|^2}\right| ^2\,\textrm{d}{\varvec{x}}\\&\le \mathcal C(d) \int _{\mathbb {R}^d} |{\varvec{x}}-{\varvec{x}}_0|^{2N} e^{-2k\beta |{\varvec{x}}-{\varvec{x}}_0|^2}\,\textrm{d}{\varvec{x}}. \end{aligned}$$

Now, by means of the change of variable $\varvec{y} = \sqrt{2k\beta }({\varvec{x}}-{\varvec{x}}_0)$, we obtain that

$$\begin{aligned} \int _{\mathbb {R}^d} |{\varvec{x}}-{\varvec{x}}_0|^{2N} e^{-2k\beta |{\varvec{x}}-{\varvec{x}}_0|^2}\,\textrm{d}{\varvec{x}}= (2k\beta )^{-\frac{d}{2}-N} \int _{\mathbb {R}^d} |\varvec{y}|^{2N} e^{-|\varvec{y}|^2}\,\textrm{d}\varvec{y}. \end{aligned}$$

Finally, by employing polar coordinates, we can compute

$$\begin{aligned} \int _{\mathbb {R}^d} |\varvec{y}|^{2N} e^{-|\varvec{y}|^2}\,\textrm{d}\varvec{y}&= \int _{\mathbb {S}^{d-1}} \int _0^{+\infty } r^{2N+d-1} e^{-r^2}\,\textrm{d}r d\sigma = \frac{1}{2}|\mathbb {S}^{d-1}| \int _0^{+\infty } \rho ^{N+\frac{d}{2}-1} e^{-\rho }\,\textrm{d}\rho \\&= \frac{d\pi ^{\frac{d}{2}}}{2\Gamma \left( \frac{d}{2}+1\right) } \int _0^{+\infty } \rho ^{N+\frac{d}{2}-1} e^{-\rho }\,\textrm{d}\rho = \frac{d\pi ^{\frac{d}{2}}}{2}\frac{\Gamma \left( \frac{d}{2}+N\right) }{\Gamma \left( \frac{d}{2}+1\right) }, \end{aligned}$$

where $\Gamma$ denotes the Euler gamma function. Putting everything together, we finally obtain that

$$\begin{aligned} \int _{\mathbb {R}^d} \left| f({\varvec{x}})e^{-k\beta |{\varvec{x}}-{\varvec{x}}_0|^2}\right| ^2\,\textrm{d}{\varvec{x}}\le \mathcal C(d) \left[ (2\beta )^{-\frac{d}{2}-N} \frac{d\pi ^{\frac{d}{2}}}{2}\frac{\Gamma \left( \frac{d}{2}+N\right) }{\Gamma \left( \frac{d}{2}+1\right) }\right] k^{-\frac{d}{2}-N}. \end{aligned}$$

$\square$

Proof of Theorem 4.1

We organize the proof in three steps, one for each statement of the theorem.

Step 1: proof of (4.7). Starting from (4.14), we have that

$$\begin{aligned} \left\| \square _c u^k\right\| _{L^2(\mathbb {R}^d)}^2 =&\,\int _{\mathbb {R}^d} |\square _c u^k|^2\,\textrm{d}{\varvec{x}}\le k^{\frac{d}{2}-2}\int _{\mathbb {R}^d} \left| e^{ik\phi }r_0\right| ^2\,\textrm{d}{\varvec{x}}+ k^{\frac{d}{2}} \int _{\mathbb {R}^d} \left| e^{ik\phi }r_1\right| ^2\,\textrm{d}{\varvec{x}}\\&+ k^{\frac{d}{2}+2}\int _{\mathbb {R}^d}\left| e^{ik\phi }r_2\right| ^2\,\textrm{d}{\varvec{x}}, \end{aligned}$$

where we recall

$$\begin{aligned}&r_0 = \square _c a \\&r_1 = a\square _c\phi + 2 a_t\phi _t - 2c \nabla a\cdot \nabla \phi \\&r_2 = \Big (c|\nabla \phi |^2-\phi _t^2\Big )a. \end{aligned}$$

Since $a,\phi \in C^\infty (\mathbb {R}^d\times \mathbb {R})$, we clearly have that also $r_0,r_1,r_2 \in C^\infty (\mathbb {R}^d\times \mathbb {R})$. Moreover, by construction, $r_1$ and $r_2$ vanish on ${\varvec{x}}={\varvec{x}}(t)$ up to the order 0 and 2. In view of that, we have that $r_0,r_1,r_2$ satisfy (A.1) with $N = 0$, $N = 1$ and $N = 3$, respectively. Then, applying Proposition A.1 with ${\varvec{x}}_0 = {\varvec{x}}(t)$ and the previous values of $N\in \mathbb {N}$, we get

$$\begin{aligned} k^{\frac{d}{2}-1}\int _{\mathbb {R}^d} \left| e^{ik\phi }r_0\right| ^2\,\textrm{d}{\varvec{x}}\le \mathcal C(a,\phi ) k^{-2} \\ k^{\frac{d}{2}} \int _{\mathbb {R}^d} \left| e^{ik\phi }r_1\right| ^2\,\textrm{d}{\varvec{x}}\le \mathcal C(a,\phi ) k^{-1} \\ k^{\frac{d}{2}+1}\int _\mathbb {R}\left| e^{ik\phi }r_2\right| ^2\,\textrm{d}{\varvec{x}}\le \mathcal C(a,\phi ) k^{-1}. \end{aligned}$$

Putting everything together, since $k\ge 1$, we finally obtain that

$$\begin{aligned} \left\| \square _c u^k\right\| _{L^2(\mathbb {R}^d)}^2 \le \mathcal C\Big (k^{-2} + k^{-1} + k^{-1}\Big ) \le \mathcal C(a,\phi ) k^{-1}, \end{aligned}$$

that is,

$$\begin{aligned} \left\| \square _c u^k\right\| _{L^2(\mathbb {R}^d)} \le \mathcal C(a,\phi ) k^{-\frac{1}{2}}. \end{aligned}$$

Step 2: proof of (4.8). Starting from (4.1), we have that

$$\begin{aligned} E_c(u^k(\cdot ,t)) =&\, \frac{1}{2} \int _{\mathbb {R}^d} \Big (|u^k_t(\cdot ,t)|^2 + c|\nabla u^k(\cdot ,t)|^2\Big )\,\textrm{d}{\varvec{x}}= \Xi _0^k(t) + \Xi _1^k(t) + \Xi _2^k(t), \end{aligned}$$

where we have denoted

$$\begin{aligned}&\Xi _0^k(t) := \frac{k^{\frac{d}{2}}}{2} \int _{\mathbb {R}^d} |a|^2\Big (|\phi _t|^2 + c|\nabla \phi |^2\Big )e^{-k({\varvec{x}}-{\varvec{x}}(t))\cdot \big [\Im (M_0)({\varvec{x}}-{\varvec{x}}(t))\big ]}\,\textrm{d}{\varvec{x}}\\&\Xi _1^k(t) := k^{\frac{d}{2}-1} \int _{\mathbb {R}^d} a\Big (a_t\phi _t + c\nabla a\cdot \nabla \phi \Big )e^{-k({\varvec{x}}-{\varvec{x}}(t))\cdot \big [\Im (M_0)({\varvec{x}}-{\varvec{x}}(t))\big ]}\,\textrm{d}{\varvec{x}}\\&\Xi _2^k(t) := \frac{k^{\frac{d}{2}-2}}{2} \int _{\mathbb {R}^d} \Big (|a_t|^2 + c|\nabla a|^2\Big )e^{-k({\varvec{x}}-{\varvec{x}}(t))\cdot \big [\Im (M_0)({\varvec{x}}-{\varvec{x}}(t))\big ]}\,\textrm{d}{\varvec{x}}. \end{aligned}$$

Notice that since $a,\phi \in C^\infty (\mathbb {R}^d\times \mathbb {R})$ and $\Im (M_0)>0$, we have that for all $t\in (0,T)$

$$\begin{aligned} |\Xi _1^k(t)| \le&\,\mathcal C(a,\phi )k^{\frac{d}{2}-1} \int _{\mathbb {R}^d} e^{-k({\varvec{x}}-{\varvec{x}}(t))\cdot \big [\Im (M_0)({\varvec{x}}-{\varvec{x}}(t))\big ]}\,\textrm{d}{\varvec{x}}\\ =&\, \mathcal C(a,\phi )\frac{d\Gamma \left( \frac{d+1}{2}\right) }{2\Gamma \left( \frac{d}{2} +1\right) }\left( \frac{\pi }{\text {det}(\Im (M_0))}\right) ^{\frac{d}{2}} k^{-1} \\ |\Xi _2^k(t)| \le&\,\mathcal C(a,\phi )k^{\frac{d}{2}-2} \int _{\mathbb {R}^d} e^{-k({\varvec{x}}-{\varvec{x}}(t))\cdot \big [\Im (M_0)({\varvec{x}}-{\varvec{x}}(t))\big ]}\,\textrm{d}{\varvec{x}}\\ =&\, \mathcal C(a,\phi )\frac{d\Gamma \left( \frac{d+1}{2}\right) }{2\Gamma \left( \frac{d}{2} +1\right) }\left( \frac{\pi }{\text {det}(\Im (M_0))}\right) ^{\frac{d}{2}} k^{-2} \end{aligned}$$

Hence,

$$\begin{aligned} \sup _{t\in (0,T)} \Big (|\Xi _1^k(t)| + |\Xi _2^k(t)|\Big ) \rightarrow 0, \quad \text { as } k\rightarrow +\infty . \end{aligned}$$

(A.3)

As for the term $\Xi _0^k(t)$, replacing in it the explicit expression (4.5) of the amplitude function a, we get that

$$\begin{aligned} \Xi _0^k(t)&= \frac{k^{\frac{d}{2}}}{2} \int _{\mathbb {R}^d} \Big (|\phi _t|^2 + c|\nabla \phi |^2\Big )e^{-({\varvec{x}}-{\varvec{x}}(t))\cdot \Big [\Big (2 I_d + k\Im (M_0)\Big )({\varvec{x}}-{\varvec{x}}(t))\Big ]}\,\textrm{d}{\varvec{x}}, \end{aligned}$$

(A.4)

where $I_d$ denotes the identity matrix in dimension $d\times d$. Moreover, using (3.7) and the explicit expression (4.6) of the phase function a, we can compute

$$\begin{aligned}&|\phi _t|^2 = \frac{c}{|{\varvec{\xi }}_0|^2}\bigg [|{\varvec{\xi }}_0|^2 + {\varvec{\xi }}_0\cdot \Big (M_0({\varvec{x}}-{\varvec{x}}(t))\Big )\bigg ]^2 \\&c|\nabla \phi |^2 = c\,\Big |{\varvec{\xi }}_0 + M_0({\varvec{x}}-{\varvec{x}}(t))\Big |^2 \end{aligned}$$

and we obtain from (A.4) that

$$\begin{aligned} \Xi _0^k(t) =&\; \frac{c}{2|{\varvec{\xi }}_0|^2} k^{\frac{d}{2}} \int _{\mathbb {R}^d} \bigg [|{\varvec{\xi }}_0|^2 + {\varvec{\xi }}_0\cdot \Big (M_0({\varvec{x}}-{\varvec{x}}(t))\Big )\bigg ]^2\\&e^{-({\varvec{x}}-{\varvec{x}}(t))\cdot \Big [\Big (2 I_d + k\Im (M_0)\Big )({\varvec{x}}-{\varvec{x}}(t))\Big ]}\,\textrm{d}{\varvec{x}}\\&+ \frac{c}{2} k^{\frac{d}{2}} \int _{\mathbb {R}^d} \Big |{\varvec{\xi }}_0 + M_0({\varvec{x}}-{\varvec{x}}(t))\Big |^2e^{-({\varvec{x}}-{\varvec{x}}(t))\cdot \Big [\Big (2 I_d + k\Im (M_0)\Big )({\varvec{x}}-{\varvec{x}}(t))\Big ]}\,\textrm{d}{\varvec{x}}. \end{aligned}$$

Now, since $2 I_d + k\Im (M_0)>0$, we can apply the change of variables

$$\begin{aligned} \Big (2 I_d + k\Im (M_0)\Big )^{\frac{1}{2}} ({\varvec{x}}-{\varvec{x}}(t)) = \varvec{y} \end{aligned}$$

and we get

$$\begin{aligned} \Xi _0^k(t) =&\; \frac{c}{2|{\varvec{\xi }}_0|^2} \frac{k^{\frac{d}{2}}}{\text {det}\Big (2 I_d + k\Im (M_0)\Big )^{\frac{d}{2}}}\\&\int _{\mathbb {R}^d} \Bigg [|{\varvec{\xi }}_0|^2 + {\varvec{\xi }}_0\cdot \bigg (\Big (2 I_d + k\Im (M_0)\Big )^{-\frac{1}{2}}M_0\varvec{y}\bigg )\Bigg ]^2 e^{-|\varvec{y}|^2}\,\textrm{d}\varvec{y} \\&+ \frac{c}{2} \frac{k^{\frac{d}{2}}}{\text {det}\Big (2 I_d + k\Im (M_0)\Big )^{\frac{d}{2}}} \int _{\mathbb {R}^d} \Big |{\varvec{\xi }}_0 + \Big (2 I_d + k\Im (M_0)\Big )^{-\frac{1}{2}}M_0\varvec{y}\Big |^2e^{-|\varvec{y}|^2}\,\textrm{d}\varvec{y}. \end{aligned}$$

Finally, the two integrals in the expression above can be computed by employing polar coordinates. In this way, we obtain that

$$\begin{aligned} \Xi _0^k(t) =&\, \mathcal C\Big (d,{\varvec{\xi }}_0,M_0\Big )\pi ^{\frac{d}{2}}\frac{k^{\frac{d}{2}}\Big (1+k^{-1}\Big )}{\text {det}\Big (2 I_d + k\Im (M_0)\Big )^{\frac{d}{2}}} \\ =&\, \frac{\mathcal C\Big (d,{\varvec{\xi }}_0,M_0\Big )\pi ^{\frac{d}{2}}}{\text {det}\Big (2 k^{-1} I_d + \Im (M_0)\Big )^{\frac{d}{2}}}\Big (1+k^{-1}\Big ). \end{aligned}$$

Hence,

$$\begin{aligned} \lim _{k\rightarrow +\infty } \Xi _0^k(t) = \mathcal C\Big (d,{\varvec{\xi }}_0,M_0\Big )\left( \frac{\pi }{\text {det}\big (\Im (M_0)\big )}\right) ^{\frac{d}{2}}. \end{aligned}$$

(A.5)

From (A.3) and (A.5), we immediately get (4.8).

Step 3: proof of (4.9). Since $a,\phi \in C^\infty (\mathbb {R}^d\times \mathbb {R})$ and $k\ge 1$, we have that

$$\begin{aligned} \int _{\mathbb {R}^d\setminus B_k(t)} \Big (|u^k_t(\cdot ,t)|^2 + c|\nabla u^k(\cdot ,t)|^2\Big )\,\textrm{d}{\varvec{x}}&\le \mathcal C\left( k^{\frac{d}{2}} + k^{\frac{d}{2}-1} + k^{\frac{d}{2}-2}\right) \\&\qquad \int _{\mathbb {R}^d\setminus B_k(t)} e^{-k({\varvec{x}}-{\varvec{x}}(t))\cdot \big [\Im (M_0)({\varvec{x}}-{\varvec{x}}(t))\big ]}\,\textrm{d}{\varvec{x}}\\&\le \mathcal Ck^{\frac{d}{2}} \int _{\mathbb {R}^d\setminus B_k(t)} e^{-k({\varvec{x}}-{\varvec{x}}(t))\cdot \big [\Im (M_0)({\varvec{x}}-{\varvec{x}}(t))\big ]}\,\textrm{d}{\varvec{x}}\\&= \mathcal Ck^{\frac{d}{2}} \int _{\mathbb {R}^d\setminus B\left( 0,k^{-\frac{1}{4}}\right) } e^{-k\varvec{y}\cdot \big [\Im (M_0)\varvec{y}\big ]}\,\textrm{d}\varvec{y}, \end{aligned}$$

with $\mathcal C= \mathcal C(a,\phi )>0$ a positive constant not depending on k. Moreover, by employing the change of variable $\varvec{y} = k^{-\frac{1}{2}}\varvec{z}$, we have that

$$\begin{aligned} \mathcal Ck^{\frac{d}{2}} \int _{\mathbb {R}^d\setminus B\left( 0,k^{-\frac{1}{4}}\right) } e^{-k\varvec{y}\cdot \big [\Im (M_0)\varvec{y}\big ]}\,\textrm{d}\varvec{y}&= \mathcal C\int _{\mathbb {R}^d\setminus B\left( 0,k^{\frac{1}{4}}\right) } e^{-\varvec{z}\cdot \big [\Im (M_0)\varvec{z}\big ]}\,\textrm{d}\varvec{z} \\&= \mathcal C\int _{\mathbb {R}^d\setminus B\left( 0,k^{\frac{1}{4}}\right) } e^{-\frac{1}{2}\varvec{z}\cdot \big [\Im (M_0)\varvec{z}\big ]}e^{-\frac{1}{2}\varvec{z}\cdot \big [\Im (M_0)\varvec{z}\big ]}\,\textrm{d}\varvec{z} \\&\le \sup _{\mathbb {R}^d\setminus B\left( 0,k^{\frac{1}{4}}\right) }\left( e^{-\frac{1}{2}\varvec{z}\cdot \big [\Im (M_0)\varvec{z}\big ]}\right) \\&\qquad \int _{\mathbb {R}^d\setminus B\left( 0,k^{\frac{1}{4}}\right) } e^{-\frac{1}{2}\varvec{z}\cdot \big [\Im (M_0)\varvec{z}\big ]}\,\textrm{d}\varvec{z} \\&= e^{-\frac{1}{2}\text {det}\big (\Im (M_0)\big )k^{\frac{1}{2}}} \int _{\mathbb {R}^d\setminus B\left( 0,k^{\frac{1}{4}}\right) } e^{-\frac{1}{2}\varvec{z}\cdot \big [\Im (M_0)\varvec{z}\big ]}\,\textrm{d}\varvec{z} \\&\le e^{-\frac{1}{2}\text {det}\big (\Im (M_0)\big )k^{\frac{1}{2}}} \int _{\mathbb {R}^d} e^{-\frac{1}{2}\varvec{z}\cdot \big [\Im (M_0)\varvec{z}\big ]}\,\textrm{d}\varvec{z} \\&= \frac{d\Gamma \left( \frac{d+1}{2}\right) }{2\Gamma \left( \frac{d}{2} +1\right) }\left( \frac{2\pi }{\text {det}(\Im (M_0))}\right) ^{\frac{d}{2}}e^{-\frac{1}{2}\text {det}\big (\Im (M_0)\big )k^{\frac{1}{2}}}. \end{aligned}$$

From this last estimate, (4.9) follows immediately. $\square$

Appendix B. Proof of Theorem 5.1

1.1 B.1. Concentration of Solutions

We give here the proof of Theorem 5.1, concerning the construction of a GB ansatz for the semi-discrete wave equation (5.2).

Proof of Theorem 5.1

We are going to split the proof into three steps, one for each point in the statement of the theorem. Moreover, in what follows, we will denote by $\mathcal C>0$ a generic positive constant independent of h. This constant may change even from line to line.

Step 1: proof of(5.10). Starting from (5.46), we have that

$$\begin{aligned} \left\| \square _{c,h} u^h_{fd}(\cdot ,t)\right\| _{\ell ^2(h\mathbb {Z})}^2 =&\; h\sum _{j\in \mathbb {Z}} \left| \square _{c,h} u^h_{fd,j}(t)\right| ^2 \\ \le&\; h^{\frac{5}{2}}\sum _{j\in \mathbb {Z}} \left| e^{\frac{i}{h}\Phi _j}\left( \mathcal R_0A_j + A_j \frac{\mathcal R_2-\mathcal R_{fd}}{h^2}\right) \right| ^2 \\&+ h^{\frac{1}{2}}\sum _{j\in \mathbb {Z}}\left( \left| e^{\frac{i}{h}\Phi _j}(\mathcal R_1A_j - \widetilde{\mathcal R}_1 A_j)\right| ^2 + \left| e^{\frac{i}{h}\Phi _j}\widetilde{\mathcal R}_1 A_j\right| ^2\right) \\&+ h^{-\frac{3}{2}}\sum _{j\in \mathbb {Z}} \left| e^{\frac{i}{h}\Phi _j}A_j\mathcal R_{fd}\right| ^2. \end{aligned}$$

Moreover, using (5.35), we see that

$$\begin{aligned} e^{\frac{i}{h}\Phi _j} =&\, e^{\frac{i}{h} \Re (\Phi _j)}e^{-\frac{1}{h} \Im (\Phi _j)} = e^{\frac{i}{h} \big (\omega (t)+\xi _0(x_j-x_{fd,j}(t)) + \frac{1}{2}\Re (M(t))(x_j-x_{fd,j}(t))^2\big )}\\&\,e^{-\frac{1}{2h}\Im (M(t))(x_j-x_{fd,j}(t))^2} \end{aligned}$$

and, therefore,

$$\begin{aligned} \left| e^{\frac{i}{h}\Phi _j}\right| ^2 = e^{-\frac{1}{2h}\Im (M(t))(x_j-x_{fd,j}(t))^2}. \end{aligned}$$

(B.1)

In view of this, we get

$$\begin{aligned} \left\| \square _{c,h} u^h_{fd}(\cdot ,t)\right\| _{\ell ^2(h\mathbb {Z})}^2 \le&\; h^{\frac{5}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| \mathcal R_0A_j + A_j \frac{\mathcal R_2-\mathcal R_{fd}}{h^2}\right| ^2 \\&+ h^{\frac{1}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\\&\left( \left| \mathcal R_1A_j - \widetilde{\mathcal R}_1 A_j\right| ^2 + \left| \widetilde{\mathcal R}_1 A_j\right| ^2 \right) \\&+ h^{-\frac{3}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| A_j\mathcal R_{fd}\right| ^2. \end{aligned}$$

Now, by means of (5.31) and (5.45), we have that

$$\begin{aligned} h^{\frac{5}{2}}&\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| \mathcal R_0A_j + A_j \frac{\mathcal R-\mathcal R_{fd}}{h^2}\right| ^2 \\&\le h^{\frac{5}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\Big (\left| \mathcal R_0A_j\right| ^2 + \mathcal C\left| A_j\right| ^2\Big ) \end{aligned}$$

and

$$\begin{aligned} h^{\frac{1}{2}}&\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| \mathcal R_1A_j - \widetilde{\mathcal R}_1 A_j\right| ^2 \\&\le \mathcal Ch^{\frac{5}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\Big (|\Delta _{c,h}\Phi _j|^2 + |\Delta _{c,h}\Phi _j|^4\Big )\\&\qquad +\mathcal Ch^{\frac{9}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}. \end{aligned}$$

Hence,

$$\begin{aligned} \left\| \square _{c,h} u^h_{fd}(\cdot ,t)\right\| _{\ell ^2(h\mathbb {Z})}^2&\le \mathcal Ch^{\frac{9}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\\ {}&\quad +\mathcal Ch^{\frac{5}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2} \\&\quad + h^{\frac{1}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| \widetilde{\mathcal R}_1 A_j\right| ^2 \\&\quad + h^{-\frac{3}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| A_j\mathcal R_{fd}\right| ^2. \end{aligned}$$

The first two terms on the right-hand side of the above inequality can be estimated by observing that

$$\begin{aligned} \sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2} = \mathcal O(h^{-\frac{1}{2}}). \end{aligned}$$

(B.2)

This can be easily seen by considering the Riemann sum approximating on the mesh $\mathcal G_h$ the integral

$$\begin{aligned} \int _\mathbb {R}e^{-\frac{1}{h} \Im (M(t))(x-x_{fd}(t))^2}\,\textrm{d}x = \mathcal Ch^{\frac{1}{2}}. \end{aligned}$$

Hence, using (B.2), we obtain that

$$\begin{aligned} \Big (h^{\frac{9}{2}} + h^{\frac{5}{2}}\Big ) \sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2} = \mathcal O(h^4) + \mathcal O(h^2) = \mathcal O(h^2) \end{aligned}$$

and, therefore,

$$\begin{aligned} \left\| \square _{c,h} u^h_{fd}(\cdot ,t)\right\| _{\ell ^2(h\mathbb {Z})}^2 \le&\; h^{\frac{1}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| \widetilde{\mathcal R}_1 A_j\right| ^2 \\&+ h^{-\frac{3}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| A_j\mathcal R_{fd}\right| ^2 + \mathcal O(h^2). \end{aligned}$$

Finally, since by construction $\widetilde{\mathcal R}_1 A_j$ and $\mathcal R_{fd}$ vanish on the discrete characteristics $x_{fd}(t)$ up to the order 0 and 2, respectively, we have that

$$\begin{aligned}&\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| \widetilde{\mathcal R}_1 A_j\right| ^2 = \mathcal O(h^{\frac{1}{2}}) \\&\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| A_j\mathcal R_{fd}\right| ^2 = \mathcal O(h^{\frac{5}{2}}). \end{aligned}$$

This is the discrete version of Proposition A.1, which can be proven once again by applying Riemann sum to approximate the corresponding integrals. In view of this, we obtain

$$\begin{aligned}&h^{\frac{1}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| \widetilde{\mathcal R}_1 A_j\right| ^2 \\&\qquad + h^{-\frac{3}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}\left| A_j\mathcal R_{fd}\right| ^2 = \mathcal O(h). \end{aligned}$$

and we can finally conclude that

$$\begin{aligned} \left\| \square _{c,h} u^h_{fd}(\cdot ,t)\right\| _{\ell ^2(h\mathbb {Z})}^2 = \mathcal O(h^2) + \mathcal O(h) = \mathcal O(h), \quad \text { as } h\rightarrow 0^+. \end{aligned}$$

Step 2: proof of(5.11). First of all, starting from (5.3), (5.4) and (5.23), we can write

$$\begin{aligned} \mathcal E_h[u_{fd}^h] =&\, \frac{h}{2}\sum _{j\in \mathbb {Z}} \Big (|\partial _t u_{fd,j}^h|^2 + c|\partial _h^+u_{fd,j}^h|^2\Big )\\ =&\, \frac{h^{\frac{5}{2}}}{2}\sum _{j\in \mathbb {Z}} \left( \left| \partial _t \left( A_j e^{\frac{i}{h} \Phi _j}\right) \right| ^2 + c\left| \partial _h^+\left( A_j e^{\frac{i}{h} \Phi _j}\right) \right| ^2\right) . \end{aligned}$$

Moreover, we can easily compute

$$\begin{aligned}&\partial _t \left( A_j e^{\frac{i}{h} \Phi _j}\right) = \frac{1}{h} e^{\frac{i}{h} \Phi _j}\Big (h\partial _t A_j + iA_j\partial _t\Phi _j\Big ) \\&\partial _h^+\left( A_j e^{\frac{i}{h} \Phi _j}\right) = \frac{1}{h} e^{\frac{i}{h} \Phi _j}\Big (h\partial _h^+A_je^{i\partial _h^+\Phi _j} + \big (e^{i\partial _h^+\Phi _j}-1\big )A_j\Big ) \end{aligned}$$

Using these identities and (B.1), we then obtain

$$\begin{aligned} \mathcal E_h[u_{fd}^h] =&\; \frac{h^{\frac{5}{2}}}{2}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2} \mathcal S_1 + h^{\frac{3}{2}}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2} \mathcal S_2 \\&+ \frac{h^{\frac{1}{2}}}{2}\sum _{j\in \mathbb {Z}} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2} \mathcal S_3, \end{aligned}$$

where we have denoted

$$\begin{aligned}&\mathcal S_1 := \Big |\partial _t A_j\Big |^2 + c\Big |\partial _h^+A_je^{i\partial _h^+\Phi _j}\Big |^2 \end{aligned}$$

(B.3a)

$$\begin{aligned}&\mathcal S_2 := \Big |iA_j\partial _t A_j\partial _t\Phi _j\Big | + c\Big |\partial _h^+A_je^{i\partial _h^+\Phi _j}\big (e^{i\partial _h^+\Phi _j}-1\big )A_j\Big | \end{aligned}$$

(B.3b)

$$\begin{aligned}&\mathcal S_3 := \Big |iA_j\partial _t\Phi _j\Big |^2 + c\Big |\big (e^{i\partial _h^+\Phi _j}-1\big )A_j\Big |^2. \end{aligned}$$

(B.3c)

Moreover, by construction of A and $\Phi$, we have that $|\mathcal S_i|\le \mathcal C(A,\phi )$ for $i\in \{1,2,3\}$. Using this and (B.2), we finally obtain

$$\begin{aligned} \mathcal E_h[u_{fd}^h] = \mathcal O(h^2) + \mathcal O(h) + \mathcal O(1) = \mathcal O(1), \quad \text { as } h\rightarrow 0^+. \end{aligned}$$

Step 3: proof of(5.12). Repeating the computations of Step 2, we have that

$$\begin{aligned} \frac{h}{2}\sum _{j\in \mathbb {Z}^\dagger (t)} \Big (|\partial _t u_{fd,j}^h|^2 + c|\partial _h^+u_{fd,j}^h|^2\Big ) =&\; \frac{h^{\frac{5}{2}}}{2}\sum _{j\in \mathbb {Z}^\dagger (t)} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2} \mathcal S_1 \\&+ h^{\frac{3}{2}}\sum _{j\in \mathbb {Z}^\dagger (t)} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2} \mathcal S_2 \\&+ \frac{h^{\frac{1}{2}}}{2}\sum _{j\in \mathbb {Z}^\dagger (t)} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2} \mathcal S_3, \end{aligned}$$

with $\mathcal S_1$, $\mathcal S_2$ and $\mathcal S_3$ defined in (B.3a), (B.3b) and (B.3c), respectively. Hence, recalling that $|\mathcal S_i|\le \mathcal C(A,\Phi )$ for $i\in \{1,2,3\}$, we obtain

$$\begin{aligned}&\frac{h}{2}\sum _{j\in \mathbb {Z}^\dagger (t)} \Big (|\partial _t u_{fd,j}^h|^2 + c|\partial _h^+u_{fd,j}^h|^2\Big ) \nonumber \\ {}&\qquad \le \mathcal C\Big (h^{\frac{5}{2}} + h^{\frac{3}{2}} + h^{\frac{1}{2}}\Big )\sum _{j\in \mathbb {Z}^\dagger (t)} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}. \end{aligned}$$

(B.4)

Now, employing the transformation

$$\begin{aligned} x_j-x_{fd,j} = y_j, \end{aligned}$$

and denoting

$$\begin{aligned} \mathbb {Z}^\ddagger (t) := \Big \{ j\in \mathbb {Z}\,:\, |y_j|>h^{\frac{1}{4}}\Big \}, \end{aligned}$$

we have that

$$\begin{aligned} \sum _{j\in \mathbb {Z}^\dagger (t)} e^{-\frac{1}{h} \Im (M(t))(x_j-x_{fd,j}(t))^2}&= \sum _{j\in \mathbb {Z}^\ddagger (t)} e^{-\frac{1}{h} \Im (M(t))y_j^2} \\&\le e^{-\frac{1}{2} \Im (M(t))h^{-\frac{1}{2}}} \sum _{j\in \mathbb {Z}^\ddagger (t)} e^{-\frac{1}{2h} \Im (M(t))y_j^2}\\&\le e^{-\frac{1}{2} \Im (M(t))h^{-\frac{1}{2}}} \sum _{j\in \mathbb {Z}} e^{-\frac{1}{2h} \Im (M(t))y_j^2}\\&= \mathcal Ch^{-\frac{1}{2}} e^{-\frac{1}{2} \Im (M(t))h^{-\frac{1}{2}}}. \end{aligned}$$

Substituting this in (B.4), we finally conclude that

$$\begin{aligned} \frac{h}{2}\sum _{j\in \mathbb {Z}^\dagger (t)} \Big (|\partial _t u_{fd,j}^h|^2 + c|\partial _h^+u_{fd,j}^h|^2\Big ) \le \mathcal C\Big (1 + h + h^2\Big ) e^{-\frac{1}{2} \Im (M(t))h^{-\frac{1}{2}}}. \end{aligned}$$

$\square$