Numerics and Control of PDEs Lecture 7. IFCAM IISc Bangalore. Feedback stabilization of a 1D nonlinear model

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1 1/3 Numerics and Control of PDEs Lecture 7 IFCAM IISc Bangalore July August, 13 Feedback stabilization of a 1D nonlinear model Mythily R., Praveen C., Jean-Pierre R.

2 /3 Plan of lecture 7 1. The nonlinear model. Eigenvalues of the linearized operator and of its adjoint 3. Stabilizability and detectability of the linearized model. Local feedback stabilization of the continuous model 5. Approximation of the closed loop nonlinear system 6. Explicit determination of unstable steady states

3 /3 1. A Burgers equation with a Dirichlet boundary control We consider the equation w t ν w x + w w x = f s in, 1,, w, t = u s + ut and ν w x 1, t = g s w, = w in, 1. for t,, We assume that f s, g s and u s are stationary data, and that w s is the solution to the equation ν w s,xx + w s w s,x = f s in, 1, w s = u s and ν w s,x 1 = g s.

4 /3 We set z = w w s. The equation satisfied by z is z t ν z x + w s z x + z w s x + z z x = in, 1,, z, t = ut and ν z 1, t = for t, T, x z, = w w s = z in, 1. The linearized model is z t ν z x + w s z x + z w s x = in, 1,, z, t = ut and ν z 1, t = for t, T, x z, = z in, 1. The linearized operator in L, 1 is defined by DA = {z H, 1 z =, z x 1 = }, Az = ν z xx w s z x z w s,x.

5 5/3 As in the case of the heat equation, it can be shown that the linearized controlled system may be written in the form z = Az + Bu, z = z, with B = λ AD for some λ >, where the Dirichlet operator D is defined by Du = w and w is the solution to the stationary equation λ w ν w xx + w s w x + w w s,x = in, 1, w = u and ν w x 1 =. The coefficient λ > is chosen so that the bilinear form aw, φ = is coercive in H{} 1, 1. λ w φ + νw x φ x + w s w x φ + w w s,x φ dx

6 6/3 Measurements We choose either a distributed measurement 1 1 H d zt = zt,, zt Y o = R No, I 1 I 1 I No I No with I j, 1 for 1 j N o, or a boundary measurement H b zt = z1, t. The adjoint of A, DA is the operator A, DA defined by DA = {φ H, 1 φ =, φ x 1 + w s 1φ1 = }, A φ = ν φ xx + w s φ x.

7 /3. Eigenvalues of A and A To study the stabilizability of the pair A + ωi, B is stabilizable and the detectability of the pairs A + ωi, H d and A + ωi, H b, we have to characterize the spectrum of A and the eigenfunctions of A and A. Since A is not selfadjoint, it may admits complex eigenvalues. Let us first show that the eigenvalues of A are real. The eigenvalue problem for the linearized model is ν z xx + w s z x + z w s,x = λ z in, 1, z = and ν z x 1 =. Since the operator is not selfadjoint, we make the following change of unknowns zx = e βx v. We have z x = β x e βx v + e βx v x z xx = β xx e βx v + β x e βx v + β x e βx v x + e βx v xx.

8 /3 Thus, the equation for v is ν v xx + ν β x + w s v x + ν β xx ν β x + w s β x + w s,x v = λ v in, 1 v = and v x 1 + β x v1 =. We choose We have β x = w s ν and ws,x ν v xx + + w s β xx = w s,x ν. v = λ v in, 1, v = and v x 1 + w s1 v1 =. ν Now, we have a selfadjoint operator. Therefore we have shown that the eigenvalues of A are real. In addition, if λ > is chosen so that the bilinear form aw, φ = λ w φ + νw x φ x + w s w x φ + w w s,x φ dx is coercive in H 1 {}, 1, we can show that λ A is an isomorphism from DA into L, 1, and its inverse λ A 1 is a compact operator in L, 1. In that case, it can be shown that the eigenvalues of A are isolated and of finite multiplicity.

9 /3 3. Stabilizability of A, B and detectability of A, H Theorem. For any ω >, the pair A + ωi, B is stabilizable and the pair A + ωi, H b is detectable. The family I j 1 j No may be chosen so that the pair A + ωi, H d is detectable. To check the stabilizability of the pair A + ωi, B, we have to verify the Hautus criterion. The adjoint of B is defined on DA by If φ H{} 1, 1 satisfies B φ = νφ x, φ DA. ν φ xx w s φ x = λ φ in, 1, φ =, and ν φ x 1 + w s 1φ1 =, and φ x =, then necessarily φ =, because φ is the solution to a second order homogeneous differential equation with a zero initial condition. Thus the pair A + ωi, B is stabilizable.

10 /3 Let us check the detectability of A + ωi, H b. If ξ H{} 1, 1 satisfies ν ξ xx + w s ξ x + ξ w s,x = λ ξ in, 1, ξ =, ν ξ x 1 = and ξ1 =, then necessarily ξ =, because ξ x 1 = and ξ1 =. Thus the pair A + ωi, H b is detectable.

11 11/3. Local feedback stabilization of the continuous model We choose a feedback gain K LZ, U such that A + ωi, B is stable and a filtering gain L LY o, Z such that A + ωi, H is stable. Here H stands either for H d or H b. It η denotes a measurement error, µ is a model error and µ is an error in the initial condition, we have to solve the system z t ν z x + w s z x + z w s x + z z x = µ z, t = Kz e t and ν z 1, t =, x z, = z + µ in, 1, in, 1,, for t,, z e t ν z e x + w z e s x + z w s e x = LHz e Hz + η z e, t = Kz e t and ν z e 1, t = x z e, = z in, 1. for t,, in, 1,,

12 1/3 Theorem Local stabilization of the nonlinear system. Let us choose s, 1/. There exist C > and c 1 >, such that if < C C and z Hs,1+ µ Hs,1+ e ω µ L, ;H 1+s,1+ e ω η L, ;Y o c 1 C, then the nonlinear system coupling z and z e admits a unique solution in the space {z, z e H 1+s,1/+s/, 1, z, z e H 1+s,1/+s/,1, C}.

13 13/3 5. Approximation of the closed loop nonlinear system For the finite dimensional approximate system, we have to solve Ez e = Az e + BKz e + LHz e Hz η, z e = z, Ez = Az + BKz e + F z + µ, z = z + µ.

14 /3 6. Explicit determination of unstable steady states We are going to show that there exist steady states of the Burgers equation for which the linearized operators about these steady states are unstable in the case of mixed boubdary conditions. We shall also consider the case of Dirichlet conditions without being able to give a positive answer. Case of mixed boundary conditions. We denote by w s a solution to ν w s,xx + w s w s,x = f s in, 1, w s = u s and ν w s,x 1 = g s. The eigenvalue problem for the linearized model is ν z xx + w s z x + z w s,x = λ z in, 1, z = and ν z x 1 =. As above, we make the following change of unknowns zx = e βx v.

15 15/3 The equation for v is if we choose We set and ν v xx + ws,x + w s v = λ v in, 1, v = and ν v x 1 + w s1 β x = w s ν and β xx = w s,x ν. v1 =. DA = {v H, 1 v =, v x 1 + w s1 ν ws,x Av = ν v xx + + w s v. v1 = },

16 16/3 Integrating by part, we obtain ws ν v xx + We have = = ν v x + ν v x + v dx 16 π and the equality holds for + w s ws,x + w s ws,x + w s v v dx v dx ν v v x x=1 v dx + w s1 v 1. vx dx for all v H{} 1, 1, vx = sin Thus for such a function v, it yields ν vx + = ws,x + w s ws,x + w s + ν π 16 πx. v dx + w s1 v 1 v dx + w s1 v 1.

17 17/3 We look for w s such that w s 1 < and w s,x + w s + ν π 16 <. We look for the solution to the equation ν w + w + ν π / = ν π ε + ε /, where < ε < 1 has to be chosen. We have Thus ν w ν π 1 + ε / + w ν π 1 + ε / + 1 =. w π1+ε/ νπ1+ε/ w νπ1+ε/ + 1 = 1. Integrating from to x, we have π1 + ε arctan w νπ1 + ε/ π1 + ε arctan w = x, νπ1 + ε/ and νπ1 + ε π1 + εx wx = tan + C.

18 18/3 For x = 1 νπ1 + ε π1 + ε w1 = tan + C. Thus, we can choose < ε < 1 so that w1 takes any prescribed value in the interval,. In particular we would like to have with vx = sin πx. v x 1 = π cos π v x 1 + w s1 ν v1 =, = π π 8, v1 = sin =, and w s 1 = πν. We have to solve νπ1 + ε π1 + ε tan + C = πν, that is π1 + ε tan + C = π π1 + ε.

19 19/3 For C =, the equation π1 + ε tan = π π1 + ε admits the solution π1 + ε = π, ε =. For π/ < C <, we have a solution ε such that π1 + ε > π. Thus we can find ε > for which w s 1 = πν. For such a function w s, we have ν vx ws,x + + w s v = ν π ε + ε dx + w s1 v dx πν v 1. v 1

20 /3 This means that the linearized Burgers operator has negative eigenvalues. Indeed, v DA, and Av v dx = ν π ε + ε v dx πν v 1 <. Since A is self-adjoint, this inequality ensures the existence of negative eigenvalues. Case of Dirichlet boundary conditions. In the case of Dirichlet boundary conditions, using the same change of unknowns as above, the eigenvalue prtoblem for v is ws,x ν v xx + + w s v = λ v in, 1, v = and v1 =.

21 1/3 Integrating by part, we obtain We have = and the equality holds for ws ν v xx + ν v x + + w s ws,x + w s v v dx v dx. v dx 1 π vx dx for all v H 1, 1, vx = sin πx. We choose vx = sin πx, which is the function giving the best constant in the above inequality. We have ν vx ws,x + + w s v dx = ws,x + w s + νπ v dx.

22 /3 If w s,x + w s + νπ <, we shall have negative eigenvalues. We look for the solution to the equation ν w + w + π = π ε + ε, where < ε < 1 has to be chosen. We have Thus ν w π 1 + ε + w π 1 + ε + 1 =. 1 w π1+ε νπ1+ε w νπ1+ε Integrating from to x, we have 1 π1 + ε arctan w νπ1 + ε and + 1 = 1. 1 π1 + ε arctan w νπ1 + ε wx = νπ1 + ε tan π1 + εx + C. We do not obtain a continuous function over [, 1]. = x,

23 3/3 Explicit expression of w s We choose C = π/8. We have to solve the equation e find the approximate solution π tan 1 + ε π = ε. ε = The corresponding w s is defined by νπ1 + ε π1 + εx wx = tan π 8.