We have already solved this partial differential equation and the general solution was:
\(\displaystyle u(r,\theta) = \sum_{n=0}^{\infty}A_nr^nP_n(\cos \theta)\)
The difference between this problem and the last one is that one of the boundary conditions is different. Let us apply it.
\(\displaystyle u'\left(r,\frac{\pi}{2}\right) = \sum_{n=0}^{\infty}A_nr^nP'_n\left(\cos \frac{\pi}{2}\right) = \sum_{n=0}^{\infty}A_nr^nP'_n(0) = 0\)
We know that \(\displaystyle A_n \neq 0\) and \(\displaystyle r^n \neq\), so it must be \(\displaystyle P'_n(0) = 0\).
We want to know when is the derivative of the Legendre function is zero. Let us find the first \(\displaystyle 5\) Legendre functions.
\(\displaystyle P_0(x) = 1\)
\(\displaystyle P_1(x) = x\)
\(\displaystyle P_2(x) = \frac{1}{2}(3x^2 - 1)\)
\(\displaystyle P_3(x) = \frac{1}{2}(5x^3 - 3x)\)
\(\displaystyle P_4(x) = \frac{1}{8}(35x^4-30x^2 + 3)\)
Let us take their derivatives at \(\displaystyle x = 0\).
\(\displaystyle P'_0(0) = 0\)
\(\displaystyle P'_1(0) = 1\)
\(\displaystyle P'_2(0) = 0\)
\(\displaystyle P'_3(0) = -\frac{3}{2}\)
\(\displaystyle P'_4(0) = 0\)
This pattern tells us that the derivative of the Legendre function is zero when \(\displaystyle n\) is even, that is when \(\displaystyle n = 0, 2, 4, \cdots\).
Or
\(\displaystyle 2n, \ \ \ n = 0, 1, 2, \cdots\)
Then, the general solution becomes:
\(\displaystyle u(r,\theta) = \sum_{n=0}^{\infty}A_{2n}r^{2n}P_{2n}(\cos \theta)\)
Now we can apply the second boundary condition which will help us find the constant \(\displaystyle A_{2n}\).
\(\displaystyle u(c,\theta) = f(\theta) = \sum_{n=0}^{\infty}A_{2n}c^{2n}P_{2n}(\cos \theta)\)
\(\displaystyle \int_{0}^{\pi} f(\theta)P_{2n}(\cos \theta) \sin \theta \ d\theta = \sum_{n=0}^{\infty}A_{2n}c^{2n}\int_{0}^{\pi}P_{2n}(\cos \theta)P_{2n}(\cos \theta) \sin \theta \ d\theta\)
\(\displaystyle \int_{0}^{\pi} f(\theta)P_{2n}(\cos \theta) \sin \theta \ d\theta = \sum_{n=0}^{\infty}A_{2n}c^{2n}\int_{0}^{\pi}P^2_{2n}(\cos \theta) \sin \theta \ d\theta\)
\(\displaystyle \int_{0}^{\pi} f(\theta)P_{2n}(\cos \theta) \sin \theta \ d\theta = \sum_{n=0}^{\infty}A_{2n}c^{2n}\frac{2}{2(2n) + 1}\)
This gives:
\(\displaystyle A_{2n} = \frac{4n + 1}{2c^{2n}}\int_{0}^{\pi} f(\theta)P_{2n}(\cos \theta) \sin \theta \ d\theta\)
Properties of Legendry function:
\(\displaystyle \bold{1.}\) When \(\displaystyle n\) is even, \(\displaystyle P_n(x)\) is even.
\(\displaystyle \bold{2.}\) When \(\displaystyle n\) is odd, \(\displaystyle P_n(x)\) is odd.
In calculus, we have learnt that \(\displaystyle \int_{-1}^{1} = 2\int_{0}^{1}\) when the function is even. This means that \(\displaystyle \int_{0}^{\pi} = 2\int_{0}^{\frac{\pi}{2}}\).
Then,
\(\displaystyle A_{2n} = \frac{4n + 1}{c^{2n}}\int_{0}^{\frac{\pi}{2}} f(\theta)P_{2n}(\cos \theta) \sin \theta \ d\theta\)
The final solution is:
\(\displaystyle u(r,\theta) = \sum_{n=0}^{\infty}A_{2n}r^{2n}P_{2n}(\cos \theta)\)
where \(\displaystyle A_{2n}\) as mentioned above.
