Chem Differential Eq HW Solutions Fall 2011 61

Chem Differential Eq HW Solutions Fall 2011 61 - Section...

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Unformatted text preview: Section 4.4 and Steady-State Temperature in a Disk 61 2π n 1 bn = a π f (θ) sin nθ dθ. 0 Hence an = 2π a nπ f (θ) cos nθ dθ and bn = 0 a nπ 2π f (θ) sin nθ dθ. 0 Note that a0 is still arbitray. Indeed, the solution of a Neumann problem is not unique. It can be determined only up to an additive constant (which does not affect the value of the normal derivative at the boundary). 21. Using the fact that the solutions must be bounded as r → ∞, we see that c1 = 0 in the first of the two equations in (3), and c2 = 0 in the second of the two equations in (3). Thus r a R(r) = Rn(r) = cnr−n = n for n = 0, 1, 2, . . . . The general solution becomes ∞ u(r, θ) = a0 + n=1 n a r (an cos nθ + bn sin nθ) , r > a. Setting r = a and using the boundary condition, we obtain ∞ f (θ) = a0 + (an cos nθ + bn sin nθ) , n=1 which implies that the an and bn are the Fourier coefficients of f and hence are given by (5). 25. The hint does it. 29. (a) Recalling the Euler formulas for the Fourier coefficients, we have ∞ u(r, θ) = a0 + n=1 = 1 2π f (φ) dφ n=1 1 2π 1 π 2π f (φ) cos nφ dφ cos nθ + 0 1 π 2π f (φ) sin nφ dφ sin nθ 0 f (φ) dφ n=1 = n 0 ∞ 1 2π r a 2π + = [an cos nθ + bn sin nθ] 0 ∞ 1 2π n 2π + = r a r a n 1 π 2π f (φ) cos nφ cos nθ + sin nφ sin nθ dφ 0 ∞ 2π f (φ) dφ + 0 n=1 ∞ 2π f ( φ) 1 + 2 0 n=1 n r a r a 1 π 2π f (φ) cos n(θ − φ) dφ 0 n cos n(θ − φ) dφ. ...
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