Explanation Given: The boundary conditions are as follows: ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 0 < x < 1 , 0 < y < 1 ∂ u ∂ x | x = 0 = 0 u ( 1 , y ) = u 0 u ( x , 0 ) = 0 and ∂ u ∂ y | y = 1 = 0 Calculation: The boundary conditions are given as follows: ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 0 < x < 1 , 0 < y < 1 ∂ u ∂ x | x = 0 = 0 u ( 1 , y ) = u 0 u ( x , 0 ) = 0 and ∂ u ∂ y | y = 1 = 0 Separate the variables with the help of separation constant and take the separation constant λ as follows: X ″ X = − Y ″ Y = λ The separated equations are written as follows: X ″ − λ X = 0 … … ( 1 ) Y ″ + λ Y = 0 … … ( 2 ) The product solution is as follows: u ( x , y ) = X ( x ) Y ( y ) … … ( 3 ) Substitute 0 for y in equation ( 3 ) . u ( x , 0 ) = X ( x ) Y ( 0 ) X ( x ) Y ( 0 ) = 0 ( u ( x , 0 ) = 0 ) Y ( 0 ) = 0 Differentiate equation ( 3 ) with respect to y . u ′ y ( x , y ) = X ( x ) Y ′ ( y ) Substitute 1 for y in the above equation. u ′ y ( x , 1 ) = X ( x ) Y ′ ( 1 ) X ( x ) Y ′ ( 1 ) = 0 ( u ′ y ( x , 1 ) = 0 ) Y ′ ( 1 ) = 0 Differentiate equation ( 3 ) with respect to x . u ′ x ( x , y ) = X ′ ( x ) Y ( y ) Substitute 0 for x in the above equation. u ′ x ( 0 , y ) = X ′ ( 0 ) Y ( y ) X ′ ( 0 ) Y ( y ) = 0 ( u ′ x ( 0 , y ) = 0 ) X ′ ( 0 ) = 0 There are three cases for the value of λ . First case is λ = 0 . Substitute 0 for λ in equation ( 2 ) . Y ″ = 0 The general; solution of the above equation is as follows: Y ( y ) = c 1 + c 2 y … … ( 4 ) Substitute 0 for y in equation ( 4 ) . Y ( 0 ) = c 1 + c 2 ⋅ 0 c 1 = 0 ( Y ( 0 ) = 0 ) Substitute 0 for c 1 in equation ( 4 ) . Y ( y ) = c 2 y Differentiate the above equation with respect to y . Y ′ ( y ) = c 2 Substitute 1 for y in the above equation. Y ′ ( 1 ) = c 2 c 2 = 0 ( Y ′ ( 1 ) = 0 ) Since, the value of c 1 and c 2 is 0 so, for λ = 0 the non-trivial solution does not exist. Second case is λ = − α 2 ( λ < 0 ) . Substitute λ = − α 2 in equation ( 2 ) . Y ″ − α 2 Y = 0 The general solution of the above equation is as follows: Y ( y ) = c 3 cos h ( α y ) + c 4 sinh ( α y ) … … ( 5 ) Substitute 0 for y in equation ( 5 ) . Y ( 0 ) = c 3 cos h ( 0 ) + c 4 sinh ( 0 ) c 3 = 0 ( Y ( 0 ) = 0 ) Substitute 0 for c 3 in equation ( 5 ) . Y ( y ) = c 4 sinh ( α y ) Differentiate the above equation with respect to y as follows: Y ′ ( y ) = α c 4 cosh ( α y ) Substitute 1 for y in the above equation. Y ′ ( 1 ) = α c 4 cosh ( α ) α c 4 cosh ( α ) = 0 ( Y ′ ( 1 ) = 0 ) c 4 = 0 Since the value of c 3 and c 4 is 0 so, for λ = − α 2 the non-trivial solution does not exist. Third case is λ = α 2 ( λ > 0 ) . Substitute λ = α 2 in equation ( 2 ) . Y ″ + α 2 Y = 0 The general solution of the above equation is as follows: Y ( y ) = c 5 cos ( α y ) + c 6 sin ( α y ) … … ( 6 ) Substitute 0 for y in equation ( 6 ) . Y ( 0 ) = c 5 cos ( 0 ) + c 6 sin ( 0 ) c 5 = 0 ( Y ( 0 ) = 0 ) Substitute 0 for c 5 in equation ( 6 ) . Y ( y ) = c 6 sin ( α y ) … … ( 7 ) Differentiate the above equation with respect to y . Y ′ ( y ) = α c 6 cos ( α y ) Substitute 1 for y in the above equation. Y ′ ( 1 ) = α c 6 cos ( α ) α c 6 cos ( α ) = 0 ( Y ′ ( 1 ) = 0 ) cos ( α ) = 0 α = ( 2 n − 1 ) π 2 Therefore, the Eigen values are as follows: λ = α n 2 = ( 2 n − 1 ) 2 π 2 4 Substitute the value of α in equation ( 7 )

9th Edition

ISBN: 9781337810906

Author: ZILL

Publisher: CENGAGE L

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Chapter 12.7, Problem 7E

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The solution u(x,y) of the differential equation ∂2u∂x2+∂2u∂y2=0 such that ∂u∂x|x=0=0, u(1,y)=u0, u(x,0)=0 and ∂u∂y|y=1=0.

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Chapter 12.7, Problem 6E

Chapter 12.7, Problem 8E

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The solution u ( x , y ) of the differential equation ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = 0 such that ∂ u ∂ x | x = 0 = 0 , u ( 1 , y ) = u 0 , u ( x , 0 ) = 0 and ∂ u ∂ y | y = 1 = 0 .

Solution Summary: The author explains the solution of the differential equation partial x,y, and y.

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The solution u(x,y) of the differential equation ∂2u∂x2+∂2u∂y2=0 such that ∂u∂x|x=0=0, u(1,y)=u0, u(x,0)=0 and ∂u∂y|y=1=0.

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Chapter 12 Solutions