Recovery of wave speeds and density of mass across a heterogeneous smooth interface from acoustic and elastic wave reflection operators

Abstract

We revisit the problem of recovering wave speeds and density across a curved interface from reflected wave amplitudes. Such amplitudes have been exploited for decades in (exploration) seismology in this context. However, the analysis in seismology has been based on linearization and mostly flat interfaces. Here, we present an analysis without linearization and allow curved interfaces, establish uniqueness and provide a reconstruction, while making the notion of amplitude precise through a procedure rooted in microlocal analysis.

A Proofs of lemmas and propositions from Section 2

These are proofs of the main statements in the acoustic case. Since they are a simpler, yet more lucid version of the elastic case, we relegate them to this appendix.

1.1 A.1 Zeroth order recovery of the parameters at the interface

Proof of Lemma 2.1

By solving (2.4), we get

$$\begin{aligned} (a_R)_0(x',\tau ,\xi ') = \frac{\mu ^{(-)}\xi _I - \mu ^{(+)}\xi _T}{\mu ^{(-)}\xi _I + \mu ^{(+)}\xi _T} = \frac{\mu ^{(-)}\xi _I/\xi _T - \mu ^{(+)}}{\mu ^{(-)}\xi _I/\xi _T + \mu ^{(+)}} = \frac{af-b}{af+b}, \end{aligned}$$

where we denote $f = \xi _I/\xi _T, a = \mu ^{(-)}, b = \mu ^{(+)}.$ Note that $f = f(|\xi ' |/\tau )$ i.e. it is a function of the parameter $|\xi ' |/\tau$ while a, b only depend on x. Now, since $(a_R)_0 = ({\tilde{a}}_R)_0$ and assuming $\mu ^{(-)} = {\tilde{\mu }}^{(-)}$ (i.e. $a=\tilde{a}$) on $\Gamma$ we obtain

$$\begin{aligned} \frac{af-b}{af+b} = \frac{a{\tilde{f}}-{\tilde{b}}}{a{\tilde{f}}+\tilde{b}},\quad \text {if and only if} \quad a{\tilde{b}} f = ab {\tilde{f}}, \quad \text {if and only if} \quad \frac{{\tilde{b}}}{ b} = \frac{\tilde{f}}{f}. \end{aligned}$$

(A.1)

Varying $|\xi ' |/\tau$ keeping everything else constant, we get

$$\begin{aligned} \frac{{\tilde{b}}}{ b} = \frac{{\tilde{f}}_1}{f_1}, \end{aligned}$$

where $f_1$ is f evaluated at different value of $|\xi ' |/\tau$. Thus,

$$\begin{aligned} \frac{{\tilde{f}}}{f}=\frac{{\tilde{f}}_1}{f_1}\quad \text {if and only if} \quad \frac{({\tilde{c}}^{(+)}_S)^{-2} - d^2}{(c^{(+)}_S)^{-2} - d^2} = \frac{({\tilde{c}}^{(+)}_S)^{-2} - d_1^2}{(c^{(+)}_S)^{-2} - d_1^2}, \end{aligned}$$

where we used $c^{(-)}_S = {\tilde{c}}^{(-)}_S$ and labelled $d = |\xi ' |/\tau$, $d_1 = |\xi '_1 |/\tau _1$. Cross multiplying we get the algebraic equation

$$\begin{aligned} (({\tilde{c}}^{(+)}_S)^{-2}- ( c^{(+)}_S)^{-2})(d^2 - d_1^2) = 0 \end{aligned}$$

Note that, as long as we pick $d_1 \ne \pm d$, we recover $c_S^{(+)} = \tilde{c}_S^{(+)}$. Then going back to (A.1) one gets $b=\tilde{b}$, that is $\mu ^{(+)} = {{\tilde{\mu }}}^{(+)}$ on $\Gamma$. $\square$

Remark A.1

In geophysical experiments, one often only has access to relative amplitudes where the amplitude is normalized to be 1 for an incident wave hitting the interface at a fixed particular angle. Concretely, for a fixed, $(x',\tau _1, \xi '_1)$ in the hyperbolic set, suppose one instead measures $R:= (a_R)_0(x',\tau ,\xi ')/ (a_R)_0(x',\tau _1,\xi '_1)$. The questions is can one recover $\mu ^{(+)}$ and $c_S^{(+)}$ at $x'$ from R at various incident angles (i.e. varying $\tau , \xi '$ in the hyperbolic set)?

In order to recover a single unknown parameter such as $\mu ^{(+)}$, then this can be done with elementary means, but disentangling two material parameters is less clear. For the uniqueness question, ignoring spacial variables, assume

$$\begin{aligned}\frac{(a_R)_0(\tau , \xi ')}{(a_R)_0(\tau _1, \xi _1)} = \frac{(\widetilde{a_R})_0(\tau , \xi ')}{(\widetilde{a_R})_0(\tau _1, \xi _1)}. \end{aligned}$$

Then one can show $\mu ^{(+)} = {{\tilde{\mu }}}^{(+)}$ and $c_S^{(+)} = {\tilde{c}}_S^{(+)}$ with a similar argument as above.

One can even obtain a partial reconstruction algorithm directly from R. Via computation, one can show that

$$\begin{aligned} L:= (R-1)/(R+1) = \frac{2ac(f-f_1)}{a^2ff_1 - b^2}, \end{aligned}$$

where $a,b,f,f_1$ are as in the lemma. By solving a quadratic equation, we compute

$$\begin{aligned} 2b = \alpha /L + \sqrt{\alpha ^2/L^2 + 4\beta } \end{aligned}$$

where $\alpha = 2a(f-f_1), \beta = a^2ff_1$ are independent of b. Since b is independent of those variable, one can vary $\tau , \xi '$ within the hyperbolic set to obtain a nonlinear equation that needs to be solved for $c_S$ only (without any terms involving $b= \mu ^{(+)}$), but it is unclear whether this can be done by elementary means. If it can, then our approach shows how one can do the recovery even with reflected amplitudes, and there is a reconstruction formula.

1.2 A.2 Recovery of the derivatives of the parameters at the interface

Proof of Lemma 2.2

Observe that for $J=0$, from (2.6) and (2.8) one obtains

$$\begin{aligned} \partial _{x_3}(a_\bullet )_0 = \frac{1}{2\rho c_S^2 \xi _{3, \bullet }}\left( P\phi _{\bullet }\right) (a_{\bullet })_0 + R_0. \end{aligned}$$

(A.2)

In order to calculate the term $P\phi _{\bullet } = (\rho \partial _t^2 - \nabla _x\cdot \mu \nabla _x)\phi _{\bullet }$, we start with

$$\begin{aligned} \partial _{x_3} e^{i\phi _\bullet } = i \partial _{x_3}\phi _\bullet e^{i\phi _\bullet } = \pm \left( \sqrt{| \partial _{x'}\phi _{\bullet } |^2 -c^{-2}_{S}| \partial _{t}\phi _{\bullet } |^2}\right) e^{i\phi _\bullet }, \qquad \text{ on } \Gamma . \end{aligned}$$

(A.3)

Therefore, $\partial _{x_3} \phi _\bullet = \pm \sqrt{c^{-2}_{S}| \partial _{t}\phi _{\bullet } |^2 - | \partial _{x'}\phi _{\bullet } |^2}$ can be recovered from $c_S$ and the tangential derivatives of $\phi _{\bullet }$ on $\Gamma$. In other words, $\partial _{x_3} \phi _\bullet$ at $\Gamma$ is a $R_0$ term. Taking one more derivative of (A.3) in the normal direction we get

$$\begin{aligned} \partial ^2_{x_3}\phi _\bullet = \frac{1}{\xi _{3, \bullet }}\left[ -c_S^{-2}( \partial _{x_3}\log c_S)| \partial _t \phi _\bullet |^2 + c_S^{-2}( \partial _t \phi _\bullet ) \partial _{x_3} \partial _t \phi _\bullet - \langle \partial _{x_3}\nabla _{x'}\phi _\bullet , \nabla _{x'}\phi _\bullet \rangle \right] . \end{aligned}$$

Here the last two terms above are determined by $c_S$, $\phi _{\bullet }$ and their tangential derivatives on $\Gamma$. Hence, we have

$$\begin{aligned} \partial ^2_{x_3}\phi _\bullet =-( \partial _{x_3}\log c_S)c_S^{-2}\xi _{3, \bullet }^{-1}| \partial _t \phi _\bullet |^2 + E_0, \quad \text{ where } E_0 \text{ is } \text{ a } R_0 \text{ term }. \end{aligned}$$

If we take one more normal derivative of $\phi _{\bullet }$, then $\partial _{x_3}E_0$ can have at most one derivative of $c_S$ as well as the term $c_S^{-2}\xi _{3, \bullet }^{-1}| \partial _t \phi _\bullet |^2$. Thus, we obtain

$$\begin{aligned} \partial _{x_3}^3 \phi _\bullet = -( \partial ^2_{x_3}\log c_S)c_S^{-2}\xi _{3, \bullet }^{-1}| \partial _t \phi _\bullet |^2 + R_1. \end{aligned}$$

In general, one obtains

$$\begin{aligned} \partial _{x_3}^k \phi _\bullet = -( \partial ^{k-1}_{x_3}\log c_S)c_S^{-2}\xi _{3, \bullet }^{-1}| \partial _t \phi _\bullet |^2 + R_{k-2}. \end{aligned}$$

(A.4)

Now we calculate

$$\begin{aligned} \nabla _x \cdot \mu \nabla _x \phi _\bullet&= ( \partial _{x_3}\mu ) \partial _{x_3} \phi _\bullet + \mu \partial ^2_{x_3}\phi _\bullet + R_0\\&= \partial _{x_3}(\rho c_S^2)\xi _{3, \bullet }- \rho c_S^2( \partial _{x_3}\log c_S)c_S^{-2}\xi _{3, \bullet }^{-1}( \partial _t \phi _\bullet )^2 +R_0 \\&= \rho ( ( \partial _{x_3}\log \rho )c_S^2\xi _{3,\bullet } + ( \partial _{x_3}\log c_S)(2c_S^2\xi _{3,\bullet } - \xi _{3, \bullet }^{-1}( \partial _t \phi _\bullet )^2)) + R_0. \end{aligned}$$

Also note that $\partial ^2_t \phi _{\bullet } = \partial ^2_t\left( -\tau t + x'\cdot \xi '\right) = 0$ on $\Gamma$. Thus, from a direct calculation, we obtain

$$\begin{aligned} (1/2\rho )P\phi _{\bullet } =&\frac{1}{2} \partial _t^2 \phi _{\bullet } - (2\rho )^{-1} \nabla _x\cdot \mu \nabla _x \phi _{\bullet }\\ =&-(1/2)( ( \partial _{x_3}\log \rho )c_S^2\xi _{3,\bullet } \!+\! ( \partial _{x_3}\log c_S)(2c_S^2\xi _{3,\bullet } - \xi _{3, \bullet }^{-1}( \partial _t \phi _\bullet )^2)) \!+\! R_0,\\&\text {on}\quad \Gamma . \end{aligned}$$

Therefore, going back to (A.2) we get

$$\begin{aligned}&c_S^2\xi _{3,\bullet } \partial _{x_3}(a_\bullet )_0\\&\quad = -(1/2)( ( \partial _{x_3}\log \rho )c_S^2\xi _{3,\bullet } + ( \partial _{x_3}\log c_S)(2c_S^2\xi _{3,\bullet } - \xi _{3, \bullet }^{-1}( \partial _t \phi _\bullet )^2))(a_\bullet )_0 + R_0,\\ \end{aligned}$$

so that

$$\begin{aligned} \partial _{x_3}(a_\bullet )_0 = -\left[ ( \partial _{x_3}\log \sqrt{\rho }) - \partial _{x_3}\log c_S\left( 1 - \frac{( \partial _t \phi _\bullet )^2}{2c_S^2 \xi ^2_{3,\bullet }}\right) \right] (a_\bullet )_0 + R_0. \end{aligned}$$

$\square$

Remark A.2

Observe that $\partial _{x_3}(a_{I})_0$ and $\partial _{x_3}(a_{R})_0$ are indeed $R_0$ terms, because $\rho ^{(-)}$ and $c_{S}^{(-)}$ are known on the $\Omega _{-}$ region and so are $\partial _{x_3}\rho ^{(-)}$, $\partial _{x_3}c_{S}^{(-)}$ on $\Gamma$. The rest of the terms in the expression of $\partial _{x_3}(a_{I})_0$ and $\partial _{x_3}(a_{R})_0$ can be determined from the 0-th order transmission condition (2.4). On the other hand $\partial _{x_3}(a_{T})_0$ is not $R_0$ but $R_1$ since it involves $\rho ^{(+)}$ and $c_{S}^{(+)}$, which cannot be determined from (2.4).

We provide the proofs of several claims made in the elastic case.

Proof of Lemma 2.3

We start with the transmission conditions for $(a_R)_{-1}$. From (2.5) for $J=-1$, we get

$$\begin{aligned} (a_R)_{-1} =&\frac{1}{\mu ^{(-)}\xi _{3, I} + \mu ^{(+)}\xi _{3, T}}\left[ \mu ^{(-)} \partial _{x_3}(a_I)_0 + \mu ^{(-)} \partial _{x_3}(a_R)_0 - \mu ^{(+)} \partial _{x_3}(a_T)_0 \right] . \end{aligned}$$

(A.5)

Note that $\mu ^{(+)}$ can be determined by the 0-th order transmission condition (see Lemma 2.1), therefore, $\left( \mu ^{(-)}\xi _{3, I} + \mu ^{(+)}\xi _{3, T}\right) ^{-1}$ is a $R_0$ quantity. Furthermore, thanks to Lemma 2.2, $\mu ^{(-)} \partial _{x_3}(a_I)_0$ and $\mu ^{(-)} \partial _{x_3}(a_R)_0$ are $R_0$, see Remark A.2. From (A.5) and (2.9) one obtains

$$\begin{aligned} (a_R)_{-1} =\ R_0 + \left[ \left( \partial _{x_3}\log \sqrt{\rho ^{(+)}}\right) - \partial _{x_3}\log c_S^{(+)}\left( 1 - \frac{( \partial _t \phi _T)^2}{2(c_S^{(+)})^2 \xi ^2_{3,T}}\right) \right] \frac{(a_T)_0}{R_0}. \end{aligned}$$

We denote $f=f(|\xi ' |/\tau ) = \left( 1 - \frac{( \partial _t \phi _T)^2}{2(c_S^{(+)})^2 \xi ^2_{3,T}}\right)$. If we have ${\tilde{\rho }}^{(-)} = \rho ^{(-)}$, ${\tilde{\mu }}^{(-)} = \mu ^{(-)}$ on $\Omega _{-}$ and $\tilde{R} = R$ on $\Gamma$, then one gets $\tilde{R}_0 = R_0$ and $(\tilde{a}_R)_{-1} = (a_R)_{-1}$ on $\Gamma$. Therefore, we obtain

$$\begin{aligned} R_0 +&\left[ \left( \partial _{x_3}\log \sqrt{\rho ^{(+)}}\right) - \partial _{x_3}\log c_S^{(+)}\left( 1 - \frac{( \partial _t \phi _T)^2}{2(c_S^{(+)})^2 \xi ^2_{3,T}}\right) \right] \frac{(a_T)_0}{R_0}\\ =&R_0 + \left[ \left( \partial _{x_3}\log \sqrt{{\tilde{\rho }}^{(+)}}\right) - \partial _{x_3}\log \tilde{c}_S^{(+)}\left( 1 - \frac{( \partial _t {\tilde{\phi }}_T)^2}{2(\tilde{c}_S^{(+)})^2 {\tilde{\xi }}^2_{3,T}}\right) \right] \frac{(a_T)_0}{R_0},\quad \text{ on }\quad \Gamma . \end{aligned}$$

Note that $(a_T)_0 \ne 0$ on the hyperbolic set when solving (2.4), which implies

$$\begin{aligned} \qquad \left( \partial _{x_3}\log \sqrt{\frac{\rho ^{(+)}}{{\tilde{\rho }}^{(+)}}}\right) =\left( \partial _{x_3}\log c_S^{(+)}\right) f - \left( \partial _{x_3}\log \tilde{c}_S^{(+)}\right) \tilde{f}\qquad \text{ on }\quad \Gamma . \end{aligned}$$

(A.6)

Observe that f is a $R_0$ quantity so that $f=\tilde{f}$. Furthermore, $\rho ^{(+)}$, ${\tilde{\rho }}^{(+)}$ depends only on x, hence by varying $(|\xi ' |/\tau )$ we obtain

$$\begin{aligned} \left( \partial _{x_3}\log \frac{c_S^{(+)}}{\tilde{c}_S^{(+)}}\right) f = \left( \partial _{x_3}\log \frac{c_S^{(+)}}{\tilde{c}_S^{(+)}}\right) f_1 \qquad \text{ on }\quad \Gamma , \end{aligned}$$

(A.7)

where f and $f_1$ are evaluated in different values of $|\xi ' |/\tau$. Note that $c_S^{(+)} = \tilde{c}_S^{(+)}$ on $\Gamma$ (see Lemma 2.1). If we take two values of $|\xi ' |/\tau$ such a way that $f \ne f_1$ on $\Gamma$, then (A.7) implies

$$\begin{aligned} \left( \partial _{x_3}\log \frac{c_S^{(+)}}{\tilde{c}_S^{(+)}}\right) = 0, \ \ \text{ so } \text{ that } \ \ \partial _{x_3}c_S^{(+)} = \partial _{x_3}\tilde{c}_S^{(+)} \quad \text{ on } \Gamma . \end{aligned}$$

Going back to (A.6) we obtain $\partial _{x_3}\rho ^{(+)} = \partial _{x_3}{\tilde{\rho }}^{(+)}$ and thus $\partial _{x_3}\mu ^{(+)} = \partial _{x_3}{\tilde{\mu }}^{(+)}$ on $\Gamma$. $\square$

Proof of Lemma 2.4

We prove this lemma via an iterative argument. First we note that for $J=0,-1$ we already have Lemma 2.1 and Lemma 2.3.

In order to prove the lemma for $J<-1$ we study the transport equation (2.6). For $J<0$, in the transport equations (2.6), we encounter the term $P(t,x,D_{t,x})(a_\bullet )_J = \rho \partial _t^2(a_\bullet )_J - \nabla \cdot \mu \nabla (a_\bullet )_J$. We calculate

$$\begin{aligned} \nabla \cdot \mu \nabla (a_\bullet )_0 =&\, ( \partial _{x_3}\mu ) \partial _{x_3}(a_\bullet )_0 + \mu \partial ^2_{x_3}(a_\bullet )_0 + R_0\\ =&\, \mu \partial ^2_{x_3}(a_\bullet )_0 + R_1 \\ =&\, -\mu \left[ ( \partial ^2_{x_3}\log \sqrt{\rho }) + \partial ^2_{x_3}\log c_S\left( 1 - \frac{( \partial _t \phi _\bullet )^2}{2c_S^2 \xi ^2_{3,\bullet }}\right) \right] (a_\bullet )_0\\&\, -\mu \left[ ( \partial _{x_3}\log \sqrt{\rho }) + \partial _{x_3}\log c_S\left( 1 - \frac{( \partial _t \phi _\bullet )^2}{2c_S^2 \xi ^2_{3,\bullet }}\right) \right] ^2(a_\bullet )_0 +R_1 \end{aligned}$$

Using the equation earlier for $\partial _{x_3}(a_\bullet )_0$, we see the second term is in fact $R_1$. Thus, we obtain

$$\begin{aligned} P(t,x,D_{t,x})(a_\bullet )_0 = \mu \left[ ( \partial ^2_{x_3}\log \sqrt{\rho }) + \partial ^2_{x_3}\log c_S\left( 1 - \frac{( \partial _t \phi _\bullet )^2}{2c_S^2 \xi ^2_{3,\bullet }}\right) \right] (a_\bullet )_0 + R_1\nonumber \\ \end{aligned}$$

(A.8)

Now, from the transport equation (2.6) and the relation (2.7) ,(A.8) we get

$$\begin{aligned} \partial _{x_3}(a_\bullet )_{-1} = -i/(2\xi _{3,\bullet }) \left[ ( \partial ^2_{x_3}\log \sqrt{\rho }) + \partial ^2_{x_3}\log c_S\left( 1 - \frac{( \partial _t \phi _\bullet )^2}{2c_S^2 \xi ^2_{3,\bullet }}\right) \right] (a_\bullet )_0 + R_1.\nonumber \\ \end{aligned}$$

(A.9)

Since the microlocal transmission conditions (2.5) helps us to connect $(a_R)_{-2}$ to $\partial _{x_3}(a_T)_{-1}$, using the same argument as in Lemma 2.3 we see that $(a_R)_{-2}$ uniquely determines $\partial ^2_{x_3} \rho ^{(+)}$ and $\partial ^2_{x_3} \mu ^{(+)}$ at $\Gamma$. Iterating the above argument gives us

$$\begin{aligned}&\partial _{x_3}(a_\bullet )_J \nonumber \\&\quad =(-i/(2\xi _{3,\bullet }))^{|J |} \left[ ( \partial ^{|J |+1}_{x_3}\log \sqrt{\rho }) + \partial ^{|J |+1}_{x_3}\log c_S\left( 1 - \frac{( \partial _t \phi _\bullet )^2}{2c_S^2 \xi ^2_{3,\bullet }}\right) \right] (a_\bullet )_0 + R_{|J |}.\nonumber \\ \end{aligned}$$

(A.10)

Then we get from the $|J |$-th order transmission conditions

$$\begin{aligned} \partial _{x_3}(a_T)_{J+1} = R_{|J+1 |}(R_{|J+1 |} - (a_R)_{J}) \text { on } \Gamma \end{aligned}$$

so that

$$\begin{aligned} (a_R)_J= & {} -\frac{ \partial _{x_3}(a_T)_{J+1}}{R_{|J+1 |}} + R_{|J+1 |} \nonumber \\= & {} -(-i/(2\xi _{3,T}))^J \left[ ( \partial ^{|J |}_{x_3}\log \sqrt{\rho ^{(+)}}) \right. \nonumber \\&\left. + \partial ^{|J |}_{x_3}\log c^{(+)}_S\left( 1 - \frac{( \partial _t \phi _T)^2}{2c_S^2 \xi ^2_{3,T}}\right) \right] \frac{(a_T)_{J+1}}{R_{|J+1 |}} + R_{|J+1 |} \end{aligned}$$

(A.11)

Using the same argument as above, and noting that the transmission conditions already determine $(a_T)_{J+1}$ from knowledge of $(a_R)_{J+1}$, shows that $(a_R)_J$ determines $\partial ^{|J |}_{x_3} \rho ^{(+)}$ and $\partial ^{|J |}_{x_3} \mu ^{(+)}$ at $\Gamma$. $\square$

This completes the proof of Theorem 1.1. The essential piece to make this work is verifying that $(a_R)_J$ at $\Gamma$ only depends on at most $|J |$ normal derivatives of the material parameters using the transmission conditions to continue unique recovery inductively. We finish this section by the following remark.

Remark A.3

Note that the recovery of the parameters on the boundary is obtained directly from the principal symbol of the reflection operator R, whereas recovering the higher order derivatives one relies on the recursive equations obtained from the interface conditions for the lower order terms of the asymptotic expansion of the geometric optics solution.

1.3 A.3 Proofs of several lemmas in the elastic case

Lastly in this appendix, we provide the remaining proof of lemma 3.1, which is similar to our previous computations and the

Remaining proof of lemma 3.1

Having $r_{11} = \tilde{r}_{11}$ we obtain

$$\begin{aligned} \begin{aligned} \tilde{\mathcal {A}}&\left[ {\hat{\tau }}^4\left( \xi _{3,I,S} + \xi _{3,T,S}\right) \left( \xi _{3,I,P} - \xi _{3,T,P}\right) \right. \\&\quad + 4{\hat{\tau }}^2\left( \mu ^{(-)} - \mu ^{(+)}\right) \left( \xi _{3,I,P}\xi _{3,I,S} - \xi _{3,T,S}\xi _{3,T,P}\right) \\&\left. \quad + 4\left( \mu ^{(-)} - \mu ^{(+)}\right) ^2\left( 1+\xi _{3,I,S}\xi _{3,I,P}\right) \left( 1-\xi _{3,T,S}\xi _{3,T,P}\right) \right] \\&= \mathcal {A}\left[ {\hat{\tau }}^4\left( \xi _{3,I,S} + \xi _{3,T,S}\right) \left( \xi _{3,I,P} - {\tilde{\xi }}_{3,T,P}\right) \right. \\&\quad + 4{\hat{\tau }}^2\left( \mu ^{(-)} - \mu ^{(+)}\right) \left( \xi _{3,I,P} \xi _{3,I,S} - \xi _{3,T,S}{\tilde{\xi }}_{3,T,P}\right) \\&\quad \left. + 4\left( \mu ^{(-)} - \mu ^{(+)}\right) ^2\left( 1+\xi _{3,I,S}\xi _{3,I,P}\right) \left( 1-{\tilde{\xi }}_{3,T,S}{\tilde{\xi }}_{3,T,P}\right) \right] . \end{aligned} \end{aligned}$$

Here we use that fact that so far we have $\rho ^{(\pm )} = {\tilde{\rho }}^{(\pm )}$, $\mu ^{(\pm )} = {\tilde{\mu }}^{(\pm )}$, $c_S^{(\pm )} = \tilde{c_S}^{(\pm )}$ on $\Gamma$ and therefore, $\xi _{3,\bullet ,S} = {\tilde{\xi }}_{3,\bullet ,S}$. Equating the coefficient of ${\hat{\tau }}^8$ on the both sides of the above equation we obtain

$$\begin{aligned} \begin{aligned}&\frac{\left( \xi _{3,I,S} + \xi _{3,T,S}\right) \left( \xi _{3,I,P} - \xi _{3,T,P}\right) }{\left( \xi _{3,I,S} - \xi _{3,T,S}\right) \left( \xi _{3,I,P} + \xi _{3,T,P}\right) } = \frac{\left( \xi _{3,I,S} + \xi _{3,T,S}\right) \left( \xi _{3,I,P} - {\tilde{\xi }}_{3,T,P}\right) }{\left( \xi _{3,I,S} - \xi _{3,T,S}\right) \left( \xi _{3,I,P} + {\tilde{\xi }}_{3,T,P}\right) }\\&\text {which implies}\quad \frac{\left( \xi _{3,I,P} - \xi _{3,T,P}\right) }{\left( \xi _{3,I,P} + \xi _{3,T,P}\right) } = \frac{\left( \xi _{3,I,P} - {\tilde{\xi }}_{3,T,P}\right) }{\left( \xi _{3,I,P} + {\tilde{\xi }}_{3,T,P}\right) }. \end{aligned} \end{aligned}$$

After a cross multiplication and simplification, on $\Gamma$ we obtain

$$\begin{aligned} \frac{\xi _{3,T,P}}{{\tilde{\xi }}_{3,T,P}}= & {} \frac{\xi _{3,I,P}}{\xi _{3,I,P}} = 1, \text { which implies }\ \frac{|\xi ' |^2 - \tau ^2(c_P^{(+)})^{-2}}{|\xi ' |^2 - \tau ^2(\tilde{c}_P^{(+)})^{-2}} = 1 \\&\text {and thus, }\ (c_P^{(+)})^{-2} = (\tilde{c}_P^{(+)})^{-2}, \quad [\because \tau \ne 0]. \end{aligned}$$

Since $c_P^{(+)}$ and $\tilde{c}_P^{(+)}$ are wave speeds and cannot be negative, hence, we obtain $c_P^{(+)} = \tilde{c}_P^{(+)}$ on $\Gamma$. $\square$

We also finish the proof of proposition 4.5 in the elastic case.

proof of proposition 4.5 in the elastic case:

First, we need the following operator appearing in the elastic transport equations (see (Rachele 2000, equation (25)))

$$\begin{aligned} P_{sc}(t,x,D) \phi = \rho ( \partial _t^2 \phi ) - \left[ ( \nabla _x \otimes \lambda \nabla _x \phi ) + (\text {div} \mu \nabla _x \phi ))I + ( \nabla _x \otimes \mu \nabla _x \phi )^t \right] . \end{aligned}$$

The divergence term is computed exactly as in the acoustic case with the mean curvature appearing, which is still an $R_0$ term.

Next, we let $\nabla$ be the Levi-Civita connection and $e_\alpha = \partial _{{\tilde{x}}_\alpha }$ a basis for $T\Omega$ near $\Gamma$ for $\alpha = 1, 2, 3$, and dual basis $\{{\hat{e}}_\alpha \}$. Note that $\partial _{{\tilde{x}}_3} = \partial _\nu$ when restricted to $\Gamma$ by fixing a unit normal $\nu$ with the correspending sign. Then $\nabla _x \phi = \partial _{{\tilde{x}}_\alpha }\phi \ e_\alpha$ with the usual summation convention and

$$\begin{aligned} \nabla _{{\tilde{x}} _j}( \lambda \nabla _x \phi ) = ( \partial _{{\tilde{x}}_j} \lambda ) \partial _{{\tilde{x}}_\alpha }\phi + \lambda \partial _{{\tilde{x}}_j,\tilde{x}_\alpha } \phi + \lambda \partial _{{\tilde{x}}_\beta }\phi \Gamma ^\alpha _{\beta j} e_\alpha , \end{aligned}$$

where in particular, the last term is the only new term we have from the flat case and it is an $R_0$ term. An analogous calculation applies to $( \nabla _x \otimes \mu \nabla _x \phi )^t$. Let $N = \nabla \phi _P/|\nabla \phi _P |$ and observe that

$$\begin{aligned} N^t( \partial _{{\tilde{x}}_\beta }\phi _P \Gamma ^\alpha _{\beta \gamma } e_\alpha \otimes {\hat{e}}_\gamma )N =|\nabla \phi _P |^{-2} \Gamma ^\alpha _{\beta \gamma } ( \partial _{{\tilde{x}}_\beta }\phi _P) ( \partial _{{\tilde{x}}_\alpha }\phi _P) ( \partial _{{\tilde{x}}_\gamma }\phi _P). \end{aligned}$$

Since $\Gamma ^3_{3 j} = 0$ in our coordinates, if we pick $\xi _{tan} = 0$ (normal incidence wave), then the quantity above vanishes at $\Gamma$. Thus,

$$\begin{aligned}&\frac{N^t(P_{sc}(t,x,D)\phi _{P,\bullet })N}{-2\rho c_P^2\tilde{\xi }_{3,\bullet }}(\alpha _\bullet )_0\nonumber \\&\quad =-\left( \frac{1}{2}( \partial _\nu \log c_P) \left[ 1+\frac{c_S^2}{c_P^2}(1-\frac{|\nabla _x \phi _{P,\bullet } |^2}{( \partial _\nu \phi _{P,\bullet }^2})\right] + ( \partial _\nu \log \sqrt{\rho }) \right) (\alpha _\bullet )_0\nonumber \\&\qquad \quad -\frac{c_S^2}{2c_P^2} H(x) (\alpha _\bullet )_0 + R_0, \end{aligned}$$

(A.12)

where for $\xi _{tan} = 0$, $R_0$ contains no curvature terms when restricted to $\Gamma$.

Following (Rachele 2000, equation (34)), we also need to compute the term

$$\begin{aligned}N^t( \partial _{\tau ,{{\tilde{\xi }}}} p)(t, {\tilde{x}}, \partial _{t,{\tilde{x}}} \phi _P) \cdot \partial _{t, {\tilde{x}}}N\end{aligned}$$

restricted to $\Gamma$ where p is the principal symbol of Q. We will also choose $\xi _{tan} = 0$ initially. Observe that $\partial _{\tilde{\xi }_1} ( {\tilde{\xi }}\otimes {\tilde{\xi }}) = {\tilde{\xi }}\otimes e_1 + e_1 \otimes {\tilde{\xi }}$ so that when restricted to ${\tilde{\xi }}= d\phi _P$, at $\Gamma$, for $\xi _{tan} =0$ (we assume all statements in this subsection are with this restriction)

$$\begin{aligned} N^t \partial _{{{\tilde{\xi }}}_1} ( {\tilde{\xi }}\otimes {\tilde{\xi }}) \cdot \partial _{ \tilde{x}_1}N= & {} (N\cdot {\tilde{\xi }}) \langle e_1, \nabla _{e_1}N\rangle = \frac{( \partial _\nu \phi _P)^2}{|\nabla \phi _P |} \langle e_1, \nabla _{e_1}[ ( \partial _\nu \phi _P/|\nabla \phi _P |)e_3]\rangle \nonumber \\= & {} \frac{( \partial _\nu \phi _P)^3}{|\nabla \phi _P |^2} \langle e_1, \nabla _{e_1}e_3\rangle = \xi _{3,P} \langle e_1, \nabla _{e_1}e_3\rangle . \end{aligned}$$

(A.13)

Using that $\langle \nu , \nabla _\nu \nu \rangle = 0$, we get

$$\begin{aligned} N^t \partial _{{{\tilde{\xi }}}_1} ( {\tilde{\xi }}\otimes {\tilde{\xi }}) \cdot \partial _{ \tilde{x}_1}N + N^t \partial _{{{\tilde{\xi }}}_2} ( {\tilde{\xi }}\otimes {\tilde{\xi }}) \cdot \partial _{ \tilde{x}_2}N = \xi _{3,P} H(x) \end{aligned}$$

(A.14)

where H(x) is from before and proportional to the mean curvature at $\Gamma$. Similarly,

$$\begin{aligned} N^t \partial _{{{\tilde{\xi }}}_3} ( {\tilde{\xi }}\otimes {\tilde{\xi }}) \cdot \partial _{ \tilde{x}_3}N= & {} \frac{( \partial _\nu \phi _P)^3}{|\nabla \phi _P |^2} \langle e_3, \partial _\nu ( N_{tan} + e_3 \partial _\nu \phi _P/|\nabla \phi _P |) \rangle \\= & {} \xi _{3,P} \partial _\nu \left( \frac{ \partial _\nu \phi _P}{|\nabla \phi _P |} \right) . \end{aligned}$$

One may also check that the quantity $N^t ( \partial _{ {\tilde{\xi }}} | {\tilde{\xi }} |^2 I) \cdot \partial _{ \tilde{x}}N$ has no curvature terms when we restrict to $\Gamma$ and $\xi _{tan} = 0$. Then from (Rachele 2000, equation (50)), we get

$$\begin{aligned}&\frac{-\left[ N_\bullet ( \partial _{\tau ,{{\tilde{\xi }}}} p)(t, {\tilde{x}}, \partial _{t,{\tilde{x}}} \phi _P) \cdot \partial _{t, {\tilde{x}}}N_\bullet \right] (\alpha _\bullet )_0}{-2\rho c_P^2 \xi _{3, \bullet , P}} \nonumber \\&\quad = ( \partial _\nu \log c_P)\left[ \frac{c_{\lambda +\mu }^2|\xi _{tan} |^2}{2c_P^2( \partial _\nu \phi _P)^2} \right] (\alpha _\bullet )_0 + \frac{c_{\lambda +\mu }^2}{2c_P^2}H(x) (\alpha _\bullet )_0 + R_0, \end{aligned}$$

(A.15)

where $R_0$ also does not contain any curvature terms and can be compute from the values of the parameters and not their normal derivative. Combining these calculations, we obtain from (Rachele 2000, lemma 3.5)

$$\begin{aligned} \partial _\nu (\alpha _\bullet )_0= & {} -\left[ \frac{1}{2}( \partial _\nu \log c_P) \left( 1 - \frac{|\xi _{tan} |^2}{( \partial _\nu \phi _P)^2} \right) \right. \nonumber \\&\left. + ( \partial _\nu \log (\sqrt{\rho }) \right] (\alpha _\bullet )_0 + b_\bullet H(x) (\alpha _\bullet )_0 + R_0 \end{aligned}$$

(A.16)

where $R_0$ does not depend on curvature when restricted to $\Gamma$ and $\xi _{tan} = 0$, and $b_\bullet$ is a fixed constant whose sign is dependent on $\bullet .$ One may derive an analogous formula for $\partial _\nu (\alpha _{k,\bullet })_0$ for $k=1,2$ using (Rachele 2000, lemma 3.5) as well. To recover the mean curvature as in the acoustic case, we use the transmission conditions (3.13) for the lower order symbols and this produces an additional term:

$$\begin{aligned}&\begin{bmatrix}(\mathcal {A}_R)_J\\ (\mathcal {A}_T)_J\end{bmatrix} =D^{-1} \begin{bmatrix}S_I\\ \mathcal {T}_I\end{bmatrix}(\mathcal {A}_I)_J \nonumber \\&\quad + D^{-1} \Big [\begin{array}{ll} &{}0\\ B_I(x,D)(a_I)_{J+1} + B_R(x,D) &{}(a_R)_{J+1} - B_T(x,D)(a_T)_{J+1} \end{array}\Big ],\qquad \qquad \end{aligned}$$

(A.17)

where we denoted by D the matrix appearing in (3.13) which does not depend on curvature. The right hand side of the above equation will contain curvature terms in the form of $\partial _{ \tilde{x}_i}N_{\bullet }$ and $\partial _{ \tilde{x}_i}N_{\bullet ,k}$. However, when $\xi _{tan} = 0$ and restricted to $\Gamma$, the inner product of these vectors with $N_\bullet$ will not have curvature terms since $\Gamma ^3_{3l} = 0$. Thus, by setting $J = -1$ and taking the inner product of the equation for $(\mathcal {A}_R)_{-1}$ in (A.17) with $N_R$, then we can recover the mean curvature from $(\mathcal {A}_R)_{-1}$ as in the acoustic case by using (A.16). After a long computation similar to the one above, we may then recover $\partial _\nu H(x)$ at the boundary from $(\mathcal {A}_R)_{-2}$ as in the acoustic case and invoke lemma 4.3, keeping in mind that H(x) and $\kappa$ are proportional. $\square$