Math 293 solution set 12

# Math 293 solution set 12 - Math 293 Solutions to Problem...

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Math 293 Solutions to Problem Set 12. § 10.5 #4 . In general : The equation u t = ku xx with boundary conditions u x (0 , t ) = 0 = u x ( L, t ) and initial condition u ( x, 0) = f ( x ) has solution u ( x, t ) = a 0 2 + X n =0 a n e - kn 2 π 2 t/L 2 cos( nπx/L ) where the a n ’s are the coefficients in the Fourier sine series for f ( x ), hence a n = 2 L Z L 0 f ( x ) cos( nπx/L ) dx In the present problem we have k = 2, L = 1, and f ( x ) = x - x 2 . To compute a n one can use integration by parts, but it is much quicker to use tabular integration, explained in Thomas & Finney on page 566. Whichever method you use, you should get a n = 2 ( - cos - 1 n 2 π 2 ) . Since cos = ( - 1) n , we see that a n is 0 for n odd and - 4 n 2 π 2 for n even. This calculation doesn’t work for n = 0 since it would involve dividing by 0, so a separate calculation gives a 0 = 2 R 1 0 x - x 2 dx = 1 3 . Now we plug these values into the general formula for u ( x, t ) above, replacing n by 2 n since we only want the even values of n , and we get u ( x, t ) = 1 6 - 4 π 2 X n =1 1 (2 n ) 2 e - 2(2 n ) 2 π 2 t cos 2 nπx #7 . In general : The equation u t = ku xx + P ( x ) with boundary conditions u (0 , t ) = A , u ( L, t ) = B , and initial condition u ( x, 0) = f ( x ), has solution u ( x, t ) = v ( x ) + w ( x, t ) where: (1) v satisfies 0 = kv

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