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

Differential Equations with Boundary-Value Problems

9th Edition

ISBN: 9781337632515

Author: Dennis G. Zill

Publisher: Cengage Learning US

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PS 9 Two films are shown on screen A and screen B at a cinema each evening. The numbers of people viewing the films on 12 consecutive evenings are shown in the back-to-back stem-and-leaf diagram. Screen A (12) Screen B (12) 8 037 34 7 6 4 0 534 74 1645678 92 71689 Key: 116|4 represents 61 viewers for A and 64 viewers for B A second stem-and-leaf diagram (with rows of the same width as the previous diagram) is drawn showing the total number of people viewing films at the cinema on each of these 12 evenings. Find the least and greatest possible number of rows that this second diagram could have. TIP On the evening when 30 people viewed films on screen A, there could have been as few as 37 or as many as 79 people viewing films on screen B.

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