Fredholm Integral Equations
4.4 Fredholm Integral Equations of the First Kind
4.4.2 The Homotopy Perturbation Method
166 4 Fredholm Integral Equations Exercises 4.4.1
Combine the regularization method with any other method to solve the Fredholm integral equations of the first kind
1. 1
2(1−e−2)e3x= 1
0 e3x−4tu(t)dt 2. 1 2e3x=
1
2
0 e3x−3tu(t)dt 3. 3
4x= 1
0 xt2u(t)dt 4. 6 5x2=
1
0 x2t2u(t)dt 5. 2
5x2= 1
−1x2t2u(t)dt 6. 1 5x=
1
0 xtu(t)dt 7. 1
6x2= 1
0 x2t2u(t)dt 8. 2 3x2=
1
−1x2t2u(t)dt 9. −1
4x= 1
0 xtu(t)dt 10. 1
4x= 1
0 xtu(t)dt 11. 1
12x= 1
0 xtu(t)dt 12. 7
12x= 1
0 xtu(t)dt 13. π
2sinx= π
0 cos(x−t)u(t)dt 14. π 2cosx=
π
0 cos(x−t)u(t)dt 15.2−π+ 2x=
π
0 (x−t)u(t)dt 16.2 +π−2x= π
0 (x−t)u(t)dt
4.4 Fredholm Integral Equations of the First Kind 167 v(x) =f(x) +
b a
K(x, t)v(t)dt. (4.297) We now define the operator
L(u) =u(x)−f(x)− b
a
K(x, t)u(t)dt= 0, (4.298) whereu(x) =v(x). Next we define the homotopyH(u, p), p∈[0,1] by
H(u,0) =F(u), H(u,1) =L(u), (4.299) whereF(u) is a functional operator. We construct a convex homotopy of the form
H(u, p) = (1−p)F(u) +pL(u) = 0. (4.300) This homotopy satisfies (4.299) forp= 0 andp= 1 respectively. The embed- ding parameterpmonotonically increases from 0 to 1 as the trivial problem F(u) = 0 continuously deformed [10] to the original problemL(u) = 0. The homotopy perturbation method admits the use of the expansion
u= ∞ n=0
pnun, (4.301)
and consequently
v= lim
p→1
∞ n=0
pnun. (4.302)
The series (4.302) converges to the exact solution if such a solution exists.
Substituting (4.301) into (4.300), usingF(x) =u(x)−f(x), and equating the terms with like powers of the embedding parameter p we obtain the recurrence relation
p0:u0(x) =f(x), pn+1:un+1= b
a
K(x, t)un(t)dt, n0. (4.303) Notice that the recurrence relation (4.303) is the same standard Adomian decomposition method as presented before in this chapter. This proves the following theorem:
Theorem 4.3The Adomian decomposition method is a homotopy perturba- tion method with a convex homotopy given by
H(u, p) =u(x)−f(x)−p b
a
K(x, t)un(t)dt= 0. (4.304)
HPM for Fredholm Integral Equation of the First Kind
In what follows we present the homotopy perturbation method for handling the Fredholm integral equations of the first kind of the form
f(x) = b
a
K(x, t)v(t)dt. (4.305)
168 4 Fredholm Integral Equations We now define the operator
L(u) =f(x)− b
a
K(x, t)u(t)dt= 0. (4.306) We construct a convex homotopy of the form
H(u, p) = (1−p)u(x) +pL(u)(x) = 0. (4.307) The embedding parameterpmonotonically increases from 0 to 1. The homo- topy perturbation method admits the use of the expansion
u= ∞ n=0
pnun, (4.308)
and consequently
v(x) = lim
p→1
∞ n=0
pnun(x). (4.309)
The series (4.309) converges to the exact solution if such a solution exists.
Substituting (4.308) into (4.307), and proceeding as before we obtain the recurrence relation
u0(x) = 0, u1(x) =f(x), un+1(x) =un(x)−
b a
K(x, t)un(t)dt, n1. (4.310) If the kernel is separable, i.e.K(x, t) =g(x)h(t), then the following condition
1−
b a
K(t, t)dt
<1, (4.311)
must be justified for convergence. The proof of this condition is left to the reader.
We will concern ourselves only on the case whereK(x, t) =g(x)h(t). The HPM will be used to solve the following Fredholm integral equations of the first kind.
Example 4.41
Use the homotopy perturbation method to solve the Fredholm integral equa- tion of the first kind
1 3ex=
13
0 ex−tu(t)dt. (4.312)
Notice that
1− 13
0 K(t, t)dt = 2
3 <1. (4.313)
Using the recurrence relation (4.310) we find
4.4 Fredholm Integral Equations of the First Kind 169 u0(x) = 0, u1(x) = 1
3ex, un+1(x) =un(x)−
13
0 ex−tun(t)dt, n1.
(4.314)
This in turn gives
u0(x) = 0, u1(x) =1 3ex, u2(x) =u1(x)−
1
3
0 ex−tu1(t)dt=2 9ex, u3(x) =u2(x)−
13
0 ex−tu2(t)dt= 4 27ex, u4(x) =u3(x)−
13
0 ex−tu3(t)dt= 8 81ex,
(4.315)
and so on. Consequently, the approximate solution is given by u(x) =ex
1 3 +2
9 + 4 27+ 8
81+· · ·
, (4.316)
that converges to the exact solution
u(x) =ex. (4.317)
Example 4.42
Use the homotopy perturbation method to solve the Fredholm integral equa- tion of the first kind
1 4e−x=
14
0
et−xu(t)dt. (4.318)
Notice that
1− 14
0 K(t, t)dt = 3
4 <1. (4.319)
Using the recurrence relation (4.310) we find u0(x) = 0, u1(x) = 1
4e−x, un+1(x) =un(x)−
14
0 et−xun(t)dt, n1.
(4.320)
This in turn gives
u0(x) = 0, u1(x) = 1 4e−x, u2(x) =u1(x)−
1
4
0
et−xu1(t)dt= 3 16e−x, u3(x) =u2(x)−
1
4
0 et−xu2(t)dt= 9 64e−x,
(4.321)
170 4 Fredholm Integral Equations u4(x) =u3(x)−
14
0 et−xu3(t)dt= 27 256e−x, and so on. Consequently, the approximate solution is given by
u(x) =e−x 1
4 + 3 16+ 9
64+ 27 256+· · ·
, (4.322)
that converges to the exact solution
u(x) =e−x, (4.323)
obtained by evaluating the sum of the infinite geometric series.
Example 4.43
Use the homotopy perturbation method to solve the Fredholm integral equa- tion of the first kind
x= 1
0 xt u(t)dt. (4.324)
Notice that
1− 1
0
K(t, t)dt =2
3 <1. (4.325)
Using the recurrence relation (4.310) we find u0(x) = 0, u1(x) =x, un+1(x) =un(x)−
1
0
xtun(t)dt, n1. (4.326) This in turn gives
u0(x) = 0, u1(x) =x, u2(x) =u1(x)−
1
0 xtu1(t)dt= 2 3x, u3(x) =u2(x)−
1
0 xtu2(t)dt= 4 9x, u4(x) =u3(x)−
1
0
xtu3(t)dt= 8 27x,
(4.327)
and so on. Consequently, the approximate solution is given by u(x) =x
1 + 2
3 +4 9+ 8
27+· · ·
, (4.328)
that converges to the exact solution
u(x) = 3x. (4.329)
Example 4.44
Use the homotopy perturbation method to solve the Fredholm integral equa- tion of the first kind
5 6x=
1
0
xt u(t)dt. (4.330)
4.4 Fredholm Integral Equations of the First Kind 171
Notice that
1− 1
0
K(t, t)dt =2
3 <1. (4.331)
Using the recurrence relation (4.310) we find u0(x) = 0, u1(x) =5
6x, un+1(x) =un(x)−
1
0 xtun(t)dt, n1.
(4.332)
This in turn gives
u0(x) = 0, u1(x) = 5 6x, u2(x) =u1(x)−
1
0
xtu1(t)dt= 5 9x, u3(x) =u2(x)−
1
0
xtu2(t)dt= 10 27x, u4(x) =u3(x)−
1
0
xtu3(t)dt= 20 81x,
(4.333)
and so on. Consequently, the approximate solution is given by u(x) =5
6x+5 9x
1 + 2
3+4 9 + 8
27+· · ·
, (4.334)
that converges to the exact solution u(x) =5
2x, (4.335)
obtained by evaluating the sum of the infinite geometric series.
It is interesting to point out that another solution to this equation is given by
u(x) = 1 +x. (4.336)
As stated before, the Fredholm integral equation of the first kind is ill-posed problem. For ill-posed problems, the solution might not exist, and if it exists, the solution may not be unique.
Example 4.45
Use the homotopy perturbation method to solve the Fredholm integral equa- tion of the first kind
1 4x=
1
0 xt u(t)dt. (4.337)
Notice that
1− 1
0 K(t, t)dt =2
3 <1. (4.338)
Using the recurrence relation (4.310) we find
172 4 Fredholm Integral Equations u0(x) = 0, u1(x) =1
4x, un+1(x) =un(x)−
1
0 xtun(t)dt, n1.
(4.339)
This in turn gives
u0(x) = 0, u1(x) = 1 4x, u2(x) =u1(x)−
1
0 xtu1(t)dt= 1 6x, u3(x) =u2(x)−
1
0
xtu2(t)dt= 1 9x, u4(x) =u3(x)−
1
0
xtu3(t)dt= 2 27x,
(4.340)
and so on. Consequently, the approximate solution is given by u(x) =1
4x
1 + 2 3 +4
9 + 8 27+· · ·
, (4.341)
that converges to the exact solution u(x) =3
4x. (4.342)
It is interesting to point out that another solution to this equation is given by
u(x) =x2. (4.343)
As stated before, the Fredholm integral equation of the first kind is ill-posed problem and the solution may not be unique. Notice that the ill-posed Fred- holm problem is linear.
Exercises 4.4.2
Use the homotopy perturbation method to solve the Fredholm integral equations of the first kind
1. 1
2(1−e−2)e3x= 1
0 e3x−4tu(t)dt 2. 1 2e3x=
1
2
0 e3x−3tu(t)dt 3. 3
4x= 1
0 xt2u(t)dt 4. 6
5x2= 1
0 x2t2u(t)dt 5. 2
5x2= 1
−1x2t2u(t)dt 6. 1 5x=
1
0 xtu(t)dt 7. 1
6x2= 1
0 x2t2u(t)dt 8. 2 3x2=
1
−1x2t2u(t)dt 9. − 1
4x= 1
0 xtu(t)dt 10. 1
4x= 1
0 xtu(t)dt
References 173 11. 1
12x= 1
0 xtu(t)dt 12. 7
12x= 1
0 xtu(t)dt
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