18.303 Midterm Solutions, Fall 2012
Problem 1: Adjoints and operators (20 points)
(a) Integrating by parts, and using the fact that u0 (L) = αu(L), have
Z
Z L
Z L
L
00
0 L
0
0
0
0
ūv = ūv |0 −
u v = ūv − u v 0 +
hu, Âvi =
u00 v
0
0
0
L
0 (0) + u0 (0)v(0)
u(0)v
= u(L)v 0 (L) − u0 (L)v(L) − + hÂu, vi
i
h
(
(
(((
(u(L)v(L)
(−
= α u(L)v(L)
+ hÂu, vi
((((
= hÂu, vi,
hence  = Â∗ .
(b) For it to have decaying solutions, as in class we want  to be negative definite (or negative semidefinite, if we
allow the solutions to decay to a nonzero asymptote). Hence, we need hu, Âui < 0 (or ≤ 0) for u 6= 0. As in
class, we look at the middle step above where we have integrated by parts once:
Z L
0 L
u0 u0
hu, Âui = ūu |0 −
0
2
Z
= α |u(L)| −
L
2
|u0 (x)| dx
0
This is ≤ 0 if α ≤ 0. It would = 0 if u0 = 0 and αu(L) = 0. However, u0 = 0 implies u(x) = constant, but since
u(0) = 0 the only constant it can be is 0. Hence, hu, Âui < 0 for u 6= 0 for any α ≤ 0 .
Problem 2: Finite differences (20 points)
Let m = M + 1 correspond to the x = L boundary, in which case we set uM +1 = 0 similar to class. The discrete form
of the left boundary condition will be something like
u1 − u0
=⇒ u0 = u1 − ∆x.
∆x
There is some subtlety about where to place the left boundary. If we want this boundary condition to be imposed with
−u0
second-order accuracy, we need u1∆x
to be a center-difference approximation, in which case we should make x = 0
L
correspond to m = 0.5. This means that (M + 1 − 0.5)∆x = L, or ∆x =
. [However, I didn’t specify that
M + 0.5
second-order accuracy was required, so it is acceptable if you make, e.g., m = 0 correspond to x = 0 similar to class,
in which case we have a first-order forward difference approximation for u0 (0) and you will get ∆x = L/(M + 1). I do
expect you to specify ∆x.]
m +um−1
Then, applying u00m ≈ um+1 −2u
for m = 1, 2, . . . , M will give us the same matrix as usual except that we
∆x2
need to be careful with the first and last rows, where:
u0 (0) = 1 ≈
u2 − 2u1 + u0
u2 − u1
1
,
=
−
∆x2
∆x2
∆x
− 2uM + uM −1
uM+1
u00M = .
∆x2
In particular, the 1/∆x term in the u001 equation needs to be moved to the right-hand side of Âu − cu =
term just gives us a diagonal term −cm um . Writing this all out in matrix form, we obtain:
u001 =
Au = b
where

−1
 1
1 

A=

∆x2 

1
−2
..
.

1
..
.
1
..
.
−2
1

 
 
 
−
 
1  
−2
1

c1
c2
..






.
cM −1
cM
∂u
∂t .
The −cu
and

1
0
..
.

1 

b=

∆x 
 0
0




.


Notice that, much like in class, the nonzero boundary condition adds a source term to the right-hand-side, and the
source term looks much like a (discretized) delta function since it has height 1/∆x and “width” ∆x (one “pixel”)
Problem 3: Green-ish functions (20 points)
(a) As in class, for x 6= y we have u00 = 0, or u is a straight line. Since it has to go through zero at the endpoints,
we have
(
αx
x<y
u(x) =
β(x − L) x > y
for some constants α and β to be determined.
The first derivative, in the distribution sense, gives
(
α
u (x) = [β(y − L) − αy] δ(x − y) +
β
0
x<y
,
x>y
where we have picked up a δ function from the discontinuity at y. The second derivative, in the distribution
sense, then gives
u00 (x) = [β(y − L) − αy] δ 0 (x − y) + (β − α)δ(x − y),
where the derivative of δ gave us δ 0 and the discontinuity β − α gave us another δ. Comparing to −u00 (x) =
δ 0 (x − y), we immediately obtain the equations
β = α,
1
.
L
Note that since y ∈ [0, L], the denominator in α is never zero and so α is always finite. Hence, we obtain the
regular distribution
(
x<y
1 x
u(x) =
,
L x−L x>y
α(y − L) − αy = −1 =⇒ α =
which is discontinuous but has continuous slope.
(b) Similar to the Green’s functions in class, we can do this by superposition:
Z
u(x) = −
L
D(x, y)f (y)dy ,
0
since
Z
Âu = −
L
Z
[AD(x, y)] f (y)dy =
0
L
[−δ 0 (x − y)]f (y)dy = −δ 0 (x − y){f (y)} = f 0 (y).
0
2