Page 1
Method of Separation of Variables
Institute of Lifelong Learning pg. 1
Subject : Maths
Lesson : Method of Separation of Variables
Lesson Developer : Chaman Singh
College/Department : Acharya Narendra Dev College ,
University of Delhi
Page 2
Method of Separation of Variables
Institute of Lifelong Learning pg. 1
Subject : Maths
Lesson : Method of Separation of Variables
Lesson Developer : Chaman Singh
College/Department : Acharya Narendra Dev College ,
University of Delhi
Method of Separation of Variables
Institute of Lifelong Learning pg. 2
Table of Contents:
Chapter : Method of Separation of Variables
? 1. Learning Outcomes
? 2. Introduction
? 3. Separation of Variables
? 4. The Vibrating String Problem
? 5. Solution of the vibrating string problem
? 5.1. Uniqueness Theorem
? 6. The Heat Conduction Problem
? 7. Solution of the heat conduction problem
? 7.1. Uniqueness Theorem
? 8. Non-homogeneous Problems
? Summary
? Exercises
? Glossary
? References/ Further Reading
1. Learning Outcomes:
After reading this chapter, the learner should be able to understand the
followings:
? Method of separation of variables to study the solutions of various
partial differential equations encountered in physical applications
in which the domain of interest are finite.
? Use of Separations of Variable Method to solve the problem of a
vibrating string of finite length l.
? Solution of the heat conduction problem.
? Solution of Non-homogeneous problems.
Page 3
Method of Separation of Variables
Institute of Lifelong Learning pg. 1
Subject : Maths
Lesson : Method of Separation of Variables
Lesson Developer : Chaman Singh
College/Department : Acharya Narendra Dev College ,
University of Delhi
Method of Separation of Variables
Institute of Lifelong Learning pg. 2
Table of Contents:
Chapter : Method of Separation of Variables
? 1. Learning Outcomes
? 2. Introduction
? 3. Separation of Variables
? 4. The Vibrating String Problem
? 5. Solution of the vibrating string problem
? 5.1. Uniqueness Theorem
? 6. The Heat Conduction Problem
? 7. Solution of the heat conduction problem
? 7.1. Uniqueness Theorem
? 8. Non-homogeneous Problems
? Summary
? Exercises
? Glossary
? References/ Further Reading
1. Learning Outcomes:
After reading this chapter, the learner should be able to understand the
followings:
? Method of separation of variables to study the solutions of various
partial differential equations encountered in physical applications
in which the domain of interest are finite.
? Use of Separations of Variable Method to solve the problem of a
vibrating string of finite length l.
? Solution of the heat conduction problem.
? Solution of Non-homogeneous problems.
Method of Separation of Variables
Institute of Lifelong Learning pg. 3
2. Introduction:
The techniques we have used to solve the PDEs of first order and second
order with variable and constant coefficients so far, seems to have certain
limitations in the sense that the solution obtained in terms of arbitrary
functions either fail to be particularised for physical/engineering problems
of interest or the algebra becomes quite involved.
In this chapter, a powerful technique is studied to solve the various partial
differential equations encountered in physical applications in which the
domain of interest are finite. In particular, we shall study the boundary
and initial conditions needed to specify the unique solutions of their
respective one-dimensional unsteady or two – dimensional steady
version. The basic philosophy of the method is to reduced the single linear
PDE into two or more (depending on the dimensions of the problem)
ODEs, each involving only one of the independent variables.
3. Method of Separation of Variables:
Here, we will describe the method of separation of variables to solve the
second order initial boundary value problems with variable and constant
coefficients separately.
Case I:
Let us a take the second order homogenous partial differential equation
0,
uu uv vv u v
Az Bz Cz Dz Ez Fz ? ? ? ? ? ? (1)
where , , , A B C D and F are functions of u and v .
by considering the transformation,
( , ), ( , ) x x u v y y u v ?? (2)
where
( , )
0
( , )
xy
uv
?
?
?
.
equation (1) can be transformed into standard/ canonical form
? ? ? ? ? ? ? ? ? ? , , , , , 0,
xx yy x y
a x y z c x y z d x y z e x y z f x y z ? ? ? ? ? (3)
Page 4
Method of Separation of Variables
Institute of Lifelong Learning pg. 1
Subject : Maths
Lesson : Method of Separation of Variables
Lesson Developer : Chaman Singh
College/Department : Acharya Narendra Dev College ,
University of Delhi
Method of Separation of Variables
Institute of Lifelong Learning pg. 2
Table of Contents:
Chapter : Method of Separation of Variables
? 1. Learning Outcomes
? 2. Introduction
? 3. Separation of Variables
? 4. The Vibrating String Problem
? 5. Solution of the vibrating string problem
? 5.1. Uniqueness Theorem
? 6. The Heat Conduction Problem
? 7. Solution of the heat conduction problem
? 7.1. Uniqueness Theorem
? 8. Non-homogeneous Problems
? Summary
? Exercises
? Glossary
? References/ Further Reading
1. Learning Outcomes:
After reading this chapter, the learner should be able to understand the
followings:
? Method of separation of variables to study the solutions of various
partial differential equations encountered in physical applications
in which the domain of interest are finite.
? Use of Separations of Variable Method to solve the problem of a
vibrating string of finite length l.
? Solution of the heat conduction problem.
? Solution of Non-homogeneous problems.
Method of Separation of Variables
Institute of Lifelong Learning pg. 3
2. Introduction:
The techniques we have used to solve the PDEs of first order and second
order with variable and constant coefficients so far, seems to have certain
limitations in the sense that the solution obtained in terms of arbitrary
functions either fail to be particularised for physical/engineering problems
of interest or the algebra becomes quite involved.
In this chapter, a powerful technique is studied to solve the various partial
differential equations encountered in physical applications in which the
domain of interest are finite. In particular, we shall study the boundary
and initial conditions needed to specify the unique solutions of their
respective one-dimensional unsteady or two – dimensional steady
version. The basic philosophy of the method is to reduced the single linear
PDE into two or more (depending on the dimensions of the problem)
ODEs, each involving only one of the independent variables.
3. Method of Separation of Variables:
Here, we will describe the method of separation of variables to solve the
second order initial boundary value problems with variable and constant
coefficients separately.
Case I:
Let us a take the second order homogenous partial differential equation
0,
uu uv vv u v
Az Bz Cz Dz Ez Fz ? ? ? ? ? ? (1)
where , , , A B C D and F are functions of u and v .
by considering the transformation,
( , ), ( , ) x x u v y y u v ?? (2)
where
( , )
0
( , )
xy
uv
?
?
?
.
equation (1) can be transformed into standard/ canonical form
? ? ? ? ? ? ? ? ? ? , , , , , 0,
xx yy x y
a x y z c x y z d x y z e x y z f x y z ? ? ? ? ? (3)
Method of Separation of Variables
Institute of Lifelong Learning pg. 4
which is hyperbolic, parabolic or elliptic according as a = - c, a = 0, a = c
respectively.
To solve, equation (3), we assume
( ) ( ) 0, z X x Y y ?? (4)
where X and Y are twice continuously differentiable functions.
Substituting equation (4) into equation (3), we get,
'' '' ' '
0, aX Y cXY d XY eXY f XY ? ? ? ? ? (5)
Let there exist a function p(x, y) such that if the equation (5) is divided
by p(x, y), we get,
? ? ? ? ? ? ? ? ? ? ? ?
'' '' ' '
1 1 2 2 3 3
0, a x X Y b y XY a x XY b y XY a x b y XY ?? ? ? ? ? ? ?
??
on dividing the obtained result by XY, we get,
? ? ? ? ? ? ? ? ? ? ? ?
'' '' ' '
1 1 2 2 3 3
0,
X Y X Y
a x b y a x b y a x b y
X Y X Y
?? ? ? ? ? ? ?
??
'' ' '' '
1 2 3 1 2 3
,
X X Y Y
a a a b b b
X X Y Y
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
(6)
The L.H.S. of this equation is the function of X only and R.H.S. of this
equation is the function of Y only.
on differentiating equation (6) w.r.t. x, we have
'' '
1 2 3
0,
d X X
a a a
d x X X
??
? ? ?
??
??
Now integrating, we get,
'' '
1 2 3
,
XX
a a a
XX
? ? ? ?
where ? is a constant.
'' '
1 2 3
,
YY
b b b
YY
? ? ? ? ?
? ?
'' '
1 2 3
0, a X a X a X ? ? ? ? ? (7)
? ?
'' '
1 2 3
0, b Y b Y b Y ? ? ? ? ? (8)
If the solutions of the equations (7) and (8) are denoted by X(x) and Y(y)
respectively, Then, ( , ) z x y is the solution of equation (3).
Case II:
If the coefficients are constants, then reduction to canonical form is not
necessary.
Consider the equation
0,
xx xy yy x y
Az Bz Cz Dz Ez Fz ? ? ? ? ? ? (9)
where A, B, C, D, E and F are constants such that at least one of them is
non-zero.
Page 5
Method of Separation of Variables
Institute of Lifelong Learning pg. 1
Subject : Maths
Lesson : Method of Separation of Variables
Lesson Developer : Chaman Singh
College/Department : Acharya Narendra Dev College ,
University of Delhi
Method of Separation of Variables
Institute of Lifelong Learning pg. 2
Table of Contents:
Chapter : Method of Separation of Variables
? 1. Learning Outcomes
? 2. Introduction
? 3. Separation of Variables
? 4. The Vibrating String Problem
? 5. Solution of the vibrating string problem
? 5.1. Uniqueness Theorem
? 6. The Heat Conduction Problem
? 7. Solution of the heat conduction problem
? 7.1. Uniqueness Theorem
? 8. Non-homogeneous Problems
? Summary
? Exercises
? Glossary
? References/ Further Reading
1. Learning Outcomes:
After reading this chapter, the learner should be able to understand the
followings:
? Method of separation of variables to study the solutions of various
partial differential equations encountered in physical applications
in which the domain of interest are finite.
? Use of Separations of Variable Method to solve the problem of a
vibrating string of finite length l.
? Solution of the heat conduction problem.
? Solution of Non-homogeneous problems.
Method of Separation of Variables
Institute of Lifelong Learning pg. 3
2. Introduction:
The techniques we have used to solve the PDEs of first order and second
order with variable and constant coefficients so far, seems to have certain
limitations in the sense that the solution obtained in terms of arbitrary
functions either fail to be particularised for physical/engineering problems
of interest or the algebra becomes quite involved.
In this chapter, a powerful technique is studied to solve the various partial
differential equations encountered in physical applications in which the
domain of interest are finite. In particular, we shall study the boundary
and initial conditions needed to specify the unique solutions of their
respective one-dimensional unsteady or two – dimensional steady
version. The basic philosophy of the method is to reduced the single linear
PDE into two or more (depending on the dimensions of the problem)
ODEs, each involving only one of the independent variables.
3. Method of Separation of Variables:
Here, we will describe the method of separation of variables to solve the
second order initial boundary value problems with variable and constant
coefficients separately.
Case I:
Let us a take the second order homogenous partial differential equation
0,
uu uv vv u v
Az Bz Cz Dz Ez Fz ? ? ? ? ? ? (1)
where , , , A B C D and F are functions of u and v .
by considering the transformation,
( , ), ( , ) x x u v y y u v ?? (2)
where
( , )
0
( , )
xy
uv
?
?
?
.
equation (1) can be transformed into standard/ canonical form
? ? ? ? ? ? ? ? ? ? , , , , , 0,
xx yy x y
a x y z c x y z d x y z e x y z f x y z ? ? ? ? ? (3)
Method of Separation of Variables
Institute of Lifelong Learning pg. 4
which is hyperbolic, parabolic or elliptic according as a = - c, a = 0, a = c
respectively.
To solve, equation (3), we assume
( ) ( ) 0, z X x Y y ?? (4)
where X and Y are twice continuously differentiable functions.
Substituting equation (4) into equation (3), we get,
'' '' ' '
0, aX Y cXY d XY eXY f XY ? ? ? ? ? (5)
Let there exist a function p(x, y) such that if the equation (5) is divided
by p(x, y), we get,
? ? ? ? ? ? ? ? ? ? ? ?
'' '' ' '
1 1 2 2 3 3
0, a x X Y b y XY a x XY b y XY a x b y XY ?? ? ? ? ? ? ?
??
on dividing the obtained result by XY, we get,
? ? ? ? ? ? ? ? ? ? ? ?
'' '' ' '
1 1 2 2 3 3
0,
X Y X Y
a x b y a x b y a x b y
X Y X Y
?? ? ? ? ? ? ?
??
'' ' '' '
1 2 3 1 2 3
,
X X Y Y
a a a b b b
X X Y Y
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
(6)
The L.H.S. of this equation is the function of X only and R.H.S. of this
equation is the function of Y only.
on differentiating equation (6) w.r.t. x, we have
'' '
1 2 3
0,
d X X
a a a
d x X X
??
? ? ?
??
??
Now integrating, we get,
'' '
1 2 3
,
XX
a a a
XX
? ? ? ?
where ? is a constant.
'' '
1 2 3
,
YY
b b b
YY
? ? ? ? ?
? ?
'' '
1 2 3
0, a X a X a X ? ? ? ? ? (7)
? ?
'' '
1 2 3
0, b Y b Y b Y ? ? ? ? ? (8)
If the solutions of the equations (7) and (8) are denoted by X(x) and Y(y)
respectively, Then, ( , ) z x y is the solution of equation (3).
Case II:
If the coefficients are constants, then reduction to canonical form is not
necessary.
Consider the equation
0,
xx xy yy x y
Az Bz Cz Dz Ez Fz ? ? ? ? ? ? (9)
where A, B, C, D, E and F are constants such that at least one of them is
non-zero.
Method of Separation of Variables
Institute of Lifelong Learning pg. 5
We assume a separable solution of the form ( , ) ( ) ( ) z x y X x Y y ? , and putting
it into equation (9), we get,
'' ' ' '' ' '
0, AX Y BXY CXY DXY EXY F XY ? ? ? ? ? ?
Dividing by AXY, we get,
'' ' ' '' ' '
0,
X B XY C Y D X E Y
F
X A XY A Y A X A Y
? ? ? ? ? ? (10)
Differentiating w. r. t. x, we get,
''
'' ' ' '
0,
X B X Y D X
X A X Y A X
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
Dividing the above eqn. by
'
'
BX
AX
??
??
??
, we get,
'
''
'
'
'
,
X
X DY
BY
BX
AX
?
??
??
??
? ? ? ?
??
??
??
where ? is a separation constant.
'
'
'
0, 0,
D B X
YY
B A X
??
??
??
? ? ? ? ?
?? ??
??
??
(11)
Integrating second eqn. w. r. t. x, we get,
'' '
,
X D B X
X B A X
??
??
??
? ? ? ?
?? ??
??
??
(12)
where ß is a constant to be determined.
Substituting equation (11) in equation (10), we get,
? ? ? ?
'' ' '' '
0,
X B X C Y D X E
F
X A X A Y A X A
??
??
? ? ? ? ? ? ? ?
??
??
'' ' ''
0,
X D B X Y E F C
X B A X Y C C A
??
? ? ? ?
??
? ? ? ? ? ? ?
? ? ? ? ??
??
? ? ? ?
(13)
From equation (12) and equation (13), we get,
''
,
Y E F C
Y C C A
??
??
? ? ? ?
??
??
If the solutions of the equations (13) and (11) are denoted by X(x) and
Y(y) respectively, Then, ( , ) z x y is the solution of equation (9).
4. The Vibrating String Problem:
To illustrate the details of the method, let us take the problem of a
vibrating string of length l, that is fixed at both ends and released from
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