temperature in a hemisphere - 3

[logistic_guy]

The steady-state temperature in a hemisphere of radius \(\displaystyle r = c\) is determined from

\(\displaystyle \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cot \theta}{r^2}\frac{\partial u}{\partial \theta} = 0\)

\(\displaystyle r > c, \ \ \ \ 0 < \theta < \frac{\pi}{2}\)

\(\displaystyle u\left(r,\frac{\pi}{2}\right) = 0, \ \ \ \ r > c\)

\(\displaystyle u(c,\theta) = f(\theta), \ \ \ \ 0 < \theta < \frac{\pi}{2}\)

Solve for \(\displaystyle u(r,\theta)\). [Hint: \(\displaystyle P_{n}(0) = 0\) only if \(\displaystyle n\) is odd.]

 

[khansaheb]

The steady-state temperature in a hemisphere of radius \(\displaystyle r = c\) is determined from

\(\displaystyle \frac{\partial^2 u}{\partial r^2} + \frac{2}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} + \frac{\cot \theta}{r^2}\frac{\partial u}{\partial \theta} = 0\)

\(\displaystyle r > c, \ \ \ \ 0 < \theta < \frac{\pi}{2}\)

\(\displaystyle u\left(r,\frac{\pi}{2}\right) = 0, \ \ \ \ r > c\)

\(\displaystyle u(c,\theta) = f(\theta), \ \ \ \ 0 < \theta < \frac{\pi}{2}\)

Solve for \(\displaystyle u(r,\theta)\). [Hint: \(\displaystyle P_{n}(0) = 0\) only if \(\displaystyle n\) is odd.]

show us your effort/s to solve this problem.

 

[logistic_guy]

We have already solved this partial differential equation twice, but before we write the general solution let us go back one step. Let us write the solution of the radial equation again.

\(\displaystyle R(r) = c_1 r^n + c_2\frac{1}{r^{n+1}}\)

The interval for \(\displaystyle r\) is now \(\displaystyle c < r < \infty\).

\(\displaystyle c_1r^n \rightarrow \infty \ \ \ \) as \(\displaystyle r \rightarrow \infty\).
\(\displaystyle c_2\frac{1}{r^{n+1}} \rightarrow 0 \ \ \ \) as \(\displaystyle r \rightarrow \infty\).

We are only interested in bounded solutions, so we have to choose \(\displaystyle c_1 = 0\) to get rid of the unbounded solution.

Then,

\(\displaystyle R(r) = c_2\frac{1}{r^{n+1}}\)

And the general solution now becomes:

\(\displaystyle u(r,\theta) = \sum_{n=0}^{\infty}A_n \frac{1}{r^{n+1}}P_n(\cos\theta)\)

Let us apply the first boundary condition.

\(\displaystyle u\left(r,\frac{\pi}{2}\right) = \sum_{n=0}^{\infty}A_n \frac{1}{r^{n+1}}P_n(\cos\frac{\pi}{2}) = \sum_{n=0}^{\infty}A_n \frac{1}{r^{n+1}}P_n(0) = 0\)

We know that \(\displaystyle A_n \neq 0\) and \(\displaystyle \frac{1}{r^{n + 1}} \neq 0\), so it must be \(\displaystyle P_n(0) = 0\).

From previous hint we know that \(\displaystyle P_n(0) = 0\) only when \(\displaystyle n\) is odd. In other words, \(\displaystyle n = 1, 3, 5, \cdots\)

Or

\(\displaystyle 2n + 1, \ \ \ n = 0, 1, 2, \cdots\)

Then, the general solution becomes:

\(\displaystyle u(r,\theta) = \sum_{n=0}^{\infty}A_{2n + 1} \frac{1}{r^{2n + 2}}P_{2n + 1}(\cos\theta)\)

We will apply the second boundary condition so that we can find the constant \(\displaystyle A_{2n+1}\).

\(\displaystyle u(c,\theta) = f(\theta) = \sum_{n=0}^{\infty}A_{2n + 1} \frac{1}{c^{2n + 2}}P_{2n + 1}(\cos\theta)\)

We have already done this step twice. This gives:

\(\displaystyle A_{2n+1} = (4n + 3)c^{2n+n}\int_{0}^{\pi/2}f(\theta)P_{2n+1}(\cos \theta)\sin \theta \ d\theta\)

 

[khansaheb]

I think @logistic_guy wants to be recognized as a teacher the worst way. Unfortunately he doesn't realize giving away xerox able answer is NOT one of the desired virtues of a teacher.

 

[logistic_guy]

I think @logistic_guy wants to be recognized as a teacher the worst way. Unfortunately he doesn't realize giving away xerox able answer is NOT one of the desired virtues of a teacher.

I know that you think I have the behavior of a bad teacher. But be frank with me, if you were my student, how would you rate my explanation style? \(\displaystyle 10\) of \(\displaystyle 10\) is the best.

 

[khansaheb]

I know that you think I have the behavior of a bad teacher. But be frank with me, if you were my student, how would you rate my explanation style? \(\displaystyle 10\) of \(\displaystyle 10\) is the best.

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