Section 5.2 Dirichlet Problems with Symmetry81Solutions to Exercises 5.21.This problem is similar to Example 2. Note thatfis its own Legendre series:f(θ)=20(P1(cosθ)+P0(cosθ)).So really there is no need to compute the Legendre coeﬃcients using integrals. Wesimply haveA0= 20 andA1= 20, and the solution isu(r, θ) = 20 + 20rcosθ.3.We haveu(r, θ)=∞±n=0AnrnPn(cosθ),withAn=2n+12²π0f(θ)Pn(cosθ) sinθdθ=2n2²π20100Pn(cosθ) sin+2n2²ππ220Pn(cosθ) sinθdθ.Letx= cosθ,dx=-sin. ThenAn=50(2n+1)²10Pn(x)dx+ 10 (2n²0-1Pn(x)dx.The casen= 0 is immediate by usingP0(x)=1,A0=50²10dx+10²0-1dx=60.Forn>0, the integrals are not straightforward and you need to refer to Exercise 10,Section 5.6, where they are evaluated. Quoting from this exercise, we have²10P2n(x)dx=0,n=1,2,...,and²10P2n+1(x)dx=(-1)n(2n)!22n+1(n!)2(n,1,2,....SinceP2n(x) is an even function, then, for
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