Solving Differential Equations in Terms of Bessel Functions

42 CHAPTER 2. TRANSFORMATIONS
R is a operator of degree one and the coefficients are a K-linear combination of
r0,r ′ 0 ,r′′ 0 ,r1,r ′ 1 and r′′ 1 . Equating these coefficients with zero yields a system of two
differential equations of order two with two variables.
Furthermore, replacing r ′ 0 and r′ 1 by two new variables ¯r0 and ¯r1 transforms this
system into a system of differential equations of order one. We add two equations
r ′ 0 − ¯r0 = 0 and r ′ 1 − ¯r1 = 0 and finally get a system of four order one equations in
four variables. In matrix representation such a system is written as Y ′ − AY = 0,
where A is a 4×4 matrix and Y is the vector including the undetermined functions
r0,r1, ¯r0 and ¯r1.
These hyperexponential solutions can be found with the cyclic vector method.
Definition 2.19 Let V be a n-dimensional vector space and ∂ : V → V an endomorphism.
A vector v ∈ V is called a cyclic vector of V if
is a basis of V .
Definition 2.20 Let
{z,∂z,∂ 2 z,...,∂ n−1 z}
L = an∂ n + an−1∂ n−1 + ··· + a0∂ 0
be a differential operator. We define the adjoint operator
L ∗ := (−1) n (−∂) 0 a0 + ... + (−∂) n−1 an−1 + (−∂) n
an .
The adjoint operator satisfies L ∗∗ = L and (L1L2) ∗ = L ∗ 2 L∗ 1 .
The cyclic vector method works as follows.
Theorem 2.21 Let A be the 4 × 4 matrix of the equation Y ′ − AY = 0 and let
∂y = y ′ − Ay. Then we can compute a hyperexponential solution by the following
steps:
1. Pick a random element v ∈ K4 .
2. Check whether v is cyclic, otherwise repeat step one and two.
3. Compute L = a0 + a1∂ + a2∂ 2 + a3∂ 3 + ∂ 4 such that Lv = 0.
4. Compute a hyperexponential solution s of L∗
(e.g. use expsols in Maple).
5. Compute R such that L = ∂ + s′
1s
s R .
6. Let R = y0 + y1∂ + y2∂ 2 + y3∂ 3 and y = y0v + y1∂v + y2∂ 2v + y3∂ 3v. Then y is a hyperexponential solution of Y ′ − AY = 0.