Download 2 - CFD

PARTIAL DIFFERENTIAL
EQUATIONS
Formation of Partial Differential equations
Partial Differential Equation can be formed either
by elimination of arbitrary constants or by the
elimination of arbitrary functions from a relation
involving three or more variables .
SOLVED PROBLEMS
1.Eliminate two arbitrary constants a and b from
2
2
2
2
x  a    y  b  z  R here R is known constant
.
(OR) Find the differential equation of all spheres
of fixed radius having their centers in x y- plane.
solution
x  a    y  b 
2
2
 z  R .......1
2
2
Differentiating both sides with respect to x and y
z
2z
  2( x  a )
x
z
2z
 2 ( y  b )
y
z
z
 p,
q
x
y
 x  a   pz , y  b   qz
By substituting all these values in (1)
2
p z
2
 z
2
q z
2

p2
2
z
2
 R
2
R2
 q2 1
or
z2 
R2
 z 


 x 
2
 z

 y





2
1
2. Find the partial Differential Equation by eliminating
2
2
z

f
(
x

y
)
arbitrary functions from
SOLUTION
z  f ( x  y )..........(1)
d .w.r.to.xandy
2
2
z
'
2
2
 f ( x  y )  2 x......(2)
x
z
'
2
2
 f ( x  y )  2 y......(3)
y
By
( 2)
(3)
 z 


 x    x
y
 z 


 y 


p
x

 py  qx  0
q
y
3.Find Partial Differential Equation
by eliminating two arbitrary functions from
z  yf ( x)  xg( y )
SOLUTION
z  yf ( x)  xg( y )......(1)
Differentiating both sides with respect to x and y
z
 yf ( x)  g ( y )........( 2)
x
z
 f ( x)  xg ( y )........(3)
y
Again d . w .r. to x and yin equation (2)and(3)
 z
 f ( x)  g ( y )
xy
2
x  (2)  y  (3)......to...get
z
z
x
y

x
y
xg( y )  yf ( x)  xy( f ( x)  g ( y ))
 z  xy f   g 
  z 
z
z

x
y
 z  xy
x
y
 xy 
2
Different Integrals of Partial Differential
Equation
1. Complete Integral (solution)
z z
Let F ( x, y, z, , )  F ( x, y, z, p, q)  0......(1)
x y
be the Partial Differential Equation.
The complete integral of equation (1) is given
by  ( x, y, z , a, b)  0..........(2)
2. Particular solution
A solution obtained by giving particular values to
the arbitrary constants in a complete integral is
called particular solution .
3.Singular solution
The eliminant of a , b between
 ( x, y , z , a , b )  0


 0,
0
a
b
when it exists , is called singular solution
4. General solution
In equation (2) assume an arbitrary relation
of the form b  f (a) . Then (2) becomes
 ( x, y, z, a, f (a))  0.........(3)
Differentiating (2) with respect to a,
 

f (a)  0..........(4)
a b
The eliminant of (3) and (4) if exists,
is called general solution
Standard types of first order equations
TYPE-I
The Partial Differential equation of the form
f ( p, q )  0
has solution
z  ax  by  c
with f (a, b)  0
TYPE-II
The Partial Differential Equation of the form
z  px  qy  f ( p, q) is called Clairaut’s form
of pde , it’s solution is given by
z  ax  by  f (a, b)
TYPE-III
If the pde is given by
f ( z , p, q )  0
then assume that
z   ( x  ay )
u  x  ay
z   (u )
z z u z
dz
p 
 .1 
x u x u
du
z z u z
dz
q 
 .a  a
y u y u
du

The given pde can be written as
dz dz
f ( z, , a )  0.And also this can
dx dy
be integrated to get solution
TYPE-IV
The pde of the form f ( x, p)  g ( y, q) can be
solved by assuming
f ( x, p )  g ( y , q )  a
f ( x, p )  a  p   ( x, a )
g ( y, q )  a  q   ( y, a )
z
z
dz 
dx  dy
x
y
dz   ( x, a )dx   ( y, a )dy
Integrate the above equation to get solution
SOLVED PROBLEMS
2
p
 q  1 and find the complete
1.Solve the pde
and singular solutions
Solution
Complete solution is given by
z  ax  by  c
with
a b 1
2
 b  a 1
2
z  ax  (a  1) y  c
2
d.w.r.to. a and c then
z
 x  2 ay
a
z
 1  0 Which is not possible
c
Hence there is no singular solution
2.Solve the pde pq  p  q  0and find the
complete, general and singular solutions
Solution
The complete solution is given by
z  ax  by  c
with
ab  a  b  0
b
a
b 1
b
z 
x  by  c.......(1)
b 1
z
1

x y 0
2
b
b  1
z
 1  0 no singular solution
c
To get general solution assume that
c  g (b)
From eq (1)
b
z 
x  by  g (b).......( 2)
b 1
z
1


x

y

g
(b).......(3)
2
c
b  1
Eliminate from (2) and (3) to get general
solution
3.Solve the pde z  px  qy  1  p  q
and find the complete and singular solutions
2
Solution
The pde
z  px  qy  1  p  q
2
is in Clairaut’s form
2
2
complete solution of (1) is
z  ax  by  1  a  b .......(2)
2
2
d.w.r.to “a” and “b”
z
a

 x
 0
2
2
a
1 a  b

........(
3
)

z
b
 y
 0 
2
2
b
1 a  b

From (3)
2
2
a
b
2
x 
,y 
2
2
2
2
1 a  b
1 a  b
2
2
a b
2
2
x y 
2
2
1 a  b
1
2
2


1

(
x

y
)
2
2
1 a  b
2
ax 
by 
a
2
1 a  b
2
b
2
1 a  b
2
2
2
0
0
ax  by  1  a  b 
2
z
1
1
2
1 a  b
2
2
0
 0  z  1 (x  y )
2
2
2
1  a2  b2
2
2
2
 x  y  z  1 is required singular solution
4.Solve the pde (1  x) p  (2  y )q  3  z
Solution
pde (1 
x) p  (2  y )q  3  z
z  px  qy  (3  p  2q )
Complete solution of above pde is
z  ax  by  (3  a  2b)
5.Solve the pde
Solution
Assume that
p q  z
2
2
z   ( x  ay )
u  x  ay
z   (u )
z z u z
dz
p 
 .1 
x u x u
du
z z u z
dz
q 
 .a  a
y u y u
du
From given pde
2
2
 dz 
2  dz 
2
p q  z  a    z
 du 
 du 
2
2
2
z
 dz 
  
2
 du  1  a
z
dz
1
 dz 


du
 
2
2
z
 du  1  a
1 a
Integrating on both sides
2
z 
2
z 
u
1 a2
x  ay
1 a2
b
b
6. Solve the pde
zpq  p  q
Solution
Assume q  ap
Substituting in given equation
zpap  p  ap
1 a
1 a
p
,q 
az
z
z
z
dz 
dx 
dy
x
y
1 a
1 a
 dz 
dx 
dy
az
z
zadz  (1  a )( dx  ady )
Integrating on both sides
a 2
z  (1  a )( x  ay )  b
2
7.Solve pde pq  xy
z z
(or) ( )( )  xy
x y
Solution
p
q

x
y
Assume that
p
y

a
x
q
y
 p  ax, q 
a
y
dz  pdx  qdy  axdx  dy
a
Integrating on both sides
2
2
x
y
za

b
2
2a
8. Solve the equation
p q  x y
2
2
Solution
p x  yq  a
2
2
p  a  x, q  y  a
dz  pdx  qdy  a  x dx  y  a dy
integrating
3
2
3
2
2
z  (a  x)  ( y  a )  b
3
Equations reducible to the standard forms
m
n
(
y
q) occur in the pde as in
(
x
p
)
(i)If
and
m
n
Or
in
F
(
z
,
x
p
,
y
q)  0
F ( x p, y q)  0
m
Case (a)
n
1 m
Put x  X and y
if m  1 ; n  1
1 n
Y
z z X z
m
p 

(1  m) x
x X x X
z z Y z
n
q 
 (1  n) y
y Y x Y
z
x p
(1  m)  P(1  m)
X
z
n
y q
(1  n)  Q(1  n)
Y
m
where
Then
z
z
 P,
Q
X
Y
F ( x m p, y n q)  0 reduces to F ( P, Q)  0
Similarly F ( z, x
m
p, y q)  0 reduces
n
to F ( z, P, Q)  0
case(b)
If m  1 or
n 1
put log x  X , log y  Y
z
p
X
z
q
Y
(ii)If
k
( z p)
and
Or in
(z
1
 px  P
x
1
 qy  Q
y
k
q) occur in pde as in
f1 ( x, z p)  f 2 ( y, z q)
k
k
k
k
F ( z p, z q)
Case(a) Put
z z

x Z
z z

y Z
z
1 k
 Z if
k  1
Z k
 `1 Z
k
 `1
 z (1  k )
 z p  (1  k ) P
x
x
Z k
 `1 Z
k
 `1
 z (1  k )
 z q  (1  k ) Q
y
y
Z
Z
 P,
Q
where
x
y
Given pde reduces to
F ( P, Q) and
f1 ( x, P)  f 2 ( y, Q)
Case(b) if k  1
z z

x Z
z z

y Z
log z  Z
Z
Z
1
z z pP
x
x
Z
Z
1
 z  z qQ
y
y
Solved Problems
1.Solve
p x q y  z
2
4
2
2
4
2
2




px
qy
Solution 
 z    z   1.......(1)

 

2
2
m  2, n  2
k  1
1
x X
1
y Y
log z  Z
z z Z X
 2 Z
2
p 
  zx
  zx P
x Z X x
X
z z Z Y
 2 Z
2
q 
  zy
  zy Q
y Z Y y
Y
where
Z
Z
 P,
Q
X
Y
2
2
px
qy
  P,
 Q
z
z
(1)becomes
 P    Q
2
2
1
P  Q 1
2
2
 Z  aX  bY  c
a  b  1, b 
2
2
1 a
2
log z  ax  1  a y  c
2
2
2
2. Solve the pde
p  q  z (x  y )
2
2
2
2
2
SOLUTION



2
2
p q
2
2
     ( x  y ).....(1)
z z
k  1 log z  Z
z z Z
Z
1

z
z pP
x Z x
x
z z Z
Z
1

z
 z qQ
y Z y
y
Eq(1) becomes
P  Q  ( x  y ).....(2)
2
2
2
2
P  x  y Q  a
2
2
2
2
2
a
x 2
1  x 
2
log z  sinh    (a  x ) 
2
a 2
2
y (y  a ) a
1  y 
 cosh    b
2
2
a
2
2
2
Lagrange’s Linear Equation
Def: The linear partial differenfial equation
of first order is called as Lagrange’s linear Equation.
This eq is of the form
Where
P, Q
and
Pp  Qq  R
R are functions x,y and z
The general solution of the partial differential
equation Pp  Qq  R
is F (u, v)  0
Where F is arbitrary function of u( x, y, z)  c1
and v( x, y, z)  c2
Here u  c1 and v  c2 are independent solutions
dx dy dz


of the auxilary equations
P
Q
R
Solved problems
2
2
x
p

y
q  ( x  y) z
1.Find the general solution of
Solution
dx dy
auxilary equations are 2  2 
x
y
dz
( x  y) z
dx dy
 2
2
x
y

1
u x y
Integrating on both sides
1
 c
1
dx  dy
dz

2
2
x y
( x  y) z
d ( x  y)
dz

( x  y )( x  y ) ( x  y ) z
d ( x  y ) dz Integrating on both sides

( x  y)
z
log( x  y )  log z  log c2
v  ( x  y) z
1
 c2
F (u, v)  0
The general solution is given by
1
1
1
F ( x  y , ( x  y) z )  0
2.solve x ( y  z )  y ( z  x)q  z ( x  y)
solution Auxiliary equations are given by
2
2
2
dx
dy
dz
 2
 2
2
x ( y  z ) y ( z  x) z ( x  y )
dy
dx
dz
2
2
2
y
x

 z
( y  z ) ( z  x) ( x  y )
dx dy dz
 2 2
2
x
y
z
( y  z )  ( z  x)  ( x  y )
dx dy dz
 2  2 0
2
x
y
z
Integrating on both sides
1
1
1
u


a
x
y
z
1
1
1
x dx
y dy
z dz


x( y  z ) y ( z  x) z ( x  y )
1
1
1
x dx  y dy  z dz
x( y  z )  y ( z  x)  z ( x  y )
dx dy dz


 0 Integrating on both sides
x
y
z
v  xyz  b
The general solution is given by
1
1
1
F ( x  y  z , xyz)  0
HOMOGENEOUS LINEAR PDE WITH
CONSTANT COEFFICIENTS
Equations in which partial derivatives
occurring are all of same order (with degree
one ) and the coefficients are constants ,such
equations are called homogeneous linear PDE
with constant coefficient
z
z
z
z
 a1 n1  a2 n2 2  ........an n  F ( x, y)
n
x
x y
x y
y
n
n
n
n


Assume that D  , D  .
x
y
th
then n order linear homogeneous equation is
given by
( D  a1D D  a2 D D  .........  an D ) z  F ( x, y )
n
n 1
n2
2
or
f ( D, D) z  F ( x, y ).........(1)
n
The complete solution of equation (1) consists
of two parts ,the complementary function and
particular integral.
The complementary function is complete
solution of equation of f ( D, D) z  0
Rules to find complementary function
Consider the equation
 z
 z
 z
 k1
 k2 2  0
2
x
xy
y
2
2
or
2
2


( D  k1DD  k2 D ) z  0.............(2)
2
The auxiliary equation for (A.E) is given by
2


D  k1DD  k2 D  0
2
And by giving
D  m, D  1
The A.E becomes
m  k1m  k2  0....(3)
2
Case 1
If the equation(3) has two distinct roots
The complete solution of (2) is given by
z  f1 ( y  m1x)  f 2 ( y  m2 x)
m1 , m2
Case 2
If the equation(3) has two equal roots i.e
m1  m2
The complete solution of (2) is given by
z  f1 ( y  m1 x)  xf2 ( y  m1 x)
Rules to find the particular Integral
Consider the equation
( D  k1DD  k2 D ) z  F ( x, y)
2
2
f ( D, D) z  F ( x, y )
F ( x, y )
Particular Integral (P.I) 
f ( D, D)
Case 1 If F ( x, y )  e
then P.I=
If
ax by
1
axby
e
f ( D, D)
1
ax by

e
, f ( a, b)  0
f ( a, b)
a
f (a, b)  0 and ( D  D) is
b
factor of f ( D, D) then
P.I
ax by
 xe
a
If f (a, b)  0 and ( D  D) 2 is
b

f
(
D
,
D
)
factor of
2
then
Case 2
x axby
e
P.I 
2
F ( x, y )  sin( mx  ny )or cos( mx  ny )
P.I
sin( mx  ny)
sin( mx  ny)


2
2
2
2
f ( D , DD, D ) f (m ,mn,n )
Case 3
P.I
F ( x, y)  x y
m
n
1
1 m n
m n

x y   f ( D, D) x y
f ( D, D)
Expand  f ( D, D) 1 in ascending powers of
m n

D
and operating on x y term by term.
D or
Case 4 when F ( x, y ) is any function of x
and y.
1
F ( x, y )
P.I=
f ( D, D)
1
F ( x, y )   F ( x, c  mx)dx
D  mD
( D  mD) is factor of
Here
Where ‘c’ is replaced by
f ( D, D)
( y  mx) after integration
Solved problems
1.Find the solution of pde
3
2
2


( D  D  3DD  3D D) z  0
3
Solution
The Auxiliary equation is given by
Solution
The Auxiliary equation is given by
m  1  3m  3m  0
3
By taking
2
D  m, D  1
 m  1,1,1.
2
Complete solution  f1 ( y  x)  xf2 ( y  x)  x f 3 ( y  x)
2. Solve the pde ( D3  4D 2 D  5DD) z  0
Solution
The Auxiliary equation is given by
m  4m  5m  0
 m  0,1,5
z  f1 ( y )  f 2 ( y  x)  f 3 ( y  5 x)
3
2
3. Solve the pde ( D 2  D2 ) z  0
Solution
the A.E is given by
m 1  0
2
m  i
 z  f1 ( y  ix)  f 2 ( y  ix)
4. Find the solution of pde
( D  3DD  4D ) z  e
2
2
2 x4 y
Solution
Complete solution =
Complementary Function + Particular Integral
The A.E is given by
m  3m  4  0
2
m  4,1
C.F  1 ( y  x)  2 ( y  4x)
2 x4 y
2 x4 y
e
e
P.I  2

2
D  3DD  4 D
 36
Complete solution
 C.F  P.I
 1 ( y  x)  2 ( y  4 x) 
e
2 x4 y
36
5.Solve
( D  3DD  2D ) z  e
3
3
2 x y
e
x y
Solution
A.E  m  3m  2
 m  1,1,2.
C.F  1 ( y  x)  x2 ( y  x)  3 ( y  2 x)
3
2 x y
2 x y
e
e
P.I1  3

2
3
2
2
D  3DD  2D ( D  D) ( D  2D)
2 x y
2 x y
e
xe
P.I1 

2
( D  D) ( D  2D)
9
x y
x y
e
e
P.I 2  3

2
3
2
2
D  3DD  2 D ( D  D) ( D  2 D)
2
x x y
 P.I 2  e
6
z  C.F  P.I1  P.I 2
2 x y
2
xe
x x y
z  1 ( y  x)  x2 ( y  x)  3 ( y  2 x) 
 e
9
6
6.Solve ( D 2  DD) z  cos x cos 2 y
Solution
1
( D  DD) z  cos( x  2 y )  cos( x  2 y )
2
2
A.E  m  m  0
2
m  0,1
C.F  1 ( y  x)  2 ( y )
cos( x  2 y) cos( x  2 y)
P.I1  2

 cos( x  2 y)
( D  DD) ((1)  (2))
cos( x  2 y) cos( x  2 y) cos( x  2 y)
P.I 2  2


( D  DD) (( 1)  (2))
3
1
z  1 ( y  x)  2 ( y  x)  cos( x  2 y )  cos( x  2 y )
3
7.Solve
2
2 2


( D  DD  6D ) z  x y
2
Solution A.E  m 2  m  6  0
m  2,3.
C.F  1 ( y  2x)  2 ( y  3x)
2
2
x y
P.I  2
2
D  DD  6 D
1
 2
D
  D
D
1    6 2
D
D


2
1
 2 2
 x y

2
2 2
  D
D   D
D  2 2
2
 D 1    6 2     6 2   x y
D  D
D  
  D

2
2

 2 2





D
D
D
2
 D 1  
 D  6 D2 
  D2  x y




2
2
2




2
x
y
2
x
2
x
2
2
2
 D x y  
 D  6 D2 
  D2 




3
4
4




2
x
y
2
x
2
x
2
2
2
 D x y  
 3  6 12 
  12 




3
4


2
x
y
2
x
2
2
2
 D x y 
8

3
12 

 x4 y2
2 x5 y
2x6 




12
60
90


7.Solve ( D 2  5DD  6D2 ) z  y sin x
Solution
A.E is
m  5m  6  0
2
m  3, m  2.
C.F  1 ( y  3x)  2 ( y  2x)
P.I 

y sin x
y sin x

2
D  5 DD  6 D
( D  3D)( D  2 D)
2
 y sin x 
1
( D  3D)  ( D  2 D) 
1
(a  2 x) sin xdx
( D  3D) 
1
 a cos x  2( x cos x  sin x)

( D  3D)
1
2 x cos x  2 sin x  ( y  2 x) cos x

( D  3D)

y sin x
y sin x
P.I  2

2
D  5 DD  6 D
( D  3D)( D  2 D)
 y sin x 
1





( D  3D )  ( D  2 D ) 
here
a  y  2 x 
1

( a  2 x) sin xdx

( D  3D)
1
 a cos x  2( x cos x  sin x)

( D  3D)
1
2 x cos x  2 sin x  ( y  2 x) cos x

( D  3D)
1
 y cos x  2 sin x

( D  3D)
  ((b  3x) cos x  2 sin x)dx
here
b  y  3x 
 b sin x  2 cos x  3( x sin x  cos x)
 ( y  3x) sin x  2 cos x  3( x sin x  cos x)
 5 cos x  y sin x
Non Homogeneous Linear PDES
If in the equation
f ( D, D) z  F ( x, y )............(1)
the polynomial expression f ( D, D) is not
homogeneous, then (1) is a non- homogeneous
linear partial differential equation
Ex
2
2 x 3 y


( D  3D  D  4D ) z  e
2
Complete Solution
= Complementary Function + Particular Integral
To find C.F., factorize f ( D, D)
into factors of the form ( D  mD  c)
If the non homogeneous equation is of the form
( D  m1 D  c1 )( D  m2 D  c2 ) z  F ( x, y )
C.F  e  ( y  m1 x)  e  ( y  m2 x)
c1 x
1.Solve
c2 x
( D  DD  D) z  x
2
2
Solution
2

f ( D, D )  D  DD  D  D( D  D  1)
C.F  e  x1 ( y  x)  2 ( y)
1
x
1  ( D  1)  2
P.I  2
 2 1 
x
D  DD  D D 
D 
2
1
 2
D
 2  ( D  1)  2  ( D  1)  2

x

x

......
x  




 D 
 D 


1
 2
D
 2 x x  x
x
x 

x      

3 12   3.4 3.4.5 12.5.6 

2
3
4
4
5
6
2.Solve ( D  D  1)( D  2 D  3) z  4
Solution
4
z  e 1 ( y  x)  e 1 ( y  2 x) 
3
x
3x