Heat equation
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Transcript of Heat equation
Heat (or Diffusion) equation in 1D*
• Derivation of the 1D heat equation• Separation of variables (refresher)• Worked examples
*Kreysig, 8th Edn, Sections 11.4b
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
heat flow
Derivation of the heat equation in 1D
),( txutxA
K
by time at point at etemperatur the Denote is area sectional Cross
is material the ofdensity The is heatspecific The
is wirethe inty conductivi thermal the that Suppose
ρσ
x xx δ+
xxuKA
xuKAxx
xuKAx
xx
x
δ
δδ
2
2
:
∂∂
∂∂
+
∂∂
−
+
: :is outflow net the So
: face the At
at face across bar intoflow Heat
xtuAx
xuKA δσρδ
∂∂
=∂∂
2
2
:gives heat of onConservati
σρKc
xuc
tu
=∂∂
=∂∂ 2
2
22 where,
Boundary and Initial Conditions
0),(),0( == tLutu
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,(
)(
===
Lffxfxu
xf
:conditionsboundary the from Evidently, :condition initial the have we
,by given is rod the in ondistributi etemperatur initial the If
Solution by Separation of Variables
1. Convert the PDE into two separate ODEs
2. Solve the two (well known) ODEs3. Compose the solutions to the two ODEs
into a solution of the original PDE– This again uses Fourier series
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 2 ODEs
)()(),( tGxFtxu = :that Assume
2
2
2
2
''
''
dxFdF
dtdGG
GFxu
GFtu
==
=∂∂
=∂∂
and where
:find weating,Differenti
&
&GFcGF ''2=&
kFF
GcG
==''
2
&
So that
equivalently
0
0''2 =−
=−
GkcG
kFF&
The two (homogeneous) ODEs are:
Step 2a: solving for F*0'' =−kFF
00
0
)(02
===+
=+
+=
>=
−
−
BABeAe
BA
BeAexFk
LL
xx
so and
conditionsboundary the ng Applyi
:is solution general the case
µµ
µµ
µ
0)(
:0
==+=
=
babaxxF
k
:conditionsboundary the Applying
case
0sin,0sin)(0)0(
sincos)(:02
=====
+=<−=
pLpLBLFAF
pxBpxAxFpk
so and
find we,conditionsboundary the ng Applyi
is solution the case
We have to solve:
xL
nxF
Lnp
npL
nπ
ππ
sin)( =
=
=
that so
:isthat
2pk −= is interest of caseonly the that follows It
* This analysis is identical to the Wave Equation case we studied earlier
Step 2b: solving for G
Lcn
Lnp n
πλπ== denote weequation, wavethe for as and, ,
tnn
n
neBtG
GG2
)(
02
λ
λ−=
=+
:is solution general whose
have We &
Step 2c: combining F&G
∑∑∞
=
−∞
=
==11
sin),(),(2
n
tn
nn x
LneBtxutxu n
πλ
:is equation diffusion 1D the to solution The
)(sin)0,(1
xfxL
nBxun
n == ∑∞
=
π :condition Initial
∫=L
n xdxL
nxfL
B0
sin)(2 π : find weequation, wavethe for As
Analysing the solution
xL
netxu
txuBtxu
tn
nnn
nπλ sin),(
),(),(
2
1
−
∞
=
=
= ∑
where
:as writtenbe can equation diffusion 1D the to solution The
∫=
==
L
n
n
n
n
xdxL
nxfL
B
BtLutu
cLu
txu
0
sin)(2
0),(),0(,
),(
π
since ,conditions initial theby determined are weightsThe (2) and case; this in
conditionsboundary the and ) constants the is, (that problemgeneric theby determined completely are functions The (1)
where functions of sum weighteda is solution the that emphasises This
We now take a closer look at the functions un
Sine term x
Ln πsin :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
158.11002
21
2
2
=⎟⎠⎞
⎜⎝⎛=
==
⎟⎠⎞
⎜⎝⎛−
Lc
cLe
tL
nc
πλ
π
Then
later) slides 2 (see and that Suppose :term lexponentia the next Consider
Gabor functions
Gaussian a ...
is term the then and denote
is term lexponentia The
2
2
2
,
σ
π
πσ
s
n
tL
nc
e
ncLst
e
−
⎟⎠⎞
⎜⎝⎛−
⎟⎠⎞
⎜⎝⎛==
optics and processing signal to lfundamenta are and functions Gabor called are These term. sine aby multiplied Gaussian a
with workto have we of instead consider weif so ),(),( txsx
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 …
∑∑∞
=
−∞
=
==11
sin),(),(2
n
tn
nn x
LneBtxutxu n
πλ
:is equation diffusion 1D the to solution The
)(sin)0,(1
xfxL
nBxun
n == ∑∞
=
π :condition Initial
∫=L
n xdxL
nxfL
B0
sin)(2 π : find weequation, wavethe for As
Example 1*
xxf80
sin100)( π= :conditions initial Sinusoidal
cms 80:bar the of Length
0...,10080
sin100)(80
sin)0,(
321
1
====
=== ∑∞
=
BBB
xxfxnBxun
nππ
xetxu
Kc
K
t
80sin100),(
001758.06400
870.9158.1
sec/158.192.8092.0
95.0
092.092.895.0
001785.0
21
2
π
λ
ρσ
σ
ρ
−=
==
=⋅
==
=
=
=
cm
soC)cal/(gm
gm/cm C)sec cal/(cm :Copper
2
o
3
o
*Kreysig, 8th Edn, page 603
min5.638850:
==⇒= st 100e 50 to reduce to etemperatur maximum the for required time
0.001785t-
o
Example 2
0)0,( Txu = :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.
∞→→>==
ttxuttLutu
as all for
:conditionsBoundary
,0),(0,0),(),0(
Solution
∑∑∞
=
−∞
=
==11
sin),(),(2
n
tn
nn x
LneBtxutxu n
πλ
:is equation diffusion 1D the to solution the that Recall
01
)(sin)0,( TxfxL
nBxun
n === ∑∞
=
π :condition Initial
∫
∫
∫
=
=
=
π
θθπ
π
π
0
0
0
0
0
sin2
sin2
sin)(2
dnT
xdxL
nLTB
xdxL
nxfL
B
L
n
L
n
: find weequation, wavethe for As
Solving for nB
)12(2)12(
02
cossin0
0
−−=
=
⎥⎦⎤
⎢⎣⎡−=∫
mmn
mnnndn
equals it odd, is if
is this then even, is if
Evidently,π
π θθθ
∑∞
=
⎟⎠⎞
⎜⎝⎛ −
−−⋅
−=
1
)12(0 )12sin(2
)12(2),(
2
m
tL
cm
xL
meTm
txu ππ
π
find weso, and
Changing the initial condition
L
x
T
To
All the other boundary conditions remain the same:
0),( →⇒∞→ txut (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)
⎪⎩
⎪⎨
⎧
≤≤−
≤≤=
LxLxLLT
LxxLT
xf
2)(2
202
)(0
0
if
ifinitially that Suppose
Solution
∑∑∞
=
−∞
=
==11
sin),(),(2
n
tn
nn x
LneBtxutxu n
πλ
:is equation diffusion 1D the to solution the that Recall
)(sin)0,(1
xfxL
nBxun
n == ∑∞
=
π :condition Initial
We have to solve for the coefficients using Fourier series.
Instead of orthogonality, we consult HLT
f(φ)
φ
A
-Aπ/2 π
which is
.....)3cos91(cosA8)(f 2 ++= φφ
πφ
Let’s shift this by π/2 in the φ direction: θ = φ-π/2 to give:
π/2
.....))2(3cos91)2(cos(A8)2(f)(g 2 ++++=+= πθπθ
ππθθ
∑∞
=
≡++=1n
n2 nsinB.....)3sin91(sinA8 θθθ
π
g(θ)
A
-Aπ
θ
This is now what we need between 0 < θ < π with Α = Το
So, 220
n )1n2(T8B
−=
π
and the final solution for this problem is
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−
−−
=∑∞
=2
22
220
1
)12(exp)12(sin)12(
8),(L
tnkL
xnn
TtxTn
πππ
Compare this with our previous example of constant initial temperature distribution:
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−
−−
=∑∞
=2
220
1
)12(exp)12(sin)12(
4),(L
tnkL
xnn
TtxTn
πππ
Derivation from electrostatics: the ‘Telegraph Equation’
δx
I(x)
V(x)
RδxI(x+δx)
V(x+δx)Cδx
R: Resistance per unit length
C: Capacitance per unit length
( ) ( )tVxCxIxxI
∂∂
−=−+ δδ
:find we, FromtVCI
∂∂
=
( ) ( ) xIRxVxxV δδ −=−+
Ohm’s law:
( ) ( ) xxIxIxxI δδ
∂∂
+=+
( ) ( ) xxVxVxxV δδ
∂∂
+=+
tVC
xI
∂∂
=∂∂
−
RIxV
=∂∂
−
We findtVRC
xV2
2
∂∂
=∂∂ The diffusion
equation
Example from electrostatics
0,0),(0,),0( 0
>=>=
ttLutVtu
all all
:conditionsBoundary
Lxxu ≤≤= 0,0)0,( for :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( == tLutu 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 2 ODEs
)()(),( tGxFtxu = :that Assume
2
2
2
2
''
''
dxFdF
dtdGG
GFxu
GFtu
==
=∂∂
=∂∂
and where
:find weating,Differenti
&
&GFcGF ''2=&
kFF
GcG
==''
2
&
So that
equivalently
0
0''2 =−
=−
GkcG
kFF&
The two (homogeneous) ODEs are:
No change here!!
Step 2a: solving for F&G .. k≥00'' =−kFF
before as whichfrom that so :is solution general the case
00)0()()0,()(02
====+=>= −
BAGxFxuBeAexFk xx µµµ
⎟⎠⎞
⎜⎝⎛ −=
−=+==
==
+==
LxVxF
LVabaLLF
bVF
baxxFk
1)(
,0)(
)0(
)(:0
0
0
0
that so
so
:conditionsboundary the Applying
case
We have to solve:
(ignore) constant is that so , have we Since GGk 0,0 == &
Step 2b: Solving for F&G …k<0
0sin,0sin)(0)0(
sincos)(:02
=====
+=<−=
pLpLBLFAF
pxBpxAxFpk
so and
find we,conditionsboundary the ng Applyi
is solution the case
xL
nxF
Lnp
npL
nπ
ππ
sin)( =
=
=
that so
:is that
Lcn
Lnp n
πλπ== denote webefore, as and, ,
tnn
n
neBtG
GG2
)(
02
λ
λ−=
=+
:is solution general whose
have We &
This case is as before
Step 2c: combining F&G
∑∞
=
−+⎟⎠⎞
⎜⎝⎛ −=
10 sin1),(
2
n
tn x
LneB
LxVtxu n
πλ
:is equation diffusion 1D the to solution the case, this In
∑
∫
∫
∑
∞
=
−
∞
=
−⎟⎠⎞
⎜⎝⎛ −=
−=
⎟⎠⎞
⎜⎝⎛ −
−=
⎟⎠⎞
⎜⎝⎛ −−=
=+⎟⎠⎞
⎜⎝⎛ −=
1
00
0
0
0
0 0
10
sin121),(
1.2
sin12
sin12
0sin1)0,(
2
n
t
L
n
nn
xL
nen
VLxVtxu
nV
dnV
xdxL
nLxV
LB
xL
nBLxVxu
nπ
π
π
θθπθ
π
π
π
λ
π
:is Solution
list) earlier on last to next HLT use (or parts)(by which
that so
:condition Initial