Nair S. Advanced Topics in Applied Mathematics: For Engineering and the Physical Sciences
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Green’s Functions 49
1.5 Convert the equation
Lu =
d
2
u
dx
2
+x
2
du
dx
+2u = f
into the Sturm-Liouville form.
1.6 Find the adjoint system for
xu
+u
+u =0, u(1) =0, u(2) +u
(2) =0.
1.7 Solve the differential system
u
+u
−2u = x
2
, u(0) =0, u
(1) =0.
1.8 Solve the differential system
(x
2
u
)
−n(n +1)u =0, u(0) =0, u(1) =1.
1.9 Convert the following system to one with homogeneous bound-
ary conditions:
(x
2
u
)
−n(n +1)u =0, u(0) =0, u(1) =1.
1.10 Obtain the Green’s function for the equation
(xu
)
−
9
x
u =f (x), u(0) =0, u(1) =0.
Using the Green’s function, find an explicit solution when f (x) =
x
n
. Check if there are any values for the integer n for which your
solution does not satisfy the boundary conditions.
1.11 Find the Green’s function for
u
+ω
2
u =f (x), u(0) =0, u(π ) =0.
Examine the special cases ω =n, n =0,1,....
50 Advanced Topics in Applied Mathematics
1.12 Express the equation
u
−2u
+u =f (x), u(0) =0, u(1) =0,
in the self-adjoint form. Obtain the solution using the Green’s
function when f (x) =e
x
.
1.13 Transform the equation
xu
+2u
=f (x); u
(0) =0, u(1) =0,
into the self-adjoint form. Find the Green’s function and express
the solution in terms of f (x). State the restrictions on f (x) for the
solution to exist.
1.14 Find the Green’s function for
x
2
u
−xu
+u =f (x), u(0) =0, u(1) =0.
1.15 Using the self-adjoint form of the differential equation
x
2
u
+3xu
−3u =f (x), u(0) =0, u(1) =0,
find the Green’s function and obtain an explicit solution when
f (x) =x.
1.16 Solve the equation
x
2
u
+3xu
=x
2
, u(1) =1, u(2) =2,
using the Green’s function.
1.17 For the problem
Lu =u
+u
=0, u(0) =0, u
(1) =0,
obtain the adjoint system. Solve the eigenvalue problems,
Lu =λu, L
∗
v = λv,
and show that their eigenfunctions are bi-orthogonal.
Green’s Functions 51
1.18 By solving the nonhomogeneous problem
u
=δ
(x −ξ), u(0) =0, u(1) =0,
where
δ
(x) =
0, |x| >,
1
2
, |x| <,
in three parts:
(a) 0 < x <ξ−,
(b) ξ −<x <ξ+,and
(c) ξ +<x < 1,
show that, in the limit → 0, we recover the Green’s function.
1.19 Expanding
g(x, ξ)=
x(ξ −1), x <ξ,
ξ(x −1), x >ξ,
in terms of u
n
=
√
2sinnπx as a Fourier series, show that
g(x, ξ)=
n=1
u
n
(x)u
n
(ξ)
λ
n
, λ
n
=−π
2
n
2
.
1.20 Obtain the Green’s function for
κ∇
2
u(x
1
,x
2
) =v
1
∂u
∂x
1
+v
2
∂u
∂x
2
,
where v
1
and v
2
are constants, by transforming the dependent
variable.
1.21 The anisotropic Laplace equation in a two-dimensional infinite
domain is given by
k
2
1
∂
2
u
∂x
2
1
+k
2
2
∂
2
u
∂x
2
2
=0.
Find the Green’s function for this equation.
1.22 In a semi-infinite medium, −∞< x < ∞,0< y < ∞, the pressure
fluctuations satisfy the wave equation
∇
2
p =
1
c
2
∂
2
p
∂t
2
,
52 Advanced Topics in Applied Mathematics
where c is the wave speed and t is time. If the boundary, y =0, is
subjected to a pressure p =P
0
δ(x)h(t), show that
p(x,y,t) =A
y
r
2
t
√
t
2
−k
2
r
2
h(t −kr),
where r =
x
2
+y
2
and k = 1/c, is a solution of the wave
equation. Evaluate the constant A using equilibrium of the
medium in the neighborhood of the applied load. Hint: Use polar
coordinates.
1.23 For the two-dimensional wave equation
∇
2
u =
1
c
2
∂
2
u
∂t
2
,
steady-state solutions are obtained using u(x,y, t) = v(x,y)e
−it
,
where v satisfies the Helmholtz equation,
Lv = 0, L =∇
2
+k
2
, k = /c.
Show that the Hankel functions H
(1)
0
(kr) and H
(2)
0
(kr) satisfy
Lg = δ(x,y) when r =
x
2
+y
2
= 0. Examine their asymptotic
forms for kr << 1 and for kr >> 1, using the results shown in
AbramowitzandStegun(1965)andselectmultiplicationcon-
stants A and B to have g = AH
(1)
0
or g = BH
(2)
0
by comparing
the asymptotic form with the Green’s function for the Laplace
operator (k → 0). Show that H
(1)
0
corresponds to an outgoing
wave and H
(2)
0
to an incoming wave.
1.24 Show that
g =−
e
±ikr
4πr
,
satisfies the Helmholtz equation
∇
2
g +k
2
g = δ(x −ξ),
in a 3D infinite domain, with r =|x −ξ|. Assuming the Helmholtz
equation is obtained from the wave equation by separating
Green’s Functions 53
the time dependence using a factor e
−it
, show that (±) signs
correspond to outgoing and incoming waves, respectively.
1.25 Consider a volume V enclosed by the surface S in 3D. From
∇
2
u +k
2
u =f , ∇
2
g +k
2
g = δ(x −ξ),
obtain the solution
u(ξ) =
V
g(ξ ,x)f (x)dV −
S
g
∂u
∂n
−u
∂g
∂n
dS.
1.26 In the previous problem, assuming V is a sphere of radius R
centered at x = 0 and x is a point on its surface, and g is the
Green’s function for the 3D infinite space, show that
u(ξ) =
1
4π
S
e
ikr
r
∂u
∂n
−u
∂
∂n
e
ikr
r
dS,
if f =0. If the included angle between x and x −ξ is ψ, show that
dr/dn =cosψ. Also show that as R →∞, r →R and R
2
(1−cos ψ)
is finite and for the surface integral to exist
r
∂u
∂r
−iku
→0, and u →0.
These are known as the Sommerfeld radiation conditions.
1.27 By reconsidering the previous problem, show that the surface
integral exists if the less restricted condition
r
∂u
∂r
−iku +
u
r
→0
is satisfied.
1.28 A wedge-shaped 2D domain has the boundaries: y =0andy =x.
Obtain the Green’s function for the Poisson equation, ∇
2
u =
f (x, y), for this domain if u(x,0) = 0andu(x,x) = 0. Use the
method of images for your solution.
54 Advanced Topics in Applied Mathematics
1.29 A unit circle is mapped into a cardioid by the transformation
w = c(1 +z)
2
,
where c is a real number. Obtain the Green’s function, g, for a
domain in the shape of a cardioid with g = 0 on the boundary.
1.30 The steady-state temperature in a semi-infinite plate satisfies
∇
2
u(x,y) = 0, −∞< x < ∞,0< y < ∞.
On the boundary, y = 0, the temperature is given as u = f (x).
Obtain the temperature distribution in the domain using the
Green’s function.
1.31 Show by substitution that the solutions of
[(1 −x
2
)u
]
−
h
2
1 −x
2
u =0
are
u =
1 −x
1 +x
±h/2
.
1.32 Obtain the Green’s function for the above operator when h = 0
if the boundary conditions are
u(−1) =finite, u(1) =finite.
1.33 If h =0 in the preceding problem, show that
U(x) =
1
√
2
is a normalized solution of the above homogeneous equation,
which satisfies both the boundary conditions. Obtain the gener-
alized Green’s function for this case.
1.34 Find the generalized Green’s function for
u
=f (x), u(1) = u(−1), u
(1) =u
(−1).
Green’s Functions 55
1.35 To solve the self-adjoint problem
Lu =f (x),
with homogeneous boundary conditions at x = a and x = b,we
use a generalized Green’s function g which satisfies
Lg = δ(x −ξ)−U(x)U(ξ ),
with homogeneous boundary conditions, where U is the normal-
ized solution of the homogeneous equation satisfying the same
boundary conditions. The solution is written as
u(x) =AU(x) +
b
a
g(x, ξ)f (ξ) dξ.
By operating on this equation using L, show that u is the required
solution.
1.36 For the equation
Lu =u
+u
=0, u
(0) =u
(1) =0,
obtain the adjoint equation and its boundary conditions. Obtain
normalized homogeneous solutions of LU = 0andL
∗
U
∗
= 0.
Construct generalized Green’s functions g and g
∗
.
2
INTEGRAL EQUATIONS
An equation involving the integral of an unknown function is called an
integral equation. The unknown function itself may appear explicitly
in an integral equation along with its integral. If the derivatives of
the unknown are also present, we have what are known as integro-
differential equations. Here are some examples of integral equations:
u(x) −
1
0
(1 +xξ)u(ξ )dξ =x
2
, (2.1)
π
0
sin(x +ξ)u(ξ )dξ =cos(x), (2.2)
x
0
u(ξ ) dξ
√
x −ξ
=f (x). (2.3)
2.1 CLASSIFICATION
One way of classifying integral equations is based on the explicit pres-
ence of the unknown function. Integral equations of the first kind do
not have the unknown function present and these equations have the
form
b(x)
a(x)
k(x,ξ)u(ξ) dξ = f (x), (2.4)
where k(x,ξ) is called the kernel of the integral equation. If
k(x,ξ)= k(ξ ,x), (2.5)
the kernel is called a symmetric kernel.
56
Integral Equations 57
An integral equation of the second kind will have the unknown
explicitly in it, for example
u(x) −
b(x)
a(x)
k(x,ξ)u(ξ) dξ = f (x). (2.6)
The integral equations of the first and second kind are further classified
as Fredholm type and Volterra type based on the nature of the limits
of integration. If the limits are constants, such as
u(x) −
b
a
k(x,ξ)u(ξ) dξ = f (x), (2.7)
we have a Fredholm equation of the second kind. If the upper or
lower limit depends on the independent variable, x, we have Volterra
equations.
Unlike differential equations, there are no separate conditions such
as the boundary conditions in the integral equation formulations. In
certain problems involving infinite domains, based on the physics,
conditions on the growth of the unknown as |x|→∞may be imposed.
We assume the kernel, k(x,ξ), is continuous in x and ξ. The Volterra
equation,
u(x) −
x
a
k(x,ξ)u(ξ) dξ = f (x), (2.8)
may be written in the Fredholm form,
u(x) −
b
a
¯
k(x,ξ)u(ξ) dξ = f (x), (2.9)
provided
¯
k(x,ξ)=
k(x,ξ), ξ<x,
0, ξ>x.
(2.10)
The continuity of the kernel now requires,
k(x,x) = 0. (2.11)
58 Advanced Topics in Applied Mathematics
Integral equations with discontinuous kernels are called singular inte-
gral equations. We will consider these later in this chapter. In the
preceding equations, if the forcing function f (x) = 0, the equations
are homogeneous. The foregoing examples are all linear integral
equations. For homogeneous, linear integral equations, if u
1
and u
2
are solutions, u
1
+u
2
and cu
1
are also solutions. Here c is any constant.
In higher dimensions, with
x = x
1
,x
2
,..., x
n
, ξ =ξ
1
,ξ
2
,..., ξ
n
, dξ =dξ
1
dξ
2
...dξ
n
, (2.12)
we can have
u(x) −
k(x,ξ)u(ξ ) dξ =f (x), (2.13)
where is the domain of integration. Using u and f as vector-valued
functions and k as a matrix function, we may also extend this to a
coupled system of integral equations.
2.2 INTEGRAL EQUATION FROM DIFFERENTIVAL
EQUATIONS
It is always possible to convert a differential equation into an integral
equation. However, in general, it is not possible to convert an integral
equation into a differential equation.
Consider a differential equation of the form
Lu =h, (2.14)
with homogeneous boundary conditions. If we know the Green’s func-
tion for the operator, L, we may formally write a solution for u in terms
of the forcing function, h. As we know, this is not always possible.
However, we may split the operator, L,as
L =L
1
+L
2
, (2.15)