Chem Differential Eq HW Solutions Fall 2011 152

# Chem Differential Eq HW Solutions Fall 2011 152 - 4 1 2 = 5...

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152 Chapter 12 Green’s Functions and Conformal Mappings Solutions to Exercises 12.2 1. The function u ( x, y )= e x cos y is harmonic for all ( x, y ) (check that 2 u =0 for all ( x, y )). Applying (1) at ( x 0 ,y 0 )=(0 , 0) with r = 1, we obtain 1= u (0 , 0) = 1 2 π ± 2 π 0 e cos t cos(sin t ) dt. 5. u ( x, y )= x 2 - y 2 , u xx =2 , u yy = - 2, u xx + u yy = 0 for all ( x, y ). Since u is harmonic for all ( x, y ) it is harmonic on the given square region and continuus on its boundary. Since the region is bounded, u attains its maximum and minimum values on the boundary, by Corollary 1. Starting with the side of the square on the x -axis and moving counterclockwise, label the sides of the square by 1, 2, 3, and 4. On side 1, 0 x 1, y = 0, and u ( x, y )= u ( x, 0) = x 2 . On this side, the maximum value is 1 and is attained at the point (1 , 0), and the minimum value is 0 and is attained at the point (0 , 0). On side 2, 0 y 1, x = 1, and u ( x, y )= u (1 ,y )=1+ y - y 2 = f ( y ). On this side, f ± ( y )= - 2 y +1, f ± ( y )=0 y =1 / 2. Minimum value f (0) = f (1) = 1, attained at the points (1 , 0) and (1 , 1). Maximum value f (1 / 2)=1 -
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Unformatted text preview: / 4 + 1 / 2 = 5 / 4, attained at the point (1 , 1 / 2). On side 3, 0 ≤ x ≤ 1, y = 1, and u ( x, y ) = u ( x, 1) = x 2 + x-1 = f ( x ). On this side, f ± ( x ) = 2 x + 1, f ± ( x ) = 0 ⇒ x =-1 / 2. Extremum values occur at the endpoints: f (0) =-1, f (1) = 1. Thus the minimum value is-1 and is attained at the point (0 , 1). Maximum value is 1 and is attained at the point (1 , 1). On side 4, 0 ≤ y ≤ 1, x = 0, and u ( x, y ) = u (0 , y ) =-y 2 . On this side, the maximum value is 0 and is attained at the point (0 , 0), and the minimum value is-1 and is attained at the point (0 , 1). Consequently, the maximum value of u on the square is 5 / 4 and is attained at the point (1 , 1 / 2); and the minimum value of u on the square is-1 and is attained at the point (0 , 1) (see Fgure). Plot3D x^2 y^2 x y, x, 0, 1 , y, 0, 1 , ViewPoint 2, 2, 1 1...
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## This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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