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Chapter_3_Solutions

# Chapter_3_Solutions - CHAPTER 1 Orthogonal Expansions 1 The...

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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 π integraldisplay π 0 f ( x ) sin nxdx 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 ncπ integraldisplay π 0 g ( x ) sin nxdx 2. Orthogonal Expansions Exercise 1. The requirement for orthogonality is integraldisplay π 0 cos mx cos nxdx = 0 , m negationslash = n 1

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2 1. ORTHOGONAL EXPANSIONS For the next part make the substitution y = πx/l to get integraldisplay l 0 cos( mπx/l ) cos( nπx/l ); dx = integraldisplay π 0 cos my cos nydy = 0 , m negationslash = n We have c n = ( f, cos( nπx/l )) || cos( nπx/l ) || 2 Thus c 0 = 1 l integraldisplay l 0 f ( x ) dx, c n = 2 l integraldisplay l 0 f ( x ) cos( nπx/l ) dx, n 1 Exercise 3. Up to a constant factor, the Legendre polynomials are P 0 ( 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 0 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 π integraldisplay π/ 2 0 cos x sin 2 nxdx 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 0 ( 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 n , v ′′ m + x 2 v m = (2 m + 1) v n Multiply the first equation by v m and the second by v n , subtract, and then integrate over R to get integraldisplay R ( v ′′ n v m + v ′′ m v n ) dx = 2( n m ) integraldisplay R v m v n dx But integrating by parts twice gives integraldisplay R v m v ′′ n dx = integraldisplay R v n v ′′ m dx

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