18.336 Mid-term Solutions
1
0.9
Problem 1 (30 points): Velocity
ei(θm−φn) ,
(a) Plugging in
0.8
0.7
and employing the
0.6
vg / a
usual trig identites, we nd:
sin(φ/2) = ±aλ sin(θ/2)
0.5
0.4
0.3
This gives a group velocity
of:
vg = dφ/dθ/λ
0.2
0.1
cos(θ/2)
vg
=q
.
a
1 − (aλ)2 sin2 (θ/2)
0
0
0.1
0.2
0.3
0.4
0.5
0.6
β∆x/π = θ/π
0.7
0.8
0.9
1
This is actually exactly the same as the
pset 3.
It clearly is
1
for
|aλ| < 1, decreases to
θ = π . If we've forgotten
for
frog wave equation, for
θ = 0 and,
0 for β∆x =
what the plot
pulse in the signal-processing community).
show that it decreases monotonically. If we
Since the gure showed a pulse with slower
dvg /dθ
) = (2u0 v −
take its derivative, the numerator of
0
uv )/2v
3/2
(u/v
oscillations at the left, and there are no
1/2 0
boundaries to change the pulse direction,
by the product rule):
we conclude that we are looking at the leftgoing pulse.
− sin(θ/2) · [1 − (aλ)2 sin2 (θ/2)]
+ cos(θ/2) · (aλ)2 sin(θ/2) cos(θ/2)
Problem 2 (30 points): Stability
= − sin(θ/2) · [1 − (aλ)2 ] ≤ 0
(a) The simplest thing is to consider a at solution (ux
ut = σu and thus
eσt . More geni(βx−ωt)
erally, we can look at a wave e
, and
plugging this in we see that −iω = −iaβ+σ ,
or β = ω/a − iσ/a, which leads to solutions
σx/a
that grow as e
. This is exponentially
growing towards the right if a is positive
and to the left if a is negative, and is there-
so the curve must be monotonically decreasing, with zero slope at
θ =0
that it must be concave downward, by tak-
d2 vg /dθ2 ,
obtaining:
− cos(θ/2) · [1 − (aλ)2 sin2 (θ/2)]
− 3 sin(θ/2) · (aλ)2 sin(θ/2) cos(θ/2)
of propagation
θ = π.
a
σ(x)
not constant via the coordinate-stretching
Rx
exp[ σ(x0 )dx0 /a]
constant σ , ω = aβ + iσ so
approach where we get
plot must look something like Fig. 1 (which
growth. Or, for
aλ = 0.9).
every Fourier component grows in time as
(b) The pulse must be travelling to the
eσt . Another way to show that all constantσ solutions are exponentially growing is to
write u(x, t) = f (x − at, t), which yields the
equation ft − afx = −afx + σf , and we see
σt
that ft = σf so that f (x, t) = f (x, 0)e .
left.
The reason is that we saw from part (a) that
faster spatial frequencies (larger
more slowly than smaller
growing in the direction
regardless of the sign of
this argument can be generalized to
Putting this together, we conclude that the
is
which gives
fore exponentially
= − cos(θ/2) · [1 + 2(aλ)2 sin2 (θ/2)] ≤ 0,
with equality (zero curvature) only at
= 0),
has solutions that grow as
and negative
slope elsewhere. Furthermore, we can show
ing the numerator of
for leap-
slower oscillations (this is called a chirped
looks like, with a little more work we can
looks like (recall that
vg /a vs. θ/π
|aλ| = 0.9.
Figure 1: Group velocity
group velocity for the leap-frog method from
θ.
θ)
travel
At the start,
the pulse has the same oscillation frequency
everywhere in its envelope, but eventually
(b) Plugging in the usual
the trailing edge of the pulse will have faster
unm = g n eiθm , we nd:
(g − 1) = −iaλg sin θ + σ∆t,
oscillations and the leading edge will have
1
or
just set the value of
quire
1 + σ∆t
|g| = q
≤ 1 + σ∆t,
that
1 + (aλ)2 sin2 θ
φ(0) = 0.
φ(2π) = 0
φ
at a point, e.g.
re-
By periodicity, this means
too, and thus we have just
changed periodic boundary conditions into
Dirichlet boundary conditions (at least in
which is a sucient condition for stability as
1d)!
we showed in class. So, it is unconditionally
stable.
Note that the problem does
to do with whether
(c) Cal is over-simplifying the denitions. Saying
it
is
stable
doesn't
mean
that
in a nite time
as
it
ing
= d˜k /k 4 , com4
pared to the exact solution ck = dk /k . As we
showed in class, the L2 error in u is exactly equal
to the L2 error in the Fourier coecients, which
is the sum of two terms: a discretization error
and a truncation error for
|c̃k − ck |2 +
|k|≤M
X
|k| > M .
|ck |2 .
|k|>M
|c̃k − ck |2 = |d˜k −
1
1
dk | /k = O( M 2` ) k8 . A sum of 1/k 8 is a convergent series and is just sum number for large M ,
1
so we get O( 2` ). The second sum is of terms
M
1
|ck |2 = |dk |2 /k 8 = O( |k|2`+8
); when summed,
1
this gives O( 2`+7 ) by the usual series bound.
M
Obviously, the rst term dominates, and so the
1
overall error is O( ` ).
M
The rst sum is of terms
2
8
Extra credit (5 points): Poisson
(a) A singular matrix means that the solution
is not unique, and indeed that is the case
here: we can add any constant to
φ
and it
still solves the same equation.
(b) To x this, we just need to specify a specic solution
φ
R
dierential equation).
The collocation solution will be c̃k
s X
ρ that does not
have
φ has a discontinuous slope
ρdx = φ0 (2π) − φ0 (0) by integrating the
will be one for which
(since
|k| ≤ M ,
If we take our formulation
zero integral, we will still get a solution, but it
t → ∞.
for
φ to zero, which is equivalent to requirφdx = 0, another possible choice (albeit
from (b) above and plug in a
Problem 3 (30 points): Accuracy
c̃k
R
dierence scheme).
converges to the exact solution, which blows
in the
φ
a slightly harder choice to enforce in a nite-
well-posed,
So, there is no contradiction: the solution
the missing Fourier coecients with
solution
For our spectral
cient of
∆t → 0.
which is a condition for Lax's theorem too).
up as
have anything
solution in class, we set the zeroth Fourier coef-
And indeed, the exact solution doesn't blow
up in a nite time either (it is
not
ρdx = 0...the
is still not unique regardless.
doesn't blow up. Rather, it means that it
doesn't blow up
R
that we want out of all the
innite possibilities. The simplest way is to
2