Explanation Given: The given boundary conditions are u ( 0 , y ) = 0 , u ( x , 0 ) = 0 , u ( π , y ) = 50 and u ( x , π ) = 0 . Calculation: The partial differential equation is, ∂ 2 u ∂ x 2 = ∂ 2 u ∂ y 2 . (1) Consider the equation, u ( x , y ) = X ( x ) Y ( y ) . (2) Separate the variables of equation (2), − X ″ X = − λ and Y ″ Y = − λ The equation of the variable X is, X ″ − λ X = 0 . (3) The equation of the variable Y is, Y ″ + λ Y = 0 . (4) Substitute λ = α 2 in equation (3), X ″ − α 2 X = 0 The general solution of the above equation is, X ( x ) = c 1 cosh α x + c 2 sinh α x . (5) At boundary condition X ( 0 ) = 0 , c 1 = 0 . For non-trivial solution c 2 ≠ 0 , α = n . Substitute c 1 = 0 and α = n in the equation (5), X ( x ) = c 2 sinh n x . (6) Substitute λ = α 2 in equation (4) as follows

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ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Chapter 13, Problem 7CR

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The steady state temperature u(x,y) in the square plate.

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Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E Ch. 13.1 - Prob. 5E Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

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The steady state temperature u ( x , y ) in the square plate.

Solution Summary: The author explains the steady state temperature in the square plate and the partial differential equation.

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The steady state temperature u(x,y) in the square plate.

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