Analogous to differentiation, linear ODEs are solved in Chapter 9. Analogous to integra- tion, there is no general method available for solving integral equations. However, certain special cases may be treated with our integral transforms (Chapter 15). For convenience these are listed here. If
ψ (x)= 1
√2π ∞
−∞
eixtϕ(t ) dt, then
ϕ(x)= 1
√2π ∞
−∞
e−ixtψ (t ) dt (Fourier). (16.35) If
ψ (x)= ∞
0
e−xtϕ(t ) dt, then
ϕ(x)= 1 2π i
γ+i∞ γ−i∞
extψ (t ) dt (Laplace). (16.36) If
ψ (x)= ∞
0
tx−1ϕ(t ) dt,
then
ϕ(x)= 1 2π i
γ+i∞ γ−i∞
x−tψ (t ) dt (Mellin). (16.37) If
ψ (x)= ∞
0
t ϕ(t )Jν(xt ) dt, then
ϕ(x)= ∞
0
t ψ (t )Jν(xt ) dt (Hankel). (16.38) Actually the usefulness of the integral transform technique extends a bit beyond these four rather specialized forms.
Example 16.2.1 FOURIERTRANSFORMSOLUTION
Let us consider a Fredholm equation of the first kind with a kernel of the general type k(x−t ),
f (x)= ∞
−∞
k(x−t )ϕ(t ) dt, (16.39)
in whichϕ(t )is our unknown function. Assuming that the needed transforms exist, we apply the Fourier convolution theorem (Section 15.5) to obtain
f (x)= ∞
−∞
K(ω)(ω)e−iωxdω. (16.40)
The functionsK(ω), (ω), andF (ω)are the Fourier transforms ofk(x), ϕ(x), andf (x), respectively. Taking the Fourier transform of both sides of Eq. (16.40), by Eq. (16.35) we have
K(ω)(ω)= 1 2π
∞
−∞
f (x)eiωxdx=F (ω)
√2π. (16.41)
Then
(ω)= 1
√2π ã F (ω)
K(ω), (16.42)
and, using the inverse Fourier transform, we have ϕ(x)= 1
2π ∞
−∞
F (ω)
K(ω)e−iωxdω. (16.43)
For a rigorous justification of this result one can follow Morse and Feshbach (see the Additional Readings) (1953) across complex planes. An extension of this transformation
solution appears as Exercise 16.2.1.
Example 16.2.2 GENERALIZEDABELEQUATION, CONVOLUTIONTHEOREM
The generalized Abel equation is f (x)=
x 0
ϕ(t )
(x−t )αdt, 0< α <1, with
f (x)known,
ϕ(t ) unknown. (16.44) Taking the Laplace transform of both sides of this equation, we obtain
L{f (x)} =L x
0
ϕ(t ) (x−t )αdt
=L' x−α(
L' ϕ(x)(
, (16.45)
the last step following by the Laplace convolution theorem (Section 15.11). Then L'
ϕ(x)(
=s1−αL{f (x)}
(−α)! . (16.46)
Dividing bys,1we obtain 1 sL'
ϕ(x)(
=s−αL{f (x)}
(−α)! =L{xα−1}L{f (x)}
(α−1)!(−α)! . (16.47) Combining the factorials (Eq. (8.32)) and applying the Laplace convolution theorem again, we discover that
1 sL'
ϕ(x)(
=sinπ α
π L
x 0
f (t ) (x−t )1−αdt
. (16.48)
Inverting with the aid of Exercise 15.11.1, we get x
0
ϕ(t ) dt=sinπ α π
x 0
f (t )
(x−t )1−αdt, (16.49) and finally, by differentiating,
ϕ(x)=sinπ α π
d dx
x 0
f (t )
(x−t )1−αdt. (16.50)
Generating Functions
Occasionally, the reader may encounter integral equations that involve generating func- tions. Suppose we have the admittedly special case
f (x)= 1
−1
ϕ(t )
(1−2xt+x2)1/2dt, −1≤x≤1. (16.51) We notice two important features:
1. (1−2xt+x2)−1/2generates the Legendre polynomials.
2. [−1,1]is the orthogonality interval for the Legendre polynomials.
1s1−αdoes not have an inverse for 0< α <1.
If we now expand the denominator (property 1) and assume that our unknownϕ(t )may be written as a series of these same Legendre polynomials,
f (x)= 1
−1
∞
n=0
anPn(t ) ∞
r=0
Pr(t )xrdt. (16.52) Utilizing the orthogonality of the Legendre polynomials (property 2), we obtain
f (x)= ∞
r=0
2ar
2r+1xr. (16.53)
We may identify theanby differentiatingntimes and then settingx=0:
f(n)(0)=n! 2
2n+1an. (16.54)
Hence
ϕ(t )= ∞
n=0
2n+1 2
f(n)(0)
n! Pn(t ). (16.55)
Similar results may be obtained with the other generating functions (compare Exer- cise 7.1.6).
• This technique of expanding in a series of special functions is always available. It is worth a try whenever the expansion is possible (and convenient) and the interval is appropriate.
Exercises
16.2.1 The kernel of a Fredholm equation of the second kind, ϕ(x)=f (x)+λ
∞
−∞
K(x, t )ϕ(t ) dt,
is of the formk(x−t ).2Assuming that the required transforms exist, show that ϕ(x)= 1
√2π ∞
−∞
F (t )e−ixtdt 1−√
2π λK(t ).
F (t )andK(t )are the Fourier transforms off (x)andk(x), respectively.
16.2.2 The kernel of a Volterra equation of the first kind, f (x)=
x 0
K(x, t )ϕ(t ) dt,
2This kernel and a range 0≤x <∞are the characteristics of integral equations of the Wiener–Hopf type. Details will be found in Chapter 8 of Morse and Feshbach (1953); see the Additional Readings.
has the formk(x−t ). Assuming that the required transforms exist, show that ϕ(x)= 1
2π i
γ+i∞ γ−i∞
F (s) K(s)exsds.
F (s)andK(s)are the Laplace transforms off (x)andk(x), respectively.
16.2.3 The kernel of a Volterra equation of the second kind, ϕ(x)=f (x)+λ
x 0
K(x, t )ϕ(t ) dt,
has the formk(x−t ). Assuming that the required transforms exist, show that ϕ(x)= 1
2π i
γ+i∞
γ−i∞
F (s)
1−λK(s)exsds.
16.2.4 Using the Laplace transform solution (Exercise 16.2.3), solve (a) ϕ(x)=x+
x 0
(t−x)ϕ(t ) dt.
ANS.ϕ(x)=sinx.
(b) ϕ(x)=x− x
0
(t−x)ϕ(t ) dt.
ANS.ϕ(x)=sinhx.
Check your results by substituting back into the original integral equations.
16.2.5 Reformulate the equations of Example 16.2.1 (Eqs. (16.39) to (16.43)), using Fourier cosine transforms.
16.2.6 Given the Fredholm integral equation, e−x2=
∞
−∞
e−(x−t )2ϕ(t ) dt,
apply the Fourier convolution technique of Example 16.2.1 to solve forϕ(t ).
16.2.7 Solve Abel’s equation, f (x)=
x 0
ϕ(t )
(x−t )αdt, 0< α <1, by the following method:
(a) Multiply both sides by(z−x)α−1and integrate with respect tox over the range 0≤x≤z.
(b) Reverse the order of integration and evaluate the integral on the right-hand side (with respect tox) by the beta function.
Note.
z t
dx
(z−x)1−α(x−t )α =B(1−α, α)=(−α)!(α−1)! = π sinπ α. 16.2.8 Given the generalized Abel equation withf (x)=1,
1= x
0
ϕ(t )
(x−t )αdt, 0< α <1, solve forϕ(t )and verify thatϕ(t )is a solution of the given equation.
ANS.ϕ(t )=sinπ α π tα−1. 16.2.9 A Fredholm equation of the first kind has a kernele−(x−t )2:
f (x)= ∞
−∞
e−(x−t )2ϕ(t ) dt.
Show that the solution is
ϕ(x)= 1
√π ∞
π=0
f(n)(0) 2nn! Hn(x), in whichHn(x)is annth-order Hermite polynomial.
16.2.10 Solve the integral equation f (x)=
1
−1
ϕ(t )
(1−2xt+x2)1/2dt, −1≤x≤1, for the unknown functionϕ(t )if
(a)f (x)=x2s, (b) f (x)=x2s+1.
ANS. (a)ϕ(t )=4s+1
2 P2s(t ), (b)ϕ(t )=4s+3
2 P2s+1(t ).
16.2.11 A Kirchhoff diffraction theory analysis of a laser leads to the integral equation v(r2)=γ
K(r1,r2)v(r1) dA.
The unknown,v(r1), gives the geometric distribution of the radiation field over one mirror surface; the range of integration is over the surface of that mirror. For square confocal spherical mirrors the integral equation becomes
v(x2, y2)=−iγ eikb λb
a
−a
a
−a
e−(ik/b)(x1x2+y1y2)v(x1, y1) dx1dy1,
in which b is the centerline distance between the laser mirrors. This can be put in a somewhat simpler form by the substitutions
kxi2
b =ξi2, kyi2
b =η2i, and ka2
b =2π a2 λb =α2. (a) Show that the variables separate and we get two integral equations.
(b) Show that the new limits,±α, may be approximated by±∞for a mirror dimen- siona≫λ.
(c) Solve the resulting integral equations.