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2. Let us fix T > 0. Consider the square N = {x € R² : 0 < x₁ < π,0 < x₂ < π} and the linear evolution problem ₁x1x1x2x2 = f,in (0,7) x u(0, x) = g(x) ,forx € Ω (1) u(t, x) = 0 , for all t = [0, T] and x = 0, where g € L²(Q) and ƒ € L²((0, T), L²(N)). u € L²([0, T], H¹ (22)) ^ H¹([0, T], H¹(Q)*) (u' (t), 4)H¹H¹ + B(u(t), y) = (f(t), 9)2, \ € H¹(N), for a.e. t € (0, T) (2) u(0) = g is a weak formulation of the linear evolution problem (1), where g in the last equation is identified through the embedding L²(Q) c H¹(Q)* with an element of H¹(Q)*. b) Consider (en)nz an orthonormal basis of L²(Q), made of Neumann-eigenvectors for ▷ on , respectively associated to the eigenvalues (An)n≥1 with λ₁ ≤ λ₂ ≤ … and limn→ An = +∞0. For any N≥ 1, set V₁ = span{en, 1 ≤ n ≤N}. Assume that g = V₁ and f(t) € VN for a.e. t. [5 (i) Show that there a unique solution u solving (2) while u(t) € VN for a.e. t. (ii) Show that for any 0 ≤ a ≤ T, [4 |u(@){};m) + 2 S " || Vu(1)||:;„di = |gl|?:) + 2 = |gl|2²) + 2 * (ƒ(1), u(t)), dt. 2(12) L²(2) (3)