Nair S. Advanced Topics in Applied Mathematics: For Engineering and the Physical Sciences
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Integral Equations 89
We define the spatial norm, spatial inner product, and the temporal
expectation value by
v
2
=
v
2
dx, v, u=
vudx, E{v}=
1
T
T
0
vdt, (2.192)
respectively. In POD, we wish to express the data, v,as
v(x, t) =
N
i
f
i
(t)u
i
(x), (2.193)
where both, f
i
(t) and the spatial basis functions u
i
(x), have to be found.
We assume all the functions involved are continuous, and the integral
over space and the integral over time commute. To start the process,
let us ask ourselves the following: What is the single term f (t)u(x),
which is the best approximation for v(x, t)?
The criterion for finding u is the requirement that the expectation
value of the square-error E has to be a minimum. The square-error E
is defined as
E =
[v(x, t) −f (t)u(x)]
2
dx, (2.194)
and its expectation value, as
E{E}=
1
T
T
0
[v(x, t) −f (t)u(x)]
2
dx dt. (2.195)
We assume our single basis function is normalized. That is,
u=1. (2.196)
The extremization of E{E} with the constraint u=1 can be done
using the modified functional
E{E
∗
}=E{E}+µ[u
2
−1], (2.197)
where µ is a Lagrange multiplier. Explicitly,
E{E
∗
}=
1
T
T
0
[v(x, t) −f (t)u(x)]
2
dx dt +µ[u
2
−1]. (2.198)
90 Advanced Topics in Applied Mathematics
The first variation of this functional using f −→ f +δf and u −→ u +δu,
gives
1
T
T
0
2[v −fu](−uδf )dx dt = 0, (2.199)
1
T
T
0
2[v −fu](−f δu)dx dt +2
uδudx =0. (2.200)
Noting that f is a function of only t,andu is that of only x,wehave
f =v ,u, (2.201)
1
T
T
0
[vf −f
2
u]dt −µu =0, (2.202)
where we have used u=1. Using Eq. (2.201)in(2.202),
E{v(x, t)v(x
,t)}u(x
)dx
=[E{f
2
}+µ]u(x). (2.203)
This relation shows that the function u gets mapped into itself through
a symmetric integral operator R(x,x
) defined by
R(x,x
) =E{v(x,t)v(x
,t)}, (2.204)
and the associated eigenvalue problem is
R(x,x
)u(x
)dx
=
1
λ
u(x). (2.205)
Equation (2.203) becomes
1
λ
=E{f
2
}+µ. (2.206)
Integral Equations 91
Let us compute the first term on the right-hand side.
E{f
2
}=E{
v(x, t)u(x)dx}
2
=E{
v(x, t)u(x)dx
v(x
,t)u(x
)dx
}
=
E{v(x, t)v(x
,t)}u(x
)dx
u(x)dx
=
u
2
(x)
λ
dx
=
1
λ
. (2.207)
Thus, the Lagrange multiplier µ turns out to be zero. Also, as E{f
2
}> 0,
the eigenvalue is positive.
From the Hilbert-Schmidt theory, our eigenvalue problem with a
real symmetric continuous kernel gives a set of N orthonormal eigen-
functions u
i
with eigenvalues λ
i
, where N may be infinite. With this,
we extend our approximation to
v(x, t) =
N
i
f
i
(t)u
i
(x), (2.208)
where f
i
=v,u
i
and the orthonormal eigenfunctions u
i
satisfy
u
i
(x) =λ
i
R(x,x
)u
i
(x
)dx
. (2.209)
The expectation value of the error becomes
E{E}=E{
[v −
f
i
u
i
]
2
dx}, (2.210)
92 Advanced Topics in Applied Mathematics
which may be broken into the three integrals:
I
1
=E{
v(x, t)v(x,t)dx},
I
2
=−2
E{
v(x, t)f
i
u
i
(x)dx},
I
3
=
E{
f
2
i
u
2
i
(x)dx},
where in I
3
, we have anticipated the orthogonality of u
i
.
With the expansion of the kernel using Mercer’s theorem stated in
Eq. (2.100), we have
E{v(x, t)v(x
,t)}=R(x,x
) =
u
i
(x)u
i
(x
)
λ
i
,
I
1
=
R(x,x)dx =
1
λ
i
,
I
2
=−2
E{f
2
i
}=−2
1
λ
i
,
I
3
=
E{f
2
i
}=
1
λ
i
. (2.211)
These integrals add up to zero, and our choice of basis functions
makes the expectation value of the error zero, provided the complete
sequence of eigenfunctions are used for the expansion.
In applications, an approximate representation of the data using
a finite number of eigenfunctions is more practical. Also, in order
to improve numerical accuracy the spatial mean value of the data is
removed before finding the low dimensional representation through
the proper orthogonal decomposition.
SUGGESTED READING
Abramowitz, M. and Stegun, I. (1965). Handbook of Mathematical Functions
(National Bureau of Standards), Dover.
Barber, J. R. (2002). Elasticity, 2nd ed., Kluwer.
Brebbia, C. A. (1978). The Boundary Element Method for Engineers, Pentech
Press, London.
Integral Equations 93
Chatterjee, A. (2000). An introduction to the proper orthogonal decomposi-
tion, Current Science, Vol. 78, No. 7, pp. 808–817.
Courant, R. and Hilbert, D. (1953). Methods of Mathematical Physics, Vol. 2,
Interscience.
Hartmann, F. (1989). Introduction to Boundary Elements, Springer-Verlag.
Hildebrand, F. B. (1965). Methods of Applied Mathematics, 2nd ed., Prentice-
Hall.
Holmes, P., Lumley, J. L., and Berkooz, G. (1996). Turbulence, Coherent
Structures, Dynamical Systems and Symmetry, Cambridge University Press.
Pozrikides, C. (2002). A Practical Guide to Boundary Element Methods,
Chapman & Hall/CRC.
Rahman, M. (2007). Integral Equations and Their Applications, WIT Press.
Sneddon, I. N. (1966). Mixed Boundary-Value Problems in Potential Theory,
North Holland.
Tricomi, F. G. (1957). Integral Equations, Interscience.
EXERCISES
2.1 Using differentiation, convert the integral equation
u(x) =
1
2
1
0
|x −ξ|u(ξ )dξ +
x
2
2
into a differential equation. Obtain the needed boundary condi-
tions, and solve the differential equation.
2.2 Convert
u(x) =λ
1
−1
e
−|x−ξ |
u(ξ )dξ
into a differential equation. Obtain the required number of
boundary conditions.
2.3 Solve the integral equation
u(x) =e
x
+λ
1
−1
e
(x−ξ)
u(ξ )dξ ,
and discuss the conditions on λ for a unique solution.
2.4 Solve
u(x) =x +
1
0
(1 −xξ)u(ξ )dξ.
94 Advanced Topics in Applied Mathematics
2.5 Solve
u(x) =sin x +
π
0
cos(x −ξ)u(ξ )dξ.
2.6 Obtain the eigenvalues and eigenfunctions of the equation
u(x) =λ
2π
0
cos(x −ξ)u(ξ )dξ.
2.7 Find the eigenvalues and eigenfunctions of
u(x) =λ
1
0
(1 −xξ)u(ξ )dξ.
2.8 From the differential equation and the boundary conditions
obtainedforExercise2.2,findthevaluesofλfortheexistence
of non-trivial solutions.
2.9 Show that the equation
u(x) =λ
1
0
|x −ξ|u(ξ )dξ
has an infinite number of eigenvalues and eigenfunctions.
2.10 Obtain the values of λ and a for the equation
u(x) =λ
1
0
(x −ξ)u(ξ )dξ +a +x
2
to have (a) a unique solution, (b) a nonunique solution, and
(c) no solution.
2.11 Obtain the resolvent kernel for
u(x) =λ
2π
0
e
in(x−ξ)
u(ξ )dξ +f (x).
2.12 Find the resolvent kernel for
u(x) =λ
π
0
cos(x −ξ)u(ξ )dξ +f (x).
2.13 Show that for a Fredholm equation
u(x) =λ
b
a
k(x,ξ)u(ξ)dξ +f (x),
Integral Equations 95
the resolvent kernel g(x, ξ) satisfies
g(x, ξ)=k(x,ξ)+λ
b
a
k(x,η)g(η,ξ)dη.
Thus, the resolvent kernel satisfies the integral equation for u
when the forcing function f is replaced by k with ξ and λ kept as
parameters.
2.14 Demonstrate the preceding result when k =1−2xξ in the domain
0 < x, ξ<1.
2.15 For the Volterra equation,
u(x) =
x
0
(1 −xξ)u(ξ )dξ +1,
starting with u
(0)
=1, obtain iteratively, u
(2)
.
2.16 For the integral equation of the first kind
2π
0
sin(x +ξ)u(ξ )dξ = sinx,
discuss the consequence of assuming
u(x) =
n=0
a
n
cosnx +
n=1
b
n
sinnx,
where the constants, a
n
and b
n
, are unknown.
2.17 With the quadratic forms,
J
1
[u]=
b
a
b
a
k(x,ξ)u(x)u(ξ)dξ dx, J
2
[u]=
b
a
u
2
dx,
where u is a smooth function in (a,b), show that the functions u,
which extremize
J[u]=
J
1
[u]
J
2
[u]
,
are the eigenfunctions of [k(x,ξ)+k(ξ ,x)]/2. Also show that the
extrema of J correspond to the reciprocal of the eigenvalues of
k(x,ξ).
96 Advanced Topics in Applied Mathematics
2.18 In the preceding problem, assuming k is symmetric, compute
J[v] if
v(x) =
n
a
n
u
n
(x),
where u
n
are normalized eigenfunctions.
2.19 The generalized Abel equation,
x
0
u(ξ )dξ
(x −ξ)
α
=f (x),
is solved by multiplying both sides by (η −x)
α−1
and integrating
with respect to x from 0 to η. Implement this procedure, and
discuss the allowable range for the index, α.
2.20 Solve the singular equation
x
0
u(ξ )dξ
(x
2
−ξ
2
)
=f (x),
using a change of the independent variable.
2.21 If an elastic, 3D half space (z > 0) is subjected to an axi-symmetric
z-displacement, w(r), where r =
x
2
+y
2
, we need to solve the
integral equation for the distributed pressure on the horizontal
surface, z =0,
r
0
g(ρ)dρ
r
2
−ρ
2
=
µ
1 −ν
w(r),0< r < a,
where µ is the shear modulus, ν is the Poisson’s ratio, and a is the
contact radius. Obtain the pressure distribution g(r) (see Barber
(2002)).
2.22 In the preceding problem, if the contact displacement is due to
a rigid sphere of radius R pressing against the elastic half space,
we assume
w(r) =d −
r
2
2R
,
where d is the maximum indentation. Obtain the pressure distri-
bution, the value of the contact radius a, and the total vertical
force.
Integral Equations 97
2.23 Consider the integral equation
u(x) =
1
0
k(x,ξ)u(ξ)dξ +x
2
,
where
k(x,ξ)=
x(ξ −1), x <ξ,
ξ(x −1), x >ξ.
Obtain an exact solution to this equation. Using the approximate
kernel Ax(1 −x), find A using the method of least square error.
Obtain an approximate solution for u. Compare the values of the
exact and approximate solutions at x =0.5.
2.24 Consider the differential equation
u
+u =x, u(0) = 0, u(1) =0.
Find an exact solution. Obtain a finite difference solution by
dividing the domain into four equal intervals.
Convert the differential equation into an integral equation using
the Green’s function for the operator L = d
2
/dx
2
. Obtain a
numerical solution of the integral equation by dividing the
domain into four intervals and using the trapezoidal rule and the
collocation method. Compare the results with the exact solution.
2.25 Consider the nonlinear integral equation
u(x) −λ
1
0
u
2
(ξ)dξ = 1.
Show that for λ<1/4 this equation has two solutions and for
λ>1/4 there are no real solutions. Also show that one of these
solutions is singular at λ =0 and two solutions coalesce at λ =1/4.
Sketchthesolutionsasfunctionsofλ.(BasedonTricomi(1957).)
3
FOURIER TRANSFORMS
The method of Fourier transforms is a powerful technique for solving
linear, partial differential equations arising in engineering and physics
when the domain is infinite or semi-infinite. This is an extension of the
Fourier series, which is applicable to periodic functions defined on an
interval −a < x < a. First, let us review the Fourier series and extend
it to infinite domains in a heuristic form.
3.1 FOURIER SERIES
If f (x) is a continuous function on −a < x < a, we expand it in terms of
the orthogonal functions,
{1,cos(π x/a),cos(2π x/a),...,sin(πx/a),sin(2πx/a), ...}
as
f (x) =A
0
+
∞
n=1
[A
n
cos(nπx/a) +B
n
sin(nπx/a)]. (3.1)
Taking the inner product of this equation with each member of the
orthogonal set (basis), we obtain
A
0
=
1
2a
a
−a
f (t)dt, (3.2)
A
n
=
1
a
a
−a
f (t) cos(nπt/a) dt, (3.3)
B
n
=
1
a
a
−a
f (t) sin(nπt/a) dt. (3.4)
98