Boundary Control of the Beam Equation by Linear Quadratic Regulation
aa r X i v : . [ m a t h . O C ] F e b BOUNDARY CONTROL OF THE BEAM EQUATION BY LINEARQUADRATIC REGULATION
ARTHUR J. KRENERA
BSTRACT . We present and solve a Linear Quadratic Regulator (LQR) for the boundarycontrol of the beam equation. We use the simple technique of completing the square to getan explicit solution. By decoupling the spatial frequencies we are able to reduce an infinitedimensional LQR to an infinte family of two two dimensional LQRs each of which can besolved explicitly.
1. I
NTRODUCTION
We consider the stabilization of the linear beam equation using a Linear Quadratic Reg-ulator (LQR). By decoupling the spatial frquencies we obtain a complete and explicit solu-tion to the LQR including the closed loop eigenvalues. The simple technique of completingthe square yields Riccati PDEs. At each spatial frequency the Riccati PDEs reduce to thealgebraic Riccati equation of a two dimensional problem which is readily solvable. Thesums of the optimal cost and optimal feedback of these two dimensional problems yieldthe optimal cost and optimal feedback of the infinite dimensional LQR. The only techni-cal issues that arise are whether these sums are convergent. We discuss which Lagrangianyield convergence.The study of optimal control of systems governed by partial differential equations goesback at least to Lions [18]. More recent treatises on this topic are the works of Curtainand Zwart [6], [7], Lasiecka and Trigiani [17] and Krstic and Smyshlyaev [15] who usebackstepping. LQR boundary control has been used by Lasiecka and Trigiani [16], Burnsand King [3], Hulsing [11], Burns and Hulsing [2], Cristofaro, DeLuca and Lanari [5].Coron, D’Andrea and Bastin [4] found Lyapunov functions for the boundary control ofhyperbolic conservation laws. Guo et al. have considered boundary control of the beamequation in the presence of disturbaces, [9], [8]. Other papers on boundary control of thebeam equation are Morgul [20], Militec and Arnold [19], Han, Li, Xu and Liu [10].More recently we introduced the Completing the Square technique to solve LQR prob-lems for partial differential equations. We solved an LQR problem for the heat equationunder distributed control in [12] and under boundary control in [13]. In both cases usingan extension of Al’brekht’s method [1] we were able to find the higher degree terms in theTaylor polynomial expansions of the optimal cost and the optimal feedback.In [14] we solved an LQR for the boundary control of the wave equation by decouplingthe spatial frequencies. This allowed us to reduce an infinite dimensional LQR problemto an infinite family of two dimensional LQR problems each of which can be explicitlysolved. In this paper we show that an LQR problem for the beam equation can also be
Mathematics Subject Classification.
Key words and phrases.
Boundary Control, Beam Equation, Linear Quadratic Regulation, Completing theSquare.This work was supported by AFOSR under FA9550-20-1-0318 . reduced to an infinite family of two dimensional LQR problems each of which can beexplicitly solved. 2. B
OUNDARY C ONTROL OF THE B EAM E QUATION
In Exercise 3.18 Curtain and Zwart consider the undamped beam equation subject toboundary control action ∂ f∂t ( x, t ) = − ∂ f∂x ( x, t ) f (0 , t ) = 0 , f (1 , t ) = 0 ∂ f∂x (0 , t ) = 0 , ∂ f∂x (1 , t ) = u ( t ) f ( x,
0) = f ( x ) , ∂f∂t ( x,
0) = f ( x ) It is convenient to intoduce vector notation, let z ( x, t ) = (cid:20) z ( x, t ) z ( x, t ) (cid:21) = (cid:20) f ( x, t ) ∂f∂t ( x, t ) (cid:21) We also allow damping ddt z ( x, t ) = A z ( x, t ) where α ≥ and A is the matrix differential operator A = (cid:20) − ∂ ∂x − α (cid:21) , The boundary conditions are z (0 , t ) = 0 , z (1 , t ) = 0 ∂ z ∂x (0 , t ) = 0 , ∂ z ∂x (1 , t ) = u ( t ) and the initial conditions are z ( x,
0) = f ( x ) , z ( x,
0) = f ( x ) If α = 0 the beam is undamped and if α > the beam is damped.The eigenvalues of the open loop system are λ n = − α + sign ( n ) √ α − n π and the corresponding eigenvectors are v n = (cid:20) λ n (cid:21) sin | n | πx for n = ± , ± , ± , . . . . If | n | is small enough the eigenvalues λ n , λ − n can be realnumbers but for large | n | the eigenvalues λ n , λ − n are complex and conjugate.We wish to find a feedback to stabilize the beam, z ( x, t ) → as t → ∞ . Anotherpossibility is that we wish to stabilize the beam to some open loop trajectory z ∗ ( x, t ) . Wedefine ˜ z ( x, t ) = z ( x, t ) − z ∗ ( x, t ) and we seek a feedback to drive ˜ z ( x, t ) → as t → ∞ .Because of linearity these are equivalent problems so we only consider the first one. OUNDARY CONTROL OF THE BEAM EQUATION BY LINEAR QUADRATIC REGULATION 3
We shall use a Linear Quadratic Regulator (LQR) to find the desired feedback. Wechoose a × nonnegative definite matrix valued function Q ( x , x ) which is symmet-ric in x , x , Q ( x , x ) = Q ( x , x ) , and a positive scalar R . Consider the problem ofminimizing Z ∞ (cid:18)Z Z S z ′ ( x , t ) Q ( x , x ) z ( x , t ) dA + Ru ( t ) (cid:19) dt (1)subject to the beam dynamics where S = [0 , and dA = dx dx .Let P ( x , x ) be any × symmetric matrix valued function which is also symmetricin x , x , P ( x , x ) = P ( x , x ) . Suppose there is a control trajectory u ( t ) such thatthe corresponding state trajectory z ( x, t ) goes to as t → ∞ then by the FundamentalTheorem of Calculus Z Z S z ′ ( x , t ) P ( x , x ) z ( x , t ) dA + Z ∞ Z Z S ddt ( z ′ ( x , t ) P ( x , x ) z ( x , t )) dA dt We expand the integrand of the time integral.
Z Z S z ′ ( x , t ) P ( x , x ) z ( x , t ) dA + Z ∞ Z Z S dz ′ dt ( x , t ) P ( x , x ) z ( x , t ) dA dt + Z ∞ Z Z S z ′ ( x , t ) Q ( x , x ) dzdt ( x , t ) dA dt Z Z S z ′ ( x , t ) P ( x , x ) z ( x , t ) dA + Z ∞ Z Z S (cid:20) z ( x, t ) − ∂ z ∂x ( x , t ) − αz ( x , t ) (cid:21) ′ (cid:20) P , ( x , x ) P , ( x , x ) P , ( x , x ) P , ( x , x ) (cid:21) (cid:20) z ( x , t ) z ( x , t ) (cid:21) dA dt + Z ∞ Z Z S (cid:20) z ( x , t ) z ( x , t ) (cid:21) ′ (cid:20) P , ( x , x ) P , ( x , x ) P , ( x , x ) P , ( x , x ) (cid:21) " z ( x, t ) − ∂ z ∂x ( x, t ) − αz ( x , t ) dA dt Z Z S z ′ ( x , t ) P ( x , x ) z ( x , t ) dA (2) + 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 ) − ∂ z ∂x ( x , t ) P , ( x , x ) z ( x , t ) − z ( x , t ) P , ( x , x ) ∂ z ∂x ( x , t ) − ∂ z ∂x ( x , t ) P , ( x , x ) z ( x , t ) − z ( x , t ) P , ( x , x ) ∂ z ∂x ( x , t ) (3) − αz ( x , t ) P , ( x , x ) z ( x , t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) dA dt ARTHUR J. KRENER
We assume that P ( x , x ) satisfies these boundary conditions P (0 , x ) = P (1 , x ) = P ( x ,
0) = P ( x ,
1) = 0 (4) ∂ P∂x (0 , x ) = ∂ P∂x (1 , x ) = ∂ P∂x ( x ,
0) = ∂ P∂x ( x ,
1) = 0 (5)We integrate by parts four times to get these identities
Z Z S − ∂ z ∂x ( x , t ) P , ( x , x ) z ( x , t ) dA = Z Z S u ( t ) ∂P , ∂x (1 , x ) z ( x , t ) − z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) dA Z Z S − z ( x , t ) P , ( x , x ) ∂ z ∂x ( x , t ) dA = Z Z S − z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂P , ∂x ( x , u ( t ) dA Z Z S − ∂ z ∂x ( x , t ) P , ( x , x ) z ( x , t ) dA = Z Z S u ( t ) ∂P , ∂x (1 , x ) z ( x , t ) − z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) dA Z Z S − z ( x , t ) P , ( x , x ) ∂ z ∂x ( x , t ) dA = Z Z S − z ( x , t ) ∂ P , ∂x ( x , x ) z ( x , t ) + z ( x , t ) ∂P , ∂x ( x , u ( t ) dA We plug these identities into (2 ) and obtain
Z Z S z ′ ( x , t ) P ( x , x ) z ( x , t ) dA (6) + 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 )+ u ( t ) ∂P , ∂x (1 , x ) z ( x , 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 , u ( t )+ u ( t ) ∂P , ∂x (1 , 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 , u ( t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) dA dt We add the the right side of (6) to the criterion (1) to get an equivalent criterion
OUNDARY CONTROL OF THE BEAM EQUATION BY LINEAR QUADRATIC REGULATION 5 Z ∞ (cid:18)Z Z S z ′ ( x , t ) Q ( x , x ) z ( x , t ) dA + Ru ( t ) (cid:19) dt (7) Z Z S z ′ ( x , t ) P ( x , x ) z ( x , t ) 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 )+ u ( t ) ∂P , ∂x (1 , x ) z ( x , 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 , u ( t )+ u ( t ) ∂P , ∂x (1 , 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 , u ( t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) − αz ( x , t ) P , ( x , x ) z ( x , t ) dA dt We would like to find a × matrix valued function K ( x ) = (cid:2) K ( x ) K ( x ) (cid:3) (8)so that the time integrand in (7) 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 Clearly the terms quadratic in u ( t ) agree so we equate terms containing the product of u ( t ) and z ( x , t ) . This yields the equation − R (cid:2) K ( x ) K ( x ) (cid:3) = h ∂P , ∂x (1 , x ) ∂P , ∂x (1 , x ) i so we set (cid:2) K ( x ) K ( x ) (cid:3) = − R − h ∂P , ∂x (1 , x ) ∂P , ∂x (1 , x ) i (9)By symmetry (cid:2) K ( x ) K ( x ) (cid:3) = − R − h ∂P , ∂x ( x , ∂P , ∂x ( x , i ARTHUR J. KRENER
Then we equate terms containing the product of z ( x , t ) and z ( x , t ) and we obtain theRiccati PDEs for the boundary control of the beam equation, ∂ P , ∂x ( x , x ) + ∂ P , ∂x ( x , x ) + Q , ( x , x ) (10) = γ ∂P , ∂x ( x , ∂P , ∂x (1 , x ) P , ( x , x ) − ∂ P , ∂x ( x , x ) − αz ( x , t ) P , ( x , x ) z ( x , t ) + Q , ( x , x ) (11) = γ ∂P , ∂x ( x , ∂P , ∂x (1 , x ) P , ( x , x ) − ∂ P , ∂x ( x , x ) − αz ( x , t ) P , ( x , x ) z ( x , t ) + Q , ( x , x ) (12) = γ ∂P , ∂x ( x , ∂P , ∂x (1 , x ) P , ( x , x ) + P , ( x , x ) − αz ( x , t ) P , ( x , x ) z ( x , t ) + Q , ( x , x ) (13) = γ ∂P , ∂x ( x , ∂P , ∂x (1 , x ) where γ = R − β .To simplify the problem we decouple the spatial frequencies by assuming that Q ( x , x ) has the expansion Q ( x , x ) = ∞ X n =1 (cid:20) Q n,n , Q n,n , Q n,n , Q n,n , (cid:21) sin nπx sin nπx (14)and Q n,n , = Q n,n , .We assume that P ( x , x ) has a similar expansion 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 (15)with P m,n , = P m,n , . Clearly any such P ( x , x ) satisfies the boundary conditions (4) and(5).Then (9) implies K ( x ) = − R − β ∞ X n =1 (cid:2) P m,n , P m,n , (cid:3) nπ sin nπx (16)and the Riccati PDEs (10, 11, 12, 13) imply that − n π P n,n , + Q n,n , − n π γ (cid:0) P n,n , (cid:1) (17) P n,n , − n π P n,n , − αP n,n , + Q n,n , − n π γ P n,n , P n,n , (18) P n,n , − n π P n,n , − αP n,n , + Q n,n , − n π γ P n,n , P n,n , (19) P n,n , − αP n,n , + Q n,n , − n π γ (cid:0) P n,n , (cid:1) (20)where γ = R − β . OUNDARY CONTROL OF THE BEAM EQUATION BY LINEAR QUADRATIC REGULATION 7
For each n = 1 , , . . . these are the Riccati equations of the two dimensional LQR withmatrices(21) 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) We use the quadratic formula to solve (17) P n,n , = − n π ± q n π + γ Q n,n , n π γ (22)then (20) implies P n,n , = − α ± q α + n π γ (cid:0) Q n,n , + 2 P n,n , (cid:1) n π γ (23)Since we want P n,n , to be nonnegative we take the positive sign. Then (18) implies P n,n , = αP n,n , + n π P n,n , − Q n,n , + n π γ P n,n , P n,n , (24)If the two dimensional LQR (21) satisfies the standard conditions then the associated Ric-cati equation has a unique nonnegative definite solution. This implies that if we take thenegative sign in (22) 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 − n π 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 (25)for 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) (26)The corresponding eigenvectors of the infinite dimensional 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 (27)Notice that at least for large | n | , 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 ARTHUR J. KRENER If µ n and µ − n are complex and conjugatethen ζ n and ζ − n must be complex conjugates for z ( x, t ) to be real valued.Notice we can control each spatial frequency independently. 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 .But we must address the questions of whether (14) and (15) converge. If there is an N > and an r > such that (cid:13)(cid:13) Q n,ni,j (cid:13)(cid:13) ∞ ≤ qn r for i, j = 1 , and n > N then clearly (14) converges. We assume that we have chosen Q n,n such that this is true.We apply the Mean Value Theorem to (22) to obtain P n,n , = 12 s / Q n,n , n π for some s between n π and q n π + γ Q n,n , n π . Since s / is monotonically decreasingon this interval and takes on its maximum value at n π we conclude that P n,n , ≤ Q n,n , n π ≤ q n r π so clearly the sum P , ( x , x ) = P , ( x , x ) = ∞ X n =0 P n,n , sin nx sin nx converges.If α = 0 then (23) implies that P n,n , = 1 nπγ q Q n,n , + 2 P n,n , ≤ cn r/ for n > N and some constant c so clearly the sum P , ( x , x ) = ∞ X n =0 P n,n , sin nx sin nx (28)converges.If α > then again by the Mean Value Theorem (23) implies that there exists an s between α and q α + n π γ (cid:0) Q n,n , + 2 P n,n , (cid:1) such that P n,n , = 12 s / (cid:0) Q n,n , + 2 P n,n , (cid:1) ≤ α / cn r so again the sum (28) converges.But because of n π P n,n , term in (23) in order for the sum P , ( x , x ) = ∞ X n =0 P n,n , sin nx sin nx (29)to converge r must be larger than when α = 0 and r must be larger than when α > .If α > then all of the closed loop eigenvalues (26) are in the open left half of thecomplex plane. In particular for large | n | the real parts of the closed loop eigenvalues (26)are more negative than α .Can we shift all the eigenvalues into the open left half of the complex plane if α = 0 ? If Q n,n > but decays like n r as n → ∞ then the term outside the square root in (26) will OUNDARY CONTROL OF THE BEAM EQUATION BY LINEAR QUADRATIC REGULATION 9 be negative but it will decay in absolute value like n r/ − . For (29) to converge r must begreater than so the term outside the square root in (26) is converging to zero faster than r . But we are more interested in the convergence of feedback (16) than the convergenceof the optimal cost(15). For (16) to converge r need only greater than . If we choose < r < then the term outside the square root in (26) will grow like n − r so the higherthe mode the higher the damping. It is intersting to note that even if the optimal cost of anLQR problem does not exist, the LQR methodogy may yield a stabilizing feedbak.3. C ONCLUSION
We have used the simple and constructive technique of completing the square to solvethe LQR problem for the stabilization of the linear beam equation using boundary control.The result is an explicit formula for the quadratic optimal cost and the linear optimal feed-back. Our approach allows us to decouple the spatial frequencies so we can damp out allor just some frequencies. R
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