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Thus, with
∞
18.12
Rλ(x, y) =
λiKi(x, y),
i=1
we have
18.13
Although 18.10, 18.12, 18.13 together give an “explicit” solution to the Fredholm
equation, this explicitness is only theoretical. For, computing Rλ is of the same order
of difficulty as solving 18.1 (in fact, even harder).
On the other hand, if the kernel K is simple enough, analytic solutions might be
possible. The following illustrates the computations for such a special case.
18.14 EXAMPLE. Suppose that
n
x, y ∈ [a, b]
for some continuous functions p1, . . . , pn and q1, . . . , qn on [a, b]. For φ continuous on
[a, b], consider the Fredholm equations 18.1. Now, if f ∈ C satisfies 18.1, then
n
18.15
where
18.16
In view of 18.15, then
zi
=
=
75
Thus, letting
18.17
we obtain
18.18
Note that the ci and aij are known. If we can solve 18.18 for the zi’s, then 18.15 gives
the solution f.
b
f(x) = φ(x) +
Rλ(x, y)φ(y)dy.
a
K(x, y) =
pj(x)qj(y)
j=1
f(x) = φ(x) + λ
zjpj(x)
j=1
b
zj =
qj(y)f(y)dy,
j = 1, . . . , n.
a
a
b
qi(x)f(x)dx
b
b
qi(x)φ(x)dx + λ
qi(x)pj(x)dx zj.
n
j=1
a
a
a
b
ci =
qi(x)φ(x)dx,
aij =
qi(x)pj(x)dx,
n
zi = ci + λ
aijzj,
i = 1, 2, . . . , n.
j=1
a
b
x ∈ [a, b].
It is called the Volterra equation. It differs from the Fredholm equation only slightly,
and in form only. If we define
76
DIFFERENTIAL AND INTEGRAL EQUATIONS
In vector-matrix notation, 18.18 becomes
z = c + λAz,
whose solution is easy to discern. We can solve it for z (for arbitrary c) as long as
I - λA is invertible, that is, as long as 1/λ is not an eigenvalue for A. Thus, we have
a solution z for arbitrary b provided that λ ∈ (-1/λ0, 1/λ0) where λ0 is the modulus
of the largest eigenvalue of A.
Volterra Equation
Let K be a continuous function on [a, b] × [a, b] and let φ be a continuous function on
[a, b]. Consider the equation
Now, for f and g in C,
|T f(x) - T g(x)|
=
≤
x
18.19
f(x) = φ(x) + λ
K(x, y)f(y)dy,
a
K(x, y)
if y ≤ x,
if y x,
ˆ
K(x, y) =
ˆ
then 18.19 becomes the Fredholm equation 18.1 with kernel K. However, it is easier
to attack 18.19 directly.
18.20 THEOREM. For each λ ∈ R, the Volterra equation 18.19 has a unique solution
f that is continuous on [a, b].
PROOF. Let C = C([a, b], R), the set of all continuous functions from [a, b] into R,
with the usual uniform metric f - g . Let c be the maximum of |K(x, y)| over all
x, y ∈ [a, b]; this number is finite since K is continuous. Define the transformation
T : f → T f on C by
x
T f(x) = φ(x) + λ
K(x, y)f(y)dy.
a
x
|λ
K(x, y)[f(y) - g(y)]dy|
a
|λ|c(x - a) f - g , x ∈ [a, b].
18. INTEGRAL EQUATIONS
We use this, next, to bound T f - T g = T (T f - T g):
x
|T f(x) - T g(x)|
=
|λ
K(x, y)[T f(y) - T g(y)]dy|
a
77
≤
≤
≤
Iterating in this manner, we see that
k
k
- g
a)k
f
for all x ∈ [a, b]. Hence,
[|λ|c(b - a)]k
T f - T g ≤
f - g .
k!
Recalling that rn/n! tends to 0 as n → ∞ for any r ∈ R, we conclude that there exists
k such that T is a contraction: simply take k large enough to have [|λ|c(b - a)]k/k!
1. Finally, the existence and uniqueness of f ∈ C satisfying f = T f follows from the
next theorem. Obviously, if f = T f, then f solves 18.19.
Generalization of the Fixed Point Theorem
18.21 THEOREM. Let E be a complete metric space and let T be a continuous trans-
formation on E. If T is a contraction for some k ≥ 1, then T has a unique fixed point.
PROOF. Fix k such that U = T is a contraction. By Theorem 16.3, then, U has
a unique fixed point x, and limn Unx0 = x for every point x0 in E. Now, by the
continuity of T ,
T x
=
lim T Unx0
n
=
lim T T
x0
n
=
lim T
T x0
=
lim UnT x0
n
=
x,
2
2
a
2
2
x
|λ|
|K(x, y)||λ|c(y - a) f - g dy
x
|λ|2c2
(y - a)dy f - g
a
|λ|2c2(x - a)2
f - g .
2
|λ|kck(x -
|T f(x) - T g(x)| ≤
k!
k
k
k
k
k
kn
kn
v(t,x(t))
1
x(t)
x0
78
DIFFERENTIAL AND INTEGRAL EQUATIONS
t0
t
Figure 14: A moving particle.
that is, x is a fixed point of T . To show that it is the only fixed point of T we note that
every fixed point of T is a fixed point of T = U, whereas U has only one fixed point,
namely x.
Exercises:
18.1 Solve the Fredholm equation 18.1 for arbitrary φ, on [a, b] = [0, 2π], with
the kernel
K(x, y) = sin(x + y).
18.2 Do the same with [a, b] = [0, 1] and K(x, y) = (x - y)2.
18.3 Let p be a continuous function of [0, b]. Show that
x
f(x) = φ(x) +
p(y)f(x - y)dy,
x ∈ [0, b],
has a unique solution f for each continuous function φ.
19
Differential Equations
We continue with applications of the fixed point theorem by discussing Picard’s method
of successive approximations for solving systems of differential equations. [ Pobierz całość w formacie PDF ]