Chapter 11 Introduction to PDE.pdf

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Chapter 11

Introduction to Partial Differential Equations

11.1 Basic concepts and Definitions

11.2 Classification of Partial Differential equations

11.2.1 Initial and Boundary Value Problems

11.2.2 Classification of second order Partial Differential Equations

11.3 Solutions of Partial Differential Equations of First Order

11.3.1. Solutions of Partial Differential Equations of First order with

constant coefficients.

11.3.2. Lagrange's Method for Partial Differential Equations of First-

order with Variable coefficients.

11.3.3. Charpit's Method for solving nonlinear Partial Differential

Equations of first-order.

11.3.4. Solutions of Special type of Partial Differential Equations of first

order.

11.3.5. Geometric concepts Related to Partial Differential Equations of

first-order.

11.4 Solutions of Linear Partial Differential Equations of Second order with

constant coefficients.

11.4.1 Homogeneous Equations

11.4.2 Non-homogeneous Equations

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11.5 Monge's Method for a special class of nonlinear Equations (Quasi

linear Equations) of the second order.

11.6 Exercises.

Partial differential equations are central theme to scientific and

technological studies besides occupying pivotal position in pure and applied

mathematics. Under appropriate conditions they represent real world systems

and for proper understanding of these problems, solutions of partial

differential equations must be understood. In this chapter we introduce basic

properties of partial differential equations and their solutions.

In the next chapter we present some well known partial differential

equations representing important problems of science and engineering.

11.1 Basic concepts and definitions

An equation containing the dependent and independent variables and

one or more partial derivatives of the dependent variable is called a partial

differential equation. In general it may be written in the form

F (x,y,.....,u,ux, uy,......,uxx, uyy,......)=0 (11.1)

involving several independent variables x,y,.....,an unknown function u

of these variables and the partial derivatives ux, uy,......,uxx, uxy, uyy ........of the

function. (11.1) is considered in a suitable subset D of Rn. For the sake of

convenience we confine our discussion for n=2. However extension of

properties discussed here to higher values of n is possible.

Here as in the case of ordinary differential equations, we define the

order of a partial differential equation to be the order of the derivative of

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highest order occurring in the equation. The power of the highest order

derivative in a differential equation is called the degree of the partial

differential equation.

Example 11.1 (a) xx

u

+ y

y

u

=0 is a first-order equation in two variables

with variable coefficients.

(b) a x

u

+ b

y

u

=c; where x,y are independent variables, a and b are

constants; is partial differential equation of first-order with constant

coefficients.

(c) x

u

+

y

u

-(x+y) u=0 is a partial differential equation of first-order.

(d) a(x) 2

2

x

u

+2b(x)

yx

u2

+c(x)

2

2

y

u

=x+y+u+

x

u

+

y

u

is a partial differential

equation of second-order.

(e) a(x) 2

2

x

u

+2b(x)

yx

u2

+c(x)

2

2

y

u

= f(x,y,u,

x

u

,

y

u

)

where a(x), b(x) and c(x) are functions of x and f(..,..,.,.,.) is a function of

x,y,u, x

u

and

y

u

, is a partial differential equation of second order.

(f) u yx

u2

+

x

u

=y is a partial differential equation of second-order.

(g) 2

2

x

u

+ 2y

yx

u2

+ 3x

2

2

y

u

= 4 sin x is a partial differential equation of

second-order and degree one.

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(h) 2

2

x

u

=

2

2

y

u

is a partial differential of second-order.

(i)

2

x

u

+

2

y

u

= 1 is a partial differential equation of first-order and

second degree.

By a solution of a partial differential equation of the type

F (x,y,u, ux,uy,uxx,uyy, uxy) =0 (11.2)

we understand functions u=(x,y) which satisfy (11.2) identically in D, that is,

if we put values of quantities on the left hand side we get right hand side.

Example 11.2.

(i) Show that sin n(x+y), cosn(x+y) and ex+y are solutions of the partial

differential equation

x

u -

y

u=0

(ii) Show that u(x,y)=(x+y)3 and u(x,y)=sin (x-y) are solutions of the partial

differential equation

2

2

x

u

-

2

2

y

u

=0

Solution (i) x

u

= n cos n (x+y) if u(x,y) = sin n(x+y)

y

u

= n cos n (x+y) if u(x,y) = sin n(x+y)

L.H.S. of the equation is x

u

-

y

u

= ncos n(x+y) – n cos n(x+y) =0=R.H.S.

x

u

= -nsin n(x+y) if u(x,y) = cos n(x+y)

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y

u

= -nsin n(x+y) if u(x,y) = cos n(x+y)

L.H.S. = [-nsin n(x+y)]-[-n sin n(x+y)]=0=R.H.S.

x

u

= ex+y if u(x,y) ex+y

y

u

= ex+y if u(x,y) = ex+y

L.H.S. = ex+y -ex+y =0 = R.H.S.

(ii) For u(x,y) = (x+y)3, x

u

=3(x+y)2,

2

2

x

u

=6 (x+y)

For u(x,y)=(x+y)3, y

u

=3(x+y)2,

2

2

y

u

= 6(x+y)

This implies that L.H.S. of the given partial differential is

2

2

x

u

-

2

2

y

u

= 6(x+y)-6(x+y)=0=R.H.S.

For u(x,y)=sin (x-y), x

u

=cos (x-y),

2

2

x

u

= - sin (x-y)

y

u

= - cos (x-y),

2

2

y

u

= - sin (x-y)

L.H.S. of the partial differential equation is

2

2

x

u

-

2

2

y

y

= - cos (x-y) + cos (x-y) = 0 = R.H.S.

Therefore (x+y)3 and sin (x-y) are solutions of

2

2

x

u

-

2

2

y

u

= 0.

A partial differential equation is said to be linear if the unknown

function u(.,.) and all its partial derivatives appear in an algebraically linear

form, 'that is, of the first degree. For example the equation

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A uxx+2Buxy+Cuyy+Dux+Euy+Fu = f (11.3)

where the coefficients A,B,C,D.E and F and the function f are functions of x

and y, is a second-order linear partial differential equation in the unknown

u(x,y).

Left hand side of (11.3) can be abbreviated by Lu, where u has

continuous partial derivatives of upto second order.

If u is a function having continuous partial derivatives of appropriate

order, say n then a partial derivative can be written as Lu=f where L is a

differential operator, that is, L carries u to the sum of scalar multiplications of

its partial derivatives of different order. An operator L is called linear

differential operator if L (u+v)= Lu+v where and are scalars and u

and v are any functions with continuous partial derivatives of appropriate

order. A partial differential equation is called homogeneous if Lu=0, that is, f

on the right hand side of a partial differential equation is zero, say f=0 in 11.3.

The partial differential equation is called non-homogeneous if f0.

(x+2y) ux +x2uy = sin (x2+y2) is a non-homogeneous partial differential

equation of first-order.

(x+2y) ux+x2uy=0 is a homogeneous linear partial differential equation

of first-order.

xuxx +yuxy+uyy=0 is a homogeneous linear partial differential equation

of second-order.

xuxx+y uxy+uyy=sin x is a non-homogeneous linear partial differential

equation of second-order.

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The general solution of a linear partial differential equation is a linear

combination of all linearly independent solutions of the equation with as many

arbitrary functions as the order of the equation; a partial differential equation

of order 2 has 2 arbitrary functions. A particular solution of a differential

equation is one that does not contain arbitrary functions or constants.

Homogeneous linear partial differential equation has an interesting property

that if u is its solution then a scalar multiple of u, that is, cu, where c is a

constant, is also its solution. Any equation of the type F(x,y,u,c1,c2)=0, where

c1 and c2 are arbitrary constants, which is a solution of a partial differential

equation of first-order is called a complete solution or a complete integral

of that equation. An equation F(,)=0 involving arbitrary function. F

connecting two known functions and of x, y and u, and providing a

solution of a first order differential equation is called a general solution or

general integral of that equation. It is clear that in some sense general

solution provides a much broader set of solutions than a complete solution.

However a general solution may be derived once a complete solution is

known.

Very often ux = x

u

, uy =

y

u

,uxx =

2

2

x

u

uxy = yx

u2

and uyy =

2

2

y

u

are respectively denoted by p, q,r, s and t.

In this notation the general form of partial differential equation of first-

order is

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F(x,y,u,p,q)=0 (11.4)

The general second-order partial differential equation is of the form

F(x,y,u,p,q,r,s,t)=0 (11.5)

A partial differential equation is said to be quasilinear if it is linear in all

the highest-order derivatives of the dependent variable. The most general

form of a quasi linear second- order equation is

A(x,y,u,p,q) uxx + B(x,y,u,p,q) uxy + C(x,y,u,p,q) uyy +f(x,y,u,p,q)=0 (11.6)

A partial differential equation of first-order is called semilinear if it is

linear in the principal part, namely the terms involving first derivatives: thus,

for A x

u

+ B

y

u

= C, these equations are defined to be such that the left hand

side, which contains all derivatives is linear in u in that A,B depend on x and y

alone; however C may depend non linearly on u. A semi linear partial

differential equation of second-order is of the form

A2

2

x

u

+ 2B

yx

u2

+C

2

2

y

u

= f(x,y,u,

x

u

,

y

u

) (11.7)

where A,B,C are functions of x and y.

11.2. Classification of Partial Differential Equations

We have seen the classification of Partial Differential equations into

linear, quasilinear, semi linear, homogeneous and non-homogeneous

categories in Section 11.1. In this section we mainly focus on the classification

of second order equations into elliptic, hyperbolic and parabolic types. Notion

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of Cauchy data (initial and boundary conditions) and characteristic for partial

differential equations are introduced.

11.2.1 Initial and Boundary Value Problems

A partial differential equation subject to certain conditions in the form of

initial or boundary condition is known as an initial-value or a boundary value

problem. The initial conditions, also known as Cauchy conditions, are the

values of the unknown function u(.,.) and an appropriate number of its

derivatives at the initial point.

Let us consider a second-order partial differential equation for the

function u(.,.) in the independent variables x and y, and suppose that this

equation can be solved explicitly for uyy, and hence can be represented in the

form

uyy = F(x,y,u,ux,uy,uxx,uxy) (11.8)

For some value y=y0, we prescribe the initial values of the unknown

function u and of the derivative with respect to y

u(x,y0)=f(x) (11.9)

uy(x,y0)=g(x) (11.10)

The problem of determining the solution of (11.8) satisfying initial

conditions (11.9)-(11.10) is known as the initial-value problem. Here initial-

value usually refer to the data assigned at y=y0. If initial values are prescribed

along some curve in the (x,y) plane, that is, finding solution of equation

(11.8) subject to prescribed value of y on some curve is called the Cauchy

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problem. These conditions are called Cauchy data. Actually two names are

synonymous.

Example 11.3 (a) ut = uxx 0<x<1, t>0

u(x,0)= cos x 0 x l

is an initial-value problem.

(b) Suppose that is a curve in the (x,y) plane; we define Cauchy data to

be the prescription of u on . It is convenient to write this boundary

condition in the parametric form

x=x0(s), y=y0(s), u=u0(s), for s1 s s2.

A(x,y,u) x

u

+B (x,y,u)

y

u

=C (11.10)

subject to (11.10) is a Cauchy problem

(c) Let us consider the equation

A(x,y) uxx + B(x,y) uxy + C uyy = F(x,y,u,ux,uy). (11.11)

Let (x0,y0) denote points on a smooth curve in the (x,y) plane. Also let

the parametric equations of this curve be

x=x0 (), y0=y0 ()

where is a parameter.

We suppose that two functions f() and g() are prescribed along the

curve . The Cauchy problem is now one of determining the solution u(x,y) of

Equation (11.11) in the neighbourhood of the curve satisfying the Cauchy

conditions

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u=f(), n

u

=g()

on the curve . n is the direction of the normal to which lies to the left of in

the counter clockwise direction of increasing arc length. The functions f() and

g(() are the Cauchy data.

The solution of the Cauchy problem is a surface, called an integral

surface, in the (x,y,u) space passing through a curve having as its

projection in the (x,y) plane and satisfying n

u

=g() which represents a

tangent plane to the integral surface along .

Types of Boundary Conditions

The boundary conditions on partial differential equation (11.6) fall into

the following three categories:

(i) Dirichlet boundary conditions (also known as boundary conditions of

the first kind), when the values of the unknown function u are prescribed

at each point of the boundary of a given domain on which (11.6) is

defined.

(ii) Neumann boundary conditions (also known as boundary conditions of

the second kind), when the values of the normal derivatives of the

unknown function u are prescribed at each point of the boundary .

(iii) Robin boundary conditions (also known as boundary conditions of the

third kind, or mixed boundary conditions), when the values of a linear

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combination of the unknown function u and its normal derivative are

prescribed at each point of the boundary .

Example 11.4 (i) t

u

= k

2

2

x

u

, 0<x<l,t>0

u(x,o)=f(x)

t

u

(x,o)=g(x), 0<x<l

u(0,t) =T1(t)

u(l,t)=T2(t), t>0

It is a Dirichlet boundary value problem.

(ii) t

u

= k

2

2

x

u

, 0<x< l, t>o

u(x,o)=f(x), t

u

(x,o)=g(x), 0<x< l

n

u

(0,t) =T3(t),

n

u

(l,t)=T4(t), t>0

It is an example of Neumann boundary value problem.

(iii) t

u

= k

2

2

x

u

, 0<x< l, t>0

u(x,o)=f(x), t

u

(x,o)=g(x), 0<x< l,

.0t

0)t,l(x

ut)u(l,

0,t)(0, n

u t)u(0,

It is an example of Robin boundary problem.

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It may be observed, a major part of scientific and technological studies

are devoted to initial and boundary value problems. Solutions of few important

initial and boundary value problems will be discussed in the next chapter.

11.2.2 Classification of Second-order partial differential Equations

For f=0 in Equation (11.3), the most general form of a second-

order homogeneous equation

A uxx + 2B uxy+C uyy + D ux+E uy +Fu=0 (11.12)

For a correspondence of this equation with an algebraic quadratic

equation, we replace ux by , uy by , uxx by 2, uxy by , and uyy by 2. The

left hand side of Equation (11.12) reduces to a second degree polynomial in

and :

P( ,)=A2+2B+C2+D+E+F=0 (11.13)

It is known from analytical geometry and algebra that the polynomial

equation P (,)=0 represents a hyperbola, parabola, or ellipse according as

its discriminant. B2-AC is positive, zero, or negative. Thus, the partial

differential equation (11.12) is classified as hyperbolic, parabolic, or elliptic

according as the quantity

B2-AC>0, B2-AC=0, or B2-AC<0.

The equation

A u2x+2B uxy + C u2

y = 0 (14.14)

is called the characteristic equation of the partial differential equation

(11.13). Solutions of (11.14) are called the characteristics

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Example 11.5 Examine whether the following partial differential equations

are hyperbolic, parabolic, or elliptic.

(i) 2

2

x

u

+ x

2

2

y

u

+ 4 = 0

(ii) 2

2

x

u

+ y

2

2

y

u

= 0

(iii) 2

22

x

uy

-

2

2

y

u

= 0

(iv) uxx + x2 uyy = 0

(v) x uxx + 2x uxy + y uyy = 0

Solution (i) A = 1, C = x, B = 0

B2-AC = 0 –x <0 for x>0

Thus the equation is elliptic if x > 0, is hyperbolic if x < 0 and it is

parabolic if x = 0.

(ii) A=1, B=0, C=y

B2-AC=0-y >0 if y<0 and so the equation is hyperbolic if y<0. It is

parabolic if y=0 and it is elliptic if y>0.

(iii) A=y2, B=0, C = -1.

B2-AC=y2>0 for all y. Therefore the equation is hyperbolic.

(iv) A=1, B=0, C=x2

B2-AC=0-x2<0 for all x. The equation is elliptic

(v) A=x, B=x, C=y

B2-AC=x2-xy=x(x-y)>0 for x>0 x>y

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In this case the equation is hyperbolic B2-AC=o if x=y. For this the

equation is parabolic. B2-AC <0 if x>y and x<0 or if x<y and x>0

In this case the equation is elliptic.

11.3 Solutions of Partial Differential Equations of First-order

11.3.1 Solution of Partial Differential Equations of first-order with

constant coefficients.

The most general form of linear partial differential equations of first

order with constant coefficients is

Aux+Buy+Ku=f(x,y) (11.15)

where A,B and K are constants

Let u(x,y) be a solution of (11.15) then

du=uxdx+uydy (11.16)

From (11.15) and (11.16) we get the auxiliary system of equations

(comparing coefficients of ux, uy and remaining terms).

A

dx=

B

dy=

Ku)y,x(f

du

(11.17)

The solution of the left pair is Bx-Ay=c or y= A

cBx , where c is an

arbitrary constant of integration

B

dy

A

dx or Bdx-Ady=0 or Bx-Ay=c by integrating both sides of

the previous equation

.

The other pair

Ku-)y,x(f

du

A

dx

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is reduced to an ordinary differential equation with u as the dependent

variable and x as the independent variable, namely

A

Ku-)y,x(f

dx

du

or A

)A

c-Bx,x(f

A

)y,x(f

A

Ku

dx

du

The integrating factor of this differential equation is eKx/A. Making

change of variable by v=ueKx/A (11.15) takes the form

Avx+Bvy = f(x,y)e Akx

=g(x,y)

The substitution v=ueKy/B in (11.15) leads to Avx+Bvy=f(x,y) eKy/B. Thus,

we need to consider only the formal reduced form

Aux+Buy=f(x,y) (11.18)

The auxiliary system of equations for (11.18) is

)y,x(f

du

B

dy

A

dx (11.19)

The solution of B

dy

A

dx is

Bx-Ay=c, which gives

x=B

cAy

Substituting this value in

)y,x(f

du

B

dy we get

)y,B

cAy(f

du

B

dy

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or du=F(y,c) dy where F(y,c)= B

)y,B

cAy(f

Solution of this equation is

u=G(y,c)+c1, where Gy (y,c)= F(y,c).

Thus, the general solution is obtained by replacing c1 by (c) and c by

Bx-Ay, thereby yielding

u(x,y)=G(y, Bx-Ay)+ ( Bx-Ay), and the solution of equation (11.15) is

u(x,y)=[G(y, Bx-Ay)+ ( Bx-Ay)]e-Kx/A (11.20)

Equations (11.17) are called the equations of the Characteristics.

These equations contain two independent equations, with two solutions of the

form F(x,y,u)=c1 and G(x,y,u)=c2. Each of these represents a family of

surfaces. The curves of intersection of these two families of surfaces are

known as the characteristics of the partial differential equation. The

projections of these curves in the (x,y)-plane are called the base

characteristics. The general solution represents a family of surfaces, and

these surfaces are called integral surfaces.

Thus, the equation Bx-Ay=c represents a family of planes. The

intersection of any one of these planes with an integral surface is a curve

whose projection in the (x,y)-plane is again given by Bx-Ay=c, but this time

this equation represents a straight line and is the base characteristic.

Therefore, the solution u on a base characteristic Bx-Ay=c is given by

u=G(y,c)+c1, and the general solution is the same as above.

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Example 11.6 Find the general solution of the first-order linear partial

differential equation with the constant coefficients:

4ux+uy=x2y

Solution: The auxiliary system of equations is

yx

du

1

dy

4

dx2

From here we get

1

dy

4

dx or dx-4dy=0. Integrating both sides

we get x-4y=c. Also yx

du

4

dx2

or x2y dx=4du

or x2 dx)4

c-x( =4du or

16

1 (x3 – cx2) dx = du

Integrating both sides we get

u=c1+ 192

cx4-x3 34

= f(c)+ 192

cx4-x3 34

After replacing c by x-4y, we get the general solution

u=f(x-4y)+ 192

x)y4-x(4-x3 34

=f(x-4y)- 12

yx

192

x 34

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11.3.2. Lagrange's Method

The general form of first-order linear partial differential equations with

variable coefficients is

P(x,y)ux+Q(x,y)uy+f(x,y)u=R(x,y) (11.21)

We can eliminate the term in u from (11.21) by substituting u=ve-(x,y),

where (x,y) satisfies the equation

P(x,y) x(x,y)+ Q (x,y) y(x,y)=f(x,y)

Hence, Eq (11.21) is reduced to

P(x,y)ux+Q (x,y) uy =R(x,y) (11.22)

where P,Q,R in (11.22) are not the same as in (11.21). The following theorem

provides a method for solving (11.22) often called Lagrange's Method.

Theorem 11.1 The general solution of the linear partial differential equation of

first order

Pp+Qq=R; (11.23)

where p=y

uq,

x

u

, P, Q and R are functions of x y and u

is F(, ) = 0 (11.24)

where F is an arbitrary function and (x,y,u) =c1 and (x,y,u)=c2 form a

solution of the auxiliary system of equations

R

du

Q

dy

P

dx (11.25)

Proof: Let (x,y,u)=c1 and (x,y,u)=c2 satisfy (11.25), then equations

xdx+y dy +udu=0

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and

R

du

Q

dy

P

dx

must be compatible, that is, we must have P x+Qy+Ru=0

Similarly we must have

Px+Qy+Ru=0

Solving these equations for P,Q, and R, we have

)y,x(/),(

R

)x,u(/),(

Q

)u,y(/),(

P

(11.26)

where (,)/(y,u)= yu- yu0 denotes the Jacobian.

Let F(,)=0. By differentiating this equation with respect to x and y,

respectively, we obtain the equations

0pux

Fp

ux

F

0quy

Fq

uy

F

and if we now eliminate

Fand

F from these equations, we obtain

the equation p)u,y(

),(

+q

)x,u(

),(

=

)y,x(

),(

(11.27)

Substituting from equations (11.26) into equation (11.27), we see that

F(,)=0 is a general solution of (11.23). The solution can also be written as

=g() or =h(),

Example 11.7 Find the general solution of the partial differential equation

y2up + x2uq = y2x

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Solution: The auxiliary system of equations is

222 xy

du

ux

dy

uy

dx (11.28)

Taking the first two members we have x2dx = y2dy which on integration

given x3-y3 = c1. Again taking the first and third members,

we have x dx = u du

which on integration given x2-u2 = c2

Hence, the general solution is

F(x3-y3,x2-u2) = 0

11.3.3 Charpit's Method for solving nonlinear Partial Differential

Equation of First-Order

We present here a general method for solving non-linear partial

differential equations. This is known as Charpit's method.

Let

F(x,y,u, p.q)=0 (11.29)

be a general non linear partial differential equation of first-order. Since

u depends on x and y, we have

du=uxdx+uydy = pdx+qdy (11.30)

where p=ux=x

u

, q = uy=

y

u

If we can find another relation between x,y,u,p,q such that

f(x,y,u,p,q)=0 (11.31)

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then we can solve (11.28) and (11.30) for p and q and substitute them in

equation (11.29). This will give the solution provided (11.29) is integrable.

To determine f, differentiate (11.28) and (11.30) w.r.t. x and y so that

0x

q

q

F

x

p

p

Fp

u

F

x

F

(11.32)

0x

q

q

f

x

p

p

fp

u

f

x

f

(11.33)

0y

q

q

F

y

p

p

Fq

u

F

y

F

(11.34)

0y

q

q

f

y

p

p

fq

u

f

y

f

(11.35)

Eliminating x

p

from, equations (11.31) and (11.32), and

y

q

from

equations (11.33) and (11.34) we obtain

0dx

q

p

F

q

f-

p

f

q

Fp

p

F

u

f-

p

f

u

F

p

F

x

f-

p

f

x

F

0dy

p

q

F

p

f-

q

f

p

Fq

q

F

u

f-

q

f

u

F

q

F

y

f-

q

f

y

F

Adding these two equations and using

y

p

yx

u

x

q 2

and rearranging the terms, we get

0q

f

u

fq

y

F

p

f

u

Fp

x

F

u

f

q

Fq-

p

Fp-

y

f

q

F-

x

f

p

F-

(11.36)

385

Following arguments in the proof of Theorem 11.1 we get the auxiliary

system of equations

0

df

u

Fq

y

F

dq

u

Fp

x

F

dp

q

Fq-

p

Fp-

du

q

F-

dy

p

F-

dx

(11.37)

An Integral of these equations, involving. p or q or both, can be taken

as the required equation (11.30). p and q determined from (11.28) and (11.30)

will make (11.29) integerable.

Example 11.8 Find the general solution of the partial differential equation.

0u-yy

ux

x

u22

(11.38)

Solution: Let p = x

u

, q =

y

u

The auxiliary system of equations is

2222 q-q

dq

p-p

dp

)yqxp(2

du

qy2

dy

px2

dx

(11.39)

which we obtain from (11.36) by putting values of

22 qy

F,1-

u

F,p

x

F,qy2

q

F,px2

p

F

and multiplying by -1 throughout the auxiliary system. From first and 4th

expression in (11.38) we get

dx = py

pxdp2dxp2 . From second and 5th expression

dy= qy

qydq2dyq2

386

Using these values of dx and dy in (11.38) we get

xp

pxdp2dxp2

2 =

yq

qydq2dyq2

2

or q

dq2

y

dydp

p

2

x

dx

Taking integral of all terms we get

ln|x| + 2ln|p| = ln|y|+2ln|q|+lnc

or ln|x| p2 = ln|y|q2c

or p2x=cq2y, where c is an arbitrary constant. (11.40)

Solving (11.37) and (11.39) for p and q we get cq2y+q2y -u=0

(c+1)q2y=u

q=2

1

y)1c(

u

p= 2

1

x)1c(

cu

(11.29) takes the following form in this case

du= dyy)1c(

udx

x)1c(

cu 21

21

or dyy

idx

x

cdu

u

c1 21

21

21

By integrating this equation we obtain 12

12

12

1

c)y()cx()u)c1((

This is a complete solution.

387

11.3.4 Solutions of special type of partial differential equations

(i) Equations containing p and q only

Let us consider a partial differential equation of the type

F(p,q)=0 (11.41)

The auxiliary system of equations of Charpit's method (Equation

(11.36)) takes the form

0

dq

0

dp

qFpF

du

F

dy

F

dx

qpqp

It is clear that p=c is a solution of these equations. Putting value of p in

(11.40) we have

F(c,q)=0 (11.42)

So that q=G(c) where c is a constant

Then observing that

du=cdx+G(c) dy

we get the solution u=cx +G(c) y+c1,

where c1 is another constant.

Example 11.9 Solve p2+q2=1

Solution: The auxiliary system of equation is

-0

dq

0

dp

q2-p2-

du

q2

dy

p2-

dx22

or 0

dq

0

dp

qp

du

q

dy

p

dx22

388

Using dp =0, we get p=c and q= 2c-1 , and these two combined with

du =pdx+qdy yield

u=cx+y 2c-1 + c1 which is a complete solution.

Using du

dx = p , we get du =

c

dx where p= c

Integrating the equation we get u = c

x + c1

Also du = q

dy, where q = 22 c-1p-1

or du = 2c-1

dy. Integrating this equation we get u =

2c-1

1 y +c2

This cu = x+cc1 and 2c-1u = y + c2 2c-1

Replacing cc1 and c2 2c-1 by - and - respectively, and eliminating

c, we get

u2 = (x-)2 + (y-)2

This is another complete solution.

This is another complete solution.

(ii) Clairaut equations

An equation of the form

u=px+qy+f(p,q)

or

F=px+qy+f(p,q)-u=0 (11.43)

is known as Clairaut equation.

389

The auxiliary system of equations for Clairaut equation takes the form

pfx

dx

=

qfy

dy

=

qp qfpfqypx

du

=

0

dp=

0

dq

From here we find that

dp=0, dq=o implying

p=c1, q=c2

If we put these values of p and q in Eq. (11.42), we get

u = c1 x +c2y +f (c1, c2)

Therefore, F(x,y,u,c1,c2) = c1x + c2y + f (c1,c2) -u=0 is a complete

solution of (11.42).

(iii) Equations not containing x and y

Consider a partial differential equation of the type

F(u,p,q) = 0 (11.44)

The auxiliary system of equations take the form

pF

dx =

qF

dy =

qq qFpF

du

=

upF-

dp =

uqF-

dq

The last two terms yield p

dp =

q

dq

i.e. p = a2q where a2 is an arbitrary constant

This equation together with 11.43 can be solved for p and q and we

proceed as in previous cases.

Example 11.10 Solve u2+pq – 4 = 0

Solution. The auxiliary system of equations is

390

q

dx =

p

dy =

pq2

du =

up2-

dp =

uq2-

dq

The last two equations yield p = a2q.

Substituting in u2+pq – 4 = 0 gives

q = 2u-4a

1 and p = + a 2u-4

Then du = pdx+qdy yields

du = +

dy

a

1adxu-4 2

or 2u-4

du = + dy

a

1adx

Integrating we get sin--1

2

u = +

cy

a

1adx

or u = + 2 sin

cy

a

1ax

which is the required complete solution.

(iv) Equations of the type

f(x,p) = g(y,q)

Then each of these functions must be constant, that is

f(x, p) = g(y, q) = C

Solving for p and q, and using du=pdx+qdy we can obtain the solution

Example 11.11 Solve p2(1-x2)-q2(4-y2) = 0

Solution Let p2(1-x2) = q2 (4-y2) = a2

391

This gives p = 2x-1

a and q =

2y-4

a

(neglecting the negative sign).

Substituting in du = pdx + q dy we have

du = 2x-1

adx +

2y-4

ady

Integration gives u = a

2

ysin' x sin' + c.

which is the required complete solution.

11.3.5. Geometric concepts related to Partial Differential Equations of

First order

We have discussed geometrical interpretation of a first order ordinary

differential equation in chapter. .........

The situation for a partial differential equation is some what

complicated. In this case the values of p=x

u

, q=

y

u

are not unique at a point

(x,y,u). If an integral surface is g(x,y,u)=0, then p and q represent the slopes

of the curves of intersection of the surface with the planes y=constant and

x=constant, respectively. Moreover, p,q,-1 represent the direction ratios of the

normals to the surface at the point (x,y,u). The derivatives p and q are

constrained by F(x,y,u,p,q)=0. Obviously, at a fixed point, p and q can be

represented by a single parameter. Hence, there are infinitely many possible

normals and consequently infinitely many integral surfaces passing through

392

any fixed point. So, unlike the case of ordinary differential equations, we

cannot determine a unique integral surface by making it pass through a point.

Cauchy established that a unique integral surface can be obtained by

making it pass through a continuous twisted space curve, also known as an

initial curve, except when the curve is a characteristic of the differential

equation.

The infinity of normals passing through a fixed point generates a cone

known as the normal cone. The corresponding tangent planes to the integral

surfaces envelope a cone known as the Monge cone. In the case of a linear

or a quasi linear equation, the normal cone degenerates into a plane since

each normal is perpendicular to a fixed line. Consider the equation ap+bq=c,

where a,b, and c are functions x,y, and u. Then the direction p,q,-1 is

perpendicular to the direction ratios a,b,c. This direction is fixed at a fixed

point. The Monge cone then degenerates into a coaxial set of planes known

as the Monge pencil. The common axis of the planes is the line through the

fixed point with direction ratios a,b,c. This line is known as the Monge axis.

11.4 Solutions of Linear Partial Differential Equation of Second Order

with Constant Coefficients

11.4.1 Homogeneous Equations

Let Dx= ,y

D,x

D,y

D,x i

ii

yi

ii

xy

We are looking for solving equations of the type

393

0y

uk

yx

uk

x

u2

2

2

2

12

2

(11.45)

where k1 and k2 are constants.

(11.44) can be written as

0uDkDDkD 2

y2yx1

2

x

or F(Dx, Dy) u=0 (11.46)

The auxiliary equation of (11.45) (compare with Section 5.5) is

0DkDDkD 2

y2yx1

2

x

Dy then equation (11.45) can be written as

Let the roots of this equation be m1 and m2, that is, Dx=m1Dy, Dx=m2Dy

(Dx-m1Dy) (Dx-m2Dy)u=0- (11.47)

This implies

(Dx-m2Dy) u=0 or p-m2q=0

The auxiliary system of equations for p-m2q=0 is of the type

0

du

m-

dy

1

dx

2

This gives us -m2dx=dy

or y+m2x=c

and u=c1= (c)

Thus, u=(y+m2x) is a solution of (11.44).

From (11.46) we also have (Dx-m1Dy) u=0

or p-m1q=0

Its auxiliary system of equations is

394

0

du

m-

dy

1

dx

1

This gives –m1dx=dy or m1x+y=c1 and u=c2 and so u=(y+m1x) is a

solution of (11.44).

Therefore u= (y+m2x) + (y+m1x) is the complete solution of (11.44)

If the roots are equal (m1 = m2) then Equation 11.44 is equivalent to

(Dx-m1Dy)2 u = 0

Putting (Dx-m1Dy) u = z, we get

(Dx-m1Dy) z=0 which gives

z= (y+m1x)

Substituting z in (Dx-m1Dy) u=z gives

(Dx-m1Dy) u = (y+m1x)

or p-m1q = (y+m1x)

Its auxiliary system of equations is

)xmy(

du

m-

dy

1

dx

11

which gives y+m1x = a & u + (a) x+b

The complete solution in this case is

u= x (y+m1x) + (y+m1x)

Example 11.12 Find the solution of the equation

2

2

2

2

y

u-

x

u

= 0

Solution: In the terminology introduced above this equation can be written as

395

(Dx2-Dy

2) u = 0.

or (Dx-Dy) (Dx+Dy)u=0

Its auxiliary equation is

(Dx-Dy)(Dx+Dy)=0,

that is, Dx - Dy =0

or Dx= -Dy. that is,

p=q or p = - q

p-q = 0 or p+q=0

Auxiliary system of equations for p-q=0 is

0

du

1-

dy

1

dx

This gives x+y = c.

The auxiliary system for p+q = 0 is

0

du

1

dy

1

dx

This gives x-y =c1

The complete solution is

u=(x+y)+ (x-y) where and are arbitrary functions.

Non-homogeneous Partial Differential Equations of the second-order

Equations of the type

2

2

2

2

12

2

y

uk

yx

uk

x

u

=f(x,y) (11.48)

are called non-homogeneous partial differential equations of the

second-order with constant coefficients.

396

Let uc be the general solution of

2

2

2

2

12

2

y

uk

yx

uk

x

u

= 0 (11.49)

and let up be a particular solution of (11.47)

Then uc+up is the solution of (11.47)

We have discussed the method for finding the general solution

(complementary function) of (11.48). In Section 5.6 we described the method

of undetermined coefficients for ordinary differential equations. That method is

applicable in finding particular solution of partial differential equations of the

type (11.47) Let f(Dx,Dy) be a linear partial differential operator with constant

coefficients, then the corresponding inverse operator is defined

as )D,D(f

1

yx

The following results hold

f(Dx,Dy) )y,x()y,x()D,D(f

1

yx

(11.50)

)y,x(

)D,Df

1

)D,D(f

1)y,x(

)D,D(f)D,D(f

1

yx2yx1yx2yx1

(11.51)

= )D,D(f

1

yx2

)y,x(

)D,D(f

1

yx1

(11.52)

397

)y,x()D,D(f

1

)y,x()D,D(f

1)y,x()y,x(

)D,D(

1

2

yx

1

yx

21

yx

(11.53)

Kasus 1,

0),(,),(

1

),(

1 bafe

bafe

DDf

byaxbyax

yx

(11.54)

Jika 𝑓(𝑎, 𝑏) = 0 , perhatikan bahwa (𝐷𝑥 −𝑎

𝑏𝐷𝑦)

𝑘

adalah factor dari

𝑓(𝐷𝑥, 𝐷𝑦). Dalam kasus ini,

1

(𝐷𝑥 −𝑎𝑏𝐷𝑦)

𝑒𝑎𝑥+𝑏𝑦 = 𝑥𝑒𝑎𝑥+𝑏𝑦

1

(𝐷𝑥 −𝑎𝑏𝐷𝑦)

𝑘 𝑒𝑎𝑥+𝑏𝑦 =

𝑥𝑟

𝑟!𝑒𝑎𝑥+𝑏𝑦

Kasus 2,

f(Dx,Dy) (x,y) eax+by=eax+by f(Dx+a, Dy+b) (x,y)

)y,x()bD,aD(f

1ee)y,x(

)D,D(f

1

yx

byaxbyax

yx

(11.55)

= )y,x(e)bD,D(f

1e)y,x(e

)D,aD(f

1e ax

yx

byby

yx

ax

(11.56)

f )D,D( 2

y

2

x cos (ax+by) = f(-a2,-b2) cos (ax+by)

)byax(cos)b-,a-(f

1)byax(cos

)D,D(f

1222

y

2

x

(11.57)

f )D,D( 2

y

2

x sin (ax+by) = f(-a2,-b2) sin (ax+by)

398

)byax(sin)b-,a-(f

1)byax(sin

)D,D(f

1222

y

2

x

(11.58)

When (x,y) is any function of x and y, we resolve )D,D(f

1

yx

into partial

fractions treating f(Dx, Dy) as a function of Dx alone and operate each partial

fraction on (x,y), remembering that

yx DmD

1

(x,y) = dx)mxc,x(

where c is replaced by y+mx after integration.

Example 11.13

Find the particular solution of the following partial differential equations

(i) y3x2

2

2

ey

u-

yx

u4

x

u3

(ii) )yxsin(ey

u-

x

u3 x

2

2

Solution: (i) The equation can be written as

)D-DD4D3( yyx

2

x u = ex-3y

up = yyx

2

x D-DD4D3

1

ex-3y

= )3-(-)3-(43

1

ex-3y by (11.53)

= 6

1- ex-3y

(ii) The equation can be written as

399

(3D2x-Dy)u=ex sin (x+y)

up = y

2

x D-D3

1 ex sin (x+y)

= ex )D-)1D(3(

1

y

2

x sin (x+y)

= ex )D3D6D3(

1

yx

2

x sin(x+y)

= ex yx D-3D6)1-(3(

1

sin(x+y)

= ex

yx D-D6

1sin (x+y) = ex

y22

x

yx

D-D36

)DD6( sin(x+y)

= ex 35-

)yx(cos7

= - 5

1 ex cos(x+y).

Example 11.14 Solve the partial differential equation

2

22

2

2

x

uc-

t

u

= e-xsin t

Solution: The equation can be written as

(D 2

t -c2Dx2) u = e-xsin t

The particular solution is

up= tsineDc-D

1 x

2

x

22

t

= x

x

2

t

x etsin)1-D(c(-D

1e tsin

c-1-

12

400

= - tsine1c

1 x

2

By proceeding on the lines of the solution of Example 11.12 we get

uc = (x-ct)+ (x+ct)

u(x,t)= (x-ct)+ (x+ct) - tsine1c

1 x

2

The solution uc is known as the d' Alembert's solution of the wave

equation

2

22

2

2

x

uc-

t

u

=0.

11.5 Monge's Method for a special class of non linear Equations

(quasi linear Equations) of the Second order.

Let u(x,y) be a function of two variables x and y

Let p = 2

222

y

ut,

yx

us,

x

ur,

y

uq,

x

u

Monge's method provides a technique for solving a special class of

partial differential equation of second order of the type

F(x,y,u,p,q,r,s,t)=0 (11.59)

Monge's method comprises in establishing one or two first integrals of

the form

= f() (11.60)

where and are known function of x,y,u, p and q and the function f is

arbitrary; that is, in finding relations of the type (11.59) such that equation

401

(11.58) can be derived from equation (11.59). The following equations are

obtained from it by partial differentiation.

x+up+pr+qs=f'() {x+up+pr+qs} (11.61)

y+uq+ps+qt=f'() {y+uq+ps+qt} (11.62)

It may be noted that every equation of the type (11.58) does not have a

first integral of the type (11.59). By eliminating f'() from equations (11.60) and

(11.61), we find that any second order partial differential equation which

possesses a first integral of the type (11.59) must be expressible in the form

R1r+S1s+T1t+U1(rt-s2)=V1 (11.63)

where R1, S1,T1,U1 and V1 are functions of x,y,u, p and q defined by the

relations

R1 = )q,u(

),(p

)q,x(

),(T,

)u,p(

),(q

)y,p(

),(1

(11.64)

S1= )u,p(

),(p

)x,p(

),(

)u,q(

),(q

)y,q(

),(

(11.65)

U1= )x,y(

),(

)u,y(

),(p

)x,u(

),(qV,

)q,p(

),(1

(11.66)

The equation (11.62) reduces to the form

R1r+S1s+T1t=V1 (11.67)

if and only if the Jacobian pp- qp=0 identically. Equation (11.66) is a

non-linear equation because the coefficients R1, S1, T1, V1 are functions of p

and q as well as of x,y, and u. Infact it is a quasi linear equation. We explain

here the method of finding solution of the equation of the type (11.66), namely

402

Rr+Ss+Tt = V (11.68)

for which a first integral of the form (11.59) exists. For any function u of

x and y we have the relations dp =rdx+sdy, dq=sdx+tdy (11.69)

Eliminating r and t from this pair of equations and equation (11.67), we

see that any solution of (11.67) must satisfy the relation

Rdpdy+Tdqdx - Vdxdy=0 (11.70)

Rdy2 +Tdx2 –Sdxdy=0 (11.71)

The method of finding solutions of (11.69) and (11.70) is explained

through the following example:

Example: 11.15

Solve the equation 0y

u

x

u

yx

u

y

u

x

u2-

x

u

y

u2

222

2

22

This equation is of the form (11.67) where

R= 0V,y

utand,

x

uT,

yx

us,

y

u

x

u2S,

x

ur,

y

u2

222

2

22

Therefore (11.69) and (11.70) become respectively

q2dpdy + p2dq dx=0 (11.72)

(pdx+qdy)2 = 0 (11.73)

By the equation du=pdx+qdy and (11.72) we get du=0, which gives

integral u=c1. From (11.71) and (11.72) we have qdp =pdq, which has solution

p=c2q. Thus, the first integral is

p=q f(u) (11.74)

403

where f(.) is arbitrary. We solve (11.73) by Lagrange's method. The

auxiliary system of equations (characteristic equations) are

0

du

)u(f-

dy

1

dx

with integral u=c1, y+x f(c1)=c2 leading to the general solution

y+x f(u)=g(u)

where the functions f and g are arbitrary.

404

11.6 Exercises

Write down the order and degree of partial differential equations in

problems 1-5.

1. 2uy

u

x

u

2. t

u

x

u2

2

3. 0y

u

x

u3

4. 0x

u100

t

u

5. 0y

u

x

u32

6. Verify that the functions u(x,y)=x2-y2 and u(x,y) = ex sin y are

solutions of the equation

0y

u

x

u 2

2

2

7. Let u=f(x,y), where f is an arbitrary differentiable function. Show

that u satisfies the equation

x ux –y uy = 0

Examine whether cos (xy), exy and (xy)3 are solutions of this partial

differential equation.

Classify the partial differential equations as hyperbolic, parabolic, or

elliptic.

405

8. 4 uxx-7 uxy + 3 uyy= 0

9. 4 uxx-8 uxy + 4 uyy= 0

10. a2 uxx+2a uxy +uyy = 0, a0

11. 4 0x

u9

tx

u12-

t

u2

22

2

2

12. 8 0y

u3-

yx

u2-

x

u2

22

2

2

For what values of x and y are the following partial differential

equations hyperbolic, parabolic, or elliptic?

13. uxx+2xuxy+(1-y2) uyy=0

14. (1+y2) uxx+(1+x2) uyy=0

15. uxx + x2 uyy = 0

16. uxx -2 sin x uxy – cos2x uy = 0

17. Find the general solution of 2 ux-3 uy = cos x

18. Solve ux+exuy=y, u(0,y) = 1+y

Find the complete solutions of the equations in problem 19-25

19. p=(u +qy)2

20. 2(u+xp+yq)=yp2

21. u2=pqxy

22. xp+3yq=2(u-x2q2)

23. pq=1

24. p2y(1+x2)=qx2

25. u=p2-q2

406

26. p2q2+x2y2=x2q2(x2-y2)

27. Discuss the method for finding a complete solution of the

equation of the type

F(u,p,q)=0

Solve partial differential equations of problems 28 to 32.

28. 02x

u12

x

u2

2

29. 0y

u15

y

u16-

x

u4

2

2

2

2

2

2

30. 0y

u-

yx

u4

x

u3

2

2

2

31.

y

u-

x

u3

2

2

sin (ax+by)

32.

2

2

2

2

y

u5-

y

u

x

u2-

x

u3 3x+y+ex-y

Solve equations in problems 33-36 using Monge's method

33. 2

2

2

2

y

u

x

u

34.

2

22

x

u

y

u-

yx

u

x

ux-

y

u

x

u

35. yx

u

y

u-

y

u

x

u

x

u

y

u

yx

u-

q

u

x

u2-

y

u

x

u 2

2

22

2

222

2

2

36. 2

2

y

u

x

u