18.303 Problem Set 3 Solutions
Problem 1: (5+5+5+10+2+5+5)
dQn
1
n
(a) The heat capacity equation tells us that dT
dt = cρa∆x dt , where dQn /dt is the rate of change
of the heat in the n-th piece. The thermal conductivity equation tells us that dQn /dt, in
turn, is equal to the sum of the rates q at which heat flows from n + 1 and n − 1 into n:
dTn
1 dQn
1
κa
=
=
[(Tn+1 − Tn ) + (Tn−1 − Tn )] = α(Tn+1 −Tn )+α(Tn−1 −Tn )
dt
cρa∆x dt
cρa∆x ∆x
where α =
κ
. The only difference for T1 and TN is that they have no heat flow
cρ(∆x)2
n − 1 and n + 1, respectively, since the ends are insulated:
α(TN −1 − TN ).
dT1
dt
= α(T2 − T1 ) and
dTN
dt
=
(b) We can obtain A in two ways. First, we can simply look directly at our equations above,
n
which give dT
dt = α(Tn+1 − 2Tn + Tn−1 ) for every n except T1 and TN , and read off the
corresponding rows of the matrix

−1
 1


A = α


1
−2
..
.

1
..
.
1
..
.
−2
1



.

1 
−1
Alternatively, we can write each of the above steps—differentiating T to get the rate of heat
flow q to the left at each of the N − 1 interfaces between the pieces, then taking the difference
of the q’s to get dT /dt, in matrix form, to write:




1
−1 1

 −1 1




−1 1


−1 1
1 1 
1 
κ




.
.
..
..
A=
κa
 = − DDT ,



.
.
.
.

 ∆x 
cρa ∆x 
cρ
.
.




−1 1

−1 1 
−1 1
−1
|
{z
}
|
{z
}
−D T : (N −1)×N
D: N ×(N −1)
in terms of the D matrix from class (except with N reduced by 1), which gives the same A
as above. As we will see in the parts below, this is indeed a second-derivative approximation,
but with different boundary conditions—Neumann conditions—than the Dirichlet conditions
in class.
By the way, it is interesting to consider −DDT , compared to the −DT D we had in class.
Clearly, −DDT is real-symmetric and negative semidefinite. It is not, however, negative definite, since DT does not (and cannot) have full column rank (its rank must be ≤ the number
of rows N − 1, and in fact in class we showed that it has rank N − 1).
dTn
dt
κ ∂2T
cρ ∂x2
(c) Ignoring the ends for the moment, for all the interior points we have
which is exactly our familiar center-difference approximation for
1
=
κ Tn+1 −2Tn +Tn−1
,
cρ
∆x2
at the point n (x =
[n − 0.5]∆x). Hence, everywhere in the interior our equations converge to
∂T
∂T
=
κ ∂2T
cρ ∂x2
, and
2
thus  =
κ ∂
.
cρ ∂x2
∂T
= 0 at x = 0, L. The easiest way to see this is to
∂x
observe that our heat flow q is really a first derivative, and zero heat flow at the ends
−Tn
is really an approximate derivative:
means zero derivatives. That is, qn+0.5 = κa Tn+1
∆x
∂T
∂T
qn+0.5 ≈ κa ∂x n+0.5 = κa ∂x n∆x , while the flows q0.5 and qN +0.5 to/from n = 0 and
n = N + 1 is zero, and hence q0.5 = qN +0.5 = 0 ≈ κa ∂T
∂x 0,L .
(d) The boundary conditions are
2
κ
Working backwards, consider ÂT = ∂∂xT2 = T 00 (setting cρ
= 1 for convenience) with these
boundary conditions and center-difference approximations. We are given Tn = T ([n −
Tn+1 −Tn
0
for n =
0.5]∆x, t) for n = 1, . . . , N . First, we compute ∂T
∂x n∆x ≈ Tn+0.5 =
∆x
T
1, . . . , N − 1 (−D T using the D above). Unlike the Dirichlet case in class, we don’t com0
pute T0.5
and TN0 +0.5 , since these correspond to ∂T /∂x at x = 0, L, which are zero by the
T0
−T 0
boundary conditions. Then, we compute our approximate 2nd derivatives Tn00 = n+0.5∆x n−0.5
0
for n = 1, . . . , N , where we let T0.5
= TN0 +0.5 = 0 (DT0 using the D from above). This
0
0
0−TN
T1.5
−0
+TN −1
T2 −T1
−0.5
00
=
=
,
T
= −TN∆x
at the endpoints, and
2
2
N
∆x
∆x
∆x
(Tn+1 −Tn )−(Tn −Tn−1 )
Tn+1 −2Tn +Tn−1
=
for
1
<
n
<
N
,
which
are
precisely the rows of
∆x2
∆x2
gives T100 =
Tn00 =
our A
matrix above.
(e) To convert T to u = cρT , we just multiply both sides of
∂T
∂t
= ÂT by cρ to get
∂u
∂t
= B̂u for
B̂ = Â . This was a silly question.
(f) This is a special case of the Sturm-Liouville operator from class so we can more-or-less quote
ˆ L
the result, adjusting for the difference in boundary conditions. Choose hu, vi =
ūv dx .
0
κ
cρ
is a real, positive number. So, integrating by parts as in class, we obtain: hu, Âvi =
L
L
´ 00
´ 0 0
´ 00
κ
κ κ
κ 0
κ
ūv = cρ
ū v = − cρ
ū v = hÂu, vi, where the boundary terms
ūv 0 − cρ
ū v + cρ
cρ
0
00
0
vanish because u and v are zero there. Hence, Â (= B̂) is self-adjoint.
To check
´ 0 2definiteness, as usual we do the same process but stop halfway through: hu, Âui =
κ
− cρ
|u | ≤ 0. So it is negative semidefinite. It is not negative definite, since these
boundary conditions allow u to be a nonzero constant, for which u0 = 0 and hence we can
have hu, Âui = 0 for u 6= 0. Hence the eigenvalues are real and ≤ 0 (and there is a zero
eigenvalue for the eigenfunction u = 1).
(g) As usual, we can write the solution as an infinite series, from class, using the orthogonality
of the eigenfunctions to give us an explicit equation for the coefficients:
u(x, t) =
∞
X
hun , f i
un (x)eλn t ,
hu
,
u
i
n
n
n=1
where Âun = λn un are the eigensolutions (and the hun , un i denominator can be omitted if we
normalize kun k = 1). For large t, most of these terms decay exponentially to zero. However,
2
the λ1 = 0 solution for u1 (x) = 1 does not decay, and hence as t → ∞ this is all that remains:
h1, f i 0t
1e =
u(x, ∞) =
h1, 1i
´L
0
f (x)dx
1
=
´L
L
dx
0
ˆ
L
f (x)dx .
0
That is, after a long time, the initial f “spreads out” and we obtain the average initial
energy density (or average initial temperature) everywhere in the rod.
Problem 2: (5+5+5+10+5)
(a) Writing out the derivation in 18.02 fashion, this is tedious but straightforward:

∇ · (u × v) =
∂
∂x
∂
∂y
∂
∂z

uy vz − uz vy
 uz vx − ux vz 
ux vy − uy vx
∂(uy vz − uz vy ) ∂(uz vx − ux vz ) ∂(ux vy − uy vx )
+
+
∂x
∂y
∂z
∂uy
∂uz
∂ux
∂uz
∂ux
∂uy
−
−
−
=
vx +
vy +
vz
∂y
∂z
∂z
∂x
∂x
∂y
∂vz
∂vx
∂vy
∂vy
∂vz
∂vx
ux
−
−
−
+ uy
+ uz
∂z
∂y
∂x
∂z
∂y
∂x
=
= (∇ × u) · v − u · (∇ × v).
(A much more compact derivation is possible using Einstein notation and the Levi–Civitá
tensor, but probably most of you haven’t seen this notation.)
(b) Given the above identity, integration by parts is straightforwards:
ˆ
ˆ
ū · (∇ × v) =
∇ · (ū × v) + ∇ × u · v
hu, ∇ × vi =
Ω
‹Ω
=
(ū × v) · dS + h∇ × u, vi,
∂Ω
applying the divergence theorem in the second line. So, the surface term
w = ū × v .
‚
∂Ω
w · dS is for
(c) We must have (ū × v) · dS = 0. Let dS = n dS, where n is the outward unit normal vector
at each point on ∂Ω. Then we must have ū × v ⊥ n, which is true if, for example, both u
and v are parallel to n at the boundary. i.e. if u × n|∂Ω = 0 . It is not necessary to require
u|∂Ω = 0 on the boundary, and that answer will not be accepted as I specifically requested
that you not constrain all the components of u on the boundary.
Another way to see this is to write (ū × v) · dS = n · (ū × v)dS = v · (n × ū)dS = ū · (v × n)dS
by elementary triple-product identities, and hence we again see that it is sufficient to have
u × n = 0 on the boundary.
Although I will accept the above answer, it is actually possible to contrive a slightly weaker
condition: u and v can have components ⊥ to n on the boundary, as long as those surfaceparallel components are in the same direction for both u and v (to obtain zero cross product).
3
That is, suppose p(x) is some surface-parallel (p ⊥ n) vector field on ∂Ω.1 Then it is sufficient for the surface-parallel component of u to be ⊥ to p everywhere on the boundary, or
equivalently u · p|∂Ω = 0.
[Actually, there are other possible conditions on u if we allow non-local boundary conditions, where u at one point on the boundary is related to u at another point. For example,
if
‚ Ω is a cube and u is periodic (i.e. u on one face equals u on the opposite face), then the
vanishes because each face of the cube cancels the opposite face, without requiring any
∂Ω
component of u to be zero.]
(d) We just “integrate by parts” twice:
ˆ
‹
ˆ
hu, ∇ × ∇ × vi =
ū · (∇ × ∇ × v) =
[ū
×
(∇×v)] · dS + (∇ × ū) · (∇ × v)
Ω
∂Ω
Ω
‹
ˆ
=
[(∇ ×
× v] · dS + (∇ × ∇ × ū) · v = h∇ × ∇ × u, vi,
ū)
∂Ω
Ω
where the surface terms cancel if either u × n|∂Ω = 0 or (∇ × u) × n|∂Ω = 0 , that is if
either u or its curl are normal to the surface. (As in the previous part, one can actually
weaken this slightly, but this is sufficient for our purposes.)
To check definiteness, carry integration by parts “halfway” through:
ˆ
hu, ∇ × ∇ × ui =
|∇ × u|2 ≥ 0,
Ω
so it is positive semidefinite. It is not positive-definite since ∇ × u = 0 for u = ∇φ 6= 0
for any non-constant scalar field φ , and we can easily choose such a φ such that ∇φ satisfies
the boundary conditions (e.g., choose φ so that it is constant in a neighborhood of ∂Ω).
∂B
∂
(e) Taking the curl of both sides of ∇×E = − ∂B
∂t , we obtain ∇×∇×E = −∇× ∂t = − ∂t ∇×B =
∂2E
− ∂t2 . That is, we have
∂2E
= ÂE
∂t2
where  = −∇ × ∇×. From the previous parts, for E ⊥ n at the surface, this  is self-adjoint
and negative semidefinite, and hence we have a hyperbolic equation.
As in class, we therefore expect orthogonal eigenfunctions √
and real λ ≤ 0, and hence oscillating “normal mode” solutions with eigenfrequencies ω = −λ.
[Technically, we also obtain λ = 0 solutions which are non-oscillatory—from above, these are
nullspace solutions E = ∇φ for some φ, which physically correspond to the time-independent
fields of fix charge distributions, where −φ is the potential and ∇ · E = ∇2 φ is the charge
density. Everything else, all of the other eigenfunctions, are oscillating solutions.]
1 By
the “hairy ball theorem” of topology, p must vanish somewhere on ∂Ω if Ω is simply connected (no holes), at
which point u must be parallel to n.
4