for conven ience, choose to specify the homogeneous solutions y
1
(x) and y
2
(x) by requiring
16
y
1
(a) = 0 , y
2
(b) = 0 . (4.265)
This choice is not obligatory; any choice of boundary conditions for y
1
(x) an d y
2
(x) will
allow us to solve the inhomogeneous equation, as long as we make sure that our boundary
conditions lead to linearly-independent solutions y
1
(x) and y
2
(x). (We know this because
we have already proved that (4.264) gives the most general possible solution of the in-
homogeneous solution, provided that y
1
and y
2
are linearly-independent solutions of the
homogeneous equ ation.) The choice in (4.265) is very convenient, as we shall now see.
In our example, we want our inhomogeneous solution u(x) to satisfy u(a) = 0 and u(b) =
0. To ensure these two conitions, we have at our disposal to choose the two integration limits
x
1
and x
2
. In view of (4.265), we can see that u(a) = 0 implies we should take x
1
= a.
Similarly, u(b) = 0 implies we should take x
2
= b. T hus f rom (4.264) we can write the
solution as
u(x) =
Z
x
a
dt f(t)
y
1
(t) y
2
(x)
∆(y
1
, y
2
)
+
Z
b
x
dt f(t)
y
2
(t) y
1
(x)
∆(y
1
, y
2
)
, (4.266)
(Note that the sign has changed on th e second term because we have reversed the order of
the limits.)
Note that (4.266) can be interpreted as the equation
u(x) =
Z
b
a
dt G(x, t) f (t) , (4.267)
where the Green function G(x, t) is given by
G(x, t) =
y
1
(x) y
2
(t)
∆(y
1
, y
2
)
if x ≤ t ,
=
y
2
(x) y
1
(t)
∆(y
1
, y
2
)
if x ≥ t . (4.268)
Here ∆(y
1
, y
2
) is a function of the integration variable, t.
We can now try comparing this result with our previous eigenfunction expansion (4.259)
for the Green function, since the two should in pr inciple agree. Doing this in general would
16
A full specification of boundary conditions that leads to a unique solution requires two cond itions, and
not just one. For example, y
1
(x) is not fully pinned down simply by specifiying y
1
(a) = 0, since we can take
any constant multiple of y
1
(x) and it again satisfies the condition of vanishing at x = a. But t his scaling
arbitrariness is completely unimportant in the present context, because, as can be seen from (4.264), y
1
appears linearly in both the numerator and the d enominator (via the Wronskian) in each term, as does y
2
.
Thus we do not need to specify the scale factor in y
1
and y
2
, and so we need only specify the one condition
on each of y
1
and y
2
.
91