Fisika Matematika III · Linearity Concept A linier operator by definition satisfies Example L(c1u1 +c2u2 ) =c1L(u1)+c2L(u2) L t u c t u c u c u c t ∂ ∂ + ∂ ∂ + = ∂ ∂

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


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 =


((λλ>0)>0)

2

=L

nπλ K,3,2,1=n

L

xncxcx

πλφ sinsin)( 22 ==

L

xnx

πφ sin)( =

Eigenvalue

Eigenfunction

12 =c


((λλ=0)=0)

xcc 21 +=φ

λφφ −=2

2

dx

d

0)0( 1 == cφ 0)( 21 =+= LccLφ02 =c

Trivial Solution


((λλ<0)<0)

λφφ −=2

2

dx

d 0)0( =φ0)( =Lφ


General Solution xcxc λλφ −+−= sinhcosh 43

Plug into BC

00sin0cosh)0( 23 =−+−= λλφ cc

03 =c

0sinh)( 4 =−= LcL λφ


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


�� 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


((λλ>0)>0)

λφφ −=2

2

dx

d 0)0( =φ0)( =Lφ


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


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

Documents

Fisika Matematika III · Linearity Concept A linier operator by definition satisfies Example L(c1u1 +c2u2 ) =c1L(u1)+c2L(u2) L t u c t u c u c u c t ∂ ∂ + ∂ ∂ + = ∂ ∂