Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The...

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11 The Fourier Transform and its Applications Solutions to Exercises 11.1 1. We have f (w) = 1 2π 1 -1 xe -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 1 w 2 sin wx - x w cos wx 1 0 = -i 2 π sin w - w cos w w 2 . 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|) cos wx dx - i 2π =0 1 -1 (1 -|x|) sin wx 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 = - 2 π cos wx w 2 1 0 = 2 π 1 - cos w w 2 .

Transcript of Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The...

Page 1: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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√

∫ 1

0

x sin wx dx

=−2i√

[ 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

= −√

cos wx

w2

∣∣∣∣∣

1

0

=

√2π

1 − cos w

w2.

Page 2: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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;

∫ ∞

−∞

cos wπ2

1 − w2cos wx dx =

cos x if |x| < π2

0 if |x| ≥ π2;

∫ ∞

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.

Page 3: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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) =

∫ ∞

−∞

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) =

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

∫ ∞

−∞

x

(1 + x2)2eiWx dx,

Page 4: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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

IR = −iwe−|w|

2.

The formula works for w ≥ 0 since it defines an odd function.

Page 5: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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|) =√

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

].

Page 6: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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) =

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)

).

Page 7: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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)

).

Page 8: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

Section 11.2 Operational Properties 231

Specializing to α = 1 and f(x) = e−|x|, f(w) =√

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

).

Page 9: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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.

Page 10: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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) .

Page 11: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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)];

Page 12: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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.

Page 13: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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

)].

Page 14: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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 .

Page 15: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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.)

Page 16: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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 .

Page 17: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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,

∫ ∞

0

w

w2 − 1sin 2πw cos wx dw =

{cos x if 0 < x < 2π,0 if x > 2π.

Page 18: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

Section 11.6 The Fourier Cosine and Sine Transforms 241

For x = 2π, the integral converges to 1/2. So

∫ ∞

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 .

Page 19: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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.

Page 20: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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)√

T0w

= 0, we find A(w) = −√

T0w

. So

us(w t) =

√2π

T0

w−√

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) +√

=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

.

Page 21: Solutions to Exercises 11 - faculty.missouri.eduasmarn/complexpde/chapter11.pdfSection 11.1 The Fourier Transform 227 which is the desired integral. So let us compute the contour integral,

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.