CHAPTER 1 PARTIAL DIFFERENTIAL EQUATIONS - … 1...1 CHAPTER 1 PARTIAL DIFFERENTIAL EQUATIONS A partial differential equation is an equation involving a function of two or more variables

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CHAPTER 1

PARTIAL DIFFERENTIAL EQUATIONS

A partial differential equation is an equation involving a function of two or

more variables and some of its partial derivatives. Therefore a partial differential

equation contains one dependent variable and one independent variable.

Here z will be taken as the dependent variable and x and y the independent

variable so that .

We will use the following standard notations to denote the partial derivatives.

The order of partial differential equation is that of the highest order derivative

occurring in it.

Formation of partial differential equation:

There are two methods to form a partial differential equation.

(i) By elimination of arbitrary constants.

(ii) By elimination of arbitrary functions.

Problems

Formation of partial differential equation by elimination of arbitrary

constants:

(1)Form the partial differential equation by eliminating the arbitrary constants

from .

Solution:

Given ……………... (1)

2

Here we have two arbitrary constants a & b.

Differentiating equation (1) partially with respect to x and y respectively we get

……………… (2)

………………. (3)

Substitute (2) and (3) in (1) we get

, which is the required partial differential equation.

(2) Form the partial differential equation by eliminating the arbitrary constants

a, b, c from .

Solution:

We note that the number of constants is more than the number of independent

variable. Hence the order of the resulting equation will be more than 1.

.................. (1)

Differentiating (1) partially with respect to x and then with respect to y, we get

Differentiating (2) partially with respect to x,

……………..(4)

Where .

From (2) and (4) , .

From (5) and (6), we get

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, which is the required partial differential

equation.

(3) Find the differential equation of all spheres of the same radius c having their

center on the yoz-plane. .

Solution:

The equation of a sphere having its centre at , that lies on the -plane

and having its radius equal to c is

……………. (1)

If a and b are treated as arbitrary constants, (1) represents the family of spheres

having the given property.

Differentiating (1) partially with respect to x and then with respect to y, we

have

…………… (2)

and …………….(3)

From (2), …………….(4)

Using (4) in (3), ……………..(5)

Using (4) and (5) in (1), we get

. i.e. , which is the required partial differential

equation.

Problems

Formation of partial differential equation by elimination of arbitrary

functions:

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(1)Form the partial differential equation by eliminating the arbitrary function ‘f’

from

solution: Given

i.e. ……………(1)

Differentiating (1) partially with respect to x and then with respect to y, we

get

………….(2)

…………….(3)

where

Eliminating f’(u) from (2) and (3), we get

i.e.

(2) Form the partial differential equation by eliminating the arbitrary function ‘ ’

Solution: Given ………………(1)

Let ,

Then the given equation is of the form .

The elimination of from equation (2), we get,

5

i.e.

i.e

i.e

(3) Form the partial differential equation by eliminating the arbitrary function ‘f’

from

Solution: Given .…………(1)

Differentiating (1) partially with respect to x,

………….(2)

Where and

Differentiating (1) partially with respect to y,

…………. (3)

Differentiating (2) partially with respect to x and then with respect to y,

…………. (4)

and ………….. (5)


………….. (6)

Eliminating and from (4), (5) and (6) using determinants, we

have

= 0

i.e.

or

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(4) Form the partial differential equation by eliminating the arbitrary function ‘ ’

from

.

Solution: Given …………...(1)

Where

Differentiating partially with respect to x and y, we get

…………(2)

………….(3)

………..(4)

…………(5)

…………..(6)


=

=

i.e.

Solutions of partial differential equations Consider the following two equations

………..(1)

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and ………..(2)

Equation (1) contains arbitrary constants a and b, but equation (2) contains only one

arbitrary function f.

If we eliminate the arbitrary constants a and b from (1) we get a partial differential

equation of the form . If we eliminate the arbitrary function f from (2) we get

a partial differential equation of the form .

Therefore for a given partial differential equation we may have more than one type of

solutions.

Types of solutions:

(a) A solution in which the number of arbitrary constants is equal to the number of

independent variables is called Complete Integral (or) Complete solution.

(b) In complete integral if we give particular values to the arbitrary constants we get

Particular Integral.

(c) The equation which does not have any arbitrary constants is known as Singular

Integral.

To find the general integral:

Suppose that ………....(1)

is a first order partial differential equation whose complete solution is

………..(2)

Where a and b are arbitrary constants.

Let , where ‘f’is an arbitrary function.

Then (2) becomes

……….(3)

Differentiating (3) partially with respect to ‘a’, we get

……….(4)

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The eliminant of ‘a’ between the two equations (3) and (4), when it exists, is called the

general integral of (1).

Methods to solve the first order partial differential equation: Type 1:

Equation of the form ………...(1)

i.e the equation contains p and q only.

Suppose that …….....(2)

is a solution of the equation

substitute the above in (1), we get

on solving this we can get , where is a known function.

Using this value of b in (2), the complete solution of the given partial differential

equation is

…………(3)

is a complete solution,

To find the singular solution, we have to eliminate ‘a’ and ‘c’ from

Differentiating the above with respect to ‘a’ and ‘c’, we get

,

and 0=1.

The last equation is absurd. Hence there is no singular solution for the equation of

Type 1.

Problems:

(1) Solve .

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Solution:

Given: ………….(1)

Equation (1) is of the form .

Assume ………….(2)

be the solution of equation (1).

From (2) we get .

(1)

……….(3)

Substitute (3) in (2) we get

……....(4)

This is a complete solution.

To find the general solution:

We put in (4),where ‘f’ is an arbitrary function.

i.e. …………(5)


……………(6)

Eliminating ‘a’ between equations (5) and (6), we get the required general solution.

To find the singular solution:

Differentiate (4) partially with respect to ‘a’ and ‘c’, we get

,

0=1.(which is absurd)

so there is no singular solution.

(2) Solve

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Solution:

Given: ………..(1)

Equation (1) is of the form

Assume ………(2)

be the solution of equation (1).

From (2) we get

(1)

…….....(3)

Substituting (3) in (2), we get

………(4)

This is a complete solution.


We put in (4), we get

……..(5)


………..(6)

Eliminating ‘a’ between equations (5) and (6), we get the required general solution


Differentiating (4) with respect to ‘a’ and ‘c’.

and 0=1 (which is absurd).

So there is no singular solution.

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Type 2: (Clairaut’s type)

The equation of the form

………(1)

is known as Clairaut’s equation.

Assume …………(2)

be a solution of (1).

Substitute the above in (1), we get

………..(3)

which is the complete solution.

Problem:

(1) Solve

Solution:

Given: ……….(1)

Equation (1) is a Clairaut’s equation

Let ………..(2)

be the solution of (1).

Put in (1), we get

……….(3)

which is a complete solution.



………(4)

Differentiate (4) partially with respect to ‘a’, we get

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………..(5)

Eliminating ‘a’ between equations (4) and (5), we get the required general solution

To find singular solution,


………..(6)

Differentiate (3) partially with respect to ‘b’, we get

………..(7)

Multiplying equation (6) and (7),we get

(2) Solve

Solution:

Given: ……….(1)

Equation (1) is a Clairaut’s equation

Let ………...(2)

be the solution of (1).

Put in (1), we get

……….(3)

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which is the complete solution.



……..(4)


……..(5)

Eliminating ‘a’ between equations (4) and (5), we get the required general

solution


Differentiate (3) partially with respect to ‘a’,

.............. (4)

Differentiate (3) partially with respect to ‘b’,

………….(5)

Substituting equation (4) and (5) in equation (3), we get

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Type 3:

Equations not containing x and y explicitly, i.e. equations of the form

……….(1)

For equations of this type ,it is known that a solution will be of the form

……….(2)

Where ‘a’ is the arbitrary constant and is a specific function to be found out.

Putting , (2) becomes

and

If (2) is to be a solution of (1), the values of p and q obtained should satisfy (1).

i.e. ……..(3)

From (3), we get

……….(4)

Now (4) is a ordinary differential equation, which can be solved by variable separable

method.

The solution of (4), which will be of the form , is the

complete solution of (1).

The general and singular solution of (1) can be found out by usual method.

Problems:

(1)Solve .

Solution:

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Given: …………(1)


Assume where, be a solution of (1).

…….(2)

……(3)

Substituting equation (2) & (3) in (1), we get

By variable separable method,

By integrating, we get

……….(4)

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This is the complete solution.



……..(5)


………..(6)


solution.


Differentiate (4) partially with respect to ‘a’ and ‘k’, we get

………..(7)

and (which is absurd)


(2)Solve .

Solution:

Given: ………(1)


Assume where , be a solution of (1).

…….(2)

……(3)

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Substituting equation (2) & (3) in (1), we get

Integrating the above, we get

………..(4)




……..(5)


……..(6)


solution.


Differentiate (4) partially with respect to ‘a’ and ‘k’, we get

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………..(7)

and (which is absurd)


Type 4:

Equations of the form

………..(1)

i.e. equation which do not contain z explicitly and in which terms containing p and x can

be separated from those containing q and y.

To find the complete solution of (1),

We assume that .where ‘a’ is an arbitrary constant.

Solving ,we can get and solving ,we can get

.

Now

i.e.

Integrating with respect to the concerned variables, we get

……….(2)

The complete solution of (1) is given by (2), which contains two arbitrary constants ‘a’

and ‘b’.


Problems:

(1)Solve .

Solution:

Given:

………..(1)


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Let (say)

………….(2)

Similarly, …………(3)

Assume be a solution of (1)

Substitute equation (2) and (3) to the above, we get

Integrating the above we get,

………..(4)



(2) Solve .

Solution:

Given:

………….. (1)


Let (say)

…………. (2)

Similarly, ……………(3)

Assume be a solution of (1)

Substitute equation (2) and (3) to the above, we get

Integrating the above we get,

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……………(4)



Equations reducible to standard types-transformations: Type A:

Equations of the form .

Where m and n are constants, each not equal to 1.

We make the transformations .

Then and

Therefore the equation reduces to .which is a

type 1 equation.

The equation reduces to .which is a type 3

equation.

Problem:

(1)Solve .

Solution:

Given:

This can be written as

.

Which is of the form , where m=2,n=2.

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Put

Substituting in the given equation,

.

This is of the form .

Let , where

Equation becomes, .

Solving for ,

is a complete solution.

The general and singular solution can be found out by usual method.

Type B:

Equations of the form .

Where k is a constant, which is not equal to -1.

We make the transformations .

Then and

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Therefore the equation reduces to which is a type 1

equation.

The equation reduces to which is a

type 4 equation.

Problems:

(1)Solve:

Solution:

Given:

The equation can be rewritten as ………(1)

Which contains .

Hence we make the transformation

Similarly

Using these values in (1), we get

………..(2)

As (2) is an equation containing P and Q only, a solution of (2) will be of the form

………….(3)

Now obtained from (3) satisfy equation (2)

i.e.

Therefore the complete solution of (2) is

i.e complete solution of (1) is

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Singular solution does not exist. General solution is found out as usual.

Type C:

Equations of the form , where

We make the transformations

Then

and

Therefore the given equation reduces to

This is of type 1 equation.

Problem:

(1)Solve

Solution:

Given:

It can be rewritten as ………..(1)

which is of the form

we make the transformations

i.e.

Then

,

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Similarly, ,

Using these in (1),it becomes

…………(2)

As (2) contains only P and Q explicitly, a solution of the equation will be of the

form

………….(3)

Therefore obtained from (3) satisfy equation (2)

i.e.



Singular solution does not exist. General solution is found out as usual.

Type D:

Equation of the form

By putting the equation reduces

to

where .

Problems:

(1)Solve .

Solution:

Given: . …………….(1)

Rewriting (1),

. …………….(2)

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As (2) contains , we make the substitutions

Then

i.e.

Similarly,

Using these in (2), it becomes

…………..(3)

which contains only P and Q explicitly. A solution of (3) is of the form

…………(4)

Therefore obtained from (4) satisfy equation (3)

i.e.


Therefore the complete solution of (1) is ……..(5)

General solution of (1) is obtained as usual.

General solution of partial differential equations: Partial differential equations, for which the general solution can be obtained

directly, can be divided in to the following three categories.

(1) Equations that can be solved by direct (partial) integration.

(2) Lagrange’s linear equation of the first order.

(3) Linear partial differential equations of higher order with constant coefficients.

Equations that can be solved by direct (partial) integration:

Problems:

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(1)Solve the equation if when when

Also show that when .

Solution:

Given: …………….(1)

Integrating (1) partially with respect to x,

…………….(2)

When in (2), we get .

Equation (2) becomes …………….(3)

Integrating (3) partially with respect to t, we get

……………(4)

Using the given condition, namely when we get

Using this value in (4), the required particular solution of (1) is

Now

i.e. when .

(2) Solve the equation simultaneously.

Solution: Given

Integrating (1) partially with respect to x,

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…………..(3)


…………...(4)

Comparing (2) and (4), we get

Therefore the required solution is

, where c is an arbitrary constant.

Lagrange’s linear equation of the first order: A linear partial differential equation of the first order , which is of the form

where are functions of is called Lagrange’s linear equation.

working rule to solve

(1)To solve , we form the corresponding subsidiary simultaneous equations

(2)Solving these equations, we get two independent solutions .

(3)Then the required general solution is .

Solution of the simultaneous equations

Methods of grouping:

By grouping any two of three ratios, it may be possible to get an ordinary

differential equation containing only two variables, eventhough P;Q;R are in

general, functions of x,y,z. By solving this equation, we can get a solution of the

simultaneous equations. By this method, we may be able to get two independent

solutions, by using different groupings.

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Methods of multipliers:

If we can find a set of three quantities l,m,n which may be constants or

functions of the variables x,y,z, such that , then the solution of the

simultaneous equation is found out as follows.

Since If is an exact

differential of some function , then we get Integrating this, we get

, which is a solution of

Similarly, if we can find another set of independent multipliers we can get

another independent solution .

Problems:

(1)Solve .

Solution:

Given: .

This is of Lagrange’s type of PDE where .

The subsidiary equations are

Taking first two members

Integrating we get

i.e. ………….(1)

Taking first and last members .

i.e. .

Integrating we get ………......(2)

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Therefore the solution of the given PDE is .

(2)Solve the equation

Solution:

Given:

This is of Lagrange’s type of PDE where .

The subsidiary equations are ……….(1)

Using the multipliers 1,1,1, each ratio in (1)= .

.

Integrating, we get ……………(2)

Using the multipliers y,x,2z, each ratio in (1)= .

.


Therefore the general solution of the given equation is .

(3)Show that the integral surface of the PDE .

Which contains the straight line .

Solution:

The subsidiary equations of the given Lagrange ‘s equation are

……………(1)

Using the multipliers . each ratio in (1)= .

.


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Using the multipliers y,x,-1, each ratio in (1)= .

.


The required surface has to pass through

……………(4)

Using (4) in (2) and (3), we have

……………(5)

Eliminating x in (5) we get,

…………….(6)

Substituting for a and b from (2) and (3) in (6), we get

, which is the equation of the required

surface.

Linear P.D.E.S of higher order with constant coefficients:

The standard form of a homogeneous linear partial differential equation of the

order with constant coefficients is

…………….(1)

where a’s are constants.

If we use the operators , we can symbolically write equation (1)

as

…………….(2)

…………….(3)

where is a homogeneous polynomial of the degree in .

The method of solving (3) is similar to that of solving ordinary linear differential

equations with constant coefficients.

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The general solution of (3) is of the form z = (complementary function)+(particular

integral),where the complementary function is the R.H.S of the general solution of

and the particular integral is given symbolically by .

Complementary function of :

C.F of the solution of is the R.H.S of the solution of

. …………(1)

In this equation, we put ,then we get an equation which is called the

auxiliary equation.

Hence the auxiliary equation of (1) is

………….(2)

Let the roots of this equation be .

Case 1:

The roots of (2) are real and distinct.

The general solution is given by

Case 2:

Two of the roots of (2) are equal and others are distinct.


Case 3:

‘r’ of the roots of (2) are equal and others distinct.


To find particular integral:

Rule (1): If the R.H.S of a given PDE is , then

Put

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If refer to Rule (4).


Replace provided the

denominator is not equal to zero.

If the denominator is zero, refer to Rule (4).


Expand by using Binomial Theorem and then operate on .

Rule (4): If the R.H.S of a given PDE is any other function [other than

Rule(1),(2) and(3)] resolve into linear factors say

etc. then the

Note: If the denominator is zero in Rule (1) and (2) then apply Rule (4).

Working rule to find P.I when denominator is zero in Rule (1) and Rule (2).

If the R.H.S of a given PDE is of the form

Then

This rule can be applied only for equal roots.

Problems:

(1) Solve

Solution:

Given:

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The auxiliary equation is

The general solution of the given equation is

(2)Solve

Solution:

Given:


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Therefore the general solution is

(3)Solve

Solution:

Given:


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Therefore the general solution is

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UNIT 1

Part A (1)Form a partial differential equation by eliminating arbitrary constants a and b from

Ans: Given ……. (1)

Substituting (2) & (3) in (1), we get

(2) Solve: Ans: Auxiliary equation

(3)Form a partial differential equation by eliminating the arbitrary constants a and b from the equation . Ans: Given: ….. (1) Partially differentiating with respect to ‘x’ and ‘y’ we get ….. (2) …... (3) (2) …… (4) (3) ….. (5) Substituting (4) and (5) in (1) we get . . (4)Find the complete solution of the partial differential equation Ans: Given:

…….. (1) Let us assume that ……… (2) be the solution of (1) Partially differentiating with respect to ‘x’ and ‘y’ we get

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…….. (3)

Substituting (3) in (1) we get

From the above equation we get,

………. (4) Substituting (5) in (2) we get

(5)Find the PDE of all planes having equal intercepts on the x and y axis. Ans: The equation of such plane is

………. (1) Partially differentiating (1) with respect to ‘x’ and ‘y’ we get

……….. (2)

……….. (3) From (2) and (3), we get (6)Find the solution of . Ans: The S.E is

Taking first two members, we get

Integrating we get

i.e

Taking last two members, we get

Integrating we get

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i.e

The complete solution is

(7)Find the singular integral of the partial differential equation Ans: The complete integral is

Therefore

(8)Solve: Ans:

…….. (1) Let us assume that …… (2) be the solution of (1) Partially differentiating with respect to ‘x’ and ‘y’ we get

…….. (3)


This is the required solution. (9)Form a partial differential equation by eliminating the arbitrary constants a and b from

Ans:

……… (1) Partially differentiating with respect to ‘x’ and ‘y’ we get

39

…….. (2)


This is the required PDE. (10)Solve:

Ans: Auxiliary equation

(11)Form a partial differential equation by eliminating the arbitrary constants a and b from



(12)Solve:

Ans: The given equation can be written as

We know that the C.F corresponding to the factors

In our problem

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(13)Form a partial differential equation by eliminate the arbitrary function f from

Ans:

From (1), we get

Substituting (3) in(2), we get

(14)Solve:


(15)Obtain partial differential equation by eliminating arbitrary constants a and b from



(16)Find the general solution of

Ans: Auxiliary equation is

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General solution is

(17)Find the complete integral of

Ans: Let us assume that ……… (1) be the solution of the given equation. Partially differentiating with respect to ‘x’ and ‘y’ we get

…….. (2)


Substituting the above in (1) we get

This gives the complete integral. (18)Solve:


(19)Find the PDE of the family of spheres having their centers on the line x=y=z. Ans: The equation of such sphere is Partially differentiating with respect to ‘x’ and ‘y’ we get

From (1),

From (2),

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This is the required PDE. (20)Solve:


Part B

(1)(i) Form a partial differential equation by eliminating arbitrary functions from

(ii) Solve: (2)(i) Solve:

(ii) Solve:

(3)(i) Solve: (ii) Solve: (4)(i) Solve: (ii) Solve: (5)(i) Solve: (ii) Solve: (6)(i) Solve: (ii) Solve: (7)(i) Solve: (ii) Solve: (8)(i) Solve: (ii) Solve:

43

(9)(i) Solve: (ii) Solve: (10)(i) Solve: (ii)Solve: (11)(i) Form the partial differential equation by eliminating from

(ii) Solve: (12)(i) Find the complete integral of (ii) Solve: (13)(i) Solve: (ii) Solve: (iii) Solve: (14)(i) Solve (ii) Solve: (15)(i) Form a partial differential equation by eliminating arbitrary functions f and g in

(ii) Solve: (16)(i) Form a partial differential equation by eliminating arbitrary functions f and g in

(ii) Solve: (17)(i) Find the singular solution of (ii) Solve: (18) (i) Solve: (ii) Find the singular integral of a partial differential equation (19)(i) Solve: (ii) Form a partial differential equation by eliminating arbitrary functions ‘f’ from

(20)(i) Solve: (ii) Solve:

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CHAPTER 1 PARTIAL DIFFERENTIAL EQUATIONS - … 1...1 CHAPTER 1 PARTIAL DIFFERENTIAL EQUATIONS A partial differential equation is an equation involving a function of two or more variables