# 18.303 Problem Set 6 Solutions Problem 1: (5+5+10)+(5+5+10) points

```18.303 Problem Set 6 Solutions
Problem 1: (5+5+10)+(5+5+10) points
(a) Solution:
Rb
(i) As pointed out in the notes, for any f (x) to define a regular distribution, it is sufficient for a |f (x)|dx
to be finite for any [a, b]. Here, even though f (x) diverges as x → 0+ , the integral of f (x) does not:
R b dx
√ b
√ = 2 x| , which is finite for any a, b ≥ 0.
a
a
x
Rb
√ b
(ii) In this case, the integral does diverge: if a, b ≥ 0, then a |g(x)|dx = −1/ x|a , which blows up as a → 0+ .
R∞
√ dx, where the limit exists since, for small x, the integrand is
(iii) As a distribution, f {φ} = lim→0 φ(x)
x
√
≈ φ(0)/ x, which as a finite integral from above. The definition of the distributional derivative is
∞
−φ0 (x)
√ dx
f {φ} = f {−φ } = lim
→0 x
Z ∞
−1 d
√
[φ(x) − φ(0)] dx
= lim
→0 x dx
∞
Z ∞
−1
−1 d
√
[φ(x) − φ(0)] + lim
= lim
[φ(x) − φ(0)] dx
3/2
→0 →0
x dx
2x
Z ∞
= lim
g(x) [φ(x) − φ(0)] dx,
0
Z
0
→0
where in the second line we have used the hint, in the third line we have integrated by parts, and in the
fourth line we have used the fact that φ(x) vanishes at ∞ (by definition of the space of test functions) and
0
√
φ()−φ(0)
√
→ φ√()
= φ0 () → 0 as → 0 (by virtue of φ being differentiable). By the same token, the
√
integrand g(x)[φ(x) − φ(0)] is asymptotically proportional to φ0 (0)/2 x in the limit of small x, and hence
has a finite integral. This is not a regular distribution—it is not an integral of φ(x) multiplied by any
function—it is a singular distribution. It is, however, the same as the naive integration against g(x) for any
φ(x) with φ(0) = 0.
(b) Solution:
∂
∂
(i) For r &gt; 0, −∇2 g(r) = − 1r ∂r
(rg 0 ) = 0, hence ∂r
(rg 0 ) = 0, hence rg 0 (r) = c for some constant c, hence
c
0
g (r) = r , hence g(r) = c ln r + C. The +C constant is just a function in the nullspace of ∇2 independent
of the right-hand side, and we will set this to zero.
R 2π R ∞
(ii) Writing the integral as g{φ} = lim→0 0 dθ c ln rφ(r, θ)r dr, it is clear that the only possible problem
could come from the r integral. By definition of the space of test functions, φ(r, θ) must be zero for
r &gt; R for some R, and it is bounded for r &lt; R (being continuous). The c ln rφ(r, θ)r integrand may look
worrying at first glance because ln r → −∞ as r → 0+ . However, application of L’Hopital’s rule shows that
1/r
ln r
= limr→0 −1/r
limr→0 r ln r = limr→0 1/r
2 = 0, so the integrand is bounded (and continuous) for r &lt; R,
and hence the integral exists. (Note that we don’t have to worry about the r → ∞ limit, since φ = 0 for
all sufficiently large r: no limit is required. This wouldn’t be the case if we had defined the test functions
differently, requiring them to → 0 at ∞ rather than having compact support—in that case, we would have
to worry about how fast they → 0.)
(iii) Taking the distributional derivative, we find:
2
(−∇ g){φ} =
Z
Z
2
g(r)[−∇ φ] = − lim
→0
2π
Z
dθ
0
∞
1 ∂
cr ln r
r ∂r
∂φ
1 ∂2φ
r
+ 2 2 dr.
∂r
r ∂θ
Now, we will perform this integral in two pieces, as in class. For the ∂ 2 φ/∂θ2 term, we will do the dθ
integral first:
2π
Z 2π
∂2φ
∂φ dθ 2 =
=0
∂θ
∂θ 0
0
1
−0.999
−0.9995
−1
σn
−1.0005
−1.001
−1.0015
−1.002
0
50
100
150
200
θ (degrees)
250
300
350
Figure 1: Point-source values σn versus angle θ for a “point charge” at (2, 0) near a unit circle with Dirichlet (0)
boundary conditions.
where we have used the fact that any function in polar coordinates must be periodic in θ. Fort the ∂φ/∂r
term, we will integrate by parts:
∞
Z ∞
Z ∞
1 ∂φ
1 ∂
∂φ
∂φ ∞
c r dr = − cφ(r, θ)|0 = cφ(0),
lim
cr ln r r
dr = lim c (ln r) r − lim
→0 →0 →0
∂r
∂r r ∂r
r ∂r
where in the second
R 2π step we have used the fact that limr→0 r ln r = 0. Combined with the other integrals
above, we get − 0 cφ(0)dθ = −2πcφ(0). Since we want this to be δ{φ} = φ(0), we immediately obtain
c = −1/2π .
Problem 2: 5+5+5+10+10 points
(a) If we simply used yn = xn , then the diagonal entries of A would blow up (since they would have ln 0). In
general, these kinds of singularities in the Green’s functions for coincident points require lots of careful handling
in integral-equation methods. (Although using yn 6= xn as we did here is a simple fix, it has the unfortunate sideeffect of making A non-symmetric. There are cleverer ways around this problem in serious Nystr&ouml;m methods.)
(b) Since we want
As =
N
X
σn ln |ym − xn | = − ln |ym − x0 | = −b
n=1
in order to satisfy the Dirichlet boundary conditions, we should solve As = −b.
(c) (There is a typo in the problem set: s should be plotted versus x_theta, not y_theta, not that it matters much;
the two differ by only about 0.2◦ .) The command plot(x_theta * 180/pi, A \ -b) yields the solution plot
shown in figure 1.
(d) We just need to compute
ln |x − x0 | +
N
X
σn ln |x − xn |
n=1
for x = (1, 1). In Matlab, this corresponds to the command log(sqrt((1-2)^2+(1-0)^2)) + sum((A \ -b)
.* log(sqrt((1-x_x).^2+(1-x_y).^2))), which gives −345.3383.
P
(e) Given s, we can easily change the total charge to zero just by subtracting the mean σ (=
σn /N ) from each
σn . (A little more cleverness is to realize that the mean σ is exactly −1, but it’s okay if you don’t notice this.)
However, we also require that the potential G be a constant on the cylinder, and for an arbitrary shape subtracting constant from all the σn0 s wouldn’t change G by a constant. Here, we are saved by symmetry: since, in
2
the case of a circle, every point ym is equivalent under rotation to every other ym , so adding a constant to all
0
the σn0 s changes G at every ym by the same amount, so the result is a constant G at the ym
s (which are the
only points we enforce the boundary conditions at in this approximation). This is easily verified by multipling A
by o (the vector of all ones) in Matlab by the command A * o’, which results in a vector of all the same values.
In fact, since the old G at the surface is zero and the mean σ was −1, we are just adding +1 to all the σn0 s and
get a new potential on the surface of Ao, which is 0.6931.
(For a more general shape, matters are not so easy because we cannot simply add a constant to all the charges
to get a constant G on the surface. In that case, we would have to set all the potentials at ym to equal the
potential at y1 , remove the equation for y1 , and add a new equation that the sum of the σn ’s is zero, to obtain
a new system of equations to solve.)
3
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