Volterra Integral Equations
3.2 Volterra Integral Equations of the Second Kind
3.2.1 The Adomian Decomposition Method
The Adomian decomposition method (ADM) was introduced and developed by George Adomian in [5–7] and is well addressed in many references. A con- siderable amount of research work has been invested recently in applying this method to a wide class of linear and nonlinear ordinary differential equations, partial differential equations and integral equations as well.
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), (3.4)
or equivalently
u(x) =u0(x) +u1(x) +u2(x) +· · ·, (3.5) where the componentsun(x), n0 are to be determined in a recursive man- ner. The decomposition method concerns itself with finding the components
3.2 Volterra Integral Equations of the Second Kind 67 u0, u1, u2, . . .individually. As will be seen through the text, 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 (3.4) into the Volterra integral equation (3.3) to obtain
∞ n=0
un(x) =f(x) +λ x
0 K(x, t) ∞
n=0
un(t)
dt, (3.6)
or equivalently
u0(x) +u1(x) +u2(x) +· · ·=f(x) +λ x
0
K(x, t) [u0(t) +u1(t) +· · ·]dt.
(3.7) The zeroth component u0(x) is identified by all terms that are not included under the integral sign. Consequently, the components uj(x), j 1 of the unknown functionu(x) are completely determined by setting the recurrence relation:
u0(x) =f(x), un+1(x) =λ
x 0
K(x, t)un(t)dt, n0, (3.8) that is equivalent to
u0(x) =f(x), u1(x) =λ x
0 K(x, t)u0(t)dt, u2(x) =λ
x 0
K(x, t)u1(t)dt, u3(x) =λ x
0
K(x, t)u2(t)dt,
(3.9)
and so on for other components.
In view of (3.9), the components u0(x), u1(x), u2(x), u3(x), . . . are com- pletely determined. As a result, the solution u(x) of the Volterra integral equation (3.3) in a series form is readily obtained by using the series assump- tion in (3.4).
It is clearly seen that the decomposition method converted the integral equation into an elegant determination of computable components. It was formally shown by many researchers that if an exact solution exists for the problem, then the obtained series converges very rapidly to that solution. The convergence concept of the decomposition series was thoroughly investigated by many researchers to confirm the rapid convergence of the resulting series.
However, for concrete problems, where a closed form solution is not obtain- able, a truncated number of terms is usually used for numerical purposes.
The more components we use the higher accuracy we obtain.
Example 3.1
Solve the following Volterra integral equation:
u(x) = 1− x
0
u(t)dt. (3.10)
68 3 Volterra Integral Equations We notice thatf(x) = 1, λ=−1, K(x, t) = 1. Recall that the solutionu(x) is assumed to have a series form given in (3.4). Substituting the decomposition series (3.4) into both sides of (3.10) gives
∞ n=0
un(x) = 1− x
0
∞ n=0
un(t)dt, (3.11)
or equivalently
u0(x) +u1(x) +u2(x) +· · ·= 1− x
0 [u0(t) +u1(t) +u2(t) +· · ·]dt. (3.12) 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) = 1, uk+1(x) =−
x
0 uk(t)dt, k0, (3.13)
so that
u0(x) = 1, u1(x) =−
x
0 u0(t)dt=− x
0 1dt=−x, u2(x) =−
x 0
u1(t)dt=− x
0
(−t)dt= 1 2!x2, u3(x) =−
x 0
u2(t)dt=− x
0
1
2!t2dt=−1 3!x3, u4(x) =−
x 0
u3(t)dt=− x
0 −1
3!t3dt=−1 4!x4,
(3.14)
and so on. Using (3.4) gives the series solution:
u(x) = 1−x+ 1 2!x2− 1
3!x3+ 1
4!x4+· · · , (3.15) that converges to the closed form solution:
u(x) =e−x. (3.16)
Example 3.2
Solve the following Volterra integral equation:
u(x) = 1 + x
0
(t−x)u(t)dt. (3.17)
We notice thatf(x) = 1, λ= 1, K(x, t) =t−x. Substituting the decompo- sition series (3.4) into both sides of (3.17) gives
∞ n=0
un(x) = 1 + x
0
∞ n=0
(t−x)un(t)dt, (3.18) or equivalently
3.2 Volterra Integral Equations of the Second Kind 69 u0(x) +u1(x) +u2(x) +· · ·= 1 +
x 0
(t−x) [u0(t) +u1(t) +u2(t) +· · ·]dt.
(3.19) Proceeding as before we set the following recurrence relation:
u0(x) = 1, uk+1(x) =
x
0 (t−x)uk(t)dt, k0, (3.20) that gives
u0(x) = 1, u1(x) =
x 0
(t−x)u0(t)dt= x
0
(t−x)dt=−1 2!x2, u2(x) =
x
0 (t−x)u1(t)dt=−1 2!
x
0 (t−x)t2dt= 1 4!x4, u3(x) =
x 0
(t−x)u2(t)dt= 1 4!
x 0
(t−x)t4dt=−1 6!x6, u4(x) =
x
0 (t−x)u3(t)dt=−1 6!
x
0 (t−x)t6dt= 1 8!x8,
(3.21)
and so on. The solution in a series form is given by u(x) = 1− 1
2!x2+ 1 4!x4− 1
6!x6+ 1
8!x8+· · · , (3.22) and in a closed form by
u(x) = cosx, (3.23)
obtained upon using the Taylor expansion for cosx.
Example 3.3
Solve the following Volterra integral equation:
u(x) = 1−x−1 2x2−
x
0 (t−x)u(t)dt. (3.24) Notice that f(x) = 1−x− 12x2, λ =−1, K(x, t) = t−x. Substituting the decomposition series (3.4) into both sides of (3.24) gives
∞ n=0
un(x) = 1−x−1 2x2−
x 0
∞ n=0
(t−x)un(t)dt, (3.25) or equivalently
u0(x) +u1(x) +u2(x) +· · ·= 1−x−1 2x2−
x
0 (t−x) [u0(t) +u1(t) +· · ·]dt.
(3.26) This allows us to set the following recurrence relation:
u0(x) = 1−x−1 2x2, uk+1(x) =−
x
0 (t−x)uk(t)dt, k0,
(3.27)
70 3 Volterra Integral Equations that gives
u0(x) = 1−x−1 2x2, u1(x) =−
x 0
(t−x)u0(t)dt= 1 2!x2− 1
3!x3− 1 4!x4, u2(x) =−
x
0 (t−x)u1(t)dt= 1 4!x4− 1
5!x5− 1 6!x6, u3(x) =−
x 0
(t−x)u2(t)dt= 1 6!x6− 1
7!x7− 1 8!x8,
(3.28)
and so on. The solution in a series form is given by u(x) = 1−(x+ 1
3!x3+ 1 5!x5+ 1
7!x7+· · ·), (3.29) and in a closed form by
u(x) = 1−sinhx, (3.30)
obtained upon using the Taylor expansion for sinhx.
Example 3.4
We consider here the Volterra integral equation:
u(x) = 5x3−x5+ x
0
tu(t)dt. (3.31)
Identifying the zeroth component u0(x) by the first two terms that are not included under the integral sign, and using the ADM we set the recurrence relation as
u0(x) = 5x3−x5, uk+1(x) =
x
0 tuk(t)dt, k0. (3.32) This in turn gives
u0(x) = 5x3−x5, u1(x) =
x 0
tu0(t)dt=x5−1 7x7, u2(x) =
x
0 tu1(t)dt= 1 7x7− 1
63x9, u3(x) =
x 0
tu1(t)dt= 1
63x9− 1 693x11,
(3.33)
The solution in a series form is given by u(x) = 5x3−x5
+
x5−1 7x7
+
1 7x7− 1
63x9
+ (1
63x9− 1
693x11) +· · · . (3.34) We can easily notice the appearance of identical terms with opposite signs.
Such terms are called noise terms that will be discussed later. Canceling the identical terms with opposite signs gives the exact solution
3.2 Volterra Integral Equations of the Second Kind 71
u(x) = 5x3, (3.35)
that satisfies the Volterra integral equation (3.31).
Example 3.5
We now consider the Volterra integral equation:
u(x) =x+x4+1 2x2+1
5x5− x
0
u(t)dt. (3.36) Identifying the zeroth componentu0(x) by the first four terms that are not included under the integral sign, and using the ADM we set the recurrence relation as
u0(x) =x+x4+1 2x2+1
5x5, uk+1(x) =−
x
0 uk(t)dt, k0.
(3.37)
This in turn gives
u0(x) =x+x4+1 2x2+1
5x5, u1(x) =−
x 0
u0(t)dt=−1 2x2−1
5x5−1 6x3− 1
30x6, u2(x) =−
x
0 u1(t)dt= 1 6x3+ 1
30x6+ 1
24x4+ 1 210x7, u3(x) =−
x 0
u2(t)dt=−1
24x4− 1
210x7+· · · .
(3.38)
The solution in a series form is given by u(x) =
x+x4+1 2x2+1
5x5
− 1
2x2+1 5x5+1
6x3+ 1 30x6
+ 1
6x3+ 1
30x6+ 1
24x4+ 1 210x7
− 1
24x4+ 1
210x7+· · ·
+· · · . (3.39) We can easily notice the appearance of identical terms with opposite signs.
This phenomenon of such terms is callednoise termsphenomenon that will be presented later. Canceling the identical terms with opposite terms gives the exact solution
u(x) =x+x4. (3.40)
Example 3.6
We finally solve the Volterra integral equation:
u(x) = 2 +1 3
x 0
xt3u(t)dt. (3.41)
Proceeding as before we set the recurrence relation
72 3 Volterra Integral Equations u0(x) = 2, uk+1(x) =1
3 x
0
xt3uk(t)dt, k0. (3.42) This in turn gives
u0(x) = 2, u1(x) = 1 3
x
0 xt3u0(t)dt=1 6x5, u2(x) = 1
3 x
0
xt3u1(t)dt= 1 162x10, u3(x) = 1
3 x
0 xt3u2(t)dt= 1 6804x15, u4(x) = 1
3 x
0
xt3u3(t)dt= 1 387828x20,
(3.43)
and so on. The solution in a series form is given by u(x) = 2 +1
6x5+ 1
6·33x10+ 1
6·34·14x15+ 1
6·35·14·19x20+· · ·. (3.44) It seems that an exact solution is not obtainable. The obtained series solution can be used for numerical purposes. The more components that we determine the higher accuracy level that we can achieve.
Exercises 3.2.1
In Exercises 1–26, solve the following Volterra integral equations by using theAdo- mian decomposition method:
1.u(x) = 6x−3x2+ x
0 u(t)dt 2.u(x) = 6x−x3+ x
0
(x−t)u(t)dt 3.u(x) = 1−1
2x2+ x
0 u(t)dt 4.u(x) =x−2 3x3−2
x
0 u(t)dt 5.u(x) = 1 +x+
x
0 (x−t)u(t)dt 6.u(x) = 1−x+ x
0 (x−t)u(t)dt 7.u(x) = 1 +x−x
0
(x−t)u(t)dt 8.u(x) = 1−x−x
0
(x−t)u(t)dt 9.u(x) = 1− x
0 (x−t)u(t)dt 10.u(x) = 1 + x
0 (x−t)u(t)dt 11.u(x) =x−
x
0
(x−t)u(t)dt 12.u(x) =x+ x
0
(x−t)u(t)dt 13.u(x) = 1 +
x
0 u(t)dt 14.u(x) = 1− x
0 u(t)dt 15.u(x) = 1 + 2
x
0 tu(t)dt 16.u(x) = 1−2 x
0 tu(t)dt 17.u(x) = 1−x2− x
0
(x−t)u(t)dt 18.u(x) =−2 + 3x−x2−x
0
(x−t)u(t)dt
3.2 Volterra Integral Equations of the Second Kind 73 19.u(x) =x2+
x
0
(x−t)u(t)dt 20.u(x) =−2 + 2x+x2+ x
0
(x−t)u(t)dt 21.u(x) = 1 + 2x+ 4
x
0 (x−t)u(t)dt 22.u(x) = 5 + 2x2− x
0 (x−t)u(t)dt 23.u(x) = 1 +x+1
2x2+1 2
x
0
(x−t)2u(t)dt 24.u(x) = 1−1
2x2+1 6
x
0
(x−t)3u(t)dt 25.u(x) = 1 +1
2x+1 2
x
0 (x−t+ 1)u(t)dt 26.u(x) = 1 +x2− x
0
(x−t+ 1)2u(t)dt
In Exercises 27–30, use theAdomian decomposition methodto find the series solution 27.u(x) = 3 +1
4 x
0 xt2u(t)dt 28.u(x) = 3 +1 4
x
0 (x+t2)u(t)dt 29.u(x) = 1 +1
2 x
0
(x2−t2)u(t)dt 30.u(x) = 1 +1 2
x
0 x2u(t)dt