Math 5440
Aaron Fogelson
Fall, 2013
Math 5440 Problem Set 7 – Solutions
1: (Logan, 3.2 # 1) Verify that the set of functions {1, cos( x ), cos(2x ), ...} form an orthogonal set on the interval [0, π ]. Next verify that the set of functions cos(nπx/L),
n = 0, 1, 2, ... form an orthogonal set on the interval [0, L]. If
∞
f (x) =
∑ cn cos(nπx/L),
n =0
in the mean-square sense on [0, L], what are the formulas for the cn ?
Let m ≥ 0 and n ≥ 0. Then
Z π
0
cos(mx ) cos(nx )dx
Z π
1
=
{cos((m + n) x ) + cos((m − n) x )} dx
2
1 sin((m + n) x ) π 1 sin((m − n) x ) π
+ 2
= 0.
2
m+n
m−n
0
0
0
if m6=n
=
(1)
(2)
To show {1, cos(πx/L), cos(2πx/L), ...} are orthogonal on [0, L], note that with the change
of variables y = πx/L,
Z L
0
cos
nπx L
cos
mπx L
L
dx =
π
Z π
0
cos(ny) cos(my)dy = 0,
from the above provided m 6= n.
If
∞
f (x) =
∑ cn cos(nπx/L),
in L2 [0, L],
n =0
then
( f , cos(mπx/L)) = cm (cos(mπx/L), cos(mπx/L)).
From (1), in case m = n, we see that (cos(mπx/L), cos(mπx/L)) = ( L/π )(π/2) = L/2,
so
Z
nπx 2 L
cn =
f ( x ) cos
dx.
L 0
L
1
2: (Logan, 3.2 # 4) For which powers of r is the function f ( x ) = xr in L2 [0, 1]? In
L2 [0, ∞]?
Z 1
0
Z 1
x2r dx
 x2r+1 1
r 6= −1/2
= lim 2r+1 a
a →0 
ln( x )|1a r = −1/2
(
1
a2r+1
− 2r
2r
+
1
+1 r 6 = −1/2 .
= lim
a→0 − ln( a )
r = −1/2
2r
x dx = lim
a →0 a

The limit is finite for r > −1/2 because then the exponent 2r + 1 > 0, so xr ∈ L2 [0, 1] for
r > −1/2.
R∞
Only for r > −1/2 could 0 x2r dx be finite because this integral is greater than the
integral of x2r over the smaller interval [0, 1]. Assume r > −1/2.
Z ∞
0
b
x2r+1 x dx = lim
b→∞ 2r + 1 0
1
= lim
b2r+1 .
b→∞ 2r + 1
2r
Only if 2r + 1 < 0 is this finite, but we’ve already set that r > −1/2, so there is no value
of r for which xr ∈ L2 [0, ∞].
2
3: (Logan, 3.2 # 5) Let f be defined and integrable on [0, L]. The orthognal expansion
∞
∑
n =1
bn sin
nπx L
bn =
,
2
L
Z L
0
f ( x ) sin
nπx L
dx,
is called the Fourier sine series for f on [0, L]. Find the Fourier sine series for f ( x ) =
cos( x ) on [0, π/2]. What is the Fourier sine series for f ( x ) = sin x on [0, π ]?
For f ( x ) = cos( x ) on [0, π/2],
bn =
=
=
=
=
=
π/2
2
cos( x ) sin(2nx )dx
π/2 0
Z
4 1 π/2
{sin(2nx + x ) + sin(2nx − x )} dx
π2 0
2 − cos((2n + 1) x ) cos((2n − 1) x ) π/2
−
π
2n + 1
2n − 1
0
2
1
cos((2n + 1)π/2) cos((2n − 1)π/2)
2
1
−
+
−
+
π
2n + 1
2n − 1
π 2n + 1 2n − 1
2
2n − 1 + 2n + 1
π (2n + 1)(2n − 1)
n
8
.
2
π 4n − 1
Z
So
∞
cos( x ) =
8n
sin(2nx ).
π (4n2 − 1)
n =1
∑
The Fourier Sine Series for sin( x ) on [0, π ] is the single term sin( x ).
3
4: (Logan, 3.2 # 8) For f , g ∈ L2 [ a, b], prove the Cauchy-Schwarz inequality
|( f , g)| ≤ || f || || g||.
Let q(t) = ( f + tg, f + tg) = ( f , f ) + 2( f , g)t + ( g, g)t2 . Because q(t) ≥ 0, q(t) has 1 or no
real roots. The roots t± are
p
−2( f , g) ± 4( f , g)2 − 4( g, g)( f , f )
t± =
,
2( g, g)
and for there to be at most one real root the discriminant must be nonpositive. Hence
( f , g)2 ≤ ( g, g)( f , f ),
or
|( f , g)| ≤ ( g, g)1/2 ( f , f )1/2 ,
4
5: (Logan, 3.3 # 2) Find the Fourier series for f ( x ) = x2 on [−π, π ]. Sketch a graph of
the function defined on all of R to which the Fourier series converges. Is the convergence
pointwise? Use the Fourier series to show
1 1
1
π2
= 1− + −
+ ....
12
4 9 16
Graph the frequency spectrum.
The Fourier coefficients for f ( x ) = x2 on [−π, π ] are
and
1
an =
π
Z π
x2 cos(nx )dx,
1
bn =
π
Z π
x2 sin(nx )dx.
x2
−π
−π
Since is even and sin(nx ) is odd, their product is odd, and the integrals defining bn are
all 0, so bn = 0 for all n. The integral for a0 can be evaluated easily and give a0 = 2π 2 /3.
The integrals for an for n ≥ 1 can be found using integration by parts twice as follows
(
)
π
Z π
1 x2 sin(nx ) sin(nx )
an =
− −π 2x n dx
π
n
−π
π
2
x sin(nx )dx
= −
πn −π
Z π 2
cos(nx ) π
− cos(nx )
= −
−x
dx
− −π
πn
n
n
−π
−π (−1)n sin(nx ) π
−π (−1)n
2
+
+
= −
πn
n
n
n2 − π
4
2
= 2 (2(−1)n ) = 2 (−1)n
n
n
Z
So
x2 =
∞
π2
4
+ ∑ 2 (−1)n cos(nx ).
3
n
n =1
Dividing both sides of this equation by 4 and evaluating the result at x = 0, we get
0=
1 1
1
π2
+ (−1 + − +
− ...)
12
4 9 16
which upon rearranging gives the desired sum. The function to which the Fourier series
for x2 on [−π, π ] is the extension of x2 from this interval to the entire real line as a 2π periodic function. The extended function is twice continuously differentiable except when
x equals an integer multiple of π. The extended function is continuous for all x and is
2π periodic by construction. The first derivative of the extended function is continuous
5
except when x is an integer multiple of π. By inspection an = O(1/n2 ) and bn = 0, so
the series converges pointwise to the extended function at points
at which that function
p
2
is continuous, that is, for all x. Because the series with terms an + bn2 converges and the
extended function is continuous and 2π-periodic, the convergence of the Fourier series is
in fact uniform.
–8
–6
–4
–2
10
10
8
8
6
6
4
4
2
2
0
2
4
10
–8
–6
–4
–2
6
–8
8
–6
–4
–2
x
0
8
6
6
4
4
2
2
2
4
4
6
8
–8
x
–6
–4
–2
0
6
8
6
8
x
10
8
0
2
2
4
x
Figure 0.1: Upper left: The extended function. Upper Right: A three term approximation.
Lower Left: A five term approximation. Lower Right: A seven term approximation.
6
4
3
2
1
0
5
10
15
Figure 0.2: Frequency Spectrum.
7
20
6: (Logan, 3.3 # 5) Let f ( x ) = −1/2 on −π < x < 0 and f ( x ) = 1/2 on 0 < x < π.
Show that the Fourier series for f is
∞
2
sin(2n − 1) x.
(2n − 1)π
n =1
∑
If s N ( x ) denotes the sum of the first N terms, sketch the graphs of s1 ( x ), s3 ( x ), s7 ( x ),
and s10 ( x ) and compare with f ( x ). Observe that the approximations overshoot f ( x ) in
the neighborhood of x = 0, and the overshoot is not improved regardless of how many
terms are taken in the approximation. This overshoot behavior of Fourier series near a
discontinuity is call the Gibbs phenomenon.
Since f ( x ) is odd, an = 0 for all n.
bn
1 π
f ( x ) sin(nx )dx
=
π −π
Z π
Z 0
1
=
sin(nx )dx −
sin(nx )dx
2π
0
−π
(
)
− cos(nx ) π − cos(nx ) 0
1
=
−
2π
n
n
0
−π
n
n
1
1 (−1)
(−1)
1
+ + −
−
=
2π
n
n n
n
n
1
2 2(−1)
=
−
2π n
n
(
0
n even
=
2
n odd.
nπ
Z
All positive odd integers can be written 2n − 1 for n = 1, 2, 3, . . . , so b2n−1 =
8
2
.
(2n−1)π
0.6
0.6
0.4
0.4
0.2
0.2
–3
–2
–1
0
1
2
3
–3
–2
–1
0
1
2
3
2
3
x
x
–0.2
–0.2
–0.4
–0.4
–0.6
–0.6
–3
–2
–1
0.6
0.6
0.4
0.4
0.2
0.2
0
1
2
3
–3
–2
–1
0
1
x
x
–0.2
–0.2
–0.4
–0.4
–0.6
–0.6
Figure 0.3: Plot of partial sums s1 ( x ), s3 ( x ), s7 ( x ), and s10 ( x ).
9
7: (Logan, 3.4 # 1) Show that the substitution of u( x, t) = g(t)y( x ) into the PDE
ut = ( p( x )u x ) x − q( x )u,
a < x < b, t > 0,
leads to the pair of differential equations
g′ = −λg,
−( p( x )y′ )′ + q( x )y = λy,
where λ is some constant.
Let u( x, t) = g(t)y( x ). Then ut ( x, t) = g′ (t)y( x ) and u x ( x, t) = g(t)y′ ( x ). Substituting
these expressions into the PDE, we find
g′ (t)y( x ) = ( p( x ) g(t)y′ ( x )) x − q( x ) g(t)y( x ) = g(t)( py′ )′ − qgy.
Rearranging this equation yields
( py′ )′ − qy
g′
=
.
g
y
Since the left side is a function of t and the right side is a function of x, both sides must
equal the same constant which we call −λ. Hence,
g′ = −λg,
−( py′ )′ + qy = λy.
10
8: (Logan, 3.4 #2) Show that the SLP
−y′′ ( x ) = λy( x ),
y(0) = 0,
0 < x < L,
y( L) = 0,
has eigenvalues λn = n2 π 2 /L2 and corresponding eigenfunctions yn ( x ) = sin(nπx/L),
n = 1, 2, ....
We know from the energy argument presented in class that the eigenvalues for this problem (with p( x ) = 1, q( x ) = 0, and Dirichlet boundary conditions) are positive. So let
λ = k2 for k > 0. Then the general solution of the ODE is
y( x ) = A cos(kx ) + B sin(kx )
where A and B are arbitrary constants. The boundary conidtions y(0) = 0 implies that
A = 0, and the boundary condition y( L) = 0 then implies that 0 = B sin(kL). We cannot
choose B = 0 because this would give us an identically zero function, so we must have
sin(kL) = 0. Therefore, kL is an integer multiple of π, i.e., kL = nπ for positive integer n.
So λn = n2 π 2 /L2 and yn ( x ) = sin(kx ) = sin(nπx/L).
11
9: (Logan, 3.4 #3) Show that the SLP
−y′′ ( x ) = λy( x ),
y′ (0) = 0,
0 < x < L,
y( L) = 0,
with mixed Dirichlet and Neumann boundary conditions has eigenvalues
λn =
(1 + 2n)π
2L
2
and corresponding eigenfunctions
yn ( x ) = cos
(1 + 2n)πx
2L
for n = 0, 1, 2....
Again, the conditions of this SLP are such that the energy argument works and we therefore conclude that all eigenvalues are positive. Let λ = k2 for k > 0. The general solution
of the ODE is
y( x ) = A cos(kx ) + B sin(kx )
where A and B are arbitrary constants. The derivative of y( x ) is y′ ( x ) = − Ak sin(kx ) +
Bk cos(kx ). The boundary conidtion y′ (0) = 0 implies that B = 0, and y( x ) = A cos(kx ),
and the boundary condition y( L) = 0 then implies that 0 = A cos(kL). Hence Lk must
(2n−1)π
(2n−1)π 2
and
y
(
x
)
=
cos
, for n =
be an odd multiple of π/2. So k n =
n
2L
2L
1, 2, 3, . . . .
12
10: (Logan, 3.4 #6) Consider the SLP
−y′′ ( x ) = λy( x ),
0 < x < 1;
y(0) + y′ (0) = 0,
y(l ) = 0.
Is λ = 0 an eigenvalue? Are there any negative eigenvalues? Show that there are infinitely many positive eigenvalues by finding an equation whose roots are those eigenvalues, and show graphically that there are infinitely many roots.
The energy argument would give us the relation
2
− y (0) +
Z 1
0
′ 2
(y ) dx = λ
Z 1
0
y2 dx,
and the sign of the left hand sign is ambiguous, so we cannot use the energy argument to
conclude anything about the eigenvalues. So we consider, in turn, the possibilities λ = 0,
λ < 0, and λ > 0.
If λ = 0, the general soluton of the ODE is y( x ) = Ax + B for constants A and B. So
′
y ( x ) = A. Both boundary conditions reduce to the condition A + B = 0, so B = − A.
Therefore, λ = 0 is an eigenvalue with eigenfunction y0 ( x ) = x − 1.
If λ < 0, let λ = −k2 . The general soluton of the ODE is y( x ) = Aekx + Be−kx for
constants A and B. Then y′ ( x ) = Akekx − Bke−kx , and the boundary conditions imply
that 0 = A + B + k ( A − B) and 0 = Aek + Be−k . So we seek A and B that satisfy the linear
system
1+k 1−k
0
A
=
0
B
ek
e−k
There is a nontrivial solution only if the determinant e−k (1 + k ) − ek (1 − k ) = 0. This is
k
equivalent to the equation e2k = 11+
−k . The question then is whether this equation has a
positive solution k. The functions on the two sides of this equation are equal at k = 0,
and so are their first and second derivatives. However, the series expansions for these
functions, convergent for k ∈ [0, 1) are
1
1
e2k = 1 + 2k + (2k )2 + (2k )3 + . . .
2
6
and
1+k
= (1 + k)(1 + k + k2 + k3 + . . . ) = 1 + 2k + 2k2 + 2k3 + . . .
1−k
+k
2k
and we see that 11−
k > e for any positive k in the interval [0, 1), because the first three
+k
terms in the two series match and the remaining terms are larger for 11−
k . For k = 1 one
side of the equation is undefined. For k > 1 one side of the equation is negative, while
the other is positive. Hence there are no positive solutions to the above equation and
therefore there are no negative eigenvalues.
So consider λ = k2 > 0 for k > 0. The general solution of the ODE is y( x ) =
A cos(kx ) + B sin(kx ). The boundary conditions imply that A + Bk = 0 and A cos(k ) +
B sin(k ) = 0. So A = −kB, and 0 = [−k cos(k ) + sin(k )] B. Hence k must satisfy
13
k = tan(k ). From the graphs of the functions z = k and z = tan(k ) we can see that there
are an infinite number of positive intersections of these curves and therefore an infinite
number of positive eigenvalues.
14
11: (Logan, 3.4 #7) Show that the SLP
−y′′ ( x ) = λy( x ),
y(0) + 2y′ (0) = 0,
0 < x < 2;
3y(2) + 2y′ (2) = 0,
has exactly one negative eigenvalue. Is zero an eigenvalue? How many positive eigenvalues are there?
We consider, in turn, the possilities, λ < 0, λ = 0, and λ > 0.
Suppose λ = −k2 for k > 0. The general solution of y′′ = k2 y is y = Aekx + Be−kx . To
satisfy the boundary conditions, we must find a nontrivial soluton to the linear system
1 + 2k
1 − 2k
0
A
.
=
0
B
(3 + 2k)e2k (3 − 2k)e−2k
Setting the determinant of the matrix equal to 0 gives us an equation for k:
3 + 4k − 4k2
= e4k .
2
3 − 4k − 4k
The graphs of the functions on the two sides of this equation shows there is one positive
intersection (between 0 and 1/2). Thus there is exactly one negative eigenvalue.
10
8
6
y
4
2
0
0
0.1
0.2
0.3
0.4
k
Consider λ = 0. The general solution of the ODE is y( x ) = Ax + B. To satisfy the
boundary conditions, we must find a nontrivial solution of the linear system
2 1
A
0
=
.
8 3
B
0
Since the determinant of the matrix in not zero, there is no nontrivial solution of this
system, so λ = 0 is not an eigenvalue.
We know that a regular SLP has an infinite number of real eigenvalues. We know
there is one negative eigenvalue and that zero is not an eigenvalue. Therefore there are
15
an infinite number of positive eigenvalues. This may also be shown directly. For λ = k2 ,
the general solution to the ODE is y( x ) = A cos(kx ) + B sin(kx ). To satisfy the boundary
conditions, we must find a nontrivial solution of the linear system
0
A
1
2k
.
=
0
B
3 cos(2k ) − 2k sin(2k ) 3 sin(2k ) + 2k cos(2k )
The determinant of this matrix is 0 if and only if k satisfies
tan(2k ) =
4k
.
+3
4k2
We see graphically that there are infinitely many positive intersections of the curves z =
tan(2k ) and z = 4k4k
2 +3 , so there are infinitely many positive eigenvalues.
1
0.8
0.6
y
0.4
0.2
0
0
5
10
k
16
15
20
12: (Logan, 3.4 #9) Consider the SLP
−( x2 y′ )′ = λy,
y(1) = y(π ) = 0.
1 < x < π,
Use an energy argument to show that any eigenvalue must be nonnegative. Find the
eigenvalues and eigenfunctions.
We try the energy argument to see if we can tell the sign of the eigenvalues. Multiply both
sides of the ODE by y( x ), integrate over the interval [1, π ], and use integration by parts
and the boundary conditions to find that
Z π
1
x2 (y′ )2 dx = λ
Z π
1
y2 dx.
So λ ≥ 0. The only way that λ = 0 could occur is if y′ ( x ) = 0 for all x ∈ [1, π ]. Then y( x )
would be constant and because of the boundary conditions, the constant would have to
be 0. Therefore λ = 0 is not an eigenvalue for this problem.
So assume λ = k2 for k > 0. The ODE becomes
x2 y′′ ( x ) + 2xy′ ( x ) + k2 y( x ) = 0
This is a Cauchy-Euler equation (discussed in Appendix to Logan). We look for solutions
of the form y( x ) = xr . Substituting this into the ODE we find that r must satisfy
r2 + r + k2 = 0.
This has two solutions
√
1 − 4k2
2
There are three possible scenarios. Case 1: If r+ , r− are real and unequal, then the general
solution to the ODE is y( x ) = Axr+ + Bxr− . Case 2: If r+ , r− are real and both equal to r,
then the general solution is y( x ) = Axr + Bxr ln( x ). Case 3: If r+ , r− are complex conjugates r± = a ± ib, then the general solution is y( x ) = Ax a sin(b ln( x )) + Bx a cos(b ln( x )).
Case 1: y( x ) = Axr+ + Bxr− with r± real and unequal. The boundary conditions are
satisfied if
0
A
1
1
.
=
0
B
π r+ π r−
r± =
−1 +
The determinant of this system is π r+ − π r− which is never zero for two different real
numbers r± , so there are no eigenvalues in case 1.
Case 2: For r+ = r− = r to occur, 1 − 4k2 = 0, and r = −1/2, so the solution in this case is
y( x ) = Ax −1/2 + Bx −1/2 ln( x ).
The boundary conditions are satisfied if
1
0
0
A
=
.
−
1/2
−
1/2
B
0
π
π
ln(π )
17
The determinant of this system is not zero,
so k2 = 1/4 is not an eigenvalue.
√
Case 3: For case 3 to occur, r± = −21 ± 2i 4k2 − 1 with k2 > 1/4. Then the general solution
is
!
!
√
√
2
2
4k − 1
4k − 1
ln( x ) + Bx −1/2 sin
ln( x ) .
y( x ) = Ax −1/2 cos
2
2
To satisfy the boundary conditions, we need to find a nontrivial solution of the linear
system
! 1
0
0
A
√ 2
√ 2
.
=
4k −1
4k −1
−
1/2
−
1/2
0
B
π
cos
ln(π ) π
sin
ln(π )
2
2
Setting the determinant of this system to zero gives us the equation
!
√
2−1
4k
π −1/2 sin
ln(π ) = 0.
2
This is satisfied if and only if
√
4k2 − 1
ln(π ) = nπ
2
for some positive integer n. That is only if
1
λ=k = +
4
2
nπ
ln(π )
2
n = 1, 2, 3, . . . .
These are the eigenvalues of the problem and the corresponding eigenfunctions are
nπx
−1/2
yn ( x ) = x
sin
.
ln(π )
18