18.303 Problem Set 10
Due Monday, 29 November 2010.
Problem 1: Energy flow and group velocity
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In class (and notes), we showed that we can turn the scalar wave equation b∇ · (a∇u) = ∂∂t2u (a > 0 and b > 0) into
∂v
two coupled first-derivative equations: ∂u
∂t = b∇ · v, ∂t = a∇u by introducing a new (vector) unknown v(x, t). By
T
defining w = (u, v) , we obtained the form
∂w
b∇·
=
w = D̂w,
a∇
∂t
where D̂ was anti-Hermitian (D̂∗ = −D̂) under the inner product hw, w0 i =
boundary conditions (e.g. u|dΩ = 0). This gave us conservation of “energy”:
hD̂∗ w, wi + hD̂w, wi = 0.
´
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ūu0 + a1 v̄ · v0 , for
Ω b
∂
∂t hw, wi = hw, D̂wi +
appropriate
hD̂w, wi =
(a) To show that D̂∗ = −D̂, we integrated by parts and relied on the boundary conditions to ensure that the surfaceintegral terms from integration by parts (the divergence theorem) vanished. Now, instead, assume that you aren’t
given any information about u or v on dΩ, so that we cannot drop the surface integrals when integrating by
parts in hw, D̂wi to move D̂ to the other side. Show that we now obtain
‹
∂
hw, wi =
F · dA
∂t
dΩ
for some vector field F in terms of u and v. We can therefore interpret F as a flow rate (“flux”) of energy per
unit area per unit time. (The electromagnetic version of this is called the “Poynting vector”.)
Reminder: the key identity for integration by parts is that,´ for any function
φ and any vector field f , we
´
have ∇ · (φf ) = (∇φ) · f + φ∇ · f . e.g. this allows us to write Ω φ∇ · f = Ω [∇ · ´(φf ) − (∇φ) · f ], where the first
term turns into a surface integral by the divergence theorem, and vice versa for Ω (∇φ) · f .
(b) Suppose that we are in 3d have a and b invariant in the z direction (e.g. in a waveguide along z), and the boundary
conditions also being independent of z (Ω being infinite in z). Then we can look for separable eigenfunctions
w(x, y, z, t) = wk (x, y)ei(kz−ωt) , similar to class, satisfying D̂w = −iωw = ∂w
∂t , which leads to the reduced 2d
eigenproblem:
D̂k wk = −iωwk .
What is D̂k ? wk is only a function of (x, y), so all we need is a 2d domain Ωxy . Suppose u|dΩ = 0, and so
uk |dΩxy = 0. Show that D̂k∗ = −D̂k under the 2d version of the inner product hwk , wk0 i, same as above but only
integrated over Ωxy .
(c) For an eigenfunction D̂k wk = −iωwk , we can write the frequency ω(k) as the Rayleigh quotient:
ω(k) =
hwk , iD̂k wk i
.
hwk , wk i
Show that:
D̂k
hwk , i ∂∂k
wk i
∂ω
=
∂k
hwk , wk i
by differentiating via the usual product rule etcetera. You should find that the
fact that iD̂k is Hermitian [(iD̂k )∗ = (−i)D̂k∗ = iD̂k ].
∂wk
∂k
terms all cancel, due to the
D̂k
(d) Show that the numerator hwk , i ∂∂k
wk i from the part (c) is exactly an integral of F · ẑ [the F from part (a)].
Hence, the group velocity ∂ω/∂k is exactly equal to the power flowing in z divided by the energy density
hwk , wk i (= energy per unit z), and can be interpreted as an energy velocity.
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