UNIT – III

PARTIAL DIFFERENTIAL EQUATION

TYPES OF SOLUTION:

(i) COMPLETE INTEGRAL:

A solution of a PDE which contains as many arbitrary
constants as the number of independent variables is
called complete integral or complete solution of the
equation.

(ii) SINGULAR INTEGRAL:

Let the complete integral of the PDE f(x,y,z,p,q) = 0 be
ϕ(x,y,z,a,b) =0 →(1)
Differntiating (1) partially w.r.t ‘a’ and ‘b’ and then
equate them to 0.
∂ϕ ∂ϕ
Eliminating ‘a’ and ‘b’ from =0 and =0 we get
∂a ∂b
singular integral.

(iii) GENERAL SOLUTION:

A solution of a PDE which contains as many arbitrary
functions as the order of the equation is called general
solution or the general integral of the equation.

NOTATIONS USED IN PDE :

p= ∂z , q = ∂z , r = ∂ ∂2z
and t = ∂
2 2
z z
, s= .
∂x ∂y ∂x 2
∂x∂y ∂y 2
 SOLVING FIRST ORDER NON LINEAR PDE( 6 TYPES)
STANDARD TYPES
Singular General Integral
Types Complete Integral
Integral

Substitute c = φ(a)
z = ax + y f(a) + φ(a)
I
-------------(1)
∂z
f(p,q) = 0
= x + y f ’(a) + φ’(a)
z = ax + y f(a) + c
NO -------------(2)
Example: ∂a
pq+p+q=0 Eliminating ‘a’ from (1) and (2), we get General integral

Equating
∂z
II
= 0 and
∂z
z =px+qy + f(p,q)
∂a
put b = φ(a) in complete integral and find
∂z
then
∂a
z = ax +by + f(a,b)
Example = 0 and
p ∂b eliminating ‘a’ between them, we get General integral
z = px + qy + -p eliminating
q
‘a’& ‘b’

III Put z = f(u) where u = x + ay.

f(z,p,q) = 0 dz
Replacing p by and q by
du Can be found
Follow Type II
Example: dz as usual
z = p2 + q2 a Separate the variables ‘z’
du
and ‘u’ then integrate.

IV Put F1(x,p) = a = F2(y,q)

F1(x,p) = F2(y,q) Find p and q interms of ‘a’.
NO Follow Type II
Example: Sub. these values in
p2 + q2 = x2 + y2 dz=pdx + qdy and integrate.
EQUATIONS REDUCIBLE TO STANDARD TYPES

TYPE V: F(xmp, ynq) = 0 and F(z, xmp , ynq) = 0

Case (i) : If m ≠ 1 and n ≠ 1

Put xmp = (1 – m)P ; ynq = ( 1 – n) Q

Substitute in given equations we get a differential equation of the form

F(P,Q) = 0 → FOLLOW TYPE I
(OR) F(z, P, Q) = 0 → FOLLOW TYPE III

Case(ii) : If m = n = 1 then log x = X ⇒ xp = P and

log y = Y ⇒ yq = Q

Substitute in given equations we get a differential equation of the form

F(P,Q) = 0 → FOLLOW TYPE I
(OR) F(z, P, Q) = 0 → FOLLOW TYPE III

TYPE VI: F(zmp, zmq) = 0 or F1(x, zmp) = F2(y,zmq)

Case (i) : If m ≠ -1,Put Z = z m+1 and zmp = P

, zmq = Q
m +1 m +1
Substitute in given equations, we get a PDE of the form
F(P,Q) = 0 → FOLLOW TYPE I
(OR) F1(x,p) = F2(y,q) → FOLLOW TYPE IV
Solving the above eqns. using their corresponding working rule.

Case(ii) : If m = -1 then Put Z = log z (i.e) P = z –1 p, Q = z-1 q

Substitute in the given equations, we get a PDE of the form
F(P,Q) = 0 → FOLLOW TYPE I
(OR) F1(x,p) = F2(y,q) → FOLLOW TYPE IV
Solving the above eqns. using their corresponding working rule.

LAGRANGE’S LINEAR EQUATION

Pp + Qq = R → Lagrange’s linear equation , where P,Q,R are functions of x,y,z

TYPE I: METHOD OF GROUPING

Working rule to solve Pp + Qq = R
dx dy dz
Step 1: Write the subsidiary equation = = .
P Q R
Step 2: Group any 2 of the three ratio’s and solve it . So that we get two
independent solutions u=a and v= b.
Step 3: Write down the general soln. f(u,v) = 0.
TYPE II: METHOD OF MULTIPLIERS
Step 1: Write the subsidiary equation dx = dy = dz .
P Q R
Step 2: Choose the set of multipliers l,m,n in such a way that
lP + mQ + nR=0,so that l dx + mdy + ndz =0 in
dx dy dz ldx + mdy + ndz
= = =
P Q R lP + mQ + nR
Step 3: Integrating l dx + mdy + ndz =0 we get u=a , which is a solution of
dx dy dz
= =
P Q R
Step 4: |||ly we can find another set of independent multipliers l ',m ',n ', so that
we get another independent solution v=b.
Step 5: Write down the general soln. f(u,v) = 0.

Note: If x,y,z are in cyclic then 1,1,1 (or) x,y,z (or) 1/x,1/y,1/z will
be the multipliers.

Homogeneous Linear PDE

General form:
(a0D n + a1D n − 1 D ' + a2D n − 2 D ' + …+ anD ' ) z = F(x,y) →(1)
2 n

(Or) f(D,D ') = F(x,y) , where D = ∂ and D ' = ∂

∂x ∂y
The complete solution of (1) is of the form
z = Complementary function + Particular Integral

Working Rule to find Complementary Function:

Put D=m and D ' =1,
we get the A.E as a0mn + a1mn-1 + a2mn-2 + …+ an = 0
Let the roots of the above eqn. be m1, m2,…,mn

Case(i): If the roots are real (or) imaginary and distinct then
z = f1(y +m1x) + f2(y +m2x) + f3(y +m3x) +…. + fn(y +mnx)

Case(ii): If any two of the roots are equal say m1=m2=m and other roots are
distinct then
z = f1(y +mx) + x f2(y +mx) + f3(y +m3x) +…. + fn(y +mnx)

Case(iii): If any three of the roots are equal say m1=m2=m3=m and other roots are
distinct then
z = f1(y +mx) + x f2(y +mx) + x2 f3(y +mx) + f4(y +m4x) +….
+ fn(y +mnx)
 Working Rule to find Particular Integral:

F(x,y)[R.H.S of (1)] Working rule

ax + by ax + by
provided f(a,b) ≠0
1 1
e = e
eax+by f(D, D' ) f(a, b)
if Denominator= 0 follow the Rule V
1
sin(ax + by ) (or) cos(ax + by)
f(D, D' )
sin(ax +by) or 2
Replacing D2 by -a2, DD ' by –ab and D ' by – b2
cos(ax +by)
provided Denominator≠0 if Denominator= 0 follow
the Rule V.
1
x m y n = [f(D,D ')]-1 xmyn
f(D, D' )
xmyn
[f(D,D’)]-1 is to be expanded using binomial
expansion.
eax+byφ(x,y) ax + by ax + by
φ ( x, y ) = e
1
where φ(x,y) is any φ ( x, y )
1
e
f(D, D' ) f(D + a, D'+ b)
function
Step 1 If the Dr. becomes zero then split the Dr. into factors.
Step 2 For each factor apply the following rule
Φ( x, y ) = ∫ Φ( x, c1 − mx)dx, where y = c1 − mx.
1
D − mD '
Step 3 Replace the constant c1 by y + mx. This process will be continued till
we apply each factor in the Dr.

SPECIAL NOTE:
If the denominator is zero, find the partial derivative of the denominator w.r.to
‘D’ after putting x in the numerator. This process may be continued until we get a
non-zero denominator.

NON – HOMOGENEOUS LINEAR PDE

General Form: f(D,D ')z = F(x,y) ---------(1)

Here the sum of the degree of each and every term is not equal.
The complete solution of (1) is of the form
z = Complementary function + Particular Integral
Working Rule to find Complementary Function:

f (D,D ') is factorisable into linear factors.

Case (i): If the given equation is of the form (D- mD’ – α)z = 0 then the
general solution of this equation is z = eαx f(y+mx), where m is
arbitrary.

Case (ii): If the given equation is of the form

(D- m1D '–α1) (D– m2D '–α2)(D–m3D '–α3)…(D–mnD '–αn) z = 0 then

α x α x α x
z= e 1 f ( y + m x) + e 2 f ( y + m x) + L + e n f ( y + m x )
1 1 2 2 n n

Case(iii): If the given equation is of the form (D- mD ' – α)r z = 0 then

z = eαxf1(y+mx) + xeαxf2(y+mx) +x2eαxf3(y+mx)+ … +xr-1eαxfr(y+mx)

Particular Integral:

The method of finding P.I is same as that of the homogeneous linear PDE.