arXiv:2203.03012v1 [math.OC] 6 Mar 2022

CONTROL OF THE LINEARIZED STEFAN PROBLEM IN APERIODIC BOX

BORJAN GESHKOVSKI AND DEBAYAN MAITY

Abstract. In this paper we consider the linearized one-phase Stefan problem withsurface tension, set in the strip T× (−1, 1), thus with periodic boundary conditionsrespect to the horizontal direction x1 ∈ T. When the support of the control is notlocalized in x1, namely, is of the form ω = T × (c, d), we prove that the systemis null-controllable in any positive time. We rely on a Fourier decomposition withrespect to x1, and controllability results which are uniform with respect to the Fourierfrequency parameter for the resulting family of one-dimensional systems. The latterresults are also novel, as we compute the full spectrum of the underlying operatorfor the non-zero Fourier modes. The zeroth mode system, on the other hand, is seenas a controllability problem for the linear heat equation with a finite-dimensionalconstraint. We extend the controllability result to the setting of controls with asupport localized in a box: ω = (a, b) × (c, d), through an argument inspired by themethod of Lebeau and Robbiano, under the assumption that the initial data areof zero mean. Numerical experiments motivate several challenging open problems,foraying even beyond the specific setting we deal with herein.

Contents

1. Introduction and main results 22. Discussion 73. The linear semigroup 114. A family of projected systems 165. Proof of Theorem 1.1 266. Proof of Theorem 1.2 277. Epilogue 33Appendix A. Numerics 35Appendix B. Linearization 37References 38

Keywords. Stefan problem, phase transitions, controllability, free boundary problem,Gibbs-Thomson correction, surface tension.AMS Subject Classification. 93B05, 35R35, 35Q35, 93C20.

Date: March 8, 2022.A major part of this work was done while B.G. was affiliated with Departamento de Matemáticas,

Universidad Autónoma de Madrid, and the Chair of Computational Mathematics, Fundación Deusto.1

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2 BORJAN GESHKOVSKI AND DEBAYAN MAITY

1. Introduction and main results

The Stefan problem is the quintessential macroscopic model of phase transitionsin liquid-solid systems. The physical setup thereof typically consists in consideringa domain Ω ⊂ Rd, which is occupied by water (the liquid phase), a part of whoseboundary is some interface Γ, describing contact with a deformable solid such as ice(the solid phase). Due to melting or freezing, the regions occupied by water and icewill change over time and, consequently, the interface Γ will also change its positionand shape. This leads to a free boundary problem. Albeit classical (see [19, 48] foran overview of the mathematical literature), the Stefan problem continues to be of usein many contemporary applications, such as additive manufacturing of alloys ([32]), icemodeling for video rendering in computer graphics ([31]), and, reaching even beyond itsoriginal fluid-mechanical nature, in the context of mathematical biology, for modelingthe spread of various infectious diseases ([14, 36]).

1.1. Setup. We shall focus on the strong formulation of the one-phase Stefan problem(i.e., where the temperature of the ice is a known constant), with surface tension effects,following [17, 25, 27, 49]. We shall focus on the problem in two spatial dimensions (d =2). To describe the geometrical setup, let T := R/(2πZ) denote the one-dimensionalflat torus, which we identify with [0, 2π]. Set

Ω := T× (−1, 1). (1.1)

The domain Ω will serve as the reference configuration. In the one-phase Stefan problem,a heat-conducting liquid fills a time-varying domain Ω(t) ⊂ R2 for t > 0. We willsuppose that the boundary ∂Ω(t) of the liquid consists of two components: an unknown,time-dependent component (the free boundary Γ(t)), and a fixed and static component.More specifically, for any t > 0, Ω(t) is assumed to have a flat, rigid bottom, while thefree boundary will be parametrized by an unknown function h(t, z1), representing theformer’s displacement away from the reference boundary T × z2 = 1 (see Figure 1),and thus described by the equation z2 = 1 + h(t, z1).

In other words,

Ω(t) :=z = (z1, z2) ∈ T× R

∣∣∣ −1 < z2 < 1 + h(t, z1),

where h = h(t, z1) is the unknown height function, while the free boundary is given by

Γ(t) :=z = (z1, z2) ∈ T× R

∣∣∣ z2 = 1 + h(t, z1).

Given a time horizon T > 0 the strong formulation of the one-phase Stefan problemtakes the form1

∂t%−∆% = 0 in (0, T )× Ω(t),

∂th+√

1 + |∂z1h|2∇%|Γ(t)· n = 0 on (0, T )× T,

%(t, z1,−1) = 0 on (0, T )× T%(t, z1, z2) = −σκ(h(t, z1)) on (0, T )× Γ(t),

(%, h)|t=0=(%0, h0

)in Ω(0)× T.

(1.2)

1We make use of the standard notation (0, T ) × Ω(t) :=⋃

06t6T t × Ω(t) and the analog for(0, T )× Γ(t).

CONTROL OF THE LINEARIZED STEFAN PROBLEM IN A PERIODIC BOX 3

This is a coupled system, where the unknown state is the pair (%, h). The initial domainΩ(0) is therefore given by

Ω(0) :=z = (z1, z2) ∈ T× R

∣∣∣ −1 < z2 < 1 + h0(z1).

The constant σ > 0 represents the surface tension coefficient2, whereas κ(h(t, z1)) de-notes the mean curvature of the free boundary Γ(t), defined as

κ(h) :=∂2z1h(

1 + |∂z1h|2)3/2

.

As seen later on, the assumption σ > 0 will be a critical part of our study. Finally,n = n(t, z1) denotes the unit normal to Γ(t) outward Ω(t) and is given by

n :=1√

1 + |∂z1h|2

[−∂z1h

1

].

Ω(t)

x2 = 1 + h(t, x1)

x2

x1(0,−1) 2π

1

control

Ω

x2

x1(0,−1) 2π

1

Figure 1. (Left) The moving domain Ω(t), and the free boundary Γ(t),parametrized by the height function h(t, z1). Our goal is to steer boththe temperature %(t), and the free boundary Γ(t), to rest in time T > 0(right).

1.2. Main results. In view of the applications presented just before, analyzing thecontrolled evolution of trajectories to (1.2) is rather natural. Our interest is the prob-lem of null-controllability for (1.2): given a time horizon T > 0, we seek to steer thetemperature %(t), as well as the interface height h(t), to the equilibrium position (0, 0),by means of some control u(t, z1, z2)1ω actuating either along the bottom boundaryT × z2 = −1 (replacing the Dirichlet boundary condition in (1.2)), or distributedwithin the fluid domain Ω(t) (as a source term in (1.2)).

2The condition involving the surface tension and the mean curvature is referred to as the Gibbs-Thomson correction. The physical reason for introducing the Gibbs-Thomson correction stems fromthe need to account for possible coarsening or nucleation effects ([43]). When σ = 0, we are dealingwith the classical Stefan problem. The mesoscopic limit σ 0 has been addressed in [25] (withoutcontrol).


Note that, due to the presence of the free boundary parametrized by h, such a con-trollability problem would amount to also controlling the domain Ω(t) to the referenceconfiguration Ω. This is seen in Figure 1. And as a first but necessary step in solvingthis problem, in this work, we shall focus solely on the system linearized around theequilibrium (0, 0).

The linearized system, obtained after fixing the domain in (1.2) and dropping thenonlinearities from the subsequent system (detailed in Appendix B), with a distributedcontrol, will take the form

∂ty −∆y = u1ω in (0, T )× Ω,

∂th(t, x1)− ∂x2y(t, x1, 1) = 0 on (0, T )× T,y(t, x1,−1) = 0 on (0, T )× T,y(t, x1, 1) = σ∂2

x1h(t, x1) on (0, T )× T,

(y, h)|t=0=(y0, h0

)in Ω× T.

(1.3)

For the time being, ω ⊂ Ω is assumed to be open and non-empty. We shall focus ondistributed controls since, as it is well known for parabolic equations, a simple extension-restriction argument allows one to obtain results for boundary controls as well.

The controllability properties of (1.3) have not been addressed in the literature, tothe best of our knowledge. In fact, even the well-posedness may appear tricky at a firstglance, due to the peculiar coupling of the states y and h. Nonetheless, we may notethe energy dissipation law (when u ≡ 0)

d

dt

∫Ω|y(t)|2 dx+ σ

∫T

∣∣∂x1h(t)∣∣2 dx1

= −2

∫Ω

∣∣∇y(t)∣∣2 dx, (1.4)

which will aid in ensuring that the system is well-posed when considered on the energyspace L2(Ω)×H1(T). As a matter of fact, we show that the governing operator of (1.3)generates an analytic semigroup on this energy space (Proposition 3.2 & Corollary 3.1).Identity (1.4) also clearly illustrates the strength of the coupling between y and h, andthe impact of σ > 0.

The main results in our paper regard the null-controllability of (1.3). We begin withthe following theorem.Theorem 1.1. Suppose T > 0 and σ > 0 are fixed, and suppose that ω = T × (c, d),where (c, d) ⊂ (−1, 1). Then for any

(y0, h0

)∈ L2(Ω)×H1(T), there exists some control

u ∈ L2((0, T ) × ω) such that the unique solution (y, h) ∈ C0([0, T ];L2(Ω) ×H1(T)) to(1.3) satisfies y(T, ·) ≡ 0 in Ω and h(T, ·) ≡ 0 in T.

The control u in Theorem 1.1 is supported everywhere in the horizontal (i.e., x1 ∈ T)direction. This is of critical importance in the strategy of proof – we touch upon thedetails and reasons behind our methodological choice in Section 1.3 just below.

By slightly tweaking our strategy, we can further localize the support of the control,under the assumption that the initial data are of zero mean.Theorem 1.2. Suppose T > 0 and σ > 0 are fixed, and suppose that ω = (a, b)× (c, d)where (a, b) ⊂ T and (c, d) ⊂ (−1, 1). Then for any

(y0, h0

)∈ L2(Ω)×H1(T) such that∫

Ty0(x1, x2) dx1 =

∫Th0(x1) dx1 = 0


for all x2 ∈ (−1, 1), there exists some control u ∈ L2((0, T ) × ω) such that the uniquesolution (y, h) ∈ C0([0, T ];L2(Ω) × H1(T)) to (1.3) satisfies y(T, ·) ≡ 0 in Ω andh(T, ·) ≡ 0 in T.

We postpone further comments to Section 2.

1.3. Strategy of proof. When one looks to prove the controllability of (1.3), theprimal instinct would be to first write the adjoint system, which reads as

−∂tζ −∆ζ = 0 in (0, T )× Ω,

−∂t`(t, x1)− ∂2x1∂x2ζ(t, x1, 1) = 0 on (0, T )× T,

ζ(t, x1,−1) = 0 on (0, T )× T,ζ(t, x1, 1) = `(t, x1) on (0, T )× T,(ζ, `)|t=T = (ζT , `T ) in Ω.

(1.5)

Note that, since the natural energy space for (1.3) is L2(Ω)×H1(T), the adjoint problem(1.5) ought to be analyzed in the dual space. Proceeding by the Hilbert UniquenessMethod (HUM, [37]), one would look to show an observability inequality of the form

‖ζ(0)‖2L2(Ω) + ‖`(0)‖2(H1(T))′ .T,ω,σ

∫ T

0

∫ω|ζ(t, x)|2 dtdx, (1.6)

for all (ζT , `T ) ∈ L2(Ω)×(H1(T))′. If ω ⊂ Ω is open, non-empty, but otherwise arbitrary,the canonical way to proceed in such an endeavor would be through the use of Carlemaninequalities. At this stage, we are not aware of existing Carleman inequalities whichcome even close to being adapted to the specific nature of the adjoint problem (1.5),due to the asymmetric nature of the coupling between both states.

There is however, a simpler way in which the problem can be tackled, as a first step.If one first supposes that ω = T × (c, d), with (c, d) ⊂ (−1, 1), we may then furtherexploit the periodicity of the control, and decompose all functions appearing in (1.3)into Fourier series with respect to x1 ∈ T. Namely, we write

y(t, x1, x2) =1√2π

∑n∈Z

yn(t, x2)einx1 (1.7)

withyn(t, x2) =

1√2π

∫Ty(t, ξ, x2)einξ dξ, (1.8)

with analogous decompositions for h and u, and the initial data (y0, h0). We then findourselves with a family of systems for the Fourier coefficients, parametrized by n ∈ Z:

∂tyn − ∂2x2yn + n2yn = un1(c,d) in (0, T )× (−1, 1),

h′n(t)− ∂x2 yn(t, 1) = 0 in (0, T ),

yn(t,−1) = 0 in (0, T ),

yn(t, 1) = −σn2hn(t) in (0, T ),(yn, hn

)|t=0

=(y0n, h

0n

)in (−1, 1).

(1.9)

Clearly, if the Fourier coefficients (yn, hn) solving (1.9) are controllable to 0 in timeT > 0 for any n ∈ Z, then by summing up all these Fourier coefficients as in (1.7),


we will deduce the desired controllability result for (1.3) as well. This is summarizedin Proposition 4.1. Such procedures have been used in the literature, typically in thecontext of hypoelliptic operators ([5, 6, 8, 9]).

Curiously, when n 6= 0, the linear operator governing (1.9) is self-adjoint, and so thecontrollability of (1.9) can be ensured by proving an appropriate observability inequalityfor the (simpler) adjoint system, which can now be done by utilizing spectral techniques.This in turn, requires the computation of the full spectrum of the operator, which weprovide in Lemma 4.1. Note that the latter is a nontrivial computation, as the operatorin (1.9) is not a linear shift of the Laplacian. On another hand, when n = 0, we seethat (1.9) becomes uncoupled, in the sense that we may control the heat component y0

independently of h0. Yet, we can view the constraint h0(T ) = 0 as a one-dimensionalconstraint on the control u0 – this can then be covered via a compactness-uniquenessargument, and is a rather classical procedure for one-dimensional free boundary prob-lems ([18, 23, 38]). Chaining all of these arguments together, one is led to the statementof Theorem 1.1.

An interesting byproduct of this strategy is the possibility of further localizing thecontrol domain – as stated in Theorem 1.2 –, all the while avoiding the use of Carlemaninequalities. Let us corroborate this claim.

First of all, note that the linear operator generating (1.3) is actually self-adjoint on

˙H :=

(y, h) ∈ L2(Ω)×H1(T)

∣∣∣∣∣∫Ty(x1, ·) dx1 =

∫Th(x1) dx1 = 0

, (1.10)

endowed with the norm∥∥(y, h)∥∥2

H:= ‖y‖2L2(Ω) + σ‖∂x1h‖2L2(T), (1.11)

and the inferred inner product 〈f1, f2〉H = 12‖f1‖2H + 1

2‖f2‖2H −12‖f1 − f2‖2H . The

adjoint problem for (1.3) taken on ˙H , for initial data in ˙H , then reads as

−∂tζ −∆ζ = 0 in (0, T )× Ω,

−∂t`(t, x1)− ∂x2ζ(t, x1, 1) = 0 on (0, T )× T,ζ(t, x1,−1) = 0 on (0, T )× T,ζ(t, x1, 1) = σ∂2

x1`(t, x1) on (0, T )× T,

(ζ, `)|t=T = (ζT , `T ) in Ω.

(1.12)

When decomposing the solution to (1.12) in Fourier series – the zeroth-mode being re-moved due to the definition of ˙H – we end up precisely with the adjoint of (1.9). Butif one uses the resulting observability inequality for the low frequency Fourier modesonly (namely, corresponding to |n| 6 µ, for any µ > 0), we can further leverage aclassical inequality for the eigenfunctions of the Laplacian3 (due to Lebeau and Rob-biano, [35]) to obtain an observability inequality for the solutions to (1.12) possessinglow-frequencies only. To cover the high-frequencies and complete the proof, followingthe Lebeau-Robbiano argument, we seek to exploit the exponentially stable characterof the semigroup for (1.12). But since λ = 0 is, a priori, an eigenvalue of the governing

3Here, we consider the Laplacian on the torus T, whose eigenfunctions are the complex exponentialswhich span L2(T).


operator in (1.12), identity (1.4) alone will not yield exponential decay of solutions.However, by taking initial data in ˙H , we mod out the constants, and thus λ = 0 aswell. We may consequently ensure the exponential decay of solutions to (1.12), whichwill lead us to the desired result (Theorem 1.2).

1.4. Scope. The remainder of the paper is organized as follows.• In Section 2, we provide an in-depth comparison of our results and techniqueswith existing works on the control of free boundary problems, as well as acommentary on the potential limitations and extensions of our results to moregeneral settings – the latter is also corroborated by numerical experiments.• In Section 3, we begin by presenting the basic functional setting, as well as thewell-posedness, and the symmetry properties in a mean-zero energy space of theoperator governing system (1.3).• In Section 4, we consider the family of systems, indexed by the Fourier fre-quency parameter, projected with respect to the horizontal direction x1 ∈ T.In particular, we show the null-controllability of these systems with a controlcost uniform in the frequency parameter. This is done by computing the fullspectrum of the underlying operator.• In Section 5, we make use of the result presented in Section 4 to immediatelyderive Theorem 1.1.• In Section 6, we provide the proof to Theorem 1.2, through the Lebeau-Robbianoinspired argument discussed just above.• Finally, in Section 7, we conclude with a selection of related open problems.

1.5. Notation. Whenever the dependence on parameters of a constant is not specified,we will make use of Vinogradov notation and write f .S g whenever a constant C > 0,depending only on the set of parameters S, exists such that f 6 Cg.

2. Discussion

2.1. Previous work. The null-controllability results (Theorem 1.1, Theorem 1.2) weprove in this work are among the first of their kind for multi-dimensional free-boundaryproblems in which the free boundary depends on the spatial variable – even in the lin-earized regime. In this sense, our setup differs from existing works on the controllabilityof multi-dimensional fluid-structure interaction models with rigid bodies ([11, 30]), andthe controllability of one-dimensional free boundary problems ([12, 18, 23, 38, 50], seealso [15]), as therein, the free boundary is parametrized by the graph of a time-onlydependent function, modeling a rigid body. In particular, the spatial regularity of theheight function h plays a crucial role in the analysis (or even well-posedness) results.

A partial controllability result for the two-dimensional classical Stefan problem (σ =0) is shown in [13] – only the temperature % is controlled to 0 without any considerationof the height function h(t, z1) defining the free boundary Γ(t). In fact, the geometricalsetting is also different, as the free boundary Γ(t) manifests as the entire boundary ofthe fluid domain Ω(t). Moreover, the Stefan law governing the velocity of the heightfunction is regularized by adding a Laplacian term, which significantly simplifies theanalysis.


Albeit for a system of different nature to ours, we also refer to [3] (and [1, 2, 53] forrelated results) for an exact-controllability result of the velocity and the free surfaceelevation of the water waves equations in two dimensions, by means of a single controlactuating along an open subset of the free surface. In the aforementioned works, thetwo-dimensional geometrical strip-like setting of the free boundary problem is the sameas ours. These results are extended to the three dimensional context in [56].

2.2. The (curious) case of σ = 0. We were unsuccessful in applying our techniquesto cover the case σ = 0. In this case, the linearized system (1.3) is uncoupled, andproceeding by writing the adjoint system directly might appear as an arid endeavor.We provide more insight into some of the possible obstacles.

• Since (1.3) is uncoupled, one can first control the heat equation for y to 0 throughHUM, and then see the null-controllability for h as a linear constraint of theform

0 = h(T, x1) = h0(x1) +

∫ T

0∂x1y(t, x1, 1) dt (2.1)

for all x1 ∈ T. The difference with respect to the one-dimensional case (see, e.g.,[16, 23]) is that (2.1) is not a finite-dimensional constraint anymore, due to thefact that h depends on the spatial variable. Hence, the compactness-uniquenessarguments of these works are not directly applicable.

• In this spirit, one can rather proceed by Fourier series decompositions to derive(1.9) with σ = 0. Now, for any fixed frequency n, the compactness-uniquenessarguments of the above-cited works can be used to derive the null-controllabilityof the full system, since (2.1) will transform into a one-dimensional constraintfor the heat control. The caveat is that, due to the compactness-uniquenessargument used for addressing this finite-dimensional constraint, the controlla-bility cost will depend on n, with an explicit dependence on n being uncertain.Consequently, we cannot paste the controls for all n to derive the controllabilityof (1.3) with σ = 0.

To motivate future work in this direction, we provide illustrations of numerical exper-iments for finding the minimal L2((0, T )×ω)-norm control for (1.3), in the cases σ > 0(specifically, σ = 2) and σ = 0. The results are displayed in Figure 2 and Figure 3 respec-tively. Numerical discretization and computing details may be found in Appendix A. Forsimplicity of the implementation, we worked on a rescaled domain Ω = (0, 2)× (−1, 1),whilst keeping periodic boundary conditions with respect to x1. In both cases, we tookT = 0.14, with h0(x1) = x1(2− x1) and y0(x1, x2) = 70 sin(πx1) sin(πx2) (up to x2 = 1when σ = 0, whereas we impose the compatibility condition y(0, x1, 1) = σ∂2

x1h0(x1)

when σ = 10). The numerical experiments depicted in Figure 3 insinuate that null-controllability might also hold when σ = 0. For the time being, a rigorous analyticalproof (or disproof) of such a result remains an open problem.

4Note that the simulations yield the expected results even when T 1. We present experimentswith T 1 since this is somewhat the "interesting" regime in the context of null-controllability ofheat-like equations.


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Figure 2. Controllability of (1.3) with σ = 10. (Top) The L2(ω)-norm, over time, of the minimal L2((0, T ) × ω)-norm control; (Middle)The temperature y(t, x1, x2); (Bottom) the height h(t, x1). The numer-ical experiment reflects Theorem 1.1.

2.3. Extensions. Let us conclude this section with a brief discussion on the currentlimitations and possible extensions of our results.

Remark 1 (Geometric setup). (1) Theorem 1.2 actually implies the controllabilityfrom any open and non-empty subset ω ⊂ Ω – given any such ω, one can alwaysfind a rectangle ω as in the statement, such that ω ⊂ ω, and then applyTheorem 1.2 to ω.

(2) We parametrize the horizontal variable x1 of any x = (x1, x2) ∈ Ω(t) (or inΓ(t)) over the torus T for convenience. This is actually standard in the literatureon free boundary problems. This choice allows us to avoid rather complicatedensuing arguments regarding the regularity of the moving domains. Moreover,our geometrical setting is also amenable to Fourier analysis, which gives us anatural blueprint, based on Fourier decomposition and spectral analysis for one-dimensional problems, for tackling the control problem.


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4

5

Figure 3. Controllability of (1.3) with σ = 0. (Top) The L2(ω)-norm of the minimal L2((0, T )×ω)-norm control; (Middle) The tempera-ture y(t, x1, x2); (Bottom) the height h(t, x1). The numerical experimentstipulates that null-controllability holds even when the components areuncoupled, and thus h(t, x1) is only controlled indirectly, through thetime-average of the Neumann trace of y, evaluated at x2 = 1. Note thathere, the scale is 10 times lower than in Figure 2, to better capture thecontrolled evolution.

(3) There is no strict need of working in the two-dimensional setting regarding thelinear problem (1.3). We do so mainly for convenience. The one-dimensionaltorus T may be replaced by Td−1, d > 3, throughout, without changing the strat-egy in the slightest. Indeed, just like when d = 2, Fourier decomposition can beperformed with respect to the first d − 1 variables (x1, . . . , xd−1), when d > 3 –only the "vertical" one (namely xd) will remain in the projected, one-dimensionalsystem, and the strategy of proof for Theorem 1.1 and Theorem 1.2 can be main-tained.

(4) The higher-dimensional (d > 3) setting may pose an obstacle in a prospectiveextension of Theorem 1.1 and Theorem 1.2 to the nonlinear setting, where the


dimension would play a role regarding the spatial regularity of solutions (in par-ticular that of the height function h(t, ·)). More specifically, one might have toensure that the controls for the linearized system are more regular than simplyL2tL

2x. We leave this open for a future study – a possible roadmap would include

using the penalized-HUM method, as done in [33].

Remark 2 (The nonlinear problem). In view of Theorem 1.1 and Theorem 1.2, onecould stipulate a local controllability result for the nonlinear problem (1.2) by means of afixed point and smallness of the initial data. This could be accomplished by first addingsource terms through the so-called source term method of [38] (see also [7, 21, 29, 33]),and then performing a Banach fixed point for small enough initial data.

Currently, there are a couple of issues vis-à-vis this strategy.

(1) First of all, with regard to the source-term method of [38], we need to ensure thatthe controllability cost in Theorem 1.1 is of the form ∼ e

1T when T 1. Due

to the compactness-uniqueness argument through which the zeroth Fourier modesystem is controlled, the explicit time-dependence of the controllability cost islost for the full system, and so the source term method is not directly applicable.In other words, further study is needed regarding the controllability cost of thesystem corresponding to the first Fourier mode.

(2) When arguing along the lines of the Lebeau-Robbiano method for Theorem 1.2,we can sharpen the proof to ensure that the control cost is of such exponentialform. But now the issue is the functional setup of ˙H , as there is no guaranteethat the quadratic nonlinear terms will be of mean zero, therefore posing animpediment regarding the invariance of the fixed point map.

All in all, the nonlinear adaptation of Theorem 1.1 and/or Theorem 1.2 remains open.

3. The linear semigroup

3.1. Functional setting. Recalling that Ω := T× (−1, 1), let us consider the Hilbertspace

H := L2(Ω)×H1(T),

which, by Plancharel’s theorem, may be endowed with the norm

∥∥(f, g)∥∥2

H:=∑n∈Z

∥∥fn∥∥2

L2(−1,1)+(

1 + σn2)∣∣gn∣∣2, (3.1)

where the Fourier coefficients (yn, gn) are defined as in (1.8). The inner product on Hcan then be inferred from (3.1).

We look to rewrite (1.3) in a canonical first-order form evolving on state space H .The well-posedness thereof will follow by studying the governing operator generatingthe semigroup of (1.3). To write (1.3) as an abstract control system evolving in H , weintroduce the unbounded operator A : D(A)→H , defined by

A(y, h)

:=(∆y, ∂x2y(·, 1)

),


with domain

D(A) =

(y, h) ∈ H2(Ω)×H7/2(T)

∣∣∣∣∣ y(x1, 1) = σ∂2x1h(x1) on T, (3.2)

y(x1,−1) = 0 on T,

∂x2y(·, 1) ∈ H1(T)

.

It can readily be seen that the operator A : D(A)→H is closed and densely defined.By using the above definitions, we can rewrite (1.3) (for a general source term f insteadof u1ω) as a first order system in the Hilbert state space H :

∂t(y, h)

= A(y, h)

+(f, 0)

in (0, T ),(y, h)|t=0

=(y0, h0

).

(3.3)

3.2. Well-posedness of (3.3). To study the well-posedness of (3.3), we shall rely ona Fourier decomposition in the horizontal (x1 ∈ T) variable. This will lead to a systemakin to (1.9). Specifically, for any n ∈ Z we consider the Hilbert space

Hn := L2(−1, 1)× R,

which we endow with the inner product⟨(f1, g1), (f2, g2)

⟩Hn

:=⟨f1, f2

⟩L2(−1,1)

+ σn2g1g2,

when n 6= 0, and the canonical inner product when n = 0. We then define, for anyn ∈ Z, the operator An : D(An)→Hn by

An

(y, h)

:=(∂2x2y − n2y, ∂x2y(1)

),

with domain

D(An) =

(y, h)∈ H2(−1, 1)× R

∣∣∣∣∣ y(−1) = 0, y(1) = −σn2h

.

We will show that the operators A and An (n ∈ Z) generate analytic semigroups onH , and Hn, respectively. The fact that An generates an analytic semigroup on Hn

actually follows directly from [40, Theorem 1.25]. To show that A generates an analyticsemigroup, we shall express its resolvent in terms of the resolvent of the operators An.Thus, we shall need resolvent estimates for An which are uniform with respect to n ∈ Z.

For θ ∈(π2 , π

)and β > 0, we define the sector

Σθ,β :=λ ∈ C \ 0

∣∣∣ |arg(λ)| < θ, |λ| > β.

The following result then holds.

Proposition 3.1 (Regarding An). Suppose n ∈ Z and σ > 0.(1) If n 6= 0, the operator An : D(An)→Hn is self-adjoint, has compact resolvents,

and its spectrum spec(An) consists only of negative eigenvalues.


(2) There exist θ ∈(π2 , π

), β > 0 and Cθ,β > 0, all independent of n, such that

|λ|∥∥∥(λ Id−An)−1

∥∥∥L (Hn)

6 Cθ,β (3.4)

holds for all λ ∈ Σθ,β.

Consequently, for any n ∈ Z, the operator An generates an analytic semigroupetAn

t>0

on Hn.

Proof. The fact both claims imply that An generates an analytic semigroup followsfrom [10, Theorem 2.11 (p. 112), Proposition 2.11 (p. 122)].

Let us begin by proving the first claim. If n 6= 0, through standard integration byparts, is readily seen that An is self-adjoint and has compact resolvents. Therefore, itsspectrum spec(An) is a discrete subset of R. Let us conclude the proof the first claimby showing that spec(An) ⊂ (−∞, 0). We argue by contradiction. Let λ ∈ spec(An)with λ > 0. Thus there exists a vector (y, h) ∈ D(An) \ 0 such that

λy − ∂2x2y + n2y = 0 in (−1, 1),

λh− ∂x2y(1) = 0,

y(−1) = 0,

y(1) = −σn2h.

We now multiply the first equation by y and integrate by parts to obtain

λ

∫ 1

−1|y|2 dx2 +

∫ 1

−1|∂x2y|2 dx2 − ∂x2y(1)y(1) + n2

∫ 1

−1|y|2 dx2 = 0.

Using the boundary conditions, this identity entails

λ

∫ 1

−1|y|2 dx2 +

∫ 1

−1|∂x2y|2 dx2 + λσn2h2 + n2

∫ 1

−1|y|2 dx2 = 0.

Since λ > 0 and n 6= 0, from this identity, we may readily conclude that y ≡ h ≡ 0.This is a contradiction, and hence λ ∈ (−∞, 0).

We now look to prove the second claim. Suppose that n ∈ Z and (fn, gn) ∈ Hn arearbitrary. Let us consider the resolvent problem

λyn − ∂2x2yn + n2yn = fn in (−1, 1),

λhn = ∂x2yn(1) + gn

yn(−1) = 0

yn(1) = −σn2hn.

(3.5)

If n = 0, the operator A0 generates an analytic semigroup (see [40, Theorem 1.25]).Therefore, there exist θ0 ∈

(π2 , π

), β0 > 0, and Cθ0,β0 > 0 such that

|λ|∥∥(y0, h0

)∥∥L2(−1,1)×C 6 Cθ0,β0

∥∥(f0, g0

)∥∥L2(−1,1)×C , (3.6)

holds for all λ ∈ Σθ,β. Now let n 6= 0 be arbitrary and fixed. Since spec(An) ⊂ (−∞, 0),we also have

Σθ0,β0 ⊂ res(An),


where res(An) denotes the resolvent set of An. Consequently, for any n 6= 0, λ ∈ Σθ,β,and (fn, gn) ∈Hn, (3.5) admits a unique solution (yn, hn) ∈ D(An). Let us now fix

λ = β0eiθ0 ∈ Σθ0,β0 .

Multiplying the first equation in (3.5) by e−iθ0/2 and taking the inner product with yn,we obtain

β0eiθ02

∫ 1

−1|yn|2 dx2 + e−i

θ02

∫ 1

−1|∂x2yn|2 dx2 − e−

iθ02 ∂x2yn(1)yn(1)

+e−iθ02 n2

∫ 1

−1|yn|2 dx2 = e−

iθ02

∫ 1

−1fnyn dx2.

Using the boundary conditions, the above identity can be rewritten as

β0eiθ02

∫ 1

−1|yn|2 dx2 + e−i

θ02

∫ 1

−1|∂x2yn|2 dx2 + β0e

iθ02 σn2 |hn|2 + e−i

θ02 n2

∫ 1

−1|yn|2 dx2

= e−iθ02

∫ 1

−1fnyn dx2 + e−i

θ02 σn2hngn

= e−iθ02

⟨(yn, hn

),(fn, gn

)⟩Hn

.

By taking the real part on both sides in the above identity, and subsequently usingCauchy-Schwarz, we find

β0

(∫ 1

−1|yn|2 dx2 + σn2 |hn|2

)+

∫ 1

−1|∂x2yn|2 + n2

∫ 1

−1|yn|2 dx2

6∥∥(yn, hn)∥∥Hn

∥∥(fn, gn)∥∥Hn

Taking into account the fact that |λ| = β0, we deduce that

|λ|∥∥(yn, hn)∥∥Hn

6∥∥(fn, gn)∥∥Hn

.

Since n 6= 0 and λ ∈ Σθ0,β0 were taken arbitrary, the above estimate in junction with(3.6) leads us to (3.4).

The above result then leads us to the following.

Proposition 3.2 (Regarding A). Suppose σ > 0. There exist θ ∈(π2 , π

), β > 0 and

Cθ,β > 0 such that|λ|∥∥∥(λ Id−A)−1

∥∥∥L (H )

6 Cθ,β (3.7)

holds for all λ ∈ Σθ,β. Consequently, the operator A : D(A)→H generates an analyticsemigroup

etAt>0

on H .

Proof of Proposition 3.2. Let θ ∈(π2 , π

)and β > 0 be the constants stemming from

Proposition 3.1. For λ ∈ Σθ,β and (f, g) ∈H , we consider the eigenvalue problemλy −∆y = f on Ω,

λh− ∂x2y(·, 1) + g on T,y(·,−1) = 0 on T,y(·, 1) = σ∂2

x1h(·) on T.

(3.8)


We decompose all functions appearing in the above eigenvalue problem in Fourier serieswith respect to the periodic, x1–variable, as in (1.7) – (1.8). We see that for any n ∈ Z,the pair (yn, hn) ∈Hn of Fourier coefficients solves (3.5), and moreover, because of theFourier series expansion of (y, h) as in (1.7),

(λ Id−A)−1 (f, g) =1√2π

∑n∈Z

(λ Id−An)−1 (fn, gn)ein·also holds. Combining the above relation with (3.4), we immediately obtain (3.7). Thiscompletes the proof.

As a consequence of the above result, we have

Proposition 3.3 (Resolvent of A). Let λ ∈ Σθ,β, where θ ∈ (π2 , π) and β > 0 stemfrom Proposition 3.2. For any (f, g) ∈H , the eigenvalue problem (3.8) admits a uniquesolution (y, h) ∈ D(A). In particular, the resolvent of A is compact in H .

Proof of Proposition 3.3. From Proposition 3.2, we know that (y, h) ∈H . We considerthe elliptic boundary value problem (3.8)1−3 satisfied by y. Note that

∂x2y(·, 1) = λh− g ∈ H1(T).

Therefore, y ∈ H2(Ω) and y(·, 1) ∈ H3/2(T). Finally, using (3.8)4 we obtain h ∈ H7/2(T).Whence (y, h) ∈ D(A), and the compactness readily follows.

Remark 3 (Fourier decomposition of the semigroup). In view of Proposition 3.1 andProposition 3.2, we have

etA(y0, h0

)=

1√2π

∑n∈Z

etAn(y0n, h

0n

)ein·, (3.9)

where(y0n, h

0n

)denote the Fourier coefficients of an initial datum

(y0, h0

), defined as in

(1.8).

Taking stock of Proposition 3.2, and using standard results from parabolic equations(see e.g. [10, Thm. 2.12, Sect. 2]), we deduce the well-posedness of the linear system(3.3).

Corollary 3.1. Suppose T > 0 and σ > 0. For every(y0, h0

)∈ L2(Ω) × H1(T) and

f ∈ L2(0, T ;L2(Ω)), there exists a unique solution (y, h) ∈ C0([0, T ];L2(Ω) × H1(T))to (3.3).

To prove Theorem 1.2, we also need to ensure the well-posedness of the linear system(3.3) on the energy space consisting of mean zero functions, namely ˙H defined in (1.10).Due to Plancharel’s theorem, the canonical norm on ˙H , induced from the inner product

〈(f1, g1), (f2, g2)〉H := 〈f1, f2〉L2(Ω) + σ 〈∂x1g1, ∂x1g2〉L2(T) ,

is clearly equivalent to ∑n∈Zn6=0

∥∥fn∥∥2

L2(−1,1)+ σn2

∣∣gn∣∣2

1/2

.


From (3.9), it is clear that the space ˙H is invariant under the action ofetAt>0

, andthe operator A may thus be restricted to ˙H .

Definition 3.1. [55, Definition 2.4.1] We define the part of A in ˙H as the restrictionof A to the domain D(A) ∩ ˙H .

Proposition 3.4. The part of A in ˙H is self-adjoint.

Proof of Proposition 3.4. First of all, by integrating by parts and using the Schwarztheorem vis à vis the symmetry of second derivatives, we may readily find⟨

A(y, h),(z, r)⟩

H=⟨(y, h),A(z, r)⟩

H

for all (y, h), (ζ, `) ∈ (D(A) ∩ ˙H )2. This shows that D(A) ∩ ˙H ⊂ D(A∗), and that A

is symmetric. To conclude, we show that D(A) ∩ ˙H = D(A∗). To this end, pick anarbitrary (ζ, `) ∈ D(A∗). Then, there exists (f, g) ∈ ˙H such that⟨

A(y, h),(ζ, `)⟩

H=⟨(y, h),(f, g)⟩

H

holds for all (y, h) ∈ D(A) ∩ ˙H . Let us now set

(ϕ, s) := A−1(f, g) ∈ D(A) ∩ ˙H .

Then from the above identities, we may infer that⟨(y, h),(f, g)⟩

H=⟨(y, h),A(ϕ, s)⟩

H=⟨A(y, h),(ϕ, s)⟩

H,

holds for all (y, h) ∈ D(A) ∩ ˙H . Therefore, we may also conclude that⟨A(y, h),(ϕ, s)−(ζ, `)⟩

H= 0,

for all (y, h) ∈ D(A) ∩ ˙H , whence (ζ, `) = (ϕ, s) ∈ D(A) ∩ ˙H . This completes theproof.

4. A family of projected systems

To prove Theorem 1.1, we will make use of the periodicity with respect to the x1

variable of the functions appearing in (1.3). We write the Fourier series expansions of(y, h, u) as in (1.7), with the corresponding Fourier coefficients (yn, hn, un) being definedas in (1.8). It is readily seen that for any n ∈ Z, these Fourier coefficients satisfy (1.9).

Our objective in this section is thus to prove the null-controllability of (1.9), with acontrollability cost which is uniform in n ∈ Z. This will allow us to simply sum up allof the coefficients as in (1.7), and deduce Theorem 1.1. We postpone the full conclusionto the next section, and solely focus on the result for the Fourier coefficients here.

Proposition 4.1. Suppose T > 0 and σ > 0 are fixed. Suppose (c, d) ⊂ (−1, 1). For anyn ∈ Z, and for any pair of initial data (y0

n, h0n) ∈ L2(−1, 1)×R, there exists some control

un ∈ L2 ((0, T )× (c, d)) such that the unique solution (yn, hn) ∈ C0([0, T ];L2(−1, 1)×R)to (1.9) satisfy yn(T, ·) ≡ 0 in (−1, 1) and hn(T ) = 0. Moreover, there exist a constantC(T, σ) > 0 independent of n such that∥∥un∥∥L2((0,T )×(c,d))

6 C(T, σ)∥∥(y0

n, h0n

)∥∥Hn

.


The full proof may be found in Section 4.2. More specifically, when n 6= 0, we usethe customary Hilbert Uniqueness Method (HUM), which renders the controllabilityproblem equivalent to a proof of an observability inequality for the adjoint system. Theobservability inequality will be shown by means of spectral arguments (based on resultspresented in Section 4.1, which come with some degree of difficulty). For the zerothmode n = 0, we shall note that the eigenfunctions of the governing linear operator arenot orthogonal (the operator is not self-adjoint), but the system is of cascade type andfalls into the setting of [23].

4.1. The spectrum of An. Recall that, when n ∈ Z \ 0, the operator An isself-adjoint due to the specific inner product we endowed to Hn (see Proposition 3.1).And since An has compact resolvents, by the Hilbert-Schmidt theorem, it may bediagonalized to find an orthonormal basis of Hn consisting of eigenfunctions of An,associated to a decreasing sequence of eigenvalues.

To prove the controllability of (1.9) when n 6= 0 by using spectral arguments, weneed to explicitly characterize the spectrum of An. This is the goal of the followingresult.

Lemma 4.1 (The spectrum of An). Let σ > 0 and n ∈ Z \ 0 be fixed. Then, thesequence λn,k+∞k=0, with λn,k < 0, of eigenvalues of An, reads as follows:

λn,k+∞k=0 =

−(

(k + 1)π

2+π

4− εk+1

)2

− n2

+∞

k=0

if σ > 2, for some εk ∈(0, π4

), and

λn,k+∞k=0 =

−(kπ

2+π

4− εk

)2

− n2

+∞

k=1

⋃ ν2

0 − n2 if σ < 2

−n2 if σ = 2

otherwise, for some ν0 ∈(0, 1

σ

)independent of n and k. (Here, we imply that the second

"set" in the union is indexed as k = 0.)Furthermore, the following properties hold.(1) The sequence λn,k+∞k=0 is separated uniformly with respect to n ∈ Z \ 0, in

the sense thatinfk>0

∣∣∣λn,k+1 − λn,k∣∣∣ > γ (4.1)

holds for some γ > 0 independent of n.(2) Moreover,

− λn,k = rk2 + n2 + Ok→+∞

(k) (4.2)

for some r > 0 independent of n and σ.(3) Suppose that (c, d) ⊂ (−1, 1) is fixed, with c < d. Then, there exists some

constant C = C(σ, c − d) > 0 such that for any n ∈ Z \ 0 and k > 0, thenormalized eigenfunctions

Φn,k :=

(ϕn,k, −

1

σn2ϕn,k(1)

)


of An are such that‖ϕn,k‖L2(c,d) > C (4.3)

holds.

Before proceeding with the proof of Lemma 4.1, let us make the following importantobservation.

Lemma 4.2 (Spectral gap). Suppose σ > 0 and n ∈ Z \ 0. Then An has a spectralgap – namely,

λ 6 −minσ

2, 1n2

for all λ ∈ spec(An).

In particular, λ 6 −n2 if σ > 2. We shall make use of this property in the proof ofLemma 4.1, in particular distinguishing the cases σ ∈ (0, 2) and σ > 2, for convenience.The factor 2 in the denominator refers to the Lebesgue measure of the interval [−1, 1],so more general intervals can be envisaged.

Proof of Lemma 4.2. Let λ < 0 be such thatλy − ∂2

x2y + n2y = 0 in (−1, 1),

λh− ∂x2y(1) = 0,

y(−1) = 0,

y(1) = −σn2h,

(4.4)

holds for some (y, h) ∈ D(An) \ (0, 0). This entails the identity

− λ(∫ 1

−1|y|2 dx2 + σn2h2

)=

∫ 1

−1|∂x2y|2 dx2 + n2

∫ 1

−1|y|2 dx2. (4.5)

Now note that from the boundary condition y(1) = −σn2h and an elementary Sobolevembedding5 for solutions to (4.4), we derive

σ2n4h2 = |y(1)|2 6 2

∫ 1

−1|∂x2y|2 dx2. (4.6)

Plugging (4.6) in (4.5), we find

−λ(∫ 1

−1|y|2 dx2 + σn2h2

)>σ

2σn4h2 + n2

∫ 1

−1|y|2 dx2

> minσ

2, 1n2

(σn2h2 +

∫ 1

−1|y|2 dx2

),

as desired.

Proof of Lemma 4.1. We recall that An : D(An) → Hn is self-adjoint, has compactresolvents, and its spectrum consists of a decreasing sequence of negative eigenvalues,namely a sequence λn,k+∞k=0 with −∞ < . . . 6 λn,k 6 . . . 6 λn,0 < 0. We shalldistinguish three different scenarios for the computation of these eigenvalues.

5We simply use y(x) = y(x)−y(−1) =∫ x−1∂zy(z) dz combined with the Cauchy-Schwarz inequality.

The factor 2 occurs as meas([−1, 1]) = 2.


Case 1: λ < −n2. Suppose that λ ∈ (−∞, 0) is an eigenvalue of An which alsosatisfies λ < −n2. So, there must exist a vector (ϕ, `) ∈ D(An) \ (0, 0) such that

∂2x2ϕ+ (−λ− n2)ϕ = 0 in (−1, 1),

∂x2ϕ(1) = λ`

ϕ(−1) = 0

ϕ(1) = −σn2`.

(4.7)

In other words, ϕ would solve the mixed Dirichlet-Robin problem∂2x2ϕ+ (−λ− n2)ϕ = 0 in (−1, 1),

ϕ(−1) = 0

ϕ(1) + σn2

λ ∂x2ϕ(1) = 0.

(4.8)

Since −λ− n2 > 0, one may readily see that the solutions to (4.8) are of the form

ϕ(x2) = C sin(ν(1 + x2)

),

with C > 0, where ν :=√−λ− n2 is the positive root of the transcendental equation(

ν2

n2+ 1

)tan(2ν) = σν. (4.9)

Locating the positive roots of this equation suggest a study of the fixed points of thefunction f(ν) =

(ν2

n2 + 1)

tan(2ν), defined and non-decreasing on the union of consec-utive intervals of the form

+∞⋃k=1

(π

4+

(k − 1)π

2,π

4+kπ

2

).

0 1 2 3 4 5 6 7

8

-15

-10

-5

0

5

10

15

20

0 1 2 3 4 5 6 7

8

-5

0

5

10

15

20

25

30

35

Figure 4. The function f(ν) = (ν2/n2 + 1) tan(2ν) (blue), with ν 7→ σxsuperposed (black), with n = 104 and σ = 1/2 (left) and σ = 5 (right).We see how the fixed points of f are localized over each subinterval.


Moreover, for k > 1,

limνπ

4+

(k−1)π2

f(ν) = −∞, f

(kπ

2

)= 0, lim

νπ4

+ kπ2

f(ν) = +∞.

Thus, (4.9) has a sequence of positive roots νk+∞k=1 of the form

νk =kπ

2+π

4− εk (4.10)

for k > 1, where εk ∈(0, π4

)may a priori depend on σ and n. Consequently, the

eigenvalues λn,k < 0 in this case are of the form

− λn,k =

(kπ

2+π

4− εk

)2

+ n2 (4.11)

for k > 1.

Case 2: λ = −n2. We note that λ = −n2 is an eigenvalue if and only if σ = 2. Indeed,should λ = −n2 be an eigenvalue, then ϕ in (4.7) would be harmonic, and thus an affinefunction: ϕ(x2) = C0x2 +C1, for some C0, C1 ∈ R. Using the boundary conditions, wemoreover find that C0 = C1, as well as 2C0 = λσ`. Since ∂x2ϕ(1) = C0 = λ`, we areled to σ = 2. Summarizing, we find that when λ = −n2,

ϕ(x2) = C0(1 + x2).

Case 3: λ ∈ (−n2, 0). Following Lemma 4.2 regarding the spectral gap, this case mayonly occur if σ < 2. So let us henceforth suppose that σ < 2, and that λ ∈ (−n2, 0) isan eigenvalue of An. Consequently, there must exist a vector (ϕ, `) ∈ D(An) \ (0, 0)such that (4.7) holds. Then ϕ would again solve the mixed Dirichlet-Robin problem(4.8). Since now −λ− n2 < 0, one may readily see that the solutions to (4.8) are nowof the form

ϕ(x2) = C(eνx2 − e−ν(2+x2)

),

for some C > 0, where ν :=√n2 + λ denotes the positive root(s) of the transcendental

equation

eν − e−3ν − σn2

n2 − ν2

(νeν + νe−3ν

)= 0

in (0, |n|). We may equivalently rewrite the above equation as(− ν2 − σn2ν + n2

)+(ν2 − σn2ν − n2

)e−4ν = 0. (4.12)

We designate (4.12) as f(ν) = 0, and we claim that f has a unique root6 in (0, |n|).Let us henceforth focus on proving this claim. Existence follows from the fact thatf is increasing and positive near 0, and decreasing and negative near |n|. To ensureuniqueness, we will look to show that f is strictly concave in

[0, |n|

](see Figure 5). We

shall designatef1(ν) := −ν2 − σn2ν + n2

6One may try to compute this root by using the Lambert W function and its generalizations ([42]).We omit this from our work as it is not needed for our analysis.


and

f2(ν) :=(ν2 − σn2ν − n2

)e−4ν ,

so that f = f1 + f2. We see that

f ′′1 (ν) = −2. (4.13)

On another hand, we also have

f ′′2 (ν) =(

2− 16ν + 8σn2 + 16ν2 − 16σn2ν − 16n2)e−4ν . (4.14)

As f = f1 + f2, and taking (4.13) into account, we see that it suffices to ensure thatf ′′2 (ν) < 2 in [0, |n|]. This may be shown without too much difficulty. Indeed, wheneverν ∈

[0, 1

2

]we see that 8σν2 − 16n2 < 0 as well as 16ν2 − 16ν < 0 in (4.14). Similarly,

whenever ν ∈[

12 , |n|

], we find that 8σn2 − 16σn2ν < 0 as well as 16ν2 − 16n2 < 0

in (4.14). This yields f ′′2 (ν) < 2 in [0, |n|], whence f is strictly concave in [0, |n|].Consequently, it follows that f has at most 2 roots in [0, |n|]. (This elementary propertyis readily shown by arguing by contradiction and Rolle’s theorem.) And since f(0) = 0,we conclude that any root of f in (0, |n|) is unique. On another hand, we may readilysee that

f2(ν) < 0 for ν ∈[0, |n|

],

as well as

f1(ν) 6 0 for ν ∈[

1

2

(√n4σ2 + 4n2 − n2σ

), |n|

).

Consequently we find that

f(ν) < 0 for ν ∈[

1

2

(√n4σ2 + 4n2 − n2σ

), |n|

),

and therefore the unique root ν0 of f must be located in the complement of the aboveinterval, namely ν0 ∈

(0, 1

2

(√n4σ2 + 4n2 − n2σ

)). Hence, in this case, the first eigen-

value λn,0 < 0 of An will have the form

λn,0 = ν20 − n2. (4.15)

Note that, moreover,

ν0 61

2

(√n4σ2 + 4n2 − n2σ

)=

4n2

2(√

n4σ2 + 4n2 + n2σ) 6 1

σ,

for all n ∈ Z \ 0.Having analyzed all the possible scenarios, we collect the sequence of eigenvalues

λn,k+∞k=0, which if σ > 2 may actually be indexed as λn,k+∞k=1 with λn,k with k > 1defined in (4.11); if σ = 2, we also have λn,0 = −n2, and if σ ∈ (0, 2), we have λn,0 < 0defined in (4.15). One thus readily sees that (4.2) holds. On another hand, since


-10 -8 -6 -4 -2 0 2 4 6 8 10

8

-50

-40

-30

-20

-10

0

10

20

30

40

50

f1(8)

f2(8)

f(8)

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

8

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1#10 11

f1(8)

f2(8)

f(8)

-4 -2 0 2 4 6 8

-1

-0.5

0

0.5

1

108

Figure 5. Plot of f1 (blue), f2 (green), and f = f1+f2 (red), for σ = 25 ,

with n = 5 (left) and n = 104 (right). We see that f has a unique rootν0 ∈ (0, |n|), and this root does not collapse to 0 as n→ +∞.

εk ∈(0, π4

), we see that for k > 1,

−λn,k+1 + λn,k =

((k + 1)π

2+π

4− εk+1

)2

−(kπ

2+π

4− εk

)2

=

(kπ + π − εk+1 − εk

)(π

2− εk+1 + εk

)

>3π

2· π

4=

3π2

8.

Furthermore, if σ ∈ (0, 2),

−λn,1 + λn,0 =

(π

2+π

4− εk

)2

+ ν20 >

π2

4,

whereas if σ = 2, we similarly find

−λn,1 + λn,0 =

(π

2+π

4− εk

)2

>π2

4.

Hence, the separation condition (4.1) holds as well (actually, also uniformly in σ > 0).Let us finally prove (4.3). We recall that the normalized eigenfunctions Φn,k have the

form

Φn,k =

(ϕn,k, −

1

σn2ϕn,k(1)

),


where ϕn,k is given by

ϕn,k(x2) = Cn,k sin

(√−λn,k − n2 (1 + x2)

)for k > 1 (following the indexing of the eigenvalues depending on where σ is located),while

ϕn,0(x2) = Cn,0

(e√n2+λn,0 x2 − e−

√n2+λn,0(2+x2)

)if σ ∈ (0, 2), and

ϕn,0(x2) = Cn,0

(1 + x2

)if σ = 2. Let us first suppose k > 1. Reusing the notation νk :=

√−λn,k − n2 > 0, we

note that in order to ensure that the eigenfunctions Φn,k are of norm 1 in Hn, namely‖Φn,k‖Hn = 1, we see that Cn,k > 0 needs to satisfy

C2n,k

(1− sin(4νk)

4νk

)+C2n,k

n2σsin2(2νk) = 1

for all k > 1. On the other hand, using elementary trigonometric identities, we mayalso find

‖ϕn,k‖2L2(c,d) =C2n,k

2

((d− c) +

sin (2νk(c+ 1))

2νk− sin (2νk(d+ 1))

2νk

). (4.16)

In view of (4.10), namely the asymptotics of νk when k → +∞, we find that there existsδ = δ(d− c) > 0 independent of n ∈ Z \ 0 and k such that

(d− c) +sin (2νk(c+ 1))

2νk− sin (2νk(d+ 1))

2νk> δ (4.17)

holds for all k > 1. Therefore, we see from (4.16) and (4.17) that, in order to obtain(4.3), it suffices to have an appropriate lower bound on Cn,k for all k > 1. To this end,we note that

C2n,k =

((1− sin(4νk)

4νk

)+

sin2(2νk)

n2σ

)−1

.

By virtue of (4.10), we see that

sin(4νk)

4νk−→k→+∞

0 and sin2(2νk) −→k→+∞

0,

henceC2n,k > C∗

for some C∗ > 0 independent of n, k and σ. This concludes the proof of (4.3) whenk > 1.

Let us now consider the case k = 0 and σ ∈ (0, 2]. First suppose that σ ∈ (0, 2). Wesee that to ensure orthonormality, Cn,0 > 0 needs to satisfy

C2n,0

sinh(2ν0)− 4ν0

ν0e−2ν0 +

C2n,0

n2σ

(eν0 − e−3ν0

)2= 1,


thus

C2n,0 =

sinh(2ν0)− 4ν0

ν0e−2ν0 +

(eν0 − e−3ν0

)2

n2σ

−1

. (4.18)

Taking (4.18) into stock, we see that since ν0 ∈(0, 1

σ

)and sinh(2x)

x is positive andcontinuous for x ∈

[0, 1

σ

],

sinh(2ν0)− 4ν0

ν0e−2ν0 +

(eν0 − e−3ν0

)2

n2σ6

sinh(2ν0)

ν0+

1

σe

1σ 6 C1(σ) +

1

σe

1σ

holds for some C1(σ) > 0 independent of n. Consequently,

C2n,0 >

1

C1(σ) + 1σe

1σ

. (4.19)

We also have

‖ϕn,0‖2L2(c,d) = C2n,0

(− 2ν0(d− c)− sinh

(2(c+ 1)ν0

)+ sinh

(2(d+ 1)ν0

))ν0

e−2ν0 .

Similarly, using the continuity and the positivity of the function

x 7→ sinh(2(d+ 1)x)− sinh(2(c+ 1)x)− 2(d− c)xx

on(0, 1

σ

), and using (4.19), we conclude that there exists C(σ, d− c) > 0 independent

of n such that‖ϕn,0‖2L2(c,d) > C(σ, d− c),

holds for all n ∈ Z \ 0. This is precisely (4.3).Finally, in the case k = 0 and σ = 2, since the (normalized) eigenfunction is an affine

function, the proof of (4.3) is straightforward. This concludes the proof.

4.2. Proof of Proposition 4.1. We may conclude the study of the family of pro-jected systems.

Proof of Proposition 4.1. We distinguish two separate cases.Case 1: n = 0. By virtue of the results in [16, 23], we know that there exists somecontrol u0 such that (1.9) with n = 0 is null-controllable in time T > 0. Moreover, thereexists a constant C(T, σ) > 0 such that∥∥u0

∥∥2

L2((0,T )×(c,d))6 C(T, σ)

∥∥∥(y00, h

00

)∥∥∥2

L2(−1,1)×R. (4.20)

The proof of this result consists in seeing that since n = 0, the resulting system (1.9) isof cascade type, with

h0(t) = h00 +

∫ t

0∂x2yn(s, 1) ds.

Controlling h0(t) to 0 in time T can then be seen as a one-dimensional constraint onthe heat control found by HUM, and may be achieved by a compactness-uniquenessargument within the global Carleman inequality for the heat operator. We refer to[16, 23] for all the necessary details.


Case 2: n 6= 0. We thus focus on the case n 6= 0. We recall that, by the customaryHUM ([37]), (1.9) is null-controllable in any time T > 0 by means of a control un ofminimal L2((0, T )× ω) norm, satisfying∥∥un∥∥2

L2((0,T )×(c,d))6M1e

M2T

∥∥∥(y0n, h

0n

)∥∥∥2

Hn

,

for some M1 > 0 and M2 > 0 independent of n and T , if and only if the functional

J (ζT , `T ) :=1

2

∫ T

0

∫ d

c|ζ|2 dx2 dt−

⟨(ζ(0), `(0)

),(y0n, h

0n

)⟩Hn

,

where (ζ, `) is the unique solution to the adjoint system−∂t

(ζ, `)

= An

(ζ, `)

in (0, T )(ζ, `)|t=T

=(ζT , `T

),

(4.21)

has a unique minimizer. (We have dropped the subscripts n to declutter the notation.)This, as is well-known, is nothing else but an integration by parts argument on one side,combined with the Euler-Lagrange equation for the functional J at its minimizer forthe converse. On another hand, by the direct method in the calculus of variations, Jhas a minimizer if the observability inequality

M1eM2T

∫ T

0‖ζ(t, ·)‖2L2(c,d) dt >

∥∥(ζ(0, ·), `(0))∥∥2

Hn(4.22)

holds for some M1 > 0 and M2 > 0 independent of n and T , and for all (ζT , `T ) ∈Hn,where (ζ, `) is the solution to (4.21). As An has an orthonormal basis of eigenfunctionsΦn,k+∞k=0 and corresponding decreasing sequence of negative eigenvalues −λn,k+∞k=0,we may write the Fourier series decomposition of ζ as

ζ(t, x2) =+∞∑k=0

e−λn,k(T−t)⟨(ζT , `T

),Φn,k

⟩Hn

ϕn,k(x2).

Denoting by ψj+∞j=0 the orthonormal basis of L2 (c, d), and via the shift T − t 7→ t, weobtain∫ T

0‖ζ(t, ·)‖2L2(c,d) dt =

+∞∑j=0

∫ T

0

∣∣∣∣∣+∞∑k=0

e−λn,kt⟨(ζT , `T

),Φn,k

⟩Hn

〈ϕn,k, ψj〉L2(c,d)

∣∣∣∣∣2

dt.

(4.23)

Now, making use of (4.1) and (4.2), we deduce from [54, Cor. 3.6] that there existM1 > 0 and M2 > 0 depending only on r > 0 and γ > 0 such that

M1eM2T

∫ T

0

∣∣∣∣∣+∞∑k=0

ake−(λn,k−n2

)t

∣∣∣∣∣2

dt >+∞∑k=0

|ak|2e−2(λn,k−n2

)T (4.24)


for any ak+∞k=0 ∈ `2(R), and hence

M1eM2T

∫ T

0

∣∣∣∣∣+∞∑k=0

ake−λn,kt

∣∣∣∣∣2

dt >M1eM2T e−n

2T

∫ T

0

∣∣∣∣∣+∞∑k=0

ake−(λn,k−n2

)t

∣∣∣∣∣2

dt

>(4.24)

e−n2T

+∞∑k=0

|ak|2e−2(λn,k−n2

)T

=+∞∑k=0

|ak|2e−2λn,kT .

The above estimate, combined with (4.23), implies that

M1eM2T

∫ T

0‖ζ(t, ·)‖2L2(c,d) dt >

+∞∑j=0

+∞∑k=0

e−2λn,kT

∣∣∣∣⟨(ζT , `T ),Φn,k

⟩Hn

∣∣∣∣2 ∣∣∣〈ϕn,k, ψj〉L2(c,d)

∣∣∣2 .After employing the Fubini theorem, we may apply (4.3) to the above estimate, revertthe time shift, and deduce that

M1eM2T

∫ T

0‖ζ(t, ·)‖2L2(c,d) dt > C

∥∥(ζ(0, ·), `(0))∥∥2

Hn,

which holds for all (ζT , `T ) ∈ Hn. Here, the constant C = C(σ, c − d) > 0 stemsfrom (4.3) in Lemma 4.1. This concludes the proof of (4.22), and thus the proof of theproposition.

5. Proof of Theorem 1.1

We may now proceed with the proof of Theorem 1.1.

Proof of Theorem 1.1. Let us define the control u ∈ L2((0, T )× ω) by

u(t, x1, x2) =1√2π

∑n∈Z

un(t, x2)einx1 in (0, T )× ω,

where, for n ∈ Z, un ∈ L2((0, T )× (c, d)) are the controls provided by Proposition 4.1,thus such that yn and hn, which solve (1.9), vanish at time T > 0: yn(T, ·) ≡ 0 in(−1, 1) and hn(T ) = 0. Defining y and h via Fourier series similarly as u above, wereadily see that (y, h) is the unique solution to (1.3). Moreover, since all the Fouriercoefficients of y and h vanish at time T , then also (y, h) vanish at time T . The estimateon the control follows by summing up the estimate of the Fourier coefficient controlsover all n, and using the fact that all the constants intervening in this estimate areindependent of n. This concludes the proof.


6. Proof of Theorem 1.2

Let us now consider ω = (a, b) × (c, d), where (a, b) ⊂ T and (c, d) ⊂ (−1, 1). Weconsider the adjoint system

−∂tζ −∆ζ = 0 in (0, T )× Ω,

−∂t`(t, x1)− ∂x2ζ(t, x1, 1) = 0 on (0, T )× T,ζ(t, x1, 0) = 0 on (0, T )× T,ζ(t, x1, 1) = σ∂2

x1`(t, x1) on (0, T )× T,

(ζ, `)|t=T = (ζT , `T ) in Ω× T.

(6.1)

We also define the filtered space of low frequencies

Eµ :=⊕|n|6µn6=0

Hn ⊗

1√2πein·.

Let Πµ : ˙H → ˙H denote the orthogonal projection from ˙H onto Eµ.

6.1. Control of the low frequencies. We first recall the following version of theLebeau-Robbiano spectral inequality for the eigenfunctions of the Laplacian on T withperiodic boundary conditions.

Lemma 6.1 (Spectral inequality). Let a, b ∈ R be such that a < b. There exists C > 0such that for every µ > 0 and ann∈Z ⊂ C, the inequality

∑|n|6µ

|an|2 6 CeCµ∫ b

a

∣∣∣∣∣∣∑|n|6µ

aneinx1

∣∣∣∣∣∣2

dx1

holds.

Proof. See [35], and also [34, Theorem 5.4], [5, Proposition 5].

Using Lemma 6.1, we may derive the following observability inequality for the lowfrequencies.

Proposition 6.1 (Filtered observability). Let σ > 0 be fixed, and suppose ω = (a, b)×(c, d), where (a, b) ⊂ T and (c, d) ⊂ (−1, 1). Then, there exists a constant C > 0 suchthat for every T > 0, for every µ > 0, and for every (ζT , `T ) ∈ Eµ, the unique solution(ζ, `) ∈ C0([0, T ];L2(Ω)×H1(T)) to (6.1) satisfies∫

Ω

∣∣ζ(0)∣∣2 dx+ σ

∫T

∣∣∂x1`(0)∣∣2 dx1 6 Ce

C( 1T

+µ)∫ T

0

∫ω|ζ(t)|2 dx dt.

Proof. By assumption, we can write(ζ0(x1, x2), `0(x1)

)=

1√2π

∑|n|6µn6=0

(ζ0n(x2), `0n

)einx1 ,


with the Fourier coefficients being defined as in (1.8). The solution (ζ, `) to (6.1) is thengiven by (

ζ(t, x1, x2), `(t, x1))

=1√2π

∑|n|6µn 6=0

(ζn(t, x2), `n(t)

)einx1 ,

where for every n 6= 0, (ζn, `n) denotes the unique solution to−∂t

(ζn, `n

)= An

(ζn, `n

)in (0, T ),(

ζn, `n)|t=T = (ζT,n, `T,n) .

Combining (4.22) with Lemma 6.1, we find that∫Ω|ζ(0)|2 dx+ σ

∫T|∂x1`(0)|2 dx1 =

∑|n|6µn6=0

∥∥∥(ζn(0), `n(0))∥∥∥2

Hn

6M1eM2T

∑|n|6µn 6=0

∫ T

0

∫ d

c

∣∣∣ζn(t, x2)∣∣∣2 dx2dt

6Lem.6.1

CeC(

1T

+µ) ∫ T

0

∫ d

c

∫ b

a

∣∣∣∣∣∣∣∣∑|n|6µn6=0

ζn(t, x2)einx1

∣∣∣∣∣∣∣∣2

dx1dx2dt

= CeC(

1T

+µ) ∫ T

0

∫ω

∣∣ζ(t, x1, x2)∣∣2 dx1dx2dt

holds for some constant C > 0 independent of T and µ. This is the desired conclusion.

By customary HUM arguments, we also deduce the following result.

Proposition 6.2 (Filtered control). Let σ > 0 be fixed, and suppose ω = (a, b)× (c, d),where (a, b) ⊂ T and (c, d) ⊂ (−1, 1). Then, there exists a constant C > 0 such that forany T > 0, for every µ > 0, and for every (y0, h0) ∈ ˙H , there exists uµ ∈ L2((0, T )×ω)

such that the unique solution (y, h) ∈ C0([0, T ]; ˙H ) to (1.3) with control uµ satisfies

Πµ (y(T ), h(T )) = 0, (6.2)

and ∥∥uµ∥∥L2((0,T )×ω)6 CeC

(1T

+µ) ∥∥∥(y0, h0

)∥∥∥H. (6.3)

Proof. We again proceed by using customary HUM arguments, as done in the proof ofProposition 4.1, and also closely following [57]. Let us define the functional

J (ζT , `T ) :=1

2

∫ T

0

∫ω|ζ|2 dx dt−

⟨(ζ(0), `(0)

),(y0, h0

)⟩H,

for (ζT , `T ) ∈ Eµ. The functional J has a unique minimizer (ζ∗T , `∗T ) ∈ Eµ by virtue

of the observability inequality of Proposition 6.1. Let (ζ∗, `∗) denote the corresponding


solution to (6.1). By writing down the Euler-Lagrange equation, one quickly finds thatthe minimizer (ζ∗T , `

∗T ) is such that⟨(

y(T ), h(T )),(ϕT , sT

)⟩H

= 0 (6.4)

for all (ϕT , sT ) ∈ Eµ. This yields (6.2), by definition of Πµ. Estimate (6.3) followssimilarly by virtue of Proposition 6.1.

6.2. Decay of the high frequencies. To go beyond the above theorem, as is commonwith the Lebeau-Robbiano method, we will need to make use of the following exponentialdecay result for the high frequency components.

Lemma 6.2 (Exponential decay). Suppose σ > 0. The following statements hold.(1) The part of A in ˙H generates an exponentially stable semigroup on ˙H . More

specifically, ∥∥∥etA(y0, h0)∥∥∥

H6 e−min

σ2,1t∥∥∥(y0, h0

)∥∥∥H

holds for all t > 0 and (y0, h0) ∈ ˙H .(2) Let µ > 0 be fixed, and suppose that

(y0, h0

)∈ ˙H is such that Πµ

(y0, h0

)= 0.

Then the solution of (1.3) with u ≡ 0 satisfies∥∥∥(y(t), h(t))∥∥∥

H6 e−min

σ2,1µ2t∥∥∥(y0, h0

)∥∥∥H

for all t > 0.

Proof. Let us prove the first claim. We may write(y0(x1, x2), h0(x1)

)=

1√2π

∑n6=0

(y0n(x2), h0

n

)einx1 ,

for (x1, x2) ∈ Ω, with the Fourier coefficients being defined as in (1.8). The solution(y, h) to (1.3) with u ≡ 0 is then given by(

y(t, x1, x2), h(t, x1))

=1√2π

∑n6=0

(yn(t, x2), hn(t)

)einx1 , (6.5)

where (yn, hn) is the unique solution to (1.9) with un ≡ 0. On another hand, sincen 6= 0, and since An then has a decreasing sequence of negative eigenvalues λn,k+∞k=0

with corresponding sequence of eigenfunctions Φn,k+∞k=0 forming an orthonormal basisof Hn, we may write(

yn(t, x2), hn(t))

=

+∞∑k=0

eλn,kt⟨(yn(t, ·), hn(t)

),Φn,k

⟩Hn

Φn,k(x2). (6.6)

Using Lemma 4.2 and n2 > 1, it follows that∥∥∥(yn(t), hn(t))∥∥∥

Hn

6 eλ0,nt∥∥∥(y0

n, h0n

)∥∥∥Hn

6 e−minσ2,1n2t∥∥∥(y0

n, h0n

)∥∥∥Hn

6 e−minσ2,1t∥∥∥(y0

n, h0n

)∥∥∥Hn


for all t > 0. Combining (6.5), (6.6), and the Plancharel theorem to sum up over n 6= 0,we arrive to the desired conclusion.

The proof of the second claim is almost identical. We provide details nonetheless.Since Πµ

(y0, h0

)= 0, we may write(y0(x1, x2), h0(x1)

)=

1√2π

∑|n|>µ

(y0n(x2), h0

n

)einx1 ,

with the Fourier coefficients being again defined as in (1.8). The solution (y, h) to (1.3)with u ≡ 0 is then given by(

y(t, x1, x2), h(t, x1))

=1√2π

∑|n|>µ

(yn(t, x2), hn(t)

)einx1 ,

where (yn, hn) is the unique solution to (1.9) with un ≡ 0. We again make use of (6.6)and Lemma 4.2, this time to find that∥∥∥(yn(t), hn(t)

)∥∥∥Hn

6 eλ0,nt∥∥∥(y0

n, h0n

)∥∥∥Hn

6 e−minσ2,1n2t∥∥∥(y0

n, h0n

)∥∥∥Hn

6 e−minσ2,1µ2t∥∥∥(y0

n, h0n

)∥∥∥Hn

holds for all t > 0. Summing up over |n| > µ yields the desired conclusion.

Combining Proposition 6.2 and Lemma 6.2, we obtain the following result.

Proposition 6.3. Let σ > 0 be fixed, and suppose ω = (a, b)× (c, d), where (a, b) ⊂ Tand (c, d) ⊂ (−1, 1). Then, there exists C > 0 such that for any T > 0, for any(y0, h0) ∈ ˙H , and for any µ > 0, there exists a control uµ ∈ L2((0, T ) × ω) such thatthe unique solution (y, h) ∈ C0([0, T ]; ˙H ) to (1.3) with control uµ satisfies∥∥uµ∥∥L2((0,T )×ω)

6 CeC(

1T

+µ)∥∥∥(y0, h0

)∥∥∥H, (6.7)

and ∥∥∥(y(T ), h(T ))∥∥∥

H6(

1 + C√T)eC(

1T

+µ)−min

σ2,1µ2 T

2

∥∥∥(y0, h0)∥∥∥

H. (6.8)

Proof. By virtue of Proposition 6.2, there exists a control uµ ∈ L2((

0, T2)× ω

)such

that the solution (y, h) to (1.3) with control uµ satisfies

Πµ

(y

(T

2

), h

(T

2

))= 0, (6.9)

and

‖uµ‖L2((0,T2 )×ω) 6 Ce

C(

1T

+µ)∥∥∥(y0, h0

)∥∥∥H. (6.10)

The constant C > 0, stemming from Proposition 6.2, is independent of both T > 0 andµ > 0. By virtue of the Duhamel formula, the semigroup contractivity per Lemma 6.2,


the Cauchy-Schwarz inequality, and (6.10), we find∥∥∥∥(y(T2), h

(T

2

))∥∥∥∥H

6∥∥∥(y0, h0

)∥∥∥H

+√T ‖uµ‖L2((0,T

2 )×ω)

6(

1 + C√T)eC(

1T

+µ)∥∥∥(y0, h0

)∥∥∥H. (6.11)

Between T2 and T, we let the system dissipate according to Lemma 6.2. Namely, let us

define

uµ(t) :=

uµ(t) if t ∈

(0, T2

)0 if t ∈

(T2 , T

).

From the above definition, we immediately have (6.7). Finally, using the second claimin Lemma 6.2 combined with (6.9), and subsequently (6.11), we deduce∥∥∥(y(T ), h(T )

)∥∥∥H6 e−min

σ2,1µ2 T

2

∥∥∥∥(y(T2), h

(T

2

))∥∥∥∥H

6(

1 + C√T)eC(

1T

+µ)e−min

σ2,1µ2 T

2

∥∥∥(y0, h0)∥∥∥

H.

This concludes the proof.

Remark 4. In the constants appearing in the estimates (6.7) and (6.8), the term Tactually indicates the length of the interval, rather than the endpoint. In other words,the interval (0, T ) may be replaced by an arbitrary interval (τ1, τ2) in the above result,in which case the T appearing in the constants of these estimates would be replaced byτ2 − τ1.

6.3. Proof of Theorem 1.2. We may now complete the proof of our second mainresult.

Proof of Theorem 1.2. Let β > 0 be fixed and to be chosen suitably later on. For j > 1,we consider the dyadic sequences

Tj := 2−jT and µj := 2jβ,

and we define the sequence

τj :=

0 for j = 0,

j∑k=1

Tk for j > 1.

We note that

(0, T ) =+∞⋃j=0

(τj , τj+1).

For j > 1, on any interval (τj−1, τj), by virtue of Proposition 6.3, with µ = µj ,we may build a control uj ∈ L2((τj−1, τj) × ω) with corresponding state (yj , hj) ∈C0([τj−1, τj ]; ˙H ) such that

‖uj‖L2((τj−1,τj)×ω) 6 CeC

(1Tj

+µj

) ∥∥∥(yj(τj−1), hj(τj−1))∥∥∥

H, (6.12)


as well as∥∥∥(yj(τj), hj(τj))∥∥∥H

6(

1 + C√Tj

)eC

(1Tj

+µj

)−minσ2 ,1µ2

j

Tj2∥∥∥(yj(τj−1), hj(τj−1)

)∥∥∥H

(6.13)

hold, with the convention (y1(0), h1(0)) =(y0, h0

). From (6.12) and (6.13), we gather

‖uj‖L2((τj−1,τj)×ω) (6.14)

6 C(

1 + C√T)j−1

exp

(j∑

k=1

C

(1

Tk+ µk

)−min

σ2, 1 j−1∑k=1

µ2k

Tk2

)∥∥∥(y0, h0)∥∥∥

H.

as well as∥∥∥(yj(τj), hj(τj))∥∥∥H

(6.15)

6(

1 + C√T)j

exp

(j∑

k=1

[C

(1

Tk+ µk

)−min

σ2, 1µ2k

Tk2

])∥∥∥(y0, h0)∥∥∥

H.

Let us now define

β0 := −(

2C

T+ 2βC −min

σ2, 1β2T

)and

β1 := −(

2C

T+ 2βC −min

σ2, 1 β2T

2

),

and we choose β > 0 sufficiently large so that β0 > 0 and β1 > 0. We may observe thatj∑

k=1

[C

(1

Tk+ µk

)−min

σ2, 1µ2k

Tk2

]=

2C

T

(2j − 1

)+ 2βC

(2j − 1

)−min

σ2, 1T(2j − 1

)= −2jβ0 + β0,

as well asj∑

k=1

C

(1

Tk+ µk

)−min

σ2, 1 j−1∑k=1

µ2k

Tk2

= −2jβ1 + β0.

Thus, from (6.14) and (6.15), we infer that

‖uj‖L2((τj−1,τj)×ω) 6C

1 + C√T

(1 + C

√T)jeβ0e−2jβ1

∥∥∥(y0, h0)∥∥∥

H. (6.16)

and ∥∥∥(yj(τj), hj(τj))∥∥∥H6(

1 + C√T)jeβ0e−β02j

∥∥∥(y0, h0)∥∥∥

H, (6.17)

We select m = m(T ) ∈ N sufficiently large so that(1 + C

√T)< 2m. (6.18)


Observe that, since β1 > 0, from (6.16) we have

‖uj‖L2((τj−1,τj)×ω) 6Ceβ0β−m1

1 + C√T

((1 + C

√T)

2−m)j

(β12j)me−β12j∥∥∥(y0, h0

)∥∥∥H

6 C1

((1 + C

√T)

2−m)j ∥∥∥(y0, h0

)∥∥∥H, (6.19)

where C1 = C1(T, σ) > 0 is independent of j. We define the control u : (0, T )× ω → Rby pasting all the uj as

u =∞∑j=1

uj1(τj−1,τj).

By virtue of (6.19) and (6.18), we get

‖u‖2L2(0,T ;L2(ω)) =∞∑j=1

‖uj‖2L2((τj−1,τj)×ω) = C21

∥∥∥(y0, h0)∥∥∥2

H

∞∑j=1

((1 + C

√T)

2−m)2j

= C2

∥∥∥(y0, h0)∥∥∥2

H,

for some7 C2 = C2(T, σ) > 0, and therefore u ∈ L2((0, T )×ω). Now consider the uniquesolution (y, h) ∈ C0([0, T ];L2(Ω)×H1(T)) to (1.3), with control u. Clearly(

y(t), h(t))

=(yj(t), hj(t)

)for t ∈ [τj−1, τj ].

In particular, (y(τj), h(τj)

)=(yj(τj), hj(τj)

),

so from (6.17) we deduce that(y(T, ·), h(T, ·)

)≡ 0 in Ω× T. This concludes the proof.

7. Epilogue

We have shown that the linearized Stefan problem with surface tension (Gibbs-Thomson correction) is null-controllable, in the sense that both the temperature andthe height function are controllable to zero.

We expect that this result should lead to a local controllability result for the nonlinearproblem, pending technical developments. There are, however, several questions andproblems which we believe merit further attention and clarity, even in the linearizedregime. In addition to addressing the case σ = 0, discussed in greater depth in Section 2,other problems may include

(1) Control on the free boundary. While not exactly perfectly clear to interpretin physical terms, one can consider the problem of controlling through the freeboundary, which would mean putting a control u1ω in the second equation in(1.3) (namely the evolution equation for h). This would be in the spirit of worksin control of water waves ([3]), and also the simplified piston problem ([12]). We

7As a matter of fact, assuming T < 1, and with suitable choice of β > 0 above, we can refine thisestimate to obtain

‖u‖2L2((0,T )×ω) 6 CeCT3∥∥(y0, h0)∥∥2

H,

for some C = C(σ) > 0 depending on σ, but independent of T.


expect this problem to be significantly more challenging to address than the onewe had considered here.

(2) Spectral optimization. The governing operator A of (1.3) appears somewhatopaque, and a variety of alternative (control) problems can be envisaged andstudied regarding (1.3). We believe that further analysis is warranted in theanalysis of the spectral properties of A, in particular with regard to extensionof various control results to general geometries (beyond strips).

As we have noted, −A is self-adjoint when considered on the space ˙H definedin (1.10). In particular, one could define the first eigenvalue λ1 > 0 of −Athrough the min-max theorem as the Rayleigh quotient

λ1 := inff∈D(A)\0

〈−Af, f〉H‖f‖2

H

.

But as is typical for the Laplacian, one seeks to use the symmetry and obtaina more tractable representation. In [28, Lemma 4.5] (see also [26, Chapter 3,Section 4, Lemma 4.5]), it also is stated that

λ1 = inff=(f1,f2)∈S\0

∫Ω|∇f1|2 dx∫

Ω|f1|2 dx+ σ

∫T|∂x1f2|2 dx1

, (7.1)

where the space S is defined as

S :=

f = (f1, f2) ∈ H1(Ω)×H5/2(T)

∣∣∣∣∣∫

Ωf1(x1, ·) dx1 +

∫Tf2(x1) dx1 = 0

.

Taking stock of (7.1), there are a variety of different spectral optimization prob-lems one could then envisage for (1.3), such as characterizing optimal actuatorand observer domains ω in the spirit of [24, 45, 46, 47], and in particular, com-paring how these designs differ from that of the classical heat equation, or thelimit of these designs as σ 0 (should controllability hold for the latter).

(3) The obstacle problem. It is by now well-known that the classical Stefanproblem (σ = 0), without source terms, is related to the parabolic obstacleproblem through the so-called Duvaut transform (see [19, 51] and the referencestherein). For control purposes, one could envisage transferring results from theStefan problem to the parabolic obstacle problem (which is actually a problemto be studied in its own right, [20, 51]). But this is highly nontrivial due tothe fact that the Duvaut transform applies to non-negative solutions of theStefan problem, and it is not clear if existing techniques on controllability underpositivity constraints ([39], or the so-called staircase method [41, 44, 52]) wouldbe applicable here. The bottom line is that the controllability properties of theparabolic obstacle problem remain widely open. (See [22, Section 1.5.1].)

Acknowledgments. This research began with discussions between both authors dur-ing the “VIII Partial Differential Equations, Optimal Design and Numerics” workshopthat took place in Benasque, in August 2019. We thank the organizing committee, andthe center "Pedro Pascual" for their hospitality.


Funding. B.G. has received funding from the European Union’s Horizon 2020 re-search and innovation programme under the Marie Sklodowska-Curie grant agreementNo.765579-ConFlex. Debayan Maity was partially supported by INSPIRE faculty fel-lowship (IFA18-MA128) and by Department of Atomic Energy, Government of India,under project no. 12-R & D-TFR-5.01-0520.

Appendix A. Numerics

A.1. Discretizing (1.3). As we did not find precisely the same scheme in the literature,for completeness and future reproducibility purposes, let us briefly discuss the numer-ical discretization we used for computing, and obtaining the simulations presented inFigure 2 and Figure 3.

(1) Setup. We shall focus on Ω = (0, 2) × (−1, 1) for simplicity. The controldomain ω is a thin neighborhood of a line ranging from x1 = 1

2 to x1 = 32 , tilted

at an angle of 45 (i.e. with slope 1) with respect to the horizontal axis (chosenfor the relative simplicity of numerical implementation, and the sparsity of theresulting matrix). We choose the same mesh-size 4x > 0 for the horizontal andvertical variables, and we set nx := 2

4x − 1. Since we are working with periodicboundary conditions in the x1-variable, and Dirichlet boundary conditions inthe x2-variables, y(t, x1, x2) will be unknown at (nx + 1)nx points, and h(t, x1)at (nx + 1) points.

(2) Finite difference semi-discretization. We define an equi-distributed gridxi1, x

j2

i∈0,...,nx+1j∈0,...,nx+1

of Ω through

xi1 = i4x and xj2 = −1 + j4x.

We discretize the the two-dimensional Laplacian ∆x1,x2 with the classical 5-pointfinite-difference stencil, and the Neumann trace ∂x2y(t, x1, 1) with a centereddifference scheme. Henceforth denoting

yi,j(t) := y(t, xi1, x

j2

),

with analog definitions for ui,j and hi, the finite-difference semi-discretizationof (1.3) reads as

yi,j −yi+1,j + yi−1,j + yi,j+1 + yi,j−1 − 4yi,j

4x2= ui,j1ωnx 1, . . . , nx + 1 × 1, . . . , nx,

hi =yi,nx+1 − yi,nx−1

24x 1, . . . , nx + 1,

y0,j = ynx+1,j 1, . . . , nx,yi,0 = 0 1, . . . , nx + 1,

yi,nx+1 = σhi+1 + hi−1 − 2hi

4x21, . . . , nx + 1,

(A.1)


for t ∈ (0, T ). With analog definitions for u[j] and h, setting

y[j] :=

y1,j

y2,j...

ynx+1,j

for j ∈ 1, . . . , nx, as well as z := (y[1], . . . , y[nx],h) and then, similarly, set-ting u := (u[1], . . . , u[nx], 0Rnx+1), we may rewrite (A.1) as a canonical finite-dimensional linear system z = Anxz + Bnxu, where

Anx :=

A0 A1

A1 A0 A1

. . . . . . . . .A1 A0 A1

A1 A0 A2

A3 A4

where A1 = 1

4x2 Idnx+1 and A3 = − 124x Idnx+1, whereas

A0 =1

4x2

−4 1 11 −4 1

. . . . . . . . .1 −4 1

1 1 −4

and A` = c`

−2 1 11 −2 1

. . . . . . . . .1 −2 1

1 1 −2

,for ` ∈ 2, 4, with c2 = σ

4x4 and c4 = 2σ4x3 .

0 20 40 60 80 100 120 140 160

nz = 845

0

20

40

60

80

100

120

140

160

0 20 40 60 80 100 120 140 160

nz = 767

0

20

40

60

80

100

120

140

160

Figure 6. The sparsity pattern of the matrix Anx when σ = 10 (left)and σ = 0 (right).


(3) Time-stepping. Given nodes tkk∈0,...,nt with tk = k4t for some 4t = Tnt,

we set zk = z(tk). Time-stepping for (1.3) is done with a Crank-Nicolson method

zk+1 − zk

4t =1

2Anx

(zk + zk+1

)+

1

2Bnx

(uk + uk+1

)for k ∈ 0, . . . , nt, which is unconditionally stable.

A.2. Computing. We solve

minu∈L2((0,T )×ω)

(y,h) solve (1.3)(y(T ),h(T ))≡0

‖u‖2L2((0,T )×ω)

by making use of the above discretization for parametrizing the PDE constraints. Thisis a convex program, which can be solved using interior point methods. We use theIpOpt solver embedded into Casadi for Matlab ([4]). For the experiments in Figure 2and Figure 3, we took 4x = 2

13 , and thus nx = 12, with nt = 200. All codes are openlymade available at https://github.com/borjanG/2022-stefan-control.

Appendix B. Linearization

B.1. Fixing the domain. A simple change of variables which allows us to pass from(1.2), set in the moving domain Ω(t), to a nonlinear problem in the time-independentreference domain Ω, consists in defining the map Ψ(t, ·) : Ω→ Ω(t) for t > 0 by

Ψ(t, x) :=(x1,(1 + h(t, x1)

)x2

), x = (x1, x2) ∈ Ω.

If h(t, ·) ∈ Hs(T), then Ψ(t, ·) ∈ Hs(Ω) – namely, the transformation Ψ(t, ·) preservesthe spatial regularity of h. This is not desirable when d > 3, since one would have toensure that h is rather smooth, and consequently, the control too.

One can exhibit a slightly different transformation which entails a gain in spatialregularity with respect to that of the height function h(t, ·). Given h ∈ C0([0, T ];Hs(T))for some s > 0, for any t > 0 we consider the solution ψ(t, ·) to the Poisson problem

∆ψ(t, ·) = 0 in Ω

ψ(t, x1, 0) = 0 on Tψ(t, x1, 1) = h(t, x1) on T,

and we define Ψ(t, ·) : Ω→ Ω(t) by

Ψ(t, x) :=(x1, x2 + ψ(t, x)

).

In this case, if h(t, ·) ∈ Hs(T) then Ψ(t, ·) ∈ Hs+1/2(Ω). Note that Ψ is similar to thetransformation defined in [25, Eq. (1.6)]. From elliptic estimates, it can be seen that

‖Ψ(t, ·)− Id‖Hs+1/2(Ω) . ‖h(t, ·)‖Hs(T)

for all t > 0, so whenever h(t, ·) is sufficiently small, Ψ(t, ·) is a diffeomorphism from Ω

onto Ω(t) by the inverse function theorem.

https://github.com/borjanG/2022-stefan-control


B.2. Linearizing. In this case, we denote by X(t, ·) = [Ψ(t, ·)]−1 the inverse of Ψ(t, ·)for all t > 0, and consider

y(t, x) = %(t,Ψ(t, x)) for (t, x) ∈ (0, T )× Ω.

In other words,

%(t, z) = y(t,X(t, z)) for (t, z) ∈ (0, T )× Ω(t).

We also introduce the standard notation

BΨ := Cof(∇Ψ), and AΨ :=1

det(∇Ψ)B>ΨBΨ,

where δΨ := det(∇Ψ) denotes the Jacobian determinant of∇Ψ, and Cof(M) denotes thecofactor matrix of M , satisfying M(Cof(M))> = (Cof(M))>M = det(M)Id. System(1.2) can then be equivalently rewritten as

∂ty −∆y = N1(y, h) in (0, T )× Ω,

∂th(t, x1) = (∇y(t, x1, 1) · e2) +(N3

(y(t, x1, 1)

)· e2

)on (0, T )× T,

y(t, x1,−1) = 0 on (0, T )× T,y(t, x1, 1) = σ∂2

x1h(t, x1) +N2(h(t, x1)) on (0, T )× T,

(y, h)|t=0=(y0, h0

)in Ω× T,

(B.1)

where y0(·) := %0(Ψ(0, ·)), e2 := (0, 1)>, with the nonlinear terms having the form

N1(y, h) := −(det(∇Ψ)− 1)∂ty + div(N3), N3(y) := (AΨ − Id)∇y.and

N2(h) := σ(κ(h)− ∂2

x1h).

The linearized problem can then be obtained simply by dropping Nj from (B.1).

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Borjan GeshkovskiDepartment of MathematicsMassachusetts Institute of TechnologySimons Building, Room 24177 Massachusetts AvenueCambridgeMA02139-4307 USAe-mail: [email protected]

Debayan MaityTIFR Centre for Applicable Mathematics560065 BangaloreKarnataka, Indiae-mail: [email protected]

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arXiv:2203.03012v1 [math.OC] 6 Mar 2022

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