1. No category

AMath 569, Spring 2013
Sample Solutions for Assignment 5.
Reading: Chapter 4 in the Notes or Secs. 4.1-4.2 in the text.
1. Exercise 1 on p. 77 in the Notes.
utt = uxx + uyy ,
0 < x < a, 0 < y < b,
u(x, y, 0) = φ(x, y), ut (x, y, 0) = ϕ(x, y),
0 ≤ x ≤ a, 0 ≤ y ≤ b,
u(0, y, t) = u(a, y, t) = 0,
0 ≤ y ≤ b,
u(x, 0, t) = u(x, b, t) = 0,
0 ≤ x ≤ a.
Assume that u(x, y, t) = X(x)Y (y)T (t). Substituting into the differential
equation, we find
XY T 00 = X 00 Y T + XY 00 T, or
X 00 Y 00
T 00
=
+
= K,
T
X
Y
where K is some constant independent of x, y, and t. The equation for T is
then T 00 = KT , while that for X and Y is
Y 00
X 00
=−
+ K = L,
X
Y
where L is another constant independent of x, y, and t since one side of the
equation depends only on x and the other side depends only on y. Thus
X 00 = LX and Y 00 = (K − L)Y . The first boundary condition is u(0, y, t) =
X(0)Y (y)T (t) = 0, 0 ≤ y ≤ b, t ≥ 0. Since we assume that Y and T are
not identically 0, we must have X(0) = 0. Applying the other boundary
conditions, we find X(a) = 0, Y (0) = Y (b) = 0.
Solutions for X and Y have already been derived in the Notes. The equation
X 00 = LX, X(0) = X(a) = 0 has nontrivial solutions only if L = −(nπ/a)2 ,
n = 1, 2, . . ., and then Xn (x) = sin((nπx)/a). Similarly, Y 00 = (K − L)Y ,
Y (0) = Y (b) = 0, has nontrivial solutions only if (K − L) = −(mπ/b)2 ,
m = 1, 2, . . ., and then Ym (y) = sin((mπy)/b). Finally, we have the equation
T 00 = KT , where K = −(nπ/a)2 − (mπ/b)2 < 0, so the general solution for
T is
p
T (t) = αmn cos(λmn t) + βmn sin(λmn t), λmn = (nπ/a)2 + (mπ/b)2 .
1
Assembling our results, solutions to the differential equation that satisfy the
given boundary conditions are of the form
u(x, y, t) =
∞ X
∞
X
sin(nπx/a) sin(mπy/b)(αmn cos(λmn t) + βmn sin(λmn t)),
m=1 n=1
for some coefficients αmn , βmn , m, n = 1, 2, . . ..
Finally, we must apply the initial conditions to determine the coefficients αmn
and βmn . Suppose φ(x, y) can be expanded in a two dimensional Fourier sine
series as
∞ X
∞
X
φ(x, y) =
Amn sin(nπx/a) sin(mπy/b).
m=1 n=1
Then since u(x, y, 0) = φ(x, y), it follows that αmn = Amn . If ϕ(x, y) is
expanded as
ϕ(x, y) =
∞ X
∞
X
Bmn sin(nπx/a) sin(mπy/b),
m=1 n=1
then since ut (x, y, 0) = ϕ(x, y), it follows that βmn = Bmn /λmn . It can
be shown that if φ and ϕ are C 2 functions on the rectangle [0, a] × [0, b]
then these expansions are valid and they converge uniformly. The solution
u(x, y, t) is then given by
u(x, y, t) =
∞ X
∞
X
sin(nπx/a) sin(mπy/b)(Amn cos(λmn t)+(Bmn /λmn ) sin(λmn t)).
m=1 n=1
2. Exercises 1 and 9 on pp. 99-100 in the text.
utt = c2 uxx ,
0 < x < L, t > 0,
u(0, t) = ux (L, t) = 0,
u(x, 0) = f (x), ut (x, 0) = g(x),
t ≥ 0,
0 ≤ x ≤ L.
Assume that u(x, t) = X(x)T (t). Substitute into the PDE to find
T 00 X = c2 X 00 T ⇒
T 00
X 00
= c2
.
T
X
Since the left-hand side is a function of t alone and the right-hand side is a
function of x alone, they both must be equal to a constant K. This leads to
two ODE’s: T 00 − c2 KT = 0 and X 00 − KX = 0. Look first at the equation
for X, together with the boundary conditions:
X 00 − KX = 0, X(0) = 0, X 0 (L) = 0.
2
Will show that K must be negative in order for this to have nontrivial solutions.
√
Suppose
K > 0. Then the general solution of X 00 − KX = √0 is c1 e√ Kx +
√
c2 e−√Kx . X(0) = 0 ⇒ c1 + c2 = 0 and X 0 (L) = 0 ⇒ K(c1 e KL −
c2 e− KL ) = 0. Since these two equations are independent, together they
imply that c1 = c2 = 0.
Suppose K = 0. Then the general solution of X 00 = 0 is c0 + c1 x. X(0) =
0 ⇒ c0 = 0 and X 0 (L) = 0 ⇒ c1 = 0.
Now suppose
K < 0. pThen the general solution of X 00 + |K|X = 0 is
p
0
c1 p
cos( |K|x)
p + c2 sin( |K|x).
p X(0) = 0 ⇒ c1 = 0. X (L) = 0 ⇒
c2 |K| cos( |K|L) = 0 ⇒ |K|L = (2n − 1)π/2, n = 1, 2, . . .. Thus
the eigenvalues are
2
(2n − 1)π
K=−
, n = 1, 2, . . . ,
2L
and the corresponding eigenvectors are
(2n − 1)πx
, n = 1, 2, . . . .
Xn (x) = sin
2L
Now look at the equation for T after substituting −((2n − 1)π/(2L))2 for K:
Tn00 +
c2 (2n − 1)2 π 2
Tn = 0.
4L2
The general solution of this is
c(2n − 1)πt
c(2n − 1)πt
Tn (t) = αn cos
+ βn sin
.
2L
2L
It follows that the solution u(x, t) has the form
∞
X
(2n − 1)πx
c(2n − 1)πt
c(2n − 1)πt
u(x, t) =
sin
αn cos
+ βn sin
.
2L
2L
2L
n=1
(1)
Now apply the initial conditions:
∞
X
(2n − 1)πx
u(x, 0) =
αn = f (x),
sin
2L
n=1
ut (x, 0) =
∞
X
n=1
sin
(2n − 1)πx
2L
βn
c(2n − 1)π
= g(x).
2L
These will be Fourier series for f and g provided
Z
2 L
(2n − 1)πx
αn =
sin
f (x) dx,
L 0
2L
3
Z L
4
(2n − 1)πx
βn =
g(x) dx,
sin
c(2n − 1)π 0
2L
and f and g satisfy the boundary conditions: f (0) = g(0) = 0, f 0 (L) =
g 0 (L) = 0, since in this case an odd extension of f and g will have a sine
series expansion in which the coefficients of even terms are zero.
Finally, we must find conditions on f and g that will ensure that term-byterm differentiation of the Fourier series for u in (1) really is justified and
therefore u is a C 2 solution of the differential equation. By the Weierstrass
M-test, it suffices to show that there is a constant C such that
2
2
C
C
(2n − 1)π
(2n − 1)π
2
|αn | ≤ 2 , and
|βn |2 ≤ 2 ,
2L
n
2L
n
P∞
since n=1 (C/n2 ) is finite. Following the approach in the Notes, we will
look at the expressions for αn and βn and use integration by parts:
Z
(2n − 1)πx
2 L
sin
αn =
f (x) dx
L 0
2L
L
(2n − 1)πx 4
= −f (x)
cos
+
(2n − 1)π
2L
0
Z L
4
(2n − 1)πx
cos
f 0 (x) dx
(2n − 1)π 0
2L
The boundary term will be 0 if f (0) = 0. In that case, using integration by
parts for the remaining integral, we find
"
#
L Z L
(2n
−
1)πx
4L
(2n
−
1)πx
−
f 0 (x) sin
αn =
sin
f 00 (x) dx .
(2n − 1)2 π 2
2L
2L
0
0
The boundary term here will be 0 provided f 0 (L) = 0. Using integration by
parts again, the next boundary term will be 0 and hence |αn | will be O(1/n3 )
if f 00 (0) = 0. Using integration by parts one more time, the next boundary
term will be 0 and |αn | will be O(1/n4 ) provided also f 000 (L) = 0. Thus,
conditions on f in order for u in (1) to be a solution are
f ∈ C 4 , f (0) = 0, f 0 (L) = 0, f 00 (0) = 0, f 000 (L) = 0.
A similar procedure with the formula for βn shows that since βn = O(1/n),
it will be O(1/n2 ) if g(0) = 0, O(1/n3 ) if also g 0 (L) = 0, and O(1/n4 ) if also
g 00 (0) = 0. Thus, conditions on g in order for u in (1) to be a solution are
g ∈ C 3 , g(0) = 0, g 0 (L) = 0, g 00 (0) = 0.
3. Consider the Dirichlet problem for the Laplacian on the unit disk D; i.e., given f on
∂D (the boundary of D), find u satisfying 4u = 0 in D, u = f on ∂D. Note that in
polar coordinates (r, θ), Laplace’s equation becomes:
1
1
urr + ur + 2 uθθ = 0.
r
r
4
(2)
(a) Assume that equation (2) has a solution of the form u(r, θ) = v(r)w(θ), and find
v and w. (Note: u should be bounded near the origin, and w must be 2π-periodic.
To solve the equation for v, you may need to look up Euler equations in an ODE
book.)
Substituting u(r, θ) = v(r)w(θ) into the PDE, we find
1
1
v 00 (r)w(θ) + v 0 (r)w(θ) + 2 v(r)w00 (θ) = 0.
r
r
Moving r variables to the left and θ variables to the right, this becomes
r2 v 00 (r) + rv 0 (r)
w00 (θ)
=−
= λ,
v(r)
w(θ)
where λ is a constant independent of r and θ.
First consider the equation for w. The function w is 2π-periodic and
satisfies
w00 = −λw.
√
If √
λ < 0, then w00 = |λ|w so w is a linear combination of e |λ|θ and
e− |λ|θ , but these functions are not 2π-periodic (and no nontrivial linear
combination of them is 2π-periodic), so in this case there are no nontrivial
solutions for w.
If λ = 0, then w = c0 + c1 θ, so the only nontrivial 2π-periodic solutions
are constant functions.
If λ > 0, then
√ solution of the differential equation is of the
√ the general
form a cos( λθ) + b sin( λθ), and this will be 2π-periodic if and only if
λ = n2 , n = 1, 2, . . ..
Thus the 2π-periodic solutions of w00 = −n2 w are linear combinations of
{1, cos(nθ), sin(nθ) : n = 1, 2, . . .}.
Since λ = n2 , the equation for v becomes
r2 v 00 (r) + rv 0 (r) − n2 v(r) = 0.
Solving this (Euler’s) equation, we find
v(r) = c1 rn + c2 r−n , n ≥ 1.
In order for this to be bounded at r = 0, we must have c2 = 0; hence
v(r) = c1 rn . For n = 0, solutions of v 00 = −(1/r)v 0 , or, of ϕ = v 0 and
ϕ0 = −(1/r)ϕ, are ϕ = c1 /r so v = c1 ln r + c2 . In order for this to be
bounded at r = 0, we need c1 = 0 hence v = c2 . Thus v is some linear
combination of the functions {rn : n = 0, 1, . . .}.
5
A solution u(r, θ) therefore has an expansion of the form
∞
a0 X n
+
r (an cos(nθ) + bn sin(nθ)).
2
n=1
(3)
R 2π
(b) Suppose f ∈ L2 (∂D) is a 2π-periodic square integrable function ( 0 |f (θ)|2 dθ <
∞) defined on the boundary of D. Write f in a Fourier series and derive a series
for u. Show that this series converges for r < 1 to a C 2 solution of 4u = 0, which
satisfies the boundary condition in the sense that ku(r, ·) − f (·)kL2 → 0 as r → 1.
If the Fourier series of f is
∞
a0 X
+
(an cos(nθ) + bn sin(nθ)),
2
n=1
then the series for u is that in (3). Assuming that we can do term-byterm differentiation, it satisfies Laplace’s equation in polar coordinates
since
ur =
∞
X
nr
n−1
(an cos(nθ)+bn sin(nθ)), urr =
∞
X
n(n−1)rn−2 (an cos(nθ)+bn sin(nθ)),
n=1
n=1
uθθ = −
∞
X
rn n2 (an cos(nθ) + bn sin(nθ)),
n=1
and hence
∞
X
1
1
urr + ur + 2 uθθ =
[n(n−1)+n−n2 ]rn−2 (an cos(nθ)+bn sin(nθ)) = 0.
r
r
n=1
We will show that, for each fixed r < 1, all of these series converge
uniformly on [0, 2π]; in fact, for any δ > 0 they converge uniformly on
[0, 1 − δ] × [0, 2π]. Hence term-by-term differentiation is justified and
the function u defined in (3) is therefore C 2 inside D. Since f ∈ L2 ,
its Fourier coefficents an and bn approach 0 as n → ∞; hence there is a
constant M such that |an cos(nθ) + bn sin(nθ)| ≤ M for all n. It follows
that each term in the series (3) is bounded in absolute value by M rn , so
for any fixed r < 1 the series in (3) converges absolutely (and uniformly
1
(M times
of P
the geometric
in θ), P
with sum bounded by M 1−r
P∞the sum
n n
n
2 n
series 0 r ). By the same argument, since n=1 nr P
and ∞
n=1 n r are
∞
2 n
2
2
finite (since, for n sufficiently large, n r < 1/n and n=1 (1/n ) < ∞),
the series for the derivatives of u converge uniformly in θ and uniformly
in r provided r ∈ [0, 1 − δ] for some δ > 0. Thus u is C 2 inside D.
Now look at ku(r, ·) − f (·)kL2 . For r < 1, the Fourier series for u(r, θ) −
f (θ) is
∞
X
(rn − 1)(an cos(nθ) + bn sin(nθ)),
n=1
6
so that, by Parseval’s identity,
ku(r, θ) − f (θ)k2L2 = π
∞
X
(rn − 1)2 (|an |2 + |bn |2 ).
n=1
We wish to show that this approaches 0 as r % 1. To see this, break the
infinite series into two pieces:
ku(r, θ)−f (θ)k2L2 = π
N
X
(rn −1)2 (|an |2 +|bn |2 )+π
n=1
∞
X
(rn −1)2 (|an |2 +|bn |2 ).
n=N +1
(4)
P∞
2
2
n
2
(|a
|
+
|b
|
)
is
finite,
and
(r
−
1)
≤
1
for
all
Since kf k2L2 =
n
n
n=1
n, given any > 0 we can choose N large enough so that the second
sum in (4) is less than /2. Since the first
we can pull the
Psum isn finite,
2
2
2
(r
−
1)
(|a
limit inside the sum to see that limr→1 N
n | + |bn | ) =
n=1
PN
n
2
2
2
n=1 limr→1 (r − 1) (|an | + |bn | ) = 0. Thus by choosing r sufficiently
close to 1 we can make this piece less than /2 as well. Therefore ku(r, ·)−
f (·)kL2 → 0 as r % 1.
7