Chem Differential Eq HW Solutions Fall 2011 150

# Chem Differential Eq HW Solutions Fall 2011 150 - ±± D...

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150 Chapter 12 Green’s Functions and Conformal Mappings Solutions to Exercises 12.1 1. We have M ( x, y )= xy , N ( x, y )= y , M y ( x, y )= x , N x ( x, y )= y . The right side of (4) is equal to ± 1 0 ± 1 0 - xdxdy = - 1 2 . 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. We have ± C ( M ( x, y ) dx + N ( x, y ) dy ) = 4 ² j =1 ± side j ( M ( x, y ) dx + N ( x, y ) dy ) = 4 ² j =1 I j . On side 1, y = 0, hence M = N = 0 and so I 1 = 0. On side 2, x = 1 and y varies from 0 to 1; hence M = y , N = y , and dx = 0, and so I 2 = ± 1 0 ydy = 1 2 . On side 3, y = 1 and x varies form 1 to 0; hence M = x , N = 1, and dy = 0, and so I 3 = ± 0 1 xdx = - 1 2 . On side 4, x =0and y varies form 1 to 0; hence M =0 , N = y , and dx = 0, and so I 4 = ± 0 1 ydy = - 1 2 . Consequently, ± C ( M ( x, y ) dx + N ( x, y ) dy ) =0+ 1 2 - 1 2 - 1 2 = - 1 2 , which veriFes Green’s theorem in this case. 5. We have M ( x, y )=0 , N ( x, y )= x , M y ( x, y )=0 , N x (
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Unformatted text preview: ±± D dxdy = (area of annular region) = π-π 4 = 3 π 4 . We have ± Γ ( M ( x, y ) dx + N ( x, y ) dy ) = ± C 2 + ± C 1 ( M ( x, y ) dx + N ( x, y ) dy ) = I 1 + I 2 . Parametrize C 1 by x = cos t , y = sin t , 0 ≤ t ≤ 2 π , dx =-sin tdt , dy = cos t . Hence I 1 = ± 2 π cos 2 tdt = π. Parametrize C 2 by x = 1 2 cos t , y = 1 2 sin t , t varies from 2 π to 0, dx =-1 2 sin tdt , dy = 1 2 cos t . Hence I 2 = 1 4 ± 2 π dt ==-± 2 π 1 4 cos 2 tdt =-π 4 . Consequently, ± Γ ( M ( x, y ) dx + N ( x, y ) dy ) = π-π 4 = 3 π 4 , which veriFes Green’s theorem in this case....
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