Identification of the diffusion coefficient in a nonlinear parabolic PDE and numerical methods

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1 ANZIAM J. 57 (E) pp.e E38, 26 E Identification of the diffusion coefficient in a nonlinear parabolic PDE and numerical methods Sameh Abidi Jamil Satouri 2 (Received 26 September 24; revised 27 April 26) Abstract This paper presents an iterative method to identify the diffusion in a semi-linear parabolic problem. This method can be generalized to other kind of problems, elliptic, parabolic and hyperbolic in twodimensional and three-dimensional case. The diffusion is obtained by solving an optimal control problem. By imposing specific conditions to the data, we build a sequence of linear problems which converge to the exact solution. We discretize our problem by a finite element method in the first case and a spectral method in the second case, using the sensibility method for approximating the gradient of the functional. Some numerical experiments prove the efficiency of this method. Keywords: Parabolic equation, diffusion, optimization, finite-elements, spectral method gives this article, c Austral. Mathematical Soc. 26. Published May 6, 26. issn (Print two pages per sheet of paper.) Copies of this article must not be made otherwise available on the internet; instead link directly to this url for this article.

2 Contents E2 Contents Introduction E2 2 The linearized problems E4 2. The Assumptions E4 2.2 Linearized problems E5 2.3 The variational Formulation of (P n ) E6 3 Continuity of the map Φ E6 4 The convergence E 5 Optimal control E5 6 The discrete problem E6 6. Finite element method E6 6.. The variational Problem E Calculating the gradient of J n E9 6.2 Spectral discretization of the problem (PV n ) E Calculating the gradient of J n E26 7 Some numerical experiments E27 References E35 Introduction In the phenomena of exchange at the sediment-water interface of the seabed or lagoons, the concentration u(x, t) where x is the spatial coordinate system and t is the time, is governed by the one-dimensional diffusion equation u t v u x x ( D u x ) = f(x, t, u),

3 Introduction E3 where D(x, t) is the diffusion coefficient that accounts both molecular diffusion and dispersion kinetics [4] and f is the production-destruction term. Defining Q T = Ω ], T[, for open interval Ω =], [ and T is a positive real number, χ T is the concentration at the time T, and v is the concentration at time t =. Neglecting the convective term, the problem in this paper is to find the coefficient of diffusion D minimizing the functional J(D) = 2 [u(x, T; D) χ T (x)] 2 dx, where for a given D, u(x, t; D) is the solution of the problem Find u(x, t) : Q T R, such that u t ( D u ) = f(x, t, u), (x, t) Q T, x x u(x, ) = v (x), x Ω, (P D ) u(, t) = u(, t) = g(t), t ], T[, where f, g and v > are given. Thus the problem min J(D) such that D U ad, u(x, t; D) solves (P D ), (Q D ) is an optimization problem. A similar problem in linear case was first studied by Bouchiba and Abidi [3]. However, here and elsewhere the formulation that is considered in this work deals with the nonlinear source term f. There are Several applications in this field, discussing parabolic and hyperbolic, linear and nonlinear systems describing numerical simulation results. These include control of distributed parameter systems [5, 6, 7], regularization for highly ill-posed distributed parameter estimation problems [4]. This performed numerical analysis based on an iterative Gauss Newton method and solved by Levenberg Marquardttype method, which applied to the output least squares formulation.

4 2 The linearized problems E4 In section 2, we study after linearizing, the existence and the uniqueness of a weak solution of the linearized problem of (P n ) for a fixed D. In section 3, we prove the continuity of the map Φ : D u(x, t, D). In section 4, we establish the convergence of the sequence (u n ) n N where u n is solution of (P n ) to u solution of (P D ). In section 5, we describe the optimal control problem under constraints (P D ) []. Indeed we prove the existence of a global optimal solution for the optimal control problem (Q) [, 2]. In section 6, we give a numerical algorithm to solve (P n ) by computing the gradient of the functional J [4], [3, linear version]. The numerical solution of the pde system is generated by both the finite elements method [7] and the spectral method [8, 9]. Finally in Section 7, we present a comparative study between the two methods (finite elements method and spectral method) to see the best convergence. 2 The linearized problems 2. The Assumptions Let O be an open set of R d, let p [, + ] and let k be a natural number. Define the Sobolev space W k,p (O) = {u L p (O) : α u L p (O) for all α k}, where α is a multi-index, and α is a partial derivative of u in the weak sense. We suppose that the nonlinear source term f in the first equation of (P D ), satisfies the following assumptions: f C ( ) 2 Ω ], ( ) T[ R W 2, Ω ], T[ R, (H) f (x, t, s), for all s R and (x, t) Ω ], T[, (H2) s s f(,, s) is concave on R, (H3) f(x, t, ) >. (H4)

5 2 The linearized problems E5 2.2 Linearized problems Using the first equation of (P D ), we denote the residual R(ψ) = ψ t ( D ψ ) f(,, ψ), x x so that u is a solution of R(u) =. By applying Newton s method linearized which has the advantage to converge rapidly towards our solution, we build a recurrent sequence of functions (u n ) n with u n : Q T R, for n N, u n = u n R(u n) R (u n ). When u is the first term, then u n t ( D u ) n f x x u (,, u n )u n = f u (,, u n )u n +f(,, u n ). We obtain a sequence of linear problems: initially u t ( D u ) = k, in Q T x x u (x, ) = v, x Ω, (P ) u (, t) = u (, t) = g(t), t ], T[, where k = f. For n, u n t ( D u ) n f x x u (x, t, u n )u n = F(x, t, u n ), in Q T, u n (x, ) = v, x Ω, (P n ) where u n (, t) = u n (, t) = g(t), t ], T[, F(x, t, u n ) = f(x, t, u n ) f u (x, t, u n )u n.

6 3 Continuity of the map Φ E6 2.3 The variational Formulation of (P n ) Let O be an open set of R d and let k be a natural number. Define the Sobolev space H k (O) = { u L 2 (O) : α u L 2 (O) for all α N d, α k } For a fixed D, the variational formulation of problem (P n ) is Find u n (, t) H (Ω), with u n (, t) ḡ(, t) H (Ω), such that d dt (u n(, t), v) + a D (u n (, t), v) = (F(, t, u n ), v), for all v H (Ω), u n (, ) = v, (PV n ) where u n (, t) : Ω R and ḡ(, t) is the lifting of the boundary condition g(t) in H (Ω) and coincides with g(t) on {, }, the boundary condition is reflected by the fact that u n (, t) ḡ(, t) H (Ω) [9, Proposition.2]. The bilinear form, for all v H (Ω), a D (u n, v) = D u n(, t) x v x dx f u (, t, u n )u n (, t)v dx. So for F(,, u n ) L 2 (, T, L 2 (Ω)) and ḡ(, t) continuous in Ω, v L 2 (Ω) and D L (Q T ) such that D α >, problem (PV n ) admits a unique solution [, 2], u n L 2 (, T, H (Ω) ) C (, T, L 2 (Ω) ). 3 Continuity of the map Φ The following Lemma is needed to prove the continuity of the map Φ. Lemma. If function u is a solution of problem (P D ), then u is bounded in L 2 (, T, H (Ω)).

7 3 Continuity of the map Φ E7 Proof: The variational formulation of problem (P D ) is Find u(, t) H (Ω); u(, t) ḡ(, t) H (Ω) such that d dt (u(, t), v) + ã D(u(, t), v) = (f(, t, u), v), for all v H (Ω), (PV D ) u(, ) = v, where ã D (u(, t), v) = D u(,t) v dx. We denote ū(, t) = u(, t) ḡ(, t). x x Then u(, t) is solution of problem (PV D ) if and only if ū(, t) is solution of the problem Find ū(, t) H (Ω) such that, for all v H (Ω), d dt (ū(, t), v) + ã D(ū(, t), v) = (f(, t, ū + ḡ(t)), v) ã D (ḡ(, t), v), u(, ) = v. () Using () in the particular case for v = ū(, t) H (Ω), such that D α > and the Cauchy Schwarz inequality combined with Poincare inequality we prove that u is bounded in L 2 (, T, H (Ω)). Let the map Φ : L (Q) L 2 (, T, H (Ω) ) D u where u is the unique solution of (P D ). Let U = {ϑ L (Q T ) : there exists α\ϑ α > }. For D, D 2 U, D D 2, we denote u i = Φ(D i ) i=,2 and f i (x, t) = f(x, t, u i ) i=,2, for all (x, t) Q T. Proposition 2. The map Φ is continuous u u 2 L2 (,T,H (Ω)) C D D 2 L (Q T ), (2) where C is a constant independent of D and D 2.

8 3 Continuity of the map Φ E8 Proof: We denote ū = ū ū 2. Using (2), after subtracting for all v H (Ω), ( ū (, t) = t ), v + (D D 2 ) ū (, t) x v x dx + D 2 ū (, t) x v x dx ( f (, t) f 2 (, t) ) v dx. (3) In particular, equation (3) is true for v = ū (, t) H (Ω) and after integration between and t ( t ) 2 ū (, t) 2 L 2 (Ω) + ū 2 (, t) D 2 x dx dσ t [ = ( f (, t) f 2 (, t) ) ] ū (, t) dx dσ t [ + (D 2 D ) ū ] (, t) ū (, t) dx dσ. x x We denote A = 2 ū (, t) 2 L 2 (Ω), B = t ( D 2 ū (, t) x 2 dx ) dσ. Then, as D α >, ( t ) ū 2 (, t) x dx dσ α t + α [ t [ ( f (, t) f 2 (, t) ) ] ū (, t) dx dσ (D 2 D ) ū (, t) x ū (, t) x dx ] dσ. (4)

9 3 Continuity of the map Φ E9 And also, 2 ū (, t) 2 L 2 (Ω) t + [ t ( f (, t) f 2 (, t) ) ] ū (, t) dx dσ [ (D 2 D ) ū ] (, t) ū (, t) dx dσ. x x The inequality (4) is true for all t ], T[. Recall the assumption (H) and using Taylor s formula, we obtain the following inequality which is related to B defined above ū 2 (, t) x L 2 (,T,L 2 (Ω)) α D 2 D L (Q T ) ū (, t) x L 2 (,T,L 2 (Ω)) + c α D 2 D L (Q T ) ū L 2 (,T,L 2 (Ω)) By Lemma, ū (, t) x 2 We obtain ū (, t) x L 2 (,T,L 2 (Ω)) 2 L 2 (,T,L 2 (Ω)) ū (, t) x α c D 2 D ū (, t) L (Q T ) x L 2 (,T,L 2 (Ω)) L 2 (,T,L 2 (Ω)) + c α D 2 D L (Q T ) ū (, t) L 2 (,T,L 2 (Ω)). C D 2 D L (Q T ) ū (, t) L2 (,T,H (Ω)). Similarly to (4), the following inequality in terms of the defined variable A, 2 ū (, t) 2 L 2 (Ω) C D 2 D L (Q T ) ū (, t) L2 (,T,H (Ω)).

10 4 The convergence E After integration between and T, ū (, t) 2 L 2 (,T,L 2 (Ω)) 2TC D 2 D L (Q T ) ū (, t) L2 (,T,H (Ω)). We set C = 2 sup(c, 2TC ), and conclude that ū (, t) L2 (,T,H (Ω)) C D 2 D L (Q T ), and finally u u 2 L2 (,T,H (Ω)) C D D 2 L (Q T ). 4 The convergence Remark 3. The symbol X Y denotes the continuous and dense embedding of X into Y. The symbol X Y denotes the weak convergence of X to Y. The symbol X Y denotes the strong convergence of X to Y. Consider now the sequence (u n ) n N where u n is solution of (P n ). Proposition 4. The sequence {u n } n= satisfies < u n+ u n u.

11 4 The convergence E Proof:. We first prove that u n+ u n. Let ω (x, t) = u (x, t) u (x, t), for all (x, t) Q T, with u is the solution of (P ) and u is the solution of P, (P n ) for n =, then u t ( D u ) x x u t ( D u ) = k. x x f u (,, u )u = F(,, u ), After subtracting (P ) and P, (P n ) for n =, we get ω t ( D ω ) f x x u (,,, u )ω = f(,, u ) k. Using the assumption (H2), the difference field ω via the maximum principle []. Subtracting P n+ and (P n ), setting ω n = u n+ u n, with u n is the solution of (P n ) and u n+ is the solution of P n+, (P n ) for n n +, ω n t x ( D ω ) n f x u (,, u n)ω n = G n, where via (H3), G n = f(,, u n ) f(,, u n ) f u (,, u n )ω n. Then, using the assumption (H2) the maximum principle gives ω n, for all n. 2. We prove now that u n > for all n. In problem (P ), k = f > and v > by the maximum principle u >. In problem P n+, (P n ) for n n +, under assumption (H3) and (H4) F(,, u n ) >. Using the assumption (H2) and v >, then via the maximum principle u n > for all n.

12 4 The convergence E2 Theorem 5. The sequence (u n ) n N where u n is solution of (P n ) converges to u, where u is a solution of (P D ). Proof: By Lemma, u n is bounded in L 2 (, T, H (Ω)) then, we can extract a subsequence still denoted u n such that u n u in L 2 (, T, H (Ω) ) weakly. (5) Using (H4), we prove that u >. But it appears difficult because from (5), it is generally not guaranteed that ( D u ) n ( D u ). x x x x Then, we increase the regularity of u using the singular perturbation method. For that we introduce a new bilinear form, continuous on H 2 (Ω), such that (ϕ, ψ) b(ϕ, ψ), b(ψ, ψ) β ψ 2 H 2 (Ω), for all ψ H2 (Ω). Using Riez s Theorem, we represent the bilinear form b by an operator B such that b(ϕ, ψ) = Bϕ, ψ H 2 (Ω),H 2 (Ω). Similarly for the bilinear form a D : it is represented by the operator A(D) such that a D (ϕ, ψ) = A(D)ϕ, ψ H (Ω),H (Ω). For ε sufficiently small and positive we denote a ε,d (ϕ, ψ) = εb(ϕ, ψ) + a D (ϕ, ψ), for all ϕ, ψ H 2 (Ω). We verify that a ε,d (ψ, ψ) ε ψ 2 H 2 (Ω) + c ψ 2 H (Ω),

13 4 The convergence E3 where c is a positive constant. The singular perturbation problem is u n,ε + (εb + A(D))u n,ε = F(,, u n ) in Q T, t u n,ε (x, ) = v (x) in Ω, (P n,ε ) where u n,ε L 2 (, T, H 2 (Ω)). The operator + (εb + A(D)) is the singular t perturbation operator of + A(D). By applying the theorem of Lions [2] t such that u n,ε H 2 (Ω) and a D (u n, v) is replaced by a ε,d (u n, v) and using the assumption of uniform coercivity, a ε,d (ψ, ψ) ε ψ 2 H 2 (Ω) + c ψ 2 H (Ω), (6) problem (P n,ε ) admits a unique solution u n,ε. Proposition 6. As ε the following strong convergences hold: u n,ε u in L 2 (, T, H (Ω)); εu n,ε in L 2 (, T, H 2 (Ω)); t u n,ε t u in L2 (, T, H 2 (Ω) ). Since ε is fixed positive, we deduce from (6) that u n is bounded in L 2 (, T, H 2 (Ω) ). (7) From the first equation of (P n,ε ), we prove that t u n,εis bounded in L 2 (, T, H 2 (Ω) ). (8) From (7), we extract a sequence, still denoted by u n, such that u n converge in a weak sense to u in L 2 (, T, H 2 (Ω)). Since the injection of H 2 (Ω) in H (Ω) is compact, there exists a subsequence, still denoted u n, which converge strongly to u in L 2 (, T, H (Ω)).

14 4 The convergence E4 From (8), we conclude that u n converge weakly to u t t then D u n x D u in Q T, x in a weak sense. Indeed, for all ψ D(Q T ), X = (x, t) Q T, D u n Q T x ψ dx = D u ( Q T x ψ dx + un D Q T x u x We conclude that because Q T D Q T D u n x ψ dx Q T D u x ψ dx, in L2 (, T, H 2 (Ω) ), ) ψ dx. ( un x u ) ψ dx x C u n x u x ψ L 2 (Q T ), L 2 (Q T ) and Since f(q T ) is C 2, then u n x u x. L 2 (Q T ) f u (X, u n) f (X, u), u and f(x, u n ) f u (X, u n )u n f(x, u) f (X, u)u. u We conclude that t u n ( D ) x x u n + f u (X, u n)u n t u (D x ) x u + f (X, u)u, u weakly in Q T. Therefore u n satisfies the two equations of (P n,ε ).

15 5 Optimal control E5 5 Optimal control In this section we are interested by the optimal control problem under constraints (P D ) [] min J(D) = [u(x, T; D) χ T (x)] 2 dx, such that 2 u t ( D u ) = f(x, t, u), (x, t) Q T, x x u(x, ) = v (x), x Ω, (Q) u(, t) = u(, t) = g(t), t ], T[, u C. The set C is a closed convex subset of C (Ω) = {ω C( Ω) : ω = on {, }}, the space of continuous functions on Ω vanishing on {, }. Remark 7. Sometimes we adopt the notation u(x, t; D) as a reminder that u implicitly depends on D. The set of admissible solutions is defined as U ad = {D L (Q T ) : u(x, t; D) satisfies the state equation in (P D )}. The function u(x, T; D), for all (x, t) Q T, is the solution of problem (P D ) at t = T, χ T is given in L 2 (Ω). We now prove the main result of this section. Theorem 8. If U ad is non-empty, then there exists a global optimal solution for the optimal control problem (Q). Proof: Because U ad is not empty, we may take a minimizing sequence D n U ad. We obtain that D n < which implies that D n is uniformly bounded in L (Q T ). Then we may extract a weakly convergent subsequence, also denoted by D n, which converge in a weak-star sense to D L (Q T ). And we denote u n (x, t) = ū(x, t; D n ), for all (x, t) Q T, by () and using

16 6 The discrete problem E6 Lemma combined with (H), u n is uniformly bounded in H (Ω). Then we extract a weakly convergent subsequence, also denoted by u n, such that u n ū H (Ω). In order to see that (ū ; D) is a solution of the (P D ) equations, the only problem is to pass to the limit in the nonlinear form a n = f(, t, u n)v dx of (). Due to the compact embedding of H (Ω) into L 2 (Ω) [2] and the continuity of the a n, and without forgetting that a n is bounded for all x, t and u n, it follows that f(, t, u n )v dx f(, t, u)v dx. Consequently, taking into account the linearity and continuity of all terms involved, the limit (ū ; D) satisfies the state equations. Since C is convex and closed, it is weakly closed, so u n ū H (Ω) and the embedding H (Ω) C (Ω) imply that ū C. Taking into consideration that J(D) is weakly lower semi continuous, then there exists a global optimal solution for the optimal control problem (Q). 6 The discrete problem 6. Finite element method In this section, to simplify, we study the problem (PV n ) in the case of homogeneous boundary conditions (g = ). 6.. The variational Problem We put H = L 2 (Ω), V = H (Ω), Ω =], [. Then V H V. We consider the variational problem (PV n ) which admits a unique solution u n L 2 (, T, H (Ω) ) C ([, T], L 2 (Ω)).

17 6 The discrete problem E7 To build an approximation u n,h of u n, we choose a subspace of V composed of affine functions over an equal number of intervals of equal size. More precisely, let N be a natural number and h =, i N +. We N+ denote x i = ih, i N +, that subdivides the interval Ω = [, ] to N + intervals K i = [x i, x i+ ], i N, each of length h and V h = {v C ( Ω) : v() = v() =, v Ki P, i N}, a finite dimensional subspace of V []. Setting F n (t) = F(, t, u n ), the approached problem is Find u n,h (, t) V h, such that for all t [, T], d dt u n,h(, t), v h + a D (u n,h (, t), v h ) = F n (t), v h for all v h V h, u n,h (, ) = v,h. (PV nh ) The problem (PV nh ) admits an unique solution u n,h (, t) V h : the proof is obtained by direct application of Lions s Theorem [5]. We introduce the basis (ϕ i ) i N of V h, for j N. Then there exists u D n,,j and ud n,j functions of [, T] in R, such that, for all (x, t) Q T, v,h (x, t) = u D n,,j(t)ϕ j (x), u n,h (x, t) = u D n,j(t)ϕ j (x). j= Then problem (PV nh ) is equivalent to the differential system (ϕ j, ϕ i ) d dt ud n,j(t) + a D (ϕ j, ϕ i )u D n,j(t) = F n,i (t), i N, j= j= u D n,i() = u D n,,i, i N, (S ) where F n,i (t) = (F n (t), ϕ i ). We denote j= R D (t) = (a D (ϕ j, ϕ i )) i,j N the stiffness matrix, M = (ϕ j, ϕ i ) i,j N the mass matrix, u D n (t) = (u D n,(t), u D n,2(t),..., u D n,n(t)) T, and F n (t) = (F n, (t), F n,2 (t),..., F n,n (t)) T.

18 6 The discrete problem E8 The system (S ) is written as M d dt u D n (t) + R D (t)u D (t) = F n (t), n u D n () = u D n,. (S 2 ) A simple calculation gives the elements of M as (ϕ i, ϕ i ) = 2h 3, i N, We put Setting K n (t) = K n,i+ 2 K + n,i 2 (ϕ i, ϕ i+ ) = h 6, (ϕ i+, ϕ i ) = h, i N, 6 (ϕ i, ϕ j ) =, i j >. D i+ 2 = xi+ x i D(t) dx. f (, t, u u n )u n (, t)v dx, then = = xi+ x i K n (t)(x i+ x) 2 dx, xi x i K n (t)(x x i ) 2 dx, xi+ K n,i+ 2 = K n (t)(x i+ x)(x x i ) dx. x i We obtain for the coefficients of R D : a D (ϕ i, ϕ i ) = h 2 (D i+ 2 + D i 2 + K n,i+ 2 + K + n,i 2 ), i N, a D (ϕ i+, ϕ i ) = h 2 (D i+ 2 + K n,i+ 2 ), a D (ϕ i, ϕ i+ ) = h (D 2 i+ + K 2 n,i+ ), i N, 2 a D (ϕ i, ϕ j ) =, i j >. (9)

19 6 The discrete problem E9 Using the modified Euler method of integration, K n (t)ϕ 2 i dx = ( ) K + + K h 2 n,i 2 n,i+ 2 = 2h f ( xi + h, t, u D f 3 u n,i) + ( u + h f 6 u ( + f u x i + h 2, t, ud n,i + ud n,i 2 x i+ + h 2, t, ud n,i+ + ud n,i+2 2 ( xi+ + h, t, u D n,i+2) ) ), () and ( K n (t)ϕ i ϕ i+ dx = h f x i + h 24 u 2, t, ud n,i + ) ud n,i+. () 2 The system (S 2 ) is solved using a fourth order Runge Kutta method Calculating the gradient of J n Consider the problem min J n (D), D U ad, u n (,.; D) solves (P n ), (Q n ) where we define the functional J n (D) = 2 [u n (x, T; D) χ T,n (x)] 2 dx. (2) As χ T,n = N i= χ n,t (x i )ϕ i, we write J n (D) = [ N ( u D 2 n,i (T) χ n,t (x i ) ) ] 2 ϕ i (x) dx. (3) i=

20 6 The discrete problem E2 Expand the expression of J, J n (D) = h ( u D 3 n,i (T) χ n,t (x i ) ) 2 i= + h 6 N ( u D n,i+ (T) χ n,t (x i+ ) ) ( u D n,i(t) χ n,t (x i ) ). i= To calculate an approximation of the gradient J, we consider a mesh of the rectangle Ω [, T] by triangles T ij with vertices and T ij with vertices where δt = (ih, jδt), ((i + )h, jδt), (ih, (j + )δt), (ih, (j + )δt), (ih, (j + )δt), ((i + )h, jδt), T is a time step. There D M+ i,j D(ih, jδt). And we denote λ i respectively (λ i ), i =, 2, 3, the barycentric coordinates of a point from the triangle T ij, respectively T A simple calculation gives ij, D ij Tij = λ D ij + λ 2 D i+,j + λ 3 D i,j+, D ij T ij = λ D i+,j + λ 2D i,j+ + λ 3D i+,j. λ (x, t) = x h + t t (i + j + ) = λ (x, t), λ 2(x, t) = x h + i + = (λ 2(x, t) ), λ 3(x, t) = t t + j + = (λ 3(x, t) ). The calculation of D i+ 2 is more delicate when x varies between x i and x i+ and t between jδt and (j + )δt the point (x, t) T ij T ij so if x [x i, x i + a(t)] then (x, t) T i, if x [x i + a(t), x i ] then (x, t) T i.

21 6 The discrete problem E2 Figure : Mesh where a(t) = h( s j) (see Figure ), with s = t/δt so D i+ 2 = h 2 [α j(t)d ij + β j (t)d i+,j + γ j (t)d i,j+ + δ j (t)d i+,j+ ], (4) with α j (t) = (+j s) 2, β j (t) = (+j s)( j+s), γ j (t) = (s j)(2+j s), and δ j (t) = (s j) 2. Thus, knowledge of D ij, i N +, j M +, approach D(x, t) on Q T and consequently reduce the problem of minimization in finite dimensions: D will equate to a vector R k where k = (N + 2)(M + 2). We denote D = (D,..., D (N+), D,..., D (N+),..., D (M+),..., D (N+)(M+) ) T. The problem (Q n ) becomes min J n (D), D R k, D α. (Q nh )

22 6 The discrete problem E22 Then we obtain J n (D) = ( Jn D, J n D,..., J n D (N+),..., J n D (N+)(M+) ) T. As and J n D ij = J n u D k= n,k u D n,k D ij (t), ( ) T u D n u D n, (t) =, ud n,2,..., ud n,n. D ij D ij D ij D ij The expression (3) gives J n = 2 u n,k 3 h [ u D k,n(t) χ T,n (x k ) ] + h [ u D 6 k,n (T) χ T,n (x k ) ] + h 6 ud k+,n(t) χ T,n (x k+ ). Knowing that χ T (x ) = χ T (x N+ ) = and u D J n D ij, we assume = u D N+ =, to calculate ( ) T v ij (t) = u D n u D n, (t) =, ud n,2,..., ud n,n. D ij D ij D ij D ij Then By using (S 2 ), it follows v ij t = D ij ( ) u D n. t v ij t = M R D u D n D ij (t) M R D (t) u D n D ij (t).

23 6 The discrete problem E23 To calculate (v ij ), it suffices to solve the following system, using the fourthorder Runge Kutta method, v ij t = M R D D ij u D n (t) M R D (t)v ij, v ij () =. The matrix coefficients R D (t) and R D(t) D ij are calculated using (). 6.2 Spectral discretization of the problem (PV n ) In this section we recall some formulas in the spectral method in a reference field Λ =], [, then we use these results to write the variational formulation in our field of study Ω. Let N be an integer 2. Denote the space P N (Λ) of polynomials with degree N and the space P N (Λ) of polynomials in P N(Λ) vanishing on the boundary of Λ [8, 9]. We introduce the space P N (, ) of restrictions to Λ of polynomials with degree N. Setting ξ = and ξ N =, we introduce the N nodes ξ j, j N, and the N + weights ρ j, j N, of the Gauss Lobatto quadrature formula, recall that φ(ζ)dζ = φ(ξ j )ρ j. (5) We also recall [6, (3.2)] the following property, which is useful in what follows: for all ϕ N P N (, ), j= N ϕ N 2 L 2 (Λ) ϕ 2 N(ζ)ρ j 3 ϕ N 2 L 2 (Λ). (6) Relying on this formula, we introduce the discrete product [8, 9] defined on continuous functions u and v by (u, v) = u(ξ j )v(ξ j )ρ j. (7) j= j=

24 6 The discrete problem E24 It follows from (6) that this discrete product is a scalar product on P N (Λ). Let finally I N denote the Lagrange interpolation operator at the nodes ξ i, i N, with values in P N (Λ). We now assume that the function F n (t) is continuous on Ω. Thus the discrete problem is constructed from (PV n ) using the Galerkin method combined with numerical integration. It reads Find u n,n (, t) P N (Ω), with u n,n (, t) I N ḡ(, t) P N(Ω), such that for all v N P N(Ω) d dt (u n,n(, t), v N ) N + a D,N (u n,n (, t), v N ) = (F n (t), v N ) N, u n,n (, ) = v,n, (8) where the bilinear form ( u n,n (, t) a D,N (u n,n, v N ) = D N, v ) N. (9) x x N It follows from (PV n ), combined with Cauchy Schwarz inequalities, that the form a D,N is continuous on P N (Ω) P N (Ω), with norm bounded independently of N. Then to investigate the well-posedness of the problem we use Lions s theorem [2, 5]. To simplify the problem, we suppose that we know the values of F n (t) at the nodes of Ω. Let l i be Lagrange polynomials at the nodes ξ i, i N [6, Remark.5]. Then we calculate the coordinates u n,i (t) of solution u n,n (, t) in the base l i, for each t in [, T]. So We note and u n,n (x, t) = u n,i (t)l i (x). i= v N = l r, r N, D N (x, t) = m=p= D mp l m (x) l p (t).

25 6 The discrete problem E25 The variational problem (8) (l i, l r ) dt u n,i(t) + i= u n,i () = v,i, a D,N (l i, l r )u n,i (t) = F n (t)( ξ r )ρ r, (2) i= (2) where u n,i (t) are the values of u n,n (t) at the nodes ξ i of the Gauss Lobatto in Ω. We obtain N equations with N unknowns. So (9) is equivalent to the linear system M d dt u D n (t) + R D (t)u D (t) = F n (t), n u D n () = ṽ, (22) where u D n (t) is the vector of coordinates u n,i (t), i N. M is the mass matrix, R D (t) is the stiffness matrix. F n (t) is the vector of coordinates 2 ( f( ξ r, t, u n ( ξ r, t)) f s ( ξ r, t, u n ( ξ r, t))u n ( ξ r ) a D,n (l, l r )u n ( ξ, t) + a D,n (l N, l r )u n ( ξ N, t), r N. ṽ is constituted with the values v,i, i N, at the nodes ξ i of the Gauss Lobatto in Ω. The coefficients of M are written as [3, Appendix] { if i < r N, (l i, l r ) = if i = r N. 2 N(N+)L 2 N ( ξ i ) And the coefficients of R D (t) are a D,N (l i, l r ) = 2 lp (t) p= ) D kp l i( ξ k )l r( ξ k )ρ k k= f 2 s ( ξ r, t, u n ( ξ r, t))δ ir ρ r. ρ r

26 6 The discrete problem E Calculating the gradient of J n Consider the problem (Q n ) as u n,n (t) (x) = u D n,jl j (x), χ T (x) = j= χ T,i l i (x), i= and setting u D n ( ξk ) (T) = u D n,k (T), it follows J n (D) = k= [ u D n,k(t) χ T ( ξ k )] 2 ρk. (23) To calculate an approximation of the gradient J n (2), we consider a mesh of rectangle Ω [, T]. Then J n (D) D mp = We then observe that, via (23), k= J n (D) u D n,k u D n,k D mp. J n (D) u n,k = 2(u D n,k(t) χ T,k )ρ k. And then find ( u D n (t) u D n, (t) =,..., ud n,n (t) ), D mp D mp D mp we assume v m,p (t) = u D n (t) D mp, and using (22) gives v m,p t = M (R Du D n (t)) D mp, v m,p () =. (24)

27 7 Some numerical experiments E27 To solve (24), R D u D n = 2 m= p= ( N D mp and finally ( R D u D n (t) N = 2 D mp i= j= i= j= ) u ij l i( ξ m ) l j (t) l r( ξ m ) l p (t) ρ m, ) u ij l i( ξ m ) l j (t) l r( ξ m ) l p (t) ρ m. 7 Some numerical experiments In this section we present some numerical experiments using matlab as a computational tool. The algorithm of resolution of the problem (Q), is divided into two parts. Firstly we solve the problem (PV n ), in the matrix form (S 2 ) and (22) in both methods. The resolution of U is done knowing D = D(, ). The method used is a fourth order Runge Kutta. Secondly, once U is calculated, it is then introduced in the minimization problem (Q D ) to recover D. The best method used in our case is a Levenberg Marquardt method based on the calculation of gradient of J(D) detailed in subsections 6..2 and We continue this process until we get the desired convergence. The algorithm requires about 7% of total time for calculating the solution u and 93% to calculate D. Runge Kutta is used three times in the program. In the Levenberg Marquardt algorithm, the iteration number to achieve maximum convergence of D varies somewhat as function of N, and it does not exceed five iterations. We need also just a few iterations to assure the convergence of the global system. Table reports the average cpu time (in

28 7 Some numerical experiments E28 Table : The cpu time in seconds required for one iteration of Runge Kutta. N fem fem2 sp.m seconds) required for one iteration of Runge Kutta using the three methods with N varying between 8 and 36. We use the bicgstabl method (gmres method is also a suitable alternative). Several preconditioners can be found in literature. In our case a preconditioner performing an incomplete lu factorization is the best appropriate to be used for all three methods. The convergence is assured in four iterations at most. Without using any preconditioner the number of iterations increases as a function of N. For example by fixing a tolerance 6 the iteration number exceeds ten for N greater than 3. Without loss of generality, we consider the domain of study Q T. Numerically we prove the convergence of the same problem in a more general case taking u(, t) = g (t) and u(, t) = g 2 (t) where g and g 2 are given. We present a complete study of the problem, using finite element method (explained in subsection 6.). We give in this method two different approaches. In the first approach (fem) we use the trapezoidal formula for approximating integrals. Regarding the second approach (fem2) we employ the Gauss Lobatto formula. Finally, we present a spectral method (sp.m), with a comparative study for each test case.

29 7 Some numerical experiments E29. We consider the following analytical functions: u(x, t) = (x 2 + 2)(t + ), D(x, t) = (x + )(t + ), (25) 24 f(x, t) = s(x, t)u 2 + h(x, t), with s(x, t) = (t 2 + )e x, and h(x, t) verifies the equation (P D ). Figures 2 and 3 plot the two quantities log ( u u N L 2 (Ω) ) and log ( D D N L 2 (Ω) ) as functions of log (δt). We apply this for both the methods. In all these calculations we take T =. and N = We consider the following functions: u(x, t) = e t (x + ), D(x, t) = 2 e t (x 2 + ), (26) f(x, t) = s(x, t)u 2 + h(x, t), with s(x, t) = tx 2, and h(x, t) verifies the equation (P D ). Figures 4 and 5 plot the two quantities log ( u u N L 2 (Ω) ) and log ( D D N L 2 (Ω) ) as functions of log (δt). We apply this for both the methods. All these calculations are made by fixing T =. and N = 35. In both the cases (first test and second test) presented, we find that the spectral method is more precise than finite elements method. The slope of the curve estimations in approach 2 (finite element method) and the spectral method proves the good convergence in time of the solutions u and D. While in approach (finite element method) the slopes of the curve estimations decreases very slowly with time, this result is due to the poor convergence space stops the evolution in time of errors. 3. For the same example as the first case, Table 2 presents the error u u N and Figure 6 the approached solution u N for times t varying between and in the nodes ξ i, i =,..., 8. The results are for T =, δt = 3 and N = 8.

30 7 Some numerical experiments E3 Figure 2: L 2 error between the exact solution u given in the first test and the experimental solution u N using the three approaches. 2 4 log u u N L 2 (Ω) f.e.m. f.e.m.2 sp.m log (δt) Figure 3: L 2 error between the exact solution D given in the first test and the experimental solution D N. log D D N L 2 (Ω) 5 f.e.m. f.e.m.2 sp.m log (δt)

31 7 Some numerical experiments E3 Figure 4: L 2 error between the exact solution u given in the second test and the experimental solution u N. 5 log u u N L 2 (Ω) f.e.m. f.e.m.2 sp.m log (δt) Figure 5: L 2 error between the exact solution D given in the second test and the experimental solution D N. log D D N L 2 (Ω) 5 f.e.m. f.e.m.2 sp.m log (δt)

32 7 Some numerical experiments E32 Table 2: The values u(ξ i,.) u N (ξ i,.) in the third test calculated at the nodes ξ i for different values of t, with N = 8 and T =. t ξ ξ 2 ξ 3 ξ 4 ξ 5 ξ 6 ξ 7 ξ 8. e 3 e 3 e 2 7e 2 5e 2 e 2e 3e.2 e 3 e 2 2e 2 7e 2 2e 5e e e.3 e 2 2e 2 6e 2 e 5e e 2e 3e.4 e 2 4e 2 e 3e e 2e 5e 6e.5 2e 2 7e 2 2e 5e e 4e 9e e 9.6 4e 2 e 3e e 2e 6e e 9 e 9.7 6e 2 e 5e e 4e e 9 2e 9 2e 9.8 9e 2 2e 8e 2e 5e e 9 2e 9 3e 9.9 e 4e e 3e 8e e 9 3e 9 4e 9 2e 6e e 4e e 9 2e 9 4e 9 5e 9 4. We study the convergence of solutions depending on the parameter n. Figures 7 and 8 show the evolution of curves as a function of N; N varying between 6 and 24. We take T =. and δt = 2, and the exact solutions are u(x, t) = (t + )(x 2 + ) /2 D(x, t) =.25 + t (x2 + ) 3/2, (27) f(x, t) = s(x, t)u 4 + h(x, t), with s(x, t) = e t2 + (x 2 + ), and h(x, t) verifies the equation (P D ). In this case u is a singular function. The maximum error of the solutions is reached at N = 24 in approach and 2 of finite element methods, and at N = 4 in the spectral method. Knowing that, if we take u as a polynomial function, then the maximum is reached from N = 4 for the two methods.

33 7 Some numerical experiments E33 Figure 6: The curves of u N (ξ i, t) as a function of t, calculated in the third test at the nodes ξ i with N = 8, T = and δt = 3..8 Estimations by Finite elements method.6 u(.,t))in the values ξ i,i=: u(ξ 8,t) u(ξ,t) t 5. In this test, we take u f(x, t) = ( (x + )(t + ) )2 + x + 5, u(x, ) = x +, (28) { (t + ) on {} g(t) = Ω, 2(t + ) on {} Ω (29), D(x, ) = x +. The solution in this case is unknown. The calculations done with the second approach (finite element methods) and spectral method are presented in Figures 9 and. Figures 9 and are almost identical, this proves the good convergence of the two algorithms.

34 7 Some numerical experiments E34 Figure 7: L 2 error between the exact solution u given in the fourth test and the experimental solution u N. 2 f.e.m. f.e.m.2 sp.m. log u u N L 2 (Ω) N Figure 8: L 2 error between the exact solution D given in the fourth test and the experimental solution D N. 2 f.e.m. f.e.m.2 sp.m. 3 log D D N L 2 (Ω) N

35 t t References E35 Figure 9: The discrete solution u N and D N using spectral method with data f and g given in the fifth test. Sp.m isovalues of u u t.5 x x.5 isovalues of D.8.8 D t.5 x x.2 Conclusion The selection of the two algorithms introduced in the paper gives a good and stable numerical convergence of the two methods. The results of the numerical simulations that we performed confirm all theoretical results found. References [] P. A. Raviart, D. J. M. Thomas, Introduction à l analyse numérique des équations aux dérivées partielles, Masson (983). E6, E, E7 [2] J.L. Lions, E. Magenes, Problèmes aux limites non homogènes et applications, Vol., Dunod (968). E6, E3, E6, E24

36 t t References E36 Figure : The discrete solution u N and D N using finite elements method with data f and g given in the fifth test. F.e.m.2 isovalues of u u t.5 x x.5 isovalues of D.8.8 D t.5 x x.2 [3] M. Bouchiba, S. Abidi, Identification of the Diffusion in a lineair Parabolic problem, International Journal of Applied Mathematics, Vol. 23, No. 3, 2, E3, E4 [4] M. Kern, Problèmes Inverses Aspects Numériques, INRIA (23). E3, E4 [5] J. E. Rakotoson, J. M. Rakotoson, Analyse Fonctionnelle appliquée aux équations aux dérivées partielles, Presse Universitaires de France, (999). E7, E24 [6] C. Bernardi, Y. Maday, Spectral Methods, in the Handbook of Numerical Analysis V, P. G. Ciarlet & J. L. Lions eds., North-Holland, (997), E23, E24

37 References E37 [7] V. Girault, P.-A. Raviart, Finite Element Methods for Navier Stokes Equations, Theory and Algorithms, Springer-Verlag (986). E4 [8] C. Bernardi, Y. Maday, F. Rapetti, Discrétisation Variationnelles de problèmes aux limites elliptiques, Springer-Verlag (24). E4, E23 [9] C. Bernardi, Y. Maday, Approximations spectrales de problèmes aux limites elliptiques, Springer-Verlag (992). E4, E6, E23 [] I. Ekeland, R. Teman, Analyse Convexe et problèmes Variationnels, Bordas (9674). E4, E5 [] J. L. Lions, Controle Optimal de systèmes gouvernés par des équations aux dérivées partielles, Dunod, Paris, (968). E4 [2] J. Henry, Étude de la contrôlabilité de certaines équations paraboliques, Thèse d état, Université Paris VI, (978). E4 [3] J. Satouri, Méthodes déléments spectraux avec joints pour des géométries axisymétriques, Thèse de l Université Pierre et Marie Curie, Paris VI (2). E25 [4] K. Van de Doel, U. M. Ascher, On level set regularization for highly ill-posed distributed parameter systems, Journal of Computational Physics, 26, 26: E3 [5] M. Böhm, M. A. Demetriou, S. Reich, and I. G. Rosen, Model Reference Adaptive Control of Distributed Parameter Systems, SIAM J. Control Optim., 36(), 33 8; doi:.2/rnc.98 E3 [6] C. Jia, A note on the model reference adaptive control of linear parabolic systems with constant coefficients. Journal of Systems Science and Complexity 24, 7; doi:.7/s E3 [7] D. B. Pietria, M. Krstic, Output-feedback adaptive control of a wave PDE with boundary anti-damping. Automatica 5, 47 45; doi:.6/j.automatica E3

38 References E38 Author addresses. Sameh Abidi, University of Tunis El Manar, Faculty of Sciences of Tunis, Department of Mathematics, 6 Tunis, Tunisia. mailto:labidisameh@yahoo.fr. 2. Jamil Satouri, IPEIT-University of Tunis, 2 street Jawaher Lel Nahra-89, Monfleury, Tunisia. mailto:jamil.satouri@yahoo.fr