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FIRST ORDER PARTIAL DIFFERENTIA L
EQUATIONS
Unit structure
1.0 Objectives
1.1 Introduction
1.2 First order partial differential equation and Cauchy problem
1.3 Semi linear and quasi linear equation in two independent variable s
1.4 First order nonlinear equations in two independent variable s
1.5 Complete Integral
1.6 Let us sum up
1.7 List of references
1.8 Bibliography
1.0 OBJECTIVES
After doing this unit, you will be able to:
identify partial differential equation of order one.
classify different types of partial differential eq uation
solve problems of semi linear and quasilinear problems
determine characteristic equations
solve Cauchy problem
find general solution
find complete integral.
1.1 INTRODUCTION
A partial differential equation for a function () of m independ ent
variables (=1,2,3……) is a relationship between the function
and its partial derivatives ,,….. We represent this relationship in
the form
,……,;; ,……; ;……=0 (1.1)
Or briefly
,, , ……=0 munotes.in
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where only a finite number of derivatives occur on the lefthand side and
the function is defined over Doma in . The orde r of the partial
derivative is the order of the highest derivatives appearing in the
function .
A gen uine solution of the partial differential equation is a function =
()defined over a domain of space such that all partial derivative
of appearing in the equation exist and are continuous in ,
,(),(),(),……∈ when ∈ and
,(),(),(),……=0 for all ∈. We also say
that the function satisfies equation (1.1). we shall refer the genuine
solution simply as a solution.
While studying partial differential equations, we shall assume that all
functions are real valued with real arguments unle ss otherwise stated.
The simplest partial differential equations to study are those of the first
order for the determination of just one unknown function. Apart from the
fact that they form the basis of the study of higher order equations called
hyperbol ic equations, they are the simplest kind of equations for which
method of solutions are available and for which the existence, uniqueness
and stability can be discussed in detail. In this chapter, we shall present
some basic result concerning first order p artial differential equation.
1.2 FIRST ORDER PARTIAL DIFFERENTIAL
EQUATION AND CAUCHY PROBLEM
In this chapter while dealing with the partial differential equation in two
independent variables, we shall denote the independent variable by
and .
A first order partial differential equation in two unknowns in its most
general form is given by
,,,,=0 (1.2)
where is a known function of its ar guments.
1.2.1 Classification of partial differential Equation
Linear equation: when the function is linear in ,and u. then the
equation of the form
(,)+(,)=(,)+(,) (1.3)
is called linear equation. Where ,,, will depend on x and y.
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For example,
+= +and +=3 +.
Semi linear equation: When the fun ction is linear in ,then the
equation of the form
(,)+(,)=(,,) (1.4)
is called semi linear equation. W herea and b depend on x and y whereas c
depends on x, y and u.
For example,
+ = and 2 +3 =5
Quasilinear equation: When the function is linear in ,then the
equation of the form
(,,)+(,,)=(,,) (1.5)
Is called quasi linear. Where ,, depend on , and .
For example,
+ =and (−)+(−) = −
Nonlinear equation: When the function is not linear in ,then the
equation (1.2) is called non linear equation .
For example,
+= and +=
The solution =(,) represents a surface in (,,) space. This
surface is called integral surface of the partial differential equation.
While dealing with partial differential equations appearing in science and
engineering, we rarely to find out or discu ss properties of a solution in its
most general form. Almost always we deal with those solution of
differential equations which satisfy certain conditions. In the case of first
order partial differential equations, the search for these specific solutions
can be formulated as a Cauchy problem.
1.2.2. The Cauchy problem
Consider an interval I on the real line and three arbitrary functions
(),() () of single variable ∈ suchthat the derivatives
() () are piecewise continuous and ()+()≠0.
A Cauchy problem for a first order equation (1.2) is to find the domain D
in (,) plane containing (),()for all ∈ and a solution
=(,) of the equation such that munotes.in
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(),()=() (1.6)
for all values of ∈ .
Geometrically, =(),=() represents a curve in (,) plane.
We call the curve datum curve. The Cauchy problem is to determine the
solution of ,,,,=0 in a neighbourhood of such that
takes prescribed values () on .
The solution of Cauchy problem also involves such questions as the
conditions on the functions ,(),() () under which the
solution exists and its unique.
1.3 SEMILINEAR AND QUASILINEAR EQUATIONS IN
TWO INDEPENDENT VARIABLES
We start with a semi linear equation instead of linear equation as the
theory of the former does not require any special treatment as compared to
that of latter.
1.3.1 Semilinear Equation
Consider a s ingle first order equation in two independent variables (,)
for a single unknown quantity:
(,)+(,)=(,,) (2.1)
We assume that ,, are continuously differential functions of their
arguments and a and b are not simultaneously zero. ,∈() and
∈(), where and are domains in (,) plane and (,,)
space respectively, such that whenever (,,)∈,(,)∈.
At a given point (,)∈, (,)+(,) represents a
derivative of (,) in the direction of vector ((,),(,)).
Therefore, if we consider a one parameter family of curves whose tangent
at each point is in the above direction i.e. the fa mily of curves defined by
ordinary differential equation
=(,)
(,) (2.2)
the variation of u along these curves is given by
=+
=
, which with the help of (2.1) gives
=(,,)
(,) (2.3)
Consider a curve represented by a solution of equation (2.2). we can
choose a variable such that the curve has a parametric representation
=(),=() and () and () satisf y a pair of ordinary
differential equations munotes.in
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=(,),
=(,) (2.4)
The variation of u along the curve is given by
=(,,) (2.5)
The equations (2.2) or (2.4) are called characteristic equations . The
solution of (2.2) can be written in the form
(,,)=0 (2.6)
Where C is a constant of integration. This equation represents one
parametric family of cu rves with as a parameter. We call these curves
the characteristiccurves of the partial differential equation. In the domain
consider another curve =(),=() such that it is nowhere
tangential to characteristic curve.
Solving (2 .4) with the condition =(),=() =0, we get a
solution of the form
=(,),=(,) (2.7)
Because of the equivalence of (2.2) and (2.4), th e equation (2.7) also
represents the one parameter family of characteristic curves of equation
(2.1). in the parametric representation of (2.7), varies along a
characteristic curve. remains constant along characteristic curves. The
equation (2.3) or (2.5) is called compatibility condition along a
characteristic curve.
Suppose that (,) is assigned an initial value at point
(,)in(,)−plane. Since (,),(,) and (,,) are
function of their a rguments, the initial value problem for the ordinary
differential equations (2.4) and (2.5) with initial values ,, has
unique solution. Therefore, through the point (,) there passes a
unique characteristic curve given by
=(,,),=(,,) (2.8)
and along this curve
=(,,,) (2.9)
is uniquely determined by the equation (2.5). This shows that, if is given
at any point, it is uniquely determined everywhere along the characteristic
curve denoted by passing through the point, as long as it does not pass
through a singular point and as long as (,,) remains in , where
(,,) is defined. This suggests the following method of solution of the
Cauchy problem .
We take an arbitrary point ((),(),)on the datum curve . The
value of at is () . solving the characteristic equations and the munotes.in
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compatibility condition with initial values =(),=(),=
() at =0, we get
=((),(),),=((),(),) (2.10)
and
=((),(),(),) (2.11)
Solving the pair of equations (2.10) for and in terms of , and
substituting in (2.11) we get a solution of the Cauchy Problem in
neighbourhood of the curve .
the method fails if the curve coincide with the characteristic curve.
From the compatibility condition (2.5) we also note that if is a
characteristic curve, the variation of the Cauchy data () on is
constrained by the relation (2.5) and so cannot be arbitrarily prescribed
on it.
Example 2.1: Solve the Cauchy problem of partial differential equation
2+3=1, (2.12)
with Cauchy data prescribed on the straight line :=−=0,
where and are constants . A parametric representation of Cauchy data
is =,= ,(,)=()
Solution: Initial values: =,= ,=0=2,=3,=1
Fig 1.1
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Characteristic Equations
=2,
=3=>=2, =3
=>=2+,
Applying Initial Conditions , =2∗0+=>=
=>=2+ (2.13)
And =3+,
Applying Initial Conditions, =3∗0+=>=
=>=3+ (2.14)
==>
=1=>==>=+
Applying Initial Condit ions, ()=0+=>=()
=>=+()
Solving and from (2.13) and (2.14)
=>=3−2
(3−2), =−
2−3
Substituting these values in =+()
We get, =
+
(2.15)
Provided w e assume that
3−2≠0 (2.16)
Equation (2.15) represents a genuine solution of the equation (2.12) if the
given function () is continuously differential. Then and are
function in the entire (,)−plane and satisfy the equation (2.12).
When the constants and are such that 3−2=0, the above method
of finding the solution breaks down. In this case the straight line is itself
a characteristic curve. Along a characteristic curve
=2. The
compatibili ty condition (2.5) shows that the function () in the above
Cauchy problem cannot be arbitrarily prescribed but must satisfy the
relation
()
=
(2.17)
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This condition completely determines the function ()expect for a
constant of integration:
()=
(2.18)
It is simple to check that the characteristic Cauchy problem with the
Cauchy data
=, =3
2, =
2
has a solution of the form
=
+(3−2) (2.19)
Where () is an arbitrary function of and satisfies (0)=0
This example verifies a general property namely, the solution of a
characteristic Cauchy problem when it exists, is non unique in that it
involves an arbitrary function.
Example 2.2: Solve the Cauchy problem of partial differential equation
+=, with initial conditions (,0)=1.
Solution: =1,=1,=,
With initial conditions, =,=0,=1,=0
Characteristic equation:
=1=>=
=>=+
Applying Initial Conditions, =0+=>=
=>=+=>=−
=1=>=+,
Applying Initial Conditions, 0=0+=>=0
=>=
==>
==> =+
Applying Initial Conditions, 1=0+=>=0
log==>=
=>=
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Example 2. 3:Find the character istic equation of the following PDE
−=0
Solution : Characteristic equation
=
=−
=> =−
=>∫ =−
=>
2=−
2+
=>+=
This represent s equation of circle with centre as origin.
Example 2. 4:Find the characteristic equation of the following PDE
2−(+)=0
Solution: =2,=−(+)
=
=>
=−+
2=>2 =−−
2 +=−
()=−
Integrating both the sides ,
∫()=−=>=−
3+
=>+
3−=0
=>3xy+−=0
EXERCISE 2.1
1. Find the characteristics of the equation (−+1)+
2=0
2. Show that characteristic of −=0 touches the branch of the
hyperbola =1 in the first quadrant of the (,)−plane at the
point (1,1). Verify that the point P divides the hyperbola into
two portions such t hat the Cauchy data prescribed on one portion
determines the value of on the other portion.
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3. Find the solution of −=0, given that (,0)= for
−∞<<∞
4. Show that if is prescribed on the interval 0≤≤1 of the
−axis, the so lution of (−+1)+2=0 is
completely determined in the first quadrant of the (,)−plane.
5. Find the solution of the partial differential equation
(+1)+(−1)=(+) satisfying the condition
(,0)=−1− for −1<<∞
6. Find the solution of the Cauchy problems and the domain in which
they are determined in (,)−plane:
(i) +=2 with (,0)=()for >0,
(ii) +=2 with (0,)=()for >0,
(iii) += with (,0)=1 −∞<<∞
1.3.2 Quasilinear Equations
Now, we pass on to the general quasilinear equation of the first order
(,,)+(,,)=(,,) (2.20 )
where the coefficients and depend on the dependent variable u also.
We assume that ,, are functions in the domain of
(,,)−space. We recall here the geometrical interpretation of a
solution =(,) as a surface in (,,)−space, called integral
surface . The direction ratio of the normal to the surface are
,,−1, so equation (2.20) can be written as (,,).,,−1=
0 (2.21)
where the left hand side is the scalar product of two vector s, we can
interpret the equation as being equivalent to a condition that the integral
surface at each point has the property that the vector (,,) is tangential
to the surface.
Monge direction: At any point (,,)in , the vecto r
((,, ),(,,),(,,)) defines a direction, called Monge
direction. Therefore, the coefficients in the equation (2.20) defines a
direction field i.e. the field of Monge directions in the domain of of
(,,)−space.
Monge curve: A surface =(,) is an integral surface if and only if,
at each point of the surface the tangent plane contains the Monge direction
at that point.Thus , at a given point (,,) the tangent plane of the
integral surface has one degree o f freedom, i.e. it can rotate about Monge
direction. A space curve whose tangent at every point coincide with
Monge direction is called a Monge curve and it determined by the
equations, munotes.in
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(,,)=
(,,)=
(,,) (2.22)
This equation also known as Lagrange equation.
In terms of parameter , such that is the common value of the three
ratios in (2.22), we can write the characteristic equation and c ompatibility
condition respectively as
=(,,),
=(,,) (2.23)
and
=(,,) (2.24)
As in §2.1, we consider a surface in given by =(,),=
(,),
=(,), such that it nowhere touches the Monge curve. Solving
the system of equations (2.23) and (2.2 4), with the condition =
(,),=(,),=(,)at =0 , we get a
representation of the Monge curve in the form
=(,,),=(,,),=(,,) (2.25)
The totality of Monge curves form a two parameter family of curves with
parameter and. The projection of a Monge curve on (,)−plane is
called characteristic curve of (2.20). Note that the characteristic eq uations
(2.4) of the semilinear equation (2.1) are not coupled with the
compatibility condition (2.5) and hence can be integrated independently.
Thus, the one parameter family of characteristic curves of a semilinear
equation can be drawn once for all without any reference to the
compatibility condition. For the quasi linear equation (2.20), the
characteristic equation and compatibility condition are coupled. Therefore,
to determine the characteristic in case of the quasi linear equation, we
have to draw th em by solving the three equations (2.23) and (2.24)
together. The totality of the characteristic curves in (,)−plane of
quasilinear equation form s a twoparameter family of curves. For a given
solution is a known function of and , and the eq uation (2.23) for
characteristics can be solve without any reference to the compatibility
condition (2.24), as in the case of semilinear equations. In this case
through any point (,), there is only one characteristic curve and the set
of all character istic curves from one characteristic curve and the set of all
characteristic curves form one  parameter family of curves in the
(,)−plane.
Example 2.5: consider the partial differential equation
+=0 munotes.in
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The Monge curve through the point (,,) is a straight line given by
the equations
−=(−),=
The characteristic curves through an arbitrary point (,) in
(,)−plane is the one parameter family of straight line passing throug h
the point and depending on the parameter .
Consider a surface generated by a one parameter subfamily of Monge
curves. The tangent plane at the point of the surface contains the Monge
direction at that point. Therefore, every surface generated by a one
parameter sub family of Monge curve is an integral surface of (2.20). the
converse of this statement is also true. Let =(,) be an integral
surface S. Let =(),=(),
=()≡((),()) be a space curve lying on S and suppose the
function (),() are so prescribed that the curve is not Monge curve.
Consider the solution of
=,,(,),
=(,,(,)) (2.26)
with =(),=() at =0 in the form =(,),=(,).
In (2.26) is known function of , from the equation of integral surface
. Then along the one parameter family of curves
=(,),=(,),=((,),(,)) (2.27)
with as parameter lying on , we have
=
+
=+=(,,) (2.28)
In view of (2.26) and (2.28), we infer that the curves (2. 27) are Monge
curves. These Monge curves generate the integral surface as varies. We
have shown that starting from a non Monge curve on an integral surface,
we can determine one parameter sub family of Monge curve that generate
the surface.Thus any int egral surface S is generated by a family of Monge
curve depending on a single parameter .
Now we have also proved that through an arbitrary point of an integral
surface there passes a Monge curve which lies entirely on the integral
surface. This wi th the uniqueness theorem of the solution of an initial
value problem of the ordinary differential equation (2.23) and (2.24)
implies that if Monge curve is tangential to an integral surface at any
point, it lies entirely on the integral surface.
We c an now present a method for the solution of a Cauchy problem for
the quasilinear equation (2.20). We first note that geometrically =
(),=(),=() represents a curve in (,,)−space. We
call this curve initial curve . The datum curve , on which the Cauchy data
is prescribed, is the projection of Γ on the (,)−plane . A geometrical munotes.in
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representation of a Cauchy problem for a first order partial differential
equation is to find an integral surface of the equation passing through
initial curve Γ. The result of the last two paragraphs show s that in the
order to solve a Cauchy problem we just have to fin d the surface generated
by the one parameter family of Monge curve s, starting from the points
(),(),(), in the form
=(,),=(,),=(,) (2.29)
This is a parametric repres entation of required integral surface. We shall
again have to exclude datum curve which are tangential to the
characteristic curves. We present here a precise formation in the following
theorem.
Theorem 2.1: Let (),(), and() be co ntinuously differential
function of in a closed interval say [0,1] and ,, be functions of
,, having continuous first order partial derivatives with respect to their
arguments in some domain of (,,)−space containing the in itial
curve
Γ∶=(),=(),=();0≤≤1 (2.30)
and satisfying the condition
()
(),(),()−()
(),(),()≠0.
(2.31)
Then there exists a solution =(,) of the quasi linear equation (2.20)
in the neighbourhood of the datum curve :=(),=(), and
satisfying the condition ()=(),(),0≤≤1 (2.32)
Proof: since ,, have continuous partial derivative with respect to
,,; the ordinary differential equation (2.23) and (2.24) have a unique
continuously differential solution of the form (2.29) satisfying the initial
condition
(0,)=(),(0,)=(),(0,)=() (2.33)
As (),(), and() be continuously differential, the solution (2.29)
is continuously differential with respect to . In view of assumption (2.31)
the Jacobian
(, )
(, )≡
=(−) (2.34)
does not vanish at =0 for 0≤≤1. Therefore, in the neighbourhood
of =0, we can uniquely solve for and in terms of and from the
first two relations in (2.29) and substitute in the third relation to get as a
function of and
i.e. (,)=((,),(,)) (2.35) munotes.in
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At any point of the datum curve, (),()=(0,)=(),
which shows that the initial condition (2.33) is satisfied.
From (2.24), i.e. =, we have += or +=
showing that the function (,) given by (2.35) satisfies the equation
(2.20).
To pr ove the uniqueness of the solution we first note that if a Monge curve
is tangential to an integral surface at any point, it lies entirely on the
surface. Let us assume now that there are two integral surfaces and ′
passing through the initial curve Γ, given by (2.30). Then for an arbitrary
given value of , the Monge curve (2.29) starting from the point
((),(),()) lies entirely on both the surfaces and .Hence
and are generated by same subfamily of Monge cu rves which implies
that the two integral surfaces are same.
Example 2. 6: Consider the equation
+=0 (2.36)
with the Cauchy data (,0)=,0≤≤1.
prescribed only on a portion of th e −axis. The Cauchy data can be put in
the form of (2.30):
=,=0,=,0≤≤1(2.37)
Solving the characteristic equations and compatibility condition
=,
=1,
=0
With the initial data we get
=(+1),=,= (2.38)
The characteristic curve passing through a point = on the axis is a
straight line =(+1). These characteristic for all admissible but
fixed value of i.e.0≤≤1 pass through the same point (0,−1) and
cover the wedged shaped portion of the (,)plane bounded by two
extreme characteristics =0 and =+1.= in (2.38) shows that
is constant in those characteristic s, being equal to the abscissa of the
point where the characteristic s intersects the axis. The solution is
determined in the wedged shaped region as shown in the Fig. 1.2
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We note two very important aspects of quasi linear equation from this
example.
(i) The domain in the (,) plane in which the solution is
determined depends on the data prescribed in the Cauchy problem.
Had we prescribed (,0)=constant =
, say, for 0≤≤1, the
characteristic would have been a family of parallel straight lines
−2=−2 and the domain would have been a family of
parallel straight lines −2=−2 and domain would have
been the infinite strip bounded by extreme characteristics −
2=0 and −2=−2 as shown in the Fig. 1.3.
(ii) Even though the coefficient in the equation (2.36) and the Cauchy
data (2.37) are regular, the solution develops a singularity at the
point (0,−1). Geometrically this is evident from the fact that the
characteristic which carry different values of all intersect at
(0,−1). Analytically, this is clear from the explicit form of
solution obtai ned (2.38) after eliminating and :
=
(2.39)
The appearance of the singularity in the solution of a Cauchy problem for
certain Cauchy data is properly associated with non linear di fferential
equations.
Fig 1.2
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1.3.3 the characteristic Cauchy problem
We have just seen that if the datum curve is such that Cauchy data
satisfies (2.31), then the unique solution of the Cauchy problem exists in a
neighbourhood of the curve. Now suppose t hat
()
(),(),()−()
(),(),()=0
(2.40)
Everywhere along the curve , i.e. is a characteristic curve for a possible
solution. Let us suppose further that a sol ution: =(,), of Cauchy
problem exists. Then from (2.40) and (2.20) it follows that
()
=
(),()=
(,)+
(,)
must be proportional to ((),(),()). Therefore the function
(),(),() satisfy the equations
((),(),())=
((),(),())
=
((),(),())
and the initial curve Γ is necessarily a Monge curve.
Fig 1.3
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Consider now another curve Γ′ in (,,) space which is not a Monge
curve and which intersects Γ at some point. Then we can obtain an integral
surface ′ passing through Γ. As on point of Γ lies on ′ the entire original
initial curve Γ will lie on ′ and hence ′ is an integral surface passing
through Γ. Consider now another curve Γ", which is not a Monge curve
and which intersects Γ, but does not lie on ′. Then we get another integral
surface " containing Γ and different from ′.
Therefore, the solution of a characteristic initial value problem, if it exists,
is nonunique.
1.3.4 General Solution
Until now we have discussed only those solution of a first order
differential equation which satisfy certain prescribed conditions (i.e.
solution of a Cauchy problem). In general, these particular solutions are
completely determined. For a single quasilinear equation of first order , it
is possible to get an explici t form of general solution which is define to be
a solution from which all particular solution can be obtained.
A relation of the form (,,)=, where is a constant is called a first
integral of first order ordinary differential equations (2.22)(or (2.23) and
(2.24)), if the function (,,) has a constant value along an integral
curve of (2.22) (i.e. along a Monge curv e). It follows, therefore, that if
(,,)= be a first integral of (2.22) and =(),=(),=
() be a solution of these equations, then ((),(),()) is
independence of .
The general solution of the ordinary differen tial equation (2.22) cons ists
of any two independent first integrals
(,,)= (,,)= (2.41)
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which together also constitute another representation of the two parameter
family of Monge cur ve of (2.20). The surface repr esented by a first
integral, say (,,)=, is generated by one parameter family of
Monge curves by varying a parameter and hence represents an integral
surface of (2.20). now it follows that each one of the two equations in
(2.41) represents a one parameter family of integral surface of (2.20).
Next, we prove a theorem which connect the two independent families of
integral surface two the quasilinear equations.
Theorem 2.2: if (,,)= and (,,)= be two independent
first integral of the ordinary differential equation (2.22), and +≠0
the general solution of the partial differential equation (2.20) is given by
ℎ((,,),(,,))=0 (2.42)
where ℎ is an arbitrary function.
Proof: since the first integral (,,)= represents an integral
surface, the equation (2.20) is satisfied by =−
,=−
. This
gives
++=0 (2.43)
Similarly
++=0 (2.44)
If (,,)=0 be the equation of an integr al surface of (2.20), we also
have
++=0 (2.45)
Since ++≠0, it follows from (2.43) −(2.45) that the Jacobian
(,,)
(,,) ≡0. This im plies that =ℎ(,) where ℎ is an arbitrary
function of its arguments, showing that the equation of any integral
surface is given by (2.42).
The twoparameter family of Monge curve in (,,)−space is
represented by the equation (2.41). Th e integral surface (2.42) is generated
by one parameter sub family of the Monge curves, obtained by restricting
the values of and by the relation
ℎ(,)=0 (2.46)
For a given Cauchy problem, it is simple to determine the one parameter
of subfamily of Monge curves which generate the integral surface passing
through the initial curve Γ represented by (2.30). The parameter and
for which the Mon ge curve intersect the curve Γ, satisf y
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((),(),())=
and
((),(),())=
Eliminating from these two, we get a relation of the form (2.46) between
and . This determines the func tion ℎ. the solution of the Cauchy
problem is obtained by solving in terms of and from (2.42).
Example 2.7: Find the general solution of differential equation
(+2)−(+2)=
(−) (2.47)
Solution: The characteristic equations and the compatibility conditions
are
=
()=
()
To get one first integral we derive from these,
+
2(−)=2
−=> +
2=2
=> + =4
Integrating both the sides we get,
(,,)≡+−4= (2.48)
For another independent first integral we derive a second co mbination
+
−=2
−=> + =−
(,,)≡+2= (2.49)
The general integral of the equation (2.47) is given by
ℎ(+−4,+2)=0 or (+2)=+−4 (2.50)
where ℎ are arbitrary functions of their arguments.
Consider a Cauchy problem in which is prescribed to be zero on the
straight line −=0. Parametrically, we can write it in the f orm
=,=,=0
From (2.48) and (2.49) we get, 2=and = which gives
=2.
Therefore, the solution of Cauchy problem is obtained, when we take
ℎ(,)=−2. This gives =
{−+1−1} (2.51) munotes.in
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We know that the solution of the Cauchy problem is determined uniquely
at all points in the (,)plane.
Example 2.8: Find the general solution of the following quasi linear
equations.
a)
+ =
Sol:
=
=
=>
=
=
Taking 1st two terms together
=
=>
=
=>
=
==>
3=
3+
=>(,,)≡−=
Taking 1st and 3rd term together,
=
=> = =>
2=
2+′
(,,)≡−=
The general solution is ℎ(−,−)=0
b) +=
Solution:
=
=
=>
=
=
Taking first two terms together,
=
=>(,,)≡1
−1
=
Taking last two terms together,
=
=>−
=+
=>(,,)≡1
+=
The general solution is ℎ
−
,
+log=0 munotes.in
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c) tan +tan =tan
Solution:
=
=
Solving,
=
=>cot=cot
=>log =log +log
=>log =logsin∗
=> = =>(,,)≡sin
=
Solving
=
=>(,,)≡sin
sin=
The general solution is ℎ
,
=0
d) +3=5+tan (−3)
Solution:
=
=
()
1=
3=>3=
=>3=+=>(,,)≡−3=
3=
5+tan(−3)
=>
3=
5+tanC
3=log(5+tan)∗1
5+′
(,,)≡5−3log(5+tan(−3))=
The general solution is
ℎ(−3,5−3log(5+tan(−3))=0or
(−3)=5−3log(5+tan(−3)
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EXERCISE 2.2
1. Show that all the char acteristic curves of the partial differential
equation (2+)+(2+)= through the point (1, 1)
are given by the straight line −=0.
2. Discuss the solution of the differential equation +=0,
>0,−∞<<∞ with Cauchy dat a
(,0)=−for≤
=0 for>0
3. Find the general solution of the equation
(2−)+8(−2)=2(4+)
and deduce the solution of the Cauchy problem when (,0)=
on a portion of the  axis.
4. Show that the result of elimination of an arbitrary function
ℎ(, ) of two arguments from the relation
ℎ(, ,),(,,)=0
1.4 FIRST ORDER NON LINEAR EQUATIONSIN TWO
INDEPENDENT VARIABLES
The most general first order equation, i.e. an equation of the form
(,,, )=0 (3.1) where is a given function of its arguments
and
=,= (3.2)
In this section we shall consider a non linear partial differential equation,
i.e. equation (3.1) where is not linear in and . we assume here that the
function possess c ontinuous second order partial derivatives over a
domain of (,,,,)space with +≠0. Let the projection of
on (,,)space be denoted by .
1.4.1 Monge strip and Charpit ’s Equation
Let =(,) represent an integral surface of (3.1) in (,,) space,
then (,,−1) are direction ratios of the normal to S.
The differential equation (3.1) states that at any point (,,) on S,
there is a relation between and. This relation (,,,,)=0
between ,& is not linear. Hence all the tangent to integral surface do
not pass through the fixed line but form a family of planes enveloping a
conical surface, called the Monge cone with P as its vertex. The differential
equation thus assign s a Monge cone at every point, i.e. a field of Monge
cones in the domain of (,,)space. The problem of solving the
differential equation (3.1) is to find the surface which fit in the field, i.e. munotes.in
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23
surfaces which touch the Monge cone at each point along a generator.
Also note that Monge cone need not to be closed.
Example3.1: Consider the partial differential
−=1 (3.3)
At every point of the (,,)space the relation (3.3) can be expressed
parametrically as
=cosh ,=sinh −∞<<∞ (3.4)
The equation of tangent planes at (,,) are
(−)cosh +(−)sinh −(−)=0 (3.5)
The envelope of these planes is  eliminant of (3.5) and
(−)sinh +(−)cosh =0 (3.6)
Which is obtained by differentiating (3.5) partially with respect to .
Therefore , the Monge cone of (3.3) is
(−)−(−)−(−)=0 (3.7)
This is the right circular cone with semi vertical angle
and whose axis is
the straight line passing through (,,) and parallel to axis.
Since an integral surface is touched by a Monge curve along a generator,
we proceed to determine the equations to a generator of the Monge cone
of (3.1). At a given point (,,), the relation between and can
be expressed parametrically in the form
=(,,,), =(,,,) (3.8)
which satisfy
,,,(,,,),(,,,)=0 (3.9)
For all values of the parameter for which and in (3.8) are defined.
The equations of the tangent planes for and + are
(,,,)(−)+(,,,)(−)=−
(3.10)
and
(,,,+)(−)+(,,,+)(−)=−
(3.11) munotes.in
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The limiting position of the line of intersection of these planes as →0
is a generator of the Monge cone at (,,). Expanding and in
(3.11) in powers of , using (3.10) and retaining only the first degree
terms, we get
(−)+
(−)=0 (3.12)
(3.10) and (3.12) are the equations to the generators in the terms of the
parameter . We can eliminate the derivatives
and
with the help of
(3.9) which gives
+
=0 (3.13)
From (3.10), (3.12) and (3.13) we get the following equations of the
generator of the Monge cone at (,,)
=
=
(3.14)
If we replace −,−,−by ,,, respectively,
corresponding finite infinitesimal moment, −=,−=
,−=, from (,,) along the generator, then (3 .14)tends to
=
=
(3.15)
We note that, for quasilinear equation (2.20), equations (3.15) reduce to
(2.22) showing that th e Monge cone degenerates into the Monge line
element.
Suppose we are given an integral surface :=(,), where (,)
has continuous second order partial derivatives with respect to x and y. At
the point of we know , as funct ion of and .Also at each
point of the surface , there exist Monge cone which touches the surface
along a generator of the cone. The line of contact between the tangent
plane of S and the corresponding cones, that is the generators along with
the sur face is touched, define a direction field on the surface, which is
called Monge direction on (Fig. 3.1). Monge direction for a quasilinear
equation and Monge direction on an integral surface for a non linear
equation has the common property that they ar e special direction
tangential to the integral surface. However, in the non linear case, they
have no exitance of their own but are defined only when an integral
surface is prescribed. munotes.in
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25
The above direction field also defines a one parameter family of cur ves on
, we call these curves Monge curves on , and these curves generates .
Denoting the ratios in (3.15) by , we notice that the Monge curves on
can be determined solving the ordinary differential equations
=(,,(,),(,),(,) ) (3.16)
and
=(,,(,),(,),(,) ) (3.17)
In the form of
=(,,), =(,,) (3.18)
and then determining from
=(,,)≡((,,),(,,)) (3.19)
Here (,,(,))is a point on the surface and the Monge curve on
given by (3.18) and (3.19) passes through the point. Since
=
+
It follows from (3.18) and (3.19) that along these curves varies
according to
=+(3.20)
Where =(,) has been substituted in the expression on the right 
hand side.
Fig 1.5
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Example 3.2: consider the function
=cos+sin,=constant. (3.21)
Which represents an integral surface of the equation
≡+−1=0 (3.22)
Then (3.16) and (3.17) give
=2=2cos
=2=2sin
Therefore, the Monge curve s of (3.22) on integral surface (3.21) are given
by
=+2cos, =+2sin
and
=cos+sin+2
Along the Monge curves on the variation of and are known from
the expressions =((,,),(,,)) and
=((,,),(,,)) respectively. Now we shall determine
the rates of change of and along a Monge curve on . Since (3.1) is
identically satisfied by =(,), differentiating with respect to x we
get the identity
+++=0 on . (3.23)
Along Monge curve on
=
+
=+
For sufficiently smooth solution, = so that from (3.23), we get
=−(+) (3.24)
Similarly, the variation of along a Monge curve on is
=−(+) (3.25)
Given an integral surface, we have shown that there exist a family of
Monge curves, which generate the surface and along which ,,,,
vary according to
= (3.26) munotes.in
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27
= (3.27)
=+ (3.28)
=−− (3.29)
and
=−− (3.30)
We have discussed Monge curves exist only on a given integral surface.
We now r everse the process by disregarding the fact that the system of
ordinary differential equations (3.26) to (3.30) was derived with the help
of integral surface. we call the first two equations (3.26) and (3.27)
characteristic equations, the last three equati ons (3.28)  (3.30)
compatibility conditions and the system formed with all the five equations
(3.26) (3.30), Charpit’s equations.
A set ((),(),(),(),()) of five differential function is said to
be a strip, if when we consid er the curve =(),=(),=(),
the planes with the normals given by ((),(),−1) are tangential to it.
A solution = (),= (),= (),= ()and =() of
the Charpit’s equations satisfied the strip condit ion
=()
+()
(3.31)
Note that not every set of five functions can be interpreted as a strip (Fig.
3.2). A strip requires that the plane with normal (,,−) be tan gent to
curve, i.e. they must satisfy the strip condition (3.31) and the normal
should vary continuously along the curve. For a solution of Charpit’s
equation (3.26) (3.30), the strip condition is guaranteed by the first three
equations.
Along a so lution of the Charpit’s equations, we have
=
+
+
+
+
(3.32)
which becomes identically equal to zero when we use (3.26)  (3.30).
Therefore, remains constant along an integral curve of Charpits
Equations in (,,,,)− space .If =0 is satisfied at an integral point
=0,=0 everywhere along the solution of Charpit’s equations. munotes.in
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The initial value for a solution of Charpit’s equations can be prescribed by
specifying ,,,, on the four dimensional surface in
(,,,,)−space. Therefore, the system of Charpit’s equations define
a four parameter family of strips. From this four parameter family we
choose a three parameter sub family of strips by imposing the condition
that =0 at =0. Which implies =0along these strips. We call this
three parameter sub family of strips Monge strips and the projection on
(,)plane of the corresponding sp ace curves in (,,)space,
characteristic curve .
We shall show that if a Monge strip, say M has one element (i.e. the values
of (),(),(),(),(), , =0 common with an
integral surface :(,), then the strip belong entirely to the integral
surface. let us suppose that at the point , the integral surface and the
strip has common values of (,,,,). Since is an integral surface,
we can find a unique Monge curve on through . This together with
and at points on this curve, gives a Monge strip ′ on .Since both strip
and ′ satisfy Charpit’s equations (3.26) – (3.30) with the same initial
condition at , it follows from the uniqueness theorem of solution of
ordina ry differential equation that and ’ are the same. As ′ belongs
entirely to the integral surface, the result follows.
EXERCISE 3.1
1. Show that the Monge cone of equation = is an open cone which
is generated by a one parameter family of straig ht lines whose one end
is fixed but the other and moves on a parabola.
2. Consider the partial differential equation ≡(+)−1=0
Fig 1.6
munotes.in
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(i) Show that the general solution of the Charpit equation is a four
parameter family of strips represented by
=+2
3(2)
cos,=+2
3(2)
sin
=2,=cos
√2 ,=sin
√2
Where ,, are the parameters.
(ii) Find the three parameter subfamily representing the totalit y of all
Monge strips.
(iii)Show that all characteristic curves consist of all straight line in the
(,)plane.
1.4.2 Solution of a Cauchy Problem
If there exits an integral surface passing through a space curve Γ:
=(),=(),=(); (3.33)
The first order partial derivatives =() and =(), evaluated
from the equation of integral surface at the point of Γ, satisfy the equation
(3.1), i.e. (),(),(),(),()=0 (3.34)
Moreover since ()=((),()), differentiating with respect to
, we find the strip condition with respect to :
()=()()+()′()=0 (3.35)
is satisfied at every point of Γ. Therefore, irrespective of choice of , we
can now solve for () and () from (3.34) and (3.35) to get an initial
strip
=(),=(),=(),=(),=() (3.36)
We solve the Charpit’s equations (3.26) (3.30) with initial values of
,,, and at =0 given by (3.36) and get the Monge strips starting
from the various points of Γ. Since , satisfy the strip condition (3.35)
with respect to , these Monge strips smoothly join to form a surface. Due
to (3.34), F is identically zero along each Monge strip, hence the surface
thus generated is integral surface of (3.1) passing through Γ. We note that
there can be more than one integral surface passing through Γ, since there
can more than one pair of function (),() satisfying the equations
(3.34) and (3.35). However, once a set of values and ae selected, we
expect to get a unique solution of Cauchy problem. In order that the
solution exists and unique, it will be necessary to impose some restriction
on the initial curve Γ.
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Theorem 3.1: suppose the function (,,,,)∈() where is a
domai n in (x, y, u, p, q) space. Further suppose that along a datum curve
=(),=() on ={:0≤≤1} the initial value =()
are assigned. Let the function (),(),() belong to (); the
functions (),(), satisfying two equations (3.34) and (3.35), belongs
to () and the set (),(),(),(),()∈ for ∈
and satisfies
(,,,,)−
(,,,,)≠0 (3.37)
Then we can find a domain in (,)plane containing the datum curve
and a unique solution in :
=(,) (3.38)
such that for ∈
(),()=() (3.39)
(),()=()and(),()=() (3.40)
Proof: since the function appearing on a right h and side of the Charpit’s
equation (3.26) (3.30) belong to () and
(),(),(),(),()
are (), there exists a unique solution of the Charpit’s equation with
initial condition (,,,,)=((),(),(),(),()) at
=0:
=(,),=(,),=(,),=(,),=(,)
(3.41) whose partial derivative with respect to and exists and
continuous.
From (3.26), (3.27) a nd (3.37) it follows that
(,)
( ,)( =0)=
(,,,,)−
(,,,,)≠
0 (3.42)
Therefore, there exists a neighbourhood (,) of a point
((),()) on the datum curve in (,)plane (corresponding to
=0, such that in (,) we can solve the first two equations of
(3.41) uniquely in the form
=(,),=(,) (3.43)
Substituting (3.43) in the expressions of , and in (3.41) we get
=(,),(,)≡(,) (3.44)
=(,),(,)≡(,) (3.45) munotes.in
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=(,),(,)≡(,) (3.46) which are continuously
differential differentiable function of and . we shall now show that
(3.44) is the solution the Cauchy problem. It is obvious that on the datum
curve =0, the function (3.44) takes the prescribed value (). Further,
on the family of Monge strip (3.41), (,,,,) has a constant value
(,,,,)which is zero i.e.
,,(,),(,),(,)=0 ∈(,) (3.47)
Therefore, the function (,) in (3.44) is a solution of the differential
equation (3.1) provided, we can show that
(,)=(,),(,)=(,) (3.48)
Consider the function
(,)=−− (3.49)
whose value, (0,), on the datum curve is zero. Differential (3.49) with
respect to
=−−−−
=
(−−)++−−
=0+++(+)+(+)
where we have used the Charpit’s equation in the result. Adding and
subtracting we get,
=++++−−−+
=−
Since identically zero along each of the Monge strips (3.14), ≡0.
The funct ion now satisfies the following linear homogeneous ordinary
differential equation
=−(,) (3.50) with solution
=(0,)exp−∫(,)
(3.51)
Since (0,)=0,(,)=0 for all values of (,) such that (,)∈
(,)
Therefore,
=+ (3.52) munotes.in
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From the Charpit’s Equation, we also have
=+ (3.53)
From (3.44) we get,
=+=(+)+(+)
=+++
=
+
=.1+.0=(,)=(,) (3.54) where we have
used the expressions of and from the first two equations (3.41).
similarly, we can show that
=(,) (3.55)
Therefore from (3.47) it foll ows that (,) given by (3.44) is a solution
of the differential equation (3.1), in the domain (,).
To prove the uniqueness of the solution, let us assume that is another
integral surface represented by the solution =′(,) of the Cauchy’s
problem. The surface ′can be covered by the family of Monge strips after
solving (3.16) and (3.17) with replaced by ′. These Monge strips
satisfy the same initial condition at their point of intersection with the
initial curve Γ, as the strips (3.14). from the uniqueness theorem for a
solution of the Charpit’s ordinary differential equations, it follows that this
family of Monge strips on the integral surface ′ must be the same as the
strips (3.41). Therefore, the integral sur face coincide with ′, i.e.=′
in (,)
Example 3.1: Consider the equation
+=1 (3.56)
And straight line in (,) plane.
=≡sincos,=≡sinsin (3.57)
On which prescribed by
=≡cos (3.58)
where and are constants.
The Monge cone at (,,) is the envelope of the planes
(−)cos+(−)sin−(−)=0
The Monge cone is therefore represented by the equation
(−)+(−)=(−) munotes.in
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which gives a right circular cone with vertex at (,,), axis parallel to
axis and semi vertical angle
.
For the initial strip we have to solve the equations
+=1 (3.59)
and
sincos+sinsin=cos (3.60)
If <
, the equations (3.59) and (3.60) do not possess a real solution for
and showing that the solution of Cauchy problem does not exist. This
can be explained from the fact that the space curve given by (3.57) and
(3.58) through which the integral surface should pass, lies in the interior of
the Monge cone at the origin. Naturally it is not possible for an integral
surface to touch the Monge cone along a generator of the cone and also to
pass through a line within it.
For \4<</2, we get two sets of values of and
=cotcos±sin(1−cot)/ (3.61)
=cotsin∓cos(1−cot)/ (3.62) which
is independent of .
The Charpit’s equations are
=2 ,
=2
Fig 1.7
munotes.in
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=2(+)=2(1)=2, using (3.56)
=0and
=0
Solving thesewith the initial values (3.57) , (3.58), (3.61) and (3.62), we
get
=2+sincos,=2+sinsin,=2+ ,
=,= (3.63)
Eliminating and from (3.63) we get the two solution of Cauchy
problem corresponding to the two sets of values of and .
=cot(cos+sin)±1−cot (sin−cos)
(3.64)
They represent two planes which pass through the initial line Γ and touch
the Monge cones along two generators.
1.4.3 Solution of a Characteristic Cauchy Problem
We have seen that when the condition (3.37) is satisfied, i.e. when the data
is such that datum curve in (,)plane is nowhere tangential to the
characteristic curve for a possible solution of the Cauchy problem exists
and unique. However, when ()−()=0 hold everywhere
along and the initial manifold :((),(),(),(),())
belongs to the integral surface , then following the arguments of §3.1 for
the derivation of Charpit’s equation (3.26) – (3.30) we can show that the
strip must be a Monge strip on with the parameter replaced by .
Hence in exception case, −=0, a necessary condition for the
existence of a solution of the Cauchy problem is that the initial strip is a
Monge strip. This condition is also sufficient. In fact, if this condition is
satisfied, there exist not only one but an infinite number of solutions of the
characteristic Cauchy problem.
If −=0 and the initial strip is not a Monge strip, then it
follows from above that there exists no solution of the Cauchy problem
having continuous derivatives up to second order in the neighbourhood of
the datum curve.
EXERCISE 3.2
1. Solve the Cauchy problems:
(i)
(+)= with Cauchy data prescri bed on the circle
+=1 by (cos,sin)=1,0≤≤2
(ii) ++−
−
−=0, with Cauchy data
prescribed on axis by (,0)=0 munotes.in
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(iii) 2−=0, with Cauchy data prescribed on axis by
(0,)=
(iv) 2+−=0, with Cauchy data (,1)=−
2. Consider two parameter family of functions =(,,,)
where is a known functions of its arguments and , are
parameters. If the rank of the matrix
is 2, show
that the result of elimination of and from the relation
(,,,)=,(,,,)=,
(,,,)= leads to a first non linear equation
,,,,=0
3. Two firs t order partial differential equations are said to be
compatible, if they have a common solution. Show that the necessary and
sufficient condition for two equations (,,, , )=0 and
(,,, , )=0 to be compatible is that
(,)
(,)+(,)
(,)+(,)
(,)+ (,)
(,)=0 is satisfied either identically or as
a consequence of relations =0 and =0.
1.5 COMPLETE INTEGRAL
In problem 2 in Exercise 3.2 we saw that the result of elimination of two
arbitrary constants and from a relation
=(,,,) (4.1)
leads to a non linear equation
,,,,=0 (4.2)
We note that (4.1) satisfied (4.2) for all values of and .
We shall show that a solution of the form (4.1) and (4.2) is sufficiently
general in the sense of all other solution of this equation can be obtained
from it me rely by simple operation of differentiation and elimination of
the constants.
Definition: A two parameter family of solution (4.1) of the equation (4.2)
is called complete integral of the equation if the rank the matrix
is 2 in an appropriate domain of the variables ,,,.
The condition that the above matrix has rank 2 assures that the
function depends on two independent parameters and elimination of
and from (4.1) and
=(,,,),=(,,,) (4.3)
leads to equation (4.2).
Note 1: If and be combined into one parameter =(,), then two
rows of the matrix become linearly dependent and its rank becomes one. munotes.in
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2: If the rank is two, and can solved from (4.3) and these can be
substituted in (4.1).
1.5.1 Determination of complete integral
It is simple to determine a complete integral for a given partial differential
equation (4.2). the pro blem of Exercise 3.2 gives the condition for the
existence of a common solution of two equations ,,,,=0
and ,,,,=0. Once these two equations have a common
solution, we first solve them simultaneously f or and in terms of ,
and .
=ℎ(,,)and=(,,)
and then the differential relation
ℎ(,,)+(,,)= (4.4)
will possess an integrating factor and can be integrated giving a relation
between , and and an arbitrary constant . Therefore, a complete
integral of (4.2) can be determined if we can determine a compatible
equation (,,,, )=0 containing an arbitrary constant . But this is
simple since the result of problem 3.2 shows that any satisfying the
equation:
+
++
−(+)
−+
=0
(4.5)
would be a compatible equation.
This is the first order linear homogeneous partial differential equation for
in five independent variables ,,, and . For the equation (4.5), the
characteristic equations and compatibility conditions are
=
=
=
()=
() (4.6)
Since the compatibility condition implies that =constant on the
characteristic curve in (,,,,)space, it follows that if we can get any
first integral, say (,,,,)= of the characteristic equations, then
≡ (,,,,)−=0 is the required equation containing an
arbitrary constant and compatible with ,,,,=0
The character istic equations of (4.5) are nothing but the Charpit’s
equations (3.26)  (3.30) of the equation (4.2)
Example 4.2: Find the complete integral of the partial differential equation
+−4=0. (4.7)
Solution: Charpit’s equation for the given PDE,
2=
2=
2(+)=
−2=
−2
We take the relation
=
which gives
≡=constant =, say. (4.8) munotes.in
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Taking one of the value of and from (4.7) and (4.8) and substituting in
(4.4), we get
=
+√4−
Integrating this we get a complete int egral
=log+4−log+
Containing two arbitrary constants and .
Example 4. 3 Find the complete integral of the PDE =+++
.
Solution: (,,,,)=0
=+++−=0
=,=,=−1,=+2,=+2
Charpit’s equation for the given PDE,
−+=
−+=
(+2)+(+2) =
+2=
+2
Taking
=
,
=0,=0=>=,=
Complete integral =+++.
EXERCISE 1.1
1. Show that the compete integral of
a) (,)=0, where involves only and and ,()=0 is
=+()+.
b) ≡(,)−(,)=0 is obtained by solving and from
(,)=,(,)= and integrating = + .
c) ≡−−−(,)=0 is =++(,)
2. If independent variable and do not appear in the equation
(,,)=0, then show that the complete integral can be obtained
by solving form (,,)=0, taking = and integrating
= + .
Note: These all are standard results and can be used to find complete
integral of any PDE satisfying the given condition.
1.5.2 Solution of a Cauchy Problem
Once we know a complete integral, we can find solution of the Cauchy
problem.
We are required to construct an integral surface of (4.2) passing through
an initial curve munotes.in
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Γ:=(),=(),=() (4.9)
At the point of intersection of Γ and any member o f (4.1) the parameter
satisfies
((),(),,)=() (4.10)
Differentiating both the sides with respect to
((),(),,)=′() (4.11)
Eliminating from thes e two equations we get a relation between and
. which is required integral surface.
Example 4.4: Solve the Cauchy problem
2+= (4.12)
with Cauchy data (,1)=−
Solution: Cauchy da ta can be put in the form
=()≡,=()=1,=()≡−
(4.13)
To derive a complete integral, the Charpit’s equations :
4=
=
4+=
−2+=
0
Which gives a compatible equation
= (4.14)
Containing an arbitrary constant. From (4.12) and (4.26), we get
=
(4.15)
The complete integral is given by
= +
=−
2+ =>−
−=
√2=>−
=
2+
=>−−
−
=2 (4.16)
Substituting (4.13) in (4.16),
(++)=2 (4.17)
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Which after differentiating with respect to gives
2(++)=2 (4.18)
Eliminating from (4.17) and (4.18), we get
=−2
Substituting this value of in (4. 16), we get the solution of Cauchy
problem
=
().
EXERCISE 1.2
1. Use the method of complete integrals to solve the following Cauchy
problems:
i) 2−=0,(,1)=
ii) −=
(+),(,)=
−∞<<∞
iii) +=,(cos,sin)=1 0≤≤2
iv) =+++−2,(,)=2 −∞<<∞
2. Given any two complete integrals =(,,,),=(,,,)
of a first order partial differential equation, show that one complete
integral can be derived from the other.
3. Find the complete integral of 4(+)(−−)=1.
1.6 LET US SUM UP:
In this unit two main types of partial differential equation s, semilinear and
quasi linear, out of four are discussed. Cauchy problem a nd its
characteristics is discussed. General solution can be determined. We have
also discussed Monge curve and Charpit’s equation. Solution of
characteristic Cauchy problem and complete integral is discussed.
1.7 REFERENCES
1. Garabedian P.R. (1964). Part ial Differential Equations, John Wiley
& Sons.
2. Gelfand I. M. (1962) some problem in the theory of quasilinear
equations.
3. Copson E.T. (1975). Partial differential Equations, Cambridge
University Press.
4. Courant R. and Hilbert D. (1962). Method of Mathemati cal
Physics, Vol II, Partial differential Equations, Interscience.
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1.8 BIBLIOGRAPHY
1. Sneddon I.N. (1957) Elements of Partial Differential Equations,
McGraw Hill.
2. Smith M.G. (1967). Introduction of theory of Partial Differential
Equations, Van Nostrand .
3. Lieberstein H.M. (1972). Theory of Partial Differential Equations,
Academic Press.
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2
SECOND ORDER PARTIAL
DIFFERENTIAL EQUATIONS
Unit structure
2.0 Objectives
2.1 Introduction
2.2 Classification of second order partial differential equation in two
independent variables.
2.3 Classification of partial differential equation in more than two
independent variables
2.4.1 The Cauchy problem
2.4.2 The Solution of Cauchy’s Problem
2.5 Method of reduction to normal form
2.6 Potential Theory and Elliptical differential Equation.
2.7 Harmonic function
2.8 Poisson’s formula
2.9 Let's sum up
2.10 List of references
2.11 Bibliography
2.1 OBJECTIVE
After doing this unit, you will be able to:
Classify the 2nd order PDE in two variables into hyperbola,
parabola and ellipse
Classify the 2nd order PDE in more than two variables
Find the characteristics equations of all three type s of PDE .
To solve the Cauchy’s problem .
To reduce the 2nd order PDE in its normal form.
To find the potential equations .
To study about harmonic function .
To derive Poisson’s integral formula.
To learn about Maximum  minimum properties
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2.1 INTRODUCTIO N
We have studies in previous chapter about first order partial differential
equation and its types, Cauchy equation and how to find the general
solution of this. Now we will learn about second order partial differential
equation in two variables and its classification also study the partial
differential equation more than two variable.
2.2 CLASSIFICATION OF SECOND ORDER PARTIAL
DIFFERENTIAL EQUATION IN TWO
INDEPENDENT VARIABLES.
Consider a general partial differential equation of second order for a
funct ion of two independent variables and in the form:
+++(,,,,)=0 (1.1)
Where ,, are continuous functions of and only possessing partial
derivatives defined in some domain D on −plane.
And =
,=
,=
, Then (1.1) is said to be
(i) Hyperbolic at a point (,) in domain D if
−4>0
(ii) Parabolic at a point (,) in domain D if
−4=0
(iii) Elliptic at a point (,) in domain D if
−4<0
Note that the type of (1.1) is determine solely by its principal part
(++, which involves the higher order derivative of ) and that
the type will generally change with the position in the −plane unless
,, are constants.
Remark: some authors use in place of . Then we will have
=
,=
, =
Examples:
i) Consider the one dimensional wave equation
=
i.e.−=0
Sol. Comparing it with (1.1), here ,=1,=0,=−1
−4=0−4(1)(−1)=4>0, so the given equation is hyperbolic .
ii) Consider the one dimensional diffusion equation munotes.in
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iii) Consider the twodimensional Laplace equation
+
=0 ..+
=0
Sol. Comparing it with (1.1), here ,=1,=0,=1
−4=0−4(1)(1)=−4<0,so the given equation is elliptic .
Ex.2. classify the following partial differential equations.
i) 2
+4
−3
=2
Sol: =2,=4,=−3
−4=16−4∗2∗−3=16+24=40>0 =>hyperbolic
ii)
+4
+4
=0
=1,=4,=4,−4=16−4∗1∗4=0=> parabolic
iii) −(−)−+−=2(−)
Sol.: =,=−(−),=−
−4={−(−)}−4∗∗−,
={(−)}+4=+−2+4
=++2=(+)>0 =>hyperbolic
iv) (−1)−(−1)+(−1)++=0
Sol.: (−1)−4(−1)∗(−1)
=(−1)−4(−1)
=(−1)(+1)−4(−1)
=(−1){(+1)−4}
=(−1){+1+2−4}
=(−1){+1−2}
=(−1)(−1)=(−1)>0=>ℎ
2.3 CLASSIFICATION OF SECOND ORDER PARTIAL
DIFFERENTIAL EQUATION IN THREE
INDEPENDENT VARIABLES.
A linear partial differential equation of the second order in three
independent variables , , is given by, munotes.in
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∑ ∑
+∑
+=0 (1.2)
Where =, are constant or some functions of independent
variables ,,and is thede pendent variable.
Since = ,=× is real and symmetric of order 3×3. The
eigen values of matrix are roots of the characte ristic equation of ,
namely −=0.
With the help of matrix , (1.2) is classified as follows
i) If the eigen values od are non zero and have same sign,
except precisely one of them then (1.2) is known as hyperbolic
type of equation.
ii) If =0,i.e. anyone of the eigen value of A is zero, then (1.2)
is known as parabolic type of equation.
iii) If all the eigen values of A are non zero and of the same sign,
then (1.2) is known as elliptic type of equation.
The matrix can be remembered as
= . . .
. . .
. . .
Ex.1. ℎ +=
Sol.: +−=0
=10 0
01 0
00−1
Characteristic equation of ,
−()+(++)−=0
−+(−1−1+1)+1=0
=>−−+1=0
=>(−1)−1(−1)=0
(−1)(−1)=0
=>( −1)(−1)(+1)=0
=>=1,1,−1
It is showing that all eigen values are non zero and have the same sign
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Ex.2.++++=0
Sol.: The given equation can be rewritten as +0.+0.+
0.+++0.++=0
The matrix A of the given equation is as follows
=100
011
011
Determinant o f A, =0
=>given equation is parabolic type.
Ex.3. Classify +2+=2+2
Sol.: the given can be rewritten as
+2+−−−−=0
=1 −1 0
−1 2 −1
0 −1 1,
=1(2−1)+1(−1−0)+0=1−1=0
=>given equation is parabolic type
Ex4.: classify the following equations.
i) ++=0
ii) +=
iii) 3+3+4+8+4=0
Sol.: Try yourself.
Ans.: i) elliptic ii) parabolic iii) hyperbolic
2.4.1 THE CAUCHY’S PROBLEM
We start with the general quasilinear second order equation for a function
(,) of two independent variables:
+2 + =(1)
where ,,, depend on ,,,,. The Cauchy problem consists in
finding a solution of ( 2.4.1 ) with given values of and its normal
derivative on a curve in the (,) plane.
Let the parametric representation of be: =(),=(),∈,
where is aninterval on the real line. We are given two functions () munotes.in
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and (),∈ The Cauchy problem consists in finding a solution
(,) of (2.4.1 ) which satisfies the following conditions:
(),()=(),∈and
(),()=(),∈(2)
where
denotes a normal derivative to .
For discussion of the Cauchy problem here, we assume that ,, and
are analyticfunctions, regular in some domain . Our aim is to examine
whether there exists a uniqu e analy tic solution of ( 2.4.1 ), which takes
given values on C. To do so, we formally construct a solution using a
Taylor's series expansion about any point of . The fir st step i n such a
solution is to show that the partial derivatives of of all orders are
uniqu ely determined at every point of Let suffix 0 denote the values of
partial derivatives of at point of i.e.(),()=(), and so
on. Then () and ()satisfy the following linear equations:
()+()=′()
and −()+()=+∗() (3)
where a prime ( ′ )denotes d ifferentiation with respect to . Except at
points where and vanish simultaneously and can be
determined uniquely .
Regarding second order derivatives, namely, () and () and
() they can be determined as solutions of the lin ear equations:
()+2()+()=
()()+()()={()}′
()()+()()={()}′ (4)
These equations determine (),() () uniquely
provided the determinant of the coefficient matrix is nonzero. This
requires that
−2+≠0
Or(−,)≠0 (5)
where is the characteristic quadratic form .Further we can show that the
derivatives of of all orders can be uniquely determined at points of
provided
(−,)≠0
In this way we can formally de velop a unique Taylor's series expansion
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conditions on C. The difficulty is to show that such an expansion is
convergent in some region around C. The Cauchy Kowalewski method
(see Garab edian, 1964) provides a majorant series ensuring convergence.
On the other hand, if (−,)=0, then the partial derivatives of on
the curve C cannot be determined uniquely. The exceptional curves C, on
which if and its normal derivati ve are prescribed, no unique solution of
(1) can be found satisfying these conditions, are called characteristic
curves. These curves satisfy the homogencous equation
(−,)=0.
If the curve :=(),=() in the (,) plane is given by the
equation (,)=constant.
By eliminating , then satisfied the PDE ,=
0 on(,)=constant (6)
Since −
=
=−
From the result it follows that there are two distinct families of
characteristic curves satisfying equation (6), if the equation is hyperbolic.
There are precisely (,)=constant and (,)=constant, and are
referred as characteristic variable or coordinates.
For the hyperbolic equatio n in its normal form, namely,
+,,,,=0 (7)
(,)=constant and (,)=constan are the characterist ic curves. If,
for exampl e in the Cauchy problem andare proscribed on a
charact eristic carv e. C: =constant, then we cannot determine
uniquely on = constant from the given equation (since the coefficient of
is zero in the linear second order equation ( 7)). Since and are
prescribed on =constant as () and (), say, respectively, and
can be computed on =constant and the equation ( 7) will reduce to the
compatibility condition
()+(,,,,)=0
on =constant. Compatibility conditions to be satisfied on charactoristic
curves are typical, as tho equation gives no additional information in this
case (like the value of use in equation (7)), but merely insists on a relation
between already known qu antities. If the compatibility condition is
satisfied there will be an infinity of solutions of the Cauchy problem
(choosing arbitrarily in (7)), or else there will be no solution. The
above discussion holds for data prescribed on = constant a s well. For a
hyperbolic equation, we have two compatibility conditions, one each on
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equation, we have one compatibility condition on th e single family of
characteristic curves.
In the canonical elliptic form, and can always be determined
Whenever and its normal derivative are prescribed on any curve in the
(,)−plane, since ,≠0 on any real curve (,)= constant.
We can always find a uniq ue solution for the. Cauchy problem in this
case.
In the case of independent variables, those surfaces
(,,…….)=0, on which, when the function and its normal
derivative are prescribed, no unique solution, exists satisfying the
prescribed conditions, are called characteristic surfaces. Following a
similar process, as in the case of two independent variables, it follows that
satisfies the equation ,namely
()==0 =0
The characteristic condition ()=0 is required to be satisfied on =0
but this does not require that satisfies the equation ()=
0identically.
EXERCISE 2.1
1. Let (,) satisfy the equation
−2++3−+1=0
in a region of t he (,) plane. Classify the equation and find its
characteristics. Construct a solution, if it is exists, for each of the
Cauchy data:
(i) =2,=0 on the line =0
(ii) =2,=0 on the line +=0
2.4.2 THE SOLUTION OF CAUCHY’S PROBLEM
Consi der the second order partial differential equation
+++(,,,,)=0(1)
In which ,, are functions of only. The Cauchy problems
consists of the problem of determining the solution of (1) such that on a
given space curv e it takes on prescribed value of
, where is
the distance measured along the normal to curve.
As an example of Cauchy’s problem for second order partial differential
equation, consider the following problem.
To determine the solution of
=
which of the following data
prescribed on the x axis. (,0)=(),(,0)=(). Observe that y 
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Characteristic equation and characte ristic curves :
Corresponding to equation (1), consider the −quadratic
++=0 (2)
Where −4≥0, (2) has real roots. Then the ordinary differential
equations
+(,)=0 (3)
are called the characteristic equ ations.
The solution of (3) are known as characteristic curves or simply the
characteristics of the second order partial differential equation (1).
Now consider the following cases:
Case 1: if −4>0 (i.e. if (1) is hyperboli c), then equation (2) has
two distinct real roots say so that we have two c haracteristic
equations
+(,)=0,
+(,)=0
Solving these we get two distinct families of characteristic curves.
Case2 . if −4=0 (i.e.if (1) parabolic),then equation (2) has two
equal real roots each , so that we have only one characteristic equation
+(,)=0
Solving these we get only one family of characteristic curve.
Case 3. if −4<0 (i.e. if (1) is elliptic), then equation (2) has no
real roots i.e., two complex roots. Hence there are no real characteristics.
Thus, we get two distinct families of complex characteristic curves when
(1) is elliptic.
Ex.1. Find the characteristics of −=0
Sol:=,=0,=−
−4=0−4()(−)=4>0 , hence the given equation is
hyperbolic everywhere except on the co ordinate axes =0 and =0.
The −quadratic is ++=0
i.e.,−=0
=>=
=>=±
are two distinct roots. Corresponding
characteristics equation are munotes.in
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+
=0
−
=0
=>
=−
and
=
=> =− =
Integrating, +=and +=, which are required family o f
Characteristic Curves .
Ex.2. Find the characteristics of +2 +=0
Sol: =,=2,=
−4=4−4()()=0, hence the given equation is
hyperbolic.
The −quadratic is ++=0
=>+2 +=0
=>(+)=0=>=−
,−
−
=0=>
=
=>log=log+log=>= this is
require d family of Characteristic Curves. Here it represents a family of
straight lines passing through the lines.
Ex.3. Find the characteristics of 4+5+++−2=0
Sol.: Try yourself. Ans.: −=,−
=
Ex.4. Find the characteristics of (sin)+(2cos)−=0
Sol.: Try yoursel f. Ans: + −cot = ,+ +
cot =
2.5 METHOD OF REDUCTION TO NORMAL FORM
Consider the second order partial dif ferential equation of the type
+++(,,,,)=0, where , , are continuous
function of x and y possessing continuous partial derivative of as high an
order as necessary. There is a certain method to solve different types of
PDE’ s which we are going to discuss in detail as follows.
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2.5.1 working rule for reducing a hyperbolic equation to its normal
form
Step1 . let the given equation +++(,,,,)=0, (1)
be hyperbolic so that −4>0
Step 2. Write −quadratic equation ++=0 (2)
Let be its two distinct roots.
Step 3. Then corresponding characteristic equations are
+=0 and
+=0.
Solving these, we get (,)= and (,)= (3)
Step4. We select , such that =(,) and =(,) (4)
Step5. Using relation (4), find ,,,, in terms of .
Step6. Substituting the value of ,,,, obtained in step 4 in equation
(1) and simplifying we shell get the following canonical form of (1):
=,,,
,
Ex.1. Write a canonical form of
−
=0
Sol: Re writing the given equation −=0  (1)
−4=0−4(1)(−1)=4>0=>hyperbolic
+0∗−1=0=>=1=>=±1
Characteristic equation
+1=0=>+=0=>+=
And
−1=0=>−=0=>−=
Let =+,=−
Jacobian form of u and v =
=1 1
−11=1+1=2≠ 0
=>u and v are independent function.
=
=
∗
+
∗
=
∗1+
∗−1=
−
=>
=
−
=>
=
−
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=
=
∗
+
∗
=
+
=>
=
+
=>
=
+
=
=
=
−
−
=
−
−
+
=
−2
+
=
=
=
+
+
=
+2
+
Using these values in eq. (1) the required canonical form is
−2
+
−
+2
+
=0
−2
+
−
−2
−
=0
=>−4
=0=>
=0which is required equation.
Ex.2.−+−1+
+
=0  (1)
Sol:−4=(−1)−0=1>0=> this is hyperbolic.
0∗−+1=0=>=1 corresponding ch eq. +=
=+,= (2)
Jacobian form of u and v =
=11
10=−1≠0
v and u are indepen dent functions.
=
=
∗
+
∗
=
∗1+
∗1
=>
=
+
=>
=
+
 (3)
=
=
∗
+
∗
=
=>
=
=>
=
 (4) munotes.in
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=
=
=
+
=
+
 (5)
=
=
=
(
)=
 (5)
Using the values ,, , in eq. (1)
−
−
+
+
−
1+1
+
=0
=
+
−
−1
+
=
−
+
which is required equation .
2.5.2 working rule for reducing a parabolic equation to its normal
form
Step1. let the given equation +++(,,,,)=0,
(1)
be hyperbolic so that −4=0
Step 2. Write −quadratic equation ++=0 (2)
Let be its root.
Step 3. Then corresponding characteristic equation is
+=0.
Solving th is, we get (,)= (3)
Step4. We select , such that =(,) and =(,) (4)
Where (,) is an arbitrary function of x and y and is independent of
(,). for this verify that Jacobian J of and given by (4) is n on –
zero.
=(,)
(,)=
=
∗
−
∗
≠0 (5)
Step5. Using relation (4), find ,,,, in terms of and .
Step6. Substit uting the value of ,,,, obtained in step 4 in equation
(1) and simplifying we get the following canonical form of (1) .
=,,,
,
=,,,
,
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Ex.1. Reduce +2+=0 to its canonical form.
Sol: +2+=0−−−−−−−−(1)
−4=4−4=0, the given equation is parabolic.
1∗+2+1=0=>(+1)=0=>=−1,−1
−1=0=>−=
Let =−,=  (2)
Jacobian form o f u and v =
=−11
0 1=−1≠0 u and v are
independent.
=
=
∗
+
∗
=−
=−
=>
=−
 (3)
=
=
∗
+
∗
=
∗1+
∗1
=>
=
+
=>
=
+
 (4)
=
=
=−
−
=
 (5)
=
=
=−
+
=−
−
(6)
=
=
=
+
+
=
+2
+
 (7)
Using the values ,, in eq uation (1), we get the required equation,
−2
−2
+
+2
+
=0
=>
=0
Ex.2. −2 +−
−
=0
Sol:−2 +−
−
=0 (1)
−4=4−4=0 => this equation is parabolic.
The −quadratic equation ++=0 reduces to munotes.in
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−2 +=0=>(−)=0=>=
,
The corresponding characteristic equation is
+
=0=>
+
=
Let =
+
,=
−
(2)
Jacobian form of u and v =
=
−=−−=−2≠
0
=
=
∗
+
∗
=
+
(3)
=
=
∗
+
∗
=
−
(4)
=
=
=
+
=
+
+
+
+
+
=
+
+
+2
+
using (2) (5)
=
=
−
=
−
+
−
, by (4)
=
−
+
−
+
−
=
−
+
−2
+
(6)
And =
=
=
−
=
−
+
−
=
−
(7)
Using (3), (4),(5), (6) and (7) in (1) and simplifying, we get
4
=0 so that
=0, which is the required canonical form.
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2.5.3 working rule for reducing elliptic equation to its normal form
Step 1. +++(,,,,)=0 (1)
be elliptic so that −4<0
Step.2. write −quadratic ++=0  (2)
Then and be two distinct roots.
Step 3. Then the corresponding characteristic equations are
+=
0
+=0
Solving these we get (,)+(,)=and(,)−(,)=
 (3)
Step 4 .We select u and v such that
=(,)+(,) and =(,)−(,) (4)
Step 5. Let are new real inde pendent variables such as =+
=−
Where =(,),=(,)
Step 6. Find values ,,,, in terms of and .
Step 7. Substitution the values of ,,,, in equation (1) and
simplifying we sh all get the following canonical form,
+
=,,,
,
Ex.1. Reduce the PDE in canonical form +=0
Sol:+=0  (1)
−4=0−4=−4<0 => which is ellipti c equation.
++=0=>+=0=>=±
Characteristic equation
+=0=>+
=
and
−=0=>−
2=
Let = +
and =−
Choose =,=
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Jacobia n of ,=
=01
0=−≠0=>α and β are
independent.
=
=
∗
+
∗
=
∗0+
∗=>
=
=
=
∗
+
∗
=
∗1+
∗0=>
=
=>
=
=
=
=
=
∗
+
=
+
=
+
∗
+
∗
=
+
∗0+
∗
=
+
=
+
=
=
=
=
Using these values of r and t in eq uation (1)
+=0=>
+
+
=0=>
+
+
=0
+
=−1
=>
+
=−
, which is required canonical form of given
equation .
Ex.2. find the canonical for of +−=0
Sol: +−=0  (1)
−4=0−4=−4<0 => given equation is elliptic.
++=0=>+=0=>=± munotes.in
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+=0=>
+ =0=> +=
and
−=0=> −=
Let = +,= −
Let =log,=
=
=
∗
+
∗
=
∗0+
∗1=
=>
=
=
>
=
=
=
∗
+
∗
=
∗1
+
∗0=1
=>
=1
=
=
=
=
=
=
=
1
=
1
+1
=−1
+1
−1
+1
1
+
∗0
+
=−1
+1
1
=−1
+1
Using these values in eq uation( 1),
+−=0=>
+−1
+1
−=0
−
+
−=0
=>
+
=
+
Ex.3. If the reduced canonical form is
=0 find its solution.
Sol.: Integrating w. r. t. u ,
=(),=arbitrary function.
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Integrating w. r. t. v, =∫()+(),where f is arbitrary function .
=()+()
=>=(−)+(+) it is the required general solution.
EXERCISE 2.2
Reduc e the following equation into their normal forms.
Q.1.−+−=
Q.2. +2+=0
Q.3. −4+4=0
Q.4. +=
2.6 POTENTIAL THEORY AND ELLIPTICAL
DIFFERENTIAL EQUATION:
Boundary data rather than i nitial data serve to fix properly the solution of
an elliptic differential equation. It is usually necessary to find an answer
"in the large," namely in the domain bounded by a closed boundary, and
this need for "global" constructions, rather than "local" treatment makes it
especially difficult to study nonlinear elliptic equations. We shall restrict
ourselves mainly to the linear potential equation or Laplace's equation in
$m$space variables. The boundary value problems of potential theory are
suggested b y physical phenomena from such varied field as electrostatics,
steady heat conduction and incompressible fluid flow.
Boundary Value Problems and Cauchy Problem
The general linear homogeneous second order partial differential equation
in −space variables ,,,……….. is
≡++=0,,=1,2,…… (1)
where the coefficients , and are continuous functions of the
independent variables ,,,……….. and=. Equation (1)
is said to be elliptic in a domain of −dimensional space, when the
quadratic form
()= (2)
can be express ed as the sum of squares with coefficients of the same sign,
or equivalently, () is either positive or negative definite in . The
simplest case is that of the Laplace equation or potential equation:
Δ= =0 (3)
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We shall first state three boundary value problems associated with Laplace
equation and then consider the Cauchy problem. Let be a domain in
(,,,………..)−space bounded by a piecewise smooth boundary
.Let continuous boundary values be prescribed on , by means of a
function .
1st order BVP (Dirichlet problem): The first boundary value problem,
also called the D irichlet problem, requires a solution of the Laplace
equation (3) in the domain , which is continuous in + and
coincides with on i.e.
= (4)
2nd order BVP (Neumann problem): It require s the determination of
solution in the domain , which is continuous with first order partial
derivatives in +, such that the normal derivative
of on takes
prescribed values , i.e. harmonic function (,)satisfies
=on (5)
where
is the directional derivative along outward normal and must
have a continuously normal.
3rd order BVP (Robin Problem): It is a modification of the first two
BVP where the solution (,) is a linear comb ination of
, takes
prescribed value of i.e.
+= (6)
where is a constant .
Before we discuss the Cauchy problem, we shall examine, in general, the
requirements to be satisfied by' a reasonable mathematical problem. There
are two requirements:
1. Existence requirement: There is at least one satisfying the equation
and the given boundary/Cauchy data.
2. Uniqueness requirement: There is utmost one such .
If the ma thematical problem is to be also physically realistic an extra
requirement has to be satisfied:
3. Stability requirement: Small changes in the boundary or Cauchy data
result in small changes in the solution .
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The first two requirements ensure the exist ence and uniqueness of the
solution of a mathematical problem, while all three requirements ensure,
further, stability or continuous dependence on given data for a physical
problem. If the three requirements are satisfied by a problem, it is said
towell po sed.
The Cauchy Kowalewski theorem shows that the solution of an analytic
Cauchy problem for an elliptic equation exists and is unique. However, a
Cauchy problem for Laplace's equation is not always well posed.
Hadamard gave an example of a Cauchy proble m, which violates the
stability requirement. Consider the Laplace equation in two independent
variables , with the following initial conditions:
(a) (,0)=0,(,0)=0
(b) (,0)=0,(,0)=
(7)
A solution satisfying condition (a) is
(,)=0 (8)
A solution satisfying condition (b) is
(,)=
sinsinh (9)
For sufficiently large , the Cauchy or initial values (a) and (b) are
arbitrarily cluse, but the sulutions are not, since sinh behaves like
for large .
Having noted that a Cauchy problem could be illposed for an elliptic
equatio n, we shall concentrate our attention hereafter only on the three
boundary value problems mentioned earlier and show that they are really
wellposed.
2.7 HARMONIC FUNCTION
A function () is called harmonic function in , if ()∈in +
∈ and ∆=0 in .
In case of two or three variable, the general solution of potential equation
can easily be obtained .For =2,(=,= ),.. +=0 ,
this is the real and imaginary part of any analytic functi on of the complex
variable +. For =3,(=,=,=), consider an
arbitrary function (,) analytic in the complex variable for fixed real
. Then, for arbitrary values of , both the real and imaginary parts of the
function:
=(+cos+sin, ) (10) munotes.in
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of the real variable ,, are solution of the equation Δ=0. Further
solutions may now be obtained by superposition.
=∫(+cos+sin, )
(11)
If (, ) is a solution of Laplace ’s equation in the domain D of
(,)plane, the function .
(,)=
,
,=+, (12)
Also satisfies the potential eq uation and is in t he domain ′ obtain from D
by inversion w ith respect to unit circle.
In general, m dimension, if (,…..) satisfies potential equation in
a bounded domain D then
=
,
………..
,++⋯………+= (13)
also satisfies the potential eq uatio n and is regular in the region ′
obtained from by inversion with respect to m dimensional unit sphere .
Therefore, except for the , the harmonic character of the function is
invariant with respect to sphere. Besides, the harmonic property is retained
completely under rotations, translations and simple reflections across
planes.
2.8 POISSON’S FORMULA
Ex. Dirichlet problem for a circle in the , −plane.
Sol: let a circle C is given by =,=+
B
A
oo 00
Fig. and
are inverse point with respect to :=
A
W
O munotes.in
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The problem is to find (,) ..∆=0=>+=0,
ℎ =() .
= .
Let () be analytical function in the region enclosed by C s.t. the real
part of () = ()
Letbe a complex number in that region. The inverse point of w.r.t. to
C is
which lies outside C. According to Cauchy integral formula
()=
∫()
(1)
0=
∫()
 (2)
Equation (1) – equation (2)
()=1
2()
−−()
−
=1
21
(−)−
(−)()
=1
2(−)−(−)
(−)(−)()
=1
2−−+
−(+)++()
=1
2−(−)
−{(+)−−}()
()=
∫()
{()}()  (3)
We know that
Let =,==>==>==>= ,<
From eq.
(3) =
∫()
()
Taking real part
(,)=()
∫()
{ ()}
 (4) munotes.in
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Where =+,tan=
Eq. (4) is called Poisson’s integral formula in 2D.
Maximum pr inciple: suppose that (,) be harmonic in a bounded
domain and continuous in =∪ then (,) attains its
maximum on the boundary .
Minimum principle: suppose that (,) be harmonic in a bounded
domain and continuous in =∪ then (,) attains its
minimum on the boundary .
2.9 LET US SUM UP:
In this unit we have learnt to identify different 2nd order PDE, to find the
characteristic curve and to solve Cauchy’s problem. We also discussed to
how to reduce the PDE’s in its normal form.
2.10 REFERENCES
1. Garabedian P.R. (1964). Partial Differential Equations, John Wiley
& Sons.
2. Gelfand I. M. (1962) some problem in the theory of quasilinear
equations.
3. Copson E.T. (1975). Partial differen tial Equations, Cambridge
University Press.
4. Courant R. and Hilbert D. (1962). Method of Mathematical
Physics, Vol II, Partial differential Equations, Interscience.
2.11 BIBLIOGRAPHY
1. Sneddon I.N. (1957) Elements of Partial Differential Equations,
McGr awHill.
2. Smith M.G. (1967). Introduction of theory of Partial Differential
Equations, Van Nostrand.
3. Lieberstein H.M. (1972). Theory of Partial Differential Equations,
Academic Press.
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3
GREEN FUNCTIONS I
Unit Structure:
3.1 Objectives
3.2 Introduction
3.3 Singularity functions and the fundamental solution,
3.4 Green functions
3.5 Greens identities
3.6 Lets sum up
3.7 Unit End exercise
3.8 Reference
3.1 OBJECTIVES
After going through this chapter students will be able to:
Singularity functions.
The fundamental solution of Laplace equation.
Definition of Green functions using fundamental solution.
Green’s first identity.
Green’s second identity
3.2 INTRODUCTION
Singularity functions are used in the solution of differential equations in
which the known terms are non smooth in the independent variable. In
particular, these functions are particularly useful in the study of bars,
shafts, and beams subjected to non smooth loading, such as poi nt loading
and distributed loading, that exhibits finite jumps .
The method of Green’s functions is an important technique for solving
boundary value and, initial and boundary value problems for partial
differential equations. We shall learn Green’s function method for finding
the solutions of partial differential equations. This is accomplished by
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differential equations. These integral theorems will then be used to show
how BVP and IBVP ca n be solved in terms of appropriately defined
Green’s functions for these problems. More precisely, we shall study the
construction and use of Green’s functions for the Laplace, the Heat and
the Wave equations.
3.3 SINGULARITY FUNCTIONS AND THE
FUNDAMENTAL SOLUTION:
Singularity functions are discontinuous functions or their derivatives are
discontinuous. A singularity is a point at which a function does not
possess a derivative. In other words, a singularity function is
discontinuous at its singular points. Hence a function that is described by
polynomial in t is thus a singularity function. The commonly used
singularity functions are:
Step Function, Ramp Function, and Impulse Function.
Step Function : One of the most common singularity functions is the
Heaviside* step function H(x), defined as
()=0,
1,<0
>0 (1)
y
x = 1
0 x
Figure 1 The Heaviside step function H(x)
Note that the Heaviside function H(x) is undefined at x = 0 , although it is
sometimes taken to be equal to
. Clearly, the Heaviside function H(x− a)
is analogous to the function plotted in Fig. 1,only shifted so as undergo the
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Ramp Function: The integral of the Heaviside step function is the ramp
function written as <>. With Eq. (1) taken into account, the ramp
function is given by
<>=0,
,≤0
>0 (2)
y
0 x
Figure 2 The ramp function
It is easy to see that the ramp function can be raised to any positive power,
with
<>=0,
,≤0
>0 >1 (3)
While H(x) does not have a derivative in the usual sense of a smooth
function, such a derivative can be defined as what in mathematics is
termed a distribution from the limit of a sequence of continuous
approximations to the discontinuous step function, as shown in F ig.3 below
Y y
11/w
0 (aw) a 0 (a w) a x
(i) (ii)
Figure 3 Ramp approximation to the step function and its derivative
The function depicted in Fig. 3(i) may be expressed as
(−)=1
[<−+>−<−>] munotes.in
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and it becomes H(x− a) in the limit as ω → 0, that is,
(−)=<−>
Impulse Function ( Dirac delta function ):The derivative of (−)
(depicted in Fig. 3(ii) is, in accordance with
(−)=1
[(−+)−(−)]
and in the limit as ω → 0 it formally becomes (by the standard definition
of the derivative) the derivative of H(x− a), that is,
(−)=(−)
This limit is known as the Dirac delta function and is usually denoted
δ(x− a).
We now turn to studying Laplace’s equation
∆u = 0
and its inhomogeneous version, Poisson’s equation,
−∆u = f.
We say a function u satisfying Laplace’s equation is a harmonic function.
3.3.1 The Fundamental Solution of Laplace’s equation:
Consider Laplace’s equation inℝ,
∆u = 0 x ∈ ℝ .
Clearly, there is a lot of functions u which satisfy this equation. In
particular, any constant function is harmonic. In addition, any function of
the form ()=++⋯..+ for constants is also a
solution. Of course, we can list a number of others. Here, however, we are
interested in finding a particular solution of Laplace’s equation which will
allow us to solve Poisson’s equation.
Given the symmetric nature of Laplace’s equation, we look for a radial
solution. That is, we look for a harmonic function u on ℝ such that u(x)
= v(x). In addition, to being a natural choice due to the symmetry of
Laplace’s equation, radial solutions are natural to look for because they
reduce a PDE to a n ODE, which is generally easier to solve. Therefore, we
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If u(x) = v(x), then
=
 ′() ℎ ≠0
⟹=1
()−
 ′() +
 ′′() ℎ 
≠0
Therefore,
In ℝ the solutions () of the potential equation ∆=0, which depend
only on the distance =≠0 of a fixed point x from a fixed point a,
given by the equation
∆=−1
()+ ′′()
Letting =≠0, we see that u(x) = v(x), is a radial solution of
Laplace’s equation implies v satisfies
−1
()+ ′′()=0
Therefore,
′′()=−1
()
⟹ ()
()=
⟹log()=(−1) +
⟹()=
Therefore,
()= + =2
(−2)+ ≥3
From these calculations, we s ee that for any constants ,, the function
()=+ =2
()+ ≥3 (I) munotes.in
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for x ∈ℝ,≠0 is a solution of Laplace’s equation in ℝ − {0}. We
notice that the function u defined in (I) satisfies ∆u(x) = 0 for ≠0, but
at x = 0 , ∆u(0) is undefined.
Therefore, these solution exhibits so called characteristic singularity at
=0. We defined as
()=
⎩⎨⎧−1
2− =2
1
(−2)− ≥3
Where is the surface area of the unit sphere in ndimensions given by
=2()
for singularity function ∆u =0.
()has the property that ∈ and ∆S =0 for ≠, with the
singularity =.
For =3, () correspond physically to the gravitational potential at the
point x of a unit mass concentrated at the point a. Every solution of a
potential equation ∆u =0 in D of the form
(,)=()+(), ∈
Define the function ()as follows. For ≠0, let
()=
⎩⎨⎧−1
2 =2
1
(−2) ≥3
As we will show in the following claim, ()satisfies −∆()= . For
this reason, we call ()the fundamental solution of Laplace’s equation .
Theorem: For () satisfies −∆ ()= in the sense of
distributions. That is , for all ∈, −∫()∆()=(0).ℝ
Proof: Let ∅be the distribution associated with the fundamental
solution ∅. That is let ∅:→ℝ be define such that
(∅ , )= ()()
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for all ∈. Recall that the derivative of a distribution F is defined as the
distribution G such that
(,)=−(,)
for all ∈. Therefore, the distributional Laplacian of is defined as the
distribution ∆∅, such that
(∆∅,)=(∅,∆)
for all ∈. We will show that
(∅,∆)=−(,)=−(0)
and, therefore, (∆∅,)=−(0).
which means −∆ ()=in the sense of distributions. By definition,
(∅,∆)=()()
ℝ
Now, we would like to apply the divergence theorem, but has a
singularity at x = 0 . We get around this, by breaking up the integral into
two pieces: one piece consisting of the ball of radius δ about the origin,
B(0, δ) and the other piece consisting of the complement of this ball in ℝ
. Therefore, we have
(∅,∆)=()()
ℝ
= ()∆()
(,)+ ()∆()
ℝ(,)=+
We look first at term I. For n = 2 , term I is bounded as follows
− 1
2∆()
(,)≤∆
(,)
≤ 
(,)
≤log.
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For n ≥ 3, term I is bounded as follows,
− 1
(−2)∆()
(,)≤∆
(,)
≤()
=()
2
Therefore, as δ →0, I → 0. Next, we look at term J. Applying the
divergence theorem, we have
()∆()
ℝ(,)
= ∆()()
ℝ(,)
− ()
()()
(ℝ(,))
+ ()()
()
(ℝ(,))
=− ()
()()
(ℝ(,))+ ()()
()
(ℝ(,))
=+
Using the fact that ∆ ∅()=0 for ∈ℝ−(0,)
We first look at term . Now, by assumption, g ∈ D, and, therefore, g
vanishes at ∞. Consequently, we only need to cal culate the integral over
∂B(0, ∈) where the normal derivative ν is the outer normal to ℝ − B(0,
δ). By a straightforward calculation, we see that
∇∅()=−
().
The outer unit normal to ℝ − B(0, δ).onB(0, δ) is given by
=−

Therefore, the normal derivative of ∅ on B(0, δ) is given by
()
=−
().−
=1
() munotes.in
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Therefore, can be written as
− ()
()()
(,)=− 1
()()()
(,)
=− ()()
(,)
Now if g is a continuous function, then
−∫ ()()→−(0)(,)as →0.
Lastly, we look at term . Now using the fact that g vanishes as x →
+∞, we only need to integrate over ∂B(0, δ). Using the fact that g ∈ D,
and, therefore, infinitely differentiable, we have
()()
()=
(,)()
(,)()()
(,)
≤()()
(,)
Now first, for n = 2,
∫ ()()(,)=∫ ()(,) ≤ log.
Next, for n ≥ 3,
()()
(,)
=1
()≤
(,)()
(,)
=
()≤.
Therefore, we conclude that term is bounded in absolute value by
logfor=2
for≥3
Therefore, →0 as →0 munotes.in
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Combining these estimates, we see that
()∆()=lim
→++=−(0).
ℝ
Hence prove.
Theorem : Assume ∈(ℝ)and has compact support. Let
()=∅(−)()
ℝ
where ∅ is the fundamental solution of Laplace’s equation. Then
a) ∈(ℝ)
b) −∆= in ℝ.
Proof: a) By a change of variables, we write
()=∅(−)()
ℝ=∅()(−)
ℝ
Let =(0,….1,0,0…)
be the unit vector in ℝ with a 1 in the ith slot. Then
(+ℎ)−()
ℎ=∅()(+ℎ−)−(−)
ℎ
ℝ
Now ∈implies
()()
→
(−)asℎ→0
uniformly on ℝ . Therefore,
()=∅()
(−)
ℝ
Similarly,
()=∅()
(−)
ℝ
This function is continuous because the right hand side is continuous.
b) By the above calculations and theorem 1, we see that
∆()=∅()∆(−)
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3.4 GREEN FUNCTIONS:
We are interested in solving the following problem. Let Ω⊂ℝ be an
open, bounded subset of ℝ .
Consider −∆= ∈ Ω
= ∈ Ω (I)
Suppose we can solve the problem,
−∆(,)= ∈ Ω
(,)=0 ∈ Ω
for each ∈ Ω. Then, formally, we can say that for u a solution of (I)
()=()
=−∆(,)()
=(,)()−
()()()
Now, we do know that the fundamental solution of Laplace’s equation
∅() satisfies
−∆ ()=and−∆ (−)=
Recalling the definition of distributional derivative, we will start by
looking at
()=(−)∆()
We would like to integrate this term by parts. However, we know that
(−) has a singularity at y = x .
We already find the fundamental solution of Laplace’s equation . i.e.
()=
⎩⎨⎧−1
2 =2
1
(−2) ≥3
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Applying the divergence theorem, we have
(−)∆()
=− (−)
()()
((,))
+ (−)()
()
((,))
Using theorem 1 we conclude that for any ∈(Ω)
()=∫ (−)()
−()
()()()−∫(−
)∆() (III)
We would now like to use the representation above formula to solve (I )
We proceed as follows. For each ∈ Ω, we introduce a corrector function
ℎ()which satisfies the following boundary value problem ,
∆ℎ()=0 ∈Ω,
ℎ()=(−) ∈Ω (IV)
Now suppose we can find such a (smooth) f unction ℎ which satisfies
(IV). Then using the same analysis as above, we have
ℎ()∆()
=∆ℎ()()
−ℎ()
()()
+ℎ()()
()
Now using the fact that ℎis a solution of (IV ), we conclude tha t
∫(−)()
−()
()()+−∫ℎ()∆()=0
(V)
Now subtracting (V) from (III), we conclude that
()=(−)
−ℎ()
()()
−[(−)−ℎ()]∆()
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Let (,)=(−)−ℎ()Then, u can be written as
()=(,)
()()
−[(,)]∆()
Definition: We define this function G as the Green’s function for Ω. That
is, the Green’s function for a domain Ω ⊂ℝ is the function defined as
(,)=(−)−ℎ(),∈ Ω, ≠
Where is the fundamental solution of Laplace’s equation and for each x
∈ Ω, h x is a solution of (V ). We leave it as an exercise to ve rify that G(x,
y) satisfies (II ) in the sense of distributions.
Polar form of Green’s function :
In this case we want t o solve
∆=, lim
→((,)−(,) )=0
In general, solutions to ∆u = f behave like u ∼ A log r + B as r → ∞ . The
condition just ensures that B = 0 . Again we look for a Green’s fu nction of
the form =(−)=()so that in polar coordinates
1
()=0 ≠0,
lim
→(()− ′())=0
The general solution is
=log+
Where = 0, we get
1=
(,)= =2
Where B is the unit disk, so that =
. Thus the Green’s function
is(,)=
, and the solution to given equation is
()=−()
2
ℝ
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It is sometimes useful to write G in polar coordinates. Using the law of
cosines for the distance −, one gets
(,; ,)=1
4log(++2cos (−))
Example : Let ℝ be the upper half plane in ℝ . If ℝ ={(,)∈
ℝ: >0} than find Green’s function.
Solution: We need to find a corrector function ℎ for each x ∈ℝ , such
that
∆ℎ()=0 ∈ℝ ,
ℎ()=(−) ∈ℝ
Fix x ∈ℝ We know ∆(−)=0 for all ≠. Therefore, if we
choose ∉ Ω, then ∆(−)=0 for all ∈ Ω. Now, if we choose
=() appropriately ∉ Ω such that
(−)=(−)for ∈Ω, then ℎ()=(−()).
Recall that for n = 2,
(−)=−1
2 −
Consequently, for =(,)∈ℝ,we see that for all ∈ℝ .
−=(,0)−(,)=(,0)−(,−)=−
Where =(,−)is the reflection of x in the plane .
Therefore, ℎ()=−we have found a corrector function for ℝ,
Therefore, a Green’s function for the upper half plane is given by
(,)=(−)− (−)=
[−−−].
3.5 GREENS IDENTITIES:
Green’s identities provide the main energy estimates for the Laplace and
Poisson equations.
Green’s first identity:
First recall the Divergence Theorem:
Let be a bounded solid region with a piecewise boundary surface
. Let n be the unit outward normal vector on . Let f be any vector
field on =∪. Then munotes.in
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∇⃗ . =.
Where is the volume element in and is the surface element on
.
By integrating the identity
∇⃗.∇⃗ =∇⃗ .∇⃗ +∆
Over and applying the divergence theorem, we gets
= ∇⃗ .∇⃗ + ∆
Where
= . ∇⃗ is the directional derivative in the outward normal
direction.
This is Green’s first identity.
Green’s second identity:
Switch and in Green’s first identity, then subtract it from the original
form of the identity. The result is
(∆−∆)=
−
This is Green’s second identity. It is valid for any pair of function and .
Special boundary conditions can be imposed on the functions to make the
right hand side of these identity zero, so that
∆ = ∆
Definition: A boundary condition is called symmetric for the operator ∆
on if ∬
−
=0 for all pairs of functions and
that satisfy the boundary condtion.
Note: Dirichlet, Neumann, and Robin BCs are symmetric.
Example: Show that Green’s functions are symmetric.
Solution: To show that Green’s functions are symmetric, i.e.
For all ,∈Ω,≠ , (,)=(,) munotes.in
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Let ()=(.) and ()=(,)
Now we will show that ()=()⟹(,)=(,).
By definition of Green’s function
(,)=(−)−ℎ(),∈ Ω, ≠
Where ℎ() satisfies ∆ℎ()=0 ∈Ω,
ℎ()=(−) ∈Ω
Therefore, for ∈Ω
()=(.)=(−)−ℎ()=(−)−(−)=0
()=(.)=(−)−ℎ()=(−)−(−)=0
Further, ∆=0 for ≠ and ∆=0 for ≠.
Now v is smooth, except near z = x , while w is smooth, except near z = y .
Define the region =Ω−[(,)−(,)]for >0.
Our functions are smooth. Therefore, integration by parts as follows,
∆ =
∆ −
()+
()
Using the fact that ∆v = 0 = ∆w on , we conclude that
()=
()
Using the fact that v = 0 = w on ∂ Ω, we conclude that
−
()=
(,)
−
()
(,)
where denotes the inward point ing unit vector field on ∂B(x, ) ∪ ∂B(y,
). Now we claim that as →0, the left hand side converges to w(x),
while the right hand side converg es to v(y).
For the terms on the left hand side, we first look at ∫
()(,)
Now w is smooth near x. Therefore,
is bounded near ∂B(x, ).
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Therefore, on ∂B(x, ),()≈
()
(,)≤
(,) ()
(,)=
(,)→0 →0
Now ∫
()(,)=∫ ∅
(−)−
() ()(,)
First, using the fact that ℎ is smooth and w is smooth near x, we see that
ℎ
() ()
(,)≤ ()
(,)≤
Therefore , ∫
() ()(,)→0 →0.
For the other term, we see that
∅
(−)() ()=1
()(,)1
−() ()
(,)
=1
() () ()
(,)
() ()
(,)→() →0
Hence the left hand side converges to ().
Similarly, the right hand side converges to v(y).
Hence prove,
Proposition: A Green funct ion has the following property . In the case n =
2 we assume diam Ω < 1 . 0<(,)<(−), ,∈Ω ,≠.
Proof. Since (,)=(−)+∅(,)
andG(y, x) = 0 if y ∈ ∂Ω and x ∈ Ω we have for y ∈ ∂Ω
∅(,)=−(−)
From the definition of (−)it follows that ∅(,)< 0 if x∈ ∂Ω. Thus,
since ∆∅=0 in Ω, the maximum minimum principle implies that
∅(,)<0 for all y, x ∈ Ω. Consequently
(,)<(−), ,∈Ω ,≠ munotes.in
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It remains to show that
. 0<(,), ,∈Ω ,≠
Fix x ∈ Ω and let Bρ(y) be a ball such that Bρ(y) ⊂ Ω for all 0 < ρ < ρ 0.
There is a sufficiently small ρ0> 0 such that for each ρ, 0 < ρ < ρ 0,
G(x, y ) > 0 for all y ∈Bρ(y ), ≠
see property (iii) of a Green function.
Since ∆G(x,y ) = 0 in Ω \ Bρ(y )
G(x, y) > 0 if x ∈ ∂Bρ(y)
G(x, y) = 0 if x ∈ ∂Ω
it follows from the maximum minimum principle that
G(x, y) > 0 on Ω \ Bρ(y).
Hence prove.
Example: consider a sphere with center at origin and radius ‘a’ apply the
divergence theorem to the sphere and show that ∇
=−4().
where () is a Dirac del ta function.
Solution: Applying divergence theorem to
1
=∇1
∇.∇1
=∇1
.
Where is an outward drawn normal. If =(,,∅), then
=
+1
+∅1
∅ ,
Hence ∬∇
. =∬
=∬
=
×4=
−4
By properties of ∇
, Its integral over any sphere with center at the
origin is −4.
Hence we say that ∇
=−4(). munotes.in
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Theorem: If G is continuous and
has discontinuity at ‘r’. than sho w that
lim
→
=1.
Proof: Let V be a sphere with radius bounded by .
We already know that G satisfies ∇=(−).
Integrating both sides over the sphere V, we get
∇ =1
Which can be written as lim
→∭ ∇ =1
Applying divergence theorem we get,
lim
→
=1.
Hence prove.
3.6 LETS SUM UP
In this chapter we have learnt the following:
Singularity functions.
The fundamental solution of Laplace equation.
Definition of Green functions using fundamental solution.
Green’s first identity.
Green’s second identity.
The use of Green’s function to solve partial differential equations.
3.7 UNIT END EXERCISE
1. Find the fundamental solution of Laplace equation.
2. State and prove Green’s first identity.
3. Find the Green’s function for the first quadrant in XY plane.
4. State and prove symmetric property of Green’s function.
5. Show that Green’s function is unique.
6. Find the Fundamental Solution of the Laplace Operator for n = 3 . munotes.in
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3. Find the Green’s function f or the first quadrant of ℝ, namely the
domain
Ω={(,)∈ℝ >0,>0}
8. Find the Green’s function for the upper half ball (0,) in ℝ.
9. Show that the Fundamental Solution of the Laplace Operator is given
by
()=
⎩⎨⎧1
2log =2
1
(2−) ≥3
10.Use the method of images to find the Green’s function for Laplace’s
equation to infinite
strip << in the (,)plane.
3.8 REFERENCE
Phoolan Prasad &Renuka Ravindran, Partial Differential
Equations, Wiley Eastern Limited, India
Yehuda Pinchover and Jacob Rubistein, An Introduction to Partial
Differential Equations, Cambridge University Press
munotes.in
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85
4
GREEN’S FUNCTION II
Unit Structure:
4.1 Objectives
4.2 Introduction
4.3 Green’s function for m dimensions sphere of radius R
4.4 Green’s functions Dirichlet problem in the plane,
4.5 Neumann’s function in the plane.
4.6 Lets sum up
4.7 Unit End exercise
4.8 Reference
4.1 OBJECTIVES:
After going through this chapter students will be able to:
To provide an understanding of, and methods of solution for, the most
important types of partial differential equations that arise in
Mathematics.
Use Green's functio ns to solve Laplace’s equation.
Use Green’s functionto solve Laplace’s equation for m dimensions
sphere of radius R.
Use Green’s functions to solve Dirichlet problem in the plane.
Use Green’s functions to solve Neumann’s problem in the plane.
4.2 INTRODUC TION:
In general the type of conditions that may be applied depends on the
applications that are involved. In practices two types of boundary
cond itions are commonly considered. The first one is known as the
homogeneous Dirichlet boundary condition which s tates that is zero on
and second one is known as the homogeneous Neumann condition
which is stats that ∇ is zero on . When satisfies these homogeneous
boundary condition. We obtained representation formula for problems on
ℝ. We now fix Ω to bean bounded open domain in ℝ with smooth
boundary Ω.We will try to build Green’s function using the ideas munotes.in
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developed so far. Later, we will check directly that the derived
representation formula gives the solution. We will also use the reflection
idea about the boundary of the domain.
4.3 GREEN’S FUNCTION FOR N DIMENSIONS
SPHERE OF RADIUS R:
Let Bn(0, 1) be the unit ball in ℝ . We look for a formula for the solution
of Laplace’s equation in Bn(0, 1) with Dirichlet boundary conditions,
∆=0 ∈(0,1)
= ∈(0,1) (I)
if u is a solution of (I), then u will have the form
()=− ()
(,)()
(,)
Now we just need to calculate
on ∂Bn(0, 1) where G is a Gree n’s
function for Bn(0, 1). As shown above,
(,)=∅(−)− ∅((−∗))
is a Green’s function for the unit ball in ℝ where
∗=

is the point dual to x. We consider the case when n ≥ 3. The case n = 2 can
be handled similarly. For n ≥ 3, we have
∅()=
()
⟹ ∇∅()=
()
Therefore, ∇∅(−)=()
()
While ∅(−∗)=
()
(∗)=
 ∅(−∗)
Therefore,
∇∅(−∗)=−1
(−∗)
()−∗
=−−
()(−∗) munotes.in
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=−−
()−
Now, the unit normal to Bn(0, 1) is given by
=
=
Therefore, the no rmal derivative of G(x, ·) on ∂Bn(0, 1) is given by
(,)=∅(−)
− ∅((−∗))
=−(−)
()− .+−
()− .
=−1
()−
Therefore, the solution f ormula for (I) is given by
()=− ()
(,)()
(,)
=1−
()()
−()
(,)
We can use this formula to derive the solution formula for Laplace’s
equation on the ball of radius r with Dirichlet boundary conditions,
∆=0 ∈(0,)
= ∈(0,) (II)
Suppose u is a solution of (II), then ()=()is a solution of (I) with
boundary data
()=(). Therefore, by our work above, we see the formula for is
given by
()=1−
()()
−()
(,)
=1−()
−()
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=1−()
−()
(,)
=(1−) ()
−()
(,)
()=(−)
()()
−()
(,)
which implies the solution formula for (II) is given by
()=(−)
()()
−()
(,)
This representation formula is called Poisson’s formula for the ball. The
function
(,)=(−)
()−
is called Poisson’s kerne l for the ball.
Example 1:Let ℝ be the upper half space in ℝ,
ℝ={(,,…,)∈: >0}
Find Poisson’s formula and Poisson’s kernel be the upper half space in ℝ.
Solution: G is a Green’s function for ℝ. As shown above,
(,)=∅(−)− ∅(−)
Where =(,,,……,−) and ∅ is the fundamental solution
of Laplace’s equation in ℝ . Our proposed solution has the form
()=−()
(,)()
ℝ+()(,)
ℝ
Now, we calculate
on { = 0} to find an explicit formula for solutions
to
∆=0 ∈Ω
= ∈Ω
Now ∅
()=
() munotes.in
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Therefore, the normal deriv ative of G on { = 0} is given by
(,)
=∅(−)
− ∅(−)
=−
()−−−
()−
=−2
()−
Therefore, if u is the solution of Laplace’s equation on the upper half 
space Ω with Dirichlet boundary conditions, then we suspect that u will
have the form
()=2
()()
−()
ℝ
This is called Poisson’s formula for the half space ℝ. The function
(,)=2
()−
is called Poisson’s kernel for the half space ℝ.
4.4 GREEN’S FUNCTIONS DIRICHLET PROBLEM IN
THE PLANE:
Consider the Dirichlet problem for the Poisson equation
∆= ∈
= ∈
where is a bounded planar domain with a smooth boundary . The
fundamental solution of the Laplace equation plays an important role in
our discussion. Recall that this fundamental solution is defined by
∅(,)=−1
2 =−1
4lo g (+)
The fundamental solution is harmonic in the punctured plane, and it is a
radially symmetric function with a singularity at the origin. Fix a point
(,)∈ℝ. Note that if u(x, y ) is harmonic, then u(x − a , y − b) is also
harmonic for every fixed pair (a, b). We use the notation
∅(,;,)=∅(−,−) munotes.in
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We call ∅(,;,) the fundamental solution of the Laplace equation
with a pole at (,).
The function ∅(,;,) is harmonic for any point (,) in the pla ne such
that (,)≠(,)For ε > 0, set.
=(,)∈,(−)+(−)<ε, =\
Let ∈(). We use the second Green identity in the domain where
the function (,)=∅(,;,)) is harmonic to ob tain
(∅∆−∆∅) =(∅−∅)
Therefore,
(∅∆) =(∅−∅)+(∅−∅)
Let ε tend to zero, recalling that the outward normal derivative ( with
respect to the domain ) on the boundary of is the inner radial
derivative pointing towards the pole ( a,b).
We obtain ∫(∅)≤εε→0 ε→0
(∅)
=1
2()→(,) →0
Therefore
(,)=∅(−,−)−∅(−,−)
−∅(−,−)∆
is called Green’s representation formula for Dirichlet problem in the
plane.
The function ∅[](,)=−∫∅(−,−)(,).is called
the Newtonian potential of f.
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Example 2 : Determine the Green’s function for Dirichlet problem for a
circle given by
∇=0, <
=() =.
Solution: Let (,) and Q( ,′) have po sition vectors and .
Let be the inverse of with respect to the circle so that .=
and has coordinate
,.
Now we construct the Green’s Function such that
=1
−+
Let =.′ so that it can be verified that ∇=0.
=.′
.
On the circle =,
=′
=lo g1=0
However =+′−2′cos (−)
′=
+′−2
′cos (−)
Replace value of and ′ in =.
. we get,
=1
2
+′−2′cos (−)
+′−2′cos (−)
=1
2 +
′−2
′cos (−)
+′−2′cos (−)
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But On the circle =,
=−(−)
[−2 (−)+]
Therefore, (,)=
∫
[ ()] .
The Eigen function method:
Consider the eigen value problem associated with the operator ∇ in the
domain ℝ. i.e.
∆∅+∅=0 inℝ
∅=0in ∂ℝ
Let be the eigen values and ∅ be the corresponding eigen
functions. Suppose we give furior series expression to and in terms of
the eigen functions ∅ in the following form:
(,,,)=(,)∅(,)
(−,−)=(,)∅(,)
Where =
‖∅‖∬ (−,−)∅(,) =∅(,)
‖∅‖ ℝ
‖∅‖=∅
ℝ
∴ ∇∅+∅=0
We obtain ∇∑∑(,)∅(,) =∑∑(,)∅(,)
−(,)∅(,)=∑∑∅(,)∅(,)
‖∅‖
From which we get
(,)=−∅(,)
‖∅‖
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Hence the required Green’s fun ction for the Dirichlet problem in the form
(,,,)=∑ ∑∅(,)∅(,)
‖∅‖ .
If eigen value =
+
and corresponding eigen function
∅=
where ,=1,2,…………
Therefore Green’s function for the Dirichlet problem in the form can be
written as,
(,,,)=−4
+
.
Example 3:Using Green’s function method Solve PDE,
∇=−s in()s in (2)in0<<1, 0<<2
With the initial boundary condition (,0)=0; (,2)=0,0<<
1
(0,)=0; (1,)=0, 0<<2.
Solution: Here =1,=2 and (,)=−s in()s in (2)
We have Green’s function for the Dirichlet problem in the form can be
written as,
(,,,)=−4
+
.
(,,,)=−2
+
1
2.
By definition of Green’s function,
(,)=−2
+
1
2
×−s in()s in (2)
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(,)
=8 1
4+s in()s in( )
s in(2)s in
2
×s in()s in (2)
(,)=1
28
4(1)+(4)s in()s in (2)
(,)=1
5s in()s in (2)
Example 4: Let Ω be the triangle on ℝ with vertices ( 1, 0), (1, 0) and (0,
√3 ). Solve the following Dir ichlet problem
−∆=2 in Ω
=0 on Ω.
Solution: We first need to find equations of sides of triangle.
Equations of sides of triangle are =0, +√3 −√3=0, −
√3 −√3=0.
Thus the solution has the following form
(,)=, +√3 −√3(, −√3 −√3)
Now we need to determine constant ′ which fulfilled the boundary
condition.
A direct calculation is given by
−∆=4√3=2
∴=1
2√3
Therefore the solution is
(,)=
2√3, +√3 −√3, −√3 −√3.
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4.5 NEUMANN’S FUNCTION IN THE PLANE:
We move on to present an integral representation for solutions of the
Neumann problem for the Poisson equation:
∆= ∈
= ∈
Where is a boun ded planar domain with a smooth boundary . The
fundamental solution of the Laplace equation plays an important role in
our discussion. Recall that this fundamental solution is defined by
∅(,)=−1
2 =−1
4lo g (+)
for any c losed curve that is fully contained in D.
The fundamental solution is harmonic in the punctured plane, and it is a
radially symmetric function with a singularity at the origin. Fix a point
(,)∈ℝ. Note that if u(x, y ) is harmonic, then u(x − a , y − b) is also
harmonic for every fixed pair (a, b). We use the notation
∅(,;,)=∅(−,−)
We call ∅(,;,) the fundamental solution of the Laplace equation
with a pole at (,).
The function ∅(,;,) is harmonic for any point (,) in the plane such
that (,)≠(,)For ε > 0, set.
=(,)∈,(−)+(−)<ε, =\
Let ∈(). We use the second Green identity in the domain where
the function (,)=∅(,;,)) is h armonic to obtain
(∅∆−∆∅) =(∅−∅)
Therefore,
(∅∆) =(∅−∅)+(∅−∅)
Let ε tend to zero, recalling that the outward normal derivative (with
respect to the domain ) on the boundary of is the inner radial
derivative pointing towards the pole ( a,b). munotes.in
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We obtain ∫(∅)≤εε→0 ε→0
(∅)
=1
2()→(,) →0
Therefore
(,)=∅(−,−)−∅(−,−)
−∅(−,−)∆
Enables us to reproduce the value of an arbitrary smooth function u at any
point ( a,b) in D provid ed that ∆ is given in D, and u and are given on
∂D. For the Neumann problem, u is not known on ∂D.
Let h(x, y; a, b ) be a solution (depending on the parameter ( a, b)) of the
following Neumann problem:
∆ℎ(,;,)=0(,)∈
ℎ(,;,)=∅(,;,)+1
(,)∈
Where L is the length of ∂D. Substituting u = 1 into the Green
representation formula we get
∅(,;,) =−1
Therefore, the above boundary condition a sufficient conditio n for the
solvability of the problem s to ∫ =0.∅
Definition: A Neumann function for a domain D and the Laplace operator
is the function
(,;,)=∅(,;,)−ℎ(,;,)(,),(,)∈,
(,)≠(,)
where ℎ(,;,) is a solution of
∆ℎ(,;,)=0(,)∈
ℎ(,;,)=∅(,;,)+1
(,)∈ munotes.in
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i.e. a Neumann function satisfies
∆(,;,)=−(−,−)(,)∈
(,;,)=−1
(,)∈
Therefore
(,)=(,;,)
(,)
−(,;,)∆(,) +1
We obtain the following representation formula for solutions of the
Neumann problem.
Note:
i) The kernel N is not called the Green function of the problem,
since N does not satisfy the corresponding homogeneous
boundary condition. There is no kernel function that satisfies
∆(,;,)=−(−,−)(,)∈
(,;,)=0 (,)
∈
ii) The Neumann function is determined up to an additive
constant. In order to uniquely define N it is convenient to use
the normalization
(,;,)
=0
iii) The third term in the representation formula ( 4.32) is
∫ , the average of u on the boundary, which is not
given. But since the solution is determined up to an additive
constant, it is convenient to add the condition
(,)=0
and then the pr oblem is uniquely solved, and the corresponding integral
representation uniquely determines the solution.
Example 5 : Consider the Neumann boundary value problem for Laplace’s
equation in the upper half plane ∇=0 in >0 with
=−
=()
on =0. munotes.in
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Solution: Draw boundary value condition and add image to make on the boundary condition:
(,,
Note that
=
(And as required Neumann boundary value problem,
=−
=Then, since (,0,,
(,)
98 Draw boundary value condition and add image to make on the boundary condition:
Figure 1
,)=−1
4lo g[(−)+(−)]
+−1
4lo g [(−)+(+)] ()()()+()
()() And as required Neumann boundary value problem, =1
4−2
(−)+()+2
(−)+()=0)=−
lo g[(−)+()] ()=()
lo g[(−)+()]
=0
0.
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Therefore we can write,
(,)=()
lo g[(−)+()]
We have not given condition on and
at infinity. For instance we can
think of the boundary of the upper half plane as a semi circle with
→+∞.
Figure 2
Green’s theorem in the half disc for and is
(∇−∇)=
−
.
Exam ple 6: Interior Neumann problem for Laplace’s equation in a disc,
∇=
+
=0in <,
=() on =.
Solution: Here we need,
∆=−(−)(−)+
with
=0.
Where = is the surface area of the disc. In order to deal with this
term we solve the equation ∇()=
=
⟹()=
4+ +
And take the particular solu tion with ==0. Then add in source at
inverse point and an arbitrary function ℎ to fix the symmetry and
boundary condition of .
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(,,,∅)=−1
4[+−2cos (
−∅)]−1
4
+
−2cos (−∅)
+
4+ℎ
So,
=
(∅)
(∅)−
(∅)
(∅)+
+
=−1
2−cos(−∅)+
−cos (−∅)
+−2cos (−∅)+1
2+ℎ
=−1
2+1
2+ℎ
And
=0 implise
=0 on the boundary ,
then put ℎ=
So,
(,,,∅)=
[+−2cos (−∅)]
+
−
2cos (−∅)+
on =.
=−1
4 [+−2cos (−∅)]+1
4
=−1
2[+−2cos(−∅)]−1
2
Then (,∅)=+∫()

(,∅)=−
2[+−2cos(−∅)]−1
2()
Now recall the Neumann problem compatibilit y condition;
()=0
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Indeed ∫∇ =∫
From divergence theorem,
()=0
So the term involving ∫()
in the solution (,∅) vanishes, hence
(,∅)=−
2[+−2cos(−∅)]()
Or
(,)=−
2[+−2cos(−∅)](∅)∅
4.6 LETS SUM UP:
In this chapter we have learnt the following:
Use Green's functions to solve Laplace’s equation.
Use Green’s function to solve Laplace’s equation for m 
dimensions sphere of radius R.
Use Green’s functions to solve Dirichlet problem in the plane.
Use Green’s functions to solve Neumann’s problem in the plane.
4.7 UNIT END EXERCISE:
1. Prove tha t the Neumann function for th e Poission equation is
symmetric.
2. Find the Green’s function for the Dirichlet problem on the rectangle
:0≤≤,0≤≤ described by the PDE (∇+)=0 in
and the initial boundary condition =0 on .
3. Use Green’s function technique to solve the Dirichlet problem for a
semi infinite space.
4. Find the Green’s function for Boundary value problem ∇= in the
quadrant
>0,>0.
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5. Prove that Exterior Neumann problem for Laplace’s equation in a
disc,
(,)=−
2[+−2cos(−∅)](∅)∅
6. Solve the Neumann problem in the quarter plane {x >0, y >0}.
7. Use the Green’s function method to find the solution of the Neumann
boundary value problem :
∇=0, 0<<1, 0<<1
(,0)=(,1)=0, 0<<1,
(0,)=(1,)=0, 0<<1.
8. Solve the following Dirichlet problem ,
−∆=0in(0,1)
(,)=+on=1
Where =(,)=( , ) and and are constants.
9. Find a bounded solution to the following Dirichlet problem outside a
unitball in ℝ:
−∆=0, <1
=2
5+4
Where =.
10. Let be the solution of
−∆=0inℝ
= onℝ
Given by the Poisson formula for the half space. Assume is
bounded an d ()=for ∈ℝ, ≤1. Show that is not
bounded near =0.
4.8 REFERENCE
Phoolan Prasad & Renuka Ravindran, Partial Differential
Equations, Wiley Eastern Limited, India
Yehuda Pinchover and Jacob Rubistein, An Introduction to Partial
Differential Equations, Cambridge University Press
munotes.in
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103
5
THE DIFFUSION EQUATION&
PARABOLIC DIFFERENTIAL
EQUATIONS
Unit Structure:
5.1 Objectives
5.2 Introduction
5.3 Existence and Uniqueness theorem for initial value problem in an
infinite domain
5.4 Existence and Uniqueness theorem for initial value problem i n semi 
infinite domain
5.5 One dimensional Heat equation
5.6 Maximum and Minimum Principle for the H eat equation
5.7 One dimensional wave equation
5.8 Lets sum up
5.9 Unit End exercise
5.10 Reference
5.1 OBJECTIVES:
After going through this chapter student s will be able to:
Existence and Uniqueness theorem for initial value problem in an
infinite domain and semi infinite domain.
One dimensional Heat equation and also solve its initial value
problem.
Maximum and Minimum principle for the heat equation.
One d imensional wave equation and also solves its initial value
problem.
Solve one dimensional PDE by method of separation of variables.
5.2 INTRODUCTION:
In this chapter we are going to look at one of the more common methods
for solving simple partial differen tial equations. The method we will be
taking a look at is that of Separation of Variables. munotes.in
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We will do a partial derivation of the heat equation that can be solved to
give the temperature in a one dimensional rod of length L. In addition, we
give several po ssible boundary conditions that can be used in this
situation. We do a partial derivation of the wave equation which can be
used to find the one dimensional displacement of a vibrating string .
Model heat flow in a one dimensional object (thin rod). Place r od along x 
axis, and let
(,)be a temperature in rod at position , time .
Under ideal conditions (e.g. perfect insulation, no external heat sources,
uniform rod material), one can show the temperature must satisfy
=∇ (theonedime nsional heat
equation )
The constant is called the thermal diff usivity of the rod.
Now we will discuss existence and uniqueness theorem for IVP in infinite
and semi infinite domain.
5.3 EXISTENCE AND UNIQUENESS THEOREM FOR
INITIAL VALUE PROBLEM IN A N INFINITE
DOMAIN:
We will start out by consider ing the temperature in a 1 D rod of length L.
What this means is that we are going to as sume that the bar starts off at
=0 and ends when we reach =. We are also going to so assume that
at any location, x the temperature will be constant at e very point in the
cross section at that x.
We have learn the Green’s function, using Green’s function for the
problem of heat flow in an infinite rod, the position of the rod coincide
with X axis and rod is homogeneous . Also heat is uniformly supply to it in
cross section area in time >0.(,) is the temperature at the point at
time without loss of heat through boundary condition. Then the problem
can be described by PDE,
=∆ −∞<<∞, >0
Initial boundary condition (,)=() −∞<<∞.
Suppose the Fourier transform of (,) is (,).
i.e. [(,)]=(,)−
√∫ (,)
Taking the Fourier transform of =∆ and summing that ,→
0, as →∞ we get
+=0. munotes.in
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Its solution is given by
(,)=()
When () is an arbitrary function to be determined from the initial
condition as follows.
(,0)=[(,0)]=1
√2(,0)
=1
√2()
=().
Hence, (,)=()
Hence by convolution theorem,
(,)=()×()
=1
2√()exp −(−)
4
= (−,) ()
is called the Green’s function for heat transfer in infinite domain of road.
Now we consider the case =1 and
()=0 <0
>0
(,)=
2√
−(−)
4
Put =
√
(,)=
√
√
=
√ +
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=
2+
√
√
=
21+
2√
Where is the error function.
5.4 EXIST ENCE AND UNIQUENESS THEOREM FOR
INITIAL VALUE PROBLEM IN SEMI INFINITE
DOMAIN:
Now using Green’s function for the problem of heat flow in semi infinite
rod, the position of the rod coincide with X axis and rod is homogeneous.
Also heat is uniformly supply t o it in cross section area in time >0. A
boundary condition at the finite end =0 and other end condition ∞. The
initial condition on the temperature distribution (,) can be described
by,
(,0)=() 0<<∞
There are various boundary conditions that can be prescrib ed at the end
=0.
Ist condition: The temperature is prescribed at =0 for all time (0,)=
().
IInd condition: The flux of heat across =0 is prescribed for all time. i.e.
(0,)=().
IIIrd condition: The flux of heat across =0 is pr opositional to the
difference between the temperature at =0 and the surrounding medium.
i.e.
(0,)+ (0,)=
We define a function (,) called the derived singularity function
(,)=
(,)=−2(,). >0, >0
The properties of (,) are given by
∫(,)=1
andlim→∫(,)=1 ,
>0
Using these propertieswe get the relation,
(,)=erf 2√=2
√
√
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Where the compl ementary error function, () is defined by
erf ()=1−erf ().
Theorem : If ()∈ for 0<<∞ and (0)=0 than (,)=
∫(,−)()
satisfies the heat equation in 0<<∞,<<
∞ and
Case I) lim→(,)=() 0<<
Case II) lim→(,)=0 0<<∞
Both the case is uniformly continuous. Where is a constant.
Proof: (,) satisfies the heat equations for >0.
We shall first prove case II:
Let ∈>0 such that we can determine a ,
()< 0<<.
Hence for 0<<,
(,)=(,−)()
≤(,)(−)< ∈(,
)=∈
For a specific value of , its holds for all . Hence proved.
To prove case I:
Let ∈>0 such that we can determine a which is independent to on
0≤≤ and 0<<,
(,)−()<∈
For any ∈>0, let be such that ()<∈
Whenever < due to uniform continuity of ()in 0≤≤,
We can choos e , so that we also have
(−)−()<∈
Whenever <, 0≤≤.
Now divide 0≤≤ into two sub interval 0≤≤and ≤≤,
we have
(,)−()<∈for all >0.
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(,)−()=(,−)()−(,)()
=(,)(−)−(,)()
=(,)[(−)−()]−()(,)
∴ (,)−()<∈
2(,)+3(,)
Where =Sup ().
We know that lim→∫(,)=1 ,
>0 and ∫(,)
<∈
But for sufficiently small ,we have the result
(,)−()<∈for0<< and 0≤≤
Hence the solution is given by
(,)=(,−)()
.
This result gives solution of Ist condition that the problem of finding the
temperature of semi infinite rod whose initial temperature =0 is
everywhere zero and whose temperature at finite end =0 is prescribed
by all the as ().
For IIndcondition : The temperature of semi infinite rod arises when initial
temperature =0 is everywhere zero and whose temperature at finite end
=0 is prescribed by all the as () and (0)=0.
Here (,)=−2 ∫(,−)()
is the solution for 0<<∞,
0<< where is a constant. Here (,) is satisfies the heat
equation.
We have lim→(, 0)=0 for 0≤≤∞
Also (,)=−2 ∫(,−)()
=∫(,−)()
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IIIrd condition: The flux of heat across =0 is propositional to the
difference between the temperature at =0 and the surrounding medium.
i.e.
(0,)+ (0,)=
Here (,) is the solution of the diffu sion equation satisfying the
condition
(,0)=0, ≥0; (0,)+(0,)= ∅(), ≥0 .
Where is positive constant and ∅() is continuous.
Therefore, −=∫ ∅()
(,−)
Till that we have consider initial tempe rature distribution is Zero.
Now we have to consider the case of non zero initial distribution and zero
boundary condition at =0. i.e.
(,0)=() 0≤<∞
()<
And one of the following boundary conditions:
IVth condition: (0,)=0 0<<∞.
Vth condition: (0,)=0 0<<∞.
VIth condition: (0,)−(0,)=0 0<<∞.
In IVth condition we extend () as an odd function, we get
()=−(−)for<0.
Then the solution of the initial value problem in −∞<<∞ is given by
(,)=(−,)()
=[(−,)−(+,)]
()
Since (0−,)−(0+,)=0 the boundary condition (0,)=0
is auto matically satisfies.
In Vth condition we extend () as an even function of for <0,
()=(−)for<0.
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Then the solution is given by
(,)=[(−,)−(+,)]
()
Since (0−,)−(0+,)=0 the boundary condition (0,)=
0 is automatically satisfies .
In VIth condition we extend () as
(−)=()+2 ∫()
for≥0.
In order to satisfies boundary condition
(0,)−(0,)=0 0<<∞.
Than the solution is given by
(,)=[(−,)−(+,)]
()
+2 (+,)
()
A linear combination of solution of one of the problem condition I, II, and
III wi th ()=0 and one of the conditions IV, V, VI lead us to the
general mixed initial boundary value problems for the heat equation for
semi infinite rod.
5.5 ONE DIMENSIONAL HEAT EQUATION:
We now begin to study finite difference method for time depende nt PDE
where variations in space are related to variation in time.
The diffusion equation is of the form
=∆
Taking =1 by suitable change in or .This is called as heat equation.
The one dimensional diffusion equation for (,) is
=
Where is diffusion constant.
Now solving heat equationby separation of variable method
Let (,)=() () ≠0 (*)
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Differentiate the separated solution (*) once with respect to t and twice
with respect to x and substitute these derivatives into the PDE. We then
obtain
=′′
Now, using the separation of variables step .
′′
= ′
=−
Where is positive constant.
Now we get following ODE’s
i) +=0 0<<
ii) +=0 >0
Solution of (i) is =cos+sin
Solution of (ii) is =
Therefore the general solution is
=(cos+sin)()
If the boundary conditions (0,)=0 ≥0
(,)=0 ≥0
With initial condition (,0)=() 0≤≤.
Then (0)=0 ⟹=0.
()=0⟹sin=0as≠0
∴ sin=0
Thus =
for =1,2,3,…..
Substituting thes e in solution we get
()=sin
()=
Hence the non trivial solution of the heat equation which satisfies the two
boundary condition.
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(,)= sin
for=1,2,3,…..
Where = is an arbitrary constant.
By the principle of superposition implies that any linear combination
(,)=(,)
= sin
Which satisfies the initial condition if
(,0)=()=∑ sin
.
This hold true if () can be represented by a Fourier Sine series with
coefficient
=
∫()
sin
.
Hence,
(,)=2
()
sin
sin
is the general solution of the heat conduction equation of the function f
with respect to the eigenfunctions of the pro blem, and , n = 1, 2 . . . are
called Fourier coefficients of the series .
Example 1: Solve the heat problem
= 0<<, >0
(0,)=(,)=0 ≥0
(,0)=()= 0≤≤2
− 2≤≤
Solution: The formal solution of heat equation is
(,)=2
()
sin
sin
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=2
()
sin
=2
()sin
=2
sin +
2
(−)sin
=4
sin
2
Thus the formal solution is
(,)=4
sin
2()sin()
But
sin
2= 0 =2
(−1) =2−1
Where =1,2,3,…..
Therefore we can write solution as
(,)
=4
(−1)
(2−1)sin(2−1)
2()sin((2
−1))
Example 2: Solve the heat problem
=3 0<<2, >0
(0,)=(2,)=0 >0
(,0)=50 0<<2.
Solution: Comparing with heat equation we get,
=√3 =2 munotes.in
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()=50=200
1
sin
2
Since,
=
=√3
2
We obtained the solution
(,)=
∑
sin
.
5.6 MAXIMUM AND MINIMUM PRINCIPLE FOR THE
HEAT EQUATION:
We shall pro ve the maximum and minimum properties of the heat
equation. These properties can be used to prove uniqueness and
continuous dependence on data of the solutions of these equations. To
begin with, we shall first prove the maximum principle for the
inhomogene ous heat equation ( F = 0).
Theorem: (The maximum principle) : Let :0≤≤,0≤≤ be a
closed region and let (,) be a solution of
−=(,), (,)∈ (I)
Which is continuous in the region R. if <0 in R, then (,) attains its
maximum valus on =0,=0 or = and not in the interior of the
region or at =. If >0 in R, then (,) attains its minimum values
on =0,=0 or = and not in the interior of the region or at =.
Proof: We shall show that if a max imum or minimum occurs at an inter ior
point 0<<and 0<<, then we will arrive at contradiction. Let
us consider the following cases:
Case I: first, consider the case with <0. Since (,) is continuous in a
closed and bounded region in R, (,) must attain its maximum in R. Let
(,) be the interior maximum point. Then we must have
(,)≤0, (,)≥0 (II)
Since (,)=0=(,), we have (,)=0if <.
If =, the point (,)=(,) is on the boundary of R, then we
claim that
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As may be increasing at (,). Substituting II in I , we find that the
left side the equatio n I is non negative while the right side is strictly
negative. This leads to a contradiction and hence, the maximum must be
assumed on the initial line or on the boundary.
Case –II : Consider the case with >0. Let there be an interior minimum
point (,) in R such that
(,)≥0, (,)≤0 ( III)
Note that the inequalities III is same as II with the signs reversed. Again
arguing as before, this leads to a contradiction, hence the minimum must
be ass umed on the line or on the boundary.
Note : when =0i.e. for homogeneous equation, the inequalities II at a
maximum or III at a minimum do not leads to a contradiction when they
are inserted into I as and may both vanish at (,).
Below, we present a proof of the maximum principle for the homogeneous
heat equation.
Theorem : (The maximum principle): Let :0≤≤,0≤≤ be a
closed region and let (,) be a solution of
=, (,)∈ (IV)
Which is continuous in the closed region R. The maximum and minimum
values of u(x, t)are assumed on the initial line t = 0or at the points on the
boundary x = 0or x = L.
Proof. Let us introduce the auxiliary function
(,)=(,)+∈ (V)
Where ∈>0 is a constant and usatisfies IV. Note that (,)is continuous
in R and
hence it has a maximum at some point (,) in the region R.
Assume that (,) is an interior point with 0<<and0<<.
Then we
find that
(,)≥0, (,)≤0 (VI)
Since u satisfies IV , we have
−=−−2∈=−2∈<0 (VII)
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Substituting VI into IV and using VII now leads to
0≤−<0,
which is a contradiction since the left side is non negative and the right
side is strictly
negative. Therefore, (,)assumes its maximum on the initial line or on
the boundary
since satisfies I with F <0.
Let=max {(,)} on =0,=0 and =.
i.e. Mis the maximum value of u on the initial line and boundary lines.
Then
(,)=(,)+∈ ≤+∈, for0≤ ≤,0≤≤ (VIII)
Since v has its maximum on t = 0, x = 0, or x = L, we obtain
(,)=(,)−∈ ≤(,)≤+∈ (IX)
Since is arbitrary, letting → 0, we conclude that
(,)≤for all (,)∈, (X)
Hence proof.
As a consequence of the maximum principle, we can show that the heat
flow problem has
a unique solution and depend continuously on the given initial and
boundary data.
Theorem : ( uniqueness) Let (,) and (,) be the solution of the
following problem
= 0<<,>0,
(0,)=(), (,)=ℎ(), (XI)
(,0)=(),
Where (),() and ℎ() are given function. then (,)=(,), for
all 0≤≤ and ≥0.
Proof: Let (,) and (,) be the solution of the given XI problem.
Set (,)=(,)−(,). Then satisfies
= 0<<,>0,
(0,)=, (,)=, (XI)
(,0)=0,
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By the maximum principal we have,
(,)≤0,⟹ (,)≤(,)for all 0≤≤ and ≥0.
Similarly we have =−, for that we get
⟹ (,)≤(,)for all 0≤≤ and ≥0
Therefore we have
⟹ (,)=(,)for all 0≤≤ and ≥0
Hence it has unique solution.
5.7 ONE DIMENSIONAL WAVE EQUATION:
We write the wave equation as
=∇for−∞<<∞.
This is the simplest seco nd order equation.
−∇=
−
+
=0.
This means that starting from a function (,) you compute +
call the result for (,) than you compute − and you get zero
function.
The general solution is
(,)=(+)+(−). (XII)
Where and are two arbitrary function of a single variable.
Initial value problem is to solve the wave equation
=∇for−∞<<∞.
With the initial conditions (,0)=∅(), (,0)=()
Where ∅ and are arbitrary functions of . There is one and only one
solution of this problem.
Proof: Let if ∅()=sin and ()=0 then (,)=sin cos.
The solution of above IVP is easily found from the general solution
formula (XII) replacing =0, we get
∅()=()+() (XIII)
Then using the chain rule we differentiate (XII) with respect to and put
=0, to get
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Lets regards XIII and XIV as two equations for the two unknown
functions and . To solve them change the name of variable as same
neutral to .
Now we differentiate XIII and divide XIV by to get
∅=+′and
=−′
Adding and subtracting the last pair of equations, we get
=
∅+
and=
∅−
Integrating we get,
()=1
2 ∅()+1
2+
And
()=1
2 ∅()−1
2+
Where A and B are constant because of XIII we have A + B = 0. This
tells us what and are in general formula XII. Substituting =+
into and =− into we get
(,)=1
2 ∅(+)+1
2
+1
2 ∅(−)−1
2
This simplifies to
(,)=1
2[∅(+)+∅(−)]+1
2()
This is called D’Alembert’s the solution for initial value problem of one
dimension wave equation.
Example 3: Consider the Cau chy problem
= −∞<<∞, >0
With boundary condition:
(,0)=()=0 −∞<<−1
+1 −1≤<0
1− 0≤≤1
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(,0)=()=0 −∞<<−1
1 −1≤≤1
0 1<<∞
a) Evaluate 1,12.
b) Discuss the smoothness of the solution .
Solution: a) Using D’Alembert formula of one dime nsion wave equation
we get
(,)=1
2[∅(+)+∅(−)]+1
2()
1,1
2=
+
2+ 1
2()
Since
>1, it follows that
=0. On other hand 0≤
≤1,
therefore
=
.
()
=1 =1−1
2=1
2.
Hence
1,1
2=0+
2+ 1
21
2=1
4+1
4=1
2 .
b) The solution is not classical, since ∉ . Yet is a generalized
solution of the problem. Note that although is not contin uous,
nevertheless the solution is a continuous function. The
singularities of the solution propagate along characteristics that
intersect the initial line =0 at the singulari ties of the initial
conditions. These are exactly the characte ristics ±= −1,0,1.
Therefore, the solution is smooth in a ne ighborhood of the point
1,
which does not intersect thesecharacteristics.
Method of Separation of Variables for one dimensional Wave
equation:
PDE −=0 0≤≤, >0
Boundary condition: (0,)=0 >0
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To obtained separation of variables solution we assume
= ′′
i.e.
=
=
Case I: When >0, taking = we get
−=0
−=0
The solution in the form
=+
=+
Therefore, (,)=(+)(+)
Now using boundary condition’s
(0,)=(+)(+)
This implies that +=0, also
(,)=0⟹ +=0.
This gives non trivial solution if and only if
1 1
=0
⟹=1 or=0
This implies that =0, since cannot be zero, which is against the
case I assumption.
Hence solution is not acceptable.
Case II: When =0, we get
=0 and =0.
Their solutions are formed to be
=+, =+.
Therefore required s olution of the PDE is
(,)=(+)(+)
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Using boundary conditions we get,
(0,)=0=(+)⟹=0.
(,)=0=(+)⟹=0.
Hence on ly trivial solution is possible.
Case III: When <0, taking =−, we get
+=0; +=0.
Their general solution is given by
(,)=(cos+sin)(cos +sin)
Using the boundary condition : (0,)=0 we get =0.
(,)=0we get sin=0 ⟹=
, =1,2,3,……….
Hence the possible solution
(,)=sin
cos
+sin
=1,2,3,………….
Using superposition principle, we get
(,)=sin
cos
+sin
The initial condition gives
(,0)=()=sin
Which is half range of Fourier sine series, where
=2
()sin
Also we get
(,0)=()= sin
Which is also half range of Fourier sine series, where
=2
()sin
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Hence the required solution is obtained.
Example 4: Solve the one dimensional wave equation =16for0<
<2, >0.
The boundary conditions: (0,)=(2,)=0.
The initial conditions: i) (,0)=6sin−3sin4. ii) (,0)=0.
Solution: The general solution of 1 dimensional wave equation is given
by
(,)=(cos+sin)(cos +sin)
Using boundary condition : (0,)=0 for all gives
(cos +sin)=0
Which implies that =0.
(2,)=0for all gives
sin(cos +sin)=0
For non trivial solution sin2=0⟹=
for some integer .
(,)=sin
2(cos(2)+sin(2))
Now insert the initial condition (,0)=0 for al l 0<<2.
=sin
2(−2sin(2)+2cos(2))
(,0)=sin
2(2)=0⟹=0.
Finally using the initial condition (,0)=6sin−3sin4
We get (,)=∑ sin
Hence we get,
(,)=6sincos4−3sin4cos16.
5.8 LETS SUM UP:
In this chapter we have learnt the following:
Definition diffusion equation.
1dimensional heat equation and its solution by separation of variable
method: munotes.in
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Existence and Uniqueness theorem for initial value problem for
infinite and semi infinite domain.
Maximum and Minimum principle for heat equation.
1 dimensional wave equation and its solution by separation of
variable method
5.9 UNIT END EXERCISE:
1. Let (,) be the solution of the Cauchy problem
=9 −∞<<∞, >0
(,0)=()=1 ≤2
0 >2
(,0)=()=1 ≤2
0 >2
a) Evaluate (0,
).
b) Discu ss the large time behavior of the solution.
c) Find the maximum value of (,) and point when this
maximum is achieved.
d) Find all the point when ∈.
2. Obtain the solution of the wave equation
=
Under the following conditions :
i) (0,)=(2,)=0.
ii) (,0)=
.
iii) (,0)=0.
3. Solve the following heat problem:
=1
4 0<<1, >0.
(0,)=(1,)=0 >0
(,0)=100(1−), 0<<1
4. Use the maximum/minimum principle to show that the solution u
of the problem:
= 0<<, >0.
(0,)=(1,)=0 >0
(,0)=sin+1
2sin2, 0<<
Satisfies 0≤(,)≤√
, >0.
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5. Solve the one dimensional wave equation =4 for 0<<
1, >0.
(0,)=(1,)=0 ≥0
(,0)=()= 0≤≤1.
(,0)=()= cos 0≤≤1.
6. Prove that t he solution we found by separation of variables for the
vibration of a free
string can be represented as a superposition of a forward and a
backward wave.
7. Show that the solution of the 1 dimensional wave problem if it is
exists, is unique.
8. State and prove maximum and minimum principle.
5.10 REFERENCE
1. Phoolan Prasad &RenukaRavindran, Partial Differential Equations,
Wiley Eastern Limited, India.
2. Yehuda Pinchover and Jacob Rubistein, An Introduction to Partial
Differential Equations, Cambridge University Pr ess.
3. T.Amaranath, AnElemetary Course in Partial Differential
Equations, Narosa.
4. F. John, Partial Differential Equations, Springer publications.
5. G.B. Folland, Introduction to partial differential equations,
Prentice Hall.
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