17.5 SUPERPOSITION OF EIGENFUNCTIONS: GREEN’S FUNCTIONS
17.5 Superposition of eigenfunctions: Green’s functions
We have already seen that if
Lyn(x)=λnρ(x)yn(x), (17.43)
whereLis an Hermitian operator, then the eigenvaluesλnare real and the
eigenfunctionsyn(x) are orthogonal (or can be made so). Let us assume that we
know the eigenfunctionsyn(x)ofLthat individually satisfy (17.43) and some
imposed boundary conditions (for whichLis Hermitian).
Now let us suppose we wish to solve the inhomogeneous differential equation
Ly(x)=f(x), (17.44)
subject to the same boundary conditions. Since the eigenfunctions ofLform a
complete set, the full solution,y(x), to (17.44) may be written as a superposition
of eigenfunctions, i.e.
y(x)=
∑∞
n=0
cnyn(x), (17.45)
for some choice of the constantscn. Making full use of the linearity ofL, we have
f(x)=Ly(x)=L
(∞
∑
n=0
cnyn(x)
)
=
∑∞
n=0
cnLyn(x)=
∑∞
n=0
cnλnρ(x)yn(x).
(17.46)
Multiplying the first and last terms of (17.46) byy∗jand integrating, we obtain
∫b
a
y∗j(z)f(z)dz=
∑∞
n=0
∫b
a
cnλny∗j(z)yn(z)ρ(z)dz, (17.47)
where we have usedzas the integration variable for later convenience. Finally,
using the orthogonality condition (17.27), we see that the integrals on the RHS
are zero unlessn=j, and so obtain
cn=
1
λn
∫b
ay
∗
n(z)f(z)dz
∫b
ay
∗
n(z)yn(z)ρ(z)dz
. (17.48)
Thus, if we can find all the eigenfunctions of a differential operator then (17.48)
can be used to find the weighting coefficients for the superposition, to give as the
full solution
y(x)=
∑∞
n=0
1
λn
∫b
ay
∗
n(z)f(z)dz
∫b
ay
∗
n(z)yn(z)ρ(z)dz
yn(x). (17.49)
If we work with normalised eigenfunctionsyˆn(x), so that
∫b
a
yˆn∗(z)yˆn(z)ρ(z)dz= 1 for alln,