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Closedform solutions

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Closedform solutions
INTEGRAL EQUATIONS
inhomogeneous Fredholm equation of the first kind may be written as
0 = f + λKy,
which has the unique solution y = −K−1 f/λ, provided that f = 0 and the inverse
operator K−1 exists.
Similarly, we may write the corresponding Fredholm equation of the second
kind as
y = f + λKy.
(23.6)
In the homogeneous case, where f = 0, this reduces to y = λKy, which is
reminiscent of an eigenvalue problem in linear algebra (except that λ appears on
the other side of the equation) and, similarly, only has solutions for at most a
countably infinite set of eigenvalues λi . The corresponding solutions yi are called
the eigenfunctions.
In the inhomogeneous case (f = 0), the solution to (23.6) can be written
symbolically as
y = (1 − λK)−1 f,
again provided that the inverse operator exists. It may be shown that, in general,
(23.6) does possess a unique solution if λ = λi , i.e. when λ does not equal one of
the eigenvalues of the corresponding homogeneous equation.
When λ does equal one of these eigenvalues, (23.6) may have either many
solutions or no solution at all, depending on the form of f. If the function f is
orthogonal to every eigenfunction of the equation
g = λ∗ K† g
(23.7)
that belongs to the eigenvalue λ∗ , i.e.
b
g|f =
g ∗ (x)f(x) dx = 0
a
for every function g obeying (23.7), then it can be shown that (23.6) has many
solutions. Otherwise the equation has no solution. These statements are discussed
further in section 23.7, for the special case of integral equations with Hermitian
kernels, i.e. those for which K = K† .
23.4 Closed-form solutions
In certain very special cases, it may be possible to obtain a closed-form solution
of an integral equation. The reader should realise, however, when faced with an
integral equation, that in general it will not be soluble by the simple methods
presented in this section but must instead be solved using (numerical) iterative
methods, such as those outlined in section 23.5.
806
23.4 CLOSED-FORM SOLUTIONS
23.4.1 Separable kernels
The most straightforward integral equations to solve are Fredholm equations
with separable (or degenerate) kernels. A kernel is separable if it has the form
K(x, z) =
n
φi (x)ψi (z),
(23.8)
i=1
where φi (x) are ψi (z) are respectively functions of x only and of z only and the
number of terms in the sum, n, is finite.
Let us consider the solution of the (inhomogeneous) Fredholm equation of the
second kind,
b
K(x, z)y(z) dz,
(23.9)
y(x) = f(x) + λ
a
which has a separable kernel of the form (23.8). Writing the kernel in its separated
form, the functions φi (x) may be taken outside the integral over z to obtain
b
n
y(x) = f(x) + λ
φi (x)
ψi (z)y(z) dz.
a
i=1
Since the integration limits a and b are constant for a Fredholm equation, the
integral over z in each term of the sum is just a constant. Denoting these constants
by
b
ψi (z)y(z) dz,
(23.10)
ci =
a
the solution to (23.9) is found to be
y(x) = f(x) + λ
n
ci φi (x),
(23.11)
i=1
where the constants ci can be evalutated by substituting (23.11) into (23.10).
Solve the integral equation
1
(xz + z 2 )y(z) dz.
y(x) = x + λ
(23.12)
0
The kernel for this equation is K(x, z) = xz + z 2 , which is clearly separable, and using the
notation in (23.8) we have φ1 (x) = x, φ2 (x) = 1, ψ1 (z) = z and ψ2 (z) = z 2 . From (23.11)
the solution to (23.12) has the form
y(x) = x + λ(c1 x + c2 ),
where the constants c1 and c2 are given by (23.10) as
1
z[z + λ(c1 z + c2 )] dz = 13 + 13 λc1 + 12 λc2 ,
c1 =
0
1
z 2 [z + λ(c1 z + c2 )] dz =
c2 =
0
807
1
4
+ 14 λc1 + 13 λc2 .
INTEGRAL EQUATIONS
These two simultaneous linear equations may be straightforwardly solved for c1 and c2 to
give
c1 =
24 + λ
72 − 48λ − λ2
and
c2 =
18
,
72 − 48λ − λ2
so that the solution to (23.12) is
y(x) =
(72 − 24λ)x + 18λ
.
72 − 48λ − λ2
In the above example, we see that (23.12) has a (finite) unique solution provided
that λ is not equal to either root of the quadratic in the denominator of y(x).
The roots of this quadratic are in fact the eigenvalues of the corresponding
homogeneous equation, as mentioned in the previous section. In general, if the
separable kernel contains n terms, as in (23.8), there will be n such eigenvalues,
although they may not all be different.
Kernels consisting of trigonometric (or hyperbolic) functions of sums or differences of x and z are also often separable.
Find the eigenvalues and corresponding eigenfunctions of the homogeneous Fredholm
equation
π
y(x) = λ
sin(x + z) y(z) dz.
(23.13)
0
The kernel of this integral equation can be written in separated form as
K(x, z) = sin(x + z) = sin x cos z + cos x sin z,
so, comparing with (23.8), we have φ1 (x) = sin x, φ2 (x) = cos x, ψ1 (z) = cos z and
ψ2 (z) = sin z.
Thus, from (23.11), the solution to (23.13) has the form
y(x) = λ(c1 sin x + c2 cos x),
where the constants c1 and c2 are given by
π
λπ
cos z (c1 sin z + c2 cos z) dz =
c1 = λ
c2 ,
2
0
π
λπ
sin z (c1 sin z + c2 cos z) dz =
c2 = λ
c1 .
2
0
(23.14)
(23.15)
Combining these two equations we find c1 = (λπ/2)2 c1 , and, assuming that c1 = 0, this
gives λ = ±2/π, the two eigenvalues of the integral equation (23.13).
By substituting each of the eigenvalues back into (23.14) and (23.15), we find that
the eigenfunctions corresponding to the eigenvalues λ1 = 2/π and λ2 = −2/π are given
respectively by
y1 (x) = A(sin x + cos x)
and
where A and B are arbitrary constants. 808
y2 (x) = B(sin x − cos x),
(23.16)
23.4 CLOSED-FORM SOLUTIONS
23.4.2 Integral transform methods
If the kernel of an integral equation can be written as a function of the difference
x − z of its two arguments, then it is called a displacement kernel. An integral
equation having such a kernel, and which also has the integration limits −∞ to
∞, may be solved by the use of Fourier transforms (chapter 13).
If we consider the following integral equation with a displacement kernel,
∞
K(x − z)y(z) dz,
(23.17)
y(x) = f(x) + λ
−∞
the integral over z clearly takes the form of a convolution (see chapter 13).
Therefore, Fourier-transforming (23.17) and using the convolution theorem, we
obtain
√
ỹ(k) = f̃(k) + 2πλK̃(k)ỹ(k),
which may be rearranged to give
ỹ(k) =
1−
f̃(k)
√
.
2πλK̃(k)
(23.18)
Taking the inverse Fourier transform, the solution to (23.17) is given by
∞
f̃(k) exp(ikx)
1
√
dk.
y(x) = √
2π −∞ 1 − 2πλK̃(k)
If we can perform this inverse Fourier transformation then the solution can be
found explicitly; otherwise it must be left in the form of an integral.
Find the Fourier transform of the function
#
1
g(x) =
0
if |x| ≤ a,
if |x| > a.
Hence find an explicit expression for the solution of the integral equation
∞
sin(x − z)
y(x) = f(x) + λ
y(z) dz.
x−z
−∞
(23.19)
Find the solution for the special case f(x) = (sin x)/x.
The Fourier transform of g(x) is given directly by
a
a
1
1 exp(−ikx)
exp(−ikx) dx = √
=
g̃(k) = √
2π −a
2π (−ik)
−a
2 sin ka
.
π k
(23.20)
The kernel of the integral equation (23.19) is K(x − z) = [sin(x − z)]/(x − z). Using
(23.20), it is straightforward to show that the Fourier transform of the kernel is
#
π/2 if |k| ≤ 1,
(23.21)
K̃(k) =
0
if |k| > 1.
809
INTEGRAL EQUATIONS
Thus, using (23.18), we find the Fourier transform of the solution to be
#
f̃(k)/(1 − πλ) if |k| ≤ 1,
ỹ(k) =
f̃(k)
if |k| > 1.
(23.22)
Inverse Fourier-transforming, and writing the result in a slightly more convenient form,
the solution to (23.19) is given by
1
1
1
−1 √
f̃(k) exp(ikx) dk
y(x) = f(x) +
1 − πλ
2π −1
1
1
πλ
√
f̃(k) exp(ikx) dk.
(23.23)
= f(x) +
1 − πλ 2π −1
It is clear from (23.22) that when λ = 1/π, which is the only eigenvalue of the
corresponding homogeneous equation to (23.19), the solution becomes infinite, as we
would expect.
For the special case f(x) = (sin x)/x, the Fourier transform f̃(k) is identical to that in
(23.21), and the solution (23.23) becomes
1
sin x
π
1
πλ
√
y(x) =
+
exp(ikx) dk
x
1 − πλ
2π −1 2
k=1
1 exp(ikx)
πλ
sin x
+
=
x
1 − πλ 2
ix
k=−1
sin x
sin x
sin x
πλ
1
=
+
=
.
x
1 − πλ
x
1 − πλ
x
If, instead, the integral equation (23.17) had integration limits 0 and x (so
making it a Volterra equation) then its solution could be found, in a similar way,
by using the convolution theorem for Laplace transforms (see chapter 13). We
would find
f̄(s)
,
ȳ(s) =
1 − λK̄(s)
where s is the Laplace transform variable. Often one may use the dictionary of
Laplace transforms given in table 13.1 to invert this equation and find the solution
y(x). In general, however, the evaluation of inverse Laplace transform integrals
is difficult, since (in principle) it requires a contour integration; see chapter 24.
As a final example of the use of Fourier transforms in solving integral equations,
we mention equations that have integration limits −∞ and ∞ and a kernel of the
form
K(x, z) = exp(−ixz).
Consider, for example, the inhomogeneous Fredholm equation
∞
exp(−ixz) y(z) dz.
y(x) = f(x) + λ
(23.24)
−∞
The integral over z is clearly just (a multiple of) the Fourier transform of y(z),
810
23.4 CLOSED-FORM SOLUTIONS
so we can write
y(x) = f(x) +
√
2πλỹ(x).
(23.25)
If we now take the Fourier transform of (23.25) but continue to denote the
independent variable by x (i.e. rather than k, for example), we obtain
√
ỹ(x) = f̃(x) + 2πλy(−x).
(23.26)
Substituting (23.26) into (23.25) we find
√
√
y(x) = f(x) + 2πλ f̃(x) + 2πλy(−x) ,
but on making the change x → −x and substituting back in for y(−x), this gives
√
√
y(x) = f(x) + 2πλf̃(x) + 2πλ2 f(−x) + 2πλf̃(−x) + 2πλ2 y(x) .
Thus the solution to (23.24) is given by
1
f(x) + (2π)1/2 λf̃(x) + 2πλ2 f(−x) + (2π)3/2 λ3 f̃(−x) .
y(x) =
2
4
1 − (2π) λ
(23.27)
√
√
Clearly, (23.24) possesses a unique solution provided λ = ±1/ 2π or ±i/ 2π;
these are easily shown to be the eigenvalues of the corresponding homogeneous
equation (for which f(x) ≡ 0).
Solve the integral equation
2
∞
x
+λ
y(x) = exp −
exp(−ixz) y(z) dz,
2
−∞
(23.28)
where λ is a real constant. Show that the solution is unique unless λ has one of two particular
values. Does a solution exist for either of these two values of λ?
Following the argument given above, the solution to (23.28) is given by (23.27) with
f(x) = exp(−x2 /2). In order to write the solution explicitly, however, we must calculate
the Fourier transform of f(x). Using equation (13.7), we find f̃(k) = exp(−k 2 /2), from
which we note that f(x) has the special property that its functional form is identical to
that of its Fourier transform. Thus, the solution to (23.28) is given by
2
1
x
1/2
2
3/2 3
y(x) =
.
λ
+
2πλ
+
(2π)
λ
1
+
(2π)
exp
−
1 − (2π)2 λ4
2
(23.29)
√
Since λ is restricted to be real, the solution to (23.28) will be unique unless λ = ±1/ 2π,
at which points (23.29) becomes infinite. In order to find whether solutions exist for either
of these values of λ we must return to equations
(23.25) and (23.26).
√
Let us first consider the case λ = +1/ 2π. Putting this value into (23.25) and (23.26),
we obtain
y(x) = f(x) + ỹ(x),
ỹ(x) = f̃(x) + y(−x).
811
(23.30)
(23.31)
INTEGRAL EQUATIONS
Substituting (23.31) into (23.30) we find
y(x) = f(x) + f̃(x) + y(−x),
but on changing x to −x and substituting back in for y(−x), this gives
y(x) = f(x) + f̃(x) + f(−x) + f̃(−x) + y(x).
Thus, in order for a solution to exist, we require that the function f(x) obeys
f(x) + f̃(x) + f(−x) + f̃(−x) = 0.
This is satisfied if f(x) = −f̃(x), i.e. if the functional form of f(x) is √
minus the form of its
Fourier transform. We may repeat this analysis for the case λ = −1/ 2π, and, in a similar
way, we find that this time we require f(x) = f̃(x).
In our case f(x) = exp(−x2 /2), for which, as we√ mentioned above, f(x) = f̃(x).
Therefore,
√ (23.28) possesses no solution when λ = +1/ 2π but has many solutions when
λ = −1/ 2π. A similar approach to the above may be taken to solve equations with kernels
of the form K(x, y) = cos xy or sin xy, either by considering the integral over y in
each case as the real or imaginary part of the corresponding Fourier transform
or by using Fourier cosine or sine transforms directly.
23.4.3 Differentiation
A closed-form solution to a Volterra equation may sometimes be obtained by
differentiating the equation to obtain the corresponding differential equation,
which may be easier to solve.
Solve the integral equation
x
y(x) = x −
xz 2 y(z) dz.
(23.32)
0
Dividing through by x, we obtain
x
y(x)
z 2 y(z) dz,
=1−
x
0
which may be differentiated with respect to x to give
d y(x)
y(x)
= −x2 y(x) = −x3
.
dx
x
x
This equation may be integrated straightforwardly, and we find
x4
y(x)
= − + c,
ln
x
4
where c is a constant of integration. Thus the solution to (23.32) has the form
4
x
,
y(x) = Ax exp −
4
(23.33)
where A is an arbitrary constant.
Since the original integral equation (23.32) contains no arbitrary constants, neither
should its solution. We may calculate the value of the constant, A, by substituting the
solution (23.33) back into (23.32), from which we find A = 1. 812
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