Differential Equations Lecture Work Solutions 126

Differential Equations Lecture Work Solutions 126 - sin m m...

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Θ m = sin cos µ m = m 2 m =0 , 1 , 2 , ··· T 0 T r 2 = r ( rR 0 ) 0 R H ± 2 r 2 m 2 T 0 T = 1 r ( rR 0 ) 0 R H ± 2 m 2 r 2 = ν T 0 + νT =0 1 r ( rR 0 ) 0 R = ν + H ± 2 + m 2 r 2 r ( rR 0 ) 0 R = νr 2 + H ± 2 r 2 + m 2 r ( rR 0 ) 0 ( ν H 2 ) r 2 R m 2 R =0 BC : | R (0) | < R ( a )=0 R nm` = I m s ν ` H ± 2 r This solution satisfes the boundedness at the origin. The eigenvalues ν ` can be Found by using the second boundary condition: I m s ν ` H ± 2 a =0 Since the Function I m ( x ) vanishes only at zero For any m =1 , 2 , ··· ( I 0 is never zero) then thereison lyone ν (For any n) satisFying s ν
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