Transcript Slide 1

TOPIC
INTEGRAL EQUATION
Definition: Integral equation
An integral equation is an
equation in which an unknown
function appear under one or
more integral signs.
.
EXAMPLE
For axb, atb,
the equations
b
K(x,t) y(t) dt=f(x) …(1)
a
b
y(x)-K(x,t) y(t) dt=f(x) ….(2)
a
where the function y(x) is the unknown function,
f(x) and K(x,t) are the known functions and ,a
and b are constants, are all integral equations.
LINEAR INTEGRAL EQUATION
An integral equation is called linear if only
linear operations are perfomed in it upon
the unknown function.
Example:
b
g(x) y(x)=f(x)+K(x,t) y(t) dt
a
NON LINEAR INTEGRAL
EQUATION
An integral equation which is not linear is
known as non linear integral equation.
Example:
y(x)=K(x,t)[y(t)]2dt
where the function y(x) is the unknown
function and K(x,t) is known function.
b
a
FREDHOLM INTEGRAL
EQUATION
A linear integral equation of the form
b
g(x)y(x)=f(x)+K(x,t) y(t) dt
a
where a and b are both constants
f(x),g(x) and K(x,t) are known functions while
y(x) is unknown function , is non zero
parameter.
KINDS OF FREDHOLM INTEGRAL
EQUATION
1.Fredholm integral equation of first kind
2.Fredholm integral equation of second kind
3.Fredholm integral equation of third kind
1. FREDHOLM INTEGRAL EQUATION OF
FIRST KIND
A linear integral equation of the form
b
f(x)+K(x,t) y(t) dt=0
a
is known as fredholm integral equation of
first kind.
2.Fredholm integral equation of second
kind
A linear integral equation of the form
y(x)= f(x)+K(x,t) y(t) dt
b
a
is known as fredholm integral equation of
second kind.
3.FREDHOLM INTEGRAL EQUATION OF
THIRD KIND
A linear integral equation of the form
b
g(x)y(x)=f(x)+K(x,t) y(t) dt
a
where a and b are both constants
f(x),g(x) and K(x,t) are known functions while y(x) is
unknown function , is non zero parameter.
HOMOGENEOUS FREDHOLM INTEGRAL
EQUATION OF SECOND KIND
A linear integral equation of the form
b
y(x)= K(x,t) y(t) dt
a
is known as homogeneous fredholm integral
equation of second kind.
VOLTERRA INTEGRAL
EQUATIONS
• A linear integral equation of the form
x
g(x)y(x)=f(x)+aK(x,t) y(t) dt
where upper limit of integration is
variable,is called Volterra Integral equation
VOLTERRA INTEGRAL
EQUATION OF THE FIRST KIND
A linear integral equation of the form
x
f(x)+ K(x,t) y(t) dt=0
a
where upper limit of integration is
variable.is called Volterra integral
equation of the first kind.
VOLTERRA INTEGRAL EQUATIONS
OF THE SECOND KIND
• A linear integral equation of the form
x
y(x)= f(x)+K(x,t) y(t) dt
a
is called Volterra integral equation of the 2nd kind
.
HOMOGENEOUS VOLTERRA
INTEGRAL EQUATIONOF 2ND KIND
• A linear integral equation of the form
x
y(x)= K(x,t) y(t) dt
a
is called homogeneous Volterra integral
equation of 2nd kind
SINGULAR INTEGRAL EQUATION
• When one or both limits of integration
becomes infinite or when the kernel
becomes infinite at one or more points
with in the range of integration , the
integral equation is called singular integral
equation
For exampal :
f(x)= (1/(x-t)α )y(t)dt, 0<α<1
x
0
Special kinds of kernels
• Symmetric kernel
A kernel K(x,t) is symmetric if
K(x,t)=K (x,t)
For exampal:sin(x+t)
Separable or degenerate kernel
A kernel K(x,t) is
called separable if it can be expressed as the
sum of finite number of terms , each of which is
the product of a function of x only and a function
of t only .
INTEGRAL EQUATIONS WITH
SEPARABLE KERNELS
• EXAMPLE
• 1 : solve the fredholm integral equation of
2nd kind
y(x)=x+(xt2+x2t)y(t)dt
Soln=the above equation can be written as
y(x) =x+xt2y(t)dt+x2ty(t)dt…………(1)
let C1 =t2y(t)dt ……………….(2)
1
0
1
1
0
0
1
0
1
C2=ty(t)dt……………………(3)
0
using (2) and (3),(1) reduces to
y(x)=x+C1x+C2x2…………………….(4)
from (4) y(t)=t+C1t+C2t2……………………..(5)
using (5), (2) reduces to
1
C1=t2(t+C1t+C2t2)dt
0
=[t4/4+C1t4/4+C2t5/5]
=1/4+C1/4+C2/5
(20-5)C1-4C2=5………………..(6)
`using (5),(2) reduces to
1
C2=t(t+C1t+C2t2)dt
= 1/3+C1/3+C2/4
-4C1+(12-3)C2=4……………(7)
Solving (6)and(7) for C1 and C2,we get
C1=(60+)/(240-120-2)
and C2=80/(240-120-2)
Putting these values of C1 and C2 in (4),the
Required solution is
y(x)=x+x(60+)/(240-120-2 )+80x2/(240120-2)
=((240-60)x+80x2)/(240-120-2)
0
Method of
Succesive
approximations
ITERATED KERNELS
• For Fredholm integral equation of the
second kind
b
y(x)=f(x)+K(x,t) y(t) dt
a
The iterated kernels Kn(x,t),n=1,2,3……………
are defined as follows:
K1(x,t)=K(x,t)
And Kn(x,t)=K(x,z)Kn-1(z,t)dz,n=2,3,………..
Or Kn(x,t)=Kn-1(x,z)K(z,t)dz,n=2,3,……..
b
b a
a
• For Volterra integral equation of the second kind
x
y(x)=f(x)+K(x,t)y(t)dt,
a
The iterated kernels Kn(x,t),n=1,2,3,…….
are defined as follows:
K1(x,t)=K(x,t)
x
And Kn(x,t)=K(x,z)K
n-1(z,t)dz,n=2,3,……….
t
x
Or Kn(x,t)=Kn-1(x,z)K(z,t)dz,n=2,3,…
t
RESOLVENT KERNEL
Suppose solution of fredholm integral
equation
y(x) =f(x)+K(x,t)y(t)dt………..(1)
takes the form
b
a
b
y(x)=f(x)+(x,t;)f(t)dt
a
Then (x,t;) is known as the resolvent
Kernel of (1).
Suppose solution of Volterra integral
equation of the second kind
y(x)=f(x)+K(x,t)y(t)dt……….(3)
takes the form
x
a
x
y(x)=f(x)+(x,t;)f(t)dt
a
then (x,t,) is known as the resolvent
kernel of (3)
SOLUTION OF FREDHOLM
INTEGRAL EQUATION OF SECOND
KIND
BY SUCCESSIVE APPROXIMATIONS
1
• Solve y(x)=f(x)+ex-ty(t)dt……..(1)
sol:comparing (1) with
y(x)=f(x)+K(x,t)y(t)dt,we have
K(x,t)=ex-t
K1(x,t)=K(x,t)=ex-t
K2(x,t)=K(x,z)K1(z,t)dz
=ex-zez-tdz
=ex-tdz
=ex-t
0
b
a
1
0 1
0
1
0
1
• K3(x,t)=K(x,z)K2(z,t)dz
=ex-t
Proceeding in this way,we find that all the
iterated kernels coincide with K(x,t)
The resolvent kernel is given by
(x,t;)=ex-t(1++2+……)
=ex-t/(1-), provided ||<1
The solution is given by
0
1
0
1
y(x)=f(x)+ (x,t;)f(t)dt
0
1
=f(x)+[ex-tf(t)/1-]dt where ||<1.
0
SOLUTION OF VOLTERRA
INTEGRAL EQUATION OF
SECOND KIND BY SUCCESSIVE
APPROXIMATIONS
• Solve the integral equation
X
y(x)=f(x)+ex-ty(t)dt and find the
0
resolvent kernel.
Solution:Here K(x,t)=ex-t
K1(x,t)=K(x,t)=ex-t
x
K2(x,t)=K(x,z)K1(z,t)dz
t
x
=ex-zez-tdz
t
x
=ex-tdz
t
=ex-t(x-t)
x
K3(x,t)=K(x,z)K2(z,t)dz
t
x
x-z(z-t)ez-tdz
=e
t
=ex-t(x-t)2/2!
Proceeding in this way,we find that
Kn(x,t)=(ex-t(x-t)n-1)/(n-1)!, n=1,2,3,……….
The resolvent kernel is
(x,t;)=K1(x,t)+K2(x,t)+2K3(x,t)+………..
=ex-te(x-t)
=e(x-t)(1+)
Hence the solution
is
X
(x-t)(1+)f(t)dt
y(x)=f(x)+e
0
TEST
• Define an integral equation equation.
How are they classified?Explain by giving an example of each type.
• State and prove Fredholm altrenative theorem.
• Invert the integral equation :
2
y(x)=f(x)+(sinx cosx)y(t)dt.
0
•
OR
0
Show that the integral
equation
2
y(x)=f(x)+(1/)sin(x+t)y(t)dt
0
possesses no solution for f(x)=x,but infinitely many solutions when
f(x)=1.