Section 7.8
Generalized Functions
131
Hence
=
i
F (f (x)
w
=
F (f (x)
1
eiw
eiw
e2iw
2
+
ieiw + 2ie2iw .
w
w
2 w w
45. You may want to draw a graph to help you visualize the derivatives. For
f (x) = sin x if |x| < and 0 otherwise, we have f (x) = cos x i
24 Chapter 2 Fourier Series
The modied steady-state is obtained by subtracting y3 from the steady-state.
Here is its graph.
modifiedsteadystate t_
steadystate t
alph 3 Cos 3 t
Plot Evaluate modifiedsteadystate t , t, 0, 4 Pi
bet 3 Sin 3 t
;
0.1
0.05
2
4
6
Section 2.7
Forced Oscillations
23
Clear a, mu, p, k, alph, bet, capa, capb, b, y
mu 1;
c 5 100;
k 1001 100;
p Pi;
a0 0;
0;
a n_
b n_
2
Pi n 1 Cos n Pi ;
alph0 a0 k;
capa n_
k mu n Pi p ^ 2
c n Pi p
capb n_
capa n a n
capb n b n
capa n ^ 2 capb n ^ 2
alph
Section 7.5
A Dirichlet Problem and the Poisson Integral Formula
123
Since limy0 ey|w| = 1, it follows that
lim F (Py f )(w) = lim ey|w| F (f )(w) = F (f )(w).
y 0
y 0
Taking inverse Fourier transforms, we see that limy0 Py f (x) = f (x).
The argument tha
122 Chapter 7 The Fourier Transform and its Applications
Solutions to Exercises 7.5
1. To solve the Dirichlet problem in the upper half-plane with the given boundary
function, we use formula (5). The solution is given by
u(x, y)
=
=
=
y
f (s)
ds
(x s)2 +
22 Chapter 2 Fourier Series
9. (a) Natural frequency of the spring is
0 =
k
= 10.1 3.164.
(b) The normal modes have the same frequency as the corresponding components
of driving force, in the following sense. Write the driving force as a Fourier series
F
Section 7.4
121
The Heat Equation and Gausss Kernel
You can easily check that this solution veries the heat equation and u(x, 0) = x3.
If n = 4,
2
u(x, t)
=
j =0
4
2j
tj (2j )! 42j
= x4 + 12tx2 + 12t2.
x
j!
Here too, you can check that this solution verie
Section 2.7
Forced Oscillations
21
Solutions to Exercises 2.7
1. (a) General solution of y + 2y + y = 0. The characteristic equation is
2 + 2 + 1 = 0 or ( + 1)2 = 0. It has one double characteristic root = 1. Thus
the general solution of the homogeneous e
120
Chapter 7
The Fourier Transform and its Applications
with
(xs)2
1
f (s) = e t
2t
and
(xs)2
1
e t
2t
(see the solution of Exercise 5). So
(w) = e(iwx+w
F
dn (iwx+w2 t)
e
dwn
u(x, t) = (i)n
2
t)
.
w =0
To compute this last derivative, recall the Taylor
20 Chapter 2 Fourier Series
fact that ez+w = ez ew ), we obtain
I1 + iI2
=
=
=
=
1
ex(a+ib) + C
a + ib
(a + ib)
eax eibx + C
(a + ib) (a + ib)
a ib ax
cos bx + i sin bx + C
e
a2 + b2
eax
a cos bx + b sin bx + i b cos bx + a sin bx
2 + b2
a
+ C.
(c) Equati
Section 7.4
The Heat Equation and Gausss Kernel
119
29. Parts (a)-(c) are obvious from the denition of gt(x).
(d) The total area under the graph of gt(x) and above the x-axis is
=
2
ez dz =
2
1
c 2t
2c t
c 2t
gt (x) dx =
2
ex
/(4c2 t)
2
ez dz
dx
(z =
x
,
124
Chapter 7
The Fourier Transform and its Applications
Solutions to Exercises 7.6
1. The even extension of f (x) is
1 if 1 < x < 1,
0 otherwise.
fe (x) =
The Fourier transform of fe (x) is computed in Example 1, Sec. 7.2 (with a = 1).
We have, for w 0,
Section 7.6
The Fourier Cosine and Sine Transforms
125
9. Applying the denition of the transform and using Exercise 17, Sec. 2.6 to
evaluate the integral,
2
Fs (e2x )(w) =
e2x sin wx dx
0
2 e2x
[w cos wx 2 sin wx]
4 + w2
=
x=0
2w
.
4 + w2
=
The inverse
Section 2.7
Forced Oscillations
25
modifiedfurther t_
modifiedsteadystate t
alph 1 Cos t
bet 1 Sin t ;
Plot
Evaluate modifiedfurther t , steadystate t , modifiedsteadystate t
, t, 0, 4 Pi
0.4
0.2
2
4
6
8
10
12
-0.2
-0.4
21. (a) The input function F (t) is
30 Chapter 3 Partial Dierential Equations in Rectangular Coordinates
17. We have
A(u) = u2, (x) =
x,
A(u(x(t), t) = A(x(0) = x(0).
So the characteristic lines are
x = tx(0) + x(0)
Solving for x(0), we nd
x(0) =
and so
u(x, t) = f
x(0)(t + 1) x = 0.
x
,
t+
130
Chapter 7
The Fourier Transform and its Applications
The formula is good at w = 0 if we take the limit as w 0. You will get
2 w cos w sin w
=i
w2
F x ( U 1 U 1 ) = lim i
w 0
2
w sin w
lim
= 0.
w0
2w
(Use lHospitals rue.) Unlike the Fourier transform
Section 3.1
Partial Dierential Equations in Physics and Engineering
29
Solutions to Exercises 3.1
1. uxx + uxy = 2u is a second order, linear, and homogeneous partial dierential
equation. ux(0, y) = 0 is linear and homogeneous.
5. ut ux + uxt = 2u is seco
28 Chapter 2 Fourier Series
Solutions to Exercises 2.10
5. The cosine part converges uniformly for all x, by the Weierstrass M -test. The
sine part converges for all x by Theorem 2(b). Hence the given series converges for
all x.
9.
(a) If limk sin kx = 0,
128
Chapter 7
The Fourier Transform and its Applications
Solutions to Exercises 7.8
1. As we move from left to right at a point x0, if the graph jumps by c units, then
we must add the scaled Dirac delta function by cx0 (x). If the jump is upward, c is
pos
Section 7.7
Problems Involving Semi-Innite Intervals
127
or
d
2
us (w, t) + c2 2 us(w, t) = c2
wT0 .
dt
Taking the Fourier sine transform of the boundary condition u(x, 0) = 0 for x > 0,
we get us (w, 0) = 0.
(b) A particular solution of the dierential eq
Section 2.9
Uniform Convergence and Fourier Series
27
Solutions to Exercises 2.9
1.
|fn(x)| =
sin nx
1
0 as n .
n
n
The sequence converges uniformly to 0 for all real x, because
independently of x.
1
n
controls its size
5. If x = 0 then fn (0) = 0 for al
126
Chapter 7
The Fourier Transform and its Applications
Solutions to Exercises 7.7
1. Fourier sine transform with respect to x:
=0
d
u (w,
dt s
2
w
t) = w2us (w, t) +
d
u (w,
dt s
u(0, t)
t) = w2 us(w, t).
Solve the rst-order dierential equation in us(w,
Section 2.6
Complex Form of Fourier Series
19
If m = n, then
p
1
2p
ei
m
px
ei
n
px
1
2p
dx =
p
p
ei
(mn)
p
x
dx
p
p
(mn)
i
ei p x
2(m n)
p
i
ei(mn) ei(mn)
2(m n)
i
(cos[(m n)] cos[(m n)]) = 0.
2(m n)
=
=
=
13. (a) At points of discontinuity, the Fourier
118
Chapter 7
The Fourier Transform and its Applications
As t increases, the expression erf
x+1
2
erf
erf
distribution
x1
2
x
+1
2 1et
x1
2 1et
erf
approaches very quickly
, which tells us that the temperature approaches the limiting
x+1
2
50 erf
x1
2
Section 7.2
The Fourier Transform
(c) With the help of Theorem 3
f g
w2
w
F 1 i e3 4
2
1 4 d 3 w2
1
F
i
e4
2 6 dw
12
d 3 w2
F 1 i
e4
dw
23
2
12
xe 3 x .
33
=
=
=
=
In computing F 1 e3
w2
4
12
w2
= xF 1 e3 4
23
, use Exercise 10(a) and (5) to obtain
F 1
Section 2.3
Fourier Series of Functions with Arbitrary Periods
Thus the Fourier series of g is
1
2
1+
4
k =0
1
sin(2k + 1)x
(2k + 1)
=
12
+
2
f x_
Which x 0, 0, 0 x 1, 1, x 1, 0
s n_, x_
1 2 2 Pi Sum 1
2 k 1 Sin 2 k
, x, 1, 1
Plot Evaluate f x , s 20, x
W
110
Chapter 7
The Fourier Transform and its Applications
By Theorem 3(i)
2
d
1
ew /4
dw
2
i w2 /4
e
.
22
2
F (xex ) =
i
=
1
41. We have F ( 1+x2 ) =
F(
|w |
e
.
2
So if w > 0
x
)(w) = i
1 + x2
d w
= i
e
2 dw
w
e.
2
If w < 0
F(
x
)(w) = i
1 + x2
d w
e
10 Chapter 2 Fourier Series
Solutions to Exercises 2.3
1. (a) and (b) Since f is odd, all the ans are zero and
bn
2p
n
sin
dx
p0
p
2
2
n
=
cos
(1)n 1
n
p0
n
0
if n is even,
4
if n is odd.
n
=
=
=
4
1
(2k + 1)
sin
x. At the points of discon
(2k + 1)
p
k =