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book   Mobk070    March 22, 2007  11:7








                                                 SOLUTIONS USING FOURIER SERIES AND INTEGRALS        83
                   or

                                                            2
                                                 r(rR) ∓ λ R = 0

                                                                                                (5.47)
                                                                 2
                                                        2

                                                 [(1 − x )X ] ± λ X = 0
                   The second of these is Legendre’s equation, and we have seen that it has bounded solutions at
                                 2
                   r = 1 when ±λ = n(n + 1). The first equation is of the Cauchy–Euler type with solution
                                                           n
                                                   R = C 1 r + C 2 r −n−1                       (5.48)
                   Noting that the constant C 2 must be zero to obtain a bounded solution at r = 0, and using
                   superposition,

                                                        ∞
                                                               n
                                                    u =    K n r P n (x)                        (5.49)
                                                        n=0
                   and using the condition at f (r = 1) and the orthogonality of the Legendre polynomial
                                      π                    π

                                                                 2            2K n
                                        f (θ)P n (cos θ)dθ =  K n P (cos θ)dθ =                 (5.50)
                                                                 n
                                                                             2n + 1
                                    θ=0                  θ=0
                                                           π
                                                   2n + 1
                                             K n =           f (θ)P n (cos θ)dθ                 (5.51)
                                                     2
                                                         θ=0

                   5.2    VIBRATIONS PROBLEMS
                   We now consider some vibrations problems. In Chapter 2 we found a solution for a vibrating
                   string initially displaced. We now consider the problem of a string forced by a sine function.
                   Example 5.6 (Resonance in a vibration problem). Equation (1.21) in Chapter 1 is

                                                        2
                                                  y tt = a y xx + A sin(ηt)                     (5.52)
                        Select a length scale as L, the length of the string, and a time scale L/a and defining
                   ξ = x/L and τ = ta/L,

                                                  y ττ = y ξξ + C sin(ωτ)                       (5.53)

                                                                      2 2
                   where ω is a dimensionless frequency, ηL/a and C = AL a .
                        The boundary conditions and initial velocity and displacement are all zero, so the bound-
                   ary conditions are all homogeneous, while the differential equation is not. Back in Chapter 2 we
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