Fisika Fisika MatematikaMatematika IIIIII
Parabolic Partial Differential EquationsParabolic Partial Differential Equations
-- Method of Separation of VariablesMethod of Separation of Variables
Irwan Ary Irwan Ary DharmawanDharmawan
http://http://phys.unpad.ac.id/jurusan/staff/dharmawan/kuliahphys.unpad.ac.id/jurusan/staff/dharmawan/kuliah
After this lecture After this lecture ……....
2
2
x
uk
t
u
∂∂=
∂∂
??),( txu
Separation of Variables
With IC and BC
Linearity ConceptLinearity Concept
�� A linier operator by definition satisfiesA linier operator by definition satisfies
�� ExampleExample
)()()( 22112211 uLcuLcucucL +=+
L
t
uc
t
ucucuc
t ∂∂+
∂∂=+
∂∂ 2
21
12211 )(
22
2
221
2
122112
2
)(x
uc
x
ucucuc
x ∂∂+
∂∂=+
∂∂
HomogenityHomogenity
�� See the first slide of this lectureSee the first slide of this lecture
Principle of SuperpositionPrinciple of Superposition
�� If and satisfy a linear homogeneous equation, If and satisfy a linear homogeneous equation,
then an arbitrary linear combination of them, then an arbitrary linear combination of them,
also satisfies the same linear homogeneous equationalso satisfies the same linear homogeneous equation
1u
2211 ucuc +2u
Heat Equation with Zero Temperatures Heat Equation with Zero Temperatures
at Finite Endsat Finite Ends
�� Consider the linear Consider the linear homegeneoushomegeneous Heat Equation in a oneHeat Equation in a one--
dimensional rod ( ) with constant thermdimensional rod ( ) with constant thermal al
coeficientscoeficients and no sources of thermal energyand no sources of thermal energy
Lx <<0
2
2
x
uk
t
u
∂∂=
∂∂
)()0,( xfxu =
0)(),(
0)(),0(
2
1
====
tTtLu
tTtu
IC
BC
PDE
Separation of VariablesSeparation of Variables
)()(),( tGxtxu φ=
dt
dGx
t
u)(φ=
∂∂ )(
2
2
2
2
tGdx
d
x
u φ=∂∂
2
2
x
uk
t
u
∂∂=
∂∂
Separation of Variables
Plug in to Heat Eqn.
)()(2
2
tGdx
dk
dt
dGx
φφ =
2
211
dx
d
dt
dG
kG
φφ
=
Equating
λ−= Separation constant
Linear homgn. PDE with
liniear homgn. BC)()0,( xfxu =
0)(),(
0)(),0(
2
1
====
tTtLu
tTtu
Separation of VariablesSeparation of Variables
λφφ −=2
2
dx
d
λ−=
kGdt
dG λ−=
2
211
dx
d
dt
dG
kG
φφ
=
ktcetG λ−=)(
?
Eigenvalue Problem
EigenvalueEigenvalue and and EigenfunctionEigenfunction
((λλ>0)>0)
λφφ −=2
2
dx
d 0)0( =φ0)( =Lφ
Boundary Value Problem
General Solution xcxc λλφ sincos 21 +=
Plug into BC
00sin0cos)0( 21 =+= λλφ cc
01 =c
0sincos)( 21 =+= LcLcL λλφ
0sin =Lλ2
=L
nπλ K,3,2,1=n
02 ≡c Trivial Solution
πλ nL =
EigenvalueEigenvalue and and EigenfunctionEigenfunction
((λλ>0)>0)
2
=L
nπλ K,3,2,1=n
L
xncxcx
πλφ sinsin)( 22 ==
L
xnx
πφ sin)( =
Eigenvalue
Eigenfunction
12 =c
EigenvalueEigenvalue and and EigenfunctionEigenfunction
((λλ=0)=0)
xcc 21 +=φ
λφφ −=2
2
dx
d
0)0( 1 == cφ 0)( 21 =+= LccLφ02 =c
Trivial Solution
EigenvalueEigenvalue and and EigenfunctionEigenfunction
((λλ<0)<0)
λφφ −=2
2
dx
d 0)0( =φ0)( =Lφ
Boundary Value Problem
General Solution xcxc λλφ −+−= sinhcosh 43
Plug into BC
00sin0cosh)0( 23 =−+−= λλφ cc
03 =c
0sinh)( 4 =−= LcL λφ
04 ≡c Trivial Solution
Since sinh never zero for a positive
argumen
SummarySummary
λφφ −=2
2
dx
d0)0( =φ 0)( =Lφ
2
=L
nn
πλ K,3,2,1=nL
xnxn
πφ sin)( =
Product SolutionProduct Solution
ktcetG λ−=)(
)()(),( tGxtxu φ=
2
=L
nn
πλL
xnxn
πφ sin)( =
tLnkeL
xnBtxu
2)/(sin),( ππ −=
2ccB =
K,3,2,1=n
Initial Condition
Initial Value Problems (example)Initial Value Problems (example)
2
2
x
uk
t
u
∂∂=
∂∂
tLkeL
xtxu
2)/3(3sin4),( ππ −=
0)(),(
0)(),0(
2
1
====
tTtLu
tTtu
L
xxu
π3sin4)0,( =
Principle of SuperpositionPrinciple of Superposition
tLnkM
nn e
L
xnBtxu
2)/(
1
sin),( ππ −
=∑=
Muuuu ,,,, 321 K
∑=
=++++M
nnnMM ucucucucuc
1332211 K
IfAre solutions of a linear
homogeneous problem
then
is also a solution
Initial Condition and Fourier SeriesInitial Condition and Fourier Series
L
xnBxfxu
M
nn
πsin)()0,(
1∑
=
==
FOURIER SERIES
• Any function f(x) (with certain very reasonable restriction, to be discussed
later) can be approximated by a finite linier combination of sin(nπx/L)
• The approximation may not be very good for small M, but gets to be a better
and better approximation as M is increased
• If we consider the limit as M tend to infinity, then not only is *) the best
approximation to f(x) using combination of eigenfunctions, but the resulting
infinite series will converge to f(x)
Fourier SeriesFourier Series
L
xnBxf
nn
πsin)(
1∑
∝
=
=
Any initial condition f(x) can be written as an infinite linear
combination of sin(nπx/L), known as a type of
Fourier Series
tLnk
nn e
L
xnBtxu
2)/(
1
sin),( ππ −∝
=∑=
What is more important is that we also claim that the
corresponding infinite series is the solutions of our heat
conduction problem
OrthogonalityOrthogonality of of SinesSines
∫
=≠
=L
nmL
nmdx
L
xm
L
xn
0 2/
0sinsin
ππ
L
xm
L
xnB
L
xmxf
nn
πππsinsinsin)(
1∑
∝
=
=
dxL
xm
L
xnBdx
L
xmxf
L
nn
L
∫∑∫∝
=
=010
sinsinsin)(πππ
dxL
xmBdx
L
xmxf
L
m
L
∫∫ =0
2
0
sinsin)(ππ
OrthogonalityOrthogonality of of SinesSines
dxL
xmxf
Ldx
L
xm
dxL
xmxf
BL
L
L
m ∫∫
∫==
0
0
2
0 sin)(2
sin
sin)(π
π
π
Final SummaryFinal Summary
Heat Equation with Zero Temperatures Heat Equation with Zero Temperatures
at Finite Endsat Finite Ends
tLnk
nn e
L
xnBtxu
2)/(
1
sin),( ππ −∝
=∑=
2
2
x
uk
t
u
∂∂=
∂∂ )()0,( xfxu =
0)(),(
0)(),0(
2
1
====
tTtLu
tTtu
IC
BC
PDE
The solution is
dxL
xnxf
LB
L
n ∫=0
sin)(2 π
Heat Equation in Rod with Insulated EndsHeat Equation in Rod with Insulated Ends
�� Consider the linear Consider the linear homegeneoushomegeneous Heat Equation in a oneHeat Equation in a one--
dimensional rod ( ) with constant thermdimensional rod ( ) with constant thermal al
coeficientscoeficients and no sources of thermal energyand no sources of thermal energy
Lx ≤≤0
2
2
x
uk
t
u
∂∂=
∂∂
)()0,( xfxu =
0),(
0),0(
=∂∂
=∂∂
tLx
u
tx
uIC
BC
PDE
Separation of VariablesSeparation of Variables
�� By using the same method as previous, we will have By using the same method as previous, we will have
the following resultsthe following results
∑∝
=
−=0
)/( 2
cos),(n
ktLnn e
L
xnAtxu ππ
∫=L
dxxfL
A0
0 )(1
∫=L
m dxL
xmxf
LA
0
cos)(2 π
EigenvalueEigenvalue and and EigenfunctionEigenfunction
((λλ>0)>0)
λφφ −=2
2
dx
d 0)0( =φ0)( =Lφ
Boundary Value Problem
General Solution xcxc λλφ sincos 21 +=
Plug into BC
00sin0cos)0( 21 =+= λλφ cc
01 =c
0sincos)( 21 =+= LcLcL λλφ
0sin =Lλ2
=L
nπλ K,3,2,1=n
02 ≡c Trivial Solution
πλ nL =
Heat Equation in a Thin Circular RingHeat Equation in a Thin Circular Ring
Lx
Lx
−==
Lx 20 ≤≤
2
2
x
uk
t
u
∂∂=
∂∂
)()0,( xfxu =
),(),(
),(),(
tLutLu
tLx
utL
x
u
=−∂∂=−
∂∂
IC
BC
PDE
0=x
EigenvalueEigenvalue and and EigenfunctionEigenfunction ((λλ>0)>0)
λφφ −=2
2
dx
d
)()( LL −= φφ
xcxc λλφ sincos 21 +=Boundary Value Problem
Plug into BC
LcLcLcLc λλλλ sincos)(sin)(cos 2121 +=−+−
0sin2 =Lc λLL λλ sin)(sin −=− LL λλ cos)(cos =−
We obtain
Plug into BC )()( Ldx
dL
dx
d φφ =−
)cossin( 21 xcxcdx
d λλλφ +−=
0sin1 =Lc λλWe obtain
*
**
EigenvalueEigenvalue and and EigenfunctionEigenfunction ((λλ>0)>0)
0sin2 =Lc λ
0sin1 =Lc λλ0sin =Lλ
2
=L
nπλ
Since there are no additional constraints that c1 and c2 must
satisfy. We say that both sin and cos are eigenfunctions
corresponding to the eigenvalue
,...3,2,1,sin,cos)( == nL
xn
L
xnx
ππφ
General solution General solution
ktLxneL
xntxu
2)/(cos),( ππ −=
ktLxneL
xntxu
2)/(sin),( ππ −=
∑∑∝
=
−−∝
=
++=1
)/()/(
10
22
sincos),(n
ktLxnn
ktLxn
nn e
L
xnbe
L
xnaatxu ππ ππ
In fact any linear combination of cos nπx/L and sin nπx/L is
an eigenfunctions. There are thus two infinite families of
product solutions of the PDE, n=1,2,3 …
∑∑∝
=
∝
=
++=11
0 sincos)(n
nn
n L
xnb
L
xnaaxf
ππ
General SolutionGeneral Solution
∫−
=L
L
dxxfL
a )(2
10
∫−
=L
L
m dxL
xnxf
La
πcos)(
1
∫=L
n dxL
xnxf
Lb
0
sin)(1 π
�� http://phys.unpad.ac.id/staff/irwanhttp://phys.unpad.ac.id/staff/irwan