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Unformatted text preview: MATH 300 Fall 2007 Advanced Boundary Value Problems I Solutions to Problem Set 2 To Be Completed by: Friday October 12, 2007 Department of Mathematical and Statistical Sciences University of Alberta Question 1. [p 95, #3.2.2 (e), (f)] For the following functions, sketch the Fourier series of f ( x ) on the interval L ≤ x ≤ L, and determine the Fourier coefficients: (a) f ( x ) = ( 1 for  x  < L/ 2 for  x  > L/ 2 (b) f ( x ) = ( 1 if 0 < x < L if L < x < Solution: (a) From Dirichlet’s theorem the Fourier series of f ( x ) converges to 1 2 f ( x + ) + f ( x ) if L ≤ x ≤ L, and the graph of the Fourier series of f ( x ) on the interval L ≤ x ≤ L is shown below. 1 y x L/2L LL/2 Since f ( x ) is an even piecewise smooth function on the interval [ L, L ] , it has a Fourier series repre sentation of the form f ( x ) ∼ a + ∞ X n =1 a n cos nπx L where a = 1 L Z L f ( x ) dx = 1 L Z L 2 1 dx = 1 2 , and a n = 2 L Z L f ( x ) cos nπx L dx = 2 L Z L 2 cos nπx L dx = 2 nπ sin nπx L L 2 = 2 nπ sin nπ 2 for n ≥ 1 . (b) Again, from Dirichlet’s theorem the Fourier series of f ( x ) converges to 1 2 f ( x + ) + f ( x ) if L ≤ x ≤ L, and the graph of the Fourier series of f ( x ) on the interval L ≤ x ≤ L is shown below. y xL L 1 Since f ( x ) 1 2 is an odd piecewise smooth function on the interval [ L, L ] , it has a Fourier series representation of the form f ( x ) 1 2 ∼ ∞ X n =1 b n sin nπx L where b n = 2 L Z L f ( x ) 1 2 sin nπx L dx = 1 L Z L sin nπx L dx = 1 nπ cos nπx L L = 1 nπ (cos nπ 1) = 1 nπ [1 ( 1) n ] for n ≥ 1 . Question 2. [p 96, #3.2.3] Show that the Fourier series operation is linear: that is, show that the Fourier series of c 1 f ( x ) + c 2 g ( x ) is the sum of c 1 times the Fourier series of f ( x ) and c 2 times the Fourier series of g ( x ) . Solution: Suppose that the Fourier series of f and g are given by f ( x ) ∼ A + ∞ X n =1 A n cos nπx L + B n sin nπx L and g ( x ) ∼ C + ∞ X n =1 C n cos nπx L + D n sin nπx L where A = 1 2 L Z L L f ( x ) dx, A n = 1 L Z L L f ( x ) cos nπx L dx, B n = 1 L Z L L f ( x ) sin nπx L dx for n ≥ 1 , and C = 1 2 L Z L L g ( x ) dx, C n = 1 L Z L L g ( x ) cos nπx L dx, D n = 1 L Z L L g ( x ) sin nπx L dx for n ≥ 1 . If c 1 and c 2 are scalars, and the Fourier series of c 1 f + c 2 g is c 1 f ( x ) + c 2 g ( x ) ∼ E + ∞ X n =1 E n cos nπx L + F n sin nπx L , then E = 1 2 L Z L L ( c 1 f ( x ) + c 2 g ( x )) dx = c 1 2 L Z L L f ( x ) dx + c 2 2 L Z L L g ( x ) dx = c 1 A + c 2 C . Also, E n = 1 L Z L L ( c 1 f ( x ) + c 2 g ( x )) cos nπx L dx = c 1 L Z L L f ( x ) cos nπx L dx + c 2 L Z L L g ( x ) cos nπx L dx = c 1 A n + c 2 C n for n ≥ 1 . Similarly, F n = 1 L Z L L ( c 1 f ( x ) + c 2 g ( x )) sin nπx L dx = c 1 L Z L L f ( x ) sin nπx L dx + c 2 L Z L L g ( x ) sin nπx L dx = c 1 B n + c 2 D n for n ≥ 1 . Therefore the Fourier series for c 1 f + c 2 g is c 1 f ( x ) + c 2 g...
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 Spring '00
 Courdurier
 Fourier Series, Cos, Partial differential equation, n=1

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