8.1: Example
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Consider a circular plate of radius c m, insulated from above and below. The temperature on the circumference is 100∘ C on half the circle, and 0∘ C on the other half.

The differential equation to solve is ρ2∂2u∂ρ2+ρ∂u∂ρ+∂2u∂ϕ2u=0, with boundary conditions u(c,ϕ)={100if 0<ϕ<π0if π<ϕ<2π.
8.1.1: Periodic BC
There is no real boundary in the ϕ direction, but we introduce one, since we choose to let ϕ run from 0 to 2π only. So what kind of boundary conditions do we apply? We would like to see “seamless behaviour”, which specifies the periodicity of the solution in ϕ, u(ρ,ϕ+2π)=u(ρ,ϕ),∂u∂ϕ(ρ,ϕ+2π)=∂u∂ϕ(ρ,ϕ). If we choose to put the seem at ϕ=−π we have the periodic boundary conditions u(ρ,2π)=u(ρ,0),∂u∂ϕ(ρ,2π)=∂u∂ϕ(ρ,0).
We separate variables, and take, as usual u(ρ,ϕ)=R(ρ)Φ(ϕ). This gives the usual differential equations Φ″−λΦ=0,ρ2R″+ρR′+λR=0. Our periodic boundary conditions gives a condition on Φ, Φ(0)=Φ(2π),Φ′(0)=Φ′(2π). The other boundary condition involves both R and Φ.