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Chem Differential Eq HW Solutions Fall 2011 56

# Chem Differential Eq HW Solutions Fall 2011 56 - 56 Chapter...

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56 Chapter 4 Partial Differential Equations in Polar and Cylindrical Coordinates Solutions to Exercises 4.3 1. The condition g ( r,θ ) = 0 implies that a * mn = 0 = b * mn . Since f ( r,θ ) is proportional to sin 2 θ , only b 2 ,n will be nonzero, among all the a mn and b mn . This is similar to the situation in Example 2. For n = 1 , 2 ,... , we have b 2 ,n = 2 πJ 3 ( α 2 ,n ) 2 1 0 2 π 0 (1 - r 2 ) r 2 sin 2 θJ 2 ( α 2 ,n r ) sin 2 θrdθdr = 2 πJ 3 ( α 2 ,n ) 2 1 0 = π 2 π 0 sin 2 2 θdθ (1 - r 2 ) r 3 J 2 ( α 2 ,n r ) dr = 2 J 3 ( α 2 ,n ) 2 1 0 (1 - r 2 ) r 3 J 2 ( α 2 ,n r ) dr = 2 J 3 ( α 2 ,n ) 2 2 α 2 2 ,n J 4 ( α 2 ,n ) = 4 J 4 ( α 2 ,n ) α 2 2 ,n J 3 ( α 2 ,n ) 2 , where the last integral is evaluated with the help of formula (15), Section 4.3. We can get rid of the expression involving J 4 by using the identity J p - 1 ( x ) + J p +1 ( x ) = 2 p x J p ( x ) . With p = 3 and x = α 2 ,n , we get =0 J 2 ( α 2 ,n ) + J 4 ( α 2 ,n ) = 6 α 2 ,n J 3 ( α 2 ,n ) J 4 ( α 2 ,n ) = 6 α 2 ,n J 3 ( α 2 ,n ) . So b 2 ,n = 24 α 3 2 ,n J 3 ( α 2 ,n ) . Thus u ( r,θ,t ) = 24 sin 2 θ n =1 J 2 ( α 2 ,n r ) α 3 2 ,n J 3 ( α 2 ,n ) cos( α 2 ,n t ) . 5. We have a mn = b mn = 0. Also, all a * mn and b * mn are zero except
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