fisika matematika iii · linearity concept a linier operator by definition satisfies example l(c1u1...
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![Page 1: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/1.jpg)
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
![Page 2: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/2.jpg)
After this lecture After this lecture ……....
2
2
x
uk
t
u
∂∂=
∂∂
??),( txu
Separation of Variables
With IC and BC
![Page 3: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/3.jpg)
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 ∂∂+
∂∂=+
∂∂
![Page 4: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/4.jpg)
HomogenityHomogenity
�� See the first slide of this lectureSee the first slide of this lecture
![Page 5: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/5.jpg)
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
![Page 6: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/6.jpg)
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
![Page 7: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/7.jpg)
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
![Page 8: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/8.jpg)
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
![Page 9: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/9.jpg)
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 =
![Page 10: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/10.jpg)
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
![Page 11: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/11.jpg)
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
![Page 12: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/12.jpg)
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
![Page 13: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/13.jpg)
SummarySummary
λφφ −=2
2
dx
d0)0( =φ 0)( =Lφ
2
=L
nn
πλ K,3,2,1=nL
xnxn
πφ sin)( =
![Page 14: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/14.jpg)
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
![Page 15: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/15.jpg)
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,( =
![Page 16: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/16.jpg)
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
![Page 17: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/17.jpg)
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)
![Page 18: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/18.jpg)
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
![Page 19: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/19.jpg)
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
![Page 20: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/20.jpg)
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
![Page 21: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/21.jpg)
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
![Page 22: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/22.jpg)
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
![Page 23: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/23.jpg)
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
![Page 24: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/24.jpg)
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 =
![Page 25: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/25.jpg)
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
![Page 26: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/26.jpg)
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
*
**
![Page 27: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/27.jpg)
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
![Page 28: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/28.jpg)
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
![Page 29: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/29.jpg)
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
![Page 30: 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 ∂ ∂ + ∂ ∂ + = ∂ ∂](https://reader033.vdocument.in/reader033/viewer/2022060718/607e920f9d6a95154a07852a/html5/thumbnails/30.jpg)
�� http://phys.unpad.ac.id/staff/irwanhttp://phys.unpad.ac.id/staff/irwan