Post on 28-Apr-2023
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Course: M.Sc. Mathematics (II Year)
Paper: II
Partial Differential Equation
&
Mechanics
BLOCK - I
UNIT 1 – Partial Differential Equation: Classification UNIT 2 – Non Linear First Order Partial Differential Equation
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UNIT 1
Partial Differential Equation: Classification
STRUCTURE
1.1 Introduction
1.2 Objective
1.3 Transport Equation
1.3.1 Initial Value Problem
1.3.2 Non homogeneous equation
1.4 Laplace’s equation
1.4.1 Fundamental Solution
1.4.2 Mean value Formula
1.4.3 Properties of solutions
1.5 Wave equation
1.5.1 Solution by Spherical Means
1.5.2 Non-homogeneous Equations
1.5.3 Energy Methods
1.6 Unit Summary/ Things to Remember
1.7 Assignments/ Activities
1.8 Check Your Progress
1.9 Points for Discussion/ Clarification
1.10 References
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1.1 Introduction
A Partial Differential Equation is a relationship between an unknown function of two or more
variables and its derivatives with respect to the variables. Partial differential equations have a
great variety of applications to mechanics, electrostatics, quantum mechanics and many other
fields of physics as well as to finance. The unit introduces basic examples arising in mechanics,
electromagnetism, complex analysis and other areas, and develops a number of tools for their
solution, in particular Fourier analysis and distribution theory. These tools are then applied to
the treatment of basic problems in linear PDE, including the linear transport, Laplace equation,
and wave equation, as well as more general elliptic, parabolic, and hyperbolic equations.
1.2 Objective
After the completion of this unit one should able to:
Identify with the behavior of Partial differential Equations and their Classifications.
Get a hold on the transport equation, Laplace‟s Equation, and the heat equation along with
their properties.
Find the various physical applications of PDEs.
Explain fundamental solution for wave equation.
Study Local Estimates for harmonic functions.
Get a hold on properties of PDEs under Energy methods.
Understand Liouville‟s Theorem.
Solve non-homogeneous wave equation.
Find Mean Value Formula for Laplace‟s Equation.
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1.3 Transport equation
A differential equation of the form ),0(0. n
t RinDubu …………..(1)
Where b is a fixed vector in RRuandbbbR n
n
n ).0[:),......(, 1 is the unknown, ),( txuu
where n
n Rxxxx ),......,( 21 denotes a typical point in space, and 0t denotes a typical time. Here
),.....,(21 nxxxx uuuuDDu for the gradient of u with respect to the spatial variables x.
1.3.1 Initial value problem
Let us consider the initial-value problem
}0{
),0(0.
tRongu
RinDubu
n
n
t ……………..(2)
Hence nRb and RRg n : are known, and the problem is to compute u.
Given ),( tx as above, the line through ),( tx with direction (b,1) is represented parametrically by
).(),( Rsstsbx This line hits the plane }0{: tRn when
s = -t, at the point ).0,( tbx since u is constant on the line and )()0,( tbxgtbxu .we deduce
).0,()(),( tRxtbxgtxu n ………………(3)
So, if (2) has a sufficiently regular solution u, it must certainly be given by (3). And conversely, it is
easy to check directly that if g is 1C , then u defined by (3) is indeed a solution of (2).
1.3.2 Non Homogeneous problem
Non-homogeneous problem for Transport equation is defined by:
0
,00.
tRongu
RinDubu
n
n
t ……………….(4)
Solution of non-homogeneous Problem-
Let 1).( nRtx and set ),(:)( stsbxusz for .Rs . Then
5
),(),().,()(.
stsbxfstsbxubstsbxDusz t
Consequently,
)()0()(),( tzztbxgtxu
=
0.
.
)(t
dssz
=
0
),(t
dsstsbxf
t
dssbtsxf0
),)((
So )0,(),)(()(),(0
tRxdssbtsxftbxgtxu n
t
………(5)
which shows the solution for the initial value problem (4).
1.4 Laplace Equation
Laplace equation or Potential equation is defined by
0u
Two dimensional equation is given by 02
2
2
2
y
u
x
u
Three dimensional equation is given by 02
2
2
2
2
2
z
u
y
u
x
u
Physical interpretation- The Laplace equation is very important in application. It appear in physical
phenomena such as
1. Steady-state heat incompressible fluid flow.
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2. Steady-state heat conduction in a homogeneous body with constant heat capacity and constant
conductivity.
3. Electrical potential of a stationary electrical field in a region without charge.
1.4.1 Fundamental Solution of Laplace Equation
Derivation of fundamental solution:
Let Solution of Laplace equation 0u in nRU , having the form )()( rvxu ,
Where 2/122
1 ........ nxxxr and v is to be selected so that 0u holds.
For ni ..,.........1
).(1
)(
....,.........1
1)()('',)(
)0(2)...........(2
1
'"
3
2
'
2
2
'
2122
1
rvn
nrvu
niFor
r
x
rrv
r
xrvu
r
xrvu
xr
xxxx
x
r
ii
xx
i
x
i
in
i
iii
Hence 0u if and only if
.0)(1
)( '"
rvr
nrv
If 0' v , we deduce ,1
)log('
''''
r
n
v
vv
and hence 1
' )(
nr
arv for some constant a . Consequently if r >0,we have
)3(
)2(log
)(
2nc
r
b
ncrb
rv
n
which is fundamental solution of Laplace Equation where b and c are constants.
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1.4.2 Mean Value Formula for Laplace equation
Consider an open set nRU and suppose u is harmonic function within U. Now we derive the important
mean-value formulas, which declare that u(x) equals both the average of u over the sphere .),( UrxB
These implicit formulae involving u generate a remarkable number of consequence.
Theorem-1- If )(2 UCu is harmonic, then ),(),(
)(rxBrxB
udyudSxu for each ball .),( UrxB
Proof-Set
),( )1,0().()()()(:)(
rxB BzdSrzxuydSyur
Then )1,0(
' ),().()(B
zzdSrzxDur
And consequently, using Green‟s formulae, we compute
),(
),(
),(
'
0)(
)(
)().()(
rxB
rxB
rxB
dyyun
r
ydSv
u
ydSr
xyyDur
Hence is constant and so
),(00).()()(lim)(lim)(
txBttxuydSyutr
Theorem-2 (Converse to mean-value property)
If )(2 UCu satisfies ),(
)(rxB
udSxu for each ball .),( UrxB then u is harmonic.
Proof- If ,0u there exists some ball .),( UrxB such that, say, 0u within B(x, r). But then for
as above, ),(
' 0)()(0rxB
dyyun
rr , which is contradiction.
Theorem-3 Mean value formula for two dimensional equation
Let u (P) be a harmonic function on Laplace equation.
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For the sphere S the normal vector at SP is
., 0000
a
zz
a
yy
a
xx
a
PPn
Let us make the change of variables
.cos
,sinsin
,sincos
0
0
0
zz
yy
xx
Then for )cos,sinsin,sincos(),,( 000 zyxuu
We have zyxS ua
zzu
a
yyu
a
xx
n
u 000|
a
zyx
u
uuu
|),,(
cossinsinsincos
Equation (1) becomes
.sin|),,(
sin|),,(
|),,(0
2
0 0
22
0 0
ddu
ddau
dSu
a
a
Sa
)1.........(.0
.
.:)(
,:)(
0
3
0
0
3
0
P
B
P
a
a
dSn
udVvfollowsit
dSn
uvdVuvudVvBy
aPPRPPSS
aPPRPPBB
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Above result is valid for every a>0 so that we can consider a as a variable r and we have
ddrurI
Then
ddrudr
d
sin),,()(
0sin),,(
2
0 0
2
0 0
Letting ,0r we get
)(4
sin)(
sin),,0()(lim
0
0
2
0 0
2
0 00
Pu
ddPu
ddurIr
Then it follows that
.)(4
1)(
)(
sin),,()(4
sin),,()(4
20
22
0 00
2
2
0 00
SP
SP
dSPua
Pu
dSPu
ddaauPua
ddauPu
NOTE: Mean value property is also valid in the two dimensional case. Namely, if u(x, y) is a
harmonic function in 2
000
2 ),(, RyxPR and }:{ 0
2 aPPRPKa
In a disk, aa KC then aC
PdSPua
Pu )(2
1)( 0
which is known as mean value formula for the
two dimensional equation.
1.4.3 Properties of Solution
a) Strong maximum principle; Uniqueness.
Suppose )()(2 UCUCu is harmonic within U.
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(i) then UU
uuxma
max
(ii) Furthermore, if U is connected and there exists a point Ux 0 such that U
uxu max)( 0
Then u is constant within U.
Assertion (i) is the maximum principle for Laplace‟s equation and (ii) is the strong maximum
principle.
Proof- Suppose there exists a point Ux 0 with uxmaMxuU
:)( 0 . Then for
),,(0 0 Uxdistr the mean-value property asserts
),(
0
0
.)(rxB
MudyxuM
As equality holds only if Mu within ),( 0 rxB we see ),()( 0 rxByMyu . Hence the set
})(|{ MxuUx is both open and relatively closed in U, and thus equals U if U is connected.
This proves assertion (ii), from which (i) follows.
b) Smoothness- If )(UCu satisfies the mean-value property
),(),(
)(rxBrxB
udyudSxu for each ball ,),( UrxB then
)(UCu
Proof- Let be a function. Set uu *: in }.),(|{ UxdistUxU
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We will prove u is smooth by demonstrating that in fact uu on U . Indeed if Ux then
)(
)(
)()()(1
)()(1
)((1
)()()(
),0(
1
0
),(0
),(
xu
dyxu
drrnnr
xu
drudSr
dyyuyx
dyyuyxxu
B
n
n
rxBn
xBn
U
Thus Uinuu and so 0)( eachforUCu
c) Local estimates for harmonic function- Assume u is harmonic in U. Then
),((0
01)(
rxBLkn
k ur
CxuD
………(1)
For each ball UrxB ),( 0 and each multiindex of order k
Here ........).1()(
)2(,
)(
1 1
0
kn
nkC
nC
kn
k
………(2)
Proof: By mathematical induction method, the case k=0 being immediate from the mean-value
formula.
For case k=1, we note upon differentiating Laplace‟s equation that ),.......2,1( niuix
is harmonic. Consequently
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)).2,((
2,(
2,(0
0
0
0
2
)(
2
)(
rxBL
rxBin
n
rxBxx
ur
n
dSuvrn
dxuxuii
……..(3)
Now if UrxBrxBthenrxBx ),()2,(),2,( 00 and so )),(( 0
1
2
)(
1)(
rxBL
n
urn
xu
By (1) & (2) for k=0
On combining the inequalities above, we deduce )),((1
1
00
1
1
)(
2)(
rxBLn
n
urn
nxuD
If 1 . This verifies (1) & (2) for k=1.
Assume now 2k and (1) & (2) is valid for all balls in U and each multiindex of order less than
or equal to k-1. Fix UrxB ),( 0 and let be multi index with k . Then
xiuDuD )( for some i={1,2,…..n}, 1 k . By calculation similar to those in (3), we
establish that
)),((
00
)(k
rxBL
uDr
nkxuD
If ),,( 0k
rxBx then UrxBr
k
kxB
),()
1,( 0 .
Thus (1), (2) for k-1 imply )),((
1
11
00
1
)1
)((
))1(2()(
rxBLkn
kn
u
rk
kn
knxuD
Combining the two previous estimates yields the bound )),((
1
00
1
)(
)2()(
rxBLkn
kn
urn
nkxuD
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This confirms (1), (2) for k .
d) Liouville’s Theorem- Suppose RRu n : is harmonic and bounded. Then u is constant.
Proof- Fix 0,0 RRx n and apply theorem of estimate of derivative on :),( 0 rxB
,)(
)(
0)(
1
)),((11
10
,0
nRL
rxBLn
ur
nCn
ur
CnxDu
As .r Thus 0Du
Hence u is constant.
e). Analyticity- Assume u is harmonic in U. Then u is analytic in U.
Proof- 1. Fix any point .0 Ux We must show u can be represented by a convergent power series
in some neighborhood of .0x
Let ).,(4
1: 0 Uxdistr Then .
)(
1:
))2,((1 0
rxBLn
urn
M
2. Since ),,()2,(),( 00 rxBxeachforUrxBrxB Then by theorem on estimates on
derivative provides the bound .2 1
).(( 0
r
nMuD
n
rxBL
Now Stirling‟s formula asserts .)2(
1
!lim
21
2
1
k
k
kek
k
Hence !
Ce
For some constant C and all multi indices . Furthermore,
k
kkn
!
!)1........1( whence
!!
n
14
Combining the previous inequalities now yields
!.2 21
).(( 0
r
enCMuD
n
rxBL ………….(1)
3. The Taylors series for u at 0x is
)(
!
)(0
0 xxxuD
The sum is taken over all multi indices. We asserts this power series converges, provided
.2 320
en
rxx
n …………(2)
To verify this, let us compute for each N the remainder term:
N
N
k k
N
xxxxtxuD
xxxuDxuxR
!
))(((
!
))(()(:)(
000
1
0
00
For some ,10 t t depend on x. we establish this formula by writing out the first N terms and the
error in the Taylors expansion about 0 for the function of one variable ))((:)( 00 xxtxutg at
t=1. From (1) & (2), we can estimate
.0
2)2(
1
2
2)(
32
21
Nas
CM
nCMn
en
r
r
enCMxR
NN
N
N
n
N
N
n
N
1.5 Wave equation
Wave equation is given by 0 uutt ……….(1)
And the non-homogeneous wave equation is given by futt . ………..(2)
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Subject to appropriate initial and boundary conditions. Here t>0 and ,Ux where nRU is open.
The unknown is ),(),,,0: txuutRUu , and the laplacian is taken with respect to the
spatial variables ...........1 nxxx
In Eq.(2) the function RUf ,0: is given. A common abbreviation is to write
uuu tt .
Physical interpretation- The wave equation is a simplified model for a vibrating string (n=1),
membrane (n=2).or elastic solid (n=3). In these physical interpretation ),( txu represents the
displacement in some direction of the point x at time .0t
Let V represent any smooth subregion of U. The acceleration within V is then
dxudxudt
d
V Vtt 2
2
And the net contact force is V
dSvF ,.
Where F denotes the force acting on V through V and the mass density is taken to be unity.
Newton‟s law asserts the mass times the acceleration equals the net force:
dxuV
tt V dSvF ,.
This identity obtains for each sub-region V and so divFutt .
For elastic bodies, F is a function of the displacement gradient Du; whence 0)( DudivFutt
For small Du, the linearization aDuDuF )( is often appropriate; and so 0 uautt
This is the wave equation if a=1.
Note: u=displacement, tu =velocity at time t=0
1.5 .1 Solution by Spherical Means –For the wave equation firstly solving 0 uutt
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For n=1 directly and then for 2n by the method of spherical.
(a) Solution for n=1, D’ Alembert’s formula.-
Initial-value problem for the one-dimensional wave equation in all of R:
}0{,
),0(0
tRonhugb
Rinuu
t
xxtt ……………..(1)
where g, h are given.
PDE in (1) can be “factored” as 0
xxtt uuu
xtxt…………….(2)
Suppose vuxt
………………(3)
From (2) ).0,(0 tRxvv xt
which is transport equation with constant coefficients.
at n=1, b=1, we find )( txav ……………….(4)
For )0,(:)( xvxa , Combining now (2) to (4), we obtain .,0)( Rintxauu xt
which is known as a non-homogeneous transport equation;(with n=1,b=-1, f(x, t)=a(x-t)) implies
)5...(.....................),()(2
1
)())((),(0
tx
tx
t
txbdyya
txbdssstxatxu
where ).0,(:)( xuxb
First initial condition in (1) gives );()()( Rxxgxb
Whereas the second initial condition and (3) implies
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).()()(
)0,()0,()0,()(
' Rxxgxh
xuxuxvxa xt
Put in equation (5) which gives
tx
txtxgdyygyhtxu ).()()(
2
1),( '
Hence
tx
txtRxdyyhtxgtxgtxu )0,()(
2
1)]()([
2
1),(
(b) Spherical means- Now suppose ),0(2,2 nm RCuandmn solves the initial
value problem:
.0,
),0(0
tRonhugb
Rinuu
n
t
n
tt
We intend to derive an explicit formula for u in terms of g, h. The plan will be to study first the
average of u over certain spheres. These averages, taken as functions of the time t and the radius r,
turn out to solve the Euler- Poisson-Darboux equation, a PDE which we can for odd convert into
the ordinary one-dimensional wave equation. Applying D‟Alembert‟s formula, or more precisely its
variant (1), eventually leads us to a formula for the solution.
1.5.2 Non-Homogeneous equation –Initial problem for the non-homogeneous wave equation is
defined by
.00,0
),0(
tRonuu
Rinfuu
n
t
n
tt
Motivated by Duhamel‟s principle we define u=u(x, t;s) to be the solution of
}{0(.;)(.;,0.;
),(0.;.;
stRonsfsusu
sRinsusu
n
t
n
tt
So t
n tRxdsstxutxu0
)0,();,(),(
Duhamel‟s principle asserts this is a solution of
00,0
),0(
tRonuu
Rinfuu
n
t
n
tt
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Theorem: Solution of non-homogeneous wave equation-
Assuming 2n and ),0(12/ nn RCf
Define u by )0,(;,:),(0
tRxdsstxutxu nt
n
t
tRx
xtx
tRx
xtx
n
tt
n
Rxpoeachfortxutxuiii
Rinfuuii
RCui
nn
0
0,
)0,(),(
0,
)0,(),(
2
int0),(lim,0),(lim)(
),,0()(
),,0()(
00
Proof- If n is odd, 2
11
2
nn,
By Solution of wave in odd dimension ),0()(.,.; 2 nRCsu
For each 0s and so ),0(2 nRCu . If n is even, 2
21
2
nn.
Hence ),0(2 nRCu .
We then compute:
t
tt
t
ttttt
t t
ttt
dsstxutxf
dsstxuttxutxu
dsstxudsstxuttxutxu
0
0
0 0
);,(),(
);,();,(),(
,);,();,();,(),(
Furthermore t t
tt dsstxudsstxutxu0 0
.);,();,(),(
Thus ),0,(),(),(),( tRxtxftxutxu n
tt
Hence n
t Rxforxuxu 0)0,()0,(
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1.5.3 Energy Methods
(a) Uniqueness- Let nRU be bounded, open set with a smooth boundary U ,and as usual set
.0,],,0( TwhereUUTTUU TTT
Theorem -1. (Uniqueness for wave equation)- There exists at most one function )(2TUCu
solving
}0{tUonhu
Tongu
Uinfuu
t
Ttt
Proof- If u is another such solution, then uuw : solves
}0{0 tUonw
Tongw
Uinfww
t
Ttt
Define the “energy” ).0(),(),(2
1:)(
22TtdxtxDwtxwte
Ut
We have to find :
0)(
.)(
dxwww
dxDwDwwwte
ttU
t
tttU
t
There is no boundary term since w=0, and hence 0tw on ].,0[ TU
Thus for all .0,,0)0()(,0 Tt UwithinDwwsoandeteTt
Since }0{0 tUonw , we conclude TUinuuw 0 .
(b) Domain of dependence.
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Let us examine again the domain of dependence of solutions to the wave equation in all of space.
For this, suppose 2Cu solves
Fix 0, 00 tRx n and consider the cone ttxxtttxC 000 ,0|),{(
Theorem2. (Finite propagation speed)- If 0 tuu on 0),,( 00 uthentxB within the cone C.
Proof-Define )0(),(),(2
1:)( 0
2
),(
2
00
ttdxtxDutxutettxB
t
Then
)1...(.....................2
1
2
1
2
1)(
2
1.)(
22
),(
2
),(
2
),(),(
2
),(
2
),(
00
000000
0000
dSDuuuv
u
dSDuudSuv
udxuuu
dSDuudxDuDuuute
ttttx
ttxBtt
ttxBtt
ttxBt
ttxBtt
ttxBttt
B
Now ,2
1
2
1 22DuuDuuu
v
uttt
……………..(2)
By the Cauchy- Schwarz and Cauchy inequalities, from (1) & (2) we, find
.00)0()(0)( 0ttallforetesoandte
Thus 0, Duut and consequently 0u within the cone C.
1.6 Unit Summary/ Things to Remember
1. Transport equation- 0. Dubut
2. Laplace‟s equation 0u
3. Wave equation 0 uutt
4. Wave equation 0 uut
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5. Initial value problem for Transport equation is given by
}0{
),0(0.
tRongu
RinDubu
n
n
t
6. Non-homogeneous problem for Transport equation is given by
0
,00.
tRongu
RinDubu
n
n
t
7. The wave equation is simplified model for a vibrating string (n = 1), membrane (n =2), or elastic
solid (n =3).
8. Solution of wave equation is obtained by directly method for n =1 and by method of spherical
means for n =2.
1.7 Assignment/Activities
1. State and prove D‟ Alembert‟s formula.
2. Find the solution of wave by spherical means for n=1.
3. Solve the Transport equation ),0(0. n
t RinDubu
4. Find the Fundamental solution of Laplace equation.
5. Define Wave equation and gives its physical interpretation.
1.8 Check Your Progress
1. Obtain the solution of wave equation in even dimension.
2. Find the solution of non-homogeneous wave equation.
3. Explain Energy method for the Wave equation.
4. State and Prove Mean-value formula for Laplace‟s equation.
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5. Write short note on Transportation problem with initial value problem. Explain non
homogeneous problem on transport equation. Hence or otherwise solve one dimensional wave
equation.
1.9 Points for discussion / Clarification
At the end of the unit you may like to discuss or seek clarification on some points. If so,
mention the same.
1. Points for discussion
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
2. Points for clarification
________________________________________________________________________
________________________________________________________________________
-
________________________________________________________________________
1.10 References
1. Jost, J. (2002), Partial Differential Equations, New York: Springer-Verlag.
2. Courant, R. & Hilbert, D. (1962), Methods of Mathematical Physics, II, New York:
Wiley-Interscience.
3. John, F. (1982), Partial Differential Equations (4th ed.), New York: Springer-Verlag,
23
4. Pinchover, Y. & Rubinstein, J. (2005), An Introduction to Partial Differential
Equations, New York: Cambridge University Press.
5. Polyanin, A. D. (2002), Handbook of Linear Partial Differential Equations for
Engineers and Scientists, Boca Raton: Chapman & Hall/CRC Press.
6. Evan L.C. (1998), Partial Differential equations, Graduate studies in Mathematics,
volume-19, AMS.
7. Partial differential equation by S. N. Sneddon.
1.
_________________________________________________________________________
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UNIT 2
Non Linear Partial Differential Equation
STRUCTURE
2.1 Introduction
2.2 Objective
2.3 Nonlinear First Order PDE
2.3.1 Complete Integrals
2.3.2 Envelopes
2.3.3 Characteristics
2.4 Hamilton-Jacobi Equations
2.4.1 Calculus of Variations
2.4.2 Euler-Lagrange equations
2.4.3 Hamilton’s ODE
2.4.4 Legendre transform
2.4.5 Hopf-Lax Formula
2.4.5 Weak Solutions
2.4.6 Uniqueness
2.5 Representation of Solutions
2.5.1 Separation of variables
2.5.2 Plane and Travelling Waves, solitons
2.5.3 Similarity under Scaling
2.6 Fourier and Laplace Transform
2.6.1 Cole-Hopf Transform,
25
2.6.2 Hodograph Transform
2.6.3 Potential Functions.
2.7 Unit Summary/ Things to Remember
2.8 Assignments/ Activities
2.9 Check Your Progress
2.10 Points for Discussion/ Clarification
2.11 References
26
2.1 Introduction
It is said that the great discovery of the 19th century was that the equations of nature are linear
whereas the great discovery of the 20th century was that they are not. According to the name,
nonlinear partial differential equations are the partial differential equations with nonlinear terms.
Nonlinear problems are of interest to mathematicians, engineers and physicists because most
physical systems are inherently nonlinear in nature. Such equations describe many different
physical systems, ranging from gravitation to fluid dynamics. Nonlinear PDEs appear for example
in non-Newtonian fluids, nonlinear elasticity, flow through a porous medium, and image
processing. Usually each individual equation has to be studied as a separate problem. Since
superposition is not available, methods needed to study nonlinear equations are quite different from
those of the linear theory.
2.2 Objective
After the completion of this unit one should be able to:
Find an intertwining relationship between mathematics and physical phenomena.
Know about Non linear Differential Equation, its characteristic and its application.
To write down some special solutions explicitly in terms of elementary functions.
Represent various transforms like Legendre, Fourier and Laplace.
Describe convergence properties of nonlinear functional of sequences.
Understand Vector valued problems in the Calculus of Variations and their relaxation
27
2.3 Non-linear First- order PDE
The general nonlinear first-order PDE of the form 0),,( xuDuF , where Ux and U is an open
subset of nR .
Here RURRF n : is given and RUu : is the unknown, u=u(x).
Let us write ),......,,.....(),,( 11 nn xxzppFxzpFF
For UxRzRp n ,, . Thus, “p” is the name of the variable for which we substitute the gradient
Du(x), and “z” is the variable for which we substitute u(x). We also assume that F is smooth and set
).,......(
),......(
1
1
n
n
xxx
zz
ppp
FFFD
FFD
FFFD
We are concerning with discovering solution u of the PDE 0),,( xuDuF in U,usually subject to
the boundary condition ongu where is some given subset of U and Rg : is
prescribed.
2.3.1 Complete Integrals
Analysis of Partial Differential equation F(Du, u, x)=0 …….(1)
Let us consider nRA is an open set. Assume for each parameter Aaaa n ),.......( 1 , we have a
2C solution u=u(x;a) of the PDE (1).
We used the notation as :
nnnn
n
axaxa
xaxa
xaa
uuu
auuu
uDuD
......
....
....
....
......
),(
121
1211 1
2
28
Definition: A 2C function );( axuu is called complete integral in AU provided
(i) u(x;a) solves the PDE (1) for each Aa
(ii) ).,(),( 2 AaUxnuDuDrank xaa
2.3.2 Envelopes
Let );( axuu be a 1C function of ,, AaUx where nRU and mRA are open sets.
Consider the vector equation
).,(0);( AaUxaxuDa …………..(1)
Suppose that we can solve (1) for the parameter a as a 1C function of x,
);(xa …………...(2)
Thus )(0))(;( UxxxuDa …………(3)
We then call )())(;(:)( Uxxxuxv …………(4)
The envelope of the functions .
);(.Aa
au
Example-1- Consider the PDE 1)1(22 Duu
Solution- A complete integral is )1()1(),( 212 axaxaxu
We compute 0)1(
)(
212
ax
axuDa
Provided .)( xxa
Thus 1v are singular integrals of (1).
2.3.3 Characteristics
29
Derivation of Characteristics ODE-
Basic nonlinear first-order PDE UinxuDuF 0),,( ………(1)
Subject to the boundary condition ongu ………(2)
Where RgandU : are given. We hereafter suppose that F, g are smooth function.
Let us suppose it is described parametrically by the function )),(...),........(()( 1 sxsxsx n
The parameter s lying in some subinterval of R. Assuming u is a 2C solution of (1), we define also
)).((:)( sxusz ……….(3)
In addition, set ));((:)( sxDusp ……….(4)
That is ))(),.....(()( 1 spspxp n
Where ),.......,1())(()( nisxuspix
i ………..(5)
So z(.) gives the values of u along the curve and p(.) records the values of the gradient Du. We must
choose function x(.) in such a way that we can compute z(.) and p(.).
For this, first differentiate (5):
)())(()(1
sxsxusp jn
j
xx
i
ji
………..(6)
This expression is not too promising, since it involves the second derivatives of u.
Now differentiate PDE (1) with respect to ix :
0),,(),,(),,(1
xuDux
FuxuDu
z
FuxuDu
p
F
i
xxx
n
j jiij
…………(7)
We are able to employ this identity to get rid of the “dangerous‟ second derivative terms in (6),
provided we first set
30
),......1())(),(),(()( njsxszspp
Fsx
j
j
………….(8)
Assuming now (8) holds, we evaluate (7) at x =x(s), obtaining thereby from (3),(4) the identity
: 0))(),(),(()())(),(),(())(())(),(),((1
sxszspx
Fspsxszsp
z
Fsxusxszsp
p
F
i
in
j
xx
jji
Substitute the expression and (8) into (6): we obtain
),.......,1()())(),(),(())(),(),(()( nispsxszspz
Fsxszsp
x
Fsp i
j
i
…..(9)
Finally we differentiate (3), we get
))(),(),(()(
)())(()(
1
1
..
sxszspp
Fsp
sxsx
usz
j
n
j
i
jn
j j
….(10)
The second equality holds by (5) and (8).
We summarize by rewriting equation (8)-(10) in vector notation:
..
.
..
.
))(),(),(()()(
)()).(),(),(()()(
)())(),(),(())(),(),(()()(
sxszspFDsxc
spsszspFDszb
spsxszspFDsszspFDspa
p
p
zx
…(11)
This important system of 2n+1 first-order ODE comprises the characteristic equations of the
nonlinear first-order PDE (1). The function (.))(.),......((.)(.),(.)),(.),......((.) 11 nn xxxzppp is
called the characteristic. We will sometimes refer to x(.) as the projected characteristic: it is the
projection of the full characteristics 12(.))(.),(.),( nRxzp onto the physical region nRu .
31
2.4 Hamilton-Jacobi Equations
The initial value problem for the Hamilton-Jacobi equation:
.0
),0(0)(
tRongu
RinDuHu
n
n
t
Here RRu n ),0[: is the unknown, u=u(x,t) and ).,......,(1 nxxx uuuDDu
We are given the Hamiltonian RRH n : and the initial function RRg n : .
2.4.1Calculus of variation
Assume that RRRL nn : is a given smooth function, hereafter called the Lagrangian.
We write ),(),......,,,......,(),( 11
n
nn RxqxxqqLxqLL
And
),.....,(
),......(
1
1
xnxx
qnqq
LLLD
LLLD
Where “q” is the name of the variable for which we substitute w(s), and “x” is the variable for
which we substitute w(s).
Now fix two points nRyx , and a time 0t , we introduce then the action functional
i
dsswswLwI0
))(),(((.)][
2.4.2 Euler-Lagrange equations
The function x(.) solves the system of Euler-Lagrange equations
)ts0(0))s(x),s(x(LD))s(x),s(x(LD(ds
d .
x
.
q≤≤=+
This is a vector equation, consisting of n coupled second-order equations.
32
Proof: Step 1. Choose a smooth function ),,....(,].0[: 1 nn vvvRtv satisfying
0)()0( tv
And define for R (.).(.):(.) vxw
Then Aw (.) and so (.)][(.)][ wIxI
Thus, the real-valued function (.)](.)[:)( vxIi is minimum at 0 and consequently
,'0)0('
d
di provided )0('i exists.
Step 2. We explicitly compute this derivative. Observe
dsvxxLvxxLi
fromthenset
dsvvxvxLvvxvxLi
soand
dssvsxsvsxLi
i
xi
t n
i
qi
i
xi
t n
i
qi
t
),(),(0)0(
)4(,0
),(),()(
,))()(),()(()(
...
01
'
.....
01
'
.
0
.
We recall (2) and then integration by parts in the first term inside the integral, to find
dsvxxLxxLds
d i
xi
t
qi
n
i
)],()),(([0..
01
This identity if valid for all smooth function ),.....,( 1 nvvv satisfying the boundary condition (2)
and so 0),()),((..
xxLxxLds
dxiqi For .,......,1,0 nits
2.4.3 Hamilton’s ODE
The Hamiltonian H associated with the Lagrangian L is:
),()),,((),(.:),( nRxpxxpqLxpqpxpH
33
Where the function q(.,.) is defined implicitly by the hypothesis :
Suppose for all nRpx , that the equation ),( xqLDp q can be uniquely solved for q as a
smooth function of p and x,q=q(p,x).
Derivation of Hamilton’s ODE- The function x(.) and p(.) satisfy Hamilton‟s equation:
tsfor
sxspHDsp
sxspHDsx
x
p
0
))(),(()(
))(),(()(
.
.
Furthermore, the mapping ))(),(( sxspHs is constant.
Proof: set ))(),(()(.
sxspqsx
Let us write (.))(.),......((.) 1 nqqq for :,.....,1 ni
),(
),(),(),(),(),(1
xqx
L
xqx
Lxp
x
qxq
q
Lxp
x
qpxp
x
H
i
i
n
k i
k
ki
k
k
i
And
),(
),(),(),(),(),(1
xpq
xpp
qxq
q
Lxp
p
qpxpqxp
p
H
i
n
k i
k
ki
k
k
i
i
Thus );())(),(())(),((.
sxsxspqsxspp
H ii
i
34
).(
)))(),(((
))().(())()),(),((())(),((
.
.
.
sp
sxsxq
L
ds
d
sxsxx
Lsxsxspq
x
Lsxsp
p
H
i
i
iii
Finally, observe
0
))(),((
1
..
1
iii
n
i i
i
i
in
i i
p
H
x
H
x
H
p
H
xx
Hp
p
HsxspH
ds
d
2.4.4 Legendre transform- Suppose the Lagrangian RRL n : satisfies these conditions:
the mapping )(qLq is convex and q
qL
q
)(lim
The convexity implies L is continuous.
Hence the Legendre transform of L is ).()(.sup:)(.
* n
Rq
RpqLqppLn
2.4.5 Hopf-Lax formula:
Statement-If nRx and 0t , then the solution ),( txuu of the minimization problem
t
xtwywygdsswLtxu0
.
)(,)0(:)())((inf:),( …….(1)
is
)(min),( yg
t
yxtLtxu
nRy
Proof: 1. Fix any nRy and define )0()(:)( tsyxt
sysw
35
Then the definition (1) of u implies
)(inf
)()()((),(0
.
ygt
yxtL
ygt
yxtLygdsswLtxu
nRy
t
2. on the other hand, if w(.) is any 1C function satisfying w(t)=x, we have
tt
dsswLt
dsswt
L0
.
0
.
))(((1
)(1
(
Thus if we write y = w(0), we find
t
ygdsswLygt
yxtL
0
.
)())(()(
And consequently
),()(inf txuygt
yxtL
nRy
3. We have so far shown
)(inf),( yg
t
yxtLtxu
nRy
And leave it as an exercise to prove the infimum above is really a minimum.
2.4.6 Weak solution
In this section we show that semi-concavity conditions of the sorts discovered for the Hopf-Lax
solution .
Definition- A Lipschitz continuous function RRu n ),0[: is a weak solution of the initial-
value problem:
0
),0(0)(
tRongu
RinDuHu
n
n
t ………….(1)
36
Provided
0,,0tan
)1
1(),(),(2),()(
),,0(),(..0)),((),()(
),()()0,()(
2
tRzxallandCtconssomefor
zt
Ctzxutxutzxuc
RtxeafortxDuHtxub
Rxxgxua
n
n
t
n
2.4.7 Uniqueness of weak solution-
Statement- Assume H is 2C and H is convex and p
pH
p
)(lim , and satisfies RRg n : is
Lipschitz continuous. Then there exists at most one weak solution of the initial-value problem (1).
Proof: 1. suppose that u and u are two weak solutions of (1) and write uuw :
Observe now at any point (y, s) where both u and u are differentiable and solve our PDE, we have
s).s).Dw(y,-b(y,:
s))(y,-s)(Du(y,.s)dr(y,r)-(1s)rDu(y,H( -
s))dr(y,r)-(1s)H(rDu(y, dr
d-
s))(y,()),((
),(),(),(
t
0
t
0
uDuDD
uD
uDHsyDuH
syusyusyw ttt
Consequently
0. Dwbwt a.e. …………(2)
2. Write ),0[:,0)(: Rwherewv is smooth function to be selected later, We multiply
(2) by )(' w to discover 0. Dvbvt …………(3)
37
3. Now choose 0 and define uuuu
:,; where is the standard mollifier in the
x and t variables.
)(),( uLipuDuLipDu …………..(4)
And
uDuDDuDu , a.e., as 0 …………(5)
Furthermore inequality (c) in the definition of weak solution implies
Is
CuDuD
11, 22 ………….(6)
For an appropriate constant C and all .2,,0 sRy n Verification is left as an exercise.
4. Write .)),(~)1(),((:),(1
0drsyuDrsyrDuDHsyb
Then (3) becomes DvbbDvbvt .)(.
Hence Dvbbvdivbvbdivvt .)()()(
5. Now,
sC
drurruuDrrDuHdivbkk
lkxx
n
lk
xxpp
11
)~)1(()~)1((11
1
01,
For some constant C, in view of (4), (6). Here we note that H convex implies 02 HD
6. Fix 0, 00 tRx n and set ))~(),(max()(max: uLipuLipppDHR
Define also the cone )(,0:),(: 000 ttRxxtttxC
38
Next write
))(,( 00
),()(ttRxB
dxtxvte
And compute for a.e. t >0:
))(,(
))(,(
))(,( ))(,(
))(,())(,(
))(,())(,(
.
00
00
00 00
0000
0000
).()()1
1(
).()(
).()().(
).()()(
)(
ttRxB
ttRxB
ttRxB ttRxB
ttRxBttRxB
ttRxBttRxBt
dxDvbbtet
C
dxDvbbvdivb
dxDvbbvdivbdSRvbv
dSvRdxDvbbvdivbvbdiv
dSvRdxvte
The last term on the right hand side goes to zero as 0 , for a.e. t>0, according to (4) & (5) and
the dominated convergence theorem,
Thus )()1
1()(.
tet
Cte for a.e. 00 tt …………(7)
7. Fix 00 tr and choose the function )(z to equal zero if
)~()( uLipuLipz and to be positive otherwise.
Since 0~ tRonuu n
.0)~()( tatuuwv
Thus 0)( e
Consequently, Gronwall‟s inequality and (7) imply 0)()()
11(
r
dss
C
eere
Hence ))(,()~()(~00 rtRxBonuLipuLipuu
This inequality is valid for all 0 and so uu ~ in ))(,( 00 rtRxB
39
Therefore, in particular, ),(~),( 0000 txutxu .
2.5 Representation of Solution
2.5.1 Separation of variables: The method of Separation of variables tries to construct a solution u
to a given partial differential equation as some sort of combination of functions of fewer variables.
This technique is best understood in examples.
Example-1. Let nRU be a bounded, open set with smooth boundary. Then find the solution of
initial-value problem for the heat equation by the Separation of variables.
0
),0[0
),0(0
tUongu
Uonu
Uinuut
………..(1)
Where RUg : is given.
Solution: suppose there exists a solution having the multiplicative form
);0,()()(),( tUxxwtvtxu ………..(2)
Then )()(),()()(),( ' xwtvtxuandxwtvtxut
Hence )()()()(),(),(0 ' xwtvxwtvtxutxut
If and only if )(
)(
)(
)('
xw
xw
tv
tv ………..(3)
for all 0)(),(0& tvxwthatsuchtUx
let us suppose that ),0()(
)(
)(
)('Uxt
xw
xw
tv
tv
Then vv ' and ………..(4)
ww ……….(5)
If is known, the solution of (4) is tdev for an arbitrary constant d. Consequently
40
We need only investigate equation (5).
Let is an eigen value of the operator on U provided there exists a function w, not identically
equal to zero, solving
Uonw
Uinww
0
Where w is corresponding Eigen function. Hence if s an Eigen value and w is corresponding
Eigen function then we set above, to find
wdeu t ………….(6)
Solves
),0[0
),0(0
Uonu
Uinuut ……………..(7)
with the initial condition u(.,0)=dw.
Thus the function u defined by (6) solves problem (1),provided g=dw.
More generally, if m ,......1 are Eigen values, mww ,......1 corresponding Eigen functions and
mdd ,.......,1 are constants, then k
m
k
t
k wedu k
1
Solves (7) with the initial condition
m
ik
kk wdu )0(.,
If we can find ,......,, 1wm etc. such that
m
ik
kk gwd
41
We can hope generalize further by trying to find a countable sequence ,........1 are Eigen
values, ,........1w corresponding Eigen functions and ..,.........1d are constants,
So that
ik
kk gwd in U
Hence k
k
t
k wedu k
1
Will be the solution of the initial-value problem(1).
This is an attractive representation formula for the solution.
2.5.2. Plane and traveling waves, solitons
Consider first a partial differential equation involving the two variables RtRx , . A solution u
of the form ),()(),( RtRxtxvtxu ……….(1)
is called a traveling wave. More generally, a solution u of a PDE in the n+1 variables
RtRxxxx n
n ,),......(( 1 having the form
),().(),( RtRxtxyvtxu n …………(2)
is called a plane wave.
Exponential solution: A complex-valued plane wave solutions of the form
……..(3)
Where n
n RyyyandC )......( 1 , being the frequency and n
iiy1 the waves numbers.
Consider the Korteweg-de Vries (KDV) equation in the form
),0(06 Rinuuuu xxxxt , …………(4)
This nonlinear dispersive equation being a model for surface waves in water..
A traveling wave solution having the structure
).(),( txyietxu
42
).0,()(),( tRxtxvtxu ………….(5)
Then u solves the KDV equation (4) ,provided v satisfies the ODE
)'(06 '''''
ds
dvvvv ………….(6)
Integrate (6) by first noting
avvv ''23 ………….(7)
a denotes some constant. Multiply this equality by 'v to obtain
''''2' '3 avvvvvvv
And so deduce
bavvvv
23
2'
22
………….(8)
where b is another arbitrary constant.
We investigate (8) by looking now only for solutions v which satisfy asvvv 0,, '''
Then by (7) & (8) imply a=b=0. Equation (8) thereupon simplifies to read
).
2(
2
2
2' vv
v
Hence .2 21' vvv
We take the minus sign above for computational convenience, and obtain then this implicit formula
for v: …………….(9)
for some constant c.
czz
dzs
sv
)(
1 21
)2(
43
Now substitute .sec2
2
hz It follows that
tanhsec 2hd
dz and
.tanhsec2
)2( 22
3
21
hzz
Hence (9) becomes
………(10)
where is implicitly given by the relation
)(sec2
2 svh
………..(11)
We lastly combine (10) and (11), to compute ).()(2
sec2
)( 2 Rscshsv
Conversely, it is routine to check v so defined actually solves the ODE (6).
The outcome is that
is a solution of the KDV equation for each 0, Rc . A solution of this form is called a
Soliton.
Traveling waves: The scalar reaction-diffusion equation
…….(12)
where RRf : has a “cubic-like” shape.
2.5.3. Similarity under scaling:
,2
cs
)0,()(2
sec2
),( 2
tRxctxhtxu
),0()( Rinufuu xxtt
44
Example: Find the scaling invariant solution for the porous medium equation.
Consider porous medium equation ),0(0)( n
t Rinuu ………...(13)
where 10 andu are constant.
From the fundamental solution of the heat equation, a solution u having the form
)0,()(1
),( tRxt
xv
ttxu n
………..(14)
Where the constants , and the function RRv n : must be determined.
Solution u of (13) invariant under the dilation scaling
);,(),( txutxu a
So that ),(),( txutxu a for all 0,,0 tRx n ,
We obtain (14) into (13),
0))(()(.)( )2()1()1( yvtyDvytyvt ………….(15)
For xty .
In order to convert (15) into an expression involving the variable y alone.
Let us suppose that 21 …………..(16)
Then (15) reduces to 0)(. vDvyav ………….(17)
At this point we have effected a reduction from n+1 to n variables. We simplify further by
supposing v is radial; that is )()( ywyv for some RRw : . Then (17) becomes
0)(1
)( ''''
wr
nwrww ………..(18)
45
Where dr
dandyr ',
Noe if we set n ………..(19)
Eq. (18) thereupon simplifies to read
0)())(( '''1 wrwr nn
Thus awrwr nn )()( '1 , for some constant a.
Assuming 0,lim '
wwr
, we conclude a=0;
Whence rww ')(
But then rw
1)( '1
Consquently 21
2
1rbw
where b is constant and so 1
1
2
2
1
rbw ………..(20)
where we took the positive part of the right hand side of (20) to ensure 0w
Recalling )()( rwyv and (14), we obtain
)0,(2
11),(
1
1
2
2
tRxt
xb
ttxu n
………(21)
where from (16),(19),
2)1(
1,
2)1(
nn
n ………..(22)
46
The formula (21), (22) are Barenblatt‟s solution to the porous medium equation.
2.6 Fourier and Laplace Transform
Definition of fourier transform on L 1 - If u 1L (R n ), we define its Fourier transform
)()()2(
1)(ˆ
2
n
R
ixy
nRydxxueyu
n
………(1)
And its inverse Fourier Transform
)()()2(
1)(
2
n
R
ixy
nRydxxueyu
n
……….(2)
Since )(1 1. nyix RLuande
Definition of fourier transform on 2L - Choose a sequence )()( 21
1
nn
kk RLRLu
With )(2 n
k RLinuu .
This sequence consequently converges to a limit, which we define to be u :
).(ˆˆ 2 n
k RLinuu
The definition of u does not depend upon the choice of approximating sequence 1kku .
Laplace Transform-
If )(1
RLu we define Laplace transform to be
)0()()(0
sdttuesu st
where, the Fourier transform is most appropriate for functions defined on all of R (or nR ), the
Laplace Transform is useful for functions defined only on R . In Practice this means that for a
47
Partial differential equation involving time, it may be useful to perform a Laplace transform in t,
holding the space variables x fixed.
2.6.1 Cole-Hopf transformation
A parabolic PDE with quadratic nonlinearity-
Initial value problem for a quasilinear parabolic equation:n
0
),0(02
tRongu
RinDubuau
n
n
t ……….(1)
where a>0. This sort of nonlinear PDE arises in stochastic optimal control theory.
Assuming for the moment u is a smooth solution of (1), we set
where RR : is a smooth function, as yet un-specified. We will try to choose
so that w solves a linear equation. We have ;)()(,)(2'''' Duuuuwuuw tt and
consequently (1) implies
wa
Duubuawa
Dubuauuuw tt
2'''
2''
)]()([
])[()(
Provided we choose to satisfy 0''' ba . We solve this differential equation by
setting a
bz
e
. Thus we see that if u solves (1), then a
bu
ew
……….(2)
Solve the initial- value problem for the heat equation (with conductivity a):
0
),0(0
tRonew
Rinwaw
na
bu
n
t
Formula (2) is the Cole-Hopf transformation.
)(: uw
48
Now the unique bounded solution of (3) is
);0,()4(
1),(
)(
2
4
2
tRxdyeeat
txw nyg
a
b
Rn n
at
yx
And since (2) implies wb
au log
We obtain thereby the explicit formula
)0,()4(
1log),(
)(
2
4
2
tRxdyeeatb
atxu n
yga
b
Rn n
at
yx
For a solution of quasi-linear initial-value problem (1).
2.6.2 Hodograph transform
The hodograph transform is a technique for converting certain quasi-linear systems of PDE into
linear systems, by reversing the roles of the dependent and independent variables.
Equation of steady, two-dimensional fluid flow:
0)(
0)()()()()(
2
1
1
2
2
2
2222
1
1
2
211
1
212
xx
xxxx
uub
uuuuuuuuuua
……….(6)
In 2R . The unknown is the velocity field 21,uuu and the function RR 2:(.) , the local
sound speed, is given.
The system (6) is quasi-linear.
Let us now, however, no longer regard 21 uandu as functions of 21 xandx
21
22
21
11 ,,, xxuuxxuu …………(7)
But rather regard 21 xandx as functions of 21 uandu ;
49
21
22
21
11 ,,, uuxxuuxx ; ………….(8)
According to the Inverse Function Theorem, we can locally at least, invert equations (7) to yield
(8), provided
0),(
),( 2
1
1
2
2
2
1
1
21
21
xxxx uuuu
xx
uuJ ………….(9)
In some region of 2R .
Assuming now (9) holds, we calculate
…………(10)
we insert (10 ) into (6) , to discover
0)(
0)()()()()(
2
1
1
2
1
1
2
2
22
1
1
221
2
2
212
uu
uuuu
xxb
xuuxxuuxuua
This is a linear system for 21, xxx as a function of 21,uuu .
2.6.3 Potential function:
Another technique is to utilize a potential function to convert a nonlinear system of PDE into a
single linear PDE. We consider as an example Euler‟s equations for in viscid, incompressible fluid
flow:1
0)(
,00)(
,0.)(
3
3
3
tRonguc
Rinudivb
RinfDpDuuua t
………..(1)
2
2
1
1
1
2
1
2
2
1
1
1
1
1
2
2
,
,
uxux
uxux
JxuJxu
JxuJxu
50
Here the unknown are the velocity field ),,( 321 uuuu and the scalar pressure p; the external
force ),,( 321 ffff and the initial velocity ),,( 321 gggg are given. Here D as usual denotes
the gradient in the spatial variables ).,,( 321 xxxx
The vector equation (1)(a) means
We will assume 0gdiv ………..(2)
If furthermore there exists a scalar function RRh ,0: 3 such that
F=Dh …………(3)
Now we have to find the solution (u,p) of (1) for which the velocity field u is also derived from a
potential, say u=Dv ……….(4)
As curl u=0 flow will be irrotational.
And so v must be harmonic as a function of x, for each time t >0.
If .2
1.
2DvDDuu consequently (1) (a) reads hpDDvvD t
2
2
1
Therefore we take hpDvvt 2
2
1 ………(5)
This is Bernoulli‟s law.
2.7 Unit Summary/ Things to Remember
1. General non linear partial differential equations of the form 0,, xuDuF
2. A 2C function );( axuu is called complete integral in AU provided
)3,2,1(3
1
ifpuuu i
xi
j
i
xj
ji
t
51
(i) u(x;a) solves the PDE for each Aa .
(ii) ).,(),( 2 AaUxnuDuDrank xaa
3. The envelope v is sometimes called a singular integral of PDE.
4. A complete integral, which depends upon n arbitrary constants, then a solution depending on an
arbitrary function h of n-1 variables.
5. The initial value problem for the Hamilton-Jacobi equation is defined as :
.0
),0(0)(
tRongu
RinDuHu
n
n
t
6. Hamilton‟s ODE- The Hamiltonian H associated with the Lagrangian L is
),()),,((),(.:),( nRxpxxpqLxpqpxpH
Where the function q(.,.) is defined implicitly by the hypothesis.
7. When the Lagrangian RRL n : satisfies these conditions:
the mapping )(qLq is convex and q
qL
q
)(lim .
8. Solution u of the form ),()(),( RtRxtxvtxu is called a traveling wave.
9. The hodograph transform is a technique for converting certain quasi-linear systems of PDE into
linear systems, by reversing the roles of the dependent and independent variables.
10. Utilizing a potential function to convert a nonlinear system of PDE into a single linear PDE.
52
2.8 Assignment/Activities
1. Find the complete integral of the following PDE 0)( DuHut .
2. Describe the characteristics method to solve non linear PDE
Uongu
RUinxuDuF n
0),,(
3. Find the solution of the porous medium equation by the method of separation of variable.
),0(0)( n
t Rinuu
4. Find the reaction-diffusion PDE for traveling waves.
5. Find the PDE of the initial-value problem for the viscous Burgers‟ equation.
6. Describe the following;
i) Hodograph Transform
ii) Cole Hopf- Transform
iii) Legendre Transform
iv) Fourier & Laplace Transform
2.9 Check Your Progress
1. Define the characteristics of PDE.
2. Find the complete integral of of PDE uDufDux )(. .
3. Find the solution of the Hamilton-Jacobi equation by the method of separation of variable.
),0(0)( n
t RinDuHu
4. What is Hopf-Lax formula & prove it.
53
2.10 Points for discussion / Clarification
At the end of the unit you may like to discuss or seek clarification on some points. If so,
mention the same.
1. Points for discussion
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
2. Points for clarification
________________________________________________________________________
________________________________________________________________________
-
________________________________________________________________________
2.11. References
1. Polyanin, A. D. & Zaitsev, V. F. (2004), Handbook of Nonlinear Partial Differential
Equations, Boca Raton: Chapman & Hall/CRC Press.
2. Pokhozhaev, S.I. (2001), Non-linear partial differential equation.
3. Polyanin, Andrei D.; Zaitsev, Valentin F. (2004), Handbook of nonlinear partial
differential equations, Boca Raton, FL: Chapman & Hall/CRC.
4. Evan L.C. (1998), Partial Differential equations, Graduate studies in Mathematics,
volume-19, AMS.
54
5. UNIT 3 PARTIAL DIFFERENTIAL EQUATIONS
- I
6. Structure
7. 3.0 Introduction
8. 3.1 Linear Partial Differential Equation of First Order
9. 3.2 Lagrange‟s Solution
10. 3.3 Some special types of equation which can be solved
easily by methods other than the general method
11. 3.4 Charpit‟s General Method of Solution
12. 3.5 Check Your Progress: The Key
13. 3.0 INTRODUCTION
14. An equation involving partial differential coefficients of a
function of two or more variables is known as a partial differential
equation. If a partial differential equation contains nth and lower
derivatives, it is said to be of nth order. The degree of such equation is
the greatest exponent of the highest order. Further, such equation will be
called linear if, it is of first degree in the dependent variables and its
partial derivatives (i.e. powers or products of the dependent variables
and its partial derivatives must be absent).An equation which is not
linear is called a non linear differential equation.
15. 3.1 LINEAR PARTIAL DIFFERENTIAL EQUATION
OF FIRST ORDER
16. If z is a function of two independent variables x and y, we write
then the differential coefficient z with respect to x and y are
55
called partial differential coefficients. These are denoted by following
notations:
17.
18. In case there are nth independent variables, we take them to be x1,
x2,….., xn and z is then regarded as the dependent variables. In this case we
use the following notations:
19.
20. Sometimes the partial differentiations are also denoted by making
use of suffixes. Thus we write
21.
22. Formation of Partial differential equations
23. The partial differential equations may be obtained in the following two
ways:
24. Rule I: Elimination of arbitrary constants:
25. Consider an equation
…….. (1)
26. Where a and b denote arbitrary constants. Let z be regarded as a function
of two independent variables x and y. differentiating (1) with respect to
x and y partially in turn, we get
27.
28.
56
29. Eliminating two constants a and b from (1), (2) and (3), we shall obtain
an equation of the form
30.
31. Which is partial differential equation of the first order.
32. Remark: In a similar manner it can be shown that there are more
arbitrary constants than the number of independent variables; the above
procedure of elimination will give rise to partial differential equation of
higher order than the first.
33. Rule II: Elimination of arbitrary function:
34. Let be two functions of x, y, z
connected by the relation
35.
36. We treat z as dependent variable and x and y as independent variables so
that
37.
38. Differentiating (1) partially with respect to x, we get
39.
40.
41.
42. Similarly, differentiating (1) partially with respect to y, we get
43.
57
44.
45.
46.
47.
48. Thus we obtain a linear partial differential equation of first order and of
first degree in p and q.
49. Let us consider the following problems:
50. Problem: 1 Find a partial differential equation by eliminating a and b
from the equation
51. Solution: Given
…..(1)
52. Differentiating (1) partially with respect to x and y , we get
53.
54. Substituting these values of a and b in (1) we see that the arbitrary
constant a and b are eliminated and we obtain,
55.
56. Which is required partial differential equation.
57. Problem 2: Eliminating arbitrary constant from
to form partial differential equation.
58. Solution: Given
…..(1)
59. Differentiating (1) partially with respect to x and y , we get
58
60.
61. Squaring and adding these equations we have
62.
63. or [using
(1)]
64. Problem 3: Form a partial differential equation by eliminating the
arbitrary function from at is the
order of this partial differential equation?
65. Solution: Given
…..(1)
66. Let
…..(2)
67. Then (1) becomes
…..(3)
68. Differentiating (3) with respect to „x‟ partially, we get
69.
70.
71. From (4) and (5),
72.
73.
74. Again differentiating (3) with respect to „y‟ partially, we get
59
75.
76. From (5),
77.
78.
79. From (6) and (7) by eliminating function , we obtain
80.
81.
82.
83. Which is the desired partial differential equation of first order.
84. you may now try the following exercises:
85. E 1: Find a partial differential equation by eliminating a, b, c from
86. E2: From a partial differential equation by eliminating the arbitrary
function f and F from where .
87. 3.2 LAGRANGE‟S SOLUTION
88. Consider the quasi-linear partial differential equation of order one
namely, , This equation was studied by French
mathematician Lagrange and is called Lagrange equation.
89. We have shown in Rule II
90.
…..(1)
91. Eliminating then we get
60
92.
…..(2)
93.
94. Hence (1) is the general (or integral) solution of (2).
95. Now we find value of u and v.
96. Let since
97. Differentiating,
98.
99. Solving the above equations for , we get
100.
101.
102. Which is called Lagrange‟s auxiliary (or subsidiary) equation for
(2).
103. Let us consider a few problems:
104. Problem 4: Solve .
105. Solution: Given
…..(1)
106. The Lagrange‟s auxiliary equations for (1), we have
107.
108. Taking the first two fractions of (2), we have
61
109.
110. Integrating, or
111. Now taking the first and last fractions of (2), we have
112.
113. Integrating, or
114. From (3) and (4), the required general integral is
115. , where being an arbitrary
function.
116. Problem 5: Solve
117. Solution: Given
…..(1)
118. The Lagrange‟s auxiliary equations of (1)
119.
120. Taking first two fractions, we get
121. Integrating,
122. Now taking the first and last fractions of (2) and using (3), we
have
123.
124. Integrating,
62
125. using (3)
126. From (3) and (4), the required general integral is
127. , where being an
arbitrary function.
128. Problem 6: Solve
129. Solution: Given
130. The Lagrange‟s auxiliary equations for (1) are
131.
132. Choosing as multipliers, each fraction of (2)
133.
134.
135. By integrating,
136. Choosing as multipliers, each fraction of (2)
137.
138.
139. By integrating,
140. From (3) and (4), the required general solution is
63
141. , where being an arbitrary
function.
142. Problem7: Solve .
143. Solution: Given
…..(1)
144. The Lagrange‟s auxiliary equations for (1), we have
145.
146. Taking the first two fractions of (2), we have
147.
148. Integrating, or
149. Choosing as multipliers, each fraction of (2)
150.
151. Combining the third fraction of (2)with fraction (4), we get
152.
153. Integrating,
154. From (3) and (5), the required general solution is
155. , where being an arbitrary
function.
156. Problem 8: Solve
157.
64
158. Solution: Given
159.
160.
161. The Lagrange‟s auxiliary equations for (1), we have
162.
163.
164. Choosing as multipliers in turn, each
fraction of (2)
165.
166.
167. Taking the first two fractions of (3), we have
168.
169. Integrating,
170.
…..(4)
171. Taking the last two fractions of (3), we have
172.
173. Integrating,
65
174.
…..(5)
175. From (4) and (5), the required general solution is
176. , where being an
arbitrary function.
177. You may try now the following exercise:
178. E3: Solve
179. E4: Solve
180. E5: Solve
181. Let us discuss, one by one, the various types of standard forms.
182. 3.3 SOME SPECIAL TYPES OF EQUATION WHICH
CAN BE SOLVED EASILY BY METHODS OTHER THAN
THE GENERAL METHOD
183. Standard Form I: Equation involving only p and q
184. Consider equations of the form
185.
186. Then its complete integral is given by
187.
188. Where a and b are connected by the relation
189.
190. Since from equation (2), we have
191.
192. These values substituted in (3) gives rise to (1).
66
193. From the relation (3), we will find b in terms of a, say
194. Putting this value of b in (2), the complete integral of (1) is
195.
196. Which contains two arbitrary constants a and c which are equal to
the number of independent variables, namely x and y.
197. In order to find the general integral of (1), we first take
in (4), being an arbitrary function and obtain
198.
199. Now we differentiate (5) partially with respect to a and get
200.
201. Eliminating a between (5) and (6), we get the general solution of
(1).
202. The singular integral of (1) is obtained by eliminating a and
c between the complete integral (4) and the equations obtained by
differentiating (4) partially with respect to a and c ; i.e. between
203.
204. Since the last equation in (7) is meaningless, we conclude that the
equations of standard form I have no singular solution.
205. We now take up some problems to illustrate this method.
206. Problem 9: Solve , where is constant.
207. Solution: Given
208. Since (1) is of the form , its solution is
209.
67
210. Where , putting a for p and b
for q in (1).
211. So from (2), the complete integral is
212. Which contains two arbitrary constants a and c.
213. For singular solution , differentiating (3) partially with respect to a
and c, we get and is absurd. Hence
there is no singular solution of (1).
214. To find the general integral put in (3),then we get
215.
216. Now we differentiate (4) partially with respect to a and get
217.
218. Eliminating a from (4) and (5), we get the required general
solution of (1).
219. Problem 10: Solve
220. Solution: The given equation can be rewritten as
221.
222. Put
223. So that
224.
68
225.
226. Using (4) and (5) in (1), we get
227.
228. Where which is of the form
229. So complete integral of (6),
230. Where , putting a for P and b for
Q in (6).
231. So the complete integral is
232. Which contains two arbitrary constants a and c.
233. Let then
234.
235. For singular solution, differentiating (7) partially with respect to
and c‟,we get
236.
237. and
238. Eliminating and c‟ from (7), (8) and (9), the singular solution is
.
69
239. To find the general integral put in (3), then we get
240.
241. Now we differentiate (10) partially with respect to and get
242.
243. Eliminating „a‟ from (10) and (11), we get the required general
solution of (1).
244. Problem 11: Solve
245. Solution: The given equation may be written as
246.
247. Now substituting
248.
249. We obtain
250.
251.
252.
253. Hence the given equation becomes
254.
255. Where which is of the form
70
256. So solution will be of the form where ab=1
257.
258.
259. Standard Form II: Equation involving p , q and z
260.
261. Let as assume the tentative solution in the form
262.
263. Where a is arbitrary constant
264. or
265.
266.
267. so the given equation is reduced to
268.
269. Which is an ordinary differential equation of first order in z and X
can be solved as follows:
270. Suppose equation (1) on solving for gives
271.
272.
71
273. or (say)
274. or
275. which is the complete integral .
276. The general and singular solution can be obtained as explained
earlier.
277. We now take up some problems to illustrate this method.
278. Problem 12: Find the complete integral of
279. Solution: Taking
280.
281. therefore the given equation reduces to
282.
283.
284.
285. on integration, we get
286.
287. Which is the required complete integral.
288. Problem 13: Solve
289. Solution: Taking
290.
72
291. therefore the given equation reduces to
292.
293.
294.
295. on integration, we get
296.
297.
298.
299. Which is the required complete integral.
300. Standard Form III: Equation of the form
301. Let as assume the tentative solution in the form
302.
303. From these equations we may obtain
304.
305.
306.
307.
308. which is the complete integral .
73
309. The general and singular solution can be obtained as explained
earlier.
310. We now take up some problems to illustrate this method.
311. Problem 14: Solve
312. Solution: The given equation may be written as
313.
314.
315. therefore
316.
317. On integration,
318.
319. Which is the required solution.
320. Problem 15: Solve
321. Solution: The given equation may be written as
322.
323.
324. Therefore equation (1) will be transformed as
325.
326.
327.
74
328. which is now of standard form III, then
329.
330. Substituting in
331. Therefore
332. On integration,
333.
334. Which is the required solution.
335. Standard Form IV: Equation of the
form
336. In this standard form we se that equations are analogous to the
Clairaut‟s form of ordinary differential equations. It can be verified that
the complete integral will be given by
337.
338.
339. In order to obtain the general solution let us set where
denotes an ordinary function, then
340.
341. Differentiating with respect to a
342.
343. Elimination of a between (2) and (3) will give the general integral.
344. In order to obtain the singular integral, we will differentiate (1)
with respect to a and b and thus we obtain
75
345.
346. Eliminating a and b between (3) and (4), one can get singular
integral. The singular integral will not exist if is linear in p and
q.
347. We now take up some problems to illustrate this method.
348. Problem16: Solve
349. Solution: The complete integral will be given by
350.
351. In order to obtain the general solution let us set where
denotes an ordinary function, then
352.
353. Differentiating with respect to a
354.
355. Elimination of a between (2) and (3) will give the general integral.
356. In order to obtain the singular integral, we will differentiate (1)
with respect to a and b and thus we obtain
357.
358. Eliminating a and b between (1) and (4), will yield singular
integral as
359.
360. Problem17: Solve
361. Solution: Let us put , so that
76
362.
363.
364. Substituting these values of p and q in the given equation , we
have
365.
366.
367. Now this has the form of standard IV.
368. Therefore its complete integral will be
369.
370.
371. Singular integral will be obtained by elimination of a and b
between (1) and its derivative with respect to a and b which will be
372. and
373. Hence the singular integral will be
374.
375. or
376. We shall now discuss Charpit‟s method of finding the complete
integral.
377. 3.4 CHARPIT‟S GENERAL METHOD OF
SOLUTION
378. This is general Method for solving equations with two
independent variables. Since the solution by this method is generally
77
more complicated, this method is applied to solve equations which
cannot be reduced to any of the standard forms.
379. Let the given equation be
380. Since z be a function of two independent variables x and y
therefore
381.
382. Let us assume that a relation
383. Such that when the values of p and q obtained by solving (1) and
(3), are substituted in (2), it becomes integrable. The integration of (2)
will give the complete integral of (1).
384. To obtain (3), differentiate (1) and (3) with respect to x, we get
385.
386.
387. Again differentiate (1) and (3) with respect to y, we get
388.
389.
390. Eliminating from (4) we get,
391.
392.
78
393.
394.
395. Similarly, Eliminating from (5) we get,
396.
397.
398. Since
399. Now Adding (6) and (7) using (8), we get
400.
401.
402. This is a linear equation of the first order to obtain the desired
function F.
403. So Auxiliary equation by Lagrange‟ method
404.
405.
406. These equations are called Characteristic equations of differential
equation (3). Since any of the integral of (10) will satisfy (9), an integral
of (10) which involves p or q or both will serve along with the given
equation to find p and q.
407. Working Procedure:
79
408. Step 1.Transfer all terms of the given equation to L.H.S. and
denote by expression „f ‟.
409. Step 2. Write down the Charpit‟s auxiliary equation (10).
410. Step 3. Using the value of „f ‟ in step 1, write down the
values of
411. etc. occurring in step 2 and put these in Charpit‟s
equations (10).
412. Step 4. Solve two proper fractions and find simplest relation
involving at least one of p and q.
413. Step 5. The simplest relation of step 4 is solved along with the
given equation to determine p and q .
414. Put these values of p and q in which on
integration gives the complete integral of the given equation.
415. The singular and general integrals may be obtained in the
usual manner.
416. Let us take up some problems.
417. Problem 18: Find a complete integral of .
418. Solution: Given
419. The Charpit‟s auxiliary equations are
420.
421.
422.
80
423. Taking the first two fractions of (2),
424. Integrating,
425. Substituting this value of p in (1), we have
426.
427. From (3) and (4) ,
428. Putting these values of p and q in , we get
429.
430.
431. Integrating, ,
432. Which is a complete integral, a and b being arbitrary constants.
433. Problem 19: Find a complete integral of .
434. Solution: Given
435. The Charpit‟s auxiliary equations are
436.
437.
438.
439. Taking the second and fourth fractions of (2),
81
440. Integrating,
441. Substituting the above value of q in (1), we have
442.
443.
444. Putting these values of p and q in , we get
445.
446.
447. Integrating, , where a and b being
arbitrary constants.
448. Problem20. Find a complete integral of .
449. Solution. Given
450. The Charpit‟s auxiliary equations are
451.
452.
453. Now each fractions of (2),
454.
82
455.
456. Integrating,
457. Solving (1) and (3) for p and q,
458.
459. From (3) and (4),
460. Putting these values of p and q in , we get
461.
462.
463. Integrating,
464. which is a complete integral , a and b are arbitrary constant.
465. You may try now the following exercise:
466. E 6: Find a complete integral of .
467. E7. Find a complete and singular integral of
468. 3.5 SOLUTIONS/ANSWERES
469. E 1: Given
…..(1)
470. Differentiating (1) partially with respect to x and y , we get
471.
83
472. Again Differentiating (2) and (3) partially with respect to x and y ,
we get
473.
474.
475.
476. Putting this values of in (4) and dividing by , we obtain
477.
478. Similarly, from (3) and (5),
479.
480. (7) and (8) are two possible forms of the required equations.
481. E2: Given
…..(1)
482. Differentiating (1) partially with respect to x and y, we get
483.
484.
485. Differentiating (2) and (3) partially with respect to x and y, we get
486.
84
487.
488.
489. Adding (4) and (5),
490. Which is the desired partial differential equation.
491. E3: Given
…..(1)
492. The Lagrange‟s auxiliary equations for (1) are
493.
494. Taking the first two fractions of (2), we have
495.
496. Integrating, or
497. Next, the second fraction of (2) implies that
498. Integrating,
499. From (3) and (4), the required general solution is
500. , where being an arbitray function.
501. E4: Given
…..(1)
502. The Lagrange‟s auxiliary equations for (1) are
503.
504. Taking the first two functions of (2),
85
505. Integrating ,
506. Now taking Second and third fractions of (2) and using (3), we get
507.
508. Integrating,
509. using (3)
510. From (4) and (6), the required general solution is
511. , where being an arbitrary
function.
512. E5: Given
…..(1)
513. The Lagrange‟s auxiliary equations for (1) are
514.
515. Choosing as multipliers, each fraction of (2)
516.
517.
518. By integrating,
519. Choosing as multipliers, each fraction of
(2)
86
520.
521.
522.
523. By integrating,
524. From (3) and (4), the required general solution is
525. , where being an
arbitrary function.
526. E 6: Given
527. The Charpit‟s auxiliary equations are
528.
529.
530.
531. Taking the first two fractions of (2),
532. Integrating,
533. Solving (1) and (3) for p and q,
534.
87
535. Putting these values of p and q in , we get
536.
537. Integrating,
538. Putting so that
539.
540.
541.
542.
543. which is a complete integral , a and b are arbitrary constant.
544. E7. Given
545. The Charpit‟s auxiliary equations are
546.
547.
548.
549. Taking the second fraction of (2),
550. Substituting the above value of q in (1), we have
551. Putting these values of p and q in , we get
88
552.
553. Integrating,
554.
555. which is complete integral, a and b being arbitrary constants.
556. Differenting (3) partially with respect to a and b ,we get
557.
558. Solving (4) for a and b,
559. Substituting the above values of a and b given by (4) in (3), we
have
560. , which is the required singular integral.
561. *****
89
UNIT 4 PARTIAL DIFFERENTIAL EQUATIONS - II
Structure
4.0 Introduction
4.1 Partial Differential Equation of Second and Higher Orders
4.2 Classification of Linear Partial Differential Equation Second Orders
4.3 Homogeneous and Non Homogeneous Equation with constant
Coefficients
4.4 Partial Differential Equation reducible to equation with constant
Coefficients
4.5 Check Your Progress: The Key
4.0 INTRODUCTION
In this unit, we shall discuss certain types of PDEs of order higher than one.
More specifically, we shall begin this unit by considering the general form of linear
PDE.
4.1 PARTIAL DIFFERENTIAL EQUATION OF SECOND AND HIGHER
ORDERS
The general form of the linear partial differential equation (PDE) of order
is
90
Where are either constants
or functions of and . This shows that equation is not merely linear with
regard to the different coefficients but also with regard to the dependent variable .
If the coefficients are constant, then it will be a partial differential equation of
order with constant coefficients.
For the sake of convenience the operator and are denoted by
and respectively and then the equation can be written as
or more briefly may be written as
where corresponds to the polynomial of operators in the left side of
relation
The solution of equation consists of complementary function and
particular integral similar to ordinary differential equations. The complementary
function is the solution of containing arbitrary functions. Particular
integral will be the solution of not involving any arbitrary function.
4.2 CLASSIFICATION OF LINEAR PARTIAL DIFFERENTIAL EQUATION
SECOND ORDERS
The linear PDE(3) with constant coefficient may be classified into the
following two types depending on the form of the function
Homogeneous Equations
91
Non-Homogeneous Equations
Depending on the form of equation (3) may be classified into
two types, namely,
Reducible Equations
Irreducible Equations
4.3 HOMOGENEOUS AND NON HOMOGENEOUS EQUATION WITH
CONSTANT COEFFICIENTS
4.3.1 Homogeneous Equations
A Partial differential equation is called homogeneous if all the derivatives
appearing in the equation are of the same order or it is of the form
The Complementary functions of the Homogeneous Equation with constant
Coefficients
The complementary function is the solution of or of
Now Let us consider that is a solution of , then
and so on.
Substituting these values in , we obtain
or
known as auxiliary equation.
Above equation is of degree and therefore will have roots.
When Auxiliary Equation has Real and distinct Roots
92
Suppose all roots (say) are distinct, then the complementary
function will be
where the functions are arbitrary.
When Auxiliary Equation has Complex Roots
In case the auxiliary equation has complex roots (say and
then the complementary function can easily be obtained by
substituting values of and in equation
When Auxiliary Equation has Repeated Roots
Let a root of the auxiliary equation be repeated twice, ,
then complementary function will be
Let us first take the part of equation corresponding to repeated roots, which is
Substituting
equation becomes
whose solution will be
Putting this value of in , we get
93
which has the form of Lagrange’s linear equation whose auxiliary
equations will be
From first two terms, we obtain
(const.)
and from the first and last member, we have
Integrating
Hence the general solution of will be
If the equation has repeated roots, then proceeding in the same fashion we
can obtain the corresponding part of the complementary function as
Let us look the following problems:
Problem 1: Solve:
Solution. We know that
Hence the given equation can be written as
94
Auxiliary equation in this case will be
which gives
Hence the solution of the given equation will be
Problem 2: Find the surface passing through the two lines and
and satisfying the differential equation
Solution. The given differential equation can be written as
Which is symbolic form will be
Auxiliary equation in this case will be
or
Hence the general solution will be
Now we will have determine the arbitrary functions and with the help
of given conditions.
We have been given that the surface passes through , which from
gives 0= , the function and so
95
Hence equation reduces to
It also passes through , so substituting and in
the above equation, we obtain
taking , we get
Thus
Putting the values of arbitrary functions in , the required solution will be
obtained as
or
You may try now the following exercise:
E 1: Find a surface satisfying the equation
and touching the elliptic paraboloid + along its section be the plane
.
96
E2: Solve :
The particular Integral (General Method)
Let us consider the differential equation
Then the particular integral will be given by
First method: we will be dealing first with the simplest case
then
Which has the form of linear partial differential equation (Lagrange’s form),
and therefore can be solved with the help of subsidiary equations.
which give
and
Method explained above can be applied to general case, for example if
then particular integral
97
Hence
[replacing by after integration]
Then
where
Proceeding in the similar manner, we find the particular integral as
Note : At each step after integration for all applicable values of is to be
replaced by
Second method : When
have factors more than one, like
then it can be decomposed into partial fractions and then finding the value
corresponding to each term (as obtained above), the particular integral may be
obtained.
98
Note : When the R.H.S.is a polynomial in x and y particular integral
can be obtained by expanding factor correspond to
in ascending powers of D and In this case 1/D will mean integration with respect
to x, while 1/D' means integration with respect to y.
Let us consider the following problems:
Problem3: Solve :
Solution. The equation can be written as
Auxiliary equation will be , which gives therefore the
complementary function will be
Particular integral will be given by
Hence the complete solution of the given equation will be
Let us take up another problem:
99
Problem4: Solve:
Solution: The given equation can be written as
or
The complementary function in this case will be
Particular integral will be given by
where
100
[Substituting
where
[as
Hence the complete integral will be
You may now try the following exercise:
E3: Solve :
E4: Solve:
Short Methods of Computing Particular Integral
101
The general method of obtaining the particular integral has been explained in
the previous section. The calculation of particular integral becomes easy, if the
function assumes the form as
We know that
…………………………………..
Similarly
……………………………………
where ….. . ..... are the derivatives of with respect to
Hence
Since is a homogeneous function of degree and so in the right
hand side of above result and have been replaced by and respectively.
In result is derived function of with respect to
being the degree of
Provided that
Working Rule:
102
From above result we can say that when is a function of
and then particular integral
is given by
is nth differential coefficient of with respect to and n is the
degree of
If we put then the desired result may be written as
Provided that
The method of obtaining particular integral in this case will be clear with the
help of following examples.
Problem 5: Solve :
Solution: The given equation can be written as
or symbolically
Auxiliary equation in this case is which gives
Hence the complementary function will be
103
Particular integral will be given by
Where
Therefore the complete solution will be
Particular Integral of when
If then the method of obtaining particular integral discussed
above article, fails.
In such cases when it is clear that should have at least
one factor We will discuss the case when there will be only one factor
of this type and later on the obtained result will be generalized.
Let
Then particular integral corresponding to the factor will
be given by
Then
which is Lagrange’s linear equation. Hence subsidiary equations will be
104
Form first two, terms, we will obtain on integration
and from the first and last term using we get
with the help of relation we can write as
Now if the factor is repeated times, then the corresponding
particular integral will be
Working rule:
While solving questions based on this article where , students
should proceed in the following manner:
(1) Differentiate with respect to partially and multiply the expression
by , so that
(2) If is also zero, differentiate with respect to and multiply
by so that
105
Proceed in the same way as long as the derivative vanishes when
and when a stage is reached where the value can be
obtained by earlier article.
To illustrate this method ,let us take up a problem:
Problem 6: Solve :
Soltion: The given equation can be written as
or
Auxiliary equation gives
Therefore the complementary function will be
Particular integral will be obtained from
Hence the complete solution will be given by
And now a few exercise for you:
E5: Solve :
106
E6: Solve:
4.3.2 Non- Homogeneous Linear Equations
A linear partial differential equation, which is not homogeneous, may be
written as
where is not necessarily homogeneous. Had it been homogeneous then it
could be easily resolved into linear factors but now it may be reducible into linear
factors as well as may be irreducible.
We will deal with these cases separately.
Case I: Complementary Function Corresponding to Linear Factors
Suppose the non-homogeneous linear equation is written as
Right hand side being taken equal to zero for C.F.
Firstly the solution corresponding to will be obtained.
we have
or
which as the form of Lagrange’s linear equation. So subsidiary equation will
be
First and second terms produce
107
Also the first and last term yield
Hence the solution of is
By the generalization of above result it can be shown that the solution of
will be
In case there are repeated roots of the type
Then the corresponding solution can be easily obtained as
Particular Integral
In case of non- homogeneous linear equations the particular integral can be
obtained in the same way as in the case of linear equations with constant
coefficients.
Here the formulae are being given for the sake of convenience of readers.
Provided that
In case of failure
108
Provided that
In case of failure, one may write
Real part in
Similar expression will be for
Let us consider the following problems:
Problem 7: Solve:
Solution: Here the solution will consist of two parts, complementary function and
particular integral.
Complementary function will be given by
Particular integral will be
109
Thus the complete solution of the given equation will be
Problem 8: Solve :
Solution: The given equation can be written as
or
Complementary function will be
Particular integral will be given by
110
Hence the required solution will be
Case II : When Can not be Resolved into Linear Factors
In case when is irreducible it can be factorised into linear
factors in and the method discussed in case I of obtaining the complementary
function fails. We are left with a trial solution which is usually taken and
when substituted in the given equation gives the relationship between and The
method will become clear to the readers by the following problem:
Problem 9: Solve :
Solution: Let us assume that the solution of the given equation be
that
and
Substituting these values in the given equations
Thus will be a solution of the given equation if
Putting for in the solution will be given as
Since all the values of k satisfy the given equation, so more general solution
will be given by
111
You may try the following exercise:
E 7: Solve :
E 8: Solve :
4.4 LINEAR PARTIAL DIFFERENTIAL EQUATION EQUATIONS REDUCIBLE TO
EQUATIONS WITH CONSTANT COEFFICIENTS
A differential equation having variable coefficients can sometimes by reduced
to equations with constant coefficients by suitable substitutions.
One such form is
In this case we substitute and ,
so that and
Now denoting corresponding by D and D' it can be shown that
and in general
112
These substituting reduce the equation into an equation having constant
coefficients, which can be easily solved.
We illustrate this method by the following problems:
Problem 9: Solve :
Solution: Let
or
the equation is transformed in
Complementary function will be given by
Problem 10: Solve :
113
Solution: Substituting
the equation is transformed in
or
which is a linear equation with constant coefficients.
Complementary function will be given by
=
=
Thus the complete solution will be
114
You may now try the following exercise:
E 9: Solve :
E10: Solve :
4.5 SOLUTIONS/ANSWERES
E1: The given differential equation is
Hence the auxiliary equation will be given by
which gives and
Thus the general solution will be given by
Now it has been given that the surfaces
and touch each other along the section by the plane This
condition will be used in the determination of arbitrary functions and In
other words the given condition requires that the values of p and form and
must be equal at
Calculating the values of and form , we get
115
while the values of and from are
and
equating values of p from and at , we have
at
or
To integrate we replace by , to get
Whose integral is easily obtained as
Hence,
Similarly equating q from and at , we get
at ,
or
or ,
using and noting that
Integrating,
Hence the required surface with the help of and is given
by
To evaluate it may be noted that for
116
the surface and must give the same section, . from
and from
This gives
Hence the required surface is
E2: The given equation can be written as
or symbolically
Here the auxiliary equation will be
or
which gives
Therefore the solution of the given equation will be
E3: The given equation can be written as
or
117
Auxiliary equation in this case will be
which gives and
Hence the complementary function will be
Particular integral with be given by
[Substituting the value of ]
[Substituting the value of ]
Therefore the complete integral will be
E4: The given equation can be written as
118
Auxiliary equation gives
Hence the complementary function will be
Particular integral will be given by
Where
Integrating by parts and then substituting the value of the particular
integral is obtained as
Therefore the complete solution will be given by
119
E5: The given equation may be written as
Auxiliary equation is which gives
Hence the complementary function is
Particular integral will be
where and
Therefore the complete solution will be
E6: Auxiliary equation in this case will be
which gives
Hence the complementary function is
120
Particular integral will be given by
Therefore the complete solution will be given by
E 7: The given equation can be written as
or
The complementary function will be given as
Particular integral will be given by
121
Therefore the required solution is
E 8: Let us assume that the solution is
then
Substituting these values in the given equation, we obtain
or
Now may be taken as and may be taken as which satisfy
the relation Thus the solution will be
122
which is satisfied for all values of α, and thus the more general solution
will be
E 9: Substituting
the equation is transformed in
or
or
which is a linear equation with constant coefficients.
Complementary function will be given by
Thus the complete solution will be
123
E10: Substituting and , so that
the equation is transformed in
or
which is a linear equation with constant coefficients.
Complementary function will be given by
Particular integral
125
UNIT 5 CALCULUS OF VARIATIONS
Structure
5.0 Introduction
5.1 Variational problem with fixed boundaries
5.2 Euler equation for functional containing First order derivative and
one independent variable
5.3 Functional dependent on higher order derivatives
5.4 Functional dependent on more than one independent variable
5.5 Variational problem in parametric form
5.6 Invariance of Eular’s equation under Coordinates Transformation
5.7 Check Your Progress: The Key
5.0 INTRODUCTION
The calculus of variation has been one of the major branches of analysis for
more than two centuries. It is a tool of great power that can be applied to a wide
variety of problems in mathematics and mathematical physics. It is primarily
concerned with the study of certain maxima and minima problems which cannot be
handled by the methods of elementary differential calculus. But in mathematical
physics we are generally encountered with the problems where one has to find the
maximal and minimal values (i.e., extreme values) of special quantities called
“functional.”
Functionals are variable quantities whose values are determined by the
choice of one or several functions. In short we may say that “functionals are
functions of functions.”
126
5.1 VARIATIONAL PROBLEM WITH FIXED BOUNDARIES
The Calculus of variations provides a method for determining maximal and
minimal values of functionals. Such problems are known as variational problems.
Before we develop a formal analysis for the calculus of variations, it may be
interesting to know that three problems of historical importance influenced the
development of this subject, which were considered and partly solved by the
ancient Greeks. The analysis was finally developed by Leonhard Euler (1707-1783) in
1744 with his discovery of the basic differential equation for maximizing curve.
Problem I: The problem of brachistochrone or the path of quickest descent.
Let and are two points in a vertical plane, but not in the
same vertical line. The question which was asked by Johann Bernoulli in 1696, was
no determine the curve, among the infinite many
possibilities, along which a bead slides under
gravity will take the shortest possible time of
descent. The common sense, as was believed
earlier, may say that the straight line joining A and B
will be the path of quickest decent as it is the
shortest distance or Galileo believed that the bead would decend more quickly
along the arc of a circle . But Bernoulli proved by an ingenious method that the path
is an arc of a cycloid. In this case although the path becomes longer but a
considerable portion of the arc will be covered at a greater speed.
Problem II: The problem of Geodesic
Geodesics are curves of shortest distance, measured along the surface,
between any two points on the surface.
In other words, we have to find a space curve
y
O x
127
Which lies on a surface such that the length s joining two
points, and , on the surface is minimum, i.e.,
is minimum.
This problem was first solved by Jakob Bernoulli in 1698. He found that the
geodesic on earth, taken as a sphere, is a smaller are of a great circle, that is, that it
lies on a plane through the centre of the sphere.
Problem III: The isoperimetric problem
The ancient Greek proposed the problem of finding the closed plane curve of given
length that encloses the largest area they called this the isoperimetric problem and
showed in a more or less rigorous manner that the obvious answer: a circle, is
correct.
Let the parametric equation of the plane curve be and if
the curve is traversed once counter clockwise as t increases from to , then the
enclosed area is given by
The problem is to maximize (find the extremum of) the functional A subject to
the condition that the length s of the given curve, i.e.,
must have a constant value.
128
General methods for solving problems with isoperimetric conditions were
handled by L. Euler.
5.2 EULER EQUATION FOR FUNCTIONAL CONTAINING FIRST ORDER
DERIVATIVE AND ONE INDEPENDENT VARIABLE
Euler-Lagrange Differential Equation for an Extremal
The calculus of variations, in the simplest form, is the method of determining
an admissible curve joining the points and such that the
integral [functional] of the variational problem,
when taken along this curve, is a maximum or a minimum ( stationary). In other
words the integral has extreme value. The curve itself often called “extremal”.
It may be noted that and and the form of the function are fixed by
given considerations, by the physics of the problem.
Remark: The unknown functions that have continuous second order
derivatives and satisfy the given boundary conditions and .
Functions of this kind are called “admissible.”
Analysis : Let and be two fixed points in the -plane and
be a curve joining and Then the integral
where is a given function of and , has a definite value for a
prescribed curve Y C
129
Now, if we vary the curve passing through the
fixed points and by some other curve through
them, the value of in general, will change and it may
happen that for some admissible curve
through these fixed points the integral has stationary
value.
Let us take one parameter family of admissible curves, say , as
close to , where is a small quantity and is an arbitrary
continuous differentiable function of , satisfying the conditions
to ensure that passes through A and B, i.e., not only are fixed but also
and
It may be noted from the adjoining figure that is called
the variation of the function and is the source of the name Calculus of variations.
If we consider the value of the integral (1) only on curves of the family
then becomes a function of and we call
The curve for which is the curve which makes stationary by
hypothesis. Therefore, the necessary condition for this is
Differentiating (4) with respect to , keeping in view that
C1
P1
A
B
P
X
130
and
we get
The stationary requirement yields
This while integrating its second term by parts, may be written as
using the boundary condition (2), it reduces to
Since is an arbitrary function subject to the condition (2), the integral (9)
can vanish for all only if
131
Hence, the curve along which the integral defined in (1), has
stationary value, must satisfy the differential equation (10),
or in short
which is called Euler’s equations or better called Euler Lagrange equation. It is a
second order non-linear differential equation.
Note : when the boundary conditions are given the solution of equation (11)
gives us only the curve for which has a stationary value it may be a
maximum or a minimum. It is only the geometry of the variational problem in many
cases may help us in deciding the nature of its extreme value.
The solutions of Euler-Lagrange equation as which are
unrestricted by the boundary conditions are called the extremals.
Other Forms of Euler’s Equations:
Form I:
Form II:
Now we shall take up few problems which will illustrate the application of
Euler-Lagrange equation.
Let us look the following problems:
132
Problem 1: On what curves can the functional
be extremized ?
Solution: Here the function under consideration is
Therefore, the Euler-Lagrange equation
for stationary value of , reduces to
This gives on integration
using the boundary conditions, we find
Therefore, an extremum can be achieved only the curve
Problem 2: [The brachistochrone problem]: Find the path on which a particle, in the
absence of friction, will slide from one fixed point to another point not in the same
vertical line in the shortest time under the action of gravity.
133
Solution: We take the origin of coordinates at one A(0
,0) of the fixed points x-axis horizontal and -axis
as downward vertical so that the particle slides from
to another fixed point not in the
P( x, y) same vertical line.
The velocity at any point on the curve
for motion of the particle under gravity is given
by
we know that the arc length is
Therefore the time taken by the particle for moving from
to is obtained from (1) and (2) as
such that
does not contain explicitly.
The functional will be stationary only when satisfies the Euler-
Lagrange equation
s
X
P(x,y)
A(0,0)
B(x1,y1) Y
134
or its corresponding form
Substituting (4) in (5), we get
or on simplification,
Separating the variables of equation (6) we may write
To integrate (7) we introduce a variable Φ by putting
so that
and
This on integration gives
Since the curve is passing through we have when
and consequently Thus
135
If we put and then equations (10) may be put in a more
recognized form
Which are the standard parametric equations of a cycloid generated by a point on
the circumference of a circle rolling along the -axis.
As stated earlier, in the geometry of the problem suggests that the time of
descent will be a minimum only. Thus the brachistochrone is a cycloid.
5.3 FUNCTIONAL DEPENDENT ON HIGHER ORDER DERIVATIVES
If the function contains higher order derivatives, say upto any order , then
then we need to extremize the integral of the form
where we consider the function differentiable times with respect to all
arguments and we assume that the boundary conditions are of the form
at the end points not only the value of the dependent variable but also of its
derivatives upto order inclusive are prescribed.
Using the same kind of analysis as we used in 5.2 and performing repeated
integration by parts, it can be shown that the required extremizing function
satisfies
136
This non-linear differential equation of order is called Euler-Poisson
equation, and its integral curves are termed extremals of the variational problem
under consideration.
Variational Problems involving several dependent variables
let us first consider a variational problem containing two dependent variables
and
we are now required to extremize the functional
for prescribed boundary conditions
That is to say that the functional is defined by the choice of the space curve
As in 5.2 we now consider one parameter family of admissible space curves as
where { is the space curve satisfying the boundary conditions, for
which is stationary, where ε is small quantity and , are arbitrary
continuous differentiable functions of satisfying the conditions
137
Now from (1)
To make stationary, we squire Therefore
Integrating the second and fourth terms by parts and using the boundary
conditions (4) at the end points, we finally get
Since this argument is applicable to any curve in a space N-
dimensions, satisfying the end points conditions
we get a system of second order non-linear differential equations
These are Euler-Lagrange equations for functionals having dependent
variables.
5.4 FUNCTIONAL DEPENDENT ON MORE THAN ONE INDEPENDENT
VARIABLES
With independent variables, we need to extremize multiple integrals of the
form
138
where z is the dependent variable.
Using the same kind of analysis as we did earlier, we find that the extermizing
function must satisfy
where stands for . This is Euler-Lagrange equation when the extremizing
curve is a functions of n independent variables.
It is also called Ostrogradsky equation.
5.5 VARIATIONAL PROBLEM IN PARAMETRIC FORM
In the investigation of certain functional
For an extremum it is sometimes more convenient to seek the solution in the
parametric form . In this case the functional reduces to the form
The corresponding Euler Lagrange equations for two dependent variable will
be
139
However, in this case these equations are not independent but one is the
consequence of the other because of the choice of the parametric representation.
For example if the arc length is taken as parameter then to the equation
we can adjoin the equation
Isoperimetric Problems and Isoperimetric Conditions [or integral side conditions]
In the strict sense of the word, as we have seen in Problem III of 5.1,
isoperimetric problems are problems in which one is required to find closed plane
curves of prescribed length (perimeter) that encloses the largest area. At present,
the term isoperimetric problem is usually extended to include the general case of
finding extremals for one integral subject to the constraints; where one or more
integral having the same integral limits and involving the same variables to take on
prescribed values, which are known as isoperimetric conditions [or integral side
conditions] are given.
Such problems are solved by the method of Lagrange multiplier parallel to the
problems studied in elementary calculus for finding out stationary values of a
function of two or more variables subject to certain constraints, which we present
here as a Lemma for a better understanding.
Lemma : Lagrange multipliers :
Suppose, we wish to determine the values that yield stationary values for
a given function where the variables are not independent but are
constrained by a side condition
140
For the extreme value of we require
Here are not independent but constrained by the condition
which gives
Multiplying (3) by an unknown number λ and adding to (2), we
get
Where λ is called Lagrange undetermined multiplier. This renders
independent and therefore for the validity of (4) we must choose λ such that
These equations together with the constraint are sufficient to
find the three unknowns λ and the values at the required stationary point.
Method of solution of the isoperimetric problem:
Let us now take up the variational problem where we wish to find the
stationary values of the functional
141
subject to the constraint that value of the functional
is held constant, and assume prescribed values at the
end points.
Following the method of Lagrange undetermined multipliers let us take a new
functional (or auxiliary functional)
Now following the analysis of 5.2 and taking
we investigate the unconstrained stationary values of at by
means of the necessary conditions
This yield the equation
Using (9), it may be written as
142
This is the Euler-Lagrange equation for isoperimetric problems, which
together with the original constraint const. and the boundary conditions, will
yield the required solution
Note: In the case of isoperimetric problems involving two or more dependent
variables, e.g. for two variables
subject to the constraint
The stationary functions must satisfy the equations
with the corresponding necessary boundary conditions.
Let us now try the following problems:
Problem 1: Find the extremals of the fuctional
with the boundary conditions
143
Solution: Here,
It contains two dependent variables and , therefore for extremals of , the
Euler-Lagrange system of equations, viz.
reduce to
Eliminating, say, from this system, we get
Its solution is easily obtained as
Now, from the first equation of the set (3)
Applying the given boundary conditions
On (4) and (5), we find
Hence the extremal is the curve of intersection of the surfaces
Let us consider another problem:
144
Problem 2: [Parametric Equation]
Show that the curve of shortest distance (geodesic) on a right circular cylinder
is a helix or a generator.
Solution: Let the parametric equations of a right circular cylinder be
Let be a curve lying on the surface of the cylinder. The length of
curve between two points and is given by the formula
The functional will be stationary only when the integrand satisfies the Euler-
Lagrange equation, viz.
This gives
On integration, we get
145
Let at then
This together with gives the parametric equation of a
helix
If then equation (2) gives const. which will correspond to a
generator of the cylinder.
You may now try the following exercise:
E1: Find the curve of fixed length that joins the points and lies
above the -axis, and encloses the maximum area between itself and -axis.
Next, we look for transformations under which the canonical Euler equations
preserve their canonical form.
5.6 INVARIANCE OF EULAR’S EQUATION UNDER COORDINATES
TRANSFORMATION
We proved the invariance of the Euler equation
Under coordinate transformations of the form
(Such transformations change to in the original functional.) The canonical
Euler equations also have this invariance property. Furthermore, because of the
symmetry between the variables in the canonical equations, they permit
146
even more general changes of variables, i.e., we can transform the variables x,
into new variables x.
In other words, we can think of letting the transform according to their
own formulas, independently of how the variables transform. However, the
canonical equations do not preserve their form under all transformations (1). We
now study the conditions which have to be imposed on the transformations (1) if
the Euler equations are to continue to be in canonical form when written in the new
variables. i.e., if the canonical equations are to transform into new equations
Where = is some new function. Transformations of the
form (1) which preserve the canonical form of the Euler equations are called
canonical transformations.
To find such canonical transformations. We use the fact that the canonical
equations
are the Euler equations of the functional
147
In which the are regarded as 2n independent functions. We want the new
variables to satisfy the equations (2) for some function . This suggests
that we write the functional which has (2) as its Euler equations. This functional is
Where are the functions of defined by (1) and is the
derivative of thus, the functional (4) and (5) represent two different variational
problems involving the same variables and the requirement that the new
system of canonical equations (2) be equivalent to the old system (3. i.e., that it be
possible to obtain (2) from (3) by variational problems corresponding to the
functional (4) and (5) be equivalent.
Since two variational problems are equivalent (i.e., have the same externals) if
the intergrades’ of the corresponding functional differ from each other by a total
differential, which in this case means that
For some function . Thus, if a given transformation (1) from the
variables to the variables is such that there exists a function
satisfying the condition (6). Then the transformation (1) is canonical. In this case, the
function defined by (6) is called the generating function of the canonical
transformation. The function is only specified to within an additive constant.
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Since, as is well known, a function in only specified by its total differential to within
an additive constant.
To justify the term ‘generating function.’ We must show how to actually find the
canonical transformation corresponding to a given generating function . this is
easily done. Writing (6) in the form
We find that
Then (7) is precisely the desired canonical transformation. In fact, the
equations (7) establish the connection between the old variables and the new
variables . and they also give an expression for the new Hamiltonian .
Moreover, it is obvious that (7) satisfies the condition (6), so that the transformation
(6) is indeed canonical. If the generating function does not depend on explicitly.
Then in this case, to obtain the new Hamiltonian , we need only replace
in H by their expressions in terms of
In writing (7). We assumed that the generating function is specified as a
function of x. the old variables and the new variables
It may be more convenient to express the generating function in terms of
instead . To this end, we rewrite (6) in the form
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Thereby obtaining a new generating function
Which is to be regarded as a function of the variables Denoting (8)
by we can write the corresponding canonical
transformation in the form
Noether’s Theorem:
The system of Euler equations corresponding to the functional.
Where F does not depend on x explicitly, has the first integral
It is clear that the statement “F does not depend on explicitly” is equivalent
to the statement “F, and hence the integral (9). Remains the same if we replace by
the new variable
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Where is an arbitrary constant.” It follows that H is a first integral of the
system of Euler equations corresponding to the functional (10) if and only if (10) is
invariant under the transformation (11).
We now show that even in the general case, there is a connection between
the existence of certain first integrals of a system of Euler equations and the
invariance of the corresponding functional under certain transformations of the
variables . We begin by defining more precisely what is meant by the
invariance of a functional under some set of transformations. Suppose we are given
a functional.
Which we write in the concise form
Where now y indicates the n-dimensional vector and ’ the n-
dimensional vector . Consider the transformation
Where The transformation (13) carries the curve y. with the
vector equation
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Into another curve . In fact, replacing y,y in (45) by y(x), y(x), and eliminating x
from resulting n+1 equations, we obtain the vector equation
For where = .
Definition. The functional (12) is said to be invariant under the transformation (13)
i.e., if
Problem 1: The functional
is invariant under the transformation
Where is an arbitrary constant. In fact, given a curve with equation
The “transformed” curve i.e., the curve obtained from by shifting it a
distance along the x-axis, has the equation
and then
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5.7 SOLUTIONS/ANSWERES
E1: (Restricted version of Isoperimetric Problem)
The area bounded by the arc AB,
, and the -axis is given by
Subject to the condition that the
length of the arc , say is fixed length
It is clearly necessary that has to greater than 1.
Now, the formulation of the problem is to determine the shape of the arc,
so as to maximize (1) under the isoperimetric condition (2).
We therefore, use the method of Lagrange multipliers and define a new
functional
Y
X
y=y(x)
(0,0) B(1,0)
L
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For the stationary value of the Euler-Lagrange equation is
which is view of (5) reduce to
Integration once, we find
Solving (7) for we get
Integrating once again, in the parametric form by the substitution
and then eliminating between the forms of and , we find
which is a circle having the center as and radius and is the required curve
maximizing the enclosed area.
The constant and are given by the boundary conditions
ad the constraint
This gives
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To get a circle of real center, from (10) it is clear that further , for admissible
values of sine function, from (11), we find
It is clear necessary that . Thus combining the three conditions, we
conclude
Because if the circular arc determined by (9) will not define as a single
valued function of . In this case the negative value of , which will correspond to
the curve below -axis will give the maximum area.
*****