EXACT NEUMANN BOUNDARY CONTROLLABILITY FOR SECOND ORDER HYPERBOLIC EQUATIONS

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1 Published in Colloq. Math , 7-4. EXAC NEUMANN BOUNDARY CONROLLABILIY FOR SECOND ORDER HYPERBOLIC EUAIONS BY Weijiu Liu and Graham H. Williams ABSRAC Using HUM, we study the problem of exact controllability with Neumann boundary conditions for second order hyperbolic equations. We prove that these systems are exactly controllable for all initial states in L H and we derive estimates for the control time. Key Words: Exact controllability; Second order hyperbolic equations; Neumann boundary condition; HUM AMS subject classification: 93B5, 35B37, 35L. Introduction and Main Result. Let be a bounded domain open, connected, and nonempty in lr n n with suitably smooth boundary Γ =. For >, set =, and Σ = Γ,. he aim of this paper is to discuss the problem of exact controllability for second order hyperbolic equations with Neumann boundary control y n i,j= a ij x, t y = in, y = y, y = y in, y = φ ν A on Σ.. In., a ij x, t are suitably smooth real valued functions and a ij x, t = a ji x, t, i, j =,,, n, n A = a ij x, t,. the co-normal derivative y ν A i,j= with respect to A is equal to n i,j= a ij x, tν i y x j, and ν = ν, ν,, ν n is the unit normal on Γ pointing towards the exterior of, y = y t, y = yx,, y = y x,, and φ is a boundary control function.

2 More precisely, the problem of exact controllability can be stated as follows. Given >, for any initial state y, y and any terminal state z, z in a suitable Hilbert space H, find a boundary control φ such that the solution y = yx, t; φ of. satisfies yx, ; φ = z, y x, ; φ = z in..3 Since system. is linear, it is sufficient to look for controls driving the system. to rest, i.e., yx, ; φ =, y x, ; φ = in..4 hroughout this paper, we will adopt the following notation. C [, ; lr n, and set Let x t mx, t = x x t = x x t,, x n x nt = m x, t,, m n x, t,.5 Σx = {x, t Σ : mx, t νx = n m k x, tν k x > },.6 k= Σ x = Σ Σx,.7 Γx = {x Γ : mx, νx > },.8 Rt = max x R t = max x Σx = Γx,,.9 mx, t = max x m x, t = max x n x k x kt,. k= n x k t,. k= R = max Rt.. t Before stating the main results of this paper, we impose certain conditions on a ij. We suppose a ij x, t, a ij x, t, a ij x, t C[, ; L, a ij x, t L,, i, j, k =,,, n. and there exists a constant α > such that.3 a ij x, tξ i ξ j α ξ, ξ lr n, x, t..4

3 Here and in the sequel, we use the summation convention for repeated indices, for example, n a ij x, tξ i ξ j = a ij x, tξ i ξ j. i,j= If Set we set = at = n α bt = n α max max i,j n x a ijx, t,.5 max a ijx, t..6 x i,j,k n max at, bt, R t L,,.7 R b, R α e a, R, α e a,,.8 where, denotes the norm of L,. Furthermore, if or a ijx, tξ i ξ j, x, t [,, ξ lr n,.9 a ijx, tξ i ξ j, x, t [,, ξ lr n,. then can be refined slightly to = R b, R α R, α e a,,. or = R b, R α e a, R, α e a,.. If we suppose at, bt, R t L,,.3 3R a, R α b, R, < α,.4 where, denotes the norm of L,. In the sequel, W s,p denotes the usual Sobolev space and s,p its norm for any s lr and p. We write H s for W s, and s for s,. We now state the main result as follows. heorem.. Let be a bounded domain in R n with the boundary Γ of class C. Suppose.3 and.4 hold and Σx Σx. If either.7 holds and > or.3 and.4 hold and is large enough so that 3R a, R α b, R, < α R,.5 3

4 then for all initial states there exists a control y, y L H, φ = { φ on Σx, φ on Σ x, with φ H Σx and φ H Σ x such that the solution y = yx, t; φ of. satisfies.4. Corollary.. Under the conditions of heorem., if Σ x =, then for all initial states y, y L H, there exists a control φ H, ; L Γ, such that the solution y = yx, t; φ of. satisfies.4. Remark.3. Σ x = if x t x and is star-shaped with respect to x see [3]. he method of proof of heorem. uses multiplier techniques and the Hilbert Uniqueness Method HUM for short introduced by Lions [9]. We now compare our result with the existing literature. he problem of exact controllability for second order hyperbolic equations for both Dirichlet and Neumann boundary controls has been extensively studied. he first work for Dirichlet boundary controls was done probably by Komornik [5], who dealt with the wave equation with variable coefficients but not depending on time by using HUM. Later the time-dependent case was considered by Apolaya [] and Miranda []. In addition, making use of the theory of pseudodifferential operators, Bardos, Lebeau and Rauch [] considered the Neumann boundary controllability with rather smooth coefficients and domains. he control considered in this paper is of Neumann type and the coefficients and domain are required to be less smooth. Generally speaking, Neumann control is more delicate than the Dirichlet one. We also allow for the case that Σx is not a cylinder of a form Σx = Γx,, where x is independent of t, and give delicate estimates for the control time as given in.8 and.5. Further, the condition.4 generalizes condition 3 of [5]. he rest of this paper is divided into four parts. Section is devoted to a discussion of the regularity of solutions of Neumann boundary value problems. We then establish an identity for the solution in section. Using the identity, we obtain an observability inequality in section 3. We prove heorem. in section 4.. Regularity of Solutions. We first give some preliminary results on solutions of the following Neumann boundary value problem u Au = f in, u = u, u = u in,. = on Σ. ν A 4

5 hroughout this paper, it is assumed that there is α > such that a ij x, tξ i ξ j α ξ, ξ R n, x, t [, ].. Let X be a Banach space. We denote by C k [, ], X the space of all k times continuously differentiable functions defined on [, ] with values in X, and write C[, ], X for C [, ], X. By example 3 of chapter XVIII of [3], we have heorem.. Let be a bounded domain in lr n with Lipschitz boundary Γ. Suppose that a ij x, t, a ijx, t C[, ], L, i, j =,,, n..3 hen, for u, u, f H L L, ; L, problem. has a unique solution with u C[, ]; H C [, ]; L..4 Moreover, there exists a constant c = c such that u C[, ];H u C[, ];L.5 c[ u u f L, ;L ]. A solution to. which satisfies.4 is called a weak solution. Set with norm and W,, ; L = {f : f, f L, ; L }.6 f W,= f L, ;L f L, ;L,.7 DA = { u H : We will need the following regularity result. } =..8 ν A heorem.. Let be a bounded domain in R n with boundary Γ of class C. Suppose that a ij x, t, a ij x, t, a ij x, t C[, ]; L, a ij x, t L, i, j, k =,,, n. Assume that {u, u } DA H. 5.9

6 i If f W,, ; L, then problem. has a unique solution with u C[, ]; DA C [, ]; H C [, ]; L.. Moreover, there exists a constant c = c such that u t u t [ c u u f L, ;L ], t [, ].. ii If f L, ; H, then problem. has a unique solution with u C[, ]; DA C [, ]; H.. Moreover, there exists a constant c = c such that ut u t [ c u u f L, ;H ], t [, ]..3 A solution satisfying. is called a strong solution. Proof. We first prove.. o this end, we first suppose that f D, ; L the space of all infinitely differentiable functions with supports in, and values in L. Set and at; ut, vt = a t; ut, vt = a t; ut, vt = x, t vx, t a ij x, t dx,.4 a x, t vx, t ijx, t dx,.5 a x, t vx, t ijx, t dx..6 Let, denote the scalar product in L. For any v H, multiplying. by v and integrating over, we obtain u t, v at; ut, v = ft, v..7 Differentiating.7 with respect to t, we obtain u t, v a t; ut, v at; u t, v = f t, v..8 Replacing v by u t in.8 gives [u t, u t at; u t, u t] a t; ut, u t a t; u t, u t = f t, u t..9 6

7 But a t; ut, u t = [a t; ut, u t] a t; ut, u t a t; u t, u t.. Integrating.9 from to t and using., we have u t at, u t, u t = u a, u, u a, u, u a t, ut, u t 3 t t a s, us, u sds a s, u s, u sds t f s, u sds.. It therefore follows from. and.9 and. that the following c s denoting various constants depending on a, α, u t α u t α [ u t c u u u ut t max s t u s us u s ds t ] f s ds, which, by adding u t to both sides of the above inequality, implies u t u t [ c u u u ut u t t us u s ds max s t u s t ] f s ds.. But ut = u t u sds and u t = u t u sds yield respectively and In addition, by. we have t ut u u s ds,.3 t u t u u s ds..4 u = Au f = Au..5 7

8 So we deduce from.-.5 that u t u t [ c u u f L, ;L t ] u s u s ds max s t u s,.6 from which, setting we deduce wt = max s t u s u s,.7 t ] wt c [ u u f L, ;L wsds..8 Gronwall s inequality see [4, p.36] shows [ wt c u u f L, ;L ]..9 his implies.. By a density argument, we can show. still holds for f W,, ; L. Now we prove.. Using the proof of heorem 8. of [, Vol.I, p.75] and.7 of [, Vol.II, p.97], we can prove u C [, ]; H C [, ]; L..3 On the other hand, by inequality 6.7 of [7, p.66] and Remark 6. of [7, p.77], we have ut c Aut c ut.3 c[ u t ut ft ]. It follows from. and.3 that ut ut c[ u t u t ut ut ft ft ]..3 hus, the continuity of u and f implies u C[, ]; DA..33 It remains to prove.3. Multiplying. by Au and integrating over, we obtain Aut, Aut at; u t, u t a t; ut, u t = at; u t, ft a.34 t; ut, ft. 8

9 Combining this and. gives [Aut, Aut at; u t, u t] = 3 a t; u t, u t a t; ut, u t [a t; ut, u t].35 at; u t, ft a t; ut, ft. Integrating.35 from to t, we have Aut at, u t, u t = Au a, u, u a, u, u a t, ut, u t 3 t t a s, u s, u sds t a s, us, u sds [as; u s, fs a s; us, fs]ds..36 It therefore follows from.,.5, and.36 that there exists a constant c = c > such that Aut u t t c [ u u f L, ;H max s t u s. ] u s ds.37 from which, as in the proof of.9, we deduce Aut u t [ ] c u u f L, ;H..38 hus.3 follows from.5,.3, and.38. Finally,. is a consequence of.3 through a density argument.. An Identity. We are now in a position to establish an identity, which is indispensable for obtaining an observability inequality in the following section. then, We define the energy of the solution u of. with f = by Et = u x, t dx at; ut, ut,. E t = a t; ut, ut,. 9

10 and Et = E t a s; us, usds,.3 where at; ut, ut and a t; ut, ut are given by.4 and.5, respectively. For the coming calculation, we introduce the notion of tangential differential operators with respect to A which are similar to those introduced in [9, p.37]. Let be a bounded domain with a Lipschitz boundary Γ. Since by. we { n } n have a ij x, tν i ν j α, the vector ν A = a ij x, tν i is not tangential to Γ j= for almost all x Γ. hus, we can define a tangential vector field {τa kx}n k= such that{ν A x, τa n x,, τa x} forms a basis in lrn for almost all x Γ. For a smooth function u, there exist β j A, γk,j A j =,,, n; k =,,, n depending on {ν A x, τa n x,, τa x} such that n = β j A x j ν A k= γ k,j A τ k A i=, on Γ, j =,,, n..4 Set then, n σj A u = k= γ k,j A τ k A, j =,,, n,.5 = β j A σj A u..6 x j ν A Evidently, σj A j =,,, n are independent of the choice of the tangential vector field {τa kx}n k=. herefore we obtain a family of first order tangential differential operators σj A j =,,, n on Γ with respect to A. We can define the tangential gradient of u on Γ by σ Au = {σj A u} n j=..7 For any subset Σ of Σ, σj A H Σ L Σ. Set j =,,, n are linear and continuous from A Σ = n σj A σj A,.8 j= where σ A j denotes the adjoint of σ A j. hen the operator A Σ is linear and continuous from H Σ H Σ and satisfies A Σ u, v = σ Au σ AvdΣ, u, v H Σ..9 Σ

11 Lemma.. Let be a bounded domain in lr n with boundary Γ of class C. Let q = q k be a vector field in [C [, ] n. Suppose u is the weak solution of.. hen the following identity holds: = Σ q k ν k u a ij x, tσi A uσj A u dσ a ij x, t q k x j x i q k u a ij x, t u t, q k a ij x, t q k q k f u q k,. where u t, q k = u tq k dx. Remark.. If n =, then. becomes qν u dσ Σ = u t, q ax, t q x x x q u ax, t x x ax, t q q x x x f u q x.. Proof. We first prove. in the case of strong solutions, that is, we assume initial conditions u, u DA H andf L, ; H. Multiplying. by q k and integrating on, we have q k u q k a ij x, t = q k f.. Integrating by parts, we obtain q k u = u t, q k x k u q k q k u q k ν k u dσ, Σ.

12 and But, q k = = a ij x, t a ij x, t x j a ij x, t x j qk x i q k x i u q k. x i a ij x, t u q k x j x i = q k a ij x, t a ijx, t x k = q k ν k a ij x, t dσ a ij x, t q k Σ q k a ij x, t. By.3 and.4, and noting that = σi A u on Σ, we have x i q k a ij x, t x i x j = Σ q k ν k a ij x, tσi A uσj A udσ q k a ij x, t a ij x, t q k x j x i q k a ij x, t It follows from.,., and.5 that u q k ν k a ij x, tσi A uσj A u dσ Σ = u t, q k a ij x, t q k x j x i q k u a ij x, t a ij x, t q k his is.. q k f u q k. We now consider the general case of weak solutions with u, u H L and f L, ; L. We take u n, u n DA H andf n L, ; H such that

13 u n, u n u, u in H L, f n f in L, ; L. Now for strong solutions u n with initial conditions u n, u n, and right hand side f n, the identity. holds. Due to heorem., we have u n u in C[, ], H, u n u in C[, ], L. hus, as in the proof of Lemma.3 of chapter 3 of [9, p.39], taking the limit in. we deduce that. still holds in this case. 3. Observability Inequality. o establish an observability inequality, we need the following lemma. Lemma 3.. Let be a bounded domain in lr n with boundary Γ of class C. hen for all weak solutions u of. with f = the following hold: i If n >, then = Σ m k ν k u a ij x, tσi A uσj A u dσ u t, m k n a ij x, t m k ut Etdt u m k. 3. If n =, then mν u dσ = u t, m Etdt Σ x ax, t m x x u m x. 3. ii If n >, then u t, m k n ut R α n Et ut dx α 8R αn m k ν k ut dγ, t [, ]. 4R Γ 3.3 3

14 If n =, then for γ, u t, m x γ ut R αγ Et ut dx α 8R α γ mν ut dγ, t [, ]. 4R Γ 3.4 Proof. We prove the lemma only in the case of n >. It is similar in the case of n =. i aking q k = m k in., we have = Σ m k ν k u a ij x, tσi A uσj A u dσ u t, m k n = u t, m k a ij x, t x j x i u a ij x, t m k a ij x, t x i n x j u a ij x, t m k a ij x, t u m k u a ij x, t x i x j u m k. 3.5 But herefore, = Σ u a ij x, t = u, u m k ν k u a ij x, tσi A uσj A u dσ u t, m k n a ij x, t m k ut Etdt u m k his shows 3. 4

15 ii From the Cauchy-Schwarz inequality we have u t, m k n ut R u t α dx α R α As shown in [6] by Komornik, we have m k n ut dx = m k n dx 4 However, m k, ut = m k utdx = m k ut dx = n ut dx Combining 3.9 and 3., we obtain m k n ut dx n = m k dx ut dx n 4 R ut dx n ut dx 4 n m k ν k ut dγ. Γ hus, by. and 3.8 we have m k n ut dx. ut dx n m k, ut. Γ m k ν k ut dγ. Γ m k ν k ut dγ u t, m k n ut R α n Et ut dx α 8R αn m k ν k ut dγ, t [, ]. 4R Γ

16 Lemma 3.. Let be a bounded domain in R n with boundary Γ of class C. Suppose.3 and.4 hold. If either.7 holds and > or.3 and.4 hold and is large enough so that 3R a, R α b, R, < α R, 3.3 then for all weak solutions u of. with f = there exists c = c > such that m k ν k u a ij x, tσi A uσj A udσ m k ν k u u dγ Σ Γ 3.4 c u u, for n >, and mν u dσ mν u u dγ Σ Γ c u u, for n =. 3.5 Furthermore, if condition.9 or. is satisfied, then can be refined slightly to. or., respectively. Proof. i Suppose.7 holds and >. Case I: n >. It follows from.4 and.5 that where at is given by.5. Let atet E t atet, t, 3.6 ht = t then e h E = e h E h e h E hus, as ds, 3.7 e h ae ae h E =. 3.8 Et Ee ht Ee a,, t. 3.9 On the other hand, it follows from Gronwall s inequality and 3.6 that Et Ee a,, t. 3. Set Z = u t, m k n ut. 3. 6

17 It follows from 3.3 and 3. that Z Z Z R E e a, α n u u dx α 8R αn m k ν k u u dγ. 4R In addition, by.4 and 3., we have a ij x, t m k R Γ bta ij x, t x i x j R btetdt R E b, e a,, where bt is given by.6. Also, u m k u R t u R t u α u α R tetdt α E R, α e a,. It therefore follows from 3., 3.9, 3., 3.3, and 3.4 that m k ν k u a ij x, tσi A uσj A u dσ Σ Ee a, R E b, e a, R E e a, α α n u u dx 8R αn 4R hus, Σ Γ m k ν k u u dγ E R, α e a,. m k ν k u a ij x, tσi A uσj A u αn dσ m k ν k u u dγ 4R Γ e a R b, e a, R e a, α R, e a, αn E u dx. α 8R

18 his implies 3.4. Case II: n =. By 3., we have mν u dσ = u t, m Σ x Etdt ax, t m x x Choose γ, such that u m x. 3.7 γ e a, R b, e a, R α e a, R, α e a, >. 3.8 We write Etdt = γ Etdt γ u ax, t x γ ax, t. x hen, mν u dσ Σ = u t, m x γ ut γ ax, t m x x u t, m x γ ut γ ax, t m x x Etdt u m x Etdt γ from which, as in the case n >, we can deduce ax, t x u m x, 3.3 Furthermore, if.9 is satisfied, then E t. Consequently, Et E, for t. 3.3 hen 3.6 becomes m k ν k u a ij x, tσi A uσj A u dσ Σ αn m k ν k u u dγ 4R Γ e a, R b, R R, E α α αn u dx. 8R 8 3.3

19 So can be refined to.. If. is satisfied, then E t. Consequently, hen 3.6 becomes m k ν k u a ij x, tσi A uσj A u dσ Σ αn m k ν k u u dγ Et E, for t R Γ R b, e a, R e a, R, e a, α α αn 8R u dx. So can be refined to.. and E 3.34 ii Suppose.3 and.4 hold and is large enough so that 3.3 holds. By.3 we deduce E E a, Etdt, 3.35 Etdt E a, Etdt It therefore follows from 3., 3.9, 3., 3.3, and 3.4 that m k ν k u a ij x, tσi A uσj A u dσ Σ R b, R a, R, Etdt R E α α α α n u u dx 8R αn m k ν k u u dγ 4R Γ R b, R a, R, E R E α α a, α α n u u dx 8R αn m k ν k u u dγ. 4R Γ

20 hus, Σ m k ν k u a ij x, tσi A uσj A u αn m k ν k u u dγ 4R Γ α R 3R a, R α b, R, E α a, αn u dx. 8R dσ 3.38 aking into account 3.3, this implies 3.4. Remark 3.3. If x t is independent of t, then R t. If a ij are independent of x, then bt. If a ij are independent of t, then at. Let Γ be any subset of Γ and Σ = Γ,. hen Γ u u dγ As a matter of fact, by calculation we have and Σ u u dσ u dγ = u dtdγ tdu dγ Γ Γ Γ u u dσ, Σ u dγ = u dtdγ t du dγ Γ Γ Γ u u dσ. Σ herefore 3.39 follows from the above. By Lemma 3., we obtain the following observability inequality. Lemma 3.4. Observability inequality Suppose Σx Σx, and suppose the conditions of Lemma 3. are satisfied. hen there exists a constant c = c > such that for all strong solutions u of. with f = Σx u u dσ Σ x σ Au dσ c[ u u ]. 3.4

21 4. Proof of heorem.. We apply HUM. o do so, we consider the problem: u Au =, in, u = u, u = u, in, 4. =, on Σ. ν A For any u, u C DA C, problem 4. has a unique strong solution due to heorem.. Define u, u H = u u dσ σ Au dσ, 4. Σx Σ x which is a norm on C DA C due to Lemma 3.4. Let H be the completion of C DA C with respect to the norm H. hen Lemma 3.4 implies that H H L. 4.3 Consequently H L H. 4.4 According to the definition of H, we have for any u, u H, u Σx, u Σx L Σx, σ Au Σ x L Σ x n. 4.5 o apply the HUM, we need to consider the backward problem: v Av =, in, v = {, v =, in, v u = t u, on Σx, ν A Σ x u, on Σ x. 4.6 he solution of 4.6 can be defined by the transposition method see [9, ] as follows. Let, denote the duality pairing between H and H. Definition 4.. v is said to be an ultraweak solution of 4.6 if there exist ρ, ρ H such that v satisfies fv ρ, ρ, θ, θ = Σx θu θ u dσ Σ x σ Aθ σ AudΣ, 4.7 for any θ, θ H, f L, ; H, and where θ is the solution of the following problem θ Aθ = f, in, θ = θ, θ = θ, in, θ 4.8 =, on Σ. ν A

22 We define v = ρ, v = ρ. 4.9 Lemma 4.. Problem 4.6 has a unique ultraweak solution in the sense of Definition 4. satisfying v L, ; H, 4. Moreover, there exists c > such that v, v H. 4. v, v H c u, u H. 4. We assume Lemma 4. for the moment. We then define a linear operator Λ by aking f = in 4.7, we find Λu, u, u, u = Λu, u = v, v. 4.3 Σx u u dσ Σ x σ Au dσ. 4.4 It therefore follows from Lemma 3.4, Lemma 4., and the Lax-Milgram heorem that Λ is an isomorphism from H onto H. his means that for all y, y H, the equation Λu, u = y, y 4.5 has a unique solution u, u. With this initial condition we solve problem 4., and then solve problem 4.6. hen set φ = { u t u, on Σx, Σ x u, on Σ x, 4.6 and yx, t; φ = vx, t; φ. 4.7 hen we have constructed a control function φ such that the solution yx, t; φ of. satisfies.4. hus, we have proved heorem. provided we can prove Lemma 4.. Proof of Lemma 4.. he solution θ of problem 4.8 can be written as θ = η w, where η and w are solutions of the following problems: η Aη =, in, η = θ, η = θ, in, η =. on Σ, ν A 4.8

23 and w Aw = f, in, w = w =, in, w =. on Σ. ν A 4.9 Since {θ, θ } H, we have {θ, θ } H = Σx η η dσ Σ x σ η dσ. 4. On the other hand, by heorems.-. and the trace theorem see [, Chap. ], we have Σx w w dσ Σ x σ w dσ c f L, ;H. 4. herefore, fv ρ, ρ, θ, θ = θu θ u dσ σ Aθ σ AudΣ Σx Σ x ηu η u dσ σ Aη σ AudΣ Σx Σ x wu w u dσ σ Aw σ AudΣ Σx Σ x c {θ, θ } H f L, ;H {u, u } H. 4. hus, there exist v L, ; H and {ρ, ρ } H such that 4.7 holds, that is, v is an ultraweak solution of 4.6 and {v, v } H. aking f =, 4. gives 4.. Acknowledgments. he authers would like to thank the anonymous referee for helpful comments and bringing the papers [5, 6] to their attention. he first auther is grateful for the financial support from the Overseas Postgraduate Research Scholarship from Australian Government, the University of Wollongong Postgraduate Research Award, and the Department of Mathematics Analysis Research Group Scholarship from the Faculty of Informatics at the University of Wollongong. 3

24 REFERENCES [] R.F. Apolaya, Exact controllability for temporally wave equations, Portugal. Math., 5 994, [] C. Bardos, G. Lebeau, and J. Rauch, Sharp sufficient conditions for the observation, control, and stabilization of waves from the boundary, SIAM J. Control Optim. 3 99, [3] R. Dautray and J.L. Lions, Mathematical Analysis and Numerical Methods for Science and echnology, Vol.5, Evolution problems I, Springer-Verlag, Berlin, 99. [4] J.K. Hale, Ordinary Differential Equations, Wiley-interscience, New York, 969. [5] V. Komornik, Exact controllability in short time for the wave equation, Ann. Inst. H. Poincaré Anal. Non Linéaire 6 989, [6] V. Komornik, Contrôlabilité exacte en un temps minimal, C. R. Acad. Sci. Paris Sér. I Math , 3-5. [7] O.A. Ladyzhenskaya, he Boundary Value Problems of Mathematical Physics, Springer-Verlag, New York, 985. [8] I. Lasiecka, J.L. Lions, and R. riggiani, Nonhomogeneous boundary value problems for second order hyperbolic operators, J. Math. Pures Appl., , [9] J.L. Lions, Contrôlabilité Exacte Perturbations et Stabilisation de Systèmes Distribués, ome, Contrôlabilité Exacte, Masson, Paris Milan Barcelone Mexico, 988. [] J.L. Lions and E. Magenes, Non-homogeneous Boundary Value Problems and Applications, Vol. I and II, Springer-Verlag, Berlin, Heidelberg, New York, 97. [] M. M. Miranda, HUM and the wave equation with variable coefficients, Asymptotic Analysis, 995, [] J.E.M. Rivera, Exact controllability: coefficient depending on the time, SIAM J. Control Optim., 8 99, [3] D. L. Russell, Exact boundary value controlability theorems for wave and heat processes in star-complemented regions, in Differential Games and Control heory, Roxin, Liu, and Sternberg, Eds., Marcel Dekker Inc., New York, 974, Department of Mathematics University of Wollongong Northfields Avenue, Wollongong NSW 5, Australia w.liu@uow.edu.au; ghw@uow.edu.au 4