Green’s function from F 1 (x, y) and F 2 (x, y), L x should be Hermitian. However, if we
had, e.g., F 1 (x, y) ¼ x À r and F 2 (x, y) ¼ y À s, G(x, y) 6 ¼ G(y, x), and so L x would not
be Hermitian.
The Green’s functions must satisfy the homogeneous BCs. That is,
B 1 G
ð Þ ¼ B 2 G
ð Þ ¼ 0:
ð10:140Þ
Also, we require continuity condition of G(x, y) at x ¼ y and discontinuity condition
of
∂G x, y
ð Þ
∂x
at x ¼ y described by (10.122). Thus, we have four conditions including
BCs and continuity and discontinuity conditions to be satisfied by G(x, y). Thus, we
can determine four constants c 1 , c 2 , d 1 , and d 2 by the four conditions.
Now, let us inspect further details about the Green’s functions by an example.
Example 10.4 Let us consider a following differential equation
d
2 u
dx 2 þ u ¼ 1:
ð10:141Þ
We assume that a domain of the argument x is [0, L]. We set boundary conditions
such that
u 0
ð Þ ¼ σ 1 and u L
ð Þ ¼ σ 2 :
ð10:142Þ
Thus, if at least one of σ 1 and σ 2 is not zero, we are dealing with an inhomogeneous
differential equation under inhomogeneous BCs.
Next, let us seek conditions that the Green’s function satisfies. We also seek a
fundamental set of solutions of a homogeneous equation described by
d
2 u
dx 2 þ u ¼ 0:
ð10:143Þ
This is obtained by putting a ¼ c ¼ 1 and b ¼ 0 in a general form of (10.5) with a
weight function being unity. The differential equation (10.143) is therefore selfadjoint according to the argument of Sect. 10.3. A fundamental set of solutions are
given by
e
ix and e
Àix
:
Then, we have
F 1 x, y
ð Þ ¼ c 1 e
ix
þ c 2 e
Àix for 0 x < y,
F 2 x, y
ð Þ ¼ d 1 e
ix
þ d 2 e
Àix for y < x L:
ð10:144Þ
The functions F 1 (x, y) and F 2 (x, y) must satisfy the following BCs such that
10.5 Construction of Green’s Functions
405