Boundary Control of the Wave Equation via Linear Quadatic Regulation
aa r X i v : . [ m a t h . O C ] F e b Boundary Control of the Wave Equationvia Linear Quadatic RegulationArthur J. KrenerDepartment of Applied MathematicsNaval Postgraduate SchoolMonterey, CA, [email protected]
Abstract
We consider the Linear Quadratic Regulation for the boundary control of theone dimensional linear wave equation under both Dirichlet and Neumann activa-tion. For each activation we present a Riccati partial differential equation that weexplicitly solve. The derivation the Riccati partial differential equations is by thesimple and explicit technique of completing the square.
The control of infinite dimensional systems is well treated in numerous works, [15], [6],[14]. [11], [7]. In particular Lasiecka and Triggiani [13] prove the existence of solutionsto the Riccati equations that arise in the boundary control of hyperbolic problems inhigher dimensions. Burns and King [3] obtain integral representations for the feedbackoperators for hyperbolic problems with Kelvin-Voight damping and non-compact inputoperators.More recently [10] we were able to explicitly solve a boundary control problemfor a nonlinear reaction diffusion equation by completing the square and extendingAl’brekht’s method [1]. In this paper we explicitly find infinite horizon Linear QuadraticRegulator (LQR) for the one dimensional linear wave equation under either Dirichletand Neumann activation. We use the simple and explicit technique of completing thesquare to derive Riccati partial differential equations for the optimal cost and optimalcontrol. By decoupling the spatial frequencies we reduce Riccati partial differentialequations to an infinite family of two dimensional algebraic Riccati equations that canbe solved explicitly.
Consider the linear wave equation subject to Dirichlet boundary control at one end. Wecould consider Dirichlet boundary control at both ends but to keep the notation simple1e do not do so. The spatial variable is x ∈ [0 , and the model is ∂ w∂t ( x, t ) = ∂ w∂x ( x, t ) − α ∂w∂t ( x, t ) w (0 , t ) = βu ( t ) , w (1 , t ) = 0 w ( x,
0) = w ( x ) , ∂w∂t ( x,
0) = w ( x, where α is a nonegative constant. If α > the wave equation is damped and if α = 0 it is undamped. For Dirichlet activation is convenient to have the control act at x = 0 .Suppose w ∗ ( x, t ) is any solution of unforced wave equation subject to zero Dirich-let boundary conditions and w ( x, t ) is a solution of wave equation subject to Dirichletboundary control. Then we define z ( x, t ) = (cid:20) z ( x, t ) z ( x, t ) (cid:21) = (cid:20) w ( x, t ) − w ∗ ( x, t ) ∂w∂t ( x, t ) − ∂w ∗ ∂t ( x, t ) (cid:21) (1)and z ( x, t ) satisfies ∂z∂t ( x, t ) = A z ( x, t ) subject to boundary and initial conditions z (0 , t ) = βu ( t ) , z (0 , t ) = 0 z ( x,
0) = w ( x ) − w ∗ ( x, , z ( x,
0) = w ( x, − ∂w ∗ ∂t ( x, where A is the matrix differential operator A = (cid:20) ∂ ∂x − α (cid:21) (2)The open loop eigenvalues of A under Dirichlet boundary conditions are λ n = − α + sign ( n ) √ α − n π (3)for n = ± , ± , . . . . The coresponding eigenvectors are φ n ( x ) = (cid:20) λ n (cid:21) sin | n | πx Notice that if the equation is undamped, α = 0 , then all the eigenvalues are imagi-nary. If α > then some of the eigenvalues might be negative real numbers but as | n | gets larger they become complex numbers with negative real part.We wish to stabilize this system to z ( x, t ) = 0 so we set up a LQR problem. Wechoose a × dimensional matrix valued function Q ( x , x ) = (cid:20) Q , ( x , x ) Q , ( x , x ) Q , ( x , x ) Q , ( x , x ) (cid:21) ( x , x ) ∈ S = [0 , × [0 , andsymmetric with respect to ( x , x ) , Q ( x , x ) = Q ( x , x ) . We also choose a positivescalar R and consider the problem of minimizing by choice of control u ( t ) the criterion Z ∞ Z Z S z ′ ( x , t ) Q ( x , x ) z ( x , t ) dA + R ( u ( t )) dt (4)where dA = dx dx .Let P ( x , x ) be a × dimensional symmetric matrix valued function P ( x , x ) = (cid:20) P , ( x , x ) P , ( x , x ) P , ( x , x ) P , ( x , x ) (cid:21) which is also symmetric in ( x , x ) , P ( x , x ) = P ( x , x ) . Suppose there exists acontrol trajectory u ( t ) such that the resulting state trajectory z ( x, t ) → as t → ∞ .Then by the Fundamental Theorem of Calculus Z Z S z ′ ( x , P ( x , x ) z ( x , dA + Z ∞ Z Z S ddt ( z ′ ( x , t ) P ( x , x ) z ( x , t )) dA dt We expand the time integrand into components
Z Z S z ′ ( x , P ( x , x ) z ( x , dA + Z ∞ Z Z S z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x , t )+ z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x , t )+ (cid:18) ∂ z ∂x ( x , t ) − αz ( x , t ) (cid:19) P , ( x , x ) z ( x , t )+ z ( x , t ) P , ( x , x ) (cid:18) ∂ z ∂x ( x , t ) − αz ( x , t ) (cid:19) + (cid:18) ∂ z ∂x ( x, t ) − αz ( x , t ) (cid:19) P , ( x , x ) z ( x , t )+ z ( x , t ) P , ( x , x ) (cid:18) ∂ z ∂x ( x, t ) − αz ( x , t ) (cid:19) dA dt We assume that P ( x , x ) also satisfies Dirichlet boundary conditions P (0 , x ) = P (0 , x ) = 0 P ( x ,
0) = P ( x ,
0) = 0 and we integrate by parts twice with respect to the x i to get3 = Z Z S z ′ ( x , P ( x , x ) z ( x , dA + Z ∞ Z Z S z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x , t )+ z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x )+ z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t )+ z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) dA dt + Z ∞ Z βu ( t ) ∂P , ∂x (0 , x ) z ( x , t ) dx dt + Z ∞ Z z ( x , t ) ∂P , ∂x ( x , βu ( t ) dx dt + Z ∞ Z βu ( t ) ∂P , ∂x (0 , x ) z ( x , t ) dx dt + Z ∞ Z z ( x , t ) ∂P , ∂x ( x , βu ( t ) dx dt We add this to the criterion (4) to get the equivalent criterion to be minimized4 Z S z ′ ( x , P ( x , x ) z ( x , dA + Z ∞ Z Z S z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x , t )+ z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x )+ z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t )+ z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t )+ z ′ ( x , t ) Q ( x , x ) z ( x , t ) dA + R ( u ( t ) dt + Z ∞ Z βu ( t ) ∂P , ∂x (0 , x ) z ( x , t ) dx dt + Z ∞ Z z ( x , t ) ∂P , ∂x ( x , βu ( t ) dx dt + Z ∞ Z βu ( t ) ∂P , ∂x (0 , x ) z ( x , t ) dx dt + Z ∞ Z z ( x , t ) ∂P , ∂x ( x , βu ( t ) dx dt We would like to choose K ( x ) = (cid:2) K ( x ) K ( x ) (cid:3) so that the sum of the integrands of the time integrals is a perfect square of the form Z Z S ( u ( t ) − K ( x ) z ( x , t )) ′ R ( u ( t ) − K ( x ) z ( x , t )) dA Obviously the terms quadratic in u ( t ) match so we equate terms containing the productof u ( t ) and z ( x , t ) . This leads to the equation − RK ( x ) = β h ∂P , ∂x (0 , x ) ∂P , ∂x (0 , x ) i (5)so we assume K ( x ) = (cid:2) K ( x ) K ( x ) (cid:3) = − R − β h ∂P , ∂x (0 , x ) ∂P , ∂x (0 , x ) i (6)By symmetry K ( x ) = (cid:2) K ( x ) K ( x ) (cid:3) = − R − β h ∂P , ∂x ( x , ∂P , ∂x ( x , i z ( x, t ) and we obtain the equations ∂ P , ∂x ( x , x ) + ∂ P , ∂x ( x , x ) + Q , ( x , x )= γ ∂P , ∂x ( x , ∂P , ∂x (0 , x ) (7) P ( x , x ) − αP , ( x , x ) + ∂ P , ∂x ( x , x ) + Q , ( x , x )= γ ∂P , ∂x ( x , ∂P , ∂x (0 , x ) (8) P ( x , x ) − αP , ( x , x ) + ∂ P , ∂x ( x , x ) + Q ( x , x )= γ ∂P , ∂x ( x , ∂P , ∂x (0 , x ) (9) P , ( x , x ) + P , ( x , x ) − αP , ( x , x ) + Q ( x , x )= γ ∂P , ∂x ( x , ∂P , ∂x (0 , x ) (10)where γ = R − β .We call these equations (7,8,9,10) the Riccati PDE for the Dirichlet LQR controlof the wave equation.We assume that P ( x , x ) can be expressed in terms of the eigenfunctions of ∂ ∂x with respect to Dirichlet boundary conditions at x = 0 and x = 1 . P ( x , x ) = ∞ X m =1 ∞ X n =1 (cid:20) P m,n , P m,n , P m,n , P m,n , (cid:21) sin mπx sin nπx (11)and P m,n , = P m,n , . Then (6) implies K ( x ) = − R − β ∞ X m =1 ∞ X n =1 (cid:2) P m,n , P m,n , (cid:3) mπ sin nπx (12)We assume that Q ( x , x ) has a similar expansion Q ( x , x ) = ∞ X m =1 ∞ X n =1 Q m,n sin mπx sin nπx (13)where Q m,n = (cid:20) Q m,n , Q m,n , Q m,n , Q m,n , (cid:21) and Q m,n , = Q m,n , .This leads to an infinite algebraic Riccati equation for P m,n . To simplify the anal-ysis we decouple the spatial frequencies by assuming Q m,n = δ m,n (cid:20) Q n,n , Q n,n , Q n,n , Q n,n , (cid:21) (14)6hen we guess P m,n = δ m,n P n,n If this is true then the Riccati PDE (7,8,9,10) implies that − n π P n,n , + Q n,n , − n π γ (cid:0) P n,n , (cid:1) (15) P n,n , − αP n,n , − n π P n,n , + Q n,n , − n π γ P n,n , P n,n , (16) P n,n , − αP n,n , − n π P n,n , + Q n,n , − n π γ P n,n , P n,n , (17) P n,n , − αP n,n , + Q n,n , − n π γ (cid:0) P n,n , (cid:1) (18)where γ = R − β .For each n = 1 , , . . . these are the Riccati equations of the two dimensional LQRwith matrices F n,n = (cid:20) − n π − α (cid:21) , G n,n = (cid:20) nπβ (cid:21) Q n,n = (cid:20) Q n,n , Q n,n , Q n,n , Q n,n , (cid:21) , R n,n = (cid:2) R (cid:3) (19)We use the quadratic formula to solve (15) P n,n , = − ± q γ Q n,n , n π γ (20)then the quadratic formula applied to (18) implies P n,n , = − α ± q α + n π γ (cid:0) Q n,n , + 2 P n,n , (cid:1) n π γ (21)Since we want P n,n , to be nonnegative we need to take the positive sign. Then (16)implies P n,n , = αP n,n , + n π (cid:0) γ P n,n , (cid:1) P n,n , − Q n,n , (22)If the two dimensional LQR (19) satisfies the standard conditions then the associatedRiccati equation has a unique nonnegative definite solution. This implies that if wetake the negative sign in (20) the resulting P n,n is not nonnegative definite.The × closed loop system is F n,n + GK n,n = (cid:20) − n π − γ P n,n , − α − γ P n,n , (cid:21) and the closed loop eigenvalues are µ n = − α + n π γ P n,n , sign n ) q ( α + n π γ P n,n , ) − n π + γ P n,n , )2 (23)7or n = ± , ± , ± , . . . . For n = 1 , , . . . the corresponding eigenvectors of the n th × closed loop system are (cid:20) µ n (cid:21) , (cid:20) µ − n (cid:21) (24)The corresponding eigenvectors of the infinite closed loop system are v n ( x ) = (cid:20) µ n (cid:21) sin | n | πx, v − n ( x ) = (cid:20) µ − n (cid:21) sin | n | πx (25)Notice µ n and µ − n are complex conjugates as are v n ( x ) and v − n ( x ) .The trajectories of the infinite closed loop system are z ( x, t ) = ∞ X n = −∞ ζ n ( t ) (cid:20) µ n (cid:21) sin | n | πx where ζ n ( t ) = e µ n t ζ n Since we want a real valued z ( x, t ) , ζ n and ζ − n must be complex conjugates.The quadratic kernel of the optimal cost is given by (11) which reduces to P ( x , x ) = ∞ X n =1 (cid:20) P n,n , P n,n , P n,n , P n,n , (cid:21) sin nπx sin nπx (26)Notice we can control each spatial frequency indepedently. If we don’t want to dampout the n th spatial frequency then we set Q n,n = 0 so that P n,n = 0 and K n,n = 0 .A critical issue is whether the series (26) is convergent. Clearly some norm of Q n,n needs to go to zero faster than n for the series (13) to converge. But probably Q n,n needs to go to zero even faster for (26) to converge. Of course if the system is dampedwe can set Q n,n = 0 for n > N and let the damping stabilize the higher spatial modes.But if the system is undamped how many spatial modes can we stabilize and howfast can we dampen them? To answer these questions we look at a simple example, weassume β = 1 R = 1 Q n,n , = Q n,n , = qn r Q n,n , = Q n,n , = 0 For (13) to converge we must take r > . 8or this system γ = 1 P n,n , = − r qπ n r P n,n , = s q/n r + 2 P n,n , n π P n,n , = n π (cid:0) P n,n , (cid:1) P n,n , First note that by the Mean Value Theorem there exists an s between and qπ n r such that P n,n , = r qπ n r +2 − s / qπ n r The function s / is montonically decreasing between and qπ n r and takes onits maximum value at s = 1 so < P n,n , ≤ qπ n r +2 Then P n,n , = s q/n r + 2 P n,n , n π ≤ c n r/ P n,n , ≤ c n − r/ for some positive constants c , c . For P , ( x , x ) term in (26) to converge we musttake r > . Then P n,n , < c n s for some s > . If we choose r a little bigger than then the term outside the square root in the closed loop eigenvalues (23), − n π γ P n,n , is decaying so the higher spatial modes are substatially less damped. Moreover thefirst term inside the the square root in (23) is going to zero while the negative term isgrowing so at least the higher eigenvalues are complex.But for the optimal feedback (12) to converge we only need to take r > . If wetake r a little larger than then the term outside the square root in the closed loopeigenvalues (23), − n π γ P n,n , , is decaying like n − s for some s a little larger than .So again the higher the spatial mode the less damping.9 Neumann Boundary Control of the Linear Wave Equa-tion via LQR
In this section we assume Neumann boundary control instead of Dirichlet boundarycontrol. The model is now ∂ w∂t ( x, t ) = ∂ w∂x ( x, t ) − α ∂w∂t ( x, t ) ∂w∂x (0 , t ) = 0 , ∂w∂x (1 , t ) = βu ( t ) w ( x,
0) = w ( x ) , ∂w∂t ( x,
0) = w ( x, Now it is convenient to have the control act at x = 1 .Let z ( x, t ) and A be as before (1, 2). The new boundary conditions on z ( x, t ) are ∂z ∂x (0 , t ) = 0 , ∂z ∂x (1 , t ) = βu ( t ) The open loop eigenvalues of A under Neumann boundary conditions are λ n = − α + sign ( n ) √ α − n π (27)for n = 0 , ± , ± , . . . . The coresponding eigenvectors are φ ( x ) = (cid:20) (cid:21) , φ n ( x ) = (cid:20) λ n (cid:21) cos nπx for n = ± , ± , . . . .Again we wish to stabilize this system to z ( x, t ) = 0 which implies w ( x, t ) = w ∗ ( x, t ) so we consider the Linear Quadratic Regulator problem of minimizing (4)with Q ( x , x ) and R as before. Let P ( x , x ) be a × dimensional symmetricmatrix valued function P ( x , x ) = (cid:20) P , ( x , x ) P , ( x , x ) P , ( x , x ) P , ( x , x ) (cid:21) which is also symmetric in ( x , x ) , P ( x , x ) = P ( x , x ) . Suppose there exists acontrol trajectory u ( t ) such that the resulting state trajectory z ( x, t ) → as t → ∞ .Then by the Fundamental Theorem of Calculus Z Z S z ′ ( x , P ( x , x ) z ( x , dA + Z ∞ Z Z S ddt ( z ′ ( x , t ) P ( x , x ) z ( x , t )) dA dt Now we assume that P ( x , x ) satisfies Neumann boundary conditions in both itsarguments ∂P∂x (0 , x ) = ∂P∂x (1 , x ) = 0 ∂P∂x ( x ,
0) = ∂P∂x ( x ,
1) = 0
Z Z S z ′ ( x , P ( x , x ) z ( x , dA + Z ∞ Z Z S z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x , t )+ z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x )+ z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t )+ z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) dA dt + Z ∞ Z βu ( t ) P , (1 , x ) z ( x , t ) dx dt + Z ∞ Z z ( x , t ) P , ( x , βu ( t ) dx dt + Z ∞ Z βu ( t ) P , (1 , x ) z ( x , t ) dx dt + Z ∞ Z z ( x , t ) P , ( x , βu ( t ) dx dt
11e add this to the criterion (4) to get the equivalent criterion to be minimized Z ∞ Z Z S z ′ ( x , t ) Q ( x , x ) z ( x , t ) dA + R ( u ( t )) dt + ZZ S z ′ ( x , P ( x , x ) z ( x , dA + Z ∞ Z Z S z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x , t )+ z ( x , t ) P , ( x , x ) z ( x , t ) + z ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x ) − αz ( x , t ) P , ( x , x ) z ( x )+ z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t )+ z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) dA dt + Z ∞ Z βu ( t ) P , (1 , x ) z ( x , t ) dx dt + Z ∞ Z z ( x , t ) P , ( x , βu ( t ) dx dt + Z ∞ Z βu ( t ) P , (1 , x ) z ( x , t ) dx dt + Z ∞ Z z ( x , t ) P , ( x , βu ( t ) dx dt We complete the square as before to obtain K ( x ) = (cid:2) K ( x ) K ( x ) (cid:3) = − R − β (cid:2) P , (1 , x ) P , (1 , x ) (cid:3) (28)and ∂ P , ∂x ( x , x ) + ∂ P , ∂x ( x , x ) + Q , ( x , x )= γ P , ( x , P , (1 , x ) (29) P ( x , x ) − αP , ( x , x ) + ∂ P , ∂x ( x , x ) + Q , ( x , x )= γ P , ( x , P , (1 , x ) (30) P ( x , x ) − αP , ( x , x ) + ∂ P , ∂x ( x , x ) + Q ( x , x )= γ P , ( x , P , (1 , x ) (31) P , ( x , x ) + P , ( x , x ) − αP , ( x , x ) + Q ( x , x )= γ P , ( x , P , (1 , x ) (32)12here γ = R − β . We call these equations (29,30,31,32) the Riccati PDE for theLQR Neumann boundary control of the wave equation.Since P ( x , x ) satisfies Neumann boundary conditions at x = 0 and x = 1 it canbe expressed in terms of the eigenfunctions of ∂ ∂x with respect to Neumann boundaryconditions. As before it is convenient to decouple the spatial frequencies. P ( x , x ) = ∞ X n =0 (cid:20) P n,n , P n,n , P n,n , P n,n , (cid:21) cos nπx cos nπx (33)and P n,n , = P n,n , . Then (28) implies K ( x ) = − R − β ∞ X n =0 (cid:2) P n,n , P n,n , (cid:3) cos nπx (34)We assume that Q ( x , x ) has a similar expansion Q ( x , x ) = ∞ X n =0 Q n,n cos nπx cos nπx (35)This leads to an infinite algebraic Riccati equation for P n,n .The Riccati PDE (29,30,31,32) implies that − n π P n,n , + Q n,n , − γ (cid:0) P n,n , (cid:1) (36) P n,n , − αP n,n , − n π P n,n , + Q n,n , − γ P n,n , P n,n , (37) P n,n , − αP n,n , − n π P n,n , + Q n,n , − γ P n,n , P n,n , (38) P n,n , − αP n,n , + Q n,n , − γ (cid:0) P n,n , (cid:1) (39)where γ = R − β .For each n these are the Riccati equations of the two dimensional LQR with matri-ces F n,n = (cid:20) − n π − α (cid:21) , G n,n = (cid:20) β (cid:21) Q n,n = (cid:20) Q n,n , Q n,n , Q n,n , Q n,n , (cid:21) , R n,n = (cid:2) R (cid:3) (40)Notice G n,n is different from before.We solve these equations and obtain P n,n , = αP n,n , + (cid:0) n π + γ P n,n , (cid:1) P n,n , − Q n,n , (41) P n,n , = − n π ± q n π + Q n,n , γ (42) P n,n , = − α ± q α + γ (cid:0) Q n,n , + 2 P n,n , (cid:1) γ (43)13ince we want P n,n , to be nonegative we need to take the positive sign in (43). If thetwo dimensional LQR (40) satisfies the standard conditions then the associated Riccatiequation has a unique nonnegative definite solution. This implies that if we take thenegative sign in (42) the resulting P n,n is not nonnegative definite.The × closed loop system is F n,n + G n,n K n,n = (cid:20) − n π − γ P n,n , − α − γ P n,n , (cid:21) and the closed loop eigenvalues are µ n = − α + γ P n,n , sign n ) q ( α + γ P n,n , ) − n π + γ P n,n , )2 (44)for n = 0 , ± , ± , . . . . For n = 0 , , , . . . the corresponding eigenvectors of the n th × closed loop system are (cid:20) µ n (cid:21) , (cid:20) µ − n (cid:21) (45)The corresponding eigenvectors of the infinite closed loop system are v n ( x ) = (cid:20) µ n (cid:21) cos nπx, (cid:20) µ − n (cid:21) cos nπx (46)The trajectories of the infinite closed loop system are z ( x, t ) = ∞ X n = −∞ ζ n ( t ) (cid:20) µ n (cid:21) cos nπx (47)where ζ n ( t ) = e µ n t ζ The quadratic kernel of the optimal cost is given by (11) which reduces to P ( x , x ) = ∞ X n =1 (cid:20) P n,n , P n,n , P n,n , P n,n , (cid:21) cos nπx cos nπx (48)Again we can control each spatial frequency indepedently.But again we face the question is (48) convergent. And if the system is undamped, α = 0 , how many spatial modes can we stabilize and how fast? To partially answerthese questions we look at our simple example again, we assume β = 1 R = 1 Q , , = Q , , = qQ n,n , = Q n,n , = qn r for n > Q n,n , = Q n,n , = 0 for n ≥ r > .For this system γ = 1 P n,n , = − n π + r n π + qn r P n,n , = q q/n r + 2 P n,n , P n,n , = (cid:0) n π + P n,n , (cid:1) P n,n , Again by the Mean Value Theorem there exists an s between n π and n π + qn r such that P n,n , = r n π + qn r − n π = q s / n r The function s / n r is montonically decreasing between n π and n π + qn r andtakes on its maximum value at s = n π so < P n,n , ≤ q πn r +1 Then P n,n , = q q/n r + 2 P n,n , = O ( n − r/ ) P n,n , = (cid:0) n π + P n,n , (cid:1) P n,n , = O ( n − r/ ) This time for P , ( x , x ) in the optimal cost (33) to converge we must take r > so − r/ < − . Then P n,n , = O ( n − r/ ) so the higher spatial modes are less damped.Moreover the first term inside the the square root in (44) is going to zero while thenegative term is growing so at least the higher eigenvalues are complex.But for the optimal feedback (34) to converge we only need to take r > . Ifwe take r a little larger than then the term outside the square root in the closedloop eigenvalues (44) is decaying so the higher spatial modes are less damped. Hencewe conclude for this simple example Dirichlet boundary control of the wave equationis preferable to Neumann boundary control. We strongly suspect that this is true ingeneral. We have used the simple and constructive technique of completing the square to solvethe LQR problems for the linear wave equation under both Dirichlet and Neumannboundary control. The results are explicit formulas for the quadratic optimal cost andthe linear optimal feedback. These formulas decouple the spatial frequencies so we candamp out all or just some frequencies. We can also use the linear optimal feedback tostabilize the wave to any desired trajectory of the open loop system. We would like tothank Miroslav Krstic for his helpful comments.15 eferences [1] E. G. Al’brekht, On the Optimal Stabilization of Nonlinear Systems, PMM-J.Appl. Math. Mech., 25:1254-1266, 1961.[2] J. Burns and K. Hulsing, Numerical methods for approximating functionalgains in LQR boundary control problems, Mathematical and Computer Mod-elling 33:89-100, 2001.[3] J. Burns and B. King, Representation of feedback operators for hyperbolicsystems, Computation and Control IV, pp. 57-73, Springer 1995.[4] J. Burns, D. Rubio and B. King, Regularity of feedback operators for bound-ary control of thermal processes, in the Proceedings of the First internationalconference on nonlinear problems in aviation and aerospace, 1994.[5] J. Crank and P. 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