Mathematical Physics Vol 1

5.10 Examples

289

By multiplying the first equation by v , and the second equation by − u , and then adding 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 in the interval a ≤ x ≤ b (see proof of the previous theorem), and thus by integrating, bearing in mind the boundary conditions (as in the previous theorem), we obtain

u 2 + v 2 p d x = r uv ′ − vu ′ b a

b Z a

− β

= 0 .

As y is a main function, it follows that y̸ ≡ 0. Further, as y and p are continuous functions, where p > 0 or p < 0 in the interval a ≤ x ≤ b , and y 2 = u 2 + v 2̸ ≡ 0, it follows that the integral on the left hand side of the last equation is different from zero. From here, 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. The theorem is thus proven.

Mittag-Leffler (ML) functions

Problem 203 Prove that E 1 ( ± x ) is an exponential function.

Solution The ML function is defined by the expression (5.183) E α ( x )= ∑ k = 0 x k Γ ( α k + 1 )

, α > 0 ,

so that, for α = 1,

∞ ∑ k = 0

∞ ∑ k = 0

( ± x ) k Γ ( k + 2 )

( ± x ) k k !

= e ± x .

E 1 ( ± x )=

=

We have used here the relation Γ ( k + 1 )= k ! and the expansion of the exponential function into the Taylor (Maclaurin) series.

Problem 204 Prove that E 1 , 2 ( x )= 1 + xE 1 , 3 ( x ) . Solution ML function of two parameters (5.184) is defined by the following series E α , β ( x )= ∑ k = 0 x k Γ ( α k + β ) .