Fredholm Integral Equations
4.2 Fredholm Integral Equations of the Second Kind
4.2.1 The Adomian Decomposition Method
The Adomian decomposition method (ADM) was introduced and developed by George Adomian in [2–5] and was used before in Chapter 3. The Adomian method will be briefly outlined.
The Adomian decomposition method consists of decomposing the un- known function u(x) of any equation into a sum of an infinite number of components defined by the decomposition series
u(x) = ∞ n=0
un(x), (4.13)
or equivalently
u(x) =u0(x) +u1(x) +u2(x) +· · · (4.14) where the components un(x), n 0 will be determined recurrently. The Adomian decomposition method concerns itself with finding the components u0, u1, u2, . . .individually. As we have seen before, the determination of these components can be achieved in an easy way through a recurrence relation that usually involves simple integrals that can be easily evaluated.
To establish the recurrence relation, we substitute (4.13) into the Fredholm integral equation (4.12) to obtain
122 4 Fredholm Integral Equations ∞
n=0
un(x) =f(x) +λ b
a
K(x, t) ∞
n=0
un(t)
dt, (4.15)
or equivalently
u0(x) +u1(x) +u2(x) +· · ·=f(x) +λ b
a
K(x, t) [u0(t) +u1(t) +· · ·]dt.
(4.16) The zeroth componentu0(x) is identified by all terms that are not included under the integral sign. This means that the components uj(x), j0 of the unknown functionu(x) are completely determined by setting the recurrence relation
u0(x) =f(x), un+1(x) =λ b
a
K(x, t)un(t)dt, n0, (4.17) or equivalently
u0(x) =f(x), u1(x) =λ
b a
K(x, t)u0(t)dt, u2(x) =λ
b a
K(x, t)u1(t)dt, u3(x) =λ
b a
K(x, t)u2(t)dt,
(4.18)
and so on for other components.
In view of (4.18), the components u0(x), u1(x), u2(x), u3(x), . . . are com- pletely determined. As a result, the solution u(x) of the Fredholm integral equation (4.12) is readily obtained in a series form by using the series as- sumption in (4.13).
It is clearly seen that the decomposition method converted the integral equation into an elegant determination of computable components. It was formally shown that if an exact solution exists for the problem, then the ob- tained series converges very rapidly to that exact solution. The convergence concept of the decomposition series was thoroughly investigated by many researchers to confirm the rapid convergence of the resulting series. How- ever, for concrete problems, where a closed form solution is not obtainable, a truncated number of terms is usually used for numerical purposes. The more components we use the higher accuracy we obtain.
Example 4.1
Solve the following Fredholm integral equation u(x) =ex−x+x
1
0
tu(t)dt. (4.19)
The Adomian decomposition method assumes that the solution u(x) has a series form given in (4.13). Substituting the decomposition series (4.13) into
4.2 Fredholm Integral Equations of the Second Kind 123 both sides of (4.19) gives
∞ n=0
un(x) =ex−x+x 1
0 t ∞ n=0
un(t)dt, (4.20) or equivalently
u0(x) +u1(x) +u2(x) +· · ·=ex−x+x 1
0
t[u0(t) +u1(t) +u2(t) +· · ·]dt.
(4.21) We identify the zeroth component by all terms that are not included under the integral sign. Therefore, we obtain the following recurrence relation
u0(x) =ex−x, uk+1(x) =x 1
0 tuk(t)dt, k0. (4.22) Consequently, we obtain
u0(x) =ex−x, u1(x) =x
1
0
tu0(t)dt=x 1
0
t(et−t)dt=2 3x, u2(x) =x
1
0 tu1(t)dt=x 1
0
2
3t2dt= 2 9x, u3(x) =x
1
0 tu2(t)dt=x 1
0
2
9t2dt= 2 27x, u4(x) =x
1
0
tu3(t)dt=x 1
0
2
27t2dt= 2 81x,
(4.23)
and so on. Using (4.13) gives the series solution u(x) =ex−x+2
3x
1 + 1 3+1
9 + 1 27 +· · ·
. (4.24)
Notice that the infinite geometric series at the right side hasa1= 1, and the ratior= 13. The sum of the infinite series is therefore given by
S= 1 1−13 = 3
2. (4.25)
The series solution (4.24) converges to the closed form solution
u(x) =ex, (4.26)
obtained upon using (4.25) into (4.24).
Example 4.2
Solve the following Fredholm integral equation u(x) = sinx−x+x
π2
0 u(t)dt. (4.27)
Proceeding as before, we substitute the decomposition series (4.13) into both sides of (4.27) to find
124 4 Fredholm Integral Equations ∞
n=0
un(x) = sinx−x+x π
2
0
∞ n=0
un(t)dt, (4.28) or equivalently
u0(x) +u1(x) +u2(x) +· · ·= sinx−x+x π
2
0
[u0(t) +u1(t) +· · ·]dt. (4.29) We identify the zeroth component by all terms that are not included under the integral sign. Therefore, we obtain the following recurrence relation:
u0(x) = sinx−x, uk+1(x) =x π2
0 uk(t)dt, k0. (4.30) Consequently, we obtain
u0(x) = sinx−x, u1(x) =x
π2
0 u0(t)dt=x−π2 8 x, u2(x) =x
π
2
0 u1(t)dt= π2 8 x−π4
64x, u3(x) =x
π
2
0
u2(t)dt= π4 64x− π6
512x, u4(x) =x
π2
0
u3(t)dt= π6
512x− π8 4096x,
(4.31)
and so on. Using (4.13) gives the series solution u(x) = sinx−x+
1−π2
8
x+ π2
8 −π4 64
x +
π4 64− π6
512
x+ π6
512− π8 4096
x+· · · . (4.32) We can easily observe the appearance of the noise terms, i.e the identical terms with opposite signs. Canceling these noise terms in (4.32) gives the exact solution
u(x) = sinx. (4.33)
Example 4.3
Solve the following Fredholm integral equation u(x) =x+ex−4
3 + 1
0
tu(t)dt. (4.34)
Substituting the decomposition series (4.13) into both sides of (4.34) gives ∞
n=0
un(x) =x+ex−4 3 +
1
0
t ∞ n=0
un(t)dt, (4.35) or equivalently
4.2 Fredholm Integral Equations of the Second Kind 125 u0(x)+u1(x)+u2(x)+· · ·=x+ex−4
3+ 1
0 t[u0(t) +u1(t) +· · ·]dt. (4.36) Proceeding as before, we set the following recurrence relation
u0(x) =x+ex−4
3, uk+1(x) = 1
0 tuk(t)dt, k0. (4.37) Consequently, we obtain
u0(x) =x+ex−4
3, u1(x) = 1
0
tu0(t)dt=2 3, u2(x) =
1
0
tu1(t)dt= 1
3, u3(x) = 1
0
tu2(t)dt=1 6, u4(x) =
1
0 tu3(t)dt= 1 12,
(4.38)
and so on. Using (4.13) gives the series solution u(x) =x+ex−4
3 +2 3
1 + 1
2+1 4 +1
8+· · ·
. (4.39)
Notice that the infinite geometric series at the right side hasa1= 1, and the ratior= 12. The sum of the infinite series is therefore given by
S= 1
1−12 = 2. (4.40)
The series solution (4.39) converges to the closed form solution
u(x) =x+ex. (4.41)
Example 4.4
Solve the following Fredholm integral equation u(x) = 2 + cosx+
π 0
tu(t)dt. (4.42)
Proceeding as before we find ∞ n=0
un(x) = 2 + cosx+ π
0 t ∞ n=0
un(t)dt, (4.43) or equivalently
u0(x) +u1(x) +u2(x) +· · ·= 2 + cosx+
π 0
t[u0(t) +u1(t) +· · ·]dt. (4.44) We next set the following recurrence relation
u0(x) = 2 + cosx, uk+1(x) = π
0
tuk(t)dt, k0. (4.45) This in turn gives
126 4 Fredholm Integral Equations u0(x) = 2 + cosx,
u1(x) = π
0 u0(t)dt=−2 +π2, u2(x) =
π 0
u1(t)dt=−π2+1 2π4, u3(x) =
π
0 u2(t)dt=−1 2π4+1
4π6, u4(x) =
π 0
u3(t)dt=−1 4π6+1
8π8,
(4.46)
and so on. Using (4.13) gives the series solution u(x) = 2 + cosx+ (−2 +π2) +
−π2+1 2π4
+
−1 2π4+1
4π6
+
−1 4π6+1
8π8
+· · · . (4.47) We can easily observe the appearance of the noise terms, i.e the identical terms with opposite signs. Canceling these noise terms in (4.47) gives the exact solution
u(x) = cosx. (4.48)
Example 4.5
Solve the following Fredholm integral equation u(x) = 1 +1
2 π4
0
sec2xu(t)dt. (4.49) Substituting the decomposition series (4.13) into both sides of (4.49) gives
∞ n=0
un(x) = 1 +1 2sec2x
π4
0
∞ n=0
un(t)dt, (4.50) or equivalently
u0(x) +u1(x) +u2(x) +· · ·= 1 +1 2sec2x
π4
0 [u0(t) +u1(t) +· · ·]dt. (4.51) Proceeding as before, we set the recurrence relation
u0(x) = 1, uk+1(x) =1 2sec2x
π4
0
uk(t)dt, k0. (4.52) Consequently, we obtain
u0(x) = 1, u1(x) = 1
2sec2x π4
0 u0(t)dt= π 8sec2x,
4.2 Fredholm Integral Equations of the Second Kind 127 u2(x) = 1
2sec2x π
4
0
u1(t)dt= π 16sec2x, u3(x) = 1
2sec2x π4
0
u2(t)dt= π 32sec2x, u4(x) = 1
2sec2x π
4
0 u3(t)dt= π 64sec2x,
(4.53)
and so on. Using (4.13) gives the series solution u(x) = 1 + π
8 sec2x
1 + 1 2+1
4 +1 8 +· · ·
. (4.54)
The sum of the infinite series at the right side is S = 2. The series solution (4.54) converges to the closed form solution
u(x) = 1 + π
4 sec2x. (4.55)
Example 4.6
Solve the following Fredholm integral equation u(x) =πx+ sin 2x+x
π
−π
tu(t)dt. (4.56)
Proceeding as before we find ∞
n=0
un(x) =πx+ sin 2x+x π
−π
t ∞ n=0
un(t)dt. (4.57) To determine the components ofu(x), we use the recurrence relation
u0(x) =πx+ sin 2x, uk+1(x) =x π
−π
tuk(t)dt, k0. (4.58) This in turn gives
u0(x) =πx+ sin 2x, u1(x) =x
π 0
u0(t)dt=−πx+2 3π4x, u2(x) =x
π
0 u1(t)dt=−2
3π4x+4 9π7x, u3(x) =x
π
0 u2(t)dt=−4
9π7x+ 8 27π10x,
(4.59)
and so on. Using (4.13) gives the series solution u(x) =πx+ sin 2x+
−π+2 3π4
x+
−2 3π4+4
9π7
x +
−4 9π7+ 8
27π10
x+· · ·. (4.60)
Canceling the noise terms in (4.60) gives the exact solution
u(x) = sin 2x. (4.61)
128 4 Fredholm Integral Equations Exercises 4.2.1
In Exercises 1–20, solve the following Fredholm integral equations by using theAdo- mian decomposition method
1. u(x) =ex+ 1−e+ 1
0 u(t)dt 2. u(x) =ex+e−1 1
0 u(t)dt 3. u(x) = cosx+ 2x+
π
0 xtu(t)dt 4. u(x) = sinx−x+ π
2
0 xtu(t)dt 5. u(x) =ex+2−2
1
0 ex+tu(t)dt 6. u(x) =ex+ ex+1−1 x+ 1 − 1
0 extu(t)dt 7. u(x) =x+ (1−x)ex+
1
0 x2et(x−1)u(t)dt 8. u(x) = 1 +1
2sin2x π
2 0 u(t)dt 9. u(x) =xex−1
2+1 2
1
0 u(t)dt 10. u(x) =xsinx− 1 2+ 1
2 π
2 0 u(t)dt 11. u(x) =xcosx+ 1 +1
2 π
0 u(t)dt 12. u(x) = sinx+ π
2 0
sinxcostu(t)dt 13. u(x) =x+ sinx− π
2
0 xu(t)dt 14. u(x) = 1− 1 15x2+
1
−1
(xt+x2t2)u(t)dt 15. u(x) = 1−19
15x2+ 1
−1(xt+x2t2)u(t)dt 16. u(x) =−x+ sinx+
π
2
0 (1 +x−t)u(t)dt 17. u(x) =e−x+e−(x+1)−1
x+ 1 + 1
0 e−xtu(t)dt 18. u(x) =3
2ex− 1 2ex+2+
1
0 ex+tu(t)dt 19. u(x) = 1
2cosx+ π
2 0
cosxsintu(t)dt20. u(x) =π
4−sec2x− π
4 0 u(t)dt