{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

soln2 - MATH 300 Fall 2007 Advanced Boundary Value Problems...

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

View Full Document Right Arrow Icon

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

View Full Document Right Arrow Icon

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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/2-L L-L/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 x-L 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...
View Full Document

{[ snackBarMessage ]}

Page1 / 18

soln2 - MATH 300 Fall 2007 Advanced Boundary Value Problems...

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

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