Mathematical Physics Vol 1

5.9 Orthogonal and normalized functions

269

In these expressions α and β are real constants, and u and v real functions. By substituting these values into equation (5.229) we obtain ru ′ + irv ′ ′ +( q + α p + i β p )( u + iv )= 0 . In order for this complex equation to be satisfied, it is necessary that both its real and imaginary parts be equal to zero, i.e. ru ′ ′ +( q + α p ) u − β pv = 0 , rv ′ ′ +( q + α p ) v + β pu = 0 . If we multiply the first equation by v , and the second equation by − u , and then add the resulting equations, we obtain − β u 2 + v 2 p = u ( rv ′ ) ′ − v ( ru ′ ) ′ = = ( rv ′ ) u − ( ru ′ ) v ′ . The expression in the square brackets is a continuous function on the interval a ≤ x ≤ b (see the proof of the previous theorem), and thus, by integration, taking into account the boundary conditions (as in the case of the previous theorem), we obtain

u 2 + v 2 p d x = r uv ′ − vu ′ b a

b Z a

− β

= 0 .

Given that y is a principal function, it follows that y̸ ≡ 0. Further, as y and p are continuous functions, where p > 0 or p < 0 on the interval a ≤ x ≤ b , and y 2 = u 2 + v 2̸ ≡ 0, it follows that the integral on the left side of the last equation is not equal to zero. From this, it follows that β must be equal to 0, i.e. β = 0. Given that λ = α + i β and β = 0, it follows that λ = α , that is, λ is a real number. Thus the theorem has been proved.