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2020-04-03, Friday

Today we consider PDE problem in cylindrical coordinate and spherical coordinate.

Cylindrical Coordinate

Recall that Laplacian in cylindrical coordinate $r, \theta, z$ is written as $$ \Delta u = \frac{1}{r} \d_r(r \d_r(u)) + \frac{1}{r^2} \d_\theta^2 u + \d_z^2 u. $$ We shall look for eigenfunctions of $\Delta$ of the following form $$ u(r,\theta, z) = R( r ) \Theta(\theta) Z(z), $$ Then we have $$ \frac{1}{u}\Delta u = \frac{1}{R} \frac{1}{r} \d_r(r \d_r(R)) + \frac{1}{\Theta} \frac{1}{r^2} \d_\theta^2 \Theta + \frac{1}{Z} \d_z^2 Z. $$ such that $$ \Theta^{-1} \d_\theta^2 \Theta = \lambda_\theta $$ $$ \frac{1}{R} \frac{1}{r} \d_r(r \d_r(R)) + r^{-2} \lambda_\theta = \lambda_r $$ $$ \frac{1}{Z} \d_z^2 Z = \lambda_z. $$

eigenvalue problem for $\Theta$.
$$\lambda_\theta = -n^2, \quad \Theta(\theta) = \cos(n \theta), \sin(n \theta) $$

eigenvalue problem for $R( r)$.
Assume $\lambda_\theta = -n^2$, then we get $$ r \d_r(r \d_r(R)) + (-\lambda_r r^2 - n^2) R = 0. $$

Compare with Bessel equation (12.2) from Boas $$ x(xy')' + (x^2 - p^2) y = 0 \Rightarrow y(x) = a J_p(x) + b Y_p(x).$$ And notice that we want $R(0)$ finite, and $|Y_p(0)| = \infty$, we see $$ R( r) = \begin{cases} J_n(\sqrt{-\lambda_r} r) & \lambda_r \neq 0 \cr r^n & \lambda_r = 0 \cr \end{cases} $$ Note that, the $\sqrt{-\lambda_r}$ has an ambiguity of $\pm 1$ sign, but $J_n(x)=(-1)^n J_n(-x)$, hence you can take either sign for the square root.

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) will oscillate as $r$ increase.

If $\lambda_r > 0$, say $\lambda_r = k^2$ for $k>0$, then we will take $$ R( r) = I_n(kr) = i^n J_n(i k r) $$ where $I_n(kr)$ is the 'hyperbolic' Bessel function. $R( r)$ looks like expoential function, with no oscillation.

Eigenvalue problem for $Z( z)$
$$ Z(z) = \begin{cases} e^{\sqrt{\lambda_Z} z} & \lambda_z \neq 0 \cr 1, z & \lambda_z = 0 \end{cases} $$

Hence, for any $\lambda_\theta = -n^2, \lambda_r \in \R, \lambda_z \in \R$, we have an eigen-function of Laplacian $$ u = R_{\lambda_r}( r) \Theta_n(\theta) Z_{\lambda_z}(z), \quad \Delta u = (\lambda_r + \lambda_z) u. $$ where the $R, \Theta, Z$ are given as above.

Solve the Steady State Temperature problem

Suppose we are given a cylinder of radius $1$ and height $1$ ($z \in [0,1]$), and suppose we specify the temperature on the boundary of the cylinder. We need to solve the equation $$ \Delta u = 0 $$ satisfying the boundary condition.

Note that $\lambda_z + \lambda_r = 0$, hence we will use $\lambda_r$ and $n$ to label the solutions.

We consider two special cases of boundary value specification.

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 $\lambda_r = -k^2 < 0$, and $$ u_{k,n}(r, \theta, z) = J_n(kr) \cos(n\theta) \sinh(k z), \quad J_n(kr) \sin(n\theta) \sinh(k z)$$ 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.

More precisely, let $k_{n,m} > 0$ be the $m$-th root of $J_n(x)$. Then the general solution is $$ u(r,\theta,z) = \sum_{n=1}^\infty \sum_{m=1}^\infty J_n(k_{n,m} r) \sinh(k_{n,m} z) (a_{n,m} \cos(n\theta) + b_{n,m} \sin(n\theta)) $$ Set $z=1$, we can use boundary condition to determine the coefficients $a_{n,m}, b_{n,m}$. Recall that $$ \int_0^1 J_n(k_{n,m_1} r) J_n(k_{n,m_2} r) rdr = 0, \quad m_1 \neq m_2 $$ Hence $$ a_{n,m} = \frac{\int_0^1 \int_{0}^{2\pi} u(r,\theta,1) J_n(k_{n,m} r) \cos(n\theta) r dr d\theta } {\sinh(k_{n,m}) \int_0^1 \int_{0}^{2\pi} J_n(k_{n,m} r)^2 \cos^2(n\theta) r dr d\theta } $$ and $b_{n,m}$ is obtained similarly, replacing $\cos(n\theta)$ by $\sin(n \theta)$.

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 $$ u(r,\theta, z) = \sum_{n=1}^\infty \sum_{m=1}^\infty I_n( m \pi r) \sin(m \pi z) (a_{n,m} \cos(n\theta) + b_{n,m} \sin(n\theta)) $$.

We figure out the coefficient $a_{n,m}, b_{n,m}$ using the boundary value at $r=1$. $$ u(1, \theta, z) = \sum_{n=1}^\infty \sum_{m=1}^\infty I_n( m \pi) \sin(m \pi z) (a_{n,m} \cos(n\theta) + b_{n,m} \sin(n\theta)) $$. Hence by integrating $\theta, z$ we get $$ a_{n,m} = \frac{ \int_0^1 \int_0^{2\pi} u(1,\theta,z) \sin(m \pi z) \cos(n\theta) d \theta d z}{I_n( m \pi) \int_0^1 \int_0^{2\pi} \sin^2(m \pi z) \cos^2(n\theta) d \theta d z} $$ and similarly for $b_{n,m}$.

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.txt · Last modified: 2020/04/03 10:39 by pzhou