Chem Differential Eq HW Solutions Fall 2011 63

Chem Differential Eq HW Solutions Fall 2011 63 - Section...

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Section 4.5 Steady-State Temperature in a Cylinder 63 Solutions to Exercises 4.5 1. Using (2) and (3), we have that u ( ρ, z )= ± n =1 A n J 0 ( λ n ρ ) sinh( λ n z ) n = α n a , where α n = α 0 ,n is the n th positive zero of J 0 , and A n = 2 sinh( λ n h ) a 2 J 1 ( α n ) 2 ² a 0 f ( ρ ) J 0 ( λ n ρ ) ρdρ = 200 sinh(2 α n ) J 1 ( α n ) 2 ² 1 0 J 0 ( α n ρ ) = 200 sinh(2 α n ) α 2 n J 1 ( α n ) 2 ² α n 0 J 0 ( s ) sds (let s = α n ρ ) = 200 sinh(2 α n ) α 2 n J 1 ( α n ) 2 [ J 1 ( s ) s ] ³ ³ ³ ³ α n 0 = 200 sinh(2 α n ) α n J 1 ( α n ) . So u ( ρ, z )=200 ± n =1 J 0 ( α n ρ ) sinh( α n z ) sinh(2 α n ) α n J 1 ( α n ) . 5. (a) We proceed exactly as in the text and arrive at the condition Z ( h )=0 which leads us to the solutions Z ( z Z n ( z ) = sinh( λ n ( h - z )) , where λ n = α n a . So the solution of the problem is u ( ρ, z ± n =1 C n J 0 ( λ n ρ ) sinh( λ n ( h - z )) , where C n = 2 a 2 J 1 ( α n ) 2 sinh( λ n h ) ² a 0 f ( ρ ) J 0 ( λ n ρ ) ρ dρ. (b) The problem can be decomposed into the sum of two subproblems, one
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