Maxwell s equations and a second order hyperbolic system: Simultaneous exact controllability

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1 Maxwell s equations and a second order hyperbolic system: Simultaneous exact controllability by B. Kapitonov 1 and G. Perla Menzala Abstract We present a result on simultaneous exact controllability for two models which describe two distinct hyperbolic dynamics. One is the system of Maxwell equations and the other a vector-wave equation with a pressure term. We obtain the main result using modified multipliers in order to generate a necessary observability estimate which allow us to use the Hilbert Uniqueness Method (HUM) introduced by J.L. Lions. Key words: distributed systems, boundary observability, simultaneous exact controllability. AMS Subject Classification: 35Q99, 74F99, 35B4. 1 Introduction We consider two models which describe two distinct hyperbolic dynamics: One of them is the system of Maxwell equations and the other is a vector wave equation with a pressure term. In solution (and vector-valued) variables {E, H, u} satisfy EE t curl H = µh t + curl E = div E =, div H = in (, T ) (1.1) E(x, ) = E (x), H(x, ) = H (x) in (1.) 1 Sobolev Institute of Mathematics, Siberian Branch of the Russian Academy of Science, Novosibirsk, Russia, borisvk@math.nsc.ru National Laboratory of Scientific Computation (LNCC/MCT), Av. Getulio Vargas 333, Quitandinha, Petrópolis, RJ, , Brasil and Institute of Mathematics, Federal University of Rio de Janeiro, RJ, P.O. Box 6853, Rio de Janeiro, Brasil, perla@lncc.br 1

2 η E = Q(x, t) on (, T ) (1.3) and ρu tt α u + grad p = div u = in (, T ) (1.4) u(x, ) = g 1 (x), u t (x, ) = g (x) in (1.5) u = P (x, t) on (, T ). (1.6) In (1.1), E = (E 1, E, E 3 ) and H = (H 1, H, H 3 ) denote the electric and magnetic field respectively and E and µ are positive constants representing the permittivity and magnetic permeability respectively. In (1.4), ρ denotes the scalar density which we will assume to be a positive constant and p = p(x, t) is the pressure term (an scalar function). Also, α is a positive constant depending on the elastic properties of the material. In (1.3), η = η(x) denotes the unit normal vector at x pointing the exterior of. Finally, grad, curl,, div and denote the usual gradient, rotational, Laplace operator, divergent and vector product in R 3. Generally speaking the problem of exact controllability (for either problem (1.1) (1.3) or (1.4) (1.6)) can be state as follows: Given a time T > and any initial data and desired terminal data, to find a corresponding control F driving the system to the terminal data at time T. One of the most usufull methods to solve such problems of controllability is the Hilbert Uniqueness Method (HUM) introduced by J.L. Lions in the middle 8 s and is based on the construction of appropiate Hilbert space structures on the space of initial data. These Hilbert structures are connected with uniqueness properties. Several authors considered the problem of exact controllability either for problem (1.1) (1.3) (see for instance D.L. Russell [16], J. Lagnese [9], M. Eller and J. Masters [], M. Eller [3], B. Kapitonov [5], N. Weck [17], K. Phong [14]) or problem (1.4) (1.6) (see for instance J.L. Lions [1] and A. Rocha dos Santos [15]). Clearly, the above results provide the existence of controls Q(x, t) and P (x, t) which are not necessarily related one to the other. In the middle 8 s, D. Russell [16] and J.L. Lions [1] raised the question if it is possible to solve the exact controllability problem for two evolution models using only one control. They named this problem simultaneous exact controllability. In the absence of dissipative effects, due to technical difficulties simultaneous exact controllability were only treated for one system with two different boundary conditions (see for instance, J.L. Lions [1], B.V. Kapitonov [6], B.V. Kapitonov, M. Raupp [7] and B.V. Kapitonov, G. Perla Menzala [8]).

3 Now, let us formulate the simultaneous exact controllability for systems (1.1) (1.3) and (1.4) (1.6): Given initial states (f 1, f, g 1, g ) and terminal data (ϕ 1, ϕ, ψ 1, ψ ) in suitable function spaces we ask if it is possible to find one vector-valued function P = P (x, t) such that the solution {E, H, u} de (1.1) (1.6) with Q in terms of P satisfies at terminal time T (E(T ), H(T ), u(t ), u t (T )) = (ϕ 1, ϕ, ψ 1, ψ ). The main purpose of this work is to prove that this is indeed the case, P serving as a control function for (1.4) (1.6) while the vector-valued function is a control function for (1.1) (1.3). Q = µη (η P t ) Let us briefly describe the sections in this paper: In Section we briefly indicate the function space where the solutions of systems (1.1) (1.3) and (1.4) (1.6) will be considered. Then, we use modified multipliers in order to obtain a boundary observability inequality valid for system (1.1) (1.6) with P and Q identically equal to zero (see inequalities (.17), (.18)). However, the right hand side of this inequality is not suitable to apply the steps of the Hilbert Uniqueness Method. That is why we assume a suitable (numerically) relation between important parameters of (1.1) (1.6). The observability inequality is obtain assuming that the region is substar-shaped. In Section 3, the simultaneous exact controllability is studied by means of the Hilbert Uniqueness Method (HUM) introduced by J.L. Lions (see [11], [1]). Systems (1.1) (1.3) and (1.4) (1.6) are not directly coupled to each other. A more interesting problem would be to study the case when they are coupled say with coupling terms γ curl E and γ curl u t (with γ > ) for system (1.1) (1.3) and (1.4) (1.6) respectively. As far as we know this remains an open problem. Our techniques seem to work when the coefficients E, µ, ρ and α are functions of x which are smooth and they and their partial derivatives of first order are bounded below by strictly positive constants. We will use standard notations which can be found in G. Duvaut and J.L. Lions book [1]. For example, H m () and H r ( ) will denote the Sobolev spaces of order m and r on and respectively. Given a real-valued function g the notation gdγ means the surface integral of g over the surface. If X is a vector space, then [X] m means X X X X m-times. For any vector v R 3, v denotes the usual norm of v in R 3. 3

4 A Boundary Observability Inequality We consider suitable function spaces where the solutions {E, H, u, u t } of problem (1.1) (1.6) (with P and Q identically equal to zero) will be considered. Let be a bounded region of R 3 with smooth boundary and E >, µ > as indicated in Section 1. We consider the Hilbert space H defined as follows with inner product given by H = [L ()] 3 [L ()] 3 v, w H = [Ev 1 w 1 + µv w ]dx for any v = (v 1, v ), w = (w 1, w ) in H. Here the dot means the usual inner product in R 3. We also consider the Hilbert space H(curl, ) = {w [L ()] 3 ; curl w [L ()] 3 } with inner product v 1, v H(curl,) = [v 1 v + curl v 1 curl v ]dx for any v 1, v H(curl, ). It is well known (see [1]) that the map Z η Z from [C()] 1 3 into [C 1 ( )] 3 extends by continuity to a continuous linear map from H(curl, ) into [H 1/ ( )] 3. This result allow us to consider the space H (curl, ) = {w H(curl, ), η w = on }. Here η = η(x) denotes the unit normal vector at x pointing the exterior of. We define the operator A: D(A) H H with domain D(A) given by D(A) = H (curl, ) H(curl, ) and A is defined as follows A(v 1, v ) = (E 1 curl v, µ 1 curl v 1 ) The skew-selfadjointness of A can be easily verified. By Stone s Theorem, the operator A generates a one-parameter group of unitary operators {U(t)} t R on H. Observe that in order U(t)f to solve problem (1.1) (1.3) with Q and given f D(A) remains to prove that the components of U(t)f are divergent free. In order to overcome this issue we consider M = {(grad ϕ 1, grad ϕ ) with ϕ 1, ϕ C ()} and M 1 = M. We observe that M is not closed 4

5 in H but M 1 is closed in H. In the distributional sense it is easy to prove that whenever w = (w 1, w ) M 1 then div w 1 = and div w =. Furthermore, we claim that U(t) takes M 1 D(A) into itself. Indeed, for any w in M and v in M 1 D(A) we have d dt U(t)v, w H = AU(t)v, w H = U(t)v, A w H = for any t R. Thus U(t)v, w H = constant for any t R. Therefore taking t = we have v, w =. Consequently U(t)v M 1 D(A) for any t R which proves our claim. Observe that for any element v = (v 1, v ) belonging to M 1 D(A) and w = (, grad ϕ ) with ϕ H () we have = v, w H = µv grad ϕ dx = µ ϕ v ηdγ. Since ϕ H () is arbitrary, the above identify say that η v = on. (.1) We conclude that problem (1.1) (1.3) with Q has a generator A 1 which applies M 1 D(A) into M 1 D(A) and for any w = (w 1, w ) M 1 D(A) the relation η w = on holds. Next, we consider problem (1.3) (1.6) with P. We can use Galerkin s method to find u and p (defined up to a constant). We choose the spaces V = {ϕ [C ()] 3, div ϕ = in }. Let Y be the closure of V with respect to the norm of [H()] 1 3 and W = Y [H ()] 3. Considering u W, u 1 Y we obtain by using the Galerkin method a unique solution of problem (1.3) (1.6) such that u C([, + ); V ) C 1 ([, + ); H) where H denotes the closure of V with respect to the norm of [L ()] 3. We can also obtain more regularity. For example, if u V [H 4 ()] 3 and u 1 W then the solution u C([, ); W ) C 1 ([, + ); V ) with p H (). Let us now concern ourselves with the simultaneous boundary observability problem. We use the multiplier method. They are conveniently modified in such a way that we can handle 5

6 the extra boundary terms appearing in the identities. Let h(x) be an auxiliary scalar smooth function on, which we will choose later on. For problem (1.1) we consider the multipliers M 1 = te + µ grad h H and M = th E grad h E Since {E, H} solves (1.1) (1.) then we have the identity = M 1 (EE t curl H) + M (µh t + curl E) + E(grad h E) div E + µ(grad h H) div H. (.) We can rearrange the terms on the right hand of (.) to obtain A t = div( B) + D (.3) where A = A(x, t) = t(e E + µ H ) + Eµ grad h (H E) (.4) and where D = B = B(x, t) = th E + grad h(e E + µ H ) i,j=1 EE(E grad h) µh(h grad h) (.5) h x i x j (EE i E j + µh i H j ) ( h 1)(E E + µ H ) (.6) E = Ej and H = j=1 Similarly, for problem (1.4) we consider the multipliers Hj. j=1 M 3 = tu t + (grad h grad)u + u and M 4 = tp + p(grad h grad) + p. t Here grad h grad = 3 h j=1 x j x j. Observe that M 4 is actually an operator. Since {u, p} is a solution of (1.4) then we have the identity = M 3 (ρu tt α u + grad p) + M 4 div u. (.7) 6

7 We can rearrange terms in identity (.7) to obtain where with and A 1 = t ( ρ u t + α A 1 t = div G + div F + D 1 (.8) ) + ρ[u t (grad h grad)u + u] x i i=1 G = (G 1, G, G 3 ) G i = [tu t + (grad h grad)u + u] α x ( i + h ρ u t α ) (.9) x i x j j=1 F = p[tu t + (grad h grad)u + u] D 1 = (3 h)ρ u t + ( h 1)α x j=1 j ( h α ) + p x i x q x i x q i,q=1 i,j=1 h x i x j j x i. Remark 1. If we choose h(x) = 1 x x for some x R 3 then we can verify that D and D 1 in (.3) and (.8) are identically zero. Therefore in the case (.3) and (.8) represent a conservation law. Let (E, H ) M 1 D(A) and (E, H) be the corresponding solution of problem (1.1) (1.3) with Q. Integration in (, T ) of identity (.3) give us T {E E + µ H }dx + Eµ grad h (H E)dx t=t t= T T = β(e, H, h)dγdt + D dxds (.1) where β(e, H, h) = tη (H E) + h (E E + µ H ) E(E η)(e grad h) µ(h η)(h grad h). (.11) and D is given as in (.6). Using the boundary conditions, (1.3) (with Q ) and the vector identities on we get η (H E) = (η E) H =, E = (E η) + E η = (E η) 7

8 E = η (η E) + η(e η) = η(e η), H = (H η) + H η = H η because η = 1 and (.1) holds for H. Thus we obtain from (.11) β(e, H, h) = h {µ H η E(E η) }. (.1) Now, we want to get appropiate estimates for the term T a convenient function h(x). We consider the elliptic problem D dxdt in (.1). Let us choose Φ = 1 Φ = in Vol() Area( ) on which admits a solution Φ C () C 1 (). Here Area( ) means the surface area of S =. Let < δ < 1 and x R 3 we define substitution of such h(x) into (.6) give us D = δ i,j=1 Let C 1 = C 1 (Φ) be the constant given by h(x) = δφ(x) + 1 x x. (.13) Φ x i x j (EE i E j + µh i H j ) δ(e E + µ H ). C 1 = We can easily verified that C 1 1/3 and max x i,j=1,,3 Φ(x) x i x j. D δc (E E + µ H ) (.14) where C = 6C 1 1 > and < δ < 1. Since the quantity 1 {E E + µ H }dx is constant for any t R for the solution (E, H) of (1.1) (1.3) (with Q ) then, from (.14) it follows that T D dxdt δc T Now, we want to estimate the term Eµ grad h (H E)dx t=t by C 3 = max{ grad Φ(x) + x x } x {E E + µ H }dx. (.15) t= in (.1). Let C 3 > given 8

9 Since HE E H and h as in (.13) we deduce Eµ grad h (H E)dx 4(1 + δ)c 3 Eµ 4(1 + δ)c 3 Eµ Eµ E H dx (E E + µ H )dx (.16) for any t T. Thus, from identity (.1) and inequalities (.1), (.15) and (.16) we get (1 δc )(T T ) {E E + µ H }dx T (.17) h {µ H η E(E η) }dγ where T = 4(1 + δ)c 3 Eµ /1 δc. Similarly, using identity (.8) and assuming some geometric condition on the region (say for instance, substar like [4], [8]) we can prove that the solution of problem (1.4) (1.6) (with P ) satisfies the inequality { (1 δ C )(T T ) ρ u t + α } dx x i=1 i T h α η dγdt for some C > and T >. (.18) Remark. By choosing δ > sufficiently small and adding inequalities (.17) and (.18) we would obtain a boundary observability provided h on. However, in order to apply the techniques to use the HUM would not help that much. We want to prove the following observability inequality Theorem 1. Let {E, H, u, u t } be the solution of problem (1.1) (1.6) with P = Q = on. Suppose there exist δ 1 > and x R 3 (δ 1 < min{c 1 1, C }) where C and C are as in (.17) and (.18) such that δ 1 Vol() Area( ) + (x x ) η > for all x (.19) and the parameters in (1.1) and (1.4) satisfy (numerically) the relation ρ = Eµα. Then, there exist constants C 5, C 6 and a T 1 > such that { ( δ 1 C 6 )(T T 1 ) E E 3 + µ H + ρ u t + α T { C 5 µh + α x i i=1 } h E(E η) } dγdt 9

10 Proof: In (.13) we choose δ = δ 1. Observe that (.19) tell us that h can easily verify the identity µh (ρu tt α u + grad p) + ρe 1 curl u (EE t curl H) > for all x. we + ρu t (µh t + curl E) + (µp αµ div u) div H + (ρe 1 αµ) curl u curl H (.) = t A div B. where and A = ρu t EH + ρ curl u E B = ρu t E + αµ(div u)h + αµh curl u µph. From (.) it follows that A t = div B. Integration over (, T ) give us {ρu t EH + ρ curl u E}dx t=t t= T = αµ (H η) curl u dγdt. We use the identity (.1) µ(h η) α curl u = µ H η αµ(h η) curl u + α curl u. Substitution into (.1) give us {ρu t EH + ρ curl u E}dx T { 1 = = T t=t t= µ(h η) α curl u 1 µ H η α curl u { 1 µ(h η) α curl u 1 µ H η α } dγdt η } dγdt (.) because u = on (, T ) (and u [H () H()] 1 3 then i x j η on (; T ). Observe that curl u ( ) i = i,j=1 x j 1 x j j=1 = η i j, therefore curl u =. (.3)

11 Using (.3) we can also obtain the inequality {ρu t EH + ρ curl u E}dx { } C 4 ρ u t + α x j + E E + µ H dx j=1 (.4) where C 4 = max{(ρ/µ) 1/ E, ρ/ E}. Consider { h C 5 = max α 1, L ( ) h L ( ) } µ 1. We multiply (.) by C 5 and obtain from (.17), (.18), (.) and (.4) { (1 δ 1 C 6 )(T T 1 ) E E + µ H + ρ u t + α } dx x i=1 i T { h µ H η E(E η) + α η } + C 5 µ(h η) α curl u C 5 µ H η α C 5 η dγdt T { C5 µ(h η) α curl u h } E(E η) dγdt (.5) where C 6 = C + C and T 1 = (T + T ) δ 1 (C T + C T ) δ 1 C 6 > We claim that the term µ(h η) α curl u on the right hand side of (.19) equals to α +µh for any (x, t) (, T ) q.t.p. In fact, using the boundary conditions we have ( ) µ(h η) α curl u = µ(h η) + α η. Using the identity v η + (v η) = v valid for any vector of v R 3 we obtain ( µh + α ) η + [( µh + α ) η] = α + µh because H η = and η = on (, T ). This proves our claim and the conclusion of Theorem 1. 11

12 Corollary 1. Let {E, H, u, u t } be the solution of problem (1.1)-(1.6) with zero boundary conditions and assume the conditions of Theorem 1. If the condition µh + α holds, then for any T > T 1 we will have = on (, T ) E(x, t) H(x, t) u(x, t) in (, T ). 3 Simultaneous exact controllability Let {E, H, u, u t } be the solution of problem (1.1) (1.6) with zero boundary conditions. In the function space of initial data (for strong solutions) we consider the Hilbert space F obtained by completing such space with respect to the norm ( T ) 1/ (f, g) F = µh + α dγdt for T > T 1 where f = (f 1, f ) and g = (g 1, g ) are the initial data of problems (1.1) (1.3) and (1.4) (1.6) respectively. From Corollary 1 it follows that F is indeed a norm. Let us denote by K the energy norm { (f, g) K = E E + µ H + ρ u t + α j=1 } dx x j then, clearly we have F K and (f, g) K C (f, g) F for some positive constant C. Let us consider the dual space of F with respect to K. We will denoted by F. Definition. Given R = R(x, t) [L ( (, T ))] 3 and (f 1, f, g 1, g ) F. We say that {E, H, u, u t } is a solution of the Maxwell/elasticity system if {E, H} solves (1.1) (1.) with boundary condition. η E = = µη (η R) on (, T ) (.6) and (u, u t ) solves (1.4) (1.5) with boundary conditions u t = R on (, T ). Furthermore a) (E(, t), H(, t), u(, t), u t (, t)) L (, T ; F ) 1

13 and b) (E(, t), H(, t), u(, t), u t (, t)), (Ẽ(, t), H(, t), ũ(, t), ũ t (, t)) K = (f 1, f, g 1, g ), ( f 1, f, g 1, g ) K t ( + R µ H + α ũ ) pη dγdτ (.7) holds for any ( f 1, f, g 1, g ) F and t (, T ) where (Ẽ, H, ũ, ũ t ) is a solution pf (1.1) (1.6) with zero boundary conditions. In (.6) (f 1, f, g 1, g ), ( f 1, f, g 1, g ) K { } = Ef 1 f 1 + µf f g 1 + α g 1 + ρg g x i x i Here p denotes the pressure term for the solution (ũ, ũ t ) of (1.4) (1.6) with zero boundary conditions. Similarly we define Definition. Given R = R(x, t) [L ( (, T )] 3 we say that {E, H, u, u t } is a solution of the Maxwell/elasticity system with zero initial data at time t = T if {E, H} solves (1.1) (1.) with boundary condition i=1 η E = µη (η R) on (, T ) and {u, u t } solves (1.4) (1.5) with boundary condition Furthermore dx. u t = R on (, T ) (.8) a) (E(, t), H(, t), u(, t), u t (, t)) L (, T ; F ) and b) (E(, t), H(, t), u(, t), u t (, t)), (Ẽ(, t), H(, t), ũ(, t), ũ t (, t)) K T ( = R α ũ ) + µ H pη dγdτ (.9) t 13

14 holds for any ( f 1, f, g 1, g ) F and t (, T ). Due to linearity and reversibility of system (1.1)-(1.6) it is clear that in order to solve the problem of exact controllability it is sufficient to prove that for any initial data (f 1, f, g 1, g ) F then the corresponding solution can be driven to the equilibrium state at time T. Theorem. Under the assumptions of Theorem 1. If T > T 1, then for any initial data (f 1, f, g 1, g ) F of problems (1.1) (1.) and (1.4) (1.5) there exist a control P H 1 (, T ; [L ( )] 3 ) such that u = P on (, T ) and the corresponding solution satisfies (u, u t ) = (, ) t=t while the vector-valued function Q = µη (η P t ) drives system (1.1) (1.) such that η E = Q on (, T ) and the corresponding solution satisfies (E, H) = (, ) t=t Proof: We use our previous discussion to apply the Hilbert Uniqueness Method (HUM). Let (h 1, h, q 1, q ) be an (arbitrary) element of F and (ϕ, ψ, v, v t ) the solution of (1.1) (1.6) with zero boundary conditions and initial data (ϕ, ψ, v, v t ) = (h 1, h, q 1, q ). (.3) t= Finally, let (E, H, u, u t ) be the solution of (1.1), (.6), (1.4), (.8) with zero initial data at t = T > T 1 where R is chosen to be R(x, t) = µψ + α v We consider the map M : F F given by M(h 1, h, q 1, q ) = (E, H, u, u t ). t= on (, T ). (.31) Our objective is to show that M is an isomorphism from F onto F. From to (.9) (with t = ) and (.31) it follows M(h 1, h, q 1, q ), ( f 1, f, q 1, q ) K = T ( = µψ + α v ) ( α ũ ) + µ H pη dγdτ (.3) 14

15 where (Ẽ, H, ũ, ũ t ) is a solution of (1.1) (1.6) with zero boundary conditions. Since we know that ψ η = and v η = on (, T ) because (.1) and v = on (, T ), then it ( ) follows that µψ + α v pη = on (, T ). Therefore from (.3) we deduce M(h 1, h, q 1, q ), ( f 1, f, q 1, q ) K = T ( = µψ + α v ) ( α ũ + µ H ) dγdτ (.33) = (h 1, h, q 1, q ), ( f 1, f, q 1, q ) F for any ( f 1, f, q 1, q ) F. Clearly (.33) implies that M is an isomorphism from F onto F. Now, we return to problems (1.1), (1.), (.6) and (1.4), (1.5), (.8). Let (f 1, f, g 1, g ) F. We set and (h 1, h, q 1, q ) = M 1 (f 1, f, g 1, g ) ( R = µψ + α v ) where (ϕ, ψ, v, v t ) is the solution of (1.1), (1.6) with zero boundary conditions and initial data at t = as in (.3). From (.9) with t = T > T 1 we deduce (E(T ), H(T ), u(t ), u t (T )), (Ẽ(T ), H(T ), ũ(t ), ũ t (T )) K = = M(h 1, h, q 1, q ), ( f 1, f, q 1, q ) K (h 1, h, q 1, q ), ( f 1, f, q 1, q ) F. (.34) Using (.33), we conclude that the right hand side of (3.34) is equal to zero. This means that (E(T ), H(T ), u(t ), u t (T )) generates the zero functional on F. Now, the conclusion of Theorem is a consequence of the above discussion: Construct R(x, t) as in (.31) and let P (x, t) = t R(x, s)ds + g 1 (x) Obviously P = u and η E = µη (η P t ) by construction. Acknowledgements: Both authors were partially supported by Research Grants of the Brazilian Government (MCT, Brasil). The first author (B.K.) by a Grant of CNPq Project and the second (G.P.M.) by a Grant of CNPq (Proc ). 15

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