Chem Differential Eq HW Solutions Fall 2011 81

# Chem Differential Eq HW Solutions Fall 2011 81 - Section...

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Section 5.2 Dirichlet Problems with Symmetry 81 Solutions to Exercises 5.2 1. This problem is similar to Example 2. Note that f is its own Legendre series: f ( θ )=20( P 1 (cos θ )+ P 0 (cos θ )) . So really there is no need to compute the Legendre coeﬃcients using integrals. We simply have A 0 = 20 and A 1 = 20, and the solution is u ( r, θ ) = 20 + 20 r cos θ. 3. We have u ( r, θ )= ± n =0 A n r n P n (cos θ ) , with A n = 2 n +1 2 ² π 0 f ( θ ) P n (cos θ ) sin θdθ = 2 n 2 ² π 2 0 100 P n (cos θ ) sin + 2 n 2 ² π π 2 20 P n (cos θ ) sin θdθ. Let x = cos θ , dx = - sin . Then A n = 50(2 n +1) ² 1 0 P n ( x ) dx + 10 (2 n ² 0 - 1 P n ( x ) dx. The case n = 0 is immediate by using P 0 ( x )=1 , A 0 =50 ² 1 0 dx +10 ² 0 - 1 dx =60 . For n> 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 have ² 1 0 P 2 n ( x ) dx =0 ,n =1 , 2 ,..., and ² 1 0 P 2 n +1 ( x ) dx = ( - 1) n (2 n )! 2 2 n +1 ( n !) 2 ( n , 1 , 2 ,.... Since P 2 n ( x ) is an even function, then, for
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