Volterra Integral Equations
3.2 Volterra Integral Equations of the Second Kind
3.2.4 The Variational Iteration Method
82 3 Volterra Integral Equations
3.2 Volterra Integral Equations of the Second Kind 83 Integration by parts is usually used for the determination of the Lagrange multiplier λ(ξ). In other words we can use
x 0
λ(ξ)un(ξ)dξ =λ(ξ)un(ξ)− x
0
λ(ξ)un(ξ)dξ, x
0 λ(ξ)un(ξ)dξ =λ(ξ)un(ξ)−λ(ξ)un(ξ) + x
0 λ(ξ)un(ξ)dξ, x
0
λ(ξ)un(ξ)dξ =λ(ξ)un(ξ)−λ(ξ)un(ξ) +λ(ξ)un(ξ)
− x
0 λ(ξ)un(ξ)dξ, x
0
λ(ξ)u(iv)n (ξ)dξ =λ(ξ)un(ξ)−λ(ξ)un(ξ) +λ(ξ)un(ξ)−λun(ξ) +
x
0 λ(iv)(ξ)un(ξ)dξ,
(3.90)
and so on. These identities are obtained by integrating by parts.
For example, ifLun(ξ) =un(ξ) in (3.89), then (3.89) becomes δun+1=δun+δ
x 0
λ(ξ) (Lun(ξ))dξ
. (3.91)
Integrating the integral of (3.91) by parts using (3.90) we obtain δun+1=δun+δλ(ξ)un(ξ)−
x 0
λ(ξ)δun(ξ)dξ, (3.92) or equivalently
δun+1=δun(ξ)(1 +λ|ξ=x)− x
0
λδundξ. (3.93) The extremum condition of un+1 requires thatδun+1= 0. This means that the left hand side of (3.93) is zero, and as a result the right hand side should be 0 as well. This yields the stationary conditions:
1 +λ|ξ=x= 0, λ|ξ=x= 0. (3.94) This in turn gives
λ=−1. (3.95)
As a second example, ifLun(ξ) =un(ξ) in (3.89), then (3.89) becomes δun+1=δun+δ
x 0
λ(ξ) (Lun(ξ))dξ
. (3.96)
Integrating the integral of (3.96) by parts using (3.90) we obtain δun+1=δun+δλ((un))x0−(λδun)x0+
x 0
λδundξ, (3.97) or equivalently
δun+1=δun(ξ)(1−λ|ξ=x) +δλ((un)|ξ=x) + x
0
λδundξ, (3.98)
84 3 Volterra Integral Equations The extremum condition of un+1 requires that δun+1 = 0. This means that the left hand side of (3.98) is zero, and as a result the right hand side should be 0 as well. This yields the stationary conditions:
1−λ|ξ=x= 0, λ|ξ=x= 0, λ|ξ=x= 0. (3.99) This in turn gives
λ=ξ−x. (3.100)
Having determined the Lagrange multiplier λ(ξ), the successive approxi- mationsun+1, n0, of the solutionu(x) will be readily obtained upon using selective function u0(x). However, for fast convergence, the function u0(x) should be selected by using the initial conditions as follows:
u0(x) =u(0), for first orderun,
u0(x) =u(0) +xu(0), for second orderun, u0(x) =u(0) +xu(0) + 1
2!x2u(0), for third orderun, ...
(3.101)
and so on. Consequently, the solution u(x) = lim
n→∞un(x). (3.102)
In other words, the correction functional (3.87) will give several approxima- tions, and therefore the exact solution is obtained as the limit of the resulting successive approximations.
The determination of the Lagrange multiplier plays a major role in the determination of the solution of the problem. In what follows, we summarize some iteration formulae that show ODE, its corresponding Lagrange multi- pliers, and its correction functional respectively:
(i)
⎧⎪
⎨
⎪⎩
u+f(u(ξ), u(ξ)) = 0, λ=−1, un+1=un−
x
0 [un+f(un, un)]dξ, (ii)
⎧⎪
⎨
⎪⎩
u+f(u(ξ), u(ξ), u(ξ)) = 0, λ= (ξ−x), un+1=un+
x
0 (ξ−x) [un+f(un, un, un)]dξ,
(iii)
⎧⎪
⎪⎨
⎪⎪
⎩
u+f(u(ξ), u(ξ), u(ξ), u(ξ)) = 0, λ=−1
2!(ξ−x)2, un+1=un− x
0
1
2!(ξ−x)2[un +f(un, . . . , un)]dξ,
(iv)
⎧⎪
⎪⎨
⎪⎪
⎩
u(iv)+f(u(ξ), u(ξ), u(ξ), u(ξ), u(iv)(ξ)) = 0, λ= 1
3!(ξ−x)3, un+1=un+
x
0
1
3!(ξ−x)3
un +f(un, un, . . . , u(iv)n ) dξ,
3.2 Volterra Integral Equations of the Second Kind 85 and generally
(v)
⎧⎪
⎪⎨
⎪⎪
⎩
u(n)+f(u(ξ), u(ξ), . . . , u(n)(ξ)) = 0, λ= (−1)n 1
(n−1)!(ξ−x)(n−1), un+1=un+ (−1)n
x
0
1
(n−1)!(ξ−x)(n−1)
un +f(un, . . . , u(n)n ) dξ, forn1.
To use the variational iteration method for solving Volterra integral equa- tions, it is necessary to convert the integral equation to an equivalent initial value problem or to an equivalent integro-differential equation. As defined before, integro-differential equation is an equation that contains differential and integral operators in the same equation. The integro-differential equa- tions will be studied in details in Chapter 5. The conversion process is pre- sented in Section 2.5.1. However, for comparison reasons, we will examine the obtained initial value problem by two methods, namely, standard methods used for solving ODEs, and by using the variational iteration method as will be seen by the following examples.
Example 3.15
Solve the Volterra integral equation by using the variational iteration method u(x) = 1 +
x
0 u(t)dt. (3.103)
Using Leibnitz rule to differentiate both sides of (3.103) gives
u(x)−u(x) = 0. (3.104)
Substituting x= 0 into (3.103) gives the initial conditionu(0) = 1.
Using the variational iteration method
The correction functional for equation (3.104) is un+1(x) =un(x) +
x
0 λ(ξ)(un(ξ)−u˜n(ξ))dξ. (3.105) Using the formula (i) given above leads to
λ=−1. (3.106)
Substituting this value of the Lagrange multiplierλ=−1 into the functional (3.105) gives the iteration formula:
un+1(x) =un(x)− x
0
(un(ξ)−un(ξ))dξ. (3.107) As stated before, we can use the initial condition to selectu0(x) =u(0) = 1.
Using this selection into (3.105) gives the following successive approximations:
u0(x) = 1, u1(x) = 1−
x
0 (u0(ξ)−u0(ξ))dξ= 1 +x, u2(x) = 1 +x−
x 0
(u1(ξ)−u1(ξ))dξ= 1 +x+ 1 2!x2,
86 3 Volterra Integral Equations u3(x) = 1 +x+ 1
2!x2− x
0
(u2(ξ)−u2(ξ))dξ= 1 +x+ 1 2!x2+ 1
3!x3, (3.108) and so on. The VIM admits the use of
u(x) = limn→∞un(x)
= limn→∞(1 +x+ 1 2!x2+ 1
3!x3+ 1
4!x4+· · ·+ 1
n!xn), (3.109) that gives the exact solution by
u(x) =ex. (3.110)
Using ODEs method
The ODE (3.104) is of first order, therefore the integrating factorμ(x) is given by
μ(x) =ex(−1)dx=e−x. (3.111) For first order ODE, we use the formula:
u(x) = 1 μ
x
μq(x)dx+C
=Cex. (3.112)
To obtain the particular solution, we use the initial condition u(0) = 1 to find that C= 1. This gives the particular solution:
u(x) =ex. (3.113)
Example 3.16
Solve the Volterra integral equation by using the variational iteration method u(x) =x+
x
0 (x−t)u(t)dt. (3.114) Using Leibnitz rule to differentiate both sides of (3.114) once with respect to xgives the integro-differential equation:
u(x) = 1 + x
0
u(t)dt, u(0) = 0. (3.115) However, by differentiating (3.115) with respect toxwe obtain the differential equation:
u(x) =u(x). (3.116)
Substitutingx= 0 into (3.114) and (3.115) gives the initial conditionsu(0) = 0 andu(0) = 1. The resulting initial value problem, that consists of a second order ODE and initial conditions is given by
u(x)−u(x) = 0, u(0) = 0, u(0) = 1. (3.117) The integro-differential equation (3.115) and the initial value problem (3.116) will be handled independently by using the variational iteration method.
Using the variational iteration method
(i) We first start using the variational iteration method to handle the integro- differential equation (3.115) given by
3.2 Volterra Integral Equations of the Second Kind 87 u(x) = 1 +
x 0
u(t)dt, u(0) = 0. (3.118) The correction functional for Eq. (3.118) is
un+1(x) =un(x) + x
0
λ(ξ)
un(ξ)−1− ξ
0
˜ un(r)dr
dξ. (3.119) Using the formula (i) forλwe find that
λ=−1. (3.120)
Substituting this value of the Lagrange multiplierλ=−1 into the functional (3.119) gives the iteration formula:
un+1(x) =un(x)− x
0
un(ξ)−1− ξ
0 un(r)dr
dξ. (3.121)
We can use the initial conditions to select u0(x) = u(0) = 0. Using this selection into (3.121) gives the following successive approximations:
u0(x) = 0, u1(x) =x−
x 0
u0(ξ)−1− ξ
0
u0(r)dr
dξ=x, u2(x) =x−
x 0
u1(ξ)−1− ξ
0
u1(r)dr
dξ=x+ 1 3!x3, u3(x) =x−
x 0
u2(ξ)−1− ξ
0 u2(r)dr
dξ=x+ 1 3!x3+ 1
5!x5, ...
un(x) =x+ 1 3!x3+ 1
5!x5+ 1
7!x7+· · ·+ 1
(2n+ 1)!x2n+1.
(3.122)
The VIM admits the use of
u(x) = lim
n→∞un(x), (3.123)
that gives the exact solution by
u(x) = sinhx. (3.124)
(ii) We can obtain the same result by applying the variational iteration method to handle the initial value problem (3.117) given by
u(x)−u(x) = 0, u(0) = 0, u(0) = 1. (3.125) The correction functional for Eq. (3.117) is
un+1(x) =un(x) + x
0
λ(ξ) (un(ξ)−u˜n(ξ))dξ. (3.126) Using the formula (ii) given above leads to
λ=ξ−x. (3.127)
88 3 Volterra Integral Equations Substituting this value of the Lagrange multiplierλ=ξ−xinto the functional (3.126) gives the iteration formula:
un+1(x) =un(x) + x
0
(ξ−x) (un(ξ)−un(ξ))dξ. (3.128) We can use the initial conditions to selectu0(x) =u(0) +xu(0) =x. Using this selection into (3.128) gives the following successive approximations
u0(x) =x, u1(x) =x+
x
0 (ξ−x) (u0(ξ)−u0(ξ))dξ=x+ 1 3!x3, u2(x) =x+ 1
3!x3+ x
0
(ξ−x) (u1(ξ)−u1(ξ))dξ=x+ 1 3!x3+ 1
5!x5, u3(x) =x+ 1
3!x3+ 1 5!x5+
x
0 (ξ−x) (u2(ξ)−u2(ξ))dξ
=x+ 1 3!x3+ 1
5!x5+ 1 7!x7, ...
un(x) =x+ 1 3!x3+ 1
5!x5+ 1
7!x7+· · ·+ 1
(2n+ 1)!x2n+1.
(3.129) The VIM admits the use of
u(x) = lim
n→∞un(x), (3.130)
that gives the exact solution by
u(x) = sinhx. (3.131)
Standard methods for solving ODEs
The initial value problem (3.125) is of second order, therefore the auxiliary equation is of the form
r2−1 = 0, (3.132)
that givesr=±1. This in turn gives the general solution by
u(x) =Asinhx+Bcoshx. (3.133) To obtain the particular solution, we use the initial conditions u(0) = 0, u(0) = 1 to find that the particular solution is given by
u(x) = sinhx. (3.134)
Example 3.17
Solve the Volterra integral equation by using the variational iteration method u(x) = 1 +x+ 1
3!x3− x
0
(x−t)u(t)dt. (3.135) Using Leibnitz rule to differentiate both sides of (3.135) once with respect to xgives the integro-differential equation:
3.2 Volterra Integral Equations of the Second Kind 89 u(x) = 1 + 1
2!x2− x
0
u(t)dt, u(0) = 1, (3.136) and by differentiating again we obtain the initial value problem
u(x) +u(x) =x, u(0) = 1, u(0) = 1. (3.137) Using the variational iteration method
(i) We first start using the variational iteration method to handle the integro- differential equation (3.136) given by
u(x) = 1 + 1 2!x2−
x
0 u(t)dt, u(0) = 1, (3.138) The correction functional for Eq. (3.138) is
un+1(x) =un(x)− x
0
λ(ξ)
un(ξ)−1−1 2ξ2+
ξ 0
˜ un(r)dr
dξ. (3.139) Using the formula (i) forλwe find that
λ=−1. (3.140)
Substituting this value of the Lagrange multiplierλ=−1 into the functional (3.139) gives the iteration formula
un+1(x) =un(x)− x
0
un(ξ)−1−1 2ξ2+
ξ
0 un(r)dr
dξ. (3.141) We can use the initial conditions to select u0(x) = u(0) = 1. Using this selection into (3.141) gives the following successive approximations
u0(x) = 1, u1(x) =x−
x 0
u0(ξ)−1−1 2ξ2+
ξ
0 u0(r)dr
dξ
= 1 +x− 1 2!x2+ 1
3!x3, u2(x) =x−
x 0
u1(ξ)−1−1 2ξ2+
ξ 0
u1(r)dr
dξ
= 1 +x− 1 2!x2+ 1
4!x4− 1 5!x5, u3(x) =x−
x 0
u2(ξ)−1−1 2ξ2+
ξ 0
u2(r)dr
dξ
= 1 +x− 1 2!x2+ 1
4!x4− 1 6!x6, ...
un(x) =x+
1− 1 2!x2+ 1
4!x4− 1
6!x6+· · ·+(−1)n (2n)!x2n
.
(3.142)
The VIM admits the use of
90 3 Volterra Integral Equations u(x) = lim
n→∞un(x), (3.143)
that gives the exact solution by
u(x) =x+ cosx. (3.144)
(ii) We next use the variational iteration method for solving the initial value problem
u(x) +u(x) =x, u(0) = 1, u(0) = 1. (3.145) The correction functional for Eq. (3.145) is
un+1(x) =un(x) + x
0
λ(ξ) (un(ξ) + ˜un(ξ)−ξ)dξ. (3.146) Using the formula (ii) given above leads to
λ=ξ−x. (3.147)
Substituting this value of the Lagrange multiplierλ=ξ−xinto the functional (3.146) gives the iteration formula
un+1(x) =un(x) + x
0 (ξ−x) (un(ξ) +un(ξ)−ξ)dξ. (3.148) We can use the initial conditions to select u0(x) =u(0) +xu(0) = 1 +x.
Using this selection into (3.148) gives the following successive approximations u0(x) = 1 +x,
u1(x) = 1 +x+ x
0 (ξ−x) (u0(ξ) +u0(ξ)−ξ)dξ
= 1 +x− 1 2!x2, u2(x) = 1 +x− 1
2!x2+ x
0
(ξ−x) (u1(ξ) +u1(ξ)−ξ)dξ
= 1 +x− 1 2!x2+ 1
4!x4, u3(x) = 1 +x− 1
2!x2+ 1 4!x4+
x 0
(ξ−x) (u2(ξ) +u2(ξ)−ξ)dξ
= 1 +x− 1 2!x2+ 1
4!x4− 1 6!x6 ...
un(x) =x+
1− 1 2!x2+ 1
4!x4− 1 6!x6+ 1
8!x8− · · ·+(−1)n (2n)!x2n
.
(3.149)
Proceeding as before, the VIM gives the exact solution by
u(x) =x+ cosx. (3.150)
Standard methods for solving ODEs
The ODE (3.169) is of second order and nonhomogeneous. The auxiliary equation for the homogeneous part is of the form
3.2 Volterra Integral Equations of the Second Kind 91
r2+ 1 = 0, (3.151)
that givesr=±i, i2=−1. The general solution is given by u(x) =uc+up,
u(x) =Acosx+Bsinx+a+bx, (3.152) whereucis the complementary solution, andupis a particular solution. Using ODE methods and initial conditions we find thatB=a= 0, andA=b= 1.
The particular solution is given by
u(x) =x+ cosx. (3.153)
Example 3.18
Solve the Volterra integral equation by using the variational iteration method u(x) = 1 +x+1
2x2+1 2
x
0 (x−t)2u(t)dt. (3.154) Using Leibnitz rule to differentiate both sides of (3.154) three times with respect to xgives the two integro-differential equations
u(x) = 1 +x+ x
0
(x−t)u(t)dt, u(0) = 1 u(x) = 1 +
x
0 u(t)dt, u(0) = 1, u(0) = 1.
(3.155)
and the third order initial value problem
u(x) =u(x), u(0) =u(0) =u(0) = 1. (3.156) Using the variational iteration method VIM
(i) We first note that we obtained two equivalent integro-differential equations (3.155). We will apply the VIM to these two equations. We first start using the VIM to handle the integro-differential equation
u(x) = 1 +x+ x
0
(x−t)u(t)dt, u(0) = 1. (3.157) The correction functional for Eq. (3.157) is
un+1(x) =un(x) + x
0
λ(ξ)
un(ξ)−1−ξ− ξ
0
(ξ−r)˜un(r)dr
dξ.
(3.158) Proceeding as before we find
λ=−1, (3.159)
that gives the iteration formula un+1(x) =un(x)−
x 0
un(ξ)−1−ξ− ξ
0
(ξ−r)un(r)dr
dξ. (3.160) We can use the initial conditions to select u0(x) = u(0) = 1. Using this selection into (3.160) gives the following successive approximations
92 3 Volterra Integral Equations u0(x) = 1,
u1(x) = 1− x
0
u0(ξ)−1−ξ− ξ
0 (ξ−r)u0(r)dr
dξ
= 1 +x+ 1 2!x2+ 1
3!x3, u2(x) = 1−
x 0
u1(ξ)−1−ξ− ξ
0
(ξ−r)u1(r)dr
dξ
= 1 +x+ 1 2!x2+ 1
3!x3+ 1 4!x4+ 1
5!x5+ 1 6!x6, ...
un(x) = 1 +x+ 1 2!x2+ 1
3!x3+ 1 4!x4+ 1
5!x5+ 1
6!x6+· · ·+ 1 n!xn.
(3.161)
This in turn gives the exact solution by
u(x) =ex. (3.162)
(ii) We next consider the integro-differential equation u(x) = 1 +
x
0 u(t)dt, u(0) = 1, u(0) = 1. (3.163) The correction functional for Eq. (3.163) is
un+1(x) =un(x) + x
0
λ(ξ)
un(ξ)−1− ξ
0
˜ un(r)dr
dξ. (3.164) Notice that the integro-differential equation is of second order. Therefore, we can show that
λ=ξ−x, (3.165)
that gives the iteration formula un+1(x) =un(x) +
x 0
(ξ−x)(un(ξ)−1− ξ
0
(ξ−r)un(r)dr)
dξ.
(3.166) We can use the initial conditions to selectu0(x) = 1 +x. Using this selection into (3.166) gives the following successive approximations
u0(x) = 1 +x, u1(x) = 1 +x+ 1
2!x2+ 1 3!x3+ 1
4!x4, u2(x) = 1 +x+ 1
2!x2+ 1 3!x3+ 1
4!x4+ 1 5!x5+ 1
6!x6+ 1 7!x7, ...
un(x) = 1 +x+ 1 2!x2+ 1
3!x3+ 1 4!x4+ 1
5!x5+ 1
6!x6+· · ·+ 1 n!xn.
(3.167)
This in turn gives the exact solution by
3.2 Volterra Integral Equations of the Second Kind 93
u(x) =ex. (3.168)
(iii) We next use the variational iteration method for solving the initial value problem
u(x)−u(x) = 0, u(0) =u(0) =u(0) = 1. (3.169) un+1(x) =un(x) +
x
0 λ(ξ) (un(ξ)−u˜n(ξ))dξ. (3.170) Using the formula (iii) given above forλleads to
λ=−1
2!(ξ−x)2. (3.171)
Substituting this value of the Lagrange multiplier into the functional (3.170) gives the iteration formula
un+1(x) =un(x)− 1 2!
x 0
(ξ−x)2(un(ξ)−un(ξ))dξ. (3.172) As stated before, we can use the initial conditions to select
u0(x) =u(0) +xu(0) +1
2u(0) = 1 +x+ 1
2!x2. (3.173) Using this selection into (3.172) gives the following successive approximations
u0(x) = 1 +x+ 1 2!x2, u1(x) = 1 +x+ 1
2!x2+ 1 3!x3+ 1
4!x4+ 1 5!x5, u2(x) = 1 +x+ 1
2!x2+ 1 3!x3+ 1
4!x4+ 1 5!x5+ 1
6!x6+ 1 7!x7+ 1
8!x8, (3.174) ...
un(x) = 1 +x+ 1 2!x2+ 1
3!x3+ 1 4!x4+ 1
5!x5+ 1 6!x6+ 1
7!x7+ 1
8!x8+· · · The VIM admits the use of
u(x) = lim
n→∞un(x), (3.175)
that gives the exact solution by
u(x) =ex. (3.176)
Using standard methods for solving ODEs
The ODE (3.169) is of third order, therefore the auxiliary equation is of the form
r3−1 = 0, (3.177)
that givesr= 1,−12±√23i, i2=−1. The general solution is given by u(x) =Aex+e−12x
Bcos
√3
2 x+Csin
√3 2 x
. (3.178)
To obtain the particular solution, we use the initial conditions to find that the particular solution
94 3 Volterra Integral Equations
u(x) =ex. (3.179)
It is interesting to point out that we need to use different approaches to solve ODEs by standard methods, whereas the variational iteration method attacks all problems directly and in a straightforward manner.
Exercises 3.2.4
Use the variational iteration methodto solve the following Volterra integral equa- tions by converting the equation to initial value problem or to an equivalent integro- differential equation:
1. u(x) = 1− x
0 u(t)dt 2. u(x) =x+x4+1 2x2+1
5x5− x
0 u(t)dt 3. u(x) = 1−1
2x2+ x
0 u(t)dt 4. u(x) = 1−x− 1 2x2+
x
0
(x−t)u(t)dt 5. u(x) = 1− x
0
(x−t)u(t)dt 6. u(x) =x+ x
0
(x−t)u(t)dt 7. u(x) = 1 + 2x+ 4
x
0 (x−t)u(t)dt 8. u(x) = 5 + 2x2− x
0 (x−t)u(t)dt 9. u(x) = 1 +x+1
2x2+1 2
x
0
(x−t)2u(t)dt 10. u(x) = 1 +1
2x+1 2
x
0
(x−t+ 1)u(t)dt 11. u(x) = 1 +x2−
x
0 (x−t+ 1)2u(t)dt 12. u(x) = 1−1
2x2+1 6
x
0
(x−t)3u(t)dt 13. u(x) = 2 +x−2 cosx−x
0
(x−t+ 2)u(t)dt 14. u(x) = 1−2 sinhx+
x
0 (x−t+ 2)u(t)dt 15. u(x) = 1 +3
5x+ 1 10x2+ 1
10 x
0
(x−t+ 2)2u(t)dt 16. u(x) = 2 + 3x+ 3
2x2+1 6x3−1
2 x
0 (x−t+ 2)2u(t)dt 17. u(x) = 1−xsinx+
x
0 tu(t)dt 18. u(x) =xcoshx− x
0 tu(t)dt 19. u(x) =−1 +ex+1
2x2ex−1 2
x
0 tu(t)dt 20. u(x) = 1−xsinx+xcosx+
x
0 tu(t)dt
3.2 Volterra Integral Equations of the Second Kind 95