Download - Chemical Engineering Mathematics
CHEMICAL ENGINEERING MATHEMATICS II
ALI ALTWAY
REFERENCESRichard G.Rice, Duong D.Do,”Applied Mathematics and Modeling for chemical Engineers”, John Wiley,New YorkMickley, Reed, Sherwood,”Applied Mathematics in Chemica Engineering”,MsGraw-HillJenson and Jeffrey,”Mathematical Methods in Chemical Engineering”,Academic Press
Course Outline1. Mathematical Formulation of Physicochemical Problems 2. Analytical Series Solution of Ordinary Differential Equation and Special Functions3. Analytical Solution of Partial Differential Equation
Course ObjectiveStudents have capability to apply mathematics to solve Physicochemical Problems.
Evaluation Assignment 20% QUIZ 40%
Exam 40% ------- 100%
MATHEMATICAL FORMULATION OF PHYSICOCHEMICAL PROBLEMS
Objective: Students have capability to develop mathematical formulation of Physicochemical problems and solve the mathematical problems using known mathematical methods.
MATHEMATICAL FORMULATION The mathematical treatment of engineering problems involves three basic steps: the expression of the problem in mathematical language, the solution of mathematical problems, and the interpretation of the results.
Physicochemical Problems
Mathematical Formulation
Solution of Mathematical Formulation
Interpretation
Assumption
Law Conservation
Rate
Equilibrium
Analytical
Numerical
FUNDAMENTAL LAWS There are three basic physical and chemical law, they are: Conservation law, Rate expression, and Equilibrium relation.
CONSERVATION LAWS
Rate of mass accumulation in system = Rate of mass in – Rate of mass out
Mass, Overall:
Mass, Component:
Rate of mass accumulation of component i in system = Rate of mass of Component i in – Rate of mass of component i out +Rate of mass generation of Component i
Rate of Energy Accumulation = Rate of Energy in – Rate of Energy out + Rate of Energy Generation
Energy
Rate of Momentum Accumulation = Rate of Momentum in – Rate of Momentum Out + Rate of Momentum Generation
Momentum
Body Force Surface Force
RATE EXPRESSION
FUNDAMENTAL LAWS
Heat Transfer
Conduction
Convection (Interface Transport)
Mass Transfer
Diffusion :
Convection (Interface Transport):
x
TkAq x
)TT(hAq fs
x
CSDN A
AAx
)CC(SkN AbASCA
Molecular (Newtonian) :
Interface Transport
x
U yxy
f.K
Chemical Reaction
Momentum Transfer
cCbBaA BAA CkCr
EQUILIBRIUM RELATION
FUNDAMENTAL LAWS
Phase Equilibrium : Vapor-liquid ------- Raoult Law Liquid-liquid Gas/vapor-solid Liquid-solid
Chemical Equilibrium
cCbBaA
bB
aA
cC
CC
CK
LUMP PARAMETER AND DISTRIBUTED PARAMETER MODELS
Mathematical models can be classified into two distinct types: Lump parameter and Distributed parameter models. The first type is characterized by the uniformity of the parameter value in the system such as mixed flow reactor, while the second type is characterized by the variability of the variable/parameter value in the system such as plug flow reactor. The distributed parameter model is usually called Transport Phenomena models, because it involves the phenomena of heat, mass or momentum transport. For the distributed parameter models, we have to consider more specifically on the boundary condition of the system.
GENERAL TYPE OF BOUNDRY CONDITION
1. The temperature at a surface may be specified2. The heat flux at a surface may be given, e.g. q=qo. When a surface is assumed completely isolated, then the heat flux at that surface is equal to zero. 3. At solid-fluid interface the heat flux may related to the difference between the temperature at the interface and that in the fluid, thus
fluidTThq
4. At solid-solid interface, the continuity of temperature and the normal component of the heat flux may be specified.5. At plane, axes or point of symmetry, the heat flux is equal to zero, except the symmetry is treated as the heat source.
h=heat transfer coefficient
HEAT TRANSFER
1. The concentration at a surface can be specified.2. The mass flux at a surface can be specified.3. If diffusion is occurring in a solid, it may happen that at the solid surface substance A is lost to a surrounding fluid stream according to the relation,
MASS TRANSPORT
AfAcA CCkN 00
4. The rate of chemical reaction at the surface can be specified. For example, if a substance A disappears at a surface by a first-order chemical reaction,
AA CkN "10
5. At the plane, axes, or point of symmetry the mass flux is equal to zero.
kc = mass transfer coefficient
GENERAL TYPE OF BOUNDRY CONDITION
1. At solid-fluid interfaces the fluid velocity equals the velocity with which the surface itself is moving.2. At liquid-gas interface, the momentum flux (hence the velocity gradient) is equal to zero.3. At liquid-liquid interfaces the momentum flux perpendicular to the interface, and the velocity, are continuous across the interface.4. At the plane, axes, or point of symmetry the momentum flux is equal to zero.
MOMENTUM TRANSPORT
GENERAL TYPE OF BOUNDRY CONDITION
Gas-liquid interface
Solid-Liquid interfaceliquid
GENERAL STEPS
1. Draw the sketch of the system to be modeled and label/define the various geometric, physical and chemical quantities.2. Carefully select the important variables, and list the parameters that are expected to be important3. Establish a ”control volume” for a differential or finite element of the system to be modeled.4. Write the conservation law on the control volume and use the necessary rate expression and equilibrium relation to derive equations describing the system. 5. Write boundary and initial condition6. Solve the equations7. Interpret the solution
ILLUSTRATIVE EXAMPLES (1)
Two tanks each contains 100 liters salt solution (20 gr/lit). A stream of water is fed into the first tank at a rate of 5 liters/min. The liquid flows from the tank to the second tank at a rate of 8 liters/min. The liquid flows from the second tank at a rate of 8 liters/min where part of it (3 liters/min) is directed to the first tank and the balance flows to some points out of the system. Determine the salt concentration (gr/lit) in the first and second tank as a function of time. Assume is constant in all streams.
5 lt/min
8 lt/min
5 lt/min
SOLUTION
0835Vdt
d
0dt
dV1
1211 830,5 CCCVdt
d
121 C8C3
dt
dC100
088Vdt
d2
02 dt
dV
2122 C8C8CVdt
d
212 C8C8
dt
dC100
dt
dC5.12CC 2
21
ILLUSTRATIVE EXAMPLES (1)
V1= const = 100V2 = const = 100
Tank I:Consevation of mass (overall)
Conservation of mass (salt)
Tank II
Consevation of mass (overall)
Conservation of mass (salt)
(1)
(2)
22
221
dt
Cd5.12
dt
dC
dt
dC
dt
dC100C8C3
dt
Cd1250
dt
dC100 2
2222
22
0C3dt
dC40
dt
Cd250 2
22
22
t129.02
t031.012 eKeKC
tt eKeKdt
dC 129.02
031.01
2 129.0031.0
tt eeC 129.031.02 33.63261.26
tt eeC 129.0031.01 875.3125.16
Initial Condition: t = 0 C1=20, C2 = 20
Eq (2) is differentiated with respected to t:
(3)
Eq. (2) is substituted into Eq. (1):
m1 = - 0.031 ; m2 = - 0.129 :
(4)
(5)
Initial Condition:
t = 0 C2=20 02 dt
dC
C2 = 20 = K1 + K2
0 = - 0.031 K1 – 0.129 K2
K1 = 26.33
K2 = - 6.33
ILLUSTRATIVE EXAMPLES (1)
ILLUSTRATIVE EXAMPLES (2)
(a) A container is maintained at a constant temperature of 800o F and is fed with a pure gas A at a steady rate of 1 lbmole/min; the gas product gas stream is withdrawn from the container at the rate necessary to keep the total pressure constant at a value of 3 atm. The container contents are vigorously agitated, and the gas mixture is always well mixed. The following irreversible second order gas phase reaction occurs in the container:
At a temperature of 800 oF, the reaction rate constant for the reaction has the numerical value of 1000 ft3 /(lbmole min ). Both A and B are perfect gases. Because of their low temperature, no reaction occurs in the lines to and from the vessel. If under steady state condition, the product stream is to contain 33 1/3 mole % B, how large ( in cubic feet ) should be the volume of the reaction container ?(b) After the steady state of a) has been attained, the valve on the exit pipe ofisothermal vessel is abruptly closed. The feed rate is controlled so that the total tank pressure is maintained at 3 atm. If the mixing is still perfect, how many minutes will it take (after the instant of closing the valve) for the tank content to be 90 mole % B. The feed rate is controlled so that the total tank pressure is maintained at 3 atm.
2A → B
P = 3 atmT = 800oC
Feed
A
Product
A, B
ILLUSTRATIVE EXAMPLES (2)Solution
VV
nkF10
2
2A
Af
VV
nk
2
1F10
2
2A
Bf
VV
nkF2
2
2A
Bf
BfAf F2F10
Mass Balance:
A:
B:
3
1
FF
F
BfAf
Bf AfBf FF 2
AfAf FF10
5.0AfF
VV
nk5.010
2
2A
RT
PVy
RT
VPn AA
A
22
223)3/2()1000(5.010
TR
V
3
2Ay 8.105
)3()3/2)(1000(
)1260( )7302.0)(5.0(22
22
V
a)
V
nkF
dt
dn 2A
AoA
V
nk
dt
dn AB2
20
V
nk
dt
dn2
2AA
dtV2
k
n
dn2A
ART
PVy
RT
VPn AA
A KtV2
k
n
1
A
0345.0)1260)(7302.0(
8.105*3*1.0An
23.0 2287.0)1260)(7302.0(
8.105*3*)3/2(
K023.0
1
348.4t)8.105(*2
1000
n
1
A
348.4726.41
tnA
348.4t726.40345.0
1
b) Mass Balance:
A:
B:
t=0
348.4K
t=?→ nA = 0.0345t=5.213
ILLUSTRATIVE EXAMPLES (2)
02
2
0 V
nkF
dt
dn AA
Solution
V
nkF AA
2
02
The apparatus shown diagrammatically in fig 1 is to be used for the continuous extraction of benzoic acid from toluene, using water as the extracting solvent the two stream are fed into a tank A where they are stirred vigorously, and the mixture is then pumped into tank B where it is allowed to settle into two layers. The upper toluene layer and the lower water layer are discharged separately and the problem is to find what proportion of the benzoic acid has passed into the solvent phase.
Toluene + benzoic acid water
Water+ benzoic acid
Toluene + benzoic acid
ILLUSTRATIVE EXAMPLES (3)
R m3/s TolueneC kg/m3 benzoic acid
R m3/s Toluene X kg/m3 benzoic acid
S m3/s water S m3/s water Y kg/m3 benzoic acid
The mixture is so efficient that the two stream leaving the stage are always in equilibrium with one another. This can be expressed mathematically : y = m x
ILLUSTRATIVE EXAMPLES (3)
Mass balance :Benzoic acid :
R . C + S . 0 = R . X + S . Y
R . C = R . X + S . m X SmR
RC
X =
SmR
mS
RC
SYE
Solution:
Proportion of benzoic acid extracted is :
Two Stages :
Stage 1 Stage 2R R R
C
S S S
X1X2
Y1 Y2
The above eample will now be reconsidered, but two stages will be used for the extraction.
Mass balance :Stage 1 : R C + S Y2 = R X1 + S Y1
R C + S m X2 = R X1 + S m X1 ……………(1)
Stage 2 : R X1 + S 0 = R X2 + S Y2
= R X2 + S m X2 …………………(2) From Eq.(1), we obtain, X2=((R+Sm)X1-RC)/Sm And from EQ.(2), we obtain , X2=RX1/(R+Sm) From the last two equations we obtain, X1=RC(R+Sm)/(R2+RSm+S2m2) And the proportion of Benzoic acid extracted is, E=Sm(R+Sm)/(R2+RSm+S2m2)
ILLUSTRATIVE EXAMPLES (3)
No gram of solid material was placed in W gram of water at time to . The liquid was continuously stirred and maintained at a constant temperature. At the end of t1 second, N1 gram of solid remained undissolved. At the end of very long period of time, N2 gram of solid remained undissolved. Set up the differential equation required to determined the rate of solution of the solid in term of No, N1, N2, and to, t1 . Do not integrate your expression.Notes:It is assumed that the rate of solution is proportional to (1) the surface area of the material and (2) the concentration driving force , where the concentration is expressed as gram of solid per gram of water. The original solid consisted of S sphere each of initial diameter Do ft.
ILLUSTRATIVE EXAMPLES (4)
Mass balance:
System solid : ]CC[A.k0]N[dt
d *c
W
NNC 2o*
W
NNC o
A = D2 S ;
6
DN
3
6
DN
3o
o 3o
o
D
N6
Solution:
3o
o3
D
Nx
6
DN
3o
3
o D
D
N
N
3oo N/NDD 3 2
o2o )N/N()D)((
ILLUSTRATIVE EXAMPLES (4)
A = D2 S = S
W
NN
W
NNS )N/N(DK
dt
dN o2o3 2o
2oc
ILLUSTRATIVE EXAMPLES (5)
Consider the case of starting the equilibrium still. The still is initially charged with 20 lb mole of feed stock of composition xF =0.32 mole fraction of benzene. Feed is supplied at the rate of 10 lb mole/hr, and the heat input is adjusted so that the total mole of liquid in the still remain constant at 20 . It is desired to estimate the time required for the composition of overhead product YD to fall to 0.4 moles fraction of benzene. Benzene and toluene may be assumed to follow Raoult’s law, and the relative volatility may be taken as constant at an average value of 2.48.
Feed
XF
Source of heat
Distillate
Solution:
ILLUSTRATIVE EXAMPLES (5)
Mass balance
Total
Benzene
0dt
dWD10 D = 10
dt
dxW
dt
x.dW WW
(10)(0.32)-DYD =
PB = PB* XB
Hub antara YD dan XW
PT = PT* XT
P = PB + PT = PB* XB + PT
* (1-XB)
*B
*T
BB*B
*B
* P
P)X1(X
P
P
P
P
B
)X1(
XP
P BB*
B
BB
B
B
*B X
)X1(X
1X
P
PY
B
BB
BB
*B X
X)1(1
XX
P
PY
BW
W
X).1(1
X.
YD =
dt
dX
X
X W
W
W 2048.11
48.2102.3
dt
dX
X
XX W
W
WW 2048.11
8.2492.52.3
tt
t
XX
X
WW
W dtdXX
XWW
W 032.0 1.202.3
)48.11(20W
W
W
W
W dXX 1.202.3
X 6.29
X 1.202.3
dX 20t
K)32.0( 1.202.3ln()1.20(
2.3
1.20
X6.29)X1.202.3ln(
1.20
20t
2W
W
for: t = 0 Xw = 0.32 K)32.0* 1.202.3ln()1.20(
2.3
1.20
32.06.29)32.0*1.202.3ln(
1.20
200
2
)32.0* 1.202.3ln(
)1.20(
2.3
1.20
32.06.29)32.0*1.202.3ln(
1.20
202K=
)32.0* 1.202.3
X 1.202.3ln(
)1.20(
2.3
1.20
32.0X6.29)
32.0*1.202.3
X1.202.3ln(
1.20
20t W
2WW
YD = 0.4 Xw = 0.21
ILLUSTRATIVE EXAMPLES (5)
At Xw = 0.21t = 1.58 hr
ILLUSTRATIVE EXAMPLES (6)
A jacketed vessel of A m2 heating surface is heated by steam condensed at TS oC. The vessel is filled with M kg liquid [heat capacity = C joule /(kg oC)] at To oC. The value of overall heat transfer coefficient is U watt/ (m2 oC). Obtain the temperature of liquid in the tank as a function of time.
Heat balance : TTUAMCTdt
dS
dt UATT
dTMC
S
t A UTT
TTln
oS
S
t A UTT
TTln
oS
S
UAt
oS
S eTT
TT
UAtoSS e )TT(TT
Solution
Disk shaped catalyst is used to accelerate the following irreversible first-order reaction :
ILLUSTRATIVE EXAMPLES (7)
The concentration of A at the catalyst surface is CAS. The effective diffusivity A trough the catalyst is DA a) Determine concentration distribution of A in the catalyst b) Derive an equation to estimate the mass transfer rate of A into the catalyst
A → B
CAS
Z
CASZZZAZN
ZZAZN
Z
Solution
ZSkCNN0 AZZZAZZZAZ
Mass Balance:a)
ZSkCZS
)NN(0 A
ZZAZZZZAZ
)kCdZ
dN
S
1(0 A
AZ
As Z→0
AA kC
dZ
dCS
dZ
d
S
AD
10 0
D
A2
2
AA C
k
dZ
CdZ
D
kZ
D
k
AAA eKeKC
2
1
ZD
k
A
ZD
k
A
A AA eD
kKe
D
kK
dZ
dC
2
1
BD
kB
D
k
AS
AA ee
CKK
21
Z Z
C AA
AA
D
k
D
k
BD
kB
D
k
ASA ee
ee
C
BZA
BZAZ dZ
dCSN AD
BZ
D
k
D
k
BD
kB
D
k
D
k
AS AA
AA
A
ee
ee
C Z Z
A
SD
B
D
k tanh
D
kC SD
AAAA S
ILLUSTRATIVE EXAMPLES (7)
B.C.1: 0,0 dZ
dCZ A
B.C.2: ASA CCBZ ,
b)
ILLUSTRATIVE EXAMPLES (8)
A copper fin L ft long is rectangular in cross section. It is W ft thick and B ft wide. The base of this metal is maintained at a constant temperature TB and the fin loses heat by convection to the surrounding air which is at a temperature To. The surface coefficient of heat transfer is h Btu/hr (ft2 oF).Determine the temperature distribution in the fin?Derive the formula to estimate heat losses from the fin to the surrounding air!
ILLUSTRATIVE EXAMPLES (8)
x qx x qx x+x
x
Solution:
Accumulation = Input – output
)TT(xhB2qq0 oxxxxx
)TT(xhB2
x
qq0 o
xxxxx
for x 0
)TT(hB2dx
dq0 o
x )TT(hB2dx
dThBW
dx
d0 o
)TT(kW
h2
dx
Td0 o2
2
= T-To
0kW
h2
dx
d2
2
ILLUSTRATIVE EXAMPLES (8)
xx KeKe 0kW
h2
dx
d2
2
General Solution
kW
h2
x = 0 T = TB = TB –To TB -To = K1 + K2
BC 1:
x = L
BC 2:
0dx
d
x
2x
1 eKeKdx
d
LL eKeK 210
L212 eKK L2
11oB eKKTT
L2oB
1 e1
TTK
L2
oBoB2 e1
TTTTK
xL2
oBoB
xL2
oB ee1
TTTTe
e1
TT
L2
)xL2(xoB
e1
eeTT
ILLUSTRATIVE EXAMPLES (8)
L
0ox )TT(hBdx2dq
dx)TT(hB2qL
0ox
dxeee1
)TT(hB2
L
0
)xL2(xL2
oB
L
0
xL2xL2
oB e1
xee1
e1
)TT(hB2
10
1e
1Lee
1
e1
)TT(hB2 LL2L
L2oB
L2LLL2
oBx Leee
1
e1
)TT(hB2q
b) Heat transfer rate
SERIES SOLUTION OF ORDINARY DIFFERENTIAL EQUATION AND SPESIAL FUNCTION
Objective:1.Students have ability to get series solution of ordinary differential equation with variable coefficients2.Student has ability to identify special functions and have ability to apply them in solving some physicochemical problems.
POWER SERIES
0
22)10 ................)(()( 000
n
nn xxAxxAAxxA
Form:
Such a series is said to “converge” if it is approaches a finite value as n approaches infinity. The simplest test of convergence is the ratio test; if the absolute value of the ratio of the (n+1)st term to the nth term in any infinite series approaches a limit J as n , then the series converges for J<1, diverges for J>1 , and the test fails for J =1.
Convergence:
1lim 001
xxLJxx
A
A
n
n
n
1. Within the interval of convergence of the original power series, series formed by termwise differentiation and integration of the original series is convergent.2. The product of two power series converges inside the common interval of convergence of the original series.3.The ratio of two power series converges inside the common interval of convergence of the original series, provided that the devisor does not vanish in common interval.
Properties:
POWER SERIES
Taylor Series
The power series may be put into the useful form known as “ Taylor series “
n
n
n
xxn
xfxfy )(
!
)()( 0
0
0
A function which can be represented by a Taylor series or the completely equivalent power series about xo is said to be “ regular “ at x = xo.
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
d y
dxa x
dy
dxa x y
2
2 1 1 0 ( ) ( )
Standard form
The behavior of the series solutions in the neighborhood of a point xo can be predicted from the behavior of the functions a1(x) and a2 (x) near xo. It is customary to classify the point xo as follows:
1. xo is termed an “ ordinary “ point of the differential equation if both a1(x) and a2 (x) can be represented by convergent power series which include x = xo in the interval of convergence, i.e., if a1(x) and a2(x) are regular at x = xo.2. xo is termed a “singular “ point of the differential equation if either a1(x) or a2(x) fails to prove regular at x = xo.3. xo is termed a “ regular singular “ point of the differential equation if 2 holds but the products (x – xo)a1(x) and (x - xo)2a2(x) both prove to be regular at x = xo4. xo is termed an “ irregular singular “ point of the differential equation if 2 holds but 3 fails.
EXAMPLE – 1 :
Locate and identify the singular points of the differential equation
x xd y
dxx x
dy
dxy2 2 2
2
21 2 1 0( ) ( )
Solution :
d y
dx
x x
x x
dy
dx x xy
2
2 2 2 2 2 2 2
2 1
1
1
10
( )
( ) ( )d y
dx
x x
x x x
dy
dx x xy
2
2 2 2 2 2 2 2
2 1
1 1
1
10
( )
( ) ( ) ( )
d y
dx x x x
dy
dx x xy
2
2 2 2 2 2
2
1 1
1
10
( )( ) ( )
)1)(1.(
2)(
21
xxxxa
222 )1(
1)(2
xxxa
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
1. x0 = 0 : a1(x) and a2(x) : non regular
)1)(1.(
2)(
21
xxxax
2222
)1(
1)(
xxax
regular
regular
x0 = 0 : regular singular point
2. x0 = 1 : a1(x) and a2(x) : non regular
2)1.(.
2)()1( 1
xxxax
regular
222
2
)1.(
1)(1
xxxax
regular
x0 = 1 : regular singular point
3. x0 =-1 : a1(x) and a2(x) : non regular
)1)(1(
2)()1( 1
xxxxax
222
2
)1(
1)(1
xxxax
non regular
regular
x0 = -1 : irregular singular point
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Solution (continued):
1.If xo is an ordinary point of (2-11), two linearly independent power-series solutions which are regular at x = xo will be obtained. Each solution will be of the form.
0
0n
nn xxAy
2. If xo is a regular singular point of (2-11), a power-series solution which is regular at x = x o cannot be guaranteed. However, the method of the next section will always generate at last one of the form.
0
00n
nn
s xxAxxy
In which s is a number whose value is determined in the course of the analysis
3. If xo is an irregular singular point of (2-11), a power-series solution may or may not exist.
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
EXAMPLE – 2 :
Obtain the general solution of the following differential equation, valid near x = 0.
d y
dxxdy
dxy
2
20
Solution : Function a1(x) = x and a2(x) = 1 are regular for xo, so point xo is ordinary, then the power series solution as
0
)()(n
nn xAxy
dy
dxnA xn
n
n
( ) 1
0
d y
dxn n A xn
n
n
2
22
0
1
( ) ( )
n n A x x n A x A xnn
nn
nn
n
nn
.( ). . . . .
1 02
0
1
00
[2A2 + 6A3.x + 12A4.x2 + 20A5.x3 + 30A6.x4 + 42A7.x5 + ...] + [A1.x + 2A2.x2 + 3A3.x3 + 4A4.x4 + 5A5.x5 + ...] + [A0 + A1.x + A2.x2 + A3.x3 + A4.x4 + A5.x5 + ...] = 0
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
From identity :2A2 + A0 = 0 A2 = -1/2 A06A3 + 2A1 = 0 A3 = -1/3 A112A4 + 3A2 = 0 A4 = -1/4 A2 = 1/8 A020A5 + 4A3 = 0 A5 = -1/5 A3 = 1/15 A130A6 + 5A4 = 0 A6 = -1/6 A4 = -1/48 A042A7 + 6A5 = 0 A7 = -1/7 A5 = -1/105 A1
AnAn
n
n2 0
1
2
( )
. !
0
7531
20 ...
105
1
15
1
3
1..
!.2
)1()`(
n
nn
n
xxxxAxn
Axy
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
METHOD OF FROBENIUS.
The method proceeds to find solution which is valid in the region of the point x = 0, Solution valid in the region of point x = xo may be obtained by transformation of the differential equation by use of the new variable z = x – xo. In the following discussion it is assumed that such a transformation has already been accomplished. The Standard Form:
0)(1
)(1
22
2
yxVxdx
dyxP
xdx
ydxR
Assumtions:
1.R(0)=1
2.R(x), P(x), and V(x) are regular at x = 0
0k
kk xRxR
0k
kk xPxP
0k
kk xVxV
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
0n
nn
s xAxy
R x n s n s A xkk
kn
n s
n
0
2
0
1( ).( ) P x n s A xkk
kn
n s
n
0
2
0
( ) 00
2
0
n
snn
k
kk xAxV
0.)()1).((0 0
2
k n
snknkkk xAVPsnRsnsn k + n = l
( ).( ) ( ). .l s k l s k R l s k P V Ak k k l kk
1 00
[s(s – 1 ) R o + sPo + Vo] Ao = 0
s2 + (Po – 1)s + Vo = 0
METHOD OF FROBENIUS.
Form of Solution:
for each fixed value of l between 0 and
0l 00 tok
Indicial Equation
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
METHOD OF FROBENIUS.
ooo
AVPsss
VVsPRssA
)1()1(
)1( 11111
A
q s n A
f s nn
k n kk
n
( ).
( )1
101 tokl
ntoknl 0
kkkkk VksRPksRsq 2
002 1 VsPssf
Recurrence Formula:
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE COEFFICIENT
Exceptional Cases
METHOD OF FROBENIUS.
1. If s1 and s2 do not differ by zero or a real integer, two independent solution obtained.2. If s1 = s2 only one solution is obtained.3. If s1 - s2 =N, where N is a a real integer, use of the larger value of s (s1) will always given one solution .
0n
nn
s xAxyForm of solution:
4. In all cases in which only one solution
0
11 )(1
nO
snn xuAxAy
can be found, the second independent solution is of the form
0
12.)ln().(.
n
snn xBxxuCy
Substitution into the original differential equation will determine the coefficient Bn in terms of the arbitrary constant C.
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENTMETHOD OF FROBENIUS.
Using method of Frobenius, obtain the general solution of the following differential equation, valid near x = 0,
2 1 2 02
2xd y
dxxdy
dxy ( )
Solution :The differential equation is changed to the Frobenius form:
R xd y
dx xP x
dy
dx xV x y( ). . ( ) . ( ).
2
2 2
1 10
d y
dx x
x dy
dx xx y
2
2 2
1 1 2
2
1 1
20
. . .
R(x) = 1 R(0) = 1 xx
xP
2
1
2
21)( xxV
2
1)(
EXAMPLE – 3 :
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
METHOD OF FROBENIUS.
1............)( 2210
0
xRxRxRxRxRn
kk
0..........,1 210 RRR
xxPxPPxPxPk
kk
2
1..........2
2100
0............,1,2
13210 PPPP
xxVxVVxVxVk
kk 2
1.............2
2100
0..........,2
1,0 3210 VVxVV
01 002 VsPs 01
2
12
ss 0,
2
121 ssIndicial Equation:
A
q s n A
f s nn
k n kk
n
( ).
( )1Recurrsion Formula
kkkkk VksRPksRsq 2 002 1 VsPssf
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENTMETHOD OF FROBENIUS.
12
1
2
1)01(01
2
31
2
3
2
3111
2
11
VRPRq
2
3
4
60
2
31
2
1
2
3
2
32
f
2
11 ss
1
2
1
12
101
1
f
Aq
A
0
01 3
2
2
3A
AA
54
200
2
51
2
1
2
5
2
52
f
22
1
2
3)01(01
2
51
2
5
2
5111
2
11
VRPRq
)5,2(
]).5,2().5,2([ 02112 f
AqAqA
02
52
q
0
2
11
2 53
2
5
2
5
]0.2[A
xA
AA
n=1
n=2
011
)12(........531
2
5
2
5
]0.2[A
nxxxxA
AA
n
n
Generalisasi:
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
)1(
]).1([ 011 f
AqA
METHOD OF FROBENIUS.
for s = s1 = 0 :
AA
A10
0
1 2
1 2
[ / . ]
/
2
1
2
10011111 111
211 VRPRq
2
1011
2
111 2
f
n=1
2
3
2
1)1)(01(012122 111
211 VRPRq
30212
122 2
f
)2(
]).2().2([ 02112 f
AqAqA
223
]0.2
3[
011
2
AAAA
n=2
Aq A q A q A
f31 2 2 1 3 03 3 3
3
[ ( ). ( ). ( ). ]
( )
AA
A A31
2 0
5 2
15 2
1
3
1
6
[ / . ]
/
AnAn
10!
.
n=3
Generalisasi:
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
METHOD OF FROBENIUS.
n
n
xn
BxxxAxy
00
22
5,00 !
1............
5.3.1
2
3.1
21.)(
0
00
5.0021 !
1
)12(......531
2)()()(
n
nn
n
n
xn
Bxnxxxx
xAxyxyxy
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
METHOD OF FROBENIUS.
Using method of Frobenius obtain the general solution of the following differential equation, valid near to x = 0
xd y
dx
dy
dxxy
2
22 0
EXAMPLE – 4 :
SOLUTION
R xd y
dx xP x
dy
dx xV x y( ). . ( ) . ( ).
2
2 2
1 10
d y
dx x
dy
dx
x
xy
2
2
2
2
20
1............)( 2210
0
xRxRxRxRxRn
kk
2..........2210
0
xPxPPxPxPk
kk
22210
0
............. xxVxVVxVxVk
kk
0..........,1 210 RRR
0............,2 3210 PPPP
0..........,1,0,0 3210 VVVV
01 002 VsPs 00122 ss 1,0 21 ss
Indicial Equation:
Recurrsion Formula A
q s n A
f s nn
k n kk
n
( ).
( )1
kkkkk VksRPksRsq 2 002 1 VsPssf
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENTMETHOD OF FROBENIUS.
for s = s1 = 0 : Aq A
f11 01
1
[ ( ). ]
( )A
A1
00
20
[ . ]
Aq A q A
f21 1 2 02 2
2
[ ( ). ( ). ]
( )A
AA2
00
0 0 1
6
1
6
[ . . ]
AA A
32 00 10 0
12
0
120
[ . . . ]
AA A A A A A
43 2 1 0 2 00 1 0 0
20 20 120
[ . . . . ]
An
An
n
2 0
1
2 1
( )
( )!.
0
20
0
20
0
)!12(
)1(..
)!12(
)1(.)(1
n
nn
n
nn
xn
AxAn
xxy
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
0
2
)!12(
)1(.)(n
nn
xn
xuy c u x x B xnn s
n2 1
0
2
. ( ).ln( ) .
Second Solution:
y c xn
x B xn
n
nn
n
n2
2
0
1
0
1
2 1
.ln( ).( )
( )!.
s2=-1
dy
dxc x
n
nx c
nx B n x
nn
n
nn
nn
n
n
2 2 1
0
2 1
0
2
0
1 2
2 1
1
2 11
.ln( ).( ) .
( )!. .
( )
( )!. .( ).
d y
dxc x
n n
nx c
n
nx
nn
n
nn
n
22
22 2
0
2 2
0
1 2 2 1
2 1
1 2
2 1
.ln( ).( ) . .( )
( )!. .
( ) .
( )!.
cn
nx B n n x
nn
nn
n
n
.( ) .( )
( )!. .( ).( ).
1 2 1
2 11 22 2
0
3
0
c xn n
nx c
n
nx
nn
n
nn
n
.ln( ).( ) . .( )
( )!. .
( ) .
( )!.
1 2 2 1
2 1
1 2
2 12 1
0
2 1
0
cn
nx B n n x
nn
nn
n
n
.( ) .( )
( )!. .( ).( ).
1 2 1
2 11 22 1
0
2
0
c xn
nx c
nx B n x
nn
n
nn
nn
n
n
.ln( ).( ) .
( )!. .
( ) .
( )!. . .( ).
1 4
2 1
1 2
2 12 12 1
0
2 1
0
2
0
c xn
x B xn
n
nn
n
n
.ln( ).( )
( )!.
1
2 102 1
0 0
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
c xn n
nx c
n
nx
nn
n
nn
n
.ln( ).( ) .( )
( )!. .
( ) .( )
( )!.
1 4 2
2 1
1 4 1
2 1
22 1
0
2 1
0
c xn
x B n n x B xn
n
nn
n
nn
n
n
.ln( ).( )
( )!.( ). . .
1
2 11 02 1
0
2
0 0
cx x x x x x
16
3
120
5
142
7
3 5. . .ln( )
!
. . .ln( )
!
. . .ln( )
!...
c xx x x1
3 515
3
9
5
13
7
. .
!
.
!
.
!...
c x x
x x x x x x.ln( )
.ln( )
!
. .ln( )
!
.ln( )
!...
3 5 7
3 5 7
0...][.......]4230201262[ 55
44
33
2210
57
46
35
2432 xBxBxBxBxBBxBxBxBxBxBB
identity : - term : x-1 c = 0- term : x.ln(x) c{-6/3! + 1] = 0 0 = 0- term : x3.ln(x) c(20/5! - 1/3!] = 0 0 = 0so c = 0- term : x0 2B2 + B0 = 0 B2 = -B0 /2- term : x1 6B3 + B1 = 0 B3 = -B1 /6- term : x2 12B4 + B2 = 0 B4 = -B2 /12 = B0 /24- term : x3 20B5 + B3 = 0 B5 = -B3 /20 = B1 /120
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
BB
nn
n
2 111
2 1
( ) .
( )!
BB
nn
n
201
2
( ) .
( )!
y Bn
x Bn
xn
n
n
nn
n2 0
2 1
01
2
0
1
2
1
2 1
( )
( )!.
( )
( )!.
y An
xn
n
n
02
0
1
2 1.
( )
( )!B
nx B
nx
nn
n
nn
n0
2 1
01
2
0
1
2
1
2 1
( )
( )!.
( )
( )!.
y Bn
x A Bn
xn
n
n
nn
n
02 1
00 1
2
0
1
2
1
2 1
( )
( )!. ( )
( )
( )!.
0
20
0
120 .
)!12(
)1(.
)!2(
)1(
n
nn
n
nn
xn
Cxn
By
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
0p p
BESSEL EQUATION
0)( 222
22 ypx
dx
dyx
dx
ydx
0
2
)!(!2
)1()(
k
pk
p pkk
x
xJ J x
x
k k pp
k p
k
( )( )
!( )!
12
2
0
0p p integer xJCxJCy PP 21
Integer=n
Y x
xJ x
n kx
k
k k n
x
k n k
n
n
k n
k
n
k
k n
k
( )
ln ( )( )!
!
( ) ( ) ( )!( )!
2 2
1
2
12
1
21
2
2
0
1
1
2
0
xYCxJCy nn 21
Bessel’s function of the second kind of order n
Bessel function of the first kind of order p
=0.5772157……
100,1
1
km
kk
m
Euler’s constant
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
0
2
)!(!
2)(
k
pk
P pkk
x
xI
xKCxICy nn 21
0)( 222
22 ypx
dx
dyx
dx
ydx
0p p integer xICxICy PP 21
0p p integer
)(.)(2
1 ixYiixJixK nnn
n
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
0p p
0p
xd y
dxx a bx
dy
dxc dx b a r x b x yr s r r2
2
22 2 22 1 0 ( ) ( ( ). . )
sp
sp
rxba xs
dZcx
s
dZcexy
r
.. 21)/.(2/)1(
ps
ac
1 1
2
2
sd /
sd /
sd /
sd /
=real integer PPPP JZJZ ,
=real
p
0p p integer=n
GENERALIZED FORM OF BESSEL’S EQUATION.
General Solution:
=imaginer
=imaginer
integer
p=0, p=integer=n
nPnP YZJZ ,
PPPP IZIZ ,
nPnP KZIZ ,
Find general solution of the following equation:
Example -5
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
016
1
2
12
22 yx
dx
dyx
dx
ydx
0,2
12
2
1 babxa r
Solution
4
1,
16
1,01
16
1 222 sdcxbxrabxdcx rrs
102
2
11
4/1
1
2
p
s
d real
1p
11 , YZJZ PP
25.012
25.011
25.04
1
124
1
110
2/2
11
4
116
1
4
116
1
xYcxJcxxYcxJcexy
Properties of Bessel Functions.
x
1.0
-1.0
1.0 2.0 3.0
Y0(x)
J0(x)
I0(x)K0(x)
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Form of Curve
ppP xp
xJ .!2
1)( p
P xp
xJ
.)!(
1)(
0,)!1(2
)(
nxn
xY nn
n xxY ln
2)(0
ppP xp
xI .!2
1)( p
p
P xp
xI
.)!(
2)(
0,)!1(2)( 1 nxnxK nnn xxK ln)(0
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Properties of Bessel Functions.
Approximation for small x value
24cos
.
2)(
px
xxJ P
24sin
.
2)(
px
xxYn
x
exI
x
P 2)( x
n ex
xK .2
)(
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Properties of Bessel Functions.
Approximation for very large x value
xx
xJ sin2
)(2/1 x
xxJ cos
2)(2/1
xx
xI sinh2
)(2/1 x
xxI cosh
2)(2/1
)()(12
)( 2/32/12/1 xJxJx
nxJ nnn
)()(12
)( 2/32/12/1 xIxIx
nxI nnn
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Bessel functions of order equal to half an odd integer
Properties of Bessel Functions.
d
dxx Z x
x Z x Z J Y I
x Z x Z Kp
p
pp
pp
( )( ); , ,
( );
1
1
d
dxx Z x
x Z x Z J Y K
x Z x Z Ip
p
pp
pp
( )
( ); , ,
( );
1
1
d
dxZ x
Z xp
xZ x Z J Y I
Z xp
xZ x Z K
p
p p
p p
( )( ) ( ); , ,
( ) ( );
1
1
d
dxZ x
Z xp
xZ x Z J Y K
Z xp
xZ x Z I
p
p p
p p
( )( ) ( ); , ,
( ) ( );
1
1
2 1 1
d
dxI x I x I xp p p( ) ( ) ( )
2 1 1
d
dxK x K x K xn n n( ) ( ) ( )
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Properties of Bessel Functions.
Differentiation and Integration Properties of Bessel Function
Z xx
pZ x Z x Z J Yp p p( ) ( ) ( ) ; , 2 1 1
I xx
pI x I xp p p( ) ( ) ( ) 2 1 1 K x
x
pK x K xn n n( ) ( ) ( ) 2 1 1
integeror 0 =n when
)x(K)x(K
)x(I)x(I
)x(J)1()x(J
nn
nn
nn
n
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Properties of Bessel Functions.
Recursion Formula
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Values of x for whichJ0(x) = 0
Value of x for which J1(x) = 0
2.4048 3.8317
5.5201 7.0156
8.6537 10.1735
11.7915 13.3237
14.9390 16.4706
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENTEXAMPLE – 7
The Wedge-shaped Fin. The differential equation resulting from the analysis of heat flow through and from a wedge-shaped fin was given below.
0ykw
L sec h2
dx
dy
dx
ydx
2
2
where y = T - TaT = local fin temperature at xTa = temperature of surrounding airh = heat transfer coefficient , Btu/hr.ft2.oFk = thermal conductivity of fin material,
Btu/hr.ft.oFL = total length of fin, (ft) = half wedge angle of fin
(1)
a. Prove equation (1)
b. Determine the temperature distribution in the fin
c. Determine total rate of heat transfer from the fin to the surrounding air
Ta =1000 FTL =2000 F, temperature at x = LL = 1 fth = 2 Btu/(hr)(ft2 )(0F)k = 220 Btu/(hr)(ft)(0F)Sec =1W = 1/12 ft
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Solution
a)
x
xxxxq
xxxq
convq
L
BW
Heat balance on an element of Δx thickness:
0sec220 axxxxxxxconvxxxxxxx TTWxhqqqqq
0sec2
axxxxxxx
TTWhx
0sec2 ax TTWh
dx
dq
dx
dTWbkqx
BL
xb
0sec2
aTTWhdx
dx
dTWB
L
xkd
0sec2
aTTBk
Lh
dx
dx
dTxd
0sec2
2
2
yBk
Lh
dx
dy
dx
ydx
aTTy
)x2(Kc)x2(IcTTy 0201a
)x2(IcTT 01a
0sec2
2
2
yBk
Lh
dx
dy
dx
ydx
02
2
ydx
dy
dx
ydx 0
2
22 yx
dx
dyx
dx
ydx
Boundary Condition:
BC1: x=0 → y=finite
BC2: x=L=1 → y = TL- Ta
=200-100=100
00 K
218.01.1.220
12.1.2.2
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
Bk
Lh sec2
b)
General Solution:
10101 230.1934.0 218.02100200 cIcIc
100934.02.81 xIT o
2.811 c
c)
SOLUTION OF LINEAR HOMOGENEOUS SECOND ORDER DIFFERENSIAL EQUATION WITH VARIABLE
COEFFICIENT
L L
aa TThWdATThq0 0
dx sec 2)()(
dxxIWhcqL
o0
1 )]2([sec2
x2z
)L2(IL
L)z(zI
2
1dz)]z(zI
2
1dx)]x2(I[ 1
L2
01
L2
0o
L
0o
)2() (sec2
11 LI
L
LWhcq
)1)(218.0(2)1)(218.0(
)1)(1)(2)(1)(2.81)(2()2(
) (sec211
1 ILIL
LWhcq
0.934
Special FunctionGamma Function
( ) ,n t e dt nn t
1
0
0
( ) ( )n n n 1
( ) ( )n t e dt n t n t e dt n t e dt n nn t n t n tt e
1 00 0 00
(3.5) = 2.5(2.5) = 2.5 x 1.5(1.5) = 2.5 x 1.5 x 0.5(0.5); (0.5) = = 1.76= 2.5 x 1.5 x 0.5 x 1.76 = 3.30; (0.5) = = 1.76 dan ; (1) = 1
(n) = n!
N 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
(n) 9.51
4.59
2.99 2.22
1.76
1.49
1.30
1.16
1.07
1.00
(n) = (n+1)/n (-0.5) = (0.5)/-0.5 = -3.5222.
For n = 0 or negative integer, the value of (n) equal to infinite: (n) = , where for n = negative even number, (n) = and for n = negative odd number : (n) = - .
Beta Function
Special Function
1
0
11 )1(, dxxxnm nm
)(
)().(,
nm
nmnm
dyyayanma
nmnm 0
111 )(,
dyyynm nmn
0
1 )1(,
dnm nm 2/
0
1212 sin.cos2,
Bentuk bentuk Fungsi Betta:
y= a x
1y
yx
2sinx
Error Function
Special Function
x
n dnexerf0
22)(
X Er. f(x) X Er. f(x)
0.0 0.0000 1.6 0.9763
0.2 0.2227 1.8 0.9891
0.4 0.4284 2.0 0.9953
0.6 0.6039 2.2 0.9981
0.8 0.7421 2.4 0.9993
1.0 0.8427 2.6 0.9998
1.2 0.9103 2.8 0.9999
1.4 0.9523
32222 )2(
5.3.1
)2(
3.1
2
111)(
2
xxxx
exerf
x
x > 2.8,
PARTIAL DIFEERENTIAL EQUATION
Au
xB
u
x yCx
u
yDu
xEu
yFu f x y
2
2
2 2
22 ( , )
Au
xB
u
x yCx
u
yDu
xEu
yFu
2
2
2 2
22 0
Partial differential equation is a differential equation containing an unknown function ( or dependent variable) and several independent variables. So, the partial differential equation involves partial derivatives
The “order” of a partial differential equation is the order of the highest derivative appearing in the equation. A partial differential equation is said to be linear if no powers or product of the dependent variable or its partial derivatives are present.
If all terms in the partial differential equation have the same order than the equation is said to be homogeneous
Linear second order partial differential equation with constant coefficients:
Homogeneous linear second order partial differential equation with constant coefficients:
The second order partial differential equation can be classified as :- Elliptic, if B2-AC < 0
- Hyperbolic, if B2 - AC > 0 - Parabolic, if B2 - AC = 0
2
2
2
20
u
x
u
y
2
22
x
uc
t
u
2
22
2
2
x
uc
t
u
Two dimensional Laplace equation or Potential equation is Elliptic
A = 1, B = 0, and C = 1, therefore B2 - AC = 0 - 1.1 = -1 < 0.
One dimensional Heat equation is Parabolic
*A = c2 , B = 0, and C = 0, therefore B2 - AC = 0 – c2 .0 = 0.
One dimensional Wave equation is Hyperbolic
A = C2, B = 0, and C = -1, therefore B2 - AC = 0 – c2 (-1) = c2 > 0.
PARTIAL DIFEERENTIAL EQUATION
CLASSIFICATION OF PARTIAL DIFFERENSIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
The solution of a partial differential equation in a region R of its independent variables is a function where all the partial derivatives are exist in this region and the function satisfies the equation at all points in this region. The function has to be continue at boundary of R. Such functions are very numerous. A unique solution is limited by boundary condition and initial condition.
Theorem 1 : If u1 , u2 , ..., uk are the solution of Eq.(2) then u = c1u1 + c2u2 + ... + ck uk , where c1 , c2 ,..., ck are constants , is also the solution.
Theorem 2 : If u1 , u2 , ..., un , ... are the solution of Eq.(2) then
1
.n
nn uCu is also the solution.
SOLUTION OF PARTIAL DIFFERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
There two kinds of problems: Initial value problems and Boundary value problems. For the Boundary value problem, the region is finite and the boundary condition is set at all part of region boundary, while for initial value problem, the region is not finite.
There are some methods to solve the partial differential equation. The following are some of the methods: Laplace Transform. Separation of Variables. Combination of Variables.
TYPES OF PROBLEMS AND METHODS OF SOLUTION
PARTIAL DIFEERENTIAL EQUATION
Laplace Transform
Generally Laplace Transform method is used to solve Initial value Problems.Steps of solution are:1) Operate Laplace Transform on P.D.E and the boundary conditions using the initial conditions. This operation results an ordinary differential equation of
dependent variable in Laplace domain.2) Solve the equation in (1) to obtain an expression of dependent variable in
Laplace domain.3) Operate inverse transform on the expression of dependent variable obtained in
(2).
Example 3.1:Heat transfer in semi infinite wall.
A very thick slab (slab with infinite thick) initially at temperature of T0 through-out. Suddenly one of the slab surface is contacted with hot liquid at temperature Ts. Determine the temperature distribution in the slab.
PARTIAL DIFEERENTIAL EQUATION Solution :The heat transfer phenomena in the slab is described by the following partial differential equation (Prove it!):
T
t
k
C
T
xp
.
2
2
T
t
T
x 2
2
2
Initial condition: T(x,0) = T0 (E1-2)Boundary conditions: 1. T(0,t) = Ts (E1-3)
2. T(,t) = T0 (E1-4)
Step 1 :Operate Laplace trasform on Equation(E1-1):
(E1-1)
LT
tL
T
x
22
2. s T T x
d T
dx. ( , ) 0 2
2
2
d T
dx
sT
T2
2 202
.
Operate Laplace transform on The boundary conditions ( Eq.(E1-3) and Eq.(E1-4) 1. L{T(0,t)} = L{Ts}
T sT
ss( , )0
PARTIAL DIFEERENTIAL EQUATION
2. L{T(,t)} = L{T0} T sT
s( , ) 0
Step 2 :The general solution of Eq.(E1-5) is:
T K e K eT
s
sx
sx
1 20. .
. .
From B.C.2 (Eq.(E1-7)) and Eq.(E1-8)) we obtain K1 = 0. Therefore Eq.(E1-8) become:
T K eT
s
sx
20.
.
From BC.1 (Eq.(E1-6)) and Eq.(E1-9) the following equation is obtained:
T
sK
T
sK
T T
ss s
2
02
0
Substitute Eq.(E1-10) into Eq.(E1-9), TT T
se
T
ss
sx
0 0..
PARTIAL DIFEERENTIAL EQUATION
Step3 : ),(),( 1 sxTLtxT
sLTe
sLTT
xs
S
1.
1 10
10
00..2
Tt
xErfTTs
using boundary condition x = 0 erf(0) = 0 and x = erf() = 1 Eq.(E1-12) become:
00..2
1, Tt
xErfTTtxT S
PARTIAL DIFEERENTIAL EQUATION SEPARATION OF VARIABLES METHOD
For the application of this method the following conditions have to be satisfied:1. The differential equation is homogeneous.2. The boundary condition is homogeneous.
General steps:1)Perform variable separation to obtain two ordinary differential equations.2)Solve these two equations satisfying boundary conditions to obtain partial solutions.3)Obtain total solution satisfying initial condition.
If the boundary conditions and differential equation are not homogeneous then before the application of this method it is necessary to perform variable transformation so that the boundary condition and the differential equation become homogeneous.For non-homogeneous boundary conditions for instance U(0,t) = U0 and U(L,t) = UL perform the following variable substitution (in case of Cartesian coordinate):
V = U + a + bxV(0,t) = U(0,t) + a
0 = U0 + a a = -U0 ,V(L,t) = U(L,t) + a + bL
0 = UL + a + bL
L
UUbLbUU L
L
0
00 xL
UUUUV L
00
xL
UUUVU L
00
(See Example 3.5 for the explanation of this)
PARTIAL DIFEERENTIAL EQUATION
For non-homogeneous differential equation, dependent variable is expressed as the summation of steady state solution ( function of only spatial coordinate) and deviation variable ( function of time and spatial coordinate ).For example consider the following partial differential equation:
T
t
T
tg x t 2
2
2( , )
with initial condition : T(x,0) = f(x) and boundary condition: T(0,t) = T(L,t) = 0. First, perform the following substitution: T(x,t) = V(x,t) - W(x,t). Then, the differential equation becomes the following differential equations:
V
t
V
t 2
2
2
where V(x,0) = f(x) and V(0,t) = V(L,t) = 0
W
t
W
tg x t 2
2
2( , )
where W(x,0) = 0 and W(0,t) = W(L,t) = 0.
PARTIAL DIFEERENTIAL EQUATION
Example 3.2 :A rod with isolated side surfaces has initial temperature distribution: T(x,0) = f(x). Suddenly (at t = 0), the two ends of the rod are contacted with iced water ( the temperature is maintained at 0 oC ).
L
X
0 oC 0 oC
Determine the temperature of the rod as function of x and t .
Solution:Heat transfer phenomena in the rod can be described as :
T
t
T
t 2
2
2
PARTIAL DIFEERENTIAL EQUATION
Initial condition: T(x,0) = f(x) (E2-2)Boundary condition: T(0,t) = 0 dan T(L,t) = 0 (E2-3)Eq.(E2-1) is homogeneous partial differential equation and the boundary conditions are also homogeneous. Then the method of separation variable can be applied.
Step 1 :T(x,t) = F(x).G(t) (E2-4)
Substitute Eq.(E2-4) into Eq.(E2-1)
F G F G. ' ". 2G
G
F
F
'
.
'
2
•It can be concluded that both sides of Eq.(E2-5) are constant:
G
G
F
FC
'
.
'
2
Eq.(E2-6) can be decomposed into 2 equations GCG 2'
0" FCF
(E2-5)
•(E2-6)
•(E2-7)
•(E2-8)
PARTIAL DIFEERENTIAL EQUATION
•The constant C may have the following values:
a. C > 0 :The solution of Eq.(E2-8) is: F x K e K eC x C x( ) . .. .
1 2
•From boundary conditions: 0 = K1 + K2
LCLC eKeK .21 ..0
•then the values of K1 and K2 are: K1 = 0 and K2 = 0, there we get trivial solution.
b. C = 0 :The solution of Eq.(E2-8) is : F x K K x( ) . 1 2
•From the boundary conditions: 0 = K1
• 0 = K1 + K2.L
•and the value of integration constants: K1 = 0 and K2 = 0, again we obtain trivial
solution.
PARTIAL DIFEERENTIAL EQUATION
c. C < 0 :The solution of Eq.(E2-8) is: F x K C x K C x( ) .cos( . ) .sin( . ) 1 2
From the boundary conditions: •0 = K1
)sin(.).cos(.0 21 LCKLCK
•Then the value of integration constants: K1 = 0 dan K2 0. In this case we obtain non-trivial solution, the solution we expected.
•Therefore the value of C must be negative,for instance C = -p2.Then the Eq.(E2-8) become:
0" 2 FpF GpG 22'
PARTIAL DIFEERENTIAL EQUATION
Step 2 :The solution of Eq.(E2-9) :The general solution of Eq.(E2-9) is: F(x) = K1.cos (px) + K2.sin (px)From boundary condition: F(0) = 0 , we obtain: 0 = K1 + 0 or K1 = 0 Then the general solution becomes:F(x) = K2.sin (px), and using the other boundary condition: 0 = K2.sin (pL), then sin (pL) = 0. so, pL = n. or p = n./L, where n = 1,2,3, ... .. Finally, the solution of Eq.(E2-9) becomes:
F x Kn x
Ln n( ) .sin. .
2
*The solution of Eq.(E2-10):
G e p t 2 2. . G t en
n
Lt
( ).
. .
22
Therefore:
11
)().(,,n
nnn
n tGxFtxTtxT
1.
..sin.),(
...
2
22
ne
L
xnKtxT
tL
n
n
1.
..sin.
... 2
2
ne
L
xnA
tL
n
n
PARTIAL DIFEERENTIAL EQUATION
Step 3 :From initial condition:
1
..sin.)()0,(
n L
xnAxfxT n
AL
f xn x
Ldxn
L
2
0
( ).sin. .
It can be seen that f(x) is Sinus Fourier series where:
Therefore the overall solution is Eq.(E2-11) with constants An obtained from Eq. (E2-12).
E2-11
E2-12
PARTIAL DIFEERENTIAL EQUATION
Kombinasi Variabel.
An illustrative example of this method is heat transfer problem in semi infinite (very thick) wall. We define new variable:
x
t2. .
Partial differential equation describing this heat transfer problem is:
T
t
T
x 2
2
2
Initial Condition: t=0 0 < x < ~ T = T0
Boundary Condition: t > 0 x = 0 T = Ts
t > 0 x = ~ T = T0
PARTIAL DIFEERENTIAL EQUATION
Solution:
T
t
T
t
t
t
x
t t
T .
. . ..
4
T
x
T
x t
T .
. ..
1
2
2
2 2
2
2
1
2 1
4
T
x
T
xx
t
T
x t
T
. ..
. ...
x
t t
T
4. . ..
2
2
22 .
...4
1.
T
t
2
2 2
2
2
T x
t
T
.
. .. .
RT
Substitution :
2
2
T R then:
RR 2. .
R
R 2. .
R K e 1
2
.
TK e
1
2
.
T K e 1
2
. T K e K 1
0
2
2
. .
PARTIAL DIFEERENTIAL EQUATION
Kondisi-kondisi batas dalam variabel adalah :
= 0 : T = Ts (8) = : T = T0 (9)
dari kondisi batas pers. (8) dan pers. (7) :
Ts = 0 + K2 K2 = Ts, sehingga : T K e Ts 1
0
2
. .
T K Erf Ts 1 2. ( )
dari kondisi batas pers. (9) dan pers. (11) :
T K Ts0 1 21 . .( )
K T Ts1 02
.
Dengan kondisi batas x = 0 erf(0) = 0 dan x = erf() = 1, sehingga pers. (11) menjadi :
00..2
1, Tt
xErfTTtxT S
PARTIAL DIFEERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION
PARTIAL DIFEERENTIAL EQUATION