Stone M., Goldbart P. Mathematics for Physics: A Guided Tour for Graduate Students

Подождите немного. Документ загружается.

8.5 Further exercises and problems 305

Example:Inn dimensions, the “l = 0” part of the Laplace operator is

(n − 1)

This is formally self adjoint with respect to the natural inner product

ψ, χ



∞

n−1

∗

χ dr. (8.206)

The zero eigenvalue solutions are ψ

(r) = 1 and ψ

(r) = r

2−n

. The second of these

ceases to be normalizable once n ≥ 4. In four space dimensions and above, therefore,

we are always in the limit-point case. No point interaction – no matter how strong –

can affect the physics. This non-interaction result extends, with slight modification, to

the quantum field theory of relativistic particles. Here we find that contact interactions

become irrelevent or non-renormalizable in more than four space-time dimensions.

8.5 Further exercises and problems

Here are some further problems involving Legendre polynomials, associated Legendre

functions and Bessel functions.

Exercise 8.8: A sphere of radius a is made by joining two conducting hemispheres

along their equators. The hemispheres are electrically insulated from one another and

maintained at two different potentials V

and V

(a) Starting from the general expression

V (r, θ) =

∞



l=0



l+1



(cos θ)

find an integral expression for the coefficients a

, b

that are relevant to the electric

field outside the sphere. Evaluate the integrals giving b

, b

and b

(b) Use your results from part (a) to compute the electric dipole moment of the sphere

as a function of the potential difference V

− V

potential. The sphere is immersed in a uniform electric field E. What is its dipole

moment now?

Problem 8.9: Tides and gravity. The Earth is not exactly spherical. Two major causes

of the deviation from sphericity are the Earth’s rotation and the tidal forces it feels from

the Sun and the Moon. In this problem we will study the effects of rotation and tides on

a self-gravitating sphere of fluid of uniform density ρ

306 8 Special functions

(a) Consider the equilibrium of a nearly spherical body of fluid rotating homogeneously

with angular velocity ω

. Show that the effect of rotation can be accounted for by

introducing an “effective gravitational potential”

eff

= ϕ

grav

(cos θ) − 1),

where R, θ are spherical coordinates defined with their origin in the centre of the

body and ˆz along the axis of rotation.

(b) A small planet is in a circular orbit about a distant massive star. It rotates about an axis

perpendicular to the plane of the orbit so that it always keeps the same face directed

towards the star. Show that the planet experiences an effective external potential

tidal

=−

(cos θ),

together with a potential, of the same sort as in part (a), that arises from the once-

per-orbit rotation. Here  is the orbital angular velocity, and R, θ are spherical

coordinates defined with their origin at the centre of the planet and ˆz pointing at

the star.

the surface is given by

R(θ, φ) = R

+ ηP

(cos θ).

(with θ being measured with respect to different axes for the rotation and tidal

effects). Show that, to first order in η, this deformation does not alter the volume of

the body. Observe that positive η corresponds to a prolate spheroid and negative η

to an oblate one.

(d) The gravitational field of the deformed spheroid can be found by approximating it

as an undeformed homogeneous sphere of radius R

, together with a thin spherical

shell of radius R

and surface mass density σ = ρ

ηP

(cos θ). Use the general

axisymmetric solution

ϕ(R, θ, φ) =

∞



l=0



l+1



(cos θ)

of Laplace’s equation, together with Poisson’s equation

∇

ϕ = 4πGρ(r)

for the gravitational potential, to obtain expressions for ϕ

shell

in the regions R > R

and R ≤ R

(e) The surface of the fluid will be an equipotential of the combined potentials of the

homogeneous sphere, the thin shell and the effective external potential of the tidal or

8.5 Further exercises and problems 307

centrifugal forces. Use this fact to find η (to lowest order in the angular velocities) for

the two cases. Do not include the centrifugal potential from part (b) when computing

the tidal distortion. We never include the variation of the centrifugal potential across

a planet when calculating tidal effects. This is because this variation is due to the

once-per-year rotation, and contributes to the oblate equatorial bulge and not to the

prolate tidal bulge.



Answer: η

rot

=−

4πGρ

, and η

tide



4πGρ



Exercise 8.10: Dielectric sphere. Consider a solid dielectric sphere of radius a and

permittivity . The sphere is placed in an electric field which takes the constant value

E = E

ˆz a long distance from the sphere. Recall that Maxwell’s equations require that

⊥

and E



be continuous across the surface of the sphere.

(a) Use the expansions





(cos θ)



out



+ C

−l−1

(cos θ)

and find all non-zero coefficients A

, B

, C

(b) Show that the E field inside the sphere is uniform and of magnitude

3

+2

polarization-induced surface charge density

induced

= 3



 − 

 + 2



cos θ.

(Some systems of units may require extra 4π ’s in this last expression. In SI units D ≡

E = 

E + P, and the polarization-induced charge density is ρ

induced

=−∇·P.)

Exercise 8.11: Hollow sphere. The potential on a spherical surface of radius a is (θ, φ).

We want to express the potential inside the sphere as an integral over the surface in a

manner analagous to the Poisson kernel in two dimensions.

(a) By using the generating function for Legendre polynomials, show that

1 − r

(1 + r

− 2r cos θ)

3/2

∞



l=0

(2l + 1)r

(cos θ), r < 1.

Our Earth rotates about its axis 365

+ 1 times in a year, not 365

times. The “+1” is this effect.

308 8 Special functions

(b) Starting from the expansion



(r, θ , φ) =

∞



l=0



m=−l

(θ, φ)





(θ, φ)



∗

(θ, φ) d cos θ dφ

and using the addition formula for spherical harmonics, show that



(r, θ , φ) =

a(a

− r

)

4π



(θ



, φ



)

+ a

− 2ar cos γ)

3/2

d cos θ



dφ



where cos γ = cos θ cos θ



+ sin θ sin θ



cos(φ − φ



local maximum or minimum.

Problem 8.12: We have several times met with the Pöschel–Teller eigenvalue problem



−

− n(n + 1)sech



ψ = Eψ ,()

in the particular case that n = 1. We now consider this problem for any positive integer n.

(a) Set ξ = tanh x in () and show that it becomes



dξ

(1 − ξ

)

dξ

+ n(n + 1) +

1 − ξ



ψ = 0.

(b) Compare the equation in part (a) with the associated Legendre equation and deduce

that the bound-state eigenfunctions and eigenvalues of the original Pöschel–Teller

equation are

(x) = P

(tanh x), E

=−m

, m = 1, ..., n,

where P

(ξ) is the associated Legendre function. Observe that the list of bound

states does not include ψ

= P

(tanh x) ≡ P

(tanh x). This is because ψ

is not

normalizable, being the lowest of the unbound E ≥ 0 continuous-spectrum states.

(x) = e

ikx

f (tanh x),

and show if we take E = k

, where k is any real number, then f (ξ ) obeys

(1 − ξ

)

dξ

+ 2(ik − ξ)

dξ

+ n(n + 1)f = 0. ()

8.5 Further exercises and problems 309

(d) Let us denote by P

(k)

(ξ) the solutions of () that reduce to the Legendre polynomial

(ξ) when k = 0. Show that the first few P

(k)

(ξ) are

(k)

(ξ) = 1,

(k)

(ξ) = ξ − ik,

(k)

(ξ) =

(3ξ

− 1 − 3ikξ − k

Explore the properties of the P

(k)

(ξ), and show that they include

(i) P

(k)

(−ξ) = (−1)

(−k)

(ξ).

(ii) (n + 1)P

(k)

n+1

(ξ) = (2n + 1)xP

(k)

(ξ) − (n + k

/n)P

(k)

n−1

(ξ).

(iii) P

(k)

(1) = (1 − ik)(2 − ik)...(n − ik)/n!.

(The P

(k)

(ξ) are the ν =−µ = ik special case of the Jacobi polynomials P

(ν,µ)

(ξ).)

Problem 8.13: Bessel functions and impact parameters. In two dimensions we can

expand a plane wave as

iky

∞



n=−∞

(kr)e

inθ

(a) What do you think the resultant wave will look like if we take only a finite segment

of this sum? For example

φ(x) =



l=10

(kr)e

inθ

Think about:

(i) The quantum interpretation of l as angular momentum = kd, where d is the

impact parameter, the amount by which the incoming particle misses the origin.

(ii) Diffraction: one cannot have a plane wave of finite width.

(b) After writing down your best guess for the previous part, confirm your understanding

by using Mathematica or another package to plot the real part of φ as defined above.

The following Mathematica code may work.

Clear[bit,tot]

bit[l_,x_,y_]:=Cos[l ArcTan[x,y]]BesselJ[l,Sqrt[xˆ2+yˆ2]]

tot[x_,y_] :=Sum[bit[l,x,y],{l,10,17}]

ContourPlot[tot[x,y],{x,-40,40},{y,-40,40},PlotPoints ->200]

Display["wave",\%,"EPS"]

Run it, or some similar code, as a batchfile. Try different ranges for the sum.

310 8 Special functions

Exercise 8.14: Consider the two-dimensional Fourier transform

;

f (k) =



ik·x

f (x) d

of a function that in polar coordinates is of the form f (r, θ) = exp{−ilθ }f (r).

(a) Show that

;

f (k) = 2πi

−ilθ



∞

(kr)f (r) rdr,

where k, θ

are the polar coordinates of k.

(b) Use the inversion formula for the two-dimensional Fourier transform to establish

the inversion formula (8.120) for the Hankel transform

F(k) =



∞

(kr)f (r) rdr.

Integral equations

A problem involving a differential equation can often be recast as one involving an

integral equation. Sometimes this new formulation suggests a method of attack or approx-

imation scheme that would not have been apparent in the original language. It is also

usually easier to extract general properties of the solution when the problem is expressed

as an integral equation.

9.1 Illustrations

Here are some examples.

A boundary-value problem: Consider the differential equation for the unknown u(x)

−u



+ λV (x)u = 0 (9.1)

with the boundary conditions u(0) = u(L) = 0. To turn this into an integral equation we

introduce the Green function

G(x, y) =

⎧

⎨

⎩

x(y − L),0≤ x ≤ y ≤ L,

y (x − L),0≤ y ≤ x ≤ L,

(9.2)

so that

−

G(x, y) = δ(x − y). (9.3)

Then we can pretend that λV (x)u(x) in the differential equation is a known source term,

and substitute it for “f (x)” in the usual Green function solution. We end up with

u(x ) + λ



G(x, y)V (y)u(y) dx = 0. (9.4)

This integral equation for u has not solved the problem, but is equivalent to the original

problem. Note, in particular, that the boundary conditions are implicit in this formulation:

if we set x = 0orL in the second term, it becomes zero because the Green function is

zero at those points. The integral equation then says that u(0) and u(L) are both zero.

311

312 9 Integral equations

An initial value problem: Consider essentially the same differential equation as before,

but now with initial data:

−u



+ V (x)u = 0, u(0) = 0, u



(0) = 1. (9.5)

In this case, we claim that the inhomogeneous integral equation

u(x ) −



(x − t)V (t)u(t) dt = x (9.6)

is equivalent to the given problem. Let us check the claim. First, the initial conditions.

Rewrite the integral equation as

u(x ) = x +



(x − t)V (t)u(t) dt, (9.7)

so it is manifest that u(0) = 0. Now differentiate to get



(x) = 1 +



V (t)u(t) dt. (9.8)

This shows that u



(0) = 1, as required. Differentiating once more confirms

that u



= V (x)u.

These examples reveal that one advantage of the integral equation formulation is

that the boundary or initial value conditions are automatically encoded in the integral

equation itself, and do not have to be added as riders.

9.2 Classification of integral equations

The classification of linear integral equations is best described by a list:

(A) (i) Limits on integrals fixed ⇒ Fredholm equation.

(ii) One integration limit is x ⇒ Volterra equation.

(B) (i) Unknown under integral only ⇒Type I.

(ii) Unknown also outside integral ⇒ Type II.

(ii) Inhomogeneous.

For example,

u(x ) =



G(x, y)u(y) dy (9.9)

9.3 Integral transforms 313

is a Type II homogeneous Fredholm equation, whilst

u(x ) = x +



(x − t)V (t)u(t) dt (9.10)

is a Type II inhomogeneous Volterra equation.

The equation

f (x) =



K(x, y)u(y) dy, (9.11)

an inhomogeneous Type I Fredholm equation, is analogous to the matrix equation

Kx = b. (9.12)

On the other hand, the equation

u(x ) =



K(x, y)u(y) dy, (9.13)

a homogeneous Type II Fredholm equation, is analogous to the matrix eigenvalue

problem

Kx = λx. (9.14)

Finally,

f (x) =



K(x, y)u(y) dy, (9.15)

an inhomogeneous Type I Volterra equation, is the analogue of a system of linear

equations involving an upper triangular matrix.

The function K(x , y) appearing in these expressions is called the kernel. The phrase

“kernel of the integral operator” can therefore refer either to the function K or the

null-space of the operator. The context should make clear which meaning is intended.

9.3 Integral transforms

When the kernel of the Fredholm equation is of the form K(x − y), with x and y taking

values on the entire real line, then it is translation invariant and we can solve the integral

equation by using the Fourier transformation

;u(k) = F(u) =



∞

−∞

u(x )e

ikx

dx (9.16)

u(x ) = F

−1

(;u) =



∞

−∞

;u(k)e

−ikx

2π

. (9.17)

314 9 Integral equations

Integral equations involving translation-invariant Volterra kernels usually succumb to a

Laplace transform

;u(p) = L(u) =



∞

u(x )e

−px

dx (9.18)

u(x ) = L

−1

(;u) =

2πi



γ +i∞

γ −i∞

;u(p)e

dp. (9.19)

The Laplace inversion formula is the Bromwich contour integral, where γ is chosen so

that all the singularities of ;u(p) lie to the left of the contour. In practice one finds the

inverse Laplace transform by using a table of Laplace transforms, such as the Bateman

tables of integral transforms mentioned in the introduction to Chapter 8.

For kernels of the form K(x/y) the Mellin transform,

;u(σ ) = M(u) =



∞

u(x )x

σ −1

dx (9.20)

u(x ) = M

−1

(;u) =

2πi



γ +i∞

γ −i∞

;u(σ )x

−σ

dσ , (9.21)

is the tool of choice. Again the inversion formula requires a Bromwich contour integral,

and so usually requires tables of Mellin transforms.

9.3.1 Fourier methods

The class of problems that succumb to a Fourier transform can be thought of as a

continuous version of a matrix problem where the entries in the matrix depend only

on their distance from the main diagonal (Figure 9.1).

Example: Consider the Type II Fredholm equation

u(x ) − λ



∞

−∞

−|x−y|

u(y) dy = f (x), (9.22)

where we will assume that λ<1/2. Here the x -space kernel operator

K(x − y) = δ(x − y) − λe

−|x−y|

(9.23)



Figure 9.1 The matrix form of the equation



∞

−∞

K(x − y)u(y) dy = f (x ).