110
An Introduction to Beam Physics
In order to solve the inhomogeneous eq. (4.37) of order j, we first determine the homogeneous solution, and then perform a so-called variation of
parameters. The homogeneous solution is exactly the same form as for the
linearized part. To obtain the inhomogeneous solution, we make the ansatz
R j (s) = ˆ
L(s) ·
T (s). Then
d
ds
R j =
d
ds
ˆ
L(s)
·
T (s) + ˆ
L(s) ·
d
ds
T (s) .
Using eq. (4.36), the first term in the right hand side is
d
ds
ˆ
L(s)
·
T (s) = ˆ
M (s) · ˆ
L(s) ·
T (s) = ˆ
M (s) ·
R j (s).
Thus, from eq. (4.37), we obtain
ˆ
L(s) ·
d
ds
T (s) =
Q j (s, ˆ
L,
R k ),
that is
T (s) =
s
0
ˆ
L
−1 (¯ s)
Q j (¯ s, ˆ
L,
R k )d¯ s,
(4.38)
where the choice of the lower integration boundary as 0 ensures that
T (0)
= 0, which agrees with the initial condition
R j (0) = 0. Altogether we have
R j (s) = ˆ
L(s) ·
s
0
ˆ
L
−1 (¯ s)
Q j (¯ s, ˆ
L,
R k )d¯ s.
The integral is often referred to as the aberration integral, and the integrand ˆ
L
−1
Q j as the driving term. The complete solution then is obtained
as
r (s) = ˆ
L(s) r i +
∞
j=2
R j (s) = ˆ
L(s) r i +
∞
j=2
ˆ
L(s) ·
s
0
ˆ
L
−1 (¯ s)
Q j (¯ s, ˆ
L,
R k )d¯ s.
So, once the linear solution is known, everything else just boils down to
quadratures. If within a piece in which it is constant, ˆ
M (s) is diagonalizable,
the linear solutions can be written as combinations of sin, cos, sinh, cosh and
s. In other important cases where ˆ
M (s) is singular, often a complete set of
linear solutions that are polynomials in s can be obtained.
In both of these cases, the insertion into the polynomials
R j (s) leads to
terms that are polynomials in sin, cos, sinh, cosh and s. By expressing such
functions in terms of exponentials times powers of s, one can show that the
result of any integration can again be expressed as a polynomial of sin, cos,
sinh, cosh and s.
For practical cases, it is worthwhile to discuss the complexity of the procedure. With each new order, the expansion of the ODE becomes more complicated; then all previous orders have to be inserted, multiplied with the linear
An Introduction to Beam Physics
In order to solve the inhomogeneous eq. (4.37) of order j, we first determine the homogeneous solution, and then perform a so-called variation of
parameters. The homogeneous solution is exactly the same form as for the
linearized part. To obtain the inhomogeneous solution, we make the ansatz
R j (s) = ˆ
L(s) ·
T (s). Then
d
ds
R j =
d
ds
ˆ
L(s)
·
T (s) + ˆ
L(s) ·
d
ds
T (s) .
Using eq. (4.36), the first term in the right hand side is
d
ds
ˆ
L(s)
·
T (s) = ˆ
M (s) · ˆ
L(s) ·
T (s) = ˆ
M (s) ·
R j (s).
Thus, from eq. (4.37), we obtain
ˆ
L(s) ·
d
ds
T (s) =
Q j (s, ˆ
L,
R k ),
that is
T (s) =
s
0
ˆ
L
−1 (¯ s)
Q j (¯ s, ˆ
L,
R k )d¯ s,
(4.38)
where the choice of the lower integration boundary as 0 ensures that
T (0)
= 0, which agrees with the initial condition
R j (0) = 0. Altogether we have
R j (s) = ˆ
L(s) ·
s
0
ˆ
L
−1 (¯ s)
Q j (¯ s, ˆ
L,
R k )d¯ s.
The integral is often referred to as the aberration integral, and the integrand ˆ
L
−1
Q j as the driving term. The complete solution then is obtained
as
r (s) = ˆ
L(s) r i +
∞
j=2
R j (s) = ˆ
L(s) r i +
∞
j=2
ˆ
L(s) ·
s
0
ˆ
L
−1 (¯ s)
Q j (¯ s, ˆ
L,
R k )d¯ s.
So, once the linear solution is known, everything else just boils down to
quadratures. If within a piece in which it is constant, ˆ
M (s) is diagonalizable,
the linear solutions can be written as combinations of sin, cos, sinh, cosh and
s. In other important cases where ˆ
M (s) is singular, often a complete set of
linear solutions that are polynomials in s can be obtained.
In both of these cases, the insertion into the polynomials
R j (s) leads to
terms that are polynomials in sin, cos, sinh, cosh and s. By expressing such
functions in terms of exponentials times powers of s, one can show that the
result of any integration can again be expressed as a polynomial of sin, cos,
sinh, cosh and s.
For practical cases, it is worthwhile to discuss the complexity of the procedure. With each new order, the expansion of the ODE becomes more complicated; then all previous orders have to be inserted, multiplied with the linear