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BITS PilaniPilani Campus
BITS Pilani
presentationDr. Padma Murali
Department of Mathematics
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Engineering Mathematics
IILecture 16
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Separable partial differential equation
Review: Partial differential equation(PDEs), like ordinary differential equations
(ODEs), are classified as linear or
nonlinear. Analogous to a linear ODE (see(6) of section 1.1) the dependent variable
and its partial derivatives appear only to
the first power in a linear PDE. In this
chapter we are concerned only with linear
partial differential equations.
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Linear Partial Differential Equation: If we
let u denote the dependent variable and x
and y the independent variables, then the
general form of a linear second-order
partial differential equation is given by
.2
22
2
2
GFuy
uE
x
uD
y
uC
yx
uB
x
uA
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Where the coefficients A,B,C,G areconstants or functions of x and y. When
G(x,y) = 0. equation (1) is said to be
homogeneous; otherwise it isnonhomogeneous. For example, the linear
equations
xyy
u
x
u
andy
u
x
u
2
2
2
2
2
2
0
are homogeneous and nonhomogeneous,
respectively.
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Separation of variables:
Although there are several methods that canbe tried to find particular solutions of a linear
PDE, in the method of separation of variables
we seek to find a particular solution of the formof a product of a function of x and a function of
y. yYxXyxu ,
With this assumption, it is sometimes possible
to reduce a linear PDE in two variables to two
ODEs. To this end we observe that
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YXy
uYX
x
uYX
y
uYX
x
u
2
2
2
2
,,,
Where the primes denote ordinarydifferentiation.
Example: Using Separation of Variables
Find product solution of yu
xu
42
2
Solution: Substituting u(x,y) = X (x) Y(y) into
the partial differential equation yields
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YXYX 4After dividing both sides by 4XY we have
separated the variables:
.4 Y
Y
X
X
Since the left hand side of the last equationis independent of y and is equal to the right
hand side, which is independent of x, we
conclude that
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both sides of the equation are independentof x and y. In other words, each side of the
equation must be a constant. As a practical
matter it is convenient to write this realseparation constant as From the two equalities.
Y
Y
X
X
4
We obtain the two linear ordinary differential
equations.
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004 YYandXX
For the three cases for
zero, negative, or positive; that is
0,00,0 22 whereand
the ODEs in (2) are, in turn
,00
YandX004 22 YYandXX
004 22 YYandXX
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Case I: 0
The DEs in (3) can be solved by integration.
The solutions are X = c1 + c2x and Y=c3.
Thus a particular product solution of the
given PDE is
xBAcxccXYu
11321
Where we have replaced c1c3 and c2c3 by
A1 and B1, respectively
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Case II 2 The general solutions of the DEs in (4) are
yecYandxcxcX2
654 2sinh2cosh
respectively. Thus another particular
product solution of the PDE is
yecxcxcXYu2
654 2sinh2cosh
OR
xeBxeAu yy 2sinh2cosh22
22
.652642 ccBandccA Where
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Case III 2
Finally, the general solutions of the DEs in
(5) are:yecYandxcxcX
2
987 2sin2cos
respectively. These results give yet anotherparticular solution
xeBxeAu yy 2sin2cos22
33
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Where 973 ccA And 983 ccB
It is left as an exercise to verify that (6), (7)
and (8) satisfy the given partial differential
equation
yxx uu 4
See problem 29 in exercises 13.1
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Theorem 13.1 Superposition Principle
If kuuu ,........, 21
are solutions of a homogeneous linear
partial differential equation, then the linearcombination.
kkucucucu .......2211
Where the kici ,....,2,1,
are constants is also a solution.
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Definition 13.1 Classification of Equations
The linear second order partial differentialequation
02
22
2
2
Fuy
u
Ex
u
Dy
u
Cyx
u
Bx
u
A
Where A,B,C,D,E and F are real
constants is said to be
Hyperbolic if 042 ACB
Parabolic if 042 ACB
Elliptic if042 ACB
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Heat Equation
Introduction
Consider a rod of length L with an initial
temperature f(x) through-out and whoseends are held at temperature zero for all
time t>0. The temperature u(x,t) in the rod is
determined from the boundary value
problem
0,0,2
2
tLx
t
u
x
uk (1)
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)3(...0,0,
)2(...0,0,,0,0
Lxxfxu
ttLutu
Solution of the BVP: Using the product
andtTxXtxu ,,as the separation constant, leads to
kT
T
X
X(4)
OR
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0 XX (5)0 TkT (6)
Now the boundary conditions in (2) becomes
0,00,0 tTLXtLuandtTXtuSince the last equalities must hold for all
time t, we must have X(0) = 0 and X(L) = 0.
These homogeneous boundary conditionstogether with the homogeneous ODE(5)
constitute a regular Sturm-Liouville
problem:
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0,00,0 LXXXX (7)
The general solution of the DEs are 0,21 ifxccxX
(8)
0,sinhcosh2
21
ifxcxcxX(9)
0,sincos 221 ifxcxcxX (10)
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Recall, when the boundary condition X(0) =0 and X (L) = 0 are applied to (8) and (9)
these solutions yield only X(x) = 0 and so
we are left with the unusable result u = 0.Applying the first boundary condition X(0) =
0 to the solution in (10) gives c1 = 0.
Therefore xcxX
sin2
The second boundary condition X(L) = 0
now implies
0sin
2 LcLX
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If c2 = 0, then X = 0 so that u=0. But (11)can be satisfied for 02 c when
.0sin L
This last equation implies that ,/LnL where n = 1,2,3,.. Hence (7) possesses
nontirivial solution when
,....3,2,1,/
2222
nLnnn The values n
and the corresponding solutions
,.....3,2,1,sin2 nxL
ncxX
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are the eigenvalues and eigenfunctions,respectively, of the problem in (7). The
general solution of (6) is
,
222 /
3
tLnk
ecT
and so xL
neAtTxXu tLnknn
sin222 / (13)
where we have replaced the constant c2c3 by
An. The products un(x,t) given in (13) satisfythe partial differential equation (1) as well as
the boundary condition (2) for each value of
the positive integer n.
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However, in order of the functions in (13)
to satisfy the initial condition (3), we
would have to choose the coefficient An
in such a manner that
xL
nAxfxu nn
sin0, (14)
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is not a solution of the problem given in (1)
(3). Now by the superposition principlethe function
In general, we would not expect
condition (14) to be satisfied for an
arbitrary, but reasonable, choice of f.
Therefore we are forced to admit that txun ,
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1 1
/ sin,222
n n
tLnk
nn xL
neAutxu
(15)
Must also, although formally, satisfy
equation (1) and the conditions in (2). If we
substitute t = 0 into (15) then
1
.sin0,n
n xL
nAxfxu
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The last expression is recognized as thehalf-range expansion of f in a sine series.
If we make the identification
........3,2,1, nbA nnit follows from (5) to section 12.3 that
L
n dxx
L
nxf
L
A0
.sin2
(16)
We conclude that a solution of the
boundary-value problem described in (1),
(2) and (3) is given by the infinite series
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1
/
0
sinsin2
,223
n
tLnkL
xL
nedxx
L
nxf
Ltxu
(17)
In the special case when the initial
temperature is 1,,1000, kandLxu you should verify that the coefficient (16)
are given by
n
A
n
n
11200
And that the series (17) is
nxen
txu tn
n
n
sin11200
,2
1
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Wave Equation
Introduction
The vertical displacement u(x,t) of a string
of length L that is freely vibrating in thevertical plane is determined from
0,0,2
2
2
2
2
tLxt
u
x
ua (1)
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0,0,0,0 tLutu (2)
.0,,0,0
Lxxgt
uxfxu
t
(3)
Solution of the BVP with the usual
assumption that u(x,t) = X (x) T(t), separating
variables in (1) gives
Ta
T
X
X2
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So that0 XX
02 TaT
(4)
(5)
As in Section 13.3 the boundary condition(2) translate into X(0) = 0 and X(L) = 0.
The ODE in (4) along with these
boundary-conditions is the regular Sturm-Liouville problem
0,00,0 LXXXX (6)
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Of the usual three possibilities for the parameter
.0,0,0: 22 and
only the last choice leads to nontirivalsolutions. Corresponding to 0,2
the general solution of (4) is
xcxcX sincos 21 X(0) = 0 and X (L) = 0 indicate that c1 = 0
and 0sin2
Lc
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The last equation again implies that./LnornL
The eigenvalues and corresponding
eigenfunctions of (6) are ,.....3,2,1,sin/ 2
222 nxL
ncxXandLnn
the general solution of the second order
equation (5) is then
.sincos 43 tL
anct
L
anctT
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By rewriting c2c3 and An and c2c4 as Bn
solutions that satisfy both the wave equation
(1) and boundary conditions (2) are
xL
n
L
anBt
L
anAu nnn
sinsincos
(7)
and
1
.sinsincos,n
nn xL
nt
L
anBt
L
anAtxu
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Setting t = 0 in (8) and using the initial
condition u(x,0) = f(x) gives
1 .sin0, nn xL
n
Axfxu
Since the last series is a half-range
expansion for f in a sine series, we can write
An = bn.
L
n dxxL
nxf
LA
0
.sin2
(9)
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To determine Bn we differentiate (8) with
respect to t and then set t = 0.
1sincossin
n
nn xL
ntL
an
L
anBtL
an
L
anAt
u
10
.sin
n
n
t
xL
n
L
anBxg
t
u
In order for this last series to be half
range sine expansion of the initial velocity g
on the interval, the total coefficient
LanBn /
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must be given by the form bn in (5) of section
12.3 that is,
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L
n xdxLnxg
LLanB
0
sin2
From which we obtain
L
n dxxL
nxgan
B0
sin2
(10)
The solution of the boundaryvalue problem
(1) (3) consists of the series (8) with
coefficients An and Bn defined by (9) and
(10), respectively.
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We note that when the string is released
from rest, then g(x) = 0 for every x in the
interval
Lx 0 and consequently
0
nB
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