Masujima M. Applied Mathematical Methods in Theoretical Physics
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186 5 Hilbert–Schmidt Theory of Symmetric Kernel
This is a statement of the completeness of the eigenfunctions; the discrete eigen-
functions {φ
n
(x), ψ
n
(y)}and the continuum eigenfunctions {φ
λ
(x), ψ
λ
(y)}together
form a complete set. We anticipate that the orthogonality of the eigenfunctions will
survive with minor modification.
We first consider the following integral:
h
0
ψ
n
(x)L
x
φ
m
(x)dx = λ
m
h
0
ψ
n
(x)φ
m
(x)dx
=
h
0
(L
T
x
ψ
n
(x))φ
m
(x)dx = λ
n
h
0
ψ
n
(x)φ
m
(x)dx,
from which, we obtain
(λ
n
− λ
m
)
h
0
ψ
n
(x)φ
m
(x)dx = 0
⇒
h
0
ψ
n
(x)φ
m
(x)dx = 0whenλ
n
= λ
m
. (5.7.31)
The eigenfunctions {φ
m
(x), ψ
n
(x)} belonging to the distinct eigenvalues are or-
thogonal to each other. Secondly, we multiply ψ
n
(x) on the completeness relation
(5.7.30) and integrate over x.Sinceλ
n
= λ
, we have by Eq. (5.7.31)
h
0
dxψ
n
(x)δ(x − y) =
m
ψ
m
(y)
h
0
dxψ
n
(x)φ
m
(x),
⇒ ψ
n
(y) =
m
ψ
m
(y)
h
0
ψ
n
(x)φ
m
(x)dx. (5.7.32)
Then we must have
h
0
ψ
n
(x)φ
m
(x)dx = δ
mn
. (5.7.33)
Thirdly, by multiplying ψ
λ
(x) on the completeness relation (5.7.30) and integrating
over x,sinceλ
= λ
n
, we have by Eq. (5.7.32)
ψ
λ
(y) =
λ
B
−∞
dλ
ψ
λ
(y)
h
0
dxψ
λ
(x)φ
λ
(x).
Then we must have
h
0
ψ
λ
(x)φ
λ
(x)dx = δ(λ
− λ
). (5.7.34)
Thus the discrete eigenfunctions {φ
m
(x), ψ
n
(x)} and the continuum eigenfunctions
{φ
λ
(x), ψ
λ
(x)} are normalized, respectively, as Eqs. (5.7.33) and (5.7.34).
5.8 Problems for Chapter 5 187
5.8
Problems for Chapter 5
5.1. (due to H. C.). Find an upper bound and a lower bound for the first
eigenvalue of
K(x, y) =
(1 − x)y,0≤ y ≤ x ≤ 1,
(1 − y )x,0≤ x ≤ y ≤ 1.
5.2. (due to H. C.). Consider the Bessel equation
(xu
)
+ λxu = 0,
with the boundary conditions
u
(0) = u(1) = 0.
Transform this differential equation to an integral equation and find
approximately the lowest eigenvalue.
5.3. Obtain an upper limit for the lowest eigenvalue of
∇
2
u + λru = 0,
where, in three dimensions,
0 < r < a and u = 0onr = a.
5.4. (due to H. C.). Consider the Gaussian kernel K(x, y)givenby
K(x, y) = e
−x
2
−y
2
, −∞ < x, y < +∞.
(a) Find the eigenvalues and the eigenfunctions of this kernel.
(b) Verify the Hilbert–Schmidt expansion of this kernel.
(c) By calculating A
2
and A
4
, we obtain the exact lowest eigenvalue.
(d) Solve the integro-differential equation
∂
∂t
φ(x, t) =
+∞
−∞
K(x, y)φ(y, t)dy,withφ(x,0)= f (x).
188 5 Hilbert–Schmidt Theory of Symmetric Kernel
5.5. Show that the boundary condition of the Sturm–Liouville system can be
replaced with
α
1
φ(0) + α
2
φ
(0) = 0andβ
1
φ(h) + β
2
φ
(h) = 0
where α
1
, α
2
, β
1
,andβ
2
are some constants and the corresponding
boundary condition on G(x, y) is replaced accordingly.
5.6. Verify the Hilbert–Schmidt expansion for the case of Direction 1 in Section
5.6, when the kernel K(x, y)isself-adjoint and square-integrable.
5.7. (due to H. C.). Reproduce all the results of Section 5.7 with Green’s
function G(x, y;λ)definedby
(L
x
− λr(x))G(x, y;λ) = δ(x − y),
with the weight function
r(x) > 0onx ∈ [0, h].
5.8. (due to H. C.). Consider the eigenvalue problem of the fourth-order
ordinary differential equation of the form
d
4
dx
4
+ 1
φ(x) =−λxφ(x), 0 < x < 1,
with the boundary conditions
φ(0) = φ
(0) = 0,
φ(1) = φ
(1) = 0.
Do the eigenfunctions form a complete set?
Hint: Check whether the differential operator L
x
defined by
L
x
≡−
d
4
dx
4
+ 1
is self-adjoint or not under the specified boundary conditions.
5.9. (due to D. M.) Show that, if
˜
λ is an eigenvalue of the symmetric kernel
K(x, y), the inhomogeneous Fredholm integral equation of the second kind,
φ(x) = f (x) +
˜
λ
b
a
K(x, y)φ(y)dy, a ≤ x ≤ b,
has no solution, unless the inhomogeneous term f (x)isorthogonaltoallof
the eigenfunctions φ(x) corresponding to the eigenvalue
˜
λ.
5.8 Problems for Chapter 5 189
Hint: You may suppose that {λ
n
} is the sequence of the eigenvalues of the
symmetric kernel K(x, y) (ordered by the increasing magnitude), with the
corresponding eigenfunctions {φ
n
(x)}. You may assume that the eigenvalue
˜
λ has the multiplicity 1. The extension to the higher multiplicity k, k ≥ 2, is
immediate.
5.10. (due to D. M.) Consider the integral equation
φ(x) = f (x) + λ
1
0
sin
2
π(x − y)
!
φ(y)dy,0≤ x ≤ 1.
(a) Solve the homogeneous equation by setting
f (x) = 0.
Determine all the eigenfunctions and eigenvalues. What is the spectral
representation of the kernel?
Hint: Express the kernel
sin
2
[π(x − y)],
which is translationally invariant, periodic, and symmetric, in terms of
the powers of
exp[π(x − y)].
(b) Find the resolvent kernel of this equation.
(c) Is there a solution to the given inhomogeneous integral equation when
f (x) = exp[imπ x], m integer,
and λ = 2?
5.11. (due to D. M.) Consider the kernel of the Fredholm integral equation of the
second kind, which is given by
K(x, y) =
3, 0 ≤ y < x ≤ 1,
2, 0 ≤ x < y ≤ 1.
(a) Find the eigenfunctions φ
n
(x) and the corresponding eigenvalues λ
n
of
the kernel.
(b) Is K(x, y) symmetric? Determine the transposed kernel K
T
(x, y), and
find its eigenfunctions ψ
n
(x) and the corresponding eigenvalues λ
n
.
190 5 Hilbert–Schmidt Theory of Symmetric Kernel
(c) Show by an explicit calculation that any φ
n
(x) is orthogonal to any
ψ
m
(x)ifm = n.
(d) Derive the spectral representation of K(x, y)intermsofφ
n
(x)and
ψ
n
(x).
5.12. (due to D. M.) Consider the Fredholm integral equation of the second kind,
φ(x) = f (x) + λ
1
0
1
2
(x + y) −
1
2
x − y
φ(y)dy,0≤ x ≤ 1.
(a) Find all nontrivial solutions φ
n
(x) and corresponding eigenvalues λ
n
for f (x) ≡ 0.
Hint: Obtain a differential equation for φ(x) with the suitable
conditions for φ(x)andφ(x)
.
(b) For the original inhomogeneous equation (f (x) = 0), will the iteration
series converge?
(c) Evaluate the series
'
n
λ
−2
n
by using an appropriate integral.
5.13. If
h
< 1, find the nontrivial solutions of the homogeneous integral
equation,
φ(x) =
λ
2π
π
−π
1 − h
2
1 − 2h cos(x − y) + h
2
φ(y)dy.
Evaluate the corresponding values of the parameter λ.
Hint: The kernel of this equation is translationally invariant. Write
cos(x − y) as a sum of two exponentials, express the kernel in terms of the
complex variable ζ = exp[i(y − x)], use the partial fractions, and then
expand each fraction in powers of ζ .
5.14. If
h
< 1, find the solution of the integral equation,
f (x) =
λ
2π
π
−π
1 − h
2
1 − 2h cos(x − y) + h
2
φ(y)dy,
where f (x) is the periodic and square-integrable known function.
Hint: Write cos(x − y) as a sum of two exponentials and express the kernel
in terms of the complex variable ζ = exp[i(y − x)].
5.15. Find the nontrivial solutions of the homogeneous integral equation,
φ(x) = λ
π
−π
1
4π
(x − y)
2
−
1
2
x − y
φ(y)dy.
5.8 Problems for Chapter 5 191
Evaluate the corresponding values of the parameter λ.Findthe
Hilbert–Schmidt expansion of the symmetric kernel displayed above.
5.16. (a) Transform the following integral equation to the integral equation
with the symmetric kernel:
φ(x) = λ
1
0
xy
2
φ(y)dy.
(b) Solve the derived integral equation with the symmetric kernel.
5.17. (a) Transform the following integral equation to the integral equation
with the symmetric kernel:
φ(x) − λ
1
0
xy
2
φ(y)dy = x + 3.
(b) Solve the derived integral equation with the symmetric kernel.
5.18. (a) Transform the following integral equation to the integral equation
with the symmetric kernel.
1
0
xy
2
φ(y)dy = 2x.
(b) Solve the derived integral equation with the symmetric kernel.
5.19. Obtain the eigenfunctions of the following kernel and establish the
Hilbert–Schmidt expansion of the kernel,
K(x, y) = 1, 0 < x, y < 1.
5.20. Obtain the eigenfunctions of the following kernel and establish the
Hilbert–Schmidt expansion of the kernel,
K(x, y) = sin x sin y,0< x, y < 2π.
5.21. Obtain the eigenfunctions of the following kernel and establish the
Hilbert–Schmidt expansion of the kernel,
K(x, y) = x + y,0< x, y < 1.
5.22. Obtain the eigenfunctions of the following kernel and establish the
Hilbert–Schmidt expansion of the kernel,
K(x, y) = exp[x + y], 0 < x, y < ln 2.
192 5 Hilbert–Schmidt Theory of Symmetric Kernel
5.23. Obtain the eigenfunctions of the following kernel and establish the
Hilbert–Schmidt expansion of the kernel,
K(x, y) = xy + x
2
y
2
,0< x, y < 1.
193
6
Singular Integral Equations of the Cauchy Type
6.1
Hilbert Problem
Suppose that rather than the discontinuity f
+
(x) − f
−
(x) across a branch cut, the
ratio of the values on either side is known. That is, suppose that we wish to find
H(z) which has branch cut on [a, b ]with
H
+
(x) = R(x)H
−
(x)onx ∈ [a, b]. (6.1.1)
Solution. Take the logarithm of both sides of Eq. (6.1.1) to obtain
ln H
+
(x) − ln H
−
(x) = ln R(x)onx ∈ [a, b]. (6.1.2)
Define
h(z) ≡ ln H(z)andr(x) ≡ ln R(x). (6.1.3)
Then we have
h
+
(x) − h
−
(x) = r(x)onx ∈ [a, b ]. (6.1.4)
Hence we can apply the discontinuity formula (1.8.5) to determine h(z):
h(z) =
1
2πi
b
a
r(x)
x − z
dx + g(z) (6.1.5)
where g(z) is an arbitrary function with no cut on [a, b], i.e.,
H (z) = G(z)exp
1
2πi
b
a
ln R(x)
x − z
dx
. (6.1.6)
We check this result. Supposing G(z) to be continuous across the cut, we have
from this result that
H
+
(x) = G(x)exp
1
2πi
P
b
a
ln R(y )
y − x
dy +
1
2
ln R(x)
, (6.1.7a)
Applied Mathematical Methods in Theoretical Physics, Second Edition. Michio Masujima
Copyright
2009 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim
ISBN: 978-3-527-40936-5
194 6 Singular Integral Equations of the Cauchy Type
H
−
(x) = G(x)exp
1
2πi
P
b
a
ln R(y)
y − x
dy −
1
2
ln R(x)
. (6.1.7b)
Here, we used the Plemelj formulas (1.8.14a) and (1.8.14b) to take the limit as
z → x from above and below. Dividing H
+
(x)byH
−
(x), we obtain
H
+
(x)H
−
(x) = exp
ln R(x)
= R(x ),
as expected.
Thus, in general, G(z) may be of any form as long as it is continuous across the
cut. In particular, we can choose G(z)soastomakeH(z) have any desired behavior
as z → a or b,bytakingitoftheform
G(z) = (a − z)
n
(b − z)
m
, (6.1.8)
wherein G(z) = exp(g(z)) has no branch points on [a, b]. In particular, we can always
take G(z)oftheformabovewithn and m integers, and choose n and m to obtain a
desired behavior in H(z)asz → a, z → b,andz →∞.
We now address the inhomogeneous Hilbert problem.
Inhomogeneous Hilbert problem:
Obtain (z) which has a branch cut on [a, b]with
+
(x) = R(x)
−
(x) + f (x)onx ∈ [a, b]. (6.1.9)
Solution. To solve the inhomogeneous problem, we begin with the solution to the
corresponding homogeneous Hilbert problem,
H
+
(x) = R(x)H
−
(x)onx ∈ [a, b], (6.1.10)
whose solution, we know, is given by
H (z) = G(z)exp
1
2πi
b
a
ln R(x)
x − z
dx
. (6.1.11)
We then seek for a solution of the form
(
z
)
= H (z)K(z); (6.1.12)
hence K(z)mustsatisfy
H
+
(x)K
+
(x) = R(x)H
−
(x)K
−
(x) + f (x) (6.1.13)
or, from Eq. (6.1.10), H
+
(x)K
+
(x) = H
+
(x)K
−
(x) + f (x). Therefore, we have
K
+
(x) − K
−
(x) = f (x)H
+
(x). (6.1.14)
6.1 Hilbert Problem 195
The problem reduces to an ordinary application of the discontinuity theorem
provided that H(z) is chosen so that f (x)H
+
(x)islesssingularthanapoleatboth
end points. With such a choice for H(z), we then get
K(z) =
1
2πi
b
a
f (x)
H
+
(x)(x − z)
dx + L(z), (6.1.15)
where L(z) has no branch points on [a, b ]. We thus fi nd the solution to the original
inhomogeneous problem to be
(
z
)
= H
(
z
)
K
(
z
)
, (6.1.16)
i.e.,
(
z
)
= G
(
z
)
exp
1
2πi
b
a
ln R(x)
x − z
dx
1
2πi
b
a
f (x)
H
+
(x)(x −z)
dx + L(z)
,
(6.1.17)
with L(z) arbitrary and G(z) chosen such that f (x)H
+
(x)isintegrableon[a, b].
Neither L(z)norG(z) can have branch points on [a, b].
Example 6.1. Find (z) s atisfying
+
(x) =−
−
(
x
)
+f (x)onx ∈ [a, b]. (6.1.18)
Solution.
Step 1. Solve the corresponding homogeneous problem first:
H
+
(x) =−H
−
(x). (6.1.19)
Taking the logarithm on both sides we obtain
ln H
+
(x) = ln H
−
(x) + ln(−1).
Setting
h(z) = ln H(z),
we have
h
+
(x) − h
−
(x) = iπ. (6.1.20)
Then we can solve for h(z) with the use of the discontinuity formula as
h(z) =
1
2πi
b
a
iπ
x − z
dx + g(z) =
1
2
ln
b − z
a − z
+ g(z),