Fredholm Integral Equations
4.4 Fredholm Integral Equations of the First Kind
4.4.1 The Method of Regularization
The method of regularization was established independently by Phillips [11]
and Tikhonov [12]. The method of regularization consists of replacing ill- posed problem by well-posed problem. The method of regularization trans- forms the linear Fredholm integral equation of the first kind
f(x) = b
a
K(x, t)u(t)dt, x∈D, (4.258) to the approximation Fredholm integral equation
μuμ(x) =f(x)− b
a
K(x, t)uμ(t)dt, x∈D, (4.259) whereμis a small positive parameter. It is clear that (4.259) is a Fredholm integral equation of the second kind that can be rewritten
uμ(x) = 1
μf(x)− 1 μ
b a
K(x, t)uμ(t)dt, x∈D. (4.260) Moreover, it was proved in [7,13] that the solution uμ of equation (4.260) converges to the solutionu(x) of (4.258) asμ→0 according to the following lemma [14]:
Lemma 4.1
Suppose that the integral operator of (4.258)is continuous and coercive in the Hilbert space where f(x), u(x), anduμ(x)are defined, then:
1.|uμ| is bounded independently ofμ, and 2.|uμ(x)−u(x)| →0 whenμ→0.
The proof of this lemma can be found in [7,13].
In summary, by combining the method of regulariztion with any of the methods used before for solving Fredholm integral equation of the second
162 4 Fredholm Integral Equations kind, we can solve Fredholm integral equation of the first kind (4.258). The method of regulariztion transforms the first kind to a second kind. The re- sulting integral equation (4.260) can be solved by any of the methods that were presented before in this chapter. The exact solutionu(x) of (4.258) can thus be obtained by
u(x) = lim
μ→0uμ(x). (4.261)
In what follows we will present five illustrative examples where we will use the method of regulariztion to transform the first kind equation to a second kind equation. The resulting equation will be solved by any appropriate method that we used before.
Example 4.36
Combine the method of regulariztion and the direct computation method to solve the Fredholm integral equation of the first kind
1 4ex=
14
0
ex−tu(t)dt. (4.262) Using the method of regularization, Equation (4.262) can be transformed to
uμ(x) = 1 4μex−1
μ 14
0
ex−tuμ(t)dt. (4.263) The resulting Fredholm integral equation of the second kind will be solved by the direct computation method. Equation (4.263) can be written as
uμ(x) = 1
4μ−α μ
ex, (4.264)
where
α= 14
0
e−tuμ(t)dt. (4.265)
To determine α, we substitute (4.264) into (4.265), integrate the resulting integral and solve to find that
α= 1
1 + 4μ. (4.266)
This in turn gives
uμ(x) = ex
1 + 4μ. (4.267)
The exact solutionu(x) of (4.262) can be obtained by u(x) = lim
μ→0uμ(x) =ex. (4.268) Example 4.37
Combine the method of regulariztion and the direct computation method to solve the Fredholm integral equation of the first kind
4.4 Fredholm Integral Equations of the First Kind 163 ex+ 1 =
1
0 (4tex+ 3)u(t)dt. (4.269) Notice that the data function f(x) =ex+ 1 contains components which are matched by the correspondingxcomponents of the kernelK(x, t) = 4tex+ 3.
This is a necessary condition to guarantee a solution.
Using the method of regularization, Equation (4.269) can be transformed to
uμ(x) = 1 μex+ 1
μ− 1 μ
1
0
(4tex+ 3)uμ(t)dt. (4.270) The resulting Fredholm integral equation of the second kind will be solved by the direct computation method. Equation (4.270) can be written as
uμ(x) = 1
μ−4α μ
ex+
1 μ−3β
μ
, (4.271)
where
α= 1
0
tuμ(t)dt, β = 1
0
uμ(t)dt. (4.272)
To determine αand β, we substitute (4.271) into (4.272), integrate the re- sulting integrals and solve to find that
α= 3(e−3−μ)
2(6e−18−7μ−μ2), β=− −2(e+ 6 +μe)
6e−18−7μ−μ2. (4.273) Substituting this result into (4.271) gives the approximate solution
uμ(x) = (1 +μ)ex+ (7−3e+μ)
6(3−e) + (7μ+μ2) . (4.274) The exact solutionu(x) of (4.269) can be obtained by
u(x) = lim
μ→0uμ(x) = 1
6(3−e)ex+ 7−3e
6(3−e). (4.275) It is interesting to point out that another solution to this equation is given by
u(x) =x2. (4.276)
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 4.38
Combine the method of regulariztion and the direct computation method to solve the Fredholm integral equation of the first kind
π 2sinx=
π 0
sin(x−t)u(t)dt. (4.277) Notice that the data function f(x) = π2sinxcontains component which is matched by the correspondingxcomponent of the kernelK(x, t). This is a necessary condition to guarantee a solution.
164 4 Fredholm Integral Equations Using the method of regularization, Equation (4.277) can be transformed to
uμ(x) = π
2μsinx− 1 μ
π
0 sin(x−t)uμ(t)dt, (4.278) that can be written as
uμ(x) = π
2μ−α μ
sinx+β
μcosx, (4.279)
where
α= π
0
costuμ(t)dt, β= π
0
sintuμ(t)dt. (4.280) To determine αand β, we substitute (4.279) into (4.280), integrate the re- sulting integrals and solve to find that
α= π3
2(π2+ 4μ2), β = π2μ
π2+ 4μ2. (4.281) Substituting this result into (4.279) gives the approximate solution
uμ(x) = 2πμ
π2+ 4μ2sinx+ π2
π2+ 4μ2cosx. (4.282) The exact solutionu(x) of (4.277) can be obtained by
u(x) = lim
μ→0uμ(x) = cosx. (4.283) Example 4.39
Combine the method of regulariztion and the Adomian decomposition method to solve the Fredholm integral equation of the first kind
1 3e−x=
13
0
et−xu(t)dt. (4.284) Using the method of regularization, Equation (4.284) can be transformed to
uμ(x) = 1
3μe−x−1 μ
13
0
et−xuμ(t)dt. (4.285) The Adomian decomposition method admits the use of
uμ(x) = ∞ n=0
uμn(x), (4.286)
and the recurrence relation uμ0(x) = 1
3μe−x, uμk+1(x) =−1
μ 13
0 et−xuμk(t)dt, k0.
(4.287)
This in turn gives the components
4.4 Fredholm Integral Equations of the First Kind 165 uμ0(x) = 1
3μe−x, uμ1(x) =− 1 9μ2e−x, uμ2(x) = 1
27μ3e−x, uμ3(x) =− 1 81μ2e−x,
(4.288) and so on. Substituting this result into (4.286) gives the approximate solution
uμ(x) = 1
1 + 3μe−x. (4.289)
The exact solutionu(x) of (4.284) can be obtained by u(x) = lim
μ→0uμ(x) =e−x. (4.290) Example 4.40
Combine the method of regulariztion and the successive approximations method to solve the Fredholm integral equation of the first kind
1 4x=
1
0
xt u(t)dt. (4.291)
Using the method of regularization, Equation (4.291) can be transformed to uμ(x) = 1
4μx−1 μ
1
0
xtuμ(t)dt. (4.292)
To use the successive approximations method, we first select uμ0(x) = 0.
Consequently, we obtain the following approximations uμ0(x) = 0,
uμ1(x) = 1 4μx, uμ2(x) = 1
4μx− 1 12μ2x, uμ3(x) = 1
4μx− 1
12μ2x+ 1 36μ3x, uμ4(x) = 1
4μx− 1
12μ2x+ 1
36μ3x− 1 108μ4x,
(4.293)
and so on. Based on this we obtain the approximate solution uμ(x) = 3
4(1 + 3μ)x. (4.294)
The exact solutionu(x) of (4.291) can be obtained by u(x) = lim
μ→0uμ(x) =3
4x. (4.295)
It is interesting to point out that another solution to this equation is given by
u(x) =x2. (4.296)
As stated before, the Fredholm integral equation of the first kind is ill-posed problem and the solution may not be unique.
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