$(a\\#b)(x,y,\xi,\eta):=(2\pi)^{-n}\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}{e^{i[\xi^{\prime}X+\eta^{\prime}\cdot
Y]}a(x,y,\xi^{\prime}+\xi,\eta^{\prime}+\eta)b(x,y,X,Y,\xi,\eta)\,d\xi^{\prime}\,d\eta^{\prime}\,dX\,dY}.$

to show

$\partial_{\eta_{j}}(a\\#b)=(\partial_{\eta_{j}}a)\\#b+a\\#(\partial_{\eta_{j}}b).$

For all $j$ we have $\partial_{\eta_{j}}a\in S^{m-1,l,k+1,j+1}\subset
S^{m,l,k,j}$ and $\partial_{\eta_{j}}b\in
S^{m^{\prime}-1,l,k^{\prime}+1,j^{\prime}+1}\subset
S^{m^{\prime},l^{\prime},k^{\prime},j^{\prime}}_{full}$, so the lemma holds
for $V=\partial_{\eta_{j}}$ (the same logic holds for $\partial_{\xi}$ as
well). Furthermore, since

$\eta_{i}(a\\#b)=a\\#(\eta_{i}b),$

it follows that for $V=\eta_{i}\partial_{\eta_{j}}$ we can write

$\eta_{i}\partial_{\eta_{j}}(a\\#b)=(\partial_{\eta_{j}}a)\\#(\eta_{i}b)+a\\#(\eta_{i}\partial_{\eta_{j}}b)=(\partial_{\eta_{j}}a)\\#(\eta_{i}b)+a\\#(Vb).$

We again have $\partial_{\eta_{j}}a\in S^{m-1,l,k+1,j+1}$, while if
$i=2,\dots,n-1$, then

$\eta_{i}=\rho^{-1}\tau_{i}\in S^{1,0,-1,-1}.$

Thus, $\eta_{i}b\in
S^{m^{\prime}+1,l^{\prime},k^{\prime}-1,j^{\prime}-1}_{full}$, so the orders
of $\partial_{\eta_{j}}a$ and $\eta_{i}b$ add to the desired quantities.
Hence, the lemma holds for $V=\eta_{i}\partial_{\eta_{j}}$ with
$i=2,\dots,n-1$. The same argument holds for
$V=\xi\partial_{\xi}$,$\eta_{i}\partial_{\xi}$, and $\xi\partial_{\eta_{j}}$.
For $V=\eta_{j}\partial_{\eta_{1}}$, we have

$V(a\\#b)=(\partial_{\eta_{1}}a)\\#(\eta_{j}b)+a\\#(Vb).$

Noting that $\partial_{\eta_{1}}a\in S^{m-1,l,k,j}$ since
$\partial_{\eta_{1}}$ is tangent to $\\{\tau=0\\}$, and $\eta_{j}b\in
S^{m^{\prime}+1,l^{\prime},k^{\prime},j^{\prime}}$, we see that the lemma
holds here as well. Finally, for $V=x^{1/2}\rho^{-1/2}\partial_{\eta_{i}}$, we
have

$V(a\\#b)=(\partial_{\eta_{i}}a)\\#(x^{1/2}\rho^{-1/2}b)+a\\#Vb.$

In the first term, we have $\partial_{\eta_{i}}a\in S^{m-1,l,k+1,j+1}\subset
S^{m-1/2,l+1/2,k,j}$ and $x^{1/2}\rho^{-1/2}b\in
S^{m^{\prime}+1/2,l^{\prime}-1/2,k^{\prime},j^{\prime}}$, so the sum of the
orders add to the desired quantities, while in the second term we have that
$Vb$ has the same orders as $b$. Hence the lemma holds here as well; the logic
applies similarly to $V=x^{1/2}\rho^{-1/2}\partial_{\xi}$. ∎

From Theorem 4.7, we obtain the following ellipticity result, following
standard arguments in microlocal analysis:

######  Proposition 4.11 (Elliptic Parametrix).

Suppose $a\in S^{m,l,k,j}$ satisfies an ellipticity estimate

$|a|\geq c\rho^{-m}x^{-l}d_{\Sigma}^{-k}d_{\Gamma}^{-j}\quad\text{near
}\partial\mathscr{X}_{2}.$

If $A$ is the corresponding quantization of $a$, then $A$ admits an elliptic
parametrix, i.e. there exists $B\in\Psi^{-m,-l,-k,-j}$ such that

$BA=\text{Id}+R,\quad R\in\Psi^{-\infty,-\infty}.$

Similarly, using Theorem 4.7 and Proposition 4.6, we obtain the
$L^{2}$-boundedness property, again following standard arguments in microlocal
analysis:

######  Proposition 4.12 ($L^{2}$-boundedness).

An operator in $\Psi^{0,0,0,0}$ is bounded as a map from
$L^{2}\left([0,\infty)\times\mathbb{R}^{n-1},\frac{dx\,dy}{x^{(n+1)}}\right)$
to itself.

Recalling that the scattering Sobolev spaces $H^{s,r}_{sc}$ are defined by the
property that scattering operators in $\Psi^{s,r}$ map it into $L^{2}$, we
obtain the following corollary of the $L^{2}$-boundedness of $bb$-operators:

######  Corollary 4.13.

An operator in $\Psi^{m,l,0,0}$ is bounded as a map from $H^{s,r}_{sc}$ to
$H^{s-m,r-l}_{sc}$ for any $s,r\in\mathbb{R}$.

######  Remark 4.14.

In general, we can conclude mapping properties of operators in
$\Psi^{m,l,k,j}$ (for $k$, $j$ not necessarily zero) by including it into a
space of the form $\Psi^{m^{\prime},l^{\prime},0,0}$: for example, we have

$\Psi^{m,l,k,j}\subset\Psi^{m+k_{+}/2,l+j_{+}/2,0,0},$

so operators in $\Psi^{m,l,k,j}$ map $H^{s,r}_{sc}$ into
$H^{s-(m+k_{+}/2),r-(l+j_{+}/2)}_{sc}$ for any $s,r\in\mathbb{R}$.

##  5\. Ellipticity of Operators from Elastic Inverse Problem

We now investigate the composition of the pseudolinearization operator $I$
with a “localized” formal adjoint of the form (1.3). Recall in this case that
we are working near some point $p$ on the boundary, with $\tilde{x}$ a
function strictly convex with respect to all dynamics considered satisfying
$\tilde{x}(p)=0$ and $d\tilde{x}(p)\neq 0$, and that we have introduced an
artificial boundary $\\{x=0\\}$, where $x=\tilde{x}+c$, with $c>0$ a small
number of our choice. We show that the composition of $I$ with such localized
formal adjoints are, after conjugation with a factor of $e^{\digamma/x}$,
scattering pseudodifferential operators, which normally are not elliptic as
scattering operators, but are elliptic in a new operator class to be defined.

###  5.1. The localized operators

We recall that our operators of interest are the pseudolinearization operators

$I[u](x_{0},\xi_{0})=\int_{\mathbb{R}}{A(X(t,x_{0},\xi_{0}),\Xi(t,x_{0},\xi_{0}))u(X(t,x_{0},\xi_{0}))\,dt},$

with $A(x,\xi)=-E^{\nu}(x,\xi)\frac{\partial\tilde{\Xi}}{\partial\xi}(x,\xi)$,
and a formal adjoint of the form

$Lv(x,y)=x^{-2}\int_{\mathbb{S}^{2}}{\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\frac{\digamma^{2}\hat{\lambda}^{2}}{2\alpha}\right)v(\gamma_{\lambda,\omega})\,d\mathbb{S}^{2}(\lambda,\omega)}$

(recalling that $x$ is as above and $y=(y_{1},y_{2})$ are tangential
coordinates such that $(x,y)$ form valid coordinates near $p$). Here
$0<\epsilon<1/2$, $\digamma>0$, $\chi\in C_{c}^{\infty}(\mathbb{R})$ is
identically $1$ in a neighborhood of $0$, $\hat{\lambda}=\lambda/x$,
$\gamma_{\lambda,\omega}$ is the trajectory passing through $(x,y)$ with
tangent vector $\lambda\partial_{x}+\omega\cdot\partial_{y}$, and
$\alpha=\alpha(x,y,\lambda,\omega)$ is an “acceleration” factor in the $x$
direction satisfying

$\gamma_{\lambda,\omega}-x=\lambda t+\alpha
t^{2}+O(t^{3})\quad(\text{equivalently
}\frac{\gamma_{\lambda,\omega}-x}{x^{2}}=\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}+O(x\hat{t}^{3})\text{
if }\hat{\lambda}=\frac{\lambda}{x},\hat{t}=\frac{t}{x}).$

Note that this differs slightly in structure from previous papers, as the
cutoff chosen is not a compact cutoff in $\hat{\lambda}$, but rather a
Gaussian. Such a cutoff works well with finite point symbol computations, but
in previous works one uses a compactly supported approximation of a Gaussian
using the Gaussian computation as justification; here we will directly use the
Gaussian. However, we will still select for $\hat{\lambda}$ in a controlled
manner, namely that $\hat{\lambda}$ is at most $O(x^{-\epsilon})$ as $x\to 0$,
i.e. $\lambda$ is $O(x^{1-\epsilon})$; as we will see below, this is good
enough for dynamics purposes.

We remark that for the pseudolinearization operator $I$, we are free to adjust
the factor $A$ so that it is supported in $\\{x<2c,|t|<T\\}$, where $T$ is
larger than the maximum travel time of a trajectory in
$\overline{\Omega}\cap\\{x\geq 0\\}$, since the function of interest, namely
the difference of parameters $r_{\nu}$, is supported in $\overline{\Omega}$.
Since $x=\tilde{x}+c$, by making $c$ arbitrarily small we can arrange for $T$
to be sufficiently small as well; this will turn out to be useful for
dynamical assumptions below.

For $N=L\circ I$, we then have

$\displaystyle N[u](x,y)$
$\displaystyle=x^{-2}\int_{\mathbb{R}\times\mathbb{S}^{2}}{\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\frac{\digamma^{2}\hat{\lambda}^{2}}{2\alpha}\right)A((X,\Xi)(x,y,t,\lambda,\omega))u(X(x,y,t,\lambda,\omega))\,dt\,d\mathbb{S}^{2}(\lambda,\omega)}.$

Here, we let $(X,\Xi)(x,y,t,\lambda,\omega)$ denote the Hamilton flow at time
$t$ (viewed in the cotangent bundle) starting at $(x,y)$ with initial
(spatial) velocity $\lambda\partial_{x}+\omega\cdot\partial_{y}$ (equivalently
with initial momentum corresponding to an initial velocity of
$\lambda\partial_{x}+\omega\cdot\partial_{y}$). We write
$A(x,y,t,\lambda,\omega)=A((X,\Xi)(x,y,t,\lambda,\omega))$ and
$(x^{\prime},y^{\prime})=X(x,y,t,\lambda,\omega)$; this now frees up the
letter $X$ to be used as a scattering coordinate. We can thus write its
Schwartz kernel as

$\displaystyle K(x,y,X,Y)$
$\displaystyle=x^{-2}\int_{\mathbb{R}\times\mathbb{S}^{2}}{\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\frac{\digamma^{2}\hat{\lambda}^{2}}{2\alpha}\right)A(x,y,t,\lambda,\omega)\delta\left(X-\frac{x^{\prime}-x}{x^{2}},Y-\frac{y^{\prime}-y}{x}\right)\,dt\,d\mathbb{S}^{2}(\lambda,\omega)}$
$\displaystyle=\int_{\mathbb{R}_{\hat{t}}\times\mathbb{R}_{\hat{\lambda}}\times\mathbb{S}^{1}_{\omega}}{\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\frac{\digamma^{2}\hat{\lambda}^{2}}{2\alpha}\right)A(x,y,x\hat{t},x\hat{\lambda},\omega)\delta\left(X-\frac{x^{\prime}-x}{x^{2}},Y-\frac{y^{\prime}-y}{x}\right)\,d\hat{t}\,d\hat{\lambda}\,d\mathbb{S}^{1}(\omega)}.$

Note that we can write

$d\mathbb{S}^{2}(\lambda,\omega)=d\lambda\,d\mathbb{S}^{1}(\omega)$

since we are in dimension $3$. It turns out that the correct operator to
consider is not $N$ itself, but rather its conjugate
$N_{\digamma}:=e^{-\digamma/x}Ne^{\digamma/x}$; its Schwartz kernel is just
the above Schwartz kernel times $e^{-\digamma/x}e^{\digamma/x^{\prime}}$.
Hence

$\displaystyle K_{\digamma}(x,y,X,Y)$
$\displaystyle=\int_{\mathbb{R}_{\hat{t}}\times\mathbb{R}_{\hat{\lambda}}\times\mathbb{S}^{1}_{\omega}}\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}-\left(\frac{1}{x^{\prime}}-\frac{1}{x}\right)\right)\right)A(x,y,x\hat{t},x\hat{\lambda},\omega)$
$\displaystyle\cdot\delta\left(X-\frac{x^{\prime}-x}{x^{2}},Y-\frac{y^{\prime}-y}{x}\right)\,d\hat{t}\,d\hat{\lambda}\,d\mathbb{S}^{1}(\omega).$

We can find the scattering symbol of $N$ by taking the $(X,Y)$-Fourier
transform of its $(X,Y)$-Schwartz kernel. Thus the symbol has the form

(5.1)
$a=\int_{\mathbb{R}_{\hat{t}}\times\mathbb{R}_{\hat{\lambda}}\times\mathbb{S}^{1}_{\omega}}{e^{i\left(\xi\frac{x^{\prime}-x}{x^{2}}+\eta\cdot\frac{y^{\prime}-y}{x}\right)}\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}-\left(\frac{1}{x^{\prime}}-\frac{1}{x}\right)\right)\right)A(x,y,\hat{\lambda},\omega,\hat{t})\,d\hat{t}\,d\hat{\lambda}\,d\omega}.$

We now comment on the Hamiltonian dynamics near $p$, which determine what
assumptions we may put on the function $A$ above.Since the function
$\tilde{x}$ (and hence $x=\tilde{x}+c$) is chosen to be convex with respect to
the Hamiltonian dynamics involved, we can make an estimate on the acceleration
component $\alpha\geq C_{0}$ for some constant $C_{0}>0$. In fact, since we
have $x=\tilde{x}+c$ and are also free to adjust $c$ to consider all
sufficiently small $c$, with the region of interest (i.e.
$\overline{\Omega_{c}}=\\{\tilde{x}>-c\\}\cap\overline{\Omega}$) shrinking to
a point as $c\to 0^{+}$, we may choose $c$ sufficiently small so that $\alpha$
also satisfies an upper bound in a neighborhood of $\overline{\Omega_{c}}$,
say $C_{0}\leq\alpha\leq 1.01C_{0}$; note that by Taylor’s theorem this would
also imply that

$C_{0}x^{2}\hat{t}^{2}\leq\frac{x^{\prime}-x}{x^{2}}-\hat{\lambda}\hat{t}\leq
1.01C_{0}x^{2}\hat{t}^{2}$

in this region, where $x^{\prime}$ is the $x$ component of
$\gamma_{\lambda,\omega}(t)$. Note that the lower bound implies the existence
of $C^{\prime}$ such that

$|\lambda|\leq\frac{C}{x^{1/2}}\implies x^{\prime}\geq 0\text{ for all }t$

(see [11] for a justification), and hence the cutoff
$\chi(x^{\epsilon}\hat{\lambda})$ guarantees that all trajectories with
parameters supported in the support of this cutoff stay in the region
$\\{x>0\\}$ for all $t$. Moreover, the function $A$ is smooth, grows at most
polynomially as $(\hat{t},\hat{\lambda})\to\infty$, and by the remarks
regarding the pseudolinearization operator $I$, we can arrange for $A$ to be
supported in a set of the form $\\{x<2c,x|\hat{t}|<T\\}$ where $c,T>0$ can be
arranged to be arbitrarily small.

###  5.2. Scattering symbol

We first show such integrals are indeed scattering symbols. If the Gaussian
$\exp\left(-\frac{\digamma^{2}\hat{\lambda}^{2}}{2\alpha}\right)$ were
replaced by a $C_{c}^{\infty}$ function approximating it, then this would be a
result of the works in [11] etc.; however we can apply similar methods even
after the Gaussian change. We first show, for each $(x,y)$, that the integral
admits an asymptotic expansion in $|(\xi,\eta)|$ as $|(\xi,\eta)|\to\infty$.
We then show that applying $b$-vector fields like $x\partial_{x}$ and
$\partial_{y}$ essentially returns the integral to the same class, for which
we also have a similar asymptotic expansion, thus showing the symbolicity of
the integral.

The idea is that the phase should have stationary points along precisely the
submanifold $\\{\hat{t}=0,\xi\hat{\lambda}+\eta\cdot\omega=0\\}$, with an
asymptotic expansion following stationary phase on that submanifold, and the
integral away from the stationary set should be suitably controlled.

To rigorously justify this, we first show that the integral, while over a non-
compact region in $\hat{\lambda}$ and $\hat{t}$, is well-controlled due to the
(real) exponential factor, which will end up decaying at least exponentially.
Specifically, we have the following:

######  Lemma 5.1.

On the support of the integrand (i.e. for $|\hat{\lambda}|\leq Cx^{-\epsilon}$
and $|\hat{t}|\leq T/x$), there exist constants $c_{1},c_{2}>0$ such that

$\frac{\hat{\lambda}^{2}}{2\alpha}+\left(\frac{1}{x}-\frac{1}{x^{\prime}}\right)\geq\begin{cases}c_{1}(\hat{\lambda}^{2}+\hat{t}^{2})&\text{for
}|\hat{t}|\leq\frac{c_{2}}{x^{1/2}},\\\
c_{1}(\hat{\lambda}^{2}+|\hat{t}|)&\text{for
}|\hat{t}|\geq\frac{c_{2}}{x^{1/2}}\end{cases}.$

###### Proof.

We take $c_{2}=\frac{1}{5\sqrt{C_{0}}}$. In the region of support we have
$|\hat{\lambda}|\leq Cx^{-\epsilon}=Cx^{1/2-\epsilon}x^{-1/2}$, so by choosing
$c$ sufficiently small, we can arrange
$|\hat{\lambda}|\leq(100c_{2})^{-1}x^{-1/2}$ on the region of support. Then in
the region $|\hat{t}|\leq c_{2}/x^{1/2}$ we have

$x^{\prime}-x\geq x^{2}(C_{0}\hat{t}^{2}-|\hat{\lambda}\hat{t}|)\geq
x^{2}(-0.01c_{2}^{-1}x^{-1/2}\cdot c_{2}x^{-1/2})=-0.01x$

i.e. $x^{\prime}\geq 0.99x$, while similarly

$x^{\prime}-x\leq x^{2}(|\hat{\lambda}\hat{t}|+1.01C_{0}\hat{t}^{2})\leq
x^{2}\left(0.01x^{-1}+1.01C_{0}\cdot\frac{1}{25C_{0}}x^{-1}\right)\leq 0.06x$

so $x^{\prime}\leq 1.06x$. Note that when $x^{\prime}\leq x$, we have
$\frac{1}{x}-\frac{1}{x^{\prime}}=\frac{x^{\prime}-x}{xx^{\prime}}\geq\frac{x^{\prime}-x}{0.99x^{2}}$
(since $x^{\prime}-x\leq 0$ there), while in $x^{\prime}\geq x$ we have
$\frac{1}{x}-\frac{1}{x^{\prime}}\geq\frac{x^{\prime}-x}{1.06x^{2}}$. Thus, we
have

$\frac{\hat{\lambda}^{2}}{2\alpha}+\left(\frac{1}{x}-\frac{1}{x^{\prime}}\right)\geq\left\\{\begin{matrix}0.99\left(\frac{\hat{\lambda}^{2}}{1.98\alpha}+\frac{x^{\prime}-x}{x^{2}}\right)&x^{\prime}\leq
x\\\
1.06\left(\frac{\hat{\lambda}^{2}}{2.12\alpha}+\frac{x^{\prime}-x}{x^{2}}\right)&x^{\prime}\geq
x\end{matrix}\right.$

It thus suffices to obtain a lower bound on
$\frac{\hat{\lambda}^{2}}{2.12\alpha}+\frac{x^{\prime}-x}{x^{2}}$ in this
region. Note that we can bound

$\displaystyle\frac{\hat{\lambda}^{2}}{2.12\alpha}+\frac{x^{\prime}-x}{x^{2}}\geq\frac{\hat{\lambda}^{2}}{2.12\alpha}+C_{0}\hat{t}^{2}-|\hat{\lambda}\hat{t}|$
$\displaystyle\geq\frac{\hat{\lambda}^{2}}{2.12\alpha}+C_{0}\hat{t}^{2}-\frac{1}{2}\left(\frac{1}{2(C_{0}-\epsilon)}\hat{\lambda}^{2}+2(C_{0}-\epsilon)\hat{t}^{2}\right)$
$\displaystyle\geq\frac{1}{2}\left(\frac{1}{1.06\alpha}-\frac{1}{2(C_{0}-\epsilon)}\right)\hat{\lambda}^{2}+\epsilon\hat{t}^{2}.$

For sufficiently small $\epsilon$, we have $2(C_{0}-\epsilon)>1.06\alpha$
(since by assumption we will always have $\alpha\leq 1.01C_{0}$), which then
gives the desired lower bound.

On the other region $|\hat{t}|\geq c_{2}x^{-1/2}$, we have

$x^{\prime}-x\geq
x^{2}(C_{0}\hat{t}^{2}-|\hat{\lambda}\hat{t}|)=x^{2}|\hat{t}|(C_{0}|\hat{t}|-|\hat{\lambda}|).$

Note that
$|\hat{\lambda}|\leq(100c_{2})^{-1}x^{-1/2}=\frac{1}{20}\sqrt{C_{0}}x^{-1/2}=\frac{C_{0}c_{2}}{4}x^{-1/2}\leq\frac{C_{0}}{4}|\hat{t}|$,
so we have

$x^{2}|\hat{t}|(C_{0}|\hat{t}|-|\hat{\lambda}|)\geq
x^{2}|\hat{t}|\frac{3}{4}C_{0}|\hat{t}|\geq\frac{3}{4}x^{2}C_{0}(c_{2}^{2}x^{-1})=0.03x,$

i.e. $x^{\prime}\geq 1.03x$ in the region. Thus
$\frac{1}{x}-\frac{1}{x^{\prime}}\geq\left(1-\frac{1}{1.03}\right)\frac{1}{x}$.
Since on the region of support we have $x|\hat{t}|\leq T$, we have
$\frac{1}{x}\geq|\hat{t}|/T$. Thus, we have

$\frac{\hat{\lambda}^{2}}{2\alpha}+\frac{1}{x}-\frac{1}{x^{\prime}}\geq\frac{1}{2\alpha}\hat{\lambda}^{2}+\left(1-\frac{1}{1.03}\right)\frac{1}{x}\geq\frac{1}{2\alpha}\hat{\lambda}^{2}+\frac{1-\frac{1}{1.03}}{T}|\hat{t}|,$

as desired. ∎

Thus, the exponential term
$\exp\left(-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}-\left(\frac{1}{x^{\prime}}-\frac{1}{x}\right)\right)\right)$
will decay at least exponentially as $(\hat{\lambda},\hat{t})\to\infty$, so
any polynomial-order growth in the other terms will not affect the
integrability of the integral. Furthermore, derivatives of the exponent will
also be at most polynomial growth in $(\hat{\lambda},\hat{t})$, so derivatives
of the exponential term will also not affect integrability.

We now cut off outside a compact neighborhood of $\hat{t}$. To do so, write
$\gamma=\left(\frac{x^{\prime}-x}{x^{2}},\frac{y^{\prime}-y}{x}\right)$, i.e.

$\gamma=(\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}+O(x\hat{t}^{3}),\omega\hat{t}+O(x\hat{t}^{2})).$

Also write $(\xi,\eta)=|(\xi,\eta)|(\tilde{\xi},\tilde{\eta})$ so
$\tilde{\xi}^{2}+|\tilde{\eta}|^{2}=1$. Then the oscillatory exponent becomes
$|(\xi,\eta)|\phi$ where $\phi=(\tilde{\xi},\tilde{\eta})\cdot\gamma$.

We now calculate derivatives of $\gamma$. Let $\partial_{\theta}$ be the
angular derivative on $\mathbb{S}^{1}$, so that
$\partial_{\theta}\omega=\omega^{\perp}:=(-\omega_{2},\omega_{1})$. We have

$\displaystyle\partial_{\hat{t}}\gamma$
$\displaystyle=(\hat{\lambda}+2\alpha\hat{t}+O(x\hat{t}^{2}),\omega+O(x\hat{t}))$
$\displaystyle\partial_{\hat{\lambda}}\gamma$
$\displaystyle=(\hat{t}+\partial_{\hat{\lambda}}\alpha\hat{t}^{2}+O(x\hat{t}^{3}),O(x\hat{t}^{2}))=\hat{t}\left(1+\partial_{\hat{\lambda}}\alpha\hat{t}+O(x\hat{t}^{2}),O(x\hat{t})\right)$
$\displaystyle\partial_{\theta}\gamma$
$\displaystyle=(\partial_{\theta}\alpha\hat{t}^{2}+O(x\hat{t}^{3}),\omega^{\perp}\hat{t}+O(x\hat{t}^{2}))=\hat{t}\left(\partial_{\theta}\alpha\hat{t}+O(x\hat{t}^{2}),\omega^{\perp}+O(x\hat{t})\right).$

We can actually be a bit more precise with the $\hat{\lambda}$ derivative:
note that we can write

$x^{\prime}-x=\lambda
t+\alpha(x,y,\lambda,\omega)t^{2}+\Gamma(x,y,\lambda,\omega,t)t^{3}$

for some smooth $\Gamma$, and thus

$\frac{x^{\prime}-x}{x^{2}}=\hat{\lambda}\hat{t}+\alpha(x,y,x\hat{\lambda},\omega)\hat{t}^{2}+\Gamma(x,y,x\hat{\lambda},\omega,x\hat{t})\hat{t}^{3}.$

Thus we can actually write

$\partial_{\hat{\lambda}}\left(\frac{x^{\prime}-x}{x^{2}}\right)=\hat{t}\left(1+x\partial_{\lambda}\alpha\hat{t}+\partial_{\lambda}\Gamma
x^{2}\hat{t}^{2}\right).$

Recall we always have $x\hat{t}=O(T)$ where we can arrange $T$ to be
arbitrarily small. It follows that if we write the vectors
$(\frac{\partial_{\hat{\lambda}}\gamma}{\hat{t}},\frac{\partial_{\theta}\gamma}{\hat{t}},\partial_{\hat{t}}\gamma)$
as a matrix, we have

$\begin{pmatrix}\frac{\partial_{\hat{\lambda}}\gamma}{\hat{t}}&\frac{\partial_{\theta}\gamma}{\hat{t}}&\partial_{\hat{t}}\gamma\end{pmatrix}=\begin{pmatrix}1+O(T)&(\partial_{\theta}\alpha+O(T))\hat{t}&\hat{\lambda}+(2\alpha+O(T))\hat{t}^{2}\\\
\vec{0}_{2\times 1}&\omega&\omega^{\perp}\end{pmatrix}+O(T).$

Note that this matrix has determinant $1+O(T)$ and is thus invertible, with an
inverse whose entries grow at most polynomially along with all derivatives. It
follows that

$\displaystyle\begin{pmatrix}\partial_{\hat{\lambda}}\gamma&\partial_{\theta}\gamma&\partial_{\hat{t}}\gamma\end{pmatrix}$
$\displaystyle=\begin{pmatrix}\frac{\partial_{\hat{\lambda}}\gamma}{\hat{t}}&\frac{\partial_{\theta}\gamma}{\hat{t}}&\partial_{\hat{t}}\gamma\end{pmatrix}\begin{pmatrix}\hat{t}&0&0\\\
0&\hat{t}&0\\\ 0&0&1\end{pmatrix}$
$\displaystyle\implies\begin{pmatrix}\partial_{\hat{\lambda}}\gamma&\partial_{\theta}\gamma&\partial_{\hat{t}}\gamma\end{pmatrix}^{-1}$
$\displaystyle=\begin{pmatrix}1/\hat{t}&0&0\\\ 0&1/\hat{t}&0\\\
0&0&1\end{pmatrix}\begin{pmatrix}\frac{\partial_{\hat{\lambda}}\gamma}{\hat{t}}&\frac{\partial_{\theta}\gamma}{\hat{t}}&\partial_{\hat{t}}\gamma\end{pmatrix}^{-1},$

i.e.
$\begin{pmatrix}\partial_{\hat{\lambda}}\gamma&\partial_{\theta}\gamma&\partial_{\hat{t}}\gamma\end{pmatrix}$
is invertible away from $\hat{t}=0$, with inverse whose entries are
$\frac{1}{\hat{t}}$ times terms growing at most polynomially along with all
derivatives. All of this leads to the following:

######  Lemma 5.2.

For each unit vector $(\tilde{\xi},\tilde{\eta})\in\mathbb{S}^{2}$ there
exists a smooth vector field $V$ on
$\mathbb{R}_{\hat{t}}\times\mathbb{R}_{\hat{\lambda}}\times\mathbb{S}^{1}_{\omega}$
such that $V\gamma=\hat{t}(\tilde{\xi},\tilde{\eta})$. The coefficients of $V$
are bounded polynomially and have derivatives also bounded polynomially, with
the bounds uniform in $(\tilde{\xi},\tilde{\eta})$. Moreover, the coefficient
of $\partial_{\hat{t}}$ in $V$ will vanish at $\hat{t}=0$. In particular, on
any region of the form $\\{|\hat{t}|>\epsilon\\}$, there exists a smooth
vector field $\tilde{V}$ such that
$\tilde{V}\gamma=(\tilde{\xi},\tilde{\eta})$, with the coefficients of
$\tilde{V}$ satisfying the same properties as above on
$\\{|\hat{t}|>\epsilon\\}$.

Thus, we can split $a=a_{1}+a_{2}$, where $a_{1}$ is the integral in (5.1)
with a cutoff $\chi_{1}(\hat{t})$ inserted where $\chi_{1}\in
C_{c}^{\infty}(\mathbb{R})$ is identically one in a neighborhood of $0$. We
then have $a_{2}$ is decreasing to order $|(\xi,\eta)|^{-\infty}$, since we
can argue as follows: note that for a fixed $(\tilde{\xi},\tilde{\eta})$, with
$V$ chosen accordingly by the lemma above, we have

$Ve^{i|(\xi,\eta)|(\tilde{\xi},\tilde{\eta})\gamma}=|(\xi,\eta)|e^{i|(\xi,\eta)|(\tilde{\xi},\tilde{\eta})\gamma}.$

Thus repeated integration by parts w.r.t. $|(\xi,\eta)|^{-1}V$ gives the
desired decay in $|(\xi,\eta)|$, noting that since the integrand in $a_{2}$
will be supported uniformly away from $\\{\hat{t}=0\\}$, the coefficients of
$V$ will be bounded polynomially (along with its derivatives), and any
additional polynomial growth will be harmless against the exponential decay in
the integral.

We can assume the cutoff $\chi_{1}$ is supported say in
$[-1/(100C_{0}),1/(100C_{0})]$, and thus we focus on

(5.2)
$a_{1}=\int_{[-1/(100C_{0}),1/(100C_{0})]_{\hat{t}}\times\mathbb{R}_{\hat{\lambda}}\times\mathbb{S}^{1}_{\omega}}{e^{i|(\xi,\eta)|(\tilde{\xi},\tilde{\eta})\cdot\gamma}\chi_{1}(\hat{t})A^{\prime}(x,y,\hat{\lambda},\omega,\hat{t})\,d\hat{t}\,d\hat{\lambda}\,d\omega}$

where
$A^{\prime}(x,y,\hat{\lambda},\omega,\hat{t})=\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}-\left(\frac{1}{x^{\prime}}-\frac{1}{x}\right)\right)\right)A(x,y,\hat{\lambda},\omega,\hat{t})$.
The stationary points will end up being at
$\\{\hat{t}=0,\tilde{\xi}\hat{\lambda}+\tilde{\eta}\cdot\omega=0\\}$. Since
the support of $\hat{\lambda}$ is not uniformly compact (in particular the
support increases to $\mathbb{R}$ as $x\to 0$), we will divide the domain into
two further regions.

We first suppose $|\tilde{\xi}|$ is small. Let
$\omega_{\parallel}=\frac{\tilde{\eta}}{|\tilde{\eta}|}\cdot\omega$. We now
divide $\mathbb{S}^{1}_{\omega}$ via smooth cutoffs, say

$1=\chi_{1,1}(\omega_{\parallel})+\chi_{1,2}(\omega_{\parallel}),$

where $\chi_{1,1}$ is supported where $|\omega_{\parallel}|$ is small (say at
most $2/3$), and $\chi_{1,2}$ is supported where $|\omega_{\parallel}|$ is
large (say at least $1/2$). Let $a_{1}=a_{1,1}+a_{1,2}$ be the corresponding
partition of the integral in (5.2). In the former region we write
$\omega=\omega_{\parallel}\frac{\tilde{\eta}}{|\tilde{\eta}|}+\sqrt{1-\omega_{\parallel}^{2}}\omega_{\perp}$,
where $\omega_{\perp}$ is one of two vectors in $\mathbb{S}^{1}$ perpendicular
to $\tilde{\eta}$, so the integral becomes

$a_{1,1}=\int_{\mathbb{R}\times\mathbb{S}^{0}}\int_{-\frac{1}{100C_{0}}}^{\frac{1}{100C_{0}}}\int_{-2/3}^{2/3}{e^{i(\xi,\eta)\cdot\gamma}\chi_{1,1}(\omega_{\parallel})\chi_{1}(\hat{t})A^{\prime}(x,y,\hat{\lambda},\omega,\hat{t})(1-\omega_{\parallel}^{2})^{-1/2}\,d\omega_{\parallel}\,d\hat{t}\,d\hat{\lambda}\,d\omega_{\perp}}.$

We will use $\hat{t}$ and $\omega_{\parallel}$ as the stationary phase
variables, and $\hat{\lambda}$ and $\omega_{\perp}$ as parameters. We also
write the phase as

$|\eta|\left(\frac{\xi}{|\eta|},\frac{\eta}{|\eta|}\right)\cdot\gamma=|\eta|\left(\frac{\xi}{|\eta|}(\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}+O(x\hat{t}^{3}))+\omega_{\parallel}\hat{t}+O(x\hat{t}^{2})\right)$

We split the integral in $\hat{\lambda}$ into two parts: one where
$|\frac{\xi}{|\eta|}\hat{\lambda}|<1$ and one where
$|\frac{\xi}{|\eta|}\hat{\lambda}|\geq 3/4$. In the former part where
$|\frac{\xi}{|\eta|}\hat{\lambda}|<1$, we can proceed with stationary phase as
usual: the stationary set is

$\left\\{\hat{t}=0,\frac{\xi}{|\eta|}\hat{\lambda}+\omega_{\parallel}=0\right\\}$

and the Hessian at the stationary set (w.r.t. $\hat{t},\omega_{\parallel}$) is

$\begin{pmatrix}\frac{\xi}{|\eta|}\alpha&1\\\ 1&0\end{pmatrix}.$

Thus stationary phase gives that the integral satisfies the asymptotic
expansion999Note that the lower order terms are not exactly as described here
since the phase is not exactly a quadratic form, but the fact that an
asymptotic expansion exists remains true.

$\displaystyle|\eta|^{-1}\sum_{j=0}^{\infty}\frac{i^{j}}{j!|\eta|^{j}}\int_{\mathbb{R}\times\mathbb{S}^{0}}$
$\displaystyle\left(\partial_{\hat{t}}\partial_{\omega_{\parallel}}-\frac{\xi}{|\eta|}\partial_{\omega_{\parallel}}^{2}\right)^{j}$
$\displaystyle\left[\chi_{1,1}(\omega_{\parallel})\chi_{1}(\hat{t})A^{\prime}(x,y,\hat{\lambda},\omega,\hat{t})(1-\omega_{\parallel}^{2})^{-1/2}\right]|_{\hat{t}=0,\omega_{\parallel}=-\frac{\xi}{|\eta|}\hat{\lambda}}\,d\hat{\lambda}\,d\omega_{\perp}.$

In the latter part where $|\frac{\xi}{|\eta|}\hat{\lambda}|\geq 3/4$, we note
that due to support properties we can arrange for the phase to be non-
stationary: note that

$\partial_{\hat{t}}\left(\left(\frac{\xi}{|\eta|},\frac{\eta}{|\eta|}\right)\cdot\gamma\right)=\frac{\xi}{|\eta|}\hat{\lambda}+\frac{\xi}{|\eta|}(2\alpha+O(x\hat{t}))\hat{t}+\omega_{\parallel}+O(x\hat{t}),$

so due to all the support properties (i.e.
$\frac{\xi}{|\eta|}\hat{\lambda}\geq 3/4$, $|\omega_{\parallel}|<2/3$,
$\hat{t}$ being sufficiently small (this is where the $1/(100C_{0})$ comes
in), etc.), we have that the above derivative is bounded below, so we can
integrate by parts without problem.

For the second integral $a_{1,2}$, we will not reparametrize $\omega$ by
$\omega_{\parallel}$, as it will be a singular parametrization in the region
of support. Instead, we use $\omega$ as a parameter and use $\hat{t}$,
$\hat{\lambda}$ as the stationary phase variables. We split up the integral
similarly as before using a cutoff in $\tilde{\xi}\hat{\lambda}$:

$1=\tilde{\chi}_{1}\left(\tilde{\xi}\hat{\lambda}\right)+\tilde{\chi}_{2}\left(\tilde{\xi}\hat{\lambda}\right),$

with the support of $\tilde{\chi}_{1}$ contained where
$|\tilde{\xi}\hat{\lambda}|$ is at least $1/3$ and the support of
$\tilde{\chi}_{2}$ contained where $|\tilde{\xi}\hat{\lambda}|$ is strictly
less than $1/2$. On the latter region one can integrate by parts as before
(noting that now we have $|\omega_{\parallel}|$ is at least $1/2$ on the
region of support). On the former region, we have

$a_{1,2}^{\prime}=\int_{\mathbb{S}^{1}}{\chi_{1,2}(\omega_{\parallel})\left(\int\int{e^{i|(\xi,\eta)|(\tilde{\xi},\tilde{\eta})\cdot\gamma}\tilde{\chi}_{1}\left(\tilde{\xi}\hat{\lambda}\right)\chi_{1}(\hat{t})A^{\prime}(x,y,\hat{\lambda},\omega,\hat{t})\,d\hat{t}\,d\hat{\lambda}}\right)\,d\omega}.$

When we try to perform stationary phase w.r.t. $(\hat{t},\hat{\lambda})$, we
obtain that the stationary set is still
$\hat{t}=0,\tilde{\xi}\hat{\lambda}+\tilde{\eta}\cdot\omega=0$, with the
Hessian now

$\begin{pmatrix}2\tilde{\xi}\alpha&\tilde{\xi}\\\ \tilde{\xi}&0\end{pmatrix}.$

One complication with stationary phase is the following: the determinant of
the Hessian is now $-\tilde{\xi}^{2}$, which may be small; however for this
integral to be nonzero at all requires $\tilde{\xi}$ to be nonzero! Thus the
stationary phase expansion would give

$a_{1,2}^{\prime}\sim\sum_{j\geq
0}{\frac{1}{|(\xi,\eta)|^{j+1}}\int_{\mathbb{S}^{1}}{\frac{\chi_{1,2}(\omega_{\parallel})}{\tilde{\xi}^{2}}\left(\frac{\partial_{\hat{t}}\partial_{\hat{\lambda}}-\alpha\partial_{\hat{\lambda}}^{2}}{\tilde{\xi}}\right)^{j}\left(\tilde{\chi}_{1}\left(\tilde{\xi}\hat{\lambda}\right)\chi_{1}(\hat{t})A^{\prime}(x,y,\hat{\lambda},\omega,\hat{t})\right)\Big{|}_{\hat{t}=0,\hat{\lambda}=(-\tilde{\eta}\cdot\omega)/\tilde{\xi}}\,d\omega}}.$

Note that any derivatives on the integrand
$\tilde{\chi}_{1}\left(\tilde{\xi}\hat{\lambda}\right)\chi_{1}(\hat{t})A^{\prime}(x,y,\hat{\lambda},\omega,\hat{t})$
equals the exponential factor
$\exp\left(-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}-\left(\frac{1}{x^{\prime}}-\frac{1}{x}\right)\right)\right)$
times terms which are smooth in $(x,y,\hat{\lambda},\omega,\hat{t})$ (aside
from possible $x^{\epsilon}$ terms from the $\chi(x^{\epsilon}\hat{\lambda})$,
which are nonetheless conormal to $x=0$). The exponential factor evaluated at
$\hat{t}=0,\hat{\lambda}=(-\tilde{\eta}\cdot\omega)/\tilde{\xi}$ gives the
term
$\exp\left(-\digamma\frac{\omega_{\parallel}^{2}}{2\alpha(\xi/|\eta|)^{2}}\right)$,
i.e. a term which is bounded by a Gaussian in $1/\tilde{\xi}$. It follows that
the powers of $1/\tilde{\xi}$ which arise from the degeneracy of the phase are
actually completely controlled by the exponential decay, and thus this term,
while not necessarily zero away from $\tilde{\xi}=0$, will vanish extremely
rapidly (in particular smoothly) as one approaches $\tilde{\xi}=0$.

Thus, in the region where $\tilde{\xi}$ is relatively small, we have that $a$
satisfies an asymptotic expansion in $(\xi,\eta)$ for each fixed $(x,y)$. In
the region where $\tilde{\xi}$ is relatively large, we can perform stationary
phase in $(\hat{t},\hat{\lambda})$ like the above case (but without needing to
split the integral into regions), noting that the determinant
$-\tilde{\xi}^{2}$ will be nondegenerate in this region.

Thus overall we have that $a$ satisfies an asymptotic expansion

$a\sim\sum_{j\geq
0}{a_{-1-j}\left(x,y,\frac{(\xi,\eta)}{|(\xi,\eta)|}\right)|(\xi,\eta)|^{-1-j}}$

for every $(x,y)$. This shows, for every fixed $(x,y)$, that $a$ is a symbol
in $(\xi,\eta)$. To show it is a scattering symbol, we can note that, by
following the stationary phase arguments above, that the coefficients will be
conormal to $x=0$, as the integrand consists of terms which are smooth, except
for $\chi(x^{\epsilon}\hat{\lambda})$ whose derivatives may introduce terms
like $x^{\epsilon}$, which are still conormal. Alternatively, we can use the
characterization that $a$ is a symbol if for all $(\alpha,\beta)$, we have
that $(xD_{x})^{\alpha}D_{y}^{\beta}a$ satisfies a similar asymptotic
expansion, i.e. taking vector fields tangent to the boundary does not affect
its symbol behavior in $(\xi,\eta)$. Indeed, for
$A(x,y,\hat{\lambda},\omega,\hat{t})$, if we let

$I[A](x,y,\xi,\eta)=\int_{\mathbb{R}_{\hat{t}}\times\mathbb{R}_{\hat{\lambda}}\times\mathbb{S}^{1}_{\omega}}{e^{i(\xi,\eta)\cdot\gamma}\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}-\left(\frac{1}{x^{\prime}}-\frac{1}{x}\right)\right)\right)A(x,y,\hat{\lambda},\omega,\hat{t})\,d\hat{t}\,d\hat{\lambda}\,d\omega}$

then

$(xD_{x})I[A]=I[A^{\prime}]$

where

$A^{\prime}=xD_{x}A+\left(\frac{\epsilon
x^{\epsilon}\hat{\lambda}\chi^{\prime}(x^{\epsilon}\hat{\lambda})}{\chi(x^{\epsilon}\hat{\lambda})}+(\xi,\eta)\cdot
x\partial_{x}\gamma+xD_{x}\left(-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}-\left(\frac{1}{x^{\prime}}-\frac{1}{x}\right)\right)\right)\right)A.$

Essentially the only possibly problematic term is the $(\xi,\eta)\cdot
xD_{x}\gamma$, given that a priori it adds an additional order to the symbol.
However, we note that

$x\partial_{x}\gamma=x\partial_{x}(\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}+O(x\hat{t}^{3}),\omega\hat{t}+O(x\hat{t}^{2}))=xO(\hat{t}^{2})$

(where $O(\hat{t}^{2})$ means $\hat{t}^{2}$ times smooth), so using the
integration by parts Lemma, we can use one factor of $\hat{t}$ to integrate by
parts, thus cancelling the additional $|(\xi,\eta)|$ order. (We note that we
can actually do so again because the coefficient of the necessary vector field
$V$ will vanish at $\hat{t}=0$, so it applied to $\hat{t}$ times smooth will
keep it that way. In other words, the term is actually one order lower, but we
will not need this fact here.) The exponential term is also fine, since

$\frac{1}{x}-\frac{1}{x^{\prime}}=\frac{X}{1+xX}=\frac{\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}+O(x\hat{t}^{3})}{1+x\hat{\lambda}\hat{t}+x\alpha\hat{t}^{2}+O(x^{2}\hat{t}^{3})}$

so

$x\partial_{x}\left(\frac{1}{x}-\frac{1}{x^{\prime}}\right)=O(\hat{\lambda}\hat{t})+O(\hat{t}^{2}).$

Thus this contributes a lower-order contribution to the symbol, but more
importantly the derivative is well-behaved as $x\to 0$.

###  5.3. Symbol computation

We will assume our symbol is in fact smooth on the unscaled variables, i.e. we
write it as $A(x,y,x\hat{\lambda},\omega,x\hat{t})$ where $A$ is smooth in
$(x,y,\lambda,\omega,t)$, and suppose that $A$ satisfies the following
property: for a fixed $\eta_{0}\in\mathbb{S}^{1}$, we have

$A\geq 0\text{ everywhere},\quad
A(x,y,\lambda,\omega,0)=0\iff\omega\cdot\eta_{0}=0.$

Note this automatically implies that all derivatives of $A$ vanish at
$(x,y,\lambda,\omega,0)$ where $\omega\cdot\eta_{0}=0$. Moreover, for $\omega$
near the equatorial sphere $\omega\cdot\eta_{0}=0$, decompose $\omega$ as
$\omega=\omega_{\parallel}\eta_{0}+\sqrt{1-\omega_{\parallel}^{2}}\omega_{\perp}$,
and assume that $\partial^{2}_{\omega_{\parallel}}A\geq C>0$ when
$\omega\cdot\eta_{0}=0$.

Note that for our operators of interest, we take $\eta_{0}=(1,0)$, i.e.
$\eta_{0}\cdot dy=dy_{1}$, as we have chosen coordinates such that $dy_{1}$
aligns with the axis of isotropy.

Furthermore, write $y^{\prime}-y=\omega t+\beta(x,y,\lambda,\omega,t)t^{2}$,
so that
$\frac{y^{\prime}-y}{x}=\omega\hat{t}+x\beta(x,y,x\hat{\lambda},\omega,x\hat{t})\hat{t}^{2}$.
Assume as well that

$(\partial_{t}\partial_{\omega_{\parallel}}-\eta_{0}\cdot\beta\partial_{\omega_{\parallel}}^{2})A(x,y,x\hat{\lambda},\omega,0)\geq
C>0\text{ for all }\omega\in\eta_{0}^{\perp}.$

Let $\Sigma=\text{span }\eta_{0}\frac{dy}{x}$. Let $a$ admit the stationary
phase expansion

$a\sim\sum_{j=0}^{\infty}{a_{-1-j}(x,y,\tilde{\xi},\tilde{\eta})|(\xi,\eta)|^{-1-j}}.$

We then have the following:

######  Lemma 5.3.

For $a$ as above, the first three terms $a_{-1}$, $a_{-2}$, and $a_{-3}$ have
the following properties:

  * •

$a_{-1}\geq 0$ and vanishes only on $\Sigma$, where it vanishes quadratically;
moreover the vanishing is non-degenerate. In particular, $a_{-1}$ satisfies a
bound of the form $|a_{-1}|\geq C|\tau|^{2}$, where $\tau=(\tau_{1},\tau_{2})$
is a boundary-defining function for $\Sigma\cap\mathbb{S}^{2}$.

  * •

$a_{-2}$ is of the form $a_{-2}=ix\tilde{a}$, where $\tilde{a}\in S^{0,0}$ and
is real-valued and non-vanishing on $\Sigma$. In particular, $|\tilde{a}|\geq
C$ in a region of the form $|\tau|<\epsilon$.

  * •

$a_{-3}$ is positive on $\Sigma\cap\\{x=0\\}$, i.e. $a_{-3}\geq C$ in a region
of the form $|\tau|+x<\epsilon$.

Moreover, all of the lower bounds above have constants which can be chosen to
vary continuously as $A$ varies in the $C^{2}$ norm and as the phase varies in
the $C^{0}$ norm101010In the following sense: the phase can always be written
as $(\hat{\lambda}\hat{t},\omega\hat{t})+O(\hat{t}^{2})$; the $C^{0}$
continuity is in the coefficient of $\hat{t}^{2}$ in the $O(\hat{t}^{2})$
term..

###### Proof.

The quadratic vanishing of $a_{-1}$ is well-known; the only comment is that
the constant in the lower bound follows from the lower bound of the second
derivative $\partial^{2}_{\omega_{\parallel}}A$. For the other quantities, we
calculate the symbol at $(\xi,\eta)\sim(0,\eta_{0})$. This simplifies
calculations, since in the decomposition $a=a_{1,1}+a_{1,2}$ in the previous
section, we only need the $a_{1,1}$ term since $\tilde{\xi}=0$. Thus, for the
other two terms, it suffices to calculate the stationary phase expansions of

$\int_{\mathbb{R}\times\mathbb{S}^{0}}\int_{-1/(100C_{0})}^{1/(100C_{0})}\int_{-2/3}^{2/3}{e^{i(\xi,\eta)\cdot\gamma}\chi_{1,1}(\omega_{\parallel})\chi_{1}(\hat{t})A^{\prime}(x,y,x\hat{\lambda},\omega,x\hat{t})(1-\omega_{\parallel}^{2})^{-1/2}\,d\omega_{\parallel}\,d\hat{t}\,d\hat{\lambda}\,d\omega_{\perp}}.$

For $(\xi,\eta)=\pm|(\xi,\eta)|(0,\eta_{0})$, we have that the phase is
$|\eta|(\omega_{\parallel}\hat{t}+x\eta_{0}\cdot\beta\hat{t}^{2})$. Thus the
term $a_{-2}$ equals

$i\int_{\mathbb{R}\times\mathbb{S}^{0}}{\chi(x^{\epsilon}\hat{\lambda})\left(\partial_{\hat{t}}\partial_{\omega_{\parallel}}-x\eta_{0}\cdot\beta\partial_{\omega_{\parallel}}^{2}\right)\left(e^{-\digamma(\hat{\lambda}^{2}+x^{-1}-(x^{\prime})^{-1})}A(x,y,x\hat{\lambda},\omega,x\hat{t})\right)|_{\hat{t}=0,\omega_{\parallel}=0}\,d\hat{\lambda}\,d\omega_{\perp}}.$

Note that $A$ vanishes on the set $\\{\hat{t}=0,\omega_{\parallel}=0\\}$, and
hence derivatives on the exponential factor times $A$ will vanish except for
the terms where both derivatives fall on $A$. Thus the term is111111Note that
this calculation implicitly assumes that $\beta$ is a constant; however we can
approximate $\beta$ by its value at $\\{\hat{t}=0,\omega_{\parallel}=0\\}$;
doing so will introduce an error of order
$O(x\omega_{\parallel}\hat{t}^{2})+O(x\hat{t}^{3})$ into the phase, and this
error vanishes to sufficiently high degree at the critical set such that it
will not affect the subprincipal behavior.

$ix\int_{\mathbb{R}\times\mathbb{S}^{0}}{\chi(x^{\epsilon}\hat{\lambda})(\partial_{t}\partial_{\omega_{\parallel}}-\eta_{0}\cdot\beta(x,y,x\hat{\lambda},\omega_{\perp},0)\partial_{\omega_{\parallel}}^{2})A(x,y,x\hat{\lambda},\omega_{\perp},0)\,d\hat{\lambda}\,d\omega_{\perp}}.$

By assumption, the integrand is bounded from below everywhere; moreover such a
bound will vary $C^{2}$ in $A$ and $C^{0}$ in the phase (due to the $\beta$
term).

For the term $a_{-3}$, we calculate the term at $x=0$ and on $\Sigma$. At
$x=0$, the phase is just $\omega_{\parallel}\hat{t}$, and the term
$x^{-1}-(x^{\prime})^{-1}$ in the exponential is
$\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}$, and the cutoff
$\chi(x^{\epsilon}\hat{\lambda})$ is $1$. Hence the term $a_{-3}$ is

$\displaystyle\frac{i^{2}}{2}\int_{\mathbb{R}\times\mathbb{S}^{0}}{(\partial_{\hat{t}}\partial_{\omega_{\parallel}})^{2}\left(e^{-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}+\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}\right)}A(0,y,0,\omega,0)\right)|_{\hat{t}=0,\omega_{\parallel}=0}\,d\hat{\lambda}\,d\omega_{\perp}}$
$\displaystyle=-\frac{1}{2}\int_{\mathbb{R}\times\mathbb{S}^{0}}{\left(\partial_{\hat{t}}^{2}\left(e^{-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}+\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}\right)}\right)\partial_{\omega_{\parallel}}^{2}A(0,y,0,\omega,0)\right)|_{\hat{t}=0,\omega_{\parallel}=0}\,d\hat{\lambda}\,d\omega_{\perp}}$
$\displaystyle=\frac{1}{2}\int_{\mathbb{R}\times\mathbb{S}^{0}}{(2\digamma\alpha-\digamma^{2}\hat{\lambda}^{2})e^{-\digamma\hat{\lambda}^{2}/(2\alpha)}\partial^{2}_{\omega_{\parallel}}A(0,y,0,\omega_{\perp},0)\,d\hat{\lambda}\,d\omega_{\perp}}.$

(Note that the second line follows since the $\hat{t}$ derivatives can only
fall on the exponential term to produce a nonzero term, and the
$\omega_{\parallel}$ derivative must fall at least twice on $A$ to produce a
nonzero term since $A$ vanishes to second order in $\omega$.) Since

$\int_{\mathbb{R}}{(2\digamma\alpha-\digamma^{2}\hat{\lambda}^{2})e^{-\digamma\hat{\lambda}^{2}/(2\alpha)}\,d\hat{\lambda}}=\sqrt{2\pi}\alpha^{3/2}\digamma^{1/2}$

it follows that

$a_{-3}=\frac{\sqrt{\pi\digamma}}{2}\sum_{\omega_{\perp}}{\alpha^{3/2}\partial^{2}_{\omega_{\parallel}}A(0,y,0,\omega_{\perp},0)}>0.$

Note again that this depends continuously in the $C^{2}$ norm for $A$ and in
the $C^{0}$ norm for the phase (via $\alpha$). ∎

Note that while this means the symbol is not elliptic when considered as a
standard scattering symbol, by Lemma 3.7 the result gives us that the symbol
is elliptic as a symbol in $S^{-1,0,-2,-2}$, at least near fiber infinity.

We still need to show that the symbol is elliptic at finite points, so we
discuss this now. Recall that at $x=0$ the symbol becomes

$a(0,y,\xi,\eta)=\int{e^{i(\xi,\eta)\cdot(\hat{\lambda}\hat{t}+\alpha\hat{t}^{2},\omega\hat{t})}e^{-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}+\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}\right)}A(0,y,0,\omega,0)\,d\hat{t}\,d\hat{\lambda}\,d\omega}.$

where $\alpha=\alpha(0,y,0,\omega)$. We can evaluate this integral as follows:

We make the substitution $X=\hat{\lambda}\hat{t}+\alpha\hat{t}^{2}$,
$Y=\omega\hat{t}$. Then
$dX\,dY=\hat{t}^{2}\,d\hat{t}\,d\hat{\lambda}\,d\omega\implies
d\hat{t}\,d\hat{\lambda}\,d\omega=|Y|^{-2}\,dX\,dY$. Furthermore,
$\hat{\lambda}=\frac{X-\alpha|Y|^{2}}{|Y|}$. Thus the integral becomes

$\displaystyle a$
$\displaystyle=\int_{\mathbb{R}_{X}\times\mathbb{R}^{2}_{Y}}{e^{i(\xi
X+\eta\cdot
Y)}e^{-\digamma\left(\frac{(X-\alpha|Y|^{2})^{2}}{2\alpha|Y|^{2}}+X\right)}A(0,y,0,\frac{Y}{|Y|},0)|Y|^{-2}\,dX\,dY}$
$\displaystyle=\int_{\mathbb{R}_{X}\times\mathbb{R}^{2}_{Y}}{e^{i(\xi
X+\eta\cdot
Y)}e^{-\digamma\left(\frac{X^{2}}{2\alpha|Y|^{2}}+\frac{\alpha|Y|^{2}}{2}\right)}A(0,y,0,\frac{Y}{|Y|},0)|Y|^{-2}\,dX\,dY}.$

Evaluating the Fourier transform in $X$, we get

$a=\sqrt{2\pi}\int_{\mathbb{R}^{2}_{Y}}{e^{i\eta\cdot
Y}e^{-\frac{\alpha\xi^{2}|Y|^{2}}{2\digamma}}e^{-\frac{\digamma\alpha|Y|^{2}}{2}}A(0,y,0,\frac{Y}{|Y|},0)\sqrt{\alpha/\digamma}|Y|^{-1}\,dY}.$

Note that all terms in the integrand aside fro $e^{i\eta\cdot Y}$ are real.
Hence, we can evaluate the real part121212If we assume $A(0,y,0,\omega,0)$ is
even in $\omega$, as one often does in practice, then we actually know $a$
must be real. In any case, we can always estimate the real part from below. of
$a$ by polar coordinates:

$\displaystyle\text{Re }a$
$\displaystyle=\sqrt{2\pi}\int_{\mathbb{R}^{2}_{Y}}{\cos(\eta\cdot
Y)e^{-\frac{\alpha\xi^{2}|Y|^{2}}{2\digamma}}e^{-\frac{\digamma\alpha|Y|^{2}}{2}}A(0,y,0,\frac{Y}{|Y|},0)\sqrt{\alpha/\digamma}|Y|^{-1}\,dY}$
$\displaystyle=\sqrt{2\pi}\int_{\mathbb{S}^{1}}{\left(\int_{0}^{\infty}{\cos(r\eta\cdot\omega)e^{-\frac{\left(\frac{\alpha}{\digamma}\xi^{2}+\digamma\alpha\right)r^{2}}{2}}\,dr}\right)A(0,y,0,\omega,0)\sqrt{\alpha/\digamma}\,d\omega}.$

Since

$\int_{0}^{\infty}{\cos(r\eta\cdot\omega)e^{-\frac{\left(\frac{\alpha}{\digamma}\xi^{2}+\digamma\alpha\right)r^{2}}{2}}\,dr}=\frac{1}{2}\int_{\mathbb{R}}{e^{ir\eta\cdot\omega}e^{-\frac{\left(\frac{\alpha}{\digamma}\xi^{2}+\digamma\alpha\right)r^{2}}{2}}\,dr}=\sqrt{\frac{\pi}{2}\frac{\digamma}{\alpha(\xi^{2}+\digamma^{2})}}e^{-\frac{\digamma}{\alpha(\xi^{2}+\digamma^{2})}(\eta\cdot\omega)^{2}}$

it follows that

$\text{Re
}a=\pi\sqrt{\frac{\digamma}{\xi^{2}+\digamma^{2}}}\int_{\mathbb{S}^{1}}{e^{-\frac{\digamma}{\alpha(\xi^{2}+\digamma^{2})}(\eta\cdot\omega)^{2}}A(0,y,0,\omega,0)\sqrt{\alpha/\digamma}\,d\omega}.$

Noting that the terms in the integrand are all strictly positive aside from
$A$, which is positive almost everywhere on $\mathbb{S}^{1}$, it follows that
the integral is positive, and in particular $a$ is nonzero, as desired.

######  Remark 5.4.

By using Laplace’s method, i.e. the analogue of the method of stationary phase
for real exponentials, one can obtain an asymptotic expansion as
$|(\xi,\eta)|\to\infty$, as predicted by stationary phase.

##  6\. The Local Recovery Problem

In this section we make the argument for the local recovery problem. Once the
local recovery result is established, a global recovery result can be obtained
via a convex foliation argument, which would follow exactly the argument used
in previous papers.

Recall we are in the setting where we assume all of the parameters were known
except for one, call it $\nu$, and we assume that the travel time data for the
two parameters $a_{\nu}$ and $\tilde{a}_{\nu}$ were the same. For simplicity
of notation, we let $I=I^{\nu}$ and $\tilde{I}=\tilde{I}^{\nu}$. Note that

$I[u](x_{0},\xi_{0})=\int_{\mathbb{R}}{A(X(t,x_{0},\xi_{0}),\Xi(t,x_{0},\xi_{0}))u(X(t,x_{0},\xi_{0}))\,dt},\quad
A(x,\xi)=-E^{\nu}(x,\xi)\frac{\partial\tilde{\Xi}}{\partial\xi}(x,\xi)$

and

$\tilde{I}=\int_{\mathbb{R}}{\tilde{A}(X(t,x_{0},\xi_{0}),\Xi(t,x_{0},\xi_{0}))u(X(t,x_{0},\xi_{0}))\,dt}$

with

$\tilde{A}(x,\xi)=-\partial_{x}E^{\nu}\frac{\partial\tilde{\Xi}}{\partial\xi}+\partial_{\xi}E^{\nu}\frac{\partial\tilde{\Xi}}{\partial
x}.$

Recall then that the pseudolinearization formula reads

$0=I[\partial_{x}r_{\nu},\partial_{y_{1}}r_{\nu},\partial_{y_{2}}r_{\nu}]+\tilde{I}[r_{\nu}].$

If we let

$Lv(x,y)=x^{-2}\int_{\mathbb{S}^{2}}{\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\frac{\digamma^{2}\hat{\lambda}^{2}}{2\alpha}\right)v(\gamma_{\lambda,\omega})\,d\mathbb{S}^{2}(\lambda,\omega)}$

then we have

(6.1)
$0=N[\partial_{x}r_{\nu},\partial_{y_{1}}r_{\nu},\partial_{y_{2}}r_{\nu}]+\tilde{N}[r_{\nu}]$

where $N=L\circ I$ and $\tilde{N}=L\circ\tilde{I}$.

We now let $\digamma>0$, and we consider the conjugation of $N$ and
$\tilde{N}$ by the function $e^{\gamma/x}$. That is, we consider

$N_{\digamma}=e^{-\digamma/x}Ne^{\digamma/x},\quad\tilde{N}_{\digamma}=e^{-\digamma/x}\tilde{N}e^{\digamma/x}.$

As shown in Section 5.1, we have that $N_{\digamma}$ is a scattering
pseudodifferential operator whose symbol is of the form

$a=\int_{\mathbb{R}_{\hat{t}}\times\mathbb{R}_{\hat{\lambda}}\times\mathbb{S}^{1}_{\omega}}{e^{i\left(\xi\frac{x^{\prime}-x}{x^{2}}+\eta\cdot\frac{y^{\prime}-y}{x}\right)}\chi(x^{\epsilon}\hat{\lambda})\exp\left(-\digamma\left(\frac{\hat{\lambda}^{2}}{2\alpha}-\left(\frac{1}{x^{\prime}}-\frac{1}{x}\right)\right)\right)A(x,y,\hat{\lambda},\omega,\hat{t})\,d\hat{t}\,d\hat{\lambda}\,d\omega}.$

Hence, from the results in Section 5, particularly Lemma 5.3, as well as Lemma
3.7, we have the following:

######  Theorem 6.1.

For any $\digamma>0$, we have $N_{\digamma}\in\Psi^{-1,0,-2,-2}$ and
$\tilde{N}_{\digamma}\in\Psi^{-1,0,-1,-1}$. Moreover, if

$(\partial_{t}\partial_{\omega_{\parallel}}-\eta_{0}\cdot\beta\partial_{\omega_{\parallel}}^{2})A(x,y,x\hat{\lambda},\omega,0)\geq
C>0\text{ for all }\omega\in\eta_{0}^{\perp}$

where $\eta_{0}\in\mathbb{S}^{1}$ is parallel to the axis of isotropy and
$\omega=\omega_{\parallel}\eta_{0}+\omega_{\perp}$, then $N_{\digamma}$ is in
fact elliptic.

Note that the subprincipal condition is satisfied under reasonable geological
assumptions for the Earth: in fact, working through the definition of $A$
above, we can see that the term in the subprincipal condition is the same as
the term

$\left(\partial_{t}\xi_{T}(x,0,\omega^{\prime})\partial_{\omega_{\parallel}}\xi_{T}(x,0,\omega)-\frac{\zeta}{|\zeta|}\cdot\alpha(\omega)(\partial_{\omega_{\parallel}}\xi_{T}(x,0,\omega))^{2}\right)$

appearing in [13] when considering the subprincipal symbol in the global
setting. (See the discussion in [13], particularly following Equation 2.10.)

We now let $r_{\nu}=e^{\digamma/x}f$. It now suffices to show, from (6.1),
that $f\equiv 0$, assuming $c$ is chosen sufficiently small. Indeed, from
(6.1) we have

$\displaystyle 0$
$\displaystyle=e^{-\digamma/x}N[\partial_{x}(e^{\digamma/x}f),\partial_{y_{1}}(e^{\digamma/x}f),\partial_{y_{2}}(e^{\digamma/x}f)]+e^{-\digamma/x}\tilde{N}[e^{\digamma/x}u]$
$\displaystyle=N_{\digamma}[(\partial_{x}-\digamma/x^{2})f,\partial_{y_{1}}f,\partial_{y_{2}}f]+\tilde{N}_{\digamma}[f].$

For a fixed $\digamma>0$, the operator $N_{\digamma}$ is elliptic, and hence
there exists a parametrix $Q\in\Psi^{1,0,2,2}\otimes\text{Mat}_{3\times 3}$
such that $QN_{\digamma}=I+R$, with
$R\in\Psi^{-\infty,-\infty}\otimes\text{Mat}_{3\times 3}$. Applying $Q$ to
both sides gives

(6.2)
$0=\left(\partial_{x}f-\frac{\digamma}{x^{2}}f,\partial_{y_{1}}f,\partial_{y_{2}}f\right)+E[f]$

where

$E[f]=Q\circ\tilde{N}[f]+R[\partial_{x}u-\frac{\digamma}{x^{2}}f,\partial_{y_{1}}f,\partial_{y_{2}}f].$

Note131313In fact, it would have sufficed for the parametrix error $R$ to be
in $\Psi^{-3,-3,3,3}\otimes\text{Mat}_{3\times
3}\subset\Psi^{-2,-2,1,1}\otimes\text{Mat}_{3\times 3}$ (as opposed to all the
way in $\Psi^{-\infty,-\infty}$), since we need to cancel one order in
derivative and two orders in decay. then that
$E\in\Psi^{0,0,1,1}\otimes\text{Mat}_{3\times 1}$.

Writing $E=(E_{0},E_{1},E_{2})$, we can rewrite (6.2) as the three scalar
equations

$\displaystyle 0$ $\displaystyle=(x^{2}\partial_{x}-\digamma)f+x^{2}E_{0}[f],$
$\displaystyle 0$ $\displaystyle=x\partial_{y_{i}}f+xE_{i}[f],\quad i=1,2.$

Note that each equation is of the form $0=Af+Ef$, where
$A\in\text{Diff}_{sc}^{1,0}$, and $E$ is at worst in $\Psi^{0,-1,1,1}$.
Moreover, the symbols of the scattering differential operators are
$i(\xi+i\digamma)$, $i\eta_{1}$, and $i\eta_{2}$, respectively, so one can
take $\Psi^{0,0}_{sc}$ combinations141414Specifically, we apply $\Psi$DOs
whose symbols are $\frac{\xi-i\digamma}{\langle(\xi,\eta)\rangle}$ and
$\frac{\eta_{i}}{\langle(\xi,\eta)\rangle}$, respectively. of the scattering
differential operators to form an elliptic element $B\in\Psi^{1,0}_{sc}$.
Applying the $\Psi^{0,0}_{sc}$ combinations to the equations above, we end up
with

$0=Bf+Ff,\quad B\in\Psi^{1,0}_{sc}\text{ elliptic},F\in\Psi^{0,-1,1,1}$

so applying an elliptic parametrix for $B$ yields

$0=f+Gf,\quad G\in\Psi^{-1,-1,1,1}\subset\Psi^{-1/2,-1/2,0,0}.$

Hence, we have

$\|f\|_{L^{2}_{sc}}\leq\|G\|_{H^{-1/2,-1/2}_{sc}\to
L^{2}_{sc}}\|f\|_{H^{-1/2,-1/2}_{sc}}.$

It suffices to show that for any $\epsilon>0$ we have
$\|f\|_{H^{-1/2,-1/2}_{sc}}\leq\epsilon\|u\|_{L^{2}_{sc}}$ if $f$ is supported
sufficiently close to the boundary. But this just follows by noting that if
$f$ is supported in $\\{x<c\\}$, then $x^{1/2}|f|\leq c^{1/2}|f|$, and hence

$\|f\|_{H^{-1/2,-1/2}_{sc}}\leq\|f\|_{H^{0,-1/2}_{sc}}=\|x^{1/2}f\|_{L^{2}_{sc}}\leq
c^{1/2}\|f\|_{L^{2}_{sc}}.$

Thus, taking $c$ sufficiently small, we can conclude that $f$ (restricted to
$\\{\tilde{x}>-c\\}$) is identically zero. This concludes the proof of Theorem
1.1, which, as mentioned in the Introduction, when combined with a convex
foliation condition proves Theorem 1.2 as well.

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