Mathematical Methods for Physics and Engineering - Matematica.NET

CALCULUS OF VARIATIONS22.5 (a) ∂x/∂t =0andsoẋ = ∑ i ˙q i∂x/∂q i ; (b) use∑( )d ∂T˙q i = d dt ∂˙q i dt (2T ) − ∑ ii22.7 Use result (22.8); β 2 =1+α 2 .¨q i∂T∂˙q i.Put ρ = uc to obtain dθ/du = β/[u(u 2 − 1) 1/2 ]. Remember that cos −1 is amultivalued function; ρ(θ) =[R cos(θ 0 /β)]/[cos(θ/β)].22.9 −λy ′ (1 − y ′2 ) −1/2 = 2gP(s), y = y(s), P (s) = ∫ s0 ρ(s′ ) ds ′ . The solution, y =−a cos(s/a), and 2P (πa/4) = M together give λ = −gM. Therequiredρ(s) isgiven by [M/(2a)] sec 2 (s/a).22.11 Note that the φ E–L equation is automatically satisfied if v ≠ v(φ). A =1/a.22.13 Circle is λ 2 x 2 +[λy +(1− λ 2 b 2 ) 1/2 ] 2 =1.Usethefactthat ∫ ydx = V/h todetermine the condition on λ.22.15 Denoting (ds) 2 /(dt) 2 by f 2 , the Euler–Lagrange equation for φ gives r 2 ˙φ = Af,where A corresponds to the angular momentum of the particle. Use the resultof exercise 22.10 to obtain c 2 − (2GM/r) =Bf, where, to first order in smallquantities,cB = c 2 − GM + 1 r 2 (ṙ2 + r 2 ˙φ2 ),which reads ‘total energy = rest mass + gravitational energy + radial andazimuthal kinetic energy’.22.17 Convert the equation to the usual form, by writing y ′ (x) =u(x), and obtainx 2 u ′′ +4xu ′ − 4u = 0 with general solution Ax −4 + Bx. Integrating a second timeand using the boundary conditions gives y(x) =(1+x 2 )/2 andJ =1;η(1) = 0,since y ′ (1) is fixed, and ∂F/∂u ′ =2x 4 u ′ =0atx =0.22.19 Using y =sinx as a trial function shows that λ 0 ≤ 2/π. The estimate must be>λ 0 since the trial function does not satisfy the original equation.22.21 Z ′′ + ρ −1 Z ′ +(ω/c) 2 Z =0,withZ(a) =0andZ ′ (0) = 0; this is an SL equationwith p = ρ, q = 0 and weight function ρ/c 2 .Estimateofω 2 =[c 2 ν/(2a 2 )][0.5 −2(ν +2) −1 +(2ν +2) −1 ] −1 , which minimises to c 2 (2 + √ 2) 2 /(2a 2 )=5.83c 2 /a 2 whenν = √ 2.22.23 Note that the original equation is not self-adjoint; it needs an integrating factorof e x .Λ(y) =[ ∫ 2(1 + 0 x)ex y ′2 dx]/[ ∫ 20 ex y 2 dx; λ 0 ≤ 3/8. Since y ′ (2) must equal 0,γ =(π/2)(n + 1 ) for some integer n.222.25 E 1 ≤ (ω/2)(8n 2 +12n +3)/(4n + 1), which has a minimum value 3ω/2 wheninteger n =0.22.27 (a) V = ∫ L(p − q) dx. (c)UseV =(a − −L b)L2 to eliminate b from the expressionfor E; now the minimisation is with respect to a alone. The values for a and bare ±V/(2L 2 ) − Vρg/(6γ).22.29 The SL equation has p =1,q =0,andρ =1.Use u(x, z) =x(4 − x)z(1 − z) as a trial function; numerator = 1088/90, denominator= 512/450. Direct solution k 2 =17π 2 /16.802