using Laplace transform to solve heat equation

using Laplace transform to solve heat equation

Along the whole

positive

x

-axis, we have an heat-conducting rod, the surface of which is .  The initial temperature of the rod is 0 .  Determine the temperature

function

u

⁢

(

x

,

t

)

when at the time

t

=

0

(a) the head

x

=

0

of the rod is set permanently to the constant temperature;

(b) through the head

x

=

0

one directs a constant heat flux.

The

heat equation

in one

dimension

reads

u

x

⁢

x

′′

⁢

(

x

,

t

)

=

1

c

2

⁢

u

t

′

⁢

(

x

,

t

)

.

(1)

In this we have

(a)

⁢

{

boundary conditions

⁢

u

⁢

(

∞

,

t

)

=

0

,

u

⁢

(

0

,

t

)

=

u

0

,

initial conditions

u

⁢

(

x

,

0

)

=

0

,

u

t

′

⁢

(

x

,

0

)

=

0

⏟

for

⁢

x

>

0

and

(b)

⁢

{

boundary conditions

⁢

u

⁢

(

∞

,

t

)

=

0

,

u

x

′

⁢

(

0

,

t

)

=

-

k

,

initial conditions

u

⁢

(

x

,

0

)

=

0

,

u

t

′

⁢

(

x

,

0

)

=

0

⏟

for

⁢

x

>

0

.

For solving (1), we first form its

Laplace transform

(see the table of Laplace transforms)

U

x

⁢

x

′′

⁢

(

x

,

s

)

=

1

c

2

⁢

[

s

⁢

U

⁢

(

x

,

s

)

-

u

⁢

(

x

,

0

)

]

,

which is a ordinary

linear differential equation

U

x

⁢

x

′′

⁢

(

x

,

s

)

=

(

s

c

)

2

⁢

U

⁢

(

x

,

s

)

of order (

http://planetmath.org/ODE

) two.  Here,

s

is only a

parametre

, and the

general solution

of the equation is

U

⁢

(

x

,

s

)

=

C

1

⁢

e

s

c

⁢

x

+

C

2

⁢

e

-

s

c

⁢

x

(see this entry (

http://planetmath.org/SecondOrderLinearODEWithConstantCoefficients

)).  Since

U

⁢

(

∞

,

s

)

=

∫

0

∞

e

-

s

⁢

t

⁢

u

⁢

(

∞

,

t

)

⁢

𝑑

t

=

∫

0

∞

0

⁢

𝑑

t

≡

0

,

we must have

C

1

=

0

.  Thus the Laplace transform of the solution of (1) is in both cases (a) and (b)

U

⁢

(

x

,

s

)

=

C

2

⁢

e

-

s

c

⁢

x

.

(2)

For (a), the second

boundary condition

implies

U

⁢

(

0

,

s

)

=

u

0

s

.  But by (2) we must have

U

⁢

(

0

,

s

)

=

C

2

⋅

1

, whence we infer that

C

2

=

u

0

s

.  Accordingly,

U

⁢

(

x

,

s

)

=

u

0

⋅

1

s

⁢

e

-

x

c

⁢

s

,

which corresponds to the solution function

u

⁢

(

x

,

t

)

:=

u

0

⁢

erfc

⁢

x

2

⁢

c

⁢

t

of the heat equation (1).

For (b), the second boundary condition says that

U

x

′

⁢

(

0

,

s

)

=

-

k

s

,  and since (2) implies that

U

x

′

⁢

(

x

,

s

)

=

-

s

c

⁢

C

2

⁢

e

-

s

c

⁢

x

,  we can infer that now

C

2

=

c

⁢

k

s

⁢

s

.

Thus

U

⁢

(

x

,

s

)

=

c

⁢

k

s

⁢

s

⁢

e

-

x

c

⁢

s

,

which corresponds to

u

⁢

(

x

,

t

)

:=

k

⁢

[

2

⁢

c

⁢

t

π

⁢

e

-

x

2

4

⁢

c

2

⁢

t

-

x

⁢

erfc

⁢

x

2

⁢

c

⁢

t

]

.

Title

using Laplace transform to solve heat equation

Canonical

name

UsingLaplaceTransformToSolveHeatEquation

Date of creation

2015-05-30 6:55:05

Last modified on

2015-05-30 6:55:05

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Example

Classification

msc 35K20

Classification

msc 35Q99

Classification

msc 35K05

Synonym

using Laplace transform to solve

partial differential equation

Related topic

LaplaceTransform

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