2018 Exam 3 Spring.pdf - Math 3363 Exam III Solutions Spring 2018 Please use a pencil and do the problems in the order in which they are listed You may

2018 Exam 3 Spring.pdf - Math 3363 Exam III Solutions...

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Math 3363 Exam III Solutions Spring 2018 Please use a pencil and do the problems in the order in which they are listed . You may use the following information without derivation. A proper listing of eigenvalues and eigenfunctions for (i) 00 ( ) =  ( ) for 0 , (ii) (0) = 0 , and (iii) ( ) = 0 is { } =1 and { } =1 where = (  ) 2 and ( ) = sin  A proper listing of eigenvalues and eigenfunctions for (i) 00 ( ) =  ( ) for 0 , (ii) 0 (0) = 0 , and (iii) 0 ( ) = 0 is { } =0 and { } =0 where = (  ) 2 and ( ) = cos  Note that 0 = 0 and 0 ( ) = 1 A proper listing of eigenvalues and eigenfunctions for (i) 00 ( ) =  ( ) for 0 , (ii) (0) = 0 , and (iii) 0 ( ) = 0 is { } =1 and { } =1 where = μ (2 1) 2 2 and ( ) = sin (2 1)  2 A proper listing of eigenvalues and eigenfunctions for (i) 00 ( ) =  ( ) for 0 , (ii) 0 (0) = 0 , and (iii) ( ) = 0 is { } =1 and { } =1 where = μ (2 1) 2 2 and ( ) = cos (2 1)  2 1
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A proper listing of eigenvalues and eigenfunctions for (i) 00 ( ) =  ( ) for , (ii) ( ) = ( ) , and (iii) 0 ( ) = 0 ( ) is { } =0 and { } =0 where 0 = 0   0 ( ) = 1 for  2 1 = 2 = (  ) 2 for = 1 2 3      2 1 ( ) = cos   and 2 ( ) = sin  for = 1 2 3     and  A proper listing of eigenvalues and eigenfunctions for (i) −∇ 2 (   ) =  (   ) for 0 and 0 and (ii) (   ) = 0 for (   ) on the boundary of [0   ] × [0   ] is {  }  =1 and {  }  =1 where  = (  ) 2 +(  ) 2 and  (   ) = sin  sin  1. Suppose that each of , , and is a positive number. Derive the solution to the following heat equation problem.   (    ) = 2 (    ) for 0  0 and 0 (1) (0    ) = 0 for 0 and 0 (2) (    ) = 0 for 0 and 0 (3) (  0   ) = 0 for 0 and 0 (4) (    ) = 0 for 0 and 0 (5) (   0) = (   ) for 0 and 0  (6) Solution. Suppose that is an elementary separated solution of the form (   ) = ( ) ( ) (7) for in [0   ] × [0   ] and 0 Putting this into (1) we have μ 2  2 ( ) + 2  2 ( ) ( ) =  ( ) 0 ( ) Assuming for now that (   ) 6 = 0
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