18.303 Midterm Solutions, Fall 2013
Problem 1: (7+8+(5+10) = 30 points)
(a) Consider an eigensolution Âu = λu. Then <hu, Âui = <hu, λui = < (λhu, ui) = hu, ui<λ < 0, hence <λ < 0
(since hu, ui is real and positive).
hu,(Â+Â∗ )ui
Âui
Âu,ui
Â∗ ui
(b) From the properties of inner products, <hu, Âui = hu,Âui+hu,
= hu,Âui+h
= hu,Âui+hu,
=
<
2
2
2
2
0 for all u 6= 0, and hence  + Â∗ is negative definite according to the definition from class, and conversely if
 + Â∗ is negative definite then <hu, Âui < 0.
∂v
∂v
(c) Consider the system of equations ∂u
∂t = ∂x − αu and ∂t =
Dirichlet boundary conditions u(0) = u(L) = 0.
u
(i) Similar to class, in terms of w =
, we have
v
∂w
=
∂t
so
 =
where D̂ =
R
ūu0 + v̄v 0 .
∂/∂x
∂/∂x
u̇
v̇
−α
=
−α
∂
∂x
∂
∂x
∂u
∂x
∂
∂x
− βv for some α(x) and β(x), for Ω = [0, L] with
∂
∂x
w = Âw,
−β
−β
= D̂ −
α
β
0 u
u
= −D̂ is the scalar-wave operator from class, using h
,
i =
v
v0
∗
(ii) There are a couple of ways to approach this.
We can look at eigensolutions. For an eigensolution Âw = λw, this has the solution eλt w(t = 0), which is
decaying if <λ < 0. If we assume (as usual) that we have a basis of eigenfunctions, then (from above) it is
sufficient for  + Â∗ to be negative-definite in order for the solutions to be exponentially decaying. Then
α
ᾱ
α
 + Â∗ = D̂ +
D̂∗ −
−
= −<
,
β
β̄
β
(where by inspection
we have seen the adjoint of α and β is just ᾱ and β̄). For this to be negative definite,
R
we need −< α|u|2 + β|v|2 < 0 for all u, v 6= 0, which implies
<α > 0,
<β > 0
almost everywhere.
∂
hw, wi for any solution w(x, t) of
Another way to derive the same sufficient condition is to consider ∂t
∂w
∂
∂
our PDE ∂t = Âw. If ∂t hw, wi < 0 for all w 6= 0, then the solution is decaying, and ∂t
hw, wi = · · · =
∗
hw, (Â + Â )wi, similar to the derivation in class of conservation of energy for wave equations, hence we
arrive again at  + Â∗ < 0.
Problem 2: (10+10+10 = 30 points)
(a) Let um denote u(m∆x), as usual. We want to be able to write u0 (0) = u00 as a center difference with spacing ∆x,
−0.5
i.e. u00 = u0.5 −u
. Hence, we need to store our unknowns at grid points um+0.5 for m = 0, 1, 2, . . . , N − 1
∆x
(for N unknowns). Similarly, we should put the other boundary at N so that we can write u0N =
Hence N ∆x = L or ∆x = L/N (which is different from the
L
N +1
uN +0.5 −uN −0.5
.
∆x
we used with Dirichlet boundary conditions!).
(b) First, we apply the boundary conditions u00 = 0 and u0N = 0 from above to obtain u−0.5 = u+0.5 and uN +0.5 =
uN −0.5 . Then we can write the second derivative, similar to class, by
u00m+0.5 =
um+1.5 − 2um+0.5 + um−0.5
,
∆x2
1
where for the first row (m = 0) and the last row (m = N − 1) of the matrix we will have
u000.5 =
u1.5 − u0.5
u1.5 − 2u0.5 + u−0.5
=
,
∆x2
∆x2
uN +0.5 − 2uN −0.5 + uN −1.5
−uN −0.5 + uN −1.5
=
2
∆x
∆x2
via the boundary conditions. Hence, A looks like
u00N −0.5 =
−1
1
1
A=
∆x2
1
−2
..
.
1
..
.
1
..
.
−2
1
.
1
−1
(c) This A is obviously real-symmetric. To show that it is definite, the easiest thing to do, similar to class, is to
factorize A as the product of two first-derivative operations. Let us construct the matrix D that computes
u01 , u02 , . . . , u0N −1 from u0.5 , u1.5 , . . . , uN −0.5 . (Since u00 = 0 and u0N = 0, we need not compute them.)
u01
u02
..
.
u0N −1
−1
1
= Du =
∆x
1
−1
1
..
.
..
.
−1
u0.5
u1.5
..
.
uN −1.5
1
uN −0.5
.
Notice that D is (N − 1) × N . Similarly, to get u000.5 , u001.5 , . . . , u00N −0.5 from u01 , u02 , . . . , u0N −1 , we do:
u000.5
u001.5
..
.
u00
N −1.5
u00N −0.5
1
−1
1
=
∆x
1
..
.
..
.
−1
u01
u02
..
.
1
u0N −1
−1
= −DT
u01
u02
..
.
u0N −1
,
where we have used the Neumann boundary conditions for the first and last rows. Hence A = −DT D, which is
at least negative semidefinite. It is not negative definite, because it is easy to see that N (A) = N (D) contains
the constant vector (1, 1, . . . , 1, 1)T .
2
0
