CALCULUS OF VARIATIONS
22.5 (a)∂x/∂t=0andso ̇x=
∑
i ̇qi∂x/∂qi; (b) use
∑i̇qid
dt(
∂T
∂ ̇qi)
=
d
dt(2T)−
∑
ïqi∂T
∂ ̇qi.
22.7 Use result (22.8);β^2 =1+α^2.
Putρ=ucto obtaindθ/du=β/[u(u^2 −1)^1 /^2 ]. Remember that cos−^1 is a
multivalued function;ρ(θ)=[Rcos(θ 0 /β)]/[cos(θ/β)].
22.9 −λy′(1−y′^2 )−^1 /^2 =2gP(s),y=y(s),P(s)=
∫s
0 ρ(s′)ds′.Thesolution,y=
−acos(s/a), and 2P(πa/4) =Mtogether giveλ=−gM.Therequiredρ(s)is
given by [M/(2a)] sec^2 (s/a).
22.11 Note that theφE–L equation is automatically satisfied ifv=v(φ).A=1/a.
22.13 Circle isλ^2 x^2 +[λy+(1−λ^2 b^2 )^1 /^2 ]^2 =1.Usethefactthat
∫
ydx=V/hto
determine the condition onλ.
22.15 Denoting (ds)^2 /(dt)^2 byf^2 , the Euler–Lagrange equation forφgivesr^2 φ ̇=Af,
whereAcorresponds to the angular momentum of the particle. Use the result
of exercise 22.10 to obtainc^2 −(2GM/r)=Bf, where, to first order in small
quantities,
cB=c^2 −
GM
r+
1
2
( ̇r^2 +r^2 φ ̇^2 ),
which reads ‘total energy = rest mass + gravitational energy + radial and
azimuthal kinetic energy’.
22.17 Convert the equation to the usual form, by writingy′(x)=u(x), and obtain
x^2 u′′+4xu′− 4 u= 0 with general solutionAx−^4 +Bx. Integrating a second time
and using the boundary conditions givesy(x)=(1+x^2 )/2andJ=1;η(1) = 0,
sincey′(1) is fixed, and∂F/∂u′=2x^4 u′=0atx=0.
22.19 Usingy=sinxas a trial function shows thatλ 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 equation
withp=ρ,q= 0 and weight functionρ/c^2 .Estimateofω^2 =[c^2 ν/(2a^2 )][0. 5 −
2(ν+2)−^1 +(2ν+2)−^1 ]−^1 , which minimises toc^2 (2 +
√
2)^2 /(2a^2 )=5. 83 c^2 /a^2 when
ν=√
2.
22.23 Note that the original equation is not self-adjoint; it needs an integrating factor
ofex.Λ(y)=[
∫ 2
0 (1 +x)exy′^2 dx]/[∫^2
0 exy (^2) dx;λ 0 ≤ 3 /8. Sincey′(2) must equal 0,
γ=(π/2)(n+^12 ) for some integern.
22.25 E 1 ≤(ω/2)(8n^2 +12n+3)/(4n+ 1), which has a minimum value 3ω/2when
integern=0.
22.27 (a)V=
∫L
−L(p−q)dx.(c)UseV=(a−b)L(^2) to eliminatebfrom the expression
forE; now the minimisation is with respect toaalone. The values foraandb
are±V/(2L^2 )−Vρg/(6γ).
22.29 The SL equation hasp=1,q=0,andρ=1.
Useu(x, z)=x(4−x)z(1−z) as a trial function; numerator = 1088/90, denomi-
nator = 512/450. Direct solutionk^2 =17π^2 /16.