Section 4.6The Helmholtz and Poisson Equations65Solutions to Exercises 4.61.Write (1) in polar coordinates:φrr+1rφr+1r2φθθ=-kφφ(a, θ)=0.Consider a product solutionφ(r, θ)=R(r)Θ(θ). Sinceθis a polar angle, it followsthatΘ(θ+2π)=Θ(θ);in other words, Θ is 2π-periodic. Plugging the product solution into the equationand simplifying, we FndR±±Θ+1rR±1r2RΘ±±=-kRΘ;(R±±+1rR±+)Θ=-1r2RΘ±±;r2R±±R+rR±R+kr2=-Θ±±Θ;hencer2R±±R+rR±R+2=λ,and-Θ±±Θ=λ⇒Θ±±+λΘ=0,whereλis a separation constant. Our knowledge of solutions of second order linearode’s tells us that the last equation has 2π-periodic solutions if and only ifλ=m2,m=0,±1,±2,....This leads to the equationsΘ±±+m2,andr2R±±R+rR±R+2=m2⇒r2R±±+rR±+(2-m2)R.These are equations (3) and (4). Note that the conditionR(a) = 0 follows fromφ(a, θ⇒R(a)Θ(θ⇒R(a) = 0 in order to avoid the constant 0 solution.5.We proceed as in Example 1 and tryu(r, θ∞±m=0∞±n=1Jm(λmnr)(Amncosmθ+Bmnsinmθ∞±m=0∞±n=1φmn(r, θ),whereφmn(r, θJm(
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