Heat (or Diffusion) equation in 1D*

Heat (or Diffusion) equation in 1D*

• Derivation of the 1D heat equation

• Separation of variables (refresher)

• Worked examples

Physical assumptions

• We consider temperature in a long thin

wire of constant cross section and

homogeneous material

• The wire is perfectly insulated laterally, so

heat flows only along the wire

insulation

Derivation of the heat equation in 1D

) , (x t u t x A K by time at point at e temperatur the Denote is area sectional Cross is material the of density The is heat specific The is wire the in ty conductivi thermal the that Suppose ρ σ

x

x

+

δ

x

x x u KA x u KA x x x u KA x x x x δ δ δ 2 2 : ∂ ∂ ∂ ∂ + ∂ ∂ − + : : is out flow net the So : face the At at face across bar into flow Heat x t u A x x u KA δ σρ δ ∂ ∂ = ∂ ∂ 2 2 : gives heat of on Conservati

σρ

K

c

x

u

c

t

u

=

∂

∂

=

∂

∂

₂ 2 2 2

where

,

Boundary and Initial Conditions

0

)

,

(

)

,

0

(

t

=

u

L

t

=

u

As a first example, we will assume that the perfectly insulated rod is of finite length L and has its ends maintained at zero temperature.

0 ) ( ) 0 ( ) ( ) 0 , ( ) ( = = = L f f x f x u x f : conditions boundary the from Evidently, : condition initial the have we , by given is rod the in on distributi e temperatur initial the If

Solution by

Separation of Variables

1. Convert the PDE into two separate

ODEs

2. Solve the two (well known) ODEs

3. Compose the solutions to the two ODEs

into a solution of the original PDE

The first step is to

assume

that the function of two variables has a very

special form: the product of two separate functions, each of one variable,

that is:

Step 1: PDE

Î

2 ODEs

)

(

)

(

)

,

(

x

t

F

x

G

t

u

=

:

that

Assume

2 2 2 2

''

''

dx

F

d

F

dt

dG

G

G

F

x

u

G

F

t

u

=

=

=

∂

∂

=

∂

∂

and

where

:

find

we

ating,

Differenti

&

&

G

F

c

G

F

&

=

2

''

k

F

F

G

c

G

=

''

=

2

&

So that

equivalently

0

0

''

2

=

−

=

−

G

kc

G

kF

F

&

Step 2a: solving for F*

0

''

−kF

=

F

0 0 0 ) ( 0 2 = = = + = + + = > = − − B A Be Ae B A Be Ae x F k L L x x so and conditions boundary the ng Applyi : is solution general the case µ µ µ µ µ 0 ) ( : 0 = = + = = b a b ax x F k : conditions boundary the Applying case 0 sin , 0 sin ) ( 0 ) 0 ( sin cos ) ( : 0 2 = = = = = + = < − = pL pL B L F A F px B px A x F p k so and find we , conditions boundary the ng Applyi is solution the case

We have to solve:

x L n x F L n p n pL n π π π sin ) ( = = = that so : is that 2 p k = − is interest of case only the that follows It

Step 2b: solving for G

L

c

n

L

n

p

=

π

,

and,

as

for

the

wave

equation,

we

denote

λ

_n

=

π

t n n n n

e

B

t

G

G

G

2

)

(

0

2 λ

λ

−

=

=

+

:

is

solution

general

whose

have

We

&

Step 2c: combining F&G

∑

∑

∞ = − ∞ =

=

=

1 1

sin

)

,

(

)

,

(

2 n t n n n

x

L

n

e

B

t

x

u

t

x

u

λn

π

:

is

equation

diffusion

1D

the

to

solution

The

)

(

sin

)

0

,

(

1

x

f

x

L

n

B

x

u

n n

=

=

∑

∞ =

π

:

condition

Initial

∫

=

L n

xdx

L

n

x

f

L

B

0

sin

)

(

2

π

:

find

we

equation,

wave

the

for

As

Analysing the solution

x L n e t x u t x u B t x u t n n n n n π λ sin ) , ( ) , ( ) , ( 2 1 − ∞ = = =

∑

where : as written be can equation diffusion 1D the to solution The

∫

= = = L n n n n xdx L n x f L B B t L u t u c L u t x u 0 sin ) ( 2 0 ) , ( ) , 0 ( , ) , ( π since , conditions initial the by determined are weights The (2) and case; this in conditions boundary the and ) constants the is, (that problem generic the by determined completely are functions The (1) where functions of sum weighted a is solution the that emphasises This

Sine term

x L n

π

sin : term sine the first Consider

Evidently, the higher the value of n the higher the frequency component of f(x) that is analysed

Exponential term

0011 . 0 158 . 1 100 2 2 1 2 2 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − L c c L e t L nc π λ π Then later) slides 2 (see and that Suppose : term l exponentia the next Consider

Gabor functions

Gaussian a ... is term the then and denote is term l exponentia The 2 2 2 , σ π π σ s n t L nc e nc L s t e − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = optics and processing signal to l fundamenta are and functions Gabor called are These term. sine a by multiplied Gaussian a with work to have we of instead consider we if so (x,s) (x,t)

Denis Gabor (born Budapest 1900, died London 1979). Left Berlin for London during the 1930s. Eventually

professor of electronics at Imperial College. Credited with invention of the hologram in 1947. Awarded the Nobel prize for physics in 1971

This appears to be a short burst of sine wave in a Gaussian shaped envelope. This is fundamental to AM/FM communications and to wavelets

Recall the solution …

∑

∑

∞ = − ∞ =

=

=

1 1

sin

)

,

(

)

,

(

2 n t n n n

x

L

n

e

B

t

x

u

t

x

u

λn

π

:

is

equation

diffusion

1D

the

to

solution

The

)

(

sin

)

0

,

(

1

x

f

x

L

n

B

x

u

n n

=

=

∑

∞ =

π

:

condition

Initial

∫

=

L n

xdx

L

n

x

f

L

B

0

sin

)

(

2

π

:

find

we

equation,

wave

the

for

As

Example 1*

x x f 80 sin 100 ) ( = π : conditions initial Sinusoidal cms 80 : bar the of Length 0 ... , 100 80 sin 100 ) ( 80 sin ) 0 , ( 3 2 1 1 = = = = = = =

∑

∞ = B B B x x f x n B x u n n π π x e t x u K c K t 80 sin 100 ) , ( 001758 . 0 6400 870 . 9 158 . 1 sec / 158 . 1 92 . 8 092 . 0 95 . 0 092 . 0 92 . 8 95 . 0 001785 . 0 2 1 2 π λ ρσ σ ρ − = = = = ⋅ = = = = = cm so C) cal/(gm gm/cm C) sec cal/(cm : Copper 2 o 3 o

*Kreysig, 8th _{Edn, page 603}

min 5 . 6 388 50 : = = ⇒ = t s 100e 50 to reduce to e temperatur maximum the for required time 0.001785t -o

Example 2

0

)

0

,

(

x

T

u

=

:

condition

Initial

Somewhat more realistically, we assume that the bar is initially entirely at a constant temperature, then at time t=0, it is insulated and quenched, that is, the temperatures at its two ends instantly reduce to zero.

∞

→

→

>

=

=

t

t

x

u

t

t

L

u

t

u

as

all

for

:

conditions

Boundary

,

0

)

,

(

0

,

0

)

,

(

)

,

0

(

Solution

∑

∑

∞ = − ∞ =

=

=

1 1

sin

)

,

(

)

,

(

2 n t n n n

x

L

n

e

B

t

x

u

t

x

u

λn

π

:

is

equation

diffusion

1D

the

to

solution

the

that

Recall

0 1

)

(

sin

)

0

,

(

x

f

x

T

L

n

B

x

u

n n

=

=

=

∑

∞ =

π

:

condition

Initial

∫

∫

∫

=

=

=

π

θ

θ

π

π

π

0 0 0 0 0

sin

2

sin

2

sin

)

(

2

d

n

T

xdx

L

n

L

T

B

xdx

L

n

x

f

L

B

L n L n

:

find

we

equation,

wave

the

for

As

Solving for

B

n

)

1

2

(

2

)

1

2

(

0

2

cos

sin

0 0

−

−

=

=

⎥⎦

⎤

⎢⎣

⎡−

=

∫

m

m

n

m

n

n

n

d

n

equals

it

odd,

is

if

is

this

then

even,

is

if

Evidently,

π π

θ

θ

θ

∑

∞ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − −

−

⋅

−

=

1 ) 1 2 ( 0

sin(

2

1

)

2

)

1

2

(

2

)

,

(

2 m t L c m

x

L

m

e

T

m

t

x

u

π

π

π

find

we

so,

and

Changing the initial condition

L

x T

T_o

All the other boundary conditions remain the same:

0

)

,

(

→

⇒

∞

→

u

x

t

t

(must eventually cool down to zero)

u(0,t) = 0 for all t > 0

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

(quenched at both ends)

⎪ ⎩ ⎪ ⎨ ⎧ ≤ ≤ − ≤ ≤ = L x L x L L T L x x L T x f 2 ) ( 2 2 0 2 ) ( 0 0 if if initially that Suppose

Solution

∑

∑

∞ = − ∞ =

=

=

1 1

sin

)

,

(

)

,

(

2 n t n n n

x

L

n

e

B

t

x

u

t

x

u

λn

π

:

is

equation

diffusion

1D

the

to

solution

the

that

Recall

)

(

sin

)

0

,

(

1

x

f

x

L

n

B

x

u

n n

=

=

∑

∞ =

π

:

condition

Initial

We have to solve for the coefficients using Fourier series.

Instead of orthogonality, we consult HLT

f(φ) φ A -A π/2 π which is ...) 3 cos 9 1 (cos A 8 ) ( f = ₂ φ + φ + π φ

Let’s shift this by

π

/2 in the

φ

direction:

θ

=

φ

-

π

/2 to give:

π/2 ...) ) 2 ( 3 cos 9 1 ) 2 (cos( A 8 ) 2 ( f ) ( g = + = ₂ θ +π + θ +π + π π θ θ

∑

∞ = ≡ + + = 1 n n 2 sin3 ...) B sinn 9 1 (sin A 8 _{θ θ θ} π g(θ) A -A π θ

This is now what we need between

0 <

θ

<

π

with

Α = Τ

_ο

So,

2 2 0 n

)

1

n

2

(

T

8

B

−

=

π

and the final solution for this problem is

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−

−

−

=

∑

∞ = 2 2 2 2 2 0 1

)

1

2

(

exp

)

1

2

(

sin

)

1

2

(

8

)

,

(

L

t

n

k

L

x

n

n

T

t

x

T

n

π

π

π

Compare this with our previous example of constant

initial temperature distribution:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

−

−

−

=

∑

∞ = 2 2 2 0 1

)

1

2

(

exp

)

1

2

(

sin

)

1

2

(

4

)

,

(

L

t

n

k

L

x

n

n

T

t

x

T

n

π

π

π

Derivation from electrostatics: the ‘Telegraph Equation’

δx I(x) V(x) Rδx I(x+δx) V(x+δx) Cδx

R: Resistance per unit length C: Capacitance per unit length

(

) ( )

t

V

x

C

x

I

x

x

I

∂

∂

−

=

−

+

δ

δ

:

find

we

,

From

t

V

C

I

∂

∂

=

(

x

x

) ( )

V

x

IR

x

V

+

δ

−

=

−

δ

Ohm’s law:

(

) ( )

x

x

I

x

I

x

x

I

δ

δ

∂

∂

+

=

+

(

) ( )

x

x

V

x

V

x

x

V

δ

δ

∂

∂

+

=

+

t

V

C

x

I

∂

∂

=

∂

∂

−

RI

x

V

=

∂

∂

−

We find

t

V

RC

x

V

2 2

∂

∂

=

∂

∂

The diffusion

equation

Example from electrostatics

0

,

0

)

,

(

0

,

)

,

0

(

₀

>

=

>

=

t

t

L

u

t

V

t

u

all

all

:

conditions

Boundary

L

x

x

u

(

,

0

)

=

0

,

for

0

≤

≤

:

conditions

Initial

Initially, a uniform conductor has zero potential throughout. One end (x=0) is then subjected to constant potential V₀ while the other end (x=L) is held at zero potential. What is the transient potential distribution?

We again use separation of variables; but we need to start from scratch because so far we have assumed that the boundary conditions were

0

)

,

(

)

,

0

(

t

=

u

L

t

=

u

but this is not the case here.

We now retrace the steps for the original solution to the heat equation, noting the differences

The first step is to

assume

that the function of two variables has a very

special form: the product of two separate functions, each of one variable,

that is:

Step 1: PDE

Î

2 ODEs

)

(

)

(

)

,

(

x

t

F

x

G

t

u

=

:

that

Assume

2 2 2 2

''

''

dx

F

d

F

dt

dG

G

G

F

x

u

G

F

t

u

=

=

=

∂

∂

=

∂

∂

and

where

:

find

we

ating,

Differenti

&

&

G

F

c

G

F

&

=

2

''

k

F

F

G

c

G

=

''

=

2

&

So that

equivalently

0

0

''

2

=

−

=

−

G

kc

G

kF

F

&

The two (homogeneous) ODEs are:

Step 2a: solving for F&G ..

k

≥

0

0

''

−

kF

=

F

before

as

which

from

that

so

:

is

solution

general

the

case

0

0

)

0

(

)

(

)

0

,

(

)

(

0

2

=

=

=

=

+

=

>

=

−

B

A

G

x

F

x

u

Be

Ae

x

F

k

µ

µx µx ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − = + = = = = + = = L x V x F L V a b aL L F b V F b ax x F k 1 ) ( , 0 ) ( ) 0 ( ) ( : 0 0 0 0 that so so : conditions boundary the Applying case

We have to solve:

(ignore) constant is that so , have we Since k = 0, G& = 0 G

Step 2b: Solving for F&G …

k<0

0 sin , 0 sin ) ( 0 ) 0 ( sin cos ) ( : 0 2 = = = = = + = < − = pL pL B L F A F px B px A x F p k so and find we , conditions boundary the ng Applyi is solution the case x L n x F L n p n pL n π π π sin ) ( = = = that so : is that

L

c

n

L

n

p

=

π

,

and,

as

before,

we

denote

λ

_n

=

π

t n n n n

e

B

t

G

G

G

2

)

(

0

2 λ

λ

−

=

=

+

:

is

solution

general

whose

have

We

&

Step 2c: combining F&G

∑

∞ = − + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = 1 0 1 sin ) , ( 2 n t n x L n e B L x V t x u λn π : is equation diffusion 1D the to solution the case, this In

∑

∫

∫

∑

∞ = − ∞ = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = = + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = 1 0 0 0 0 0 0 0 1 0 sin 1 2 1 ) , ( 1 . 2 sin 1 2 sin 1 2 0 sin 1 ) 0 , ( 2 n t L n n n x L n e n V L x V t x u n V d n V xdx L n L x V L B x L n B L x V x u n π π π θ θ π θ π π π λ π : is Solution list) earlier on last to next HLT use (or parts) (by which that so : condition Initial