Upload
votruc
View
271
Download
8
Embed Size (px)
Citation preview
1
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
3
, 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:
4
(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)
Differentiating (3) partially with respect to y,
………….. (6)
Eliminating and from (4), (5) and (6) using determinants, we
have
= 0
i.e.
or
6
(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)
From (4) and (6), we get
=
=
i.e.
Solutions of partial differential equations Consider the following two equations
………..(1)
7
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)
8
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 .
9
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)
Differentiating (5) partially with respect to ‘a’, we get
……………(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
10
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.
To find the general solution:
We put in (4), we get
……..(5)
Differentiating (5) partially with respect to ‘a’, we get
………..(6)
Eliminating ‘a’ between equations (5) and (6), we get the required general solution
To find the singular solution:
Differentiating (4) with respect to ‘a’ and ‘c’.
and 0=1 (which is absurd).
So there is no singular solution.
11
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.
To find the general solution:
We put in (3), we get
………(4)
Differentiate (4) partially with respect to ‘a’, we get
12
………..(5)
Eliminating ‘a’ between equations (4) and (5), we get the required general solution
To find singular solution,
Differentiate (3) partially with respect to ‘a’, we get
………..(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)
13
which is the complete solution.
To find the general solution:
We put in (3), we get
……..(4)
Differentiate (4) partially with respect to ‘a’, we get
……..(5)
Eliminating ‘a’ between equations (4) and (5), we get the required general
solution
To find the singular 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
14
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:
15
Given: …………(1)
Equation (1) is of the form
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)
16
This is the complete solution.
To find the general solution:
We put in (4), we get
……..(5)
Differentiate (5) partially with respect to ‘a’, we get
………..(6)
Eliminating ‘a’ between equations (4) and (5), we get the required general
solution.
To find the singular solution:
Differentiate (4) partially with respect to ‘a’ and ‘k’, we get
………..(7)
and (which is absurd)
So there is no singular solution.
(2)Solve .
Solution:
Given: ………(1)
Equation (1) is of the form
Assume where , be a solution of (1).
…….(2)
……(3)
17
Substituting equation (2) & (3) in (1), we get
Integrating the above, we get
………..(4)
This is the complete solution.
To find the general solution:
We put in (4), we get
……..(5)
Differentiate (5) partially with respect to ‘a’, we get
……..(6)
Eliminating ‘a’ between equations (4) and (5), we get the required general
solution.
To find the singular solution:
Differentiate (4) partially with respect to ‘a’ and ‘k’, we get
18
………..(7)
and (which is absurd)
So there is no singular solution.
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’.
The general and singular solution of (1) can be found out by usual method.
Problems:
(1)Solve .
Solution:
Given:
………..(1)
Equation (1) is of the form
19
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)
This is the complete solution.
The general and singular solution of (1) can be found out by usual method.
(2) Solve .
Solution:
Given:
………….. (1)
Equation (1) is of the form
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,
20
……………(4)
This is the complete solution.
The general and singular solution of (1) can be found out by usual method.
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.
21
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
22
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
23
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
,
24
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.
Therefore the complete solution of (2) is
Therefore the complete solution of (1) is
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)
25
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 (3) is
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:
26
(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,
27
…………..(3)
Differentiating (1) partially with respect to y,
…………...(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.
28
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)
29
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)= .
.
Integrating, we get ……………(3)
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)= .
.
Integrating, we get ……………(2)
30
Using the multipliers y,x,-1, each ratio in (1)= .
.
Integrating, we get ……………(3)
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.
31
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.
The general solution is given by
Case 3:
‘r’ of the roots of (2) are equal and others distinct.
The general solution is given by
To find particular integral:
Rule (1): If the R.H.S of a given PDE is , then
Put
32
If refer to Rule (4).
Rule (2): If the R.H.S of a given PDE is , then
Replace provided the
denominator is not equal to zero.
If the denominator is zero, refer to Rule (4).
Rule (3): If the R.H.S of a given PDE is , then
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:
33
The auxiliary equation is
The general solution of the given equation is
(2)Solve
Solution:
Given:
The auxiliary equation is
34
Therefore the general solution is
(3)Solve
Solution:
Given:
The auxiliary equation is
35
Therefore the general solution is
36
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
37
…….. (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
38
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)
Substituting (3) in (1) we get
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)
Substituting (2) in (1) we get
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
Ans: Given ……. (1)
Substituting (2) & (3) in (1), we get
(12)Solve:
Ans: The given equation can be written as
We know that the C.F corresponding to the factors
In our problem
40
(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:
Ans: Auxiliary equation
(15)Obtain partial differential equation by eliminating arbitrary constants a and b from
Ans: Given ……. (1)
Substituting (2) & (3) in (1), we get
(16)Find the general solution of
Ans: Auxiliary equation is
41
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 (2) in (1) we get
Substituting the above in (1) we get
This gives the complete integral. (18)Solve:
Ans: Auxiliary equation
(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),
42
From (3) and (4), we get
This is the required PDE. (20)Solve:
Ans: Auxiliary equation
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: