Chem Differential Eq HW Solutions Fall 2011 59

Chem Differential Eq HW Solutions Fall 2011 59 - Section...

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Section 4.4 Steady-State Temperature in a Disk 59 Exercises 4.4 1. Since f is already given by its Fourier series, we have from (4) u ( r, θ )= r cos θ = x. 5. Let us compute the Fourier coefficients of f . We have a 0 = 50 π ± π/ 4 0 = 25 2 ; a n = 100 π ± π/ 4 0 cos nθ dθ = 100 sin ² ² ² ² π 0 = 100 sin 4 ; b n = 100 π ± π/ 4 0 sin nθ dθ = - 100 cos ² ² ² ² π 0 = 100 (1 - cos 4 ) . Hence f ( θ 25 2 + 100 π ³ n =1 1 n ´ sin 4 cos +(1 - cos 4 ) sin µ ; and u ( r, θ 25 2 + 100 π ³ n =1 1 n ´ sin 4 cos - cos 4 ) sin µ r n . 9. u ( )=2 r 2 sin θ cos θ =2 xy .So u ( xy T if and only if 2 xy = T if and only if y = T 2 x , which shows that the isotherms lie on hyperbolas centered at the origin. 13. We follow the steps in Example 4 (with α = π 4 ) and arrive at the same equation in Θ and R . The solution in Θ is Θ n ( θ ) = sin(4 ) ,n =1 , 2 ,..., and the equation in R is r 2 R ±±
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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