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Controllability of surface gravity waves and the sloshing problem

M. A. Fontelos (Corresponding Author) ICMAT-CSIC, C/Nicolás Cabrera, no 13-15 Campus de Cantoblanco, UAM, 28049 Madrid, Spain. [email protected] and J. López-Ríos Universidad Industrial de Santander, Escuela de Matemáticas, A.A. 678, Bucaramanga, Colombia [email protected]

Abstract.

We study the problem of controlling the free surface for a two dimensional solid container in the context of the gravity waves and the sloshing problem. By using conformal maps and the Dirichlet–Neumann operator, the problem is formulated as a second order evolutionary equation on the free surface involving a self-adjoint operator. We present then the appropriate Sobolev spaces where having solutions for the system and study the exact controllability through an observability inequality for the adjoint problem.

Key words and phrases:

Controllability; gravity waves; the sloshing problem; Hilbert transform.

2010 Mathematics Subject Classification:

35Q31; 35J57; 44A15

M. A. Fontelos was partially supported by Ministerio de Economía, Industria y Competitividad, Gobierno de España (Grant No. MTM2017–89423–P)

1. Introduction

Finding and controlling the frequencies for the oscillations of the free boundary in the context of gravity waves and studying related spectral problems is a classical, well known problem in the literature [11, 14, 15, 31]. Roughly speaking, controlling a system consist not only in testing that its behavior is satisfactory, but also in putting things in order to guarantee that it behaves as desired. In mathematical terms, controlling the state $y$ , ruled by the state equation

A(y)=f(v),

where $v$ is the control, consists in finding $v\in U_{ad}$ , the set of admissible controls, such that the solution to the equation gets close to a desired prescribed state, $\bar{y}$ .

In this paper, we study controllability of a Partial Differential Equation (PDE), in the context of controlling the oscillations of a liquid free surface in a two-dimensional bounded container and the so-called sloshing problem. We formulate the geometrical problem in terms of an integrodifferential equation by using the Hilbert transform, then we establish the appropriate Sobolev spaces to study existence of solutions for the eigenvalue problem and, finally, we set up an observability inequality for the homogeneous adjoint problem. The sloshing problem adduces to an important difference with respect to the classical water-waves formulation: the presence of vertical walls and the contact with the free surface. Inspired in the developments for the classical wave equation, we introduce analytical tools to prove that it is possible to control the oscillations of the free surface, by injecting fluid on the rigid side walls.

The main strengths of our method lie in the use of the Hilbert transform to formulate the problem as an evolutionary equation involving a self-adjoint operator. This is known as the boundary integral method and has proved to be very fruitful in the study of water waves problems (see [13] and references therein, for instance). Moreover, the use of Tchebyshev polynomials provides an explicit orthogonal basis which allows to study, analytically, the associated eigenvalue problem. Then, an observability inequality arises naturally.

Generally speaking, the water-waves problem for an ideal liquid consists of describing the motion of a layer of incompressible, inviscid fluid, delimited below by a solid bottom, and above by a free surface under the influence of gravity. In mathematical terms, if $\boldsymbol{u}(x,y)$ is the fluid velocity and $\varphi$ is the velocity potential such that $\boldsymbol{u}=\nabla\varphi$ , by the conservation laws [19]

\left\{\begin{aligned} &\Delta\varphi=0,&&\Omega_{t},\\ &\eta_{t}+\eta_{x}\varphi_{x}=\varphi_{y},&&y=\eta,\\ &\varphi_{t}+\frac{1}{2}(\varphi_{x}^{2}+\varphi_{y}^{2})+g\eta=0,&&y=\eta,\\ &\partial_{n}\varphi=0,&&y=b,\end{aligned}\right.

(1)

where $\eta$ and $b$ are the free boundary and bottom parametrization, respectively, $g$ is the acceleration due to gravity and $\Omega_{t}=\{(x,y)\in\mathbb{R}^{2}:b(x)<y<\eta(t,x)\}$ . We put ourselves in the general situation described in (1). However, we consider the case of a bounded domain and the sloshing problem of describing the contact line between the free surface and the solid walls [14]. Two conditions are customary, the pinned–end boundary condition where the contact line is always pinned to the solid surface, as considered in [7, 16], and the free–end condition where the contact angle between the fluid–air interface and the side walls is fixed and the contact line is allowed to move [24]. We will consider both and will impose conditions on the Cauchy problem to deal with it, accordingly.

Controlling the surface by different methods is of practical interest in oceanography, controllability and inverse problems theory. We mention, for example, the work by Reid and Russell [26] where the authors dealt with the linear conservation laws and the null-controllability in infinite time of the free surface, by a source control, in a two dimensional domain with flat side walls. Also, the work by Reid, [25], where the capillary version and the control in finite time is considered. Concerning nonlinear water-waves, there is the recent work by Alazard [3], for a two dimensional rectangular domain, where the stabilization through an external pressure acting on a small part of the free surface is studied. Also [4], where the author studied the boundary observability problem in a three dimensional rectangular domain; namely, an estimate for the energy of the system in terms of the surface velocity at the contact line with a vertical wall. Finally, in [5], Alazard et al. addressed the local exact controllability of the two dimensional full water-waves system, by controlling a localized portion of the free surface, through the external pressure. On the literature concerning the generation of waves by wave-makers, controllability and stability properties in the water-waves context, we refer to [22, 27, 28]. From the optimal control point of view, we mention [23], where the authors designed the ‘best’ moving solid bottom generating a prescribed wave under the context of a BBM-type equation. We mention also the possibility of studying the inverse problem of detecting the source where jets originate, denoted as $J$ , by measuring the free surface as in [20].

When we talk about the controllability by fluid injection, for instance, we mean the condition

\frac{\partial\varphi}{\partial n}=J,

for a given function $J=J(t,x)$ , with $n$ being the outward normal vector. This boundary condition lead to think of a boundary control of the gravity waves problem; nevertheless, we will restate the problem as an integrodifferential equation on the free surface and the boundary condition becomes a source term. Then, we may use the classical approach of interior controllability by means of the adjoint problem and the observability inequality [21, 30].

By addressing this problem, we give an answer to a practical question raised in [9] and numerically studied in [15]; namely, the problem of controlling undesirable splashing appearing in a cooper converter when air is injected into the molten matte. In [15], the authors studied the problem by using triangular finite elements to mesh a half-ball bounded domain, on a damped linear gravity waves model. Our approach allows to consider any general simply connected two dimensional domain, through a conformal mapping into the lower half-plane. Moreover, if $f$ represents such a conformal mapping, the geometry is characterized explicitly by the term $1/|f^{\prime}|$ appearing as a factor in the evolution problem (see (47) below). On this matter, see [14], where oscillations are numerically computed for bottoms with rectangular even distributions.

Following the methods introduced in [14], after linearizing (1), the problem is restated through a conformal map into the lower half–plane. We link the normal derivative to the specific conformal map and rewrite the problem as a second order evolution equation on the interval $[-1,1]$ , where the Hilbert transform is involved. In [14], we used this approach to propose an efficient computational method to find the sloshing frequencies on general 2d domains. We mention [18] where the capillary version of the problem is developed, under the context of the oscillations in a nozzle of an inkjet printer. In these works, two possibilities for the contact line were considered: the ‘pinned-end edge condition’, where the contact line is always pinned to the solid surface, and the ‘free-end condition’ where the contact angle between the fluid–air interface and the side walls is fixed and the contact line is allowed to move, with contact angle $\pi/2$ .

With the present work, we have completed the numerical analysis started in [14]. Namely, we establish a Sobolev frame where the Cauchy problem is well-posed and study the eigenvalue associated problem. We also prove an observability inequality for the adjoint problem and explore the possibility of finding explicit controls taking the oscillations at the free surface to zero.

The rest of the paper is organized as follows: In section 2 we formulate the general equations to be considered, the linearization approach and corresponding formulation on the half-plane by the conformal mapping. In section 3 we use the Hilbert transform to state a second order evolutionary PDE modeling the dynamics of the fluid interface on the bounded domain $[-1,1]$ . In section 4 we make use of the Tchebyshev polynomials to study the stationary adjoint problem in suitable Sobolev spaces. In section 5 we prove an observability inequality for the adjoint problem. Finally, in section 6 we establish the controllability of the problem and explore the possibility of determining possible control functions, explicitly.

2. Formulation

Let us consider a two dimensional container, filled with water, bounded from above by a free surface. In this context, the motion is governed by the incompressible Euler equations with zero surface tension:

	$\displaystyle\nabla\cdot\mathbf{u}$	$\displaystyle=0,$
	$\displaystyle\rho\left[\frac{\partial\mathbf{u}}{\partial t}+(\mathbf{u}\cdot\nabla)\mathbf{u}\right]$	$\displaystyle=-\nabla p-ge_{2},$

with $-ge_{2}$ being the constant acceleration of gravity, $g>0$ , and $e_{2}$ the unit upward vector in the vertical direction, $\rho$ is the (constant) density of the fluid and $p$ is the pressure inside. We use the classical notation $(x,y)\in\mathbb{R}^{2}$ and $z=x+iy$ for complex numbers.

By considering the potential function $\varphi$ of $\mathbf{u}$ , so that $\mathbf{u}=\nabla\varphi$ :

	$\displaystyle\Delta\varphi$	$\displaystyle=0,$		(2)
	$\displaystyle\frac{\partial\varphi}{\partial t}+\frac{1}{2}\|\nabla\varphi\|^{2}+\frac{1}{\rho}p+gy$	$\displaystyle=\text{const}.$		(3)

We complement system above with the nonzero Neumann condition at the solid walls, $\mathbf{u}\cdot n=J(t,x)$ , and a kinematic condition on the free boundary. In terms of $\varphi$ :

	$\displaystyle\frac{\partial\varphi}{\partial n}$	$\displaystyle=J(t,x),\quad\text{at the solid boundary }$		(4)
	$\displaystyle\eta_{t}$	$\displaystyle=\frac{\partial\varphi}{\partial n},\quad\|x\|<1.$		(5)

2.1. Linearized equations

Next, we are going to linearize Euler’s system around the zero state. This will allow us to obtain a single, explicit, evolutionary conservation law on the free surface modeling the fluid dynamics of the free surface side bounded by vertical solid walls. This equation must be complemented with boundary conditions at the contact line, as explained in section 3.

Even if we consider the reference domain $\Omega=\{(x,y)\in\mathbb{D}:y^{\prime}<0\}$ , other domains can be considered though. Let

\eta(t,x)=\epsilon\zeta(t,x),

and

J(t,x)=\epsilon j(t,x).

Then if

\varphi=const.+\epsilon\phi,

conditions (2)-(4) in terms of $\phi$ , at the first order for $\epsilon<<1$ , become (after re-scaling to make $g=1$ )

$\displaystyle\Delta\phi$	$\displaystyle=0,$	(6)
$\displaystyle\phi_{t}+\zeta$	$\displaystyle=0,\quad\text{at }y=0,\ \|x\|\leq 1,$	(7)
$\displaystyle\zeta_{t}$	$\displaystyle=\frac{\partial\phi}{\partial n},\quad\text{at }y=0,\ \|x\|\leq 1,$	(8)
$\displaystyle\frac{\partial\phi}{\partial n}$	$\displaystyle=j(t,x),\quad\text{on the solid walls.}$	(9)

System above is complemented with initial conditions $\phi_{0}$ , $\zeta_{0}$ such that the following mass conservation is satisfied:

\int_{-1}^{1}\zeta_{0}(x)dx=0.

(10)

Therefore, from (7)-(8), on the free boundary and for $|x|\leq 1$ ,

\phi_{tt}+\frac{\partial\phi}{\partial n}=0.

(11)

Refer to caption — Figure 1. Geometry of the problem.

We remark that the motion of the fluid interface may also be affected by external forces such as electric and magnetic fields (cf. [10]), vibrational forces of the container (cf. [2]), etc. In those cases, the pressure $p$ at the interface is not constant, and this results in a nontrivial right hand side of equation (7), and hence in a nonhomogeneous version of (11):

\phi_{tt}+\frac{\partial\phi}{\partial n}=h(t,x).

(12)

2.2. Conformal transplants

As is explained in [6, 14], let $\Psi$ be a real-valued function written as

\Psi:(x,y)\to\Psi(x,y)=\Psi(z)

be defined in a domain $D$ . Also, let $\widetilde{\Psi}$ be defined in $\widetilde{D}$ as follows: for any $\omega\in\widetilde{D}$ we have

\widetilde{\Psi}(\omega):=\Psi(f^{[-1]}(\omega))=\Psi(x(x^{\prime},y^{\prime}),y(x^{\prime},y^{\prime})).

Then the following relation between the normal derivatives holds:

\frac{\partial\widetilde{\Psi}}{\partial\widetilde{n}}=\left|\frac{dz}{d\omega}\right|\frac{\partial\Psi}{\partial n}=\frac{1}{|f^{\prime}(z)|}\frac{\partial\Psi}{\partial n}.

(13)

2.3. Reformulation as an integrodifferential PDE

In the particular case of the half-cylinder geometry, $\Omega$ , we use the conformal map

\omega=f(z)=\frac{1}{\frac{1}{2}\left(z+\frac{1}{z}\right)},

into the half plane $\Omega^{\prime}:=\{(x^{\prime},y^{\prime})\in\mathbb{\ R}^{2}:y^{\prime}<0\}$ .

From (13), equations (6), (9), and (11), in variables $\omega=x^{\prime}+iy^{\prime}$ become (see Figure 1)

$\displaystyle\Delta\widetilde{\phi}$	$\displaystyle=0,\quad\text{for}\ y^{\prime}<0,$	(14)
$\displaystyle\widetilde{\phi}_{tt}+\|f^{\prime}(x^{\prime})\|\frac{\partial\widetilde{\phi}}{\partial\widetilde{n}}$	$\displaystyle=0,\quad\text{at }y^{\prime}=0,\ \|x^{\prime}\|\leq 1,$	(15)
$\displaystyle\frac{\partial\widetilde{\phi}}{\partial\widetilde{n}}$	$\displaystyle=\frac{\widetilde{j}(t,x^{\prime})}{\|f^{\prime}(x^{\prime})\|},\quad\text{at }y^{\prime}=0,\ \|x^{\prime}\|>1,$	(16)

where $j=\widetilde{j}\circ f$ . That is, for $|x^{\prime}|>1$ , $x^{\prime}=f(x)$ .

By relation (13) and since $|f^{\prime}(z)|\underset{z\to-i}{\longrightarrow}\infty$ , we complement system above with the following boundary conditions at infinity:

\partial_{x^{\prime}}\widetilde{\phi},\partial_{y^{\prime}}\widetilde{\phi}\to 0,\quad\text{as }y^{\prime}\to-\infty\ \text{or }|x^{\prime}|\to\infty.

By taking the Fourier transform in (14), in the variable $x^{\prime}$ ; using notation $\widetilde{\Phi}$ and the convention

\Phi{(k)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\phi(x^{\prime})e^{-ikx^{\prime}}dx^{\prime}},

we have

\widetilde{\Phi}_{y^{\prime}y^{\prime}}-k^{2}\widetilde{\Phi}=0,

which implies

	$\displaystyle\widetilde{\Phi}(t,k,y^{\prime})$	$\displaystyle=\widetilde{\Phi}(t,k,0)e^{\|k\|y^{\prime}}$		(17)
		$\displaystyle=\widetilde{\Phi}(t,k,0)\widehat{\frac{-\sqrt{2}y^{\prime}}{\sqrt{\pi}(x^{\prime 2}+y^{\prime 2})}}.$

By taking inverse Fourier transform

\widetilde{\phi}(t,x^{\prime},y^{\prime})=-\frac{y^{\prime}}{\pi}\int_{-\infty}^{+\infty}\frac{\widetilde{\phi}(t,\xi,0)}{(x^{\prime}-\xi)^{2}+y^{\prime 2}}d\xi.

Since we want to establish properties on the normal derivative, $\widetilde{\phi}_{y^{\prime}}$ , taking the $y^{\prime}$ derivative in (17) instead, and evaluating at $y^{\prime}=0$ we find

	$\displaystyle\widetilde{\Phi}_{y^{\prime}}(t,k,0)$	$\displaystyle=\frac{1}{i}\operatorname{sgn}(k)(ik)\widetilde{\Phi}(t,k,0),$
		$\displaystyle=\widehat{\frac{\sqrt{2}}{\sqrt{\pi}x^{\prime}}}\widetilde{\Phi}_{x^{\prime}}(t,k,0).$		(18)

Then, taking inverse Fourier transform

\left.\widetilde{\phi}_{y^{\prime}}\right|_{y^{\prime}=0}=\frac{1}{\pi}P.V.\int_{-\infty}^{+\infty}\frac{\widetilde{\phi}_{x^{\prime}}(t,\xi,0)}{x^{\prime}-\xi}d\xi=H\left(\left.\widetilde{\phi}_{x^{\prime}}\right|_{y^{\prime}=0}\right).

(19)

Since $HH=-I$ (see [17]):

\left.\widetilde{\phi}_{x^{\prime}}\right|_{y^{\prime}=0}=-\frac{1}{\pi}P.V.\int_{-\infty}^{\infty}\frac{\widetilde{\phi}_{y^{\prime}}(t,\xi,0)}{x^{\prime}-\xi}d\xi=-\frac{1}{\pi}P.V.\int_{-\infty}^{\infty}\frac{\frac{\partial\widetilde{\phi}}{\partial\widetilde{n}}(t,\xi,0)}{x^{\prime}-\xi}d\xi.

(20)

2.4. The Dirichlet-Neumann operator

Let us consider in this section the basic situation when $\frac{\partial\phi}{\partial n}=0$ , on the solid walls of the domain $\Omega$ . The value of $\frac{\partial\phi}{\partial n}$ at the fluid interface may be viewed as the result of an operator (the so-called Dirichlet-Neumann operator) acting on the function $\phi$ restricted to the interface. We are going to deduce a few simple consequences obtained from the properties of this operator. Keeping in mind the equation (8), we can easily deduce the following mass conservation relation:

\frac{d}{dt}\int_{-1}^{1}\zeta dx=\int_{-1}^{1}\frac{\partial\phi}{\partial n}dx=\int_{\partial\Omega}\frac{\partial\phi}{\partial n}=\int_{\Omega}\nabla\cdot\left(\nabla\phi\right)=0.

(21)

Notice then that, in the mapped coordinates, by (13)

0=\int_{-1}^{1}\frac{\partial\phi}{\partial n}dx=\int_{-1}^{1}\left|f^{\prime}(x^{\prime})\right|\frac{\partial\widetilde{\phi}}{\partial\widetilde{n}}\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}=\int_{-1}^{1}\frac{\partial\widetilde{\phi}}{\partial\widetilde{n}}dx^{\prime},

(22)

and hence, from (21) and (13)

\frac{d}{dt}\int_{-1}^{1}\widetilde{\zeta}\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}=0.

Next, since $\frac{\partial\phi}{\partial n}=0$ at the solid boundaries of $\Omega$ , we have

\int_{-1}^{1}\psi\frac{\partial\phi}{\partial n}dx=\int_{\partial\Omega}\psi\frac{\partial\phi}{\partial n}dx

for any harmonic function $\psi$ also vanishing at the solid boundaries of $\Omega$ , and therefore by Green’s identity we deduce

\int_{\partial\Omega}\psi\frac{\partial\phi}{\partial n}dx-\int_{\partial\Omega}\phi\frac{\partial\psi}{\partial n}dx=\int_{\Omega}(\psi\Delta\phi-\phi\Delta\psi)=0,

which implies

\int_{\partial\Omega}\psi\frac{\partial\phi}{\partial n}dx=\int_{\partial\Omega}\phi\frac{\partial\psi}{\partial n}dx,

or, equivalently,

\int_{-1}^{1}\widetilde{\psi}\frac{\partial\widetilde{\phi}}{\partial\widetilde{n}}dx^{\prime}=\int_{-1}^{1}\widetilde{\phi}\frac{\partial\widetilde{\psi}}{\partial\widetilde{n}}dx^{\prime},

showing that the Dirichlet-Neumann operator, mapping $\widetilde{\phi}$ at the free surface into $\frac{\partial\widetilde{\phi}}{\partial\widetilde{n}}$ , is formally selfadjoint. The details on the appropriate functional space where the operator is self-adjoint will be given in section 4.

Finally, from (7), (21), and the mass conservation property (10):

\int_{-1}^{1}\phi_{t}dx=-\int_{-1}^{1}\zeta dx=0,

(23)

which implies

\frac{d}{dt}\int_{-1}^{1}\phi(t,x)dx=0.

Therefore, by choosing $\phi(0,x)$ such that $\int_{-1}^{1}\phi(0,x)dx=0$ (this can always be achieved by adding a suitable constant to a given $\phi$ ), we have

\int_{-1}^{1}\phi(t,x)dx=0,

implying

\int_{-1}^{1}\widetilde{\phi}(t,x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}=0.

(24)

3. An associated Cauchy problem

As in the last section, let $\frac{\partial\phi}{\partial n}=0$ on the solid walls of the domain $\Omega$ . Then, from the integrodifferential formulation (20), we obtain

\left.\widetilde{\phi}_{x^{\prime}}\right|_{y^{\prime}=0}=-\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\widetilde{\phi}_{y^{\prime}}(t,\xi,0)}{x^{\prime}-\xi}d\xi.

(25)

We observe two properties of $\widetilde{\phi}_{x^{\prime}}$ . Firstly, by making use of the identity (see section 4.3 of [29])

\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-x^{\prime 2}}}\frac{dx^{\prime}}{x^{\prime}-\xi}=0,\ \text{for }\xi\in\left(-1,1\right),

(26)

we prove

$\displaystyle\int_{-1}^{1}\frac{\widetilde{\phi}_{x^{\prime}}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}dx^{\prime}$	$\displaystyle=-\frac{1}{\pi}\int_{-1}^{1}\frac{1}{\sqrt{1-x^{\prime 2}}}\left(P.V.\int_{-1}^{1}\frac{\widetilde{\phi}_{y^{\prime}}(\xi)}{x^{\prime}-\xi}d\xi\right)dx^{\prime}$	(27)
	$\displaystyle=-\frac{1}{\pi}\int_{-1}^{1}\widetilde{\phi}_{y^{\prime}}(\xi)\left(P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-x^{\prime 2}}(x^{\prime}-\xi)}dx^{\prime}\right)d\xi$
	$\displaystyle=0.$

Secondly, by the mass conservation $\int_{-1}^{1}\widetilde{\phi}_{y^{\prime}}(\xi)d\xi=0$ (see (22)), and using (see (51), for $r=1$ )

\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{x^{\prime}}{\sqrt{1-x^{\prime 2}}}\frac{dx^{\prime}}{x^{\prime}-\xi}=1,\ \text{for }\xi\in(-1,1),

we also have

$\displaystyle\int_{-1}^{1}\frac{x^{\prime}\widetilde{\phi}_{x^{\prime}}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}dx^{\prime}$	$\displaystyle=-\frac{1}{\pi}\int_{-1}^{1}\frac{x^{\prime}}{\sqrt{1-x^{\prime 2}}}\left(P.V.\int_{-1}^{1}\frac{\widetilde{\phi}_{y^{\prime}}(\xi)}{x^{\prime}-\xi}d\xi\right)dx^{\prime}$	(28)
	$\displaystyle=-\frac{1}{\pi}\int_{-1}^{1}\widetilde{\phi}_{y^{\prime}}(\xi)\left(P.V.\int_{-1}^{1}\frac{x^{\prime}}{\sqrt{1-x^{\prime 2}}(x^{\prime}-\xi)}dx^{\prime}\right)d\xi$
	$\displaystyle=-\int_{-1}^{1}\widetilde{\phi}_{y^{\prime}}(\xi)d\xi=0.$

By using the inverse transform instead, from (25) we have, for $|x^{\prime}|<1$ ,

\left.\widetilde{\phi}_{y^{\prime}}\right|_{y^{\prime}=0}=\frac{1}{\sqrt{1-x^{\prime 2}}}\frac{1}{\pi}P.V.\int_{-1}^{1}\sqrt{1-\xi^{2}}\frac{\widetilde{\phi}_{x^{\prime}}(t,\xi,0)}{x^{\prime}-\xi}d\xi+\frac{C}{\sqrt{1-x^{\prime 2}}},

(29)

where $C$ is an arbitrary constant (see [17], Chapter 5.2, Example 13 on the airfoil equation). In order for (29) to be the general solution to the integral equation (25) it suffices to require $\widetilde{\phi}_{x^{\prime}}(t,x^{\prime},0)$ to satisfy (see [17])

\|\widetilde{\phi}_{x^{\prime}}(t,x^{\prime},0)\|_{L_{(1-x^{\prime 2})^{1/2}}^{2}}^{2}\equiv\int_{-1}^{1}\sqrt{1-x^{\prime 2}}|\widetilde{\phi}_{x^{\prime}}(t,x^{\prime},0)|^{2}dx^{\prime}<\infty\text{.}

(30)

Finally, due to mass conservation and using (28), the constant must be chosen to be zero:

	$\displaystyle 0$	$\displaystyle=\int_{-1}^{1}\widetilde{\phi}_{y^{\prime}}(x^{\prime})dx^{\prime}$
		$\displaystyle=\int_{-1}^{1}\sqrt{1-\xi^{2}}\widetilde{\phi}_{x^{\prime}}(\xi)\left(\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-x^{\prime 2}}}\frac{1}{x^{\prime}-\xi}dx^{\prime}\right)d\xi+\pi C$
		$\displaystyle=\pi C.$

If, instead of (30), one assumes the stronger condition

\|\widetilde{\phi}_{x^{\prime}}(t,x^{\prime},0)\|_{L_{(1-x^{\prime 2})^{-1/2}}^{2}}^{2}\equiv\int_{-1}^{1}\frac{|\widetilde{\phi}_{x^{\prime}}(t,x^{\prime},0)|^{2}}{\sqrt{1-x^{\prime 2}}}dx^{\prime}<\infty\text{,}

(31)

together with (27), then (cf. [17], Chapter 5.2, Example 13)

\left.\widetilde{\phi}_{y^{\prime}}\right|_{y^{\prime}=0}=\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-\xi^{2}}}\frac{\widetilde{\phi}_{x^{\prime}}(t,\xi,0)}{x^{\prime}-\xi}d\xi.

As a consequence of the computations above, it is possible to state the following.

Lemma 1.

Given $\widetilde{\phi}_{x^{\prime}}\in L_{\sqrt{1-x^{\prime 2}}}^{2}(-1,1)$ , let $\widetilde{\phi}_{y^{\prime}}$ be satisfying (25) and the mass conservation condition $\int_{-1}^{1}\widetilde{\phi}_{y^{\prime}}(\xi)d\xi=0$ . Then, the following relations hold,

	$\displaystyle\left.\widetilde{\phi}_{y^{\prime}}\right\|_{y^{\prime}=0}$	$\displaystyle=\frac{1}{\sqrt{1-x^{\prime 2}}}\frac{1}{\pi}P.V.\int_{-1}^{1}\sqrt{1-\xi^{2}}\frac{\widetilde{\phi}_{x^{\prime}}(t,\xi,0)}{x^{\prime}-\xi}d\xi$		(32)
		$\displaystyle=\partial_{x^{\prime}}\left(\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-\xi^{2}}}\frac{\widetilde{\phi}(t,\xi,0)}{x^{\prime}-\xi}d\xi\right).$		(33)

Moreover, if $\widetilde{\phi}_{x^{\prime}}\in L_{(1-x^{\prime 2})^{-1/2}}^{2}(-1,1)\subset L_{\sqrt{1-x^{\prime 2}}}^{2}(-1,1)$ then $\widetilde{\phi}_{y^{\prime}}$ may also be given by the equivalent expression

\left.\widetilde{\phi}_{y^{\prime}}\right|_{y^{\prime}=0}=\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-\xi^{2}}}\frac{\widetilde{\phi}_{x^{\prime}}(t,\xi,0)}{x^{\prime}-\xi}d\xi.

(34)

Proof.

We present a proof based on expansions in terms of Tchebyshev polynomials in the Appendix A. ∎

Expression (34) was used in our previous article [14], while expression (33) will be more convenient in the present work. We present next an alternative deduction of (33). We introduce the function $\chi$ defined by

\widetilde{\phi}=\chi_{x^{\prime}}

so that

\chi=\int_{-\infty}^{x^{\prime}}\widetilde{\phi}(s,y^{\prime})ds,

(note that integrability at $-\infty$ is guarantied by the decay of $\widetilde{\phi}$ from (20) after integration by parts and the mass conservation (22)) and

\Delta\chi=0.

We have then

\chi_{y^{\prime}}=\frac{\partial}{\partial y^{\prime}}\int_{-\infty}^{x^{\prime}}\widetilde{\phi}dx^{\prime}=\int_{-\infty}^{x^{\prime}}(\chi_{y^{\prime}})_{x^{\prime}}dx^{\prime}=\begin{cases}0,&x^{\prime}\leq-1,\\ \chi_{y^{\prime}},&-1<x^{\prime}\leq 1,\\ 0,&x^{\prime}>1.\end{cases}

Since $\Delta\chi=0$ , by (25) we have

\chi_{x^{\prime}}(x^{\prime})=-\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\chi_{y^{\prime}}(\xi)}{x^{\prime}-\xi}d\xi.

(35)

Therefore, inverting Hilbert transform

\chi_{y^{\prime}}(x^{\prime})=\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-\xi^{2}}}\frac{\chi_{x^{\prime}}(\xi)}{x^{\prime}-\xi}d\xi,

(36)

and taking $x^{\prime}$ derivative

\widetilde{\phi}_{y^{\prime}}(x^{\prime})=\partial_{x^{\prime}}\left(\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-\xi^{2}}}\frac{\widetilde{\phi}(\xi)}{x^{\prime}-\xi}d\xi\right)

(37)

which, by (15), yields the following evolution problem

\widetilde{\phi}_{tt}=-\left|f^{\prime}(x^{\prime})\right|\partial_{x^{\prime}}\left(\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\widetilde{\phi}(t,\xi)}{\sqrt{1-\xi^{2}}(x^{\prime}-\xi)}d\xi\right).

(38)

The equation (38) may be rewritten as an identical equation for $\widetilde{\zeta}$ after taking an additional time derivative and writing $\widetilde{\phi}_{t}=-\widetilde{\zeta}$ . We remark again that the formulation (38) is equivalent to the formulation given in [14] (see Appendix A).

We need to complement (38) with suitable initial and boundary conditions, namely

	$\displaystyle\widetilde{\phi}(0,x^{\prime})$	$\displaystyle=\widetilde{\phi}_{0}(x^{\prime}),$
	$\displaystyle\widetilde{\phi}_{t}(0,x^{\prime})$	$\displaystyle=\widetilde{\phi}_{1}(x^{\prime}),$

where, by equation (23)

\int_{-1}^{1}\frac{\widetilde{\phi}_{1}(x^{\prime})}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}={\color[rgb]{0,0,1}-}\int_{-1}^{1}\frac{\widetilde{\zeta}(x^{\prime})}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}=0.

(39)

Moreover, from (27), (28), we impose on the initial data:

\int_{-1}^{1}\frac{\widetilde{\phi}_{0,x^{\prime}}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}dx^{\prime}=0,

(40)

\int_{-1}^{1}\frac{x^{\prime}\widetilde{\phi}_{0,x^{\prime}}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}dx^{\prime}=0,

(41)

which are automatically fulfilled by defining an initial normal derivative $\widetilde{\phi}_{0,y^{\prime}}(x^{\prime})$ with vanishing mean value in $\left[-1,1\right]$ and the corresponding tangential derivative $\widetilde{\phi}_{0,x^{\prime}}(x^{\prime})$ defined by (25).

In the case when $\left|f^{\prime}(x^{\prime})\right|$ is a symmetric function (corresponding to a symmetric domain $\Omega$ ) it is useful to think of $\widetilde{\phi}(t,x^{\prime})$ as decomposed into symmetric and antisymmetric part; that is

\widetilde{\phi}(t,x^{\prime})=S(t,x^{\prime})+N(t,x^{\prime})

with

S(t,x)=\frac{\widetilde{\phi}(t,x)+\widetilde{\phi}(t,-x)}{2},\quad N(t,x)=\frac{\widetilde{\phi}(t,x)-\widetilde{\phi}(t,-x)}{2},

and the initial data decomposed accordingly

	$\displaystyle S_{0}(x)$	$\displaystyle=\frac{\widetilde{\phi}_{0}(x)+\widetilde{\phi}_{0}(-x)}{2},\quad N_{0}(x)=\frac{\widetilde{\phi}_{0}(x)-\widetilde{\phi}_{0}(-x)}{2},$
	$\displaystyle S_{1}(x)$	$\displaystyle=\frac{\widetilde{\phi}_{1}(x)+\widetilde{\phi}_{1}(-x)}{2},\quad N_{1}(x)=\frac{\widetilde{\phi}_{1}(x)-\widetilde{\phi}_{1}(-x)}{2}.$

Then, one can consider the evolution problem for symmetric and antisymmetric functions separately. For symmetric functions, the initial data need to satisfy (39), (41) while for antisymmetric functions the conditions are (39), (40). Finally, we remark that (39), (40), (41) do not only hold initially, but for any time by replacing $(\widetilde{\phi}_{0}(x^{\prime}),\widetilde{\phi}_{1}(x^{\prime}))$ by $(\widetilde{\phi}(t,x^{\prime}),\widetilde{\phi}_{t}(t,x^{\prime}))$ .

We discuss now on boundary conditions for (38). Two kind of conditions are customary: pinned end and free end boundary conditions. In pinned end conditions one imposes $\widetilde{\zeta}(t,\pm 1)=0$ , and since $\widetilde{\phi}_{t}=-\widetilde{\zeta}$ , this implies

\widetilde{\phi}(t,\pm 1)=0\ \text{(pinned-end boundary condition)}.

(42)

In the case of free-end boundary conditions one imposes $\widetilde{\zeta}_{x^{\prime}}(t,\pm 1)=0$ and since $\widetilde{\phi}_{x^{\prime}t}=\widetilde{\zeta}_{x^{\prime}}$ this translates into the condition

\widetilde{\phi}_{x^{\prime}}(t,\pm 1)=0\ \ \text{(free-end boundary condition)}.

(43)

We discuss now the case with fluid injection, i.e. with the condition (16) where the flux $j$ is such that $\ \int\frac{\widetilde{j}(t,x^{\prime})}{|f^{\prime}(x^{\prime})|}dx^{\prime}=0$ (in order to preserve the total fluid mass). Then, by (20), equation (25) needs to be replaced by

\left.\widetilde{\phi}_{x^{\prime}}\right|_{y^{\prime}=0}=-\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\widetilde{\phi}_{y^{\prime}}(t,\xi,0)}{x^{\prime}-\xi}d\xi+\frac{1}{\pi}\int_{\mathbb{R}\setminus[-1,1]}\frac{\widetilde{j}(t,z)}{|f^{\prime}(z)|(\xi-z)}dz.

Inverting as in (32), (33) we get the following formula for the normal derivative:

	$\displaystyle\left.\widetilde{\phi}_{y^{\prime}}\right\|_{y^{\prime}=0}$	$\displaystyle=\partial_{x^{\prime}}\left(\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-\xi^{2}}}\frac{\widetilde{\phi}(t,\xi,0)}{x^{\prime}-\xi}d\xi\right)$
		$\displaystyle\phantom{=}-\frac{1}{\sqrt{1-x^{\prime 2}}}\frac{1}{\pi^{2}}P.V.\int_{-1}^{1}\frac{\sqrt{1-\xi^{2}}}{(x^{\prime}-\xi)}\int_{\mathbb{R}\setminus[-1,1]}\frac{\widetilde{j}(t,z)}{\|f^{\prime}(z)\|(\xi-z)}dzd\xi,$		(44)

which leads to the evolution equation

\widetilde{\phi}_{tt}=-\left|f^{\prime}(x^{\prime})\right|\partial_{x^{\prime}}\left(\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\widetilde{\phi}(t,\xi)}{\sqrt{1-\xi^{2}}(x^{\prime}-\xi)}d\xi\right)+\widetilde{h}(t,x^{\prime}),

(45)

where

\widetilde{h}(t,x^{\prime})=\frac{\left|f^{\prime}(x^{\prime})\right|}{\sqrt{1-x^{\prime 2}}}\frac{1}{\pi^{2}}P.V.\int_{-1}^{1}\frac{\sqrt{1-\xi^{2}}}{(x^{\prime}-\xi)}\int_{\mathbb{R}\setminus[-1,1]}\frac{\widetilde{j}(t,z)}{|f^{\prime}(z)|(\xi-z)}dzd\xi.

(46)

Notice that $\chi_{y^{\prime}}=\int_{-\infty}^{x^{\prime}}\widetilde{\phi}_{y^{\prime}}dx^{\prime}$ implies expression (46) can be obtained by replacing (35) with

\chi_{x^{\prime}}(x^{\prime})=-\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\chi_{y^{\prime}}(\xi)}{x^{\prime}-\xi}d\xi-\frac{1}{\pi}P.V.\int_{\mathbb{R}\setminus[-1,1]}\frac{J(\xi)}{x^{\prime}-\xi}d\xi

where $J$ is the primitive of $\widetilde{j}/|f^{\prime}|$ , inverting the Hilbert transform (restricted to $\left[-1,1\right]$ ) in the first term at the right hand side as in (36), (37) and using finally (32)–(33).

As mentioned above, equation (45), for given $\widetilde{h}(t,x^{\prime})$ , is also valid in situations where the interface is actuated by means of external forces such as electric and magnetic ones, external container vibration, etc. Hence, we will present a general discussion on controllability for general $\widetilde{h}(t,x^{\prime})$ and will only specify for boundary injection in the final section.

4. Spectrum of the sloshing problem

From now on, from equation (45) and to summarize the computations from the preceding sections, we are concerned with the following Initial Value Problem,

\begin{cases}\frac{\widetilde{\phi}_{tt}}{|f^{\prime}|}+\mathcal{A}\widetilde{\phi}=\widetilde{h}(t,x^{\prime}),\quad&(t,x^{\prime})\in(0,\infty)\times(-1,1),\\ \widetilde{\phi}(0,x^{\prime})=\widetilde{\phi}_{0}(x^{\prime}),&x^{\prime}\in(-1,1),\\ \widetilde{\phi}_{t}(0,x^{\prime})=\widetilde{\phi}_{1}(x^{\prime}),&x^{\prime}\in(-1,1),\end{cases}

(47)

where $\mathcal{A}$ is the integral, non-local operator

\mathcal{A}\widetilde{\phi}\equiv\partial_{x^{\prime}}\left(\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\widetilde{\phi}(\xi)}{\sqrt{1-\xi^{2}}(x^{\prime}-\xi)}d\xi\right).

Following [21], to study the interior controllability problem (47), we need to consider the homogeneous (backward in time) adjoint version in $(0,T)$ as that given in (38). For that purpose, let us consider first the eigenvalue problem

\lambda\frac{\widetilde{\phi}}{\left|f^{\prime}(x^{\prime})\right|}=\partial_{x^{\prime}}\left(\sqrt{1-x^{\prime 2}}P.V.\frac{1}{\pi}\int_{-1}^{1}\frac{\widetilde{\phi}(\xi)}{\sqrt{1-\xi^{2}}(x^{\prime}-\xi)}d\xi\right).

(48)

Let $T_{n}(x)$ , $U_{n}(x)$ , with $n\in\mathbb{N}\cup\{0\}$ , be the Tchebyshev polynomials of the first and second kind respectively. They are defined as the polynomial solutions of the equations (see [1] for details)

	$\displaystyle T_{n}(\cos\theta)$	$\displaystyle=\cos(n\theta),$
	$\displaystyle U_{n}(\cos\theta)$	$\displaystyle=\frac{\sin((n+1)\theta)}{\sin\theta}.$

They satisfy the following orthogonality relations in $L^{2}(-1,1)$ with the corresponding inner products $\langle f,g\rangle=\int_{-1}^{1}\frac{fg}{\sqrt{1-x^{2}}}dx$ , $\langle f,g\rangle=\int_{-1}^{1}fg\sqrt{1-x^{2}}dx$ :

\int_{-1}^{1}T_{n}(x)T_{m}(x)\frac{dx}{\sqrt{1-x^{2}}}=\begin{cases}0,\ \text{if }n\neq m,\\ \pi\ \text{if }n=m=0,\\ \frac{\pi}{2}\ \text{if }n=m\neq 0,\end{cases}

(49)

\int_{-1}^{1}U_{n}(x)U_{m}(x)\sqrt{1-x^{2}}dx=\begin{cases}0,\ \text{if }n\neq m,\\ \frac{\pi}{2}\ \text{if }n=m.\end{cases}

(50)

Moreover, for $n\geq 1$ , we have relations

	$\displaystyle\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{T_{n}(\xi)}{\sqrt{1-\xi^{2}}(x^{\prime}-\xi)}d\xi$	$\displaystyle=-U_{n-1}(x^{\prime}),$		(51)
	$\displaystyle\frac{d}{dx^{\prime}}\left(\sqrt{1-x^{\prime 2}}U_{n-1}(x^{\prime})\right)$	$\displaystyle=-r\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}.$

By writing

\widetilde{\phi}(\xi)=\sum_{n=0}^{\infty}a_{n}T_{n}(\xi),

(52)

and using the identities above, we find

\partial_{x^{\prime}}\left(\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\widetilde{\phi}(\xi)}{\sqrt{1-\xi^{2}}(x^{\prime}-\xi)}d\xi\right)=\sum_{n=1}^{\infty}na_{n}\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}.

(53)

We use now the fact that $\int_{-1}^{1}\widetilde{\phi}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}=0$ (see (24)), to deduce

a_{0}\int_{-1}^{1}T_{0}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}+\sum_{n=1}^{\infty}a_{n}\int_{-1}^{1}T_{n}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}=0.

So that

a_{0}=-\frac{1}{\int_{-1}^{1}T_{0}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}}\sum_{n=1}^{\infty}a_{n}\int_{-1}^{1}T_{n}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|},

and conclude the estimate

a_{0}^{2}\leq\frac{\sum_{n=1}^{\infty}\left(\int_{-1}^{1}T_{n}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}\right)^{2}}{\left(\int_{-1}^{1}T_{0}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}\right)^{2}}\sum_{n=1}^{\infty}a_{n}^{2}.

Now, let $c_{n}$ be such that

\frac{\sqrt{1-x^{\prime 2}}}{|f^{\prime}(x^{\prime})|}=\sum_{n=0}^{\infty}c_{n}T_{n}(x^{\prime}).

Then, by the orthogonality of the Tchebyshev polynomials,

c_{0}=\frac{1}{\pi}\int_{-1}^{1}\frac{T_{0}(x^{\prime})}{|f^{\prime}(x^{\prime})|}dx^{\prime},\quad c_{n}=\frac{2}{\pi}\int_{-1}^{1}\frac{T_{n}(x^{\prime})}{|f^{\prime}(x^{\prime})|}dx^{\prime}.

Therefore

\displaystyle\int_{-1}^{1}\frac{1-x^{\prime 2}}{|f^{\prime}(x^{\prime})|^{2}}\frac{1}{\sqrt{1-x^{\prime 2}}}dx^{\prime}=\sum_{n,m=0}^{\infty}c_{n}c_{m}\int_{-1}^{1}\frac{T_{n}T_{m}}{\sqrt{1-x^{\prime 2}}}dx^{\prime}=\pi c_{0}^{2}+\frac{\pi}{2}\sum_{n=1}^{\infty}c_{n}^{2},

namely

\frac{1}{\pi}\left(\int_{-1}^{1}T_{0}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}\right)^{2}+\frac{2}{\pi}\sum_{n=1}^{\infty}\left(\int_{-1}^{1}T_{n}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}\right)^{2}=\int_{-1}^{1}\sqrt{1-x^{\prime 2}}\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|^{2}}<\infty,

which implies

a_{0}^{2}\leq C\sum_{n=1}^{\infty}a_{n}^{2}.

(54)

Let

\widetilde{\psi}=\sum_{n=0}^{\infty}b_{n}T_{n}(x^{\prime}).

(55)

We consider now the scalar product

(\mathcal{A}\widetilde{\phi},\widetilde{\psi})_{L^{2}}=\int_{-1}^{1}\widetilde{\psi}\frac{\partial\widetilde{\phi}}{\partial\widetilde{n}}dx^{\prime},

and find

(\mathcal{A}\widetilde{\phi},\widetilde{\psi})_{L^{2}}=\int_{-1}^{1}\left(\sum_{m=0}^{\infty}b_{m}T_{m}(x^{\prime})\right)\left(\sum_{n=1}^{\infty}na_{n}\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}\right)dx^{\prime}=\frac{\pi}{2}\sum_{n=1}^{\infty}na_{n}b_{n},

(56)

so that, $(\widetilde{\psi},\mathcal{A}\widetilde{\phi})_{L^{2}}=(\mathcal{A}\widetilde{\psi},\widetilde{\phi})_{L^{2}}$ , implying the selfadjoint character of the operator $\mathcal{A}$ , a fact already shown in a previous section.

We study the problem

\mathcal{A}\widetilde{\phi}=u,

(57)

where $u\in L_{w}^{2}$ with $w=\sqrt{1-x^{\prime 2}}$ and

L_{w}^{2}\equiv\left\{u:\int_{-1}^{1}\sqrt{1-x^{\prime 2}}\left|u\right|^{2}dx^{\prime}<\infty\right\}.

Since $\left\{T_{n}(x^{\prime}){/\sqrt{1-x^{\prime 2}}}\right\}_{n=0}^{\infty}$ form an orthogonal base for $L_{w}^{2}$ , given $u\in L_{w}^{2}$ we can expand

u=\sum_{n=0}^{\infty}u_{n}\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}},

and hence, assuming

u_{0}=\int_{-1}^{1}u(x^{\prime})dx^{\prime}=0,

(58)

write the following weak version of (57)

(\mathcal{A}\widetilde{\phi},\widetilde{\psi})_{L^{2}}=(u,\widetilde{\psi})_{L^{2}},

(59)

to be satisfied for any $\widetilde{\psi}$ in

H_{w^{-1}}^{\frac{1}{2}}\equiv\left\{\widetilde{\psi}:\left\|\widetilde{\psi}\right\|_{H_{w^{-1}}^{\frac{1}{2}}}^{2}=\left\|\widetilde{\psi}\right\|_{L_{w^{-1}}^{2}}^{2}+\sum_{n=1}^{\infty}n\left(\int_{-1}^{1}\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}\widetilde{\psi}(x^{\prime})dx^{\prime}\right)^{2}<\infty\right\}

(60)

such that, in addition,

\int_{-1}^{1}\widetilde{\psi}(x^{\prime})\frac{dx^{\prime}}{\left|f^{\prime}(x^{\prime})\right|}=0.

(61)

By using the preceding functional framework, we can state a result on existence of weak solutions for problem (57).

Lemma 2.

Let $u\in L^{2}_{w}$ satisfying (58). Then, there exists a unique weak solution $\widetilde{\phi}\in H_{w^{-1}}^{1/2}$ for problem (57).

Proof.

We can write the equation (59) in the form

\sum_{n=1}^{\infty}na_{n}b_{n}=\sum_{n=1}^{\infty}u_{n}b_{n},

(62)

making it clear that $(\mathcal{A}\widetilde{\phi},\widetilde{\psi})_{L^{2}}$ defines a continuous bilinear form on $H_{w^{-1}}^{\frac{1}{2}}$ . Indeed,

Remark 1.

By using the orthogonality property (49) on (55) we have, for $n\geq 1$ ,

\sqrt{n}\int_{-1}^{1}\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}\widetilde{\psi}(x^{\prime})dx^{\prime}=\frac{\pi}{2}\sqrt{n}b_{n},

or, equivalently,

\sum_{n=1}^{\infty}n\left(\int_{-1}^{1}\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}\widetilde{\psi}(x^{\prime})dx^{\prime}\right)^{2}=\frac{\pi^{2}}{4}\sum_{n=1}^{\infty}n|b_{n}|^{2}.

Notice, from this last equality and (56), that $H_{w^{-1}}^{1/2}$ can be written as

H_{w^{-1}}^{\frac{1}{2}}\equiv\left\{\widetilde{\psi}:\left\|\widetilde{\psi}\right\|_{H_{w^{-1}}^{\frac{1}{2}}}^{2}=\left\|\widetilde{\psi}\right\|_{L_{w^{-1}}^{2}}^{2}+\frac{\pi}{2}(\mathcal{A}\widetilde{\psi},\widetilde{\psi})_{L^{2}}<\infty\right\}.

Thus from (56) and the last remark, we have

	$\displaystyle\|(\mathcal{A}\widetilde{\phi},\widetilde{\psi})_{L^{2}}\|$	$\displaystyle=\frac{\pi}{2}\left\|\sum_{n\geq 1}na_{n}b_{n}\right\|\leq\frac{\pi}{2}\left(\sum_{n\geq 1}n\|a_{n}\|^{2}\right)^{1/2}\left(\sum_{n\geq 1}n\|b_{n}\|^{2}\right)^{1/2}$
		$\displaystyle\leq\frac{2}{\pi}\\|\widetilde{\phi}\\|_{H_{w^{-1}}^{1/2}}\\|\widetilde{\psi}\\|_{H_{w^{-1}}^{1/2}}.$

$(\mathcal{A}\widetilde{\phi},\widetilde{\psi})_{L^{2}}$ is also coercive (the $L_{w^{-1}}^{2}$ part of the norm of $\widetilde{\phi}$ is trivially bounded by $\sum_{n=1}^{\infty}na_{n}^{2}$ , except for $a_{0}^{2}$ that is also bounded by (54)):

Remark 2.

By using the orthogonality property (49) on (52),

\|\widetilde{\phi}\|_{L_{w^{-1}}^{2}}^{2}=\int_{-1}^{1}\frac{|\widetilde{\phi}|^{2}}{\sqrt{1-x^{\prime 2}}}dx^{\prime}=\pi a_{0}^{2}+\frac{\pi}{2}\sum_{n=1}^{\infty}a_{n}^{2}.

Moreover,

a_{n}=\begin{cases}\frac{1}{\pi}\int_{-1}^{1}\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}\widetilde{\phi}(x^{\prime})dx^{\prime},\quad&n=0,\\ \frac{2}{\pi}\int_{-1}^{1}\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}\widetilde{\phi}(x^{\prime})dx^{\prime},&n\geq 1.\end{cases}

Then, from the last remark and (54), coerciveness holds as follows:

$\displaystyle(\mathcal{A}\widetilde{\phi},\widetilde{\phi})_{L^{2}}$	$\displaystyle=\frac{\pi}{2}\sum_{n\geq 1}n\|a_{n}\|^{2}\geq C\left[\pi\sum_{n=1}^{\infty}a_{n}^{2}+\frac{\pi}{2}\sum_{n=1}^{\infty}a_{n}^{2}+\frac{\pi^{2}}{4}\sum_{n=1}^{\infty}na_{n}^{2}\right]$	(63)
	$\displaystyle\geq C\left[\pi a_{0}^{2}+\frac{\pi}{2}\sum_{n=1}^{\infty}a_{n}^{2}+\frac{\pi^{2}}{4}\sum_{n=1}^{\infty}na_{n}^{2}\right]$
	$\displaystyle=C\left[\\|\widetilde{\phi}\\|_{L_{w^{-1}}^{2}}^{2}+\frac{\pi}{2}(\mathcal{A}\widetilde{\phi},\widetilde{\phi})\right]$
	$\displaystyle=C\\|\widetilde{\phi}\\|_{H_{w^{-1}}^{1/2}}^{2},$

for a constant $C>0$ .

Finally, $u\in L_{w}^{2}$ defines a linear continuous functional on $H_{w^{-1}}^{1/2}$ :

	$\displaystyle\left\|\int_{-1}^{1}u\widetilde{\phi}dx^{\prime}\right\|$	$\displaystyle\leq\int_{-1}^{1}\sqrt[4]{1-x^{\prime 2}}\|u\|\frac{\widetilde{\phi}}{\sqrt[4]{1-x^{\prime 2}}}dx^{\prime}$
		$\displaystyle\leq\\|u\\|_{L_{w}^{2}}\\|\widetilde{\phi}\\|_{L_{w^{-1}}^{2}}$
		$\displaystyle\leq\\|u\\|_{L_{w}^{2}}\\|\widetilde{\phi}\\|_{H_{w^{-1}}^{1/2}}.$

Hence, by Lax-Milgram’s theorem, there exists a unique weak solution $\widetilde{\phi}$ to (57) that belongs to $H_{w^{-1}}^{\frac{1}{2}}$ . ∎

If we consider now $u=\frac{v}{\left|f^{\prime}(x^{\prime})\right|}$ then the problem

\mathcal{A}\widetilde{\phi}=\frac{v}{\left|f^{\prime}(x^{\prime})\right|}

where $v$ is such that

\int_{-1}^{1}\sqrt{1-x^{\prime 2}}\frac{v^{2}(x^{\prime})}{\left|f^{\prime}(x^{\prime})\right|^{2}}dx^{\prime}<\infty

(64)

has also a unique solution $\widetilde{\phi}\in H_{w^{-1}}^{\frac{1}{2}}$ . Note that (64) is satisfied if $v\in L_{\left|f^{\prime}(x^{\prime})\right|^{-1}}^{2}$ provided there exists a constant $C$ such that

\left|f^{\prime}(x^{\prime})\right|\geq C\sqrt{1-x^{\prime 2}},\quad x^{\prime}\in[-1,1],

(65)

because

\int_{-1}^{1}\sqrt{1-x^{\prime 2}}\frac{v^{2}(x^{\prime})}{\left|f^{\prime}(x^{\prime})\right|^{2}}dx^{\prime}\leq\frac{1}{C^{2}}\int_{-1}^{1}\frac{v^{2}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}dx^{\prime},

\left\|u\right\|_{L_{|f^{\prime}|^{-1}}^{2}}^{2}\leq\frac{1}{C}\left\|u\right\|_{L_{w^{-1}}^{2}}^{2}.

We define now the operator $\mathcal{T}$ such that $\widetilde{\phi}=\mathcal{T}v$ . If we view $\mathcal{T}$ as an operator from $L_{\left|f^{\prime}\right|^{-1}}^{2}$ to $L_{\left|f^{\prime}\right|^{-1}}^{2}$ , since $H_{w^{-1}}^{\frac{1}{2}}\subset L_{w^{-1}}^{2}\subset L_{\left|f^{\prime}\right|^{-1}}^{2}$ , we can see that $\mathcal{T}$ is a selfadjoint operator (due to the selfadjoint character of $\mathcal{A}$ ) and also a compact operator (due to the compact embedding $H_{w^{-1}}^{\frac{1}{2}}\subset L_{w^{-1}}^{2}$ (see Appendix B) and the continuous inclusion $L_{w^{-1}}^{2}\subset L_{\left|f^{\prime}\right|^{-1}}^{2}$ ) . Moreover, $\ker(\mathcal{T})=\left\{0\right\}$ and

(\mathcal{T}v,v)_{L_{|f^{\prime}|^{-1}}^{2}}=(\widetilde{\phi},|f^{\prime}|\mathcal{A}\widetilde{\phi})_{L_{|f^{\prime}|^{-1}}^{2}}=(\widetilde{\phi},\mathcal{A}\widetilde{\phi})_{L^{2}}\geq 0,

for any $v\in L_{\left|f^{\prime}\right|^{-1}}^{2}$ . Therefore, by the spectral decomposition theorem, $L_{\left|f^{\prime}\right|^{-1}}^{2}$ admits a Hilbert basis $\left\{e_{n}\right\}$ formed by eigenvectors of $\mathcal{T}$ , with eigenvalues $\mu_{n}$ such that $\mu_{n}>0$ and $\mu_{n}\rightarrow 0$ as $n\rightarrow\infty$ . We have then $\mathcal{T}e_{n}=\mu_{n}e_{n}$ , $e_{n}\in H_{w^{-1}}^{\frac{1}{2}}$ , and thus

{(\mathcal{A}e_{n},\widetilde{\psi})_{L^{2}}=\frac{1}{\mu_{n}}(\mathcal{A}(\mathcal{T}e_{n}),\widetilde{\psi})_{L^{2}}=\frac{1}{\mu_{n}}\left(\frac{e_{n}}{|f^{\prime}|},\widetilde{\psi}\right)_{L^{2}},}\quad\text{for any }\widetilde{\psi}\in H_{w^{-1}}^{\frac{1}{2}}.

(66)

This implies $e_{n}$ is a weak solution to the eigenvalue problem (48) with $\lambda=\lambda_{n}=\mu_{n}^{-1}$ . The base can be made orthonormal, i.e.

\int_{-1}^{1}\frac{e_{i}e_{j}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}=\delta_{ij}.

Let us remark that condition (65) is satisfied provided the angles $\alpha_{1,2}$ between the fluid interface and the solid container satisfy

\alpha_{1,2}\geq\frac{\pi}{2},

since $|f^{\prime}(x^{\prime})|$ has a singularity weaker (i.e. with larger exponent) than $|x\pm 1|^{1/2}$ otherwise [6].

We summarize the results above in the following theorem.

Theorem 1.

Let $\Omega$ be a domain such that the interior angles between the free liquid interface and the solid are greater or equal than $\pi/2$ , so that (65) is fulfilled. There exist a Hilbert basis $\{e_{n}\}_{n\geq 1}$ of $L_{\left|f^{\prime}\right|^{-1}}^{2}$ and a sequence $\{\lambda_{n}\}_{n\geq 1}$ of real numbers with $\lambda_{n}>0$ $\forall n$ and $\lambda_{n}\rightarrow+\infty$ such that

e_{n}\in H_{w^{-1}}^{\frac{1}{2}},

\mathcal{A}e_{n}=\lambda_{n}\frac{e_{n}}{\left|f^{\prime}(x^{\prime})\right|}.

We will show next that the eigenfunctions $e_{n}$ possess further regularity. From (52)-(53), if $e_{n}(x^{\prime})=\sum_{m=0}^{\infty}a_{m}T_{m}(x^{\prime})$ , by using (49) we have

\int_{-1}^{1}\sqrt{1-x^{\prime 2}}\left|\mathcal{A}e_{n}\right|^{2}dx^{\prime}=\frac{\pi}{2}\sum_{m=1}^{\infty}m^{2}a_{m}^{2}.

Therefore, by using condition (65),

	$\displaystyle\int_{-1}^{1}\sqrt{1-x^{\prime 2}}\left\|\mathcal{A}e_{n}\right\|^{2}dx^{\prime}$	$\displaystyle=\lambda_{n}^{2}\int_{-1}^{1}\sqrt{1-x^{\prime 2}}\frac{\left\|e_{n}\right\|^{2}}{\left\|f^{\prime}(x^{\prime})\right\|^{2}}dx^{\prime}$
		$\displaystyle\leq C_{n}\int_{-1}^{1}\frac{\left\|e_{n}\right\|^{2}}{\sqrt{1-x^{\prime 2}}}dx^{\prime}=C_{n}\\|e_{n}\\|_{L_{w^{-1}}^{2}}^{2},$

we have that $\sum_{m=1}^{\infty}m^{2}a_{m}^{2}$ is bounded by $\|e_{n}\|_{L^{2}_{w^{-1}}}^{2}$ . On the other hand, since $\frac{dT_{m}(x^{\prime})}{dx^{\prime}}=mU_{m-1}(x^{\prime})$ , from the orthogonality property (50) we deduce

\int_{-1}^{1}\sqrt{1-x^{\prime 2}}\left|\frac{de_{n}}{dx^{\prime}}\right|^{2}dx^{\prime}=\frac{\pi}{2}\sum_{m=1}^{\infty}m^{2}a_{m}^{2},

which is bounded by $\|e_{n}\|_{L_{w^{-1}}^{2}}^{2}$ . Hence, $e_{n}\in H_{w^{-1}}^{1}$ implying, by Sobolev embeddings, that $e_{n}(x^{\prime})$ is a continuous function in $(-1,1)$ . Moreover, it is bounded at $x=\pm 1$ since:

e_{n}(x^{\prime})-e_{n}(-1)=\int_{-1}^{x^{\prime}}e_{n}^{\prime}(x)dx\leq\left(\int_{-1}^{x^{\prime}}\frac{1}{\sqrt{1-x^{2}}}dx\right)^{\frac{1}{2}}\left(\int_{-1}^{x^{\prime}}\sqrt{1-x^{2}}\left|e_{n}^{\prime}(x)\right|^{2}dx\right)^{\frac{1}{2}}.

So that

\sup_{-1\leq x^{\prime}\leq-1+\delta}\frac{\left|e_{n}(x^{\prime})-e_{n}(-1)\right|^{2}}{\left|1+x^{\prime}\right|^{\frac{1}{2}}}\leq C\int_{-1}^{1}\sqrt{1-x^{2}}\left|e_{n}^{\prime}\right|^{2}dx,

and hence

\left|e_{n}(x^{\prime})-e_{n}(-1)\right|\leq C\left|1+x^{\prime}\right|^{\frac{1}{4}}.

Identical estimate may be obtained in the neighborhood of $x^{\prime}=1$ .

Note that no condition has been imposed on the eigenfunctions at $x^{\prime}=\pm 1$ . The eigenfunctions correspond to free-end boundary conditions. We can also consider the pinned-end boundary condition by writing the same weak formulation (59) but assuming that $\widetilde{\psi}$ belongs to the closure of $C_{c}^{\infty}(-1,1)$ in the topology defined by the $H_{w^{-1}}^{\frac{1}{2}}$ norm and satisfying (61). We denote such space as $H_{w^{-1},0}^{\frac{1}{2}}$ . The same arguments as for the free-end case lead to the existence of a complete set of eigenfunctions as in the Theorem above.

As a final remark, notice that for bounded symmetric $\left|f^{\prime}(x^{\prime})\right|$ , one has $L_{\left|f^{\prime}(x^{\prime})\right|^{-1}}^{2}\subset L^{2}$ and hence vectors in $L_{\left|f^{\prime}\right|^{-1}}^{2}$ may be expanded in trigonometric basis of $L^{2}$ . Using the base $\left\{\sin(n\pi x^{\prime})\right\}_{n=1}^{\infty}$ one can approximate antisymmetric eigenfunctions (solutions to (48)) with pinned contact lines while the set $\left\{\cos\left(\left(n+\frac{1}{2}\right)\pi x^{\prime}\right)\right\}_{n=0}^{\infty}$ (with the extra mass conservation condition also imposed) allows to approximate symmetric eigenfunctions. Likewise, the sets $\left\{\sin\left(\left(n+\frac{1}{2}\right)\pi x^{\prime}\right)\right\}_{n=0}^{\infty}$ and $\left\{\cos(n\pi x^{\prime})\right\}_{n=1}^{\infty}$ allow to find antisymmetric and symmetric eigenfunctions with free-end condition. This approach was followed in [14], in order to compute eigenvalues and eigenfunctions numerically for various domains.

5. Observability and energy estimates

We will consider next the control problem consisting of finding $\widetilde{h}(t,x^{\prime})$ in equation (45) such that for given initial data the solution vanishes at some time $T$ . As it is customary in controllability theory, we need to consider first the homogeneous adjoint problem and obtain observability estimates.

The analysis presented above allows us to consider the solution in $(0,T)$ of the following homogeneous adjoint problem (see (38))

\frac{\widetilde{\phi}_{tt}}{\left|f^{\prime}\right|}+\mathcal{A}\widetilde{\phi}=0,

(67)

with suitable initial conditions $\widetilde{\phi}(0)=\widetilde{\phi}_{0}$ , $\widetilde{\phi}^{\prime}(0)=\widetilde{\phi}_{1}$ , whose solution we can write as

\widetilde{\phi}=\sum_{n=1}^{\infty}\left(A_{n}\cos(\theta_{n}t)+B_{n}\sin(\theta_{n}t)\right)e_{n},

(68)

where $\theta_{n}=\sqrt{\lambda_{n}}$ and

\widetilde{\phi}_{0}=\sum_{n=1}^{\infty}A_{n}e_{n},\quad\widetilde{\phi}_{1}=\sum_{n=1}^{\infty}\theta_{n}B_{n}e_{n},

with

A_{n}=\int_{-1}^{1}\frac{\widetilde{\phi}_{0}e_{n}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}=(\widetilde{\phi}_{0},e_{n})_{|f^{\prime}|^{-1}},\quad B_{n}=\frac{1}{\theta_{n}}\int_{-1}^{1}\frac{\widetilde{\phi}_{1}e_{n}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}=\frac{1}{\theta_{n}}(\widetilde{\phi}_{1},e_{n})_{|f^{\prime}|^{-1}}.

Hence $\|\widetilde{\phi}\|_{L^{2}(L_{|f^{\prime}|^{-1}}^{2})}^{2}=\int_{0}^{T}\int_{-1}^{1}\frac{\widetilde{\phi}^{2}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}dt$ can be bounded from below by

	$\displaystyle\int_{0}^{T}\sum_{n}\left(A_{n}^{2}\cos^{2}(\theta_{n}t)+B_{n}^{2}\sin^{2}(\theta_{n}t)+2A_{n}B_{n}\sin(\theta_{n}t)\cos(\theta_{n}t)\right)dt$		(69)
	$\displaystyle=\sum_{n}\left[\frac{A_{n}^{2}}{2}\left(T+\frac{\sin(2\theta_{n}T)}{2\theta_{n}}\right)+\frac{B_{n}^{2}}{2}\left(T-\frac{\sin(2\theta_{n}T)}{2\theta_{n}}\right)+A_{n}B_{n}\frac{1-\cos(2\theta_{n}T)}{2\theta_{n}}\right]$
	$\displaystyle\geq\sum_{n}\left[\frac{A_{n}^{2}}{2}\left(T+\frac{\sin(2\theta_{n}T)}{2\theta_{n}}\right)+\frac{B_{n}^{2}}{2}\left(T-\frac{\sin(2\theta_{n}T)}{2\theta_{n}}\right)-\frac{A_{n}^{2}+B_{n}^{2}}{2}\frac{1-\cos(2\theta_{n}T)}{2\theta_{n}}\right]$
	$\displaystyle=\frac{T}{2}\sum_{n}\left[A_{n}^{2}\left(1-\frac{1}{2\theta_{n}T}+\frac{\sin(2\theta_{n}T)}{2\theta_{n}T}+\frac{\cos(2\theta_{n}T)}{2\theta_{n}T}\right)\right.$
	$\displaystyle\phantom{\ =}\left.+B_{n}^{2}\left(1-\frac{1}{2\theta_{n}T}-\frac{\sin(2\theta_{n}T)}{2\theta_{n}T}+\frac{\cos(2\theta_{n}T)}{2\theta_{n}T}\right)\right]$
	$\displaystyle\geq C\frac{T}{2}\sum_{n}\left[A_{n}^{2}+B_{n}^{2}\right],$

for some positive $C>0$ and $T>2.42/2\min(\theta_{n})=2.42/2\theta_{1}$ .

From the orthogonality of the Hilbert basis $\{e_{n}\}$ ,

\|\widetilde{\phi}_{0}\|_{L_{|f^{\prime}|^{-1}}^{2}}^{2}=\int_{-1}^{1}\frac{\widetilde{\phi}_{0}^{2}}{|f^{\prime}(x^{\prime})|}dx^{\prime}=\sum_{n=1}^{\infty}A_{n}^{2}.

(70)

Moreover, from (66), the decomposition given in (68), and (67) we conclude $\lambda_{n}=\theta_{n}^{2}$ and thus

\sum_{n=1}^{\infty}B_{n}^{2}=\sum_{n}\left(\frac{1}{\theta_{n}}\int_{-1}^{1}\frac{\widetilde{\phi}_{1}e_{n}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}\right)^{2}=\sum_{n}\frac{1}{\lambda_{n}}\left(\int_{-1}^{1}\frac{\widetilde{\phi}_{1}e_{n}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}\right)^{2}.

(71)

Let us define, accordingly, the space

\widetilde{H}^{-\frac{1}{2}}=\left\{\widetilde{\phi}:\sum_{n}\frac{1}{\lambda_{n}}\left(\int_{-1}^{1}\frac{\widetilde{\phi}e_{n}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}\right)^{2}<\infty\right\},

and its dual space

\widetilde{H}^{\frac{1}{2}}=\left\{\widetilde{\phi}:\sum_{n}\lambda_{n}\left(\int_{-1}^{1}\frac{\widetilde{\phi}e_{n}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}\right)^{2}<\infty\right\}.

Note that, by (56) and definition of $\lambda_{n}$ and $e_{n}$ ,

{\ \left\|\widetilde{\phi}\right\|_{\widetilde{H}^{\frac{1}{2}}}^{2}=\sum\left(\int\frac{\widetilde{\phi}\lambda_{n}e_{n}}{\left|f^{\prime}(x)\right|}\right)\left(\int\frac{\widetilde{\phi}e_{n}}{\left|f^{\prime}(x)\right|}\right)=\int\widetilde{\phi}\mathcal{\ A}\widetilde{\phi}=\frac{\pi}{2}\sum_{n\geq 1}na_{n}^{2}\sim\left\|\widetilde{\phi}\right\|_{H_{w^{-1}}^{\frac{1}{2}}}^{2}.}

From (69)–(71), we conclude then for some $C>0$ and $T$ sufficiently large,

C^{-1}T\left(\left\|\widetilde{\phi}_{1}\right\|_{\widetilde{H}^{-\frac{1}{2}}}^{2}+\left\|\widetilde{\phi}_{0}\right\|_{L_{|f^{\prime}|^{-1}}^{2}}^{2}\right)\geq\int_{0}^{T}\int_{-1}^{1}\frac{\widetilde{\phi}^{2}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}dt\geq CT\left(\left\|\widetilde{\phi}_{1}\right\|_{\widetilde{H}^{-\frac{1}{2}}}^{2}+\left\|\widetilde{\phi}_{0}\right\|_{L_{|f^{\prime}|^{-1}}^{2}}^{2}\right).

This proves the following theorem:

Theorem 2.

The equation (67) is observable in time $T>T_{0}$ for some $T_{0}$ sufficiently large. That is, there exists a positive constant $C>0$ such that for $T$ sufficiently large

\int_{0}^{T}\int_{-1}^{1}\frac{\widetilde{\phi}^{2}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}dt\geq CT\left(\left\|\widetilde{\phi}_{1}\right\|_{\widetilde{H}^{-\frac{1}{2}}}^{2}+\left\|\widetilde{\phi}_{0}\right\|_{L_{|f^{\prime}|^{-1}}^{2}}^{2}\right).

We consider next the nonhomogeneous problem (as in (45))

\frac{\widetilde{\psi}_{tt}}{\left|f^{\prime}\right|}+\mathcal{A}\widetilde{\psi}=\frac{\widetilde{h}}{|f^{\prime}|}.

(72)

It will be convenient the following energy estimate:

\frac{1}{2}\frac{d}{dt}\int_{-1}^{1}\frac{\widetilde{\psi}_{t}^{2}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}+\int_{-1}^{1}\widetilde{\psi}_{t}\mathcal{A}\widetilde{\psi}dx^{\prime}=\int_{-1}^{1}\frac{\widetilde{h}\widetilde{\psi}_{t}}{|f^{\prime}(x^{\prime})|}dx^{\prime}.

(73)

If we let

\widetilde{\psi}=\sum_{n=1}^{\infty}c_{n}e_{n},

then

	$\displaystyle c_{n}=\int_{-1}^{1}$	$\displaystyle\frac{\widetilde{\psi}e_{n}}{\left\|f^{\prime}(x^{\prime})\right\|}dx^{\prime},$
	$\displaystyle\widetilde{\psi}_{t}=\sum_{n=1}^{\infty}c_{n,t}e_{n}\quad\implies\quad$	$\displaystyle\int_{-1}^{1}\frac{\psi_{t}^{2}}{\left\|f^{\prime}(x^{\prime})\right\|}dx^{\prime}=\sum_{n=1}^{\infty}c_{n,t}^{2}.$

Noticing that

	$\displaystyle\int_{-1}^{1}\widetilde{\psi}_{t}\mathcal{A}\widetilde{\psi}dx^{\prime}$	$\displaystyle=\int_{-1}^{1}\left(\sum_{n}c_{n,t}e_{n}\right)\left(\sum_{m}\lambda_{m}c_{m}\frac{e_{m}}{\|f^{\prime}\|}\right)dx^{\prime}$
		$\displaystyle=\sum_{n}\lambda_{n}c_{n,t}c_{n}$
		$\displaystyle=\frac{1}{2}\frac{d}{dt}\sum_{n}\lambda_{n}c_{n}^{2},$

(73) is equivalent to the following inequality

\frac{dE}{dt}\leq\|\widetilde{h}\|_{L_{|f^{\prime}|^{-1}}^{2}}E^{\frac{1}{2}},

for the energy defined as

E=\frac{1}{2}\sum_{n}c_{n,t}^{2}+\frac{1}{2}\sum_{n}\lambda_{n}c_{n}^{2}.

Hence, the natural initial data for which the problem is well-posed is $\left(\widetilde{\psi}_{0},\widetilde{\psi}_{1}\right)\in\widetilde{H}^{\frac{1}{2}}\times L_{|f^{\prime}|^{-1}}^{2}$ and, the energy is bounded provided $\int_{0}^{T}\|\widetilde{h}\|_{L_{|f^{\prime}|^{-1}}^{2}}^{2}dt<\infty$ , i.e. $\widetilde{h}\in L^{2}(0,T;L_{|f^{\prime}|^{-1}}^{2})$ . Namely, we have the following:

Theorem 3.

Given $f$ a conformal mapping that satisfies condition (65), for any $\widetilde{h}\in L^{2}(0,T;L_{|f^{\prime}|^{-1}}^{2})$ and $\left(\widetilde{\psi}_{0},\widetilde{\psi}_{1}\right)\in\widetilde{H}^{\frac{1}{2}}\times L_{|f^{\prime}|^{-1}}^{2}$ equation (72) has a unique weak solution

(\widetilde{\psi},\widetilde{\psi}^{\prime})\in C([0,T];H_{w^{-1}}^{\frac{1}{2}}\times L_{|f^{\prime}|^{-1}}^{2}).

6. Controllability

Once we have studied in the previous sections the forward evolution equation and the backward homogeneous system, given by (72) and (67) respectively, we are in position of comparing them to get the controllability condition on the system.

That is, if we assume the control drives the initial data of system (72) to zero, by multiplying the source term in (72) by the solution to the adjoint problem (67), we obtain

	$\displaystyle\int_{0}^{T}\int_{-1}^{1}\frac{\widetilde{h}\widetilde{\phi}}{\|f^{\prime}\|}dx^{\prime}dt$	$\displaystyle=\int_{0}^{T}\int_{-1}^{1}\left[\frac{\widetilde{\psi}_{tt}}{\|f^{\prime}\|}+\mathcal{A}\widetilde{\psi}\right]\widetilde{\phi}dx^{\prime}dt$
		$\displaystyle=\int_{-1}^{1}\frac{\widetilde{\psi}_{t}\widetilde{\phi}\|_{0}^{T}}{\|f^{\prime}\|}dx^{\prime}-\int_{0}^{T}\int_{-1}^{1}\frac{\widetilde{\psi}_{t}\widetilde{\phi}_{t}}{\|f^{\prime}\|}dx^{\prime}dt+\int_{0}^{T}\int_{-1}^{1}\widetilde{\psi}\mathcal{A}\widetilde{\phi}dx^{\prime}dt$
		$\displaystyle=\int_{-1}^{1}\frac{\widetilde{\psi}_{t}\widetilde{\phi}\|_{0}^{T}}{\|f^{\prime}\|}dx^{\prime}-\int_{-1}^{1}\frac{\widetilde{\psi}\widetilde{\phi}_{t}\|_{0}^{T}}{\|f^{\prime}\|}dx^{\prime}+\int_{0}^{T}\int_{-1}^{1}\widetilde{\psi}\left[\frac{\widetilde{\phi}_{tt}}{\|f^{\prime}\|}+\mathcal{A}\widetilde{\phi}\right]dx^{\prime}dt$
		$\displaystyle=-\int_{-1}^{1}\frac{\widetilde{\psi}_{1}\widetilde{\phi}_{0}}{\|f^{\prime}\|}dx^{\prime}+\int_{-1}^{1}\frac{\widetilde{\psi}_{0}\widetilde{\phi}_{1}}{\|f^{\prime}\|}dx^{\prime},$

where $(\widetilde{\psi}_{0},\widetilde{\psi}_{1})=(\widetilde{\psi}(0),\widetilde{\psi}_{t}(0))\in\widetilde{H}^{\frac{1}{2}}\times L_{|f^{\prime}|^{-1}}^{2}$ . So that $\widetilde{h}$ can be chosen as the minimizer of the functional

J[\widetilde{\phi}_{0},\widetilde{\phi}_{1}]=\frac{1}{2}\int_{0}^{T}\int_{-1}^{1}\frac{\widetilde{\phi}^{2}}{|f^{\prime}|}dx^{\prime}dt+\int_{-1}^{1}\frac{\widetilde{\psi}_{1}\widetilde{\phi}_{0}}{|f^{\prime}|}dx^{\prime}-\int_{-1}^{1}\frac{\widetilde{\psi}_{0}\widetilde{\phi}_{1}}{|f^{\prime}|}dx^{\prime}.

Notice that

\left|\int_{-1}^{1}\frac{\widetilde{\psi}_{1}\widetilde{\phi}_{0}}{|f^{\prime}|}dx^{\prime}-\int_{-1}^{1}\frac{\widetilde{\psi}_{0}\widetilde{\phi}_{1}}{|f^{\prime}|}dx^{\prime}\right|\leq\|\widetilde{\phi}_{0}\|_{L_{|f^{\prime}|^{-1}}^{2}}\|\widetilde{\psi}_{1}\|_{L_{|f^{\prime}|^{-1}}^{2}}+\|\widetilde{\phi}_{1}\|_{\widetilde{H}^{-\frac{1}{2}}}\|\widetilde{\psi}_{0}\|_{\widetilde{H}^{\frac{1}{2}}}.

As it is customary in controllability theory, coerciveness of the functional $J$ is guaranteed if the adjoint problem is observable in time $T$ , that is:

\int_{0}^{T}dt\int_{-1}^{1}\frac{\widetilde{\phi}^{2}}{|f^{\prime}|}dx^{\prime}\geq C\left(\|\widetilde{\phi}(0)\|_{L_{|f^{\prime}|^{-1}}^{2}}^{2}+\|\widetilde{\phi}_{t}(0)\|_{\widetilde{H}^{-\frac{1}{2}}}^{2}\right),

(74)

a fact that was proved in the previous section. Hence, we have proved the following controllability theorem:

Theorem 4.

The system (72) is exactly controllable in time $T$ . That is, for any initial data $\left(\widetilde{\psi}_{0},\widetilde{\psi}_{1}\right)\in\widetilde{H}^{\frac{1}{2}}\times L_{|f^{\prime}|^{-1}}^{2}$ , there exist $\widetilde{h}\in L^{2}(0,T;L^{2}_{|f^{\prime}|^{-1}})$ and $T>0$ such that

\|\widetilde{\psi}\|_{H_{w^{-1}}^{\frac{1}{2}}}=0,\quad\text{for }t>T.

The fact that the control $\widetilde{h}\in L^{2}((0,T);L_{|f^{\prime}|^{-1}}^{2})$ implies that one can write

\widetilde{h}(t,x^{\prime})=\sum_{n}\widetilde{h}_{n}(t)e_{n},

with

\int_{0}^{T}\sum_{n}|\widetilde{h}_{n}(t)|^{2}dt<\infty.

We are going to discuss next on how to approach the control function $\widetilde{h}$ , defined at the fluid interface by means of a function $j$ defined at the solid boundaries. This function $j$ will represent a fluid injection at certain points $\left\{x_{1}^{\prime},x_{2}^{\prime},...,x_{N}^{\prime}\right\}$ of the solid boundary with a flow rate $\left\{j_{1}(t),j_{2}(t),...,j_{N}(t)\right\}$ and under the mass conservation condition

\sum_{i=1}^{N}j_{i}(t)=0.

More precisely,

\frac{\partial\widetilde{\phi}}{\partial\widetilde{n}}=\sum_{i}j_{i}(t)\delta(x^{\prime}-x_{i}^{\prime}).

(75)

We replace the expression for $\widetilde{j}(t,z,0)$ in (46) by the right hand side of (75) at the solid side walls and obtain

$\displaystyle\frac{\|f^{\prime}(x^{\prime})\|}{\sqrt{1-x^{\prime 2}}}\frac{1}{\pi^{2}}$	$\displaystyle P.V.\int_{-1}^{1}\frac{\sqrt{1-\xi^{2}}}{(x^{\prime}-\xi)}\int_{\mathbb{\ R}\setminus[-1,1]}\frac{\widetilde{j}(t,z,0)}{\|f^{\prime}(z)\|(\xi-z)}dzd\xi$
	$\displaystyle=\sum_{i}\frac{j_{i}(t)}{\|f^{\prime}(x_{i}^{\prime})\|}\frac{\|f^{\prime}(x^{\prime})\|}{\sqrt{1-x^{\prime 2}}}\frac{1}{\pi^{2}}P.V.\int_{-1}^{1}\frac{\sqrt{1-\xi^{2}}}{(x^{\prime}-\xi)(\xi-x_{i}^{\prime})}d\xi$
	$\displaystyle=\sum_{i}\frac{j_{i}(t)}{\|f^{\prime}(x_{i}^{\prime})\|}\frac{\|f^{\prime}(x^{\prime})\|}{\sqrt{1-x^{\prime 2}}}\frac{1}{x^{\prime}-x_{i}^{\prime}}\frac{1}{\pi^{2}}P.V.\int_{-1}^{1}\sqrt{1-\xi^{2}}\left[\frac{1}{x^{\prime}-\xi}+\frac{1}{\xi-x_{i}^{\prime}}\right]d\xi$
	$\displaystyle=\frac{\|f^{\prime}(x^{\prime})\|}{\sqrt{1-x^{\prime 2}}}\sum_{i}j_{i}(t)\frac{G(x_{i}^{\prime})-G(x)}{\pi^{2}\|f^{\prime}(x_{i}^{\prime})\|}\frac{1}{x^{\prime}-x_{i}^{\prime}},$	(76)

with

G(x_{i}^{\prime})=P.V.\int_{-1}^{1}\frac{\sqrt{1-\xi^{2}}}{(\xi-x_{i}^{\prime})}d\xi.

Hence, the challenge is to approximate the control function $\widetilde{h}$ by (6). One possibility is to approximate the first $N-1$ modes so that

\int_{-1}^{1}\frac{\widetilde{h}e_{i}}{\left|f^{\prime}(x^{\prime})\right|}dx^{\prime}=h_{i}(t),\quad i=1,...,N-1,

or, denoting $m_{ij}=\frac{1}{\pi^{2}|f^{\prime}(x_{j}^{\prime})|}\int_{-1}^{1}\frac{G(x_{j}^{\prime})-G(x)}{\sqrt{1-x^{\prime 2}}(x^{\prime}-x_{j}^{\prime})}e_{i}dx^{\prime}$ and including the mass conservation condition, solving the system

	$\displaystyle\sum_{j=1}^{N}m_{ij}j_{j}(t)$	$\displaystyle=h_{i}(t),\quad i=1,...,N-1,$
	$\displaystyle\sum_{i=1}^{N}j_{i}(t)$	$\displaystyle=0.$

The fact that $h_{i}(t)\in L^{2}\left(0,T\right)$ implies $j_{i}(t)\in L^{2}\left(0,T\right)$ .

Let us summarize what we have achieved so far in connection with the injection of fluid problem and the control of splashing appearing in a cooper converter [9, 15]. First, the main relation between the inner source $\widetilde{h}$ , and the injection of fluids $\widetilde{j}$ at the solid boundary of the half-plane, is given by (46). Second, by the computations above, we found the source $\widetilde{h}$ given in (75), which corresponds to the fluid injection of jets $\widetilde{j}$ on a finite number of points. Third, once $\widetilde{j}$ is known on the half-plane, we can recover $j$ , on the cylindrical container, from (13). Of course this procedure is directly connected to the controllability, through Theorem 4.

Finally, it is important to highlight that, although we used the case of the cylinder as a reference, all the computations are valid for any simply connected domain through the conformal mapping term $|f^{\prime}|$ , which appears as a factor on the main operator for the Cauchy problem.

As far as the three-dimensional case is concerned, the conformal mapping strategy cannot be replicated directly. While it is true that the integral equations can be formulated in an equivalent frame in the 3d case (see [18] for more details).

Acknowledgments

We would like to thank the anonymous referees for their comments and suggestion which helped to significantly improve this work.

Appendix A Equivalent expressions of the operator $\mathcal{A}$

We discuss in this appendix the relation between different expressions of the operator $\mathcal{A}$ based on expansions in Tchebyshev polynomials. We remind that the Tchebyshev polynomials $\left\{T_{n}(x^{\prime})\right\}_{n=0}^{\infty}$ form an orthogonal basis in $L_{(1-x^{\prime 2})^{-1/2}}^{2}(-1,1)$ and the Tchebyshev polynomials $\left\{U_{n}(x^{\prime})\right\}_{n=0}^{\infty}$ form an orthogonal basis in $L_{(1-x^{\prime 2})^{1/2}}^{2}(-1,1)$ . Remind that tangential and normal derivatives of $\widetilde{\phi}$ in the interval $(-1,1)$ are related by

\widetilde{\phi}_{x^{\prime}}(x^{\prime})=-\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\widetilde{\phi}_{y^{\prime}}(\xi)}{x^{\prime}-\xi}d\xi.

(77)

Let

\widetilde{\phi}=\sum_{n=0}^{\infty}a_{n}T_{n},

(78)

and observe the following well-known identities

$\displaystyle\frac{dT_{n}}{dx^{\prime}}$	$\displaystyle=nU_{n-1},$	(79)
$\displaystyle-U_{n-1}(x^{\prime})$	$\displaystyle=\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{T_{n}(\xi)}{\sqrt{1-\xi^{2}}(x^{\prime}-\xi)}d\xi,$	(80)
$\displaystyle T_{n}(x^{\prime})$	$\displaystyle=\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{\sqrt{1-\xi^{2}}U_{n-1}(\xi)}{(x^{\prime}-\xi)}d\xi,$	(81)
$\displaystyle\frac{d}{dx^{\prime}}\left(\sqrt{1-x^{\prime 2}}U_{r-1}(x^{\prime})\right)$	$\displaystyle=-r\frac{T_{r}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}.$	(82)

Since, by (79),

\widetilde{\phi}_{x^{\prime}}=\sum_{n=1}^{\infty}na_{n}U_{n-1}(x^{\prime}),

(83)

we have

\int_{-1}^{1}\sqrt{1-x^{\prime 2}}|\widetilde{\phi}_{x^{\prime}}(x^{\prime})|^{2}dx^{\prime}=\frac{\pi}{2}\sum_{n=1}^{\infty}n^{2}a_{n}^{2}.

By writing

\widetilde{\phi}_{y^{\prime}}=\sum_{n=0}^{\infty}b_{n}\frac{T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}},

and using

\int_{-1}^{1}\widetilde{\phi}_{y^{\prime}}(x^{\prime})dx^{\prime}=\sum_{n=0}^{\infty}b_{n}\int\frac{T_{0}(x^{\prime})T_{n}(x^{\prime})}{\sqrt{1-x^{\prime 2}}}=b_{0}=0,

together with (77), (80) we conclude $b_{n}=na_{n}$ . By using (81), one can write

\widetilde{\phi}_{y^{\prime}}(x^{\prime})=\frac{1}{\sqrt{1-x^{\prime 2}}}\frac{1}{\pi}P.V.\int_{-1}^{1}\sqrt{1-\xi^{2}}\frac{\widetilde{\phi}_{x^{\prime}}(\xi)}{x^{\prime}-\xi}d\xi,

which is equivalent, by (78), (80) and (82), to

\widetilde{\phi}_{y^{\prime}}(x^{\prime})=\partial_{x^{\prime}}\left(\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-\xi^{2}}}\frac{\widetilde{\phi}(\xi)}{x^{\prime}-\xi}d\xi\right).

If we expand, instead of (83), in the form

	$\displaystyle\widetilde{\phi}_{x^{\prime}}$	$\displaystyle=\sum_{n=1}^{\infty}a_{n}T_{n}(x^{\prime}),$		(84)
	$\displaystyle\widetilde{\phi}_{y^{\prime}}$	$\displaystyle=\sum_{n=1}^{\infty}b_{n}\sqrt{1-x^{\prime 2}}U_{n-1}(x^{\prime}),$		(85)

with $a_{n}$ such that

\int_{-1}^{1}\frac{|\widetilde{\phi}_{x^{\prime}}(x^{\prime})|^{2}}{\sqrt{1-x^{\prime 2}}}dx^{\prime}=\pi a_{0}^{2}+\frac{\pi}{2}\sum_{n=1}^{\infty}a_{n}^{2}<\infty,

then, $b_{n}=a_{n}$ and using (80) one can write

\widetilde{\phi}_{y^{\prime}}(x^{\prime})=\sqrt{1-x^{\prime 2}}\frac{1}{\pi}P.V.\int_{-1}^{1}\frac{1}{\sqrt{1-\xi^{2}}}\frac{\widetilde{\phi}_{x^{\prime}}(\xi)}{x^{\prime}-\xi}d\xi.

Note that the mass conservation condition $\int_{-1}^{1}\widetilde{\phi}_{y^{\prime}}(x^{\prime})dx^{\prime}=0$ implies $b_{1}=0$ which yields $a_{1}=0$ implying

\int_{-1}^{1}\frac{x^{\prime}\widetilde{\phi}_{x^{\prime}}(x^{\prime})dx^{\prime}}{\sqrt{1-x^{\prime 2}}}=0\text{,}

and the absence of $T_{0}(x^{\prime})$ term in (84) implies

\int_{-1}^{1}\frac{\widetilde{\phi}_{x^{\prime}}(x^{\prime})dx^{\prime}}{\sqrt{1-x^{\prime 2}}}=0\text{.}

Appendix B The compact embedding of $H_{w^{-1}}^{\frac{1}{2}}$ into $L_{w^{-1}}^{2}$

Since the spaces $H_{w^{-1}}^{\frac{1}{2}}$ and $L_{w^{-1}}^{2}$ can be defined in terms of an orthogonal basis, following [8], let us prove the more general result $h^{1/2}\overset{c}{\hookrightarrow}\ell^{2}$ for sequence spaces

h^{1/2}:=\left\{(a_{n}):\sum_{n=1}^{\infty}n|a_{n}|^{2}<\infty\right\},\quad\ell^{2}:=\left\{(a_{n}):\sum_{n=1}^{\infty}|a_{n}|^{2}<\infty\right\}.

Consider $M\subseteq h^{1/2}$ a bounded set in $h^{1/2}$ . Then, given $\mathbf{a\in}\mathbb{\ell}^{2}$ ,

\left\|\mathbf{a}\right\|_{\mathbb{\ell}^{2}}^{2}=\sum_{n=1}^{\infty}a_{n}^{2}\leq\sum_{n=1}^{\infty}na_{n}^{2}=\left\|\mathbf{a}\right\|_{h^{\frac{1}{2}}}^{2}<K^{2},

where $K$ is such that $\left\|\mathbf{a}\right\|_{h^{\frac{1}{2}}}<K$ , $\forall\mathbf{a}\in M$ . We take now $Y_{\varepsilon}=span\left\{\mathbf{e}_{1},\mathbf{e}_{2},...,\mathbf{e}_{N}\right\}$ where $N$ will be chosen later as a function of $\varepsilon$ , and $\mathbf{e}_{i}$ is the vector in $\mathbb{\ell}^{2}$ with all components zero except the $i$ -th component that is $1$ . We write $\mathbf{x}=a_{1}\mathbf{e}_{1}+a_{2}\mathbf{\ e}_{2}+...+a_{N}\mathbf{e}_{N}$ so that $\mathbf{x}\in Y_{\varepsilon}\subset\mathbb{\ell}^{2}$ , $\mathbf{x}=(a_{1},a_{2},...,a_{N-1},0,0,...)$ and

\left\|\mathbf{x}-\mathbf{a}\right\|_{\mathbb{\ell}^{2}}^{2}=\sum_{n=N}^{\infty}a_{n}^{2}=\sum_{n=N}^{\infty}\frac{1}{n}na_{n}^{2}\leq\frac{1}{N}\sum_{N}^{\infty}na_{n}^{2}\leq\frac{K^{2}}{N}<\frac{\varepsilon}{2},

which implies $N=O(1/\varepsilon)$ . Let $\mathbf{x\in}\overline{M}$ . There exists $\mathbf{x}_{n}\in M$ such that $\mathbf{x}_{n}\rightarrow\mathbf{x}$ in the $\mathbb{\ell}^{2}$ topology. For $n$ sufficiently large, $\left\|\mathbf{x}_{n}-\mathbf{x}\right\|_{\mathbb{\ell}^{2}}<\frac{\varepsilon}{2}$ and all $\mathbf{x}_{n}$ are at distance $\frac{\varepsilon}{2}$ of $Y_{\varepsilon}$ ; then by the triangle inequality $\mathbf{x}$ is within $\varepsilon$ of $Y_{\varepsilon}$ and the closure of $M$ in $\mathbb{\ell}^{2}$ is compact by Proposition 7.4 in [12] or Proposition 2.1 in [8]. Hence $h^{1/2}$ is compactly embedded in $\mathbb{\ell}^{2}$ .

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	$\displaystyle\left\|\int_{-1}^{1}u\widetilde{\phi}dx^{\prime}\right\|$	$\displaystyle\leq\int_{-1}^{1}\sqrt[4]{1-x^{\prime 2}}\|u\|\frac{\widetilde{\phi}}{\sqrt[4]{1-x^{\prime 2}}}dx^{\prime}$
		$\displaystyle\leq\\|u\\|_{L_{w}^{2}}\\|\widetilde{\phi}\\|_{L_{w^{-1}}^{2}}$
		$\displaystyle\leq\\|u\\|_{L_{w}^{2}}\\|\widetilde{\phi}\\|_{H_{w^{-1}}^{1/2}}.$

	$\displaystyle\int_{0}^{T}\int_{-1}^{1}\frac{\widetilde{h}\widetilde{\phi}}{\|f^{\prime}\|}dx^{\prime}dt$	$\displaystyle=\int_{0}^{T}\int_{-1}^{1}\left[\frac{\widetilde{\psi}_{tt}}{\|f^{\prime}\|}+\mathcal{A}\widetilde{\psi}\right]\widetilde{\phi}dx^{\prime}dt$
		$\displaystyle=\int_{-1}^{1}\frac{\widetilde{\psi}_{t}\widetilde{\phi}\|_{0}^{T}}{\|f^{\prime}\|}dx^{\prime}-\int_{0}^{T}\int_{-1}^{1}\frac{\widetilde{\psi}_{t}\widetilde{\phi}_{t}}{\|f^{\prime}\|}dx^{\prime}dt+\int_{0}^{T}\int_{-1}^{1}\widetilde{\psi}\mathcal{A}\widetilde{\phi}dx^{\prime}dt$
		$\displaystyle=\int_{-1}^{1}\frac{\widetilde{\psi}_{t}\widetilde{\phi}\|_{0}^{T}}{\|f^{\prime}\|}dx^{\prime}-\int_{-1}^{1}\frac{\widetilde{\psi}\widetilde{\phi}_{t}\|_{0}^{T}}{\|f^{\prime}\|}dx^{\prime}+\int_{0}^{T}\int_{-1}^{1}\widetilde{\psi}\left[\frac{\widetilde{\phi}_{tt}}{\|f^{\prime}\|}+\mathcal{A}\widetilde{\phi}\right]dx^{\prime}dt$
		$\displaystyle=-\int_{-1}^{1}\frac{\widetilde{\psi}_{1}\widetilde{\phi}_{0}}{\|f^{\prime}\|}dx^{\prime}+\int_{-1}^{1}\frac{\widetilde{\psi}_{0}\widetilde{\phi}_{1}}{\|f^{\prime}\|}dx^{\prime},$

$\displaystyle\frac{\|f^{\prime}(x^{\prime})\|}{\sqrt{1-x^{\prime 2}}}\frac{1}{\pi^{2}}$	$\displaystyle P.V.\int_{-1}^{1}\frac{\sqrt{1-\xi^{2}}}{(x^{\prime}-\xi)}\int_{\mathbb{\ R}\setminus[-1,1]}\frac{\widetilde{j}(t,z,0)}{\|f^{\prime}(z)\|(\xi-z)}dzd\xi$
	$\displaystyle=\sum_{i}\frac{j_{i}(t)}{\|f^{\prime}(x_{i}^{\prime})\|}\frac{\|f^{\prime}(x^{\prime})\|}{\sqrt{1-x^{\prime 2}}}\frac{1}{\pi^{2}}P.V.\int_{-1}^{1}\frac{\sqrt{1-\xi^{2}}}{(x^{\prime}-\xi)(\xi-x_{i}^{\prime})}d\xi$
	$\displaystyle=\sum_{i}\frac{j_{i}(t)}{\|f^{\prime}(x_{i}^{\prime})\|}\frac{\|f^{\prime}(x^{\prime})\|}{\sqrt{1-x^{\prime 2}}}\frac{1}{x^{\prime}-x_{i}^{\prime}}\frac{1}{\pi^{2}}P.V.\int_{-1}^{1}\sqrt{1-\xi^{2}}\left[\frac{1}{x^{\prime}-\xi}+\frac{1}{\xi-x_{i}^{\prime}}\right]d\xi$
	$\displaystyle=\frac{\|f^{\prime}(x^{\prime})\|}{\sqrt{1-x^{\prime 2}}}\sum_{i}j_{i}(t)\frac{G(x_{i}^{\prime})-G(x)}{\pi^{2}\|f^{\prime}(x_{i}^{\prime})\|}\frac{1}{x^{\prime}-x_{i}^{\prime}},$	(76)

Controllability of surface gravity waves and the sloshing problem

Abstract.

Key words and phrases:

2010 Mathematics Subject Classification:

1. Introduction

2. Formulation

2.1. Linearized equations

2.2. Conformal transplants

2.3. Reformulation as an integrodifferential PDE

2.4. The Dirichlet-Neumann operator

3. An associated Cauchy problem

Lemma 1.

Proof.

4. Spectrum of the sloshing problem

Lemma 2.

Proof.

Remark 1.

Remark 2.

Theorem 1.

5. Observability and energy estimates

Theorem 2.

Theorem 3.

6. Controllability

Theorem 4.

Acknowledgments

Appendix A Equivalent expressions of the operator 𝒜\mathcal{A}

Appendix B The compact embedding of Hw−112H_{w^{-1}}^{\frac{1}{2}} into Lw−12L_{w^{-1}}^{2}

References

Appendix A Equivalent expressions of the operator $\mathcal{A}$

Appendix B The compact embedding of $H_{w^{-1}}^{\frac{1}{2}}$ into $L_{w^{-1}}^{2}$