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math121b:04-03 [2020/04/03 06:01]
pzhou 		math121b:04-03 [2026/02/21 14:41] (current)
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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. 
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-** Case 1: Bottom and Top face boundary value = 0. $u(r=1,\theta,z)$ is given **+** Case 2: Bottom and Top face boundary value = 0. $u(r=1,\theta,z)$ is given **
 This forces $\lambda_z < 0$. In particular, $Z(z) = \sin(m \pi z)$, hence $\lambda_z = -(m\pi)^2$ and $\lambda_r = (m \pi)^2$.  The general solution is written as  This forces $\lambda_z < 0$. In particular, $Z(z) = \sin(m \pi z)$, hence $\lambda_z = -(m\pi)^2$ and $\lambda_r = (m \pi)^2$.  The general solution is written as 
 $$ u(r,\theta, z) = \sum_{n=1}^\infty \sum_{m=1}^\infty I_n( m \pi r) \sin(m \pi z) (a_{n,m} \cos(n\theta) + b_{n,m} \sin(n\theta)) $$.  $$ u(r,\theta, z) = \sum_{n=1}^\infty \sum_{m=1}^\infty I_n( m \pi r) \sin(m \pi z) (a_{n,m} \cos(n\theta) + b_{n,m} \sin(n\theta)) $$. 
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 $$ a_{n,m} = \frac{ \int_0^1 \int_0^{2\pi} u(1,\theta,z) \sin(m \pi z) \cos(n\theta) d \theta d z}{I_n( m \pi) \int_0^1 \int_0^{2\pi} \sin^2(m \pi z) \cos^2(n\theta) d \theta d z} $$ $$ a_{n,m} = \frac{ \int_0^1 \int_0^{2\pi} u(1,\theta,z) \sin(m \pi z) \cos(n\theta) d \theta d z}{I_n( m \pi) \int_0^1 \int_0^{2\pi} \sin^2(m \pi z) \cos^2(n\theta) d \theta d z} $$
 and similarly for $b_{n,m}$.  and similarly for $b_{n,m}$. 
 +
 +** General case **
 +Decompose the boundary condition into three parts, top face, bottom face and side face. Break the original problem into 3 smaller problems in which only one part is non-zero. Then add up the solution from the three smaller problems. You end up with a solution that satisfies $\Delta u =0$ and has the desired boundary value. 
  
  
  
math121b/04-03.1585893691.txt.gz · Last modified: 2026/02/21 14:44 (external edit)

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