Author Topic: Domain of the theta function in section 6.4  (Read 1365 times)

Shentao YANG

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Domain of the theta function in section 6.4
« on: November 05, 2016, 07:14:10 PM »
Can any explain why we need the $\Theta $ function in section 6.4 (and onward) defined on $[0,2\pi ]$ instead of on $[0,2\pi )$ so that we can remove the periodic assumption of the $\Theta $ function and the boundary conditions related to $\theta $.

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter6/S6.4.html

From Wikipedia, a standard convention for defining polar coordinate system to achieve Uniqueness of polar coordinates is restrict the domain to $[0, 2\pi)$ or $(−\pi, \pi]$.

https://en.wikipedia.org/wiki/Polar_coordinate_system
« Last Edit: November 05, 2016, 07:22:07 PM by Shentao YANG »

Victor Ivrii

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Re: Domain of the theta function in section 6.4
« Reply #1 on: November 05, 2016, 09:17:35 PM »
It is a really good question. In fact, in the "standard" settings $\theta$ runs $(-\infty,\infty)$ but $\Theta$ must be $2\pi$-periodic. So problem  is
\begin{align}
&\Delta u=0,\label{eq1}\\
&u|_{r=a}=g(\theta),\label{eq2}\\
&u(2\pi)=u(0), \ u_\theta(2\pi)=u_\theta(0).\label{eq3}
\end{align}

On the other hand, we can consider membrane also in the same shape, but it is "clumped" along $\{\theta=0\}$, in which case $\theta$ runs $(0,2\pi)$ and (\ref{eq3}) is replaced by
\begin{align}
&u(2\pi)=u(0)=0.\label{eq4}
\end{align}
We can also consider membrane also in the same shape, but it is "cut" along $\{\theta=0\}$ and both sides of cut are left free  in which case $\theta$ runs $(0,2\pi)$ and (\ref{eq3}) is replaced by
\begin{align}
&u_\theta(2\pi)=u_\theta (0)=0.\label{eq5}
\end{align}
And so on. These are different problems. Separation of variables leads to different decompositions.
« Last Edit: November 06, 2016, 12:01:31 AM by Victor Ivrii »

Shentao YANG

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Re: Domain of the theta function in section 6.4
« Reply #2 on: November 05, 2016, 11:14:41 PM »
Just to make sure, in the "membrane" case you describe, we basically do not need the periodic assumption of $\Theta$ (Except the boundary condition), right?


Victor Ivrii

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Re: Domain of the theta function in section 6.4
« Reply #3 on: November 06, 2016, 12:03:57 AM »
Just to make sure, in the "membrane" case you describe, we basically do not need the periodic assumption of $\Theta$ (Except the boundary condition), right?
There is only one $2\pi$-periodic case and it is the "standard" one. In this problem condition (\ref{eq3}) is equivalent to periodicity