化工應用數學 授課教師： 郭修伯 lecture 8 solution of partial differentiation equations
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化工應用數學
授課教師： 郭修伯
Lecture 8 Solution of Partial Differentiation Equations
Solution of P.D.E.s
– To determine a particular relation between u, x, and y, expressed as u = f (x, y), that satisfies
• the basic differential equation• some particular conditions specified
– If each of the functions v1, v2, …, vn, … is a solution of a linear, homogeneous P.D.E., then the function
is also a solution, provided that the infinite series converges and the dependent variable u occurs once and once only in each term of the P.D.E.
1
nvv
Method of solution of P.D.E.s
• No general formalized analytical procedure for the solution of an arbitrary partial differential equation is known.
• The solution of a P.D.E. is essentially a guessing game.• The object of this game is to guess a form of the
specialized solution which will reduce the P.D.E. to one or more total differential equations.
• Linear, homogeneous P.D.E.s with constant coefficients are generally easier to deal with.
Example, Heat transfer in a flowing fluid
An infinitely wide flat plate is maintained at a constant temperature T0. The plate is immersed in an infinately wide the thick stream of constant-density fluid originally at temperature T1. If the origin of coordinates is taken at the leading edge of the plate, a rough approximation to the true velocity distribution is:
Turbulent heat transfer is assumed negeligible, and molecular transport of heat is assumed important only in the y direction. The thermal conductivity of the fluid, k is assumed to be constant. It is desired to determine the temperature distribution within the fluid and the heat transfer coefficient between the fluid and the plate.
00 zyx VVyV
B.C.T = T1 at x = 0, y > 0T = T1 at x > 0, y = T = T0 at x > 0, y = 0T0
x
y
T1 T1dx
dy
Input - output = accumulation
y
Tk
yx
CTVx const. properties
2
2
y
T
CV
k
x
T
x
yVx
2
2
y
T
y
A
x
T
C
kA
T = T1 at x = 0, y > 0T = T1 at x > 0, y = T = T0 at x > 0, y = 0
Heat balance on a volume element of length dx and height dy situated in the fluid :
Input energy rate: y
TkdxCTdyVx
Output energy rate:
dyy
Tkdx
yy
TkdxdxCTdyV
xCTdyV xx
10
1
TT
TT
2
2
yy
A
x
= 0 at x = 0, y > 0 = 0 at x > 0, y = = 1 at x > 0, y = 0
2
2
yy
A
x
B.C. = 0 at x = 0, y > 0 = 0 at x > 0, y = = 1 at x > 0, y = 0
Assume: fx
yf
n
Compounding the independent
variables into one variable
= 0 at = = 1 at = 0
Replace y and x in the P.D.E by
dyy
dxxd
dd
d
dd
d
d
x
n
d
d
x
ny
xd
d
x n
1
nx
y
d
d
xyd
d
y n
1
dy
d
d
d
xd
d
yxd
d
xyy nnn
2
2
2
2 111
2
2
yy
A
x
2
2
3
1
d
dA
xd
d
x
nn
In order to eliminate x and y, we choose n = 1/3
03
2
2
2
d
d
Ad
d
AB
d
d
9exp
3
10
13
0 9exp
TT
TTd
ABd
= 0 at = = 1 at = 0
dA
dA
dB
0
3
0
3
1
0
9exp
1
9exp
dA
dA
TT
TT
0
3
3
10
1
9exp
9exp
= 1 at = 0
0
10 )(
y
TkTThLocal heat transfer coefficient
y
TTy
T
10
d
d
xyd
d
y n
1
10
1
TT
TT
yd
dT
y
T
d
d
x
TT
y
T
31
10
d
A
B
0
3
9exp
1
AB
d
d
9exp
3
dA
A
x
TT
y
T
0
3
3
31
10
9exp
9exp
0
10 )(
y
TkTTh
d
Ax
kh
0
33
1
9exp
= 0 at y = 0
31
43.0
kx
Ckh
C
kA
Separation of variables: often used to determine the solution of a linear P.D.E.
Suppose that a slab (depending indefinately in the y and z directions) at an initial temperature T1 has its two faces suddenly cooled to T0. What is the relation between temeprature, time after quenching, and position within the slab?
2R
x
dx
Since the solid extends indefinately in the y and z direction, heat flows only in the x direction. The heat-conduction equation:
2
2
x
T
t
T
Boundary condition:
0,2
0,0
0,
0,0
0
0
0
1
tRxatTT
txatTT
xtatTT
xtatTT
Dimensionless:2
2
x
T
t
T
0,2
0,0
0,
0,0
0
0
0
1
tRxatTT
txatTT
xtatTT
xtatTT 01
0
TT
TT
2
2
xt
0,20
0,00
0,0
0,01
tRxat
txat
xtat
xtat
Assume: )(tgxf Separation of variables
)(
)(
2
2
tgxfx
tgxft
2
2
xt
)()()()( tgxftgxf
)(
)(
)(
)(
tg
tg
xf
xf
independent of t independent of x
0)()(
0)()(
tgtg
xfxf
)(
)(
)(
)(
tg
tg
xf
xf
when 0
tCetg
xBxAxf
)(
cossin)(
when = 0
0
00
)(
)(
Ctg
BxAxf
)(tgxf texBxA cossin
)(tgxf00 BxA
A0, B0, A, B, and have to be chosen to satisfy the boundary conditions.
0,20
0,00
0,0
0,01
tRxat
txat
xtat
xtat
texBxABxA cossin00Superposition:
000 BA
0B
R
n
2
n is an integer
texBxABxA cossin00
tR
n
eR
xnA
2
2sin
The constant has to be determined.But no single value can satisfy the B.C.
0,01 xtatB.C.
More general format of the solution (by superposition):
1
2
2sin
n
tR
n
n eR
xnA
0,01 xtat
1 2sin1
nn R
xnA
dxR
xnA
R
xmdx
R
xm R
nn
R
2
01
2
0 2sin
2sin
2sin
dxR
xmAdx
R
xn
R
xmAdx
R
xm R
nn
R
n
R
2
0
2
1
2
0
2
0 2sin
2sin
2sin
2sin
Orthogonality property
ndx
R
xn
RA nR
n
2])1(1[
2sin
1 2
0
nA n
n
2])1(1[
1
2
2sin
n
tR
n
n eR
xnA
01
0
TT
TT
1
4
10
1 2
22
2sin
)1(12
n
tR
nn
eR
xn
nTT
TT
The representation of a function by means of an infinite series of sine functions is known as a “Fourier sine series”.
More about the “Orthogonal Functions”Two functions m(x) and n(x) are said to be “orthogonal” with respect to the weighting function r(x) over interval a, b if:
b
a nm dxxxxr 0)()()(
dxR
xmAdx
R
xn
R
xmAdx
R
xm R
nn
R
n
R
2
0
2
1
2
0
2
0 2sin
2sin
2sin
2sin
R
xm
2sin
R
xn
2sin
and are orthogonal with respect to the weight function
(i.e., unity) over the interval 0, 2R when m n.
Each term is zero except when m = n.
Back to our question, we had two O.D.E.s and the solutions are :
0)()(
0)()(
tgtg
xfxf
tCetg
xBxAxf
)(
cossin)( where shows!
R
xn
2sin
0)()( xfxf 00 xat
xCxf sin)( Rxat 20
R
nn 2
These values of are called the “eigenvalues” of the equation, and the
correponsing solutions, are called the “eigenfunctions”.
n
xsin
Sturm-Liouville Theory
• A typical Sturm-Liouville problem involves a differential equation defined on an interval together with conditions the solution and/or its derivative is to satisfy at the endpoints of the interval.
• The Strum-Liouville differential equation:
• In Strum-Liouville form:
0)]()([)( yxPxQyxRy
0)]()([])([ xpxqyxr eigenvalue
• The regular problem on [a,b]
• The periodic problem on [a,b]
• The singular problem on [a,b]
0)()(
0)()(
21
21
byBbyB
ayAayA
)()(
)()(
byay
byay
0)()(0)( 21 byBbyBar
0)()(0)( 21 ayAayAbr0)()( brar
A Strum-Liouville differential equation with boundary conditions at each end point x = a and x = b which satisfy one of the following forms:
has solutions, the eigenfunctions m(x) and n(x) which are orthogonal provided that the eigenvalues, m and n are different.
If the eigenfunctions, (x) result from a Strum-Liouville differential equation and nemce be orthogonal. The formal expansion of a general solution f(x) can be written in the form:
0
)()(n
nn xAxf
The value of An can be obtained by making use of the orthogonal properties of the functions (x)
b
a
b
an
nnmm dxxAxxrdxxfxxr0
)()()()()()(
b
a
b
a nmn
nm dxxxxrAdxxfxxr )()()()()()(0
Each term is zero except when m = n.
b
a
n
b
a
n
n
dxxxr
dxxfxxr
A2)]()[(
)()()(
0, 1, 2…… are eigenfunctions
Steady-state heat transfer with axial symmetry
0sinsin
12
T
r
Tr
r
0cot22
2
2
22
TT
r
Tr
r
Tr
Assume: )(grfT
)(
)(
grfT
grfr
T
0)()(cot)()()()(2)()(2 grfgrfgrfrgrfr
)1(cot22
llg
gg
f
frfr Dividing by fg and separate variables
)1(cot22
llg
gg
f
frfr 0)1(cot
0)1(22
gllgg
fllfrfr
0)1(22 fllfrfr
nArf set
0)1(2)1( nnn ArllnArArnn 0)1(2)1( llnnn
1 ll BrArf
0)1(cot gllgg cosmset
ddm sin
0)1(2)1( 2 gllgmgmLegendre’s equation of order l
Solved by the method of Frobenius
0
)(n
cnnmamgset
0)1(2)1( 2 gllgmgm
0
)(n
cnnmamg
0)1()(2)1)(()1(00
1
0
22
n
cnn
n
cnn
n
cnn mallmacnmmacncnm
比較係數2cm 0)1( 0 acc 10 orc
1cm 0)1( 1 cac 00 1 aorc
and
ss alcslcsacscs )1)(()1)(2( 2
)()()( mBQmAPmg ll where Pl(m) is the “Legendre polynomial”
)(grfT 1 ll BrArf
)()()( mBQmAPmg ll
cosm
)(cos)( 1 ll
ll
l PrBrAT
superposition
0
1 )(cos)(l
ll
ll
l PrBrAT
Unsteady-state heat transfer to a sphere
t
TT
12
t
T
r
T
rr
T
12
2
2
A sphere, initially at a uniform temperature T0 is suddenly placed in a fluid medium whose temperature is maintained constant at a value T1. The heat-transfer coefficient between the medium and the sphere is constant at a value h. The sphere is isotropic, and the temperature variation of the physical properties of the material forming the sphere may be neglected. Derive the equation relating the temperature of the sphere to the radius r and time t.
independent of and
Boundary condition:
0,00
0,
0,0
1
0
tratr
T
rtatTT
rtatTT
022
1 44)(0
rratrr
TkrTThq rrs
Assume: )(tgrfT
t
T
r
T
rr
T
12
2
2
gfgfr
gf 12
g
g
f
f
rf
f 12
0)()(
0)()(2)( 22
tgtg
rfrrfrrfr
g
g
f
f
rf
f 12
0)()(2)( 22 rfrrfrrfr Bessel’s equation see next slide...
rJrcrJrcrf 2
12
1
22
12
1
1)(
if 0
rccrf 1)( 43 if = 0
• Bessel’s equation of order
– occurs in studies of radiation of energy and in other contexts, particularly those in cylindrical coordinates
– Solutions of Bessel’s equation• when 2 is not an integer
• when 2 is an integer
– when = n + 0.5– when = n + 0.5
0)( 222 yxyxyx
0
22 )1(!2
)1()(
n
nn
n
xnn
xJ )()()( 21 xJcxJcxy
)()()( 5.025.01 xJcxJcxy nn
0)()( tgtg
tectg 5)( if 0
if = 0
6)( ctg
rcrcr
rJrcrJrcrf
cossin21
)( 212
12
1
22
12
1
1
tectg 5)(
)(tgrfT
Dr
CerBrAr
T t
1
cossin21
rccrf 1)( 43
6)( ctg
0,00
0,
0,0
1
0
tratr
T
rtatTT
rtatTTB.C.
B = C = 0
D = T1
0)( 1 rrr
TkTTh
1sin21
TerAr
T t
0,00 rtatTTB.C.
ter
rA
r
rA
r
T
2
sincos2
0)( 1 rrr
TkTTh
0)( 1 rrr
TkTTh
tt er
rA
r
rAkerA
rh
2
0
0
0
00
0
sincos2sin
21
hrk
krr
0
00tan
1sin21
TerAr
T t
hrk
krr
0
00tan
0,00 rtatTT
B.C.
0,00 rtatTT
rAr
TT n
n
sin21
10
rr
TTA
n
n
sin21
10
11
10 sin
sin21
21
n
tn
n
n
n
Ter
rr
TT
rT n
More general format of the solution (by superposition):
or
11sin
21
n
tnn
n
TerAr
T n
11 sin
21
n
tnn
n
nerAr
TT
If the constants An can be determined by making use of the properties of orthogonal function?
0)()(2)( 22 rfrrfrrfr solution of the form rr
r n
n
n
sin21
)(
orthogonal
100
00
0
0
2
2
0 102
2
cossin2
sin21
sin21
)]()[(
)()()(
)(0
0
TTr
rr
r
drrr
r
drTTrr
r
dxxxr
dxxfxxr
rA
n
n
n
nn
r
n
n
r
n
n
b
a
n
b
a
n
n
11
sin21
TerAr
Tn
tnn
n
n
where
100
00
0
cossin2)( TT
rr
r
rrA
n
n
n
nnn
and
hrk
krr n
n0
00tan
Equations involving three independent variables The steady-state flow of heat in a cylinder is governed by Laplace’s equation in cylindrical polar coordinates:
011
2
2
2
2
22
2
z
TT
rr
T
rr
T
There are three independent variables r, , z.
Assume: )(, zgrfT
011
2
2
22
2
gfgf
rg
r
f
rg
r
f
22
2
22
2 111v
g
gf
rr
f
rr
f
f
Separation of variables
0)()(
0
2
222
2
2
22
zgvzg
fvrf
r
fr
r
fr
OK,
vzv
vzv eBeAzg )(
Two independent variable P.D.E.
0222
2
2
22
fvrf
r
fr
r
fr
Assume: )()(, GrFrf
0222 FGvrGFGFrGFr
2222
kG
Gvr
F
FrFr
Separation of variables
0)()(
0)(2
2222
GkG
FkvrFrFr
OK, kBkAG kk sincos)(
Bessel’s equation
The solution of the Bessel’s equation: )()()( rvBYrvAJrF kk
vz
vvz
vkkkk eBeAkBkArvBYrvAJ
zgGrFzgrfT
sincos)()(
)()()()(,
In the study of flow distribution in a packed column, the liquid tends to aggregate at the walls. If the column is a cylinder of radius b m and the feed to the column is distributed within a central core of radius a m with velocity U0 m/s, determine the fractional amount of liquid on the walls as a function of distance from the inlet in terms of the parameters of the system.
z
U0
a
b
r
U
r
UDV
horizontal component of fluid velocity
Material balance:
Input: zrr
UDrrU 22
Output: rzrr
UD
rzr
r
UDzrrU
zrrU
2222
022
rzrr
UD
rzrrU
z
022
rzrr
UD
rzrrU
z
r
U
rr
UD
z
U 12
2
B.C.at z = 0, if r < a, U = U0
at z = 0, if r > a, U = 0at r = 0, U = finite
at r = b, z
Ukbb
r
UD
22
Assume: )(zgrfU
)1
( gfr
gfDgf 2
1
f
fr
f
Dg
g
0)()(
0
2
222
22
zDgzg
frdr
dfr
dr
fdr
DzeAzg2
)(
Bessel’s equation
The solution of the Bessel’s equation: )()()( 00 rYBrJArf if 0
rBArf ln)( 00 if = 0
)(zgrfU
DzeAzg2
)(
Dznn
nerJAAzgrfU2
)()()( 00
0)( Arf
)()( 0 rJArf
General form:
1
00
2
)(n
Dznn
nerJAAU
The Laplace Transform
• It is defined of an improper integral and can be used to transform certain initial value problems into algebra problems.
• Laplace Transform table!
0
)()( dttfesf st
)0()()( fssFsf
)0()0()()( 2 fsfsFssf
The Laplace transform method for P.D.E.
• The Laplace transform can remove the derivatives from an O.D.E.
• The same technique can be used to remove all derivatives w.r.t. one independent variable from a P.D.E. provided that it has an open range.
• A P.D.E has two independent variables can use “the Laplace transform method” to remove one of them and yields an O.D.E..
• The boundary conditions which are not used to transform the equation must themselves be transformed.
For unsteady-state one-dimensional heat conduction:
0)( TsTst
T
2
2
2
2 )(
dx
sTd
x
T
and
2
2
x
T
t
T
Laplace transform
2
2
0
)()(
dx
sTdTsTs
Second order linear O.D.E.
x and t are independent variables
x
dx
2
2
x
T
t
T
Boundary condition:
0
0
1
0
xatTT
tatTT The initial condition can use the Laplace transform method
00 TTsTs
TBeAeT
xs
xs
0
s regards as constant
The boundary condition: 01 xatTT
Laplace transform
0)( 1 xats
TsT
s
TBeAeT
xs
xs
0
0)( 1 xats
TsTand
when x , T remains finite remains finite B = 0T
B.C.
s
TAeT
xs
0
0)( 1 xat
s
TsT
s
TTA 01
s
Te
s
TTT
xs
001
inverse transform 001
2T
t
xerfcTTT
For unsteady-state one-dimensional heat conduction:
)(sTst
T
2
2
2
2 )(
dx
sTd
x
T
and
2
2
x
T
t
T
Laplace transform
2
2 )()(
dx
sTdsTs
Second order linear O.D.E.
x
dx
2
2
x
T
t
T
Boundary condition:
onstantcH
xtatT
0,00
0 TsT xs
xs
BeAeT
s regards as constant
the heat is concentrated at the surface initially
x and t are independent variables
xs
AeT
The boundary condition:
0
TdxCHtantconsH p
Laplace transform
0
dxTCs
Hp
x and t are independent variables
0
dxTCs
Hp
0dxAe
Cs
H xs
p
s
A
Cs
H
p
s
e
C
HT
xs
p
inverse transform
t
e
C
HT
t
x
p
4
2
For unsteady-state one-dimensional heat conduction:
)(sTst
T
2
2
2
2 )(
dx
sTd
x
T
and
2
2
x
T
t
T
Laplace transform
2
2 )()(
dx
sTdsTs
Second order linear O.D.E.
x
dx
2
2
x
T
t
T
Boundary condition:
0
0,00
xatx
TkQ
xtatT
0 TsT xs
xs
BeAeT
s regards as constant
the heat is supplied at a fixed rate
x and t are independent variables
xs
AeT
The boundary condition: 0
xatx
TkQ
Laplace transform
s
ks
QA
xs
esk
QT 2
3 inverse transform
0
xatx
Tk
s
Q
x and t are independent variables
0
x
xs
es
Aks
Q
t
x
dek
QT
0
4
2
1
Heat conduction between parallel planes
Consider the flow of heat between parallel planes maintained at different temperatures:
T0
T1
x Boundary condition:
LxatTT
xatTT
xtatTT
1
0
0
0
0,0 The initial condition can use the Laplace transform method
2
2
x
T
t
T
0)( TsTst
T
2
2
2
2 )(
dx
sTd
x
T
and
2
2
x
T
t
T
Laplace transform
2
2
0
)()(
dx
sTdTsTs
Second order linear O.D.E.
x and t are independent variables
00 TTsTs
TBeAeT
xs
xs
0
s regards as constant
The boundary condition: 00 xatTT
Laplace transform
0)( 0 xats
TsT
LxatTT 1
Laplace transform
Lxats
TsT 1)(
s
TBeAeT
xs
xs
0
0)( 0 xats
TsT
Lxats
TsT 1)( s
TBeAe
s
T
BA
Ls
Ls
01
0
s
T
ee
ee
s
TTT
Ls
Ls
xs
xs
001
inverse transform
001
0 2
22
sin)1(2
n
L
tnn e
L
xn
nL
x
TT
TT
Symmetrical heat conduction between parallel planes
Consider a wall of thickness 2L with a uniform initial temperature throughout, and let both faces be suddenly raised to the same higher temperature.
Boundary condition:
Lxatx
T
xatTT
xtatT
0
0
0,00
0
The initial condition can use the Laplace transform method
2
2
x
T
t
T
)(sTst
T
2
2
2
2 )(
dx
sTd
x
T
and
2
2
x
T
t
T
Laplace transform
2
2 )()(
dx
sTdsTs
Second order linear O.D.E.
x and t are independent variables
0 TsT xs
xs
BeAeT
s regards as constant
x
The boundary condition: 00 xatTT
Laplace transform
0)( 0 xats
TsT
Lxatx
T
0
Laplace transform
Lxatx
sT
0)(
xs
xs
BeAeT
0)( 0 xats
TsT
0
0
Ls
Ls
es
Bes
A
s
TBA
Ls
Ls
xs
Ls
xs
Ls
ee
eeee
s
TT
22
22
0
11
inverse transform
00
2
)22(
2
)2()1(
n
n
t
xLnLerfc
t
nLxerfcTT
Lxatx
sT
0)(
ExampleAn extensive shallow oilfield is to be exploited by removing product at a constant rate from one well. How will the pressure distribution in the formation vary with time?Taking a radial coordinate r measured from the base of the well system, it is known that the pressure (p) follows the normal diffusion equation in the r direction:
t
p
r
p
rr
p
11
2
2
where is the hydraulic diffusivity
If the oil is removed at a constant rate q:r
prkhq
r
2lim
0
where k is the permeability; h is the thickness of the formation; and is the coefficient of viscosity
0,00 rtatpp
0
1)(
1psp
s
t
p
2
2
2
2 )(
dr
spd
r
p
and
r and t are independent variables
dr
spd
r
p )(
t
p
r
p
rr
p
11
2
2 Laplace Transform
0
2
2 1 pp
s
dr
pd
rdr
pd
Second O.D.E
0
2
2 1 pp
s
dr
pd
rdr
pd 0022
p
ps
rprpr
Modified Bessel’s equation
s
pxBKxAIp 0
00 )()(
s
rx
0022
s
ppxpxpx
)(; 0 xIx
s
pxBKp 0
0 )(
The boundary condition: r
prkhq
r
2lim
0
Laplace transform
dr
pdrkh
s
qr
2lim
0
rkhs
qB
2
xxKx ln)(; 0
s
pxBKp 0
0 )( srkhs
qB
2
s
psrK
sp 0
0 )(
inverse transform
00
2
)4
exp(4
pdtt
r
rkh
qp
t
0
2
44p
t
rEi
rkh
qp
The restriction on the use of Laplace transform to solve P.D.E. problems
• The problem must be of initial value type.• Th dependent variable and its derivative remain
finite as the transformed variable tends to infinity• The Laplace transform should be tried whenever a
variable has an open range and the method of separation should be used in all other cases.
• There are many P.D.E.s which cannot be solved by either method, and the numerical methods are recommended.
The Laplace Transform• A particular “operational method” of solving differential
equations.
• An O.D.E. is converted into an equivalent algebraic form which can be solved by the laws of elementary algebra.
• If f(t) is a continuous function of an independent variable t for all values of t greater than zero, then the integral with respect to t of the product of f(t) with e-st between the limits 0 and is defined as the Laplace transform of f(t).
)()()(0
sFtfdttfe st
The parameter s must be large enough to make the integral convergent at the upper limit and t must be positive.
s
KKdtKe st
0
)(
11
0
0
)1()(
nn
zn
nnst
s
n
s
dzeztdtte
as
edtee atatst
1
)(0
)0()()()0()(
)]([)()()]([0
0
0
fssFtfsftf
dttfesetfdttfe ststst
)0()0()()(
)0()()0()()0()(
)]([)()()]([
2
0
0
0
fsfsFstf
fssFsftfsftf
dttfesetfdttfe ststst
)0()0(...)0()0()()( )1()2(21 nnnnnn fsffsfssFstf
Only valid for continuous functions
Usually given as boundary conditions
The shifting theorem
22
)sin(
s
t
(1) Boundary conditions are introduced into the problem before solution of the equation.(2) The differential equation is reduced to an algebraic equation in terms of the operator s.Note:The operation described is only applicable to “initial value” problems. (i.e., the value of the function and its derivatives must be known when the independent variable is zero)
Advantages about Laplace Transform:
The inverse transformation
• It must be convenient to convert the transform back to a function of the independent variable:
• Using partial fraction
• Laplace transform table
)()(1 tfsF
22 )())((
1
bs
C
bs
B
as
A
bsas
Example:
Solve , where y(0) = y’(0) = 1 342
2
ydt
yd
342
2
ydt
yd
Laplace Transform
342
2
y
dt
yd
s
sYysysYs3
)(4)0()0()(2
)0()0()()( 2 fsfsFssf s
KK
B.C.
s
sYssYs3
)(41)(2 )4(
3
4
1
4)(
222
ssss
ssY
)4(
3
4
1
4)(
21
21
211
ssss
ssY
Inverse Laplace Transform
)2cos1(4
3sin
2
12cos)( tsttty
Properties of the Laplace transform
• Differentiation of the transform
• Integral of a function
00
)()()()( dttftedttfeds
dsF stst
)(1
)(1
)()(
)()()(
00
00
0
00
sFs
dtetfss
edttfdtedttf
dttftfdttfe
ststtstt
tst
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