MATH 333: Partial Differential Equations
Problem Set 6, Final version
Due Date: Fri., Oct. 22, 2010
Read Sections 4.1–4.3 from the Olver text. Some of the material in 4.1 should be quite
familiar. For the rest you need not read to get every detail, but just a general feel for what
is being done (particularly the application of separation of variables to non-heat problems).
There is accompanying material to be found in Lessons 7, 20, and 31–33 from Farlow’s
book.
?18 (a) Suppose v is an element of a Hilbert space H, and {φ1 , . . . , φN } are mutually
orthogonal nonzero elements in H. Let
D
E
v, φn
cn =
,
n = 1, 2, . . . , N.
kφn k2
Show that
2
N
N
X
X
2
v −
|cn |2 kφn k2 .
cn φn = kvk −
n=1
n=1
(1)
(b) Now suppose S = {φ1 , φ2 , . . .} is a collection (finite or infinite) of mutually
orthogonal nonzero elements in H. Use the previous result to explain Bessel’s
Inequality:
D
E2
X v, φn kvk2 ≥
,
kφn k2
n
where there is a term in the right-hand sum for each element in S.
(c) Show that if S (see part (b)) is a complete orthogonal basis of H, Parseval’s
Equality holds:
D
E2
X v, φn 2
.
(2)
kvk =
2
kφ
k
n
n
[Note: In fact, the following tautology may be proved: S is a complete orthogonal basis of H if and only if (2) holds for every v ∈ H.]
?19 Consider the series solution
u(t, x) =
∞
X
n=1
cn e
−n2 π2 t/`2
nπx
sin
,
`
with
cn =
f, sin(nπ · /`)
k sin(nπ · /`)k22
,
(the subscript 2 in k · k2 denotes size is being measured by the L2 -norm, or 2-norm for
short) of the heat problem with Dirichlet BCs
(
BCs: u(t, 0) = 0 = u(t, `),
ut = uxx , 0 < x < `, t > 0, subject to
IC:
u(0, x) = f (x).
In this problem we investigate the error incurred by truncating the series solution.
MATH 333
Problem Set 6
2
(a) Suppose N is a fixed positive integer and we define
TN (t, x) :=
N
X
cn e−n π t/` sin
2 2
2
n=1
nπx
,
`
and a remainder
RN (t, x) := u(t, x) − TN (t, x) =
∞
X
cn e−n π t/` sin
2 2
n=N+1
2
nπx
,
`
so that u(t, x) = TN (t, x) + RN (t, x) is the sum of the series truncated at the Nth
term and this remainder. At any fixed t > 0, the square of the L2 -norm (i.e., the
2-norm) of RN is defined to be
∞
2
X
nπ· 2 2
2
cn e−n π t/` sin
kRN (t, ·)k22 = ` n=N+1
2
*X
!+
∞
∞
X
nπ·
kπ·
−n2 π2 t/`2
−k2 π2 t/`2
=
cn e
,
ck e
sin
sin
`
`
n=N+1
k=N+1
Employ the orthogonality of the functions {sin(nπ · /`)}∞
in L2 (0, `) to simplify
n=1
this expression for kRN (t, ·)k22 , and then use Bessel’s inequality to conclude that,
for t ≥ t0 > 0,
2 2
2
kRN (t, ·)k2 ≤ e−(N+1) π t0 /` k f k2 .
(b) Now take ` = π and
(
f (x) =
0, 0 < x < π/2,
1, π/2 < x < π.
If we want the 2-norm of the remainder kRN (t, ·)k2 < 0.01 for all t ≥ 0.1, use your
estimate from part (b) to determine a sufficiently large N. That is, determine
N so that the truncated series TN (t, x) approximates u(t, x) this accurately (in
2-norm) whenever t ≥ 0.1. For this N, produce plots on [0, π] for f (x), TN (0, x),
TN (0.1, x) and TN (0.2, x).
?20 (a) Consider the BVP
L[v] = 0,
0 < x < `,
subject to
v0 (0) = 0 = v0 (`),
where L is the differential operator L = d2 /dx2 operating on appropriate functions v(x) of one variable. Find all solutions to this problem.
(b) Now consider the inhomogeneous problem
(L[u])(x) = f (x),
0 < x < `,
subject to
u0 (0) = 0 = u0 (`),
where L is a self-adjoint linear operator. Show that solvability of this inhomogeneous problem is sufficient to conclude
f, v = 0
whenever
L[v] = 0.
(3)
PS6—Final version
MATH 333
Problem Set 6
3
Said another way, the Equation (3) is a necessary compatability condition for the
solvability of the inhomogeneous problem.
(c) Determine what the compatability condition (3) says about f when L = d2 /dx2 .
?21 In class, I set out to demonstrate that the Sturm-Liouville operator
(
d α1 u(a) + α2 u0 (a) = 0,
0
(K[u]) (x) := −
p(x)u (x) +q(x)u(x), a < x < b, subj. to BCs
β1 u(b) + β2 u0 (b) = 0,
dx
with |α1 | + |α2 | > 0 and |β1 | + |β2 | > 0, is self-adjoint. Here, in blue text, is how I did it
in my notes:
Assuming everything is real-valued here, then for sufficiently smooth functions u, v
satisfying the boundary conditions we have
b
!
Z b
d
du
v(x)
p(x)
dx +
q(x)u(x)v(x) dx
hK[u], vi = −
dx
dx
a
a
!
Z b
b Z b
dv
du
0
p(x)
dx +
= −p(x)u (x)v(x) +
q(x)u(x)v(x) dx
a
dx
a
a dx
Z
(after an integration by parts on first integral from previous line)
!
#
"
h
ib h
ib Z b
dv
d
p(x)
+ q(x)v(x) u(x) dx
= − p(x)u0 (x)v(x) + p(x)u(x)v0 (x) +
−
a
a
dx
dx
a
=
h
(after a second integration by parts, again on first integral)
ib h
ib
− p(x)u0 (x)v(x) + p(x)u(x)v0 (x) + hu, K[v]i .
a
a
For K to be self-adjoint, we need these boundary terms to vanish. In the case where
each α j , β j is nonzero (j = 1, 2), we have
h
ib
ib h
− p(x)u0 (x)v(x) + p(x)u(x)v0 (x)
a
a
= p(a)u (a)v(a) − p(b)u (b)v(b) + p(b)u(b)v0 (b) − p(a)u(a)v0 (a)
0
=
0
p(a)
p(b)
[α2 u0 (a) α1 v(a) − α1 u(a) α2 v0 (a)] +
[β1 u(b) β2 v0 (b) − β2 u0 (b) β1 v(b)]
α1 α2
β1 β2
= ...
= 0,
(details)
giving that K is self-adjoint.
Your task, for this exercise, is to fill in the missing ”details”; that is, show
p(a)
p(b)
[α2 u0 (a) α1 v(a) − α1 u(a) α2 v0 (a)] +
[β1 u(b) β2 v0 (b) − β2 u0 (b) β1 v(b)] = 0.
α1 α2
β1 β2
PS6—Final version
MATH 333
Problem Set 6
4
4.1.2 Consider the IBVP
ut = uxx ,
and
0 < x < 10,
t > 0,









IC: u(0, x) = f (x) := 







subject to BCs: u(t, 0) = 0 = u(t, 10),
x − 1,
11 − 5x,
5x − 19,
5 − x,
0, otherwise.
1 ≤ x ≤ 2,
2 ≤ x ≤ 3,
3 ≤ x ≤ 4,
4 ≤ x ≤ 5,
Discuss what happens to the solution as t increases. You do not need to write
down an explicit formula for u(t, x) (in particular, you do not need to give the paper
calculations of the Fourier coefficients), but for full credit you must explain (sketches
can help) at least 3–4 interesting things that happen to the solution as time progresses.
The Octave file fourierCoeffs.m may well be of use to you here, as well as some
modified version of the file fss approx.m. An Octave version of the function f
which carries out pointwise evaluation on vectors is
function y = f(x)
y = zeros(size(x));
y = (x >= 1 & x <= 2)
y += (x > 2 & x <= 3)
y += (x > 3 & x <= 4)
y += (x > 4 & x <= 5)
end
.*
.*
.*
.*
(x - 1);
(11 - 5*x);
(5*x - 19);
(5 - x);
?22 Rectangular coordinates (x, y) and polar coordinates
(r, θ) have the following relationships:
p
x = r cos θ
r = x2 + y2
y = r sin θ
θ = arctan (y/x)
r
y
θ
x
(a) The chain rule says ∂u/∂x = (∂u/∂r)(∂r/∂x) + (∂u/∂θ)(∂θ/∂x). Use the chain
rule along with the relationships above to conclude that
∂
∂
sin θ ∂
= cos θ
−
.
r ∂θ
∂x
∂r
(b) Working as in part (a), it is possible to conclude
∂
∂
cos θ ∂
= sin θ
+
.
r ∂θ
∂y
∂r
Use this and the result from (a) to write the two-dimensional Laplacian operator
∆ = ∂2 /∂x2 +∂2 /∂y2 in polar coordinates. (Note: The polar expressions for ∂2 /∂x2
and ∂2 /∂y2 are a good deal uglier than that for ∆; there are a lot of cancellations
when the two are combined.)
PS6—Final version
MATH 333
Problem Set 6
5
{?23} {Optional exercise, but suggested for those interested in Schrödinger’s equation.}
Hermite functions. Consider the differential equation
−y00 + x2 y = Ey,
x ∈ R,
E = constant,
and the functions
2
Hn (x) = (−1)n ex
dn −x2
e ,
dxn
n = 0, 1, 2, . . . .
(a) Find Hn (x), for n = 0, 1, . . . , 4. Why is Hn (x) always a polynomial (known as an
Hermite polynomial)?
(b) Verify that vn (x) = Hn (x)e−x /2 is a solution of the differential equation when
E = 2n + 1. Hint: First, note that
2
dn+1 −x2
dn
2
(e
)
=
(−2xe−x )
n
n+1
dx
dx
!
dn−1
dn −x2
n d
2
(−2x) n−1 (e−x )
= (−2x) n (e ) +
dx
1 dx
dx
)
(
dn−k
dk
+ terms of form k (−2x) n−k , where k ≥ 2
dx
dx
n
n−1
d
d
2
2
= −2x n (e−x ) − 2n n−1 (e−x ),
dx
dx
and use this to show Hn0 = 2nHn−1 .
R∞
(c) Show that −∞ vn vm dx = 0, m , n, and thus the vn are orthogonal on the interval
(−∞, ∞). Hint: Use the fact that both vn and vm satisfy the differential equation
for appropriate E to show that v00m vn − v00n vm = 2(n − m)vm vn .
P
(d) If a function f (x) can be represented by f (x) = ∞
would you
n=0 cn vn (x), how √
expect to find the cn ? Assume uniform convergence. Take f (x) = 1/ 1 + x4 and
use software to find cn , n = 0, . . . 4.
PS6—Final version