18.303 Midterm Solutions, Fall 2014
Problem 1:
R
R
R
Define the inner product hu, vic = Ω cūv = Ω cuv = hcu, vi where hu, vi = Ω ūv . Then hu, Âvic = hcu, (cv)00 i =
h(cu)00 , cvi = hcÂu, vi = hÂu, vic , where we have used the self-adjointness of d2 /dx2 under h·, ·i from class. Therefore,
 = Â∗ under the hu, vic inner product (which is a proper inner product for real c > 0).
Problem 2:
We need −∇2 g = δ(x), and we determine this by evaluating both sides with an arbitrary test function ψ, using the
distributional derivative (−∇2 g){ψ} = g{−∇2 ψ} as in class. In cylindrical coordinates:
Z 2π
Z ∞
1 ∂
∂ψ
1 ∂2ψ
2
dφ c ln r
r dr
g{−∇ ψ} = − lim+
r
+ 2
r ∂r
∂r
r ∂φ2
→0
0
2π
Z ∞
Z 2π
Z ∞
∂ψ
c ln r
∂
∂ψ r
− lim
dr
= −c
dφ lim
dr ln r
∂r
∂r
r
∂φ 0
→0+ →0+ 0
"
#
Z 2π
Z
∞
∞
∂ψ
∂ψ ∂(ln
r)
= −c
dφ lim+ r ln r −
dr r ∂r
∂r →0
∂r
0
Z 2π
r=∞
=c
dφ lim+ ψ|r= = −2πcψ(0).
0
→0
To get δψ = ψ(0), therefore, we need c = −1/2π .
Problem 3:
It is convenient to write D̂σ = D̂ − σI, where I is the 2 × 2 identity matrix. Then it follows from D̂∗ = R−D̂ and
(σI)∗ = σI (since σ is a real scalar and I is obviously self-adjoint) under the usual inner product hw, w0 i = Ω w∗ w0
that we have D̂σ∗ = −D̂ − σI and D̂σ + D̂σ∗ = −2σI.
(a) For a solution w of D̂σ w = ∂w/∂t , we have
∂hw, wi/∂t = h∂w/∂t, wi + hw, ∂w/∂ti
= hD̂σ w, wi + hw, D̂σ wi
= hw, D̂σ∗ wi + hw, D̂σ wi = hw, D̂σ∗ + D̂σ wi
Z
= −2hw, σwi = −2 σ(x)kw(x)k2 < 0
and hence kwk2 = hw, wi is decreasing in time.
If σ(x) ≥ σ0 > 0 for some σ0 , then we can go further and say that E(t) = kwk2 is decaying at least exponentially fast in time, since in that case dE/dt ≤ −2σ0 E and from this one can show that E(t) ≤ E(0)e−2σ0 t .
(i) Given an eigensolution D̂σ wn = λn wn , we can consider
hwn , (D̂σ + D̂σ∗ )wn i = −2hwn , σwn i
= hwn , D̂σ wn i + hD̂σ wn , wn i
= hwn , λn wn i + hλn wn , wn i
= λn + λn hwn , wn i = hwn , wn i2 Re λn .
Note that we moved D̂σ∗ to act on the left via its adjoint. It is not in general true that D̂σ∗ wn = λn wn .
Then we have:
hwn , σwn i
Re λn = −
<0
hwn , wn i
since σ > 0. Hence the eigensolutions are decaying exponentially in time (while they oscillate via the
imaginary part of λn ), from their time dependence eλn t .
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Problem 4:
We will have ∂u/∂t = ∂v/∂x − σu and ∂v/∂t = ∂u/∂x − σv, so the only new terms are the σ terms. In the discretized
un +un+1
∂u/∂t equation, the left-hand side is evaluated at point m and time n+0.5, so we have to get un+0.5
= m 2 m +O(∆t2 )
m
by averaging (similarly to how we handled the Crank-Nicolson discretization in class). Similarly for the ∂v/∂t equation.
Hence, we obtain:
n+0.5
v n+0.5 − vm−0.5
un+1
− unm
un+1 + unm
m
= m+0.5
−σ m
,
∆t
∆x
2
n+0.5
n−0.5
n−0.5
vm+0.5
v n+0.5 + vm+0.5
− vm+0.5
un
− unm
= m+1
− σ m+0.5
.
∆t
∆x
2
n+0.5
Solving for un+1
and vm+0.5
, we obtain the “leap-frog” equations:
m
un+1
m
=
n+0.5
vm+0.5
σ∆t
1+
2
=
−1 σ∆t
∆t n+0.5
n+0.5
n
1−
um +
v
− vm−0.5 ,
2
∆x m+0.5
σ∆t
1+
2
−1 σ∆t
∆t n
n−0.5
n
1−
vm+0.5 +
u
− um .
2
∆x m+1
Note that σ > 0, so we are never dividing by zero in 1 + σ∆t/2, regardless of ∆t, which is comforting.
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