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Chapter 13 - 13 Boundary-Value Problems in Rectangular...

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13 Boundary-Value Problems in Rectangular Coordinates Exercises 13.1 1. If u = XY then u x = X Y, u y = XY , X Y = XY , and X X = Y Y = ± λ 2 . Then X λ 2 X = 0 and Y λ 2 Y = 0 so that X = A 1 e ± λ 2 x , Y = A 2 e ± λ 2 y , and u = XY = c 1 e c 2 ( x + y ) . 2. If u = XY then u x = X Y, u y = XY , X Y = 3 XY , and X 3 X = Y Y = ± λ 2 . Then X ± 3 λ 2 X = 0 and Y λ 2 Y = 0 so that X = A 1 e 3 λ 2 x , Y = A 2 e ± λ 2 y , and u = XY = c 1 e c 2 ( y 3 x ) . 3. If u = XY then u x = X Y, u y = XY , X Y = X ( Y Y ) , and 605
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Exercises 13.1 X X = Y Y Y = ± λ 2 . Then X λ 2 X = 0 and Y (1 λ 2 ) Y = 0 so that X = A 1 e ± λ 2 x , Y = A 2 e (1 λ 2 ) y , and u = XY = c 1 e y + c 2 ( x y ) . 4. If u = XY then u x = X Y, u y = XY , X Y = X ( Y + Y ) , and X X = Y + Y Y = ± λ 2 . Then X λ 2 X = 0 and Y ( 1 ± λ 2 ) Y = 0 so that X = A 1 e ± λ 2 x , Y = A 2 e ( 1 ± λ 2 ) y , and u = XY = c 1 e y + c 2 ( x + y ) . 5. If u = XY then u x = X Y, u y = XY , xX Y = yXY , and xX X = yY Y = ± λ 2 . Then X 1 x λ 2 X = 0 and Y 1 y λ 2 Y = 0 so that X = A 1 x ± λ 2 , Y = A 2 y ± λ 2 , and u = XY = c 1 ( xy ) c 2 . 606
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Exercises 13.1 6. If u = XY then u x = X Y, u y = XY , yX Y = xXY , and X xX = Y yY = ± λ 2 . Then X λ 2 xX = 0 and Y ± λ 2 yY = 0 so that X = A 1 e ± λ 2 x 2 / 2 , Y = A 2 e λ 2 y 2 / 2 , and u = XY = c 1 e c 2 ( x 2 y 2 ) . 7. If u = XY then u xx = X Y, u yy = XY , u yx = X Y , and X Y + X Y + XY = 0 , which is not separable. 8. If u = XY then u yx = X Y , yX Y + XY = 0 , and X X = Y yY = ± λ 2 . Then X λ 2 X = 0 and ± λ 2 yY Y = 0 so that X = A 1 e λ 2 x , Y = A 2 y ± 1 2 , and u = XY = c 1 e c 2 x y 1 /c 2 . 9. If u = XT then u t = XT , u xx = X T, kX T XT = XT , and we choose T T = kX X X = 1 ± 2 607
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Exercises 13.1 so that T ( 1 ± 2 ) T = 0 and X ( ± λ 2 ) X = 0 . For λ 2 > 0 we obtain X = A 1 cosh λx + A 2 sinh λx and T = A 3 e ( 1+ 2 ) t so that u = XT = e ( 1+ 2 ) t ( c 1 cosh λx + c 2 sinh λx ) . For λ 2 < 0 we obtain X = A 1 cos λx + A 2 sin λx and T = A 3 e ( 1 2 ) t so that u = XT = e ( 1 2 ) t ( c 3 cos λx + c 4 sin λx ) . If λ 2 = 0 then X = 0 and T + T = 0 , and we obtain X = A 1 x + A 2 and T = A 3 e t . In this case u = XT = e t ( c 5 x + c 6 ) 10. If u = XT then u t = XT , u xx = X T, kX T = XT , and X X = T kT = ± λ 2 so that X λ 2 X = 0 and T λ 2 kT = 0 . For λ 2 > 0 we obtain X = A 1 cos λx + A 2 sin λx and T = A 3 e λ 2 kt , so that u = XT = e c 2 3 kt ( c 1 cos c 3 x + c 2 sin c 3 x ) . For λ 2 < 0 we obtain X = A 1 e λx + A 2 e λx , T = A 3 e λ 2 kt , and u = XT = e c 2 3 kt ( c 1 e c 3 x + c 2 e c 3 x ) . For λ 2 = 0 we obtain T = A 3 , X = A 1 x + A 2 , and u = XT = c 1 x + c 2 . 608
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Exercises 13.1 11. If u = XT then u xx = X T, u tt = XT , a 2 X T = XT , and X X = T a 2 T = ± λ 2 so that X λ 2 X = 0 and T a 2 λ 2 T = 0 . For λ 2 > 0 we obtain X = A 1 cos λx + A 2 sin λx, T = A 3 cos aλt + A 4 sin aλt, and u = XT = ( A 1 cos λx + A 2 sin λx )( A 3 cos aλt + A 4 sin aλt ) . For λ 2 < 0 we obtain X = A 1 e λx + A 2 e λx , T = A 3 e aλt + A 4 e aλt , and u = XT = ( A 1 e λx + A 2 e λx )( A 3 e aλt + A 4 e aλt ) .
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