Exercises

23.8 EXERCISES
thus Hermitian. In order to solve this inhomogeneous equation using SH theory, however,
we must first find the eigenvalues and eigenfunctions of the corresponding homogeneous
equation.
In fact, we have considered the solution of the corresponding homogeneous equation
(23.13) already, in subsection 23.4.1, where we found that it has two eigenvalues λ1 = 2/π
and λ2 = −2/π, with eigenfunctions given by (23.16). The normalised eigenfunctions are
1
y1 (x) = √ (sin x + cos x)
π
and
1
y2 (x) = √ (sin x − cos x)
π
(23.58)
and are easily shown to obey the orthonormality condition (23.49).
Using (23.54), the solution to the inhomogeneous equation (23.57) has the form
y(x) = a1 y1 (x) + a2 y2 (x),
(23.59)
where the coefficients a1 and a2 are given by (23.53) with f(x) = sin(x + α). Therefore,
using (23.58),
√
π
1
π
1
√ (sin z + cos z) sin(z + α) dz =
a1 =
(cos α + sin α),
1 − πλ/2 0
2 − πλ
π
√
π
1
1
π
√ (sin z − cos z) sin(z + α) dz =
a2 =
(cos α − sin α).
1 + πλ/2 0
2 + πλ
π
Substituting these expressions for a1 and a2 into (23.59) and simplifying, we find that
the solution to (23.57) is given by
1
y(x) =
sin(x + α) + (πλ/2) cos(x − α) . 1 − (πλ/2)2
23.8 Exercises
23.1
Solve the integral equation
∞
cos(xv)y(v) dv = exp(−x2 /2)
0
23.2
for the function y = y(x) for x > 0. Note that for x < 0, y(x) can be chosen as
is most convenient.
Solve
∞
a
f(t) exp(−st) dt = 2
.
a + s2
0
23.3
Convert
x
f(x) = exp x +
(x − y)f(y) dy
0
into a differential equation, and hence show that its solution is
(α + βx) exp x + γ exp(−x),
23.4
where α, β and γ are constants that should be determined.
Use the fact that its kernel is separable, to solve for y(x) the integral equation
π
sin(x + z)y(z) dz.
y(x) = A cos(x + a) + λ
0
[ This equation is an inhomogeneous extension of the homogeneous Fredholm
equation (23.13), and is similar to equation (23.57). ]
819
INTEGRAL EQUATIONS
23.5
Solve for φ(x) the integral equation
1 n n x
y
φ(y) dy,
φ(x) = f(x) + λ
+
y
x
0
where f(x) is bounded for 0 < x < 1 and − 12 < n < 12 , expressing your answer
1
in terms of the quantities Fm = 0 f(y)y m dy.
(a) Give the explicit solution when λ = 1.
(b) For what values of λ are there no solutions unless F±n are in a particular
ratio? What is this ratio?
23.6
Consider the inhomogeneous integral equation
b
f(x) = g(x) + λ
K(x, y)f(y) dy,
a
for which the kernel K(x, y) is real, symmetric and continuous in a ≤ x ≤ b,
a ≤ y ≤ b.
(a) If λ is one of the eigenvalues λi of the homogeneous equation
b
fi (x) = λi
K(x, y)fi (y) dy,
a
prove that the inhomogeneous equation can only a have non-trivial solution
if g(x) is orthogonal to the corresponding eigenfunction fi (x).
(b) Show that the only values of λ for which
1
f(x) = λ
xy(x + y)f(y) dy
0
has a non-trivial solution are the roots of the equation
λ2 + 120λ − 240 = 0.
(c) Solve
1
f(x) = µx2 +
2xy(x + y)f(y) dy.
0
23.7
The kernel of the integral equation
b
ψ(x) = λ
K(x, y)ψ(y) dy
a
has the form
K(x, y) =
∞
hn (x)gn (y),
n=0
where the hn (x) form a complete orthonormal set of functions over the interval
[a, b].
(a) Show that the eigenvalues λi are given by
|M − λ−1 I| = 0,
where M is the matrix with elements
b
gk (u)hj (u) du.
Mkj =
a
If the corresponding solutions are ψ (i) (x) =
for an(i) .
820
∞
n=0
an(i) hn (x), find an expression
23.8 EXERCISES
(b) Obtain the eigenvalues and eigenfunctions over the interval [0, 2π] if
∞
1
cos nx cos ny.
n
n=1
K(x, y) =
23.8
By taking its Laplace transform, and that of xn e−ax , obtain the explicit solution
of
x
(x − u)eu f(u) du .
f(x) = e−x x +
0
23.9
Verify your answer by substitution.
For f(t) = exp(−t2 /2), use the relationships of the Fourier transforms of f (t) and
tf(t) to that of f(t) itself to find a simple differential equation satisfied by f̃(ω),
the Fourier transform of f(t), and hence determine f̃(ω) to within a constant.
Use this result to solve the integral equation
∞
2
e−t(t−2x)/2 h(t) dt = e3x /8
−∞
23.10
for h(t).
Show that the equation
f(x) = x−1/3 + λ
∞
f(y) exp(−xy) dy
0
has a solution of the form Axα + Bxβ . Determine the values of α and β, and show
that those of A and B are
1
1−
23.11
and
λ2 Γ( 31 )Γ( 32 )
λΓ( 32 )
,
1 − λ2 Γ( 13 )Γ( 32 )
where Γ(z) is the gamma function.
At an international ‘peace’ conference a large number of delegates are seated
around a circular table with each delegation sitting near its allies and diametrically
opposite the delegation most bitterly opposed to it. The position of a delegate is
denoted by θ, with 0 ≤ θ ≤ 2π. The fury f(θ) felt by the delegate at θ is the sum
of his own natural hostility h(θ) and the influences on him of each of the other
delegates; a delegate at position φ contributes an amount K(θ − φ)f(φ). Thus
2π
f(θ) = h(θ) +
K(θ − φ)f(φ) dφ.
0
Show that if K(ψ) takes the form K(ψ) = k0 + k1 cos ψ then
f(θ) = h(θ) + p + q cos θ + r sin θ
23.12
and evaluate p, q and r. A positive value for k1 implies that delegates tend to
placate their opponents but upset their allies, whilst negative values imply that
they calm their allies but infuriate their opponents. A walkout will occur if f(θ)
exceeds a certain threshold value for some θ. Is this more likely to happen for
positive or for negative values of k1 ?
x
By considering functions of the form h(x) = 0 (x − y)f(y) dy, show that the
solution f(x) of the integral equation
1
|x − y|f(y) dy
f(x) = x + 12
0
satisfies the equation f (x) = f(x).
821
INTEGRAL EQUATIONS
By examining the special cases x = 0 and x = 1, show that
f(x) =
23.13
2
[(e + 2)ex − ee−x ].
(e + 3)(e + 1)
The operator M is defined by
Mf(x) ≡
∞
K(x, y)f(y) dy,
−∞
where K(x, y) = 1 inside the square |x| < a, |y| < a and K(x, y) = 0 elsewhere.
Consider the possible eigenvalues of M and the eigenfunctions that correspond
to them; show that the only possible eigenvalues are 0 and 2a and determine the
corresponding eigenfunctions. Hence find the general solution of
∞
K(x, y)f(y) dy.
f(x) = g(x) + λ
−∞
23.14
For the integral equation
y(x) = x−3 + λ
b
x2 z 2 y(z) dz,
a
23.15
show that the resolvent kernel is 5x2 z 2 /[5 − λ(b5 − a5 )] and hence solve the
equation. For what range of λ is the solution valid?
Use Fredholm theory to show that, for the kernel
K(x, z) = (x + z) exp(x − z)
over the interval [0, 1], the resolvent kernel is
R(x, z; λ) =
exp(x − z)[(x + z) − λ( 21 x + 12 z − xz − 13 )]
,
1 2
1 − λ − 12
λ
and hence solve
1
(x + z) exp(x − z) y(z) dz,
y(x) = x2 + 2
0
23.16
1
expressing your answer in terms of In , where In = 0 un exp(−u) du.
This exercise shows that following formal theory is not necessarily the best way
to get practical results!
(a) Determine the eigenvalues λ± of the kernel K(x, z) = (xz)1/2 (x1/2 + z 1/2 ) and
show that the corresponding eigenfunctions have the forms
√
√
y± (x) = A± ( 2x1/2 ± 3x),
√
where A2± = 5/(10 ± 4 6).
(b) Use Schmidt–Hilbert theory to solve
1
K(x, z)y(z) dz.
y(x) = 1 + 52
0
(c) As will have been apparent, the algebra involved in the formal method used
in (b) is long and error-prone, and it is in fact much more straightforward
to use a trial function 1 + αx1/2 + βx. Check your answer by doing so.
822