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A Simple Introduction to
Integral EQUATIONS
Glenn Ledder
Department of MathematicsUniversity of [email protected]
Some Integral Equation Examples
2 20
( )t y s dst
t s
y t t t s y s ds( ) ( ) 20
1
Volterra
Volterra
Volterra
Fredholm
second kind
second kind
second kind
first kind
0
2 ( )( )
1 ( )
t s y sy t ds
y s
y t t t s y s dst
( ) ( ) 20
Integral Equations from Differential Equations
First-Order
Initial Value Problem
y t F t g t y t y a y' ( ) ( ) ( , ( )), ( ) 0
Volterra Integral EquationOf the Second Kind
Used to prove theorems about differential equations.
Used to derive numerical methods for differential equations.
y t y F s ds g s y s dsa
t
a
t
( ) ( ) ( , ( )) 0
• 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 Equationsof the Second Kind (VESKs)
y t f t g t s y s dsa
t
( ) ( ) ( , , ( ))
Given g : R3→R and f: R→R , find y: R→R such that
[equivalent to an initial value problem if g is independent of t]
y t y F s ds g s y s dsa
t
a
t
( ) ( ) ( , ( )) 0
Volterra Integral Equationsof the Second Kind (VESKs)
y t f t g t s y s dsa
t
( ) ( ) ( , , ( ))
y t f t k t s y s dsa
t
( ) ( ) ( , ) ( ) Linear:
Separable:
Convolved:
k t s p t q s( , ) ( ) ( )
k t s r t s( , ) ( )
Given g : R3→R and f: R→R , find y: R→R such that
Symbolic Solution of Separable VESKs
y t t t s y s dst
( ) ( ) 20
Y t s y s dst
( ) ( ) 0
Y t t y t t Y t Y' ( ) ( ) , ( ) 2 0 0
y t t t Y t( ) ( ) 2
y t t e t( ) 2
Solve
Let
Then
Solution:
New Problem:
y t t s y s dst
( ) ( ) ( ) 10
Y t y s dst
( ) ( ) 0
Y Y Y Y
Z Y tY
0 0 0 0 1
1
, ( ) , ( ) ,
Z t s y s dst
( ) ( ) 0
y t t( ) co s
Solve
Let
and
Solution:
New Problem:
Volterra Sequences
y t f t g t s y s dsa
t
1 0( ) ( ) ( , , ( )) y t f t g t s y s ds
a
t
2 1( ) ( ) ( , , ( ))
Given f, g, and y0, define y1, y2, … by
and so on.
Volterra Sequences
y t f t g t s y s ds nn na
t
( ) ( ) ( , , ( )) , , , 1 1 2
Given f, g, and y0, define y1, y2, … by
•Does yn converge to some y ?
•If so, does y solve the VESK?
Volterra Sequences
y t f t g t s y s ds nn na
t
( ) ( ) ( , , ( )) , , , 1 1 2
Given f, g, and y0, define y1, y2, … by
•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 t s y s dst
( ) ( ) ( ) 10
y t t s ds tt
1 0
12
21 1( ) ( )
y t t s s ds t tt
212
2
0
12
2 12 4
41 1 1( ) ( )( )
y t t t t 1 12
2 12 4
4 16
6! co s
Let y0 = 1. Then
↓↓
The sequence converges to the known solution.
Convergence of the Sequence
y0
y1
y2
y3
y4
y5
y6
y
A Nonlinear Example
y ts y s
y sds
t
( )( )
( )
2
10
y t s ds tt
1 0
22( )
y ts s s
s s sds
t
3 0
1 3 1 2
1 3 1 2 2( )
ln ( ) ln ( )
ln ( ) ln ( )
Let y0 = 0. Then
231
2 2 220
2( ) ln(1 ) ln(1 )
1
t s sy t ds t t t
s
Convergence of the Sequence
y3
y2
y1
Lemma 1
y t k t s y s ds nn na
t
( ) ( , ) ( ) , , , 1 1 2
| ( , ) | , | ( ) | , ,k t s K y t Y t s T 0 0
Homogeneous linear Volterra sequences
decay to 0 on [a, T ] whenever k is bounded.
Choose constants K, Y and T such that
| ( ) | ( , ) ( )y t k t s y s ds KY ds YK tt t
1 00 0
Then
Proof: (a=0)
y t k t s y s ds nn n
t
( ) ( , ) ( ) , , , 101 2
| ( , ) | , | ( ) | , ,k t s K y t Y t s T 0 0
| ( ) |y t K Y ds YK tt
1 0
| ( ) | ( )y t K YK s ds YK tt
2 0
2 21
2
| ( ) | ( )!
y t K YK s ds YK tt
312
2 2
0
3 31
3
| ( ) |!
y tnYK t nn
n n 1
Given
we have
In general,
so limn
ny 0
Theorem 2Homogeneous linear VESKs
have the unique solution y ≡ 0.
y t k t s y s dsa
t
( ) ( , ) ( )
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 3Linear VESKs have at most one solution.
( ) ( ) ( , ) ( )t f t k t s s dsa
t
( ) ( ) ( , ) ( )t f t k t s s dsa
t
y t k t s y s dsa
t
( ) ( , ) ( )
Proof:
Suppose
and
Let y = φ – ψ. Then
By Theorem 2, y ≡ 0; hence, φ = ψ.
Lemma 4Linear Volterra sequences with y0=f converge.
y t f t k t s y s ds y fn na
t
( ) ( ) ( , ) ( ) , 1 0
| |y y gn m n n
Define
Sketch of Proof:
g y yn n m n mm
| |11
1) gn → 0
2)
Together, these properties prove the Lemma.
Theorem 5Linear VESKs have a unique solution.
y t f t k t s y s ds y fn na
t
( ) ( ) ( , ) ( ) , 1 0
By Lemma 4, the sequence
y t f t k t s y s dsa
t
( ) ( ) ( , ) ( )
Proof:
converges to some y. Taking a limit as n → ∞:
The solution is unique by Theorem 3.
Fredholm Integral Equations of the Second Kind (linear)
• 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.
Given g : R3→R and f: R→R , find y: R→R such that
y t f t k t s y s dsa
b
( ) ( ) ( , ) ( )
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
Given An initial population of newborns Age-dependent birthrate Age-independent death rate
Find Total birthrate Total population
Simplified Version
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
Let f (t) be the number of births per mother
of age t per unit time.
b(t) = m(t) + d(t)
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 dxt
( ) ( ) ( ) 0
The Renewal Equation
b t m t m t x b x dxt
( ) ( ) ( ) ( ) 0
b t m t m x b t x dxt
( ) ( ) ( ) ( ) 0
or
This is a linear VESK with convolved kernel.It can be solved by Laplace transform.
Solution of the Renewal Equation
F s L f t r t LF s
F s( ) ( ) , ( )
( )
( )
1
1
p t e e r x dxt tt
( ) ( ) 0
Given the fecundity function f, define
where L is the Laplace transform. Then
b t e r tt( ) ( )
A Specific Example
p t e e eaa
a ta
t aa
a t( ) ( )( )
( _ )( )
2 2
3 44 2 2
3
Let f(x) = a x e -2x.
b t e ea a t a a t( ) ( ) ( ) 2
32
3Then
If a = 9, then p(t) →1.5. Exponential growth for a>9.The population dies if a<9.
Fecundity
a = 10a = 9a = 8
Birth Rate
a = 10a = 9a = 8
Population
a = 10a = 9a = 8
