time-dependent example of heat equation

time-dependent example of heat equation

The initial temperature (at

t

=

0

) of a plate

A

=

{

(

x

,

y

)

∈

ℝ

2

⋮

0

<

x

<

a

,

0

<

y

<

b

}

in the

x

⁢

y

-plane is given by the function

f

=

f

⁢

(

x

,

y

)

.  The faces of the plate are supposed completely isolating.  After the

t

=

0

the boundaries of

A

are held in the temperature

0

.  Determine the temperature function

u

=

u

⁢

(

x

,

y

,

t

)

on

A

(where

t

is the time).

Since it’s a question of a two-dimensional heat , the

heat equation

gets the form

∇

2

⁡

u

≡

u

x

⁢

x

′′

+

u

y

⁢

y

′′

=

1

c

2

⁢

u

t

′

.

(1)

One have to find for (1) a solution function

u

which satisfies the

initial condition

u

⁢

(

x

,

y

,

0

)

=

f

⁢

(

x

,

y

)

⁢

in

⁢

A

(2)

and the

boundary condition

u

⁢

(

x

,

y

,

t

)

=

0

⁢

on

⁢

boundary

⁢

of

⁢

A

⁢

for

⁢

t

>

0

.

(3)

For finding a simple solution of the

differential equation

(1) we try the form

u

⁢

(

x

,

y

,

t

)

:=

X

⁢

(

x

)

⁢

Y

⁢

(

y

)

⁢

T

⁢

(

t

)

,

(4)

whence the boundary condition reads

X

⁢

(

0

)

=

Y

⁢

(

0

)

=

X

⁢

(

a

)

=

Y

⁢

(

b

)

=

0

.

(5)

Substituting (4) in (1) and dividing this equation by

X

⁢

Y

⁢

T

give the form

X

′′

X

+

Y

′′

Y

=

1

c

2

⋅

T

′

T

.

(6)

It’s easily understood that such a condition requires that the both addends of the left side and the right side ought to be constants:

X

′′

X

=

-

k

1

2

,

Y

′′

Y

=

-

k

2

2

,

1

c

2

⋅

T

′

T

=

-

k

2

,

(7)

where

k

2

=

k

1

2

+

k

2

2

.  We soon explain why these constants are negative.
Because the equations (7) may be written

X

′′

=

-

k

1

2

⁢

X

,

Y

′′

=

-

k

2

2

⁢

Y

,

T

′

=

-

k

2

⁢

c

2

⁢

T

,

the

general solutions

of these

ordinary differential equations

are

{

X

=

C

1

⁢

cos

⁡

k

1

⁢

x

+

D

1

⁢

sin

⁡

k

1

⁢

x

,

Y

=

C

2

⁢

cos

⁡

k

2

⁢

y

+

D

2

⁢

sin

⁡

k

2

⁢

y

,

T

=

C

⁢

e

-

k

2

⁢

c

2

⁢

t

.

(8)

Now we remark that if the right side of the third equation (7) were

+

k

2

, then we had

T

=

C

⁢

e

k

2

⁢

c

2

⁢

t

which is impossible, since such a

T

and along with this also the temperature

u

=

X

⁢

Y

⁢

T

would ascend infinitely when

t

→

∞

.  And since, by symmetry, the right sides the two first equations (7) must have the same sign, also they must by (6) be negative.

The two first boundary conditions (5) imply by (8) that

C

1

=

C

2

=

0

,  and then the two last conditions (5) require that

D

1

⁢

sin

⁡

k

1

⁢

a

=

0

,

D

2

⁢

sin

⁡

k

2

⁢

b

=

0

.

If we had

D

1

=

0

or

D

2

=

0

,  then

X

or

Y

would vanish identically, which cannot occur.  Thus we have

sin

⁡

k

1

⁢

a

=

0

⁢

and

⁢

sin

⁡

k

2

⁢

b

=

0

,

whence only the eigenvalues

{

k

1

=

m

⁢

π

a

(

m

=

1

,

2

,

3

,

…

)

k

2

=

n

⁢

π

b

(

n

=

1

,

2

,

3

,

…

)

are possible for the obtained

X

and

Y

.  Considering the equation

k

2

=

k

1

2

+

k

2

2

we may denote

q

m

⁢

n

:=

k

2

⁢

c

2

=

[

(

m

⁢

π

a

)

2

+

(

n

⁢

π

b

)

2

]

⁢

c

2

(9)

for all

m

,

n

∈

ℤ

+

.

Altogether we have infinitely many solutions

u

m

⁢

n

=

X

⁢

Y

⁢

T

=

C

⁢

D

1

⁢

D

2

⁢

e

-

q

m

⁢

n

⁢

t

⁢

sin

⁡

m

⁢

π

⁢

x

a

⁢

sin

⁡

n

⁢

π

⁢

y

b

=

c

m

⁢

n

⁢

e

-

q

m

⁢

n

⁢

t

⁢

sin

⁡

m

⁢

π

⁢

x

a

⁢

sin

⁡

n

⁢

π

⁢

y

b

of the equation (1), where the

coefficients

c

m

⁢

n

are, for the present, arbitrary constants.  These solutions fulfil the boundary condition (3).  The sum of the solutions, i.e. the

double series

u

⁢

(

x

,

y

,

t

)

:=

∑

m

=

1

∞

∑

n

=

1

∞

c

m

⁢

n

⁢

e

-

q

m

⁢

n

⁢

t

⁢

sin

⁡

m

⁢

π

⁢

x

a

⁢

sin

⁡

n

⁢

π

⁢

y

b

,

(10)

provided it converges, is also a solution of the linear differential equation (1) and fulfils the boundary condition.  In order to fulfil also the initial condition (2), one must have

∑

m

=

1

∞

∑

n

=

1

∞

c

m

⁢

n

⁢

e

-

q

m

⁢

n

⁢

t

⁢

sin

⁡

m

⁢

π

⁢

x

a

⁢

sin

⁡

n

⁢

π

⁢

y

b

=

f

⁢

(

x

,

y

)

.

But this equation presents the

Fourier double sine series

of

f

⁢

(

x

,

y

)

in the rectangle

A

, and therefore we have the expression

c

m

⁢

n

:=

4

a

⁢

b

⁢

∫

0

a

∫

0

b

f

⁢

(

x

,

y

)

⁢

sin

⁡

m

⁢

π

⁢

x

a

⁢

sin

⁡

n

⁢

π

⁢

y

b

⁢

d

⁢

x

⁢

d

⁢

y

(11)

for the coefficients.

The result of calculating the solution of our problem is the temperature function (10) with the formulae (9) and (11).

References

1

K. Väisälä:

Matematiikka IV

.  Hand-out Nr. 141. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967).

Title

time-dependent example of heat equation

Canonical name

TimedependentExampleOfHeatEquation

Date of creation

2013-03-22 16:32:47

Last modified on

2013-03-22 16:32:47

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Example

Classification

msc 35Q99

Synonym

solving a time-dependent heat equation

Related topic

ExampleOfSolvingTheHeatEquation

Related topic

EigenvalueProblem

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