Explanation Given: The given wave equation is a 2 ( ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 ) = ∂ 2 u ∂ t 2 and the given boundary conditions are u ( 0 , y , t ) = 0 , u ( π , y , t ) = 0 , u ( x , 0 , t ) = 0 , u ( x , π , t ) = 0 , u ( x , y , 0 ) = x y ( x − π ) ( y − π ) and ∂ u ∂ t | t = 0 = 0 . Calculation: The given wave equation is, a 2 ( ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 ) = ∂ 2 u ∂ t 2 (1) Consider the equation, u ( x , y , t ) = X ( x ) Y ( y ) T ( t ) (2) Separate the variables of equation (2), X ″ X = − α 2 and − Y ″ Y + T ″ a 2 T = − α 2 The equation of the variable X is, X ″ + α 2 X = 0 (3) The equation of the variable Y and T is, Y ″ Y = T ″ a 2 T + α 2 (4) Separate the variables of equation (4), Y ″ Y = − β 2 and T ″ a 2 T + α 2 = − β 2 The equation of the variable Y is, Y ″ + β 2 Y = 0 (5) The equation of the variable T is, T ″ + a 2 ( α 2 + β 2 ) T = 0 (6) The general solution of the equation (3) is, X ( x ) = c 1 cos α x + c 2 sin α x (7) The general solution of the equation (5) is, Y ( y ) = c 3 cos β y + c 4 sin β y (8) The general solution of the equation (6) is, T ( t ) = c 5 cos a α 2 + β 2 t + c 6 sin a α 2 + β 2 t (9) At boundary condition u ( 0 , y , t ) = 0 , X ( 0 ) = 0 X ( x ) = c 1 cos 0 + c 2 sin 0 c 1 = 0 At boundary condition u ( x , 0 , t ) = 0 , Y ( 0 ) = 0 Y ( y ) = c 3 cos 0 + c 4 sin 0 c 3 = 0 At boundary conditions u ( π , y , t ) = 0 and u ( x , π , t ) = 0 , X ( π ) = 0 and Y ( π ) = 0 α = m and β = n Substitute c 1 = 0 and α = m in equation (7), X ( x ) = c 2 sin m x (10) Substitute c 3 = 0 and β = n in equation (8), Y ( y ) = c 4 sin n y (11) Substitute α = m and β = n in equation (9), T

Differential Equations with Boundary-Value Problems (MindTap Course List)

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ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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