Linear Integral Equation of the First and Second Kind of Fredholm and Volterra Type (Part - 1) | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Section 3
Integral Equations
Integral Operators and Linear Integral Equations
As we saw in Section 1 on operator notation, we work with functions de?ned in some
suitable function space. For example, f (x), g(x) may live in the space of continuous real-
valued functions on [a,b], i.e. C(a,b). We also saw that it is possible to de?ne integral as
well as di?erential operatorsacting on functions. Theorem 2.8 isan example ofan integral
operator:
u(x) =
Z
b
a
G(x,y)f (y)dy,
where G(x,y) is a Green’s function.
De?nition 3.1: An integral operator, K, is an integral of the form
Z
b
a
K(x,y)f (y)dy,
where K(x,y) is a given (real-valued) function of two variables, called the kernel.
The integral equation
Kf = g
maps a function f to a new function g in the interval [a,b], e.g.
K : C[a,b]? C[a,b], K : f 7? g.
Theorem 3.2: The integral operator K is linear
K(af +ßg)= aKf +ßKg.
Example 1: The Laplace Transform is an integral operator
Lf =
Z
8
0
exp(-sx)f (x)dx =
¯
f (s),
with kernel K(x,s) = K(s,x) = exp(-sx). Let us recap its important properties.
Theorem 3.3:
(i) L

df
dx

= s
¯
f (s)-f (0),
(ii) L

d
2
f
dx
2

= s
2
¯
f (s)-sf (0)-f
'
(0),
(iii) L
Z
x
0
f (y)dy

=
1
s
¯
f (s).
(iv) Convolution: let
f (x)*g(x) =
Z
x
0
f (y)g(x-y)dy
then
L(f (x)*g(x)) =
¯
f (s)¯ g(s).
Page 2


Section 3
Integral Equations
Integral Operators and Linear Integral Equations
As we saw in Section 1 on operator notation, we work with functions de?ned in some
suitable function space. For example, f (x), g(x) may live in the space of continuous real-
valued functions on [a,b], i.e. C(a,b). We also saw that it is possible to de?ne integral as
well as di?erential operatorsacting on functions. Theorem 2.8 isan example ofan integral
operator:
u(x) =
Z
b
a
G(x,y)f (y)dy,
where G(x,y) is a Green’s function.
De?nition 3.1: An integral operator, K, is an integral of the form
Z
b
a
K(x,y)f (y)dy,
where K(x,y) is a given (real-valued) function of two variables, called the kernel.
The integral equation
Kf = g
maps a function f to a new function g in the interval [a,b], e.g.
K : C[a,b]? C[a,b], K : f 7? g.
Theorem 3.2: The integral operator K is linear
K(af +ßg)= aKf +ßKg.
Example 1: The Laplace Transform is an integral operator
Lf =
Z
8
0
exp(-sx)f (x)dx =
¯
f (s),
with kernel K(x,s) = K(s,x) = exp(-sx). Let us recap its important properties.
Theorem 3.3:
(i) L

df
dx

= s
¯
f (s)-f (0),
(ii) L

d
2
f
dx
2

= s
2
¯
f (s)-sf (0)-f
'
(0),
(iii) L
Z
x
0
f (y)dy

=
1
s
¯
f (s).
(iv) Convolution: let
f (x)*g(x) =
Z
x
0
f (y)g(x-y)dy
then
L(f (x)*g(x)) =
¯
f (s)¯ g(s).
Recall that a di?erential equation is an equation containing an unknown function under
a di?erential operator. Hence, we have the same for an integral operator.
De?nition 3.4: An integral equation is an equation containing an unknown function
under an integral operator.
De?nition 3.5: (a) A linear Fredholm integral equation of the ?rst kind has the
form
Kf = g,
Z
b
a
K(x,y)f (y)dy = g(x);
(b) A linear Fredholm integral equation of the second kind has the form
f-?Kf = g, f (x)-?
Z
b
a
K(x,y)f (y)dy = g(x),
where the kernel K(x,y) and forcing (or inhomogeneous) term g(x) are known functions
and f (x) is the unknown function. Also, ? (?R orC) is a parameter.
De?nition 3.6: If g(x)= 0 the integral equation is called homogeneous and a value of
? for which
?Kf = f
possesses a non-trivial solution f is called aneigenvalue ofK corresponding to theeigen-
function f.
Note: ? is the reciprocal of what we have previously called an eigenvalue.
De?nition 3.7 Volterra integral equations of the ?rst and second kind take the
forms
(a)
Z
x
a
K(x,y)f (y)dy = g(x),
(b)
f (x)-?
Z
x
a
K(x,y)f (y)dy = g(x),
respectively.
Note: Volterra equations may be considered a special case of Fredholm equations. Sup-
pose a= y= b and put
K
1
(x,y) =

K(x,y) for a= y= x,
0 for x < y= b,
then
Z
b
a
K
1
(x,y)f (y)dy =
Z
x
a
+
Z
b
x

K
1
(x,y)f (y)dy =
Z
x
a
K(x,y)f (y)dy+0.
Page 3


Section 3
Integral Equations
Integral Operators and Linear Integral Equations
As we saw in Section 1 on operator notation, we work with functions de?ned in some
suitable function space. For example, f (x), g(x) may live in the space of continuous real-
valued functions on [a,b], i.e. C(a,b). We also saw that it is possible to de?ne integral as
well as di?erential operatorsacting on functions. Theorem 2.8 isan example ofan integral
operator:
u(x) =
Z
b
a
G(x,y)f (y)dy,
where G(x,y) is a Green’s function.
De?nition 3.1: An integral operator, K, is an integral of the form
Z
b
a
K(x,y)f (y)dy,
where K(x,y) is a given (real-valued) function of two variables, called the kernel.
The integral equation
Kf = g
maps a function f to a new function g in the interval [a,b], e.g.
K : C[a,b]? C[a,b], K : f 7? g.
Theorem 3.2: The integral operator K is linear
K(af +ßg)= aKf +ßKg.
Example 1: The Laplace Transform is an integral operator
Lf =
Z
8
0
exp(-sx)f (x)dx =
¯
f (s),
with kernel K(x,s) = K(s,x) = exp(-sx). Let us recap its important properties.
Theorem 3.3:
(i) L

df
dx

= s
¯
f (s)-f (0),
(ii) L

d
2
f
dx
2

= s
2
¯
f (s)-sf (0)-f
'
(0),
(iii) L
Z
x
0
f (y)dy

=
1
s
¯
f (s).
(iv) Convolution: let
f (x)*g(x) =
Z
x
0
f (y)g(x-y)dy
then
L(f (x)*g(x)) =
¯
f (s)¯ g(s).
Recall that a di?erential equation is an equation containing an unknown function under
a di?erential operator. Hence, we have the same for an integral operator.
De?nition 3.4: An integral equation is an equation containing an unknown function
under an integral operator.
De?nition 3.5: (a) A linear Fredholm integral equation of the ?rst kind has the
form
Kf = g,
Z
b
a
K(x,y)f (y)dy = g(x);
(b) A linear Fredholm integral equation of the second kind has the form
f-?Kf = g, f (x)-?
Z
b
a
K(x,y)f (y)dy = g(x),
where the kernel K(x,y) and forcing (or inhomogeneous) term g(x) are known functions
and f (x) is the unknown function. Also, ? (?R orC) is a parameter.
De?nition 3.6: If g(x)= 0 the integral equation is called homogeneous and a value of
? for which
?Kf = f
possesses a non-trivial solution f is called aneigenvalue ofK corresponding to theeigen-
function f.
Note: ? is the reciprocal of what we have previously called an eigenvalue.
De?nition 3.7 Volterra integral equations of the ?rst and second kind take the
forms
(a)
Z
x
a
K(x,y)f (y)dy = g(x),
(b)
f (x)-?
Z
x
a
K(x,y)f (y)dy = g(x),
respectively.
Note: Volterra equations may be considered a special case of Fredholm equations. Sup-
pose a= y= b and put
K
1
(x,y) =

K(x,y) for a= y= x,
0 for x < y= b,
then
Z
b
a
K
1
(x,y)f (y)dy =
Z
x
a
+
Z
b
x

K
1
(x,y)f (y)dy =
Z
x
a
K(x,y)f (y)dy+0.
Conversion of ODEs to integral equations
Example 2: Boundary value problems
Let
Lu(x) = ??(x)u(x)+g(x)
where L is a linear ODE operator as in Section 1 (of at least second order), ?(x) > 0 is
continuous, g(x)ispiecewise continuousandtheunknown u(x)issubject tohomogeneous
boundary conditions at x = a,b.
Suppose L has a known Green’s function under the given boundary conditions, G(x,y)
say, then applying Theorem 2.10,
u(x) =
Z
b
a
G(x,y)[??(y)u(y)+g(y)]dy
= ?
Z
b
a
G(x,y)?(y)u(y)dy+
Z
b
a
G(x,y)g(y)dy
= ?
Z
b
a
G(x,y)?(y)u(y)dy+h(x).
The latter is a Fredholm integral equation of the second kind with kernel G(x,y)?(y)and
forcing term
h(x) =
Z
b
a
G(x,y)g(y)dy.
The ODE and integral equation are equivalent: solving the ODE subject to the BCs is
equivalent to solving the integral equation.
Example 3: Initial value problems
Let
u
''
(x)+a(x)u
'
(x)+b(x)u(x) = g(x),
where u(0) = a and u
'
(0) = ß and a(x), b(x) and g(x) are known functions.
Change dummy variable from x to y and then Integrate with respect to y from 0 to z :
Z
z
0
u
''
(y)dy+
Z
x
0
a(y)u
'
(y)dy+
Z
z
0
b(y)u(y)dy =
Z
z
0
g(y)dy
and using integration by parts, with u = a, v
'
= u
'
, on the second term on the left hand
side
[u
'
(y)]
x
0
+[a(y)u(y)]
z
0
-
Z
z
0
a
'
(y)u(y)dy+
Z
z
0
b(y)u(y)dy =
Z
z
0
g(y)dy,
yields
u
'
(z)-ß +a(z)u(z)-a(0)a-
Z
z
0
[a
'
(y)-b(y)]u(y)dy =
Z
z
0
g(y)dy.
Integrating now with respect to z over 0 to x gives
[u(z)]
x
0
-ßx+
Z
x
0
a(z)u(z)dz-a(0)ax (3.1)
-
Z
x
0
Z
z
0
[a
'
(y)-b(y)]u(y)dydz
=
Z
x
0
Z
z
0
g(y)dydz.
Page 4


Section 3
Integral Equations
Integral Operators and Linear Integral Equations
As we saw in Section 1 on operator notation, we work with functions de?ned in some
suitable function space. For example, f (x), g(x) may live in the space of continuous real-
valued functions on [a,b], i.e. C(a,b). We also saw that it is possible to de?ne integral as
well as di?erential operatorsacting on functions. Theorem 2.8 isan example ofan integral
operator:
u(x) =
Z
b
a
G(x,y)f (y)dy,
where G(x,y) is a Green’s function.
De?nition 3.1: An integral operator, K, is an integral of the form
Z
b
a
K(x,y)f (y)dy,
where K(x,y) is a given (real-valued) function of two variables, called the kernel.
The integral equation
Kf = g
maps a function f to a new function g in the interval [a,b], e.g.
K : C[a,b]? C[a,b], K : f 7? g.
Theorem 3.2: The integral operator K is linear
K(af +ßg)= aKf +ßKg.
Example 1: The Laplace Transform is an integral operator
Lf =
Z
8
0
exp(-sx)f (x)dx =
¯
f (s),
with kernel K(x,s) = K(s,x) = exp(-sx). Let us recap its important properties.
Theorem 3.3:
(i) L

df
dx

= s
¯
f (s)-f (0),
(ii) L

d
2
f
dx
2

= s
2
¯
f (s)-sf (0)-f
'
(0),
(iii) L
Z
x
0
f (y)dy

=
1
s
¯
f (s).
(iv) Convolution: let
f (x)*g(x) =
Z
x
0
f (y)g(x-y)dy
then
L(f (x)*g(x)) =
¯
f (s)¯ g(s).
Recall that a di?erential equation is an equation containing an unknown function under
a di?erential operator. Hence, we have the same for an integral operator.
De?nition 3.4: An integral equation is an equation containing an unknown function
under an integral operator.
De?nition 3.5: (a) A linear Fredholm integral equation of the ?rst kind has the
form
Kf = g,
Z
b
a
K(x,y)f (y)dy = g(x);
(b) A linear Fredholm integral equation of the second kind has the form
f-?Kf = g, f (x)-?
Z
b
a
K(x,y)f (y)dy = g(x),
where the kernel K(x,y) and forcing (or inhomogeneous) term g(x) are known functions
and f (x) is the unknown function. Also, ? (?R orC) is a parameter.
De?nition 3.6: If g(x)= 0 the integral equation is called homogeneous and a value of
? for which
?Kf = f
possesses a non-trivial solution f is called aneigenvalue ofK corresponding to theeigen-
function f.
Note: ? is the reciprocal of what we have previously called an eigenvalue.
De?nition 3.7 Volterra integral equations of the ?rst and second kind take the
forms
(a)
Z
x
a
K(x,y)f (y)dy = g(x),
(b)
f (x)-?
Z
x
a
K(x,y)f (y)dy = g(x),
respectively.
Note: Volterra equations may be considered a special case of Fredholm equations. Sup-
pose a= y= b and put
K
1
(x,y) =

K(x,y) for a= y= x,
0 for x < y= b,
then
Z
b
a
K
1
(x,y)f (y)dy =
Z
x
a
+
Z
b
x

K
1
(x,y)f (y)dy =
Z
x
a
K(x,y)f (y)dy+0.
Conversion of ODEs to integral equations
Example 2: Boundary value problems
Let
Lu(x) = ??(x)u(x)+g(x)
where L is a linear ODE operator as in Section 1 (of at least second order), ?(x) > 0 is
continuous, g(x)ispiecewise continuousandtheunknown u(x)issubject tohomogeneous
boundary conditions at x = a,b.
Suppose L has a known Green’s function under the given boundary conditions, G(x,y)
say, then applying Theorem 2.10,
u(x) =
Z
b
a
G(x,y)[??(y)u(y)+g(y)]dy
= ?
Z
b
a
G(x,y)?(y)u(y)dy+
Z
b
a
G(x,y)g(y)dy
= ?
Z
b
a
G(x,y)?(y)u(y)dy+h(x).
The latter is a Fredholm integral equation of the second kind with kernel G(x,y)?(y)and
forcing term
h(x) =
Z
b
a
G(x,y)g(y)dy.
The ODE and integral equation are equivalent: solving the ODE subject to the BCs is
equivalent to solving the integral equation.
Example 3: Initial value problems
Let
u
''
(x)+a(x)u
'
(x)+b(x)u(x) = g(x),
where u(0) = a and u
'
(0) = ß and a(x), b(x) and g(x) are known functions.
Change dummy variable from x to y and then Integrate with respect to y from 0 to z :
Z
z
0
u
''
(y)dy+
Z
x
0
a(y)u
'
(y)dy+
Z
z
0
b(y)u(y)dy =
Z
z
0
g(y)dy
and using integration by parts, with u = a, v
'
= u
'
, on the second term on the left hand
side
[u
'
(y)]
x
0
+[a(y)u(y)]
z
0
-
Z
z
0
a
'
(y)u(y)dy+
Z
z
0
b(y)u(y)dy =
Z
z
0
g(y)dy,
yields
u
'
(z)-ß +a(z)u(z)-a(0)a-
Z
z
0
[a
'
(y)-b(y)]u(y)dy =
Z
z
0
g(y)dy.
Integrating now with respect to z over 0 to x gives
[u(z)]
x
0
-ßx+
Z
x
0
a(z)u(z)dz-a(0)ax (3.1)
-
Z
x
0
Z
z
0
[a
'
(y)-b(y)]u(y)dydz
=
Z
x
0
Z
z
0
g(y)dydz.
This can be simpli?ed by appealing to the following theorem.
Theorem 3.8: For any f (x),
Z
x
0
Z
z
0
f (y)dydz =
Z
x
0
(x-y)f (y)dy.
Proof: Interchanging the order of integration over the triangular domain in the yz-plane
reveals that the integral equals
Z
x
0
Z
z
0
f (y)dydz =
Z
x
0
f (y)
Z
x
y
dzdy
which is trivially integrated over z to give
Z
x
0
(x-y)f (y)dy
asrequired. Alternatively,integratingtherepeatedintegralbypartswithu(z) =
R
z
0
f (y)dy,
v
'
(z) = 1, i.e. u
'
(z) = f (z), v(z) = z, gives
Z
x
0
Z
z
0
f (y)dydz =

z
Z
z
0
f (y)dy

z=x
z=0
-
Z
x
0
zf (z)dz
= x
Z
x
0
f (y)dy-
Z
x
0
zf (z)dz
=
Z
x
0
xf (y)dy-
Z
x
0
yf (y)dy.

Returning to equation (3.1), it may now be written as
u(x)-a-ßx-a(0)ax+
Z
x
0
a(y)u(y)dy
-
Z
x
0
(x-y)[a
'
(y)-b(y)]u(y)dy
=
Z
x
0
(x-y)g(y)dy.
Thus,
u(x)+
Z
x
0
{a(y)-(x-y)[a
'
(y)-b(y)]}u(y)dy
=
Z
x
0
(x-y)g(y)dy+[ß +a(0)a]x+a.
This is a Volterra integral equation of the second kind.
Notes:
(1) BVPs correspond to Fredholm equations.
(2) Initial value problems (IVPs) correspond to Volterra equations.
(3) The integral equation formulation implicitly contains the boundary/initial conditions;
in the ODE they are imposed separately.
Page 5


Section 3
Integral Equations
Integral Operators and Linear Integral Equations
As we saw in Section 1 on operator notation, we work with functions de?ned in some
suitable function space. For example, f (x), g(x) may live in the space of continuous real-
valued functions on [a,b], i.e. C(a,b). We also saw that it is possible to de?ne integral as
well as di?erential operatorsacting on functions. Theorem 2.8 isan example ofan integral
operator:
u(x) =
Z
b
a
G(x,y)f (y)dy,
where G(x,y) is a Green’s function.
De?nition 3.1: An integral operator, K, is an integral of the form
Z
b
a
K(x,y)f (y)dy,
where K(x,y) is a given (real-valued) function of two variables, called the kernel.
The integral equation
Kf = g
maps a function f to a new function g in the interval [a,b], e.g.
K : C[a,b]? C[a,b], K : f 7? g.
Theorem 3.2: The integral operator K is linear
K(af +ßg)= aKf +ßKg.
Example 1: The Laplace Transform is an integral operator
Lf =
Z
8
0
exp(-sx)f (x)dx =
¯
f (s),
with kernel K(x,s) = K(s,x) = exp(-sx). Let us recap its important properties.
Theorem 3.3:
(i) L

df
dx

= s
¯
f (s)-f (0),
(ii) L

d
2
f
dx
2

= s
2
¯
f (s)-sf (0)-f
'
(0),
(iii) L
Z
x
0
f (y)dy

=
1
s
¯
f (s).
(iv) Convolution: let
f (x)*g(x) =
Z
x
0
f (y)g(x-y)dy
then
L(f (x)*g(x)) =
¯
f (s)¯ g(s).
Recall that a di?erential equation is an equation containing an unknown function under
a di?erential operator. Hence, we have the same for an integral operator.
De?nition 3.4: An integral equation is an equation containing an unknown function
under an integral operator.
De?nition 3.5: (a) A linear Fredholm integral equation of the ?rst kind has the
form
Kf = g,
Z
b
a
K(x,y)f (y)dy = g(x);
(b) A linear Fredholm integral equation of the second kind has the form
f-?Kf = g, f (x)-?
Z
b
a
K(x,y)f (y)dy = g(x),
where the kernel K(x,y) and forcing (or inhomogeneous) term g(x) are known functions
and f (x) is the unknown function. Also, ? (?R orC) is a parameter.
De?nition 3.6: If g(x)= 0 the integral equation is called homogeneous and a value of
? for which
?Kf = f
possesses a non-trivial solution f is called aneigenvalue ofK corresponding to theeigen-
function f.
Note: ? is the reciprocal of what we have previously called an eigenvalue.
De?nition 3.7 Volterra integral equations of the ?rst and second kind take the
forms
(a)
Z
x
a
K(x,y)f (y)dy = g(x),
(b)
f (x)-?
Z
x
a
K(x,y)f (y)dy = g(x),
respectively.
Note: Volterra equations may be considered a special case of Fredholm equations. Sup-
pose a= y= b and put
K
1
(x,y) =

K(x,y) for a= y= x,
0 for x < y= b,
then
Z
b
a
K
1
(x,y)f (y)dy =
Z
x
a
+
Z
b
x

K
1
(x,y)f (y)dy =
Z
x
a
K(x,y)f (y)dy+0.
Conversion of ODEs to integral equations
Example 2: Boundary value problems
Let
Lu(x) = ??(x)u(x)+g(x)
where L is a linear ODE operator as in Section 1 (of at least second order), ?(x) > 0 is
continuous, g(x)ispiecewise continuousandtheunknown u(x)issubject tohomogeneous
boundary conditions at x = a,b.
Suppose L has a known Green’s function under the given boundary conditions, G(x,y)
say, then applying Theorem 2.10,
u(x) =
Z
b
a
G(x,y)[??(y)u(y)+g(y)]dy
= ?
Z
b
a
G(x,y)?(y)u(y)dy+
Z
b
a
G(x,y)g(y)dy
= ?
Z
b
a
G(x,y)?(y)u(y)dy+h(x).
The latter is a Fredholm integral equation of the second kind with kernel G(x,y)?(y)and
forcing term
h(x) =
Z
b
a
G(x,y)g(y)dy.
The ODE and integral equation are equivalent: solving the ODE subject to the BCs is
equivalent to solving the integral equation.
Example 3: Initial value problems
Let
u
''
(x)+a(x)u
'
(x)+b(x)u(x) = g(x),
where u(0) = a and u
'
(0) = ß and a(x), b(x) and g(x) are known functions.
Change dummy variable from x to y and then Integrate with respect to y from 0 to z :
Z
z
0
u
''
(y)dy+
Z
x
0
a(y)u
'
(y)dy+
Z
z
0
b(y)u(y)dy =
Z
z
0
g(y)dy
and using integration by parts, with u = a, v
'
= u
'
, on the second term on the left hand
side
[u
'
(y)]
x
0
+[a(y)u(y)]
z
0
-
Z
z
0
a
'
(y)u(y)dy+
Z
z
0
b(y)u(y)dy =
Z
z
0
g(y)dy,
yields
u
'
(z)-ß +a(z)u(z)-a(0)a-
Z
z
0
[a
'
(y)-b(y)]u(y)dy =
Z
z
0
g(y)dy.
Integrating now with respect to z over 0 to x gives
[u(z)]
x
0
-ßx+
Z
x
0
a(z)u(z)dz-a(0)ax (3.1)
-
Z
x
0
Z
z
0
[a
'
(y)-b(y)]u(y)dydz
=
Z
x
0
Z
z
0
g(y)dydz.
This can be simpli?ed by appealing to the following theorem.
Theorem 3.8: For any f (x),
Z
x
0
Z
z
0
f (y)dydz =
Z
x
0
(x-y)f (y)dy.
Proof: Interchanging the order of integration over the triangular domain in the yz-plane
reveals that the integral equals
Z
x
0
Z
z
0
f (y)dydz =
Z
x
0
f (y)
Z
x
y
dzdy
which is trivially integrated over z to give
Z
x
0
(x-y)f (y)dy
asrequired. Alternatively,integratingtherepeatedintegralbypartswithu(z) =
R
z
0
f (y)dy,
v
'
(z) = 1, i.e. u
'
(z) = f (z), v(z) = z, gives
Z
x
0
Z
z
0
f (y)dydz =

z
Z
z
0
f (y)dy

z=x
z=0
-
Z
x
0
zf (z)dz
= x
Z
x
0
f (y)dy-
Z
x
0
zf (z)dz
=
Z
x
0
xf (y)dy-
Z
x
0
yf (y)dy.

Returning to equation (3.1), it may now be written as
u(x)-a-ßx-a(0)ax+
Z
x
0
a(y)u(y)dy
-
Z
x
0
(x-y)[a
'
(y)-b(y)]u(y)dy
=
Z
x
0
(x-y)g(y)dy.
Thus,
u(x)+
Z
x
0
{a(y)-(x-y)[a
'
(y)-b(y)]}u(y)dy
=
Z
x
0
(x-y)g(y)dy+[ß +a(0)a]x+a.
This is a Volterra integral equation of the second kind.
Notes:
(1) BVPs correspond to Fredholm equations.
(2) Initial value problems (IVPs) correspond to Volterra equations.
(3) The integral equation formulation implicitly contains the boundary/initial conditions;
in the ODE they are imposed separately.
Example 4
Write the Initial Value Problem
u
''
(y)+yu
'
(y)+2u(y)= 0
subject to u(0) = a,u
'
(0) = ß as a Volterra integral equation.
Integrate with respect to y between 0 and z:
u
'
(z)-ß +
Z
z
0
yu
'
(y)dy+2
Z
z
0
u(y)dy = 0
Integrate second term by parts:
Z
z
0
yu
'
(y)dy = [yu(y)]
z
0
-
Z
z
0
u(y)dy,
= zu(z)-
Z
z
0
y(y)dy
and substitute back to get
u
'
(z)-ß +zu(z)+
Z
z
0
u(y)dy = 0.
Integrate again, with respect to z between 0 and x:
u(x)-a-ßx+
Z
x
0
zu(z)dz +
Z
x
0
Z
z
0
u(y)dydz = 0.
Change dummy variable in ?rst integral term to y and use Theorem 3.8 in last term to
get
u(x)-a-ßx+
Z
x
0
yu(y)dy+
Z
x
0
(x-y)u(y)= 0
Simpli?cation gives
u(x)+x
Z
x
0
u(y)dy = a+ßx.