- #1
Athenian
- 143
- 33
- Homework Statement
- Original Problem:
Solve the Laplace’s equation in ##2## dimensions in polar coordinates and solve the ##r## and ##\theta## equation. Solve the problem of the steady-state temperature in a circular plate if the upper semicircular boundary is held at ##100## and the lower at ##0##.
[Full + Correct Solution Process Shown BELOW].
My Question:
Rather than having upper/lower semicircular regions (i.e. boundaries) that have two different temperatures (i.e. 0 and 100), let's say that the upper half of the circle's circumference is held at ##100## degrees whereas the lower half is held at ##0## degree. How would one then find the steady-state temperature distribution for the given circular plate in this case?
[Note that both sides of the circle are insulated against heat]
- Relevant Equations
- Refer Below ##\longrightarrow##
I learned about Bessel functions and steady-state temperature distributions in the past. Recently, I was searching online for some example problems on the topic and found the "original question" along with the solution online as a PDF file.
While I am unsure will it be appropriate for me to post these kinds of questions under the "homework" section in Physicsforum, I decided to set up a situation where rather than the two halves of the circular regions holding different temperatures, I made it into two halves of the circular perimeter that hold different temperatures. That said, I am curious about how one would go about calculating this kind of problem.
Below is the solution process of the "original question" written verbatim here by re-typing the text from http://scipp.ucsc.edu/~haber/archives/physics116C11/mathphysC6sol_11.pdf.
Solution Process:
The Laplacian in polar coordinates is
$$\nabla^2 = \frac{1}{r} \frac{d}{dr} \bigg( r \frac{d}{dr} \bigg) + \frac{1}{r^2} \frac{d^2}{d \theta^2}$$
so that Laplace’s equation ##\nabla^2 u = 0## becomes (with ##u = R(r)\Theta(\theta)##)
$$\frac{r}{R} (rR')' = -\frac{\Theta''}{\Theta} = n^2$$
The angular dependence is ##\Theta = e^{\pm i n \theta}## and ##n## must be an integer for the function to be single-valued, ##\Theta(\theta + 2\pi) = \Theta(\theta)##. The radial equation is
$$r^2 R'' + rR' - n^2 R = 0$$
To solve this we try a solution ##R = r^{\alpha}## and substituting we find ##\alpha = \pm n##. The solution to Laplace’s equation is then
$$u = \sum_{n=0} [r^n (A_n \sin(n \theta) + B_n \cos (n\theta)) + r^{-n} (A_n ' \sin (n\theta) + B_n ' \cos (n\theta)) ]$$
To solve the requested problem, first we put ##A^′ = B^′ = 0## because the temperature would diverge in the center of the plate. Then we apply the boundary condition ##u(a, \theta) = 100H(−\theta + \pi)## where ##H## is the Heaviside step function:
$$u(a, \theta) = 100H (-\theta + \pi) = \sum_{n=0} a^n (A_n \sin (n\theta) + B_n \cos (n\theta) )$$
Now we must multiply this equation by ##\sin(m\theta)## or ##\cos (m\theta)## and integrate between ##(0, 2\pi)## in order to find respectively ##A_n## and ##B_n##. Because ##\int_0^\pi \cos (m\theta) d\theta = \pi \delta_{m0}##, only the constant term in the cosine series is non-zero. It is given by
$$100 \int_0^\pi d\theta = B_0 \int_0^{2\pi} d\theta = 2\pi \Longrightarrow B_0 = 50$$
$$a^n A_n = \frac{1}{\pi} \int_0^\pi 100 \sin (n\theta) d\theta = \frac{200}{n\pi} \text{ for odd } n, \text{and 0 otherwise}$$
We can rewrite the solution as
$$u = 50 + \frac{200}{\pi} \sum_{odd \; n} \bigg( \frac{r}{a}\bigg)^n \frac{\sin (n\theta)}{n}$$
My Question:
When changing the boundary condition as explained in the homework statement (i.e. constant temperatures are located at the circumference rather than the circular region itself), how would it affect the entire solution process?
If anyone can answer the question, thank you very much!
While I am unsure will it be appropriate for me to post these kinds of questions under the "homework" section in Physicsforum, I decided to set up a situation where rather than the two halves of the circular regions holding different temperatures, I made it into two halves of the circular perimeter that hold different temperatures. That said, I am curious about how one would go about calculating this kind of problem.
Below is the solution process of the "original question" written verbatim here by re-typing the text from http://scipp.ucsc.edu/~haber/archives/physics116C11/mathphysC6sol_11.pdf.
Solution Process:
The Laplacian in polar coordinates is
$$\nabla^2 = \frac{1}{r} \frac{d}{dr} \bigg( r \frac{d}{dr} \bigg) + \frac{1}{r^2} \frac{d^2}{d \theta^2}$$
so that Laplace’s equation ##\nabla^2 u = 0## becomes (with ##u = R(r)\Theta(\theta)##)
$$\frac{r}{R} (rR')' = -\frac{\Theta''}{\Theta} = n^2$$
The angular dependence is ##\Theta = e^{\pm i n \theta}## and ##n## must be an integer for the function to be single-valued, ##\Theta(\theta + 2\pi) = \Theta(\theta)##. The radial equation is
$$r^2 R'' + rR' - n^2 R = 0$$
To solve this we try a solution ##R = r^{\alpha}## and substituting we find ##\alpha = \pm n##. The solution to Laplace’s equation is then
$$u = \sum_{n=0} [r^n (A_n \sin(n \theta) + B_n \cos (n\theta)) + r^{-n} (A_n ' \sin (n\theta) + B_n ' \cos (n\theta)) ]$$
To solve the requested problem, first we put ##A^′ = B^′ = 0## because the temperature would diverge in the center of the plate. Then we apply the boundary condition ##u(a, \theta) = 100H(−\theta + \pi)## where ##H## is the Heaviside step function:
$$u(a, \theta) = 100H (-\theta + \pi) = \sum_{n=0} a^n (A_n \sin (n\theta) + B_n \cos (n\theta) )$$
Now we must multiply this equation by ##\sin(m\theta)## or ##\cos (m\theta)## and integrate between ##(0, 2\pi)## in order to find respectively ##A_n## and ##B_n##. Because ##\int_0^\pi \cos (m\theta) d\theta = \pi \delta_{m0}##, only the constant term in the cosine series is non-zero. It is given by
$$100 \int_0^\pi d\theta = B_0 \int_0^{2\pi} d\theta = 2\pi \Longrightarrow B_0 = 50$$
$$a^n A_n = \frac{1}{\pi} \int_0^\pi 100 \sin (n\theta) d\theta = \frac{200}{n\pi} \text{ for odd } n, \text{and 0 otherwise}$$
We can rewrite the solution as
$$u = 50 + \frac{200}{\pi} \sum_{odd \; n} \bigg( \frac{r}{a}\bigg)^n \frac{\sin (n\theta)}{n}$$
My Question:
When changing the boundary condition as explained in the homework statement (i.e. constant temperatures are located at the circumference rather than the circular region itself), how would it affect the entire solution process?
If anyone can answer the question, thank you very much!