Mathematical Methods for Physics and Engineering - Matematica.NET

22.9 EXERCISESpath of a small test particle is such as to make ∫ ds stationary, find two firstintegrals of the equations of motion. From their Newtonian limits, in whichGM/r, ṙ 2 and r 2 ˙φ 2 are all ≪ c 2 , identify the constants of integration.22.16 Use result (22.27) to evaluateJ =∫ 1−1(1 − x 2 )P ′ m(x)P ′ n(x) dx,where P m (x) is a Legendre polynomial of order m.22.17 Determine the minimum value that the integralJ =∫ 10[x 4 (y ′′ ) 2 +4x 2 (y ′ ) 2 ] dxcan have, given that y is not singular at x =0andthaty(1) = y ′ (1) = 1. Assumethat the Euler–Lagrange equation gives the lower limit, and verify retrospectivelythat your solution makes the first term on the LHS of equation (22.15) vanish.22.18 Show that y ′′ − xy + λx 2 y = 0 has a solution for which y(0) = y(1) = 0 andλ ≤ 147/4.22.19 Find an appropriate, but simple, trial function and use it to estimate the lowesteigenvalue λ 0 of Stokes’ equation,d 2 y+ λxy =0, with y(0) = y(π) =0.dx2 Explain why your estimate must be strictly greater than λ 0 .22.20 Estimate the lowest eigenvalue, λ 0 , of the equationd 2 ydx − 2 x2 y + λy =0, y(−1) = y(1) = 0,using a quadratic trial function.22.21 A drumskin is stretched across a fixed circular rim of radius a. Small transversevibrations of the skin have an amplitude z(ρ, φ, t) thatsatisfies∇ 2 z = 1 ∂ 2 zc 2 ∂t 2in plane polar coordinates. For a normal mode independent of azimuth, z =Z(ρ)cosωt, find the differential equation satisfied by Z(ρ). By using a trialfunction of the form a ν − ρ ν , with adjustable parameter ν, obtain an estimate forthe lowest normal mode frequency.[ The exact answer is (5.78) 1/2 c/a.]22.22 Consider the problem of finding the lowest eigenvalue, λ 0 , of the equation(1 + x 2 ) d2 y dy+2x + λy =0, y(±1) = 0.dx2 dx(a) Recast the problem in variational form, and derive an approximation λ 1 toλ 0 by using the trial function y 1 (x) =1− x 2 .(b) Show that an improved estimate λ 2 is obtained by using y 2 (x) =cos(πx/2).(c) Prove that the estimate λ(γ) obtained by taking y 1 (x) +γy 2 (x) asthetrialfunction isλ(γ) = 64/15 + 64γ/π − 384γ/π3 +(π 2 /3+1/2)γ 2.16/15 + 64γ/π 3 + γ 2Investigate λ(γ) numerically as γ is varied, or, more simply, show thatλ(−1.80) = 3.668, an improvement on both λ 1 and λ 2 .799