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math121b:04-03 [2020/04/02 23:04]
pzhou
math121b:04-03 [2020/04/03 10:39] (current)
pzhou [Cylindrical Coordinate]
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 If $\lambda_r < 0$, say $\lambda_r = -k^2$ for $k>0$, then   If $\lambda_r < 0$, say $\lambda_r = -k^2$ for $k>0$, then  
-$$ R(r) = J_n(\sqrt{-\lambda_r} r) = J_n(kr) $$ +$$ R( r) = J_n(\sqrt{-\lambda_r} r) = J_n(kr) $$ 
-$R(r) will oscillate as $r$ increase. +$R( r) will oscillate as $r$ increase. 
  
 If $\lambda_r > 0$, say $\lambda_r = k^2$ for $k>0$, then we will take  If $\lambda_r > 0$, say $\lambda_r = k^2$ for $k>0$, then we will take 
-$$ R(r) =  I_n(kr) = i^n J_n(i k r) $$ +$$ R( r) =  I_n(kr) = i^n J_n(i k r) $$ 
-where $I_n(kr)$ is the 'hyperbolic' Bessel function. $R(r)$ looks like expoential function, with no oscillation. +where $I_n(kr)$ is the 'hyperbolic' Bessel function. $R( r)$ looks like expoential function, with no oscillation. 
  
 ** Eigenvalue problem for $Z( z)$ ** \\ ** Eigenvalue problem for $Z( z)$ ** \\
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 ** Case 1: Bottom and Side boundary value = 0. $u(r,\theta,z=1)$ is given ** ** Case 1: Bottom and Side boundary value = 0. $u(r,\theta,z=1)$ is given **
-Since $R(r)$ will have a zero at $r=1$, we should have $\lambad_r = -k^2 < 0$, and+Since $R( r)$ will have a zero at $r=1$, we should have $\lambda_r = -k^2 < 0$, and
 $$ u_{k,n}(r, \theta, z) = J_n(kr) \cos(n\theta) \sinh(k z), \quad J_n(kr) \sin(n\theta) \sinh(k z)$$ $$ u_{k,n}(r, \theta, z) = J_n(kr) \cos(n\theta) \sinh(k z), \quad J_n(kr) \sin(n\theta) \sinh(k z)$$
 For each $n$, we need to choose those $k$ such that $J_n(k 1) = 0$, hence the choice of $k$ is discrete as well.  For each $n$, we need to choose those $k$ such that $J_n(k 1) = 0$, hence the choice of $k$ is discrete as well. 
math121b/04-03.1585893863.txt.gz · Last modified: 2020/04/02 23:04 by pzhou