# [SOLVED]Checking Laplace soln

#### dwsmith

##### Well-known member
Solve Laplace's equation $\nabla^2u = 0$ on a $90^{\circ}$ wedge of radius $a$ subject to the boundary conditions
$$u(r,0) = 0\quad u_{\theta}\left(r,\frac{\pi}{2}\right) = 0\quad u(a,\theta) = f(\theta).$$
The standard form of the solutions for $R(r)$ and $\Theta(\theta)$ are
$$\Theta(\theta) = A\cos\lambda\theta + B\sin\lambda\theta\quad\text{and}\quad R(r) = r^{\pm\lambda}.$$
Using the first boundary condition, we have $A = 0$, i.e.
$$\Theta(\theta) = B\sin\lambda\theta.$$
Using the second boundary condition, we have
$$\lambda B\cos\frac{\pi\lambda}{2} = 0.$$
Therefore, $\lambda = n$ where $n\in\mathbb{Z}$.
The general solution is
$$u(r,\theta) = \sum_{n = 1}^{\infty}B_nr^{n}\sin n\theta.$$
With the last condition, we have
$$u(a,\theta) = \sum_{n = 1}^{\infty}B_na^{n}\sin n\theta = f(\theta)$$
where the Fourier coefficients are $B_n = \frac{1}{a^n\pi}\int_{-\pi}^{\pi}f(\theta)\sin n\theta d\theta$.
Then the general solution with the defined conditions is $u(r,\theta) = \sum\limits_{n = 1}^{\infty}B_nr^{n}\sin n\theta$ where $B_n$ is defined as above.

Let's take the case where $f(\theta) = \theta$.
\begin{alignat*}{3}
B_n & = & \frac{1}{a^n\pi}\int_{-\pi}^{\pi}f(\theta)\sin n\theta d\theta\\
& = & \frac{1}{a^n\pi}\left[\left.-\frac{\theta\cos n\theta}{n}\right|_{-\pi}^{\pi} + \left.\frac{\sin n\theta}{n^2}\right|_{-\pi}^{\pi}\right]\\
& = & \frac{2(-1)^{n + 1}}{a^nn}
\end{alignat*}
The solution to this Laplace equation when $f(\theta) = \theta$ is
$$u(r,\theta) = 2\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}r^{n}}{a^nn}\sin n\theta.$$

#### Sudharaka

##### Well-known member
MHB Math Helper
Using the second boundary condition, we have
$$\lambda B\cos\frac{\pi\lambda}{2} = 0.$$
Therefore, $\lambda = n$ where $n\in\mathbb{Z}$.
This is incorrect. The cosine function becomes zero only when $$\lambda$$ takes odd values. That is, $$\lambda=2n+1$$ where $$n\in\mathbb{Z}$$.

#### dwsmith

##### Well-known member
This is incorrect. The cosine function becomes zero only when $$\lambda$$ takes odd values. That is, $$\lambda=2n+1$$ where $$n\in\mathbb{Z}$$.
So after I made the correction, I ended up with

$$u(r,\theta) = 2\sum\limits_{n = 1}^{\infty}\frac{r^{n}}{a^{2n - 1}(2n - 1)}\sin (2n - 1)\theta.$$