Section 11: Eigenfunction Expansion of Green Functions in This Lecture

Section 11: Eigenfunction Expansion of Green Functions In this lecture we see how to expand a Green function in terms of eigenfunctions of the underlying Sturm-Liouville problem. First we review Hermitian matrices

11. 1. Hermitian matrices

∗ † † Hermitian matrices satisfy Hij = Hji = Hij where H is the Hermitian conjugate of H. You should recall that Hermitian matrices have real eigenvalues λn such that

H |ni = λn |ni

(where we use ‘bra ket’ notation). |ni is the eigenvector and (nondegenerate) eigenvectors are orthogonal.

X |nihn| |xi = |bi + A |0i λn n6=0 where A is an arbitrary constant.

11. 2. Hermitian Operators

We now consider the Sturm-Liouville eigenvalue problem

L(x)u(x) = λu(x)

48 with some boundary conditions imposed. An operator L is Hermitian if Z Z ∗ dx u∗(x)L(x)v(x) = dx v∗(x)L(x)u(x)

∗ † c.f Luv = Lvu = Luv In the same way as for Hermitian matrices we can show that Hermitian operators have real eigenvalues and the eigenfunctions φn(x) are orthogonal Z Z ∗ ∗ if L(x)φm(x) = λmφm(x) dx φn(x)L(x)φm(x) = λm dx φn(x)φm(x) Z Z ∗ ∗ and L(x)φn(x) = λnφn(x) dx φm(x)L(x)φn(x) = λn dx φm(x)φn(x)

Z ∗ ∗ since L Hermitian ⇒ (λm − λn) dx φn(x)φm(x) = 0 Z ∗ ⇒ λn are real and dx φn(x)φm(x) = δnm

d2 u Example: = 0 b.c. u(0) = u(L) = 0 dx2 d2 u First verify that is an Hermitian operator: dx2

Z L 2 L Z L ∗ L ∗ L Z L 2 ∗ d v ∗ dv du dv ∗ dv du ∗ d u u 2 dx = u − dx = u − v + dx v 2 0 dx dx 0 0 dx dx dx 0 dx 0 0 dx and the boundary conditions ensure that the operator is Hermitian. In this example the eigenfunctions and eigenvalues are of d2/dx2 obeying the b.c.s are

2 1/2 nπx nπ 2 φ (x) = sin λ = − n L L n L and form a discrete (i.e. countable) spectrum and that λ is bounded from above but not below. Note how the b.c.s impose the spectrum. We can check orthogonality of the eigenfunctions and completeness:

2 Z L πmx πnx dx sin sin = δmn for m, n > 0 L 0 L L ∞ 2 X mπx0 mπx sin sin = δ(x − x0) L L L m=1

11. 3. General Sturm-Liouville problem

Consider the general Sturm Liouville problem

L(x)φn(x) = λnρ(x)φn(x) a ≤ x ≤ b d Bφ (x) = 0 at x = a x = b e.g. B(a) = α + β n dx

49 d d where = L(x) = p(x) + q(x). ρ(x) is known as a ‘weight function’. dx dx It is easy to show that L(x) is Hermitian and

The analogy with the bra and ket vectors is to think of the eignfunctions φn(x) as basis vectors for a vector space. Then we can expand an arbitrary function f in terms of them. (N.B. The notation diﬀers slightly from quantum mechanics; there |φi is the eigenvector in Hilbert space of an abstract operator and ψ(x) denotes the wavefunction in position space. Here our S.L. operators are always written down in position space as are our eigenfunctions.) Consider the Green function which satisﬁes

L(x)G(x, x0) = δ(x − x0) BG(x, x0) = 0 at x = a x = b

0 X 0 We expand G(x, x ) = Cm(x )φm(x) m

X 0 0 ⇒ λmρ(x)φm(x)Cm(x ) = δ(x − x ) m

∗ Now multiply by φn(x) and integrate

X 0 ∗ 0 λmCm(x )δmn = φn(x ) m ∗ 0 X φ (x )φn(x) ⇒ G(x, x0) = λ n n This is sometimes known as the bilinear expansion of the Green function and should be compared to the expression in section 11.1 for H−1 We deduce that the Green function is basically the inverse of the Sturm Liouville operator. Example: Green Function for Finite stretched string with periodic forcing ∂2u 1 ∂2u − = f(x)e−iω0t b.c. u(0) = u(L) = 0 ∂x2 c2 ∂t2 Here u is the displacement of a stretched string and the rhs gives the external forcing which here is periodic in time. We guess the time dependence as u(x, t) = y(x)e−iω0t which leads to

d2y ω2 + k2y = f(x) where k2 ≡ 0 and the b.c.s are y(0) = y(L) = 0 ♦ dx2 0 0 c2

50 d2 2 This is ODEis the Helmholtz equation and involves a Hermitian operator dx2 + k0 for which the eigenfunctions of the Sturm-Liouville problem ♦ are r 2 n2π2 φ (x) = sin(nπx/L) λ = k2 − n L n 0 L2 The Green function obeys d2G(x, x0) + k2G = δ(x − x0) G(0, x0) = G(L, x0) = 0 dx2 0 We assume a Fourier sine series solution to this equation i.e. we insert ∞ ∞ X X 2 G = γ (x0) sin(nπx/L) δ(x − x0) = sin(nπx0/L) sin(nπx/L) n L n=1 n=1 Then equating coeﬃcients of sin nπx/L gives n2π2 2 2 sin(nπx0/L) − γ (x0) + k2γ (x0) = sin(nπx0/L) ⇒ γ (x0) = L2 n n L n L k2 − n2π2/L2

∞ 2 X sin(nπx0/L) sin(nπx0/L) Thus G(x, x0) = L k2 − n2π2/L2 n=1 0 Also note the symmetry of the Green function G(x, x0) = G(x0, x) and that the eigenfunction 2 2 2 2 expansion fails if k0 = n π /L . To understand this latter point, note that λn = 0 implies 2 2 that the external forcing (k0) coincides with an eigenfrequency of the unforced syste (nπ/L) . Thus the system we will have resonance and the oscillations will grow unboundedly (if there is no dissipation). Compare with the matrix case in 11.1 where there is one eigenvalue zero but there may still be a solution if the rhs does not overlap with eigenvector 0. In the present case we may yet have a solution if f(x) has zero overlap with the zero eigenfunction i.e. the spatial shape of the external forcing does not excite the resonance.

11. 4. Continuous spectrum

The boundary conditions of the problem may allow a continuous spectrum of eigenvalues. Example: Helmholtz equation on inﬁnite domain d2y + k2y = f(x) y(±∞) bounded dx2 0 d2y The underlying eigenvalue problem is + k2y = λy where y(±∞) bounded. The dx2 0 2 2 eigenfunctions are y = exp(±ikx) and eigenvalues λ = k0 − k with k cts. We construct G by taking the Fourier transform of d2G(x, x0) + k2G(x, x0) = δ(x − x0) dx2 0 2 2 0 −ikx0 yielding −k + k0 g(k, x ) = e Z ∞ Z ∞ ik(x−x0) 0 1 0 ikx 1 e ⇒ G(x, x ) = g(k, x )e dk = 2 2 dk 2π −∞ 2π −∞ k0 − k This formula is just the generalisation of the sum, in the bilinear expansion for the discrete case, to an integral.