Solution to Ex. 6.12
of Turbulent Flows by Stephen B. Pope, 2000
Yaoyu Hu
March 23, 2017
Show that, for integer n,

L
0
 0, for n  0
e2 inx / L dx  
 L, for n  0
(1)
and hence establish the orthonormality property Eq. (6.112).
Solution
The integral could be written as sines and cosines

L
0
L
 2 nx 
 2 nx 
e2 inx / L dx   cos 
+i sin 

 dx
0
 L 
 L 
L
L
 2 nx 
 2 nx 
  cos 
dx  i  sin 

 dx
0
0
 L 
 L 
(2)
It is obvious that Eq. (1) holds.
Eq. (6.112) is
eiκ x e iκ x
L
1
L3
1
 3
L

L
L
L
0
0
0
L
L
L
0
0
0
  
  
κ  κ κ 

1
L3
eiκx e iκ x dx1dx2dx3
e
i κ  κ   x
L
L
L
0
0
0
  
dx1dx2dx3
(3)
eiκ x dx1dx2dx3
The wavenumber vector κ” could be written as
κ   0n 
2
n
L
where n is an integer vector. Substitute Eq. (4) into Eq. (3) we have
(4)
e
iκ x  iκ x
e
L
1 L L L i 2L nx
 3   e
dx1dx2 dx3
L 0 0 0
1 L L L i 2 n1x1 i 2 n2 x2 i 2 n3 x3
 3    e L e L e L dx1dx2 dx3
L 0 0 0
1 L i 2L n1x1
1 L i 2L n2 x2
1 L i 2L n3 x3
  e
dx1  e
dx2  e
dx3
L 0
L 0
L 0
When κ = κ’, κ” = 0, then Eq. (5) equals 1. When κ ≠ κ’, then Eq. (5) is 0.
(5)