12장

12장 - Homework 1 § 12 . 1 1. Solve...

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Unformatted text preview: Homework 1 § 12 . 1 1. Solve the PDEs. (a) u xx = 4 y 2 u , (b) u yy = 4 xu y Sol. (a) If u = u ( x ), then u ( x ) = Ae 2 yx + Be- 2 yx . Thus the solution of this PDE is u ( x, y ) = A ( y ) e 2 yx + B ( y ) e- 2 yx , where A ( y ) and B ( y ) are arbitrary. (b) Setting u y = p , we have p y = 4 xp , and p = A ( x ) e 4 xy . Thus u ( x, y ) = A ( x ) 4 x e 4 xy + B ( x ) = A 1 ( x ) e 4 xy + B ( x ) , where A 1 ( x ) and B ( x ) are arbitrary. 2. Solve u xx = 0 , u xy = 0. Sol. Setting u x = p , we have p x = 0 , p y = 0, and p = c = constant. Thus u ( x, y ) = cx + A ( y ) , where A ( y ) are arbitrary. § 12 . 3 1. Find u ( x, t ) for the string of length L = 1 and c 2 = 1 when the initial velocity is zero and the initial deflection f ( x ) is as follows. (a) f ( x ) = 0 . 01(sin πx- 1 3 sin 3 πx ) (b) f ( x ) = ‰ . 2 x if ≤ x ≤ . 5- . 2 x + 0 . 2 if . 5 ≤ x ≤ 1 . Sol. We want to find a solution of the wave equation ∂ 2 u ∂t 2 = ∂ 2 u ∂x 2 satisfying the condi- tions u (0 , t ) = 0 , u (1 , t ) = 0 for all t , u ( x, 0) = f ( x ) , u t ( x, 0) = 0 for all 0 ≤ x ≤ 1 . By the method of separating variables, we have u ( x, t ) = ∞ X n =1 ( B n cos nπt + B * n sin nπt ) sin nπx. (Give the details of your answer.) From the Initial Conditions u ( x, 0) = ∞ X n =1 B n sin nπx = f ( x ) , ∂u ∂t fl fl fl fl t =0 = ∞ X n =1 B * n nπ sin nπx = 0 . Using Fourier Sine Series, B * n = 0 B n = 2 Z 1 f ( x ) sin nπxdx, n = 1 , 2 , ··· . (a) B 1 = 1 100 , B 3 =- 1 300 and B 2 = B 4 = B 5 = ··· = 0 The solution u is u ( x, t ) = 1 100 cos πt sin πx- 1 300 cos 3 πt sin 3 πx. Computational Science & Engineering (CSE) C. K. Ko Homework 2 (b) B n = 2[ Z 1 / 2 . 2 x sin nπxdx + Z 1 1 / 2 . 2(1- x ) sin nπxdx ] = . 8 sin nπ 2 ( nπ ) 2 . The solution u is u ( x, t ) = ∞ X n =1 . 8 sin nπ 2 ( nπ ) 2 cos nπt sin nπx = . 8 π 2 [cos πt sin π- 1 3 2 cos 3 πt sin 3 πx + 1 5 2 cos 5 πt sin 5 πx + ··· ] ....
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12장 - Homework 1 § 12 . 1 1. Solve...

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