examples of lamellar field

examples of lamellar field

In the examples that follow, show that the given

vector field

U

→

is

lamellar

everywhere in

ℝ

3

and determine its

scalar potential

u

.

Example 1.

Given

U

→

:=

y

⁢

i

→

+

(

x

+

sin

⁡

z

)

⁢

j

→

+

y

⁢

cos

⁡

z

⁢

k

→

.

For the rotor (

http://planetmath.org/NablaNabla

) (curl) of the we obtain

∇

×

U

→

=

|

i

→

j

→

k

→

∂

∂

⁡

x

∂

∂

⁡

y

∂

∂

⁡

z

y

x

+

sin

⁡

z

y

⁢

cos

⁡

z

|

=

(

∂

⁡

(

y

⁢

cos

⁡

z

)

∂

⁡

y

-

∂

⁡

(

x

+

sin

⁡

z

)

∂

⁡

z

)

⁢

i

→

+

(

∂

⁡

y

∂

⁡

z

-

∂

⁡

(

y

⁢

cos

⁡

z

)

∂

⁡

x

)

⁢

j

→

+

(

∂

⁡

(

x

+

sin

⁡

z

)

∂

⁡

x

-

∂

⁡

y

∂

⁡

y

)

⁢

k

→

,

which is identically

0

→

for all

x

,

y

,

z

.  Thus, by the

definition

given in the parent (

http://planetmath.org/LaminarField

) entry,

U

→

is lamellar.

Since

∇

⁡

u

=

U

→

,  the scalar potential

u

=

u

⁢

(

x

,

y

,

z

)

must satisfy the conditions

∂

⁡

u

∂

⁡

x

=

y

,

∂

⁡

u

∂

⁡

y

=

x

+

sin

⁡

z

,

∂

⁡

u

∂

⁡

z

=

y

⁢

cos

⁡

z

.

Thus we can write

u

=

∫

y

⁢

𝑑

x

=

x

⁢

y

+

C

1

,

where

C

1

may depend on

y

or

z

. Differentiating this result with respect to

y

and comparing to the second
condition, we get

∂

⁡

u

∂

⁡

y

=

x

+

∂

⁡

C

1

∂

⁡

y

=

x

+

sin

⁡

z

.

Accordingly,

C

1

=

∫

sin

⁡

z

⁢

d

⁢

y

=

y

⁢

sin

⁡

z

+

C

2

,

where

C

2

may depend on

z

.  So

u

=

x

⁢

y

+

y

⁢

sin

⁡

z

+

C

2

.

Differentiating this result with respect to

z

and comparing to the third condition yields

∂

⁡

u

∂

⁡

z

=

y

⁢

cos

⁡

z

+

∂

⁡

C

2

∂

⁡

z

=

y

⁢

cos

⁡

z

.

This means that

C

2

is an arbitrary . Thus the form

u

=

x

⁢

y

+

y

⁢

sin

⁡

z

+

C

expresses the required potential function.

Example 2.

This is a particular case in

ℝ

2

:

U

→

⁢

(

x

,

y

,

0

)

:=

ω

⁢

y

⁢

i

→

+

ω

⁢

x

⁢

j

→

,

ω

=

constant

Now,

∇

×

U

→

=

|

i

→

j

→

k

→

∂

∂

⁡

x

∂

∂

⁡

y

∂

∂

⁡

z

ω

⁢

y

ω

⁢

x

0

|

=

(

∂

⁡

(

ω

⁢

x

)

∂

⁡

x

-

∂

⁡

(

ω

⁢

y

)

∂

⁡

y

)

⁢

k

→

=

0

→

,  and so

U

→

is lamellar.

Therefore there exists a potential

u

with

U

→

=

∇

⁡

u

.  We deduce successively:

∂

⁡

u

∂

⁡

x

=

ω

⁢

y

;

u

⁢

(

x

,

y

,

0

)

=

ω

⁢

x

⁢

y

+

f

⁢

(

y

)

;

∂

⁡

u

∂

⁡

y

=

ω

⁢

x

+

f

′

⁢

(

y

)

≡

ω

⁢

x

;

f

′

⁢

(

y

)

=

0

;

f

⁢

(

y

)

=

C

Thus we get the result

u

⁢

(

x

,

y

,

0

)

=

ω

⁢

x

⁢

y

+

C

,

which corresponds to a particular case in

ℝ

2

.

Example 3.

Given

U

→

:=

a

x

i

→

+

b

y

j

→

-

(

a

+

b

)

z

)

k

→

.

The rotor is now

∇

×

U

→

=

|

i

→

j

→

k

→

∂

∂

⁡

x

∂

∂

⁡

y

∂

∂

⁡

z

a

⁢

x

b

⁢

y

-

(

a

+

b

)

⁢

z

|

=

0

→

.

From

∇

⁡

u

=

U

→

we obtain

∂

⁡

u

∂

⁡

x

=

a

⁢

x

⟹

u

=

a

⁢

x

2

2

+

f

⁢

(

y

,

z

)

(

1

)

∂

⁡

u

∂

⁡

y

=

b

⁢

y

⟹

u

=

b

⁢

y

2

2

+

g

⁢

(

z

,

x

)

(

2

)

∂

⁡

u

∂

⁡

z

=

-

(

a

+

b

)

⁢

z

⟹

u

=

-

(

a

+

b

)

⁢

z

2

2

+

h

⁢

(

x

,

y

)

(

3

)

Differentiating (1) and (2) with respect to

z

and using (3) give

-

(

a

+

b

)

⁢

z

=

∂

⁡

f

⁢

(

y

,

z

)

∂

⁡

z

⟹

f

⁢

(

y

,

z

)

=

-

(

a

+

b

)

⁢

z

2

2

+

F

⁢

(

y

)

(

1

′

)

;

-

(

a

+

b

)

⁢

z

=

∂

⁡

g

⁢

(

z

,

x

)

∂

⁡

z

⟹

g

⁢

(

z

,

x

)

=

-

(

a

+

b

)

⁢

z

2

2

+

G

⁢

(

x

)

(

2

′

)

.

We substitute

(

1

′

)

and

(

2

′

)

again into (1) and (2) and deduce as follows:

u

=

a

⁢

x

2

2

-

(

a

+

b

)

⁢

z

2

2

+

F

⁢

(

y

)

;

∂

⁡

u

∂

⁡

y

=

F

′

⁢

(

y

)

=

b

⁢

y

;

F

⁢

(

y

)

=

b

⁢

y

2

2

+

C

1

;

f

⁢

(

y

,

z

)

=

b

⁢

y

2

2

-

(

a

+

b

)

⁢

z

2

2

+

C

1

(

1

′′

)

;

u

=

b

⁢

y

2

2

-

(

a

+

b

)

⁢

z

2

2

+

G

⁢

(

x

)

;

∂

⁡

u

∂

⁡

x

=

G

′

⁢

(

x

)

=

a

⁢

x

;

G

⁢

(

x

)

=

a

⁢

x

2

2

+

C

2

;

g

⁢

(

z

,

x

)

=

a

⁢

x

2

2

-

(

a

+

b

)

⁢

z

2

2

+

C

2

(

2

′′

)

;

putting

(

1

′′

)

,

(

2

′′

)

into (1), (2) then gives us

u

=

a

⁢

x

2

2

+

b

⁢

y

2

2

-

(

a

+

b

)

⁢

z

2

2

+

C

1

,

u

=

a

⁢

x

2

2

+

b

⁢

y

2

2

-

(

a

+

b

)

⁢

z

2

2

+

C

2

,

whence, by comparing,

C

1

=

C

2

=

C

,  so that by (3), the

expression

h

⁢

(

x

,

y

)

and

u

itself have been found, that is,

u

=

a

⁢

x

2

2

+

b

⁢

y

2

2

-

(

a

+

b

)

⁢

z

2

2

+

C

.

Unlike Example 1, the last two examples are also

solenoidal

, i.e.

∇

⋅

U

→

=

0

,  which physically may be interpreted as the continuity equation of an

incompressible

fluid flow.

Example 4.

An additional example of a lamellar field would be

U

→

:=

-

a

⁢

y

x

2

+

y

2

⁢

i

→

+

a

⁢

x

x

2

+

y

2

⁢

j

→

+

v

⁢

(

z

)

⁢

k

→

with a

differentiable function

v

:

ℝ

→

ℝ

;  if

v

is a constant, then

U

→

is also solenoidal.

Title

examples of lamellar field

Canonical

name

ExamplesOfLamellarField

Date of creation

2013-03-22 17:39:25

Last modified on

2013-03-22 17:39:25

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Example

Classification

msc 26B12

Synonym

example of scalar potential

Synonym

determining the scalar potential

Related topic

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