PDES: SEPARATION OF VARIABLES AND OTHER METHODS
solution is immediate:
̄u(x, s)=Aexp
(√
s
κ
x
)
+Bexp
(
−
√
s
κ
x
)
,
where the constantsAandBmay depend ons.
We requireu(x, t)→0asx→∞and so we must also have ̄u(∞,s) = 0; consequently
we require thatA= 0. The value ofBis determined by the need foru(0,t)=u 0 and hence
that
̄u(0,s)=
∫∞
0
u 0 exp(−st)dt=
u 0
s
.
We thus conclude that the appropriate expression for the Laplace transform ofu(x, t)is
̄u(x, s)=
u 0
s
exp
(
−
√
s
κ
x
)
. (21.72)
To obtainu(x, t) from this result requires the inversion of this transform – a task that is
generally difficult and requires a contour integration. This is discussed in chapter 24, but
for completeness we note that the solution is
u(x, t)=u 0
[
1 −erf
(
x
√
4 κt
)]
,
where erf(x) is the error function discussed in the Appendix. (The more complete sets of
mathematical tables list this inverse Laplace transform.)
In the present problem, however, an alternative method is available. Letw(t)bethe
amount of salt that has diffused into the tube in timet;then
w(t)=
∫∞
0
u(x, t)dx,
and its transform is given by
̄w(s)=
∫∞
0
dtexp(−st)
∫∞
0
u(x, t)dx
=
∫∞
0
dx
∫∞
0
u(x, t)exp(−st)dt
=
∫∞
0
̄u(x, s)dx.
Substituting for ̄u(x, s) from (21.72) into the last integral and integrating, we obtain
w ̄(s)=u 0 κ^1 /^2 s−^3 /^2.
This expression is much simpler to invert, and referring to the table of standard Laplace
transforms (table 13.1) we find
w(t)=2(κ/π)^1 /^2 u 0 t^1 /^2 ,
which is thus the required expression for the amount of diffused salt at timet.
The above example shows that in some circumstances the use of a Laplace
transformation can greatly simplify the solution of a PDE. However, it will have
been observed that (as with ODEs) the easy elimination of some derivatives is
usually paid for by the introduction of a difficult inverse transformation. This
problem, although still present, is less severe for Fourier transformations.