FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS
1.1. INTRODUCTION
If a differential equation contains one or more partial derivatives, it
is called a partial differential equation.
Generally, independent variables are taken as x and y and dependent
variable is taken as z. We denote the partial derivatives of first and
higher orders as
z z  2 z  2 z  2 z
, , 2,
, 2
x y x xy y
throughout this book we shall use symbols p, q, r, s, t respectively for
z z  2 z  2 z  2 z
, , 2,
, 2
x y x xy y
In this unit, we shall confine ourself to the study of some of the
partial differential equations.
1.2. ORDER AND DEGREE OF PARTIAL DIFFERENTIAL
EQUATIONS
If a partial differential equation contains nth and lower order
derivatives, it is said to be of nth order.
Degree of a partial differential equation is the greatest exponent of
the highest order. An equation is said to be linear if the dependent
variable and its partial derivatives occur in first degree and are not
multiplied in the differential equation.
1.3. FORMATION OF EQUATION BY THE ELIMINATION OF
ARBITRARY CONSTANTS
Let
f(x,y,z,a,b) =0
…(1)
be a relation, where z is regarded as a function of two independent
variables x andy, and a and b denote the arbitrary constants.
Differentiating (1) partially w.r.t. x, we have
f f z
 . 0
x z x
i.e.,
f f
 .p  0
x z
…(2)
Differentiating (1) partially w.r.t. y, we have
f f z
 . 0
x z x
f f
 .q  0
i.e.,
…(3)
x z
Eliminating a and b from (1), (2) and (3), we obtain a partial
differential equation of the form
f (x, y, z, p, q) =0
…(4)
which is a partial differential equation of first order.
If in the above relation (1), there are more arbitrary constants than
the number of independent variables, then elimination of constants will
give us the partial differential equation of higher order than the first.
SOLVED EXAMPLES
Example 1. Form the partial differential equation by eliminating
arbitrary constants a and b from z=ax+by+a
Solution. The given equation is
z = ax + by + a2 + b2
…(1)
Differentiating(1) partially w.r.t. x, we have
z
a
…(2)
x
Differentiating (1) partially w.r.t. y, we have
z
b
…(3)
x
To eliminate a and b, putting their values from (2) and (3) in (1), we have
2
zx
2
z
z  z   z 
 y       ,
x
y  x   y 
which is the required differential equation.
Example 2. Form the partial differential equation by eliminating h
and k from the equation
(x – h)2 + (y – k)2 + z2 = 2.
Solution. The given equation is
(x – h)2 + (y – k)2 + z2 = 2.
…(1)
Differentiating (1) partially w.r.t. x, we have
z
2( x  h )  2 z  0
…(2)
x
Differentiating (1) partially w.r.t. y, we have
z
0
y
z
x  h  z
y
z
x  k  z
y
2( y  k )  2 z
Form (2),
Form (3),
…(3)
Substituting the values of (x – h) and (y – k) in (1), the required
differential equation is
2  z 
2
2  z 
2
z    z    z 2  2 .
 x 
 y 
Example 3. Find a partial differential equation by eliminating a, b, c
from
x2 y 2 z 2
 2  2  1.
2
a
b
c
x2
Solution. The given equation is a 2

y2
b
2

z2
c
2
1
…(1)
Differentiating (1) partially w.r.t. x, we have
2x
a2
i.e.,
2 z z
. 0
2 x
c
[Differentiating partially w.r.t. x implies is a constant]
x
z z
 2
0
2
a
c x
z
c xa z
0
x
2
or

2
…(2)
Differentiating (1) partially wr.t. y, we have
2 y 2 z z
2
2 z


0

c
y

b
z
0
2
2 x
y
b
c
…(3.)
In this example, the number of arbitrary constants is three more than
the number of independent variables. Therefore, more equations are
required to eliminate a, b, c.
Differentiating (2) w.r.t. x and (3) w.r.t. y, we have
2 z
2
2
2
2  z 
c a z 2 a   0
x
 x 
…(4)
2
 z 
c 2  b2 z 2  b2    0
y
 y 
2 z
…(5)
a 2 z z
Putting the value of c   x x from (2) in (4), we get
2
2
2
a 2 z z
 z 
2  z

 a z 2  a2    0
x x
x
 x 
Dividing by a2 and multiplying by x, we get
2 z
2
z
 z 
xz 2  x   z
0

x

x
x
 
…(6)
which is the required differential equation.
Another form of the differential equation can also be obtained which
is as follows:
b 2 z z
c 
.
y y
2
Form (3),
Substituting the value
of c2 in (5), we get
2
2
b2 z z
2  z
2  z 

 b z 2  b    0
y x
y
 y 
Multiplying by y and dividing by b2 we get
2
 z 
z
yz 2  y   z
 0.

y

y
y
 
2 z
Example 4. Find the differential equation by eliminating the
arbitrary constants  and A from the equation z = Ae
Solution. The given equation is
 2t
cos x.
z = Ae
Differentiating (1) partially w.r.t. x, we have
z
 2t
 Ae
( sin x)
x
Differentiating (1) partially w.r.t. t, we have
z
 2t 2
  Ae
 cos x
t
Differentiating (2) partially w.r.t x, we have
2 z
x 2
2t
  Ae
. 2 cos x
 2t
cos x. .
....(1)
…(2)
…(3)
…(4)
From (3) and (4), we have
2z
x 2

z
,
t
which is the required differential equation.
Example 5. Find the differential equation of all spheres of fixed
radius having centre in xy-plane.
Solution. Equation of given family of spheres having centre in xy
plane i.e., at (h, k, 0) and fixed radius r is
x  h2   y  k 2  z 2  r 2
…(1)
where h, k are arbitrary constants.
Differentiating (1) partially w.r.t. x, we have
z
2(x  h)  2 z
0
x
or
z
(x  h)  z
0
x
Differentiating (1) partially w.r.t. y, we have
2(y  h)  2 z
z
0
y
…(2)
(y  h)  z
or
z
0
y
x  h  z
From (2),
y  k  z
From (3),
…(3)
z
x
z
y
Substituting these values in (1), the required differential equation is
2
2
 z 
 z 
z 2    z 2    z 2  r 2
 x 
 y 
or
 z 2  z 2 
z 2       1  r 2 .
 x   y 



Example 6. Find the differential equations of the set of all right
circular cones whose axes coincide with z axis.
Solution. The family of all right circular cones whose axes coincide
with z-axis, having semi-vertical angle  and vertex at (0, 0) is
( x2  y 2 )  ( z  c)2 tan 2 
where c and  are arbitrary constants.
…(1)
Differentiating (1) partially w.r.t. x and y, we have
2 x  2( z  c)
z
tan 2 
x
z
2 y  2( z  c)
tan 2 
and
y
z
x x

Dividing (2) by (3), we get y z
y
or
x
…(2)
…(3)
z
z
y
, which is the required differential equation.
y
x
Example 7. Find the differential equations of all planes which are at
a constant distance ‘a’ from the origin.
Solution. The equation of given plane is lx + my + nz = a …(1)
where l, m, n are the direction cosines of the normal from the origin to
the plane.
Thus,
12 + m2 + n2 = 1
n  1  l 2  m2
Substituting the value of n in (1), we have
lx  my  1  l 2  m 2 z  a
Differentiating (2) partially w.r.t. x, we have
2
2 z
l  1 l  m
0
x
Differentiating (2) partially w.r.t. y, we have
m  1  l 2  m2
z
0
y
…(2)
…(3)
…(4)
Now we eliminate 1, m between (2), (3) and (4).
z
l   1 l  m
x
z
m   1  l 2  m2
y
2
From (3),
From (4),
2
Squaring (5) and (6) and adding, we get
 z 2  z 2 
l  m  (1  l  m )      
 x   y  


2
2
2
2
…(5)
…(6)
2
or
or
 z   z 
(l 2  m2 )      
 x   y 
or
 z 2  z 2 
 (l 2  m2 )      
 x   y  


  z 2  z 2   z 2  z 2
(l 2  m 2 ) 1             
  x   y    x   y 


2
2
or
2
 z   z 
    
 x   y 
l 2  m2 
2
2



z

z
 
1      
 x   y 
  z 2  z 2 
      
  x   y  
1  l 2  m2  1  
2
2
1   z    z  
  x   y  



1
2
 z   z 
1      
 x   y 
2
…(7)
l
From (5),
z
x

2
 z   z 
1      
 x   y 
m
From (6),
z
x
1  l 2  m2
[Using (7)]
2
z
x
z

y
1  l 2  m2
2
 z   z 
1      
 x   y 
2
Substituting these values in (2), we got
x
z
x
2
 z   z 
1      
 x   y 
2

y
z
x
2
 z   z 
1      
 x   y 
2
or
2

1
2
 z   z 
1      
 x   y 
2
z
z
 z   z 
z  a 1        x  y
x
y
 x   y 
which is the required differential equation.
2
za
1.4 FORMATION OF EQUATION BY THE ELIMINATION OF
ARBITRARY FUNCTIOHNS
Suppose u and v are two functions of x, y, z which are connected by
the relation
f(u, v) = 0
…(1)
Differentiating (1) w.r.t. x, we get
f  u u z  f  v v z 

. 

  . 0
u  x z x  v  x z x 
or
f  u u  f  v v 
  .p    .p  0
u  x z  v  x z 
…(2)
[When we differentiate u partially w.r.t. x we regard y as constant
and treat z as dependent variable.]
Now, differentiating (1) w.r.t. y (regarding x as constant),
f  u u z  f  v v z 
  .     .   0
u  y z y  v  y z y 
or
f  u u  f  v v 
  . q     . q   0
u  y z  v  y z 
…(3)
f
f
Now we shall eliminate u and v from (2) and (3).
From (2),
f  u u 
f  v v 

.
p




  .p  0
u  x z 
v  x z 
…(4)
From (3),
f  u u 
f  v v 
  . q      . q   0
u  y z 
v  y z 
…(5)
Dividing (5) by (4) and cross multiplying, we get
 u u   v v
 
q   
 y z   x z
  u u   v v
p   
p   
  x z   y z

p 

 u v u v 
u v u v
 u v u v 

 p  


q

.  .

or  y z z y 
x y y x
 z x x z 
or
Pp + Qq = R
…(6)
where
u v u v
 (u , v)

P = y z z y written as  ( y, z )
v u u v  (u, v)


Q = x z x z  ( z , x)
u v u v  (u, v)


R = x y y x  ( x, y )
Equation (6) is the required differential equation of first order and first
degree.
Solved Examples
Example1. Find a partial differential equation by eliminating the
1

2
z

y

2
f

log
y

.
function f from the relation
x


1

2
z

y

2
f

log
y


Solution. The given equation is
x

Differentiating (1) partially w.r.t. x, we got
z
1
  1 
 2 f '   log y  .   2 
x
x
  x 
…(1)
or
 x2
z
1
 2 f '   log
x
x

y

…(2)
Differentiating (1) partially w.r.t. y, we got
z
1
 1
 2 y  2 f '   log y  .
y
x
 y
or
 z

1

 y  2 y 2   2 f '   log y 
x

 y

…(3)
 x2
Dividing (2) by (3), we have
z
x
 z
2
 y  2y 
 x

z
z 


p

,
q


x
y 

 x 2 p  ya  2 y 2
or
x 2 p  ya  2 y 2 ,
or
which is the differential equation.
Example 2. Form a partial differential equation by eliminating the
function f from f ( x  y  z, x 2  y 2  z 2 )  0 .
f ( x  y  z, x 2  y 2  z 2 )  0
Solution. Here
or
f(u, v) = 0
…(1)
where
u=x+y+z
…(2)
and
v = x2 + y2 – z2 (say)
…(3)
Differentiating (1) partially w.r.t., x, we have
f  u u z  f  v v z 
  .    .  0
u  x z x  v  x z x 
…(4)
From (2) and (3), we have
u
v
u
v
 1,
 2 x,
 1,
 2 z
x
x
z
z
u
v
 1,
 2y
y
y
Substituting these values in (4), we get
f
f
(1  p)  (2 x  2 zp)  0
u
v
f
f
(1  p)  2 ( x  pz)  0
u
v


p
z 
x 
…(5)
Similarly differentiating (1) partially w.r.t. y, we have
f  u u z  f  v v z 
  .    .  0
u  y z y  v  y z y 
f
f
(1  q)  (2 y  2 zq)  0
u
v
f
f
(1  q)  2 ( y  zq)  0
u
v
or
From (5),
From (6),
f
f
(1  p)  2 ( x  pz)
u
v
f
f
(1  q)  2 ( y  zq)
u
v
Dividing these and cross multiplying, we get
(1 + p) (y – zq) = (1 + q) (x – pz)
or
p(y + z) – (x + z) q = x – y,
which is the required differential equation.



p
z 
y 
…(6)
Example 3. Eliminate the arbitrary functions and hence obtain the
partial differential equations z = f (x cos  + y sin  - at) +  (x cos  +
y sin  + at).
Solution. Here z  f ( x cos  y sin   at )   ( x cos  y sin   at )

z
 f ' ( x cos  y sin   at ) . cos   ' ( x cos  y sin   at ) . cos
x
2z
2
2

f
'
'
(
x
cos


y
sin


at
)
cos



'
'
(
x
cos


y
sin


at
)
cos

2
x
2

Similarly, 2z  f ' ' ( x cos  y sin   at) sin 2   ' ' ( x cos  y sin   at) sin 2 
y
…(1)
…(2)
z
 f ' ( x cos   y sin   at ) . (a)   ' ( x cos   y sin   at ) . a
t
2z
2
2

f
'
'
(
x
cos


y
sin


at
)
.
(

a
)


'
'
(
x
cos


y
sin


at
)
.
a
t 2
or
1 2z
 f ' ' ( x cos   y sin   at )   ' ' ( x cos   y sin   at )
2
2
a t
…(3)
Adding (1) and (2), we get
2z 2z
 2  f ' ' ( x cos  y sin   at )   ' ' ( x cos  y sin   at )
2
x
y
i.e.,
2z 2z 1 2z
 2  2 2,
2
x
y
a t
[Using (3)]
which is the required differential equation.
Example 4. Form the partial differential equation by eliminating
xy
arbitrary function from z  f   .
Solution.
Here
 z 
 xy 
z  f 
 z 
…(1)
Differentiating (1) partially w.r.t. x, we have
z 

z
.
1

x


z
 xy  

x

 f '  y
2
z
z

 z  




…(2)
Differentiating (1) partially w.r.t. y, we have
z 

z.1  y 

z
xy
y
 
 f '  x 

2
z
z
 z  



…(3)
Dividing (2) by (3) and writing p for
or
or
or
or
z
z
and q for , we have
x
x
 z  xp 
y 2 
p
z 
 
q
 z  xq 
x 2 
 z 
p yz  xyp

q xz  xyq
pxz – xypq = qyz – xypq
pzx = qyz
px – qy= 0, which is the required differential equation.
Example 5. Eliminate the arbitrary function to form a partial
differential equation for the equation z  e ax by f (ax  by ) .
Solution. Here z  e ax by f (ax  by )
Differentiating (1) partially w.r.t. x and y, we have
z
 e axby a f (ax  by)  e axby f ' (ax  by)a
x
…(1)
…(2)
z
 e ax by (b) f (ax  by)  e ax by f ' (ax  by)b
y
…(3)
Multiplying (2) by b and (3) by a and subtracting, we have
z
z
b  a  2ab e ax by f (ax  by)
x
y
z
z
b a
x
y
 e ax by f (ax  by)
or
2ab
z
z
Then from (1), 2ab z  b  a , which is the required differential
x
y
equation.