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pset_7sol - ACM 95/100b Problem Set 7 Solutions Faisal...

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ACM 95/100b Problem Set 7 Solutions Faisal Amlani March 16, 2009 Problem 1 Consider the ODE d dx bracketleftbigg (1 - x 2 ) dy dx bracketrightbigg + λy = x 2 over the interval - 1 x 1 with boundary conditions y ( - 1) finite y (1) finite Solve this problem using a series of Legendre polynomials. That is represent y ( x ) = summationdisplay n =0 A n P n ( x ) and solve for the A n . Does the solution exist for all values of λ ? SOLUTION : We know that the Legendre polynomials are the complete eigenfunctions of the homogeneous equation with these boundary conditions, each associated with eigenvalues λ n = n ( n + 1) . It is of Sturm- Liouville form with r ( x ) = 1 . So from the beginning of the Week 7 Notes, we have that for y ( x ) = A n φ n ( x ) = A n P n ( x ) , the solution to the inhomogeneous equation is y ( x ) = summationdisplay n =0 f n λ - λ n P n ( x ) where f n are the coefficients of the expansion of the right hand side f ( x ) = x 2 in terms of the eigen- functions (Legendre polynomials.) That is, f n = integraltext 1 1 r ( x ) f ( x ) φ n ( x ) dx integraltext 1 1 r ( x ) φ 2 n ( x ) dx = integraltext 1 1 x 2 P n ( x ) dx integraltext 1 1 P 2 n ( x ) dx = 2 n + 1 2 integraldisplay 1 1 x 2 P n ( x ) dx from the orthogonality relation of P n . But from the Rodrigues formula, we know that P 0 ( x ) = 1 1 ( x 2 - 1) 0 = 1 , P 1 ( x ) = 1 2 d dx ( x 2 - 1) 1 = x, P 2 ( x ) = 1 2 2 2! d 2 dx 2 ( x 2 - 1) 2 = 1 8 (12 x 2 - 4) = 1 2 (3 x 2 - 1) , . . . 1
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We easily suspect that we can express x 2 in terms of just P 0 and P 2 . Indeed, from f 0 = 1 2 integraldisplay 1 1 x 2 dx = 1 2 23 = 1 3 , f 1 = 3 2 integraldisplay 1 1 x 2 ( x ) dx = 3 2 (0) = 0 , f 2 = = 5 2 integraldisplay 1 1 x 2 1 2 (3 x 2 - 1) = 5 4 8 15 = 2 3 , we have that 1 3 P 0 + 2 3 P 2 = 1 3 + 2 3 1 2 (3 x 2 - 1) = x 2 . So f 3 , f 4 , ... must be zero and we have the solution y ( x ) = summationdisplay n =0 f n λ - λ n P n ( x ) = f 0 λ - λ 0 P 0 ( x ) + f 2 λ - λ 2 P 2 ( x ) = 1 / 3 λ + 2 / 3 λ - 2(2 + 1) ( 1 2 (3 x 2 - 1)) = 1 3 λ + 1 3 1 λ - 6 (3 x 2 - 1) = λx 2 - 2 λ ( λ - 6) . Clearly, the solution doesn’t exist for λ = 0 , 6 . Problem 2 Consider the Chebyshev ODE given by d dx bracketleftbigg radicalbig 1 - x 2 dy dx bracketrightbigg + λy ( x ) 1 - x 2 = 0 - 1 x 1 (a) Classify the singular points at x = ± 1 (b) Using a Taylor series about x = 0 generate Taylor series for two linearly independent solutions. Hint: it may be advantageous to rewrite the ODE so that the leading coefficient of y ′′ is a poly- nomial. (c) Show that it is only possible to get bounded solutions at x = ± 1 if λ = n 2 with n = 0 , 1 , 2 , . . ..
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