Some control results for simplified one-dimensional models of fluid-solid interaction

May 4, 2005 13:29 WSPC/103-M3AS 00052 Mathematical Models and Methods in Applied Sciences Vol. 15, No. 5 (2005) 783–824 c World Scientific Publishing Company SOME CONTROL RESULTS FOR SIMPLIFIED ONE-DIMENSIONAL MODELS OF FLUID-SOLID INTERACTION ANNA DOUBOVA∗and ENRIQUE FERN´ ANDEZ-CARA† Dpto. E.D.A.N., University of Sevilla, Aptdo. 1160, 41080 Sevilla, Spain ∗[email protected] † [email protected] Received 10 October 2003 Revised 10 December 2004 Communicated by Zuazua We analyze the null controllability of a one-dimensional nonlinear system which models the interaction of a fluid and a particle. This can be viewed as a first step in the control analysis of fluid-solid systems. The fluid is governed by the Burgers equation and the control is exerted at the boundary points. We present two main results: the global null controllability of a linearized system and the local null controllability of the nonlinear original model. The proofs rely on appropriate global Carleman inequalities, observability estimates and fixed point arguments. Keywords: Controllability; fluid-solid interaction; Burgers equation; Carleman estimates. AMS Subject Classification: 93C20, 35Q35 1. Introduction and Main Results Let us set T>0andQ=(−1,1)×(0,T). We will consider a (simplified) nonlinear system that models the interaction of a one-dimensional fluid evolving in (−1,1) and a solid particle. It will be assumed that the velocity of the fluid is governed by the viscous Burgers equation at both sides of the point mass location y=h(t). For simplicity, we will also assume that the fluid density is constant and the solid particle has unit mass. The system is thus the following:            ut−uyy +uuy=0,(y,t)∈Q, y =h(t), u(−1,t)=α(t),u(1,t)=β(t),t∈(0,T), u(h(t),t)=h(t),[uy](h(t),t)=h(t),t∈(0,T), u(y,0) = u0(y),h(0) = h0,h (0) = h1. (1.1) 783 Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 784 A. Doubova & E. Fern´andez-Cara Here, u(y,t) is the velocity of the fluid particle located at yat time t,h(t)isthe position occupied by the particle at time t,αand βare the controls (two functions at least in L∞(0,T)) and the initial data satisfy u0∈H1(−1,1),h 0∈(−1,1) and h1=u0(h0).(1.2) In (1.2) and the sequel, [f](y) denotes the jump of the function fat point y. We see that two conditions are required at y=h(t)fort∈(0,T). The first condition means that the velocity of the fluid and the solid mass coincide at these points. The second one is a transmission condition indicating that the force exerted by the fluid on the particle is equal to the product of the particle mass by its acceleration. The modelling and analysis of fluid-solid interaction have attracted a lot of attention in recent years. In particular, in the case of twoand three-dimensional Navier–Stokes fluids in contact with one or more rigid or elastic bodies, this has been the goal of Refs. 3–7, 13, 23 and 24 and the references therein. In this paper, we will be concerned with the controllability of the previous simplified one-dimensional model, already considered and analyzed in Refs. 25 and 26; more precisely, we shall investigate the null controllability properties of (1.1). It will be said that (1.1) is null controllable (at time T) if, for each u0∈ H1(−1,1) and h0∈Rwith |h0|<1, there exist controls α, β andanassociated solution (u, h)∈C0([0,T]; L2(−1,1)) ×C1([0,T]) such that u(y,T)=0 in(−1,1),h(T)=0 and h(T)=0.(1.3) Of course, we are dealing here with a simplified version of other much more complicated and realistic models. For example, for a system governed by the Navier–Stokes equations with a solid sphere inside the fluid, a related question is how to act on the fluid particles on the outer boundary to get the sphere completely stopped at t=T. This justifies the relevance of the controllability analysis of (1.1). The controllability of partial differential equations has also been the object of extensive research during the last years. Since the pioneering papers,18,19,21,22 where systems governed by linear wave and heat equations were considered, a lot of work has been done in this area. See for instance Refs. 9, 12, 15, 27 and 28 for the approximate, exact and null controllability of semilinear hyperbolic and parabolic equations; see also Ref. 16 and the more recent paper10 for the local exact controllability of the Navier–Stokes equation, etc. In (1.1), the spatial domain depends on t. Assuming that |h(t)|≤1−bwhere b is a positive small constant, we can introduce the following change of variable: For any y∈(−1,h(t)) ∪(h(t),1), we put x=(y−h)/(1 −κh), where κis the sign of x. Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 785 This leads to a reformulation of (1.1) in a domain that is independent of t:                    (1 −κh)ut−1 1−κhuxx −(1 −κx)hux+uux=0,(x, t)∈Q, x =0, u(−1,t)=α(t),u(1,t)=β(t),t∈(0,T), u(0,t)=h(t),1 1−κhux(0,t)=h(t),t∈(0,T), u(x, 0) = u0(x),h(0) = h0,h (0) = h1. (1.4) Remark 1.1. Notice that the genuine unknowns in (1.4) are uand h,thatis to say, the velocity of the fluid and the velocity of the particle. Indeed, if we set p(t)=h(t), then (1.4) can be written in the form                    (1 −κh)ut−1 1−κhuxx −(1 −κx)pux+uux=0,(x, t)∈Q, x =0, u(−1,t)=α(t),u(1,t)=β(t),t∈(0,T), u(0,t)=p(t),1 1−κhux(0,t)=p(t),t∈(0,T), u(x, 0) = u0(x),p(0) = h1, where h(t)=h0+t 0 p(s)ds =h0+T 0 u(0,s)ds. For any u∈L∞(Q)andh∈W1,∞(0,T)with|h(t)|≤1−b, we will also consider the following linearized system, where the unknowns are vand k:                    (1 −κh)vt−1 1−κhvxx −(1 −κx)hvx+1 2(uv)x=0,(x, t)∈Q, x =0, v(−1,t)=α(t),v(1,t)=β(t),t∈(0,T), v(0,t)=k(t),1 1−κhvx(0,t)=k(t),t∈(0,T), v(x, 0) = v0(x),k(0) = k0,k (0) = k1. (1.5) The first main result of this paper is the following: Theorem 1.1. For any u∈L∞(Q)and h∈W1,∞(0,T)with |h(t)|≤1−b<1, the linear system (1.5) is (globally)null controllable. More precisely,for any Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 786 A. Doubova & E. Fern´andez-Cara v0∈L2(−1,1),k 0∈(−1,1) and k1∈R,there exist controls α, β ∈L∞(0,T) and associated states (v,k)∈C0([0,T]; L2(−1,1)) ×C1([0,T]) such that v(x, T )=0 in (−1,1),k(T)=0 and k(T)=0.(1.6) For the proof of this result, we will first deduce a global Carleman inequality and some observability estimates for an associated adjoint system. Then, we will adapt standard arguments. In fact, we will first work with an extended system, similar to (1.5), where the control is distributed and its support does not intersect (−1,1) ×(0,T), see (4.1). We will prove that (4.1) is null controllable and, then, we will deduce the same property for (1.5) after restriction to (−1,1) ×(0,T). The reason is that it is not clear that a direct application to (1.5) of the techniques in Secs. 2 to 4 provides sufficient regularity to obtain controls αand βin L∞(0,T). By analogy with the boundary controllability of the linear heat equation, this may be an obstacle to have (v,k)∈C0([0,T]; L2(−1,1)) ×C1([0,T]). The second main result of this paper is the following: Theorem 1.2. The nonlinear system (1.4) is locally null controllable. More precisely,there exists ε>0depending on Tsuch that,whenever the initial data u0∈H1(−1,1),h 0and h1satisfy u0(0) = h1and u0H1(−1,1) +|h0|≤ε, we can find controls α, β ∈L∞(0,T)and associated states (u, h)in the space C0([0,T]; L2(−1,1)) ×C1([0,T]) satisfying (1.3). For the proof, we will use Theorem 1.1, a fixed point argument and some additional regularity properties of the solution of (1.4). Remark 1.2. It would be interesting to know whether Theorems 1.1 and 1.2 are still true when the control is exerted only at x=−1 (for example, taking β(t)≡0 in (1.5)). To our knowledge, this is unknown. In order to clarify the situation and trying to guess what can be expected, let us consider a very particular case:            vt−vxx =0,(x, t)∈Q, x =0, v(−1,t)=α(t),v(1,t)=0,t∈(0,T), v(0,t)=k(t),[vx](0,t)=k(t),t∈(0,T), v(x, 0) = v0(x),k (0) = k1. (1.7) This is just (1.5) with u(x, t)≡0andh(t)≡0. The previous system can be rewritten in the form      1+δ(x=0)vt−vxx =0,(x, t)∈Q, v(−1,t)=α(t),v(1,t)=0,t∈(0,T), v(x, 0) = v0(x),x∈(−1,1), (1.8) Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 787 where δ(x=0) is the Dirac distribution supported by {x=0,0<t<T}.Itcan also be written as follows:          d dt v k−Av k=Bα, t ∈(0,T), v k(0) = v0 k1, for an appropriate self-adjoint and positive linear operator A:D(A)⊂H→ H in the Hilbert space H=L2(−1,1) ×Rwith compact inverse and an appropriate (unbounded) linear operator B. Thus, according to a classical result by Russell,21 the null controllability of (1.8) at any positive time T>0 is a consequence of the fact that the similar second-order in time problem      1+δ(x=0)wtt −wxx =0,(x, t)∈(−1,1) ×(0,T ∗), w(−1,t)=α(t),w(1,t)=0,t∈(0,T ∗), w(x, 0) = w0(x),w t(x, 0) = w1(x),x∈(−1,1), is exactly controllable in H1 0(−1,1) ×L2(−1,1) at some T∗. In turn, this is a consequence of the observability inequality (p, pt)(·,0)2 L2(−1,1)×H−1(−1,1) ≤C(T∗)T∗ 0|px(−1,t)|2dt, (1.9) that holds for any T∗>2 and for all the solutions of the adjoint problem      1+δ(x=0)ptt −pxx =0,(x, t)∈(−1,1) ×(0,T ∗), p(−1,t)=p(1,t)=0,t∈(0,T ∗), p(x, T∗)=p0(x),p t(x, T∗)=p1(x),x∈(−1,1), with final data (p0,p 1)∈L2(−1,1) ×H−1(−1,1). This fundamental result of Russell (that exact controllability for a hyperbolic equation at some time implies null controllability at any time for the similar parabolic equation) has already been used in other contexts; see for instance Ref. 11, which deals with one-dimensional linear heat equations with non-smooth time-independent coefficients. The proof of (1.9) can be achieved by arguing as in Ref. 14. Hence, in the particular case (1.7), we get the null controllability of the simplified linear system with one single control. Nevertheless, this argument does not work for (1.5) (where the coefficients depend on xand t). On the other hand, the techniques used in the proof of Theorem 1.1 do not work when we have only one control (see Remarks 2.1 and 3.3 for some explanations). Remark 1.3. Of course, it is also unknown whether the nonlinear system (1.4) can be driven exactly to zero for small initial data with only one control. Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 788 A. Doubova & E. Fern´andez-Cara Remark 1.4. It would also be interesting to know whether we can get a local controllability result similar to Theorem 1.2 for a model with two or more particles. For instance, in the case of two particles, the original state equation is:              ut−uyy +uuy=0,(y,t)∈Q, y =h1(t),h 2(t), u(−1,t)=α(t),u(1,t)=β(t),t∈(0,T), u(hi(t),t)=h i(t),[uy](hi(t),t)=h i(t),t∈(0,T),i=1,2, u(y,0) = u0(y),h i(0) = h0 i,h  i(0) = h1 ii=1,2, (1.10) where u0∈H1(−1,1), −1<h 0 1<h 0 2<1andu0(h0 i)=h1 ifor i=1,2. Now, the goal is to find controls αand β,atleastinL∞(0,T), such that u(y,T)=0 in(−1,1),h 1(T)=−1 2,h 2(T)=1 2 and h 1(T)=h 2(T)=0. (1.11) To our knowledge, the null controllability of (1.10) is an open question. Notice that, arguing as in Remark 1.2, we can consider a simplified (linear) version of (1.10) to which Russell’s principle can be applied. Consequently, it is reasonable to expect a positive answer for any T>0. Remark 1.5. Let u0∈H1(−1,1) and h0∈Rbe given. It is a relevant question whether there exist a (possibly long) time T>0 depending on u0H1(−1,1) and |h0|, controls α, β ∈L∞(0,T) and associated states (u, h)∈C0([0,T]; L2(−1,1)) × C1([0,T]) such that (1.3) holds. In order to answer this question, a natural strategy would be to combine the global existence and exponential decay results in Ref. 26 and Theorem 1.2 as follows: starting from the initial data u0,h0and h1with |h0|<1andu(0) = h1,wefirsttake α(t)≡0andβ(t)≡0 and we let the system (1.4) evolve until t=T0,withT0such that (u(·,T 0),h (T0)) is small in the H1×Rnorm; then, we apply Theorem 1.2 with initial data u(·,T 0), h(T0)andh(T0) for instance in the time interval (T0,T 0+1). There is however a nontrivial difficulty to apply these ideas: unfortunately, in our local result Theorem 1.2, not only u(·,T 0)H1(−1,1) but also |h(T0)|has to be small; but this is not ensured by the natural evolution of the uncontrolled system for large T0. Consequently, the null controllability of (1.4) in long-time intervals starting from arbitrarily large initial states is an open question. Remark 1.6. An alternative approach to long-time controllability would be the following. Again, starting from the initial data u0,h0and h1=u0(0), we let the system evolve freely up to t=T0,withT0such that (u(·,T 0),h (T0)) is small in the H1(−1,1) ×Rnorm. Then, we try to apply the partial null controllability result indicated in Remark 5.1 with initial data u(·,T 0), h(T0)andh(T0)inthe Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 789 time interval (T0,T 0+ 1). If this were possible, we would have found in this way controls αand βsuch that u(x, T0+1)=0 in(−1,1),|h(T0+1)|<1andh(T0+1)=0. In this case, h(T0) does not need to be small. Nevertheless, a nontrivial difficulty appears again. Indeed, in view of Remark 5.1, in order to perform the second previous step, we need a time T0such that the uncontrolled solution (u, h)satisfies u(·,T 0)H1(−1,1) ≤ε(1,|h(T0)|), where εis a small parameter satisfying (5.20). But the possibility that |h(t)| approaches 1 (and therefore ε(1,|h(t)|) approaches 0) as t→+∞is not excluded. Consequently, it is not clear that such a T0exists for every u0and h0. Remark 1.7. In a recent paper, S. Anita and D. Tataru1have proved that dissipative semilinear parabolic equations cannot be null controllable in uniform time. More precisely, they have shown that, if the nonlinear term satisfies a “good-sign condition” and behaves superlinearly at infinity and we denote by T(R) the minimal time of exact controllability to zero of all initial data with L2norm ≤R,then T(R)≥Cif R≥1, C|log R|−1otherwise, for some C>0. It is reasonable to expect a similar result for the Boussinesq equation and also for (1.1). But for the moment this is also an open question. The rest of this paper is organized as follows. In Sec. 2, we deduce a global Carleman inequality that we need for the proof of Theorem 1.1. Section 3 deals with some observability inequalities. In Sec. 4, we give the proof of Theorem 1.1. Section 5 is devoted to the proof of Theorem 1.2. 2. A Carleman Inequality In this section, we will deduce the global Carleman inequality we need for the proof of Theorem 1.1. In the sequel, we will denote by C0,C 1,... various positive constants. We will frequently indicate the data on which they depend. On the other hand, we will denote by Ka generic constant whose value can change from line to line. Let δ>0 be given and let us set Q=(−1−δ, 1+δ)×(0,T), Q−=(−1−δ, 0) ×(0,T),Q +=(0,1+δ)×(0,T) and ω=(−1−δ, −1−δ/2) ∪(1 + δ/2,1+δ)≡ω−∪ω+. Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 790 A. Doubova & E. Fern´andez-Cara Let us introduce the functions β∗and ˜ β,with      β∗∈C2([−1−δ, 0]), β∗ x(−1−δ)=0,β ∗ xx ≤0, β∗(x)=−x+β0for x∈(−1−δ/2,0), with β0≥5/3 (2.1) and ˜ β(x)=β∗(−x)forx∈(0,1+δ).(2.2) We set β(x)=β∗(x)forx∈(−1−δ, 0], ˜ β(x)forx∈[0,1+δ). (2.3) Let us put ¯ β=5 4β∞=5 4(1 + β0). Let λ≥1 large enough to be fixed below and let us introduce the weights ϑand ϕ,with ϑ(x, t)=eλ¯ β−eλβ(x) t(T−t),ϕ(x, t)= eλβ(x) t(T−t)for (x, t)∈Q. (2.4) Notice that |β∗ x|≤1, |˜ βx|≤1, ϑx=−λβxϕand ϕx=λβxϕ. (2.5) In the sequel, we will use the following notation: C2 ±(Q) is the space of functions z∈C0(¯ Q) such that z|¯ Q−∈C2(¯ Q−)andz|¯ Q+∈C2(¯ Q+); the one-sided spatial derivatives at x= 0 of a function z∈C2 ±(Q) will be denoted by zx(0+,t)andzx(0−,t). Let us also put Lz:= (1 −κh)zt+1 1−κhzxx . Then we have the following Carleman estimate: Theorem 2.1. Assume that h∈W1,∞(0,T)and |h(t)|≤1−b<1for all t∈ (0,T). There exist positive constants C0,λ 0and s0such that Q e−2sϑ 1 sϕ(|zt|2+|zxx|2)+sλ2ϕ|zx|2+s3λ4ϕ3|z|2dx dt +T 0 e−2sϑ(0,t)sλϕ(0,t)(|zx(0+,t)|2+|zx(0−,t)|2)dt +T 0 e−2sϑ(0,t)s3λ3ϕ(0,t)3|z(0,t)|2dt Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 791 ≤C0 b4Q e−2sϑ|Lz|2dx dt +T 0 e−2sϑ(0,t)1 1−κhzx(0,t)+γ(t) 2 dt +ω×(0,T) e−2sϑs3λ4ϕ3|z|2dx dt(2.6) for any (z,γ)∈C2 ±(Q)×C1([0,T]) such that z(0,t)≡γ(t)and z(−1−δ, t)≡ z(1 + δ, t)≡0and any λ≥λ0and s≥s0. Furthermore,C 0can be chosen only depending on β∗;λ0and s0can be chosen of the form λ0=K(1 + b−2)and s0=K(T(1 + hL∞(0,T ))+T2(1 + b−2)) respectively,where Kis independent of hand T. Remark 2.1. In order to get (2.6), we need previous estimates of zand its derivatives in Q−and Q+; see (2.9) and (2.10). We do not know how to arrive at an estimate in the whole Qotherwise. But the right-hand sides of (2.9) and (2.10) contain local integrals of zin ω−×(0,T)andω+×(0,T), respectively. This means in practice that, unfortunately, we are able to solve the control problem for (1.5) only if we put controls at both sides of the line x= 0. To get the null controllability with β(t)≡0, we would need an estimate like (2.6) with ωreplaced by ω−in the right. Let us give the proof of Theorem 2.1. It consists of three steps. First, we deduce in Lemma 2.1 a global Carleman estimate for the function zin Q−. Then, another Carleman inequality is obtained in Q+. Finally, combining both estimates, we will be able to get the inequality (2.6). More precisely, our proof starts with some computations in Q−which are standard by now and provide the inequality Q− e−2sϑ1 sϕ(|zt|2+|zxx|2)+sλ2ϕ|zx|2+s3λ4ϕ3|z|2dx dt +T 0 e−2sϑ(0,t)(sλ ϕ(0,t)|zx(0,t)|2+s3λ3ϕ(0,t)3|z(0,t)|2)dt +T 0 e−2sϑ(0,t)zx(0,t)γ(t)dt ≤K1 b2Q− e−2sϑ|Lz|2dx dt +1 sT 0 e−2sϑ(0,t)ϕ(0,t)−1|γ(t)|2dt +1 b4ω−×(0,T ) e−2sϑs3λ4ϕ3|z|2dx dt(2.7) for large λand s. The possibly nonpositive last term on the left-hand side and the third term on the right-hand side of this estimate appear because zdoes not necessarily vanish at x=0 −. The structure of the former is a consequence of the choice of the function ϑ; Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 798 A. Doubova & E. Fern´andez-Cara that zn(resp. γn) is also uniformly bounded in L2(0,T−ε;H1 0)(resp.H2(0,T−ε)) for all ε>0. Hence, it can also be assumed that zn→zweakly in L2(0,T −ε;H1 0), γn→γweakly in H2(0,T −ε) for all ε>0. Obviously, the couple (z,γ)satisfies              −((1 −κh)z)t−1 1−κhzxx + ((1 −κx)hz)x−1 2˜uzx=0,(x, t)∈Q ∗, z(−1−δ, t)=0,z(1 + δ, t)=0,t∈(0,T), z(0,t)=γ(t),1 1−κhzx(0,t)=−γ(t),t∈(0,T). (3.15) In ω×(0,T)wehavezn=(zn−Pznw)+Pznw. Therefore, zn→P∗zstrongly in L2(ω×(0,T)).(3.16) Butthisimpliesthatz=P∗win ω×(0,T). Taking into account that the partial differential equation in (3.15) has the unique continuation property and wis, together with j, the solution of (3.9), we easily deduce that z=P∗win Q.But thiscanonlybetrueifz(x, t)≡0andP∗=0. From (3.16), we now see that zn→0 strongly in L2(ω×(0,T)).(3.17) Coming back to (3.3), the following holds Q e−2sϑϕ3|zn|2dx dt +T 0 e−2sϑ(0,t)ϕ(0,t)3|(γn)(t)|2dt →0asn→∞. But this is in contradiction with (3.12). Consequently, (3.11) holds. From (3.11), arguing as in the proof of Proposition 3.1, the following is found: (z(·,0),γ(0))2 L2×R≤C8ω×(0,T)|z−Pzw|2dx dt (3.18) for some C8=C8(T,b,R). On the other hand, we also have ω×(0,T ) z w dx dt 2 ≤C9ω×(0,T )|z−Pzw|2dx dt, (3.19) with C9=C9(T,b,R). This can also be proved by contradiction, arguing as above (for simplicity, we omit the details in this case). Finally, in account of (3.18) and (3.19), we see that (3.10) holds. This ends the proof. Remark 3.2. The previous proof relies on the compactness of the sequence {Pznw} and the uniqueness of (3.15). As usual, it provides the existence of a constant C5 such that an improved inequality is satisfied, but it gives no information on the Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 799 size of this constant. In other words, contrarily to C4, no explicit estimate can be given of C5in terms of T,band R. This is a price we have to pay to get the null controllability of v,kand k. Remark 3.3. At present, it is unknown whether (3.2) or (3.10) remain true with ω replaced (for instance) by ω−=(−1−δ, −1−δ/2). If this were the case, we would be able to prove null controllability properties of (1.5) with β(t)≡0, i.e. with only one control located at x=−1; see Remark 2.1. 4. Null Controllability of the Linearized System Let us give the proof of Theorem 1.1. We will consider the following auxiliary distributed controlled system:                    (1 −κh)vt−1 1−κhvxx −(1 −κx)hvx+1 2(˜uv)x=f1ω,(x, t)∈Q ∗, v(−1−δ, t)=0,v(1 + δ, t)=0,t∈(0,T), v(0,t)=k(t),1 1−κhvx(0,t)=k(t),t∈(0,T), v(x, 0) = ˜v0(x),k(0) = k0,k (0) = k1. (4.1) Here, ˜u(resp. ˜v0) is an extension of u(resp. v0) satisfying ˜uL∞(Q)≤uL∞(Q) (resp. ˜v0L2≤Cv0L2(−1,1))and1 ωdenotes the characteristic function of ω= (−1−δ, −1−δ/2) ∪(1 + δ/2,1+δ). Obviously, the distributed null controllability of (4.1) with controls f∈L2(ω×(0,T)) implies the boundary null controllability of (1.5). We have the following result: Proposition 4.1. Assume that uL∞(Q)≤R, hL∞(0,T)≤Rand |h(t)|≤ 1−b<1. For any v0∈L2(−1,1),k 0∈(−1,1) and k1∈R,there exists a control f∗∈L2(ω×(0,T)) such that the corresponding solution (v∗,k∗)of (4.1) verifies v∗(x, T )=0 in (−1−δ, 1+δ),k ∗(T)=0,(k∗)(T)=0.(4.2) Moreover,f ∗can be chosen satisfying the estimate f∗L2(ω×(0,T)) ≤C10(v0,k0,k1)L2(−1,1)×R×R,(4.3) where C10 is a positive constant,only depending on T, b and R. Proof. This controllability result is implied by the observability estimate (3.10). In fact, there are several ways to show this. Let us indicate one of them. For every ε>0, let us consider the functional Jε,givenby Jε(z0,γ1)=1 2ω×(0,T )|z−Pzw|2dx dt +ε(z0,γ1)L2×R +((z(·,0) −Pzw(·,0),γ(0) −Pzj(0)),((1 −κh(0))˜v0,k1))L2×R −Pzk0(4.4) Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 800 A. Doubova & E. Fern´andez-Cara for any (z0,γ1)∈L2×R,where(z,γ) is the solution of (3.1) associated to (z0,γ1)∈ L2×R. It is not difficult to see that Jεis continuous and strictly convex. Furthermore, in view of (3.11), the following unique continuation property holds for (3.1): If z−Pzw=0inω×(0,T), then z≡0. Therefore, we can argue as in Ref. 9 and see that lim inf (z0,γ1)L2×R→∞ Jε(z0,γ1) (z0,γ1)L2×R≥ε. This implies that the functional Jεis coercive. Consequently, Jεachieves its minimum at a unique point (z0 ε,γ1 ε)∈L2×R. Let (zε,γ ε) be the solution of (3.1) associated to (z0 ε,γ1 ε). Let us introduce the control fε,with fε=(zε−Pzεw)+ω×(0,T)|w|2dx dt−1 m(v0,k0,k1)w(4.5) in ω×(0,T),wherewehaveusedthenotation m(v0,k0,k1)=−((w(·,0),j(0)),((1 −κh(0))˜v0,k1))L2×R−k0. Let us denote by (vε,k ε) the associated solution of (4.1). We will now see that kε(T) = 0 (4.6) and (vε(·,T),k ε(T))L2×R≤ε √b.(4.7) Indeed, first we have kε(T)=T 0 vε(0,t)dt +k0 =−T 0 vε(0,t)1 1−κhwxx=0 (t)dt −T 0 k ε(t)j(t)dt +k0 =−T 01 1−κh(vε)xx=0 (t)w(0,t)dt +T 0 k ε(t)j(t)dt +ω×(0,T ) fεwdxdt+((w(·,0),j(0)),((1 −κh(0))˜v0,k1))L2×R+k0 =T 0k ε(t)−1 1−κh(vε)xx=0 (t)j(t)dt +m(v0,k0,k1) +((w(·,0),j(0)),((1 −κh(0))˜v0,k1))L2×R+k0 =0. This proves (4.6). Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 801 On the other hand, for any small ε>0, we have ω×(0,T) (zε−Pzεw)zdxdt+ε (z0,γ1)L2×R ((z0 ε,γ1 ε),(z0,γ1))L2×R +((z(·,0) −Pzw(·,0),γ(0) −Pzj(0)),((1 −κh(0))˜v0,k1))L2×R−Pzk0 =0 (4.8) for any (z0,γ1)∈L2×R.Takingintoaccountthat ω×(0,T ) (zε−Pzεw)zdxdt=ω×(0,T ) fεzdxdt−Pzm(v0,k0,k1), it is not difficult to deduce from (4.8) that ((vε(·,T),k ε(T)),((1 −κh(T))z0,γ1))L2×R =−ε (ˆz0 ε,ˆγ1 ε)L2×R ((z0 ε,γ1 ε),(z0,γ1))L2×R for all (z0,γ1)∈L2×R. Recalling that |h(t)|≤1−bfor all t, we obtain (4.7). Notice that the controls fεare uniformly bounded. Indeed, since Jε(z0 ε,γ1 ε)≤ Jε(0,0) ≤0, one has ω×(0,T )|zε−Pzεw|2dx dt ≤2(zε(·,0),γ ε(0))L2×R(v0,k1)L2(−1,1)×R+|Pzε||k0|.(4.9) Combining (4.9) and (3.10), we obtain ω×(0,T )|zε−Pzεw|2dx dt ≤4C5(v0,k0,k1)L2×R×R and, in view of (4.5), we finally have fεL2(ω×(0,T)) =ω×(0,T )|zε−Pzεw|2dx dt +m(v0,k0,k1)2 ω×(0,T )|w|2dx dt1/2 ≤C10(T,b,R)(v0,k0,k1)L2(−1,1)×R×R.(4.10) Hence, we deduce that, at least for a subsequence, we have fε→f∗weakly in L2(ω×(0,T)) and (vε(·,T),k ε(T),k ε(T)) converges to (v∗(·,T),k∗(T),(k∗)(T)) weakly in L2(−1,1)×R×R,where(v∗,k∗) is the solution of (4.1) for f=f∗. From (4.6) and (4.7), we obviously have (4.2). This ends the proof. Theorem 1.1 is an immediate consequence of Proposition 4.1. Consequently, the null controllability of (1.5) is established. Remark 4.1. The estimate (4.3) provides a bound for the cost of controlling (4.1) to zero. However, since the constant C10 is of the form KC5,whereC5is the Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 802 A. Doubova & E. Fern´andez-Cara observability constant arising in (3.10), the way C10 depends on T,band Ris unknown; see Remark 3.2. Remark 4.2. Using the observability inequality (3.2), we can establish the null controllability of vand kwith an explicit estimate of the cost of controlling. More precisely, in the conditions of Proposition 4.1, there exist controls ˆ f∈L2(ω×(0,T)) such that the corresponding solutions (ˆv, ˆ k)of(4.1) verify ˆv(x, T )=0 in(−1−δ, 1+δ)andˆ k(T)=0,(4.11) with ˆ fL2(ω×(0,T)) ≤C11(v0,k1)L2(−1,1)×R(4.12) and C11 ≤eK1+b−2+T−1(1+R)+(1+T)R2. The proof is similar (and even easier) than the proof of Proposition 4.1. In this case, it suffices to consider the functional ˆ Jε,with ˆ Jε(z0,γ1)=1 2ω×(0,T )|z|2dx dt +ε(z0,γ1)L2×R +((z(·,0),γ(0)),((1 −κh(0))˜v0,k1))L2×R(4.13) for all (z0,γ1)∈L2×R. Of course, this leads to a partial boundary null controllability result for (1.5). 5. Local Null Controllability of the Nonlinear System Let us now present the proof of theorem 1.2. As in Section 4, we introduce an auxiliary system:                  (1 −κh)ut−1 1−κhuxx −(1 −κx)hux+uux=f1ω,(x, t)∈Q ∗, u(−1−δ, t)=0,u(1 + δ, t)=0,t∈(0,T), u(0,t)=h(t),1 1−κhux(0,t)=h(t),t∈(0,T), u(x, 0) = ˜u0(x),h(0) = h0,h (0) = h1, (5.1) where ˜u0is an extension of u0to (−1−δ, 1+δ)with ˜u0H1≤u0H1(−1,1) . Recall that u0(0) = h1,Q=(−1−δ, 1+δ)×(0,T), Q ∗={(x, t)∈Q:x=0} and ω=(−1−δ, −1−δ/2) ∪(1 + δ/2,1+δ). Again, the null controllability of (5.1) with controls f∈L2(ω×(0,T)) implies the boundary null controllability of (1.4). Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 803 To prove that (5.1) is null controllable, we proceed as follows. Let us set Z=L2(Q)×W1,∞(0,T) (5.2) and let ZRbe the closed convex set ZR={(w,)∈Z:wL∞(Q)≤R, L∞(0,T)≤R}. We will denote by ·Zthe usual norm in Z: (w,)Z=(w,)L2(Q)×W∞,1(0,T )∀(w,)∈Z. Let us fix (a small) b∈(0,1) and (a large) R>0. We will use the following notation: for any K>0, TKis the real function TK(s)=   Kif s>K, sif −K≤s≤K, −Kif s<−K. Assume that (w, ) is given in ZRand consider the linear system (4.1), with ˜v0, k0,k1,h,hand ˜urespectively replaced by ˜u0,h0,h1,T1−b(), (T1−b())and w. That is to say, consider the following system:                          (1 −κT1−b())vt−1 1−κT1−b()vxx −(1 −κx)(T1−b())vx+1 2(wv)x=f1ω,(x, t)∈Q ∗, v(−1−δ, t)=0,v(1 + δ, t)=0,t∈(0,T), v(0,t)=k(t),1 1−κT1−b()vx(0,t)=k(t),t∈(0,T), v(x, 0) = ˜u0(x),k(0) = h0,k (0) = h1. (5.3) In view of Proposition 4.1, there exists a control f∈L2(ω×(0,T)) and an associated state (v,k)solving(5.3), which satisfies v(x, T )=0 in(−1−δ, 1+δ),k(T)=0 and k(T)=0.(5.4) Furthermore, the control fcanbechosensatisfying fL2(ω×(0,T)) ≤C10(T,b,R)(u0,h 0,h 1)L2(−1,1)×R×R.(5.5) Indeed, since (w, )∈ZR,wehave (T1−b())L∞(0,T)≤R, wL∞(Q)≤R. In view of (4.3), it is then clear that (5.5) holds. Using (5.5) and Lemma 7.1, we deduce that the state (v,k) associated to this control verifies (v,k)E×H2(0,T)≤C12(T,b,R)(u0,h 0)L2(−1,1)×R,(5.6) where Eis given by E={w∈L2(0,T;H1 0):wt∈L2(0,T;H−1)} Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 804 A. Doubova & E. Fern´andez-Cara and C12(T,b,R) is independent of (w,). In particular, we have (v,k)Z≤C13(T,b,R)(u0,h 0)L2(−1,1)×R.(5.7) Let us denote by A(w,)(resp.Λ(w, )) the family of all these controls (resp. the family of the associated states). In other words, let us set A(w,)={f∈L2(ω×(0,T)) : fL2(ω×(0,T )) ≤C10(u0,h 0,h 1)L2(−1,1)×R×R, v(x, T )=0in(−1−δ, 1+δ),k(T)=0and k(T)=0} and Λ(w,)={(v,k):f∈A(w,),(v,k)Z≤C13(u0,h 0)L2(−1,1)×R}. We will consider the set-valued mapping Λ: ZR→ Zand we will try to apply Kakutani’s theorem to Λ in order to deduce the existence of a fixed point. For clarity, let us recall the precise statement of this result: Theorem 5.1. Let Zbe a Banach space,let ZR⊂Zbe a nonempty closed convex set and let Λ: ZR→ ZRbe a set-valued mapping satisfying the following assumptions: •Λ(z)is a nonempty closed convex set of Zfor every z∈ZR. •There exists a nonempty convex compact set K⊂Zsuch that Λ(z)⊂Kfor every z∈ZR. •Λis upper-hemicontinuous in ZR,i.e. for each σ∈Zthe single-valued mapping z→ sup y∈Λ(z)σ, yZ,Z (5.8) is upper-semicontinuous. Then Λpossesses a fixed point in the set K, i.e. there exists z∈Ksuch that z∈Λ(z). For the proof, see for instance Ref. 2. First, it is easy to see that, for each (w, )∈ZR,Λ(w, ) is a nonempty closed convex set of Z. Furthermore, since the embeddings E→L2(Q)andH2(0,T)→ W1,∞(0,T)arecompactandwehave(5.6), there exists a compact set K⊂Z such that Λ(w,)⊂K∀(w,)∈ZR.(5.9) Let us now prove that mapping (w, )→ Λ(w, ) is upper hemicontinuous. More precisely, let us see that, for each bounded linear form (µ, λ)∈Z, the real-valued function (w,)∈Z→ sup (v,k)∈Λ(w,)(µ, λ),(v,k) Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 805 is upper semicontinuous. In other words, let us check that Bν,µ =(w, )∈Z:sup (v,k)∈Λ(w,)(µ, λ),(v,k)≥ν is a closed set of Zfor all ν∈Rand (µ, λ)∈Z. Thus, let {(wn, n)}be a sequence in Bν,µ such that (wn, n)→(w,)inZ. Our goal is to prove that (w,)∈Bν,µ.SinceallsetsΛ(wn, n)arecompactand satisfy (5.9), we deduce that ν≤sup (v,k)∈Λ(wn,n)(µ, λ),(v,k)=(µ, λ),(vn,k n)(5.10) for some (vn,k n)∈Λ(wn, n). Using the definitions of Λ(wn, n)andA(wn, n)we obtain the existence of controls fn∈L2(ω×(0,T)) such that                              (1 −κT1−b(n))vn,t −1 1−κT1−b(n)vn,xx −(1 −κx)(T1−b(n))vn,x +1 2(wnvn)x=fn1ω,(x, t)∈Q ∗, vn(−1−δ, t)=0,v n(1 + δ, t)=0,t∈(0,T), vn(0,t)=k n(t),1 1−κT1−b(n)vn,x(0,t)=k n(t),t∈(0,T), vn(x, 0) = ˜u0(x),k n(0) = h0,k  n(0) = h1. Furthermore, fnL2(ω×(0,T)) ≤C10(u0,h 0,h 1)L2(−1,1)×R×R and (vn,k n)Z≤C13(u0,h 0)L2(−1,1)×R, whence (vn,k n)(resp.fn) is uniformly bounded in Z(resp. L2(ω×(0,T))). Since (5.9) holds, we have, at least for a subsequence, that (vn,k n)→(v,k) strongly in Z and fn→fweakly in L2(ω×(0,T)). Now, it is not difficult to check that f∈A(w, )and(v,k)∈Λ(w, ). We can thus take limits in (5.10) and deduce that ν≤(µ, λ),(v,k)≤ sup (v,k)∈Λ(w,)(µ, λ),(v,k). Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 806 A. Doubova & E. Fern´andez-Cara So,wehave(w,)∈Bν,µ . This proves that (w, )→ Λ(w,) is upper semicontinuous. Let us now see that, for any R>0andb>0, there exists ε1, only depending on T,band R, such that, whenever (u0,h 0)H1(−1,1)×R≤ε1,wehave Λ(w,)⊂ZR∀(w,)∈ZR.(5.11) Indeed, if (w,)∈ZRand (v,k)∈Λ(w, ), we first have from (5.7) that kL∞(0,T)≤C13(u0,h 0)H1(−1,1)×R.(5.12) On the other hand, vL∞(Q)≤C14(u0,h 0)H1(−1,1)×R(5.13) for some C14 =C14(T,b,R). This can be justified as follows. Let us consider the equation in (5.3) for instance for (x, t)∈Q ∗,x<0. Introducing the new variable z,with v=z+1+δ+x 1+δk(t), we see that zsatisfies          (1 + T1−b())zt−1 1+T1−b()zxx =F, (x, t)∈(−1−δ, 0) ×(0,T), z(−1−δ, t)=0,z(0,t)=0,t∈(0,T), z(x, 0) = ˜u0(x)−(1 + δ+x)(1 + δ)−1h1,x∈(−1−δ, 0), (5.14) where F=f1ω+(1+x)(T1−b())zx−1 2(wz)x −1+δ+x 1+δ(1 + T1−b())k +1+x 1+δ(T1−b())k−1 2w1+δ+x 1+δkx . Let us check that z∈L∞(0,T;H1 0(−1−δ, 0)) and zL∞(0,T;H1 0(−1−δ,0)) ≤C15(T,b,R)(u0,h 1)H1(−1,0)×R.(5.15) This will suffice to prove that vis essentially bounded in (−1−δ, 0) ×(0,T). In a similar way, we will able to prove that vis L∞in (0,1+δ)×(0,T) and this will lead to (5.13). From (5.6), it is immediate that z∈L2(0,T;H1 0(−1−δ, 0)), with a norm in this space bounded in the same way. Therefore, Fcanbewrittenintheform F=P+Qx,where P∈L2((−1−δ, 0) ×(0,T)),Q∈L∞(0,T;L2(−1−δ, 0)) Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 807 and the corresponding norms are again bounded as in (5.6). Now, from standard parabolic regularity results, taking into account that the initial data in (5.14) vanishes at x=−1−δand x= 0, we deduce that z∈ L∞(0,T;H1 0(−1−δ, 0)) and (5.15) holds; see for instance Ref. 17. In view of (5.12) and (5.13), we see that the inequality (u0,h 0)H1(−1,1)×R≤ε1:= R max(C13,C 14) implies (5.11). Thus, if (u0,h 0)H1(−1,1)×R≤ε1, we can apply Kakutani’s fixed point theorem to Λ. This leads to the existence of a control fand a solution (u, h) to the nonlinear system                          (1 −κT1−b(h))ut−1 1−κT1−b(h)uxx −(1 −κx)(T1−b(h))ux+uux=f1ω,(x, t)∈Q ∗, u(−1−δ, t)=0,u(1 + δ, t)=0,t∈(0,T), u(0,t)=h(t),1 1−κT1−b(h)ux(0,t)=h(t),t∈(0,T), u(x, 0) = ˜u0(x),h(0) = h0,h (0) = h1, (5.16) such that (5.5) holds, (u, h)E×H2(0,T)≤C12(T,b,R)(u0,h 0)L2(−1,1)×R,(5.17) (u, h)L∞(Q)×L∞(0,T)≤C16(T,b,R)(u0,h 0)H1(−1,1)×R,(5.18) u(x, T )=0 in(−1−δ, 1+δ),h(T)=0 and h(T)=0.(5.19) Notice that, in particular, hL∞(0,T)≤C17(T,b,R)(u0,h 0)H1(−1,1)×R. Hence, if we set ε2=1−b C17(T,b,R) and we choose (u0,h 0)H1(−1,1)×R≤ε:= min(ε1,ε 2), we have T1−b(h)≡hand (u, h) is in fact a solution of (5.1) satisfying (5.19). This ends the proof of Theorem 1.2. Remark 5.1. Combining Remark 3.1 and a similar fixed point argument, the local null controllability of uand hcan be established. More precisely, it can be proved that, for any h0∈(−1,1), there exists ε>0 only depending on Tand |h0|such that, whenever the initial data u0and h1satisfy u0(0) = h1and u0H1(−1,1) ≤ε, Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 814 A. Doubova & E. Fern´andez-Cara Then, we can write Q− 1 sϕ|ψxx|2dx dt ≤4Q− 1 sϕ  1 1−κhψxx 2 dx dt ≤2Q− |M1ψ|2 sϕ dx dt +Ks3λ4Q− 1 (1 −κh)2ϕ3|ψ|2dx dt +KsQ− (1 −κh)2ϕ−1|ϑt|2|ψ|2dx dt. (A.25) Using (A.19) and the fact that s≥s2, we obtain: Q− 1 sϕ|ψxx|2dx dt ≤KM1ψ2 2+Ks3λ4 b2+sT2Q− ϕ3|ψ|2dx dt ≤KM1ψ2 2+Ks3λ4 b2Q− ϕ3|ψ|2dx dt. (A.26) So, (A.24) holds. On the other hand, we also have Q− 1 sϕ|ψxx|2dx dt +s3λ4Q− ϕ3|ψ|2dx dt +s3λ3T 0 e−2sϑ(0,t)ϕ(0,t)3|γ(t)|2dt ≥Kb2sλ2Q− ϕ|ψx|2dx dt. (A.27) In order to justify this assertion, notice that, after integrating by parts, we can write the following: sλ2Q− ϕ|ψx|2dx dt =−sλ2Q− ϕψxxψdxdt−sλ3Q− βxϕψψxdx dt +sλ2T 0 ϕ(0,t)ψx(0,t)ψ(0,t)dt. We deduce that sλ2Q− ϕ|ψx|2dx dt ≤sλ2 2Q− ϕ|ψx|2dx dt +1 2Q− 1 sϕ|ψxx|2dx dt +s3λ4 2Q− ϕ3|ψ|2dx dt +Ksλ4Q− ϕ|ψ|2dx dt +s2λ3 2T 0 ϕ(0,t)|γ(t)|2dt +λ 2T 0 ϕ(0,t)|ψx(0,t)|2dt which together with (A.24), gives (A.27). Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 815 From (A.23), taking into account (A.24) and (A.27), we get M2ψ2 2+Q−1 sϕ|ψxx|2+sλ2ϕ|ψx|2+s3λ4ϕ3|ψ|2dx dt +sλ T 0 ϕ(0,t)|ψx(0,t)|2dt +s3λ3T 0 e−2sϑ(0,t)ϕ(0,t)3|γ(t)|2dt +2T 0 ψx(0,t)ψt(0,t)dt ≤Kb−2e−sϑg2 2+s3λ4ω−×(0,T) ϕ3|ψ|2dx dt(A.28) for λ≥λ2and s≥s2. Observe that we also have M2ψ2 2+sλ2Q− ϕ|ψx|2dx dt ≥Kb4Q− 1 sϕ|ψt|2dx dt (A.29) for all λ≥λ2and s≥s2.Indeed,wehavefrom(A.7)that (1 −κh)ψt=M2ψ+2 1−κhsλβxϕψx. Thus, we can write Q− (1 −κh)2 sϕ |ψt|2dx dt ≤2Q− |M2ψ|2 sϕ dx dt +8sλ2Q− β2 xϕ (1 −κh)2|ψx|2dx dt. With λ≥λ2and s≥s2,weobtain b2Q− 1 sϕ|ψt|2dx dt ≤KM1ψ2 2+K b2sλ2Q−|ψx|2dx dt, (A.30) which trivially gives (A.29). From (A.30) and (A.29), we finally deduce that Q−1 sϕ(|ψt|2+|ψxx|2)+sλ2ϕ|ψx|2+s3λ4ϕ3|ψ|2dx dt +sλ T 0 ϕ(0,t)|ψx(0,t)|2dt +s3λ3T 0 e−2sϑ(0,t)ϕ(0,t)3|γ(t)|2dt +2T 0 ψx(0,t)ψt(0,t)dt ≤Kb−4e−sϑg2 2+s3λ4ω−×(0,T) ϕ3|ψ|2dx dt(A.31) for all λ≥λ2and s≥s2. Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 816 A. Doubova & E. Fern´andez-Cara Let us deduce from (A.31) that (2.9) holds for all s≥s3=K(T(1 + hL∞(0,T))+T2(1 + b−2)). Recall that ψ=e−sϑz.Wehave ψx=e−sϑ(zx−sϑxz)=e−sϑ(zx+sλβxϕz),(A.32) ψxx =e−sϑ(zxx +2sλβxϕzx+sλ(ϕβxx +βxϕx+sλβ2 xϕ2)z).(A.33) Thanks to (A.32), we can write e−sϑzx=ψx−sλβxϕe−sϑz=ψx−sλβxϕψ. Consequently, we find the following: sλ2Q− e−2sϑϕ|zx|2dx dt =sλ2Q− ϕ|ψx−sλβxϕψ|2dx dt ≤Ksλ2Q− ϕ|ψx|2dx dt +s3λ4Q− ϕ3|ψ|2dx dt. (A.34) From (A.33), we have e−sϑzxx =ψxx −2sλe−sϑβxϕzx−sλ(ϕβxx +βxϕx+sλβ2 xϕ2)ψ. Then, we obtain Q− e−2sϑ 1 sϕ|zxx|2dx dt =Q− 1 sϕ|ψxx −2sλe−sϑβxϕzx−sλ(ϕβxx +βxϕx+sλβ2 xϕ2)ψ|2dx dt ≤KQ− 1 sϕ|ψxx|2dx dt +Ksλ2Q− ϕ|zx|2dx dt +s3λ4Q− ϕ3|ψ|2dx dt ≤KQ− 1 sϕ|ψxx|2dx dt +sλ2Q− ϕ|ψx|2dx dt +s3λ4Q− ϕ3|ψ|2dx dt. (A.35) Here, we have used (A.34). On the other hand, using (A.32) we have that ψx(0,t)=e−sϑ(0,t)(zx(0,t)−sλϕ(0,t)z(0,t)) =e−sϑ(0,t)(zx(0,t)−sλϕ(0,t)γ(t)) (A.36) for all t, whence the following is found: sλ T 0 e−sϑ(0,t)ϕ(0,t)|zx(0,t)|2dt ≤KsλT 0 ϕ(0,t)|ψx(0,t)|2dt +s3λ3T 0 e−2sϑ(0,t)ϕ(0,t)3|γ(t)|2dt.(A.37) Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 817 Also, we can write that ψt=e−sϑ(zt−sϑtz),(A.38) ψt(0,t)=e−sϑ(0,t)(γ(t)−sϑt(0,t)γ(t)).(A.39) Then, from (A.36) and (A.37), we obtain T 0 ψx(0,t)ψt(0,t)dt =T 0 e−2sϑ(0,t)zx(0,t)(γ(t)−sϑt(0,t)γ(t))dt −sλ T 0 e−2sϑ(0,t)ϕ(0,t)γ(t)(γ(t)−sϑt(0,t)γ(t))dt, whichcanalsobewrittenintheform T 0 e−2sϑ(0,t)zx(0,t)γ(t)dt −sT 0 e−2sϑ(0,t)ϑt(0,t)zx(0,t)γ(t)dt =T 0 ψx(0,t)ψt(0,t)dt +sλ T 0 e−2sϑ(0,t)ϕ(0,t)γ(t)γ(t)dt −s2λT 0 e−2sϑ(0,t)ϕ(0,t)ϑt(0,t)|γ(t)|2dt. (A.40) Using (A.38), we deduce that e−sϑzt=ψt+sϑte−sϑz≡ψt+sϑtψ. Then, using (A.19) we get Q− e−2sϑ 1 sϕ|zt|2dx dt =Q− 1 sϕ|ψt+sϑtψ|2dx dt ≤KQ− 1 sϕ|ψxx|2dx dt +KsT2Q− ϕ|ψ|2dx dt. (A.41) From (A.31), (A.34), (A.35), (A.37), (A.40) and (A.41), we deduce that Q− e−2sϑ1 sϕ(|zt|2+|zxx|2)+sλ2ϕ|zx|2+s3λ4ϕ3|z|2dx dt +sλ T 0 e−2sϑ(0,t)ϕ(0,t)|zx(0,t)|2dt +s3λ3T 0 e−2sϑ(0,t)ϕ(0,t)3|γ(t)|2dt +2T 0 e−2sϑ(0,t)zx(0,t)γ(t)dt ≤Kb−4e−sϑg2 2+s3λ4ω−×(0,T ) ϕ3|ψ|2dx dt Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 818 A. Doubova & E. Fern´andez-Cara +s2λT 0 e−2sϑ(0,t)ϕ(0,t)ϑt(0,t)|γ(t)|2dt +2sT 0 e−2sϑ(0,t)ϑt(0,t)zx(0,t)γ(t)dt −2sλ T 0 e−2sϑ(0,t)ϕ(0,t)γ(t)γ(t)dt (A.42) for all λ≥λ2and s≥s2. The last three terms in (A.42) can be controlled by the second and third terms on the left-hand side. Indeed, taking s≥s3= K(T(1 + hL∞(0,T ))+T2(1 + b−2)), we have Q− e−2sϑ1 sϕ(|zt|2+|zxx|2)+sλ2ϕ|zx|2+s3λ4ϕ3|z|2dx dt +T 0 e−2sϑ(0,t)sλ ϕ(0,t)|zx(0,t)|2+s3λ3ϕ(0,t)3|z(0,t)|2dt +T 0 e−2sϑ(0,t)zx(0,t)γ(t)dt ≤K1 b2Q− e−2sϑ|Lz|2dx dt +1 sT 0 e−2sϑ(0,t)ϕ(0,t)−1|γ(t)|2dt +1 b4ω−×(0,T ) e−2sϑs3λ4ϕ3|z|2dx dt.(A.43) Here, we have also used the estimate (A.20). Finally, let us note that T 0 e−2sϑ(0,t)1 1+κh zx(0,t)γ(t)dt =T 0 e−2sϑ(0,t)zx(0,t)γ(t)dt +T 0 e−2sϑ(0,t)1 1+κh −1zx(0,t)γ(t)dt ≤T 0 e−2sϑ(0,t)zx(0,t)γ(t)dt +K bsλ T 0 e−2sϑ(0,t)ϕ(0,t)−1|γ(t)|2dt +Ksλ bT 0 e−2sϑ(0,t)ϕ(0,t)|zx(0,t)|2dt. (A.44) The inequality (A.44), together with (A.43), leads to (2.9) and this is what we wanted to prove. This ends the proof of Lemma 2.1. Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 819 Appendix B: A Technical Result For any u∈L∞(Q)andanyh∈W1,∞(0,T)with|h(t)|≤1−b<1, let us consider the linear system                  (1 −κh)vt−1 1−κhvxx −(1 −κx)hvx+1 2(uv)x=g, (x, t)∈Q, x =0, v(−1,t)=0,v(1,t)=0,t∈(0,T), v(0,t)=k(t),1 1−κhvx(0,t)=k(t),t∈(0,T), v(x, 0) = v0(x),k(0) = k0,k (0) = k1. (B.1) Then we have the following result: Lemma 7.1. For any v0∈L2(−1,1) and k1∈Rand for any u∈L∞(Q),h∈ W1,∞(0,T)with |h(t)|≤1−b<1and g∈L2(Q),there exists exactly one solution (v,k)to (B.1) that satisfies v∈C0([0,T]; L2(−1,1)) ∩L2(0,T;H1 0(−1,1)),(B.2) vt∈L2(0,T;H−1(−1,1)) (B.3) and k∈H2(0,T).(B.4) Furthermore,there exists a positive constant C7,only depending on T, b, hL∞(0,T)and uL∞(Q),such that kH1(0,T)+vL2(0,T ;H1 0(−1,1)) +vtL2(0,T;H−1(−1,1)) ≤C7(v0,k1)L2(−1,1)×R+gL2(Q).(B.5) Proof. We will apply Galerkin’s method. We will only present the main steps in this proof, since some arguments are standard and well known. Let us set V={(w,): w∈H1 0(−1,1),∈R,w(0) = } and H={(w,): w∈L2(−1,1),∈R}. The (natural) norms in the spaces Vand Hare respectively given by (w,)V=1 −1|wx|2dx1/2 and (w,)H=(w,)L2(−1,1)×R. Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 820 A. Doubova & E. Fern´andez-Cara •The choice of the special basis. We have that Vis a separable Hilbert space and V→H≡H→Vwith compact and dense embeddings. We can then choose a special basis of V.More precisely, let the couples (wj, j) and the real positive numbers λjbe the following eigenfunctions and eigenvalues:      −wj,xx =λjwj,x∈(−1,1),x=0, wj(−1) = 0,w j(1) = 0, wj(0) = j,[wj,x](0)=−λjj. Then, {(w1,l 1),...,(wp,l p),...}is an orthogonal basis of Vcharacterized by      (wj,x,w x)=λj((wj, j),(w,))L2×R, ∀(w,)∈V, (wj, j)∈V, (wj, j)H=1,λ j+∞, where (·,·) denotes the usual scalar product in L2(−1,1). We have that {(wj, j)}is orthonormal in Hand orthogonal in V. •Definition of the approximated solutions. Let Vpbe the subspace of Vspanned by the couples (wj, j)with1≤j≤p. Let us introduce the following problem: Find (vp,k p): [0,T]→ Vp,with vp= p  j=1 mjp(t)wj,k  p= p  j=1 mjp(t)j, such that              d dtk p(t)j+ ((1 −κh)vp,t,w j)+1 1−κhvp,x,w j,x −((1 −κx)hvp,x,w j)−1 2(uvp,w j,x)=(g,wj),1≤j≤p, (vp(x, 0),k p(0))L2×R=(v0 p,k1 p). (B.6) Here (v0 p,k1 p)=Pp(v0,k1)andPp:H→ Vpis the usual orthogonal projector. Since (B.6) is a Cauchy problem for a linear ordinary differential system of dimension p, it possesses exactly one solution (vp,k p) defined in the whole time interval [0,T]. •A priori estimates of vpand k p. Multiplying the differential equation in (B.6) by mjp and adding in jfor 1 ≤ j≤p, we obtain: d dtk pk p+ ((1 −κh)vpt,v p)+1 1−κhvp,x,v p,x −((1 −κx)hvp,x ,v p)−1 2(uvp,v p,x)=(g,vp). Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 821 Then, we can write 1 2 d dt |k p|2+1 −1 (1 −κh)|vp|2dx+1 −1 1 1−κh|vp,x|2dx =1 21 −1 uvpvp,x dx +(g,vp).(B.7) We deduce from (B.7) that 1 2 d dt |k p|2+1 −1 (1 −κh)|vp|2dx+1 −1 1 1−κh|vp,x|2dx ≤1 2ε1 −1 1 1−κh|vp,x|2dx +2u2 L∞(Q)1 −1 (1 −κh)|vp|2dx +1 21 −1 (1 −κh)−1|g|2dx +1 21 −1 (1 −κh)|vp|2dx. Hence, d dt|k p|2+1 −1 (1 −κh)|vp|2dx+1 −1 1 1−κh|vp,x|2dx ≤(1 + 4u2 L∞(Q))1 −1 (1 −κh)|vp|2dx +1 −1 (1 −κh)−1|g|2dx. From this inequality, we easily obtain that d dte−C(1+4u2 L∞(Q))t|k p|2+1 −1 (1 −κh)|vp|2dx+1 −1 1 1−κh|vp,x|2dx ≤C(T,uL∞(Q))1 −1 (1 −κh)−1|g|2dx. (B.8) Integrating (B.8) with respect to time in [0,t]witht∈[0,T], we also get the following: |k p|2+1 −1|vp|2dx +T 01 −1|vp,x|2dx dt ≤C(T,b,uL∞(Q))(v0 p,k1 p)2 L2(−1,1)×R+g2 L2(Q). Therefore, k p∈L∞(0,T), vp∈L∞(0,T;L2(−1,1)) ∩L2(0,T;H1 0(−1,1)) and k pL∞(0,T)+vpL∞(0,T ;L2(−1,1)) +vpL2(0,T ;H1 0(−1,1)) ≤C(T,b,uL∞(Q))v0 p,k1 p)L2(−1,1)×R+gL2(Q).(B.9) Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 822 A. Doubova & E. Fern´andez-Cara •A priori estimates of vp,t and k p. Let us introduce Mp=Mp(t)withMp(t)∈Vfor all tand      Mp(t),(w,)=−1 1−κh(t)vp,x(t)−1 2u(·,t)vp(t),w j,x −((1 −κx)h(t)vp,x(t),w)+(g,w)∀(w, )∈V. (B.10) Here, ·,· stands for the duality pairing associated to Vand V.Then ((vp,t,k p),(wj, j))L2×R=Mp+κh(vp,t,0),(wj, j)∀j=1,...,p. Consequently, for each t,wehave (vp,t,k p)(t)= ˜ Pp(Mp(t)+κh(t)(vp,t(t),0)),(B.11) where ˜ Pp:V→ Vpis the projector defined by ˜ Pp(ζ)= p  j=1ζ,(wj, j)(wj, j)∀ζ∈V. From the fact that Vpis the space spanned by the couples (wj, j)(1≤j≤p) and {(wj, j)}is the special basis of V, we deduce that ˜ Pp(ζ,m)V≤(ζ,m)V∀(ζ,m)∈V. Therefore, in view of the inequalities |h(t)|≤1−b<1, we see from (B.11) that (vp,t,k p)(t)V≤1 bMp(t)V for all t. This, the definition (B.10) of Mp(t) and (B.9) imply that vp,t ∈ L2(0,T;H−1(−1,1)), k p∈L2(0,T)and k pL2(0,T)+vp,tL2(0,T ;H−1(−1,1)) ≤C(T,b,uL∞(Q))v0 p,k1 p)L2(−1,1)×R+gL2(Q).(B.12) •Conclusions. In view of (B.9) and (B.12), we can pass to the limit (at least for a subsequence) in all the terms of (B.6). We deduce the existence of a solution (v,k)to(B.1). Furthermore, the estimates (B.9) and (B.12) imply that (B.5) holds. •Uniqueness. Let us assume that there exist two solutions (v1,k 1)and(v2,k 2) of (B.1) and let us set v=v1−v2,k=k1−k2. Math. Models Methods Appl. Sci. 2005.15:783-824. Downloaded from www.worldscientific.com by UNIVERSITY OF SEVILLE on 12/05/24. Re-use and distribution is strictly not permitted, except for Open Access articles. May 4, 2005 13:29 WSPC/103-M3AS 00052 Some Control Results for Simplified One-Dimensional Models of Fluid-Solid Interaction 823 Then (v,k)satisfies                    (1 −κh)vt−1 1−κhvxx −(1 −κx)hvx+1 2(uv)x=0,(x, t)∈Q, x =0, v(−1,t)=0,v(1,t)=0,t∈(0,T), v(0,t)=k(t),1 1−κhvx(0,t)=k(t),t∈(0,T), v(x, 0) = 0,k(0) = 0,k (0) = 0. 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