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math121b:04-03 [2020/04/02 23:01] 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 ' | + | where $I_n(kr)$ is the ' |
** Eigenvalue problem for $Z( z)$ ** \\ | ** Eigenvalue problem for $Z( z)$ ** \\ | ||
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** Case 1: Bottom and Side boundary value = 0. $u(r, | ** Case 1: Bottom and Side boundary value = 0. $u(r, | ||
- | Since $R(r)$ will have a zero at $r=1$, we should have $\lambad_r | + | Since $R( r)$ will have a zero at $r=1$, we should have $\lambda_r |
$$ u_{k,n}(r, \theta, z) = J_n(kr) \cos(n\theta) \sinh(k z), \quad J_n(kr) \sin(n\theta) \sinh(k z)$$ | $$ u_{k,n}(r, \theta, z) = J_n(kr) \cos(n\theta) \sinh(k z), \quad J_n(kr) \sin(n\theta) \sinh(k z)$$ | ||
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, | + | ** Case 2: Bottom and Top face boundary value = 0. $u(r=1, |
This forces $\lambda_z < 0$. In particular, $Z(z) = \sin(m \pi z)$, hence $\lambda_z = -(m\pi)^2$ and $\lambda_r = (m \pi)^2$. | This forces $\lambda_z < 0$. In particular, $Z(z) = \sin(m \pi z)$, hence $\lambda_z = -(m\pi)^2$ and $\lambda_r = (m \pi)^2$. | ||
$$ u(r,\theta, z) = \sum_{n=1}^\infty \sum_{m=1}^\infty I_n( m \pi r) \sin(m \pi z) (a_{n,m} \cos(n\theta) + b_{n,m} \sin(n\theta)) $$. | $$ u(r,\theta, z) = \sum_{n=1}^\infty \sum_{m=1}^\infty I_n( m \pi r) \sin(m \pi z) (a_{n,m} \cos(n\theta) + b_{n,m} \sin(n\theta)) $$. | ||
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$$ a_{n,m} = \frac{ \int_0^1 \int_0^{2\pi} u(1, | $$ a_{n,m} = \frac{ \int_0^1 \int_0^{2\pi} u(1, | ||
and similarly for $b_{n, | and similarly for $b_{n, | ||
+ | |||
+ | ** 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. | ||