2. PARTIAL DIFFERENTIAL EQUATIONS
• The relationship between a
dependent variable and two or more
independent variables and partial
differential coefficients of a
dependent variable with respect to
these independent variables is called
as a partial differential equation.
3. FORMATION OF PDEs
• A PDEs can be formed in Two
ways:
By eliminating arbitrary
constants from the given
relation .
By eliminating arbitrary
functions from the given
relation .
4. FIRST ORDER LINEAR PDEs
• A PDE of the form ;
P p + Q q = R
• Where P,Q,R are functions of
x,y,z or constant is called the
Lagrange's linear equation of
first order.
5. LAGRANGE’S METHOD
GROUPING METHOD:
In this method ,we compare any two fractions
which makes the integration possible.
MULTIPLIERS METHOD:
In this method ,we find two sets of multipliers
l,m,n and l’,m’,n’ such that;
lP+mQ+nR=0 and l’P+m’Q+n’R=0
From auxiliary equations dx/P=dy/Q=dz/R
It can be solved using the grouping or multiplier
method.Solution denoted by u(x,y)=c,v(x,y)=c’
Complete solution write as F(u,v)=0
6. FIRST ORDER NON - LINEAR PDEs
• A partial differential equation of order one
which is not linear is called a non linear partial
differential equation of first order.
• There are four forms of the FIRST ORDER NON -
LINEAR PDEs.
General form :f(x,y,z,p,q)=0
I. F(p,q)=0
II. F(z,p,q)=0
III. F(x,p)=G(y,q)
IV. Z=px+qy+F(p,q)
7. LINEAR PDEs WITH CONSTANT
COEFFICIENTS
Homogeneous linear PDEs:
If the order of all partial derivatives involved
in linear PDE are same ,then the equation
called a homogeneous linear PDE.
Non homogeneous linear PDEs:
If the order of all partial derivatives involved
in linear PDE are distinct ,then the equation
called Non homogeneous linear PDE.
8. ONE DIMENSIONAL WAVE EQUATION
• Where u(x,t) is deflection of string.
The wave equation is mainly used to
determined the motion of vibrating string.
if the initial deflection u(x,0) is f(x) and initial
velocity is g(x),then the solution of wave
equation is given by
L
nx
L
ctn
Bn
L
ctn
Bntxu
n
sin]sin*cos[),(
1
2
2
2
2
2
x
u
c
t
u
)0,(x
t
u
9. ONE DIMENSIONAL HEAT EQUATION
• The one dimensional heat equation is given by
where
• here (x,t ) is the temperature , is the thermal
diffusivity ,k the thermal conductivity , the
specific heat and the density of the material of
the body.
• It is mainly used to determine the conduction of
heat in wires or bars .
k
c 2
2
c
k
c 2
10. SOLUTION OF HEAT EQUATION
• If the ends are kept at temperature 0*c or the
boundary condition are u(0,t)=u(L,t)=0,then the
solution of the heat equation is given by
• If the ends are insulated or the boundary
conditions are
then solution of the heat equation is given by
t
L
cn
n
e
L
xn
Bntxu
2
222
sin),(
1
0),(),0(
tL
x
u
t
x
u
1
2
cos),(
n
t
L
cn
e
L
xn
AnAotxu
11. EXAMPLE:
Find u(x,t) of the string of length L= when
=1 the initial velocity is zero, and the
initial deflection is 0.01sinx.
The deflection u(x,t) is given by the wave
equation,
………….(1)
Length= ,initial velocity=g(x)=0 …..(2)
2
c
2
2
2
2
2
x
u
c
t
u
x0
12. We know that,
u(0,t)=u( ,t)=0 for all
u(x,0)=0.01sin3x
We know that the solution of (1) together
with (2) is given by………
Here g(x)=0 so , Bn*=0
…..(3)
0t
x0
L
nx
L
ctn
Bn
L
ctn
Bntxu
n
sin]sin*cos[),(
1
L
ctn
Bntxu
n
cos),(
1
L
nx
sin
13. Using u(x,0)=0.01sin3x,we get
We get,
Now put L=
0.01sin3x=B1sinx+B2sin2x+B3sin3x+…….(4)
L
nx
sin
1
)0,(
n
Bnxu
)sin()0,(
1
nxBnxu
n
14. From equation (4) we get,
By equating terms,
B3=0.01, Bn=0…… where n 3
Thus equation (3) becomes,
Here c=1 ; so by putting value of c in equation (5)
we get general solution of deflection,
L
ctn
Bntxu
n
cos),(
1
L
nx
sin
)5......(cos3sin01.0 ctx
.cos3sin01.0 tx),( txu