Homogeneização em Equações Diferenciais

Transcrição

1 Homogeneização em Equações Diferenciais Domínios finos com fronteira oscilante Marcone C. Pereira 1 2 PROGRAMA DE VERÃO EDP S E ANÁLISE FUNCIONAL INSTITUTO DE MATEMÁTICA E ESTATÍSTICA UNIVERSIDADE DE SÃO PAULO SÃO PAULO - BRASIL 1 Escola de Artes, Ciências e Humanidades - Universidade de São Paulo - São Paulo - Brasil 2 Partially supported by FAPESP 2008/ , CAPES DGU 127/07 and CNPq /

2 Referências: 1 J. M. Arrieta, A. N. Carvalho, M. C. Pereira and R. Parreira Silva; Nonlinear parabolic problems in thin domains with a highly oscillatory boundary, submitted. 3 2 J. M. Arrieta and M. C. Pereira; Homogenization in a thin domain with an oscillatory boundary, to appear in J. Math. Pures and Appl J. K. Hale and G. Raugel; Reaction-diffusion equation on thin domains, J. Math. Pures and Appl. (9) 71, no. 1, (1992). 4 Prizzi and Rybakowski; The effect of domain squeezing upon the dynamics of reaction-diffusion equations, J. of Diff. Equation 173, no. 2, (2001)

3 We are interested in studying the asymptotic behavior of w + w = f R w N = 0 R where R is a thin domain R = {(x, y) : 0 < x < 1; 0 < y < G (x)}. We suppose G(x, y) of class C 1, L-periodic in y, G (x) = G(x 1, x 1 /), 0 < G 0 G (x) G 1.

4 We divide the analysis of the problem in three cases: 1 the purely periodic case, G(x, y) = g(y); 2 the piecewise periodic case, G(x, y) is piecewise constant in x; 3 the general case.

5 Purely periodic case. G (x) = g(x/) and f (x 1, x 2 ) = f (x 1 ). We need the following ingredients: The Multiple Scale method. 5 Extension operators. Oscillatory test functions method of Tartar 6 5 Bensoussan, Lions, Papanicolaou. Asymptotic Analysis for Periodic Structures. North-Holand (1978). 6 D. Cioranescu and J. Saint Jean Paulin, Homogenization of reticulated structures, Springer-Verlag (1998).

6 Multiple Scale Method We look for a formal asymptotic expansion of the form w (x 1, x 2 ) = w 0 (x 1, x 1, x 2 ) + w 1(x 1, x 1, x 2 ) + 2 w 2 (x 1, x 1, x 2 ) +... Here, we suppose w i (x, y, z) defined for x I = (0, 1) and (y, z) Y, L-periodic on the variable y where Y is the representative basic cell Y = {(y, z) R 2 : 0 < y < L, 0 < z < g(y)}. Note that R degenerates to a line segment when 0. It suggests that w tends not to depend on the macroscopic variable x 2. Hence, we are assuming that w i is an independent function of x 2.

7 We call x = x 1, y = x 1 /, z = x 2 /. Thus, we have 2 = x + 1 y, 2 = xx + 2 xy yy, = 1 z 2 2 = 1 2 zz. Also, taking into account the geometry of the thin domain R, we have N (x 1, x 2 ) = N(x 1 /, x 2 /), (x 1, x 2 ) R where N and N are the unit outward normal to R and Y respectively.

8 Using this approach, we obtain the following homogenized equation q d 2 w 0 dx 2 (x) + w 0(x) = f (x), x (0, 1) w 0(0) = w 0(1) = 0 where q = 1 { Y 1 X } Y y (y, z) dydz and X(y, z) is the unique solution (up to an additive constant) of y,z X(y, z) = 0 in Y (y, g(y)) = g (y) on B 1 1+(g (y)) 2 X N X N (y, 0) = 0 on B 2 X L periodic in variable y. We denote by B 0,B 1 y B 2 the lateral, inferior and superior boundary, respectively.

9 Convergence result To obtain ours results, we consider the equivalent problem 2 u 2 x 1 2 u x + u = f 2 in u N1 + 1 u 2 N2 = 0 on where is the rescaled oscillating domain = {(x 1, x 2 ) R 2 0 < x 1 < 1, 0 < x 2 < g(x 1 /)} and we use an extension operator P : H 1 ( ) H 1 () uniformly bounded in > 0 with = (0, 1) (0, G 1 ).

10 Its variational formulation is: find u H 1 ( ) such that { u ϕ + 1 u ϕ } x 1 2 +u ϕ dx 1 dx 2 = f ϕdx 1 dx 2, ϕ H 1 ( ). Taking ϕ = u, we get the following apriori estimates for u u 2 L 2 ( ) u 2 L 2 ( ) + u 2 L 2 ( ) f L 2 ( ) u L 2 ( ). Thus u L2 ( ), u L and 1 u L f 2 ( ) x L 2 2 ( ) 2 ( ) > 0.

11 Let be the standard extension by zero on = (0, 1) (0, G 1 ), then { u ϕ + 1 u ϕ } 2 +ũ x ϕ dx 1 dx 2 = χ f ϕ dx 1 dx 2 ϕ H 1 () 2 where χ is the characteristic function of. If f (x 1, x 2 ) = f (x 1 ), we can extract a subsequence such that ũ u u ξ u 0 w L 2 () w L 2 () and s L 2 ().

12 Now, let χ be the characteristic function of the representative cell Y extended periodically on variable y. ( ) χ x1 kl ( x1 ) (x 1, x 2 ) = χ, x 2 = χ, x 2 for some k N such that (y, z) = ( x 1 kl, x 2 ) Y. Hence χ (x 1, x 2 ) θ(x 2 ) := 1 L L 0 χ(s, x 2 )ds w L (I) as 0 for all x 2 (0, G 1 ). Then, we can get by Lebesgue s Dominated Convergence Theorem that χ θ w L ().

13 Now, we can pass to the limit { ξ ϕ } + u ϕ dx 1 dx 2 = θ f ϕ dx 1 dx 2 ( ) for all ϕ H 1 (0, 1). Can we establish a relation between u, ξ and θ? { u ϕ + 1 u ϕ 2 + ũ x ϕ } dx 1 dx 2 = χ f ϕ dx 1 dx 2 2 with ϕ H 1 () for all > 0.

16 Oscillatory test functions method of Tartar We seek to test functions ϕ = ϕ ω, ϕ(x 1 ) C0 (0, 1) and ω H 1 (), that allow us to pass the limit. = = χ f (ϕ ω )dx 1 dx 2 { u (ϕ ω ) u } (ϕ ω ) + ũ x (ϕ ω ) dx 1 dx 2 2 { u ϕ ω + u ω ϕ + 1 u ω 2 ϕ + ũ x (ϕ ω ) 2 Todavía existen algunos términos malos". Here, we will take a suitable zero, and we use a extension operator η η 2 = 0 en P L(H 1 ( ), H 1 ()) L(L 2 ( ), L 2 ()). } dx 1 dx 2

17 Oscillatory test functions method of Tartar We seek to test functions ϕ = ϕ ω, ϕ(x 1 ) C0 (0, 1) and ω H 1 (), that allow us to pass the limit. = = χ f (ϕ ω )dx 1 dx 2 { u (ϕ ω ) u } (ϕ ω ) + ũ x (ϕ ω ) dx 1 dx 2 2 { u ϕ ω + u ω ϕ + 1 u ω 2 ϕ + ũ x (ϕ ω ) 2 Todavía existen algunos términos malos". Here, we will take a suitable zero, and we use a extension operator η η 2 = 0 en P L(H 1 ( ), H 1 ()) L(L 2 ( ), L 2 ()). } dx 1 dx 2

18 = = χ f (ϕ ω )dx 1 dx 2 { u (ϕ ω ) u } (ϕ ω ) + χ P u (ϕ ω ) dx 1 dx 2 { u (φ ω ) + 1 u } 2 (φ ω ) + χ P u (φ ω ) dx 1 dx 2 { η 1 (φ P u ) + 1 η } 2 2 (φ P u ) dx 1 dx 2 (a) η i = ω x i L 2 ( ), with i = 1, 2; (b) ω (x 1, x 2 ) = x 1 P X ( x 1 ), x 2 H 1 () (c) for i = 1, 2. { u x i ω x i ϕ η i P u } ϕ dx 1 dx 2 = 0 x i

19 We can show that the solution u H 1 ( ) satifies P u u w H 1 () where u(x 1, x 2 ) = u(x 1 ) is the unique solution of { q uxx + u = f in (0, 1) u (0) = u (1) = 0, q = 1 { Y 1 X } Y y (y, z) dydz > 0 is the homogenized coefficient, Y is the representative basic cell, and X(y, z) is the unique auxiliary in Y.

20 La existencia de un operador de extensión Sean O y O dominios definidos por O = {(x 1, x 2 ) R 2 x 1 I y 0 < x 2 < G 1 } O = {(x 1, x 2 ) R 2 x 1 I e 0 < x 2 < G (x 1 )} donde I R es un intervalo abierto, G : I R es una función C 1 satisfaciendo 0 < G 0 G (x 1 ) G 1 para todo x I y > 0. Entonces, existe un operador de extensión P L(L p (O ), L p (O)) L(W 1,p (O ), W 1,p (O)) y una constante K independiente de y p tal que P ϕ L p (O) K ϕ L p (O ) Pϕ { L ϕ L K + η() ϕ } L p (O) p (O ) p (O ) Pϕ L K ϕ L p (O) p (O ) para todo ϕ W 1,p (O ) donde 1 p y η() = sup{ G (x) }. x I

21 Using the estimates P u L 2, P u L x and 1 P u L 1 2 x M for all > where M is independent of > 0. Hence, we can get a subsequence P u such that P u u 0 w H 1 () P u u 0 s L 2 () P u 0 s L 2 ().

22 A consequence of the limits is that u 0 (x 1, x 2 ) = u 0 (x 1 ), ie. u 0 (x 1, x 2 ) = 0 a.e.. Indeed, for all ϕ C0 () ϕ u 0 dx 1 dx 2 = lim P u ϕ dx 1 dx 2 0 x 2 P u = lim ϕ dx 1 dx 2 = 0. 0 Observe that ũ = χ P u a.e.. Hence, we can pass to the limit to obtain u (x 1, x 2 ) = θ(x 2 ) u 0 (x 1 ) a.e..

23 To obtain the relation between u 0 y ξ, we consider the auxiliar function ( ( ω x1 ) ) (x 1, x 2 ) = x 1 PX, x 2 (x 1, x 2 ). X X = 0 in Y X N = 0 on B 2 X N = N 1 on B 1 L periodic in the variable y X dy 1 dy 2 = 0. Y

24 ω x 1 s L 2 () and w H 1 () ω 0 s L 2 (). Also, we define η = (η 1, η 2 ) in by η i (x 1, x 2 ) = ω x i (x 1, x 2 ). Since = 1 y 1 and = y 2, we have η1(x 1, x 2 ) = 1 X ( ) x1 k, x 2 y 1 η2(x 1, x 2 ) = X ( ) x1 k, x 2 y 2 for (x 1, x 2 ). = 1 X ( x1 ) y 1, x 2 := η 1 (y 1, y 2 ), ) = X y 2 ( x1, x 2 := η 2 (y 1, y 2 )

25 We have η 1(x 1, x 2 ) 1 L L 0 ( 1 X ) (s, x 2 ) ds := ˆq(x 2 ) y 1 w L () η η 2 = 0 in.

26 Hence, we can pass to the limit in χ f (ϕ ω )dx 1 dx 2 = { u ϕ ω + u ω ϕ + 1 u ω 2 ϕ + ũ x ϕω 2 ϕ } η 1 P u η 1 P u ϕ 1 2 η 2 P u ϕ dx 1 dx 2 to obtain { ξ (x 1, x 2 ) ϕ x 1 ˆq ϕ } u 0 (x 1 ) + u 0 (x 1 ) θ(x 2 ) ϕ x 1 dx 1 dx 2 x 1 = θ(x 2 ) f (x 1 ) ϕ(x 1 ) x 1 dx 1 dx 2 ϕ(x 1 ) C0 (0, 1).

27 Using ξ ϕ x 1 = ξ ϕ x 1 + ξ ϕ, the expression ( ) and integrating by parts, we obtain ξ (x 1, x 2 ) ϕ(x 1 ) dx 1 dx 2 = ˆq u 0 (x 1 ) ϕ(x 1 ) dx 1 dx 2 ϕ C0 (0, 1). Since u 0 (x 1, x 2 ) = u 0 (x 1 ) 1 0 {( G 1 0 ) ( G 1 ) ξ u0 } (x 1, x 2 )dx 2 ˆq(x 2 )dx 2 (x 1 ) ϕ(x 1 ) dx 1 = 0. 0 Hence ( ( G 1 ) ) G1 ξ (x 1, x 2 )dx 2 = ˆq(x 2 ) dx u 0 (x 1 ) a.e. I.

28 Also, it follows from ( ) and L G 1 0 θ(x 2 ) dx 2 = Y that 1 0 {( G 1 0 ) ϕ ξ (x 1, x 2 ) dx 2 + Y } 1 L u Y 0 ϕ dx 1 = 0 L f ϕ dx 1. Then 1 for all ϕ H 1 (I) where q = 0 { q ϕ u 0 + ϕ u 0 } dx 1 = 1 0 f ϕ dx 1 L G1 ˆq(s)ds Y = 1 { 0 Y 1 X } (y 1, y 2 ) dy 1 dy 2. Y y 1

29 Teorema Sea u la solución de 2 u x 1 2 u x + 2 u = f en 2 u N1 + 1 u N 2 2 = 0 sobre y sea u 0 la solución de { q uxx(x) + u(x) = f (x) x I u (0) = u (1) = 0 q = 1 { 1 X } (y Y 1, y 2 ) dy 1 dy 2. Y y 1 Entonces para cualquier operador de extensión a P L(L 2 ( ), L 2 ()) L(H 1 ( ), H 1 ()) tenemos que P u u 0 w H 1 () donde u 0 (x 1, x 2 ) = u 0 (x 1 ) para todo (x 1, x 2 ). a dado por el lema de extensión