Teste 1 - Análise Numérica Funcional e Optimização Instituto Superior Técnico, 8 de Novembro de 2012, 12h00-13h30

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1 Teste - Análise Numérica Funcional e Optimização Instituto Superior Técnico, 8 de Novembro de 22, 2h-3h3 (MMA,PDEC [4.] Mostre que existe pelo menos uma solução (x, y [, ] 2 para o sistema não linear (m, p N { x m 4x + y m + = x p + y p + = 4y e que se m p, há pelo menos duas soluções. 2 [2.] Mostre que em H (a, b, a norma u = u L 2 (a,b + u L 2 (a,b é uma norma equivalente à induzida pelo produto interno. 3 [.] Considere o operador A denido em C[, ]. (Au(x = u(t 2 dt u(x 2. a Determine a derivada de Fréchet de A, para qualquer u C[, ]. b Mostre que dado > 4, e f C[, ] : f 2, a equação u(x 2 + u(x = f(x + u(t 2 dt tem uma única solução u C[, ] : u. Apresente uma sucessão convergente u n u, começando com u =, e um majorante para o erro u u n. c Considere a aplicação do Método de Newton-Kantorovich à equação denida em b. Dena a equação linear que por esse método permite obter a iterada u n+ a partir da iterada u n, e obtenha u começando com u =. 4 [4.] Considere a função u(x = x a x b. 2 Calcule ( 2 u(v = u, v L 2 (R, v D(R.

2 Teste - Análise Numérica Funcional e Optimização Instituto Superior Técnico, 8 de Novembro de 22, 2h-3h3 (MMA,PDEC [4.] Show that there exists at least one solution (x, y [, ] 2 for the nonlinear system (m, p N { x m 4x + y m + = x p + y p + = 4y and that if m p, there are at least two solutions. 2 [2.] Show that in H (a, b, the norm u = u L 2 (a,b + u L 2 (a,b is equivalent to the norm induced by the inner product. 3 [.] Consider the operator A dened in C[, ] by. (Au(x = u(t 2 dt u(x 2. a Determine the Fréchet derivative of A, for any u C[, ]. b Show that given > 4, and f C[, ] : f 2, the equation u(x 2 + u(x = f(x + u(t 2 dt has an unique solution u C[, ] : u. Present a convergent sequence u n u, starting with u =, and a bound for the error u u n. c Consider Newton-Kantorovich method applied to the equation in b. Dene the linear equation given by that method to obtain u n+ from u n, and obtain u starting with u =. 4 [4.] Consider the function u(x = 2 x a x b. Calculate ( 2 u(v = u, v L 2 (R, v D(R.

3 χ- Resolution Consider the function g(x, y = 4 (xm + y,m +, x p + y p +. Taking the convex closed set S = [, ] 2, homeomorphic to the unit ball, we show the invariance, g(s S, and since g is continuous in S, we apply the Brouwer xed point theorem to conclude the existence of at least one xed point (x, y = g(x, y, which corresponds to a solution of the system. In fact, if x, y [, ], we have ( 4, 4 = 4 ( + +, + + g(x, y = 4 (xm + y,m +, x p + y p + 4 ( + +, + + = (3 4, 3 4, meaning that g(x, y [ 4, 3 4 ]2 [, ] 2 = S. This proves the invariance of g over S, and we conclude on the existence. In fact we can state that the xed point lays in [ 4, 3 4 ]2. If m p then x y. This means that (x, y and (y, x are both solutions and dierent. Otherwise, if x = y, then 2x m + = 4x = 2x p + and this implies x m = x p, and as m p, we obtain x = or x =, which are not solutions. Note: If m = p the equations imply x m + y m + 4x = x m + y m + 4y = this gives x = y and 2x m + = 4x. 2. : H (a, b [, + [ is a norm (it was not asked to prove. We can simply form the vector U = ( u L 2, u L 2 and argue that u = U + U 2, is the l norm of U in R 2. Also u 2 H (, = U 2 + U 2 2 is thel 2 norm of U in R 2, and all the norms are equivalent in R 2. We could also repeat the proof. On one hand, and on the other hand, u 2 H = u 2 + u 2 u 2 + u u u 2 = ( u + u 2 = u 2, 2 u 2 H = 2 u u 2 = ( u u 2 + ( u + u 2 ( u + u 2 = u 2. Thus u H u 2 u H. 3a We calculate the F-derivative (A(u + h A(u(x = = ( (u + h 2 (tdt (u + h 2 (x u(t 2 dt u(x 2 2u(th(tdt 2u(xh(x + } {{ } linear h 2 (tdt h 2 (x } {{ } =O( h 2 ( we conclude that A u(h(x = 2 u(th(tdt u(xh(x = 2 u, h 2(uh(x, because A(u + h A(u A u(h = max x [,] 3b The equation can be written as the xed point of a G operator: u(x 2 + u(x = f(x + h 2 (tdt h 2 (x 2 h 2 = o( h. u(t 2 dt u(x = (Au + f(x = G(u(x.

4 Then we have G u(h = A u = 2 ( u, h uh G G u L(C[,] = sup u(h 2 = sup ( u, h uh 2 h h h h sup uh + uh 4 h h u and as u we obtain contractivity with L = 4 <, because > 4. The set S = {u C[, ] : u } = B(, is convex, closed, non-void and G is invariant in S u = G(u = (Au + f ( Au + f (2 u , by the xed point theorem, we conclude existence and uniqueness. The sequence u n+ (x = ( f(x + u n (t 2 dt u 2 n(x converges to the xed point verifying u u n ( 4 n 4 u u n = ( 4 n 4 f, because u = f, as u =. 3c The equation is u = Gu and we now consider it in the form F u =, to apply Newton's method: u = G(u = (Au + f F (u = f + Au u =, By the Fréchet derivative properties F u(h = A u(h h. Therefore the Newton iteration u n+ (F u n F (u n can be written with h = u n+ u n : = u n F u n (h = F (u n A u n (h h = u n Au n f As A u n (h = 2 u, h 2(uh, we obtain a linear equation on h : When u =, we obtain 2 u n, h 2u n (xh(x h(x = u n (x 2, h 2h(x h(x = h(x = u n (t 2 dt + u n (x 2 f(x. dt + f(x (f(x + 2, h. 2 + The constant H =, h is unknown, but it can be determined, multiplying both sides:, h = H = (, f, + 2, h, (, f + 2H = H =, f The solution is then h(x = 2+ (f(x + 2(, f = 2+ (f(x + 2, f, and u (x = u (x + h(x = ( f(x f(tdt.

5 4 Using the formal calculus for the derivative u = 2 ( x a x b = (sgn(x a x b + sgn(x b x a 2 = (2δ(x a x b + 2sgn(x asgn(x b + 2δ(x b x a 2 = 2 (2δ a(x a b + 2sgn(x asgn(x b + 2δ b (x b a = (δ a (x + δ b (x a b + sgn(x asgn(x b { +, x (a, b abreviating sgn [a,b] (x = sgn(x asgn(x b =, we obtain for v D(R,, x / (a, b ( 2 u(v = u, v L 2 (R = (δ a + δ b a b sgn [a,b], v L 2 (R = (v(a + v(b a b sgn [a,b] (xv(xdx. R