A Simple Introduction to
Integral EQUATIONS
Glenn Ledder
Department of Mathematics
University of Nebraska-Lincoln
gledder@math.unl.edu
Some Integral Equation Examples
2s  y ( s)
y (t )  
ds
0 1  y( s)
Volterra
second kind
t

2 t s y( s) ds
Volterra
second kind
t

2 t s y( s) ds
Fredholm second kind
t
y (t ) 
y (t ) 

t
0
t
0
1
0
y ( s ) ds
t s
2
2
t
Volterra
first kind
Integral Equations from
Differential Equations
First-Order
Initial Value Problem

y '(t )  F (t )  g (t , y (t )),
y (t )  y 0 
Volterra Integral Equation
Of the Second Kind
y (a )  y 0
 F (s) ds   g(s, y(s)) ds
t
t
a
a
Used to prove theorems about differential equations.
Used to derive numerical methods for differential equations.
• Volterra IEs of the second kind (VESKs)
• Symbolic solution of separable VESKs
• Volterra sequences and iterative solutions
• Theory of linear VESKs
• Brief mention of Fredholm IEs of 2nd kind
• An age-structured population model
Volterra Integral Equations
of the Second Kind (VESKs)
Given g : R3→R and f: R→R , find y: R→R such that
y(t )  f (t )   g(t , s, y( s)) ds
t
a
[equivalent to an initial value problem if g is independent of t]
y (t )  y 0 
 F (s) ds   g(s, y(s)) ds
t
t
a
a
Volterra Integral Equations
of the Second Kind (VESKs)
Given g : R3→R and f: R→R , find y: R→R such that
y(t )  f (t )   g(t , s, y( s)) ds
t
a
Linear:
y (t )  f (t ) 
 k (t , s) y(s) ds
t
a
Separable:
k (t , s)  p(t ) q ( s)
Convolved:
k (t , s)  r (t  s)
Symbolic Solution of
Separable VESKs
Solve
Let
y (t ) 
Y (t ) 

t

t
0
2 t s y( s) ds
t
0
sy( s) ds
Then
y (t ) 
t  2 t Y (t )
New Problem:
Y '(t ) 
Solution:
t y  t  2t Y ( t ) ,
y (t ) 
te
t 2
Y (0)  0
y(t )  1   (t  s) y( s) ds
t
Solve
Let
Y (t ) 
0

t
0
y( s) ds
and
Z (t ) 
 sy(s) ds
t
0
New Problem:
Y   Y  0 , Y (0)  0, Y  (0)  1,
Z  Y   tY  1
Solution:
y(t )  cos t
Volterra Sequences
Given f, g, and y0, define y1, y2, … by
y1 (t )  f (t ) 

t
y 2 (t )  f ( t ) 

t
and so on.
a
a
g(t , s, y0 ( s)) ds
g(t , s, y1 ( s)) ds
Volterra Sequences
Given f, g, and y0, define y1, y2, … by
y n (t )  f (t ) 

t
a
g(t , s, yn1 ( s)) ds , n  1, 2,
•Does yn converge to some y ?
•If so, does y solve the VESK?
Volterra Sequences
Given f, g, and y0, define y1, y2, … by
y n (t )  f (t ) 

t
a
g(t , s, yn1 ( s)) ds , n  1, 2,
•Does yn converge to some y ?
•If so, does y solve the VESK?
•If so, is this a useful iterative method?
A Linear Example
y(t )  1   (t  s) y( s) ds
t
0
Let y0 = 1. Then
y1 (t )  1   (t  s) ds    1  21 t 2
t
0
y 2 (t )  1 

t
0
(t  s)(1  s ) ds    1  t 
2
1
2
1
2
2
1
24
t
4
↓↓
y  1  t 
1
2
2
1
24
t 
4
1
6!
t    cos t
6
The sequence converges to the known solution.
Convergence of the Sequence
y0
y1
y2
y3
y4
y5
y6
y
A Nonlinear Example
2 s  y ( s)
y (t )  
ds
0 1  y ( s)
t
Let y0 = 0. Then
y1 (t ) 
 2sds  t
t
2
0
2s  s
3
1
y2 (t )  
ds

t

ln(1

t
)

ln(1  t )
2
2
2
0 1 s
2
t
ln(1  s)  3 ln(1  s)  2s
y 3 (t )  
ds
0 ln(1  s)  3 ln(1  s)  2 s  2
t
Convergence of the Sequence
y3
y2
y1
Lemma 1
Homogeneous linear Volterra sequences
 k (t , s) y
t
y n (t ) 
a
n 1
( s) ds , n  1, 2,
decay to 0 on [a, T] whenever k is bounded.
Proof: (a=0)
Choose constants K, Y and T such that
| k (t , s)|  K , | y0 (t )|  Y ,  0  t , s  T
Then
| y1 (t )| 

t
0
k (t , s) y0 ( s) ds 
 KY ds  YKt
t
0
Given
| k (t , s)|  K , | y0 (t )|  Y ,  0  t , s  T
y n (t ) 
we have
 k (t , s) y
t
0
| y1 (t )| 
n 1
( s) ds , n  1, 2,
 KY ds  YKt
t
0
1
| y2 (t )|   K (YKs) ds  YK 2 t 2
0
2
t
1
3 3
| y3 (t )|   K ( YK s ) ds  YK t
0
3!
t
In general,
1
n n
| yn (t )|  YK t  n
n!
2 2
1
2
so
lim yn  0
n 
Theorem 2
Homogeneous linear VESKs
y (t ) 
 k (t , s) y(s) ds
t
a
have the unique solution y ≡ 0.
Proof:
Let φ be a solution.
Choose y0 = φ.
Then yn = φ for all n, so yn→φ.
But the lemma requires yn→0.
Therefore φ ≡ 0.
Theorem 3
Linear VESKs have at most one solution.
Proof:
Suppose
 (t )  f (t )  ak (t , s)  ( s) ds
and
 (t )  f (t )  ak (t , s)  ( s) ds
t
t
Let y = φ – ψ. Then
y (t ) 
 k (t , s) y(s) ds
t
a
By Theorem 2, y ≡ 0; hence, φ = ψ.
Lemma 4
Linear Volterra sequences with y0=f converge.
Sketch of Proof:
Define yn (t )  f (t ) 
gn 
 k (t , s) y
t
a

 |y
m 1
n m
n 1
( s) ds , y0  f
 y n  m 1 |
1) gn → 0
2)
| yn m  yn |    gn
Together, these properties prove the Lemma.
Theorem 5
Linear VESKs have a unique solution.
Proof:
By Lemma 4, the sequence
yn (t )  f (t )   k (t , s) yn1 ( s) ds , y0  f
t
a
converges to some y. Taking a limit as n → ∞:
y(t )  f (t )   k (t , s) y( s) ds
t
a
The solution is unique by Theorem 3.
Fredholm Integral Equations
of the Second Kind (linear)
Given g : R3→R and f: R→R , find y: R→R such that
y(t )  f (t )   k (t , s) y( s) ds
b
a
• Separable FESKs can be solved symbolically.
• Fredholm sequences converge (more slowly than
Volterra sequences) only if ||k|| is small enough.
• The theorems hold if and only if ||k|| is small
enough.
An Age-Structured Population
Given
An initial population of known age distribution
Age-dependent birthrate
Age-dependent death rate
Find
Total birthrate
Total population
Age distribution
An Age-Structured Population
Simplified Version
Given
An initial population of newborns
Age-dependent birthrate
Age-independent death rate
Find
Total birthrate
Total population
The Founding Mothers
Assume a one-sex population.
Starting population is 1 unit.
Life expectancy is 1 time unit.
p0′ = -p0, p0(0) = 1
Result:
p0(t) = e-t
Basic Birthrate Facts
Total Birthrate =
Births to Founding Mothers
+
Births to Native Daughters
b(t) = m(t) + d(t)
Let f (t) be the number of births per mother
of age t per unit time.
m(t) = f (t) e-t
Births to Native Daughters
Consider daughters of ages x to x+dx.
All were born between t-x-dx and t-x.
The initial number was b(t-x)dx.
The number at time t is b(t-x)e-x dx.
The rate of births is f(x)b(t-x)e-x dx
=m(x)b(t-x)dx.
d (t ) 
 m( x) b(t  x) dx
t
0
The Renewal Equation
b(t )  m(t )   m( x) b(t  x) dx
t
0
or
b(t )  m(t )   m(t  x) b( x) dx
t
0
This is a linear VESK with convolved kernel.
It can be solved by Laplace transform.
Solution of the Renewal Equation
Given the fecundity function f, define
 F ( s) 
F ( s)  L f (t ), r (t )  L 

 1  F ( s) 
1
where L is the Laplace transform. Then
t
b(t )  e r (t )
t
p(t )  e  e
t
 r( x) dx
t
0
A Specific Example
Let f(x) = a xe -2x.
Then
b(t ) 
p(t ) 
a
2
e
( a  3) t
a
( a  3) t
2( a  2)
e


4 t
a4
If a=9, then p(t) →1.5.
Exponential growth for a>9.
The population dies if a<9.
a
2
e 
e
(  a  3) t
a
(  a  3) t
2( a _ 2)
e
Fecundity
a = 10
a=9
a=8
Birth Rate
a = 10
a=9
a=8
Population
a = 10
a=9
a=8