Differential Equations Lecture Work Solutions 118

Differential Equations Lecture Work Solutions 118 - 1 c = 0...

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1 c. Θ + µ Θ = 0 Z λ Z = 0 r ( r R ) + ( λ r 2 µ ) R = 0 Z ( H ) = 0 | R (0) | < R ( a ) = 0 Solution as in 1a exactly ! But u z ( r, θ, 0) = α ( r, θ ) = m =0 n =1 ( a nm cos m θ + b nm sin m θ ) J m ( λ nm r ) λ nm cosh λ nm ( H ) a nm λ nm cosh λ nm ( H ) = 2 π 0 a 0 α ( r, θ ) cos m θ J m ( λ nm r ) rd rdθ 2 π 0 a 0 cos 2 m θ J 2 m ( λ nm r ) rd rdθ a nm = 2 π 0 a 0 α ( r, θ ) cos m θ J m ( λ nm r ) rd rdθ λ nm cosh λ nm ( H ) 2 π 0 a 0 cos
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