math121b:04-03
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| Both sides previous revision Previous revision Next revision | Previous revision Last revision Both sides next revision | ||
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math121b:04-03 [2020/04/02 23:04] pzhou |
math121b:04-03 [2020/04/03 09:55] pzhou |
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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 ' | ||
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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. | ||
math121b/04-03.txt · Last modified: 2020/04/03 10:39 by pzhou
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