Comparing (10.191) with (10.183), we recognize that the sign of RHS of (10.191)
has been reversed relative to RHS of (10.183). This is also the case after exchanging
arguments x and y. Note, however, Θ(x, y) ¼ À 1 in the present case. As a result, two
minus signs cancel and (10.191) takes exactly the same expression as (10.183).
Proceeding with calculations similarly, for both Cases I and II we arrive at a unified
solution represented by (10.190) throughout a domain (À1, +1).
10.6.4 Examples
To deepen understanding of Green’s functions, we deal with tangible examples of
the IVP below.
Example 10.5 Let us consider a following inhomogeneous differential equation
d
2 u
dx 2 þ u ¼ 1:
ð10:192Þ
Note that (10.192) is formally the same differential equation of (10.141). We may
encounter (10.192) when we are observing a motion of a charged harmonic oscillator
that is placed under a static electric field. We assume that a domain of the argument
x is a whole range of real numbers. We set boundary conditions such that
u 0
ð Þ ¼ σ 1 and u
0 0
ð Þ ¼ σ 2 :
ð10:193Þ
As in the case of Example 10.4, a fundamental set of solutions are given by
e
ix and e
Àix
:
ð10:194Þ
Therefore, following (10.173), we get
F x, y
ð Þ ¼ sin x À y
ð
Þ:
ð10:195Þ
Also following (10.190) we have
u x
ð Þ ¼
Z x
0
dy sin x À y
ð
Þþσ 2 sin x À σ 1 À cos x À y
ð
Þ
½
Š
y¼0
¼ 1 À cos x þ σ 2 sin x þ σ 1 cos x:
ð10:196Þ
In particular, if we choose σ 1 ¼ 1 and σ 2 ¼ 0, we have
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10 Introductory Green’s Functions