MATH 212 Useful File, Fall 2011

FORMULA SHEET a) The solution of the following problem :            ∂2u ∂t2 = c2∂2u ∂x2 u(0, t) = u(L, t) = 0, ∀ t u(x, 0) = f(x) ut(x, 0) = g(x), ∀ x ∈ [0, L] for a string of length L is u(x, t) = � n≥1 � Bn cos(λnt) + B∗ n sin(λnt) � sin �nπx L � , with λn = cnπ L , Bn = 2 L � L 0 f(x) sin �nπx L � dx , n = 1, 2, . . . and B∗ n = 2 cnπ � L 0 g(x) sin �nπx L � dx, n = 1, 2, . . . b) The solution of the following problem :        ∂u ∂t = c2∂2u ∂x2 u(0, t) = u(L, t) = 0 u(x, 0) = f(x) for a thin metal bar of length L is u(x, t) = � n≥1 Bn sin �nπx L � e−λn2t, with λn = cnπ L Bn = 2 L � L 0 f(x) sin �nπx L � dx, n = 1, 2, . . . c) The solution of the following equation : ∂2u ∂x2 + ∂2u ∂y2 = 0 in the following rectangle R 1 2 FORMULA SHEET such that u(x, y) = f(x) on the upper side and 0 on the other sides of R , is : u(x, y) = � n≥1 Bn sin �nπx a � sinh �nπy a � , with Bn = 2 a sinh �nπb a � � a 0 f(x) sin �nπx a � dx, n=1,2,. . . d) The solution of the following problem              ∂2u ∂t2 = c2�∂2u ∂x2 + ∂2u ∂y2 � u = 0 on the boundary u(x, y, 0) = f(x, y) ut(x, y, 0) = g(x, y) considered in the following rectangular membrane : FORMULA SHEET 3 is u(x, y, t) = � m≥1 � n≥1 � Bmn cos(λmnt)+B∗ mn sin(λmnt) � sin �mπx a � sin �nπy b � , with Bmn = 4 ab � b 0 � a 0 f(x, y) sin �mπx a � sin �nπy b � dxdy B∗ mn = 4 abλmn � b 0 � a 0 g(x, y) sin �mπx a � sin �nπy b � dxdy and λmn = cπ � m2 a2 + n2 b2 , m = 1, 2, . . . and n = 1, 2, . . . Laplace transform a) L{1} = 1 s b) L{tn} = n! sn+1, n = 1, 2, 3, . . . c) L{eat} = 1 s − a d) L{sin(kt)} = k s2 + k2 e) L{cos(kt)} = s s2 + k2 f) L{f(t − a)U(t − a)} = e−asF(s), where F(s) = L{f(t)}. g) If f, f ′,. . .,f (n−1) are continuous on [0, ∞) and if f (n)(t) is piece- wise continuous on [0, ∞), then L{f (n)(t)} = snF(s) − sn−1f(0) − sn−2f ′(0) − . . . − f (n−1)(0), where F(s) = L{f(t)}.