Chem Differential Eq HW Solutions Fall 2011 55

Chem Differential Eq HW Solutions Fall 2011 55 - u ( r, t )...

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Section 4.2 Vibrations of a Circular Membrane: Symmetric Case 55 Solutions to Exercises 4.2 1. We appeal to the solution (5) with the coefficients (6). Since f ( r ) = 0, then A n = 0 for all n . We have B n = 1 α n J 1 ( α n ) 2 ± 2 0 J 0 ( α n r 2 ) rdr = 4 α 3 n J 1 ( α n ) 2 ± α n 0 J 0 ( s ) sds (let s = α n 2 r ) = 4 α 3 n J 1 ( α n ) 2 [ sJ 1 ( s )] ² ² ² ² α n 0 = 4 α 2 n J 1 ( α n ) for all n 1 . Thus u ( r, t )=4 ³ n =1 J 0 ( α n r 2 ) α 2 n J 1 ( α n ) sin( α n t 2 ) . 5. Since g ( r ) = 0, we have B n = 0 for all n . We have A n = 2 J 1 ( α n ) 2 ± 1 0 J 0 ( α 1 r ) J 0 ( α n r ) rdr = 0 for n ± = 1 by orthogonality . For n =1 , A 1 = 2 J 1 ( α 1 ) 2 ± 1 0 J 0 ( α 1 r ) 2 rdr =1 , where we have used the orthogonality relation (12), Section 4.8, with p = 0. Thus u ( r, t )= J 0 ( α 1 r ) cos( α 1 t ) . 9. (a) Modifying the solution of Exercise 3, we obtain u ( r, t )= ³ n =1 J 1 ( α n / 2) α 2 n cJ 1 ( α n ) 2 J 0 ( α n r ) sin( α n ct ) . (b) Under suitable conditions that allow us to interchange the limit and the sum- mation sign (for example, if the series is absolutely convergent), we have, for a given ( r, t ), lim c
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Unformatted text preview: u ( r, t ) = lim c n =1 J 1 ( n / 2) 2 n cJ 1 ( n ) 2 J ( n r ) sin( n ct ) = n =1 lim c J 1 ( n / 2) 2 n cJ 1 ( n ) 2 J ( n r ) sin( n ct ) = , because lim c J 1 ( n / 2) 2 n cJ 1 ( n ) 2 = 0 and sin( n ct ) is bounded. If we let u 1 ( r, t ) denote the solution corresponding to c = 1 and u c ( r, t ) denote the solution for arbitrary c > 0. Then, it is easy t check that u c ( r, t ) = 1 c u 1 ( r, ct ) . This shows that if c increases, the time scale speeds proportionally to c , while the displacement decreases by a factor of 1 c ....
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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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