
6.4 Field Diffusion in the Two-sided Infinite Space 367
.
Applying the Ansatz
yields
,
i.e., the coefficient function results in the expanded function itself, which is nothing
more than an unfamiliar, however useful, interpretation of the defining property of
the δ-function. If the problem is solved for a basis function, then it is solved for any
function that can be expanded as a series of this base. This is the reason why the δ-
function allows one to find the Green’s function for the problem.
There is another approach, that is more systematic. We just gave the specific
solution (6.69), which is not very satisfying. Now, we want to derive a solution
starting from initial values and boundary conditions. For that purpose, we now
analyze , rather than . Then we use (6.51) and (6.68) to obtain
,
(6.79)
i.e., the partial differential equation in and becomes an ordinary differential
equation in and the initial condition is already included. The task is to find the
solution that remains finite for both and . Those are the
necessary boundary conditions (which are oftentimes implicitly assumed) to make
the solution unique. The general solution to (6.79) results from adding the solution
of the related homogeneous equation to a specific solution of the inhomogeneous
equation.
The special solution is derived from the solution of the homogeneous equation by
means of method variation of parameters. Leaving out the details, the general
solution of (6.79) is
(6.80)
The integral is a special solution of the inhomogeneous equation (6.79). The lower
boundary may be replaced by an arbitrary constant. Such a new integral is, again, a
solution of the inhomogeneous equation. On the other hand, the difference between
the two integrals is a solution of the homogeneous ordinary differential equation.
This means that it can be expressed by a suitable superposition of the two integrals.
In other words, the solution (6.80) remains unchanged when changing the lower
boundary of the integral, when simultaneously changing the coefficients and
in a suitable manner.
In particular, if
δξ ξ
0
–()δξξ'
0
–()ξd
∞–
+∞
∫
δξ
0
ξ'
0
–()=
h ξ() g ξ
0
()δξξ
0
–()ξ
0
d
∞–
+∞
∫
=
g ξ
0
() h ξ
0
()=
B
˜
z
ξ p,() B
z
ξτ,()
ξ
2
2
∂
∂
B
˜
z
ξ p,()pB
˜
z
ξ p,()h ξ()–=
ξτ
ξ
ξ∞–→ξ+∞→
B
˜
z
ξ p,()A
1
pξ–[]exp A
2
+ pξ[]exp h ξ
0
()
p ξξ
0
–()sinh
p
-------------------------------------
ξ
0
d
∞–
ξ
∫
–+=
A
1
A
2