The Chemistry Maths Book, Second Edition

12.4 Particular solutions 347

or by the equivalent trigonometric form (12.17),

y(x) 1 = 1 d

1

cos 1 ωx 1 + 1 d

2

1 sin 1 ωx (12.22)

Ifωis a given constant then two initial or boundary conditions are normally sufficient

to specify a particular solution. This is exemplified in Section 12.5 for the classical

harmonic oscillator as an initial value problem. However, ifωis a parameter to be

determined, then an additional condition is required. This is the case for the quantum-

mechanical ‘particle in a box’, as a boundary value problem, discussed in Section 12.6.

A third possibility arises when the physical situation requires the imposition of a

periodic (cyclic) boundary condition(see Section 8.6),

y(x 1 + 1 λ) 1 = 1 y(x) (12.23)

whereλis the period. In this case, replacement ofxbyx 1 + 1 λin the general solution

(12.21) gives

y(x 1 + 1 λ) 1 = 1 c

1

e

iω(x+λ)

1 + 1 c

2

e

−iω(x+λ)

1 = 1 c

1

e

iωx

e

iωλ

1 + 1 c

2

e

−iωx

e

−iωλ

The condition (12.23) is then satisfied if bothe

iωλ

1 = 11 ande

−iωλ

1 = 11. As discussed in

Section 8.6, this is the case whenωλis a multiple of 2 π,

ωλ 1 = 12 πn, n 1 = 1 0, ±1, ±2, ±3,1= (12.24)

The possible values ofωare then (using nto label the values) ω

n

1 = 12 πn 2 λ, and the

corresponding solutions are

y

n

(x) 1 = 1 c

1

e

(2πnx 2 λ)i

1 + 1 c

2

e

−(2πnx 2 λ)i

, n 1 = 1 0, ±1, ±2, ±3,1= (12.25)

The trigonometric form of these solutions is

(12.26)

The constants c

1

and c

2

, or d

1

and d

2

, are not determined. The case of the periodic

boundary condition is exemplified in Section 12.7 for the quantum-mechanical

problem of the ‘particle in a ring’.

EXAMPLE 12.9Solve the equation for the cyclic boundary

conditionsy(θ 1 + 1 π) 1 = 1 y(θ).

The general solution is

y(θ) 1 = 1 c

1

e

i

ω

θ

1 + 1 c

2

e

−i

ω

θ

.

dy

d

y

2

2

2

0

θ

+=ω

yd

nx

d

nx

n

=+

12

22

cos sin

ππ

λλ