PDE Laplace equation Calculate $u(x,0)$.

Question

Let $u(x,y)$ be a harmonic function in the domain $0\le x^2 +y^2\le 1$ that satisfy the following terms at the rim of the circle: $$\begin{cases} u(x,y)=3y & \text{if y>0} \\ u(x,y)=0 &\text{if y<0} \end{cases}$$

Calculate $u(x,0)$.

My idea is:

we know that the solution to the equation is in the form of $$u(r,\theta)=\frac{a_0}{2}+\sum_{n=1}^\infty {r^n(a_ncos{(n\theta)}+b_nsin{(}n\theta))}$$ we need to find $u(x,0)$ it's enough to calculate only the series $a_n$ but I Don't sure how we suppuse to do it :(

thanks !

Answer 1

You are on the right track. Let $f : [0,2 \pi] \to \mathbb{R}$, $f(\theta) = 3\sin(\theta)$ if $\theta \in [0,\pi]$, $f(\theta) = 0$ otherwise. Note that $u(x,y) = f(\theta)$ when $x^2 + y^2 = 1$. Substitute $r = 1$ in the solution, and impose $$ \frac{a_0}{2} + \sum_{n = 1}^\infty [a_n \cos(n\theta) + b_n \sin(n\theta)] = f(\theta). $$

Then expand $f(\theta)$ in Fourier series and impose the equality between coefficients.

Answer 2

For your problem, you only have half of a disk because $y >0$ and it is $0$ below there.

When you solve the eigenvalue problem it is a dirichlet problem

$$ \frac{d^{2}\phi}{d\theta^{2}} = -\lambda \theta \tag{1}$$ $$ \theta(0) = \theta(\pi) \tag{2} $$

$$ u(r,\theta) = \sum_{n=1}^{\infty} r^{n} B_{n} \sin(n \theta) \tag{3}$$

$$ f(\theta) = 3\sin(\theta) =\sum_{n=1}^{\infty} B_{n} \sin(n\theta) \tag{4} $$

what are the coefficients that make that true? In other words, what is $B_{n}$

$$ B_{n}a^{n} = \frac{2}{\pi} \int_{0}^{\pi} 3 \sin(\theta) \sin(n\theta) d\theta \tag{5}$$ $$ B_{n} = \frac{2}{\pi} \int_{0}^{\pi} 3 \sin(\theta) \sin(n\theta) d\theta \tag{6}$$