Explanation Given: The given boundary value problem is ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = − 2 , 0 < x < π , 0 < y < π and the given boundary conditions are u ( 0 , y ) = 0 , u ( π , y ) = 0 , u ( x , 0 ) = 0 and u ( x , π ) = 0 . Calculation: The given boundary value problem is, ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = − 2 . (1) Consider the equation, u ( x , y ) = v ( x , y ) + ψ ( x ) . (2) Partially differentiate both sides of the equation (2) with respect to x as follows. ∂ u ∂ x = ∂ v ∂ x + ψ ′ ( x ) Partially differentiate both sides of the above equation with respect to x as follows. ∂ 2 u ∂ x 2 = ∂ 2 v ∂ x 2 + ψ ″ ( x ) Substitute ∂ 2 u ∂ x 2 = ∂ 2 v ∂ x 2 + ψ ″ ( x ) and ∂ 2 u ∂ y 2 = ∂ 2 v ∂ y 2 in equation (1), ∂ 2 v ∂ x 2 + ψ ″ ( x ) + 2 = − ∂ 2 v ∂ y 2 . (3) Second derivative of ψ is zero in the equation (3) by homogeneous system, ψ ″ ( x ) = − 2 Integrate the above equation with respect to x as follows. ψ ′ ( x ) = − 2 x + c 1 Integrate the above equation with respect to x as follows. ψ ( x ) = − x 2 + c 1 x + c 2 . (4) At boundary conditions u ( 0 , y ) = 0 , u ( π , y ) = 0 , u ( x , 0 ) = 0 and u ( x , π ) = 0 , ψ ( 0 ) = 0 and ψ ( π ) = 0 . Substitute ψ ( 0 ) = 0 and ψ ( π ) = 0 in equation (4), c 1 = π and c 2 = 0 Substitute c 1 = π and c 2 = 0 in equation (4), ψ ( x ) = − x 2 + π x . (5) Consider the equation, v ( x , y ) = X ( x ) Y ( y ) . (6) Separate the variables of equation (2), X ″ X = − α 2 and Y ″ Y = − α 2 The equation of the variable X is, X ″ + α 2 X = 0 . (7) The equation of the variable Y is, Y ″ + α 2 Y = 0 . (8) The general solution of the equation (7) is, X ( x ) = c 1 cos α x + c 2 sin α x . (9) The general solution of the equation (8) is, Y ( y ) = c 3 cosh α y + c 4 sinh α y . (10) At boundary conditions u ( 0 , y ) = 0 and u ( π , y ) = 0 , X ( 0 ) = 0 X ( x ) = c 1 cos 0 + c 2 sin 0 c 1 = 0 At boundary conditions u ( 0 , y ) = 0 and u ( π , y ) = 0 , α = n Substitute c 1 = 0 and α = n in equation (9), X ( x ) = c 2 sin n x

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9th Edition

ISBN: 9781337604918

Author: Dennis G. Zill

Publisher: Cengage Learning

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