Section 17.2

3. Problem Prove that
(a) f is both even and odd if and only if it is identically zero.
(b) if f is even (and integrable), then F(x) = ∫0x f(t)dt is odd.
(c) if f is odd (and integrable), then F(x) = ∫0x f(t)dt is even.
(d) if f is even (and differentiable), then F(x) = df/dx is odd.
(e) if f is odd (and differentiable), then F(x) = df/dx is even.
Solution
(a) f(x) = f(−x) (even) = −f(−x) (odd) which implies that f(−x) = −f(−x), i.e. f(−x) = 0.
(b)
F(−x)=⌠
⌡ 		−x

0 
 f(t) dt.
Change the variable from t to s by setting t = −s so that dt = −ds and
⌠
⌡ 		−x

0 
 f(t) dt = ⌠
⌡ 		x

0 
 f(−s) (−ds) = − ⌠
⌡ 		x

0 
 f(t) dt = −F(x).
(c) Almost same as (b).
(d)
F(−x)= df(−x)
d(−x)
 = − df(−x)
dx
 = − F(x).

(e) Similar to (d).
5. Problem Determine fe(x) and fo(x). Is f even ? Odd ? Neither ?
(a) 2 − 5 x
(e) x/(x + 2)
Solution
(a)
fe = f(x) + f(−x)
2
 = (2 − 5x) + (2 + 5x)
2
 = 2
and
fo = f(x) − f(−x)
2
 = −5 x.
(e)
fe =
⎛
⎝ 		x
x + 2
 + −x
−x + 2
⎞
⎠
2
 = x2
x2 − 4

fo =
⎛
⎝ 		x
x + 2
 − −x
−x + 2
⎞
⎠
2
 = −2 x
x2 − 4
 .

6. Problem If F is even and G is odd, show that
(c) F(G(x)) is even
Solution
(c) Let H(x) ≡ F(G(x)), then, H(−x) = F(G(−x)) = F(−G(x)) = F(G(x))=H(x).
12. Problem Determine whether or not the given function (defined on −∞ < x < ∞) is periodic. If it is, find its fundamental period (if it has one).
(a) x4
(g) tan x
(m) esin 3x
Solution
(a) No. (g) Yes, π. (m) Yes, 2 π/3.
15. Problem Show that if f(x) is periodic with period T, then
(a) so is its derivative f′(x)
(b) so is its integral ∫0x f(t) d t, if and only if ∫0T f(t) dt = 0.
Solution
(a)
f′(x)=
lim
h→ 0 
 f(x + h) − f(x)
h
so
f′(x+T)=
lim
h→ 0 
 f(x + h + T) − f(x + T)
h
 =
lim
h→ 0 
 f(x + h) − f(x)
h
 = f′(x).

(b) Let
G(x) ≡ ⌠
⌡ 		x

0 
 f(t) dt,
then,
G(x+T) = ⌠
⌡ 		x+T

0 
 f(t) dt = ⌠
⌡ 		T

0 
 f(t) dt + ⌠
⌡ 		T+x

T 
 f(t) dt = ⌠
⌡ 		T+x

T 
 f(t) dt
(from the given condition). With the change of variable, t−T ≡ τ,
⌠
⌡ 		T+x

T 
 f(t) dt = ⌠
⌡ 		x

0 
 f(τ)dτ = G(x).

Section 17.3

1. Problem Are these functions piecewise continuous on [0, π] ? Explain.
(a)
sin2  x
(d)
cos  1
x
Solution
(a) Yes. In fact, f(x) is continuous everywhere.
(d) No. cos (1/x) fails to exist at x = 0.
2. Problem (a) Derive (24a), i.e.

⌠
⌡ 		l

−l 
 cos  m πx
l
 cos  n πx
l
 dx = ⎧
⎪
⎨
⎪
⎩
0
m ≠ n
l
m = n ≠ 0.
HINT:
cos A cos B = 1
2
 [cos (A+B) + cos (A−B)]
(b) Derive (24b), i.e.
⌠
⌡ 		l

−l 
 sin  m πx
l
 sin  n πx
l
 dx = ⎧
⎪
⎨
⎪
⎩
0
m ≠ n
l
m = n ≠ 0.
HINT:
sin A sin B = 1
2
 [cos (A−B) − cos (A+B)]
Solution
⌠
⌡ 		l

−l 
 cos  m πx
l
 cos  n πx
l
 dx = 1
2
 ⌠
⌡ 		l

−l 
 ⎛
⎝ 		cos  (m + n)πx
l
 +cos  (m − n)πx
l
⎞
⎠ 		dx = l
2 π
 ⎛
⎜
⎝
sin  (m + n)πx
l
m + n
 +
sin  (m − n)πx
l
m − n
⎞
⎟
⎠ 		=0
if m ≠ n. If m = n,
⌠
⌡ 		l

−l 
 cos  m πx
l
 cos  n πx
l
 dx = ⌠
⌡ 		l

−l 
 cos2 nπx
l
 dx = l.
For m = n = 0,
⌠
⌡ 		l

−l 
 cos  m πx
l
 cos  n πx
l
 dx = ⌠
⌡ 		l

−l 
 dx = 2l.

(b) Similar to (a).
4. Problem Work out the Fourier series of f, given over one period as follows. At which values of x, if any, does the series fail to converge to f(x) ? To what values does it converge at those points?
(a) x on [−π, π]
(g) | sin x | (for all x)
Solution
(a) Since f(x) is an odd function,
an = 0.

bn = 1
π
 ⌠
⌡ 		π

−π 
 x sin nx dx = 2 (−1)n+1
n
 .

f(x) = 2 ∞
∑
n=1 
 (−1)n+1
n
 sin n x
for all x except at ±π, ±3π… where f(x) = ±π but the series converges to 0.
(g) The period is π.
a0= 1
π
 ⌠
⌡ 		π/2

−π/2 
 |sin x| dx = 2
π
 .

an= 2
π
 ⌠
⌡ 		π/2

−π/2 
 |sin x| cos 2 n x dx = 4
π
 ⌠
⌡ 		π/2

0 
 |sin x| cos 2 n x dx = 2
π
 ⌠
⌡ 		π/2

0 
 (sin (2 n + 1) x − sin (2 n − 1) x) dx = − 4
π
 1
4 n2 − 1
 .

bn=0
as f(x) is an even function. Therefore,

f(x) = 2
π
 − 4
π
 ∞
∑
1 
 1
4n2 − 1
 cos 2 nx
which converges to f(x) for all x.
7. Problem (a) Use (11), i.e.,

f(x) = 2 + 8
π
 ∞
∑
1, 3, … 
 1
n
 sin n x
hw_ch17_prob7.jpg
to show that
π
4
 = ∞
∑
n = 1, 3, … 
 1
n
 sin  nπ
2
 = 1 − 1
3
 + 1
5
 − 1
7
 + …
Solution
(a) Setting x = π/2 in equation (11) yields
4 = 2 + 8
π
∑
1, 3, 5, … 
 sin (n π/2)
n
or
π
4
 =
∑
1, 3, 5, … 
 sin (n π/2)
n
 = 1 − 1
3
 + 1
5
 − 1
7
 + ….

(b) Requires many many terms. 1/(2n)  ∼ 10−6 yields n  ∼ 106. This series converges extremely slowly and is not recommended at all to compute numerical values of π.
9. Problem If the Fourier coefficients of a periodic function f(x) are an (n = 0, 1, 2, …) and bn (n = 1, 2, …), what are the Fourier coefficients An, Bn, say, of the shifted periodic function f(x − c) ?
Solution

f(x) = a0 + ∞
∑
n=1 
 ⎛
⎝ 		an cos  nπx
l
 + bn sin  nπx
l
⎞
⎠ 		.
Replacing x by x − c yields

f(x − c)
=
a0 + ∞
∑
n=1 
 ⎛
⎝ 		an cos  nπ(x − c)
l
 +bn sin  nπ(x − c)
l
⎞
⎠
=
a0 + ∞
∑
n=1 
 ⎛
⎝ 		an cos  n πx
l
 cos  n πc
l
 +an sin  n πx
l
 sin  n πc
l
 +bn sin  n πx
l
 cos  n πc
l
 −bn cos  n πx
l
 sin  n πc
l
⎞
⎠
≡
A0 + ∞
∑
n=1 
 (An cos  nπx
l
 + Bn sin  nπx
l
 )
where
A0
=
a0
An
=
an cos  n πc
l
 − bn sin  n πc
l
Bn
=
an sin  n πc
l
 + bn cos  n πc
l
 .
Note

cos (A − B)
=
cos A cos B + sin A sin B
sin (A − B)
=
sin A cos B − sin B cos A.
11. Problem (Polynomials) It is a useful fact that if p(x) is an even polynomial (i.e.,containing only even-integer powers of x) on (−l, ell), then the bn's are zero and

an = 2 l
n2 π2
 (−1)n ⎡
⎣ 		p′(l) − l2
n2 π2
 p"′(l) + l4
n4 π4
 p(v)(l) − …⎤
⎦
(1)
for n ≥ 1 : and if p(x) is an odd polynomial on (−l, l), then an's are zero and

bn = − 2
n π
 (−1)n ⎡
⎣ 		p(l) − l2
n2 π2
 p"(l) + l4
n4 π4
 p(iv)(l) − …⎤
⎦
(2)
for n ≥ 1.
(a) Derive (1) through the p"′ term.
(b) Derive (2) through the p" term.
Solution
(a) Use integration by parts repeatedly !!

an
=
1
l
 ⌠
⌡ 		l

−l 
 p(x) cos  n πx
l
 dx
=
⎡
⎣ 		1
l
 p(x) l
nπ
 sin  n πx
l
⎤
⎦ 		l

−l 
 − 1
l
 ⌠
⌡ 		l

−l 
 p′(x) l
nπ
 sin  nπx
l
 dx
=
0 + ⎡
⎣ 		− 1
nπ
 p′(x) ⎛
⎝ 		− l
nπ
 cos  nπx
l
⎞
⎠ 		⎤
⎦ 		l

−l 
 + 1
nπ
 ⌠
⌡ 		l

−l 
 p"(x) (− l
nπ
 cos  nπx
l
 ) dx
=
2l
n2 π2
 (−1)n p′(l) −⎡
⎣ 		l
n2 π2
 p"(x) ( l
nπ
 sin  nπx
l
 ) ⎤
⎦ 		l

−l 
 + l
n2 π2
 ⌠
⌡ 		l

−l 
 p"′(x) l
nπ
 sin  n πx
l
 dx
=
2l
n2 π2
 (−1)n p′(l) +⎡
⎣ 		l2
n3 π3
 p"′(x) (− l
nπ
 cos  nπx
l
 )⎤
⎦ 		l

−l 
 − l2
n3 π3
 ⌠
⌡ 		l

−l 
 p""(x) (− l
nπ
 cos  n πx
l
 ) dx
=
2l
n2 π2
 (−1)n p′(l)− 2l3
n4 π4
 p"′(l) + …
=
2l
n2 π2
 (−1)n ⎛
⎝ 		p′(l)− l2
n2 π2
 p"′(l)+…⎞
⎠ 		.
Here we used the fact that p, p", p"" … are even functions and p′, p"′, p""′, … are odd functions.
(b) Similar.
12. Problem Use (1) and/or (2) in Exercise 11 to obtain the Fourier series of the given periodic function. NOTE:(1) does not hold for n = 0, so you need to compute a0 in the usual way.
(a) f(x) = x on (−3, 3)
(d) f(x) = x2 (x2−8) on (−2, 2).
Solution
(a) p(x) = x is an odd function, p(3) = 3, p′(x) = 1, p"(x) = 0, so
bn = 6
n π
 (−1)n+1
so
f(x) = 2l
π
 ∑
 (−1)n+1
n
 sin  nπx
l
 .

(d) p(x) = −8 x2 + x4 is even and p′(2) = 0, p"(2) = 32, p"′(2) = 48 p""(2) = 24 so
an = 4
n2 π2
 (−1)n ⎛
⎝ 		0− 4
n2 π2
 48 + 0 …⎞
⎠
and
a0 = 2
4
 ⌠
⌡ 		2

0 
 (−8 x2 + x4) dx = −112/15.

f(x) = − 112
15
 + 768
π4
 ∑
 (−1)n+1
n4
 cos  nπx
2
 .
16. Problem (Complex from) Work out the complex exponential form of the Fourier series (55), i.e.
FS f = ∞
∑
n=−∞ 
 cn ei n πx/l

cn = 1
2 l
 ⌠
⌡ 		l

−l 
 f(x) e− i n πx/l dx.
for the given periodic function, defined over one period as follows.
(a) f(x) = 50 on | x | ≤ 1, and 0 on 1 < | x | ≤ 2
Solution
(a) l = 2 so

cn
=
1
4
 ⌠
⌡ 		2

−2 
 f(x) e−inπx/2 dx
=
1
4
 ⌠
⌡ 		1

−1 
 50 e−inπx/2 dx
=
50
π
 sin (n π/2)
n
 .

Section 17.9 (Fourier Integrals)

Note: Just like the Fourier series has real and complex representations, the Fourier integral has real and complex representations.

f(x)
=
⌠
⌡ 		∞

−∞ 
 c(ω) eiωxdω
cω
=
1
2π
 ⌠
⌡ 		∞

−∞ 
 f(x)e−iωx dx.
So

f(x)
=
⌠
⌡ 		∞

−∞ 
 c(ω) eiωxdω
=
⌠
⌡ 		0

−∞ 
 c(ω) eiωx dω+ ⌠
⌡ 		∞

0 
 c(ω) eiωx dω
=
⌠
⌡ 		0

∞ 
 c(−ω′) e−i ω′x (−d ω′) + ⌠
⌡ 		∞

0 
 c(ω) eiωx dω
=
⌠
⌡ 		∞

0 
 c(−ω) e−iωx dω+ ⌠
⌡ 		∞

0 
 c(ω) eiωx dω
=
⌠
⌡ 		∞

0 
 (c(−ω)e−iωx + c(ω) eiωx) dω
=
⌠
⌡ 		∞

0 
 (a(ω) cosωx + b(ω) sinωx )dω,
where

a(ω)
=
c(−ω) + c(ω)
=
1
2π
 ⌠
⌡ 		∞

−∞ 
 f(x) eiωx dx + 1
2π
 ⌠
⌡ 		∞

−∞ 
 f(x) e−iωx dx
=
1
π
 ⌠
⌡ 		∞

−∞ 
 f(x) cosωx dx
b(ω)
=
i (c(ω) − c(−ω))
=
…
=
1
π
 ⌠
⌡ 		∞

−∞ 
 f(x) sinωx dx.
These are eqs. (2a, 2b) on Page 915.
2. Problem Derive the Fourier integral representations of the following functions. At which points, if any, does the Fourier integral fail to converge to f(x) ? To what values does the integral converge at those points?
(a)
f(x) = ⎧
⎪
⎨
⎪
⎩
100
0 ≤ x ≤ 2
0
x < 0     , x > 2
(g)
f(x)=⎧
⎪
⎨
⎪
⎩
e−x
x ≥ 0
0
x < 0
Solution
(a)
a(ω) = 1
π
 ⌠
⌡ 		∞

−∞ 
 f(x) cos ωx dx = 100
π
 ⌠
⌡ 		2

0 
 cos ωx dx = 100
πω
 sin 2ω,

b(ω) = 100
π
 ⌠
⌡ 		2

0 
 sin ωx  dx = 100
πω
 (1−cos 2 ω),
so

f(x)
=
100
π
 ⌠
⌡ 		∞

0 
 (sin 2 ωcos ωx + (1 − cos 2 ω) sin ωx) dω
ω
=
100
π
 ⌠
⌡ 		∞

0 
 (sin(2−x) ω+(1 − cos 2ω) sin ωx) dω
ω
 .
This equality holds for all x except for x = 0, 2. At these points, f(x) equals to 100 whereas the integral representation converges to the average value of 50.
(g)
a(ω) = 1
π
 ⌠
⌡ 		∞

0 
 e−x cos ωx dx = 1
π
 1
1 + ω2

b(ω) = 1
π
 ⌠
⌡ 		∞

0 
 e−x sin ωx dx = 1
π
 ω
1 + ω2
so

f(x) = 1
π
 ⌠
⌡ 		∞

0 
 cos ωx +ωsin ωx
1 + ω2
 dω
for all x except x = 0 where the LHS = 1 and the RHS = 1/2.

Section 17.10

4. Problem Given ∧f(ω), use (6b), i.e.
F−1 ⎧
⎨
⎩
^
f
 
 (ω)⎫
⎬
⎭ 		= f(x) = 1
2 π
 ⌠
⌡ 		∞

−∞ 
^
f
 
 (ω) ei ωx dω.
to evaluate the inverse, f(x).
(a) e−a |ω|    (a > 0)
(b) H(ω+ a) − H(ω− a)     (a > 0)
(c) [H(ω) − H(ω− 1)] ω
(d) H(ω) e−a ω (Re     a > 0)
(e) [H(ω+ 1) − H(ω− 1)]|ω|
(f) δ(ω− a)
Solution
(a)
f(x)
=
1
2π
 ⌠
⌡ 		∞

−∞ 
 e−a |ω| ei ωx dω
(3)
=
1
2 π
 ⎛
⎝ 		⌠
⌡ 		0

−∞ 
 ea ω eiωx dω+⌠
⌡ 		∞

0 
 e−a ω eiωx dω⎞
⎠
(4)
=
(change ω = −ω′    in the first integral)
(5)
=
1
2 π
 ⎛
⎝ 		⌠
⌡ 		∞

0 
 e− a ω e− iωx dω+⌠
⌡ 		∞

0 
 e−a ω eiωx dω⎞
⎠
(6)
=
1
2 π
 ⌠
⌡ 		∞

0 
 e− aω (ei ωx + e− i ωx) dω
(7)
=
2
2 π
 ⌠
⌡ 		∞

0 
 e−aω cos ωx dω
(8)
=
(integration by parts)
(9)
=
1
π
 a
a2 + x2
 .
(10)

(b) Note that H(x) represents the step function, i.e.

H(x) = ⎧
⎪
⎨
⎪
⎩
1
x > 0
0
x < 0
so H(x − a) − H(x − b) represents a function that takes 1 for a < x < b otherwise 0.
hw_ch17_sol_1.gif
Therefore,

f(x) = 1
2π
 ⌠
⌡ 		a

−a 
 eiωx dω = sin a x
πx
 .

(c)
f(x) = 1
2π
 ⌠
⌡ 		1

0 
 ωei ωx dω = 1
2 πx2
 (eix(1 − ix) − 1).

(d)
f(x) = 1
2π
 ⌠
⌡ 		∞

−∞ 
 H(ω) e− a ωei ωx dω = 1
2 π
 ⌠
⌡ 		∞

0 
 e−(a − i x)ω dω = 1
2 π
 1
a − i x
 .

(e)
f(x)
=
1
2 π
 ⌠
⌡ 		1

−1 
 |ω| (cos x ω+ isin x ω) dω
=
1
π
 ⌠
⌡ 		1

0 
 ωcos x ωd ω
=
1
πx2
 (cos x + x sin x − 1).

(f)
f(x)= 1
2π
 ⌠
⌡ 		∞

−∞ 
 δ(ω− a)eiωx dω = ei a x
2 π
 .

6. Problem Evaluate the following using Appendix D. You may need to use more than one entry. Cite, by number , any entries that you use.
(a)
F{4 x2 e− 3 |x|}
(g)
F−1⎧
⎨
⎩ 		4 sin ω
ω
 − 1

√
|ω|
⎫
⎬
⎭
(h)
F−1{e− 2 |ω− 3|}
Solution
(a)
Note that by differentiating the both sides of

F(ω) = ⌠
⌡ 		∞

−∞ 
 f(x) e−i ωx dx
with respect to ω, one gets

d
dω
 F(ω)
=
⌠
⌡ 		∞

−∞ 
 (−ix) f(x) e− i ωx dx
d2
dω2
 F(ω)
=
⌠
⌡ 		∞

−∞ 
 (− i x)2 f(x) e−i ωx dx
=
− ⌠
⌡ 		∞

−∞ 
 x2 f(x) e−i ωx dx.
From Eq.4 in Appendix D (Page 1274), i.e.,
F( e−a |x|) = 2 a
ω2 + a2
one finds
F{ 4e− 3 |x| } = 4 F{ e− 3 |x| } = 4 6
ω2 + 9
 = 24
ω2 + 9
 .
So

F{ x2 4 e− 3 |x| } = i2 d2
dω2
 24
ω2 + 9
 = 144 3 − ω2
(ω2 + 9)3
 .

(g) From eqs.7 and 9 in Appendix D, i.e.
F( 1
√x
 ) =

√
2 π
|ω|

F(H(x + a) − H(x − a)) = 2 sin ωa
ω
 ,

F−1{ 4 sin ω
ω
 } = 2 (H(x + 1) − H(x − 1))

F−1{ 1

√
|ω|
 } = 1

√
2π
 1

√
|x|
So

F−1{ 4 sin ω
ω
 − 1

√
|ω|
 } = 2 (H(x + 1) − H(x − 1)) − 1

√
2 π|x|

(h) From eq.1 in Appendix D, i.e.
F( 1
x2 + a2
 ) = π
a
 e− a |ω|,

F−1{ e− 2 |ω| } = 2
π
 1
x2 + 4
 .
Also note that

F(ω) = 1
2π
 ⌠
⌡ 		∞

−∞ 
 f(x) e− i ωx dx
so
F(ω− a)
=
1
2 π
 ⌠
⌡ 		∞

−∞ 
 f(x) e− i (ω− a) x dx
=
1
2 π
 ⌠
⌡ 		∞

−∞ 
 (f(x) ei a x) e− i ωxdx
=
Fourier Transform of     (f(x) ei a x ).
So

F−1{ e−2 |ω− 3| } = 2
π
 e3 i x
x2 + 4
 .

10. Problem (Extension of transform tables) It follows from the results in Exercise 5 that the transform of the even and odd parts of f(x) are the even and odd part of ∧f(ω), respectively. This result can be used to obtain more information from a give transform table. For example, consider entry 2 in Appendix D, i.e.
F(H(x) e− a x) = 1
a + i ω
 .
Breaking f and ∧f into even and odd parts, show that
F{ e− a |x| } = 2a
ω2 + a2
and
F{ (sgn  x) e−a |x| } = − i ω
ω2 + a2
 ,
where
sgn  x ≡ ⎧
⎪
⎨
⎪
⎩
+1,
x > 0
−1,
x < 0,
which is read as "sign of x". Of these two results, observe that (10.1) is identical to entry 4, and that (10.2) is not contained in Appendix D.
Solution

f(x)
=
H(x)e− a x = 1
2
 (H(x) e− a x + H(−x) ea x) + 1
2
 (H(x) e− a x − H(−x) ea x)
=
1
2
 e−a |x| + 1
2
 (sgn x) e− a |x|

^
f(ω)
 
=
1
a + iω
 a − iω
a − iω
=


a
a2 + ω2
even
 
 −i

ω
a2 + ω2
odd
 
so

F{ e− a |x|} = 2a
a2 + ω2
 ,
and

F{ (sgn x) e− a |x|} = − iω
a2 + ω2
 .

11. Problem (Extension of transform tables) Another idea that enables us to extend a given Fourier transform table is that of reciprocity, namely, the reciprocity relations
F{
^
f
 
 (x)} = 2 πf(−ω)
(11)
and
F−1{ f(−ω)} =
^
f
 
 (x)
2π
 .
(12)
(a) Derive the relation (11) and (12).
(b) To illustrate, use (11) and entry 4, in Appendix D, to show that
F⎧
⎨
⎩ 		2 a
x2 + a2
⎫
⎬
⎭ 		= 2 πe− a |ω| ,     (a > 0)
or equivalently
F⎧
⎨
⎩ 		1
x2 + a2
⎫
⎬
⎭ 		= π
a
 e− a |ω| ,     (a > 0)
(In this case the result does not extend our table since it already appears as Entry 1.)
(c) Use (11) and entry 8 to show that
F⎧
⎨
⎩ 		sin a x
x
⎫
⎬
⎭ 		= π[ H(ω+ a) − H(ω− a) ].    (a > 0)
(d) Use (11) and entry 3 to show that
F⎧
⎨
⎩ 		1
a − i x
⎫
⎬
⎭ 		= 2 πH(ω) e−aω.     (Re a > 0)
(a) From Item 3, if f(x) = e−a |x|, then,
^
f
 
 (ω) = 2a
ω2 + a2
 ,
Thus, (a) in Problem 6 gives
F{ 2a
x2 + a2
 } = 2πe− a |−ω| = 2 πe− a |ω|

(c) If
f(x) = ⎧
⎪
⎨
⎪
⎩
ea x
x < 0
0
x > 0

^
f(ω)
 
 = 1
a − i ω
thus, from (a),

F{ 1
a − i x
 } = 2 π⎧
⎪
⎨
⎪
⎩
ea (− ω)
− ω < 0
0
− ω > 0
 = ⎧
⎪
⎨
⎪
⎩
2 πe−a ω
ω > 0
0
ω < 0

12. Problem (Deflection of loaded string) Related to the problem of a beam on an elastic foundation is the analogous problem for a flexible string. Imagine a string (of negligible mass per unit length ) stretched along x axis, over −∞ < x < ∞, by a tension T newtons, and let w(x) be an applied load distribution (newtons/meter), as sketched in the figure. If the displacement u(x) of the string is resisted by a distributed spring of stiffness k (newtons per meter per meter), and the slope u′(x) is sufficiently small over −∞ < x < ∞, then u(x) is accurately governed by the differential equation T u"− k u = − w, or,

u" − α2 u = − f(x),     (−∞ < x < ∞)
(13)
where α2=k/T and f(x)=w(x)/T. (Here we consider the static deflection. In Chapter 20 we will derive the governing differential equation for the not-necessarily-static case.) Assume that w(x) is sufficiently localized for u(x) to satisfy boundary conditions

u→ 0     and     u′→ 0
(14)
as x→±∞, as well.
(a) Solving (13) and (14) using a Fourier transform, derive the solution

u(x) = − 1
2α
 ⌠
⌡ 		∞

−∞ 
 e− α|x − ξ|f(ξ) dξ
(15)
(b) Evaluate u(x) from (15) for the case where f(x) is a point unit load at the origin, f(x) = δ(x), and sketch the graph of u.
(c) Verify, by formally using the Leibniz rule, that (15) satisfies the differential equation (13).
Solution
(a) By transferring the equation to the frequency domain (ω), the original differential equation becomes (iω)2U(ω) − α2 U(ω) = −F(ω) where U(ω) and F(ω) are the Fourier transforms of u(x) and f(x), respectively. Solving for U(ω) yields
U(ω) = F(ω)
ω2 + α2
By the Fourier convolution theorem,

u(x) = F−1( 1
ω2 + α2
 ) * f(x) = e− α|x|
2 α
 * f(x) = 1
2 α
 ⌠
⌡ 		∞

−∞ 
 e− α|x−ξ| f(ξ) dξ.

(I think that the sign of eq.(12.3) on Page 933 is wrong.)
(b) Note that for an arbitrary function, f(x),
⌠
⌡ 		∞

−∞ 
 f(x − ξ) δ(ξ) dξ = f(x).
So
u(x)
=
1
2α
 ⌠
⌡ 		∞

−∞ 
 e− α|x − ξ| δ(ξ) dξ
=
1
2α
 e− α|x|.

hw_ch17_sol_2.gif
(c) (skipped) Don't worry.


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On 20 Nov 2022, 18:03.