lec12-3 - .. so we can go to c 5 c 5 = 100 11 2 Z 1 1 8 (...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Steady-state temperature in a sphere Consider a sphere of radius r = 1, with the temperature T = 100 on the top half ( z > 0 or 0 < θ < π/ 2) and T = 0 on the bottom half ( z < 0 or π/ 2 < θ < π ) We know that our solution is a solution to Laplace equation 2 T = 0 most conveniently in spherical coordinates T ( r , θ, φ ) = l X m = - l X l =0 r l P m l (cos θ ) [ a lm cos m φ + b lm sin m φ ] Based on the problem, we see there is no φ dependence, so we only require m = 0 and cos m φ = 1, so we simplify T ( r , θ ) = X l =0 c l r l P m =0 l (cos θ )
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Steady-state temperature in a sphere Solution is a series in the Legendre polynomials! At r = 1 we know T ( r = 1 , θ ) T ( r = 1 , θ ) = X l =0 c l P l (cos θ ) Find the coefficents c l ! With T ( r = 1 , x = cos θ ) = 100 f ( x ), we have f ( x ) = 0 for - 1 < x < 0 and f ( x ) = 1 for 0 < x < 1, and then c l = 100 ± 2 l + 1 2 ²Z 1 - 1 f ( x ) P l ( x ) dx c 0 = 100 ± 1 2 ²Z 1 0 dx = 100 ± 1 2 ² c 1 = 100 ± 3 2 ²Z 1 0 xdx = 100 ± 3 4 ²
Background image of page 2
Steady-state temperature in a sphere, continued c 2 = 100 ± 5 2 ²Z 1 0 ± 3 2 x 2 - 1 2 ² dx = 0 c 3 = 100 ± 7 2 ²Z 1 0 ± 5 2 x 3 - 3 2 x ² dx = 100 ± - 7 16 ² The integral for c 4 = 0. .. we can even see this by symmetry.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: .. so we can go to c 5 c 5 = 100 11 2 Z 1 1 8 ( 63 x 5-70 x 3 + 15 x ) dx = 100 11 32 Steady-state temperature in a sphere, continued Now that we have our Legendre coecients, c = 100 ( 1 2 ) , c 1 = 100 ( 3 4 ) , c 2 = 0, c 3 = 100 (-7 16 ) , c 4 = 0, c 5 = 100 ( 11 32 ) , etc., we can write a series solution T ( r , ) = X l =0 c l r l P l (cos ) T ( r , ) = 100 1 2 + 3 4 r cos -7 16 r 3 5 2 cos 3 -3 2 cos + ... Notice that we write T ( r , ) since we determined from the beginning that the solution is independent of It is crucial to remember that we solved this only for r < 1! Our solution does not apply outside of this region...
View Full Document

This note was uploaded on 07/30/2011 for the course PHZ 3113 taught by Professor Staff during the Spring '03 term at University of Central Florida.

Page1 / 4

lec12-3 - .. so we can go to c 5 c 5 = 100 11 2 Z 1 1 8 (...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online