Gockenbach M.S. Partial Differential Equations. Analytical and Numerical Methods

508

Appendix

B.

Shifting

the

data

in two

dimensions

We

can

then

define

Having

computedp,

we

define

v =

u—p,

so

that

and

We

then

see

that

where

We

then obtain

the

following

formula

for

u:

The

solution

is

shown

in

Figure

B.I.

B.0.2

Inhomogeneous

Neumann

conditions

on a

rectangle

We

can

also apply

the

technique

of

shifting

the

data

to

transform

a BVP

with

inhomogeneous Neumann conditions

to a

with homogeneous Neumann

conditions. However,

the

details

are

somewhat more involved

than

in the

Dirichlet

case.

Suppose

the

Neumann conditions

are

We

first

note

that

this

is

equivalent

to

508

Appendix

B.

Shifting the data

in

two dimensions

We

can then define

P(XI,X2)

=

Xl

+

X2

-

2XIX2

+

xi

-

Xl

+

o·

X2

+ 0 +

(X2

-

X~)XI

2 2

=

X2

-

XIX2

+

Xl

-

XIX2·

Having computed p,

we

define v = u - p,

so

that

-Llv

=

-Llu

+ Llp = 0 + Llp = 2 - 2XI, X

En

and

v(x) = 0, x E

an.

We

then see that

where

We then obtain the following formula for u:

The solution is shown

in

Figure B.1.

8.0.2

Inhomogeneous Neumann conditions

on

a rectangle

We

can also apply the technique of shifting the

data

to

transform a

BVP

with

inhomogeneous Neumann conditions

to

a related

BVP

with homogeneous Neumann

conditions. However,

the

details are somewhat more involved

than

in

the

Dirichlet

case.

Suppose the Neumann conditions are

We

first note

that

this

is

equivalent to

x E r

l

,

x E r

2

,

x E r

3

,

x E r

4

.

Appendix

B.

Shifting

the

data

in two

dimensions

509

Figure

B.I.

The

solution

to the BVP

(B.2),

approximated

by a

Fourier

series

with

400

terms.

We

make

the

following

observation:

If

there

is a

twice-continuously

differentiable

function

u

satisfying

the

given Neumann conditions, then, since

we

have

Appendix

B.

Shifting the data

in

two

dimensions

509

o 0

Figure

B.t.

The solution to the

BVP

(B.2), approximated

by

a Fourier

series with 400 terms.

We

make the following observation:

If

there

is

a twice-continuously differentiable

function

u satisfying the given Neumann conditions, then, since

8

2

u 8

2

u

8:lh

8X2

8X28xl

'

we

have

510

Appendix

B.

Shifting

the

data

in two

dimensions

and

which

together imply

that

By

similar reasoning,

we

have

all of the

following

conditions:

We

will

assume

that

(B.3) holds.

We

now

explain

how to

compute

a

function

p

that

satisfies

the

desired Neu-

mann conditions.

The

method

is

similar

to

that

used

to

shift

the

data

in a

Dirichlet

problem:

we

will

"interpolate" between

the

Neumann conditions

in

each dimension

and

arrange things

so

that

the two

interpolations

do not

interfere with each other.

We

use the

fact

that

satisfies

The first

step

is to

transform

the

boundary

data

g\

to a

function

h\

satisfying

and

similarly

for

92,93,94

and

/i2,/is,/i4.

Since these derivatives

of the

boundary

data

at the

corners

are

(plus

or

minus)

the

mixed

partial

derivatives

of the de-

sired

function

at the

corners,

it

suffices

to find a

function

q(xi,xz)

satisfying

the

conditions

510

Appendix

B.

Shifting

the

data

in

two dimensions

and

which together imply

that

dg

1

(0)

=

dg4

(0).

dx

1

dx

2

By similar reasoning,

we

have all of the following conditions:

dg

1

(0)

=

dg

4

(0),

dX1

dX2

_

dg

1

(£1)

=

dg

2

(0),

dx

1

dX2

(B.3)

dg

2

(£

) =

dg

3

(£

)

dx

2 d

1,

2

Xl

_

dg

4

(£2)

=

dg

3

(0).

dx

2

dx

1

We

will assume

that

(B.3) holds.

We

now explain how

to

compute a function p

that

satisfies

the

desired Neu-

mann conditions. The method

is

similar

to

that

used

to

shift

the

data

in a Dirichlet

problem:

we

will "interpolate" between

the

Neumann conditions in each dimension

and

arrange things so

that

the two interpolations do not interfere with each other.

We

use

the

fact

that

(BA)

satisfies

(B.5)

The first step is

to

transform the boundary

data

gl

to

a function

h1

satisfying

dh1

(0) =

dh1

(£1)

= 0,

dX1 dX1

and

similarly for

g2, g3,

g4

and h

2

, h3,

h4.

Since these derivatives of the boundary

data

at

the

corners are (plus or minus) the mixed partial derivatives of

the

de-

sired function

at

the

corners, it suffices

to

find a function

Q(X1,

X2)

satisfying

the

conditions

Appendix

B.

Shifting

the

data

in two

dimensions

511

We

can

satisfy

these conditions with

a

function

of the

form

The

reader

can

verify

that

the

necessary

coefficients

are

If

p is to

satisfy

the

desired Neumann conditions, then

where

We can now

define

p

—

q by the

interpolation described

by

(B.4),

(B.5):

Then

p

satisfies

the

original Neumann conditions,

as the

interested reader

can

verify

directly.

We

will

illustrate this computation with

an

example.

Example

B.2.

We

will

solve

the BVP

by

the

above

method,

where

f)

is the

unit

square:

Appendix

B.

Shifting

the

data

in

two

dimensions

We can satisfy these conditions with a function of

the

form

The

reader can verify

that

the

necessary coefficients are

If

p

is

to

satisfy

the

desired Neumann conditions,

then

where

hl(xd

= gl(Xl) +

aXl

+ bxi,

h

2

(X2)

=

g2(X2)

-

(a

+

2i

1

b)X2

-

(c

+

2ild)x~,

ha(xd

= ga(xd -

(a

+

2i2C)Xl

-

(b

+

2i

2

d)xi,

h4(X2)

=

g4(X2)

+

aX2

+

cx~.

We can now define p - q by

the

interpolation described by (BA), (B.5):

511

Then

p satisfies

the

original Neumann conditions, as

the

interested reader can verify

directly.

We

will illustrate this computation with

an

example.

Example

B.2.

We will solve the

BVP

by

the

above

method, where n is the

unit

square:

x

E

fl'

X E

f2'

X E

fa,

x E

f4

(B.6)

512

Appendix

B.

Shifting

the

data

in two

dimensions

To

compute

p,

we

define

We

can

then

define

Having

computed

p, we

define

v =

u

—

p, so

that

and

We

can

write

the

solution

v(x\,xi}

in the

form

With

we

have

512

Appendix

B.

Shifting the data

in

two

dimensions

To

compute p,

we

define

dg

l

a=

--(0)

=

0,

dx

l

b = !

(d

9l

(0)

_

dg

l

(1)) =

0,

2

dXl

C = !

(d

ga

(0)

+

dg

l

(0))

=

0,

2

dXl

d = !

(d

9

2

(1)

+

dg

l

(1)

_

dg

a

(0)

_

dg

l

(0))

= !,

4

dX2

dXl dXl

dXl

2

hI

(xd

=

gl(Xl)

=

-1,

h

2

(X2)

=

g2(X2)

-

x~

=

0,

ha(Xl) =

ga(Xl)

-

xr

= 1,

2

h4(X2)

=

g4(X2)

=

-a'

We

can

then define

ha(xd

+

hl(xl)

2 h

2

(X2)

+ h

4

(x2)

2

P(Xl,X2) = q(Xl,X2) - hl

(Xl)X2

+

2£2

X

2

- h

4

(X2)Xl

+ 2£1

Xl

1 2 2 2 1 2

= 2XlX2 +

X2

+

aXl

-

aXl'

Having computed p,

we

define v = U - p,

so

that

and

2

-Av=-Au+Ap=O+Ap=xr+x~--,

XEO

. 3

ov

on

(x) = 0, x E

80.

We

can

write the solution V(Xl,X2) in the form

00

U(Xl,

X2)

=

aOO

+ L

amo

cos

(mnxd

+ L

aO

n

cos (

nnx

2)

m=l

n=l

00 00

+ L L a

mn

cos

(mnxl)

cos (

nnx

2).

m=ln=l

With

we

have

00 00

f(xl,x2)

=

Coo

+ L

CmO

cos

(mnxd

+ L Con cos (

nnx

2)

m=l

n=l

00

+ L L C

mn

cos

(mnxd

cos (

nnx

2),

m=l

n=l

Appendix

B.

Shifting

the

data

in two

dimensions

513

where

and

u(xi,X2)

=

^(^1,^2)

+

p(a?i,#2)-

-^

graph

of the

solution

u,

with

aoo

= 0, is

shown

in

Figure

B.2.

Figure

B.2.

The

solution

to the BVP in

Example

B.2.

This

graph

was

produced

using

a

total

of

40

terms

of the

(double)

Fourier cosine series.

It

then

follows

that

Appendix

B.

Shifting the data

in

two dimensions

513

where

Coo

=

10

1

10

1

f(X1,X2)dx2dx1

=0,

1

111

4(-1)m

CmO = 2

f(x1,x2)

cos (m1fx1)dx2

dx1

= 2 2 ' m =

1,2,3,

...

,

o 0 m

1f

1

111

4(-1)n

Con

= 2

f(x1,x2)

cos

(n1fX2)

dX2dx1 = 2 2 ' n =

1,2,3,

...

,

o

0 n

1f

C

mn

=

410

1

10

1

f(x1,

X2)

COS

(m1fxd

COS

(mrX2)

dX2

dX1

= 0,

m,

n =

1,2,3,

....

It

then follows that

and

U(X1'

X2)

=

V(X1,

X2)

+

P(X1'

X2).

A

graph

of

the solution u, with

aOO

=

0,

is

shown in Figure

B.2.

2

1.5

0.5

o

- 0.5

1

o 0

Figure

B.2.

The solution to the

BVP

in Example B.2. This

graph

was

produced using a total

of

40

terms

of

the (double) Fourier cosine series.

This page intentionally left blank

Appendix

C

Solutions

to

exercises

Chapter

1

1.

(a)

First-order, homogeneous, linear,

nonconstant-coefficient,

scalar ODE.

(b)

Second-order,

inhomogeneous,

linear, constant-coefficient, scalar PDE.

(c)

First-order, nonlinear, scalar PDE.

3. (a)

Second-order, nonlinear, scalar ODE.

(b)

First-order, inhomogeneous, linear, constant-coefficient, scalar PDE.

(c)

Second-order, inhomogeneous, linear, nonconstant-coefficient, scalar PDE.

5.

(a) No.

(b)

Yes.

(c)

No.

7.

There

is

only

one

such

/:

f(t)

=

tcos

(t).

9.

For any

constant

C

^

1, the

function

w(t)

=

Cu(t)

is a

different

solution

of the

differential

equation.

Section

2.1

1.

The

units

of

K

are

energy

per

length

per

time

per

temperature,

for

example,

J/(cm

s K) =

W/(cmK).

3.

The

units

of

Apcu&x

are

5-

&&*)

=

£,«>

*>.

7.

Measuring

x in

centimeters,

the

steady-state temperature distribution

is

u(x]

=

O.lx

+ 20, the

solution

of

515

odd-numbered

Appendix C

ions

to

bered

•

exercises

Chapter 1

1. (a) First-order, homogeneous, linear, nonconstant-coefficient, scalar ODE.

(b) Second-order, inhomogeneous, linear, constant-coefficient, scalar

PDE.

(c) First-order, nonlinear, scalar

PDE.

3. (a) Second-order, nonlinear, scalar ODE.

(b) First-order, inhomogeneous, linear, constant-coefficient, scalar

PDE.

(c) Second-order, inhomogeneous, linear, nonconstant-coefficient, scalar

PDE.

5.

(a) No.

(b) Yes.

(c) No.

7.

There

is only one such f:

f(t)

= t cos (t).

9.

For any

constant

C

f:.

1,

the

function w(t) = Cu(t) is a different solution

of

the

differential equation.

Section 2.1

1.

The

units

of

'"

are

energy

per

length

per

time

per

temperature,

for example, J / (cm s K) =

W/(cmK).

3.

The

units

of

Apcu~x

are

5.

~~(f,t)=ll<,t>to.

2 g J

cm

--3

-Kcm

= J.

cm

gK

7.

Measuring x in centimeters,

the

steady-state

temperature

distribution is

u(x)

0.1x +

20,

the

solution

of

d

2

u

- dx

2

= 0,

u(O)

= 20, u(100) = 30.

515

516

Appendix

C.

Solutions

to

odd-numbered

exercises

The

heat

flux is

Since

the

area

of the

bar's cross-section

is

vr

cm

2

,

the

rate

at

which

energy

is flowing

through

the bar is

0.08027T

=

0.252

W (in the

negative direction).

9.

Consider

the

right endpoint

(x =

I).

Let the

temperature

of the

bath

be

ue,

so

that

the

difference

in

temperature between

the

bath

and the end of the bar is

u(i,

t}

—

ut.

The

heat

flux at x = i is,

according

to

Fourier's law,

so

the

statement

that

the

heat

flux is

proportional

to the

temperature

flow is

written

where

a > 0 is a

constant

of

proportionality. This

can be

simplified

to

The

condition

at the

right

end is

similar:

(The sign change appears because,

at the

left

end,

a

positive heat

flux

means

that

heat

flows

into

the

bar,

while

at the

right

end the

opposite

is

true.)

11. The

IBVP

is

13. (a) The

rate

is

(b)

The

rate

of

mass transfer varies inversely with

t and

directly with

the

square

of

r.

15.

We

have

and

516

Appendix

C.

Solutions

to

odd-numbered exercises

The

heat flux is

du 2

-K,

dx

(x) =

-0.802

·0.1 = -0.0802 W

/cm

.

Since

the

area of

the

bar's cross-section

is

1r

cm

2

,

the

rate

at

which energy

is

flowing

through

the

bar

is

0.08021r

~

0.252 W (in

the

negative direction).

9.

Consider

the

right endpoint (x = f). Let

the

temperature of

the

bath

be

Ul,

so

that

the

difference in temperature between

the

bath

and

the

end

of

the

bar

is

u(f, t) -

Ul.

The

heat flux

at

x = f is, according to Fourier's law,

so

the

statement

that

the

heat

flux

is

proportional

to

the

temperature

flow

is

written

where

a > 0

is

a constant of proportionality. This can

be

simplified

to

The

condition

at

the

right end is similar:

au

au(O, t) - K,

ax

(0,

t) = auo·

(The sign change appears because,

at

the

left end, a positive heat flux means

that

heat

flows

into

the

bar, while

at

the

right end

the

opposite is true.)

11.

The

IBVP

is

13. (a)

The

rate

is

au

a

2

u

at

- D

ax

2

= lex, t), 0 < x < f, t > to,

u(x,

to) = tf;(x), ° < x <

f,

au

ax

(0,

t) =

0,

t > to,

au

ax

(f, t) =

0,

t > to.

2D

(Ul

- u

o

)

-1rT

--f-.

(b)

The

rate

of mass transfer varies inversely with f

and

directly with

the

square

of

T.

15.

We have

and

1

1 cos

(xy)dy

=

[sin~Xy)]1

o

y=O

sin x

x

.!:...

[SinX] =

xcosx

- sin x =

cosx

_

sinx

dx

x x

2

X

x

2

•

Appendix

C.

Solutions

to

odd-numbered

exercises

517

On

the

other hand,

Section

2.2

1.

The sum of the

forces

on the

cross-section originally

at x = t

must

be

zero.

As

derived

in the

text,

the

internal (elastic)

force

acting

on

this cross-section

is

while

the

external traction results

in a

total

force

of pA

(force

per

unit area times

area). Therefore,

we

obtain

or

The

other boundary condition,

of

course,

is

u(Q)

— 0.

3.

(a) The

stiffness

is the

ratio

of the

internal restoring

force

to the

relative change

in

length. Therefore,

if a bar of

length

t and

cross-sectional area

A is

stretched

to a

length

of 1 +p (0 < p < 1)

times

I,

then

the bar

will

pull back with

a

force

of

195pA

gigaNewtons. Equivalently, this

is the

force

that

must

be

applied

to

the end of the bar to

stretch

the bar to a

length

of (1

+p)i.

(b)

196/195

m,

or

approximately 1.005

m.

(c)

The BVP is

It is

easy

to

show

by

direct integration

that

the

solution

is

u(x]

=

o;/195,

and

therefore

u(l)

=

1/195.

Since

u

is the

displacement

(in

meters),

the final

position

of the end of the bar is 1 +

1/195

m,

that

is, the

stretched

bar is

196/195

m.

5.

The

wave equation

is

Appendix

C.

Solutions

to

odd-numbered exercises 517

On

the

other

hand,

1

d

11

o

dx

[cos (xy)]

dy=

- 0

ysin(xy)dy

=

[YCOS

(X

Y

)]

1

-11

cos

(xy)

dy

x

y=o

0 x

_

cosx

[Sin

(X

y

)]

1

-

-x-

-

-x-

2

-

y=o

cos x

sinx

=

-x--7·

Section 2.2

1.

The

sum

of

the

forces on

the

cross-section originally

at

x = £ must

be

zero. As

derived in

the

text,

the

internal (elastic) force acting on

this

cross-section is

-Ak(£)

~:

(£),

while

the

external

traction

results in a

total

force of

pA

(force

per

unit

area times

area). Therefore,

we

obtain

du

-Ak(£)

dx

(£)

+

pA

=

0,

or

du

k(£)

dx

(£)

= p.

The

other

boundary

condition, of course,

is

u(O)

=

O.

3.

(a)

The

stiffness is

the

ratio of

the

internal restoring force

to

the

relative change

in length. Therefore, if a

bar

of

length £

and

cross-sectional area A is

stretched

to

a length of 1

+p

(0 < p < 1) times

f,

then

the

bar

will pull back with a force

of

195pA gigaNewtons. Equivalently,

this

is

the

force

that

must be applied

to

the

end

of

the

bar

to

stretch

the

bar

to

a length of

(1

+ p)£.

(b) 196/195 m,

or

approximately 1.005 m.

(c)

The

BVP

is

-195

.

10

9

d

2

u = 0 0 < x <

1,

dx

2

'

u(O)

=

0,

195.

10

9

du

(1) =

10

9

.

dx

It

is easy

to

show by direct integration

that

the

solution is

u(x)

=

x/195,

and

therefore

u(l)

= 1/195. Since u is

the

displacement (in meters),

the

final

position of

the

end

of

the

bar

is 1 + 1/195 m,

that

is,

the

stretched

bar

is

196/195

m.

5.

The

wave equation is

(

3)

a

2

u (

11)

a

2

u

7.9 . 10

8t

2

-

1.95·

10

ax

2

=

!(x,

t), 0 < x <

1.

Gockenbach M.S. Partial Differential Equations. Analytical and Numerical Methods

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