18.303 Problem Set 10 Solutions
Problem 1: (10+10+10+10 points)
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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 = Ω 1b ūu0 + a1 v̄ · v0 , for appropriate
∂
boundary conditions (e.g. u|dΩ = 0). This gave us conservation of “energy”: ∂t
hw, wi = hw, D̂wi + hD̂w, wi =
∗
hD̂ w, wi + hD̂w, wi = 0.
(a) Suppose we don’t have any boudnary conditions; in this case integration by parts gives
ˆ
hw, D̂wi =
(ū∇ · v + v̄ · ∇u)
ˆΩ
=
(∇ · [ūv] − v · ∇ū + ∇ [uv̄] − u∇ · v̄)
Ω
‹
= −hD̂w, wi + 2<
(ūv)dA,
dΩ
where the surface integral comes from the divergence theorem and < denotes the real part (since z + z̄ = 2<z).
Therefore, we obtain:
‹
‹
∂
hw, wi = hw, D̂wi + hD̂w, wi = −hD̂w, wi +
F · dA + hD̂w, wi =
F · dA,
∂t
dΩ
dΩ
where F = 2<(ūv).
Thus, −F can be interpreted as a flux of energy per unit area per unit time. (Note the sign: the loss of
∂
energy from Ω is − ∂t
hw, wi.)
(b) Plugging w(x, y, z, t) = wk (x, y)ei(kz−ωt) into the eigenequation D̂w = −iωw, cancelling e−iωt from both sides,
and moving eikz to the left side, we obtain:
D̂k wk = e−ikz D̂eikz wk = −iωwk ,
where as in class this transformation of D̂ turns each ∇ into ∇xy + ik, where ∇xy is just the x and y derivatives
and k = kẑ, so that
b [∇xy + ik] ·
b∇xy ·
bk·
D̂k =
=
+i
.
a [∇xy + ik]
a∇xy
ak
Plugging this into the 2d inner product hw, w0 i =
´
Ωxy
1
0
b ūu
+ a1 v̄ · v0 , we obtain two terms from hw, D̂k w0 i,
from the two terms above. The first, from the ∇xy terms, works exactly like D̂ (since we analyzed D̂ in any
number of dimensions, and this is just the 2d case), and so is anti-Hermitian as before. The second, from the k
terms, is:
ˆ
bk·
hw, i
w0 i =
(ūik · v + v̄ · iku)
ak
Ωxy
ˆ
=−
iku · v + ik · vu
Ωxy
= h−i
and so this part is anti-Hermitian too. Hence D̂k = −D̂k∗ .
1
bk·
ak
w, w0 i,
(c) Here, we just apply the ordinary quotient/product rules of differentiation:
∂ D̂k
k
h ∂wk , iD̂k wk i + hwk , iD̂k ∂w
hwk , iD̂k wk i ∂wk
∂wk
∂ hwk , iD̂k wk i
∂k i + hwk , i ∂k wk i
h
= ∂k
−
,
w
i
+
hw
,
i
k
k
∂k hwk , wk i
hwk , wk i
hwk , wk i2
∂k
∂k
∂wk
∂wk
∂ D̂k
h
, iD̂k wk i + hiD̂k wk , ∂k i + hwk , i ∂k wk i hwk , iD̂k wk i ∂wk
∂wk
= ∂k
h
−
,
w
i
+
hw
,
i
k
k
hwk , wk i
hwk , wk i2
∂k
∂k
(
(
∂w
(
(
∂ D̂k
k
k
k
,(
wk(
i(
+(
hwk , ∂w
ω h ∂k(
ω
∂wk
∂w
∂k i + hwk , i ∂k wk i
−
h
,w
+hwk ,
i
= ((
k i
hwk , wk i
hwk , wk i ∂k
∂k
D̂k
hwk , i ∂∂k
wk i
=
,
hwk , wk i
where in the second line we have used the fact that iD̂k is Hermitian to move it from the right to the left in one
term, and in the third line we have used the eigenequation iD̂k wk = ωwk .
(d) Since k only appears in the k terms of D̂k , we obtain:
i
∂ D̂k
=−
∂k
and hence
hwk , i
∂ D̂k
wk i = −
∂k
bẑ·
aẑ
,
ˆ
ˆ
(ūvz + uv̄z ) dx dy = −
Fz dx dy,
Ωxy
Ωxy
where Fz is the z component of F from part (a). Note the minus sign: this is consistent with the interpretation
from (a) of −F as the flux of energy.
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