Differential Equations Lecture Work Solutions 125

Differential Equations Lecture Work Solutions 125 - 5. u t...

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5. u t = 1 r ( ru r ) r + 1 r 2 u θθ + u zz BC: u ( r, θ, 0) = 0 u ( r, θ, H )=0 u ( a, θ, z )=0 u ( r, θ, z, t )= R ( r )Θ( θ ) Z ( z ) T ( t ) R Θ ZT 0 = 1 r Θ ZT ( rR 0 ) 0 + 1 r 2 RZT Θ 0 + R Θ TZ 0 T 0 T = 1 r ( rR 0 ) 0 R + 1 r 2 Θ 0 Θ + Z 0 Z Z 0 Z = T 0 T 1 r ( rR 0 ) 0 R 1 r 2 Θ 0 Θ = λ Z 0 + λZ =0 BC : Z (0) = 0 Z ( H )=0 Z n =s in H z λ n = H ± 2 n =1 , 2 , ··· 1 r 2 Θ 0 Θ = T 0 T 1 r ( rR 0 ) 0 R + H ± 2 Θ 0 Θ = T 0 T r 2 r ( rR 0 ) 0 R +
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