Chapter 13

# 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 0 Y, u y = 0 , X 0 Y = 0 , and X 0 X = Y 0 Y = ± λ 2 . Then X 0 λ 2 X = 0 and Y 0 λ 2 Y =0 so that X = A 1 e ± λ 2 x , Y = A 2 e ± λ 2 y , and u = = c 1 e c 2 ( x + y ) . 2. If u = then u x = X 0 u y = 0 , X 0 Y = 3 0 , and X 0 3 X = Y 0 Y = ± λ 2 . Then X 0 ± 3 λ 2 X = 0 and Y 0 λ 2 Y so that X = A 1 e 3 λ 2 x , Y = A 2 e ± λ 2 y , and u = = c 1 e c 2 ( y 3 x ) . 3. If u = then u x = X 0 u y = 0 , X 0 Y = X ( Y Y 0 ) , and 605

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Exercises 13.1 X 0 X = Y Y 0 Y = ± λ 2 . Then X 0 λ 2 X = 0 and Y 0 (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 = then u x = X 0 Y, u y = 0 , X 0 Y = X ( Y + Y 0 ) , and X 0 X = Y + Y 0 Y = ± λ 2 . Then X 0 λ 2 X = 0 and Y 0 ( 1 ± λ 2 ) Y so that X = A 1 e ± λ 2 x , Y = A 2 e ( 1 ± λ 2 ) y , and u = = c 1 e y + c 2 ( x + y ) . 5. If u = then u x = X 0 u y = 0 , xX 0 Y = yXY 0 , and xX 0 X = yY 0 Y = ± λ 2 . Then X 1 x λ 2 X = 0 and Y 0 1 y λ 2 Y so that X = A 1 x ± λ 2 , Y = A 2 y ± λ 2 , and u = = c 1 ( xy ) c 2 . 606
Exercises 13.1 6. If u = XY then u x = X 0 Y, u y = 0 , yX 0 Y = xXY 0 , and X 0 xX = Y 0 yY = ± λ 2 . Then X λ 2 xX = 0 and Y 0 ± λ 2 =0 so that X = A 1 e ± λ 2 x 2 / 2 , Y = A 2 e λ 2 y 2 / 2 , and u = = c 1 e c 2 ( x 2 y 2 ) . 7. If u = then u xx = X 0 u yy = 0 ,u yx = X 0 Y 0 , and X 0 Y + X 0 Y 0 + 0 , which is not separable. 8. If u = then u = X 0 Y 0 , 0 Y 0 + , and X 0 X = Y 0 = ± λ 2 . Then X 0 λ 2 X = 0 and ± λ 2 0 Y so that X = A 1 e λ 2 x , Y = A 2 y ± 1 2 , and u = = c 1 e c 2 x y 1 /c 2 . 9. If u = XT then u t = 0 , u xx = X 0 T, kX 0 T = 0 , and we choose T 0 T = 0 X X = 1 ± 2 607

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Exercises 13.1 so that T 0 ( 1 ± 2 ) T = 0 and X 0 ( ± 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 ) . 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 = = e ( 1 2 ) t ( c 3 cos λx + c 4 sin λx ) . If λ 2 = 0 then X 0 = 0 and T 0 + T , and we obtain X = A 1 x + A 2 and T = A 3 e t . In this case u = = e t ( c 5 x + c 6 ) 10. If u = then u t = 0 , u xx = X 0 T, kX 0 T = 0 , and X 0 X = T 0 kT = ± λ 2 so that X 0 λ 2 X = 0 and T 0 λ 2 . 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 = = e c 2 3 kt ( c 1 cos c 3 x + c 2 sin c 3 x ) . λ 2 < 0 we obtain X = A 1 e λx + A 2 e λx , T = A 3 e λ 2 kt , and u = = e c 2 3 kt ( c 1 e c 3 x + c 2 e c 3 x ) . λ 2 = 0 we obtain T = A 3 ,X = A 1 x + A 2 , and u = = c 1 x + c 2 . 608
Exercises 13.1 11. If u = XT then u xx = X 0 T, u tt = 0 , a 2 X 0 T = 0 , and X 0 X = T 0 a 2 T = ± λ 2 so that X 0 λ 2 X = 0 and T 0 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 = =( A 1 cos λx + A 2 sin λx )( A 3 cos aλt + A 4 sin aλt ) . 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 = = ( A 1 e λx + A 2 e λx )( A 3 e aλt + A 4 e aλt ) . λ 2 = 0 we obtain X = A 1 x + A 2 , T = A 3 t + A 4 , and u = A 1 x + A 2 )( A 3 t + A 4 ) . 12. If u = then u t = 0 , u tt = 0 , u xx = X 0 a 2 X 0 T = 0 +2 kXT 0 , and X 0 X = T 0 kT 0 a 2 T = ± λ 2 so that X 0 λ 2 X = 0 and T 0 0 a 2 λ 2 T .

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## This note was uploaded on 02/13/2008 for the course ENG 342 taught by Professor Delahanty during the Fall '07 term at TCNJ.

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

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