TITLE: $M/\Gamma$ is orientable iff the elements of $\Gamma$ are orientation-preserving
QUESTION [9 upvotes]: The probem is:

Suppose $M$ is connected, orientable, smooth manifold and $\Gamma$ is
  a discrete group acting smoothly, freely, and properly on $M$. We say
  that the action is orientable-preserving if for each $\gamma \in \Gamma$, the diffeomorphism $x \rightarrow \gamma \cdot x$ is
  orientation-preserving. Show that $M/ \Gamma $ is orientable if only
  if the action of $\Gamma$ is orientable-preserving.

Notes: I tried using that $\pi\colon M\to M/\Gamma$ is a covering map. I also tried to use that for any $\gamma$ we have two disjoint open sets $U,V\subset M$ such that $\pi|_{U}$ and $\pi|_{V}$ are diffeomorphisms and $(\pi|^{-1}_{U} \circ  \pi|_V)(x)=\gamma.x$ but I didn't prove this result too.

REPLY [9 votes]: If the action is orientation-preserving, construct explicitly an orientation on the quotient. Fix an orientation on $M$. If $\bar x\in M/\Gamma$ is a point and $x\in M$ is such that $\pi(x)=\bar x$, the differential $d\pi_x:T_xM\to T_{\bar x}(M/\Gamma)$ is an isomorphism, so you can push the orientation of $T_xM$ given by the orientation of $M$ to one on $T_{\bar x}(M/\Gamma)$. Check that this depends only on $\bar x$ and not on the preimage $x$ chosen: this is where the hypothesis comes in. Finally, check that this way of orienting the tangent spaces to $M/\Gamma$ is in fact an orientation of $M/\Gamma$.
Can you do the converse?


TITLE: How to compute maximum modulus on unit circle?
QUESTION [0 upvotes]: Let $p(z) = a_0 + a_1z + \cdots + a_{n-1}z^{n-1}$ and $|a_i|\leq\frac{1}{2}$ for all $i$.
I want to know maximum modulus
$|p(z)|$ on $|z|=1$.
By the maximum modulus theorem and triangle inequality, $|p(z)| < \frac{n^2}{2}$.
But it's very trivial upper bound.
How to compute a maximum modulus or nice upper bound??

REPLY [0 votes]: Not entirely sure what you're asking, but if you are asking for, say, how to apply the maximum modulus principle for the contour integration of your $p(z)$ on the contour that is the unit circle (or anything else), then the upper bound on the modulus of your integral is ML, where $M$ is the maximum of $p(z)$ on the contour, and $L$ is the length of the contour, which, for a unit circle, is just the circumference, $2\pi $.
This ML-bound comes up a lot in complex integration problems.


TITLE: When does $V = \ker(f) \oplus \operatorname{im}(f)$?
QUESTION [16 upvotes]: For a vector space $V$ and a linear operator $f : V \to V$, under what conditions does $V = \ker(f) \oplus \operatorname{im}(f)$?
Is it always true, or only in special cases?
Edit: $V$ is finite dimensional.

REPLY [2 votes]: In Hilbert spaces (spaces with scalar product) you have that
$$\ker (B^*) = (\operatorname{ran} B)^\perp,$$
Where $\operatorname{ran}$ is the range, i.e., what you call image;
the $B^*$ is the adjoint operator, defined by the prescription $$(Bx,y)=(x,B^*y);$$
and $U^\perp$ is the orthogonal complement to $U$, i.e.
$$U^\perp=\{v:(\forall u\in U)((u,v)=0)\}.$$
There is a wide class of Hermitean operators, i.e. such that $B=B^*$. All these satisfy $\ker B = (\operatorname{ran} B)^\perp$ therefore $\ker B\oplus \operatorname{ran} B=V$.
In the finite-dimensional case, every Hermitean matrix (i.e., $a_{ij}=\overline{a_{ji}}$) satisfies this property.


TITLE: Another Number Theory Question about Polynomial
QUESTION [1 upvotes]: We know following theorem by Schur:
Suppose that $f(x) \in \mathbb{Z}[x]$ is a polynomial such that exists an integer $m$ such that $f(n) = m^2$ for
every integer $n$. Then $f(x) = g(x)^2$ for some $g(x) \in \mathbb{Z}[x]$.
Is this true if I using $3,4,5 \cdots$ instead of $2$?

REPLY [2 votes]: I think the proposer tried to say that if for every integer $n$ there is an integer $m$ such $f(n)=m^2$, then $f(x)=g(x)^2\dots$.
The generalization from exponent $2$ to higher exponents $e$ is an exercise in the second volume of the classical Polya/Szegö problem book. I don't have it at hand now, so I cannot give a more precise reference.
If one allows more sophisticated tools, then it follows immediately from Hilbert's irreducibility theorem: Let $t$ be another variable, and factorize $f(x)-t^e$ over $\mathbb Q(t)$ into irreducible factors $P_i(t,x)$. By Hilbert, there are infinitely many integers $n$ such that $P_i(n,x)$ is irreducible for all $i$. On the other hand, $f(x)-n^e$ has an integral root for each $n$. Thus there is an index $i$ such that $P_i(t,x)$ has degree $1$ with respect to $x$, so $f(x)-t^e$ has a root in $\mathbb Q(t)$. Gauss' lemma yields that this root is in $\mathbb Q[t]$, so $f(t)=g(t)^e$ for a polynomial $g$ with rational coefficients. One quickly gets that the coefficients of $g$ then are indeed integers.

REPLY [1 votes]: Yes.  See e.g. http://www.isibang.ac.in/~sury/polys.pdf Corollary 3.3.


TITLE: Space of Extensions of Holomorphic Line Bundles
QUESTION [3 upvotes]: On a compact complex manifold $X$, fix two holomorphic line bundles $L$ and $L'$. Consider a holomorphic vector bundle $V$ of rank 2 which fits in an exact sequence
$$0\to L\to V\to L'\to0$$
I would like to understand why these $V$'s are parametrized by $H^1(X, (L')^\ast\otimes L)$.
Up to tensoring with $(L')^\ast$  the short exact sequence above, we can equivalently ask why, for fixed $L$, holomorphic vector bundles $V$ of rank 2 sitting in exact sequences of the form
$$0\to L\to V\to \mathcal{O}_X\to0$$
are parametrized by $H^1(X,L)$.
I was thinking about reasoning with Cech cohomology. With respect to some covering $U_i$ of $X$, the transition functions of $V$ can be written in a upper-triangular matrix with the transition functions $\ell_{ij}$ of $L$ and $1$ on the diagonal and some $g_{ij}\in\mathcal{O}(U_i\cap U_j)$ at the top right entry. Thus, all the information should be encoded in the $g_{ij}$. The transition relations for the bundle then yield the following relations

$g_{ii}=0$ (on each $U_i$)
$g_{ij}=-\ell_{ij}g_{ji}$ (on each $U_i\cap U_j$)
$g_{ij}=\ell_{ik}g_{kj}+g_{ik}$ (on each $U_i\cap U_j\cap U_k$)

However, when interpreting $g=(g_{ij})$ as an element of the group $C^1(\mathcal{L})$ in the Cech complex, by using the above relations I get
$$(dg)_{ijk}:=g_{jk}-g_{ik}+g_{ij}=g_{jk}(1-\ell_{ij})$$
and therefore it seems that my $g$ does not even define a class in $H^1$ in general (apart from the trivial case when $L$ is trivial). Any suggestion?

REPLY [2 votes]: (This should be a comment rather than an answer, but I do not have the right to make comments, since my rep is not yet established in this town.)
I personally found Atiyah's 1957 paper 'Complex analytic connections in fibre bundles' (see the proof of Proposition 2) very enlightening with respect to the problem you posed. You may want to take a look, it's always a good idea to learn from the masters.


TITLE: Are there specific properties defining the kernel of a quotient ring?
QUESTION [1 upvotes]: I'm asking because the correction of an exercise of a course I'm following (commutative algebra) has the following conclusion:

Let I be an ideal of a commutative ring A, and $I[X]=\{\sum_{k=0}^{n}i_{k}X^{k}\mid i_{k}\in I\}\subseteq A[X]$ ideal in A[X]. Show that A[X]/I[X] is isomorphic to the ring (A/I)[X].
Let $\pi_{I}:A\rightarrow A/I$ be the natural application$^{(1)}$ and $\phi:A[X]\rightarrow(A/I)[X]\ :\ \sum_{k=0}^{n}a_{k}X^{k}\mapsto\sum_{k=0}^{n}\pi_{I}(a_{k})X^{k}$.
$\phi$ is a surjective morphism and $\ker(\phi)=I[X]$.
$\Rightarrow$ $I[X]\subseteq \ker(\phi)$ therefore by the universal quotient property there exists a natural$^{(1)}$ isomorphism $f:A[X]/I[X]\rightarrow(A/I)[X]$.

The part which causes me trouble is:

and $\ker(\phi)=I[X]$

I'm wondering where that conclusion comes from; if it is a general property of quotient rings or if it actually comes from that specific space's structure, if it comes from that space's structure in particular.
(1) translated from french: "application canonique", maybe canonical? I am unsure.

REPLY [1 votes]: This statement $\ker\phi=I[X]$ is specific to this problem, and it is immediate:
A polynomial $p(X)=\sum_{i\le k}a_iX^i$ is in the kernel of $\phi$, by definition, iff $\phi(p)=0$, iff each coefficient of $\phi(p)$ is zero, iff each $\pi_I(a_i)=0$ by definition of $\phi$.
But $\pi_I(a_i)=0\in A/I$ means exactly $a_i\in I$.
So we get that $p\in\ker\phi\iff\forall i:a_i\in I\iff p\in I[X]$.


TITLE: Classical Faraday or stress energy tensor for blackbody radiation
QUESTION [0 upvotes]: Clearly, the blackbody spectrum is derived using quantum considerations; however  It should be entirely feasible that given the blackbody distribution and considering isotropic radiation, one can calculate the classical stress energy tensor (or actually more useful for me the Faraday tensor).
Strange thing is, I can't figure out how to begin the calculation since the Fields are zero statistically at some given time. Any idea of where to start?
In cosmology it's pretty typical to consider radiation as a fluid with energy density and pressure, I was just curious is there was something a bit more quantitative.

REPLY [1 votes]: The energy-momentum tensor of the black-body radiation is the same as that of a perfect fluid, i.e.
\begin{equation}\tag{1}
T^{\mu\nu} = (\rho_0 + p_0)\frac{dx^\mu}{ds}\frac{dx^{\nu}}{ds} + g^{\mu\nu}p_0,
\end{equation}
with
\begin{equation}\tag{2}
                  \rho_0 = 3p_0.
\end{equation}
where $\rho_0$ is the proper macroscopic density of the radiation at the point of interest, $p_0$ its proper pressure, and the velocities $dx^\mu/ds$ correspond to the  macroscopic motion of the radiation.
The  applicability  of  the  energy-momentum  tensor  of  a  perfect  fluid  to  the  black-body radiation is justified by the fact that in relativistic mechanics any system whose local properties can be specified by the two scalars $\rho_0$ and $p_0$ has such an energy-momentum tensor.  And this is surely the case for the black-body radiation.
A more rigorous derivation is based on the electromagnetic nature of the black-body radiation, using the covariant form of the Maxwell’s equations.
A very good treatment of this you can find in the Tolman and Ehrenfest paper:
  Tolman R.C., Ehrenfest P., *Phys. Rev.*, **35**, 1791, (1930)


TITLE: $cos(ax)=a\cdot cos(x)$ Which are all the values of $a$?
QUESTION [0 upvotes]: How do I know for which values of $a$ is true the following?$\space$
$cos(ax)=a\cdot cos(x)$ for all $x\in \mathbb{R}$

It is trivial that it is true for $a=1$. But are the more values of $a$ for which it is true?

REPLY [1 votes]: By no means a real proof, but if you series-expand teach trig term and re-combine powers in $x$, it's pretty clear:
\begin{align*}\cos\left(ax\right) &= a \cos x \\ 1 - \frac{a^2x^2}{2!} + \frac{a^4x^4}{4!} - \cdots &= a - a\frac{x^2}{2!} + a\frac{x^4}{4!} - \cdots \\ \left(1-a\right) - \frac{x^2}{2!}\left(a^2-a\right) + \frac{x^4}{4!}\left(a^4 - a\right) - \cdots&= 0\end{align*}
For the above to hold for any $x$, the terms in parentheses must each be zero. This only works with $a=1$.


text:
\begin{document}
\maketitle

\begin{abstract}
\noindent

The problem of determining the manner in which an incoming acoustic wave is scattered by an elastic body immersed in a fluid is one of central importance in detecting and identifying submerged objects.  The problem is generally referred  to as a fluid-structure interaction and is  mathematically formulated as a time-dependent transmission problem.  In  this paper,  we
consider a typical fluid-structure interaction problem by  using a coupling procedure which
reduces  the problem to a nonlocal initial-boundary problem in the elastic  body with a system of  integral equations on the interface between the  domains occupied  by the elastic body and the fluid. We analyze this nonlocal problem by the Lubich  approach via the Laplace transform,  an essential  feature of which is that it works directly on data in the time domain rather than in the transformed domain.  Our results may serve as a mathematical foundation for  treating  time-dependent
fluid-structure interaction problems by  convolution quadrature  coupling  of FEM
and  BEM.
\end{abstract}

{\bf key words}: Fluid-structure interaction,  Coupling procedure,  Kirchhoff representation formula,
Retarded potential, Laplace transform,  Boundary integral equation,  Variational formulation, Sobolev space.

\noindent
{\bf Mathematics Subject Classifications(1991)}: 35J20, 35L05, 45P05, 65N30.65N38

\section{Introduction}
The problem of determining the manner in which an incoming acoustic wave is scattered by an elastic body immersed in a fluid is one of central importance in detecting and identifying submerged objects.  The problem is generally referred  to  as a fluid-structure interaction and is  mathematically formulated as an initial-boundary  transmission problem.   However,  most  of the investigations  study typical fluid-structure interaction problems confined to  the
 time-harmonic setting;  various  numerical methods, sometimes competitive, sometimes complementary,  have been developed.  In this regard, the governing system of partial differential equations is usually replaced by integral equations and it is these formulations upon which most numerical approximations are based. The acoustic equation is replaced by a boundary integral equation while the elastic body is treated in various ways; sometimes using an integral equation, either a  boundary or domain equation, or  alternatively using a weak or variational formulation leading to finite element approximations
(see e.g., \cite{HJ:1985, HKS:1988, EvHe:1990, BiMa:1991, LuMa:1995,  Sc:2008},  to name a few).

In  this paper,  we study and analyze a typical fluid-structure interaction problem in the time domain. Motivated by the time-harmonic fluid-structure interaction problems, we apply  a coupling procedure which is a combination of  a field equation and a boundary  integral equation. The essence of the procedure is to reduce the problem to a nonlocal initial- boundary problem in the elastic  body with integral equations on the interface between the  domains occupied  by the elastic body and the fluid.  However, in contrast to the time-harmonic setting,  the integral equations which are derived from the Kirchhoff formula are not only nonlocal in space but also nonlocal in time. This makes the analysis complicated,  in particular with respect to the choice of  appropriate solution function spaces.  To circumvent   this difficulty, we analyze this nonlocal  initial-boundary problem by the Lubich  approach via the Laplace transform as in  \cite{LaSa:2009b, Sa:2012}. The Lubich approach has been employed in the development of  numerical approximations  for  some fluid-structure interaction problems in the engineering literature (see, e.g., \cite{EsAn:1991, Sch:2001, PeBe:2009}),  but no rigorous justifications are provided.

The paper is organized as follows:   In the next section, we start with the formulation of the problem
as an initial-boundary transmission problem and reduce it to a nonlocal initial-boundary  problem.
In Section 3,   we give a brief review of the Lubich  approach  and introduce the appropriate classes of operators from \cite{LaSa:2009a} and state and prove the crucial result concerning the inversion of the Laplace transform of the classes of operators introduced in this section.  Section 4 deals with the variational formulation of the nonlocal initial- boundary problem in Section 2.  Theorems 4.1 and 4.2 are the main existence and uniqueness results. The last section, Section 5,  states the main results  in the time domain.

\section{Formulation of the problem}
\subsection{An initial -boundary transmission problem}
We are concerned with a time-dependent direct scattering problem in fluid-structure interaction, which can be simply described as follows: an acoustic wave propagates in a
fluid domain of infinite extent in which a bounded elastic body is
immersed. The problem is to determine the scattered
pressure and velocity fields in the fluid domain as well as the
displacement field in the elastic body at any time.  Throughout the paper, let $\Omega$ be the bounded domain in $\mathbb{R}^3$ occupied by the elastic body with a Lipschitz boundary $\Gamma$ and let  $\Omega^c = \mathbb{R}^3  \setminus  \overline\Omega$  its exterior occupied by a compressible fluid.  In the elastic domain $\Omega$, the  elastic displacement $\mathbf{u}(x,t)$ is governed by the
dynamic linear elastic equation:
\begin{equation}\label{eq:2.1}
\rho_e \frac{\partial^2\mathbf{u}} {\partial t^2} - \Delta^{*} \mathbf{u} = \mathbf{0},  \quad (x,t) \in
\Omega \times (0,T),
\end{equation}
where $T$ is a given positive constant and
where $\rho_e$ is the constant density of the elastic body, and $\Delta^*$ is the Lam\'e operator
\begin{eqnarray*}
 \Delta^* \mathbf{u} &:=&  \mu \Delta \mathbf{u} + (\lambda + \mu)  \nabla~div \, \mathbf{u}\\
 &=& div \; \bm{\sigma}(\bf u).
 \end{eqnarray*}
 Here $\bm {\sigma}({\bf u})$ and $\bm \varepsilon({\bf u}) $ are  the stress  and strain tensors,
 respectively,
 $$ \bm \sigma({\bf u}) =(\lambda\;div\; {\bf u}){\bf I} + 2 \mu \bm \varepsilon({\bf u}) \quad \mbox{and }\quad
 \quad \bm \varepsilon({\bf u}) = \frac{1}{2}( \nabla{\bf u} + (\nabla{\bf u})^\prime) .$$
We assume that the elastic body is homogeneous  and  isotropic  with $\mu$ and $\lambda$  the
corresponding Lam\'e constants, which are required to satisfy the constraints: $ \mu \geq 0,$ and  $3\lambda + 2 \mu \geq 0.$

In the  fluid domain  $\Omega^c$,  we consider a barotropic flow of an inviscid and compressible  fluid.  Let {\bf v}={\bf v} (x,t) be the velocity field,  and $p=p(x,t)$ and  $\rho= \rho(x,t)$ be respectively the pressure and the density of the fluid. We assume that ${\bf v},
 p $ and  $\rho$  are small perturbations of the static state ${ \bf v_0} = {\bf 0},
p_0 = constant $ and $ \rho_0 = constant. $  Then the  governing equations
may be linearized to yield  the linearized Euler equation
\begin{eqnarray}\label{eq:2.2}
\rho_0\frac{\partial {\bf v}}{\partial  t} +  \nabla\; p &=&{ \bf 0},
\end {eqnarray}
the linearized equation of continuity
\begin{eqnarray}\label{eq:2.3}
\frac{\partial \rho}{\partial t} + \rho_0 \, div \, {\bf v} &=& 0,
\end{eqnarray}
and the linearizecd state equation
\begin{eqnarray}\label{eq:2.4}
p = c^2 \rho
\end{eqnarray}
in $\Omega^c \times (0, T)$, where $c$ is the sound speed defined by
$ c^2 = f '( \rho_0) $ and $f $ is a function depending on the nature of the fluid (see e.g., \cite{Ac:1990, Se:1958}).

For an irrotational flow, this  formulation can be simplified in terms of a velocity potential $\varphi = \varphi(x,t)$  such that
$$
{\bf v} =  - \nabla\; \varphi, \quad\mbox{and}\quad
p= {\rho_0} \frac{\partial \varphi}{\partial t}.
$$
Then it follows from (\ref{eq:2.3}) and (\ref{eq:2.4}),  the velocity potential $\varphi$ satisfies the wave equation
\begin{equation}\label{eq:2.5}
\frac{\partial^2 \varphi}{\partial  t^2} - c^2 \Delta \varphi = 0 \quad\mbox{in}\quad \Omega^c \times (0, T).
\end{equation}
The time-dependent scattering problem can be formulated as {\em an initial-boundary transmission problem} consisting of  the partial differential equation (\ref{eq:2.1}) for the elastic displacement field  $\bf u$ and (\ref{eq:2.5}) for velocity
potential $\varphi$ together with the homogeneous initial conditions
\begin{equation}\label{eq:2.6}
 {\bf u}(x,0) = \frac{\partial {\bf u}(x,0)}{\partial t} = {\bf 0},\;  x \in \Omega \quad \mbox{and}
 \quad \varphi (x,0) = \frac{\partial \varphi} {\partial t} (x,0) = 0,\; x\in \Omega^c
  \end{equation}
 and the transmission conditions  on $\Gamma \times (0, T]$
 \begin{eqnarray}
 \bm \sigma({\bf u})^- {\bf n} &= & - {\rho_0}\;( \frac{\partial \varphi}{\partial t} +
  \frac{\partial \varphi^{inc}}{\partial t})^+{\bf n},  \label{eq:2.7}\\
 \frac{ \partial {\bf u}^-} {\partial t} \cdot {\bf n}& = &- (\frac{\partial  \varphi }{\partial n} +
  \frac{\partial  \varphi^{inc}}{\partial n})^+ , \label{eq:2.8}
 \end{eqnarray}
where ${\bf n}$ is the exterior unit normal for $\Omega$, and $\varphi^{inc} $ denotes the given incident field.
Here and in the sequel, we adopt the notation that $q^{\mp}$ denotes the limit of the function $q$
on $\Gamma$ from inside and outside, respectively.

\subsection{Reduction to a nonlocal initial- boundary problem}
Motivated by  time-harmonic   fluid-structure interaction problems \cite{HKR:2000}, we intend to apply the coupling of boundary integral and field equation methods to the
 transmission  problem defined by (\ref{eq:2.1}) and  (\ref{eq:2.5} ) together with
 (\ref{eq:2.6}) --(\ref{eq:2.8}).  The main idea here is  to convert the problem to a nonlocal problem
 in a bounded computational domain such as $\Omega$  by a reduction of the solution in the fluid domain to appropriate boundary integral equations on the interface boundary $\Gamma$.
 For the solution of the wave equation (\ref{eq:2.5}), we begin with the Kirchhoff formula (see e.g., \cite{HsWe:2011a, LaSa:2009a})
 \begin{equation}  \label{eq:REP}
 \varphi  = \mathcal{D} * \phi - \mathcal{S} *\lambda \quad  \mbox{in } \Omega^c \times (0, T),
\end{equation}
where $\phi:=\varphi^+$ and $\lambda:=\partial \varphi^+/ \partial n$ are the Cauchy data of $\varphi$ on $\Gamma$ and
\begin{eqnarray}
(\mathcal{S} * \lambda)(x,t) &:=& \int_\Gamma E(x,y)  \lambda(y,t-\frac{|x-y|}{c} d\Gamma_y  \label{eq:2.10}\\
(\mathcal{D}* \phi)(x,t)  &:=& -\int_\Gamma \nabla_x \Big(E(x,y) \phi (y,t-\frac{|x-y|}{c}\Big) \cdot {\bf n}_y d\Gamma_y,  \label{eq:2.11}\\
&=&   \int_{\Gamma} \Big\{\frac{\partial E(x,y)}{\partial n_y } \phi(y,t-\frac{|x-y|}{c})  -\frac{1}{c} E(x,y)
 \frac{\partial |x-y|}{\partial n_y}\phi(y,t-\frac{|x-y|}{c})\Big\}d\Gamma_y, \nonumber\\
 \label{eq:2.12}
\end{eqnarray}
are the retarded simple and double layer potentials, written in terms of the fundamental solution of the three dimensional Laplacian $E(x,y)=1/(4\pi |x-y|)$.
One may show  as in classical potential theory \cite{HsWe:2008} that  the Cauchy data $\phi$ and $\lambda$ at  smooth points  of $\Gamma$ are related   by the system of boundary integral equations form the form (see, e.g., \cite{ BaHa:1986a, BaHa:1986b, Co:2004,  Ha:1989})
\begin{eqnarray}
\phi  &=&\Big( \frac{1}{2}\phi  + \mathcal{K} * \phi  \Big)  - \mathcal{V}* \lambda
 \quad \mbox{on } \Gamma\times (0,T]  \label{eq:2.13},\\
\lambda  &=&- \mathcal{W} * \phi + \Big(\frac{1}{2} \lambda  - \mathcal{K}^{\prime} *\lambda  \Big)  \quad \mbox{on }\Gamma \times (0,T]\label{eq:2.14}.
\end{eqnarray}
The four retarded integral operators in \eqref{eq:2.13} and \eqref{eq:2.14} are called (in the order they appear in the formulas) double layer, simple layer, hypersingular, and transpose double layer operators. For instance, the explicit formulas for the operators in \eqref{eq:2.13} is
\begin{eqnarray}
(\mathcal{V} *\lambda)(x,t) &:=& \int_{\Gamma} E(x,y) \lambda (y,t-\frac{|x-y|}{c}) d\Gamma_y,  \label{eq:2.15}  \\
(\mathcal{K} *\phi)(x,t)  &:=&  \int_{\Gamma} \frac{\partial E(x,y)}{\partial n_y} \phi(y,t-\frac{|x-y|}{c}) d\Gamma_y -
\frac{1}{c}(\mathcal{V}_r * \phi_t)(x,t)
\end{eqnarray}
where
\[
( \mathcal{V}_r *\psi)(x,t):= \int_{\Gamma}E(x,y) \frac{\partial}{\partial n_y}|x-y|
\psi(y, t -\frac{|x-y|}{c}) d\Gamma_y.
\]
The operator matrix
defined by  the right-hand side of (\ref{eq:2.13}) and (\ref{eq:2.14}) resembles the familiar form of the Calder\'on  projector for the Laplacian in  potential theory (see e.g., \cite{HsWe:2008}).

 In view of the transmission condition  (\ref{eq:2.8}),  we make a  substitution:
$$
\lambda = - \big( \frac{ \partial {\bf u}^-} {\partial t} \cdot {\bf n}
+ \frac{\partial  \varphi^{inc +}}{\partial n} \big)
$$
in (\ref{eq:2.13}). This leads to the following nonlocal boundary problem  reads : Given $\varphi^{inc}$, find ${\bf u}$ in $\Omega \times (0,T] $ and
$ \phi$ on $\Gamma \times (-\infty,T]$ satisfying  the following equations and conditions :
\begin{gather}
 \rho_e \frac{\partial^2\mathbf{u}} {\partial t^2} - \Delta^{*} \mathbf{u} = \mathbf{0},  \quad \mbox{in }
\Omega \times (0,T),  \nonumber\\
 \bm \sigma({\bf u})^- {\bf n} =  - {\rho_0}\;( \phi_t +
  \frac{\partial \varphi^{inc\;+}}{\partial t}){\bf n}\quad \mbox{on} \quad \Gamma \times (0,T],
  \nonumber \\
 {\bf u}(x,0) = \frac{\partial {\bf u}(x,0)}{\partial t} = {\bf 0},\;  x \in \Omega,
\nonumber \\
 \quad \phi (x,t) =  0,\; x\in \Gamma, t \le 0, \nonumber \\
- \frac{1}{2}\phi + \mathcal K*\phi  +  \mathcal{V} * (\frac{ \partial {\bf u}^-} {\partial t}\cdot {\bf n_y})
=  - \mathcal{V} * \frac{\partial  \varphi^{inc +}}{\partial n}  \quad \mbox{on } \Gamma\times (0,T].
\label{eq:2.17}
\end{gather}
Note that the initial condition for $\phi$ has to be stated for negative values of $t$, given the fact that delays appear in the definition of the retarded integral operators.
We may also replace (\ref{eq:2.17}) by (\ref{eq:2.13}) and (\ref{eq:2.14}) in the form:
\begin{gather}
\Big( \frac{1}{2}\phi  - \mathcal{K}* \phi  \Big) + \mathcal{V} * \lambda  =0
 \quad \mbox{on } \Gamma\times (0,T]  \label{eq:2.18},\\
-  \frac{ \partial {\bf u}^-} {\partial t} \cdot {\bf n}  + \mathcal{W} *\phi - \Big(\frac{1}{2} \lambda  - \mathcal{K}^{\prime}*\lambda \Big)
= \frac{\partial  \varphi^{inc +}}{\partial n}  \quad \mbox{on } \Gamma \times (0,T]\label{eq:2.19}.
\end{gather}
In this case,  ${\bf u}$ in $\Omega \times (0,T)$,
$ \phi$ and $\lambda$  on $\Gamma \times (-\infty,T]$ are the unknown for the solutions of the nonlocal boundary problem. In view of the definition of the boundary integral operators of
$\mathcal{K}^\prime$ and $\mathcal{W}$, in addition to the homogeneous initial condition for $\phi$, we require that $ {\phi_t} (x,t) = 0$ and $\lambda(x,t) = 0,  \; x\in \Gamma, t\le 0$.

We notice that in the above formulations,   equations (\ref{eq:2.17}), (\ref{eq:2.18})
 and (\ref{eq:2.19}) are all nonlocal differential boundary integral equations. They are not only nonlocal in space but also nonlocal in time.  As pointed out in \cite{Co:2004},  it is not clear how to choose appropriate function spaces because of the retarded argument.  On the other hand,  it is known that  for the long time behavior of the solution, one may replace the nonlocal differential boundary  integral equation by an  appropriate approximated transparent  boundary condition (see, e.g. \cite{HsWe:2011a}). However, in general we prefer to employ the approach originally introduced by  Lubich in his study of convolution quadrature techniques for hyperbolic problems \cite{Lu:1994} (see also \cite{LuSc:1992} in the parabolic case). This approach has been  recently extended  systematically to  treating retarded potentials by Laliena and Sayas \cite{LaSa:2009a} by means  of   properties of  the operators in the frequency domain.  We remark  that  we will see  this technique  does not mean we are solving the problems in the transformed domain and then  applying  the inverse  Laplace transform to obtain the solutions in the time domain.  To illustrate the essence of this concept, in the next section  we begin with some preliminary results concerning the Laplace transforms of
functions and operators with causality properties.
\section{ Lubich's approach}
In this section, we give a brief review of the Lubich approach for treating time dependent boundary integral equations which has been advanced  by the work of  Laliena and Sayas. The presentation of this section follows their work in \cite{LaSa:2009b}.

\subsection{The Laplace transform}
We consider the Laplace transform for causal distributions or operator-valued functions. Throughout the paper let the complex  plane be denoted  by $\mathbb{C}$ and its positive half-plane denoted by
$$\mathbb{C}_+:= \{ s \in \mathbb{C} : Re~s > 0\}. $$
We begin with the Laplace transform for an ordinary complex-valued function. Let $f \colon [0, \infty) \to \mathbb{C}$ be a complex-valued function with limited growth at infinity.  The  Laplace transform of $f$ is defined by
$$F(s)= \mathcal{L}\{f\}(s) := \int_0^\infty e^{-st} f(t) dt. $$
A common criterion for limited growth at infinity is that $f$ be of exponential  order, but this is much too restrictive for the kind of problems we are treating here.
As in \cite{DaLi:1992, LaSa:2009b} one can define the Laplace transform for the case of causal continuous  linear maps  $f \colon \mathcal{S}(\mathbb{R}) \to X $ with limited growth at infinity  which  concept is defined as fellows for  tempered distributions defined on  $\mathcal{S}(\mathbb{R})$ with values in a complex  Banach space $X$. Here causal is taken as in the sense of $\mathcal{S}' (\mathbb{R})$ that $\langle f, \phi\rangle$ is zero element of $X$ for every $\phi$ in $\mathcal{S} (\mathbb{R})$ with support in $[0, \infty)$. Indeed
for fixed $a$ and $b$ such that $-\infty < b <a <0 $ and a $C^{\infty}(\mathbb{R})$  function $\alpha$ which vanishes identically for $t \leq b$ and is identically equal to   $1$  for $t \geq a $ the function $\phi^*$,
$$ \phi^* (t) := \alpha(t) e^{-st}$$
is in $\mathcal{S}(\mathbb{R}) $. Thus we define the Laplace transform of the casual continuous linear map $f$ as
$$
\mathcal{L}\{f\}(s) := \langle f, \phi^*\rangle.$$
Moreover, this definition is independent of the choice of $a$ and $b$.

We also want to consider the Laplace transform  of  a causal operator-valued function
$f \colon [0, \infty) \to {L}(X,Y)$,  where  $X$ and $Y$ are two complex Hilbert spaces and
  $L(X,Y)$ is  the space of bounded linear operators from $X$ to $Y$.    The Laplace transform of $f$  is  defined by Bochner's integral
$$F(s)= \mathcal{L}\{f\}(s):= \int_0^\infty e^{-st} f(t) dt, $$
if the integral exists.   We assume that $F(s)$ exists for all $s\in \mathbb{C}_+$ and decays fast enough at infinity so that inversion formula
$$ f(t) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{st} F(s) ds $$
holds for all $\sigma = Re~ s > 0$. Now let $g\colon [0, \infty)  \to  X$ and let
$$ (f*g) (t) := \int_0^t f(\tau) g(t-\tau) d \tau$$
denote the convolution  $f * g \colon [0, \infty)  \to Y$.   For appropriate  $f$ and $g$,
we see that formally
\begin{eqnarray}
(f*g) (t)& :=& \int_0^t f(\tau) g(t-\tau) d \tau\nonumber\\
 &=& \int_0^t \Big(\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{s \tau} F(s) ds\Big)
 g(t - \tau) d \tau \nonumber \\
 &=& \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} F(s) \Big( \int_0^t e^{s\tau} g( t- \tau) \Big) ds, \label{eq:3.1}
\end{eqnarray}
provided the changing orders of integrations can be adjusted by Fubini's theorem.  In fact, as we will see,  the relation in (\ref{eq:3.1}) is the essential idea  behind  the convolution  quadrature method introduced by Lubich since the late 80's   and has been employed for  treating time-dependent boundary integral equations in the early 90's (see, e.g., \cite{LuSc:1992, Lu:1994}).  We note that in the relation
(\ref{eq:3.1}), the property of the convolution integral  $f*g$ in the time domain  depends upon  g in the time domain but on $f$  only in the transformed domain. The latter is more accessible.   In  the following, we  shall summarize some of the results in \cite{LaSa:2009b}) concerning  the precise conditions for the class of operators and functions for which relation (\ref{eq:3.1}) holds.  We begin with classes of operators.

\subsection{Classes of operators $\mathcal{A}(\mu, X,Y)$ and $\mathcal{E}(\mu, \theta, X)$ }
\begin{itemize}
\item {$\mathcal{A}(\mu, X,Y)$ }: For a given $\mu \in \mathbb{R} $, the elements of the class
 $\mathcal{A}(\mu, X,Y)$ are the analytic functions $F\colon \mathbb{C}_+ \to L(X,Y)$ for which
 there exists a real number $\mu$ such that for all $\sigma > 0$ there is $C_0(\sigma)$ such that
 $$ ||F(s)|| \leq C_0(\sigma) |s|^{\mu}, \;\; \forall s \; \; s.t \; \; Re  ~ s > \sigma. $$
\item{$\mathcal{E}(\mu, \theta, X)$}: For given $\mu \in \mathbb{R}$ and a function
  $\theta \colon \mathbb{C}_+  \to  \mathbb{R}$,  we write $F \in \mathcal{E}(\mu, \theta, X) $  when
$F  \colon \mathbb{C}_+ \to L(X,X^\prime) $ is analytic (where $X^\prime$ is the dual of $X$), and there exists a non-decreasing function
 $c \colon (0, \infty) \to (0, \infty) $ such that
 $$ Re~ \Big( e^{i \theta(s)} <F(s) \psi, \overline\psi>\Big) \geq \frac{c(Re~s)}{|s|^\mu} ||\psi||^2, \;\;\forall \psi \in X, \;\;\forall s \in \mathbb{C}_+. $$
 \end{itemize}
The following theorems and  proposition are stated in \cite{LaSa:2009b}. The detailed proof of the first two appears in \cite{Sa:2012}, while the last one is a simple consequence of the Lax-Milgram lemma.

\begin{theorem} let $F \in \mathcal{A}(\mu, X, Y) $ with $\mu < -1$. Then there exists a continuous function $f \colon \mathbb{R} \to L(X,Y)$ such that supp $f \subset [0, \infty)$ such that its Laplace transform, defined in $\mathbb{C}_+$, is $F$. If $\mu < -k -1 $ with $k$ positive integer, then
$f \in C^k (\mathbb{R}, L(X, Y)) $.
\end{theorem}
\begin{theorem}
Let  $ F \in \mathcal{A}(\mu, X, Y) $  with $\mu  \geq  1 .$ Take $k$ such that $1 + \mu < k \leq 2 + \mu$. Then there exits $g \in C(\mathbb{R}, L(X,Y))$ such that $supp \;g \subset [0, \infty)$ and $F$ is the
Laplace transform of $ g^{(k)}$  in $ \mathbb{C}_+$, where the derivative is understood in the sense of distributions in $\mathbb{R}$.
\end{theorem}
\begin{proposition} If $ F\in \mathcal{E}(\mu, \theta, X)$, then $F^{-1} \in \mathcal{A} (\mu, X^\prime, X)$.
\end{proposition}
For the inversion formula, we introduce the class of $X$-space-valued functions
\begin{itemize}
\item {$\mathcal{A}(\mu, X)$}: Let  $X$ be a complex  Banach space and $\mu \in \mathbb{R}$.  We write $F \in \mathcal{A} (\mu, X)$ when $F$ is an analytic function
$$ F\colon \mathbb{C}_+ \to X
$$
satisfying
$$ ||F(s)||_X \leq C_F(Re~s) |s|^\mu, \quad \forall s \in \mathbb{C}_+,
$$
where $C_F\colon (0, \infty) \to (0, \infty)$ is a non-increasing function such that
$$ C_F(\sigma) \leq \frac{C}{\sigma^m}, \quad \forall \sigma \in (0, 1]
$$
with $C$ independent of $\sigma$.
\end{itemize}
Since $ 1 \leq |s|/(Re~s),$ it is clear that $\mathcal{A}(\mu, X) \subset \mathcal{A}(\mu + \varepsilon, X)$ for all $\varepsilon >0$.
\subsection{The inversion formula}
Let $F \in \mathcal{A}(\mu, X)$ with $\mu <-1$. For any $\sigma > 0$, we define
\begin{equation} \label{eq:3.2}
 f(t) =\mathcal{L}^{-1}\{F\}:= \frac{1}{2 \pi i} \int_{\sigma -i \infty}^{\sigma +i \infty} e^{st} F(s) ds.
\end{equation}
as the inverse of the Laplace transformed function $F(s)$.
We can see that $f$  is well defined, since
\begin{eqnarray}
|| f(t) ||_{X} & =& || \frac{1}{2 \pi} \int_0^{\infty} e^{(\sigma + i\omega)t} F(\sigma + i \omega)
d \omega ||_X \nonumber \\
&\leq& \frac{1}{2 \pi}C_{F}(\sigma) e^{\sigma t}  \int_0^{\infty} \frac{2} {|\sigma + i \omega|^{-\mu}} d \omega
, \quad - \mu >1, \nonumber\\
&=& \frac{1}{2 \pi}C_{F}(\sigma) e^{\sigma t} \sigma^{1+\mu}\int_0^{\infty} \frac{\zeta^{-1/2}}{(1 + \zeta)^{-\mu/2}} d \zeta,  \quad
\mbox{with}\quad \zeta= \omega^2/\sigma^2, \nonumber\\
&=&  \frac{1}{2 \pi}C_{F}(\sigma) e^{\sigma t} \sigma^{1 + \mu} B(\frac{1}{2}, \frac{-(1+\mu)}{2}),
\label{eq:3.3}
\end{eqnarray}
where the Euler Beta function $B$ is defined by
$$B(z_1,z_2):= \int_0^\infty t^{(z_1 -1)} /(1+t)^{(z_1 + z_2)}dt $$
 with $Re~z_1, Re~z_2 >0$.

As a consequence of ({\ref{eq:3.3}), we have
\begin{proposition} If $F  \in \mathcal{A}(\mu, X)$ with $\mu <-1$, then $F$ is the Laplace transform of a continuous causal function $f\colon \mathbb{R} \to X$ with polynomial growth.
\end{proposition}
Finally, we include here the most crucial result for  the Lubich approach related to causal time convolutions \cite{Lu:1994}.  For the benefit of the reader,  we give a brief sketch of the proof. A slight improvement of this result  can be found in \cite{DoSa:2013} and \cite{Sa:2012}: it eliminates $t^\varepsilon$ in the right-hand side of \eqref{eq:3.4} and substitutes $g^{(k)}$ by a linear differential operator of order $k$ and constant coefficients acting on $g$.
\begin{theorem}\label{th:3.5}
Let $A= \mathcal{L}\{a\} \in \mathcal{A} (\mu, X,Y) $  with $\mu \geq 0$. Let
$$ k:=\lfloor \mu +2 \rfloor,  \quad
 \varepsilon:= k - (\mu +1) \in (0, 1].$$
If $g \in C^{k-1}(\mathbb{R}, X)$ is causal and $||g^{(k)}||_X$  is integrable, then $a*g \in C(\mathbb{R}, Y)
$ is causal and
\begin{equation} \label{eq:3.4}
||a*g(t)||_Y \leq c_{\varepsilon} ~ t^{\varepsilon} ~C_A(t^{-1}) ~\int_0^t ||g^{(k)} (\tau)||_X d \tau.
\end{equation}
\end{theorem}
\begin{proof}
Let $\mathcal{L}\{g\} = G(s)$.  Since $ a*g = \mathcal{L}^{-1}( A(s)G(s))$, we see that
$\mu - k = -1 -\varepsilon < -1$ and
\begin{eqnarray*}
||a*g(t)||_Y& =& ||\mathcal{L}^{-1}\{ (s^{-k}A(s) \} * g^{(k)}||_Y  \\
&=&|| \int_0^t  \mathcal{L}^{-1} \{ (s^{(-k)} A(s) \} (\tau) g^{(k)}(t- \tau) d\tau ||_Y\\
&\leq& \int_0^t  \Big( \frac {1} {2 \pi} \int_0^\infty\frac  {2 C_A} {| \sigma +i \omega |^{1+ \varepsilon} }           d\omega  \Big)  e^{\sigma \tau}  || g^{(k)}  (t -\tau)  ||_X  d  \tau \\
&=& \frac{e^{\sigma t}}{2 \pi} C_A(\sigma) \sigma^{-\varepsilon} B(\frac{1}{2},  \frac{1}{\varepsilon})
\int_0^t || g^{(k)}  (t -\tau)  ||_X  d  \tau \\
&=& \frac{e^{\sigma t}}{2 \pi} C_A(\sigma)  \sigma^{-\varepsilon}  B(\frac{1}{2},  \frac{1}{\varepsilon})
\int_0^t || g^{(k)} (\tau) ||_X  d  \tau,
\end{eqnarray*}
 By taking $\sigma =t^{-1}$ and $c_{\varepsilon} =  1/2 \pi~  e~ B(1/2,  1/\varepsilon)$,
 this gives the estimate
(\ref{eq:3.4}).
\end{proof}
\subsection{An example}\label{s:3.4}
In order to illustrate the applicability of Theorem \ref{th:3.5}, and the concepts introduced in the section, we end this section by considering  a specific example.  Let us consider the simple retarded
boundary integral operator, namely $\mathcal{V}$ in (\ref{eq:2.15})
\begin{equation*}
  (\mathcal{V} * \lambda)(x,t) := \int_{\Gamma} E(x,y)  \lambda(y, t- \frac{|x-y|}{c})d\Gamma_y,
 \quad (x,t) \in \Gamma\times (0,T].
\end{equation*}
Note that the convolutional notation will be fully justified with this approach. It is also customary to write (at least formally)
\begin{eqnarray}
 (\mathcal{V}*\lambda) (x,t) &= &\int_0^t \int_{\Gamma} E(x,t; y,\tau)  \lambda(y, \tau)d\Gamma_y d\tau,\label{eq:3.5}
 \end{eqnarray}
where $E(x,t;y,\tau)$ is the fundamental solution of  the wave operator
$
\Box_{x,t}\varphi:= -  \Delta \varphi + \frac{1}{c^2} \partial^2 \varphi  / \partial
t^2, $ namely,
\begin{equation}\label{eq:3.6}
E(x,t; y, \tau):= \delta\Big((t - \tau) - \frac{|x-y|}{c})\Big) E(x,y),
\end{equation}
where $\delta$ is the Dirac delta.
Hence, we have
\begin{eqnarray*}
\mathcal{L}\{\mathcal{V} * \lambda \} &= &V(s) \Lambda(s) , \quad x \in \Gamma,\; s \in \mathbb C_+,  \; i.e., \\
( \mathcal{V} *\lambda)(x,t)&=&\Big( \mathcal{L}^{-1}\{V(s)\}* \lambda \Big)(x,t),
\quad x \in \Gamma,\; s \in \mathbb C_+.
\end{eqnarray*}
It can be shown that $V(s) $ is  just the simple - layer boundary integral operator for
the Laplace transform  of the wave operator,
\begin{equation}\label{eq:3.7}
\mathcal{L} \{ \Box_{x,t} \varphi\}: = -\Delta \Phi + \frac{s^2}{c^2} \Phi,
 \quad \Phi=\mathcal{L}\{\varphi\},
\end{equation}
which is defined explicitly as
$$
V(s) \psi:= \int_{\Gamma} E_{s/c}(x,y)  \psi(y) d\Gamma_y,  \quad x\in \Gamma
$$
and
$$E_{s/c}(x,y)= E(x,y) exp\{- {s |x,y| /c }\}  = \frac{ 1 }  {4 \pi |x -y|} exp\{ - s |x,y| / c  \} $$
is the fundamental solution of the transformed  wave operator  in (\ref{eq:3.7}).

We now summarize the properties of the operator $V(s)$ as follows. Note that in our notation the angled bracket is linear in both components and thus symmetry is not to be confused with self-adjointness.
\begin{itemize}
\item{Symmetry}:
$$\langle \chi, V(s) \psi\rangle = \langle \psi, V(s) \chi \rangle, \quad \forall~ \chi, \psi
\in H^{-1/2}(\Gamma). $$\\
\item{Positivity}:
$$Re \Big(e^{i \theta} \langle\overline{\psi}, V(s) \psi\rangle\Big) =\frac{\sigma}{|s|} |||u_{\psi}|||^2_{|s|, \mathbb{R}^3\setminus \Gamma}, \quad \forall \psi \in H^{-1/2}(\Gamma),$$
where  $\theta = Arg \;s$, the principal argument of $~s, \; u_{\psi} :=S(s)\psi$ in $\mathbb{R}^3 \setminus \Gamma$, and  $S(s)$ is the simple-layer potential
corresponding to $V(s)$ with norm defined by
$$ |||u_{\psi}|||^2_{|s|, \mathbb{R}^3\setminus \Gamma}:= \int_{\mathbb{R}^3 \setminus \Gamma}
\Big\{|\nabla u_{\psi} |^2
+ \frac{|s|^2}{c^2} |u_{\psi}|^2 \Big\} dx.$$
\item{Coercivity}: $$Re \Big(e^{i \theta} \langle\overline{\psi}, V(s) \psi\rangle\Big) \geq C \frac{\sigma \underline{\sigma}}{|s|^2}\; ||\psi||^2_{H^{-1/2}(\Gamma)}. $$
Here and in the sequel,  $C$ is a generic constant independent of $s$.
\item{Bounds}:
$$
||V(s)||_{L(X,X')} \leq  C \frac{|s|}{\sigma \underline\sigma^2}\quad , \quad
||V^{-1}(s) ||_{L(X', X)}\leq  C \frac{|s|^2}{\sigma \underline\sigma}.
$$
where $\sigma = Re~ s, \underline{\sigma}:= min \{1, \sigma\},  X=H^{-1/2}(\Gamma)$, and
$X^\prime = H^{1/2}(\Gamma).$
\end{itemize}
We shall return to these properties later.  Most proofs of them are readily  available (see, e.g. \cite{BaHa:1986a, Lu:1994, LaSa:2009b} and \cite{Sa:2012}). As consequences of these properties, we see that
$$ V  \in \mathcal{E}(2, \theta,
H^{-1/2}(\Gamma)) \cap \mathcal{A}(1, H^{-1/2}(\Gamma), H^{1/2}(\Gamma)),\; \mbox{and} \; V^{-1} \in \mathcal{A} (2, H^{1/2}(\Gamma), H^{-1/2}(\Gamma)). $$
Moreover, from Theorem\;\ref{th:3.5},  we have  the estimate for the simple retarded boundary integral operator
\begin{equation} \label{eq:3.8}
||\mathcal{V}*\lambda(t)||_{H^{1/2}(\Gamma)} \leq c_{\varepsilon} ~ t^{\varepsilon} ~C_V(t^{-1}) ~\int_0^t ||\lambda^{(k)} (\tau)||_{H^{-1/2}(\Gamma)} d \tau,
\end{equation}
where $ \mu =1, k=\lfloor 1+2 \rfloor=3,  \varepsilon  =3 - (1 +1)=1,\,$ and $ C_V(t^{-1}) =  C~  t\; max\{1, t^2\}$.

\section{Variational solutions}
We now return to the initial-boundary transmission problem defined by the
partial  differential equations (\ref{eq:2.1} ), (\ref{eq:2.5}), the  initial conditions
(\ref{eq:2.6}), and the transmission conditions (\ref{eq:2.7}) and (\ref{eq:2.8}). Our first step is to consider the problem in the Laplace transformed domain.

\subsection{Formulation in the transformed domain}
In the following we let $\mathbf{U}(s) :=  \mathbf{U}(x,s)= \mathcal{L}\{ {\bf u}(x,t)\},  \Phi (s)
:=\Phi(x,s) = \mathcal{L}\{\varphi(x,t)\}$. Then the initial-boundary transmission problem consisting of
(\ref{eq:2.1}), (\ref{eq:2.5}),( \ref{eq:2.6}), (\ref{eq:2.7}) and (\ref{eq:2.8}) in the Laplace transformed domain becomes  the following transmission boundary value problem:
\begin{eqnarray}
 - \Delta^{*} \mathbf{U}(s)
 +  \rho_e s^2 \mathbf{U}(s)  &=& \mathbf{0} \quad \mbox{in} \quad
\Omega, \label{eq:4.1}\\
-\Delta \Phi(s) + \frac{s^2}{c^2} \Phi(s) &=& 0 \quad \mbox{in}\quad  \Omega^c \label{eq:4.2} \\
\bm{\sigma} (\mathbf{U} )^- \mathbf{n} = - \rho_0\; s \Big( \Phi(s) + \Phi^{inc}  \Big) ^+ \mathbf{n},
&\mbox{and}&
s \mathbf{U}^- \cdot \mathbf{n} = - \Big( \frac{ \partial \Phi}{\partial n} + \frac {\partial \Phi^{inc}}{\partial n}\Big)^+ \; \mbox{on}\; \; \Gamma  \label{eq:4.3}
\end{eqnarray}
for $s\in \mathbb C_+$.  We remark that \eqref{eq:4.1}--\eqref{eq:4.3}  is an exterior scattering  problem,  and normally  a radiation condition is needed  in order to guarantee  the uniqueness  of the solution of the problem. In the present case, the radiation condition is substituted by the assumption that $\Phi\in H^1(\Omega^c)$, which is a Laplace-transform version of the weak Huygens principle.

To derive the proper nonlocal boundary problem, as usual, we begin via Green's third identity with
the representation of the solutions of (4.2) in the form:
\begin{equation}\label{eq:4.4}
 \Phi = D(s) \hat{\phi} - S(s)\hat{\lambda} \quad \mbox{in} \quad \Omega^c,
 \end{equation}
where $\hat{\phi}:= \Phi^+(s)$ and $\hat{\lambda}:= \partial \Phi^+ /\partial n$ are the Cauchy data for the operator in (\ref{eq:4.2}) and $S(s)$ and $D(s)$ are the simple-layer and duble-layer potentials
\begin{eqnarray}
S(s) \hat{\lambda} (x) &:=& \int_\Gamma  E_{s/c}(x,y) \hat{\lambda}(y) d\Gamma_y,
 \quad x \in \Omega^c,
 \label{eq:4.5}\\
D(s) \hat{\phi} (x)  &:=& \int_\Gamma \frac{\partial}{\partial n_y} E_{s/c}(x,y) \hat{\phi}(y)  d\Gamma_y, \quad x \in \Omega^c
\label{eq:4.6}.
\end{eqnarray}
Here
$$E_{s/c}(x,y) =  \frac{exp\{- s |x-y|/c\}} {4 \pi |x-y|}$$
 is the fundamental solution of the  operator  in  (\ref{eq:4.2}). By  standard arguments in potential
 theory,  we have the relations for the the Cauchy data $\hat{\lambda}$ and $\hat{\phi} $:
 \begin{equation}\label{eq:4.7}
\begin{pmatrix}
\hat{\phi} \\[3mm]
\hat{\lambda}\\
\end{pmatrix}
= \left (
\begin{matrix}
      \frac{1}{2}I + K(s) & -V(s) \\[3mm]
       -W(s)  &  ( \frac{1}{2}I - K(s))'  \\
    \end{matrix}
    \right )\begin{pmatrix}
\hat{\phi} \\[3mm]
\hat{\lambda}\\
\end{pmatrix}
\quad \mbox{on} \quad \Gamma.
 \end{equation}
Here $V, K, K^\prime $ and $W$  are the four  basic boundary integral operators familiar from potential theory such that
\begin{eqnarray*}
V(s) \hat \lambda (x) &:=& \int_\Gamma  E_{s/c}(x,y) \hat{\lambda}(y) d\Gamma_y,
 \quad x \in \Gamma\\
K(s) \hat{\phi} (x)  &:=& \int_\Gamma \frac{\partial}{\partial n_y} E_{s/c}(x,y) \hat{\phi}(y)  d\Gamma_y, \quad x \in \Gamma\\
K^\prime(s) \hat \lambda (x) &:=& \int_\Gamma  \frac{\partial}{\partial n_x} E_{s/c}(x,y) \hat{\lambda}(y) d\Gamma_y,
 \quad x \in \Gamma,\\
W(s) \hat{\phi} (x)  &:=&- \frac{\partial}{\partial n_x}  \int_\Gamma \frac{\partial}{\partial n_y} E_{s/c}(x,y) \hat{\phi}(y)  d\Gamma_y, \quad x \in \Gamma
\end{eqnarray*}
 By using the second transmission condition in (\ref{eq:4.3}), we obtain  from the second  boundary integral equation in  (\ref{eq:4.7}),
\begin{equation}\label{eq:4.8}
-s \mathbf{U}^-  \cdot \mathbf{n} - ( \frac{1}{2} I -  K)'  \hat{\lambda}  + W \hat{\phi} =
\Big(\frac{\partial \Phi^{inc }}{\partial n}\Big)^+ \;\quad \mbox{on}\quad \Gamma
\end{equation}
while  the  second boundary integral equation in (\ref{eq:4.7}) is simply
\begin{equation} \label{eq:4.9}
( \frac{1}{2}I - K) \hat{\phi} + V \hat{\lambda} = 0 \quad \mbox{on} \quad \Gamma.
\end{equation}

On the other hand, the weak solution of (\ref{eq:4.1}) in $\Omega$  leads to the operator equation in
of the form
\begin{equation}\label{eq:4.10}
{\mathbf A}_{ \Omega} \mathbf{U}  - \gamma^{\prime}(\bm{\sigma}(\mathbf {U})^-\mathbf{n}) = \mathbf{0}
\end{equation}
in the dual of $\mathbf{H}^1(\Omega)$.  Here   $\mathbf{A}_{\Omega}(s)  \colon {\mathbf H}^1( \Omega)  \to ({\mathbf H}^1(\Omega))'$ is defined by
\begin{eqnarray*}
\langle \mathbf{A}_{ \Omega}(s)\mathbf{U},\mathbf{V}\rangle: = a( \mathbf{U},
 \mathbf{V})_{s, \Omega}
=
\int_{\Omega} \Big(
\lambda (div \,\mathbf {U})(div\,  \mathbf{V} )
+ 2 \mu \bm \varepsilon(\mathbf{U}) : \varepsilon ( \mathbf{V})  + \rho_{e} \,s^2 \mathbf{U}\cdot  \mathbf{V} \Big)\,  dx  && \\
\quad \mathbf U, \mathbf V \in \mathbf H^1(\Omega), &&
\end{eqnarray*}
where $ \lambda, \mu$  the Lam\'e constants. Also in \eqref{eq:4.10} we use $\gamma^\prime$, defined as the adjoint of the trace operator. Then by using the  first transmission  condition  in (\ref{eq:4.3}),
we substitute $ - \rho_0\; s \Big( \Phi(s) + \Phi^{inc}  \Big) ^+ \mathbf{n}$
for  $\bm{\sigma} (\mathbf{U} )^- \mathbf{n} $  into (\ref{eq:4.10}) which leads to the equation
\begin{equation} \label{eq:4.11}
\tilde{\mathbf A}_{  \Omega}(s) \mathbf{U} +s \gamma^{\prime} ( \hat \phi\mathbf n) = -s \gamma' ((\Phi^{inc})^+ \mathbf{n}),
\end{equation}
where we have replaced $\tilde {\mathbf A}_{ \Omega}:=\rho^{-1}_0 {\mathbf A}_{ \Omega}$.
Collecting (\ref{eq:4.11}), (\ref{eq:4.8}) and (\ref{eq:4.9}), we arrive the following nonlocal problem, which reads: Given data $(\hat d_1, \hat d_2, \hat d_3)  \in  X' $, find $(\mathbf{U}, \hat\phi, \hat\lambda)
\in X$ such that
\begin{equation}\label{eq:4.12}
\pmb{\mathscr{A}}   \begin{pmatrix}
      \mathbf{U}   \\
      \hat\phi\\
      \hat\lambda \\
   \end{pmatrix}:=
\left (   \begin{matrix}
   \tilde{\mathbf A}_{ \Omega}&s  ({\mathbf n}^\top \gamma)^\prime &0\\
   - s (\mathbf{n}^\top\gamma) & W & - ( \frac{1}{2} I -  K)' \\
   0 & ( \frac{1}{2}I - K) & V\\
   \end{matrix}
\right )
   \begin{pmatrix}
    \mathbf{U}   \\
      \hat\phi\\
      \hat\lambda \\
    \end{pmatrix}
= \begin{pmatrix}
\hat d_1\\
\hat d_2\\
\hat d_3\\
   \end{pmatrix},
\end{equation}
where $(\hat d_1, \hat d_2, \hat d_3) : =( -s \gamma' ((\Phi^{inc})^+ \mathbf{n}), \;(\partial \Phi^{inc }/\partial n)^+,  \;0)$. The product spaces
$$ X ={\mathbf H}^1( \Omega)  \times H^{1/2}(\Gamma) \times H^{-1/2}(\Gamma) \quad \mbox{and}\quad
X^{\prime} =  ({\mathbf H}^1(\Omega))'\times H^{-1/2}(\Gamma) \times H^{1/2}(\Gamma)$$
are reciprocally dual Hilbert spaces.
Our aim is to show that Equation (\ref{eq:4.12}) has a unique solution in $X$. We will do this in the next subsection. However, before we do so, we will first show that
$\pmb{\mathscr{A} }$
 is invertible.

Using Gaussian elimination (as in  \cite{LaSa:2009b}), a simple computation shows that the matrix $\pmb{\mathscr{A}} $
 of operators can be decomposed in the form:
\begin{eqnarray}
\pmb{\mathscr{A}} &=& \left (
\begin{matrix}
I & 0 &0 \\
 0& I & -(\frac{1}{2} I -  K)' V^{-1}\\
 0 & 0& I\\
  \end{matrix}\right )
\left ( \begin{matrix}
 \tilde{\mathbf A}_{ \Omega}&s  ({\mathbf n}^\top \gamma)^\prime&0 \\
- s\, ({\mathbf n}^\top \gamma)& B& 0\\
 0 & 0& V\\
 \end{matrix}\right )
 \left ( \begin{matrix}
I&0&0 \\
0&I&0\\
0 &  V^{-1}(\frac{1}{2} I -  K)& I\\
 \end{matrix} \right )  \nonumber \\
&=:& P^{\prime} C P^{-1} ,
\label{eq:4.13}
\end{eqnarray}
where  $B := W + (\frac{1}{2} I -  K)' V^{-1}(\frac{1}{2} I -  K)$ .  We note that the operator matrix $C$  is {\em strongly elliptic}  (\cite{HsWe:2008, Mi:1970})  in the sense that
\begin{eqnarray}\label{eq:4.14}
 Re \Big\{ \langle \Theta C (v, \psi, \chi) , \overline{ (v, \psi, \chi) } \rangle \Big\}
 &\geq & c(Re \, s)\;  |s|^{-2} || (v, \psi, \chi)||^2_X
 \end{eqnarray}
for all $(\mathbf{v}, \psi, \chi) \in X $, where  $ \Theta$ is the matrix defined by
$$ \Theta:= \left (\begin{matrix}
e^{- i \theta}&0 &0\\
0& e^{-i \theta} &0\\
0&0& e^{i \theta}\\
\end{matrix} \right ).
$$
Since both $P$ and $P^{\prime}$ are invertible, it follows from (\ref{eq:4.14})  that
$\pmb{\mathscr{A}}$  is invertible.
As for   the proof of  (\ref{eq:4.14}), we only want to point out  that
$$ Re \Big\{s e^{-i\theta}  (  \langle \gamma^{\prime}(\psi  \mathbf{n}),  \bar{\mathbf{v}} \rangle   -
\langle  \gamma  \mathbf{v} , \bar{\psi} \mathbf {n} \rangle ) \Big\} = 0, $$
so that
\begin{eqnarray*}
 Re \Big\{ \langle \Theta C (\mathbf{v}, \psi, \chi) , \overline{ (\mathbf{v}, \psi, \chi) } \rangle \Big\}
 &= & Re\Big\{ e^{-i \theta}\langle \tilde{\mathbf A}_{  \Omega} \mathbf{v}, \bar{\mathbf{v}}\rangle +
 e^{-i \theta} \langle B\phi, \bar{\phi}\rangle + e^{i \theta} \langle V \chi,  \overline \chi \rangle\Big\}.\\
 \end{eqnarray*}
 It is clear that  what remans to be done is  to show that the operators  $\tilde{\mathbf A}_{ \Omega}, B$
 and $V$  belong to the appropriate class $\mathcal{E}(\mu, \theta, X)$  as $V$  in the example
 (see \S \ref{s:3.4}).
The details of  the proof will  be omitted  here.  However, in order to show that
$\pmb{\mathscr{A}}^{-1}$
 belongs to the appropriate  class $ \mathcal{A}(\mu, X, X^{\prime}) $ so that we may apply Theorem  \ref{th:3.5} to $\pmb{\mathscr{A}}^{-1}$
 for  obtaining   desired results in the time domain, we follow \cite{Sa:2012} in considering the existence and unique results of  a problem equivalent to the nonlocal problem defined by (\ref{eq:4.12}).

Suppose that $(\mathbf{U}, \hat\phi, \hat\lambda)\in X$ is a solution of (\ref{eq:4.12}).  Let
\begin{equation}\label{eq:4.15}
u:= D \hat\phi -  S \hat\lambda  \quad \mbox{in} \quad \mathbb{R}^3 \setminus \Gamma.
\end{equation}
Then  $u\in H^1(\mathbb{R}^3 \setminus \Gamma)$  is the solution of the transmission problem:
\begin{equation}\label{eq:4.16}
 - \Delta u + \frac{s^2}{c^2} u = 0 \quad \mbox{in}\quad  \mathbb{R}^3 \setminus \Gamma
 \end{equation}
satisfying  the following jump relations across $\Gamma$,
$$[\gamma u]:=  \gamma^+ u - \gamma u=\hat\phi \in H^{1/2}(\Gamma), \quad  [\partial_n u]  :=\partial_n^+ u - \partial_n u= \hat\lambda \in H^{-1/2}(\Gamma). $$

First, from (\ref{eq:4.12}) we see that
\begin{gather}
 \tilde{\mathbf A}_{  \Omega}(s)\mathbf{U}+s \gamma^{\prime}([\gamma u]\mathbf{n})= \hat d_1 \quad \mbox{in} \quad \Omega.\label{eq:4.17}\\
-s \gamma\mathbf{U}\cdot \mathbf{n} - \partial^{+}_ n u = \hat d_2 \quad\mbox{on} \quad \Gamma, \label{eq:4.18}\\
- \gamma u = \hat d_3 \quad \mbox{on} \quad \Gamma. \label{eq:4.19}
 \end{gather}
 Since $\hat d_3 =0 $, this means that $u$ is a solution of the homogeneous Dirichlet problem
 for the partial differential equation (\ref{eq:4.16}) in $\Omega$. Hence by the uniqueness of the the solution, we obtain  $u \equiv 0$ in  $\Omega$. Consequently, we have
 \begin{equation} \label{eq:4.20}
 [\gamma u] = \gamma^+ u = \hat \phi \quad \mbox{and} \quad [\partial_n u] = \partial_n^+ u = \hat \lambda.
 \end{equation}

Next, we  consider the variational formulation of the problem for equations  (\ref{eq:4.16}) and (\ref{eq:4.17})  together with the boundary condition: (\ref{eq:4.18}).   We will seek  a  solution
\begin{equation*}
(\mathbf{U}, u) \in  \pmb{\mathscr{H}}  =  {\mathbf H}^1(\Omega) \times H^1(\Omega^c)
\end{equation*}
 with the corresponding  test functions  $(\mathbf{V}, v)$ in the same function space.
 To derive the variational equations, we should keep in mind that the variational formulation should be formulated not in terms of  the Cauchy data $\hat\phi$ and $\hat \lambda$ directly but only in directly through the jumps of $u$ as indicted.

We begin with the first  Green formula for the equation (\ref{eq:4.16}).  Let  $(u,v) \in  H^1(\Omega^c) \times  H^1(\Omega^c)$. Then
\[
 - \langle \partial^+_nu, \overline{\gamma^+ v} \rangle = \int_{\Omega^c} \Big\{\nabla u \cdot \overline{\nabla v} + \frac{s^2}{c^2} u\overline{ v}\Big\} dx
 =  a_{s,\, \Omega^c} (u, v)=: \langle A_{\Omega^c}(s)\; u,\overline{ v}\rangle.
\]
From condition  (\ref{eq:4.18}), we obtain
\begin{equation} \label{eq:4.21}
 \langle A_{\Omega^c }(s)\, u,\overline{ v}\rangle  = - \langle \partial^+_n u, \overline{\gamma^+ v}
 \rangle
= \langle \hat d_2, \overline{\gamma^+ v}\rangle  + \langle  s \gamma\mathbf{U}\cdot \mathbf{n}, \overline{\gamma^+ v} \rangle.
\end{equation}
Together with the weak formulation of (\ref{eq:4.17}) we arrive at the following variational formulation:
 Find $(\mathbf{U}, u) \in  \pmb{\mathscr{H}} $    satisfying
 \begin{equation} \label{eq:4.22}
\langle \tilde{\mathbf A}_{ \Omega}(s)\mathbf{U},\overline{ \mathbf{V}}\rangle
 + \langle A_{ \Omega^c}(s)\,u,\overline{ v}\rangle
+s\Big\{ \langle \gamma^{\prime}((\gamma^+ u)\mathbf{n}), \overline{V}\rangle - \langle   \gamma\mathbf{U}\cdot \mathbf{n}, \overline{\gamma^+ v} \rangle\Big\} =\langle \hat d_1, \overline{\mathbf{V}}\rangle + \langle \hat d_2, \overline{\gamma^+ v}\rangle
\end{equation}
for all $(\mathbf{V}, v) \in  \pmb{\mathscr{H}}$.
 We remark that by the construction, it can be shown that  as in \cite{Sa:2012} this variational problem is equivalent  to   the transmission problem defined by (\ref{eq:4.17}), (\ref{eq:4.16}),
 and  (\ref{eq:4.18}). The later is equivalent to the nonlocal problem defined by (\ref{eq:4.12}).  Consequently, the variational problem (\ref{eq:4.22}) is equivalent to the nonlocal problem (\ref{eq:4.12}).  Hence
for the existence of the solution of  (\ref{eq:4.12})},  it is sufficient to show the existence of the solution of (\ref{eq:4.22}).
\subsection{Existence and uniqueness results }
We recall that  $|||u|||_{|s|, \Omega^c}$
is  the  norm of  $u$ defined by
\begin{equation}
 |||u|||_{|s|, \Omega^c} :=  \Big\{ ||\nabla u ||^2_{0, \Omega^c}
+ \frac{|s|^2}{c^2} ||u||^2_{0, \Omega^c}\Big\}^{1/2} =\langle A_{ \Omega^c}(|s|)\, u,\overline{ u}\rangle^{1/2}. \label{eq:4.23}
\end{equation}
 Similarly, we define the norm of $\mathbf{U}$
 \begin{eqnarray}
 |||\mathbf{U}|||_{|s|, \Omega} &:=& \langle \mathbf{A}_{  \Omega}(|s|)\mathbf{U},\overline{\mathbf{U}}\rangle^{1/2}. \label{eq:4.24}
 \end{eqnarray}
 We also need the following  inequalities for the equivalent norms
 \begin{gather}
\underline{\sigma } ||| \mathbf{U}|||_{1, \Omega} \leq  |||\mathbf{U}|||_{|s|, \Omega}\leq
\frac{|s|}{\underline{\sigma}}  |||\mathbf{U}|||_{1, \Omega} ,\label{eq:4.25}\\
\underline{\sigma} ||| u|||_{1, \Omega^c} \leq  |||u|||_{|s|, \Omega^c}  \leq
\frac{|s|}{\underline {\sigma} }  |||u|||_{1,\Omega},   \label{eq:4.26}
 \end{gather}
which can be obtained  from the  inequalities:
 $$\min\{1, \sigma\} \leq \min\{1, |s|\},\quad \mbox{and} \quad  \max\{1, |s|\} \min\{1, \sigma\}
 \leq |s|, \;\forall  s \in \mathbb{C}_+.$$
 We remark that the norm $ |||{ u}|||_{1, \Omega^c}$ is equivalent to $ ||{ u}||_{H^1(\Omega^c)}$ and  so is the energy norm $|||{ \mathbf{U}}|||_{1, \Omega}$  equivalent to the $\mathbf{H}^1(\Omega)$-norm of ${ \mathbf{U}} $ by the second Korn  inequality \cite{Fi:1972}.  In the following,  the ${c_j}'s $ are generic constants independent of $s$ which may or not not be the same at different places.

 Then we have the following basic results.
\begin{theorem}\label{th:4.1}
The variational problem \em{ (\ref{eq:4.22})} has a unique solution $(\mathbf{U}, u) \in
\pmb{\mathscr{H}} $. Moreover,  the following estimates hold:
 \begin{eqnarray}
 \Big\{ |||\mathbf{U}|||^2_{|s|, \Omega} + |||u|||^2_{|s|, \Omega^c}\Big\}^{1/2}
&\leq& c(\sigma, \underline{\sigma}) |s|  ||(\hat d_1, \hat d_2,  0)||_{X^{\prime}}, \label{eq:4.27}
\end{eqnarray}
where  $c(\sigma, \underline\sigma)$ is a constant depending only on  $\sigma$= {Re\, s} and $\underline{\sigma} = min\{1, \sigma\}$.
\end{theorem}
 \begin{proof}
 The existence and uniqueness results follow immediately from the identity
 \begin{eqnarray}
Re \Big\{e^{-i\theta}\Big( \langle \tilde{\mathbf A}_{ \Omega}(s)\mathbf{U},\overline{ \mathbf{U}}\rangle & + & \langle A_{ \Omega^c}(s)\,u,\overline{ u}\rangle
 + s\{ \langle \gamma^{\prime}((\gamma^+ u)\mathbf{n}), \overline{\mathbf U}\rangle - \langle
 \gamma\mathbf{U}\cdot \mathbf{n}, \overline{\gamma^+ u} \rangle\} \Big)\Big\}  \nonumber\\
&=& Re \Big\{e^{-i\theta}\Big( \langle \tilde{\mathbf A}_{  \Omega}(s)\mathbf{U},\overline{ \mathbf{U}}\rangle  +  \langle A_{ \Omega^c}(s)\,u,\overline{ u}\rangle  \Big)\Big\}   \nonumber \\
&=&\frac{\sigma}{|s|} \Big(|||\mathbf{U} |||^2_{|s|, \Omega} +|||u|||^2_{|s|, \Omega^c}\Big). \label{eq:4.28}
\end{eqnarray}
For the estimate (\ref{eq:4.27}),  it follows from  (\ref{eq:4.28}) and (\ref{eq:4.22}) that
\begin{eqnarray*}
\frac{\sigma}{|s|} \Big(|||\mathbf{U} |||^2_{|s|, \Omega} +|||u|||^2_{|s|, \Omega^c}\Big)
&\leq& \Big|\langle \hat d_1, \overline{{\mathbf{U}}}\rangle +
\langle \hat d_2, \overline{\gamma^+ {u}}\rangle\Big|\\
 &\leq&c_1 \Big\{ ||\hat d_1||_{({H^1(\Omega)})^\prime} |||{ \mathbf{U}}|||_{1, \Omega}
 + ||\hat d_2||_{H^{-1/2}( \Gamma)} |||{ u}|||_{1, \Omega^c}\Big\}\\
&\leq& c_1\Big\{ |||{ \mathbf{U}}|||^2_{1, \Omega} + |||{ u}|||^2_{1, \Omega^c} \Big\}^{1/2}
 \Big\{ ||\hat d_1||^2_{(H^1(\Omega))^\prime} + ||\hat d_2||^2_{H^{-1/2}(\Gamma)}\Big\}^{1/2}.
 \end{eqnarray*}
 Consequently, we have the estimate
 \begin{equation}\label{eq:4.29}
\Big\{|||\mathbf{U}|||^2_{|s|, \Omega} + |||u|||^2_{|s|, \Omega^c} \Big\}^{1/2}
\leq c(\sigma, \underline{\sigma})\; |s| \;||(\hat d_1,\hat d_2,0)||_{X^\prime},
\end{equation}
where $c(\sigma, \underline\sigma)= c_0/{\sigma \underline\sigma}$ with a constant $c_0$
independent of $s$ and $\sigma$.  In deriving the estimate (\ref{eq:4.29}), we
 have tacitly applied the relations (\ref{eq:4.25}) and (\ref{eq:4.26}).
\end{proof}

As we will see  the  estimate (\ref{eq:4.29}) will lead us to show that the inverse of the operator
$\pmb{\mathscr{A}} $ in (\ref{eq:4.12}) belongs to the appropriate class  $\mathcal{A}(\mu, X,Y)$.
In fact,  the following theorem holds for the operator $\pmb{\mathscr{A}}$ of (\ref{eq:4.13}).

\begin{theorem} \label{th:4.2}
Let $ X =\mathbf{ H}^1(\Omega) \times H^{1/2}(\Gamma) \times H^{-1/2}(\Gamma) $ with its dual $ X^\prime =(\mathbf{ H}^1(\Omega))^\prime  \times H^{-1/2}(\Gamma) \times H^{1/2}(\Gamma) $
and with  $X^\prime_0 := \{ (\hat d_1, \hat d_2 ,\hat d_3) \in X^\prime\,  \big | \,  \hat  d_3 = 0\}$.  Then we have
$$ \pmb{\mathscr{A}}  \in \mathcal{A}(2, X,X^\prime) \quad \mbox{and} \quad
 \pmb{\mathscr{A}}^{-1}|_{X_0'} \in  \mathcal{A}(3/2 , X_0^\prime, X) .$$
 \end{theorem}
\begin{proof}
We will only prove $ \pmb{\mathscr{A}}^{-1}|_{X_0'} \in  \mathcal{A}(3/2 , X_0^\prime, X)$.  We remark that in the proof,  for simplicity, we will replace  the norm  $|| \cdot ||_{H^1(\Omega)} $ of $\mathbf{ H}^1(\Omega)$   by its equivalent norm  $ ||| \cdot|||_{1, \Omega}$.
It was pointed out in  (\ref{eq:4.20}) that
$$
\gamma^+u=[\gamma u] =\hat\phi \in H^{1/2}(\Gamma), \quad \partial_n^+ u= [\partial_n u] =\hat\lambda \in H^{-1/2}(\Gamma).
 $$
 Then we see that
\begin{equation}\label{eq:4.30}
||\hat\phi ||^2_{H^{1/2}(\Gamma)} = ||\gamma^+ u||^2_{H^{1/2}(\Gamma)}
\leq c_1 |||u|||^2_{1, \Omega^c}
\leq c_1 \frac{1}{\underline\sigma^2} |||u|||^2_{|s|,  \Omega^c}
\end{equation}
Similarly, we have
$$
||\hat\lambda||^2_{H^{-1/2}(\Gamma)} =  || \partial^+_n u ||^2_{H^{-1/2}(\Gamma)} . $$
For $v \in H^{1/2}(\Gamma)$,  let $\tilde v $ be its extension as an $ H^1(\Omega^c)$ solution of
$ - \Delta \tilde v + |s|^2/ c^2\, \tilde v = 0 \quad \mbox{in}\quad  \Omega^c$.
Then  by the generalized Green's first theorem (see , e.g., \cite{HsWe:2008} ) we have
\begin{eqnarray*}
\langle \partial^+_n u, v\rangle &=&- \langle  A_{\Omega^c}(s)\; u, \tilde v \rangle \leq  |||u|||_{|s|, \Omega^c}  |||\tilde v|||_{|s|, \Omega^c}\\
&\leq& c_2 \;max\{ 1, |s|^{1/2} \} \;||v||_{H^{1/2}(\Gamma)} |||u|||_{|s|, \Omega^c}\\
&\leq& c_2 \Big( \frac{|s|}{\underline\sigma} \Big)^{1/2}\, ||v||_{H^{1/2}(\Gamma)} |||u|||_{|s|, \Omega^c},
\end{eqnarray*}
where on the right hand side, we have tacitly used the result in  \cite{BaHa:1986a}
(see also \cite{LaSa:2009a})   for  estimating $\tilde v$.
Hence,
\begin{equation*}
||\partial^+_n u ||_{H^{-1/2}(\Gamma)} : = sup  \frac{\langle \partial^+_n u, v\rangle }{  ||v||_{H^{1/2}(\Gamma) } }
\leq c_2\Big( \frac{|s|} {\underline\sigma} \Big)^{1/2} |||u|||_{|s|, \Omega^c}.
\end{equation*}
Thus, we have the estimates
\begin{equation} \label{eq:4.31}
||\hat\lambda||^2_{H^{-1/2}(\Gamma)} \leq  c_2^2 \frac{|s|} {\underline\sigma}
 |||u|||^2_{|s|, \Omega^c}.
\end{equation}
From (\ref{eq:4.30}) and (\ref{eq:4.31}),   we obtain the estimates
\begin{equation}\label{eq:4.32}
\frac{1}{2} \Big\{ c_1 \underline{\sigma}^2  ||\hat\phi ||^2_{H^{1/2}(\Gamma} +
 \frac {\underline\sigma}{c_2^2|s|}  |\hat\lambda||^2_{H^{-1/2}(\Gamma)}\Big\}
 \leq
 |||u|||^2_{|s|, \Omega^c}.
 \end{equation}
 As a consequence of (\ref{eq:4.29}),  it follows that
 \begin{eqnarray*}
 \Big\{ \underline\sigma^2 ||| \mathbf{U}|||^2_{1, \Omega} +
 \frac{1}{2} \Big(c_1\, \underline{\sigma}^2  ||\hat\phi ||^2_{H^{1/2}(\Gamma} +
 \frac {\underline\sigma}{c_2^2|s|}  ||\hat\lambda||^2_{H^{-1/2}(\Gamma)}\Big)\Big\}&\leq&
\Big\{  c(\sigma, \underline{\sigma}) |s|  ||(\hat d_1, \hat d_2,  0)||_{X^{\prime}}\Big\}^2.
 \end{eqnarray*}
However,  we see that
\begin{eqnarray*}
LHS &\geq&\frac{1}{2} \frac{\underline\sigma}{|s|} \Big\{ \underline\sigma^2 ||| \mathbf{U}|||^2_{1, \Omega} + c_1
 \underline{\sigma}^2  ||\hat\phi ||^2_{H^{1/2}(\Gamma)} + \frac{1}{c_2^2} ||\hat\lambda||^2_{H^{-1/2}(\Gamma)}\Big\}\\
 &\geq&\frac{1}{2} \frac{\underline\sigma^3}{|s|} \Big\{ ||| \mathbf{U}|||^2_{1, \Omega} + c_1
  ||\hat\phi ||^2_{H^{1/2}(\Gamma)} + \frac{1}{c_2^2} ||\hat\lambda||^2_{H^{-1/2}(\Gamma)}\Big\}\\
\end{eqnarray*}
This implies that
$$ \Big\{ ||| \mathbf{U}|||^2_{1, \Omega} +
  ||\hat\phi ||^2_{H^{1/2}(\Gamma)} +  ||\hat\lambda||^2_{H^{-1/2}(\Gamma)}\Big\}^{1/2}
  \leq c_0(\sigma, \underline\sigma) |s|^{3/2} ||(\hat d_1,\hat d_2, 0)||_{X^\prime},
 $$
 where
 $$c_0(\sigma, \underline\sigma)= \frac{C}{\sigma \underline\sigma^{5/2}}$$
 with constant $C$ independent of $s$ and $\sigma$.
Or, we have
\begin{equation}\label{eq:4.33}
||(\mathbf{U}, \hat\phi, \hat\lambda)||_X \leq c_0(\sigma, \underline\sigma)\; |s|^{3/2} ||(\hat d_1,\hat d_2, 0)||_{X^\prime}
\end{equation}
which is  the desired result:   $\pmb{\mathscr{A}}^{-1}|_{X_0'} \in  \mathcal{A}(3/2 , X_0^\prime, X)$ .
\end{proof}

In view of (\ref{eq:4.4}), we see that $\mathbf{U}$ and $\Phi$ are solutions of the system
 \begin{equation}\label{eq:4.34}
\begin{pmatrix}
\mathbf{U} \\[3mm]
\Phi\\
\end{pmatrix}
= \left (
\begin{matrix}
      I & 0 & 0\\[3mm]
      0 & D & -S  \\
    \end{matrix}
    \right )
\pmb{\mathscr{A}}^{-1}
    \begin{pmatrix}
\hat{d_1} \\[3mm]
\hat{d_2}\\[3mm]
{0}\\
\end{pmatrix}.
 \end{equation}
As a consequence of  Theorems~\ref{th:4.1} and \ref{th:4.2} ,  we have the following corollary.
\begin{corollary}\label{co:4.3} The matrix of operators
\begin{equation}\label{eq:4.35}
 \left (
\begin{matrix}
      I & 0 & 0\\[3mm]
      0 & D & -S  \\
    \end{matrix}
    \right )
 \pmb{\mathscr{A}}^{-1}|_{X_0'}  \quad \mbox{belongs to the class} \quad  \mathcal{A}(1 , X_0^\prime, \mathbf{H}^1(\Omega) \times H^1(\Omega^c)).
\end{equation}
\end{corollary}
\begin{proof}
We note that from (\ref{eq:4.27}), we have
 \[
\Big\{ |||\mathbf{U}|||^2_{|s|, \Omega} + |||\Phi|||^2_{|s|, \Omega^c}\Big\}^{1/2}
= \Big\{ |||\mathbf{U}|||^2_{|s|, \Omega} + |||u|||^2_{|s|, \Omega^c}\Big\}^{1/2}
\leq c(\sigma, \underline{\sigma}) |s|\; ||(\hat d_1, \hat d_2,  0)||_{X^{\prime}}\;.
\]
\end{proof}

It should be mentioned  that for $\hat\phi \in H^{-1/2}(\Gamma)$, if  $ u= D(s) \hat\phi $ in $\mathbb{R}^3 \setminus \Gamma$, then
\begin{eqnarray*}
\frac{\sigma}{|s|} |||u|||^2_{|s|, \mathbb{R}^3\setminus \Gamma}&=& Re\; \Big\{ e^{-i \theta} \langle W\hat\phi, \overline{\hat\phi} \rangle \Big\}\\
&\leq& ||W\hat\phi||_{H^{-1/2}(\Gamma)} ||\hat\phi||_{H^{1/2}(\Gamma)} \\
&\leq &c_1\Big(\frac{|s|}{\underline\sigma}\Big)^{1/2} |||u|||_{|s|, \mathbb{R}^3 \setminus \Gamma}  ||\hat\phi||_{H^{1/2}(\Gamma)} .
\end{eqnarray*}
Hence from (\ref{eq:4.26}) we obtain that
$$ ||D(s) \hat\phi||_{H^1( \mathbb{R}^3 \setminus \Gamma)} \leq  c_1 \frac{|s|^{3/2}}{\sigma {\underline\sigma}^{3/2}} ||\hat\phi||_{H^{1/2}(\Gamma)},$$
which implies $D \in
 \mathcal{A}(3/2 , H^{/1/2}(\Gamma), H^1(\mathbb{R}^3 \setminus \Gamma)).
$
Similarly, we may show as in \cite{LaSa:2009b} that for $\hat \lambda$, if we set $ u= S(s) \hat\lambda $ in $\mathbb{R}^3 \setminus \Gamma$, then we can show that
$$||S(s) \hat\lambda||_{H^1(\mathbb{R}^3 \setminus \Gamma)} \leq c_2 \frac{|s|} {\sigma \underline\sigma^2}  ||\hat\lambda||_{H^{-1/2}(\Gamma)}, $$
and hence $S \in
 \mathcal{A}(1 , H^{-1/2}(\Gamma), H^1(\mathbb{R}^3 \setminus \Gamma)).
$
This means
\begin{equation*}
 \left (
\begin{matrix}
      I & 0 & 0\\[3mm]
      0 & D & -S  \\
    \end{matrix}
    \right ) \in  \mathcal{A} (3/2 , X,  \mathbf{H}^1(\Omega)
\times H^1(\Omega^c)).
\end{equation*}
Following \cite{LaSa:2009b}, if we apply the composition rule and Theorem~\ref{th:4.2}, we find  the matrix of operators in (\ref{eq:4.34}) ended  with an index $\mu = 3/2 + 3/2 = 3$. However, this only gives an upper bound for the actual index as in the Corollary\;\ref{co:4.3}.
\section{Main results in the time domain }
With the properties of the solutions in the transformed domains available, we now return to the solutions in the time domain.  As the example in \S \ref{s:3.4}  properties of the solutions in the time domains may be  obtained by applying  the inversion formula( \ref{eq:3.2}) to the  relevant   operators in the  transformed domain.

We recall that  the matrix of operators  $\pmb{\mathscr{A}} $
 belongs to the class $\mathcal{A}(2, X,X^\prime)$ with  $ X =\mathbf{ H}^1(\Omega) \times H^{1/2}(\Gamma) \times H^{-1/2}(\Gamma) $
(see Theorem \ref{th:4.2}).  Indeed, we see that
\begin{eqnarray}\label{eq:5.1}
||\pmb{\mathscr{A}}(s)||  &= & 3^2 \max_{1 \leq i,j  \leq 3}  ||\mathbb{A}_{ij}(s)||  \nonumber\\
&=& 3^2\; || W(s)||
\leq c_0 \frac{|s|^2}{\sigma \underline\sigma},
\end{eqnarray}
where $c_0$ is a constant  independent of $s$ and $\sigma$.
We now apply  Theorem  \ref{th:3.5} to
$\pmb{\mathscr{A}}$,
 with $\mu =2, k=4$  and $ \varepsilon =1$, and writing $\mathbf a:=\mathcal L^{-1}\{\pmb{\mathscr{A}}\}$.

\begin{theorem}\label{th:5.1}
Let  $ X =\mathbf{ H}^1(\Omega) \times H^{1/2}(\Gamma) \times H^{-1/2}(\Gamma) $  and
$\mathbf{g}(t) := \mathcal{L}^{-1} \{ ( {\mathbf U}, \hat \phi, \hat \lambda)^{\top} \}$.
If $\mathbf{g} \in C^3([0, T], X)$ and $\mathbf{g}^{(4)}$  is integrable, then $
\mathbf a*\mathbf{g}$ belongs to $C([0, T], X^{\prime}) $  and
\begin{equation} \label{eq:5.2}
||\mathbf a*\mathbf{g}(t)||_{X^{\prime}} \leq c_1 ~ t^2\, max\{1, t\} ~\int_0^t ||g^{(4)} (\tau)||_X\; d \tau.
\end{equation}
\end{theorem}
\noindent
Similarly, by applying Theorem  \ref{th:3.5} to $\pmb{\mathscr{A}}^{-1}|_{X_0'}$
with $\mu =3/2,\; k=3$  and $ \varepsilon =1/2$, the following theorem holds.
\begin{theorem}\label{th:5.2}
Let  $ X =\mathbf{ H}^1(\Omega) \times H^{1/2}(\Gamma) \times H^{-1/2}(\Gamma) $  and $\mathbf{d}(t) := \mathcal{L}^{-1} \{ ( \hat{d}_1, \hat{d}_2, 0)^{\top} \}$.
If $\mathbf{d} \in C^2([0, T], X^\prime)$ and $||\mathbf{d}^{(3)}||_{X^\prime}$  is integrable, then $
(\mathbf{u}, \phi, \lambda)^{\top}$ belongs to $C([0, T], X)
$  and
\begin{equation}  \label{eq:5.3}
||\begin{pmatrix}
\mathbf{u} \\
\phi\\
\lambda
\end{pmatrix}(t)||_X \leq c_{1/2} ~ t^{ \frac{1}{2} +1}\; max \{1, t^{2 \frac{1}{2} }\} ~\int_0^t ||\mathbf{d}^{(3)} (\tau)||_{X^{\prime}}\; d \tau.
\end{equation}
\end{theorem}

Finally in view of the Corollary \ref{co:4.3}, applying Theorem \ref{th:3.5} with $\mu =1, k = 3$ and $ \varepsilon =1 $,  the elastic and potential fields, $\mathbf{u}(x,t)$ and $\varphi(x,t)$ of the fluid-structure interaction  satisfy the estimates:
\begin{theorem}\label{th:5.3}
Let  $ X =\mathbf{ H}^1(\Omega) \times H^{1/2}(\Gamma) \times H^{-1/2}(\Gamma) $  and
$\mathbf{d}(t) := \mathcal{L}^{-1} \{ ( \hat{d}_1, \hat{d}_2, 0)^{\top} \}$.
If $\mathbf{d} \in C^2([0, T], X^\prime)$ and $||\mathbf{d}^{(3)}||_{X\prime}$  is integrable, then $
(\mathbf{u}, \varphi)^{\top}$ belongs to $C([0, T], \mathbf{H}^1(\Omega) \times H^1(\Omega^c) )
$  and
\begin{equation} \label{eq:5.4}
||\begin{pmatrix}
\mathbf{u} \\[3mm]
\varphi\\
\end{pmatrix}
(t)||_{\mathbf{H}^1(\Omega) \times H^1(\Omega^c)} \leq c_1 ~ t^2 \max \{1, t\} ~\int_0^t ||\mathbf{d}^{(3)} (\tau)||_{X^{\prime}}\; d \tau.
\end{equation}
\end{theorem}

We remark that Theorems \ref{th:5.1}-\ref{th:5.3} are mathematical foundations for the semi- and full-discretization  schemes based on the boundary element method and convolution quadrature method. We will pursue these investigations  in a separate communication.

config: arxiv


TITLE: I would like to know what variables to consider when I experiment with pouring from bottles and the "glug" effect?
QUESTION [3 upvotes]: I am a middle school science teacher and want to model a science fair project for my students. Specifically I want to investigate using a vent in the top of the handle of a gallon milk jug to alleviate the air glugging when pouring and achieve faster pouring rates. So far I believe the following factors are in play (or would at least seem to be worthy of consideration as possible variables):
Initial angle of tilt.
Opening size, both the pouring opening and the vent opening.
Stiffness of the container.
Viscosity of the liquid.
Atmospheric pressure.
At this point in time I intend to make assumptions that the last three are relatively constant given that I will be using standard plastic milk jugs and water from the tap. I do not believe the changes in atmospheric pressure will be significant but I do think that will make for a good thinking/discussion point for my students.
I am particularly interested in the variables aspect of experimentation as I know from experience this is something my 6th graders are not good with. I am mostly looking for the obvious things, factors that even with their limited experimental and life experiences they can relate to.
Any input would be much appreciated!

REPLY [1 votes]: I'll just throw some ideas out.

Initial angle of tilt.

Higher the angle the higher the back pressure has to get before air will hold the fluid from flowing until it equalizes (glugs).  It will have longer time between "glugs" but this will also tend to slow the flow down.
At a very low angle air can enter while fluid exists at the same time even without a vent.

Opening size, both the pouring opening and the vent opening.

At some point the vent will be large enough to allow all the air to flow through at a high enough rate to prevent the back pressure from building and stopping the fluid flow.  Perhaps the experiment might start with a large vent that you partially cover.  The challenge is for angle X, how do you get the bottle to drain fast (adjust vent size so laminar flow occurs).  But this could be tedious and depending on the shape of the bottle it could he hard to get an obvious laminar flow.  Also the idea of turbulence and friction might be hard to convey.
The opening will help larger amounts of fluid to exit faster but equalization between air pressures will still have to occur.

Viscosity of the liquid.

A thicker fluid will take longer to pour but the same principles apply.
The other aspect is the shape of the opening because as fluid flows it will either create a laminar flow or a turbulent flow.  The ideal rate will balance the air vent inflow with the fluid outflow so that the fluid remains laminar because that will have the least friction and the fastest smoothest flow.

I would suggest looking into another experiment that shows the exchange between air and fluid in the form of a tornado.  This is a classic demonstration.  You can find many examples and construction ideas but here's one:

If you just flip the bottles upside down and let one drain into the other it is very slow.  However if you circulate the fluid with a little swirl a "tornado" forms and the air flows upwards rapidly and the fluid will spin and fall into the lower bottle.
There are many things you could learn from this and tease the kids with. Challenge them to find a fast way to move the fluid down (don't show them the swirl trick).  Then discuss why it works and what doesn't work, and perhaps prepare a little side material about how this happens in nature (low pressure/high pressure etc.).


text:
\begin{definition}[Definition:Negative Categorical Statement]
A '''negative categorical statement''' is either one of:
=== [[Definition:Universal Negative|Universal Negative]] ===
{{:Definition:Universal Negative}}
=== [[Definition:Particular Negative|Particular Negative]] ===
{{:Definition:Particular Negative}}
[[Category:Definitions/Categorical Statements]]
i4tlzj2k8juuoz2pt0zforyz6b8lyw6
\end{definition}

config: wiki


TITLE: Functions with $ f \bigg( p \Big( f \big( p ( x ) \big) \Big) \bigg) = p \bigg( f \Big( p \big( f ( x ) \big) \Big) \bigg) $ for all polynomials $ p $
QUESTION [1 upvotes]: Find all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f \bigg( p \Big( f \big( p ( x ) \big) \Big) \bigg) = p \bigg( f \Big( p \big( f ( x ) \big) \Big) \bigg) $$
for all $ x \in \mathbb R $ and all nonconstant polynomials $ p : \mathbb R \to \mathbb R $.

Obviously, the identity function is a solution.
If constant $ p $ was allowed, identity would be the only solution, as for any $ x \in \mathbb R $, we could take $ p $ to be the constant function with the value $ x $, and the functional equation would give $ f ( x ) = x $.
Source:
Number five on the list at the end of this page.

REPLY [0 votes]: Not only it can be shown that identity is the only solution, but also one can only require the functional equation to hold for nonconstant affine functions, and still get the same answer.
For any $ y , z \in \mathbb R $ with $ z \ne 0 $, the function $ p : \mathbb R \to \mathbb R $ defined with $ p ( x ) = z x + y $ for all $ x \in \mathbb R $ is a nonconstant polynomial. Considering only the functions of this form, the functional equation reads
$$ f \big( z f ( z x + y ) + y \big) = z f \big( z f ( x ) + y \big) + y $$
for all $ x , y , z \in \mathbb R $ with $ z \ne 0 $.
Assuming $ f ( 0 ) \ne 0 $, we can substitute $ 0 $ for both $ x $ and $ y $ and $ \frac z { f ( 0 ) } $ for $ z $ to get $ f ( 0 ) f ( z ) = z f ( z ) $, which shows that for all $ z \in \mathbb R \setminus \{ 0 , f ( 0 ) \} $ we have $ f ( z ) = 0 $. But setting $ y = 0 $ and $ x = z = | f ( 0 ) | + 1 $, we will have $ x , z x + y \in \mathbb R \setminus \{ 0 , f ( 0 ) \} $, which gives $ f ( x ) = f ( z x + y ) = 0 $, and thus $ f ( 0 ) = \big( | f ( 0 ) | + 1 \big) f ( 0 ) $, which contradicts $ f ( 0 ) \ne 0 $. Therefore, we must have $ f ( 0 ) = 0 $.
Setting $ x = 0 $ and $ y = z = 1 $ we get $ f \big( f ( 1 ) + 1 \big) = f ( 1 ) + 1 $, which show that there is some $ a \in \mathbb R $ (equal to either $ 1 $ or $ f ( 1 ) + 1 $) such that $ f ( a ) \ne 0 $. Hence, substituting $ 0 $ for $ x $, $ a $ for $ y $ and $ \frac { z - a } { f ( a ) } $ for $ z $, we get $ f ( z ) = z $ for all $ z \in \mathbb R \setminus \{ a \} $. Now, we can choose some $ b \in \mathbb R \setminus \{ a \} $ with $ f ( b ) \ne 0 $, repeat the argument to get $ f ( z ) = z $ for all $ z \in \mathbb R \setminus \{ b \} $, and thus conclude that $ f $ must be the identity function.


TITLE: Inductive step for graph proofs: reduce from n+1 to n or from n to n-1?
QUESTION [1 upvotes]: I was reading a practice problem set for a discrete maths course. It says:

By far, the most common mistake in this homework was using induction incorrectly for graph
problems. When proving something about graphs by induction, you want to reduce from $n$ to $n − 1$.
This is because you are proving a statement for all graphs of size $n$, and not just a specific graph. By
proving something by constructing an $n + 1$ size graph using an $n$ size graph, you have only shown it
for “one particular” graph of size $n + 1$, whereas you wanted to show it for all graphs of size $n + 1$. (By
size, I mean either the number of vertices or the number of edges, whatever you are inducting on.)

Then they give an example:

We prove by induction on $n$ that if $|V| = n$ and $G$ is acyclic and $|E| = n − 1$, then $G$ is connected.
Base case ($n = 1$): This is trivial; $G$ consists of a single vertex and no edges.
Induction step ($n \geq 2$): First we will prove that $G$ has a leaf. Since $E \neq \emptyset$, we can pick a vertex $v \in V$
with $deg(v) > 0$. Start walking from $v$ until you reach a leaf, say $u$. This will happen because the graph is acyclic.
Now, $G − u$ has $n − 1$ vertices and $n − 2$ edges, so we can use the induction hypothesis to conclude that it is connected. Adding $u$ to $G$ with the original edge keeps it connected.
Common Mistake: Constructing $n + 1$ vertex graph using an $n$ vertex graph in induction step. If you did this, you missed showing existence of a leaf, which is crucial.

I would have done it a bit different (rough outline):
Induction step $(n + 1)$: $G$ has $n+1$ nodes. We can find a leaf node $v$ with $deg(v) = 1$. $v$ has to exists because $G$ is acyclic. We obtain a subgraph $G'$ by removing $v$ from $G$. By induction we can assume $G'$ is connected. Adding $v$ again to $G'$ keeps the resulting graph $G$ connected, which proves the theorem for $n + 1$.

Is mine the wrong approach and I should have started from $n$? I thought in the inductive step for graph proofs, you usually start from a $n+1$ graph and for example reduce it to $n$ by removing a certain node/edge/...

REPLY [3 votes]: The "common mistake" described in the problem set instructions has
nothing to do with whether your inductive step is based on $n$ and $n-1$
or $n+1$ and $n.$
It has everything to do with which number (the smaller one or the larger one) is the number of nodes in the arbitrary graph that you invoke in the inductive step.
If you go from $n$ to $n+1$ (or if you go from $n-1$ to $n,$ for that matter!) by adding a node to an arbitrary graph, the proof is wrong.
You must remove a node instead, as indeed you did.
Another vital thing to make sure of is that the inductive step can use the base case as its "input." In the problem set, this is done by making the inductive step use the assumption $P(n-1)$ to prove $P(n)$ for $n\geq 2.$
This means that when $n = 2$ in the inductive step, we use $P(1)$ (the base case) to prove $P(2).$
In your proof you need to specify that the inductive step applies
for all $n \geq 1,$ not only for $n\geq 2,$ so that when $n=1$
you use $P(1)$ to prove $P(2).$
There are some classic fake proofs that rely on inductive steps that don't "connect" with the base case.
One thing I do not like about the proof style in either the problem set or your attempt is that you both launch into the inductive step without saying what the inductive assumption is, and simply mention later that something is
true because of the inductive assumption.
But both proofs are simple enough that I can infer what the inductive assumption was supposed to be, and that it "connects" with the base case.