Post on 04-Aug-2020
11 The Fourier Transform and its Applications
Solutions to Exercises 11.1
1. We have
f (w) =1√2π
∫ 1
−1
x e−ixw dx
=1√2π
∫ 1
−1
x(cos wx − i sin wx
)dx
=−i√2π
∫ 1
−1
x sin wx dx
=−2i√
2π
∫ 1
0
x sin wx dx
=−2i√
2π
[ 1w2
sin wx − x
wcos wx
]∣∣∣1
0
= −i
√2π
sin w − w cos w
w2.
5. Use integration by parts to evaluate the integrals:
f (w) =1√2π
∫ ∞
−∞f(x)e−iwx dx
=1√2π
∫ 1
−1
(1 − |x|)(cos wx− i sin wx) dx
=1√2π
∫ 1
−1
(1 − |x|) coswx dx− i√2π
=0︷ ︸︸ ︷∫ 1
−1
(1 − |x|) sinwx dx
=2√2π
∫ 1
0
u︷ ︸︸ ︷(1 − x)
dv︷ ︸︸ ︷cos wx dx
=2√2π
((1 − x)
sin wx
w
)∣∣∣∣1
0
+2√2πw
∫ 1
0
sin wx dx
= −√
2π
cos wx
w2
∣∣∣∣∣
1
0
=
√2π
1 − cos w
w2.
Section 11.1 The Fourier Transform 225
9. We have
f (w) =1√2π
∫ π/2
−π/2
cos x e−ixw dx
=2√2π
∫ π/2
0
cos x cos wx dx
=1√2π
∫ π/2
0
[cos[(w + 1)x] + cos[(w − 1)x]
]dx
=1√2π
[sin[(w + 1)x]
w + 1+
sin[(w − 1)x]w − 1
]∣∣∣∣π/2
0
(w 6= ±1)
=1√2π
[sin[(w + 1)π/2]
w + 1+
sin[(w − 1)π/2]w − 1
](w 6= ±1)
=1√2π
[cos wπ
2
w + 1+
cos wπ2
w − 1
](w 6= ±1)
=
√2π
cos wπ2
1 − w2(w 6= ±1).
Treat the case w = ±1 separately and you will find
f (±1) =2√2π
∫ π/2
0
cos x cos x dx
=π/2√
2π=
√2π
4.
13. Apply the inverse Fourier transform to the transform of Exercise 9, then you will get the functionback; that is,
1√2π
∫ ∞
−∞
√2π
cos wπ2
1 − w2eiwx dx =
cos x if |x| < π2
0 if |x| ≥ π2;
1π
∫ ∞
−∞
cos wπ2
1 − w2cos wx dx =
cos x if |x| < π2
0 if |x| ≥ π2;
2π
∫ ∞
0
cos wπ2
1 − w2cos wx dx =
cos x if |x| < π2
0 if |x| ≥ π2.
In the last two steps, we used the fact that the integral of an odd function over a symmetric intervalis 0 and that the integral of an even function over a symmertic interval is twice the integral over thepostive half of the interval.
11 The Fourier Transform and its Applications
17. (a) Let 0 < α < 1. Applying the definition of the Fourier transform, we find, for w > 0,
F(
1|x|α
)(w) =
1√2π
∫ ∞
−∞
1|x|α
e−iwxdx
=1√2π
∫ ∞
−∞
1|x|α cos wx dx
=1√2π
∫ ∞
−∞
1|x|α
cos wx dx =2√2π
∫ ∞
0
1xα
cos wx dx
=
√2π
wα−1
∫ ∞
0
1tα
cos t dt (wx = t ⇒ x = t/w, dx = dt/w)
=
√2π
wα−1Γ(1 − α) sinαπ
2.
If w = 0, both sides are infinite. If w < 0, the integral does not change, because cos wx is even.Hence the given formula follows.(b) Take α = 1
2 , then
F(
1|x|1/2
)(w) =
√2π
w−1/2Γ(1/2) sinπ
4
1w1/2
,
where we have used Γ(1/2) =√
π (Exercise 25(a), Section 4.3).
21. We have
F
( √2 x√
π (1 + x2)2
)(w) =
1π
∫ ∞
−∞
x
(1 + x2)2e−iwx dx
=−i
π
∫ ∞
−∞
x sin wx
(1 + x2)2dx.
Note that the Fourier transform is odd in the sense that f(−w) = −f (w). For the sake of the residuemethod, we take w < 0 and write the Fourier transform as
F
( √2 x√
π (1 + x2)2
)(w) =
1π
∫
γR
z
(1 + z2)2e−iwz dz,
=1π
∫
γR
z
(1 + z2)2eiWz dz (W = −w, W > 0),
where γR is the contour in Figure 8, with R large (in fact, it is enough to take R > 1). To justifythis equality, we note that the contour integral is equal to 2πi times the sum of the residues insidethe contour. Inside the contour we have only one residue at z = i and so the integral is constant forall R > 1. Letting R → ∞, the integral on the semi-circle converges to 0 (by Jordan’s lemma) andthe integral on the real line becomes
1π
∫ ∞
−∞
x
(1 + x2)2eiWx dx,
Section 11.1 The Fourier Transform 227
which is the desired integral. So let us compute the contour integral, IR, using residues. Let
F (z) =z
(1 + z2)2eiWz
, then F has one pole of order 2 at z = i inside the contour γR. The residue at z = i is equal to
Res (F, i) =d
dz(z − i)2
z eiWz
(1 + z2)2
∣∣∣∣∣z=i
=d
dz
z eiWz
(z + i)2
∣∣∣∣∣z=i
=(eiWz + iWeiWzz)(z + i)2 − 2(z + i)zeiWz
(z + i)4
∣∣∣∣∣z=i
=e−W W
4= −w
e−|w|
4.
Thus the contour integral is equal to
−iπwe−|w|
2and the Fourier transform for w < 0 is equal to
1π
IR = −iwe−|w|
2.
The formula works for w ≥ 0 since it defines an odd function.
11 The Fourier Transform and its Applications
Solutions to Exercises 11.2
1. We have F(e−x2) = 1√
2e−w2/4. Applying Theorem 1((ii) (with n = 2), we obtain
F(x2e−x2) = −
d2
dw2
[1√2e−w2/4
]
= − 1√2
d
dw
[−w
2e−w2/4
]
=e−w2/4
4√
2
[2 − w2
].
5. We have F(e−|x|) =√
2π
11+w2 . So
F(e−|x| + 6xe−|x|) =
√2π
(1
1 + w2+ 6i
d
dw
[1
1 + w2
])
=
√2π
(1
1 + w2+ 6i
−2w
(1 + w2)2
)
=
√2π
1 − 12iw + w2
(1 + w2)2.
9. Let
g(x) =
{1 if 0 < x < 1,
0 otherwise,
and note that f(x) = x g(x). Now F(g(x)) = i√2π
e−iw−1w
(Exercise 3, Section 11.1); so, by Theo-rem 1,
F(f(x)) = F(xg(x)) = id
dw
i√2π
e−iw − 1w
=−1√2π
−iwe−iw − e−iw + 1w2
=1√2π
(1 + iw) cos w + (w − i) sin w
w2.
13. We have F(e−x2) = 1√
2e−w2/4. Using Exercise 1, we obtain
F((1 − x2)e−x2) = F(e−x2
) − F(x2e−x2)
=1√2e−w2/4 − e−w2/4
4√
2
[2 − w2
]
=e−w2/4
2√
2
[1 +
w2
2
].
Section 11.2 Operational Properties 229
17. Let f(x) = e−x2and g(x) = xe−x2
. We have F(e−x2) = 1√
2e−w2/4 and
F(xe−x2) =
i√2
d
dwe−w2/4 = − i
2√
2we−w2/4.
SoF(f ∗ g) = F(f) · F(g) =
−i
4we−w2/2 =
14i
d
dw
[e−w2/2
]=
14F(xe−x2/2).
Hencef ∗ g(x) =
14xe−x2/2.
21. Let f(x) = U−a − Ua(x) and g(x) = U−b − Ub(x), where 0 < a ≤ b. We have
F(f) =
√2π
sin aw
wand F(g) =
√2π
sin bw
w;
soF(f ∗ g) =
2π
sin(aw) sin(bw)w2
.
Instead of inverting the Fourier transform to find f ∗ g, we will compute f ∗ g by using the methodof Example 10. (This is an interesting Fourier transform that is not in the table of transforms atthe end of the book.) We have
f ′(x) = δ−a(x) − δa(x);
g′(x) = δ−b(x) − δb(x);
d2
dx2(f ∗ g)(x) =
d
dxf ∗ d
dxg(x)
=(δ−a(x) − δa(x)
)∗(δ−b(x) − δb(x)
)
=1√2π
(− δb−a(x) + δ−(a+b)(x) + δa+b(x) − δa−b(x)
).
So f ∗ g is the second antiderivative of this sum of Dirac deltas, that vanishes at infinity; that is,f ∗g(x) = 0 for large |x| (in fact, for |x| > a+b). The reason for the last assertion is that both f andg have boounded support, so f ∗ g will have bounded support. To compute the antiderivatives it isbest to do it on a graph, as we did in Example 10. We can also proceed as follows. Antidifferentiateonce and use (17), then
d
dx(f ∗ g)(x) =
1√2π
(− Ub−a(x) + U−(a+b)(x) + Ua+b(x) − Ua−b(x)
).
An antiderivative of Uα is the function (x − α)Uα or
d
dx(x − α)Uα = Uα(x).
To see this, just draw the graph of (x − α)Uα; it is continuous and equal to 0 if x < α and x − α ifx > α. Thus its derivative is 0 if x < α and 1 if x > α; thus the derivative formula is true. So andantiderivative of f ∗ g is
1√2π
(−(x − b + a
)Ub−a(x) +
(x + b + a
)U−(a+b)(x) +
(x − b − a
)Ua+b(x) −
(x + b − a
)Ua−b(x)
).
11 The Fourier Transform and its Applications
If this function vanishes for large |x|, then it would be the desired antiderivative. Let us check: Takex > a + b, then
1√2π
(−(x− b + a
)Ub−a(x) +
(x + b + a
)U−(a+b)(x) +
(x − b − a
)Ua+b(x) −
(x + b − a
)Ua−b(x)
)
= 1√2π
(−(x − b + a
)+(x + b + a
)+(x − b − a
)−(x + b − a
)= 0,
as desired. Similarly, if x < −a − b, then
1√2π
(−(x− b + a
)Ub−a(x) +
(x + b + a
)U−(a+b)(x) +
(x − b − a
)Ua+b(x) −
(x + b − a
)Ua−b(x)
)
= 1√2π
(− 0 + 0 + 0 − 0
)= 0.
This proves that
f ∗g(x) =1√2π
((x+b+a
)U−(a+b)(x)−
(x+b−a
)Ua−b(x)−
(x−b+a
)Ub−a(x)+
(x−b−a
)Ua+b(x)
).
More explicitely, we have
f ∗ g(x) =1√2π
0 if |x| > a + b;
x + b + a if − a − b < x < −b + a;(x + b + a
)−(x + b − a
)if − b + a < x < b − a;
(x + b + a
)−(x + b − a
)−(x − b + a
)if b − a < x < a + b;
so, after simplifying,
f ∗ g(x) =1√2π
0 if |x| > a + b;
x + b + a if − a − b < x < −b + a;
2a if − b + a < x < b − a;
−x + b + a if b − a < x < a + b.
The graph of f ∗ g is a nice tent.
25. (a) Use the definition of convolutions:
eiαx ∗ f(x) =1√2π
∫ ∞
−∞f(y)eiα(x−y)dy
= eiαx
=f (α)︷ ︸︸ ︷1√2π
∫ ∞
−∞f(y)e−iαydy = eiαxf (α).
(b) Using (a) and cos αx = eiαx+e−iαx
2 , we find
cos(αx) ∗ f(x) =12(eiαx ∗ f(x) + e−iαx ∗ f(x)
)
=12
(eiαxf(α) + e−iαxf(−α)
).
Section 11.2 Operational Properties 231
Specializing to α = 1 and f(x) = e−|x|, f(w) =√
2π
11+w2 , we find
cos x ∗ e−|x| =12
(eix
√2π
11 + 1
+ e−ix
√2π
11 + 1
)
=12
(eix
√2π
11 + 1
+ e−ix
√2π
11 + 1
)=
1√2π
cos x.
33. Apply (23):
F(U2(x) − U4(x)) = F(U2(x)) −F(U4(x))
=−i√2πw
(e−2iw − e−4iw
)
=−i√2πw
(e−2iw − e−4iw
).
11 The Fourier Transform and its Applications
Solutions to Exercises 11.3
1.
∂2u
∂t2=
∂2u
∂x2,
u(x, 0) =1
1 + x2,
∂u
∂t(x, 0) = 0.
Follow the solution of Example 1. Fix t and Fourier transform the problem with respect to thevariable x:
d2
dt2u(w, t) = −w2u(w, t),
u(w, 0) = F( 11 + x2
)=√
π
2e−|w|,
d
dtu(w, 0) = 0.
Solve the second order differential equation in u(w, t):
u(w, t) = A(w) cos wt + B(w) sin wt.
Using ddt u(w, 0) = 0, we get
−A(w)w sin wt + B(w)w cos wt∣∣∣t=0
= 0 ⇒ B(w)w = 0 ⇒ B(w) = 0.
Henceu(w, t) = A(w) cos wt.
Using u(w, 0) =√
π2 e−|w|, we see that A(w) =
√π2 e−|w| and so
u(w, t) =√
π
2e−|w| cos wt.
Taking inverse Fourier transforms, we get
u(x, t) =∫ ∞
−∞e−|w| cos wt eixw dw.
5.
∂2u
∂t2= c2 ∂2u
∂x2,
u(x, 0) =
√2π
sin x
x,
∂u
∂t(x, 0) = 0.
Fix t and Fourier transform the problem with respect to the variable x:
d2
dt2u(w, t) = −c2w2u(w, t),
u(w, 0) = F(√ sinx
x
)(w) = f(w) =
{1 if |w| < 10 if |w| > 1,
d
dtu(w, 0) = 0.
Section 11.3 The Fourier Transform Method 233
Solve the second order differential equation in u(w, t):
u(w, t) = A(w) cos cwt + B(w) sin cwt.
Using ddt u(w, 0) = 0, we get
u(w, t) = A(w) cos cwt.
Using u(w, 0) = f (w), we see that
u(w, t) = f (w) cos wt.
Taking inverse Fourier transforms, we get
u(x, t) =1√2π
∫ ∞
−∞f (w) cos cwt eixw dw =
1√2π
∫ 1
−1
cos cwt eixw dw.
9.
t2∂u
∂x− ∂u
∂t= 0,
u(x, 0) = 3 cos x .
The solution of this problem is very much like the solution of Exercise 7. However, there is a difficultyin computing the Fourier transform of cos x, because cos x is not integrable on the real line. Onecan make sense of the Fourier transform by treating cos x as a generalized function, but there is noneed for this in this solution, since we do not need the exact formula of the Fourier transform, asyou will see shortly.
Let f(x) = 3 cos x and Fourier transform the problem with respect to the variable x:
t2iwu(w, t) − d
dtu(w, t) = 0,
u(w, 0) = F(3 cos x)
)(w) = f (w).
Solve the first order differential equation in u(w, t):
u(w, t) = A(w)e−i w3 t3 .
Using the transformed initial condition, we get
u(w, t) = f (w)e−i w3 t3 .
Taking inverse Fourier transforms, we get
u(x, t) =1√2π
∫ ∞
−∞f (w)e−i w
3 t3 eixw dw
=1√2π
∫ ∞
−∞f (w)eiw(x+ t3
3 ) dw
= f(x +t3
3) = 3 cos(x +
t3
3).
13.
∂u
∂t= t
∂2u
∂x2,
u(x, 0) = f(x) .
11 The Fourier Transform and its Applications
Fix t and Fourier transform the problem with respect to the variable x:
d
dtu(w, t) + tw2u(w, t) = 0,
u(w, 0) = f (w).
Solve the first order differential equation in u(w, t):
u(w, t) = A(w)e−t2w2
2 .
Use the initial condition: A(w) = f (w). Hence
u(w, t) = f (w)e−t2w2
2 .
Taking inverse Fourier transforms, we get
u(x, t) =1√2π
∫ ∞
−∞f (w)e−
t2w22 eixw dw.
21. (a) To verify that
u(x, t) =12[f(x − ct) + f(x + ct)] +
12c
∫ x+ct
x−ct
g(s) ds
is a solution of the boundary value problem of Example 1 is straightforward. You just have to plugthe solution into the equation and the initial and boundary conditions and see that the equationsare verified. The details are sketched in Section 3.4, following Example 1 of that section.(b) In Example 1, we derived the solution as an inverse Fourier transform:
u(x, t) =1√2π
∫ ∞
−∞
[f (w) cos cwt +
1cw
g(w) sin cwt]eiwx dx.
Using properties of the Fourier transform, we will show that
1√2π
∫ ∞
−∞f (w) cos cwteiwx dw =
12[f(x − ct) + f(x + ct)];(1)
1√2π
∫ ∞
−∞
1w
g(w) sin cwt eiwx dw =12
∫ x+ct
x−ct
g(s) ds.(2)
To prove (1), note that
cos cwt =eicwt + e−icwt
2,
so
1√2π
∫ ∞
−∞f (w) cos cwteiwx dw
=1√2π
∫ ∞
−∞f (w) (
eicwt + e−icwt
2)eiwx dw
=12
[1√2π
∫ ∞
−∞f (w) eiw(x+ct) dw
1√2π
∫ ∞
−∞f (w) eiw(x−ct) dw
]
=12[f(x + ct) + f(x − ct)];
Section 11.3 The Fourier Transform Method 235
because the first integral is simply the inverse Fourier transform of f evaluated at x + ct, and thesecond integral is the inverse Fourier transform of f evaluated at x − ct. This proves (1). To prove(2), we note that the left side of (2) is an inverse Fourier transform. So (2) will follow if we can showthat
(3) F{∫ x+ct
x−ct
g(s) ds
}=
2w
g(w) sin cwt.
Let G denote an antiderivative of g. Then (3) is equivalent to
F (G(x + ct) − G(x − ct)) (w) =2w
G′(w) sin cwt.
Since G′ = iwG, the last equation is equivalent to
(4) F (G(x + ct)) (w) − F (G(x − ct)) (w) = 2iG(w) sin cwt.
Using Exercise 19, Sec. 7.2, we have
F (G(x + ct)) (w) −F (G(x − ct)) (w) = eictwF(G)(w) − e−ictwF(G)(w)= F(G)(w)
(eictw − e−ictw
)
= 2iG(w) sin cwt,
where we have applied the formula
sin ctw =eictw − e−ictw
2i.
This proves (4) and completes the solution.
11 The Fourier Transform and its Applications
Solutions to Exercises 11.4
1. Repeat the solution of Example 1 making some adjustments: c = 12 , gt(x) =
√2√te−
x2t ,
u(x, t) = f ∗ gt(x)
=1√2π
∫ ∞
−∞f(s)
√2√te−
(x−s)2
t ds
=20√tπ
∫ 1
−1
e−(x−s)2
t ds (v =x − s√
t, dv = − 1√
tds)
=20√π
∫ x+1√t
x−1√t
e−v2ds
= 10(
erf(x + 1√
t) − erf(
x − 1√t
))
.
9. Fourier transform the problem:
du
dtu(w, t) = −e−tw2u(w, t), u(w, 0) = f (w).
Solve for u(w, t):u(w, t) = f (w)e−w2(1−e−t).
Inverse Fourier transform and note that
u(x, t) = f ∗ F−1(e−w2(1−e−t)
).
With the help of Theorem 5, Sec. 7.2 (take a = 1 − e−t), we find
F−1(e−w2(1−e−t)
)=
1√2√
1 − e−te− x2
4(1−e−t ) .
Thus
u(x, t) =1
2√
π√
1 − e−t
∫ ∞
−∞f(s)e−
(x−s)2
4(1−e−t ) ds.
13. If in Exercise 9 we take
f(x) ={
100 if |x| < 1,0 otherwise,
then the solution becomes
u(x, t) =50
√π√
1 − e−t
∫ 1
−1
e− (x−s)2
4(1−e−t ) ds.
Let z = x−s2√
1−e−t, dz = −ds
2√
1−e−t. Then
u(x, t) =50
√π√
1 − e−t2√
1 − e−t
∫ x+1
2√
1−e−t
x−1
2√
1−e−t
e−z2dz
=100√
π
∫ x+1
2√
1−e−t
x−1
2√
1−e−t
e−z2dz
= 50[erf(
x + 12√
1 − e−t
)− erf
(x − 1
2√
1 − e−t
)].
Section 11.4 The Heat Equation and Gauss’s Kernel 237
As t increases, the expression erf(
x+12√
1−e−t
)− erf
(x−1
2√
1−e−t
)approaches very quickly erf
(x+12
)−
erf(
x−12
), which tells us that the temperature approaches the limiting distribution
50[erf(
x + 12
)− erf
(x − 1
2
)].
You can verify this assertion using graphs.
17. (a) If
f(x) ={
T0 if a < x < b,0 otherwise,
then
u(x, t) =T0
2c√
πt
∫ b
a
e−(x−s)2
4c2t ds.
(b) Let z = x−s2c
√t, dz = −ds
2c√
t. Then
u(x, t) =T0
2c√
πt2c√
t
∫ x−a
2c√
t
x−b
2c√
t
e−z2dz
=T0
2
[erf(
x − a
2c√
t
)− erf
(x − b
2c√
t
)].
25. Let u2(x, t) denote the solution of the heat problem with initial temperature distributionf(x) = e−(x−1)2 . Let u(x, t) denote the solution of the problem with initial distribution e−x2
.Then, by Exercise 23, u2(x, t) = u(x − 1, t)
By (4), we have
u(x, t) =1
c√
2te−x2/(4c2t) ∗ e−x2
.
We will apply Exercise 24 with a = 14c2t and b = 1. We have
ab
a + b=
14c2t
× 11
4c2t+ 1
=1
1 + 4c2t1√
2(a + b)=
1√2( 1
4c2t+ 1)
=c√
2t√4c2t + 1
.
So
u(x, t) =1
c√
2te−x2/(4c2t) ∗ e−x2
=1
c√
2t· c
√2t√
4c2t + 1e− x2
1+4c2t
=1√
4c2t + 1e− x2
1+4c2t ,
and henceu2(x, t) =
1√4c2t + 1
e− (x−1)2
1+4c2 t .
11 The Fourier Transform and its Applications
29. Parts (a)-(c) are obvious from the definition of gt(x).(d) The total area under the graph of gt(x) and above the x-axis is
∫ ∞
−∞gt(x) dx =
1c√
2t
∫ ∞
−∞e−x2/(4c2t) dx
=2c√
t
c√
2t
∫ ∞
−∞e−z2
dz (z =x
2c√
t, dx = 2c
√t dz)
√2∫ ∞
−∞e−z2
dz =√
2π,
by (4), Sec. 7.2.(e) To find the Fourier transform of gt(x), apply (5), Sec. 7.2, with
a =1
4c2t,
1√2a
= 2c√
2t,14a
= c2t.
We get
gt(w) =1
c√
2tF(e−x2/(4c2t)
)dx
=1
c√
2t× 2c
√2te−c2tω2
= e−c2tω2.
(f) If f is an integrable and piecewise smooth function, then at its points of continuity, we have
limt→0
gt ∗ f(x) = f(x).
This is a true fact that can be proved by using properties of Gauss’s kernel. If we interpret f(x)as an initial temperature distribution in a heat problem, then the solution of this heat problem isgiven by
u(x, t) = gt ∗ f(x).
If t → 0, the temperature u(x, t) should approach the initial temperature distribution f(x). Thuslimt→0 gt ∗ f(x) = f(x).
Alternatively, we can use part (e) and argue as follows. Since
limt→0
F (gt) (ω) = limt→0
e−c2tω2= 1,
Solimt→0
F (gt ∗ f) = limt→0
F (gt)F (f) = F (f) .
You would expect that the limit of the Fourier transform be the transform of the limit function. Sotaking inverse Fourier transforms, we get limt→0 gt ∗ f(x) = f(x). (Neither one of the argumentsthat we gave is rigorous.)
Section 11.5 A Dirichlet Problem and the Poisson Integral Formula 239
Solutions to Exercises 11.5
1. To solve the Dirichlet problem in the upper half-plane with the given boundary function, we useformula (5). The solution is given by
u(x, y) =y
π
∫ ∞
−∞
f(s)(x − s)2 + y2
ds
=50y
π
∫ 1
−1
ds
(x − s)2 + y2
=50π
{tan−1
(1 + x
y
)+ tan−1
(1 − x
y
)},
where we have used Example 1 to evaluate the definite integral.
5. Appealing to (4) in Section 7.5, with y = y1, y2, y1 + y2, we find
F(Py1)(w) = e−y1|w|, F(Py2)(w) = e−y2|w|, F(Py1+y2 )(w) = e−(y1+y2)|w|.
HenceF(Py1)(w) · F(Py2)(w) = e−y1|w|e−y2|w| = e−(y1+y2)|w| = F(Py1+y2 )(w).
ButF(Py1)(w) · F(Py2)(w) = F(Py1 ∗ Py2)(w),
HenceF(Py1+y2 )(w) = F(Py1 ∗ Py2)(w);
and so Py1+y2 = Py1 ∗ Py2 .
11 The Fourier Transform and its Applications
Solutions to Exercises 11.6
1. The even extension of f(x) is
fe(x) ={
1 if −1 < x < 1,0 otherwise.
The Fourier transform of fe(x) is computed in Example 1, Sec. 7.2 (with a = 1). We have, forw ≥ 0,
Fc(f)(w) = F(fe)(w) =
√2π
sin w
w.
To write f as an inverse Fourier cosine transform, we appeal to (6). We have, for x > 0,
f(x) =
√2π
∫ ∞
0
Fc(f)(w) cos wx dw,
or2π
∫ ∞
0
sin w
wcos wx dw =
1 if 0 < x < 1,0 if x > 1,12
if x = 1.
Note that at the point x = 1, a point of discontinuity of f , the inverse Fourier transform is equal to(f(x+) + f(x−))/2.
5. The even extension of f(x) is
fe(x) ={
cos x if −2π < x < 2π,0 otherwise.
Let’s compute the Fourier cosine transform using definition (5), Sec. 7.6:
Fc(f)(w) = =
√2π
∫ 2π
0
cos x cos wx dx
=
√2π
∫ 2π
0
12
[cos(w + 1)x + cos(w − 1)x] dx
=12
√2π
[sin(w + 1)x
w + 1+
sin(w − 1)xw − 1
]∣∣∣∣2π
0
(w 6= 1)
=12
√2π
[sin 2(w + 1)π
w + 1+
sin 2(w − 1)πw − 1
](w 6= 1)
=12
√2π
[sin 2πw
w + 1+
sin 2πw
w − 1
](w 6= 1)
=
√2π
sin 2πww
w2 − 1(w 6= 1).
Also, by l’Hospital’s rule, we have
limw→0
√2π
sin 2πww
w2 − 1=
√2π,
which is the value of the cosine transform at w = 1.To write f as an inverse Fourier cosine transform, we appeal to (6). We have, for x > 0,
2π
∫ ∞
0
w
w2 − 1sin 2πw cos wx dw =
{cos x if 0 < x < 2π,0 if x > 2π.
Section 11.6 The Fourier Cosine and Sine Transforms 241
For x = 2π, the integral converges to 1/2. So
2π
∫ ∞
0
w
w2 − 1sin 2πw cos 2πw dw =
12.
9. Applying the definition of the transform and using Exercise 17, Sec. 2.6 to evaluate the integral,
Fs(e−2x)(w) =
√2π
∫ ∞
0
e−2x sin wx dx
=
√2π
e−2x
4 + w2[−w cos wx − 2 sinwx]
∣∣∣∣∞
x=0
=
√2π
w
4 + w2.
The inverse sine transform becomes
f(x) =2π
∫ ∞
0
w
4 + w2sin wx dw.
13. We have fe(x) = 11+x2 . So
Fc
(1
1 + x2
)= F
(1
1 + x2
)=√
π
2e−w (w > 0),
by Exercise 11, Sec. 7.2.
17. We have fe(x) = cosx1+x2 . So
Fc
(cos x
1 + x2
)= F
(cos x
1 + x2
)=√
π
2
(e−|w−1| + e−(w+1)
)(w > 0),
by Exercises 11 and 20(b), Sec. 7.2.
21. From the definition of the inverse transform, we have Fcf = F−1c f. So FcFcf = FcF−1
c f = f .Similarly, FsFsf = FsF−1
s f = f .
The Fourier Transform and its Applications
Solutions to Exercises 11.7
1. Fourier sine transform with respect to x:
ddt us(w, t) = −w2us(w, t) +
√2π w
=0︷ ︸︸ ︷u(0, t)
ddt
us(w, t) = −w2us(w, t).
Solve the first-order differential equation in us(w, t) and get
us(w, t) = A(w)e−w2t.
Fourier sine transform the initial condition
us(w, 0) = A(w) = Fs(f(x))(w) = T0
√2π
1 − cos bw
w.
Hence
us(w, t) =
√2π
1 − cos bw
we−w2t.
Taking inverse Fourier sine transform:
u(x, t) =2π
∫ ∞
0
1 − cos bw
we−w2t sin wx dw.
5. If you Fourier cosine the equations (1) and (2), using the Neumann type condition
∂u
∂x(0, t) = 0,
you will get
ddt uc(w, t) = c2
[− w2uc(w, t) −
√2π
=0︷ ︸︸ ︷d
dxu(0, t)
]
ddt uc(w, t) = −c2w2uc(w, t).
Solve the first-order differential equation in uc(w, t) and get
uc(w, t) = A(w)e−c2w2t.
Fourier cosine transform the initial condition
uc(w, 0) = A(w) = Fc(f)(w).
Henceus(w, t) = Fc(f)(w)e−c2w2t.
Taking inverse Fourier cosine transform:
u(x, t) =
√2π
∫ ∞
0
Fc(f)(w)e−c2w2t cos wx dw.
Section 11.7 Problems Involving Semi-Infinite Intervals 243
9. (a) Taking the sine transform of the heat equation (1) and using u(0, t) = T0 for t > 0, we get
d
dtus(w, t) = c2
[− w2us(w, t) +
√2π
wu(0, t)];
ord
dtus(w, t) + c2ω2us(w, t) = c2
√2π
wT0.
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 differential equation can be guessed easily: us(w, t) =
√2π
T0w . The
general solution of the homogeneous differential equation:
d
dtus(w, t) + c2ω2us(w, t) = 0
is us(w, t) = A(w)e−c2w2t. So the general solution of the nonhomogeneous differential equation is
us(w, t) = A(w)e−c2w2t
√2π
T0
w.
Using us(w, 0) = A(w)√
2π
T0w
= 0, we find A(w) = −√
2π
T0w
. So
us(w t) =
√2π
T0
w−√
2π
T0
we−c2w2t.
Taking inverse sine transforms, we find
u(x, t) =2π
∫ ∞
0
(T0
w− T0
we−c2w2t
)sin wx dw
= T0
=sgn(x)=1︷ ︸︸ ︷2π
∫ ∞
0
sinwx
wdw−2T0
π
∫ ∞
0
sin wx
we−c2w2t dw
= T0 −2T0
π
∫ ∞
0
sin wx
we−c2w2t dw
13. Proceed as in Exercise 11 using the Fourier sine transform instead of the cosine transform andthe condition u(x 0) = 0 instead of uy(x, 0) = 0. This yields
d2
dx2 us(x, w) − w2us(x, w) +√
2π
=0︷ ︸︸ ︷u(x, 0) = 0
d2
dx2 us(x, w) = w2us(x, w).
The general solution isus(x, w) = A(w) cosh wx + B(w) sinh wx.
Using
us(0, w) = 0 and us(1, w) = Fs(e−y) =
√2π
w
1 + w2,
we get
A(w) = 0 and B(w) =
√2π
w
1 + w2·
1sinh w
.
The Fourier Transform and its Applications
Hence
us(x, w) =
√2π
w
1 + w2
sinh wx
sinh w.
Taking inverse sine transforms:
u(x, y) =2π
∫ ∞
0
w
1 + w2
sinh wx
sinh wsin wy dw.