Superposition of eigenfunctions Greens functions

17.5 SUPERPOSITION OF EIGENFUNCTIONS: GREEN’S FUNCTIONS
17.5 Superposition of eigenfunctions: Green’s functions
We have already seen that if
Lyn (x) = λn ρ(x)yn (x),
(17.43)
where L is an Hermitian operator, then the eigenvalues λn are real and the
eigenfunctions yn (x) are orthogonal (or can be made so). Let us assume that we
know the eigenfunctions yn (x) of L that individually satisfy (17.43) and some
imposed boundary conditions (for which L is 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 of L form a
complete set, the full solution, y(x), to (17.44) may be written as a superposition
of eigenfunctions, i.e.
y(x) =
∞
cn yn (x),
(17.45)
n=0
for some choice of the constants cn . Making full use of the linearity of L, we have
∞
∞
∞
cn yn (x) =
cn Lyn (x) =
cn λn ρ(x)yn (x).
f(x) = Ly(x) = L
n=0
n=0
n=0
(17.46)
Multiplying the first and last terms of (17.46) by yj∗ and integrating, we obtain
b
∞ b
yj∗ (z)f(z) dz =
cn λn yj∗ (z)yn (z)ρ(z) dz,
(17.47)
a
n=0
a
where we have used z as the integration variable for later convenience. Finally,
using the orthogonality condition (17.27), we see that the integrals on the RHS
are zero unless n = j, and so obtain
b ∗
yn (z)f(z) dz
1
.
(17.48)
cn =
b a
∗
λn
yn (z)yn (z)ρ(z) dz
a
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
b ∗
∞
yn (z)f(z) dz
1
yn (x).
(17.49)
y(x) =
b a
∗
λn
yn (z)yn (z)ρ(z) dz
n=0
a
If we work with normalised eigenfunctions ŷn (x), so that
b
ŷn∗ (z)ŷn (z)ρ(z) dz = 1
for all n,
a
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EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONS
and we assume that we may interchange the order of summation and integration,
then (17.49) can be written as
.
b #
∞ 1
∗
ŷn (x)ŷn (z)
f(z) dz.
y(x) =
λn
a
n=0
The quantity in braces, which is a function of x and z only, is usually written
G(x, z), and is the Green’s function for the problem. With this notation,
b
G(x, z)f(z) dz,
(17.50)
y(x) =
a
where
G(x, z) =
∞
1
ŷn (x)ŷn∗ (z).
λn
(17.51)
n=0
We note that G(x, z) is determined entirely by the boundary conditions and the
eigenfunctions ŷn , and hence by L itself, and that f(z) depends purely on the
RHS of the inhomogeneous equation (17.44). Thus, for a given L and boundary
conditions we can establish, once and for all, a function G(x, z) that will enable
us to solve the inhomogeneous equation for any RHS. From (17.51) we also note
that
G(x, z) = G∗ (z, x).
(17.52)
We have already met the Green’s function in the solution of second-order differential equations in chapter 15, as the function that satisfies the equation
L[G(x, z)] = δ(x − z) (and the boundary conditions). The formulation given
above is an alternative, though equivalent, one.
Find an appropriate Green’s function for the equation
y + 14 y = f(x),
with boundary conditions y(0) = y(π) = 0. Hence, solve for (i) f(x) = sin 2x and (ii)
f(x) = x/2.
One approach to solving this problem is to use the methods of chapter 15 and find
a complementary function and particular integral. However, in order to illustrate the
techniques developed in the present chapter we will use the superposition of eigenfunctions,
which, as may easily be checked, produces the same solution.
The operator on the LHS of this equation is already Hermitian under the given boundary
conditions, and so we seek its eigenfunctions. These satisfy the equation
y + 14 y = λy.
This equation has the familiar solution
1
1
y(x) = A sin
− λ x + B cos
− λ x.
4
4
570
17.5 SUPERPOSITION OF EIGENFUNCTIONS: GREEN’S FUNCTIONS
1
Now, the boundary conditions require that B = 0 and sin
− λ π = 0, and so
4
1
− λ = n,
where n = 0, ±1, ±2, . . . .
4
Therefore, the independent eigenfunctions that satisfy the boundary conditions are
yn (x) = An sin nx,
where n is any non-negative integer, and the corresponding eigenvalues are λn =
The normalisation condition further requires
1/2
π
2
A2n sin2 nx dx = 1
⇒
An =
.
π
0
1
4
− n2 .
Comparison with (17.51) shows that the appropriate Green’s function is therefore given
by
G(x, z) =
∞
2 sin nx sin nz
.
1
π n=0
− n2
4
Case (i). Using (17.50), the solution with f(x) = sin 2x is given by
∞
∞
2 π sin nx sin nz
2 sin nx π
sin nz sin 2z dz.
y(x) =
sin 2z dz =
1
1
2
2
π 0
π n=0 4 − n 0
−n
4
n=0
Now the integral is zero unless n = 2, in which case it is
π
π
sin2 2z dz = .
2
0
Thus
y(x) = −
2 sin 2x π
4
= − sin 2x
π 15/4 2
15
is the full solution for f(x) = sin 2x. This is, of course, exactly the solution found by using
the methods of chapter 15.
Case (ii). The solution with f(x) = x/2 is given by
π ∞
∞
2
sin nx sin nz z
1 sin nx π
z sin nz dz.
y(x) =
dz
=
1
1
π n=0
2
π n=0 4 − n2 0
− n2
0
4
The integral may be evaluated by integrating by parts. For n = 0,
π π
π
z cos nz
cos nz
z sin nz dz = −
+
dz
n
n
0
0
0
π
−π cos nπ
sin nz
+
=
n
n2 0
π(−1)n
=−
.
n
For n = 0 the integral is zero, and thus
y(x) =
∞
sin nx
,
(−1)n+1 1
n
− n2
4
n=1
is the full solution for f(x) = x/2. Using the methods of subsection 15.1.2, the solution
is found to be y(x) = 2x − 2π sin(x/2), which may be shown to be equal to the above
solution by expanding 2x − 2π sin(x/2) as a Fourier sine series. 571