The Navier-Stokes problem in velocity-pressure formulation :convergence and Optimal Control

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1 The Navier-Stokes problem in velocity-pressure formulation :convergence and Optimal Control A.Younes 1 A. Jarray 2 1 Faculté des Sciences de Tunis, Tunisie. younesanis@yahoo.fr 2 Faculté des Sciences de Tunis, Tunisie. abdennaceur.jarray@gmail.com Abstract : In this paper,we study the nonlinear Navier-Stokes problem in velocity-pressure formulation. We construct a sequence of a Newton-linearized problems and we show that the sequence of weak solutions converges towards the solution of the nonlinear one in a quadratic way.a control problem on the homogeneous problem is considered. AMS Subject Classification :35J20,49J96,. Key Words : Navier-Stokes equations,newton s algorithm, Approximation Convergence, Control. 1 Introduction Our work is based an the first part on the approximation of the solution a second order elliptic problem with Dirichlet boundary conditions [4][2]. Indeed, the flow of an incompressible viscous fluid in a domain of R 2, is characterized by two variables velocity u and pressure p, and a given function f = (f1, f2) in (L 2 ()) 2 and a constant ν>0, called kinematic viscosity. The problem is to find functions u = (u1, u2) and p defined in satisfying : The equation of momentum : ν u + (u. )u + p = f in (1) The equation of incompressibility (which reflects the conservation of mass) : div u = 0 dans (2) 1

2 We usually added to this system a Dirichlet boundary condition : u = g on. What gives the homogeneous problem for g = 0 : ν u + (u. )u + p = f in (Q) : div u = 0 in u = 0 on Γ = (Q) : describes the motion of an incompressible fluid contained in and subjected to an outside forces f, u is the velocity of fluid flow, p is the pressure and ν its viscosity. 2.1 Linearization 2 The Linearized Problems Let a bounded domain of R 2 with Lipschitz-continuous boundary Γ, and let V = {v X, div v = 0}, where X = (H0()) 1 2. We set M = L 2 0(), Y = X M. The nonlinear term (u. ) u can be written : (u. )u = 1 2 u 2 + rotu u. In order to solve (Q) we construct a sequence of Newton-linearized problems. Starting from an arbitrary u 0 X and p 0 M we consider the iterative scheme [1] : (Q n+1 ) : ν u n+1 + (u n+1. )u n + (u n. )u n+1 + p n+1 = f n in div u n+1 = 0 in u n+1 = 0 on Γ = where f n = f + (u n. )u n. Problem (Q n+1 ) is linear. 2.2 The Variational Formulation The variational formulation[6] of (Q n+1 ) is : Find (u n+1, p n+1 ) Y such that : (QV n+1 ) : a 0 (u n+1, v) + a n (u n+1, v) + a n (u n+1, v) + b(p n+1, v) = L n (v) v X b(q, u n+1 ) = 0 q M where the bilinear forms a 0, a n, a n are given for v, u X and p M by : a 0 (u, v) = ν u v dx, a n (u, v) = (u. u n ) v dx, a n (u, v) = (u n. u) v dx, b(p, v) = p v dx = p div v dx and L n (v) = f n, v where f (H 1 ()) 2 Using Green formula and div v = 0 we have : b(p, v) = 0. 2

3 We associate then at problem (QV n+1 ) the following problem : (P V n+1 ) : { Find un+1 V such that : a 0 (u n+1, v) + a n (u n+1, v) + a n (u n+1, v) = L n (v) v V Let us now show that, for each n problem (P V n+1 ) has one solution.we need for this : Lemme 1 : For fixed u n V the form (u, v) a n (u, v) and (u, v) a n (u, v) are continuous on X. According to the Sobolev Imbedding Theorem, the space H 1 () is continuously imbedded in L 4 () then by Holder s inequality we have : C > 0 such that a n (u, v) C u X v X u n X. (3) a n (u, v) C u X v X u n X. (4) To show the coercivity of the form a = a 0 + a n + a n we have for all u V : Lemme 2 : We have a n (u, u) = 0 for all u V. a n (u, v) = where a(u, u) = ν u a n (u, u) + a n (u, u). (u. )u n u dx = t J(u).u = 1 2 <J(u).u n /u>dx = ( (u 2 1 +u 2 2 ) x (u 2 1 +u2 2 ) y ) = 1 2 ( u 2 ) <u n / t J(u)u>dx (5) Using Green formula and div u n = 0 we have : 2a n (u, u) = u 2 u n dx = div u n u 2 dx = 0. (6) Let for α > 0, B α = {v V/ v X α}. Lemme 3 : We have : a(u, u) (νc 1 αc) u 2 X u B α (7) 3

4 By (6) we get a(u, u) = a 0 (u, u) + a n (u, u).by (4) we have : then (8) gives : which with (9) gives (7). a n (u, u) Cα u 2 X u, u n B α (8) a 0 (u, u) = ν u 2 (L 2 ) 2 νc 1 u 2 X (9) a n (u, u) Cα u 2 X u, u n B α (10) Lemme 4 : For ν large enough or f (L 2 ()) 2 small enough there is a α ]α 0, α 1 [,independente of n, such that : u n X α n N We must show that n N, u n X α and (νc 1 Cα ) > 0. We can choose α so that α < νc 1. C Remains to be demonstrated by induction that u n X α, n N. Indeed let u 0 B α and assume that u n B α. We note u = u n+1 and f 2 = f (L 2 ()) 2, we have for u n B α : a(u, u) = L n (u) = (f + (u n )u n )u dx (11) then a(u, u) ( f 2 + Cα 2 ) u X with (7) we obtain : (νc 1 Cα ) u X ( f 2 + Cα 2 ) which gives : u X f 2 + Cα 2 (νc 1 Cα ) So to deduce the result we must have : then we put : f 2 + Cα 2 (νc 1 Cα ) α P (α ) = 2Cα 2 νc 1 α + f 2 0 and α < νc 1 C ( ) Therefore it is necessary that the discriminant of the polynomial P (α ) verify : = ν 2 C f 2 C > 0 ie f 2 < ν2 2 C 1 and hence it has two roots 8C x 1 = νc 1 ν 2 C f 2 C 4C 4

5 x 2 = νc 1 + ν 2 C f 2 C 4C Since x 1 > 0 so just take Thus we have, the Theorem : α 0 = x 1 α α 1 = min( νc 1 C, x 2) Theorem 1 : For f (L 2 ()) 2 problem (P V n+1 ) has a unique solution u n+1 V B α. Since u n B α we have L n (v) ( f 2 + Cα 2 ) v X, which gives the continuity of L n and using Lemma 1, Lemma 2 and Lemma 3 with Lax- Milgram Theorem we obtain the result. 2.3 The Convergence The sequence (u n ) n N obtained in the preceding subsection verify : u n X α n 0 (12) which implies that the sequence (u n ) n N is bounded in X = (H 1 0()) 2. Then there exist a sub-sequence that converges weakly to φ in X. Since the injection of X in (L 2 ()) 2 is compact, there exists a subsequence still noted u n which converge strongly to φ in (L 2 ()) 2. On the other hand Lemme 5 : We have : lim u n φ 2 = 0. We put ω = u n+1 φ ν ω 2 2 = a 0 (u n+1 φ, u n+1 φ) (13) With and and a 0 (ω, ω) = a 0 (u n+1 φ, u n+1 ) a 0 (u n+1 φ, φ) (14) a 0 (ω, u n+1 ) + a n (ω, u n+1 ) + a n (ω, u n+1 ) = L n (u n+1 ) (15) a 0 (ω, φ) + a n (ω, φ) + a n (ω, φ) = L n (φ) (16) 5

6 As a result we have by subtracting (15) and (16) and using Lemma 1 : which gives using (4) and (6) : and then ν ω 2 2 = a 0 (ω, ω) = L n (ω) a n (ω, ω) (17) ν ω 2 2 ( f 2 + Cα 2 ) ω 2 + Cα ω 2 ω 2 (18) ω ν {( f 2 + Cα 2 ) ω 2 + C2 α 2 Since u n converge strongly in (L 2 ()) 2 we obtain : We need this result : 2ν ω 2 2} (19) lim u n+1 φ 2 = 0. (20) Lemme 6 : 1) We have : lim a 0 (u n+1, v) = a 0 (φ, v). 2) We have : lim a n (u n+1, v) = a (φ, v) = (φ. )φvdx. 3) We have : lim a n (u n+1, v) = a (φ, v) = (φ. )φvdx. 4) We have lim L n (v) = L (v) = [f + (φ. )φ]vdx. 1) a 0 (u n+1, v) a 0 (φ, v) ν Lemma 5 gives the result. 2) On the other hand we have : a n (u n+1, v) a (φ, v) = We can write u n+1 φ v dx ν u n+1 φ 2 v 2 {(u n+1. )u n (φ. )φ}v dx (21) (u n+1. )u n (φ. )φ = ((u n+1 φ). )u n + (φ. )(u n φ) (22) which gives using Lemma 1 : a n (u n+1, v) a (φ, v) C( u n+1 φ X u n X + φ X u n φ X ) v X (23) 3) Since a n (u n+1, v) a (φ, v) = {(u n. )u n+1 (φ. )φ}v dx we get : a n (u n+1, v) a (φ, v) C( u n φ X u n+1 X + φ X u n+1 φ X ) v X (24) 6

7 4)We have : L n (v) L (v) (u n u n (φ φ) v dx which gives : L n (v) L (v) C( u n φ 2 u n 2 + u n φ 2 φ 2 ) v 2. (25) Lemma5 gives the result. We need Rham s Theorem : Let a bounded regular domain of R 2 and L a continuous linear form on (H0()) 1 2. Then the linear form L vanishes on V if and only if there exists a unique function ϕ L 2 ()/R such that : v (H0()) 1 2, L(v) = ϕ div v dx Theorem 2 : We have lim u n = φ in V then φ is a solution of (Q). It follows from Lemma 6 that lim a 0(u n+1, v) + a n (u n+1, v) + a n (u n+1, v) = a 0 (φ, v) + 2a (φ, v) = L (v) Let now L(v) = a 0 (φ, v) + a (φ, v) + a (φ, v) L (v), therefore the form L(v) = 0 for all v V, then Rham s theorem implies there exist a unique function p L 2 ()/R such that : a 0 (φ, v) + 2a (φ, v) L (v) = p div v dx v X (26) which gives : ν φ v dx + and then in D () : (φ. )φv dx pdiv v dx = f v dx v X (27) ( ν φ + (φ. )φ + p f)v dx = 0 v X (28) ν φ + (φ. )φ + p f = 0 (29) Since φ V we can conclude that φ is solution of (Q). Remark 1 : We can clearly see that φ satisfy all the condition of u n and consequently it verify : φ X α (30) 7

8 Theorem 3 : Let u n+1 the solution of (QV n+1 ) and φ solution of (Q). Then convergence of the sequence (u n+1 ) n N towards φ is quadratic ie : u n+1 φ X C u n φ 2 X (31) Let ω n = u n φ and χ n = p n p. Subtracting problem (Q n+1 ) and (Q) we get : (D n+1 ) ν ω n+1 + (ω n+1 )u n + (u n )ω n+1 + χ n+1 = (ω n )ω n in div ω n+1 = 0 in ω n+1 = 0 on Γ Since div ω n+1 = 0 the variational formulation of (D n+1 ) is : Find (ω n+1, χ n+1 ) Y such that : (DV n+1 ) : a(ω n+1, v) + b(χ n+1, v) = F n (v) v X b(q, ω n+1 ) = 0 q M where a = a 0 + a n + a n and F n (v) = (ω n )ω n v dx. Using then Lemma 1 and Lemma 3 we obtain for u n B α and v = ω n+1 : (νc 1 Cα ) ω n+1 2 X a(ω n+1, ω n+1 ) = F ( ω n+1 ) C ω n 2 X ω n+1 X (32) which gives (30) and the convergence is quadratic. 3 Nonhomogeneous Problem we are concerned now with the following general case of nonhomogeneous problem : ν u + (u. )u + p = f in (P) div u = 0 in u = g on Γ where the state u is sought in the space (H 1 ()) 2 V. Throughout this section denotes a bounded domain in R 2, with Lipschitzcontinuous boundary Γ = Γ i i = 1,..., 4. We assume in this section that : Γ i g.n i dσ = 0 with g H = (H 1/2 (Γ)) 2 and f K = (H 1 ()) 2. (33) We assume too that for a given g H satisfying (31), there exists for any c > 0, a function w 0 (H 1 ()) 2 such that : div w 0 = 0, w 0 Γ = g. (34) 8

9 a n (w 0, u n ) c u n 2 X u n V. (35) Remark 2 : The existence of w 0 satisfying (32) and (33) is a technical result due to Hopf [3]. Theorem 4 : Given (g, f) K H satisfying (31),there exists one pair (u, p) (H 1 ()) 2 L 2 0() solution of (P). Let y 0 = w 0 + u 0 where w 0 verify (32), (33) and an arbitrary u 0 V. We consider the sequence of linear problems (F (n+1) ) : ν y n+1 + (y n+1. )y n + (y n. )y n+1 + p n+1 = f n in div y n+1 = 0 in y n+1 = g on Γ with y n+1 = w 0 + u n+1, u n+1 X and f n = f + (u n )u n. y n+1 is solution du problem variational : { Find (F V (n+1) yn+1 V such that : ) : a(y n+1, v) == L n (v) v V where a(y, v) = a 0 (y, v) + a n (y, v) + a n (y, v) + a (y, v) and where a (y, v) = (y. )w 0 v dx and L n (v) = (f n, v) a 0 (w 0, v) a (w 0, v). Taking in (33) c > ν we obtain using (6) : a(y n+1, y n+1 ) (ν c) y n+1 2 X (36) thus we have the coercivity and L is obviously continuous on V.We observe that problem (F V (n+1) ) fits into the framework of section 1 and therefore the sequence y n converge towards a solution of (P). 4 Optimal Control In this section we are concerned with the following state-constrained optimal control problem : Find (u, f ) (H 2 ()) 2 (L 2 ()) 2 which solves : min J(u, f) = 1 2 u u d 2 dx + γ 2 (S) : subject to (P u,f ) f 2 dx ν u + (u. )u + p = f in div u = 0 in u = 0 on Γ u C 9

10 where the state u is sought in the space W = (H 2 ()) 2 V. Throughout this section denotes a bounded domain in R 2, with Lipschitzcontinuous boundary Γ = Γ i i = 1,..., 4. Is maintained in this section the assumptions (33), (34) and (35). C is a closed convex subset of C 0 () = {ω C( ); ω Γ }, the space of continuous functions on () vanishing on Γ. f is a distributed control function. The function u d (L 2 ()) 2 denotes the desired state. The parameter γ > 0 stand for the control cost coefficient. State constraints are relevant in practical applications in order to suppress backward flow in channels. Below,we derive necessary optimality conditions for (S). We have in mind two types of constraint sets C.The first one is : C 1 = {v C 0 ; y a (x) v(x) y b (x), x } covers pointwise constraints on each component of the velocity vector field, i.e. v(x) y b (x) gives v i (x) y b,i (x) for i = 1,..., d, on a subdomain. This is motivated for instance by the desire to avoid recirculations and backward flow by restricting the vertical or horizontal velocity components in some parts of the domain. The set of admissible solutions is defined as : T ad = {(u, f) W (L 2 ()) 2 ; u satisfies the state equation in(p u,f ) and u C} Before we study the problem of optimal control we start with the following proposition : Proposition 1 [5] let a bounded domain of R 2 and f L 2 ().Then every solution of (Q) satisfies u H 2 () and p L 2 0() H 1 ()for the corresponding pressure. Moreover, there exists a constant c(ν, ) > 0 such that the estimate : u H 2 () + p L 2 () c (1+ f 3 L 2 () ) Theorem 5 : If T ad is non-empty, then there exists a global optimal solution for the optimal control problem S. Proof : Since there is at least one feasible pair for the problem and J is bounded below 10

11 by zero, we may take a minimizing sequence (u k, f k ) in T ad. We obtain that γ f 2 k 2 J(u k, f k ) <, which implies that {f k } is uniformly bounded in (L 2 ()) 2. Then we may extract a weakly convergent subsequence, also denoted by {f k }, such that f k f (L 2 ()) 2. From estimate the proposition 1, it follows that the sequence {u k } is also uniformly bounded in W and, consequently, we may extract a weakly convergent subsequence, also denoted by {u k } such that u k u W. In order to see that (u, f ) is a solution of the Navier-Stokes equations, the only problem is to pass to the limit in the nonlinear form (u k. u k ) v dx. Due to the compact embedding W V and the continuity of (u k. u k ) v dx, it follows that (u k. u k ) v dx (u. u ) v dx Consequently, taking into account the linearity and continuity of all terms involved, the limit (u, f ) satisfies the state equations. Since C is convex and closed, it is weakly closed, so u k u in W and the embedding (H 2 ()) (H 1 0() C 0 () imply that u C. Taking into consideration that J(u, f) is weakly lower semi continuous, the result follows. Références [1] A. Younes, S.Abidi The Dirichlet Navier-Stokes Problem in the Velocity- Pressure Formulation. International Journal of Applied Mathematics Volume 24 No , [2] D. Gilbarg N.S. Trudinger Elliptic Partial Differential equation of second order Springer-Verlag Heidelberg (1983) [3] E. Hopf On Non-Linear Partial Differential Equations. Lecture Series of the Symp. on partial Diff. Equations Berkeley. (1955). [4] H.Amann,M.G.Crandall On some existence theorems for semi-linear elliptic equations. M.R.C. Tech. Report 1772, Madison Wisconsin [5] J.C.De Los Reyes and R. Griesse state-constrained optimal control of the stationary Navier-Stokes equations, J. Math. Anal. Appl. 343 (2008) [6] Ivart Ekeland Formulation variationnelle et optimisation, book (1974). 11