Introductory Concepts of Integral Equations
2.5 Converting IVP to Volterra Integral Equation
In this section, we will study the technique that will convert an initial value problem (IVP) to an equivalent Volterra integral equation and Volterra integro-differential equation as well [3]. For simplicity reasons, we will apply this process to a second order initial value problem given by
y(x) +p(x)y(x) +q(x)y(x) =g(x) (2.48) subject to the initial conditions:
y(0) =α, y(0) =β, (2.49) whereαandβ are constants. The functionsp(x) andq(x) are analytic func- tions, and g(x) is continuous through the interval of discussion. To achieve our goal we first set
y(x) =u(x), (2.50)
where u(x) is a continuous function. Integrating both sides of (2.50) from 0 to xyields
y(x)−y(0) = x
0 u(t)dt, (2.51)
2.5 Converting IVP to Volterra Integral Equation 43 or equivalently
y(x) =β+ x
0 u(t)dt. (2.52)
Integrating both sides of (2.52) from 0 toxyields y(x)−y(0) =βx+
x 0
x
0 u(t)dtdt, (2.53) or equivalently
y(x) =α+βx+ x
0
(x−t)u(t)dt, (2.54) obtained upon using the formula that reduce double integral to a single inte- gral that was discussed in the previous chapter. Substituting (2.50), (2.52), and (2.54) into the initial value problem (2.48) yields the Volterra integral equation:
u(x) +p(x)
β+ x
0
u(t)dt
+q(x)
α+βx+ x
0
(x−t)u(t)dt
=g(x).
(2.55) The last equation can be written in the standard Volterra integral equation form:
u(x) =f(x)− x
0 K(x, t)u(t)dt, (2.56) where
K(x, t) =p(x) +q(x)(x−t), (2.57) and
f(x) =g(x)−[βp(x) +αq(x) +βxq(x)]. (2.58) It is interesting to point out that by differentiating Volterra equation (2.56) with respect to x, using Leibnitz rule, we obtain an equivalent Volterra integro-differential equation in the form:
u(x) +K(x, x)u(x) =f(x)− x
0
∂K(x, t)
∂x u(t)dt, u(0) =f(0). (2.59) The technique presented above to convert initial value problems to equiva- lent Volterra integral equations can be generalized by considering the general initial value problem:
y(n)+a1(x)y(n−1)+· · ·+an−1(x)y+an(x)y=g(x), (2.60) subject to the initial conditions
y(0) =c0, y(0) =c1, y(0) =c2, . . . , y(n−1)(0) =cn−1. (2.61) We assume that the functionsai(x),1inare analytic at the origin, and the function g(x) is continuous through the interval of discussion. Letu(x) be a continuous function on the interval of discussion, and we consider the transformation:
y(n)(x) =u(x). (2.62)
Integrating both sides with respect toxgives
44 2 Introductory Concepts of Integral Equations y(n−1)(x) =cn−1+
x 0
u(t)dt. (2.63)
Integrating again both sides with respect toxyields y(n−2)(x) =cn−2+cn−1x+
x 0
x 0
u(t)dtdt
=cn−2+cn−1x+ x
0
(x−t)u(t)dt, (2.64) obtained by reducing the double integral to a single integral. Proceeding as before we find
y(n−3)(x) =cn−3+cn−2x+1
2cn−1x2+ x
0
x 0
x 0
u(t)dtdtdt
=cn−3+cn−2x+1
2cn−1x2+1 2
x
0 (x−t)2u(t)dt. (2.65) Continuing the integration process leads to
y(x) =
n−1
k=0
ck
k!xk+ 1 (n−1)!
x
0 (x−t)n−1u(t)dt. (2.66) Substituting (2.62)–(2.66) into (2.60) gives
u(x) =f(x)− x
0
K(x, t)u(t)dt, (2.67) where
K(x, t) = n
k=1
an
(k−1)!(x−t)k−1, (2.68) and
f(x) =g(x)− n j=1
aj
j
k=1
cn−k
(j−k)!xj−k
. (2.69)
Notice that the Volterra integro-differential equation can be obtained by dif- ferentiating (2.67) as many times as we like, and by obtaining the initial conditions of each resulting equation. The following examples will highlight the process to convert initial value problem to an equivalent Volterra integral equation.
Example 2.1
Convert the following initial value problem to an equivalent Volterra integral equation:
y(x)−2xy(x) =ex2, y(0) = 1. (2.70) We first set
y(x) =u(x). (2.71)
Integrating both sides of (2.71), using the initial condition y(0) = 1 gives y(x)−y(0) =
x 0
u(t)dt, (2.72)
2.5 Converting IVP to Volterra Integral Equation 45 or equivalently
y(x) = 1 + x
0 u(t)dt, (2.73)
Substituting (2.71) and (2.73) into (2.70) gives the equivalent Volterra inte- gral equation:
u(x) = 2x+ex2+ 2x x
0 u(t)dt. (2.74)
Example 2.2
Convert the following initial value problem to an equivalent Volterra integral equation:
y(x)−y(x) = sinx, y(0) = 0, y(0) = 0. (2.75) Proceeding as before, we set
y(x) =u(x). (2.76)
Integrating both sides of (2.76), using the initial condition y(0) = 0 gives y(x) =
x 0
u(t)dt. (2.77)
Integrating (2.77) again, using the initial conditiony(0) = 0 yields y(x) =
x 0
x 0
u(t)dtdt= x
0
(x−t)u(t)dt, (2.78) obtained upon using the rule to convert double integral to a single integral.
Inserting (2.76)–(2.78) into (2.70) leads to the following Volterra integral equation:
u(x) = sinx+ x
0
(x−t)u(t)dt. (2.79) Example 2.3
Convert the following initial value problem to an equivalent Volterra integral equation:
y−y−y+y= 0, y(0) = 1, y(0) = 2, y(0) = 3. (2.80) We first set
y(x) =u(x), (2.81)
where by integrating both sides of (2.81) and using the initial condition y(0) = 3 we obtain
y= 3 + x
0 u(t)dt. (2.82)
Integrating again and using the initial conditiony(0) = 2 we find y(x) = 2 + 3x+
x 0
x 0
u(t)dtdt= 2 + 3x+ x
0
(x−t)u(t)dt. (2.83) Integrating again and usingy(0) = 1 we obtain
46 2 Introductory Concepts of Integral Equations y(x) = 1 + 2x+3
2x2+ x
0
x 0
x 0
u(t)dtdtdt
= 1 + 2x+3 2x2+1
2 x
0
(x−t)2u(t)dt. (2.84) Notice that in (2.83) and (2.84) the multiple integrals were reduced to single integrals as used before. Substituting (2.81) – (2.84) into (2.80) leads to the Volterra integral equation:
u(x) = 4 +x+3 2x2+
x
0 [1 + (x−t)−1
2(x−t)2]u(t)dt. (2.85) Remark
We can also show that ify(iv)(x) =u(x), then y(x) =y(0) +
x 0 u(t)dt y(x) =y(0) +xy(0) +
x
0 (x−t)u(t)dt y(x) =y(0) +xy(0) +1
2x2y(0) +1 2
x 0
(x−t)2u(t)dt y(x) =y(0) +xy(0) +1
2x2y(0) +1
6x3y(0) +1 6
x 0
(x−t)3u(t)dt.
(2.86) This process can be generalized to any derivative of a higher order.
In what follows we summarize the relation between derivatives ofy(x) and u(x):
Table 2.1 The relation between derivatives ofy(x) andu(x)
y(n)(x) Integral Equations
y(x) =u(x) y(x) =y(0) + x
0 u(t)dt y(x) =u(x) y(x) =y(0) +
x
0 u(t)dt y(x) =y(0) +xy(0) +
x
0
(x−t)u(t)dt y(x) =y(0) +
x
0 u(t)dt y(x) =u(x) y(x) =y(0) +xy(0) +
x
0 (x−t)u(t)dt y(x) =y(0) +xy(0) +1
2x2y(0) +1 2
x
0 (x−t)2u(t)dt
2.5 Converting IVP to Volterra Integral Equation 47
2.5.1 Converting Volterra Integral Equation to IVP
A well-known method for solving Volterra integral and Volterra integro- differential equation, that we will use in the forthcoming chapters, converts these equations to equivalent initial value problems. The method is achieved simply by differentiating both sides of Volterra equations [6] with respect to xas many times as we need to get rid of the integral sign and come out with a differential equation. The conversion of Volterra equations requires the use of Leibnitz rule for differentiating the integral at the right hand side. The initial conditions can be obtained by substituting x = 0 into u(x) and its derivatives. The resulting initial value problems can be solved easily by using ODEs methods that were summarized in Chapter 1. The conversion process will be illustrated by discussing the following examples.
Example 2.4
Find the initial value problem equivalent to the Volterra integral equation:
u(x) =ex+ x
0
u(t)dt. (2.87)
Differentiating both sides of (2.87) and using Leibnitz rule we find
u(x) =ex+u(x). (2.88)
It is clear that there is no need for differentiating again because we got rid of the integral sign. To determine the initial condition, we substitute x= 0 into both sides of (2.87) to findu(0) = 1. This in turn gives the initial value problem:
u(x)−u(x) =ex, u(0) = 1. (2.89) Notice that the resulting ODE is a linear inhomogeneous equation of first order.
Example 2.5
Find the initial value problem equivalent to the Volterra integral equation:
u(x) =x2+ x
0
(x−t)u(t)dt. (2.90) Differentiating both sides of (2.90) and using Leibnitz rule we find
u(x) = 2x+ x
0
u(t)dt. (2.91)
To get rid of the integral sign we should differentiate (2.91) and by using Leibnitz rule we obtain the second order ODE:
u(x) = 2 +u(x). (2.92)
To determine the initial conditions, we substitute x= 0 into both sides of (2.90) and (2.91) to find u(0) = 0 andu(0) = 0 respectively. This in turn gives the initial value problem:
u(x)−u(x) = 2, u(0) = 0, u(0) = 0. (2.93)
48 2 Introductory Concepts of Integral Equations Notice that the resulting ODE is a second order inhomogeneous equation.
Example 2.6
Find the initial value problem equivalent to the Volterra integral equation:
u(x) = sinx−1 2
x 0
(x−t)2u(t)dt. (2.94) Differentiating both sides of the integral equation three times to get rid of the integral sign to find
u(x) = cosx− x
0
(x−t)u(t)dt, u(x) =−sinx−
x 0
u(t)dt, u(x) =−cosx−u(x).
(2.95)
Substitutingx= 0 into (2.94) and into the first two integro-differential equa- tions in (2.95) gives the initial conditions:
u(0) = 0, u(0) = 1, u(0) = 0. (2.96) In view of the last results, the initial value problem equivalent to the Volterra integral equation (2.94) is a third order inhomogeneous ODE given by
u(x) +u(x) =−cosx, u(0) = 0, u(0) = 1, u(0) = 0. (2.97)
Exercises 2.5
Convert each of the following IVPs in 1–8 to an equivalent Volterra integral equation:
1.y−4y= 0, y(0) = 1 2.y+ 4xy=e−2x2, y(0) = 0
3.y+ 4y= 0, y(0) = 0, y(0) = 1 4.y−6y+ 8y= 1, y(0) = 1, y(0) = 1 5.y−y= 0, y(0) = 2, y(0) =y(0) = 1
6.y−2y+y=x, y(0) = 1, y(0) = 0, y(0) = 1 7.y(iv)−y= 1, y(0) =y(0) = 0, y(0) =y(0) = 1 8.y(iv)+y+y=x, y(0) =y(0) = 1, y(0) =y(0) = 0
Convert each of the following Volterra integral equation in 9–16 to an equivalent IVP:
9.u(x) =x+ 2 x
0 u(t)dt 10.u(x) = 1 +ex− x
0 u(t)dt 11.u(x) = 1 +x2+
x
0 (x−t)u(t)dt 12.u(x) = sinx−x
0 (x−t)u(t)dt 13.u(x) = 1−cosx+ 2
x
0
(x−t)2u(t)dt14.u(x) = 2 + sinhx+ x
0
(x−t)2u(t)dt 15.u(x) = 1 + 2
x
0
(x−t)3u(t)dt 16.u(x) = 1 +ex+ x
0
(1 +x−t)3u(t)dt