Laplace Differential Equation of a Half-Annulus

[mrkevelev]

Here is the DE:
Δu(r,θ)=0, 1 ≤ r ≤ 2, 0 ≤ θ ≤ pi
and here are the Boundary Conditions:
u(1,θ)=sin(θ), u(2,θ)=0, u(r,0)=0, u(r,pi)=0

Based on the Boundary Conditions I believe this is half of an annulus.
Using the 2D Laplace equation for polar coordinates, find the solution u(r,θ).

I've begun to separate variables, R(r) and Theta(θ), but I'm confused with whether I should use as +,-, or 0. The steps after that I am stuck on as well. Also, what kind of Laplace equation is this (Dirichlet, Neumann, etc.)?

 

[pasmith]

Here is the DE:
Δu(r,θ)=0, 1 ≤ r ≤ 2, 0 ≤ θ ≤ pi
and here are the Boundary Conditions:
u(1,θ)=sin(θ), u(2,θ)=0, u(r,0)=0, u(r,pi)=0

Based on the Boundary Conditions I believe this is half of an annulus.
Using the 2D Laplace equation for polar coordinates, find the solution u(r,θ).

I've begun to separate variables, R(r) and Theta(θ), but I'm confused with whether I should use as +,-, or 0. The steps after that I am stuck on as well. Also, what kind of Laplace equation is this (Dirichlet, Neumann, etc.)?

You have \Theta'' = C\Theta where in principle the constant of separation C can be anything, because the domain restricts \theta to only half a circle. Thus u need not be periodic and you are not for that reason compelled to take C = -n^2 for n \in \mathbb{Z}^{+}.

However you do need to satisfy the boundary conditions \Theta(0) = \Theta(\pi) = 0, and the boundary condition at r = 1 certainly suggests that \Theta = \sin \theta would be a good choice.