Mathematical Physics Vol 1

Chapter 5. Series Solutions of Differential Equations. Special functions

270

5.10 Examples

Problem 186 Prove that the function set

1 , sin x , sin2 x , ..., sin nx ,...,

(5.241)

is orthogonal on the interval [ − ℓ, ℓ ] .

Solution We shall use here the following, well known trigonometric relations sin α · sin β = 1 2 ( cos ( α − β ) − cos ( α + β )) , 1 2 ( sin ( α − β )+ sin ( α + β )) . It can be easily proved that the values of the following integrals are ℓ Z − ℓ sin k π x ℓ · sin m π x ℓ d x =   0 , k̸ = m , ℓ, k = m̸ = 0 , 0 , k = m = 0 cos α · cos β = sin α · cos β = 1 2 ( cos ( α − β )+ cos ( α + β )) ,

d x =   d x = 0

0 , k̸ = m , ℓ, k = m̸ = 0 , 2 ℓ, k = m = 0

ℓ Z

k π x

m π x ℓ

cos

cos

ℓ ·

− ℓ

ℓ Z

k π x

m π x ℓ

cos

sin

ℓ ·

− ℓ

Let us show this on the first example, only. For x ∈ [ − ℓ, ℓ ] , if k̸ = m ℓ Z − ℓ sin k π x ℓ · sin m π x ℓ d x = 1 2 ℓ Z − ℓ cos ( k − m ) π x ℓ − cos ℓ ( k + m ) π x

d x =

ℓ Z

ℓ Z

( k − m ) π x ℓ

( k + m ) π x ℓ

1 2

1 2

cos

d x −

cos

d x =

=

− ℓ

− ℓ

ℓ

ℓ

( k − m ) π x ℓ

( k + m ) π x ℓ

1 2

1 2

ℓ ( k − m ) π

ℓ ( k + m ) π

sin

sin

−

=

=

− ℓ

− ℓ

( k − m ) π ℓ

( k + m ) π ℓ

ℓ 2 ( k − m ) π ℓ 2 ( k − m ) π

ℓ 2 ( k + m ) π

sin

2 ℓ −

sin

2 ℓ =

=

ℓ 2 ( k + m ) π

sin2 ( k − m ) π −

sin2 ( k + m ) π =

=

= 0 .