We can now notice that $\varepsilon\kappa^{2}\big{(}\langle q_{\rm
i}\rangle\llbracket\underline{\mathtt{c}}_{1}\zeta\rrbracket-2\ell\dot{\delta}\langle\underline{\mathtt{c}}_{1}\zeta\rangle\big{)}$
is a lower order term in the sense that

$\varepsilon\kappa^{2}|\langle q_{\rm i}\rangle\llbracket
c_{1}\zeta\rrbracket-2\ell\dot{\delta}\langle
c_{1}\zeta\rangle|\leq\varepsilon^{1/2}\kappa^{1/2}\underline{C}\big{(}\underline{{\mathfrak{E}}}_{\rm
int}+\underline{{\mathfrak{E}}}_{\rm trace}\big{)},$

so that it can be absorbed by the sum of the three energies when
$\varepsilon\kappa$ is small enough to have
$\varepsilon^{1/2}\kappa^{1/2}\underline{C}<1$. For instance, if
$\varepsilon^{1/2}\kappa^{1/2}\underline{C}<1/2$, and denoting

$\widetilde{\underline{{\mathfrak{E}}}}:=\underline{\mathfrak{E}}_{\rm
ext}+\underline{\mathfrak{E}}_{\rm int}+\underline{\mathfrak{E}}_{\rm trace},$

one obtains after a Gronwall estimate

$\widetilde{\underline{{\mathfrak{E}}}}(t)\leq
3\big{[}\widetilde{\underline{{\mathfrak{E}}}}(0)+\varepsilon\underline{C}\int_{0}^{t}\big{(}|f|_{2}^{2}+|(g_{1},g_{2})|^{2}\big{)}\big{]}\exp(\sqrt{\varepsilon}\underline{C}t).$

Since moreover there exists a constant $C_{0}=C_{0}(\frac{1}{h_{\rm
min}},\frac{1}{c_{\rm min}})$ such that

$\widetilde{\underline{\mathfrak{E}}}\leq
C_{0}\big{(}|{\mathtt{Z}}|^{2}+|(\zeta,q,\kappa\partial_{x}q)|_{2}^{2}+\varepsilon\kappa^{3}|(\zeta_{-},\zeta_{+})|^{2}+\varepsilon\kappa^{5}|(\dot{\zeta}_{-},\dot{\zeta}_{+})|^{2}$

and

$|{\mathtt{Z}}|^{2}+|(\zeta,q,\kappa\partial_{x}q)|_{2}^{2}+\varepsilon\kappa^{3}|(\zeta_{-},\zeta_{+})|^{2}+\varepsilon\kappa^{5}|(\dot{\zeta}_{-},\dot{\zeta}_{+})|^{2}\leq
C_{0}\widetilde{\underline{\mathfrak{E}}},$

one deduces the estimate stated in the theorem.

Step 5. Well-posedness. By a straightforward adaptation of the proof of
Theorem 3.3, one can observe that (3.33)-(3.36) can be reformulated as an ODE
for $(\zeta,q,{\mathtt{Z}})\in H^{1}\times
H^{2}({\mathcal{E}})\times{\mathbb{R}}^{3}$ and prove existence and uniqueness
of a solution in this space by Cauchy-Lipschitz’s theorem. For data in
$(\zeta,q,{\mathtt{Z}})\in L^{2}\times
H^{1}({\mathcal{E}})\times{\mathbb{R}}^{3}$, this strategy does not work
directly because the traces $\zeta_{\pm}$ that appear in the component of the
ODE (3.53) for $\langle q_{\rm i}\rangle$ and $\dot{\delta}$ cannot be
controlled by the $L^{2}$ norm of $\zeta$. However, the energy estimate just
proved provides such a control and one can obtain the result by a classical
density argument (as used for instance in the proof of Theorem 3.1.1 in [36],
for hyperbolic initial boundary value problems where the control on the trace
is furnished by using a Kreiss symmetrizer). ∎

##  4\. Return to equilibrium

We now deal with a specific kind of wave-structure interaction that was called
the return to equilibrium problem in [26] and is commonly referred to as ”free
decay test” in engineering. This a situation where the solid is released at
zero speed from an out of equilibrium position ($\delta(t=0)\neq 0$), in a
fluid that is at rest. The solid then oscillates vertically and its motion
sends waves outwards; by this process, the solid loses energy and its
oscillations are damped so that the solid asymptotically stabilizes to its
equilibrium position. Engineers use this free decay test because by measuring
the oscillations of the object, they deduce some buoyancy properties of the
object. More precisely, assuming that the motion of the object satisfies the
phenomenological Cummins equation [11, 30]

(4.1) $M\ddot{\delta}+k\ast\dot{\delta}+a\delta=0$

with $M,a\in\mathbb{R}^{+}$ and $k\in L^{1}_{\text{loc}}(\mathbb{R}^{+})$,
they calibrate these coefficients with experimental measurements. These
measurements are also used to propose nonlinear extensions to (4.1) (by
fitting coefficients with ad hoc nonlinear terms) [39].

Our goal in this section is to study this problem from a mathematical
viewpoint, by proposing a qualitative analysis of the solutions to the
transmission problem (3.2)-(3.5) in the particular configuration corresponding
to the return to equilibrium problem. This approach is expected to lead in
some cases to an equation of the form (4.1), which would provide an analytic
description of the coefficients involved, and also to nonlinear extensions
that could be of interest to engineers.

This program was initiated and achieved in [26] for the (non dispersive)
nonlinear shallow water equations, where it was found that $\delta$ solves a
nonlinear second order ODE without integro-differential term. Still working
with the shallow water equations but in horizontal dimension $d=2$, assuming
radial symmetry and neglecting the nonlinear effects in the exterior region,
it was shown in [6] that the equation on $\delta$ should contain an integro-
differential term. Such a term is also necessary for the nonlinear shallow
water equations in dimension $d=1$ if viscosity is taken into account [33].
The goal of this section is to investigate the contribution of the dispersive
terms of the Boussinesq system to the equation satisfied by $\delta$ in this
specific configuration of the return to equilibrium problem.

From now, we assume that the initial data correspond to the configuration of
the return to equilibrium problem, namely,

(4.2)
$q(t=0)=\zeta(t=0)=0\quad\text{and}\quad\delta(t=0)=\delta_{0},\quad\dot{\delta}(t=0)=0.$

######  Notation 2.

We use throughout this section the same notations as in Section 3, namely, we
write $\kappa=\sqrt{\mu/3}$ and denote by ${\mathfrak{f}}_{\rm sw}$ the
momentum flux of the nonlinear shallow water equations,

${\mathfrak{f}}_{\rm
sw}=\frac{h^{2}-1}{2\varepsilon}+\varepsilon\frac{q^{2}}{h}=\zeta+\varepsilon\big{(}\frac{1}{2}\zeta^{2}+\frac{q^{2}}{h}\big{)}.$

We also recall that the buoyancy frequency $\tau_{\rm buoy}$ is defined in
Appendix A.

We introduce in §4.1 two Cummins operators that allow us to derive an abstract
evolution equation for the solid. We then investigate two specific cases where
it is possible to derive an explicit expression of these operators. The non
dispersive case ($\varepsilon\neq 0$, $\mu=0$) is considered in §4.2 where it
is shown that the motion of the object can be found by solving a simple
nonlinear second order scalar ODE. Waves can then be described by solving an
initial boundary value problem for a scalar Burgers equation. The opposite
case, namely, the linear dispersive case ($\varepsilon=0$, $\mu\neq 0$) is
addressed in §4.3; here again, it is possible to derive an explicit expression
for the Cummins operators leading us to an integro-differential Cummins-type
equation for the motion of the solid; qualitative properties of the solutions,
such as their decay rate are then investigated. Finally, it is shown that the
motion of the waves can be found by solving a nonlocal (in space) perturbation
of the transport equation.

###  4.1. The general Cummins equation

Quite obviously, any smooth solution of the transmission problem (3.2)-(3.5)
with initial condition (4.2) is such that $\zeta$ is an even function while
$q$ is odd – such solutions will be called “symmetric”. This implies that
$\langle q_{\rm i}\rangle=0$ and that the transmission problem can be reduced
into a simpler boundary value problem stated in the following direct corollary
of Theorem 3.1.

######  Corollary 4.1.

Any smooth symmetric solution to the transmission problem (3.2)-(3.5) solves
the following boundary value problem on the half-line $(\ell,\infty)$,

(4.3) $\begin{cases}\partial_{t}\zeta+\partial_{x}q=0\\\
(1-\kappa^{2}\partial_{x}^{2})\partial_{t}q+\partial_{x}{\mathfrak{f}}_{\rm
sw}=0,\end{cases}\quad\mbox{ for }\quad t\geq 0,\quad x\in{\mathcal{E}}^{+},$

with boundary condition

(4.4) $\displaystyle q_{|_{x=\ell}}$ $\displaystyle=-\ell\dot{\delta},$

where $\delta$ solves the ODE

(4.5)
${(\tau_{\mu}(\varepsilon\delta)^{2}+\ell\kappa\frac{1}{h_{+}})}\ddot{\delta}+\delta=\varepsilon\beta(\varepsilon\delta)\dot{\delta}^{2}+{\mathfrak{H}}_{+},$

where $h_{+}=h_{|_{x=\ell}}$, ${\mathfrak{H}}_{+}={\mathfrak{H}}_{|_{x=\ell}}$
and we recall that ${\mathfrak{H}}={\mathfrak{H}}(\zeta,q)$ with

${{\mathfrak{H}}(\zeta,q)=\frac{1}{2}\varepsilon\big{(}\frac{1}{h}\zeta^{2}-\frac{q^{2}}{h^{2}}\big{)}+\frac{1}{h}R_{1}{\mathfrak{f}}_{\rm
sw}},$

and that $\tau_{\mu}(\varepsilon\delta)$ and $\beta(\varepsilon\delta)$ are
defined in Proposition 2.4, namely,

$\displaystyle{{{\tau}_{\mu}(\varepsilon\delta)^{2}}}$
$\displaystyle=\tau_{\rm
buoy}^{2}+\frac{1}{\ell}\int_{0}^{\ell}\frac{x^{2}}{h_{\rm
eq}(x)+\varepsilon\delta}{\rm d}x+{\frac{\kappa^{2}}{h_{\rm
eq}(\ell)+\varepsilon\delta}},$ $\displaystyle\beta(\varepsilon\delta)$
$\displaystyle=\frac{1}{2}\frac{1}{\ell}\int_{0}^{\ell}\frac{x^{2}}{(h_{\rm
eq}(x)+\varepsilon\delta)^{2}}{\rm d}x.$

We know by Proposition 3.2 that if $f$ is a given $C^{1}$ function of time
then there is a unique solution $(\zeta,q)$ to (4.3) with boundary condition
$q_{|_{x=\ell}}=-\ell f$ with initial condition corresponding to the return to
equilibrium problem, namely, $(\zeta,q)(t=0)=(0,0)$. It is in particular
possible to compute the trace of $\zeta$ at $x=\ell$, so that the following
definition makes sense.

######  Definition 4.1 (Cummins operators).

Let $\varepsilon\in\mathbb{R}^{+}$, $\mu=\kappa^{2}/3>0$. Let also $f\in
C^{1}({\mathbb{R}}^{+})$ and $T>0$, and $(\zeta,q)\in
C^{1}\big{(}[0,T;H^{1}({\mathcal{E}}^{+})\times
H^{2}({\mathcal{E}}^{+})\big{)}$ be a solution to (4.3) with boundary
condition $q_{|_{x=\ell}}=-\ell f$ and initial condition
$(\zeta,q)(t=0)=(0,0)$. We define the Cummins operators
${\mathfrak{c}}_{\varepsilon,\mu}$ and ${\mathfrak{C}}_{\varepsilon,\mu}$ as

${\mathfrak{c}}_{\varepsilon,\mu}[f]:=\zeta_{|_{x=\ell}}\quad\mbox{ and
}\quad{\mathfrak{C}}_{\varepsilon,\mu}[f]:={{-}}{{\mathfrak{H}}(\zeta,q)_{|_{x=\ell}}}.$

######  Remark 4.1.

The Cummins operators can be defined for more general cases, for instance, the
solution $(\zeta,q)$ to the initial boundary value problem needs only to be
regular near the boundary $x=\ell$ (regular enough for the trace to make
sense). This allows one to extend the definition of the Cummins operators in
the case $\mu=0$, as done in §4.2 below.

######  Corollary 4.2.

The ODE (4.5) can be reformulated in a compact form as what we shall refer to
as the Cummins equation

(4.6)
${{\big{(}\tau_{\mu}(\varepsilon\delta)^{2}+\ell\kappa\frac{1}{1+\varepsilon{\mathfrak{c}}_{\varepsilon,\mu}[\dot{\delta}]}\big{)}}}\ddot{\delta}+\delta{{{+\mathfrak{C}}_{\varepsilon,\mu}[\dot{\delta}]}}=\varepsilon\beta(\varepsilon\delta)\dot{\delta}^{2},$

with initial conditions $\delta(0)=\delta_{0}$ and $\dot{\delta}(0)=0$.

The equation (4.6) is compact but not simple since the Cummins operators are
nonlinear nonlocal operators which require the resolution of the equations for
the fluid in the exterior domain. In order to get some qualitative insight on
the Cummins equation, we describe it in two limiting cases: in the nonlinear
non dispersive case ($\varepsilon>0$, not necessarily small, and $\mu=0$), and
in the linear, dispersive case ($\varepsilon=0$ and $\mu>0$, not necessarily
small). Note that in both cases, it is not necessary to compute the first
Cummins operator ${\mathfrak{c}}_{\varepsilon,\mu}[\dot{\delta}]$ and that it
is possible to provide an explicit expression of the second one
${\mathfrak{C}}_{\varepsilon,\mu}[\dot{\delta}]$.

###  4.2. The nonlinear non dispersive case

Neglecting the dispersive effects is equivalent to setting
$\mu=\kappa^{2}/3=0$ in the equations (4.3)-(4.6) ; in particular, the model
considered for the propagation of the waves is now the shallow water equations

(4.7) $\begin{cases}\partial_{t}\zeta+\partial_{x}q=0\\\
\partial_{t}q+\varepsilon\partial_{x}\left(\frac{1}{h}q^{2}\right)+h\partial_{x}\zeta=0,\end{cases}\quad\mbox{
for }\quad t\geq 0,\quad x\in{\mathcal{E}}^{+},$

the boundary condition is unchanged

(4.8) $\displaystyle q_{|_{x=\ell}}$ $\displaystyle=-\ell\dot{\delta},$

and the ODE solved by $\delta$ is simplified into

(4.9)
${{\tau_{0}(\varepsilon\delta)^{2}}}\ddot{\delta}+\delta{{{+\mathfrak{C}}_{\varepsilon,0}[\dot{\delta}]}}=-\varepsilon\tau_{0}(\varepsilon\delta)\tau_{0}^{\prime}(\varepsilon\delta)\dot{\delta}^{2},$

where we used the fact that
$\beta(\varepsilon\delta)=-2\tau_{0}(\varepsilon\delta)\tau_{0}^{\prime}(\varepsilon\delta)$
when $\mu=0$ (see (2.28) and (2.29)), and where the definition of the second
Cummins operator has been extended to the case $\mu=0$ as

${\mathfrak{C}}_{\varepsilon,0}[\dot{\delta}]:={{-}}\Big{(}\zeta+\varepsilon\frac{1}{2}\frac{q^{2}}{h^{2}}\Big{)}_{|_{x=\ell}};$

the fact that this definition makes sense follows from the decomposition of
the shallow water invariants into Riemann invariants, as shown in the proof of
the following theorem where an explicit expression of the Cummins operator is
provided. This theorem is a reformulation of Corollary 1 in [26], but with a
slight difference in the function $\gamma$, so that we reproduce a sketch of
the proof111The difference comes from the fact that in [26], the choice of the
boundary condition for the interior pressure was made by assuming that the
jump of pressure at the contact point was purely hydrostatic; as in [33, 5],
we rather use here a choice of the boundary condition on the pressure which is
consistent with the approach used throughout this paper and motivated by the
conservation of total energy, as explained in Corollary 2.1. With the choice
of [26], one would have
${\mathfrak{C}}_{\varepsilon,0}[\dot{\delta}]=-\zeta_{|_{x=\ell}}$ and
consequently
$-\ell\dot{\delta}-\varepsilon\dot{\delta}^{2}\gamma(\varepsilon\dot{\delta})=\frac{1}{\varepsilon}(\sigma_{0}(\varepsilon\frac{\ell}{2}\dot{\delta})^{2}-1)$..

######  Theorem 4.1.

Let $T>0$, $\delta\in C^{2}([0,T])$ and $(\zeta,q)$ be a continuous, piecewise
$C^{1}$ solution of (4.7)-(4.9) on $[0,T]\times(\ell,\infty)$ satisfying the
non vanishing depth condition

$\inf_{[0,T]\times{\mathcal{E}}}h>0\quad\mbox{ and
}\quad\inf_{[0,T]\times{\mathcal{I}}}h_{\rm eq}+\varepsilon\delta>0.$

If moreover $\ell\,\varepsilon\dot{\delta}<2r_{0}$, with
$r_{0}:=\frac{4}{27}$, we have
$\sqrt{h}=\sigma_{0}(\varepsilon\frac{\ell}{2}\dot{\delta})$ with the real
function

$\sigma_{0}(r)=\frac{1}{3}\Big{(}1+C_{-}(r)+C_{+}(r)\Big{)},\quad
C_{\pm}(r)=\frac{3}{2}\Big{(}-4r+2r_{0}\pm 4\sqrt{r(r-r_{0})}\Big{)}^{1/3},$

and the Cummins operator ${\mathfrak{C}}_{\varepsilon,0}$ is given explicitly
by

(4.10)
${\mathfrak{C}}_{\varepsilon,0}[\dot{\delta}]=-\varepsilon^{-1}\Big{(}\sigma_{0}(\varepsilon\frac{\ell}{2}\dot{\delta})-1\Big{)}\Big{(}3\sigma_{0}(\varepsilon\frac{\ell}{2}\dot{\delta})-1\Big{)}=:\ell\dot{\delta}{{+}}\varepsilon\dot{\delta}^{2}\gamma(\varepsilon\dot{\delta}),$

where $\gamma:(-\infty,2r_{0})\to{\mathbb{R}}$ is a smooth function such that
$\gamma(0)=\frac{1}{4}\ell^{2}$ and whose exact expression is given in (4.11)
below.

###### Proof.

The proof of Corollary 1 of [26] is based on the fact that the shallow water
equations can be put in diagonal form,

$\partial_{t}R+(\sqrt{h}+\varepsilon\frac{q}{h})\partial_{x}R=0\quad\mbox{ and
}\quad\partial_{t}L-(\sqrt{h}-\varepsilon\frac{q}{h})\partial_{x}L=0,$

where $R$ and $L$ are respectively the right and left Riemann invariants

$R=\frac{q}{h}+\frac{2}{\varepsilon}(\sqrt{h}-1)\quad\mbox{ and }\quad
L=\frac{q}{h}-\frac{2}{\varepsilon}(\sqrt{h}-1).$

One then notices that with the initial and boundary conditions considered
here, $L$ vanishes identically on $(\ell,\infty)$, which allows one to find
$\sqrt{h}$ in terms of $q$ as a root of the third order polynomial equation in
$\sigma$,

$\sigma^{3}-\sigma^{2}-\varepsilon\frac{1}{2}q=0.$

If $-\varepsilon\frac{1}{2}q<r_{0}$, then, as discussed in [26], the relevant
root is $\sigma_{0}(-\varepsilon\frac{1}{2}q)$.
Recalling that $q_{|_{x=\ell}}=-\ell\dot{\delta}$, we have
$\sqrt{h}=\sigma_{0}(\varepsilon\frac{\ell}{2}\dot{\delta})$. Moreover, $L=0$
implies

$\varepsilon\frac{1}{2}\frac{q^{2}}{h^{2}}_{|_{x=\ell}}=2\,\varepsilon^{-1}\Big{(}\sigma_{0}(\varepsilon\frac{\ell}{2}\dot{\delta})-1\Big{)}^{2}.$

Remarking that
$\zeta_{|_{x=\ell}}=\frac{1}{\varepsilon}(\sigma_{0}(\varepsilon\frac{\ell}{2}\dot{\delta})^{2}-1)$,
one gets

$\displaystyle\Big{(}\zeta+\varepsilon\frac{1}{2}\frac{q^{2}}{h^{2}}\Big{)}_{|_{x=\ell}}$
$\displaystyle=\varepsilon^{-1}(\sigma_{0}-1)(3\sigma_{0}-1)$ (4.11)
$\displaystyle=:-\ell\dot{\delta}{{-}}\varepsilon\dot{\delta}^{2}\gamma(\varepsilon\dot{\delta}),$

where we used the fact that $\sigma_{0}(0)=-\sigma_{0}^{\prime}(0)=1$. The
fact that $\gamma(0)=\frac{1}{4}\ell^{2}$ follows from the observation that
$\sigma_{0}^{\prime\prime}(0)=-4$. ∎

A first corollary is that the motion of the solid can be reduced to a simple
nonlinear ODE, provided that the initial displacement satisfies an upper bound
ensuring that the velocity of the object does not become too big.

######  Corollary 4.3.

Under the assumptions of the theorem, and with the same notations, let us
assume moreover that

$\varepsilon^{2}\delta_{0}^{2}<{\tau_{0}(\varepsilon|\delta_{0}|)^{2}}\big{(}\frac{2r_{0}}{\ell}\big{)}^{2}.$

Then, using the notations of the theorem, the motion of the solid is found by
solving the nonlinear second order ODE

(4.12)
${{\tau_{0}(\varepsilon\delta)^{2}}}\ddot{\delta}+\ell\dot{\delta}+\delta+\varepsilon{{\left(\tau_{0}(\varepsilon\delta)\tau^{\prime}_{0}(\varepsilon\delta)+\gamma(\varepsilon\dot{\delta})\right)}}\dot{\delta}^{2}=0,$

with initial condition $\delta(0)=\delta_{0}$, $\dot{\delta}(0)=0$.

######  Remark 4.2.

In the linear case ($\varepsilon=0$), this equation is almost the same as
(3.2.12) in [20], the only difference being that the author neglected the
buoyancy frequency $\tau_{\rm buoy}$ in the expression for $\tau_{0}(0)$.

###### Proof.

One just needs to check that the condition
$\ell\varepsilon\dot{\delta}<2r_{0}$, which ensures by Theorem 4.1 that the
Cummins operator takes the form (4.10), is satisfied for all times. Since at
$t=0$, one has $\dot{\delta}=0$, we now that this condition is satisfied for
small times. Since moreover one can deduce from Proposition 3.1 (by setting
$\mu=0$) that

(4.13)
$\tau_{0}(\varepsilon\delta)^{2}\dot{\delta}^{2}+\delta^{2}\leq\delta_{0}^{2},$

one deduces that $|\delta|\leq|\delta_{0}|$ and therefore that
$\dot{\delta}^{2}\leq\tau_{0}(\varepsilon|\delta_{0}|)^{-2}\delta_{0}^{2}$.
The assumption made in the statement of the corollary therefore grants the
result. ∎

The interest of reducing the motion of the solid to an ODE on the surface
displacement is that it is possible to solve it even in situations when
singularity arise in the exterior domain (typically, when shock happen). It is
in particular possible to obtain a global existence result for the ODE (4.12),
while such a result cannot be expected for strong solutions to the full
transmission problem (4.7)-(4.9) due to shock formation. Note that the first
condition on $\delta_{0}$ means that at $t=0$, the solid neither touches the
bottom nor is lifted from a height greater than the height of the water column
under the object when it is at equilibrium.

######  Proposition 4.1.

Let $\delta_{0}\in{\mathbb{R}}$ be such that

$\inf_{{\mathcal{I}}}h_{\rm eq}-\varepsilon|\delta_{0}|>0\quad\mbox{ and
}\quad\varepsilon|\delta_{0}|<\tau_{0}(\varepsilon|\delta_{0}|)\frac{2r_{0}}{\ell}.$

Then there exists a unique global solution $\delta\in
C^{\infty}({\mathbb{R}}^{+})$ to the ODE (4.12) with initial condition
$(\delta,\dot{\delta})_{|_{t=0}}=(\delta_{0},0)$.

######  Remark 4.3.

A byproduct of the proof is that
${\mathfrak{C}}_{\varepsilon,0}[\dot{\delta}]\,\dot{\delta}\geq 0$ and that
this quantity corresponds to the energy transferred at each instant to the
exterior fluid domain, that is, with the notations of Proposition 3.1, one has

$\frac{d}{dt}{\mathfrak{E}}_{\rm
ext}={\mathfrak{C}}_{\varepsilon,0}[\dot{\delta}]\dot{\delta}\geq 0.$

######  Remark 4.4.

The second condition of the proposition is a smallness condition on
$\delta_{0}$, but this condition is not restrictive as it allows $\delta_{0}$
to be of size $O(\varepsilon^{-1})$. As communicated to us by the author it is
possible, under stricter smallness conditions, to prove exponential decay of
the solution of ODEs related to (4.12) using techniques developed in [24].

###### Proof.

There exists a positive time $T>0$ such that on $[0,T)$, there is a solution
$\delta$ such that $\inf_{\mathcal{I}}h_{\rm eq}+\varepsilon\delta>0$ and
$\ell\varepsilon\dot{\delta}<2r_{0}$. We want to show that one can take
$T=+\infty$. As in the proof of Corollary 4.3, this follows from (4.13). We
therefore need to prove that (4.13) holds, without appealing to Proposition
3.1 as in the proof of Corollary 4.3, but by direct manipulations on the
solution to the ODE (4.12). We need the following two lemmas.

######  Lemma 4.1.

The function $\sigma_{0}$ is decreasing on $(-\infty,r_{0})$.

###### Proof of the lemma.

By construction, one has for all $r<r_{0}$,

$\sigma_{0}(r)^{3}-\sigma_{0}(r)^{2}+r=0;$

differentiating this identity yields

$\sigma^{\prime}_{0}(r)(3\sigma_{0}(r)^{2}-2\sigma_{0}(r))=-1,$

so that $\sigma_{0}(r)$ and $3\sigma_{0}(r)^{2}-2\sigma_{0}(r)$ have opposite
sign. It is therefore enough to prove that
$3\sigma_{0}(r)^{2}-2\sigma_{0}(r)>0$ for all $r<r_{0}$. Since
$\sigma_{0}(0)=1$, this quantity is positive at $r=0$ and must therefore
vanish if it changes sign. This means that for some $r_{1}<r_{0}$, one must
have $\sigma_{0}(r_{1})=0$ or $\sigma_{0}(r_{1})=\frac{2}{3}$. Using the cubic
equation solved by $\sigma_{0}(r_{1})$, this implies that $r_{1}=0$ or
$r_{1}=r_{0}$. Both cases have to be excluded because $\sigma_{0}(0)=1\neq 0$
and $r_{1}<r_{0}$ by assumption. The result follows. ∎

######  Lemma 4.2.

If $\dot{\delta}\neq 0$ and $\varepsilon\ell\dot{\delta}<2r_{0}$, the Cummins
operator satisfies

${\mathfrak{C}}_{\varepsilon,0}[\dot{\delta}]\,\dot{\delta}>0.$

###### Proof of the lemma.

Recalling that from (4.10), one has

${\mathfrak{C}}_{\varepsilon,0}[\dot{\delta}]\,\dot{\delta}=-\varepsilon^{-1}(\sigma_{0}(\varepsilon\frac{\ell}{2}\dot{\delta})-1)(3\sigma_{0}(\varepsilon\frac{\ell}{2}\dot{\delta})-1)\dot{\delta},$

the conclusion follows from the previous lemma and the observation that
$\sigma_{0}(0)=1$ and $\sigma_{0}(r_{0})=2/3$. ∎

We can now use the second lemma to conclude: multiplying (4.9) by
$\dot{\delta}$ and integrating in time yields

$\displaystyle\tau_{0}(\varepsilon\delta)^{2}\dot{\delta}^{2}+\delta^{2}$
$\displaystyle=\delta_{0}^{2}-\int_{0}^{t}{\mathfrak{C}}_{\varepsilon,0}[\dot{\delta}]\dot{\delta}<\delta_{0}^{2},$

which implies (4.13); the proposition is therefore proved. ∎

The following Corollary then shows that, once the ODE (4.12) has been solved,
the solution in the exterior domain reduces to a simple initial boundary value
problem for a scalar Burgers-type equation. This is a simple byproduct of the
proof of Theorem 4.1 where it was shown that the nonlinear shallow water
equations were reduced to the scalar equation on the right-going Riemann
invariant.

######  Corollary 4.4.

Under the assumptions of Corollary 4.3, $q$ is found in the exterior domain by
solving the initial boundary value problem

$\begin{cases}\partial_{t}q+\big{(}-\sigma_{0}^{\prime}(-\frac{\varepsilon}{2}q)\sigma_{0}(-\frac{\varepsilon}{2}q)\big{)}^{-1}\partial_{x}q&=0\qquad(t>0,\quad
x>\ell),\\\ q_{|_{t=0}}&=0,\\\ q_{|_{x=\ell}}&=-\ell\dot{\delta},\end{cases}$

with $\delta$ furnished by Proposition 4.1, while $\zeta$ is given in terms of
$q$ by the algebraic expression

(4.14) $\zeta=\frac{1}{\varepsilon}(\sigma_{0}(-\varepsilon q/2)^{2}-1).$

######  Remark 4.5.

More generally, if one wants to compute the waves created by an object in
forced motion, one must solve the same equations as in the corollary, but with
$\delta$ corresponding to this forced motion rather than given by Proposition
4.1.

###  4.3. The linear dispersive case

We have studied in the previous section the situation where dispersive effects
could be neglected ($\mu=\kappa^{2}/3=0$) in front of the nonlinear effects.
We consider here the opposite situation where nonlinear effects are negligible
($\varepsilon=0$) but the dispersive effects are taken into account. That is,
we consider the linear approximation to (4.3)-(4.5). The model considered for
the propagation of the waves is therefore

(4.15) $\begin{cases}\partial_{t}\zeta+\partial_{x}q=0\\\
(1-\kappa^{2}\partial_{x}^{2})\partial_{t}q+\partial_{x}\zeta=0,\end{cases}\quad\mbox{
for }\quad t\geq 0,\quad x\in{\mathcal{E}}^{+},$

the boundary condition is unchanged

(4.16) $\displaystyle q_{|_{x=\ell}}$ $\displaystyle=-\ell\dot{\delta},$

and the ODE solved by $\delta$ is simplified into

(4.17)
${(\tau^{2}_{\mu}+\ell\kappa)}\ddot{\delta}+\delta+{\mathfrak{C}}_{0,\mu}[\dot{\delta}]=0,$

where we recall that according to the definition of the Cummins operator (see
Definition 4.1 and (3.18)),

${\mathfrak{C}}_{0,\mu}[\dot{\delta}]:=-(R_{1}\zeta)_{|_{x=\ell}},$

and where, for the sake of clarity, we simply write throughout this section

$\tau_{\mu}^{2}=\tau_{\mu}(0)^{2}=\tau_{\rm
buoy}^{2}+\frac{1}{\ell}\int_{0}^{\ell}\frac{x^{2}}{h_{\rm
eq}}+\kappa^{2}\frac{1}{h_{\rm eq}}.$

We know by Theorem 3.3 that for all $n\in{\mathbb{N}}$ and $T>0$, there exists
a unique solution $(\zeta,q,\delta)\in
C^{\infty}([0,T];{\mathbb{H}}^{n}\times{\mathbb{R}})$ of (4.15)-(4.17) with
initial conditions (4.2); we want here to analyze the behavior of this
solution. As for the nonlinear non dispersive case in the previous section, we
first provide an explicit expression for the Cummins operator, from which we
are able to derive an uncoupled scalar equation for the evolution of $\delta$,
whose solution can be used to find $\zeta$ and $q$ in the exterior domain
through the resolution of a simpler scalar initial boundary value problem. All
the equations involved in this section are linear, the difficulty coming from
their nonlocal nature.

####  4.3.1. Preliminary material

In order to give an explicit representation of the Cummins operator
${\mathfrak{C}}_{0,\mu}$, we first need to recall the definition of the Bessel
functions $J_{n}$ (§8.41 in [17])

$J_{n}(t)=\frac{1}{\pi}\int_{0}^{\pi}\cos\big{(}n\theta-t\sin\theta\big{)}{\rm
d}\theta;$

we also define the causal convolution kernels ${\mathcal{K}}^{0}_{\mu}$ and
${\mathcal{K}}^{1}_{\mu}$ as

(4.18)
${\mathcal{K}}^{0}_{\mu}(t)=\frac{1}{\kappa}J_{0}(\frac{t}{\kappa})\quad\mbox{
and
}\quad{\mathcal{K}}^{1}_{\mu}(t)=\frac{1}{t}J_{1}(\frac{t}{\kappa}),\quad\mbox{
for all }t\geq 0,$

and use the following standard notation for the convolution of time causal
functions,

$\forall t\geq 0,\qquad f*g(t)=\int_{0}^{t}f(t-s)g(s){\rm d}s.$

We also need to use the Laplace transform with respect to the time variable,
which we define as

$\mathcal{L}:q\mapsto\hat{q},$

where

$\mathcal{L}[q](s)=\displaystyle\int_{0}^{\infty}q(t)e^{-st}{\rm
d}t\quad\text{ with }\quad
s\in\mathbb{C}_{0}:=\\{s\in\mathbb{C}\,|\,\mathfrak{Re}(s)>0\\}.$

We shall in particular use the following properties on Bessel functions [17]

(4.19)
${\mathcal{L}}^{-1}\big{(}\frac{1}{\sqrt{1+\kappa^{2}s^{2}}}\big{)}={\mathcal{K}}^{0}_{\mu}(t)\quad\mbox{
and }\quad{\mathcal{L}}^{-1}\big{(}\frac{1}{\sqrt{1+\kappa^{2}s^{2}}+\kappa
s}\big{)}={\mathcal{K}}^{1}_{\mu}(t),$

with ${\mathcal{K}}^{0}_{\mu}$ and ${\mathcal{K}}^{1}_{\mu}$ as defined in
(4.18).

####  4.3.2. Analysis of the equations

Using the linear structure of the equations, one can obtain an explicit
expression for the Cummins operator ${\mathfrak{C}}_{0,\mu}$.

######  Theorem 4.2.

The Cummins operator ${\mathfrak{C}}_{0,\mu}$ is given explicitly by

${\mathfrak{C}}_{0,\mu}[\dot{\delta}]:=\ell\,{\mathcal{K}}^{1}_{\mu}*\dot{\delta},$

where ${\mathcal{K}}^{1}_{\mu}$ is defined in (4.18).

###### Proof.

Applying the Laplace transform to the equations (4.15) and (4.16), which is
possible since all the functions are continuous and bounded in time (as a
consequence of Proposition 3.1), and taking into account that
$\zeta_{|_{t=0}}=q_{|_{t=0}}=0$, this yields

(4.20) $\begin{cases}s\hat{\zeta}+\partial_{x}\hat{q}=0\\\
(1-\kappa^{2}\partial_{x}^{2})s\hat{q}+\partial_{x}\hat{\zeta}=0,\end{cases}\quad\mbox{
and }\quad\widehat{q}_{|_{x=\ell}}=-\ell\widehat{\dot{\delta}}.$

This is an ODE for $(\widehat{\zeta},\widehat{q})$ on the half-line
$(\ell,\infty)$ that can be explicitly solved in terms
$\widehat{\dot{\delta}}$ (note that a representation of the solution in terms
of the Laplace transform in space is also possible [21] but not adapted to our
purpose here; see also [3] for other types of linear dispersive equations);
the formula of the lemma below provides ”right-going” solutions to the linear
Boussinesq equations and it is therefore no surprise that the relationship
between $\widehat{\zeta}$ and $\widehat{q}$ is the same as the one that arises
when imposing transparent boundary conditions as in [23].

######  Lemma 4.3.

There is one and only one solution $(\widehat{\zeta},\widehat{q})$ to (4.20)
that does not grow exponentially at infinity; it is given by

$\begin{cases}\hat{q}(s,x)=\displaystyle-\ell\,\widehat{\dot{\delta}}(s)e^{-\frac{s}{\sqrt{1+\kappa^{2}s^{2}}}(x-\ell)},\\\
\displaystyle\widehat{\zeta}(s,x)=\frac{1}{\sqrt{1+\kappa^{2}s^{2}}}\widehat{q}(s,x),\end{cases}$

where the square root is taken in order to have positive real part.

###### Proof of the lemma.

From (4.20), one deduces

(4.21)
$\partial_{x}^{2}\hat{q}(s)-\frac{s^{2}}{1+\kappa^{2}s^{2}}\hat{q}(s)=0,$

and there are therefore two constants $A(s)$ and $B(s)$ such that

$\displaystyle\hat{q}(s,x)=A(s)e^{-\frac{s}{\sqrt{1+\kappa^{2}s^{2}}}x}+B(s)e^{\frac{s}{\sqrt{1+\kappa^{2}s^{2}}}x}.$

Since exponentially increasing functions are not allowed, we have $B(s)=0$ and
thus

$\hat{q}(s,x)=A(s)e^{-\frac{s}{\sqrt{1+\kappa^{2}s^{2}}}x}.$

Then using the boundary condition on $\widehat{q}$ at $x=\ell$, we find the
expected formula for $\hat{q}$. By using the first equation of (4.20), we get
the formula for $\widehat{\zeta}$. ∎

Let us now remark that for all $f\in L^{2}({\mathcal{E}}^{+})$, one has

$(R_{1}f)_{|_{x=\ell}}=\kappa^{-1}\int_{{\mathcal{E}}^{+}}e^{-\kappa^{-1}(x-\ell)}f(x){\rm
d}x$

so that, using the lemma,

$\displaystyle\widehat{{\mathfrak{C}}_{0,\mu}[\dot{\delta}]}(s)$
$\displaystyle=-(R_{1}\widehat{\zeta})_{|_{x=\ell}}$
$\displaystyle=\displaystyle\frac{\ell\widehat{\dot{\delta}}(s)}{\kappa\sqrt{1+\kappa^{2}s^{2}}}\,\int_{{\mathcal{E}}^{+}}e^{-\big{(}\frac{1}{\kappa}+\frac{s}{\sqrt{1+\kappa^{2}s^{2}}}\big{)}(x-\ell)}{\rm
d}x.$

It follows that

$\widehat{{\mathfrak{C}}_{0,\mu}[\dot{\delta}]}(s)=\frac{\ell}{\sqrt{1+\kappa^{2}s^{2}}+\kappa
s}\,\widehat{\dot{\delta}}(s).$

Using (4.19), this yields

${\mathfrak{C}}_{0,\mu}[\dot{\delta}]=\ell{\mathcal{K}}^{1}_{\mu}*\dot{\delta};$

note also for future use that we also get from the lemma that

$\displaystyle\zeta(t,x)$ $\displaystyle={\mathcal{K}}^{0}_{\mu}*q.$

∎

As in Corollary 4.3 in the non dispersive case, it is possible to determine
the motion of the solid by the resolution of a single scalar equation on
$\delta$; due to the presence of the dispersive terms however, this equation
is no longer an ordinary differential equation but an integro-differential
equation.

######  Corollary 4.5.

The motion of the floating object for the problem (4.15)-(4.17) can be found
directly by solving the linear second order integro-differential equation,

(4.22)
$\big{(}\tau^{2}_{\mu}+\ell\kappa\big{)}\ddot{\delta}+\ell{\mathcal{K}}^{1}_{\mu}*\dot{\delta}+\delta=0,$

with initial conditions $\delta(0)=\delta_{0}$ and $\dot{\delta}(0)=0$.

######  Remark 4.6.

In [33], the authors consider the linearized shallow-water equations with some
viscosity $\nu$. More precisely, they consider (4.15) with
$-\nu\partial_{x}^{2}q$ instead of $-\kappa^{2}\partial_{x}^{2}\partial_{t}q$
in the second equation and find the following Cummins equation

(4.23)
$\tau_{\mu}^{2}\ddot{\delta}+\sqrt{\nu}\ell\delta^{\left(\frac{3}{2}\right)}+\nu\dot{\delta}+\ell{\mathcal{F}}_{\nu}\ast\dot{\delta}+\delta=0\quad\text{with}\quad{\mathcal{F}}_{\nu}:=\mathcal{L}^{-1}\left[\frac{1}{\sqrt{1+\nu
s}+\sqrt{\nu s}}\right],$

and where $\delta^{(\frac{3}{2})}$ stands for the fractional derivative of
order $3/2$ of $\delta$. This equation shares some similarities with (4.22),
in particular the convolution term, although with a different kernel (note
that one gets $\widehat{\mathcal{K}}^{1}_{\mu}(s)$ by replacing $\nu s$ by
$\kappa^{2}s^{2}$ in $\widehat{{\mathcal{F}}}_{\mu}(s)$). One the contrary,
there is in (4.23) a viscous damping term $\nu\dot{\delta}$ that has no
equivalent in (4.22). Note finally that the fractional derivative term
$\sqrt{\nu}\ell\delta^{\left(\frac{3}{2}\right)}$ in (4.23) can be related to
the added mass term ${\ell\kappa\ddot{\delta}}$ in (4.22). Indeed, in the
analysis of [33], this fractional derivative is the leading order term of a
convolution term $\ell F*\dot{\delta}$ with $\widehat{F}(s)=\sqrt{1+\nu s}$.
In the dispersive case, the same analysis would give a symbol
$\sqrt{1+\kappa^{2}s^{2}}$ and the leading order term of the same convolution
would be the dispersive added mass term $\ell\kappa\ddot{\delta}$.

In the linear non dispersive case ($\varepsilon=\mu=0$), Corollary 4.3 shows
that the motion of the object is governed by the same equation as a damped
harmonic oscillator; the return to equilibrium occurs therefore at an
exponential rate. In the presence of dispersion, Corollary 4.5 states that the
motion of the solid is now governed by the integro-differential equation
(4.22) and numerical simulations (see Figure 2) suggest that the decay gets
slower as the dispersion parameter $\kappa=\sqrt{\mu/3}$ increases.

.

Figure 2. Return to equilibrium: evolution of $\delta(t)$ with increasing
value of $\kappa$ and $\delta_{0}=1,\tau_{\rm buoy}=1/6,h_{\rm eq}=1$.

This issue is addressed in the following proposition. In particular, the fact
that $\delta$ belongs to $H^{2}({\mathbb{R}}^{+})$ implies that $\delta$ and
$\dot{\delta}$ tend to zero at infinity, but the third point of the
proposition shows that the decay cannot be stronger than $O(t^{-3/2})$ (as
opposed to the exponential convergence rate in the linear non dispersive
case), bringing a theoretical confirmation to the above numerical
observations.

######  Proposition 4.2.

i. There is a unique solution $\delta\in C^{2}({\mathbb{R}}^{+})\cap
W^{1,\infty}({\mathbb{R}}^{+})$ to (4.22) with initial data
$\delta(0)=\delta_{0}$ and $\dot{\delta}(0)=0$.
ii. Moreover, $\delta\in H^{2}({\mathbb{R}}^{+})$, but for $k\in\\{0,1,2\\}$,
$t\delta^{(k)}\not\in L^{2}({\mathbb{R}}^{+})$.
iii. For all $\alpha>0$ and $k\in\\{0,1,2\\}$ and for all $c>0$ and $T_{0}>0$,
there exists $t>T_{0}$ such that

$|\delta^{(k)}(t)|>ct^{-\frac{3}{2}-\alpha}.$

######  Remark 4.7.

The dispersive delay (convolution) term in (4.22) is responsible for the slow
decay of the solution. Indeed, in the non dispersive limit case $\kappa=0$,
the branching points $s=\pm i\kappa^{-1}$ disappear from the transfer function
$\hat{H}_{\mu}$ derived in (4.24) below, which then becomes

${\hat{H}}_{0}(s):=\frac{\tau^{2}_{0}s+\ell}{{\tau^{2}_{0}\,s^{2}+s\ell+1}},$

whose poles have a strictly negative real part, hence an exponential decay for
$\delta$.

###### Proof.

Since the kernel ${\mathcal{K}}_{1}$ belongs to $L^{1}({\mathbb{R}}^{+})$
(recall that the Bessel function $J_{1}(t)$ decays like $O(t^{-1/2})$), the
proof of the first part of the proposition does not raise any particular
problem. To prove the second part of the proposition, we need a careful
analysis of the transfer function $\widehat{H}_{\mu}$ defined by the relation
$\widehat{\delta}=\widehat{H}_{\mu}\delta_{0}$. Since
$(\delta,\dot{\delta})\in C\cap L^{\infty}(\mathbb{R}^{+})$, the Laplace
transform of $\delta$ and $\dot{\delta}$ are well defined on $\mathbb{C}_{0}$
an after remarking that

$\kappa s+\frac{1}{\sqrt{1+\kappa^{2}s^{2}}+\kappa
s}=\sqrt{1+\kappa^{2}s^{2}},$

the Laplace transform of $\delta$ is, owing to (4.22),

(4.24) $\hat{\delta}={\hat{H}}_{\mu}(s)\delta_{0}\quad\text{ where
}\quad{\hat{H}}_{\mu}(s):=\frac{\tau^{2}_{\mu}s+\ell\sqrt{1+\kappa^{2}s^{2}}}{\tau^{2}_{\mu}\,s^{2}+s\ell\sqrt{1+\kappa^{2}s^{2}}+1}.$

######  Lemma 4.4.

The transfer function ${\hat{H}}_{\mu}$ defined in (4.24) is holomorphic on
$\mathbb{C}_{0}$ and admits only two branching points at $\pm i\kappa^{-1}$.
Moreover all the zeros of the denominator in (4.24) have strictly negative
part.

###### Proof of the lemma.

We denote by $P$ the holomorphic function on $\mathbb{C}_{0}$

$P(s):=\tau^{2}_{\mu}\,s^{2}+s\ell\sqrt{1+\kappa^{2}s^{2}}+1,$

where the square root stands for the square root with positive real part.
Since $P$ has only two singularities which are the branching points $\pm
i\kappa^{-1}$, we can extend it analytically on all the complex plane except
on the cuts $i(-\infty,-\kappa^{-1})$ and $i(\kappa^{-1},\infty)$, and extend
it continuously by a function $P^{*}$ on the imaginary axis by

$P^{*}(i\omega):=\begin{cases}1-\tau^{2}_{\mu}\,\omega^{2}+i\omega\ell\sqrt{1-\kappa^{2}\omega^{2}}&|\omega|\leq\kappa^{-1},\\\
1-\tau^{2}_{\mu}\,\omega^{2}-|\omega|\ell\sqrt{\kappa^{2}\omega^{2}-1}&|\omega|>\kappa^{-1}.\end{cases}$

We first show that $s$ cannot be purely imaginary, then that if a zero is
real, it must be strictly negative, and finally that if a zero is not a real
number, it must satisfy $\Re(s)<0$.

Step 1. The zeros of $P$ cannot be purely imaginary. Indeed, if $\omega$ were
we solution of $P^{*}(i\omega)=0$, then, from the expression of $P(i\omega)$
given above, $\omega^{2}$ would be a real root of the second order polynomial

$(\tau_{\mu}^{4}-\ell^{2}\kappa^{2})X^{2}+(\ell^{2}-2\tau^{2}_{\mu})X+1.$

But the discriminant of this polynomial is
$\Delta=\ell^{4}+4\ell^{2}(\kappa^{2}-\tau^{2}_{\mu})<-\ell^{2}/3$ (since
$\tau^{2}_{\mu}>\kappa^{2}+\ell^{2}/3$), which is negative, implying that the
polynomial cannot have any real root.

Step 2. The zeros of $P$ cannot belong to ${\mathbb{R}}^{+}$ from the simple
observation that $P(\eta)>0$ for all $\eta\in{\mathbb{R}}^{+}$.

Step 3. The zeros of $P$ cannot have positive real part. In order to prove
this, we show here that $\Im(P(s))\neq 0$ for all $s=\eta+i\omega$ with
$s\notin i\mathbb{R}\cup\mathbb{R}$ (the case $s\in{\mathbb{R}}^{+}$ having
been dealt with in Step 2). The imaginary part of $P(s)$ is given by

$\Im[P(s)]=2\tau^{2}_{\mu}\,\eta\,\omega+\eta\,\ell\Im[\sqrt{1+\kappa^{2}s^{2}}]+\omega\,\ell\Re[\sqrt{1+\kappa^{2}s^{2}}].$

By definition of the square root, $\Re[\sqrt{1+\kappa^{2}s^{2}}]\geq 0$ and
the sign of $\Im[\sqrt{1+\kappa^{2}s^{2}}]$ is the same as the sign of the
product $\eta\,\omega$. The following table summarizes the sign of some
quantities in different cases (where $+$ stands for strictly positive, $-$
stands for strictly negative and ind. signifies that the sign is
indeterminate).

$\eta$ | $\omega$ | $2\tau^{2}\,\eta\,\omega$ | $\eta\,\Im[\sqrt{1+\kappa^{2}s^{2}}]$ |  $\omega\Re[\sqrt{1+\kappa^{2}s^{2}}$] | $\Im[P(s)]$
---|---|---|---|---|---
- | - | + | - | - | ind.
- | + | - | + | + | ind.
+ | - | - | - | - | -
+ | + | + | + | + | +

Therefore, if $\eta>0$ then $\Im[P(s)]$ is either strictly positive or
strictly negative, so that it does not vanish.

∎

Using Lemma 4.4, we can extend continuously the transfer function
${\hat{H}}_{\mu}$ on the imaginary axis by

${\hat{H}}^{*}_{\mu}(i\omega)=\begin{cases}\displaystyle\frac{\tau_{\mu}^{2}i\omega+\ell\sqrt{1-\kappa^{2}\omega^{2}}}{-\tau^{2}_{\mu}\,\omega^{2}+i\omega\ell\sqrt{1-\kappa^{2}\omega^{2}}+1}\delta_{0},&|\omega|\leq\kappa^{-1},\\\
\displaystyle\frac{\tau^{2}_{\mu}i\omega+i\ell{\rm
sign}(\omega)\sqrt{\kappa^{2}\omega^{2}-1}}{-\tau^{2}_{\mu}\,\omega^{2}-|\omega|\ell\sqrt{\kappa^{2}\omega^{2}-1}+1}\delta_{0},&|\omega|>\kappa^{-1}.\end{cases}$

Integrating $|{\hat{H}}^{*}_{\mu}(i\omega)|^{2}$ over $\mathbb{R}$ we get

$\displaystyle\int_{\mathbb{R}}|{\hat{H}}^{*}_{\mu}(i\omega)|^{2}{\rm
d}\omega=\displaystyle\int_{-\kappa^{-1}}^{\kappa^{-1}}|{\hat{H}}^{*}_{\mu}(i\omega)|^{2}{\rm
d}\omega+\displaystyle\int_{|\omega|>\kappa^{-1}}|{\hat{H}}^{*}_{\mu}(i\omega)|^{2}{\rm
d}\omega.$

The first integral is obviously finite (the denominator in (4.24) does not
vanish on $i{\mathbb{R}}$). The second integral is also finite since
$|{\hat{H}}^{*}_{\mu}(i\omega)|^{2}{\underset{|\omega|\to\infty}{\sim}}\omega^{-2}$.
Then ${\hat{H}}_{\mu}$ belongs to the standard Hardy space
$\mathcal{H}^{2}(\mathbb{C}_{0})$ and by the Paley-Wiener theorem (see Theorem
5.1 below), one has $\delta\in L^{2}({\mathbb{R}}^{+})$.
The same reasoning can be applied to

$\widehat{\dot{\delta}}=\left(\frac{-1}{\tau^{2}_{\mu}\,s^{2}+s\ell\sqrt{1+\kappa^{2}s^{2}}+1}\right)\delta_{0}$

and

$\widehat{\ddot{\delta}}=\left(\frac{-s}{\tau^{2}_{\mu}\,s^{2}+s\ell\sqrt{1+\kappa^{2}s^{2}}+1}\right)\delta_{0}$

so that $\dot{\delta}$ and $\ddot{\delta}$ also belong to
$L^{2}({\mathbb{R}}^{+})$.
Let us now prove that $u(t):=t\delta(t)$ does not belong to
$L^{2}({\mathbb{R}}^{+})$. Denoting by $U$ the Laplace transform of $u$, one
has

$U(s):=(-1)\left(\frac{d}{ds}\right)H_{\mu}(s)\delta_{0}\quad\mbox{ on
}\quad{\mathbb{C}}_{0},$

and the following extension to the imaginary axis holds,

$U^{*}(i\omega)=\begin{cases}\displaystyle(-1)\left(\frac{d}{d\omega}\right)\frac{\tau^{2}_{\mu}i\omega+\ell\sqrt{1-\kappa^{2}\omega^{2}}}{-\tau^{2}_{\mu}\,\omega^{2}+i\omega\ell\sqrt{1-\kappa^{2}\omega^{2}}+1}\delta_{0},&|\omega|\leq\kappa^{-1},\\\
\displaystyle(-1)\left(\frac{d}{d\omega}\right)\frac{\tau^{2}_{\mu}i\omega+i\ell{\rm
sign}(\omega)\sqrt{\kappa^{2}\omega^{2}-1}}{-\tau^{2}_{\mu}\,\omega^{2}-|\omega|\ell\sqrt{\kappa^{2}\omega^{2}-1}+1}\delta_{0},&|\omega|>\kappa^{-1}.\end{cases}$

But $U$ is no longer bounded on the imaginary axis as it contains two non
isolated singularities (of order $-1/2$ is the Puiseux series expansion) at
$\pm i\kappa^{-1}$; the integral

$\displaystyle\int_{-\kappa^{-1}}^{\kappa^{-1}}|U^{*}(i\omega)|^{2}d\omega$

is not finite and thus $U\notin\mathcal{H}^{2}(\mathbb{C}_{0})$. By the Paley-
Wiener theorem, this implies that $u\notin L^{2}(\mathbb{R}^{+})$. The same
reasoning can be applied to

$V(s)=(-1)\left(\frac{d}{ds}\right)\left(\frac{-1}{\tau^{2}_{\mu}\,s^{2}+s\ell\sqrt{1+\kappa^{2}s^{2}}+1}\right)\delta_{0}$

and

$W(s)=(-1)\left(\frac{d}{ds}\right)\left(\frac{-s}{\tau^{2}_{\mu}\,s^{2}+s\ell\sqrt{1+\kappa^{2}s^{2}}+1}\right)\delta_{0},$

which are respectively the Laplace transforms of $t\dot{\delta}$ and
$t\ddot{\delta}$. This completes the proof of the second point of the
proposition.
Let us now prove the third point by contradiction. Assuming that there exists
$C>0$ such that for $t$ large enough

$|\delta^{(k)}(t)|\leq Ct^{-\frac{3}{2}-\alpha},$

one gets

$|t\delta^{(k)}(t)|^{2}\leq Ct^{-1-2\alpha}$

which implies $t\delta^{(k)}\in L^{2}(\mathbb{R}^{+})$, which contradicts the
second point. ∎

In the non dispersive case, we showed in Corollary 4.4 that once the motion of
the object is known, it is possible to find $q$ in the exterior domain by
solving an initial boundary value problem for a Burgers-type scalar equation.
This remains true in the present dispersive linear case, but the initial
boundary value problem one has to solve is now nonlocal in time. Note that as
in Remark 4.5, the corollary can easily be generalized to describe the waves
created by an object in forced motion.

######  Corollary 4.6.

The return to equilibrium problem for the linear Boussinesq equations
(4.15)-(4.17) with initial condition (4.2) can be equivalently formulated as a
scalar nonlocal initial boundary value problem on $q$

(4.25)
$\begin{cases}\partial_{x}q+{\mathcal{K}}^{0}_{\mu}\ast\partial_{t}q&=0\qquad(t>0,\quad
x>\ell),\\\ q_{|_{t=0}}&=0,\\\ q_{|_{x=\ell}}&=-\ell\dot{\delta},\end{cases}$

where ${\mathcal{K}}^{0}_{\mu}$ is defined in (4.18) while $\zeta$ is given in
terms of $q$ by a convolution in time

(4.26) $\zeta={\mathcal{K}}^{0}_{\mu}*q,$

with $\delta$ furnished by Proposition 4.2.

######  Remark 4.8.

The nonlocal initial boundary value problem (4.25) is not standard. The most
convenient way to handle it is to see it as an evolution equation with respect
to $x$ rather than $t$; it then becomes a particular case of the nonlocal
initial boundary value problems considered in Section 5. It is in particular a
consequence of Theorem 5.2 below that (4.25) admits a unique solution $q\in
C({\mathbb{R}}^{+}_{x};H^{1}({\mathbb{R}}^{+}_{x}))\cap
C^{1}({\mathbb{R}}^{+}_{x};L^{2}({\mathbb{R}}^{+}_{x}))$. Moreover,
Proposition 5.2 and Corollary 5.1 imply that the solution if actually of class
$C^{2}({\mathbb{R}}^{+}\times{\mathbb{R}}^{+})$ and infinitely regular with
respect to time, showing that the dispersive terms induce a smoothing effect.
Indeed, when $\kappa=0$, the first equation in (4.25) becomes

(4.27) $\partial_{t}q+\partial_{x}q=0$

and the solution to the initial boundary value problem, explicitly given by

$q(x,t)=\begin{cases}-\ell\dot{\delta}(t-(x-\ell))&\text{ for }t-(x-\ell)\geq
0\\\ 0&\text{ for }t-(x-\ell)<0,\end{cases}$

does not belong to $C^{1}(\mathcal{E}_{+}\times\mathbb{R}^{+})$ because
$\ddot{\delta}(0)=-\frac{1}{\tau^{2}_{\mu}}\neq 0$.

##  5\. The initial boundary value problem for a class of nonlocal transport
equations

As shown in the previous section, the analysis of the return to equilibrium
problem in the linear dispersive case leads to a nonlocal generalization of
the transport equation. The analysis of the initial value problem for such
equations is not standard and we address it in this section. Since this
subject is of interest in its own, we work here with more standard notations.
More precisely, we consider an evolution with respect to the time variable and
a nonlocal term with respect to the space variable (this is the reverse in
§4.3). The domain of consideration is the quadrant $\\{x\geq 0,t\geq 0\\}$.
The typical initial boundary value we shall consider is therefore of the form,

$\begin{cases}\partial_{t}u+{\mathcal{K}}*_{x}\partial_{x}u&=f,\\\
u_{|_{x=0}}&={\underline{u}},\\\ u_{|_{t=0}}&=u^{\rm in}.\end{cases}$

for some convolution kernel ${\mathcal{K}}$ to be made precise later.

After presenting some technical material in §5.1.1 for the functional setting
and the Laplace transform, we remind in §5.2 some very classical facts on the
initial and/or boundary value problems for the standard transport equation,
making a distinction between the case of a positive and a negative velocity.
The nonlocal generalizations of these transport problems, in which
$\partial_{x}u$ is replaced by a nonlocal term
${\mathcal{K}}*_{x}\partial_{x}u$, are addressed in §5.3; in particular
similarities and differences (such as the presence of an additional
compatibility condition and a smoothing effect) with their local counterparts
are commented.

NB. To avoid confusions with the computations performed in §4.3 where the
Laplace transform $\widehat{u}$ was taken with respect to time (with dual
variable $s$) we denote throughout this section by $\widetilde{u}$ the Laplace
transform with respect to $x$ (with dual variable $p=\alpha+i\xi$).

###  5.1. Functional setting and a brief reminder on the Laplace transform

We gather here some definition of functional spaces that play an important
role in the analysis of initial boundary value problems, as well as some
classical facts on the Laplace transform.

####  5.1.1. Functional setting

In the study of initial boundary value problems for hyperbolic systems of
equations, the space ${\mathbb{X}}^{n}$ plays a central role; it is defined
for all $n\in{\mathbb{N}}$ as

${\mathbb{X}}^{n}=\bigcap_{j=0}^{n}C^{j}({\mathbb{R}}^{+}_{t};H^{n-j}({\mathbb{R}}^{+}_{x}));$

in particular, for all $u\in{\mathbb{X}}^{n}$, one can define for all $t\geq
0$ the quantity

$|||u(t,\cdot)|||_{n}:=\sup_{j+k\leq
n}|\partial_{t}^{j}\partial_{x}^{k}u(t,\cdot)|_{L^{2}({\mathbb{R}}^{+})}.$

Let us also define ${\mathbb{Y}}^{n}$ as

${\mathbb{Y}}^{n}=\bigcap_{j=0}^{n}W^{j,1}_{\rm
loc}({\mathbb{R}}^{+}_{t};H^{n-j}({\mathbb{R}}^{+}_{x})).$

When working with nonlocal transport equations, it is convenient to introduce
weighted versions of these spaces. For any $a\in{\mathbb{R}}$, and
$k\in{\mathbb{N}}$, we introduce therefore

$\displaystyle L^{2}_{a}({\mathbb{R}}^{+})$ $\displaystyle:=\\{u\in L^{2}_{\rm
loc}({\mathbb{R}}^{+}),\,|u|_{L^{2}_{a}}:=\Big{(}\int_{{\mathbb{R}}^{+}}e^{-2ax}|u(x)|^{2}{\rm
d}x\Big{)}^{1/2}<\infty\\},$ $\displaystyle H^{k}_{a}({\mathbb{R}}^{+})$
$\displaystyle:=\\{u\in
L^{2}_{a}({\mathbb{R}}^{+}),\,|u|_{H^{k}_{a}}:=\sum_{l=0}^{k}|\partial_{x}^{l}u|_{L^{2}_{a}}<\infty\\},$

and denote by ${\mathbb{X}}^{n}_{a}$ and ${\mathbb{Y}}^{n}_{a}$ the weighted
version of the spaces ${\mathbb{X}}^{n}$ and ${\mathbb{Y}}^{n}$ obtained by
replacing all $L^{2}({\mathbb{R}}^{+}_{x})$ based spaces by their
$L^{2}_{a}({\mathbb{R}}^{+}_{x})$ analogue: we also write

$|||u(t,\cdot)|||_{a,n}:=\sup_{j+k\leq
n}|\partial_{t}^{j}\partial_{x}^{k}u(t,\cdot)|_{L_{a}^{2}({\mathbb{R}}^{+})}.$

####  5.1.2. Some results on the Laplace transform

For all $u\in L^{1}_{\rm loc}({\mathbb{R}}^{+})$, the Laplace transform is
defined by

$\widetilde{u}(p)=\int_{0}^{\infty}e^{-px}u(x){\rm d}x,$

for all $p=\alpha+i\xi\in{\mathbb{C}}$ such that this integral converges
absolutely. Using for all $a\in{\mathbb{R}}$ the notation

${\mathbb{C}}_{a}=\\{p\in{\mathbb{C}},\Re p>a\\},$

we can define the Hardy space

$\mathcal{H}^{2}(\mathbb{C}_{a}):=\Big{\\{}U\mbox{\rm{ holomorphic on
}}\mathbb{C}_{a}\,;\,||U||^{2}_{\mathcal{H}^{2}(\mathbb{C}_{a})}:=\sup_{\alpha>a}\int_{\mathbb{R}}|U(\alpha+i\xi)|^{2}d\xi<\infty\Big{\\}}.$

Every function $U\in\mathcal{H}^{2}(\mathbb{C}_{a})$ admits a boundary trace
denoted $U^{*}$ on $a+i{\mathbb{R}}$, that belongs to
$L^{2}(a+i{\mathbb{R}})$, and $\mathcal{H}^{2}(\mathbb{C}_{a})$ is a Hilbert
space for the scalar product

$\langle
F,G\rangle_{\mathcal{H}^{2}(\mathbb{C}_{a})}=\frac{1}{2\pi}\int_{-\infty}^{\infty}F^{*}(a+i\xi)\overline{G^{*}(a+i\xi)}{\rm
d}\xi.$

Recalling that the weighted space $L^{2}_{a}({\mathbb{R}}^{+})$ is defined in
the previous section, we can state the well-known Paley-Wiener theorem.

######  Theorem 5.1.

Let $a\in{\mathbb{R}}$. The Laplace-transform

$\mathcal{L}:\begin{array}[]{lcl}L_{a}^{2}({\mathbb{R}}^{+})\to\mathcal{H}^{2}(\mathbb{C}_{a})\\\
u\mapsto\widetilde{u}\end{array}$

is an isometry between Hilbert spaces.

Recalling that

$\widetilde{\frac{du}{dx}}(p)=p\widetilde{u}(p)-u(0)$

(whenever these quantities make sense), we also have the following
characterization of the weighted Sobolev spaces $H^{k}_{a}({\mathbb{R}}^{+})$.

######  Proposition 5.1.

Let $k\in{\mathbb{N}}$ and
$({\underline{u}}_{0},\dots,{\underline{u}}_{k-1})\in{\mathbb{R}}^{k}$. The
following assertions are equivalent,

  * i.

One has $u\in H^{k}_{a}({\mathbb{R}}^{+})$ and for all $0\leq j\leq k-1$,
$\lim_{x\to 0^{+}}\partial_{x}^{j}u(x)={\underline{u}}_{j}$.

  * ii.

For all $0\leq j\leq k$, the mapping $p\mapsto
p^{j}\widetilde{u}(p)-\sum_{i=0}^{j-1}p^{j-1-i}{\underline{u}}_{i}$ belongs to
${\mathcal{H}}^{2}({\mathbb{C}}_{a})$ (with the sum taken to be zero if
$j=0$).

Moreover, for all $0\leq j\leq k$, one has
$\widetilde{\partial_{x}^{j}u}=p^{j}\widetilde{u}(p)-\sum_{i=0}^{j-1}p^{j-1-i}{\underline{u}}_{i}$.

###  5.2. Reminder on the standard transport equation

Let us start with some considerations on the standard initial boundary value
problem for the transport equations $\partial_{t}u+\partial_{x}u=f$ (referred
to as right-going case) and $\partial_{t}u-\partial_{x}u=f$ (left-going case).

####  5.2.1. The right-going case

We consider here the following initial boundary value problem

(5.1) $\begin{cases}\partial_{t}u+\partial_{x}u&=f,\\\
u_{|_{x=0}}&={\underline{u}},\\\ u_{|_{t=0}}&=u^{\rm in},\end{cases}$

with $f\in{\mathbb{Y}}^{1}$, $u^{\rm in}\in H^{1}({\mathbb{R}}^{+})$ and
${\underline{u}}\in H^{1}_{\rm loc}({\mathbb{R}}^{+})$. In order for (5.1) to
admit a solution
$u\in{\mathbb{X}}^{1}=C({\mathbb{R}}^{+}_{t};H^{1}({\mathbb{R}}^{+}_{x}))\cap
C^{1}({\mathbb{R}}^{+}_{t};L^{2}({\mathbb{R}}^{+}_{x}))$, and therefore
continuous on $[0,\infty)\times[0,\infty)$, it is necessary that

${\underline{u}}(t=0)=u^{\rm in}(x=0).$

This compatibility condition is actually sufficient to ensure the existence
and uniqueness of such a solution. Even if the data are more regular, i.e. if
$f\in{\mathbb{Y}}^{n}$, $u^{\rm in}\in H^{n}({\mathbb{R}}^{+})$ and
${\underline{u}}\in H^{n}_{\rm loc}({\mathbb{R}}^{+})$ for some $n>1$, one
cannot expect in general the solution to be in ${\mathbb{X}}^{n}$. It is a
general feature of first order hyperbolic systems that such a regularity is
achieved if and only if $n$ algebraic compatibility conditions are satisfied
(see for instance [4, 36, 37, 18]). Of course, the situation is the same if we
choose to work in the weighted space ${\mathbb{X}}^{n}_{a}$ since the presence
of the weight changes the integrability properties at infinity, but not local
regularity. In the present case, this can easily be checked on the following
explicit representation of the solution

(5.2) $u(t,x)=u^{\rm
in}(x-t)+{\underline{u}}(t-x)+\int_{0}^{t}f(t^{\prime},x-t+t^{\prime}){\rm
d}t^{\prime},$

where $u^{\rm in}$, ${\underline{u}}$ and $f(t,\cdot)$ are extended by zero in
order to be considered as functions defined on the full line ${\mathbb{R}}$
instead of ${\mathbb{R}}^{+}$.

####  5.2.2. The left-going case

It is well-known that an initial boundary value problem similar to (5.1) is
ill-posed for the left-going transport equation
$\partial_{t}u-\partial_{x}u=f$. Indeed, the initial value problem (without
boundary condition)

(5.3) $\begin{cases}\partial_{t}u-\partial_{x}u&=f,\\\ u_{|_{t=0}}&=u^{\rm
in},\end{cases}$

is well posed, and the solution can be explicitly written as

(5.4) $u(t,x)=u^{\rm in}(x+t)+\int_{0}^{t}f(t^{\prime},x+t-t^{\prime}){\rm
d}t^{\prime};$

in particular, the boundary value ${\underline{u}}$ is given in terms of
$u^{\rm in}$ and $f$ through the relation

${\underline{u}}(t)=u^{\rm in}(t)+\int_{0}^{t}f(t^{\prime},t-t^{\prime}){\rm
d}t^{\prime}$

and therefore cannot be freely prescribed. Note that using this relation in
(5.4), one can express the solution in terms of the boundary data instead of
the initial data, namely,

(5.5)
$u(t,x)={\underline{u}}(x+t)-\int_{0}^{x}f(x+t-x^{\prime},x^{\prime}){\rm
d}x^{\prime}.$

This proves in particular that the following boundary value problem (without
initial condition)

(5.6) $\begin{cases}\partial_{t}u-\partial_{x}u&=f,\\\
u_{|_{x=0}}&={\underline{u}},\end{cases}$

is also well-posed for the left-going transport equation.
We note finally that for the initial value problem (5.3) as well as for the
boundary value problem (5.6) (which are essentially the same by switching the
variables $t$ and $x$) the solution $u$ belongs to ${\mathbb{X}}^{1}$ if the
data are smooth enough without having to impose any compatibility condition,
contrary to what we saw for the right-going case.

###  5.3. The nonlocal transport equation

The aim of this section is to investigate the behavior of nonlocal
perturbations of the right-going and left-going transport equations
respectively given by

(5.7) $\partial_{t}u+{\mathcal{K}}^{0}_{\mu}*_{x}\partial_{x}u=f\quad\mbox{
and }\quad\partial_{t}u-{\mathcal{K}}^{0}_{\mu}*_{x}\partial_{x}u=f,$

where $*_{x}$ stands for the causal convolution with respect to the space
variable,

$\forall x\in{\mathbb{R}}^{+},\qquad
f*_{x}g(x)=\int_{0}^{x}f(x-x^{\prime})g(x^{\prime}){\rm d}x^{\prime}$

and with the Bessel kernel ${\mathcal{K}}^{0}_{\mu}$ as in (4.18); in
particular, we recall that

$\widetilde{{\mathcal{K}}^{0}_{\mu}}(p)=\frac{1}{\sqrt{1+\kappa^{2}p^{2}}}\qquad(\kappa^{2}=\mu/3).$

######  Remark 5.1.

Though we consider here the Bessel kernel ${\mathcal{K}}^{0}_{\mu}$, the
results of this section can easily be adapted to other kernels.

An important feature of the family $({\mathcal{K}}^{0}_{\mu})_{\mu>0}$ is that
it formally converges to the Dirac mass at $x=0$ as $\mu\to 0$, so that the
nonlocal transport equations (5.7) formally converge to the standard right-
going and left-going transport equations respectively, namely,

$\partial_{t}u+\partial_{x}u=f\quad\mbox{ and
}\quad\partial_{t}u-\partial_{x}u=f.$

A natural question is therefore to ask whether the nonlocal initial and/or
boundary value problems have a similar behavior to the behavior of their local
counterpart described in §5.2.

####  5.3.1. The right-going case

We want to address in this section the same kind of initial boundary value
problem as (5.1), but where the space derivative is now replaced by a nonlocal
term, namely, we consider

(5.8)
$\begin{cases}\partial_{t}u+{\mathcal{K}}^{0}_{\mu}*_{x}\partial_{x}u&=f,\\\
u_{|_{x=0}}&={\underline{u}},\\\ u_{|_{t=0}}&=u^{\rm in}.\end{cases}$

As for (5.1), if there exists a solution $u\in{\mathbb{X}}^{1}$ (or more
generally in the weighted version ${\mathbb{X}}^{1}_{a}$ with $a\geq 0$) to
(5.8), then it is continuous at $x=t=0$ and the data must therefore satisfy
the same compatibility condition

(5.9) ${\underline{u}}(t=0)=u^{\rm in}(x=0)$

as for the standard transport equation.
There is however a new compatibility condition that arises here. Indeed, since
${\mathcal{K}}^{0}_{\mu}\in L^{1}_{\rm loc}({\mathbb{R}}^{+})$, the trace of
${\mathcal{K}}^{0}_{\mu}*_{x}\partial_{x}u$ at $x=0$ is well defined if
$\partial_{x}u\in C({\mathbb{R}}^{+}_{t};L^{2}_{\rm
loc}({\mathbb{R}}^{+}_{x}))$, and it must be equal to zero by definition of
the convolution. Taking the trace of the first equation in (5.8), one
therefore finds the following additional compatibility condition for the
existence of solutions with the aforementioned regularity,

(5.10) $\forall
t\in{\mathbb{R}}^{+},\qquad\partial_{t}{\underline{u}}(t)=f(t,0).$

######  Remark 5.2.

The similar procedure applied to the standard transport problem (5.8) yields
the relation

$\partial_{x}u(t,0)=-\partial_{t}{\underline{u}}(t)+f(t,0),$

which is not a compatibility condition but an information on the behavior of
the trace of $\partial_{x}u$ at the boundary.

If these two compatibility conditions are satisfied, the theorem below shows
the well-posedness of the nonlocal initial boundary value problem (5.8). We
recall that the functional spaces have been defined in §5.1.1; note also that
we have to work in weighted spaces here in order to compensate the slow decay
of ${\mathcal{K}}^{0}_{\mu}$ at infinity (which is of order $O(|x|^{-1/2}$)
and that more information on the regularity of the solution is given in
Corollary 5.1 below.

######  Theorem 5.2.

Let $a>0$ and $f\in{\mathbb{Y}}^{1}_{a}$, $u^{\rm in}\in
H_{a}^{1}({\mathbb{R}}_{x}^{+})$ and ${\underline{u}}\in W^{1,1}_{\rm
loc}({\mathbb{R}}^{+}_{t})$. Assume moreover that the compatibility conditions
(5.9) and (5.10) hold. Then there exists a unique solution
$u\in{\mathbb{X}}^{1}_{a}$ to the nonlocal initial boundary value problem
(5.8), and there exists $c_{a}>0$ such that, for all $t\in{\mathbb{R}}^{+}$,

$|u(t,\cdot)|_{H_{a}^{1}({\mathbb{R}}^{+}_{x})}\leq e^{-c_{a}t}|u^{\rm
in}|_{H_{a}^{1}({\mathbb{R}}^{+}_{x})}+\int_{0}^{t}e^{-c_{a}(t-t^{\prime})}\big{[}|f(t^{\prime},\cdot)|_{H_{a}^{1}({\mathbb{R}}^{+}_{x})}+|{\mathcal{K}^{0}_{\mu}}|_{L_{a}^{2}}|{\underline{u}}(t^{\prime})|\big{]}{\rm
d}t^{\prime}.$

If moreover ${\underline{u}}=0$ then the result still holds with $a=c_{a}=0$.

###### Proof.

For the sake of clarity, we simply write ${\mathcal{K}}$ instead of
${\mathcal{K}}^{0}_{\mu}$. Taking the Laplace transform of (5.8) with respect
to space, one gets that

$\partial_{t}\widetilde{u}+\widetilde{{\mathcal{K}}}(p)(p\widetilde{u}-{\underline{u}})=\widetilde{f}\quad\mbox{
on }\quad{\mathbb{R}}^{+}.$

Solving this ODE with initial condition
$\widetilde{u}_{|_{t=0}}=\widetilde{u^{\rm in}}$, one gets the following
expression for $\widetilde{u}$, for all $p\in{\mathbb{C}}_{a}$ and
$t\in{\mathbb{R}}^{+}$,

$\displaystyle\widetilde{u}(t,p)$
$\displaystyle=e^{-p\widetilde{{\mathcal{K}}}(p)t}\widetilde{u^{\rm
in}}(p)+\int_{0}^{t}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}\widetilde{f}{\rm
d}t^{\prime}+\int_{0}^{t}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}\widetilde{{\mathcal{K}}}(p){\underline{u}}(t^{\prime}){\rm
d}t^{\prime}$ (5.11)
$\displaystyle=:\widetilde{u_{1}}+\widetilde{u_{2}}+\widetilde{u_{3}}.$

Since the Paley-Wiener Theorem 5.1 states that the Laplace transform is an
isometry between $L_{a}^{2}({\mathbb{R}}^{+})$ and
${\mathcal{H}}^{2}({\mathbb{C}}_{a})$, the following lemma shows that both
$u_{1}$ and $u_{2}$ belong to
$C({\mathbb{R}}^{+}_{t};L^{2}_{a}({\mathbb{R}}^{+}_{x}))$ if $u^{\rm in}\in
L^{2}_{a}({\mathbb{R}}^{x})$ and $f\in L^{1}_{\rm
loc}({\mathbb{R}}^{+}_{t};L^{2}_{a}({\mathbb{R}}^{+}_{x}))$.

######  Lemma 5.1.

Let $a\geq 0$. For all $U\in{\mathcal{H}}^{2}({\mathbb{C}}_{a})$, the mapping

$\begin{array}[]{lcl}{\mathbb{R}}^{+}&\to&{\mathcal{H}}^{2}({\mathbb{C}}_{a})\\\
t&\mapsto&\big{(}p\mapsto
e^{-p\widetilde{{\mathcal{K}}}(p)t}U(p)\big{)}\end{array}$

is well defined and continuous, and for all $t\in{\mathbb{R}}^{+}$,
$\|e^{-p\widetilde{{\mathcal{K}}}(p)t}U\|_{{\mathcal{H}}^{2}({\mathbb{C}}_{a})}\leq\|U\|_{{\mathcal{H}}^{2}({\mathbb{C}}_{a})}$.
If moreover $a>0$ then there exists $c_{a}>0$ such that for all
$t\in{\mathbb{R}}^{+}$,

$\|e^{-p\widetilde{{\mathcal{K}}}(p)t}U\|_{{\mathcal{H}}^{2}({\mathbb{C}}_{a})}\leq
e^{-c_{a}t}\|U\|_{{\mathcal{H}}^{2}({\mathbb{C}}_{a})}.$

###### Proof of the lemma.

Except for the last assertion, we consider only the case $a=0$ since the case
$a>0$ can easily be deduced from it. From the definition of
${\mathcal{H}}^{2}({\mathbb{C}}_{0})$ and Lebesgue’s dominated convergence
theorem, it is sufficient to prove that $e^{-p\widetilde{{\mathcal{K}}}(p)t}$
is holomorphic and bounded on ${\mathbb{C}}_{0}$. The fact that it is
holomorphic directly stems from the explicit expression
$\widetilde{{\mathcal{K}}}(p)=(1+\kappa^{2}p^{2})^{-1/2}$. For the
boundedness, this is a consequence of the fact that
$\Re(p\widetilde{{\mathcal{K}}}(p))\geq 0$ on ${\mathbb{C}}_{0}$, as we now
prove. For all $p=\alpha+i\xi\in\mathbb{C}_{0}$, one computes

(5.12)
$\Re(p\hat{{\mathcal{K}}}(p))=\frac{\alpha\,\Re({\sqrt{1+\kappa^{2}p^{2}}})+\xi\,\Im({\sqrt{1+\kappa^{2}p^{2}}})}{|\sqrt{1+\kappa^{2}p^{2}}|^{2}}.$

Since $\Re(\sqrt{1+\kappa^{2}p^{2}})$ is positive (by definition of the square
root) and the sign of $\Im({\sqrt{1+\kappa^{2}p^{2}}})$ is the same as the
sign of the product $\alpha\,\xi$, one gets the result.
Since we have proved that $\Re(p\widetilde{{\mathcal{K}}}(p))\geq 0$ on
${\mathbb{C}}_{0}$, the last assertion follows if we can prove that
$\Re(p\widetilde{{\mathcal{K}}}(p))$ does not vanish on ${\mathbb{C}}_{a}$ if
$a>0$. Since both terms in the numerator in (5.12) are positive, both must
vanish if $\Re(p\widetilde{{\mathcal{K}}}(s))$ vanishes. Since $\alpha>0$ on
${\mathbb{C}}_{0}$, this implies that there should be
$p=\alpha+i\xi\in{\mathbb{C}}_{a}$ such that $\Re(\sqrt{1+\kappa^{2}p^{2}})=0$
and $\xi\,\Im({\sqrt{1+\kappa^{2}p^{2}}})=0$, which is obviously not possible.
∎

Remarking that for any $a>0$, one has
$\widehat{\mathcal{K}}\in{\mathcal{H}}^{2}({\mathbb{C}}_{a})$, it is also a
direct consequence of the lemma that there is $c_{a}>0$ such that

$\displaystyle\|{\widetilde{u}_{3}}\|_{{\mathcal{H}}^{2}({\mathbb{C}}_{a})}$
$\displaystyle\leq\|{\widetilde{{\mathcal{K}}}}\|_{{\mathcal{H}}^{2}({\mathbb{C}}_{a})}\int_{0}^{t}e^{-c_{a}(t-t^{\prime})}|{\underline{u}}(t^{\prime})|{\rm
d}t^{\prime}.$

Together with the results already proved on $\widetilde{u}_{1}$ and
$\widetilde{u_{2}}$, we deduce (see the Paley-Wiener Theorem 5.1 below) that

$|u(t,\cdot)|_{L_{a}^{2}({\mathbb{R}}^{+}_{x})}\leq e^{-c_{a}t}|u^{\rm
in}|_{L_{a}^{2}({\mathbb{R}}^{+}_{x})}+\int_{0}^{t}e^{-c_{a}(t-t^{\prime})}\big{[}|f(t^{\prime},\cdot)|_{L_{a}^{2}({\mathbb{R}}^{+}_{x})}+|{\mathcal{K}}|_{L_{a}^{2}}|{\underline{u}}(t^{\prime})|\big{]}{\rm
d}t^{\prime}.$

In order to conclude the proof of the theorem, we still need to control
$\partial_{x}u$ and $\partial_{t}u$.

  * •

Control of $\partial_{x}u$. We want to show that $\partial_{x}u\in
C({\mathbb{R}}^{+}_{t};L^{2}_{a}({\mathbb{R}}^{+}_{x}))$, or equivalently that
$\widetilde{\partial_{x}u}\in
C({\mathbb{R}}^{+}_{t};{\mathcal{H}}^{2}({\mathbb{C}}_{a}))$. Since
$\widetilde{\partial_{x}u}=p\widetilde{u}-{\underline{u}}$, we consider

$p\widetilde{u}(t,p)=e^{-p\widetilde{{\mathcal{K}}}(p)t}p\widetilde{u^{\rm
in}}(p)+\int_{0}^{t}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}p\widetilde{f}{\rm
d}t^{\prime}+\int_{0}^{t}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}p\widetilde{{\mathcal{K}}}(p){\underline{u}}(t^{\prime}){\rm
d}t^{\prime}.$

Writing $p\widetilde{u^{\rm in}}=\widetilde{\partial_{x}u^{\rm in}}+u^{\rm
in}(0)$, $p\widetilde{f}(t,p)=\widetilde{\partial_{x}f}(t,p)+f(t,0)$, we can
remark that

$\displaystyle\int_{0}^{t}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}p\widetilde{{\mathcal{K}}}(p){\underline{u}}(t^{\prime}){\rm
d}t^{\prime}$
$\displaystyle=\int_{0}^{t}\partial_{t^{\prime}}\big{(}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}\big{)}{\underline{u}}(t^{\prime}){\rm
d}t^{\prime}$
$\displaystyle={\underline{u}}(t)-e^{-p\widetilde{{\mathcal{K}}}(p)t}{\underline{u}}(0)-\int_{0}^{t}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}\partial_{t}{\underline{u}}(t^{\prime}){\rm
d}t^{\prime},$

from which we deduce that

$\displaystyle\widetilde{\partial_{x}u}(t,p)=$ $\displaystyle
e^{-p\widetilde{{\mathcal{K}}}(p)t}\widetilde{\partial_{x}u^{\rm
in}}(p)+\int_{0}^{t}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}\widetilde{\partial_{x}f}{\rm
d}t^{\prime}$ $\displaystyle+e^{-p\widetilde{{\mathcal{K}}}(p)t}\big{(}u^{\rm
in}(0)-{\underline{u}}(0)\big{)}+\int_{0}^{t}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}\big{(}f(t^{\prime},0)-\partial_{t}{\underline{u}}(t^{\prime})\big{)}{\rm
d}t^{\prime}.$

While the first two components of the right-hand side belong to
$C({\mathbb{R}}^{+}_{t};{\mathcal{H}}^{2}({\mathbb{C}}_{a}))$ by Lemma 5.1,
the last two ones do not, unless the compatibilty conditions given in the
statement of the Theorem are satisfied, in which case these two components
cancel and the result follows together with the upper bound

$|\partial_{x}u(t,\cdot)|_{L^{2}_{a}}\leq|\partial_{x}u^{\rm
in}|_{L^{2}_{a}}+\int_{0}^{t}e^{-c_{a}(t-t^{\prime})}|\partial_{x}f(t^{\prime},\cdot)|_{L^{2}_{a}}{\rm
d}t^{\prime}.$

  * •

Control of $\partial_{t}u$. Using the equations, one has

$\displaystyle|\partial_{t}u|_{L^{2}_{a}}$
$\displaystyle\leq|{\mathcal{K}}*_{x}\partial_{x}u|_{L^{2}_{a}}+|f|_{L^{2}_{a}}$
$\displaystyle\leq|{\mathcal{K}}|_{L^{1}_{a}}|\partial_{x}u|_{L^{2}_{a}}+|f|_{L^{2}_{a}},$

with $L^{1}_{a}=L^{1}({\mathbb{R}}^{+},e^{-ax}{\rm d}x)$, showing as needed
that $\partial_{t}u\in
C({\mathbb{R}}^{+}_{t};L^{2}_{a}({\mathbb{R}}^{+}_{x}))$.

The theorem follows easily. ∎

######  Remark 5.3.

As explained above in Remark 5.1, the initial boundary value problem (5.8) can
be seen as a nonlocal perturbation of the standard transport problem (5.1)
toward which it formally converges when $\mu\to 0$. There seems however to be
some discrepancy because two compatibility conditions, namely, (5.9) and
(5.10), are needed to ensure the existence of solutions
$u\in{\mathbb{X}}^{1}_{a}$ to (5.8), while the sole compatibility condition
(5.9) is sufficient to get a similar result for the standard transport problem
(5.1). One should explain why the second compatibility condition (5.10)
disappears in the formal limit $\mu=0$.
The reason is that (5.10) is here to ensure continuity of the solution at the
boundary $x=0$. Indeed, by the initial value theorem, we know that $\lim_{x\to
0^{+}}u(t,x)=\lim_{p\in{\mathbb{C}}_{a},|p|\to\infty}p\widetilde{u}(t,p)$, and
we therefore get from the Laplace representation formula (5.11) that

$\lim_{x\to 0^{+}}u(t,x)=e^{-\frac{t}{\kappa}}u^{\rm
in}(0)+\int_{0}^{t}e^{-\frac{t-t^{\prime}}{\kappa}}f(t^{\prime},0){\rm
d}t^{\prime}+\int_{0}^{t}e^{-\frac{t-t^{\prime}}{\kappa}}\frac{1}{\kappa}{\underline{u}}(t^{\prime}){\rm
d}t^{\prime}$

where we used the fact that
$\lim_{p\in{\mathbb{C}}_{a},|p|\to\infty}p\widetilde{{\mathcal{K}}}(p)=\kappa^{-1}$;
after an integration by parts, the right-hand side can be written

${\underline{u}}(t)+e^{-\frac{t}{\kappa}}\big{(}u^{\rm
in}(0)-{\underline{u}}(0)\big{)}+\int_{0}^{t}e^{-\frac{t-t^{\prime}}{\kappa}}\big{(}f(t^{\prime},0)-\partial_{t}{\underline{u}}(t^{\prime})\big{)}{\rm
d}t^{\prime},$

so that, if the first compatibility condition (5.9) is satisfied, one has

$\lim_{x\to
0^{+}}u(t,x)-{\underline{u}}(t)=\int_{0}^{t}e^{-\frac{t-t^{\prime}}{\kappa}}\big{(}f(t^{\prime},0)-\partial_{t}{\underline{u}}(t^{\prime})\big{)}{\rm
d}t^{\prime},$

which is nonzero if the second compatibility condition is not satisfied, hence
a lack of continuity at $x=0$ (there would therefore be a Dirac mass at $x=0$
is the expression for $\partial_{x}u(t,\cdot)$ that would therefore not be in
$L^{2}_{a}({\mathbb{R}}^{+}_{x})$ as seen in the proof). However, one readily
observes that

$\lim_{\mu\to
0}\int_{0}^{t}e^{-\frac{t-t^{\prime}}{\kappa}}\big{(}f(t^{\prime},0)-\partial_{t}{\underline{u}}(t^{\prime})\big{)}{\rm
d}t^{\prime}=0\qquad(\kappa^{2}=\mu/3),$

so that this discontinuity shrinks to zero in the limit $\mu\to 0$, explaining
why the second compatibility condition is no longer necessary in the endpoint
case $\mu=0$.

Before going further, we recall that there are two possibilities to define
fractional derivatives of order $\alpha\in(0,1)$ on ${\mathbb{R}}^{+}$ using
the convolution kernel
${\mathfrak{K}}_{\alpha}(x)=x^{-\alpha}/{\Gamma(1-\alpha)}$ with
$\alpha\in(0,1)$ and $\Gamma$ the Euler Gamma function, namely, the Riemann-
Liouville and Caputo derivatives, defined respectively as

$D^{\alpha}_{\rm
RL}u=\partial_{x}\big{(}{\mathfrak{K}}_{\alpha}*_{x}u)\quad\mbox{ and }\quad
D^{\alpha}_{\rm C}u={\mathfrak{K}}_{\alpha}*_{x}\partial_{x}u.$

In the nonlocal initial boundary value problem (5.8), the space derivative
$\partial_{x}u$ in the standard transport equation has been replaced by the
nonlocal term ${\mathcal{K}}^{0}_{\mu}*\partial_{x}u$ which can be considered
as a generalized derivative of Caputo type, with the kernel
${\mathfrak{K}}_{\alpha}$ replaced by the Bessel kernel
${\mathcal{K}}^{0}_{\mu}$. It is noteworthy that working with the Riemann-
Liouville version of this operator, namely
$\partial_{x}\big{(}{\mathcal{K}}^{0}_{\mu}*_{x}u)$, the situation is
drastically different. Indeed, as shown in the following proposition, it is
not possible to impose a boundary data anymore since the knowledge of the
initial data suffices to fully determine the solution; in other words, the
initial value problem

(5.13)
$\begin{cases}\partial_{t}u+\partial_{x}\big{(}{\mathcal{K}}^{0}_{\mu}*_{x}u)&=f,\\\
u_{|_{t=0}}&=u^{\rm in},\end{cases}$

is well posed on ${\mathbb{R}}^{+}_{t}\times{\mathbb{R}}^{+}_{x}$. In
particular, the trace of the solution at the boundary $x=0$ is determined by
$f$ and $u^{\rm in}$ and therefore cannot be imposed. We also show that if the
data $u^{\rm in}$ and $f$ are smoother, then the solution is in
${\mathbb{X}}_{a}^{2}$, but generally not in ${\mathbb{X}}_{a}^{3}$ or higher
in the absence of additional compatibility condition (but we show however that
the regularity in time can be higher).

######  Proposition 5.2.

Let $a>0$, $n=1$ or $2$, and $f\in{\mathbb{Y}}^{n}_{a}$ and $u^{\rm in}\in
H_{a}^{n}({\mathbb{R}}_{x}^{+})$. Then there exists a unique solution
$u\in{\mathbb{X}}_{a}^{n}$ to the nonlocal initial boundary value problem
(5.13). Moreover, one has $u(t,\cdot)_{|_{x=0}}={\underline{u}}(t)$ for all
$t\in{\mathbb{R}}^{+}$, with ${\underline{u}}(t)$ given by

${\underline{u}}(t)=e^{-\frac{t}{\kappa}}u^{\rm
in}(0)+\int_{0}^{t}e^{-\frac{t-t^{\prime}}{\kappa}}f(t^{\prime},0){\rm
d}t^{\prime}.$

If in addition $f\in
C^{q}({\mathbb{R}}^{+}_{t};H^{n}_{a}({\mathbb{R}}^{+}_{x}))$ for some
$q\in{\mathbb{N}}$ then one also has $u\in
C^{q+1}({\mathbb{R}}^{+}_{t};H^{n}_{a}({\mathbb{R}}^{+}_{x}))$.

######  Remark 5.4.

Comparing the representation of the solution given in (5.14) below to the
representation of the solution to the initial boundary value problem (5.8)
given in (5.11), one can check that they are both the same if
${\underline{u}}=0$, which is not surprising since one can compute

$\partial_{x}({\mathcal{K}}^{0}_{\mu}*_{x}u)(t,x)=({\mathcal{K}}^{0}_{\mu}*_{x}\partial_{x}u)(t,x)+{\mathcal{K}}^{0}_{\mu}(x){\underline{u}}(t),$

so that the Caputo and Riemann-Liouville nonlocal initial boundary value
problem coincide when ${\underline{u}}=0$.

###### Proof.

As previously done, we simply write ${\mathcal{K}}={\mathcal{K}}_{\mu}^{0}$.
Taking the Laplace transform of (5.13) one readily gets

(5.14)
$\widetilde{u}(t,p)=e^{-p\widetilde{{\mathcal{K}}}(p)t}\widetilde{u^{\rm
in}}(p)+\int_{0}^{t}e^{-p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}\widetilde{f}(t^{\prime},p){\rm
d}t^{\prime};$

by the initial value theorem, one gets that $\lim_{x\to
0^{+}}u(t,x)={\underline{u}}(t)$, with ${\underline{u}}$ as in the statement
of the theorem.
For all $j$ and $l$, one deduces from the above formula for $\widetilde{u}$
that

$p^{j}\widetilde{\partial_{t}^{l}u}=(-p\widehat{{\mathcal{K}}}(p))^{l}\big{[}e^{-p\widehat{{\mathcal{K}}}(p)t}p^{j}\widetilde{u^{\rm
in}}+\int_{0}^{t}e^{-p\widetilde{\mathcal{K}}(p)(t-t^{\prime})}p^{j}\widetilde{f}\big{]}+\sum_{m=1}^{l}(-p\widehat{{\mathcal{K}}}(p))^{l-m}p^{j}\widetilde{\partial_{t}^{m-1}f}.$

Replacing in this expression

$p^{j}\widetilde{v}=\widetilde{\partial_{x}^{j}v}+\sum_{i=0}^{j-1}p^{j-1-i}(\partial_{x}^{i}v)_{|_{x=0}}$

for $v=u^{\rm in}$, $\widetilde{f}$, $\widetilde{\partial_{t}^{m-1}f}$, we
obtain

$p^{j}\widetilde{\partial_{t}^{l}u}=\sum_{i=0}^{j-1}p^{j-1-i}U_{li}(p)+F_{lj}(t,p)$

(using the convention that the summation is zero if $j-1<0$) with

$\displaystyle U_{li}(p):=$
$\displaystyle(-p\widetilde{{\mathcal{K}}}(p))^{l}\big{(}\partial_{x}^{i}u^{\rm
in}(0)+\int_{0}^{t}e^{-p\widetilde{\mathcal{K}}(p)(t-t^{\prime})}(\partial_{x}^{i}f)(t^{\prime},0){\rm
d}t^{\prime}\big{)}$
$\displaystyle+\sum_{m=1}^{l}(-p\widetilde{{\mathcal{K}}}(p))^{l-m}(\partial_{t}^{m-1}\partial_{x}^{i}f)_{|_{x=0}}$

and

$F_{lj}(t,p):=(-p\widehat{{\mathcal{K}}}(p))^{l}\big{[}\widetilde{\partial_{x}^{j}u^{\rm
in}}+\int_{0}^{t}e^{-p\widetilde{\mathcal{K}}(p)(t-t^{\prime})}\widetilde{\partial_{x}^{j}f}\big{]}+\sum_{m=1}^{l}(-p\widehat{{\mathcal{K}}}(p))^{l-m}\widetilde{\partial_{t}^{m-1}\partial_{x}^{j}f}.$

Remarking that $\lim_{|p|\to\infty}p\widehat{{\mathcal{K}}}(p)=\kappa^{-1}$,
and introducing ${\underline{u}}_{li}=\lim_{|p|\to\infty}U_{li}(p)$, namely,

${\underline{u}}_{li}=(-\kappa)^{-l}\big{(}\partial_{x}^{i}u^{\rm
in}(0)+\int_{0}^{t}e^{-\frac{t-t^{\prime}}{\kappa}}(\partial_{x}^{i}f)(t^{\prime},0){\rm
d}t^{\prime}\big{)}\\\
+\sum_{m=1}^{l}(-\kappa)^{-l+m}(\partial_{t}^{m-1}\partial_{x}^{i}f)_{|_{x=0}}$

(of course, ${\underline{u}}_{00}={\underline{u}}$), we can write

$p^{j}\widetilde{\partial_{t}^{l}u}-\sum_{i=0}^{j-1}p^{j-1-i}{\underline{u}}_{li}(p)=\sum_{i=0}^{j-1}p^{j-1-i}\big{(}U_{li}(p)-{\underline{u}}_{li}\big{)}+F_{lj}(t,p).$

From Proposition 5.1, we can deduce that $\partial_{x}^{j}\partial_{t}^{l}u$
belongs to $C({\mathbb{R}}^{+}_{t};L^{2}_{a}({\mathbb{R}}^{+}_{x}))$ if the
right-hand side of the above equality is in
$C({\mathbb{R}}^{+}_{t};{\mathcal{H}}^{2}({\mathbb{C}}_{a}))$. This is obvious
for $F_{lj}$ under the assumptions made in the statement of the proposition
(see Lemma 5.1); for the summation, the problem reduces to determine whether
the mapping $p\mapsto
p^{j-1}\big{(}p\widehat{{\mathcal{K}}}(p)-\kappa^{-1}\big{)}$ belongs to
${\mathcal{H}}^{2}({\mathbb{C}}_{a})$ or not. This mapping being holomorphic
on ${\mathbb{C}}_{a}$, we just need to check that it is square integrable on
$a+i{\mathbb{R}}$. Recalling that
$\widehat{{\mathcal{K}}}(p)=\frac{1}{\sqrt{1+\kappa^{2}p^{2}}}$, and using the
fact that for all $p\in{\mathbb{C}}_{a}$ one has $\sqrt{p^{2}}=p$, one has
$p^{j-1}\big{(}p\widehat{{\mathcal{K}}}(p)-\kappa^{-1}\big{)}\sim-\frac{1}{2\kappa^{3}}p^{j-3}$
at infinity; the mapping is therefore square integrable on $a+i{\mathbb{R}}$
if and only if $j\leq 2$, hence the results. ∎

As a corollary, we can exhibit a smoothing effect for the nonlocal transport
problem (5.8) that does not exist for the standard transport problem (5.1).
Indeed, as one can easily check on the explicit expression (5.2), even if the
data $u^{\rm in}$, ${\underline{u}}$ and $f$ are very smooth, the solution is
not $C^{1}({\mathbb{R}}^{+}\times{\mathbb{R}}^{+})$ if the additional
compatibility condition $\partial_{t}{\underline{u}}(0)=-\partial_{x}u^{\rm
in}(0)+f(0,0)$ is not imposed. There is a smoothing effect for the nonlocal
problem in the sense that the solution constructed in Theorem 5.2 actually
belongs to ${\mathbb{X}}_{a}^{3}\subset
C^{2}({\mathbb{R}}^{+}\times{\mathbb{R}}^{+})$ without any additional
compatibility condition if the data are smooth enough. Note that using the
last statement of Proposition 5.2, the proof shows that additional regularity
in time on $\partial_{x}f$ would yield additional regularity in time on
$\partial_{x}u$.

######  Corollary 5.1.

Under the assumptions of Theorem 5.2, if moreover $f\in{\mathbb{Y}}^{n}_{a}$,
$u^{\rm in}\in H_{a}^{n}({\mathbb{R}}_{x}^{+})$ and ${\underline{u}}\in
W^{n,1}_{\rm loc}({\mathbb{R}}^{+}_{t})$ for $n=2$ or $3$, then the solution
$u$ provided by the theorem belongs to ${\mathbb{X}}^{n}_{a}$.

###### Proof.

Taking the space derivative of the nonlocal transport equation in (5.8), it is
easy to see that $v=\partial_{x}u$ solves the initial boundary value problem

$\begin{cases}\partial_{t}v+\partial_{x}\big{(}{\mathcal{K}}*_{x}v)=\partial_{x}f,\\\
v_{|_{t=0}}=\partial_{x}u^{\rm in}.\end{cases}$

It follows therefore from Proposition 5.2 that
$\partial_{x}u\in{\mathbb{X}}^{n-1}$. We are therefore left to prove that
$\partial_{t}^{j}u\in C({\mathbb{R}}_{t}^{+};L^{2}_{a}({\mathbb{R}}^{+}_{x}))$
for $1\leq j\leq n$; this easily follows from the observation that

$\partial_{t}^{j}u=-{\mathcal{K}}*_{x}\partial_{t}^{j-1}\partial_{x}u+\partial_{t}^{j-1}f$

and from the fact that ${\mathcal{K}}\in L^{1}_{a}({\mathbb{R}}^{+})$. ∎

####  5.3.2. The left-going case

As for the right-going case in the previous section, we want to consider a
nonlocal perturbation of the standard transport problem in which the space
derivative $\partial_{x}$ is replaced by a nonlocal term
${\mathcal{K}}^{0}_{\mu}*\partial_{x}$. As recalled in §5.2, for the standard
left-going transport equation, one has to consider either the initial value
problem or the boundary value problem. While both cases are symmetric in the
case of the standard transport equation, this is no longer the case and, as we
shall see, the boundary value problem leads simpler expressions. We therefore
consider here its nonlocal analogue (see Remark 5.6 below for the nonlocal
analogue of the initial value problem),

(5.15)
$\begin{cases}\partial_{t}u-{\mathcal{K}}^{0}_{\mu}*_{x}\partial_{x}u&=f,\\\
u_{|_{x=0}}&={\underline{u}}.\end{cases}$

As for the boundary value problem (5.6) for the standard left-going transport
equation, there is no compatibility condition like (5.9) since $u^{\rm in}$ is
not prescribed. On the other hand, the analysis leading to the second
compatibility condition (5.10) remains valid, and it is still necessary to
have

(5.16) $\forall
t\in{\mathbb{R}}^{+},\qquad\partial_{t}{\underline{u}}(t)=f(t,0)$

in order to expect a solution $u$ that belongs to ${\mathbb{X}}^{1}$. In the
statement below, we use the notation

$H^{1}_{a}({\mathbb{R}}^{+}_{t}\times{\mathbb{R}}^{+}_{x}):=H^{1}({\mathbb{R}}^{+}_{t};L^{2}_{a}({\mathbb{R}}^{+}_{x}))\cap
L^{2}({\mathbb{R}}^{+}_{t};H^{1}_{a}({\mathbb{R}}^{+}_{x}))$

(note that the assumptions on the time dependence of $f$ and ${\underline{u}}$
are chosen in order to ensure the convergence of the integral term over the
range $(t,+\infty)$ and that they could easily be weakened).

######  Theorem 5.3.

Let $a>0$, $f\in H^{1}_{a}({\mathbb{R}}^{+}_{t}\times{\mathbb{R}}^{+}_{x})$,
and ${\underline{u}}\in H^{1}({\mathbb{R}}^{+}_{t})$. Assume moreover that the
compatibility condition (5.16) holds. Then there exists a unique solution
$u\in{\mathbb{X}}^{1}_{a}$ to the nonlocal boundary value problem (5.15), and
there exists $c_{a}>0$ such that, for all $t\in{\mathbb{R}}^{+}$,

$|u(t,\cdot)|_{H_{a}^{1}({\mathbb{R}}^{+}_{x})}\leq\int_{t}^{\infty}e^{c_{a}(t-t^{\prime})}\big{[}|f(t^{\prime},\cdot)|_{H_{a}^{1}({\mathbb{R}}^{+}_{x})}+|{\mathcal{K}^{0}_{\mu}}|_{L_{a}^{2}}|{\underline{u}}(t^{\prime})|\big{]}{\rm
d}t^{\prime}.$

###### Proof.

Still denoting ${\mathcal{K}}={\mathcal{K}}_{\mu}^{0}$ and following the same
procedure as for the proof of Theorem 5.2, one readily finds that

$\widetilde{u}(t,p)=-\int_{t}^{\infty}e^{p\widetilde{{\mathcal{K}}}(p)(t-t^{\prime})}\big{(}\widetilde{f}(t^{\prime},p)-\widetilde{{\mathcal{K}}}(p){\underline{u}}(t^{\prime})\big{)}{\rm
d}t^{\prime};$

as for the right-going case, one can check that the compatibility condition
(5.16) is necessary for the continuity of the solution at $x=0$. We omit the
proof which is an easy adaptation of the proof of Theorem 5.2. ∎

######  Remark 5.5.

As for the standard boundary transport problem (5.6), the initial data is
determined in terms of the source term $f$ and the boundary data
${\underline{u}}$ by evaluating the Laplace representation formula given in
the proof at $t=0$, namely,

(5.17) $\widetilde{u^{\rm
in}}(p)=-\int_{0}^{\infty}e^{-p\widetilde{{\mathcal{K}}_{\mu}^{0}}(p)t^{\prime}}\big{(}\widetilde{f}(t^{\prime},p)-\widetilde{{\mathcal{K}}_{\mu}^{0}}(p){\underline{u}}(t^{\prime})\big{)}{\rm
d}t^{\prime}.$

In the limit case $\mu=0$ (and therefore
$\widehat{{\mathcal{K}}}_{\mu}(p)=1$), one can check that the representation
formula of the proof is equivalent to (5.5); the additional compatibility
condition (5.16) that is not necessary for (5.6) also disappears at the limit
along a mechanism similar to the one described in Remark 5.3.

######  Remark 5.6.

For the standard left-going transport equation, the initial value problem
(5.4) and the boundary value problem (5.5) can be treated in a totally
symmetric case by switching the variables $t$ and $x$. The presence of the
nonlocal term breaks this symmetry, and the nonlocal initial value problem
would be more delicate to deal with than the boundary value problem addressed
above. In particular, one would need to find ${\underline{u}}$ in terms of $f$
and $u^{\rm in}$ by solving the nonlocal equation (5.17).

##  Appendix A Non dimensionalization of the equations

We show here how to derive the dimensionless equations of motion used
throughout this paper. To begin with, the Boussinesq-Abbott system describing
the propagation of weakly nonlinear waves in a fluid of mean depth $h_{0}$ and
with a pressure $P_{\rm atm}+\underline{P}$ exerted at the surface ($P_{\rm
atm}$ is a constant reference value for the atmospheric pressure) is given by

(A.1) $\begin{cases}\partial_{t}\zeta+\partial_{x}q=0,\\\
(1-\frac{h_{0}^{2}}{3}\partial_{x}^{2})\partial_{t}q+\partial_{x}\big{(}\frac{1}{h}q^{2}\big{)}+gh\partial_{x}\zeta=-\frac{h}{\rho}\partial_{x}\underline{P}\qquad(h=h_{0}+\zeta).\end{cases}$

######  Remark A.1.

Introducing the hydrodynamic pressure $\Pi$ as

(A.2) $\Pi=\underline{P}+\rho g\zeta,$

and alternative formulation of (A.1) is

(A.3) $\begin{cases}\partial_{t}\zeta+\partial_{x}q=0,\\\
(1-\frac{h_{0}^{2}}{3}\partial_{x}^{2})\partial_{t}q+\partial_{x}\big{(}\frac{1}{h}q^{2}\big{)}=-\frac{h}{\rho}\partial_{x}\Pi;\end{cases}$

we shall sometimes use this alternative formulation under the floating object.

Let us now consider the equations for the solid. We recall that we consider
here a floating object with vertical lateral walls located at $x=\pm\ell$
($\ell>0$) and allowed to move only vertically (heave motion). There is
therefore only one degree of freedom for the motion of the solid which can be
fully deduced from the signed distance $\delta(t)$ between the center of mass
$G=\big{(}x_{G},z_{G}(t)\big{)}$ and its equilibrium position $G_{\rm
eq}=(x_{G},z_{G,{\rm eq}})$, namely, $\delta=z_{G}(t)-z_{G,{\rm eq}}$.
Let us also assume that the water depth below the object is given at
equilibrium by a nonnegative single valued function $x\mapsto h_{\rm eq}(x)$;
the part of the bottom of the object in contact with the water (the wetted
surface) is therefore given at all time $t$ by the graph of the function
${\zeta}_{\rm w}$ defined as

(A.4) $\zeta_{\rm w}(t,x)=\delta(t)+h_{\rm eq}(x)-h_{0}.$

Newton’s equation for a body of mass $m$ that only moves vertically and
subject to gravity and hydrodynamic forces is given by

(A.5) $m\ddot{\delta}+mg=\displaystyle\int_{-\ell}^{\ell}\underline{P}_{\rm
i}(t,x){\rm d}x,$

where $\underline{P}_{\rm i}(t,x)$ is the pressure exerted by the fluid on the
object at the point $(x,\zeta_{\rm w}(t,x))$. Note that at equilibrium, the
pressure is hydrostatic, $\underline{P}_{\rm i}=-\rho g(h_{\rm eq}-h_{0})$, so
that

$m=\rho\int_{-\ell}^{\ell}(h_{0}-h_{\rm eq}(x)){\rm
d}x\qquad\mbox{(Archimedes' principle)},$

and we can rewrite Newton’s equation under the form

$m\ddot{\delta}=\displaystyle\int_{-\ell}^{\ell}\big{(}\underline{P}_{\rm
i}(t,x)+\rho g(h_{\rm eq}-h_{0})\big{)}{\rm d}x.$

By definition of the hydrodynamic pressure, its value $\Pi_{\rm i}$ in the
interior domain $(-\ell,\ell)$ is given by

$\Pi_{\rm i}=\underline{P}_{\rm i}+\rho g\zeta_{\rm w},$

from which we infer, using (A.4),

(A.6) $\tau_{\rm buoy}^{2}\ddot{\delta}+\delta=\displaystyle\frac{1}{2\rho
g\ell}\int_{-\ell}^{\ell}\Pi_{\rm i}(t,x){\rm d}x,$

where $2\pi\tau_{\rm buoy}$ is the buoyancy period defined through

$\tau_{\rm buoy}^{2}=\frac{m}{2\ell\rho g}.$

We now proceed to derive dimensionless versions of (A.1), (A.4), (A.5). We
recall that $h_{0}$ denotes the water depth at rest, and also denote by $a$
and $L$ the typical amplitude of the waves and a typical horizontal scale
respectively. For the Boussinesq-Abbott equations (A.1), we use the following
scalings

$\widetilde{x}=\frac{x}{L},\quad\widetilde{z}=\frac{z}{h_{0}},\quad\widetilde{t}=\frac{t}{L/\sqrt{gh_{0}}},\quad\widetilde{\zeta}=\frac{\zeta}{a},\quad\widetilde{q}=\frac{q}{a\sqrt{gh_{0}}},\quad\widetilde{\underline{P}}=\frac{\underline{P}}{\rho
gh_{0}}$

and consequently $\widetilde{h}=1+\varepsilon\widetilde{\zeta}$. We also
introduce the nonlinearity and shallowness parameters $\varepsilon$ and $\mu$
as

$\varepsilon=\frac{a}{h_{0}},\qquad\mu=\frac{h_{0}^{2}}{L^{2}}.$

For the sake of clarity the tildes used to denote dimensionless quantities are
omitted throughout this paper. The system (A.1) thus becomes

(A.7) $\begin{cases}\partial_{t}\zeta+\partial_{x}q=0,\\\
(1-\frac{1}{3}\mu\partial_{x}^{2})\partial_{t}q+\varepsilon\partial_{x}\big{(}\frac{1}{h}q^{2}\big{)}+h\partial_{x}\zeta=-\frac{1}{\varepsilon}h\partial_{x}\underline{P}\qquad(h=1+\varepsilon\zeta).\end{cases}$

######  Remark A.2.

The dimensionless form of the hydrodynamic pressure is naturally

$\widetilde{\Pi}=\frac{\Pi}{\rho
gh_{0}}=\widetilde{\underline{P}}+{\varepsilon}\widetilde{\zeta},$

so that the dimensionless version of the alternative formulation (A.3) is
(omitting the tildes)

(A.8) $\begin{cases}\partial_{t}\zeta+\partial_{x}q=0,\\\
(1-\frac{\mu}{3}\partial_{x}^{2})\partial_{t}q+\varepsilon\partial_{x}\big{(}\frac{1}{h}q^{2}\big{)}=-\frac{h}{\varepsilon}\partial_{x}\Pi;\end{cases}$

In order to derive the dimensionless versions of (A.4), (A.5) and (A.5), we
also need the following scalings

$\widetilde{\zeta}_{\rm w}=\frac{\zeta_{\rm
w}}{a},\qquad\widetilde{\delta}=\frac{\delta}{a},\qquad\widetilde{h}_{\rm
eq}=\frac{h_{\rm eq}}{h_{0}},\qquad\widetilde{m}=\frac{m}{2\ell\rho
h_{0}},\qquad\widetilde{\tau}_{\rm buoy}=\frac{\tau_{\rm
buoy}}{L/\sqrt{gh_{0}}},\qquad\widetilde{\ell}=\frac{\ell}{L}$

so that, omitting again the tildes for the sake of readability, we can rewrite
(A.4) and (A.5) as

(A.9) $\zeta_{\rm w}(t,x)=\delta(t)+\frac{1}{\varepsilon}\big{(}h_{\rm
eq}(x)-1\big{)}.$

and

(A.10) $\tau_{\rm
buoy}^{2}\ddot{\delta}+\frac{1}{\varepsilon}m=\displaystyle\frac{1}{\varepsilon}\frac{1}{2\ell}\int_{-\ell}^{\ell}\underline{P}_{\rm
i}(t,x){\rm d}x;$

note that in these dimensionless coordinates, the coordinates of the vertical
sides of the object are $x=\pm\ell$ and that Archimedes’ principle reads in
dimensionless form as

$m=\frac{1}{2\ell}\int_{-\ell}^{\ell}(1-h_{\rm eq}).$

Finally, the dimensionless version of A.6 is

$\tau_{\rm
buoy}^{2}\ddot{\delta}+\delta=\displaystyle\frac{1}{\varepsilon}\frac{1}{2\ell}\int_{-\ell}^{\ell}\Pi_{\rm
i}(t,x){\rm d}x.$

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