Error analysis of thermal equation with moving ends

Available online at www.sciencedirect.com Mathematics and Computers in Simulation 80 (2010) 2200–2215 Error analysis of thermal equation with moving ends M.A. Rincon a,∗ , M.O. Lacerda b a Instituto de Matemática, Universidade Federal do Rio de Janeiro, Caixa Postal 68530, Rio de Janeiro 21945-970, Brazil b Universidade Federal da Paraíba, Campus I, Cidade Universitária, CEP 58.051-900, Jo ao Pessoa, PB, Brazil Available online 24 April 2010 Abstract A mathematical model of the linear thermodynamic equations with moving ends, based on the Stefan Problem is considered. In this work, we are interested in obtaining existence, uniqueness and regularity using the Galerkin method and mainly to establish an error estimates of solutions in Sobolev spaces for the semi-discrete problem, with discretization of space variable, continuous time and for the completely discrete problem with discretization of space variable and time. © 2010 IMACS. Published by Elsevier B.V. All rights reserved. AMS Classification: 35L05; 35F30 Keywords: Stefan Problem; Moving boundary; Error estimates 1. Introduction One wide variety of problems with moving ends has been studied by many years. The thermal equations with moving ends give origin to various physical problems, such as, Stefan Problem, who can be interpreted as flame propagation problem, study of atom movements, study of combustion theory. Flame propagation problems have been established in [3] that studied a mathematical formulation suitable in the analyze for the theory of diffusion flames. Studies the mathematical model of moving ends arising in combustion theory was developed by Schmidt-Lainé and co-workers [2]. The discharge or absorption of thermal energy during the change in atom arrangement have been studied in [12] and [9]. The objective of this paper is to obtain existence, uniqueness and regularity of solutions and mainly to establish an error estimates of solutions in Sobolev spaces for the semi-discrete for the continuous and semi-discrete solutions of the linear thermodynamic model with moving ends, whose model is based on the Stefan Problem (see [14]) given by, ⎧ ⎪ ∂u ∂2 u ⎪ ⎪ − 2 = 0, ⎨ ∂t ∂x u(x, 0) = ϕ(x), ⎪ ⎪ ⎪ ⎩ u(s(t), t) = ρ(t), ∗ 0 < x < s(t) u(0, t) = ψ(t), ∀t > 0 Corresponding author. E-mail addresses: [email protected] (M.A. Rincon), [email protected] (M.O. Lacerda). 0378-4754/$36.00 © 2010 IMACS. Published by Elsevier B.V. All rights reserved. doi:10.1016/j.matcom.2010.04.013 M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 2201 where u(x, t) is the temperature and s(t) is the solution of equation:  −λ(t)s′ (t) + ux (s(t), t) = h(t) s(0) = s0 , must be determined and ψ, ρ, h, λ, ϕ, s0 are initial date of the problem defined for t ≥ 0 and ϕ(x) for 0 ≤ x ≤ s0 . 2. Formulation problem Our objective is study the approximated solutions of the following thermodynamics system, ⎧ ⎪ ∂u ∂2 u ⎪ ⎪ − k 2 = f (x, t), (x, t) ∈ Qt ⎨ ∂t ∂x (I) u(α(t), t) = 0, u(β(t), t) = 0, ∀x ∈ (α(t), β(t)), ∀t ∈ [0, T ] ⎪ ⎪ ⎪ ⎩ u(x, 0) = u (x), α(0) < x < β(0) 0 (1) where k > 0 is the thermal conductivity. We consider the domain Qt ⊂ R2 defined by Qt = {(x, t) ∈ R × (0, T ); α(t) < x < β(t)} and the horizontal length of the interval is defined by γ(t) = β(t) − α(t) > 0. Let us study the existence, uniqueness, and the approximated solution of the Problem (1). For this we consider the following hypothesis: H1: α, β ∈ C2 (0, ∞) in 0 < γ0 < γ(t) < γ1 , H2: α′ , β′ ∈ L∞ (0, ∞), ∀t ≥ 0 ∀t ≥ 0, where γ0 and γ1 are positive constants. Consider now the change of variable to transform the domain Qt in a cylindric domain Q, given by τ : Qt → Q = (0, 1)×]0, T [   x − α(t) (x, t) → (y, t) = ,t . γ(t) Note that τ ∈ C2 (Qt ) and by inverse function Theorem, the application τ −1 also is C2 (Q) (see for instance [7,10,11]). The change of variable v(y, t) = u(α(t) + γ(t)y, t), implies that u(x, t) = v (x − α(t)/γ(t), t). Using the application τ, we obtain the following cylindrical problem: ⎧ ∂2 v ∂v ⎪ ⎪ ⎪ v′ − a(t) 2 − b(y, t) = g(y, t), in Q ⎨ ∂y ∂y (2) (II) v(0, t) = 0, v(1, t) = 0, t ≥ 0 ⎪ ⎪ ⎪ ⎩ v(y, 0) = v0 (y), 0 < y < 1. where a(t) = k/γ 2 , b(y, t) = (α′ + γ ′ y)/γ and f (x, t) = f T−1 (y, t) = g(y, t), u(x, t) = u T−1 (y, t) = v(y, t). ′ For convenience, we have also used the prime ( ) to denote the derivative with respect to time t and ((.)), (.), . and |.|, the scalar product and the norms in H01 (0, 1) and L2 (0, 1), respectively. We shall show the existence and uniqueness from the solution of the cylindrical Problem (2) who implies in the existence and uniqueness of the solution of the noncylindrical Problem (1), since they are equivalent. 2202 M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 3. Existence and uniqueness We will investigate global existence and uniqueness solution for (2) and consequently for (1). For this, consider the notation = (0, 1), t = (α(t), β(t)) and 0 = (α(0), β(0)). Theorem 1. Under the hypotheses (H1) and (H2) and the initial data v0 ∈ H01 ( ) and g ∈ L1 (0, ∞; L2 ( )) ∩ L2 (0, T ; L2 ( )), there exists only one strong solution v : Q → R of the problem (I I), that is, v′ − a(t) ∂v ∂2 v − b(y, t) = g(y, t) ∂y2 ∂y in L2 (0, T ; L2 ( )), satisfying the following conditions: (i) v ∈ L2 (0, T ; H01 ( ) ∩ H 2 ( )) ∩ L∞ (0, T ; H01 ( )) (ii) v′ ∈ L2 (0, T ; L2 ( )). And consequently we obtain the following result. Theorem 2. Under the hypotheses (H1) and (H2) and the initial data u0 ∈ H01 ( 0 ) and f ∈ L1 (0, ∞; L2 ( L2 (0, ∞; L2 ( t )), there exists only one strong solution u : Qt → R of the problem (I), that is, u′ − k ∂2 u = f (x, t) ∂x2 in L2 (0, ∞; L2 ( t) ∩ t )) satisfying the following conditions: (i) u ∈ L∞ (0, ∞; H01 ( t ) ∩ H 2 ( (ii) u′ ∈ L2 (0, ∞; L2 ( t )). t )), 3.1. Proof of the Theorem 1 To prove the theorem, we introduce the approximate solutions. Let T > 0 and denote by Vm the subspace spanned by {w1 , w2 , . . . , wm }, where {wν ; nu = 1, · · ·m} are the first m eigenvectors of the space H01 ( ), solution of the spectral problem −∂2 wi /∂x2 = λi wi . If vm (t) ∈ Vm then it can be represented by m vm (y, t) = gνm (t)wν (y), ν=1 where vm is the solution of the system of ordinary differential equations     ⎧ ⎨ (v′ , w) + a(t) ∂vm , ∂w − b(y, t) ∂vm , w = (g, w), ∀w ∈ V m m ∂y ∂y ∂y ⎩ vm (0) = v0m → v0 in H01 ( ). (3) The system (3) has local solution in the interval (0, Tm ), by Carathéodory Theorem. To extend the local solution to the interval (0, T ) independent of m the following a priori estimate is needed. Estimate I: Taking w = vm ∈ Vm in (3), we obtain     ∂vm ∂vm ∂vm , , vm = (g, vm ). − b(y, t) (v′m , vm ) + a(t) (4) ∂y ∂y ∂y From the first term of (4) we have (v′m , vm ) = 1 d |vm |2 . 2 dt (5) M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 Moreover,   ∂vm ∂vm ∂vm , = a(t) a(t) ∂y ∂y ∂y 2203 2 = a(t) vm 2 , (6) where we have used the equivalence of norms in the space H01 ( ). Integrating the third term of (4), by parts and using the boundary conditions, we have,  1 ∂ 1  γ′ 1 ∂vm ∂vm vm = − (b(y, t)vm )vm = − vm + b(y, t) vm b(y, t) ∂y γ ∂y 0 ∂y 0 0 =− 1 γ′ γ 0 (vm )2 − 1 b(y, t) 0 ∂vm 1 γ′ vm = − |vm |2 . ∂y 2γ We also have by Cauchy–Schwarz inequality  1 1 2 |g| + |vm |2 (g, vm ) = gvm ≤ 2 0 (7) (8) Substituting (5)–(8) into (4), we have 1 d |vm |2 + a(t) vm 2 dt 2 + 1 γ′ 1 |vm |2 ≤ (|g|2 + |vm |2 ). 2γ 2 Integrating from [0, t), with t ∈ [0, Tm ), we get 1 |vm (t)|2 + a(t) 2 t vm 0 2 ≤ 1 1 |vm (0)|2 + |g|2 + 2 2 t 0 1 2  γ ′ ∈ L∞ (0, ∞), From the hypotheses (H1) and (H2), then    t 1 t ′ ′ 1 γ γ 1+ 1+ |vm |2 ≤ c |vm |2 ≤ γ γ0 0 2 0 2 where c is a positive constant. Let c1 = min{1/2, a(t)}, then |vm (t)|2 + t vm 2 0 ≤ 1 c 1 |vm (0)|2 + |g|2 + 2c1 2c1 c1 t 0 T 0 1+ γ′ γ  |vm |2 . |vm |2 . |vm |2 . By hypothesis, vm (0) = v0m → v0 ∈ H01 ( ) ⊂ L2 ( ), g ∈ L2 (0, T ; L2 ( )). Then |vm (t)|2 + t vm 0 2 T ≤ c2 + c2 0 |vm |2 , where c2 = 2c11 max{v0 + |g|2 , 2c}. By the use of Gronwall’s Lemma in the last inequality, we obtain the following estimates vm is bounded in L∞ (0, T ; L2 ( )), vm is bounded in L2 (0, T ; H01 ( )). Estimate II: Taking w = v′m in (3), we obtain     ∂vm ∂v′m ∂vm ′ ′ ′ , ,v (vm , vm ) + a(t) − b(y, t) = (g, v′m ), ∂y ∂y ∂y m The second term on left hand side in (10) implies that   a(t) d ∂vm ∂v′m a(t) d ∂vm 2 = a(t) , vm 2 , = ∂y ∂y 2 dt ∂y 2 dt (9) ∀v′m ∈ Vm . (10) (11) 2204 M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 by equivalence of norms in H01 ( ). As 0 < y < 1 and by hypothesis (H1) γ0 < γ(t), we obtain b(y, t) = α′ + γ ′ y |α′ | + |γ ′ | ≤ . γ γ0 Thus the third term in (10) can be estimated as, b(y, t) ∂vm ∂y |α′ | + |γ ′ | ∂vm γ0 ∂y |v′m | ≤ |v′m | ≤ |α′ | + |γ ′ | vm |v′m |. γ0 (12) Using the Cauchy–Schwarz inequality and substituting (11) and (12) in the Eq. (10), we get |v′m |2 + a(t) d vm 2 dt 2 |α′ | + |γ ′ | vm |v′m | + |g||v′m |. γ0 ≤ Consider the Young inequality  1 b2 β √ 2ab ≤ βa2 + , β β ∀a, b ∈ R, |α′ | + |γ ′ | vm γ0 |v′m | ∀β > 0. For β = 4, we have |v′m | + |g| ≤4  Therefore, 1 ′ 2 a(t) d |v | + vm 2 m 2 dt 2 ≤2  (13) |α′ |2 + |γ ′ |2 |γ0 |2 |α′ |2 + |γ ′ |2 |γ0 |2  vm 2  2 vm 1 + 2|g|2 + |v′m |2 . 2 + 2|g|2 . We observe that, from the hypothesis (H1), γ(t) > γ0 , ∀t ∈ [0, T ]. Integrating in 0 < t < Tm , using the hypothesis (H1) and the definition a(t), we obtain t 1 2 0 |v′m |2 + k 2γ12 t 0 d vm dt 2 |α′ |2 + |γ ′ |2 vm |γ|2 T ≤4 0 2 T +2 0 |g|2 . Note that, from hypothesis (H2), T 4 0 |α′ |2 + |γ ′ |2 vm |γ|2 2 T ≤ ĉ vm 2 , 0 where ĉ is a positive constant. t 0 |v′m |2 + vm (t) 2 ≤ k vm (0) 2c3 γ12 where c3 = max{1/2, k/(2γ12 )}. By hypothesis g ∈ L2 (0, T ; L2 ( t 0 |v′m |2 + vm (t) 2 ≤c+c 2 + 2 c3 T 0 |g|2 + ĉ c3 T vm 2 . 0 )), then there is a positive constant c, such that T vm 2 . 0 By Gronwall inequality we have vm is bounded in L∞ (0, T ; H01 ( )), v′m is bounded in L2 (0, T ; L2 ( )). Estimate III: Taking w = −∂2 vm /∂y2 in (3), and after to integrate the second therm by parts we have    2      ∂ vm ∂ 2 vm ∂vm ∂2 vm ∂2 vm ∂2 vm − v′m , , , + a + b(y, t) = − g, . ∂y2 ∂y2 ∂y2 ∂y ∂y2 ∂y2 (14) (15) M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 The first term on left hand side in (15) implies that   1 ∂v′ ∂v 1 d 1 ∂ 2 vm ∂ 2 vm m m v′m 2 = vm 2 . =− − v′m , 2 ∂y ∂y 0 ∂y ∂y 2 dt 0 2205 (16) The second term on left hand side in (15) give us a(t)  ∂ 2 vm ∂ 2 vm , ∂y2 ∂y2  ∂ 2 vm = a(t) ∂y2 2 2 k ∂ 2 vm ≥ 2 . γ1 ∂y2 (17) As 0 < y < 1 and from the hypothesis (H1), we have  α′ + γ ′ y γ  ∂vm ∂2 vm , ∂y ∂y2  1 α′ + γ ′ y ∂vm ∂2 vm ≤ γ ∂y ∂y2 ≤ 0  |α′ | + |γ ′ | γ0  vm ∂2 vm . ∂y2 Taking β = 4γ12 /k in the inequality (13) we obtain,  |α′ | + |γ ′ | γ0  vm 4γ12 ∂ 2 vm ≤ ∂y2 k  |α′ | + |γ ′ | γ0 2 2 2 vm + k ∂2 vm . 4γ12 ∂y2 (18) Similarly we have ∂2 vm − g, ∂y  2  ≤ |g| 4γ12 2 ∂ 2 vm k ∂ 2 vm |g| ≤ . + ∂y2 k 4γ12 ∂y2 (19) Substituting (16)–(19) in the Eq. (15), we obtain d vm dt 2 + k ∂ 2 vm γ12 ∂y2  8γ12 k 2 ≤ 8γ12 k  |α′ | + |γ ′ | γ0 2 vm 2 +  8γ12 2 |g| ≤ C vm k 2  + |g|2 , where C = max  |α′ | + |γ ′ | γ0 2 8γ 2 , 1 k  Integrating from 0 to t < Tm we get t vm (t) 2 k + 2 γ1 ∂ 2 vm ∂y2 T 2 ≤C T vm 2 0 0 |g|2 + vm (0) 2 . +C 0 As g ∈ L2 (0, T ; L2 ( )) and vm (0) → v0 in H01 ( ) we have that t vm (t) 2 + ∂ 2 vm | 2 |2 ≤ c̄ + c̄ ∂y 0 T vm 2 . 0 By Gronwall inequality we conclude that vm ∂2 vm ∂y2 is bounded in L∞ (0, T ; H01 ( )) is bounded in L2 (0, T ; L2 ( )). (20) 2206 M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 From the estimates (9)2 , (14)2 and (20)2 we conclude the existence of a subsequence of (vm ), yet denoted by (vm ) such that ∂vm ∂v → weak in L2 (0, T ; L2 ( )) ∂y ∂y v′m → v′ weak in L2 (0, T ; L2 ( )) (21) ∂2 vm ∂2 (v) → 2 ∂y ∂y2 weak in L2 (0, T ; L2 ( )). The passage to the limit and the uniqueness follows by standard way. 3.2. Proof of the Theorem 2 Let v the solution of the Problem (2), with the following initial data v0 (y) = u0 (α(0) + γ(0)y). Consider the function u(x, t) = v(y, t), where x = α(t) + γ(t)y. To verify that u(x, t), under the hypotheses of Theorem 2, is the solution of Problem (1), it is sufficient to observe that the mapping τ : Qt → Q is of class C2 . Since the Problems (1) and (2) are equivalently, then u satisfy the Problem (1). The regularity of v(y, t) given by Theorem 1, implies the regularity of u(x, t) solution of the Problem (1) and the uniqueness of solutions of the Problem (1) is a direct consequence of uniqueness of the Problem (2). 4. Regularity In the next theorem, we shall also prove the following regularity result, under further requirements of initial conditions and the regularity for α(t) and β(t), given by β ∈ W k,∞ (0, T ), H3 : α, 0 < γ0 < γ(t) < γ1 , ∀t ≥ 0 So we establish the following regularity result: Theorem 3. Under the hypotheses (H2) and (H3) and given the initial data d k v(0)/dt k ∈ (H01 (0, 1) ∩ H 2 (0, 1)) and g ∈ H k (0, T ; H k (0, 1)), then there exists a unique solution v : Q → R, satisfying the following conditions v ∈ L2 (0, T ; H0k+1 (0, 1) ∩ H k+2 (0, 1)), for k = 0, 1, · · · . Proof. Consider the cylindrical heat equation (2)1 , v′ − a(t) ∂v ∂2 v − b(y, t) = g(y, t) ∂y2 ∂y (22) Differentiating the Eq. (22) with relation to t, we have v′′ − av′yy − bv′y = g′ + b′ vy + a′ vyy (23) Here, we suppress the dependence of the variables (y, t). Multiplying (23) by (−v′yy ) and integrating in y from 0 to 1, we obtain −(v′′ , v′yy ) + a|v′yy |2 = (−g′ , v′yy ) + (bv′y , v′yy ) + (b′ vy , v′yy ) + a′ (vyy , v′yy ). Then, follows that 1 d ′ 2 a |v | + a|v′yy |2 ≤ |v′yy |2 + C1 |g′ |2 + C2 |v′y |2 + C3 |vy |2 + C4 |vyy |2 2 dt y 2 that is |v′y |2 + a 2 T 0 |v′yy |2 ≤ C0 + C1 T 0 |g′ |2 + C2 T 0 |v′y |2 (24) 2207 M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 where C0 = C0 (|v′y (0)|2 , T |vyy |2 , 0 T |vy |2 ). 0 We conclude by Gronwall inequality that is bounded in L2 (0, T ; L2 (0, 1)) v′yy v′y is bounded in L∞ (0, T ; L2 (0, 1)) its follows that v′ is bounded in L2 (0, T ; H 2 (0, 1)) ∩ L∞ (0, T ; H01 (0, 1)). Differentiating the Eq. (22) with relation the variable y, we have −avyyy = bvyy + gy − v′y + by vy ∈ L2 (0, T ; L2 (0, 1)) (25) therefore vyyy ∈ L2 (0, T ; L2 (0, 1)), and we conclude that v ∈ L2 (0, T ; H 3 (0, 1)). (26) ′′ Differentiating (23) with relation to t and after multiplying by −vyy and integrating from 0 to 1,with relation the variable y, we obtain ′′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ −(v , vyy ) + a|vyy |2 = (−g′′ , vyy ) + (bvy , vyy ) + (b′ v′y , vyy ) + (b′′ vy , vyy ) + a′ (v′yy , vyy ) + a′′ (vyy , vyy ). Its follows that a ′′ 1 d ′′ 2 ′′ ′′ |vy | + a|vyy |2 ≤ |vyy |2 + C1 |g′′ |2 + C2 |vy |2 + C3 |v′y |2 + C4 |vy |2 + C5 |v′yy |2 + C6 |vyy |2 . 2 dt 2 Then, after integration in [0, T ], becomes ′′ |vy |2 + a 2 T 0 T ′′ |vyy |2 ≤ C0 + C1 0 |g′′ |2 + C2 T 0 ′′ |vy |2 (27) where C0 = C0  |v′y (0)|2 , T 0 |v′y |2 , T 0 T 2 |vy | , 0 |v′yy |2 , T 0 |vyy | 2  . We conclude by Gronwall inequality that ′′ vyy ′′ vy is bounded in L2 (0, T ; L2 (0, 1)) is bounded in L∞ (0, T ; L2 (0, 1)) then v′′ is bounded in L2 (0, T ; H 2 (0, 1)) ∩ L∞ (0, T ; H01 (0, 1)). Differentiating the Eq. (23) with relation the variable y, we have ′′ −av′yyy = bv′yy + by v′y − vy + gy′ + b′ v′yy + by′ v′y + a′ vyyy ∈ L2 (0, T ; L2 ). Then v′ ∈ L2 (0, T ; H 3 (0, 1)). Now, differentiating the Eq. (25) with relation the variable y, we have −avyyyy = bvyyy + 2by vyy + gyy − v′yy ∈ L2 (0, T ; L2 ), where we have used that byy = 0. Then v ∈ L2 (0, T ; H 4 (0, 1)), (28) we also have −avyyyyy = bvyyyy + 3by vyyy + gyyy − v′yyy + v′y ∈ L2 (0, T ; L2 (0, 1)) (29) 2208 M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 therefore v ∈ L2 (0, T ; H 5 (0, 1)). (30) By induction we can see that −a ∂k+1 v ∂b ∂k v ∂k g ∂k v′ ∂k+2 v = b k+1 + k + k + k ∈ L2 (0, T, L2 (0, 1)) k+2 ∂y ∂y ∂y ∂yk ∂y ∂y so v ∈ L2 (0, T ; H k+2 (0, 1)), this prove the theorem.  In the next two sections, in order to achieve an error estimates for the continuous and discrete times, we will require that γ(t) is increasing, i.e, H4 : γ ′ (t) > 0, ∀t ∈ [0, T ]. 5. Error estimates—continuous time In this section, we consider error bounds for finite element approximations for the cylindrical heat equation, when the time remains continuously variable. The cylindrical heat equation is given by, ⎧ 2 ⎪ ⎪ v′ − a(t) ∂ v − b(y, t) ∂v = g(y, t), in Q = (0, 1) × (0, T ) ⎪ ⎨ ∂y2 ∂y (31) v(0, t) = 0, v(1, t) = 0, ∀t ∈ [0, T ] ⎪ ⎪ ⎪ ⎩ v(y, 0) = v0 (y), 0<y<1 which the variational formulation of the problem, as before considered, is given by ⎧ ∂v ∂v ∂w 1 ′ ⎪ ⎪ ⎨ (v , w) + a(t)( ∂y , ∂y ) − (b(y, t) ∂y , w) = (g(y, t), w), ∀w ∈ H0 (0, 1) v(0, t) = 0, v(1, t) = 0, ⎪ ⎪ ⎩ (v(0), w) = (v0 , w). ∀t ∈ [0, T ] (32) 5.1. The semi-discrete problem We now present a semi-discrete formulation for problem (32), using the Galerkin finite element method to discretize the space variable. Let {Th } be a family of polygonalization Th = {K} of , satisfying the minimum angle condition (see for instance [5]) and indexed by the parameter h representing the maximum diameter of elements K ∈ Th . For a given integer l ≥ 1, we introduce the finite element space   Vmk = qh ∈ C0 (0, 1); qh |K ∈ Pl (K), ∀ K ∈ Th , where Pl (K) is the set of polynomials on K of degree less than or equal to l, i.e., Vmk is the space of piecewise continuous polynomial functions of degree l. By standard interpolation theory (see for instance [5]), it follows that given a function w : (0, T ) → H k+1 (0, 1) there exists an interpolator ŵh : (0, T ) → Vmk such that w(t) − ŵh (t) m ≤ C1 hk+1−m w(t) k+1 , m≤k where · m denotes the usual semi-norm in the Hilbert space H m (0, 1). The Galerkin finite element semi-discrete approximation of problem (32) in Vmk is given by Find vh : (0, T ) → Vmk such that (33) M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 ⎧     ∂vh ∂wh ∂vh ⎪ ′ ⎪ ⎪ ⎨ (vh , wh ) + a(t) ∂y , ∂y − b(y, t) ∂y , wh = (g(y, t), wh ) , vh (0, t) = vh (1, t) = 0, ∀t ∈ [0, T ] ⎪ ⎪ ⎪ ⎩ (vh (0), wh ) = (v0h , wh ) ∀wh ∈ Vmk Subtracting (32) and (34) we obtain, ⎧       ∂v ∂vh ∂v ∂vh ⎪ ′ ′ ⎪ ⎪ ⎨ (v − vh , wh ) + a(t) ∂y − ∂y , wh − b(y, t) ∂y − ∂y , wh = 0 v(0, t) − vh (0, t) = 0, v(1, t) − vh (1, t) = 0 ⎪ ⎪ ⎪ ⎩ (v(0) − vh (0), wh ) = (v0 − v0h , wh ) Let â(v, w) be bilinear forms in H01 (0, 1) defined by     ∂v ∂w ∂v â(v, w) = a(t) − b(y, t) , w , , ∂y ∂y ∂y 2209 (34) (35) (36) then we have the following elliptic problem associated, ∀w ∈ H01 (0, 1). â(v, w) = (g, w), Note that 1 â(v, v) = a(t) 0 2 ∂v ∂y 1 dy − b(y, t) 0 ∂v v dy ∂y From (7), the second therm can be written in the form 1 b(y, t) 0 ∂v v dy = − ∂y 1 0 γ′ 2 |v| dy, 2γ it follows that 1 â(v, v) = a(t) 0 2 ∂v ∂y dy + γ′ 2γ 1 0 |v|2 dy. (37) Using the hypothesis (H4) we conclude that â(v, v) is coercive, i.e, â(v, v) ≥ a(t) ∂v 2 . ∂y On other hand â(v, w) is continuous. By Lax–Milgram Theorem, there exists only one v ∈ V = H01 (0, 1), such that â(v, w) = (g, w), ∀w ∈ V . Considerer, see for instance, Wheeler [15], the Rayleigh–Ritz Projection: P : V → Vmk , for each fixed t ∈ (0, T ), where Pv(t) = ṽ(t) satisfying, â(v(t) − ṽ(t), wh ) = 0, ∀wh ∈ Vmk . (38) The function ṽ(t) can be viewed as an interpolator of v(t). Since that, ṽ(t) ∈ Vmk then it can be written in the form m ṽ(t) = vim (t)ϕi (y), where (ϕi ) is the basis of Vmk . i=1 Let e(t) = v(t) − vh (t), t ∈ [0, T ], (39) 2210 M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 be the error associated with the semi-discrete problem (34). We will establish the error estimates in the H 1 (0, 1) and L2 (0, 1)-norms, due to the space-discretization. For this, we split the Eq. (39) into two parts, e(t) = v(t) − vh (t) ≤ v(t) − ṽ(t) + ṽ(t) − vh (t) = ρ(t) + θ(t), (40) where ρ(t) = v(t) − ṽ(t) and θ(t) = ṽ(t) − vh (t). From the inequality (33) and equality (38) it is possible to make the error estimates for ρ(t) ρ(t) m ≤ C1 hk+1−m v(t) (41) k+1 . Douglas and Dupont [6], has proved that the results still are valid for ρ′ (t) and ρ′′ (t), when v(t), v′ (t) and v′′ (t) belongs to H k+1 (0, 1) space. In order to achieve a convergence rate according to the error estimate for H k+1 for the problem (II), we require the following additional hypotheses for the interpolators of the initial data, i.e, H5 : vh (0) − v̂0 m ≤ C̄hk+1−m v(0) k+1 ; v′h (0) − v̂′0 m ≤ Ĉhk+1−m v′ (0) k+1 , where v̂0 and v̂′0 are the interpolators of the initial conditions v0 and v′0 and C̄ and Ĉ are positive constants independent of h. 5.2. L2 (0, 1) error estimate—time continuous Theorem 4. Under the hypotheses (H2)–(H5) and given the initial data d k v(0)/dt k ∈ (H01 (0, 1) ∩ H 2 (0, 1)) and g ∈ H k (0, T ; H k (0, 1)), then the approximate solution uh obeys the following error estimate: |v − vh |L∞ (0,T ;L2 (0,1)) ≤ Ch(k+1) (42) where C is positive constants independent of h. Proof. Summing and subtracting ṽ = ṽ(t) in (35) and using (40), we obtain:  (ρ′ + θ ′ , wh ) + a(ρ + θ, wh ) = 0 (ρ(0) + θ(0), wh ) = (v0 − v0h , wh ). (43) From (38) we have that â(ρ(t), wh ) = 0. Hence we have, (θ ′ , wh ) + (ρ′ , wh ) + â(θ(t), wh ) = 0. (44) In particular, taking wh = θ in (44), we obtain; (θ ′ , θ) + (ρ′ , θ) + â(θ, θ) = 0. But, â(θ, θ) ≥ k ∂θ 2 , γ12 ∂y (θ ′ , θ) = 1 d 2 |θ| 2 dt and | − (ρ′ , θ)| ≤ where, we have used the Cauchy–Schwarz inequality in last term. Integrating from 0 to t we have that |θ|2 + 2 k γ12 t 0 ∂θ ∂y By Poincaré inequality, |θ|2 ≤ Cp ∂θ ∂y 2 2 ≤ |θ(0)|2 + 2β T 0 |θ|2 + 1 2β T 0 |ρ′ |2 1 ′2 |ρ | + β|θ|2 . 4β M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 2211 then choosing β = k/(2γ12 Cp ) > 0, we have 2β|θ|2 ≤ 2 k ∂θ γ12 ∂y Therefore k |θ| + 2 γ1 t 2 0  2 ∂θ ∂y γ 2 Cp |θ(0)| + 1 k ≤ T 2 ′ 2 |ρ | 0  . Since that the second term on left hand side is positive, then   2C T γ p |θ|2 ≤ |θ(0)|2 + 1 |ρ′ |2 . k 0 (45) From (33) and hypothesis (H5), with m = 0, we have that |θ(0)| = |v̂(0) − vh (0)| ≤ Chk+1 v(0) k+1 |ρ′ (t)| = |v′ (t) − ṽ′ (t)|2 ≤ C1 hk+1 v(t) Substituting in (45), we obtain,  |θ(t)|2 ≤ C2 h2(k+1) v(0) 2 k+1 k+1 . 2 L2 (0,T ;H k+1 ) + v  , where C2 is a positive constant independent of h. Now, combining the error definition (40), the interpolation results (33) and using the triangle inequality we arrive at |e|L∞ (0,T ;L2 (0,1)) = |v − vh |L∞ (0,T ;L2 (0,1)) ≤ C h(k+1) .  5.3. H01 (0, 1) error estimate—time continuous Consider now in (44), wh = θ ′ ∈ Vmk . Then |θ ′ |2 + â(θ, θ ′ ) + (ρ′ , θ ′ ) = 0. (46) For the second term, using the definition of â, we obtain â(θ, θ ′ ) = 1 0 a(t) ∂θ ′ 2 ∂y 2 d dt 1 1 − b(y, t) 0 ∂θ ′ θ. ∂y From a(t) definition, 1 0 a(t) ∂θ ′ 2 ∂y 2 = 0 k ∂θ 2γ 2 ∂y 2 1 − 0 kγ ′ ∂θ γ 3 ∂y 2 Substituting the last two terms in (46), we have, 1 d dt 0 k ∂θ 2γ 2 ∂y 2 + |θ ′ |2 = 1 0 kγ ′ ∂θ γ 3 ∂y 2 1 + |θ ′ |2 + b(y, t) 0 ∂θ ′ θ − (ρ′ , θ ′ ). ∂y Integrating the 0 to T, using hypotheses (H3) and Young inequality conveniently, as before (see (13)), we have 1 0 k ∂θ γ 2 ∂y 2 T + 0 |θ ′ |2 ≤ k γ02 1 0 ∂θ(0) ∂y 2 1 T +C 0 0 ∂θ ∂y 2 + 1 2 T 0 |θ ′ |2 + T 0 |ρ′ |2 2212 M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 where C = C(|α′ |∞ , |γ ′ |∞ , γ0 ). Then we obtain 1 k γ12 0 ∂θ ∂y 2 + T 1 2 0 |θ ′ |2 ≤ k γ02 1 ∂θ(0) ∂y 0 2 1 T +C 0 0 ∂θ ∂y 2 T + 0 |ρ′ |2 It follows that by Gronwall inequality that   1 ∂θ 2 γ12 ∂θ(0) 2 γ12 T ′ 2 (γ 2 /k)CT |ρ | e 1 ≤ + k 0 γ02 ∂y 0 ∂y (47) and 1 2 T 0 1 k |θ ′ |2 ≤ 2 γ0 2 ∂θ(0) ∂y 0 T + 0 |ρ′ |2 + CT  γ12 ∂θ(0) γ02 ∂y 2 γ2 + 1 k T 0 |ρ′ |2  2 e(γ1 /k)CT (48) Using the fact that ∂v(0) ∂θ(0) ≤ chk ∂y ∂y and |ρ′ | ≤ chk+1 v(t) ≤ C1 h k k+1 k+1 , then, we conclude that θ L2 (0,T ;H01 ) + θ ′ L2 (0,T ;L2 )  ∂v(0) ∂y k+1 + v L2 (0,T ;H k+1  (49) Finally, combining the error definition (40), the interpolation results (33) and using the triangle inequality we arrive at   ∂v(0) ′ k e L2 (0,T ;H 1 ) + e L2 (0,T ;L2 ) ≤ C1 h (50) + v L2 (0,T ;H k+1 . 0 ∂y k+1 6. Error estimates—discrete time In this section, we consider error bounds for finite element approximations for the cylindrical heat equation, when the time varies discretely. The fixed continuous time interval [0, T ] considered before, is divided into N = T/t discrete time steps of uniform size t. Then the system of ordinary differential equations governing the finite element approximations with continuous time is replaced by a system of algebraic equations by using Crank–Nicolson method. We denote the discrete time in the interval [0, T ] by t n = nt, where n is an integer ranging from 0 to N. In general, let W be a function depending on the time variable t ∈ [0, T ]. Then we denote the values of W at the discrete times t n by W n . Using this notation we define wn+1/2 = (wn+1 + wn )/2 and 1 δwn+ 2 = (wn+1 − wn )/t. Suppose that w is a time-dependent vector field on norms of w; by . Then when the time varies discretely we define the following N−1 w 2 L̂2 (0,T ;H) w = t L∞ (0,T ;H) L2 ( N−1 wn 2H , w 2 L̃2 (0,T ;H) n=0 n = max{ w H 1( H, H 2( (51) = t 1 wn+ 2 2 H, n=0 n = 0, 1, · · · , N − 1}. (52) where H may be ), ) or ). As before we decompose the finite element error e = v − vh into two parts: e = v − vh = (v − ṽ) + (ṽ − vh ) = ρ + θ, (53) where ρ = v − ṽ, θ = ṽ − vh and ṽ denotes the projection of the exact solution v into Vmk with respect to the energy inner product, i.e, (v − ṽ) satisfies (38). We prove first the following preliminary estimate:   θ L∞ (0,T ;H 1 ) ≤ C hk ( v0 k+1 + v′ L2 (0,T ;H k+1 ) ) + t 2 v′′ L2 (0,T ;H k+1 ) (54) M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 2213 1 We evaluate the weak formulation problem (32) at time step t n+ 2 , obtaining n+ 21 1 1 ∀w ∈ H01 (0, 1) , w)0 + â(vn+ 2 , w) = (gn+ 2 , w)0 , (vt 1 Summing and subtracting the term δvn+ 2 1 1 1 ∀w ∈ H01 (0, 1) (δvn+ 2 , w)0 + â(vn+ 2 , w) = (gn+ 2 + f n , w)0 ,   1 n+ 1 where f n = δvn+ 2 − vt 2 . (55) Consider the discrete approximated system (34), given by n+ 21 (δvh n+ 21 , wh ) + â(vh 1 ∀wh ∈ Vmk . , wh ) = (gn+ 2 , wh ), (56) Since Vmk is a subspace of H01 (0, 1), setting w = wh in (55) and subtracting (56), we get 1 1 (δen+ 2 , wh )0 + â(en+ 2 , wh ) = (f n , wh )0 , 1 ∀wh ∈ Vmk (57) 1 Summing and subtracting δṽn+ 2 and ṽn+ 2 ∈ Vmk in first and second term respectively, and decomposing e into ρ and θ as before, we get 1 1 1 (δθ n+ 2 , wh )0 + (δρn+ 2 , wh )0 + â(θ n+ 2 , wh ) = (f n , wh )0 , 1 n+ where we have used the fact that; â(ρ 2 , wh ) = 0. 1 Taking in particular wh = θ n+ 2 in (58), we get 1 1 1 1 1 ∀wh ∈ Vmk 1 1 (δθ n+ 2 , θ n+ 2 ) + â(θ n+ 2 , θ n+ 2 ) = −(δρn+ 2 , θ n+ 2 ) + (f n , θ n+ 2 ) (58) (59) The first term on left hand side implies that 1 1 n+1 (θ − θ n , θ n+1 + θ n ) = ( θ n+1 2t 2t Since that â(.,.) is coercive, then 1 1 1 1 (δθ n+ 2 , θ n+ 2 ) = 1 â(θ n+ 2 , θ n+ 2 ) ≥ α θ n+ 2 where α = k/γ12 2 2 − θn 2 ) ≥ 0, (60) (61) ≥ a(t), by hypotheses (H3). We have by Young inequality that 1 1 1 1 n+ n+ 1 α n+ 2 (δρ 2 , θ 2 ) ≤ δρ 2 + θ 2 2 2α 2 1 1 n+ α n+ 1 n 2 n f 0+ θ 2 2 (f , θ 2 ) = 2α 2 Substituting (60)–(62) in (59), after to multiply by 2t, we obtain n+ 1 t n 2 t δρn+ 2 2 + f . α α Summing from n = 0 to n = N − 1 we have N−1  N−1 t N 2 0 2 n+ 21 2 n 2 θ ≤ θ + δρ + . f α θ n+1 2 − θn 2 ≤ n=0 (62) (63) (64) n=0 The interpolation results (33) with m = 1, and choosing the initial condition v0h and ṽ0 satisfying the error estimates, we have; θ 0 ≤ ρ0 + e0 ≤ C0 hk v0 k+1 . (65) 2214 M.A. Rincon, M.O. Lacerda / Mathematics and Computers in Simulation 80 (2010) 2200–2215 The second term on right hand side can be estimate by t α N−1 1 2 δρn+ 2 n=0 ≤ 1 α tN ρ′ (s) t0 2 2 L2 (0,T ;H 1 ) ds = C1 ρ′ ≤ C2 h2k v′ 2 L2 (0,T ;H k+1 ) (66) Now, we estimate the norm of the vector field f:   1 1 n+1 1 1 n + v ) = (v − vn ) − (vn+1 + vnt ) f n = δvn+ 2 − (vn+1 t 2 t t 2 t   tn+1 1 ′′′ = − (s − tn )(tn+1 − s)v (s) ds 2t tn Since that tn − s ≤ t and tn+1 − s ≤ t, we obtain N−1 2 L̂2 (0,T ;L2 ( )) f = t fn n=0 2 0 ≤ t 4 v ′′′ 2 L2 (0,T ;H k ) (67) Substituting (65)–(67) in (64), we obtain θN 2 ≤ h2k {C0 v0 2 k+1 + C 1 v′ 2 4 L2 (0,T ;H k+1 ) } + C3 t v ′′′ 2 L2 (0,T ;H k ) (68) where the constants C0 , C1 , C2 , C3 are independent of t. As usual, by using the equivalence of normas on a finite dimensional space, we can remove the square of norms   ′′′ θ N ≤ C4 hk v0 k+1 + hk v′ L2 (0,T ;H k+1 ) + t 2 v L2 (0,T ;H k+1 ) (69) Now since the right-hand side of (69) is independent of N, we obtain the preliminary estimate (54). Since that v, v′ , v′′ ∈ L1 (0, T, H k+1 ) then θ L∞ (0,T ;H 1 ) ≤ C(hk + t 2 ) From (33), we have for m = 1, ρ 1 ≤ C6 hk v In this manner, we can conclude the estimate e L∞ (0,T ;H 1 ) = v − vh = ρ L∞ (0,T ;H 1 ) L∞ (0,T ;H 1 ) + θ k+1 . = v − ṽ L∞ (0,T ;H 1 ) L∞ (0,T ;H 1 ) + ṽ − vh L∞ (0,T ;H 1 ) ≤ C(hk + t 2 ). Acknowledgement The author (MAR) acknowledges the partial support for research from CNPq of Brazil. References [2] M. Berstch, D. Hilhorst, C. Schmidt-Lainé, The Well-posedness of a Free-boundary Problem Arising in Combustion Theory, Publ. 21 Dép. Math. Ecc. Norm. Sup. Lyon, France, 1989. [3] J.D. Buckmaster, G.S.S. Ludford, Theory of Laminar Flames, Cambridge Univ. Press, Cambridge, 1982. 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