21.4 INTEGRAL TRANSFORM METHODS
An infinite metal bar has an initial temperature distributionf(x)along its length. Find
the temperature distribution at a later timet.We are interested in values ofxfrom−∞to∞, which suggests Fourier transformation
with respect tox. Assuming that the solution obeys the boundary conditionsu(x, t)→ 0
and∂u/∂x→0as|x|→∞, we may Fourier-transform the one-dimensional diffusion
equation (21.71) to obtain
κ
√
2 π∫∞
−∞∂^2 u(x, t)
∂x^2exp(−ikx)dx=1
√
2 π∂
∂t∫∞
−∞u(x, t)exp(−ikx)dx,where on the RHS we have taken the partial derivative with respect totoutside the
integral. Denoting the Fourier transform ofu(x, t)by ̃u(k, t), and using equation (13.28) to
rewrite the Fourier transform of the second derivative on the LHS, we then have
−κk^2 ̃u(k, t)=∂ ̃u(k, t)
∂t.
This first-order equation has the simple solution
̃u(k, t)= ̃u(k,0) exp(−κk^2 t),where the initial conditions give
̃u(k,0) =1
√
2 π∫∞
−∞u(x,0) exp(−ikx)dx=
1
√
2 π∫∞
−∞f(x)exp(−ikx)dx= ̃f(k).Thus we may write the Fourier transform of the solution as̃u(k, t)= ̃f(k)exp(−κk^2 t)=√
2 π ̃f(k)G ̃(k, t), (21.73)where we have defined the function ̃G(k, t)=(
√
2 π)−^1 exp(−κk^2 t). Since ̃u(k, t)canbe
written as the product of two Fourier transforms, we can use the convolution theorem,
subsection 13.1.7, to write the solution as
u(x, t)=∫∞
−∞G(x−x′,t)f(x′)dx′,whereG(x, t) is the Green’s function for this problem (see subsection 15.2.5). This function
is the inverse Fourier transform of ̃G(k, t) and is thus given by
G(x, t)=1
2 π∫∞
−∞exp(−κk^2 t)exp(ikx)dk=
1
2 π∫∞
−∞exp[
−κt(
k^2 −ix
κtk)]
dk.Completing the square in the integrand we find
G(x, t)=1
2 πexp(
−
x^2
4 κt)∫∞
−∞exp[
−κt(
k−ix
2 κt) 2 ]
dk=
1
2 πexp(
−
x^2
4 κt)∫∞
−∞exp(
−κtk′^2)
dk′=
1
√
4 πκtexp(
−
x^2
4 κt)
,
where in the second line we have made the substitutionk′=k−ix/(2κt), and in the last