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Hints and answers

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Hints and answers
22.10 HINTS AND ANSWERS
total energy (per unit depth) of the film consists of its surface energy and its
gravitational energy, and is expressed by
L L
2
2
2
2
E = ρg
(1 + p )1/2 + (1 + q )1/2 dx.
(p
−
q
)
dx
+
γ
2
−L
−L
(a) Express V in terms of p and q.
(b) Show that, if the total energy is minimised, p and q must satisfy
p 2
q 2
−
= constant.
2 1/2
(1 + p )
(1 + q 2 )1/2
(c) As an approximate solution, consider the equations
p = a(L − |x|),
q = b(L − |x|),
where a and b are sufficiently small that a3 and b3 can be neglected compared
with unity. Find the values of a and b that minimise E.
22.28
A particle of mass m moves in a one-dimensional potential well of the form
V (x) = −µ
2 α2
sech 2 αx,
m
where µ and α are positive constants.
As in exercise 22.26, the expectation value
E of the energy of the system is ψ ∗ Hψ dx, where the self-adjoint operator
H is given by −(2 /2m)d2 /dx2 + V (x). Using trial wavefunctions of the form
y = A sech βx, show the following:
(a) for µ = 1, there is an exact eigenfunction of H, with a corresponding E of
half of the maximum depth of the well;
(b) for µ = 6, the ‘binding energy’ of the ground state is at least 102 α2 /(3m).
22.29
[ You will find it useful to note that for u, v ≥ 0, sech u sech v ≥ sech (u + v). ]
The Sturm–Liouville equation can be extended to two independent variables, x
and z, with little modification. In equation (22.22), y 2 is replaced by (∇y)2 and
the integrals of the various functions of y(x, z) become two-dimensional, i.e. the
infinitesimal is dx dz.
The vibrations of a trampoline 4 units long and 1 unit wide satisfy the equation
∇2 y + k 2 y = 0.
By taking the simplest possible permissible polynomial as a trial function, show
that the lowest mode of vibration has k 2 ≤ 10.63 and, by direct solution, that the
actual value is 10.49.
22.10 Hints and answers
22.1
22.3
Note that the integrand, 2πρ1/2 (1 + ρ 2 )1/2 , does not contain z explicitly.
I = n(r)[r2 + (dr/dφ)2 ]1/2 dφ. Take axes such that φ = 0 when r = ∞.
If β = (π − deviation angle)/2 then β = φ at r = a, and the equation reduces to
∞
dr
β
=
,
2
2 1/2
(a2 + α2 )1/2
−∞ r(r − a )
which can be evaluated by putting r = a(y + y −1 )/2, or successively r = a cosh ψ,
y = exp ψ to yield a deviation of π[(a2 + α2 )1/2 − a]/a.
801
CALCULUS OF VARIATIONS
22.5
22.7
22.9
22.11
22.13
22.15
22.17
22.19
22.21
22.23
22.25
22.27
22.29
(a) ∂x/∂t = 0 and so ẋ = i q̇i ∂x/∂qi ; (b) use
d ∂T ∂T
d
= (2T ) −
.
q̇i
q̈i
dt
∂q̇
dt
∂q̇i
i
i
i
Use result (22.8); β 2 = 1 + α2 .
Put ρ = uc to obtain dθ/du = β/[u(u2 − 1)1/2 ]. Remember that cos−1 is a
multivalued function; ρ(θ) = [R cos(θ0 /β)]/[cos(θ/β)].
s
−λy (1 − y 2 )−1/2 = 2gP (s), y = y(s), P (s) = 0 ρ(s ) ds . The solution, y =
−a cos(s/a), and 2P (πa/4) = M together give λ = −gM. The required ρ(s) is
given by [M/(2a)] sec2 (s/a).
Note that the φ E–L equation is automatically satisfied if v = v(φ).
A = 1/a.
Circle is λ2 x2 + [λy + (1 − λ2 b2 )1/2 ]2 = 1. Use the fact that y dx = V /h to
determine the condition on λ.
Denoting (ds)2 /(dt)2 by f 2 , the Euler–Lagrange equation for φ gives r 2 φ̇ = Af,
where A corresponds to the angular momentum of the particle. Use the result
of exercise 22.10 to obtain c2 − (2GM/r) = Bf, where, to first order in small
quantities,
GM
1
cB = c2 −
+ (ṙ 2 + r2 φ̇2 ),
r
2
which reads ‘total energy = rest mass + gravitational energy + radial and
azimuthal kinetic energy’.
Convert the equation to the usual form, by writing y (x) = u(x), and obtain
x2 u + 4xu − 4u = 0 with general solution Ax−4 + Bx. Integrating a second time
and using the boundary conditions gives y(x) = (1 + x2 )/2 and J = 1; η(1) = 0,
since y (1) is fixed, and ∂F/∂u = 2x4 u = 0 at x = 0.
Using y = sin x as a trial function shows that λ0 ≤ 2/π. The estimate must be
> λ0 since the trial function does not satisfy the original equation.
Z + ρ−1 Z + (ω/c)2 Z = 0, with Z(a) = 0 and Z (0) = 0; this is an SL equation
2
2
2
with p = ρ, q = 0 and weight function ρ/c2 . Estimate
√ of ω = [c ν/(2a )][0.5 −
2(ν +√2)−1 + (2ν + 2)−1 ]−1 , which minimises to c2 (2 + 2)2 /(2a2 ) = 5.83c2 /a2 when
ν = 2.
Note that the original equation is not self-adjoint; it needs an integrating factor
2
2
of ex . Λ(y) = [ 0 (1 + x)ex y 2 dx]/[ 0 ex y 2 dx; λ0 ≤ 3/8. Since y (2) must equal 0,
1
γ = (π/2)(n + 2 ) for some integer n.
E1 ≤ (ω/2)(8n2 + 12n + 3)/(4n + 1), which has a minimum value 3ω/2 when
integer n= 0.
L
(a) V = −L (p − q) dx. (c) Use V = (a − b)L2 to eliminate b from the expression
for E; now the minimisation is with respect to a alone. The values for a and b
are ±V /(2L2 ) − V ρg/(6γ).
The SL equation has p = 1, q = 0, and ρ = 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
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