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--- math121b:04-03 [2020/04/02 19:49]
pzhou created
+++ math121b:04-03 [2020/04/03 10:39] (current)
pzhou [Cylindrical Coordinate]
@@ Line 11: / Line 11: @@
 $$ \frac{1}{u}\Delta u =  \frac{1}{R} \frac{1}{r} \d_r(r \d_r(R)) + \frac{1}{\Theta} \frac{1}{r^2} \d_\theta^2 \Theta + \frac{1}{Z} \d_z^2 Z. $$
 such that
-$$ \d_\theta^2 \Theta(\theta) = \lambda_\theta \Theta(\theta) $$
+$$ \Theta^{-1} \d_\theta^2 \Theta  = \lambda_\theta  $$
 $$  \frac{1}{R} \frac{1}{r} \d_r(r \d_r(R))  + r^{-2} \lambda_\theta = \lambda_r $$
-$$ \frac{1}{Z} \d_z^2 Z = \lambda_Z. $$
+$$ \frac{1}{Z} \d_z^2 Z = \lambda_z. $$
-Step 1: Solve the eigenvalue problem for $\Theta$. Since $\Theta(\theta) = \Theta(\theta+2\pi)$, the eigenvalues of $\theta$ are $\sin(n\theta), \cos(n\theta)$ with eigenvalues $\lambda_\theta = -n^2$ for integer $n \geq 0$. More precisely, for $\lambda_\theta=0$, the eigen-space is one-dimensional, and is generated by the constant function $1$; for $\lambda_\theta = -n^2$ for positive integer $n$, we have 2-dimensional eigenspaces $V_{\lambda_\theta}$, generated by $\sin(n\theta), \cos(n\theta)$.
+** 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
+$$ r \d_r(r \d_r(R)) + (-\lambda_r r^2 - n^2) R = 0. $$
-Step 2: Solve the eigenvalue problem for $R( r)$. The eigenvalue problem is
-$$   \frac{1}{R} \frac{1}{r} \d_r(r \d_r(R))  + r^{-2} (-n^2) = \lambda_r $$
-where we plugged in $\lambda_\theta$'s possible values. Re-arranging terms, 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
-$$ x(xy')' + (x^2 - p^2) y = 0 $$
+$$ x(xy')' + (x^2 - p^2) y = 0  \Rightarrow y(x) = a J_p(x) + b Y_p(x).$$
+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 $Z( z)$ ** \\
+$$ Z(z) = \begin{cases}
+ e^{\sqrt{\lambda_Z} z} & \lambda_z \neq 0 \cr
+, z & \lambda_z = 0
+\end{cases}
+$$
+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.

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