tailieunhanh - Ebook Student solutions manual for mathematical methods for physics and engineering (3/Ed): Part 2
Part 2 book “Student solutions manual for mathematical methods for physics and engineering” has contents: Series solutions of ODEs, special functions, quantum operators, calculus of variations, integral equations, complex variables, group theory, numerical methods, representation theory, and other contents. | 16 Series solutions of ordinary differential equations Find two power series solutions about z = 0 of the differential equation (1 − z 2 )y − 3zy + λy = 0. Deduce that the value of λ for which the corresponding power series becomes an Nth-degree polynomial UN (z) is N(N + 2). Construct U2 (z) and U3 (z). If the equation is imagined divided through by (1 − z 2 ) it is straightforward to see that, although z = ±1 are singular points of the equation, the point z = 0 is an ordinary point. We therefore expect two (uncomplicated!) series solutions with indicial values σ = 0 and σ = 1. n (a) σ = 0 and y(z) = ∞ n=0 an z with a0 = 0. Substituting and equating the coefficients of z m , (1 − z 2 ) ∞ n(n − 1)an z n−2 − 3 n=0 ∞ n=0 nan z n + λ ∞ an z n = 0, n=0 (m + 2)(m + 1)am+2 − m(m − 1)am − 3mam + λam = 0, gives as the recurrence relation am+2 = m(m − 1) + 3m − λ m(m + 2) − λ am = am . (m + 2)(m + 1) (m + 1)(m + 2) Since this recurrence relation connects alternate coefficients am , and a0 = 0, only the coefficients with even indices are generated. All such coefficients with index higher than m will become zero, and the series will become an Nth-degree polynomial UN (z), if λ = m(m + 2) = N(N + 2) for some (even) m appearing in the series; here, this means any positive even integer N. 269 SERIES SOLUTIONS OF ODES To construct U2 (z) we need to take λ = 2(2 + 2) = 8. The recurrence relation gives a2 as 0−8 a0 = −4a0 ⇒ (0 + 1)(0 + 2) n (b) σ = 1 and y(z) = z ∞ n=0 an z with a0 = 0. a2 = U2 (z) = a0 (1 − 4z 2 ). Substituting and equating the coefficients of z m+1 , (1 − z 2 ) ∞ (n + 1)nan z n−1 − 3 n=0 ∞ (n + 1)an z n+1 + λ n=0 ∞ an z n+1 = 0, n=0 (m + 3)(m + 2)am+2 − (m + 1)mam − 3(m + 1)am + λam = 0, gives as the recurrence relation am+2 = m(m + 1) + 3(m + 1) − λ (m + 1)(m + 3) − λ am = am . (m + 2)(m + 3) (m + 2)(m + 3) Again, all coefficients with index higher than m will become zero, and the series will become an Nth-degree .
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