14 EE 2402 Engineering Mathematics III Solutions to Tutorial 3 1. For n =0; 1; 2; 3; 4; 5 verify that P n (x) is a solution of Legendre's equation wit
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1 EE 0 Engineering Mathematics III Solutions to Tutorial. For n =0; ; ; ; ; verify that P n (x) is a solution of Legendre's equation with ff = n. Solution: Recall the Legendre's equation from your text or lecture notes ( x )y 00 xy 0 ff(ff )y =0 For ff = n =0,we have P 0 (x) ==) P 0 0(x) =P 00 0 (x) =0and 0 (x) xp 0 0(x)0 (0 )P 0 (x) =0 Hence P 0 (x) =isthe solution for Legendre's equation with ff =0. For ff = n =,we have P (x) =x =) P 0 (x) =,P 00 (x) =0and (x) xp 0 (x) ()P (x) = x x =0 Hence P (x) =x is the solution for Legendre's equation with ff =. For ff = n =,we have P (x) =(x )= =) P 0 (x) =x, P 00 (x) =and (x) xp 0 (x) ()P (x) = ( x ) x (x )= =0 Hence P (x) is the solution for Legendre's equation with ff =.. For ff =,wehave P (x) = (x 0x x)=8, P 0 (x) = (x 0x )=8, P 00 (x) =(x 0x)= and (x) xp 0 (x) ()P (x) =( x )(x 0x)= x(x 0x )=80(x 0x x)=8 = x =0 x 0 0 x x x 0 x 9 x x x x Hence P (x) is the solution for Legendre's equation with ff =.
2 . Expand each of the following in a series of Legendre's polynomials: (a) x x (b) x x x (c) x x Solution: Using a bit of matrix notation and the formulas given in the text or your lecture notes for P 0 (x), P (x),, P (x), we can write P 0 (x) P (x) P (x) P (x) = P (x) Inverting the above matrix gives x x x x = = 0 = = 0 = 0 =8 0 0=8 0 = = 0 = = 0 = 0 = 0 = 0 8= P 0 (x) P (x) P (x) P (x) P (x) which easily allows us to express powers of x through x in terms of Legendre polynomials. Then (a) x x = P 0(x)P (x) P (x). (b) x x x = P 0(x) P (x) P (x) P (x) x x x x (c) x x = P 0(x) P (x) P (x)
3 . Let n be a nonnegative integer. Use the fact that P n (x) is one solution of Legendre's equation with ff = n to obtain a second, linearly independent solution Q n (x) =P n (x) P n (x) ( x ) dx Hint: Let Q n (x) =P n (x)z(x) and then substitute it to the Legendre's equation to find z(x). Solution: Let Q n (x) =P n (x)z(x). We have Q 0 n(x) =P 0 n(x)z(x)p n (x)z 0 (x) and Q 00 n(x) =Pn 00 (x)z(x)pn(x)z 0 0 (x)p n (x)z 00 (x) Substituting into Legendre's equation, we have ( x )[P 00 n z P 0 n z0 P n z 00 ] x[p 0 n z P nz 0 ]n(n )P n z = z[ n xp 0 n n(n )P n ]z 00 ( x )P n z 0 [( x )]P 0 n xp n ] =0z 00 ( x )P n z 0 [( x )]P 0 n xp n ] Thus, we see that Q n (x) is a solution of Legendre's equation if and only if we choose z(x) such that z 00 ( x )P n z 0 [( x )]P 0 n xp n ]=0 z 00 P 0 n x z 0 P n =0 x Integrating this equation, we have ln jz 0 j lnj x j lnjp n j = c ln fi fi fiz 0 ( x )(P n ) fi fifi = c z 0 ( x )(P n ) = K z 0 (x) = ( x )[P n (x)] Integrate again the above equation to get z(x) =K Q n (x) =P n (x) K ( x )[P n (x)] dx ( x )[P n (x)] dx which is a second linearly independent solution. Note that we have dropped K in Q n (x) as it plays no role at all.
4 . Use the result in Problem to obtain Q 0 (x) = ln x x Q (x) = x x ln x Q (x) = x (x ) ln x x Solution: From Problem we get Q 0 (x) = for <x<. Similarly, x dx = = fi fififi fi ln xfi fi fi = ln x Q (x) =x Q (x) = (x ) x = (x ) x ( x ) dx dx x x x» = x x dx x x = x ln x x (x ) ( x ) dx x x (x p ) (x p ) = x (x ) ln x x for <x<. dx
5 8. Show the following properties of Legendre polynomials: (a) P n ( x) =( ) n P n (x) for» x» and n =0; ; ; (b) For any integer n>0, P n (x)dx =0 Hint: P n (x) =P n (x)p 0 (x). Solution: (a) From the lecture notes or the text, we have and also note that P n (x) = [n=] ( ) k (n k)! n k!(n k)!(n k)! xn k ( x) n k =( ) n k x n k =( ) n x n k (b) P n ( x) = P n ( x) = [n=] [n=] ( ) k (n k)! n k!(n k)!(n k)! ( x)n k ( ) k (n k)! n k!(n k)!(n k)! ( )n x n k =( ) n P n (x) P n (x)dx = P n (x)p 0 (x) =0 since P n (x) and P 0 (x) = are orthogonal on [ ; ] for each n.
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