Chapter_3_Solutions

Chapter_3_Solutions - CHAPTER 1 Orthogonal Expansions 1....

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 DocumentRight Arrow Icon

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

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

Unformatted text preview: CHAPTER 1 Orthogonal Expansions 1. The Fourier Method Exercise 1. Form the linear combination u ( x,t ) = summationdisplay n =1 a n cos nct sin nx Then u ( x, 0) = f ( x ) = summationdisplay n =1 a n sin nx Using the exactly same calculation as in (3.5)(3.7) in the text, we obtain a n = 2 integraldisplay f ( x ) sin nx dx Observe that u t ( x, 0) = 0 is automatically satisfied. When the initial conditions are changed to u ( x, 0) = 0 , u t ( x, 0) = g ( x ) then a linear combination of the fundamental solutions u n ( x,t ) = cos nct sin nx does not suffice. But, observe that u n ( x,t ) = sin nct sin nx now works and form the linear combination u ( x,t ) = summationdisplay n =1 b n sin nct sin nx Now u ( x, 0) = 0 is automatically satisfied and u t ( x, 0) = g ( x ) = summationdisplay n =1 ncb n sin nx Again using the argument in (3.5)(3.7), one easily shows the b n are given by b n = 2 nc integraldisplay g ( x ) sin nx dx 2. Orthogonal Expansions Exercise 1. The requirement for orthogonality is integraldisplay cos mx cos nx dx = 0 , m negationslash = n 1 2 1. ORTHOGONAL EXPANSIONS For the next part make the substitution y = x/l to get integraldisplay l cos( mx/l ) cos( nx/l ); dx = integraldisplay cos my cos ny dy = 0 , m negationslash = n We have c n = ( f, cos( nx/l )) || cos( nx/l ) || 2 Thus c = 1 l integraldisplay l f ( x ) dx, c n = 2 l integraldisplay l f ( x ) cos( nx/l ) dx, n 1 Exercise 3. Up to a constant factor, the Legendre polynomials are P ( x ) = 1 , P 1 ( x ) = x, P 2 ( x ) = 3 2 x 2 1 2 , P 3 ( x ) = 5 2 x 3 3 2 x The coefficients c n in the expansion are given by the generalized Fourier coefficients c n = 1 || P n || 2 ( f,P n ) = integraltext 1 1 e x P n ( x ) dx integraltext 1 1 Pn ( x ) 2 dx The pointwise error is E ( x ) = e x 3 summationdisplay n =0 c n P n ( x ) The mean square error is E = parenleftBigg integraldisplay 1 1 [ e x 3 summationdisplay n =0 c n P n ( x )] 2 dx parenrightBigg 1 / 2 The maximum pointwise error is max 1 x 1 | E ( x ) | . Exercise 4. Use the calculus facts that integraldisplay b 1 x p dx < , p < 1 and integraldisplay a 1 x p dx < , p > 1 ( a > 0) Otherwise the improper integrals diverge. Thus x r L 2 [0 , 1] if r > 1 / 2 and x r L 2 [0 , ] if r < 1 / 2 and r > 1 / 2, which is impossible. Exercise 5. We have cos x = summationdisplay n =1 b n sin 2 nx, b n = 4 integraldisplay / 2 cos x sin 2 nx dx Also sin x = summationdisplay n =1 b n sin nx clearly forces b 1 = 1 and b n = 0 for n 1. Therefore the Fourier series of sin x on [0 , ] is just a single term, sin x . 2. ORTHOGONAL EXPANSIONS 3 Exercise 6. (a) We find H ( x ) = 1 , H 1 ( x ) = 2 x, H 2 ( x ) = 2(2 x 2 1) H 3 ( x ) = 8 x 3 12 x, H 4 ( x ) = 16 x 4 48 x 2 + 12 (b) This is a straightforward verification. (c) To verify orthogonality, note that v n + x 2 v n = (2 n + 1) v...
View Full Document

This note was uploaded on 03/08/2012 for the course MATH 1010 taught by Professor Mihai during the Fall '08 term at UOIT.

Page1 / 12

Chapter_3_Solutions - CHAPTER 1 Orthogonal Expansions 1....

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