Mathematical Methods for Physics and Engineering : A Comprehensive Guide

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

E=ρg 2

∫L

−L

(p^2 −q^2 )dx+γ

∫L

−L

[

(1 +p′^2 )^1 /^2 +(1+q′^2 )^1 /^2

]

dx.

(a) ExpressVin terms ofpandq.

(b) Show that, if the total energy is minimised,pandqmust satisfy

p′^2

(1 +p′^2 )^1 /^2

−

q′^2

(1 +q′^2 )^1 /^2

=constant.

(c) As an approximate solution, consider the equations

p=a(L−|x|),q=b(L−|x|),

whereaandbare sufficiently small thata^3 andb^3 can be neglected compared

with unity. Find the values ofaandbthat minimiseE.

22.28 A particle of massmmoves in a one-dimensional potential well of the form

V(x)=−μ

^2 α^2

m

sech^2 αx,

whereμandαare 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

His given by−(
^2 / 2 m)d^2 /dx^2 +V(x). Using trial wavefunctions of the form

y=Asechβx, show the following:

(a) forμ= 1, there is an exact eigenfunction ofH, with a corresponding〈E〉of

half of the maximum depth of the well;

(b) forμ= 6, the ‘binding energy’ of the ground state is at least 10
^2 α^2 /(3m).

[ You will find it useful to note that foru,v≥0, sechusechv≥sech (u+v). ]
22.29 The Sturm–Liouville equation can be extended to two independent variables,x
andz, with little modification. In equation (22.22),y′^2 is replaced by (∇y)^2 and
the integrals of the various functions ofy(x, z) become two-dimensional, i.e. the
infinitesimal isdx 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 hask^2 ≤ 10 .63 and, by direct solution, that the

actual value is 10.49.

22.10 Hints and answers

22.1 Note that the integrand, 2πρ^1 /^2 (1 +ρ′^2 )^1 /^2 , does not containzexplicitly.
22.3 I=

∫

n(r)[r^2 +(dr/dφ)^2 ]^1 /^2 dφ. Take axes such thatφ= 0 whenr=∞.

Ifβ=(π−deviation angle)/2thenβ=φatr=a, and the equation reduces to

β

(a^2 +α^2 )^1 /^2

=

∫∞

−∞

dr

r(r^2 −a^2 )^1 /^2

,

which can be evaluated by puttingr=a(y+y−^1 )/2, or successivelyr=acoshψ,

y=expψto yield a deviation ofπ[(a^2 +α^2 )^1 /^2 −a]/a.