18.303 Midterm Solutions, Fall 2015
November 12, 2015
Problem 1: Adjoint (33 points)
´
As in class, let’s define the inner product as hu, vi = Ω ūv/c, where the division by c(x) will cancel the
asymmetrical-looking c factor in Â. Then we have:
ˆ
ˆ
˛
hu, Âvi = −
ū∇2 v = +
∇u · ∇v −
ū∇v · n̂da
∂Ω
˛Ω
ˆΩ
c∇2 uv/c +
[v∇ū − ū∇v] · n̂da
= −
∂Ω
Ω
˛
·
= hÂu, vi +
[v∇ū
−
ū∇v]
n̂da
∂Ω
where in the last line we have used the boundary conditions to cancel the integrand:
[v∇ū − ū∇v] · n̂ = α [vū − ūv] = 0.
Problem 2: Finite differences (34 points)
Since we have a boundary condition u0 (L) = αu(L) that requires us to know the slope at x = L, the most
convenient thing is to put x = L halfway between two grid points so that we can use a center-difference
approximation. In particular, we set up the grid so that um ≈ u(m∆x) as usual, but (M + 0.5)∆x = L .
Then we have the discretized boundary conditions:
u0 = 0
uM +1 − uM
uM + uM +1
=α
,
∆x
2
in which the left-hand side is the 2nd-order accurate center-difference approximation for u0 (L) and the righthand side is the 2nd-order accurate approximation for αu(L). (The 2nd-order accuracy of both of these was
shown in class.) We have the M unknowns u1 , . . . , uM , and we can solve the second equation above for
uM +1 =
1 + α∆x/2
uM = βuM ,
1 − α∆x/2
defining the number β.
If we use the usual center-difference approximation for the second derivative, we have
−u00m =
−um−1 + 2um − um+1
,
∆x2
which for the specific cases of m = 1 and m = M (the first and last matrix rows) gives (from the boundary
conditions):
2um − um+1
−u001 =
,
∆x2
−um−1 + (2 − β)um
−u00M =
.
∆x2
1
Hence, we have the matris

−1
2
..
.
2
 −1
1 

A=

∆x2 


−1
..
.
−1
..
.
2
−1
−1
2−β



.


Since this matrix is obviously real-symmetric, we have self-adjointness under the usual inner product hu, vi =
u∗ v.
Problem 3: Green (33 points)
We want to solve the nonlinear equation
Âu + αu2 = f
for some small |α(x)| 1. We begin by writing write u = u0 + u1 + u2 + · · · as a power series in α, where
un denotes all the terms proportional to αn . If we plug this in to our equation, we get Âu0 + Âu1 + Âu2 +
· · · + αu20 + 2αu0 u1 + αu21 + · · · = f . If we collect terms by power of α, we get the equations:
Âu0 = f,
Âu1 + αu20 = 0,
Âu2 + 2αu0 u1 = 0,
and so on. From the first equation, we get u0 = Â−1 f , i.e.
ˆ
u0 (x) =
G0 (x, x0 )f (x0 ) d3 x0 .
Ω
1. We want u ≈ u0 + u1 . From the second equation above, we get u1 = −Â−1 αu20 , i.e.
ˆ
u1 (x) = −
G0 (x, x0 )α(x0 )u0 (x0 )2 d3 x0 .
Ω
Note that if we substitute the u0 integral, we would have two nested integrals.
2. We want u ≈ u0 + u1 + u2 . From the third equation above, we get u2 = −2Â−1 αu0 u1 , i.e.
ˆ
u2 (x) = −2
G0 (x, x0 )α(x0 )u0 (x0 )u1 (x0 ) d3 x0 .
Ω
Note that if we substituted the u0 and u1 integrals, we would have a nested integral for u0 and another
two nested integrals for u1 .
It would be an amusing exercise to devise a Feynman-diagram notation for these integrals. Graphically, the
nonlinear term αu2 appears as a source term that arises when two Green’s functions “collide”.
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