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15 Integral Transform Method

Exercises 15.1

1. (a) The result follows by letting = u2 or u =

in erf(

t ) =2

t0

eu2

du.

(b) Using {t1/2} =

s1/2and the first translation theorem, it follows from the convolution theorem that

erf(

t)

=1

t0

e

d

=

1

{1}

t1/2et

=1

1

s

t1/2

ss+1

=

1

1

s

s + 1 =1

ss + 1 .

2. Since erfc(

t ) = 1 erf(t ) we have

erfc(

t )

= {1}

erf(

t )

=1

s 1

s

s + 1=

1

s

1 1

s + 1

.

3. By the first translation theorem,

et erf(

t )

=

erf(

t )

ss1

=1

s

s + 1

ss1

=1

s (s

1)

.

4. By the first translation theorem and the result of Problem 2,

et erfc(

t )

=

erfc(

t )

ss1=

1

s 1

s

s + 1

ss1

=1

s 1 1

s (s 1)

=

s 1

s (s 1) =

s 1s (

s + 1)(

s 1) =1

s (

s + 1).

5. From table entry 3 and the first translation theorem we have

eGt/Cerfx2RCt = eGt/C1 erfcx2RCt =

eGt/C

eGt/Cerfc

x

2

RC

t

=1

s + G/C e

xRC

s

s

ss+G/C

=1

s + G/C e

xRC

s+G/C

s + G/C=

C

Cs + G

1 ex

RCs+RG

.

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Exercises 15.1

6. We first compute

sinh a

s

s sinh

s=

eas ea

s

s(es es) =

e(a1)s e(a+1)

s

s(1 e2s )

=

e(a1)s

s 1 + e2s + e4s + e(a+1)s

s 1 + e2s + e4s + =

e(1a)

s

s+

e(3a)s

s+

e(5a)s

s+

e(1+a)s

s+

e(3+a)s

s+

e(5+a)s

s+

=n=0

e(2n+1a)

s

s e

(2n+1+a)s

s

.

Then

sinh a

s

s sinhs =

n=0

e(2n+1a)s

s e(2n+1+a)s

s =

n=0

erfc

2n + 1 a

2

t

erfc

2n + 1 + a

2

t

=n=0

1 erf

2n + 1 a

2

t

1 erf

2n + 1 + a

2

t

=n=0

erf

2n + 1 + a

2

t

erf

2n + 1 a

2

t

.

7. Taking the Laplace transform of both sides of the equation we obtain

{y(t)} = {1} t0

y()t d

Y(s) =

1

s Y(s)

s

s +

sY(s) =

1

s

Y(s) =1

s (

s +

).

Thus

y(t) = 1

s (

s +

) = et erfc(

t ). By entry 5 in the table

8. Using entries 3 and 5 in the table, we haveeabeb2t erfc

b

t +a

2

t

+ erfc

a

2

t

=

eabeb2t erfc

b

t +a

2

t

+

a

2

t

= eas

s (

s + b)+

eas

s

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-10 -5 5 10x

-2

-1

1

2

y

erfHxLerfcHxL

Exercises 15.1

= eas

1

s 1

s (

s + b)

= ea

s

1

s

s

s (

s + b)

= eas

s + b s

s (s + b) =bea

s

s (s + b).

9.

ba

eu2

du =

0a

eu2

du +

b0

eu2

du =

b0

eu2

dua

0

eu2

du

=

2erf(b)

2erf(a) =

2[erf(b) erf(a)]

10. Since f(x) = ex2

is an even function, aa

eu2

du = 2

a0

eu2

du.

Therefore,

aa

eu2

du = erf(a).

11. The function erf(x) is symmetric with respect to the origin,

while erfc(x) appears to be symmetric with respect to the point

(0, 1). From the graph it appears that limx erf(x) = 1and limx erfc(x) = 2.

Exercises 15.2

1. The boundary-value problem is

a22u

x2=

2u

t2, 0 < x < L, t > 0,

u(0, t) = 0, u(L, t) = 0, t > 0,

u(x, 0) = A sin

Lx,

u

t

t=0

= 0.

Transforming the partial differential equation gives

d2U

dx2 s

a

2U = s

a2A sin

Lx.

Using undetermined coefficients we obtain

U(x, s) = c1 coshs

ax + c2 sinh

s

ax +

As

s2 + a22/L2sin

Lx.

The transformed boundary conditions, U(0, s) = 0, U(L, s) = 0 give in turn c1 = 0 and c2 = 0. Therefore

U(x, s) =As

s2 + a22/L2sin

Lx

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Exercises 15.2

and

u(x, t) = A

s

s2 + a22/L2

sin

Lx = A cos

a

Lt sin

Lx.

2. The transformed equation isd2U

dx2 s2U = 2sin x 4sin3xand so

U(x, s) = c1 cosh sx + c2 sinh sx +2

s2 + 2sin x +

4

s2 + 92sin3x.

The transformed boundary conditions, U(0, s) = 0 and U(1, s) = 0 give c1 = 0 and c2 = 0. Thus

U(x, s) =2

s2 + 2sin x +

4

s2 + 92sin3x

and

u(x, t) = 2

1

s2 + 2

sin x + 4

1

s2 + 92

sin3x

=2

sin t sin x +

4

3sin3t sin3x.

3. The solution of

a2d2U

dx2 s2U = 0

is in this case

U(x, s) = c1e(x/a)s + c2e(x/a)s.

Since limx u(x, t) = 0 we have limx U(x, s) = 0. Thus c2 = 0 and

U(x, s) = c1e(x/a)s.

If {u(0, t)} = {f(t)} = F(s) then U(0, s) = F(s). From this we have c1 = F(s) and

U(x, s) = F(s)e(x/a)s

.

Hence, by the second translation theorem,

u(x, t) = f

t xa

t x

a

.

4. Expressing f(t) in the form (sin t)[1 (t 1)] and using the result of Problem 3 we find

u(x, t) = f

t xa

t x

a

= sin

t x

a

1

t x

a 1

t x

a

= sin t

x

a t x

a tx

a t x

a 1

= sin

t xa

t x

a

t x

a 1

Now

t xa

t x

a 1

=

0, 0 t < x/a1, x/a t x/a + 10, t > x/a + 1

=

0, x < a(t 1) or x > at1, a(t 1) x at

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Exercises 15.2

so

u(x, t) =

0, x < a(t 1) or x > atsin (t x/a), a(t 1) x at.

The graph is shown for t > 1.

5. We use

U(x, s) = c1e(x/a)s g

s3.

Now

{u(0, t)} = U(0, s) = As2 + 2

and so

U(0, s) = c1 gs3

=A

s2 + 2or c1 =

g

s3+

A

s2 + 2.

Therefore

U(x, s) =A

s2 + 2e(x/a)s +

g

s3e(x/a)s g

s3

and

u(x, t) = A

e(x/a)s

s2 + 2

+ g

e(x/a)s

s3

g

1

s3

= A sin

t xa

t x

a

+

1

2g

t xa

2 t x

a

1

2gt2.

6. Transforming the partial differential equation gives

d2U

dx2 s2U =

s2 + 2 sin x.


U(x, s) = c1 cosh sx + c2 sinh sx +

(s2 + 2)(s2 + 2)sin x.

The transformed boundary conditions U(0, s) = 0 and U(1, s) = 0 give, in turn, c1 = 0 and c2 = 0. Therefore

U(x, s) =

(s2 + 2)(s2 + 2)sin x

and

u(x, t) = sin x 1(s2 + 2)(s2 + 2)=

2 2 sin x

1

s2 + 2 1

s2 + 2

=

(2 2) sin t sin x 1

2 2 sin t sin x.

7. We use

U(x, s) = c1 coshs

ax +2 sinh

s

ax.

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Exercises 15.2

Now U(0, s) = 0 implies c1 = 0, so U(x, s) = c2 sinh(s/a)x. The condition EdU

dx

x=L

= F0 then yields

c2 = F0a/Es cosh(s/a)L and so

U(x, s) =aF0Es

sinh(s/a)x

cosh(s/a)L=

aF0Es

e(s/a)x e(s/a)xe(s/a)L + e(s/a)L

=aF0Es

e(s/a)(xL) e(s/a)(x+L)1 + e2sL/a

=aF0

E

e(s/a)(Lx)

s e

(s/a)(3Lx)

s+

e(s/a)(5Lx)

s

aF0E

e(s/a)(L+x)

s e

(s/a)(3L+x)

s+

e(s/a)(5L+x)

s

=aF0

E

n=0

(1)n

e(s/a)(2nL+Lx)

s e

(s/a)(2nL+L+x)

s

and

u(x, t) = aF0

En=0

(1)n e(s/a)(2nL+Lx)s

e(s/a)(2nL+L+x)s

=aF0

E

n=0

(1)n

t 2nL + L xa

t 2nL + L x

a

t 2nL + L + xa

t 2nL + L + x

a

.

8. We use

U(x, s) = c1e(x/a)s + c2e(x/a)s v0

s2.

Now limxdU

dx = 0 implies c2 = 0, and U(0, s) = 0 then gives c1 = v0/s2. Hence

U(x, s) =v0s2

e(x/a)s v0s2

and

u(x, t) = v0

t x

a

t x

a

v0t.


d2U

dx2 s2U = sxex.


U(x, s) = c1esx + c2esx 2s

(s2

1)2

ex +s

s2

1

xex.

The transformed boundary conditions limx U(x, s) = 0 and U(0, s) = 0 give, in turn, c2 = 0 andc1 = 2s/(s2 1)2. Therefore

U(x, s) =2s

(s2 1)2 esx 2s

(s2 1)2 ex +

s

s2 1 xex.

From entries (13) and (26) in the table we obtain

u(x, t) =

2s

(s2 1)2 esx 2s

(s2 1)2 ex +

s

s2 1 xex

= 2(t x)sinh(t x) (t x) tex sinh t + xex cosh t.

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0 2 4 6

8 10x

510

15

t

0

25

50

75

100

uHx,tL

2 4

5 00 10 00 15 00 2 00 0t

20

40

60

80

100

uH10,tL

Exercises 15.2

10. We use

U(x, s) = c1exs + c2exs +

s

s2 1 ex.

Now limx u(x, t) = 0 implies limx U(x, s) = 0, so we define c2 = 0. Then

U(x, s) = c1exs +s

s2 1 ex.Finally, U(0, s) = 1/s gives c1 = 1/s s/(s2 1). Thus

U(x, s) =1

s s

s2 1 exs +

s

s2 1 ex

and

u(x, t) =

s

s2 1 e(x/a)s

+

s

s2 1

ex

= cosh

t xa

t x

a

+ ex cosh t.

11. (a) We useU(x, s) = c1e

s/k x + c2es/k x.

Now limx u(x, t) = 0 implies limx U(x, s) = 0, so we define c2 = 0. Then

U(x, s) = c1es/k x.

Finally, from U(0, s) = u0/s we obtain c1 = u0/s. Thus

U(x, s) = u0es/k x

s

and

u(x, t) = u0 es/k x

s = u0 e(x/

k )s

s = u0 erfc x2kt .

Since erfc(0) = 1,

limt

u(x, t) = limt

u0 erfc(x/2

kt ) = u0.

(b)

12. (a) Transforming the partial differential equation and using the initial condition gives

kd2U

dx2 sU = 0.

Since the domain of the variable x is an infinite interval we write the general solution of this differential

equation as

U(x, s) = c1es/k x + c2e

s/k x.

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0 2 4 6 8 10x

510

15

t

0

10

20

30

40

uHx,tL

2 4

5 00 1 00 0 1 50 0 2 00 0t

100

200

300

400

500

uH10,tL

Exercises 15.2

Transforming the boundary conditions gives U(0, s) = A/s and limx

U(x, s) = 0. Hence we find c2 = 0

and c1 = A

k/s

s . From

U(x, s) = A

kes/k x

s

s

we see that

u(x, t) = A

k

es/k x

s

s

.

With the identification a = x/

k it follows from entry 49 of the table in Appendix III that

u(x, t) = A

k

2

t

ex

2/4kt xk

x/2

kt

= 2A

kt

ex

2/4kt Ax erfc

x/2

kt

.

Since erfc(0) = 1,

limt

u(x, t) = limt

2Akt

ex

2/4kt Ax erfc x2

kt

= .

(b)

13. We use

U(x, s) = c1es x + c2e

sx + u1s

.

The condition limx u(x, t) = u1 implies limx U(x, s) = u1/s, so we define c2 = 0. Then

U(x, s) = c1es x +

u1s

.

From U(0, s) = u0/s we obtain c1 = (u0 u1)/s. Thus

U(x, s) = (u0 u1) es x

s+

u1s

and

u(x, t) = (u0 u1)

exs

s

+ u1

1

s= (u0 u1) erfc

x

2

t+ u1.

14. We use


sx +

u1x

s.

The condition limx u(x, t)/x = u1 implies limx U(x, s)/x = u1/s, so we define c2 = 0. Then

U(x, s) = c1es x +

u1x

s.

From U(0, s) = u0/s we obtain c1 = u0/s. Hence

U(x, s) = u0e

s x

s+

u1x

s

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Exercises 15.2

and

u(x, t) = u0

ex

s

s

+ u1x

1

s

= u0 erfc

x

2

t

+ u1x.

15. We use

U(x, s) = c1es x + c2esx +u

0s .

The condition limx u(x, t) = u0 implies limx U(x, s) = u0/s, so we define c2 = 0. Then

U(x, s) = c1es x +

u0s

.

The transform of the remaining boundary conditions gives

dU

dx

x=0

= U(0, s).

This condition yields c1 = u0/s(s + 1). Thus

U(x, s) =

u0

es x

s(s + 1)+

u0

sand

u(x, t) = u0

exs

s(

s + 1)

+ u0

1

s

= u0ex+t erfc

t +

x

2

t

u0 erfc

x

2

t

+ u0 By (5) in the table in 15.1.

16. We use


s x.

The condition limx u(x, t) = 0 implies limx U(x, s) = 0, so we define c2 = 0. Hence

U(x, s) = c1es x.

The remaining boundary condition transforms into

dU

dx

x=0

= U(0, s) 50s

.

This condition gives c1 = 50/s(

s + 1). Therefore

U(x, s) = 50e

s x

s(

s + 1)

and

u(x, t) = 50 ex

s

s(s + 1) = 50ex+t erfc

t +

x

2t + 50 erfcx

2t .17. We use


s x.

The condition limx u(x, t) = 0 implies limx U(x, s) = 0, so we define c2 = 0. Hence

U(x, s) = c1es x.

The transform of u(0, t) = f(t) is U(0, s) = F(s). Therefore

U(x, s) = F(s)es x

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Exercises 15.2

and

u(x, t) =

F(s)exs

=x

2

t0

f(t )ex2/43/2

d.

18. We use

U(x, s) = c1es x

+ c2e

s x

.

The condition limx u(x, t) = 0 implies limx U(x, s) = 0, so we define c2 = 0. Then U(x, s) = c1esx.

The transform of the remaining boundary condition gives

dU

dx

x=0

= F(s)

where F(s) = {f(t)}. This condition yields c1 = F(s)/s . Thus

U(x, s) = F(s)e

s x

s

.

Using entry (44) of the table and the convolution theorem we obtain

u(x, t) =

F(s) es x

s

=

1

t0

f()ex2/4(t)

t d.


d2U

dx2 sU = 60.

Using undetermied coefficients we obtain


s x +

60

s.

The condition limx u(x, t) = 60 implies limx U(x, s) = 60/s, so we define c2 = 0. The transform of the

remaining boundary condition gives

U(0, s) =60

s+

40

se2s.

This condition yields c1 =40

se2s. Thus

U(x, s) =60

s+ 40e2s

es x

s.

Using entry (46) of the table in Appendix III and the second translation theorem we obtain

u(x, t) =

60

s+ 40e2s

es x

s

= 60 + 40erfc

x

2

t 2

(t 2).

20. The solution of the transformed equationd2U

dx2 sU = 100

by undetermined coefficients is

U(x, s) = c1esx + c2e

s x +100

s.

From the fact that limx

U(x, s) = 100/s we see that c1 = 0. Thus

U(x, s) = c2es x +

100

s. (1)

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Exercises 15.2

Now the transform of the boundary condition at x = 0 is

U(0, s) = 20

1

s 1

ses

.

It follows from (1) that

20s 20

ses = c2 + 100

sor c2 = 80

s 20

ses

and so

U(x, s) =

80

s 20

ses

e

sx +

100

s

=100

s 80

se

s x 20

se

s xes.

Thus

u(x, t) = 100

1

s

80

e

s x

s

20

e

s x

ses

= 100 80 erfcx/2t 20 erfcx/2t 1 (t 1).21. Transforming the partial differential equation gives

d2U

dx2 sU = 0

and so


s x.

The condition limx u(x, t) = 0 implies limx U(x, s) = 0, so we define c1 = 0. The transform of the

remaining boundary condition givesdU

dx x=1=100

s U(1, s).

This condition yields

c2

s es =

100

s c2e

s

from which it follows that

c2 =100

s(

s + 1)e

s.

Thus

U(x, s) = 100e(1x)

s

s(

s + 1).

Using entry (49) of the table in Appendix III we obtain

u(x, t) = 100e(1x)s

s(

s + 1)

= 100

e1x+t erfc

t + 1 xt

+ erfc

1 x2

t

.


kd2U

dx2 sU = r

s.

Using undetermined coefficients we obtian

U(x, s) = c1es/k x + c2e

s/k x +

r

s2.

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Exercises 15.2

The condition limx

u

x= 0 implies lim

xdU

dx= 0, so we define c2 = 0. The transform of the remaining boundary

condition gives U(0, s) = 0. This condition yields c1 = r/s2. Thus

U(x, s) = r 1s2 es/k x

s2 .Using entries (3) and (46) of the table in Appendix III and the convolution theorem we obtain

u(x, t) = r

1

s2 1

s e

s/k x

s

= rt r

t0

erfc

x

2

k

d.

23. The solution ofd2U

dx2 sU = u0 u0 sin

Lx

is

U(x, s) = c1 cosh(

s x) + c2 sinh(

s x) +u0s

+u0

s + 2/L2sin

Lx.

The transformed boundary conditions U(0, s) = u0/s and U(L, s) = u0/s give, in turn, c1 = 0 and c2 = 0.

Therefore

U(x, s) =u0s

+u0

s + 2/L2sin

Lx

and

u(x, t) = u0

1

s

+ u0

1

s + 2/L2

sin

Lx = u0 + u0e

2t/L2 sin

Lx.

24. The transform of the partial differential equation is

kd2U

dx2 hU + h um

s= sU u0

or

kd2U

dx2 (h + s)U = h um

s u0.

By undetermined coefficients we find

U(x, s) = c1e

(h+s)/k x + c2e

(h+s)/k x +hum + u0s

s(s + h).

The transformed boundary conditions are U(0, s) = 0 and U(L, s) = 0. These conditions imply c1 = 0 and

c2 = 0. By partial fractions we then get

U(x, s) =hum + u0s

s(s + h)=

ums um

s + h+

u0s + h

.

Therefore,

u(x, t) = um

1

s

um

1

s + h

+ u0

1

s + h

= um umeht + u0eht.

25. We use

U(x, s) = c1 cosh

s

kx + c2 sinh

s

kx +

u0s

.

The transformed boundary conditionsdU

dx

x=0

= 0 and U(1, s) = 0 give, in turn, c2 = 0 and

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Exercises 15.2

c1 = u0/s cosh

s/k . Therefore

U(x, s) =u0s u0 cosh

s/k x

s cosh

s/k=

u0s u0 e

s/k x + e

s/k x

s(es/k + e

s/k )

= u0s u0 es/k (x1) + es/k (x+1)

s(1 + e2s/k )

=u0s u0

es/k (1x)

s e

s/k (3x)

s+

es/k (5x)

s

u0

es/k (1+x)

s e

s/k (3+x)

s+

es/k (5+x)

s

=u0s u0

n=0

(1)n

e(2n+1x)s/k

s+

e(2n+1+x)s/k

s

and

u(x, t) = u0

1

s

u0

n=0

(1)n

e(2n+1x)s/k

s

e(2n+1+x)

s/k

s

= u0 u0n=0

(1)n

erfc

2n + 1 x

2

kt

erfc

2n + 1 + x

2

kt

.

26. We use

c(x, s) = c1 cosh

s

Dx + c2 sinh

s

Dx.

The transform of the two boundary conditions are c(0, s) = c0/s and c(1, s) = c0/s. From these conditions we

obtain c1 = c0/s and

c2 = c0(1 coshs/D )/s sinhs/D .Therefore

c(x, s) = c0

cosh

s/D x

s+

(1 cosh

s/D )

s sinh

s/Dsinh

s/D x

= c0

sinh

s/D (1 x)

s sinh

s/D+

sin

s/Dx

s sinh

s/D

= c0

es/D (1x) e

s/D (1x)

s(es/D e

s/D )

+es/Dx e

s/D x

s(es/D e

s/D )

= c0es/D x es/D (2x)

s(1 e2s/D )

+es/D (x1) es/D (x+1)

s(1 e2s/D )

= c0(e

s/Dx e

s/D (2x))

s

1 + e2

s/D + e4

s/D +

+ c0(es/D (x1) e

s/D (x+1)

s

1 + e2

s/D + e4

s/D +

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Exercises 15.2

= c0

n=0

e(2n+x)

s/D

s e

(2n+2x)s/D

s

+ c0

n=0

e(2n+1x)s/D

s e(2n+1+x)

s/D

s and

c(x, t) = c0

n=0

e

(2n+x)D

s

s

e

(2n+2x)D

s

s

+ c0

n=0

e

(2n+1x)D

s

s

e

(2n+1+x)D

s

s

= c0

n=0

erfc

2n + x

2

Dt

erfc

2n + 2 x

2

Dt

+ c0

n=0

erfc

2n + 1 x

2

Dt

erfc

2n + 1 + x

2

Dt

.

Now using erfc(x) = 1 erf(x) we get

c(x, t) = c0

n=0

erf

2n + 2 x

2

Dt

erf

2n + x

2

Dt

+ c0

n=0

erf

2n + 1 + x

2

Dt

erf

2n + 1 x

2

Dt

.

27. We use

U(x, s) = c1eRCs+RGx + c2eRCs+RG + Cu0Cs + G

.

The condition limx u/x = 0 implies limx dU/dx = 0, so we define c2 = 0. Applying U(0, s) = 0 to

U(x, s) = c1eRCsRGx +

Cu0Cs + G

gives c1 = Cu0/(Cs + G). Therefore

U(x, s) = Cu0 eRCs+RGx

Cs + G+

Cu0Cs + G

and

u(x, t) = u0 1s + G/C u0 ex

RCs+G/C

s + G/C = u0e

Gt/C u0eGt/Cerfc

x

RC

2

t

= u0eGt/C

1 erfc

x

2

RC

t

= u0eGt/Cerf

x

2

RC

t

.

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x

cHx,tL

t=0.1

t=0.5

t=1t=2t=5

Exercises 15.2

28. We use

U(x, s) = c1es+hx + c2e

s+hx.

The condition limx u(x, t) = 0 implies limx U(x, s) = 0, so we take c2 = 0. Therefore

U(x, s) = c1es+hx.

The Laplace transform of u(0, t) = u0 is U(0, s) = u0/s and so

U(x, s) = u0e

s+hx

s

and

u(x, t) = u0

e

s+hx

s

= u0

1

se

s+hx

.

From the first translation theorem,e

s+hx

= eht {ex

s} = eht x

2

t3ex

2/4t.

Thus, from the convolution theorem we obtain

u(x, s) =u0x

2

t0

ehx2/4

3/2d.

29. (a) Letting C(x, s) = {c(x, t)} we obtaind2C

dx2 s

kC = 0 subject to

dC

dx

x=0

= A.

The solution of this initial-value problem is

C(x, s) = A

ke(x/

k )s

s

,

so that

c(x, t) = Ak

tex2/4kt.

(b)

(c)

0

c(x, t)dx = Ak erfx

2

kt

0

= Ak(1 0) = Ak

30. (a) We use

U(x, s) = c1e(s/a)x + c2e(s/a)x +

v20F0(a2 v20)s2

e(s/v0)x.

The condition limx u(x, t) = 0 implies limx U(x, s) = 0, so we must define c2 = 0. Consequently

U(x, s) = c1e(s/a)x +

v20 F0(a2 v20 )s2

e(s/v0)x.

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Exercises 15.3

The remaining boundary condition transforms into U(0, s) = 0. From this we find

c1 = v20F0/(a2 v20)s2.

Therefore, by the second translation theorem

U(x, s) = v20F0(a2 v20)s2

e(s/a)x +v20F0

(a2 v20 )s2e(s/v0)x

and

u(x, t) =v20 F0

a2 v20

e(x/v0)s

s2

e(x/a)s

s2

=v20 F0

a2 v20

t x

v0

t x

v0

t xa

t x

a

.

(b) In the case when v0 = a the solution of the transformed equation is

U(x, s) = c1e(s/a)x + c2e(s/a)x

F0

2as

xe(s/a)x.

The usual analysis then leads to c1 = 0 and c2 = 0. Therefore

U(x, s) = F02as

xe(s/a)x

and

u(x, t) = xF02a

e(x/a)s

s

= xF0

2a

t x

a

.

Exercises 15.3

1. From formulas (5) and (6) in the text,

A() =

01

(1)cos xdx +1

0

(2)cos xdx = sin

+ 2sin

=

sin

and

B() =

01

(1) sin xdx +1

0

(2)sin xdx

=1 cos

2 cos 1

=

3(1 cos )

.

Hencef(x) =

1

0

sin cos x + 3(1 cos )sin x

d.


A() =

2

4cos xdx = 4sin2 sin

and

B() =

2

4sin xdx = 4cos cos2

.

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Exercises 15.3

Hence

f(x) =4

0

(sin2 sin )cos x + (cos cos2)sin x

d

=4

0

sin2 cos x cos2 sin x sin cos x + cos sin x

d

=4

0

sin (2 x) sin ( x)

d.


A() =

30

x cos xdx =x sin x

30

1

30

sin xdx

=3sin3

+

cos x

2

30

=3 sin3 + cos 3 1

2

and

B() =3

0

x sin xdx = x cos x

30

+1

30

cos xdx

= 3cos3

+sin x

2

30

=sin3 3 cos3

2.

Hence

f(x) =1

0

(3 sin3 + cos 3 1) cos x + (sin 3 3 cos3)sin x2

d

=1

0

3(sin3 cos x cos3 sin x) + cos 3 cos x + sin 3 sin x cos x2

d

= 1

0

3 sin (3 x) + cos (3 x) cos x2

d.


A() =

f(x)cos xdx

=

0

0 cos xdx +

0

sin x cos xdx +

0 cos xdx

=1

2

0

[sin(1 + )x + sin(1 )x] dx

=1

2

cos(1 + )x

1 + cos(1 )x

1

0

= 12

cos(1 + ) 1

1 + +

cos(1 ) 11

= 12

cos(1 + ) cos(1 + ) + cos(1 ) + cos(1 ) 2

1 2

=1 + cos

1 2 ,

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Exercises 15.3

and

B() =

0

sin x sin xdx =1

2

0

[cos(1 )x cos(1 + )] dx

=1

2 sin(1 )

1

sin(1 + )

1 + =sin

1

2.

Hence

f(x) =1

0

cos x + cos x cos + sin x sin

1 2 d

=1

0

cos x + cos (x )1 2 d.

5. From formula (5) in the text,

A() =

0

ex cos xdx.

Recall {cos kt} = s/(s2 + k2). If we set s = 1 and k = we obtain

A() =1

1 + 2.

Now

B() =

0

ex sin xdx.

Recall {sin kt} = k/(s2 + k2). If we set s = 1 and k = we obtainB() =

1 + 2.

Hence

f(x) =1

0

cos x + sin x

1 + 2d.


A() =11

ex cos xdx

=e(cos + sin ) e1(cos sin )

1 + 2

=2(sinh1) cos 2(cosh 1) sin

1 + 2

and

B() =

11

ex sin xdx

=e(sin cos ) e1( sin cos )

1 + 2

=2(cosh 1) sin 2(sinh 1) cos

1 + 2.

Hence

f(x) =1

0

[A()cos x + B()sin x] d.

7. The function is odd. Thus from formula (11) in the text

B() = 5

10

sin xdx =5(1 cos )

.

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Exercises 15.3

Hence from formula (10) in the text,

f(x) =10

0

(1 cos )sin x

d.

8. The function is even. Thus from formula (9) in the text

A() =

21

cos xdx =

sin2 sin

.


f(x) = 2

0

(sin2 sin )cos x

d.

9. The function is even. Thus from formula (9) in the text

A() =

0

x cos xdx =x sin x

0

1

0

sin xdx

=

sin

+

1

2 cos x

0 =

sin + cos

1

2 .

Hence from formula (8) in the text

f(x) =2

0

( sin + cos 1) cos x2

d.


B() =

0

x sin xdx = x cos x

0

+1

0

cos xdx

= cos

+1

2sin x

0

= cos + sin

2.


f(x) =2

0

( cos + sin )sin x2

d.


B() =

0

(ex sin x)sin xdx

=1

2

0

ex[cos(1 )x cos(1 + )x] dx

=1

2

0

ex cos(1 )x dx 12

0

ex cos(1 + )x, dx.

Now recall

{cos kt} =

0

est cos ktdt = s/(s2 + k2).

If we set s = 1, and in turn, k = 1 and then k = 1 + , we obtain

B() =1

2

1

1 + (1 )2 1

2

1

1 + (1 + )2=

1

2

(1 + )2 (1 )2[1 + (1 )2][1 + (1 + )2] .

Simplifying the last expression gives

B() =2

4 + 4.

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Exercises 15.3


f(x) =4

0

sin x

4 + 4d.


B() =

0

xex sin xdx.

Now recall

{t sin kt} = dds

{sin kt} = 2ks/(s2 + k2)2.If we set s = 1 and k = we obtain

B() =2

(1 + 2)2.


f(x) =4

0

sin x

(1 + 2)2d.

13. For the cosine integral,

A() =

0

ekx cos xdx =k

k2 + 2.

Hence

f(x) =2

0

k cos x

k2 + 2=

2k

0

cos x

k2 + 2d.

For the sine integral,

B() =

0

ekx sin xdx =

k2 + 2.

Hence

f(x) =2

0

sin x

k2 + 2d.

14. From Problem 13 the cosine and sine integral representations of ekx, k > 0, are respectively,

ekx =2k

0

cos x

k2 + 2d and ekx =

2

0

sin x

k2 + 2d.

Hence, the cosine integral representation of f(x) = ex e3x is

ex e3x = 2

0

cos x

1 + 2d 2(3)

0

cos x

9 + 2d =

4

0

3 2(1 + 2) (9 + 2)

cos xd.

The sine integral representation of f is

ex e3x = 2

0

sin x

1 + 2d 2

0

sin x

9 + 2d =

16

0

sin x

(1 + 2) (9 + 2)d.


A() =

0

xe2x cos x dx.

But we know

{t cos kt} = dds

s

(s2 + k2)=

(s2 k2)(s2 + k2)2

.

If we set s = 2 and k = we obtain

A() =4 2

(4 + 2)2.

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Exercises 15.3

Hence

f(x) =2

0

(4 2)cos x(4 + 2)2

d.


B() = 0

xe2x sin x dx.

From Problem 12, we know

{t sin kt} = 2ks(s2 + k2)2

.

If we set s = 2 and k = we obtain

B() =4

(4 + 2)2.

Hence

f(x) =8

0

sin x

(4 + 2)2d.


A() =

0

ex cos x cos xdx

=1

2

0

ex[cos(1 + )x + cos(1 )x] dx

=1

2

1

1 + (1 + )2+

1

2

1

1 + (1 )2

=1

2

1 + (1 )2 + 1 + (1 + )2[1 + (1 + )2][1 + (1 )2]

= 2 + 2

4 + 4.

Hence

f(x) =2

0

(2 + 2)cos x

4 + 4d.


B() =

0

ex cos x sin xdx

=1

2

0

ex[sin(1 + )x sin(1 )x] dx

= 12

1 + 1 + (1 + )2

12

1 1 + (1 )2

=1

2

(1 + )[1 + (1 )2] (1 )[1 + (1 + )2]

[1 + (1 + )2][1 + (1 )2]

=3

4 + 4.

Hence

f(x) =2

0

3 sin x

4 + 4d.

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Exercises 15.4

17. By formula (8) in the text

f(x) = 2

0

e cos xd =2

1

1 + x2, x > 0.

18. From the formula for sine integral of f(x) we have

f(x) =2

0

0

f(x)sin xdx

sin xdx

=2

10

1 sin xd +

1

0 sin xd

=2

( cos x)x

10

=2

1 cos xx

.

19. (a) From formula (7) in the text with x = 2, we have

1

2

=2

0

sin cos

d =1

0

sin2

d.

If we let = x we obtain 0

sin2x

xdx =

2.

(b) If we now let 2x = kt where k > 0, then dx = (k/2)dt and the integral in part (a) becomes0

sin kt

kt/2(k/2) dt =

0

sin kt

tdt =

2.

20. With f(x) = e|x|, formula (16) in the text is

C() =

e|x|eixdx =

e|x| cos xdx + i

e|x| sin xdx.

The imaginary part in the last line is zero since the integrand is an odd function of x. Therefore,

C() =

e|x| cos xdx = 2

0

ex cos xdx =2

1 + 2

and so from formula (15) in the text,

f(x) =1

cos x

1 + 2d =

2

0

cos x

1 + 2d.

This is the same result obtained from formulas (8) and (9) in the text.

Exercises 15.4

For the boundary-value problems in this section it is sometimes useful to note that the identities

ei = cos + i sin and ei = cos i sin imply

ei + ei = 2cos and ei ei = 2i sin .

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Exercises 15.4

1. Using the Fourier transform, the partial differential equation becomes

dU

dt+ k2U = 0 and so U(, t) = cek

2t.

Now

{u(x, 0)} = U(, 0) = e|x| .We have

e|x|

=

e|x|eixdx =

e|x|(cos x + i sin x) dx =

e|x| cos x dx.

The integral

e|x| sin xdx = 0

since the integrand is an odd function of x. Continuing we obtain

e|x|

= 2

0

ex cos xdx =2

1 + 2.

But U(, 0) = c =2

1 + 2gives

U(, t) =2ek

2t

1 + 2

and so

u(x, t) =2

2

ek2teix

1 + 2d =

1

ek2t

1 + 2(cos x i sin x)d

=1

ek2t cos x

1 + 2d =

2

0

ek2t cos x

1 + 2d.

2. Since the domain of x is (,) we transform the differential equation using the Fourier transform:k2U(, t) = du

dtdu

dt+ k2U(, t) = 0

U(, t) = cek2t. (1)

The transform of the initial condition is

{u(x, 0)} =

u(x, 0)eixdx =

01

(100eix)dx +1

0

100eixdx

= 1001

ei

i + 100

ei

1

i = 100

ei + ei

2

i

= 1002cos 2

i= 200

cos 1i

.

Thus

U(, 0) = 200cos 1

i,

and since c = U(, 0) in (1) we have

U(, t) = 200cos 1

iek

2t.

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Exercises 15.4

Applying the inverse Fourier transform we obtain

u(x, t) = 1{U(, t)}

=1

2

200cos 1

iek

2teix d

=100

200cos 1

iek

2t(cos x i sin x) d

=100

cos x(cos 1)i

ek2t

odd function

d 100

sin x(cos 1)

ek2t

even function

d

=200

0

sin x(1 cos )

ek2td.

3. Using the Fourier transform, the partial differential equation equation becomes

dUdt

+ k2U = 0 and so U(, t) = cek2t.

Now

{u(x, 0)} = U(, 0) = e2/4

by the given result. This gives c =

e2/4 and so

U(, t) =

e(14+kt)

2

.

Using the given Fourier transform again we obtain

u(x, t) =

1

{e(1+4kt)

2/4

}=

1

1 + 4ktex

2/(1+4kt).

4. (a) We use U(, t) = cek2t. The Fourier transform of the boundary condition is U(, 0) = F(). This gives

c = F() and so U(, t) = F()ek2t. By the convolution theorem and the given result, we obtain

u(x, t) = 1{F() ek2t} = 12

kt

f()e(x)2/4kt d.

(b) Using the definition of f and the solution is part (a) we obtain

u(x, t) =u0

2

kt

11

e(x)2/4ktd.

If u = x 2

kt, then d = 2ktdu and the integral becomes

u(x, t) =u0

(x+1)/2t(x1)/2

kt

eu2

du.

Using the result in Problem 9, Exercises 15.1, we have

u(x, t) =u02

erf

x + 1

2

kt

erf

x 12

kt

.

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-4-2

02

4x

2

4

6

t

0

20

40uHx,tL

4-2

02

-4 -2 2 4t

20

40

60

80

100

u

t=0.05

t=15

Exercises 15.4

5.

Since erf(0) = 0 and limx erf(x) = 1, we have

limt

u(x, t) = 50[erf(0) erf(0)] = 0

and

limx

u(x, t) = 50[erf() erf()] = 50[1 1] = 0.

6. (a) Using the Fourier sine transform, the partial differential equation becomes

dU

dt+ k2U = ku0.

The general solution of this linear equation is

U(, t) = cek2t +

u0

.

But U(, 0) = 0 implies c = u0/ and soU(, t) = u0

1 ek2t

and

u(x, t) =2u0

0

1 ek2t

sin x d.

(b) The solution of part (a) can be written

u(x, t) =2u0

0

sin x

d 2u0

0

sin x

ek

2t d.

Using

0

sin x

d = /2 the last line becomes

u(x, t) = u0 2u0

0

sin x

ek

2t d.

7. Using the Fourier sine transform we find

U(, t) = cek2t.

Now

S{u(x, 0)} = U(, 0) =1

0

sin xdx =1 cos

.

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Exercises 15.4

From this we find c = (1 cos )/ and so

U(, t) =1 cos

ek

2t

and

u(x, t) =

2

0

1

cos

ek2t

sin xd.

8. Since the domain of x is (0,) and the condition at x = 0 involves u/x we use the Fourier cosine transform:

k2U(, t) kux(0, t) = dUdt

dU

dt+ k2U(, t) = kA

U(, t) = cek2t +

A

2.

Since

{u(x, 0)} = U(, 0) = 0

we find c = A/2, so thatU(, t) = A

1 ek2t2

.

Applying the inverse Fourier cosine transform we obtain

u(x, t) = C1{U(, t)} = 2A

0

1 ek2t2

cos xd.

9. Using the Fourier cosine transform we find

U(, t) = cek2t.

Now

C{u(x, 0)} = 10

cos xdx =sin

= U(, 0).

From this we obtain c = (sin )/ and so

U(, t) =sin

ek

2t

and

u(x, t) =2

0

sin

ek

2t cos xd.

10. Using the Fourier sine transform we find

U(, t) = cek2t +

1

.

NowS{u(x, 0)} = S

ex

=

0

ex sin xdx =

1 + 2= U(, 0).

From this we obtain c = /(1 + 2) 1/. Therefore

U(, t) =

1 + 2 1

ek

2t +1

=

1

e

k2t

(1 + 2)

and

u(x, t) =2

0

1

e

k2t

(1 + 2)

sin xd.

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Exercises 15.4

11. (a) Using the Fourier transform we obtain

U(, t) = c1 cos at + c2 sin at.

If we write

{u(x, 0)} = {f(x)} = F()and

{ut(x, 0)} = {g(x)} = G()

we first obtain

c1 = F() from U(, 0) = F() and then c2 = G()/a fromdU

dt

t=0

= G(). Thus

U(, t) = F()cos at +G()

asin at

and

u(x, t) =

1

2

F()cos at + G()a sin at eixd.(b) If g(x) = 0 then c2 = 0 and

u(x, t) =1

2

F()cos ateixd

=1

2

F()

eati + eati

2

eixd

=1

2

1

2

F()ei(xat)d +1

2

F()ei(x+at)d

=

1

2 [f(x at) + f(x + at)] .12. Using the Fourier sine transform we obtain

U(, t) = c1 cos at + c2 sin at.

Now

S{u(x, 0)} =

xex

=

0

xex sin xdx =2

(1 + 2)2= U(, 0).

Also,

S{ut(x, 0)} = dUdt

t=0

= 0.

This last condition gives c2 = 0. Then U(, 0) = 2/(1 + 2

)2

yields c1 = 2/(1 + 2

)2

. Therefore

U(, t) =2

(1 + 2)2cos at

and

u(x, t) =4

0

cos at

(1 + 2)2sin x d.

13. Using the Fourier cosine transform we obtain

U(x, ) = c1 cosh x + c2 sinh x.

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Exercises 15.4

Now the Fourier cosine transforms of u(0, y) = ey and u(, y) = 0 are, respectively, U(0, ) = 1/(1 + 2) andU(, ) = 0. The first of these conditions gives c1 = 1/(1 +

2). The second condition gives

c2 = cosh (1 + 2) sinh

.

Hence

U(x, ) =cosh x

1 + 2 cosh sinh x

(1 + 2)sinh =

sinh cosh cosh sinh x(1 + 2) sinh

=sinh ( x)

(1 + 2) sinh

and

u(x, t) =2

0

sinh ( x)(1 + 2)sinh

cos y d.

14. Since the boundary condition at y = 0 now involves u(x, 0) rather than u(x, 0), we use the Fourier sine

transform. The transform of the partial differential equation is then

d2U

dx2 2U + u(x, 0) = 0 or d

2U

dx2 2U = .

The solution of this differential equation is

U(x, ) = c1 cosh x + c2 sinh x +1

.

The transforms of the boundary conditions at x = 0 and x = in turn imply that c1 = 1/ and

c2 =cosh

sinh 1

sinh +

(1 + 2)sinh .

Hence

U(, x) =1

cosh x

+

cosh

sinh sinh x sinh x

sinh +

sinh x

(1 + 2)sinh

= 1 sinh ( x) sinh sinh x(1 + 2) sinh .Taking the inverse transform it follows that

u(x, y) =2

0

1

sinh ( x)

sinh sinh x

(1 + 2) sinh

sin y d.

15. Using the Fourier cosine transform with respect to x gives

U(, y) = c1ey + c2ey.

Since we expect u(x, y) to be bounded as y we define c2 = 0. Thus

U(, y) = c1ey.

Now

C{u(x, 0)} =1

0

50 cos xdx = 50sin

and so

U(, y) = 50sin

ey

and

u(x, y) =100

0

sin

ey cos xd.

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Exercises 15.4

16. The boundary condition u(0, y) = 0 indicates that we now use the Fourier sine transform. We still have

U(, y) = c1ey, but

S{u(x, 0)} =1

0

50sin xdx = 50(1 cos )/ = U(, 0).

This gives c1 = 50(1 cos )/ and so

U(, y) = 501 cos

ey

and

u(x, y) =100

0

1 cos

ey sin xd.

17. We use the Fourier sine transform with respect to x to obtain

U(, y) = c1 cosh y + c2 sinh y.

The transforms of u(x, 0) = f(x) and u(x, 2) = 0 give, in turn, U(, 0) = F() and U(, 2) = 0. The first

condition gives c1 = F() and the second condition then yields

c2 = F()cosh2sinh2

.

Hence

U(, y) = F() cosh y F()cosh2 sinh ysinh2

= F()sinh2 cosh y cosh2 sinh y

sinh2

= F()sinh (2 y)

sinh2

and

u(x, y) =2

0

F()sinh (2 y)

sinh2sin x d.

18. The domain of y and the boundary condition at y = 0 suggest that we use a Fourier cosine transform. The

transformed equation is

d2U

dx2 2U uy(x, 0) = 0 or d

2U

dx2 2U = 0.

Because the domain of the variable x is a finite interval we choose to write the general solution of the latter

equation as


Now U(0, ) = F(), where F() is the Fourier cosine transform of f(y), and U(, ) = 0 imply c1 = F()

and c2 = F() sinh / cosh . Thus

U(x, ) = F() cosh x F() sinh cosh

sinh x = F()cosh ( x)

cosh .

Using the inverse transform we find that a solution to the problem is

u(x, y) =2

0

F()cosh ( x)

cosh cos y d.

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Exercises 15.4

19. We solve two boundary-value problems:

Using the Fourier sine transform with respect to y gives

u1(x, y) =2

0

ex

1 + 2sin y d.

The Fourier sine transform with respect to x yields the solution to the second problem:

u2(x, y) =2

0

ey

1 + 2sin xd.

We define the solution of the original problem to be

u(x, y) = u1(x, y) + u2(x, y) = 2

0

1 + 2

ex sin y + ey sin x d.

20. We solve the three boundary-value problems:

Using separation of variables we find the solution of the first problem is

u1(x, y) =n=1

Aneny sin nx where An = 2

0

f(x)sin nxdx.

Using the Fourier sine transform with respect to y gives the solution of the second problem:

u2(x, y) =200

0

(1 cos )sinh ( x) sinh

sin y d.

Also, the Fourier sine transform with respect to y gives the solution of the third problem:

u3(x, y) =2

0

sinh x

(1 + 2) sinh sin y d.

The solution of the original problem is

u(x, y) = u1(x, y) + u2(x, y) + u3(x, y).

21. Using the Fourier transform with respect to x gives

U(, y) = c1 cosh y + c2 sinh y.

The transform of the boundary conditionu

y

y=0

= 0 isdU

dy

y=0

= 0. This condition gives c2 = 0. Hence

U(, y) = c1 cosh y.

Now by the given information the transform of the boundary condition u(x, 1) = ex2

is

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Exercises 15.4

U(, 1) =

e2/4. This condition then gives c1 =

e

2/4 cosh . Therefore

U(, y) =

e

2/4 cosh y

cosh

and

U(x, y) = 12

e2

/4 cosh ycosh

eixd

=1

2

e2/4 cosh y

cosh cos xd

=1

0

e2/4 cosh y

cosh cos xd.

22. Entries 42 and 43 of the table in Appendix III imply0

estsin at

tdt = arctan

a

s

and 0

estsin at cos bt

tdt =

1

2arctan

a + b

s+

1

2arctan

a bs

.

Identifying = t, x = a, and y = s, the solution of Problem 16 is

u(x, y) =100

0

1 cos

ey sin xd

=100

0

sin x

eyd

0

sin x cos

eyd

=100

arctanx

y 1

2arctan

x + 1

y 1

2arctan

x 1y .

Exercises 15.5

1. We show that 14 F4F4 = I:

1

4F4F4 =

1

4

1 1 1 1

1 i 1 i1 1 1 11 i 1 i

1 1 1 1

1 i 1 i1 1 1 11 i 1 i

=1

4

4 0 0 0

0 4 0 0

0 0 4 0

0 0 0 4

= I.

Thus F14 =14

F4.

2. We have

f(x)(x a)dx = 12

a+a

f(x)dx =1

2f(c)(2) = f(c)

by the mean value theorem for integrals.

3. By the sifting property,

{(x)} =

(x)eixdx = ei0 = 1.

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Exercises 15.5

4. We already know that f = f. Then, by the sifting property,

(f )(x) =

f()(x )d =

f()( x)d = f(x).

5. Using integration by parts with u = f(x) and dv = (x

a) we find

f(x)(x a)dx =

f(x)(x a)dx = f(a)

by the sifting property.

6. Using a CAS we find

{g(x)} = 12

[sign(A ) + sign(A + )]where sign(t) = 1 if t > 0 and sign t = 1 if t < 0. Thus

{g(x)} =

1, A < < A0, elsewhere.

7. Using

8 =

2

2+ i

2

2

28 = i

38 =

2

2+ i

2

2

48 = 1

58 =

2

2 i

2

2

68 = i

78 =

2

2 i

2

2

88 = 1

we have

F8 =

1 1 1 1 1 1 1 1

1 22 + i2

2i 22 + i22 1 22 i22 i 22 i22

1 i 1 i 1 i 1 i1

2

2 + i

22 i

2

2 + i

22 1

2

2 i

22

i

22 i

2

2

1 1 1 1 1 1 1 11

2

2 i

22

i

22 i

2

2 1

22 + i

2

2 i

22 + i

2

2

1 i 1 i 1 i 1 i1

2

2 i

22 i

2

2 i

22 1

2

2 + i

22

i

22 + i

2

2

.

In factored form

F8 = I4 D4I4 D4

F4 00 F4

P,where I4 is the 4 4 identity matrix.

D4 =

1 0 0 0

0

2/2 + i

2/2 0 0

0 0 i 0

0 0 0 2/2 + i2/2

,

and P is the 8 8 matrix with 1 in positions (1, 1), (2, 3), (3, 5), (4, 7), (5, 2), (6, 4), (7, 6), and (8, 8).

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-1 -0.5 0.5 1 x

-1

-0.5

0.5

1

y

-2 -1 1 2

0.2

0.4

0.6

0.8

Exercises 15.5

8. The 8th roots of unity, 18, 28, . . .,

88 are shown in the solution

of Problem 7 above. The points in the complex plane are equally

spaced on the perimeter of the unit circle.

9. The Fourier transform of g(x) = (sin 2x)/x is

G() =

1, 2 < < 20, elsewhere.

This implies that (f g)(x) = 1{F()G()} is band-limited.The graph ofF()G(), which is identical to the graph of (f g), is shown.

10. For N = 6,

F6 =

1 1 1 1 1 1

1 1/2 +

3i/2 1/2 + 3i/2 1 1/2 3i/2 1/2 3i/21 1/2 + 3i/2 1/2 3i/2 1 1/2 + 3i/2 1/2 3i/21 1 1 1 1 11 1/2 3i/2 1/2 + 3i/2 1 1/2 3i/2 1/2 + 3i/21 1/2 3i/2 1/2 3i/2 1 1/2 + 3i/2 1/2 + 3i/2

.

If, for example, f = (2, 0, 1, 6, 2, 3), then

c =1

6F6f =

7/3

2/3 +

3i/3

5/6 3i/62/3

5/6 3i/62/3 i/3

.

Chapter 5 Review Exercises

1. The partial differential equation and the boundary conditions indicate that the Fourier cosine transform is

appropriate for the problem. We find in this case

u(x, y) =2

0

sinh y

(1 + 2)cosh cos xd.

2. We use the Laplace transform and undetermined coefficients to obtain

U(x, s) = c1 cosh

s x + c2 sinh

s x +50

s + 42sin2x.

The transformed boundary conditions U(0, s) = 0 and U(1, s) = 0 give, in turn, c1 = 0 and c2 = 0. Hence

U(x, s) =50

s + 42sin2x

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and

u(x, t) = 50 sin 2x

1

s + 42

= 50e4

2t sin2x.

3. The Laplace transform gives

U(x, s) = c1e

s+hx

+ c2e

s+hx

+

u0

s + h .The condition limx u/x = 0 implies limx dU/dx = 0 and so we define c2 = 0. Thus

U(x, s) = c1es+hx +

u0s + h

.

The condition U(0, s) = 0 then gives c1 = u0/(s + h) and so

U(x, s) =u0

s + h u0 e

s+hx

s + h.

With the help of the first translation theorem we then obtain

u(x, t) = u0 1

s + h u0

e

s+hx

s + h

= u0e

ht u0eht erfcx

2

t= u0e

ht

1 erfc

x

2

t

= u0e

hterf

x

2

t

.

4. Using the Fourier transform and the result

e|x|

= 1/(1 + 2) we find

u(x, t) =1

2

1 e2t2(1 + 2)

eixd

=1

2

1 e2t2(1 + 2)

cos xd

=1

0

1 e2t

2

(1 + 2

)

cos xd.



s x.

The condition limx u(x, t) = 0 implies limx U(x, s) = 0 and so we define c2 = 0. Thus

U(x, s) = c1es x.

The transform of the remaining boundary condition is U(0, s) = 1/s2. This gives c1 = 1/s2. Hence

U(x, s) =e

s x

s2and u(x, t) =

1

s

es x

s

.

Using 1

s

= 1 and

es x

s

= erfc

x

2

t

,

it follows from the convolution theorem that

u(x, t) =

t0

erfc

x

2

d.

6. The Laplace transform and undetermined coefficients gives

U(x, s) = c1 cosh sx + c2 sinh sx +s 1

s2 + 2sin x.

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The conditions U(0, s) = 0 and U(1, s) = 0 give, in turn, c1 = 0 and c2 = 0. Thus

U(x, s) =s 1

s2 + 2sin x

and

u(x, t) = sin x ss2 + 2 1 sin x s2 + 2= (sin x)cos t 1

(sin x)sin t.

7. The Fourier transform gives the solution

u(x, t) =u02

ei 1

i

eixek

2td

=u02

ei(x) eixi

ek2td

=u02

cos ( x) + i sin ( x) cos x + i sin xi

ek2td.

Since the imaginary part of the integrand of the last integral is an odd function of , we obtain

u(x, t) =u02

sin ( x) + sin x

ek2td.

8. Using the Fourier cosine transform we obtain


The condition U(0, ) = 0 gives c1 = 0. Thus

U(x, ) = c2 sinh x.

Now

C{u(, y)} = 2

1 cos ydy =

sin2

sin

= U(, ).

This last condition gives c2 = (sin 2 sin )/ sinh . Hence

U(x, ) =sin2 sin

sinh sinh x

and

u(x, y) =2

0

sin2 sin sinh

sinh x cos y d.

9. We solve the two problems

2u1x2

+2u1y2

= 0, x > 0, y > 0,

u1(0, y) = 0, y > 0,

u1(x, 0) =

100, 0 < x < 1

0, x > 1

and2u2x2

+2u2y2

= 0, x > 0, y > 0,

u2(0, y) =

50, 0 < y < 1

0, y > 1

u2(x, 0) = 0.

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Using the Fourier sine transform with respect to x we find

u1(x, y) =200

0

1 cos

ey sin x d.

Using the Fourier sine transform with respect to y we find

u2(x, y) =100

0

1 cos

ex sin y d.

The solution of the problem is then

u(x, y) = u1(x, y) + u2(x, y).


U(x, s) = c1 cosh

s x + c2 sinh

s x +r

s2.

The conditionu

x

x=0

= 0 transforms intodU

dx

x=0

= 0. This gives c2 = 0. The remaining condition u(1, t) = 0

transforms into U(1, s) = 0. This condition then implies c1 = r/s2 coshs . Hence

U(x, s) =r

s2 r cosh

s x

s2 cosh

s.

Using geometric series and the convolution theorem we obtain

u(x, t) = r

1

s2

r

cosh

s x

s2 cosh

s

= rt rn=0

(1)nt

0

erfc

2n + 1 x

2

d +

t0

erfc

2n + 1 + x

2

d

.

11. The Fourier sine transform with respect to x and undetermined coefficients give

U(, y) = c1 cosh y + c2 sinh y +A

.

The transforms of the boundary conditions are

dU

dy

y=0

= 0 anddU

dy

y=

=B

1 + 2.

The first of these conditions gives c2 = 0 and so

U(, y) = c1 cosh y +A

.

The second transformed boundary condition yields c1 = B/(1 +

2

) sinh . Therefore

U(, y) =B cosh y

(1 + 2) sinh +

A

and

u(x, y) =2

0

B cosh y

(1 + 2) sinh +

A

sin xd.

12. Using the Laplace transform gives

U(x, s) = c1 cosh

s x + c2 sinh

s x.

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The condition u(0, t) = u0 transforms into U(0, s) = u0/s. This gives c1 = u0/s. The condition u(1, t) = u0

transforms into U(1, s) = u0/s. This implies that c2 = u0(1 cosh

s )/s sinh

s . Hence

U(x, s) =u0s

cosh

s x + u0

1 coshs

s sinh

s

sinh

s x

= u0 sinhs coshs x cosh sinhs sinhs x + sinhs x

s sinh

s

= u0

sinh

s (1 x) + sinhs x

s sinh

s

= u0

sinh

s (1 x)

s sinh

s+

sinh

s x

s sinh

s

and

u(x, t) = u0

sinh

s (1 x)

s sinh

s

+

sinh

s x

s sinh

s

= u0

n=0

erf2n + 2 x2t erf2n + x2t + u0

n=0

erf

2n + 1 + x

2

t

erf

2n + 1 x

2

t

.

13. Using the Fourier transform gives

U(, t) = c1ek2t.

Now

u(, 0) =

0

exeix dx =e(i1)x

i 1

0

= 0 1i 1 =

1

1 i = c1so

U(, t) =

1 + i

1 + 2 ek2t

and

u(x, t) =1

2

1 + i

1 + 2ek

2teix d.

Since1 + i

1 + 2(cos x i sin x) = cos x + sin x

1 + 2+

i( cos x sin x)1 + 2

and the integral of the product of the second term with ek2t is 0 (it is an odd function), we have

u(x, t) =1

2

cos x + sin x

1 + 2ek

2t d.

AEMC15

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